Math

[linkspast]

I got this problem, and figured some of you may be able to help me.

so I am a freshman in college, and i tested in to math 111 which they call college algebra. no problem right?
well... last year my math teacher did not realy teach us any thing, he was more of a supivisor. so I dont know half of the things we are talking about in math 111.
I am trying to decide if I should just stick it out, or retake the test to see if i just got "lucky" and guessed the right answers to get in to math 111.
Any help would appreciated.

[Bigfoot]

Definitely get out if you feel like you don't belong. Have you guys had a first test yet in class? You might want to wait until then before you decide upon getting out of it or not. I'm in ma111 too btw heh.

[Gollgagh]

Those numbers don't mean a thing to me. MATH111 here is calculus.

But yeah, I'm having the same trouble in spanish. I'm having to withdraw from it entirely.

[Bahamut]

Wtf, I was gonna make a math thread. Fuck.

So, what is this college algebra anyway? In many colleges, it is the lowest math class offered in a school (if you want to exclude statistics).

[Dhsu]

Yeah, you should consider dropping the class if you feel like you can't keep up. Your other option is to hit the books hard and try to get up to speed. Your school math lab and/or tutor would help a lot in that case.

[The Xyco]

I'm taking the same course as well. It's essentially High Shcool Precalculus. We covered polynomial functions, complex numbers, and other sorts of related material for our first section which we were just tested on last week. For the second section we're starting now, we're going into graphical representation and functions of the previous material.

Granted, I haven't had a serious math course since my sophmore year in high school, but of the courses I'm in now, MAT 111 is my biggest concern. So far, I've managed to already enormously screw up in the class already (I got a 49% on the aforementioned test, but that's mostly because I barely touched the homework and didn't study), however, I would hesitate to say the course is difficult. After speaking with my professor, he advised me to work on extra problems in my studies to understand the types of questions on the exams better.

[Bigfoot]

I'm taking the same course as well. It's essentially High Shcool Precalculus. We covered polynomial functions, complex numbers, and other sorts of related material for our first section which we were just tested on last week. For the second section we're starting now, we're going into graphical representation and functions of the previous material.

Granted, I haven't had a serious math course since my sophmore year in high school, but of the courses I'm in now, MAT 111 is my biggest concern. So far, I've managed to already enormously screw up in the class already (I got a 49% on the aforementioned test, but that's mostly because I barely touched the homework and didn't study), however, I would hesitate to say the course is difficult. After speaking with my professor, he advised me to work on extra problems in my studies to understand the types of questions on the exams better.


Damn, you beat me by 2% on the first test :P I just didn't study as hard as I should have. I made stupid mistakes like thinking 7-(-5)=14 haha. What the fuck was wrong with me? I also left a few questions that were so easy after going over the test the next class. I just totally forgot how to do them.


Basically, I suck at test taking.

[Paratha]

I think it depends on your major.

If you're planning on doing something that will require math, don't take more than you can handle, since you'll need a solid understanding of algebra for future classes.
If you're not going down that road, try to get math out of the way as quickly as possible.

EDIT: Whoops. Totally misred the OP post.

[Ceiphied]

I got this problem, and figured some of you may be able to help me.

so I am a freshman in college, and i tested in to math 111 which they call college algebra. no problem right?
well... last year my math teacher did not realy teach us any thing, he was more of a supivisor. so I dont know half of the things we are talking about in math 111.
I am trying to decide if I should just stick it out, or retake the test to see if i just got "lucky" and guessed the right answers to get in to math 111.
Any help would appreciated.

Talk to your professor or someone else in the math department, whichever is more competent for answering this question. No one here will know the exact specifics of what is taught in your class, while a professor will know.

Trust me on this. I was a math/econ/stat major in college; talk to someone who can give you an informed opinion. If it -truly- isn't an option, find a TA or an upper-classman that's willing to give you advice.

[Steben]

Get a tutor. The vast majority of math professors range from mediocre to crappy teachers (at least at my school), and don't understand how to explain math to someone who's not naturally mathematically inclined. I'm a math major, yet for over half my math classes so far I've pretty much just taught myself everything I need to know from books. (Granted, that's getting a little bit harder now that I'm taking Senior-level classes, many of which don't have books.)

However, if you get a good tutor (or make a friend in the class who knows what's going on), you can get that person to explain things to you in a way that you can understand. Of course, if you don't need math for your major, and you're struggling, then get out. No need to sweat out a difficult class that you'll never use that'll only hurt your GPA.

[Dragonmaster_Alex]

Those numbers don't mean a thing to me. MATH111 here is calculus.

But yeah, I'm having the same trouble in spanish. I'm having to withdraw from it entirely.

Spanish sucks.

[Global-Trance]

Math 111 here doesn't exist, though anything above 100 is considered upper division math.

[Bobwillis]

So... who here is a member of the Abelian group?

/raises hand

[neminem]

So... who here is a member of the Abelian group?

Nope - I've never had to commute. All my work has been right on campus. :lol:

Someone shoot me already, that was pretty horrible. :lol:

[Ceiphied]

So... who here is a member of the Abelian group?

Nope - I've never had to commute. All my work has been right on campus. :lol:

Someone shoot me already, that was pretty horrible. :lol:

So you're saying that it's never been the case that right on campus, all your work has been?

[Ace-Of-War]

In the same boat, except this is my second time taking College Algebra. I had to withdraw a couple of semesters ago because I just couldn't keep up. Like linkspast said, our high school teachers basically gave us a calculator and showed us how to get it to think for us.

I'm just not a math kind of guy either. When I see math, I see the same problem people have done for hundreds of years, meaningless numbers that have been solved countless times. It's so much different from writing a paper or reading about an issue or event, you don't really get anything out of it. You figure out x and you throw the paper away. The next kid figures it out, does the exact same thing, and he throws it away. I can't get over that mindset, that it is just the same thing over and over again.

[Bahamut]

Well, it takes time to get to a point of mathematical maturity to do something original. Math is one of the few fields where it really is necessary to have things taught in a simple way so that you absorb the concepts before you can really dig into more complicated stuff. So you're taught stuff with polynomials and solving equations early for example, but in higher math, you get to define the notion of a polynomial as a general concept and do all sorts of crazy stuff with them (i.e. Galois Theory).

[Ace-Of-War]

Well, it takes time to get to a point of mathematical maturity to do something original. Math is one of the few fields where it really is necessary to have things taught in a simple way so that you absorb the concepts before you can really dig into more complicated stuff. So you're taught stuff with polynomials and solving equations early for example, but in higher math, you get to define the notion of a polynomial as a general concept and do all sorts of crazy stuff with them (i.e. Galois Theory).

I'm sure there is, and for those that can stomache it, more power to you. It's just that a world where math can actually be fun and creative instead of the execise in repetition that I know it to be only exists for people who like doing it to begin with.

[Bahamut]

Well, it takes time to get to a point of mathematical maturity to do something original. Math is one of the few fields where it really is necessary to have things taught in a simple way so that you absorb the concepts before you can really dig into more complicated stuff. So you're taught stuff with polynomials and solving equations early for example, but in higher math, you get to define the notion of a polynomial as a general concept and do all sorts of crazy stuff with them (i.e. Galois Theory).

I'm sure there is, and for those that can stomache it, more power to you. It's just that a world where math can actually be fun and creative instead of the execise in repetition that I know it to be only exists for people who like doing it to begin with.

Well, the creativity is usually seen outside the classroom. For an example, have you ever looked at math competition questions for various levels of schooling? Anyone with a high school level of knowledge is capable of answering a lot of them, yet they require thinking outside the box.

[LegendarySword]

First advice I can give is strengthen your weaknesses. What specifically is giving you trouble? What you do to one side of the equation do to the other, slope-intercept form for graphing, and factoring are really the main topics of basic Algebra. Having a firm grasp of those topics would be a recommended start, and then move on to more problematic areas. Keep doing the type of problems that give you the most problems.

[Omatic]

One thing that is very important here is to NOT let the first exam let you down. It's always demoralizing to have a bad start in a class, but it's not the end of the class.

My suggestion is to study hard to make up for the failed test. On top of that, this class probably allows for a dropped exam, right? (most engineering-type classes have some kind of system like that).

However if the class is not like that, then chances are you can still pull at least a B out of the class if you get a decent grade in everything else. I know this is hard, but the absolute best way to pass a class is to set aside a daily, regular schedule for the class. Since it's a entry class, an hour a day should do it, and maybe 1.5 hours the week of a test.

You sound like a smart guy (after all, you made it into college), so don't give up on the class! Stick it out until you're sure you're doomed (your college should also have a class drop deadline, try and wait until just before then to make your drop decision).

[neminem]

So... who here is a member of the Abelian group?

Nope - I've never had to commute. All my work has been right on campus. :lol:

Someone shoot me already, that was pretty horrible. :lol:

So you're saying that it's never been the case that right on campus, all your work has been?

That sentence doesn't even parse, so I have no idea what joke you were going for.

On the other hand, a joke you could have made would have involved parsing my sentence as stating that while I've been on campus, I've never turned in work that hasn't been right. Which I certainly wish were true...

[Steben]

This topic is back and badder than ever.

Here's the question I have for you kids. In Intro to Topolgy, we were asked if the intervals (0,1) and (0,1] were homeomorphic, that is, if there is a bijective continuous function f:(0,1)->(0,1] whose inverse was also continuous.

The problem was solved by talking about continuity - assuming such a function exists, and coming to a contradiction from its continuity. Therefore, they weren't homeomorphic. But what I'm wondering is, is there a bijection between the two spaces, regardless of continuity? I want to say no, but perhaps someone here can think of one?

[Bummerdude]

complex numbers
I'll have a test about that next week. ;_;

[Ab56 v2 aka Ash]

I think for math students who aren't imbued with ungodly math ability (read: everyone except Bahamut), this would be a good resource:

http://tutorial.math.lamar.edu/

It covers some Algebra, but mostly Calculus 1, 2, and 3 and Linear Algebra. It's a more useful resource than my textbook, anyway.

[Arek the Absolute]

Math 111 here doesn't exist, though anything above 100 is considered upper division math.
He speaks the truth.

[Bigfoot]

Took my 2nd test this morning...I felt 2x as confident as I did taking the first one :P so I'll find out on Wed how I did.

[Bahamut]

This topic is back and badder than ever.

Here's the question I have for you kids. In Intro to Topolgy, we were asked if the intervals (0,1) and (0,1] were homeomorphic, that is, if there is a bijective continuous function f:(0,1)->(0,1] whose inverse was also continuous.

The problem was solved by talking about continuity - assuming such a function exists, and coming to a contradiction from its continuity. Therefore, they weren't homeomorphic. But what I'm wondering is, is there a bijection between the two spaces, regardless of continuity? I want to say no, but perhaps someone here can think of one?

