Using the convergence of Fourier Series Theorem to estimate the number of terms for Fourier Series $f(x)$

Question

Attached are scans from my book. One of my homework problems requires me to let $f(x)=(x^2-1)^2$ for $-1 \leq x \leq 1$. I am using the book's example (Example 5) as a guideline, but it is driving me crazy because it is skipping steps.

I need to verify that $f(x)$ satisfies the hypotheses of Theorem 2. How many terms of Fourier Series $f(x)$ suffice to approximate $f(x)$ to within an error of $.001$ according to theorem 2.

First I need to check that my $f(x)$ satisfies the theorem, by taking second order derivatives and finding my max.

$f(x) = (x^2-1)^2$

$f(1) = (1^2-1)^2 \rightarrow 0^2 = 0$

$f(-1) = ((-1)^2-1)^2 \rightarrow (1-1)^2 \rightarrow 0^2=0$

$f'(x) = (x^2-1)4x$

$f'(1) = (0)4(1)=0$

$f'(-1) = (1-1)(-4) =0$

$f''(x) = (x^2-1)(4)+4x(2x)$

$f''(x) = 4x^2-4+8x^2 \rightarrow 12x^2-4$

$f''(1) = 12-4 = 8$

$f''(-1) = 12(-1)^2 - 4 = 12(1)-4=12-4=8$

My max is 8, but I don't understand how the book calculated $N$. I want to use the second notion where the book has $N >$ some number but I don't see how they approximate it to be 244 in the example.

Can someone please explain what the book is doing?

Answer 1

So you are clear about the displayed equation that says $$ {4(1)^2\cdot 6\over \pi^2 N} <0.01 $$ ?

If so, they are simply solving the inequality for the smallest integer $N$ which makes it true:

\begin{align*} {4(1)^2\cdot 6\over \pi^2 N} <0.01 &\implies {24\over \pi^2 N}<0.01\\ &\implies {\pi^2 N\over 24}>100\\ &\implies N>{2400\over \pi^2}\approx 243.17\text{ so take } N>244. \end{align*}

Do the same with your inequality: \begin{align*} {4\cdot 1^2\cdot 8\over \pi^2 N}<0.001&\implies {32\over \pi^2 N}<0.001\\ &\implies {\pi^2 N\over 32}>1000\\ &\implies N>{32000\over \pi^2}\approx 3242.28\\ &\implies N>3243. \end{align*}