I have to integrate pic related and I got to the infinite series at the bottom. I already know that the limit is pi/2, but how do I prove it using this fucking Taylor series? I can't use double integrals, only this shit and Wolfram Alpha redirects me to Si(x) and I can't do anything with that. My calculus exam is at stake. Pls help /sci/
>>7808066
Oh, I have to integrate it from 0 to infinity.
>>7808069
I would do it for integers and try to prove the series is cauchy, maybe try that
>>7808076
I'll try that, thanks
Fuck this, I'm going to bed, I'm gonna fail that shit anyway.
>>7808105
Yup
>>7808066
we know the integral of dx/(x^2+1) = arctanx + C
if we integrate from 0 to inf we obtain pi/2.
There's probably a way to link with the series expansion of arctan(x).
I may give it a try later tonight.
>>7808066
>>7808338
Using arctan[x] isn't a bad idea. Compare the two series:
[eqn]0<\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+1}}{(2n+1)*(2n+1)!)}<\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+1}}{2n+1}[/eqn]
The middle series represents the integral of Sinc[x] and the right series represents Arctan[x]. The equation above is valid for x > 0.
As x approaches infinity, Arctan[x] approaches Pi/2. You then get left with this:
[eqn]0<\lim_{x\rightarrow \infty }\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+1}}{(2n+1)*(2n+1)!)}<\frac{\pi}{2}[/eqn]
So this shows that the middle series *might* approach Pi/2. It isn't exactly a proof though. From what we have shown alone, there is nothing stopping the series converging to a different value like Pi/3. There might be a way to turn this into a squeeze theorem problem. I don't know though.
>>7808814
I swear to fucking god I fucking hate latex on this fucking board.
Check pic for equations.
>>7808827
Thanks, this actually helps. I just have to find a series that is minoring the original and it's limit is pi/2, so I can prove that it's pi/2 too.
>>7809045
Yes, in theory, all you have to do is find a series that is smaller than the series in the middle and that you know converges to Pi/2 when X goes to infinity. Then, via squeeze theorem, you know the middle goes to Pi/2.
I'm not sure how easy that is going to be though. It's easy finding a smaller series. Finding a series that you absolutely know converges to Pi/2 sounds like a bitch. No obvious series spring to my mind but that's just me.
>>7808827
First, I will attempt to re-write your above expressions with simple edits.
[math] \displaystyle 0 < \sum_{n=0}^{ \infty } \frac{ (-1)^{n} x^{2n+1} }{ (2n+1)(2n+1)!} < \sum_{n=0}^{ \infty } \frac{ (-1)^n x^{2n+1} } {2n+1}[/math]
[math] \displaystyle 0 < \lim_{x \rightarrow \infty } \sum_{n=0}^{ \infty } \frac{ (-1)^n x^{2n+1} }{ (2n+1)(2n+1)! } < \frac{ \pi }{2} [/math]
Now, assuming these display properly (and are along the lines of what you meant to write), I can reasonably dispense advice on how to format TeX in the 4chan dialect. I've also had personal frustrations with it, but I've also been persistent, and kept up with the solution trends here.
-I replaced the eqn tags with math tags,
-I invoked "\displaystyle" at the front of both for nice formatting of large operators (the sigams, and small letters)
-This is the single most important one. I LIBERALLY ADDED WHITESPACE, especially between function header- backslashes, and preceding matter. ALWAYS WRITE " x \frac{... ", " > \sum{... " , NOT NOT NOT! "x\frac{...", >\sum{...". ADD WHITESPACE ESP BETWEEN FUNCTION SLASHES, BRACES, ETC. You and I both know you're going to want ot attempt to write LaTeX in 4chan's current solution again at some point. This is how you do it properly at the moment.
-I also substantively changed the denominators in the larger expressions. It is possible I altered the substance of your math/argument in the process (or else fixed them).
>>7809076
Latex on /sci/ is fucking asinine. I don't understand how the previewer works fine but the posts constantly fuck up.