Well, I can't explicitly construct one, but one can talk about cardinality. The cardinality of [0, 1] = the cardinality of the real numbers (which is uncountably infinite). The cardinality of (0, 1) = [0, 1] - {0, 1} is also equal to the cardinality of [0, 1] by cardinal arithmetic. Similarly, the cardinality of (0, 1] = [0, 1] - {0} is also the cardinality of [0, 1]. Now, by definition of cardinality, two sets have the same cardinality if there exists a bijection between the two.

[Steben]

This topic is back and badder than ever.

Here's the question I have for you kids. In Intro to Topolgy, we were asked if the intervals (0,1) and (0,1] were homeomorphic, that is, if there is a bijective continuous function f:(0,1)->(0,1] whose inverse was also continuous.

The problem was solved by talking about continuity - assuming such a function exists, and coming to a contradiction from its continuity. Therefore, they weren't homeomorphic. But what I'm wondering is, is there a bijection between the two spaces, regardless of continuity? I want to say no, but perhaps someone here can think of one?

Well, I can't explicitly construct one, but one can talk about cardinality. The cardinality of [0, 1] = the cardinality of the real numbers (which is uncountably infinite). The cardinality of (0, 1) = [0, 1] - {0, 1} is also equal to the cardinality of [0, 1] by cardinal arithmetic. Similarly, the cardinality of (0, 1] = [0, 1] - {0} is also the cardinality of [0, 1]. Now, by definition of cardinality, two sets have the same cardinality if there exists a bijection between the two.
Fair enough. I've never heard cardinality be defined that way, but it certainly makes sense. And now that I'm thinking about it, I guess we want there to be a bijection intuitively anyway. We'll let f-inverse(1)=1/2, and let the inverse function map the rest of (0,1) evenly on either side of 1/2. If that makes any sense... :)

[Bahamut]

Well, I kind've skirted some set theory by saying that by definition of cardinality there exists a bijection between the two sets - the equality comes as a consequence of the Cantor-Schroeder-Bernstein Theorem.

[Steben]

Never heard of it. But now I know. Thank God for Bahamut and Wikipedia.

[Zombie]

Math makes me sad. :(

[Volt]

This topic is back and badder than ever.

Here's the question I have for you kids. In Intro to Topolgy, we were asked if the intervals (0,1) and (0,1] were homeomorphic, that is, if there is a bijective continuous function f:(0,1)->(0,1] whose inverse was also continuous.

The problem was solved by talking about continuity - assuming such a function exists, and coming to a contradiction from its continuity. Therefore, they weren't homeomorphic. But what I'm wondering is, is there a bijection between the two spaces, regardless of continuity? I want to say no, but perhaps someone here can think of one?

Well, I can't explicitly construct one, but one can talk about cardinality. The cardinality of [0, 1] = the cardinality of the real numbers (which is uncountably infinite). The cardinality of (0, 1) = [0, 1] - {0, 1} is also equal to the cardinality of [0, 1] by cardinal arithmetic. Similarly, the cardinality of (0, 1] = [0, 1] - {0} is also the cardinality of [0, 1]. Now, by definition of cardinality, two sets have the same cardinality if there exists a bijection between the two.
Fair enough. I've never heard cardinality be defined that way, but it certainly makes sense. And now that I'm thinking about it, I guess we want there to be a bijection intuitively anyway. We'll let f-inverse(1)=1/2, and let the inverse function map the rest of (0,1) evenly on either side of 1/2. If that makes any sense... :)

AAAHHH JESUS CHRIST MAKE IT STOP :(:(:(

[Jodo Kast]

From my point of view, math is entirely tolerable with the exception of proofs. I enjoyed calculus and differential equations, but when I took advanced calculus (which was all proofs), I had to drop the course. I can do proofs, much like I can clean outhouses. In other words, proofs and outhouse cleaning are equally pleasurable.

[Bahamut]

From my point of view, math is entirely tolerable with the exception of proofs. I enjoyed calculus and differential equations, but when I took advanced calculus (which was all proofs), I had to drop the course. I can do proofs, much like I can clean outhouses. In other words, proofs and outhouse cleaning are equally pleasurable.

You do not understand math then. Everything before the introduction of proofs is meant to prepare for actual mathematics - in otherwords, with proofs. So, if you know how to do proofs, would you care to share with us a proof of the Fundamental Theorem of Algebra?

[Captain Huge!]

I hate math in all forms, although imaginary numbers are easy, THEY ARE POINTLESS AND MAKE ME ANGRY AND TYPE IN CAPLSOCK

[Steben]

I hate math in all forms, although imaginary numbers are easy, THEY ARE POINTLESS AND MAKE ME ANGRY AND TYPE IN CAPLSOCK
Never become an electrician. (http://en.wikipedia.org/wiki/Imaginary_numbers#Applications_of_imaginary_number s)

[Garde]

The thing that makes math so hard is that a lot of people aren't willing to do the work to learn it (doing homework to master the applications of something you learn in class), or don't want to admit they don't understand something. When the latter happens, you get screwed. If you don't know something, ask your teacher. If they can't explain in some way that helps you, you should see if they can give you examples of it in action. If that doesn't help you, you should find someone who understands it that can explain it in a simplified way.

I know that I had a lot of trouble trying to learn Calculus in high school. Once I hit college and had a Discrete Math class, all my Calculus classes became incredibly easy. My Discrete Math teacher taught me various things, including some very important pieces of math language and their exact meaning. My high school Calc teacher expect us to know what the universal quantifier was (upside down capital A), the "There exists" symbol, "such that" symbol, and many others. He would write definitions of things up on the board and somewhere between half and three-quarters of the class ended up dropping because he never explained what the symbols meant.

So I guess what I'm trying to say is that if you don't understand something, don't pretend like you know it. If you do, you'll just end up falling farther behind as everything builds on top of what you didn't understand. Also, practice with homework, it really helps in seeing when and how problem-solving strategies can be applied.

[BlueMage]

complex numbers
I'll have a test about that next week. ;_;

Complex numbers are simple.

Or so I tell everyone who asks me.

Also, yeah, NEVER become an electrician. Or do anything involving electrics.

[vega12]

From my point of view, math is entirely tolerable with the exception of proofs. I enjoyed calculus and differential equations, but when I took advanced calculus (which was all proofs), I had to drop the course. I can do proofs, much like I can clean outhouses. In other words, proofs and outhouse cleaning are equally pleasurable.

You do not understand math then. Everything before the introduction of proofs is meant to prepare for actual mathematics - in otherwords, with proofs. So, if you know how to do proofs, would you care to share with us a proof of the Fundamental Theorem of Algebra?

Would you like a purely algebraic proof, or would complex analysis be allowed. Because man, residues make that proof almost too easy :D

(also, a good example of how complex analysis - in other words, "imaginary" numbers, are useful)

[Bahamut]

From my point of view, math is entirely tolerable with the exception of proofs. I enjoyed calculus and differential equations, but when I took advanced calculus (which was all proofs), I had to drop the course. I can do proofs, much like I can clean outhouses. In other words, proofs and outhouse cleaning are equally pleasurable.

You do not understand math then. Everything before the introduction of proofs is meant to prepare for actual mathematics - in otherwords, with proofs. So, if you know how to do proofs, would you care to share with us a proof of the Fundamental Theorem of Algebra?

Would you like a purely algebraic proof, or would complex analysis be allowed. Because man, residues make that proof almost too easy :D

(also, a good example of how complex analysis - in other words, "imaginary" numbers, are useful)

Heh, that's why I said "a" proof. I do know of the complex analysis proof, as well as a proof by Galois Theory.

Edit: Actually, the proof by complex analysis I know just quotes Rouche's Theorem.

[Captain Huge!]

I hate math in all forms, although imaginary numbers are easy, THEY ARE POINTLESS AND MAKE ME ANGRY AND TYPE IN CAPLSOCK
Never become an electrician. (http://en.wikipedia.org/wiki/Imaginary_numbers#Applications_of_imaginary_number s)

Then its a good thing im training to be a computer technician :D

[Bobwillis]

Math proof on teh internetz:

1. CTRL + C
2. CTRL + V

A.W.D.

[Rainman DX]

From my point of view, math is entirely tolerable with the exception of proofs. I enjoyed calculus and differential equations, but when I took advanced calculus (which was all proofs), I had to drop the course. I can do proofs, much like I can clean outhouses. In other words, proofs and outhouse cleaning are equally pleasurable.

You do not understand math then. Everything before the introduction of proofs is meant to prepare for actual mathematics - in otherwords, with proofs. So, if you know how to do proofs, would you care to share with us a proof of the Fundamental Theorem of Algebra?

Heh. That's easy. :P


Actually, I find mathematical proofs troublesome as well. I enjoy math, related concepts, and computation comes really easy to me. But taking this Abstract Vector Spaces course at Georgia Tech taught me one thing above all else: the math that I learned (attending a sub-par public school in North Georgia) was not sufficient in any sense of the word to prepare me for later math. AVS is a proofs-based study of linear algebra, and is the first required UG proofs-based course at GT. Frankly, it was probably the major inspiration for me to switch from a Math major to a Math minor - the only difference in scheduling is I have two 4XXX level classes left to take instead of ten.

So while I understand what Bahamut is saying, defending upper level math and its need for proofs before anything can be built upon it, it doesn't mean that even mathematicians like myself have some sort of natural bent for it either.

But, for the record, I am steadily getting better at writing and thinking in terms of proofs!

[Bigfoot]

So I got a 65 on the math test. That's a lot better than a 47 haha...


The teacher said that's the kind of things he looks for: if you're improving or doing better. He said sometimes he won't count the first test if you show improvements like that.

[Bahamut]

So here's a question that involves basic complex variables that is annoying me that (hopefully) someone here could answer, although I probably will figure out sometime anyway.

So suppose we have |1 - f(z)|/|1 + f(z)| <= |z|. (<= is less than or equal to)

Prove that |f(z)| <= (1 + |z|)/(1 - |z|)

I'm fairly certain that I gave all the info needed to solve this.

Edit: Oh and Re f(z) >= 0

[Julio Jose]

I've grown to hate math. The day I entered my Abstract Algebra 3 class, and my Linear Algebra 4 class, I came to the conclusion that, holy shit math is a pain in the ass.

[Bummerdude]

I've grown to hate math. The day I entered my Abstract Algebra 3 class, and my Linear Algebra 4 class, I came to the conclusion that, holy shit math is a pain in the ass.
Only in small doses.

I had my math test about complex numbers and polynom divisions yesterday. Since I answered every question and didn't have that much doubt about my solutions, I guess I'll get a good grade on it.

[Bahamut]

I've grown to hate math. The day I entered my Abstract Algebra 3 class, and my Linear Algebra 4 class, I came to the conclusion that, holy shit math is a pain in the ass.
Only in small doses.

