Steben Posted October 27, 2006 Report Share Posted October 27, 2006 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. Quote Link to post Share on other sites
Ab56 v2 aka Ash Posted October 27, 2006 Report Share Posted October 27, 2006 Could you power series this sucker? Quote Link to post Share on other sites
Bahamut Posted October 27, 2006 Report Share Posted October 27, 2006 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. Quote Link to post Share on other sites
Bahamut Posted October 27, 2006 Report Share Posted October 27, 2006 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. Quote Link to post Share on other sites
Hadriel Posted October 28, 2006 Report Share Posted October 28, 2006 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. Quote Link to post Share on other sites
Bahamut Posted October 28, 2006 Report Share Posted October 28, 2006 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). Quote Link to post Share on other sites
Bahamut Posted October 29, 2006 Report Share Posted October 29, 2006 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. Quote Link to post Share on other sites
Bahamut Posted October 30, 2006 Report Share Posted October 30, 2006 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? Quote Link to post Share on other sites
Fionn Posted October 30, 2006 Report Share Posted October 30, 2006 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. Because I'd express that as: d(x^2)/dx [(x^2)] = d(x)/dx + ... + d(x)/dx (x times) Quote Link to post Share on other sites
zarvoxthezadvook Posted October 30, 2006 Report Share Posted October 30, 2006 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 Quote Link to post Share on other sites
Ab56 v2 aka Ash Posted October 30, 2006 Report Share Posted October 30, 2006 ^Pretty close. Answer's on wikipedia Quote Link to post Share on other sites
Steben Posted October 30, 2006 Report Share Posted October 30, 2006 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. Quote Link to post Share on other sites
Bahamut Posted October 31, 2006 Report Share Posted October 31, 2006 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. Quote Link to post Share on other sites
Bahamut Posted November 1, 2006 Report Share Posted November 1, 2006 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. Quote Link to post Share on other sites
Steben Posted November 1, 2006 Report Share Posted November 1, 2006 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^nn=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. Quote Link to post Share on other sites
Bahamut Posted November 1, 2006 Report Share Posted November 1, 2006 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^nn=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* Quote Link to post Share on other sites
Steben Posted November 1, 2006 Report Share Posted November 1, 2006 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. Quote Link to post Share on other sites
Steben Posted November 2, 2006 Report Share Posted November 2, 2006 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... Quote Link to post Share on other sites
CHz Posted November 2, 2006 Report Share Posted November 2, 2006 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? Quote Link to post Share on other sites
Bahamut Posted November 2, 2006 Report Share Posted November 2, 2006 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. Quote Link to post Share on other sites
Bahamut Posted November 3, 2006 Report Share Posted November 3, 2006 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. Quote Link to post Share on other sites
Steben Posted November 3, 2006 Report Share Posted November 3, 2006 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. Quote Link to post Share on other sites
Bahamut Posted November 3, 2006 Report Share Posted November 3, 2006 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!) Quote Link to post Share on other sites
esperz Posted November 3, 2006 Report Share Posted November 3, 2006 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 Quote Link to post Share on other sites
Bahamut Posted November 3, 2006 Report Share Posted November 3, 2006 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. Quote Link to post Share on other sites
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