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So my function is $y(x)=x^{3} - x\pi^{2}$ for $|x|\le\pi$.

I found out the Fourier coefficients, and now I am asked to compute a specific sum with Parseval's formula.

The issue for me lies on when computing $\frac{1}{2\pi}\int_{-\pi}^{\pi} |x^{3} - x\pi^{2}|^{2} dx$.

How do I handle the absolute value?

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    In fact the absolute value disappears $|a|^2=a^2 \ \implies \ |x^{3} - x\pi^{2}|^2= (x^{3} - x\pi^{2})^2$ – 2017-02-16
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    Ok thanks , can I ask you something more? – 2017-02-16
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    Ask, but as it is late for me (CET midnight) I am not sure I can answer if it takes a long time. – 2017-02-16
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    $y(x)=x^{2}$ for $|x|\le\pi$ What choice of $a_{0}$ minimises this integral $\int_{-\pi}^{\pi} |y(x)-a_{0}| dx $ – 2017-02-16
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    No square over the absolute value ? – 2017-02-16
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    Yes, no square.. – 2017-02-16
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    I am not in agreement, I find $a_0=(\pi/2)^2$ – 2017-02-16
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    Yes my bad that is correct..but how does one find it? Whats the method.. I know that when we have a square it is the fourier coefficients..but now there is no square.. – 2017-02-16
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    I have given below a detailed explanation as an "answer" (it was too long for a comment). – 2017-02-16

2 Answers 2

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As user JeanMarie said in the comments, if $a \in \Bbb{R}$ then $$ |a|^2 = a^2 \tag{1} $$ Indeed, if $a\geq0$ then $|a|=a$ so certainly $(1)$ holds. If $a<0$ then $|a|=-a$ so $|a|^2=(-a)^2=a^2$ and $(1)$ holds also. So we conclude that $(1)$ holds for all $a \in \Bbb{R}$.

Now, as for your integral, making use of $(1)$, we get \begin{align} \| y\|_2^2 &= \frac{1}{2\pi} \int_{-\pi}^{\pi} |x^{3} - x\pi^{2}|^{2} \,dx \\ &= \frac{1}{2\pi}\int_{-\pi}^{\pi} (x^{3} - x\pi^{2})^{2} dx \\ &= \frac{1}{\pi}\int_{0}^{\pi}(x^6-2x^4\pi^2+x^2\pi^4) dx \quad \text{(even integrand)} \\ &= \frac{1}{\pi} \left[\frac{x^7}{7}-\frac{2\pi^2 x^5}{5} + \frac{\pi^4x^3}{3}\right]_{0}^{\pi} \\ &= \frac{1}{\pi} \left(\frac{\pi^7}{7}-\frac{2\pi^7}{5}+ \frac{\pi^7}{3}\right) \\ &= \frac{8 \pi^6}{105} \end{align}

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    Could you help me with this one? How do I think here : $y(x)=x^{2}$ for $|x|\le\pi$ What choice of $a_{0}$ minimises this integral $\int_{-\pi}^{\pi} |y(x)-a_{0}| dx $ – 2017-02-16
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    @fejz1234 Well, you could tackle that problem with simple calculus. Just consider the cases $a_0 \leq 0$, $0 < a_0 \leq \pi^2$ and $a_0 > \pi^2$. Note that your problem is equivalent to minimizing $\displaystyle\int_0^{\pi} |x^2-a_0|\,dx$ – 2017-02-16
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Answer to the added question:

By symmetry, it is equivalent to minimize $I(a):=\int_{0}^{\pi} |y(x)-a| dx$ (I have dropped index $0$)

$I(a)=\int_0^{\sqrt{a}}(a-x^2)dx+\int_{\sqrt{a}}^{\pi}(x^2-a)dx$

$I(a)=[ax-x^3/3]_0^{\sqrt{a}}+[x^3/3-ax]_{\sqrt{a}}^{\pi}$

$I(a)=4a\sqrt{a}/3-a\pi+\pi^3/3$

whose minimum is obtained for a value of $a$ such that $I'(a)=0$ giving

$2\sqrt{a}=\pi \ \iff a=(\pi/2)^2$

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    I think your argument works for $0\leq a \leq \pi^2$. The value of your integral when $a=(\pi/2)^2$ is $\pi^3/4$. If $a\leq0$ then the integral equals $\pi^3/3-a\pi$ which is minimized when $a=0$ with value $\pi^3/3$. Finally if $a\geq\pi^2$ then the integral equals $\pi a-\pi^3/3$ which is minimized when $a=\pi^2$ with value $2\pi^3/3$. In the end $\pi^3/4$ wins. – 2017-02-17
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    @NeedForhelp It is almost evident that the value of $a$ that minimizes the integral is in the range of values of $x^2$, for $x \in[0,\pi]$, i.e., $[0,\pi^2]$. – 2017-02-17
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    @NeedForHelp Thus I maintain that the value of $a$ that minimizes the integral is the one I have given. – 2017-02-17
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    It's OK. You are correct. If it is evident you should at least say so at the beginning of your argument. Also what is evident for someone maybe isn't for another one. Saying "it is almost evident that..." rarely constitutes a proof... – 2017-02-17
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    @NeedForHelp I agree with your sense of rigor ; but for me, it can be a burden to over-explain, for example in such a case that $a$ cannot be negative, etc. – 2017-02-17
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    Why is it by symmetry that is the same thing as ingegrating from 0 to pi? – 2017-02-17
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    The consideration of symmetry is before beginning the minimisation process: whatever the value of $a$, $x\mapsto |x^2-a|$ is an **even function** thus $\int_{-\pi}^{\pi}....=2\int_0^{\pi}...$ – 2017-02-17
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    Ah true...and if i want to find $a$ so that I can minimize $max_{x\in[-\pi,\pi]} |x^{2}-a| $ ? This should be equivalent with minimizing $max_{x\in[0,\pi]} |x^{2}-a| $? But im stuck here..the correct answer is $\frac{\pi^2}{2}$ – 2017-02-18