He sees why its a pain. Sometimes the methods of proofs just completely elude you even though it looks as though it should be simple. Sometimes it's just a lot of busy work too.

[Bahamut]

Doublepost!

But no, I figured out how to do the question I was stuck with earlier today - it just involves the Triangle Inequality & the Negative Triangle Inequality.

[Steben]

I've been working on the same stupid theorem in Analysis for the past three weeks. I'm probably just being dumb and missing something obvious, but it's been really difficult... I'll type it below for you guys, at the risk of revealing my relative inexperience at proof-based mathematics.

Show that if H is a bounded infinite set, then there exists some sequence made of distinct points of H which is always increasing or always decreasing.

[Rainman DX]

Show that if H is a bounded infinite set, then there exists some sequence made of distinct points of H which is always increasing or always decreasing.

Aaaaahhhhhhhhhh! Make the voices in my head stop! Make the voices go away!

[Bahamut]

I've been working on the same stupid theorem in Analysis for the past three weeks. I'm probably just being dumb and missing something obvious, but it's been really difficult... I'll type it below for you guys, at the risk of revealing my relative inexperience at proof-based mathematics.

Show that if H is a bounded infinite set, then there exists some sequence made of distinct points of H which is always increasing or always decreasing.

Is H a subset of the real numbers? If so, then with a bounded set, you have the least upper bound and greatest lower bound. There must exist sequences in H that converge to the least upper bound and greatest lower bound. Then, there exist subsequences of each sequences such that the subsequences are monotone and converge to the least upper bound and greatest lower bound, respectively.

[Bobwillis]

Anyone else taking the Putnam this year?

I'm nervous. It's going to be my first.

[Steben]

I've been working on the same stupid theorem in Analysis for the past three weeks. I'm probably just being dumb and missing something obvious, but it's been really difficult... I'll type it below for you guys, at the risk of revealing my relative inexperience at proof-based mathematics.

Show that if H is a bounded infinite set, then there exists some sequence made of distinct points of H which is always increasing or always decreasing.

Is H a subset of the real numbers? If so, then with a bounded set, you have the least upper bound and greatest lower bound. There must exist sequences in H that converge to the least upper bound and greatest lower bound. Then, there exist subsequences of each sequences such that the subsequences are monotone and converge to the least upper bound and greatest lower bound, respectively.

Yeah, H is a subset of the reals.

I don't follow your argument, though. Let's say H is the union of {1-1/n|n is a natural number} and {1+1/n|n is a natural number}. It's bounded by 0 and 2, but I need a sequence that is always increasing or always decreasing, rather, x_n>x_n+1 always or x_n<x_n+1 always. No such sequence converges to 0 or 2... the only way I can find such a sequence is if it converges to 1... I can't see how talking about the least upper bound or greatest lower bound helps me..

Anyone else taking the Putnam this year?

I'm nervous. It's going to be my first.

I've looked over various Putnam problems (with their solutions handy), but I've never actually taken the test. I should look into it. I probably wouldn't solve any of them, but it'd be fun to try.

[Bahamut]

I've been working on the same stupid theorem in Analysis for the past three weeks. I'm probably just being dumb and missing something obvious, but it's been really difficult... I'll type it below for you guys, at the risk of revealing my relative inexperience at proof-based mathematics.

Show that if H is a bounded infinite set, then there exists some sequence made of distinct points of H which is always increasing or always decreasing.

Is H a subset of the real numbers? If so, then with a bounded set, you have the least upper bound and greatest lower bound. There must exist sequences in H that converge to the least upper bound and greatest lower bound. Then, there exist subsequences of each sequences such that the subsequences are monotone and converge to the least upper bound and greatest lower bound, respectively.

Yeah, H is a subset of the reals.

I don't follow your argument, though. Let's say H is the union of {1-1/n|n is a natural number} and {1+1/n|n is a natural number}. It's bounded by 0 and 2, but I need a sequence that is always increasing or always decreasing, rather, x_n>x_n+1 always or x_n<x_n+1 always. No such sequence converges to 0 or 2... the only way I can find such a sequence is if it converges to 1... I can't see how talking about the least upper bound or greatest lower bound helps me..

Yes - I'm just talking about the existence, guaranteed by theorems. If you don't have the particular theorems that I'm referring to, then just prove them - if you have an element that is not the least upper bound or greatest lower bound, you can find another element closer, and keep constructing it that way.

[Steben]

I've been working on the same stupid theorem in Analysis for the past three weeks. I'm probably just being dumb and missing something obvious, but it's been really difficult... I'll type it below for you guys, at the risk of revealing my relative inexperience at proof-based mathematics.

Show that if H is a bounded infinite set, then there exists some sequence made of distinct points of H which is always increasing or always decreasing.

Is H a subset of the real numbers? If so, then with a bounded set, you have the least upper bound and greatest lower bound. There must exist sequences in H that converge to the least upper bound and greatest lower bound. Then, there exist subsequences of each sequences such that the subsequences are monotone and converge to the least upper bound and greatest lower bound, respectively.

Yeah, H is a subset of the reals.

I don't follow your argument, though. Let's say H is the union of {1-1/n|n is a natural number} and {1+1/n|n is a natural number}. It's bounded by 0 and 2, but I need a sequence that is always increasing or always decreasing, rather, x_n>x_n+1 always or x_n<x_n+1 always. No such sequence converges to 0 or 2... the only way I can find such a sequence is if it converges to 1... I can't see how talking about the least upper bound or greatest lower bound helps me..

Yes - I'm just talking about the existence, guaranteed by theorems. If you don't have the particular theorems that I'm referring to, then just prove them - if you have an element that is not the least upper bound or greatest lower bound, you can find another element closer, and keep constructing it that way.

Um. I need the sequence to be strictly increasing or strictly decreasing, not monotone. I think that's our delimma.

[Bahamut]

I've been working on the same stupid theorem in Analysis for the past three weeks. I'm probably just being dumb and missing something obvious, but it's been really difficult... I'll type it below for you guys, at the risk of revealing my relative inexperience at proof-based mathematics.

Show that if H is a bounded infinite set, then there exists some sequence made of distinct points of H which is always increasing or always decreasing.

Is H a subset of the real numbers? If so, then with a bounded set, you have the least upper bound and greatest lower bound. There must exist sequences in H that converge to the least upper bound and greatest lower bound. Then, there exist subsequences of each sequences such that the subsequences are monotone and converge to the least upper bound and greatest lower bound, respectively.

Yeah, H is a subset of the reals.

I don't follow your argument, though. Let's say H is the union of {1-1/n|n is a natural number} and {1+1/n|n is a natural number}. It's bounded by 0 and 2, but I need a sequence that is always increasing or always decreasing, rather, x_n>x_n+1 always or x_n<x_n+1 always. No such sequence converges to 0 or 2... the only way I can find such a sequence is if it converges to 1... I can't see how talking about the least upper bound or greatest lower bound helps me..

Yes - I'm just talking about the existence, guaranteed by theorems. If you don't have the particular theorems that I'm referring to, then just prove them - if you have an element that is not the least upper bound or greatest lower bound, you can find another element closer, and keep constructing it that way.

Um. I need the sequence to be strictly increasing or strictly decreasing, not monotone. I think that's our delimma.

If the least upper bound and the greatest lower bound isn't in your sets (i. e. if it's an open bounded set), then my method works in constructing a strictly increasing or decreasing sequence. Otherwise, that statement would just be plain false.

[Bobwillis]

I've looked over various Putnam problems (with their solutions handy), but I've never actually taken the test. I should look into it. I probably wouldn't solve any of them, but it'd be fun to try.

I wouldn't feel too bad about not solving any of the problems. I've heard that some years the median score is a zero.

[Bahamut]

I've looked over various Putnam problems (with their solutions handy), but I've never actually taken the test. I should look into it. I probably wouldn't solve any of them, but it'd be fun to try.

I wouldn't feel too bad about not solving any of the problems. I've heard that some years the median score is a zero.

That's usually because many high school students take the test.

[Bobwillis]

That's usually because many high school students take the test.

I thought it was for undergraduate mathematics and physics students, only.

EDIT: I just read the participation rules. Damn teenagers skewing the results :evil: .

[Bahamut]

That's usually because many high school students take the test.

I thought it was for undergraduate mathematics and physics students, only.

Some advanced public high schools will get their students to take the test I hear.

Edit: Yeah...I never actually took the Putnam myself. Some of those tests looked like I could've done well on though.

[Steben]

I've been working on the same stupid theorem in Analysis for the past three weeks. I'm probably just being dumb and missing something obvious, but it's been really difficult... I'll type it below for you guys, at the risk of revealing my relative inexperience at proof-based mathematics.

Show that if H is a bounded infinite set, then there exists some sequence made of distinct points of H which is always increasing or always decreasing.

Is H a subset of the real numbers? If so, then with a bounded set, you have the least upper bound and greatest lower bound. There must exist sequences in H that converge to the least upper bound and greatest lower bound. Then, there exist subsequences of each sequences such that the subsequences are monotone and converge to the least upper bound and greatest lower bound, respectively.

Yeah, H is a subset of the reals.

I don't follow your argument, though. Let's say H is the union of {1-1/n|n is a natural number} and {1+1/n|n is a natural number}. It's bounded by 0 and 2, but I need a sequence that is always increasing or always decreasing, rather, x_n>x_n+1 always or x_n<x_n+1 always. No such sequence converges to 0 or 2... the only way I can find such a sequence is if it converges to 1... I can't see how talking about the least upper bound or greatest lower bound helps me..

Yes - I'm just talking about the existence, guaranteed by theorems. If you don't have the particular theorems that I'm referring to, then just prove them - if you have an element that is not the least upper bound or greatest lower bound, you can find another element closer, and keep constructing it that way.

Um. I need the sequence to be strictly increasing or strictly decreasing, not monotone. I think that's our delimma.

If the least upper bound and the greatest lower bound isn't in your sets (i. e. if it's an open bounded set), then my method works in constructing a strictly increasing or decreasing sequence.
That's true. But I need it in general. For instance, the example above. It's not open bounded, so that proof doesn't help me.


I've looked over various Putnam problems (with their solutions handy), but I've never actually taken the test. I should look into it. I probably wouldn't solve any of them, but it'd be fun to try.

I wouldn't feel too bad about not solving any of the problems. I've heard that some years the median score is a zero.

I need to go back and look at some of those problems now that I've taken a math classes where we do actual proofs... maybe I could halfway get one right, if I try real hard.

Of course, the real problem with Putnam is that it's on a Saturday during football season. With any luck, I'll be at the SEC tournament that weekend.

