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March 5th, 2016, 04:27 PM   #1
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Calculus on functions with domain of natural numbers

I am going through an analysis book to prepare for grad school and am working on problem:

find a sequence integral(f(x), 0, N) that converges as N goes to infinity but the limit of integral(f(x), 0 , infinity) does not exist. The super easy solution is f(x) = sin(2 * pi * x). Sequence being 0,0,0,0,0,0,0,0,0,0,0,0,... , so it converges, but I think this problem is wanting something more. So I came up with f(x) = sin(2 * pi * ( x + (1/x))). I know it converges and playing with it on the calculator by evaluating the integral from n-1 to n shows this portion of it going to zero very fast. Faster than 1/x^2. To prove it I showed the derivatives ---> [sin(2 * pi (n + (1/n))) - sin(2 * pi ((n-1) + (1/(n-1))))]/[(-2/n^3)] goes to infinity as n goes to infinity using L'Hospital's Rule.

My question is, can I do this? If so, why? I know the series converges faster than 1/x^2 but is this the way to prove it? Also, why can we do this, if we can, indeed, do this? Thanks
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March 5th, 2016, 05:24 PM   #2
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It's not clear to me what you have done, or why. But a good way to show the rate of convergence would be to look at $\sin{(A+B)}=\sin A\,\cos B + \cos A\,\sin B$ and then us the series expansion of $\sin$ to show the rate of convergence.
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March 6th, 2016, 04:52 AM   #3
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I want to prove that integral(sin(2*pi*(x + (1/x))), x-1,x) converges to zero faster than (1/x^2) as x goes to infinity.
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March 6th, 2016, 06:00 AM   #4
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I don't think that is exactly what you mean, but to illustrate my approach from above:
$\displaystyle \sin 2\pi\left(x+\tfrac1x\right)=\sin 2\pi x \, \cos \tfrac{2\pi}{x} + \cos 2\pi x \, \sin \tfrac{2\pi}{x} $
And when evaluated at an integer value of $\displaystyle x$ we get
$\displaystyle \sin \tfrac{2\pi}{x} $
In series expansion, this is
$\displaystyle \tfrac{2\pi}{x} - \tfrac1{3!}\left(\tfrac{2\pi}{x}\right)^3 + \ldots$
But the integral expression doesn't converge. As you noted in your first post the integrand doesn't converge to anything, let alone the zero it would need to for the integral to converge.

Last edited by v8archie; March 6th, 2016 at 06:03 AM.
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March 6th, 2016, 06:33 AM   #5
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the integral converges evaluated for x-1 to x as x goes to infinity. it goes to zero because sin(2*pi*x) has a period of one. Only way to show it is by graphing it. Really, can you not understand what I am asking?
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March 6th, 2016, 06:34 AM   #6
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Quote:
Originally Posted by tnsettlemo View Post
the integral converges evaluated for x-1 to x as x goes to infinity. it goes to zero because sin(2*pi*x) has a period of one. Only way to show it is by graphing it. Really, can you not understand what I am asking?
Also, I am only evaluating it at integers, n-1 to n, which I don't think matters anyways. Thanks anyways
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March 6th, 2016, 06:50 AM   #7
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I finally figured out how to plot it on my ti 84 vs 1/x^2 so it works. I saved the data into a list and plotted the list points. it goes to zero much faster than 1/x^2.
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March 6th, 2016, 11:53 AM   #8
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if you are familiar with texas instrument calculators, here is what I did and maybe then you will understand what I achieved.

seq(fnint(sin(2*pi*(x+(1/x))),x,x-1,x),x,2,100,.25)-->L1

seq(2+.25x,x,0,392)-->L2

Now I do the scatter plot with L2 as x-values and L1 as y-values. I plotted this graph with the function +-1/x^2 and the plotted points from 2 to 100 are within these two curves, so it converges as a sequence going to zero, but not only that, if you look at my original post you will see I found a series that converges to some number. Let me know if you get what I am doing. Thanks, it was a problem in my book. The above sequence actually converges much faster if only using integers. I was wanting to prove it by hand but because of the periodic nature am unable to do it.

Last edited by tnsettlemo; March 6th, 2016 at 11:56 AM. Reason: add info
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March 6th, 2016, 07:17 PM   #9
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Quote:
Originally Posted by tnsettlemo View Post
if you are familiar with texas instrument calculators
I think I was confused in part by your re-use of the dummy variable in the integral. You are looking at
$$\int_{x-1}^{x} \sin\big(2\pi(t+\tfrac1t)\big)\,\mathrm d t$$

The best I can see for that is to expand $$\sin\big(2\pi(t+\tfrac1t)\big) = 2\pi(t+\tfrac1t) - \tfrac1{3!}\big(2\pi(t+\tfrac1t)\big)^\tfrac13 + \ldots$$
and then integrate. I'm not convinced that it would be very useful though.
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March 7th, 2016, 04:59 AM   #10
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Thanks for getting back. I am glad you understand. As you can see, it does converge and the original series from my first post converges, but yes, it would be very hard to show by hand. Pretty neat I found a way to do first order differential equations on the ti 84. You don't need the 89 after all! I guess the ti 89 would be a lot more user friendly and a lot faster though, with Euler's method and RK method. Thanks again.
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