Zer0's Conundrum #2 has been solved. No additional prizes will be given out. Stay tuned for future conundrums... Welcome to... Zer0's Conundrum #2 So, in Zer0's conundrum, Zer0 will post a problem (usually math or computer science related) and you will have to solve it! First user who solves the problem will win a prize! If no one gets it within a week, there will be no winner for that week. Post your answer here! Rules One entry per person per contest You MUST provide justification (proofs, etc) along with your answer otherwise it WILL NOT count! No collaboration (I don't understand why you would want to...) No looking up the answer or asking someone for the answer No computer programs unless otherwise specified This Week's Prize... 15 forum cash! - courtesy of some mod! This Week's Conundrum... Evaluate the following expression in closed form:
Re: Zer0's Conundrum #2 Gahhhh...why calculus?...I don't remember calculus anymore. >__< More combinatorics/logic!
Re: Zer0's Conundrum #2 summation...series...topics learned in calculus anyways...what do you mean by closed form...as an expression in terms of x?
Re: Zer0's Conundrum #2 Well its not technically calculus. I mean, there isn't really any higher math involved in solving this expression. Closed form: http://en.wikipedia.org/wiki/Closed-form_expression So yeah, some function of x
Re: Zer0's Conundrum #2 I thoroughly do not understand this question. :S Common sense tells me that the answer depends on the value of x. Ratio and root tests tell me that if x>=3, the series will diverge...meaning it's impossible to express the answer in closed form...unless you consider infinity to be closed form?
Re: Zer0's Conundrum #2 lol yes, the series diverges for -3 < x < 3 (I don't feel like checking end points, but you get the gist) Let me define a function f that equals that expression. Its domain will be (-3,3). N is just a dummy variable in the summation. So f is really just a function of x. However, what closed-form means is get rid of the summation and the dummy variable N, and explicitly define the expression as a function of x. For example: Consider the following sum That's a geometric series, and so the closed-form for that sum is:
Re: Zer0's Conundrum #2 but...but...but that only works for x<1 infinite series that diverge don't have closed forms so are we just assuming that the series will converge?
Re: Zer0's Conundrum #2 Just because it diverges for certain values of x doesn't mean its not a function... We are only interested in the portion of the function that DOES converge (i.e. the domain of the function) If the series diverges for some value of x, then the function is undefined at that point. The function 1/(1-x) IS the closed-form of that series on the domain (-1,1). We don't care about divergent series because that would be outside the domain of the function!
Re: Zer0's Conundrum #2 Domain of a function = set of values of x for which the function is defined e.g. domain of sqrt(x) = {x | x >= 0}
Re: Zer0's Conundrum #2 the function part of that is definitely defined for all values of x...it's just that just some values of x the sum is infinity/the series diverges
Re: Zer0's Conundrum #2 Yeah ok, whatever you learned in school about domains and stuff was all wrong then. The function is NOT defined for all values of x. Let's look at x = 5. By evaluating the series, you see that it diverges to infinity. Infinity is NOT a real number and so the function is not defined at x = 5. http://en.wikipedia.org/wiki/Domain_of_ ... definition for the formal definition So for example, let's say f(x) = tan(x) The domain is not all real numbers. Its R-{x+pi/2 | x = k*pi} R = real numbers k = integers
Re: Zer0's Conundrum #2 ..........................................................>____< i can't believe I didn't see it earlier (sorry for the crappy equations...I don't have latex on this computer): oh and I guess I'm just not used to thinking of functions as sums :maha:
Re: Zer0's Conundrum #2 lol, don't worry, math is all about learning new things :arf: But yeah, school math curriculums never teach this kind of stuff (and they should!). Otherwise students get the wrong idea of what a function really is. And this is why I promote math competitions! Oh and yeah, your answer is correct. Though your method is extremely over-complicated. There's a much simpler and elegant solution that doesn't require calculus at all. ^ which is equivalent to what you had