[Bahamut]

I've been working on the same stupid theorem in Analysis for the past three weeks. I'm probably just being dumb and missing something obvious, but it's been really difficult... I'll type it below for you guys, at the risk of revealing my relative inexperience at proof-based mathematics.

Show that if H is a bounded infinite set, then there exists some sequence made of distinct points of H which is always increasing or always decreasing.

Is H a subset of the real numbers? If so, then with a bounded set, you have the least upper bound and greatest lower bound. There must exist sequences in H that converge to the least upper bound and greatest lower bound. Then, there exist subsequences of each sequences such that the subsequences are monotone and converge to the least upper bound and greatest lower bound, respectively.

Yeah, H is a subset of the reals.

I don't follow your argument, though. Let's say H is the union of {1-1/n|n is a natural number} and {1+1/n|n is a natural number}. It's bounded by 0 and 2, but I need a sequence that is always increasing or always decreasing, rather, x_n>x_n+1 always or x_n<x_n+1 always. No such sequence converges to 0 or 2... the only way I can find such a sequence is if it converges to 1... I can't see how talking about the least upper bound or greatest lower bound helps me..

Yes - I'm just talking about the existence, guaranteed by theorems. If you don't have the particular theorems that I'm referring to, then just prove them - if you have an element that is not the least upper bound or greatest lower bound, you can find another element closer, and keep constructing it that way.

Um. I need the sequence to be strictly increasing or strictly decreasing, not monotone. I think that's our delimma.

If the least upper bound and the greatest lower bound isn't in your sets (i. e. if it's an open bounded set), then my method works in constructing a strictly increasing or decreasing sequence.
That's true. But I need it in general. For instance, the example above. It's not open bounded, so that proof doesn't help me.

Oh wait you have an infinite set. The real numbers have the well-ordering property, so you can take that set and find a sequence and order it so that it is always increasing or decreasing then. This argument is quite strong though...(uses the Axiom of Choice)

[Steben]

Oh wait you have an infinite set. The real numbers have the well-ordering property, so you can take that set and find a sequence and order it so that it is always increasing or decreasing then. This argument is quite strong though...(uses the Axiom of Choice)
But what if the sequence I chose was {1/4, 1/1, 1/6, 1/1930, 1/205, -4, 1/356!, ...}. I can't order it so that it is always decreasing, because of that -4. Eventually I'll have to use that -4, and then the next number will be greater than it. All that to say, the axiom of choice only lets me pick arbitrary numbers out of the sequence (I think?), and that -4 makes life difficult.

That, and, we haven't talked about the Axiom of Choice in class, so I can't use it. Dang.

The method I've been attempting to use is to prove that any infinite bounded set has a limit point, and constructing a sequence from that. Unfortunately, I'm having trouble proving any infinite bounded set has a limit point without using the fact that any infinit bounded set has a sequence which is increasing or decreasing. Whoops.

[Luna child]

I only made it to Geometry.. :(

[Bahamut]

Oh wait you have an infinite set. The real numbers have the well-ordering property, so you can take that set and find a sequence and order it so that it is always increasing or decreasing then. This argument is quite strong though...(uses the Axiom of Choice)
But what if the sequence I chose was {1/4, 1/1, 1/6, 1/1930, 1/205, -4, 1/356!, ...}. I can't order it so that it is always decreasing, because of that -4. Eventually I'll have to use that -4, and then the next number will be greater than it. All that to say, the axiom of choice only lets me pick arbitrary numbers out of the sequence (I think?), and that -4 makes life difficult.

Edit: Hmm, I'd have to think on it more. That -4 should be irrelevant though.

The method I've been attempting to use is to prove that any infinite bounded set has a limit point, and constructing a sequence from that. Unfortunately, I'm having trouble proving any infinite bounded set has a limit point without using the fact that any infinit bounded set has a sequence which is increasing or decreasing. Whoops.

Any subset of the real numbers has a limit point - each element is a limit point. That doesn't help you at all though. I'd think more on this, but I got a test tomorrow so maybe tomorrow night I can think on that more.

[Julio Jose]

I've grown to hate math. The day I entered my Abstract Algebra 3 class, and my Linear Algebra 4 class, I came to the conclusion that, holy shit math is a pain in the ass.
Only in small doses.

He sees why its a pain. Sometimes the methods of proofs just completely elude you even though it looks as though it should be simple. Sometimes it's just a lot of busy work too.
It's a little of both. Sometimes in a proof, you will spend so much time working backwards to move forwards, it's not even funny.

Other times, it's because there is just a ridiculus amount of work required by these classes. Assignments really do take 2 weeks at a time. You can't just bs around and do them 1 or 2 days before they are do. It takes a lot of commitment, and as someone who isn't even majoring in the pmat field, it's kind of crazy.

[Steben]

The method I've been attempting to use is to prove that any infinite bounded set has a limit point, and constructing a sequence from that. Unfortunately, I'm having trouble proving any infinite bounded set has a limit point without using the fact that any infinit bounded set has a sequence which is increasing or decreasing. Whoops.

Any subset of the real numbers has a limit point - each element is a limit point. That doesn't help you at all though. I'd think more on this, but I got a test tomorrow so maybe tomorrow night I can think on that more.

The definition of a limit point we're using is that "x is a limit point of a set H iff any segment about x contains an element of H distinct from x". Any subset of the real numbers contains a limit point of the real numbers, but I need a limit point of the set itself.

[Steben]

Here's some calculus, because I know it'll make your day. An engineering friend asked me if this problem was possible, I said I didn't think so, but I'm also incredibly dumb, so maybe you can? Here's what he asked.

Is it possible to simplify int[(u')^2]dx to an expression that doesn't use an integral symbol, where u is a function of x and u' is its first derivative?

[Sahaquiel Delta]

Bleah. I'm only in Calculus I. (For the second time. I hate the public high school system. I mean, I got a damn A the first time...)

All you advanced college peoples make me sad (that I'm not in Calc II and such).

[Ab56 v2 aka Ash]

I don't think calc 2 would teach you how to integrate that function; it doesn't look like integration by parts would simplify it. I'm intrigued though :D

[Jabe]

Is it possible to simplify int[(u')^2]dx to an expression that doesn't use an integral symbol, where u is a function of x and u' is its first derivative?

I don't think so because u is a generic function. Without any specifics, you can't determine the output of int[(u')^2]dx without resorting to just evaluating on a case-to-case basis. For example, there is no way to explain the following:
If u(x)=x then the function (int[(u')^2]dx) = x+C or possibly u'(x)*u(x)
If u(x)=ln(x) then the function = -1/x+C or possibly -u'(x)

Not really a proof, but it's what I say.

[Steben]

Is it possible to simplify int[(u')^2]dx to an expression that doesn't use an integral symbol, where u is a function of x and u' is its first derivative?

I don't think so because u is a generic function. Without any specifics, you can't determine the output of int[(u')^2]dx without resorting to just evaluating on a case-to-case basis. For example, there is no way to explain the following:
If u(x)=x then the function (int[(u')^2]dx) = x+C or possibly u'(x)*u(x)
If u(x)=ln(x) then the function = -1/x+C or possibly -u'(x)

Not really a proof, but it's what I say.

That's the reasoning I used.

I don't think calc 2 would teach you how to integrate that function; it doesn't look like integration by parts would simplify it. I'm intrigued though :D

It doesn't. Here's what happened when I tried integration by parts.

int((u')^2)dx = int(u'u')dx
= uu' - int(u''u)dx
= uu' - (uu' - int(u'u')dx)
= int(u'u')dx
= int((u')^2)dx

[Bahamut]

Here's some calculus, because I know it'll make your day. An engineering friend asked me if this problem was possible, I said I didn't think so, but I'm also incredibly dumb, so maybe you can? Here's what he asked.

Is it possible to simplify int[(u')^2]dx to an expression that doesn't use an integral symbol, where u is a function of x and u' is its first derivative?

Think of differential equations. In this case, the equation would be f' = (u')^2. I can't think of any generalized solution to this, can you?

[Steben]

Here's some calculus, because I know it'll make your day. An engineering friend asked me if this problem was possible, I said I didn't think so, but I'm also incredibly dumb, so maybe you can? Here's what he asked.

Is it possible to simplify int[(u')^2]dx to an expression that doesn't use an integral symbol, where u is a function of x and u' is its first derivative?

Think of differential equations. In this case, the equation would be f' = (u')^2. I can't think of any generalized solution to this, can you?
Aggh.. diff eq was so long ago. I'm gonna pull out my text book from that class now..

And still no dice. Ah well.

[Ab56 v2 aka Ash]

Could you power series this sucker?

[Bahamut]

Power series looks too ugly for this to work, since you'd have to multiply two power series, and that doesn't go well for solving differential equations.

Edit: Alright so here's a math problem for you people, and if you're able to solve it by Sunday @ midnight, then I will paypal you $20. Let m be a fixed natural number. Find the infinite sum of 1/(m * n)^s, where s is just a complex exponent (it's not important), and n is all such numbers such that (m, n) = 1, or in otherwords, m & n are relatively prime/coprime.

Edit #2: Expressions with the infinite sum of 1/k^s for k = 1, 2, .... = zeta(s) btw, so the infinite sum of 1/k^(2s) = zeta(2s) for example, where zeta(s) is the Riemann zeta function.

[Bahamut]

Another doublepost!

But I figured out how to do this problem, so it is solvable. The offer is still out there I guess for the ambitious people, but I highly doubt anyone on the forums will figure out how to do it, and if you do, you might want to consider going into math.

[Hadriel]

The Zeta function seems like something we should have studied during all the hoopla about sequences and series back in Calc II. Yet we never so much as touched it. :(

Isn't it supposed to represent the distribution of prime numbers?

Nggh, this sounds like something that'd be simple if I had another year of schooling in me.

[Bahamut]

It does have to do with the distribution of prime numbers - in fact, the Riemann zeta function is used in several proofs of the Prime Number Theorem, including the first proof. The PNT says that pi(x), the number of prime numbers less than or equal to x, is asymptotic to x/log x, also written as pi(x) ~ x/log x (since being asymptotic is an equivalence relation). Two functions f(x) & g(x) are asymptotic if lim f(x)/g(x) = 1, where the limit is as x is approaching infinity. It turns out that 1/zeta(1) = 0 is equivalent to the Prime Number Theorem. The zeta function isn't discussed in Calc 2 because it is impractical for most students in Calc 2, such as engineers & pre-med students. Also, the uses of the zeta function are considered fairly advanced, since its definition involves complex numbers and so anything involving the zeta function requires complex analysis (i.e. concepts of analytic continuation and the theory of Dirchlet series).

[Bahamut]

Power series looks too ugly for this to work, since you'd have to multiply two power series, and that doesn't go well for solving differential equations.

