Convergence of Fourier Series for Approximating an Odd 2pi Periodic Function

[standardflop]

A odd 2pi periodic function, for which x \in [0;\pi] is given by f(x)=\frac \pi{96}(x^4-2\pi<br /> x^3+\pi^3x)
was found to have the Fourier series
f(x) = \sum_{n=1}^\infty \frac{\sin(2n-1)x}{(2n-1)^5}, \ x \in \mathbb{R}
The problem is now: prove that |f(x) - \sin x| \leq 0.01, \forall x \in \mathbb{R}. The hint given was: Use integral test.

 

[shmoe]

What have you tried so far? What is |f(x) - \sin x|?

 

[standardflop]

|f(x) - \sin x| is the maximum pointwise difference between the two. I've tried looking at the integral \int_1^\infty \frac{\sin(2x-1) \tau }{(2x-1)^5} \ dx
but i couldn not interpret the result. I don't see how the integral test otherwise can be used to compare two different function... Normally i would use it to see how many terms i need for a good approximation.

 

[kleinwolf]

Since it's periodic, just integrate ¦f(x)-sin(x)| over 0->pi..then if ¦f(x)-sin(x)¦<.01 this means the integral should be smaller than .01*pi.or something like that

 

[standardflop]

Since it's periodic, just integrate ¦f(x)-sin(x)| over 0->pi..then if ¦f(x)-sin(x)¦<.01 this means the integral should be smaller than .01*pi.or something like that

I don't think that would work... That would not proove that | f(x) - sin(x) | is less than 0.01 for all x.

 

[benorin]

Isn't there a threefold inequality giving the bounds for the integral as sums related to the integral test? Googled it. Here's a http://archives.math.utk.edu/visual.calculus/6/series.7/1.html (there is a text version with flash and java examples http://archives.math.utk.edu/visual.calculus/6/series.7/). It goes like this, suppose \sum a_{n} satisfies the hypotheses of the integral test, then

\int_{N+1}^{\infty} a_{n} dn \leq \sum_{n=N+1}^{\infty} a_{n} \leq \int_{N}^{\infty} a_{n} dn

perhaps that may help you in estimating the error bounds desired above.

 

[shmoe]

|f(x) - \sin x| is the maximum pointwise difference between the two.

I mean explicitly in terms of your Fourier series.

Using crude and trivial bounds, plus the integral bounds benorin has pointed out you should be able to get the result.

 

[kleinwolf]

Well, your right, you could just integrate from 0 to y ¦f(x)-sin(x)¦dx...this should be <.01*y forall y.is this better ?...but you could also use maxima tests (take care of boundaries...)

 

[standardflop]

I mean explicitly in terms of your Fourier series.

I am not sure about what you mean.

 

[shmoe]

You want to estimate |f(x)-\sin x| and you know

f(x) = \sum_{n=1}^\infty \frac{\sin(2n-1)x}{(2n-1)^5}

Substitute this for f(x). What's the first term of this Fourier series?

You should now have an infinite sum for |f(x)-\sin x|. Bound it in the simplest way you can think of.

 

[standardflop]

Thank you, now i get it... |f(x)-sin x|= \sum_{n=2}^\infty \frac{\sin(2n-1)x}{(2n-1)^5}. But i cannot bound this series with the integral test, because the function, f(n)=\frac{\sin(2n-1)x}{(2n-1)^5} , dosent obey the hypothesis in the test (of being continuous and decreasing)?

 

[shmoe]

Thank you, now i get it... |f(x)-sin x|= \sum_{n=2}^\infty \frac{\sin(2n-1)x}{(2n-1)^5}

Where did the absolute value go?

 

[standardflop]

I got it now, thanks for the patience.