But yeah, your edits to my original latex is how I intended it to look. I will be interested to know if there is a lower bound that converges to Pi/2.
>>7809076
How the fuck can you say the last sum converges that doesn't make any fucking sense.
>>7808066
Assuming that you can integrate the terms of an infinite series term by term is a nontrivial assumption m8.
You're using the wrong method you need the squeeze theorem I think
another idea
x^n / (n*n!)
= x^n /n - (n!-1)x^n /(n*n!)
using that, we can express each term of the series in 2 components, the first of which we know converges to pi/2 as we take the limit. There must be a way to show the second terms sum to 0.
>>7808066
The standard method for the integral is contour integration. Bottom of the article:
http://mathworld.wolfram.com/SincFunction.html
>>7809433
A careful reader will understand that the purpose of the above post was not to enter the mathematical argument itself, but to advise on formatting a few lines of code (although admittedly I did edit the expression itself, and the other guy sounded agreeable to that edit). Take it up with the other guy.
>>7809433
You are right. That arctan[x] series does not converge for |x|>=1. That can be remedied by replacing that Taylor series with it's Laurent Series.
http://www.wolframalpha.com/input/?i=arctan%28x%29+series
However, now I can't be sure the center sum converges either for terms greater than 1 so it gets all dicey. Might be best to abandon this method. Laurent Series are outside the scope of Calc 2 anyway which OP appears to be hinting at taking.
>>7810540
Ok, thinking about this based on what >>7808338 said, here is a shady way of doing it because it runs on the assumption below for x >= 0:
[math] ArcTan(x) < \int_{0}^{x }\frac{Sin(t)}{t}dt < \pi-ArcTan(x) [/math]
Now you do the following:
[math] \lim_{x\rightarrow \infty }ArcTan(x) < \lim_{x\rightarrow \infty }\int_{0}^{x }\frac{Sin(t)}{t}dt <\lim_{x\rightarrow \infty } \pi-ArcTan(x) [/math]
And then you get:
[math] \frac{\pi}{2} < \int_{0}^{\infty}\frac{Sin(t)}{t}dt <\frac{\pi}{2} [/math]
I also attached an image in case the latex fucks up again.
The assumption is the first line. I’m not sure how you would prove that the integral is *really* bounded by the arctans. But yeah if you run with the assumption then you can use the squeeze theorem to prove what value the integral converges to. I’d much be interested if someone can prove that the first line really is true. It isn't obvious to me how you would prove it.
Maybe this is a bit of an overkill, but you can use the following trick:
[math] \int_0^{\infty} \frac{\sin x}{x} \ dx=\lim_{a \to \infty} \int_0^a \frac{\sin x}{x} \ dx [/math]
So since [0,a] is bounded and closed, the Riemann and Lebesgue integrals agree, so consider the function
f(x,y)=e^{-xy} sin x. Integrate this function over 0<x<a, and 0<y<\infty. Then use Fubini to deduce that
[math] \left| \frac{\pi}{2}-\int_0^a \frac{\sin x}{x} \ dx \right| \leq \frac{1}{a} [/math]
You will see that this gives you the result, as you let a go to infinity.
It's a very Standard example of Contour integration using cauchys integral theorem.
Take a contour going from epsilon <<1 to R >>1, a big semicircle over the top of the complex plane to - R then back to - epsilon, and a semicircle over 0 back to epsilon, and integrate e^(i*z) / z along the contour.
Sin x / x is differentiable inside the contour, so the whole integral is 0.
The - epsilon to epsilon bit is -pi*i, for any epsilon, so let epsilon go small. The R to - R bit goes to 0 as R gets large.
Then
Integrate e^iz/z dz from 0 to infinity - i*pi =0
Taking imaginary parts gives the answer.
(taking real parts gives cos x / x between 0 and infinity =0)
May have made some small mistakes here and there, I'm doing it off the top of my head on my phone. Just Google contour integration. This problem is always the first example.