Edit: Alright so here's a math problem for you people, and if you're able to solve it by Sunday @ midnight, then I will paypal you $20. Let m be a fixed natural number. Find the infinite sum of 1/(m * n)^s, where s is just a complex exponent (it's not important), and n is all such numbers such that (m, n) = 1, or in otherwords, m & n are relatively prime/coprime.

Edit #2: Expressions with the infinite sum of 1/k^s for k = 1, 2, .... = zeta(s) btw, so the infinite sum of 1/k^(2s) = zeta(2s) for example, where zeta(s) is the Riemann zeta function.

Bumptity bump!

So I will spoil this ahead of time, but the answer actually is just 2 * zeta(s) - 1. Sorry, no $20 for anyone.

[Bahamut]

Oo I get a triple post!

But here's a fun question for people: What's wrong in the following work?

x^2 = x * x = x + x + ... + x (x times)
d(x^2)/dx = d(x)/dx + ... + d(x)/dx (x times)
2x = 1 + 1 + ... + 1 (x times)
2x = x

So what went wrong here?

[Fionn]

d(x^2)/dx = d(x)/dx + ... + d(x)/dx (x times)


So is that the second derivative? Or the derivative of x squared?

Whatever. It's too late for me to think right now. I'll sleep on it.


EDIT: It's the derivative of x squared. Unless that assumption is the problem with the math. :idea:

Because I'd express that as:

d(x^2)/dx [(x^2)] = d(x)/dx + ... + d(x)/dx (x times)

[zarvoxthezadvook]

here's my best guess:

you seem to be treating x as an integer constant, in which case, d(x)/dx = 0, so x* d(x)/dx = 2x* d(x)/dx = 0

EDIT: scratch that.

d(x*x) != x * d(x)/dx

rather,
d(x*x) = x * d(x)/dx + x * d(x)/dx = 2x*d(x)/dx

[Ab56 v2 aka Ash]

^Pretty close.

Answer's on wikipedia

[Steben]

Before I check it against Wikipedia...

x^2 != x+x+...+x in general.

This only holds when x is a natural number. You cannot take the derivative of a function whose domain isn't some interval.

EDIT: Yeah, I have the same argument as the Wikipedians.

[Bahamut]

Before I check it against Wikipedia...

x^2 != x+x+...+x in general.

This only holds when x is a natural number. You cannot take the derivative of a function whose domain isn't some interval.

EDIT: Yeah, I have the same argument as the Wikipedians.

That is not the complete explanation. Also, any continuous integer-valued function must be constant, so if you restrict x^2 to just integers, it cannot be continuous, so you cannot differentiate it.

[Bahamut]

It's that double post time again! So here's another fun question of a more competition variety:

(USAMO 2003) Show that for each n we can find an n-digit number with all its digits odd which is divisible by 5^n.

I'll probably start posting a problem every few days, and I'll post solutions too. For this problem, I think I have the solution, so we'll see.

[Steben]

I constructed the appropriate numbers for 1<=n<=10 if people want to see. x is the appropriate n-digit number for each n.

n=1 5^n=5 x=5=5^n
n=2 5^n=25 x=75=3*5^n
n=3 5^n=125 x=375=3*5^n
n=4 5^n=625 x=9375=15*5^n
n=5 5^n=3125 x=59375=19*5^n
n=6 5^n=15625 x=359375=23*5^n
n=7 5^n=78125 x=3359375=43*5^n
n=8 5^n=390625 x=93359375=239*5^n
n=9 5^n=1953125 x=193359375=99*5^n
n=10 5^n=9765625 x=3193359375=327*5^n

I smell a pattern.

[Bahamut]

I constructed the appropriate numbers for 1<=n<=8 if people want to see. x is the appropriate n-digit number for each n.

n=1 5^n=5 x=5=5^n
n=2 5^n=25 x=75=3*5^n
n=3 5^n=125 x=375=3*5^n
n=4 5^n=625 x=9375=15*5^n
n=5 5^n=3125 x=59375=19*5^n
n=6 5^n=15625 x=359375=23*5^n
n=7 5^n=78125 x=3359375=43*5^n
n=8 5^n=390625 x=93359375=239*5^n

I smell a pattern.

*Whistles*

[Steben]

Yeah, so it's easy to brute force the number given the previous number that works. But how do I show that it works for any n?

I'm sleeping on it.

[Steben]

Well GEE DEE, guys. This is proving harder than I'd like.

The general logic I'm trying to use is this: Let x_n be the sequence of n-digit numbers that fit the hypothesis. Then, I want to inductively generate x_n+1 from x_n. If there's any justice in the world, x_n+1 should equal either x_n+1*10^n, x_n+3*10^n, x_n+5*10^n, x_n+7*10^n, or x_n+9*10^n. If x_n is divisible by 5^n+1, then x_n+1=x_n+10^n*5. But if not, I'm having difficulty figuring out which of the other four numbers to use. I tried looking at x_n/5^n mod 5, with no luck at a pattern...

[CHz]

Well GEE DEE, guys. This is proving harder than I'd like.

The general logic I'm trying to use is this: Let x_n be the sequence of n-digit numbers that fit the hypothesis. Then, I want to inductively generate x_n+1 from x_n. If there's any justice in the world, x_n+1 should equal either x_n+1*10^n, x_n+3*10^n, x_n+5*10^n, x_n+7*10^n, or x_n+9*10^n. If x_n is divisible by 5^n+1, then x_n+1=x_n+10^n*5. But if not, I'm having difficulty figuring out which of the other four numbers to use. I tried looking at x_n/5^n mod 5, with no luck at a pattern...
Well, 10^n = 2^n * 5^n, so 5^n divides x_n + 1 * 10^n, x_n + 3 * 10^n, ..., x_n + 9 * 10^n because 5^n|x_n by inductive hypothesis hypothesis. Divide all those by 5^n, and let w = x_n / 5^n, and we get w + 1 * 2^n, w + 3 * 2^n, w + 5 * 2^n, w + 7 * 2^n, and w + 9 * 2^n. 2^n is relatively prime to 5, so {w + 1 * 2^n, w + 3 * 2^n, w + 5 * 2^n, w + 7 * 2^n, w + 9 * 2^n} is a residue system module 5 because 1, 3, 5, 7, and 9 mod 5 are 1, 3, 0, 2, and 4 respectively. Thus, one of them has to be congruent to 0 mod 5 and therefore divisible by 5, which would make one of x_n + 1 * 10^n, ..., x_n + 9 * 10^n divisible by 5 * 5^n = 5^(n + 1). So, this is our x_(n + 1).

I think that works?

[Bahamut]

Well GEE DEE, guys. This is proving harder than I'd like.

The general logic I'm trying to use is this: Let x_n be the sequence of n-digit numbers that fit the hypothesis. Then, I want to inductively generate x_n+1 from x_n. If there's any justice in the world, x_n+1 should equal either x_n+1*10^n, x_n+3*10^n, x_n+5*10^n, x_n+7*10^n, or x_n+9*10^n. If x_n is divisible by 5^n+1, then x_n+1=x_n+10^n*5. But if not, I'm having difficulty figuring out which of the other four numbers to use. I tried looking at x_n/5^n mod 5, with no luck at a pattern...
Well, 10^n = 2^n * 5^n, so 5^n divides x_n + 1 * 10^n, x_n + 3 * 10^n, ..., x_n + 9 * 10^n because 5^n|x_n by inductive hypothesis hypothesis. Divide all those by 5^n, and let w = x_n / 5^n, and we get w + 1 * 2^n, w + 3 * 2^n, w + 5 * 2^n, w + 7 * 2^n, and w + 9 * 2^n. 2^n is relatively prime to 5, so {w + 1 * 2^n, w + 3 * 2^n, w + 5 * 2^n, w + 7 * 2^n, w + 9 * 2^n} is a residue system module 5 because 1, 3, 5, 7, and 9 mod 5 are 1, 3, 0, 2, and 4 respectively. Thus, one of them has to be congruent to 0 mod 5 and therefore divisible by 5, which would make one of x_n + 1 * 10^n, ..., x_n + 9 * 10^n divisible by 5 * 5^n = 5^(n + 1). So, this is our x_(n + 1).

I think that works?

Almost correct. The truth is, you don't know which is divisible by 5. You just know that one of them is divisible by 5, and that is all you need.


Alright, next problem:

Prove e, the infinite sum from n = 0 to infinity of 1/n!, is irrational.

[Bahamut]

Bumpty bump - no takers on this question? I'll give one hint - show that the finite sum from n = 0 to N of 1/n! < e < 1/(n * N!) + the finite sum from n = 0 to N of 1/n!

From there the nifty fun occurs. :)

[Steben]

Bumpty bump - no takers on this question? I'll give one hint - show that the finite sum from n = 0 to N of 1/n! < e < 1/(n * N!) + the finite sum from n = 0 to N of 1/n!

From there the nifty fun occurs. :)
Did you type that out right? Because I'm not sure what 1/(n * N!) means, since you're using n as the place holder for the summation expressions... unless I'm just being dumb.

[Bahamut]

Bumpty bump - no takers on this question? I'll give one hint - show that the finite sum from n = 0 to N of 1/n! < e < 1/(n * N!) + the finite sum from n = 0 to N of 1/n!

From there the nifty fun occurs. :)
Did you type that out right? Because I'm not sure what 1/(n * N!) means, since you're using n as the place holder for the summation expressions... unless I'm just being dumb.

Oh yeah, the 1/(n * N!) should be part of the summation, so think of it as a summation of 1/n! + 1/(n * N!)

[esperz]

Bumpty bump - no takers on this question? I'll give one hint - show that the finite sum from n = 0 to N of 1/n! < e < 1/(n * N!) + the finite sum from n = 0 to N of 1/n!

From there the nifty fun occurs. :)
Did you type that out right? Because I'm not sure what 1/(n * N!) means, since you're using n as the place holder for the summation expressions... unless I'm just being dumb.

Oh yeah, the 1/(n * N!) should be part of the summation, so think of it as a summation of 1/n! + 1/(n * N!)

If n = 0

1/(n * N!) = Invalid

[Bahamut]

Bumpty bump - no takers on this question? I'll give one hint - show that the finite sum from n = 0 to N of 1/n! < e < 1/(n * N!) + the finite sum from n = 0 to N of 1/n!

From there the nifty fun occurs. :)
Did you type that out right? Because I'm not sure what 1/(n * N!) means, since you're using n as the place holder for the summation expressions... unless I'm just being dumb.

Oh yeah, the 1/(n * N!) should be part of the summation, so think of it as a summation of 1/n! + 1/(n * N!)

If n = 0

1/(n * N!) = Invalid

Err...fuck, it's 1/(N * N!), and this is outside of the summation.

[Bahamut]

Alright bump time - time for me to spoil this problem.

