# Approximation problem

1. Nov 24, 2005

### standardflop

A odd 2pi periodic function, for wich $x \in [0;\pi]$ is given by $f(x)=\frac \pi{96}(x^4-2\pi x^3+\pi^3x)$
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.

Last edited: Nov 24, 2005
2. Nov 24, 2005

### shmoe

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

3. Nov 25, 2005

### 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 dont 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.

4. Nov 25, 2005

### 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

5. Nov 25, 2005

### standardflop

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

6. Nov 25, 2005

### 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 nifty flash presentation of it (there is a text version with flash and java examples here). 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.

Last edited: Nov 25, 2005
7. Nov 25, 2005

### shmoe

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.

8. Nov 25, 2005

### 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...)

9. Nov 27, 2005

### standardflop

Im not sure about what you mean.

10. Nov 27, 2005

### 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.

11. Nov 27, 2005

### 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)?

12. Nov 27, 2005

### shmoe

Where did the absolute value go?

13. Nov 27, 2005

### standardflop

I got it now, thanks for the patience.