Bahamut Posted November 5, 2006 Report Share Posted November 5, 2006 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. Quote Link to post Share on other sites
CHz Posted November 16, 2006 Report Share Posted November 16, 2006 Love that proof, Baha. 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. Quote Link to post Share on other sites
Bobwillis Posted November 23, 2006 Report Share Posted November 23, 2006 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. Quote Link to post Share on other sites
Bahamut Posted November 24, 2006 Report Share Posted November 24, 2006 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. Quote Link to post Share on other sites
Steben Posted November 24, 2006 Report Share Posted November 24, 2006 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... Quote Link to post Share on other sites
Bahamut Posted November 24, 2006 Report Share Posted November 24, 2006 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 Quote Link to post Share on other sites
Steben Posted November 24, 2006 Report Share Posted November 24, 2006 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. Quote Link to post Share on other sites
Bahamut Posted November 24, 2006 Report Share Posted November 24, 2006 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. Quote Link to post Share on other sites
Bobwillis Posted November 24, 2006 Report Share Posted November 24, 2006 If you plot the graphs they do not intersect. And using x = 0 will not help. Quote Link to post Share on other sites
Steben Posted November 25, 2006 Report Share Posted November 25, 2006 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. Quote Link to post Share on other sites
Psychonaut Posted November 25, 2006 Report Share Posted November 25, 2006 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. Quote Link to post Share on other sites
Bahamut Posted December 10, 2006 Report Share Posted December 10, 2006 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 is abelian. Quote Link to post Share on other sites
Effector Posted December 10, 2006 Report Share Posted December 10, 2006 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?) Quote Link to post Share on other sites
Bahamut Posted December 10, 2006 Report Share Posted December 10, 2006 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. Quote Link to post Share on other sites
Effector Posted December 10, 2006 Report Share Posted December 10, 2006 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: But I'm not sure. Why does that work, anyway? Is it because the difference between the two approach ±(π/2)? Quote Link to post Share on other sites
AbyssWyrm Posted December 10, 2006 Report Share Posted December 10, 2006 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. Quote Link to post Share on other sites
Bahamut Posted December 10, 2006 Report Share Posted December 10, 2006 See, minor mistakes in writing what I have in mind like that is what costs me huge on tests . Quote Link to post Share on other sites
HalcyonSpirit Posted December 10, 2006 Report Share Posted December 10, 2006 A problem for all you Math people out there. I need to find the sum of the following series: 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? Quote Link to post Share on other sites
AbyssWyrm Posted December 10, 2006 Report Share Posted December 10, 2006 A problem for all you Math people out there.I need to find the sum of the following series: 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? Quote Link to post Share on other sites
HalcyonSpirit Posted December 10, 2006 Report Share Posted December 10, 2006 A problem for all you Math people out there.I need to find the sum of the following series: 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. Quote Link to post Share on other sites
Rainman DX Posted December 10, 2006 Report Share Posted December 10, 2006 A problem for all you Math people out there.I need to find the sum of the following series: 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. Quote Link to post Share on other sites
HalcyonSpirit Posted December 10, 2006 Report Share Posted December 10, 2006 A problem for all you Math people out there.I need to find the sum of the following series: 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. Quote Link to post Share on other sites
Rainman DX Posted December 10, 2006 Report Share Posted December 10, 2006 Thanks guys, much appreciated. Anything to help a fellow GA Jedi Knight. Quote Link to post Share on other sites
Steben Posted December 11, 2006 Report Share Posted December 11, 2006 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 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 . 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. Quote Link to post Share on other sites
Bahamut Posted December 11, 2006 Report Share Posted December 11, 2006 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 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 . 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 and y(A intersect be in G/(A intersect (coset representation). Then xy(A intersect = yx(A intersect B since xyA = yxA and xyB = yxB, and the intersection of xyA and xyB is xy(A intersect and similarly yxA intersect yxB = yx(A intersect . 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. Quote Link to post Share on other sites
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