So, the sum from n = 0 to N of 1/n! is always < e for all N. Now, 1/(N * N!) * sum(1/n!) - e (where sum(1/n!) is from n = 0 to N) = 1/(N * N!) - sum(1/n!, n > N) = 1/N! * (1/N - 1/(N+1) - 1/((N + 1)(N + 2)) - ...) = 1/N! * (1/(N * (N + 1)) - 1/((N + 1)(N + 2)) - 1/((N + 1)(N + 2)(N + 3) - ...), which can be shown to be > 0.

Now, the fun part is suppose e is rational. Then e = p/q, where p is an integer and q is a natural number. Let N = q

Then the sum from n = 0 to q of 1/n! < p/q < sum from n = 0 to q of 1/n! + 1/(q * q!)

Then 0 < p/q - sum from n = 0 to q of 1/n! < 1/(q * q!)

Then 0 < p * q! - (q + q * (q - 1) + ... + q!) < 1

However, the middle part is an integer, so that's a contradiction.


Next question:

(AIME 2001 Version II Problem 3) The sequence a1, a2, a3, ... is defined by a1 = 211, a2 = 375, a3 = 420, a4 = 523, an = an-1 - an-2 + an-3 - an-4. Find a531 + a753 + a975.

[CHz]

Love that proof, Baha. 8)

Next question:

(AIME 2001 Version II Problem 3) The sequence a1, a2, a3, ... is defined by a1 = 211, a2 = 375, a3 = 420, a4 = 523, an = an-1 - an-2 + an-3 - an-4. Find a531 + a753 + a975.
I'm rather surprised no one went for this, since this one just falls right out when you start calculating values. =O

a_5 = a_4 - a_3 + a_2 - a_1 = 523 - 420 + 375 - 211 = 267
a_6 = a_5 - a_4 + a_3 - a_2 = 267 - 523 + 420 - 375 = -211 = -a_1
a_7 = a_6 - a_5 + a_4 - a_3 = -211 - 267 + 523 - 420 = -375 = -a_2
a_8 = a_7 - a_6 + a_5 - a_4 = -375 - (-211) + 267 - 523 = -420 = -a_3
a_9 = a_8 - a_7 + a_6 - a_5 = -420 - (-375) + (-211) - 267 = -523 = -a_4
a_10 = a_9 - a_8 + a_7 - a_6 = -(a_4 - a_3 + a_2 - a_1) = -a_5
a_11 = a_10 - a_9 + a_8 - a_7 = -(a_5 - a_4 + a_3 - a_2) = -a_6 = a_1
a_12 = -a_7 = a_2
.
.
.
a_(10k + i) = a_i if 1 <= i <= 5, -a_(i - 5) if 6 <= i <= 10, for k >= 0.

So, a_531 + a_753 + a_975 = a_1 + a_3 + a_5 = 211 + 420 + 267 = 898.

[Bobwillis]

cos(cos(cos(cosx)))=sin(sin(sin(sinx)))

I saw this in my school's newspaper yesterday....I suck at trig :(

Apparently my simplification skills are shitty.

[Bahamut]

cos(cos(cos(cosx)))=sin(sin(sin(sinx)))

I saw this in my school's newspaper yesterday....I suck at trig :(

Apparently my simplification skills are shitty.

Eww that looks disgusting. I'm too lazy to do that one.

[Steben]

cos(cos(cos(cosx)))=sin(sin(sin(sinx)))

I saw this in my school's newspaper yesterday....I suck at trig :(

Apparently my simplification skills are shitty.

Eww that looks disgusting. I'm too lazy to do that one.
I'm too lazy to do that one, because I'm fairly certain it's not true.

Actually, scratch that. Positive.

sin(sin(sin(sin(0))))=0
cos(cos(cos(cos(0))))=0.65428979...

[Bahamut]

cos(cos(cos(cosx)))=sin(sin(sin(sinx)))

I saw this in my school's newspaper yesterday....I suck at trig :(

Apparently my simplification skills are shitty.

Eww that looks disgusting. I'm too lazy to do that one.
I'm too lazy to do that one, because I'm fairly certain it's not true.

Actually, scratch that. Positive.

sin(sin(sin(sin(0))))=0
cos(cos(cos(cos(0))))=0.65428979...

How does plugging in 0 prove that it is not true?

Edit: After plugging in numbers, I don't think it's true since cos cos cos cos x is close to 1 for almost all x (and greater than 0.5 for all x), and sin sin sin sin x is always close to 0. Edit: oops I was using degrees

[Steben]

cos(cos(cos(cosx)))=sin(sin(sin(sinx)))

I saw this in my school's newspaper yesterday....I suck at trig :(

Apparently my simplification skills are shitty.

Eww that looks disgusting. I'm too lazy to do that one.
I'm too lazy to do that one, because I'm fairly certain it's not true.

Actually, scratch that. Positive.

sin(sin(sin(sin(0))))=0
cos(cos(cos(cos(0))))=0.65428979...

How does plugging in 0 prove that it is not true?

Edit: After plugging in numbers, I don't think it's true since cos cos cos cos x is close to 1 for almost all x (and greater than 0.5 for all x), and sin sin sin sin x is always close to 0. Edit: oops I was using degrees
Um. Proof by counterexample? If sin(sin(sin(sinx))))=cos(cos(cos(cosx)))) is true, it must certainly be true when x=0. However, they aren't equal when x=0, so it can't be true for all x.

[Bahamut]

cos(cos(cos(cosx)))=sin(sin(sin(sinx)))

I saw this in my school's newspaper yesterday....I suck at trig :(

Apparently my simplification skills are shitty.

Eww that looks disgusting. I'm too lazy to do that one.
I'm too lazy to do that one, because I'm fairly certain it's not true.

Actually, scratch that. Positive.

sin(sin(sin(sin(0))))=0
cos(cos(cos(cos(0))))=0.65428979...

How does plugging in 0 prove that it is not true?

Edit: After plugging in numbers, I don't think it's true since cos cos cos cos x is close to 1 for almost all x (and greater than 0.5 for all x), and sin sin sin sin x is always close to 0. Edit: oops I was using degrees
Um. Proof by counterexample? If sin(sin(sin(sinx))))=cos(cos(cos(cosx)))) is true, it must certainly be true when x=0. However, they aren't equal when x=0, so it can't be true for all x.

But isn't the question to find such an x? If it is not true for x = 0, there is no reason why it can't be true for another x.

[Bobwillis]

If you plot the graphs they do not intersect.

And using x = 0 will not help.

[Steben]

cos(cos(cos(cosx)))=sin(sin(sin(sinx)))

I saw this in my school's newspaper yesterday....I suck at trig :(

Apparently my simplification skills are shitty.

Eww that looks disgusting. I'm too lazy to do that one.
I'm too lazy to do that one, because I'm fairly certain it's not true.

Actually, scratch that. Positive.

sin(sin(sin(sin(0))))=0
cos(cos(cos(cos(0))))=0.65428979...

How does plugging in 0 prove that it is not true?

Edit: After plugging in numbers, I don't think it's true since cos cos cos cos x is close to 1 for almost all x (and greater than 0.5 for all x), and sin sin sin sin x is always close to 0. Edit: oops I was using degrees
Um. Proof by counterexample? If sin(sin(sin(sinx))))=cos(cos(cos(cosx)))) is true, it must certainly be true when x=0. However, they aren't equal when x=0, so it can't be true for all x.

But isn't the question to find such an x? If it is not true for x = 0, there is no reason why it can't be true for another x.

Oh. Crap. I've spent too much time in proofs-based math... I thought it was stating an identity to prove, not an equation.

I'm dumb. :P

[Psychonaut]

One thing that is very important here is to NOT let the first exam let you down. It's always demoralizing to have a bad start in a class, but it's not the end of the class.

I haven't read most of the thread, but I have to say "amen" to this.

I got a 20% on my first exam (by far my lowest exam score EVER) in Introduction to Scientific Computing. While that basically nuked my chances of an A in the class, I still managed to acquit myself admirably, and since I had exactly, nail-on-the-head enough credits to graduate a year early the year following, it was good that I stuck with it. Because I graduated at that specific time, I'm now going to be able to have a master's degree in four years.

[Bahamut]

Alright, so here's a math question for math people.

Suppose A, B are normal subgroups of a group G.
Suppose G/A and G/B are abelian.

Prove that G/(A intersect B) is abelian.

[Effector]

On Bahamut's topic, I was thinking about groups/subgroups the other day.

For example, it is possible to have both the group and the subgroup to have infinite possibilities, right? (I was thinking along the lines of secants and diameters of circles. There are an infinite number of possible secants, and an infinite number of possible diameters. However, all diameters are secants, but not all secants diameters.)

So, how does having infinity as a subset work out? It almost seems to indicate that inf > inf in the secant-diameter problem, and therefore inf != inf? Is that technically true? (I'm not well versed in the abstract qualities of infinity - but don't things like "inf - 2*inf = inf" work out?)

[Bahamut]

Well, think of a simple example, like the integers Z. Z, 2Z, 3Z, etc. are all subgroups of Z, and in those cases, they are isomorphic to Z sub n, the cyclic group under addition modulo n. Also, there exists other types of infinite groups, such as the infinite alternating group.

Mathematicians treat infinity differently though - there are different levels of infinity. Set size is denoted by the cardinality of a set, and two sets have the same cardinality if there exists a bijection between them. The partial order that is assigned to cardinality is the less than or equal to symbol, in which a set A has smaller cardinality than a set B if there exists an injection from A to B. When talking about infinite sets though, you can have more interesting things happen. For example, take the set of natural numbers, and take the subset of all even natural numbers. Surely there exists a bijection between the two (notably mapping every natural number n to 2n), so they have the same cardinality, even though the set of all even natural numbers is surely a subset of the natural numbers. So, the cardinality of the natural numbers is the smallest cardinality that a set can have.

Also noteworthy is that the real numbers have a strictly larger cardinality than the natural numbers, since the interval (0, 1) has a bijection into the real numbers, and the cardinality of (0, 1) is not equal to the cardinality of the natural numbers since there does not exist a bijection between the two (the famous Cantor diagonalization argument is used to prove this).


Also, I figured out that question I just posed, it's quite easy.

[Effector]

Oh! I understand it now. Between you and wikipedia, I think I comprehend it a bit better. The cardinality was what was confusing me, I think.

Oh, so anyway. A few years ago, a few students had to find the center of a circle or something. Basically, they wrote a arctan program, and I remember a little bit of it, but it didn't make a bit of sense. I think it is this, from wiki:

http://upload.wikimedia.org/math/7/9/a/79a2a1a2beefcbe3c6556528fc976d79.png

But I'm not sure. Why does that work, anyway? Is it because the difference between the two approach ±(π/2)?

[AbyssWyrm]

Sorry to nitpick...

Well, think of a simple example, like the integers Z. Z, 2Z, 3Z, etc. are all subgroups of Z, and in those cases, they are isomorphic to Z sub n, the cyclic group under addition modulo n.

Hmm? nZ is isomorphic to Z, but Z/nZ is isomorphic to Z_n.

So, the cardinality of the natural numbers is the smallest cardinality that a set can have.

Smallest cardinality an infinite set can have.

[Bahamut]

See, minor mistakes in writing what I have in mind like that is what costs me huge on tests :( .

[HalcyonSpirit]

A problem for all you Math people out there.

I need to find the sum of the following series:
http://img.photobucket.com/albums/v37/Galactic_Alliance_Jedi_Knight/Miscellaneous/Problem1.png
I know that if the starting condition was n=0, the sum would be equal to e^5. However, it has a starting point of n=1. So how do you find the sum of it?

[AbyssWyrm]

A problem for all you Math people out there.

I need to find the sum of the following series:
http://img.photobucket.com/albums/v37/Galactic_Alliance_Jedi_Knight/Miscellaneous/Problem1.png
I know that if the starting condition was n=0, the sum would be equal to e^5. However, it has a starting point of n=1. So how do you find the sum of it?

...maybe try subtracting 1?

[HalcyonSpirit]

A problem for all you Math people out there.

I need to find the sum of the following series:
http://img.photobucket.com/albums/v37/Galactic_Alliance_Jedi_Knight/Miscellaneous/Problem1.png
I know that if the starting condition was n=0, the sum would be equal to e^5. However, it has a starting point of n=1. So how do you find the sum of it?

...maybe try subtracting 1?

Unless I'm misunderstanding what you're saying, no... subtracting 1 from n wouldn't do anything. For the series to be equal to e^5, the starting condition must be n=0, and the n's can't be (n-1), (n+1), or anything else except for n by itself. That's why I'm asking about the problem. Either there's a way to get it into a form that would allow it to be equated to e^x (x being some number in the summation), or there's a test that has to be implemented to find the sum. I can't figure out either of them.

[Rainman DX]

A problem for all you Math people out there.

I need to find the sum of the following series:
http://img.photobucket.com/albums/v37/Galactic_Alliance_Jedi_Knight/Miscellaneous/Problem1.png
I know that if the starting condition was n=0, the sum would be equal to e^5. However, it has a starting point of n=1. So how do you find the sum of it?

...maybe try subtracting 1?

Actually, I'm pretty sure that's correct.

The sum, as you mentioned, looks a lot like the infinite sum representation of e^x. As such, it behaves the same way, but the one you have shown here is missing the initial term of n=0. So you can merely subtract (5^0)/0!, which is 1. Therefore, the sum converges at e^5 - 1.

[HalcyonSpirit]

A problem for all you Math people out there.

I need to find the sum of the following series:
http://img.photobucket.com/albums/v37/Galactic_Alliance_Jedi_Knight/Miscellaneous/Problem1.png
I know that if the starting condition was n=0, the sum would be equal to e^5. However, it has a starting point of n=1. So how do you find the sum of it?

...maybe try subtracting 1?

Actually, I'm pretty sure that's correct.

The sum, as you mentioned, looks a lot like the infinite sum representation of e^x. As such, it behaves the same way, but the one you have shown here is missing the initial term of n=0. So you can merely subtract (5^0)/0!, which is 1. Therefore, the sum converges at e^5 - 1.

...

You know, this is exactly why I hate this Sequence and Series stuff we've been learning. I just can't grasp how to do it myself, but when someone explains a problem to be, it seems so freakin' simple! Yet when I go to do a similar problem, I still can't see the obvious answer.

Thanks guys, much appreciated.

[Rainman DX]

Thanks guys, much appreciated.

Anything to help a fellow GA Jedi Knight.

[Steben]

Alright, so here's a math question for math people.

Suppose A, B are normal subgroups of a group G.
Suppose G/A and G/B are abelian.

Prove that G/(A intersect B) is abelian.

Just finished Intro to Abstract Algebra I. My professor freaking sucked. Fortunately his tests were ridiculously easy, so I got an A without learning a single thing.

Let's see if I can figure this out... I'll let C=(A intersect B).

To show G/C is abelian, I must show that ab=ba for a,b in G/C. a=gC and b=hC for g,h in G. ab=gChC=gCh as C is normal. gCh=(gAh intersect gBh)=(hAg intersect hBg)=hCg=hCgC=ba. Done!

Eh. Does that work? I've never had to write a solid proof for Abstract Algebra before, so I'm not really confident in it.

[Bahamut]

Alright, so here's a math question for math people.

Suppose A, B are normal subgroups of a group G.
Suppose G/A and G/B are abelian.

Prove that G/(A intersect B) is abelian.

Just finished Intro to Abstract Algebra I. My professor freaking sucked. Fortunately his tests were ridiculously easy, so I got an A without learning a single thing.

Let's see if I can figure this out... I'll let C=(A intersect B).

To show G/C is abelian, I must show that ab=ba for a,b in G/C. a=gC and b=hC for g,h in G. ab=gChC=gCh as C is normal. gCh=(gAh intersect gBh)=(hAg intersect hBg)=hCg=hCgC=ba. Done!

Eh. Does that work? I've never had to write a solid proof for Abstract Algebra before, so I'm not really confident in it.

Alright, so first you have to show that A intersect B is normal in G. Let x be in A intersect B. When you conjugate x by any element g in G, then x will be in A by normality of A and in B by normality of B, so x is in the intersection of A and B.

Next, let x(A intersect B) and y(A intersect B) be in G/(A intersect B) (coset representation). Then xy(A intersect B) = yx(A intersect B since xyA = yxA and xyB = yxB, and the intersection of xyA and xyB is xy(A intersect B) and similarly yxA intersect yxB = yx(A intersect B).

Your argument fails because gC * hC = (gh)C by definition of coset multiplication in a quotient group.


Another problem!

Suppose U is a finite multiplicative subgroup of a finite field F.

Prove that U must be cyclic.

[Rainman DX]

If I were sticking with my math major, I'd be taking Abstract Algebra I next semester, along with Numerical Analysis I...

But a math minor'll have to do instead. :oops:

[Steben]

Just got out of my Analysis exam. Here's a problem that was pretty easy, but I thought it was fun anyway.

Let f:H->K be a function. Show that if x is in H but isn't a limit point of H, then f is continuous at x.

[Bahamut]

Just got out of my Analysis exam. Here's a problem that was pretty easy, but I thought it was fun anyway.

Let f:H->K be a function. Show that if x is in H but isn't a limit point of H, then f is continuous at x.

f is any old function? And what properties does H have? Connectedness?

[Steben]

Just got out of my Analysis exam. Here's a problem that was pretty easy, but I thought it was fun anyway.

Let f:H->K be a function. Show that if x is in H but isn't a limit point of H, then f is continuous at x.

f is any old function? And what properties does H have? Connectedness?
You shouldn't need anything else. You're just considering f's continuity at a point of H which is not a limit point of H.

[Bahamut]

Just got out of my Analysis exam. Here's a problem that was pretty easy, but I thought it was fun anyway.

Let f:H->K be a function. Show that if x is in H but isn't a limit point of H, then f is continuous at x.

f is any old function? And what properties does H have? Connectedness?
You shouldn't need anything else. You're just considering f's continuity at a point of H which is not a limit point of H.

I guess since x is in the interior of H, one can take a ball around it that is contained in the interior of H. Using the fact that a function can only have a countable number of discontinuities, this means that the function must be piecewise continuous, and so must be continuous about x. This argument seems stronger than needed though.

[Steben]

Just got out of my Analysis exam. Here's a problem that was pretty easy, but I thought it was fun anyway.

Let f:H->K be a function. Show that if x is in H but isn't a limit point of H, then f is continuous at x.

f is any old function? And what properties does H have? Connectedness?
You shouldn't need anything else. You're just considering f's continuity at a point of H which is not a limit point of H.

I guess since x is in the interior of H, one can take a ball around it that is contained in the interior of H. Using the fact that a function can only have a countable number of discontinuities, this means that the function must be piecewise continuous, and so must be continuous about x. This argument seems stronger than needed though.
x isn't in the interior of H. For example, 0 is not a limit point of the integers, as the open set (-.5,.5) about 0 does not contain a point of the integers distinct from 0. The interior of the integers is empty.

Here's some useful characterizations to apply for this problem.

x is a limit point of a set H iff for every open set S about x, there is a point y in S distinct from x such that y is in H. (It follows that x is not a limit point of a set H iff there exists an open set S about x such that S does not intersect any point of H besides x.)

A fuction f:H->K is said to be continuous at x in H iff for every open set T about f(x), there is an open set S about x such that f[S] is a subset of T.



On an unrelated note, it occurs to me that we need to have some problems that require more cleverness than upper-level math experience. I should look for some of my high school math competition materials so that more people can participate.

[Bahamut]

Hm...I'm so rusty with the undergrad real analysis :( . Guess I'll get to review it in the first 3 weeks of next semester in the actual grad real analysis.

So, here's a math question for anyone with knowledge of complex analysis - does there exist a non-constant function that is harmonic on the complex plane and is 0 everywhere on the real and imaginary axis?

Oh and math problems are very easy to find - one can search for any AMC, AIME, USAMO, and IMO problems and post them here. There is a good chance I will be able to answer them.

[Steben]

Just got out of my Analysis exam. Here's a problem that was pretty easy, but I thought it was fun anyway.

Let f:H->K be a function. Show that if x is in H but isn't a limit point of H, then f is continuous at x.

f is any old function? And what properties does H have? Connectedness?
You shouldn't need anything else. You're just considering f's continuity at a point of H which is not a limit point of H.

I guess since x is in the interior of H, one can take a ball around it that is contained in the interior of H. Using the fact that a function can only have a countable number of discontinuities, this means that the function must be piecewise continuous, and so must be continuous about x. This argument seems stronger than needed though.
x isn't in the interior of H. For example, 0 is not a limit point of the integers, as the open set (-.5,.5) about 0 does not contain a point of the integers distinct from 0. The interior of the integers is empty.

Here's some useful characterizations to apply for this problem.

x is a limit point of a set H iff for every open set S about x, there is a point y in S distinct from x such that y is in H. (It follows that x is not a limit point of a set H iff there exists an open set S about x such that S does not intersect any point of H besides x.)

A fuction f:H->K is said to be continuous at x in H iff for every open set T about f(x), there is an open set S about x such that f[S] is a subset of T.



On an unrelated note, it occurs to me that we need to have some problems that require more cleverness than upper-level math experience. I should look for some of my high school math competition materials so that more people can participate.

Alrighty, here's the proof.

x is not a limit point of H, so let S be an open set such that it intersects no point of H distinct from x. Consider an open set T about f(x). f[S]=f[{x}]={f(x)} which is a subset of T. Therefore, f is continuous at x.

[Bahamut]

Once you gave all that info, I figured it out quite easily. It's not nice handing over the tools if the tools are only like 2 things :P .

[Red Shadow]

im finding myself too dumb to prove this, send help:

Tan^2x(sec^2x)=tan^2x+cos^2x-1

[Bahamut]

So you know (sin x)^2 + (cos x)^2 = 1, so just divide everything by (cos x)^2.

[Red Shadow]

for some reason im still not seeing it

aggravates the hell out of me because these are damned easy

i end up getting sin^2xcos^2x on the right wth

[Bahamut]

Oh wait I misread the problem ha. Working on it atm

Edit: I seem to be getting that this is impossible.

So, by moving the (tan x)^2 terms to the left side, you get (tan x)^2 * [ (sec x)^2 - 1 ] = (cos x)^2 - 1

==> (tan x)^4 = - (sin x)^2

==> (sin x)^4 = - (sin x)^2 * (cos x)^4 = -(sin x)^2 * (1 - (sin x)^2 )^2

==> (forget about the possible division by 0, since we're looking backwards and for arbitrary x) (sin x)^2 = - (1 - 2 * (sin x)^2 + (sin x)^4 )

==> (sin x)^4 - (sin x)^2 + 1 = 0

So, set t = sin x. Note that t^4 - t^2 + 1 = 0 is an even function, and the range of t is [-1, 1], so it is enough to look at [0, 1]. However, this equation is always greater than 0 for the possible values of t, so I think there's a typo here or your teacher gave a false problem.

[Hadriel]

Linear algebra is making me want to shoot myself in the face. One of my relatives died, so I missed several class periods and horribly bombed the first exam. The really depressing part is that there's only that exam, one more in April, and a comprehensive final, with our homework counting as one collective exam grade. Not only that, he hasn't even told us the grading scheme, although he's mentioned that the final will probably count for more than the other tests. Unless the final counts for like 40% of my grade, I think a B is the highest possible grade I can get right now; I need better if I want to transfer to UT next spring. Considering dropping so it doesn't go on my transcript. Also considering taking a differential equations class this summer at another nearby university, though the costs and driving times behind that would be troublesome. If people would have signed up for the damn class over here they wouldn't have axed it.

[Red Shadow]

yeah it should have been sin instead of sec >:[

[Bahamut]

Oh then that makes sense.

So, (tan x)^2 * (sin x)^2 = (tan x)^2 + (cos x)^2 - 1 is equivalent to

(tan x)^2 * (sin x)^2 = (tan x)^2 - (sin x)^2 (using (sin x)^2 + (cos x)^2 = 1)

(sin x)^2 + (tan x)^2 * (sin x)^2 = (sin x)^2 * (1 + (tan x)^2 ) = (tan x)^2

(sin x)^2 * (sec x)^2 = (sin x)^2 / (cos x)^2 = (tan x)^2, which is certainly true (Note: I used (tan x)^2 + 1 = (sec x)^2 here)

[Red Shadow]

yeah i went ahead on mine and changed tan^2 on the left to sec^2-1 and worked from there etc etc

[Steben]

Oh then that makes sense.

So, (tan x)^2 * (sin x)^2 = (tan x)^2 + (cos x)^2 - 1 is equivalent to

(tan x)^2 * (sin x)^2 = (tan x)^2 - (sin x)^2 (using (sin x)^2 + (cos x)^2 = 1)

(sin x)^2 + (tan x)^2 * (sin x)^2 = (sin x)^2 * (1 + (tan x)^2 ) = (tan x)^2

(sin x)^2 * (sec x)^2 = (sin x)^2 / (cos x)^2 = (tan x)^2, which is certainly true (Note: I used (tan x)^2 + 1 = (sec x)^2 here)

Usually teachers prefer starting with the left side and manipulating it to get the right.

tan^2*sin^2
= (sec^2 - 1)*sin^2 [by tan^2 = sec^2 - 1]
= tan^2 - sin^2 [by distributive law]
= tan^2 + cos^2 - 1 [by -sin^2 = cos^2 - 1]

[Escariot]

Those numbers don't mean a thing to me. MATH111 here is calculus.

But yeah, I'm having the same trouble in spanish. I'm having to withdraw from it entirely.

We don't have a Math 111, It's Math 110, which is Basic Calc, and Math 140, which is Calc w/ Analytical Geometry

[Broken]

Will someone help me with this Trig. identity, please?

cos(A)/\4 + 2cos(A)/\2 sin(A)/\2 + sin(A)/\4.

and to clarify, that's cosine A to the 4th plus two times cosine A squared times sine A squared plus sine A to the 4th.

The furthest I could get on this was to re-arrange the expression so that it was:

cosA/\4 + sinA/\4 + 2cosA/\2 sinA/\2

And then apply the pythag. identity to the terms cosA/\4 + SinA/\4, making that expression equal to 1.

So, then I had:

1 + 2cosA/\2 sinA/\2.

Am I going about this the right way?
If anyone could help me out, that would be excellent.

[Bahamut]

So, (cos A)^4 + 2 * (cos A * sin A)^2 + (sin A)^4 = ( (cos A)^2 + (sin A)^2 )^2 = 1 ^ 2 = 1

[Broken]

Oh, so the whole problem was just a factored polynomial basically?

And the two factors end up being (sinA^2 + cosA^2) and (sinA^2 + cosA^2) or 1X1.

Did I understand that correctly?

[Steben]

Oh, so the whole problem was just a factored polynomial basically?

And the two factors end up being (sinA^2 + cosA^2) and (sinA^2 + cosA^2) or 1X1.

Did I understand that correctly?

Exactly.

I'm tutoring for Precal with Trig, and that's something they like to pull - requiring you to factor trig functions in order to simplify them.

Here's a similar example: (note that I use sin^n as shorthand for sin^n(x))
(sin^3 + cos^3)/(sin + cos) = 1 - sin*cos

(sin^3 + cos^3)/(sin + cos)
= (sin + cos)(sin^2 - sin*cos + cos^2)/(sin + cos) {by the factorization of the sum of cubes}
= sin^2 - sin*cos + cos^2 {by the cancellation of (sin + cos) over itself - note that this does not hold true for angle values of 3pi/4 or 7pi/4, which cause sin(x) + cos(x) to equal zero}
= 1 - sin*cos {by the Pythagorean identity}

[Bahamut]

Well, there is a long way to do the problem too I think, but it's more trouble than it's worth (by making use of (cos x)^2 = 1 - (sin x)^2 )

[Red Shadow]

hey kids pay attention somebody is probably going to be in here with a question soon

[Native Jovian]

postcount +1
I see what you did there.

[Bleck]

nerdiest thread on the boards

[Bahamut]

I'll take on any (legit) math question.

[The Pezman]

http://imgs.xkcd.com/comics/e_to_the_pi_times_i.png

Prove it in a way we can reasonably understand.

[Bahamut]

http://imgs.xkcd.com/comics/e_to_the_pi_times_i.png

Prove it in a way we can reasonably understand.

Read this: http://en.wikipedia.org/wiki/Exponential_function#On_the_complex_plane

Basically it boils down to the Taylor series representations for sin, cos, and exp adding up correctly.

[Reaif]

Read this: http://en.wikipedia.org/wiki/Exponential_function#On_the_complex_plane

Basically it boils down to the Taylor series representations for sin, cos, and exp adding up correctly.

Wow Taylor series... It's been a while. Though I am definitely going to have to refresh on it, because I have to take calc 3 next semester, and I hear they use a lot of series' in that class... *Sigh*

But seriously, I am probably screwing myself over next semester. So far I am signed up for Linear Algebra, Calc 3, Engineering Dynamics, and Chemistry (one of the earliest ones). Sounds like a death trap to me.

[Steben]

Read this: http://en.wikipedia.org/wiki/Exponential_function#On_the_complex_plane

Basically it boils down to the Taylor series representations for sin, cos, and exp adding up correctly.

Breaking it down a little more...

http://www.auburn.edu/~clontsc/e(xi).jpg

Take e^(xi) and turn it into its Taylor series expansion. By computing the i-part of each term (i^0=1, i^1=i, i^2=-1, i^3=-i, etc.), you can then rearrange stuff like above. You get the Taylor expansion of cos(x) plus i times the Taylor expansion of sin(x).

Plug in pi for x and you get cos(pi)+i*sin(pi) = -1 + i*0 = -1.

[Bahamut]

Breaking it down a little more...

http://www.auburn.edu/%7Eclontsc/e%28xi%29.jpg

Take e^(xi) and turn it into its Taylor series expansion. By computing the i-part of each term (i^0=1, i^1=i, i^2=-1, i^3=-i, etc.), you can then rearrange stuff like above. You get the Taylor expansion of cos(x) plus i times the Taylor expansion of sin(x).

Plug in pi for x and you get cos(pi)+i*sin(pi) = -1 + i*0 = -1.

Yeah, I was too lazy to TeX up the proof...but that's the specifics.

[Red Shadow]

let's try this again:

[angcdixon]

someone help please!!
can you show me how to do these problems:
f(x)= square root of x
g(x)= 1-2x

1) (fog)(x)

2) (gof)(x)

3) (fog)(-4)

4) (gof)(-4)

5) 2g(-1) - 3g(-1)

6) 2f(-9)

the answers are as follows:
1. square root of (1-2x)
2. 1 - (2 sqrt of x)
3. 3
4. 1-4i
5. -3
6. 6i
i just need to know HOW to get the answers!!
please and thank you

[Bahamut]

let's try this again:

Nothing shows up.

someone help please!!
can you show me how to do these problems:
f(x)= square root of x
g(x)= 1-2x

1) (fog)(x)

2) (gof)(x)

3) (fog)(-4)

4) (gof)(-4)

5) 2g(-1) - 3g(-1)

6) 2f(-9)

the answers are as follows:
1. square root of (1-2x)
2. 1 - (2 sqrt of x)
3. 3
4. 1-4i
5. -3
6. 6i
i just need to know HOW to get the answers!!
please and thank you

(f o g)(x) = f(1 - 2x) = (1 - 2x)^(1/2) (square root of 1 - 2x)

(g o f)(x) = g(x^(1/2)) = 1 - 2(x)^(1/2)

3 & 4 you just plug in x = whatever number

2g(-1) - 3g(-1) = -g(-1) = - (1 - 2(-1)) = - ( 1 + 2 ) = -3

2f(-9) = 2 * (-9)^(1/2) = 2 * (9)^(1/2) * (-1)^(1/2) = 2 * 3 * i = 6i

[Red Shadow]

yeah i directed her question here before realizing i could help her with it, which was done

she'll likely ask other stuff that ive decided to forget, however