We define the functions, $f_n$, $n=0,1,2,3,...,$ on $[-1,1]$ by $$f_n(x)=x^n$$

Then I have to find the inner product for $\langle f_n,f_m \rangle$ for all values of $n, m$ and I have in particular, to show that: $$||f_n||=(n+\frac{1}{2})^{-\frac{1}{2}}$$

But I'm not sure how to do that? I'm confused, what is m? I think to show the norm I have to use $||f_n||=\sqrt{\langle f_n|f_n\rangle}$ ? But how do I find the inner product and use it finding the norm? I hope anyone can help me?


1 Answer 1


Assuming you are using the standard $L^2$-inner product $\langle f, g\rangle = \int_{-1}^1 f(x)g(x)\, dx$, we have that \begin{align*} \|f_n\|^2 = \langle f_n, f_n\rangle = \int_{-1}^1 x^{2n}\, dx = \frac{x^{2n + 1}}{2n + 1}\bigg|_{-1}^1 = \frac{2}{2n + 1} = \frac{1}{(n + 1/2)}. \end{align*}

  • $\begingroup$ Nice, that makes sense. But with the expression for $\langle f_n,f_m \rangle$ then be if for example n and m is different? $\endgroup$
    – Lifeni
    Dec 27, 2021 at 20:40
  • $\begingroup$ @Lifeni That is correct (: $\endgroup$
    – Chee Han
    Dec 27, 2021 at 20:50
  • $\begingroup$ Yeah but will the integral then be $\langle f_n,f_m \rangle=\int_{-1}^1x^nx^mdx$ or what will the expression for $\langle f_n,f_m \rangle$ be when n and m are different? $\endgroup$
    – Lifeni
    Dec 27, 2021 at 20:55
  • 1
    $\begingroup$ @Lifeni It is exactly what you written down. You would then have to analyse different cases of $n$ and $m$. Since $n, m$ are both nonnegative, there is only two cases to consider: $n + m$ odd or $n + m$ even. $\endgroup$
    – Chee Han
    Dec 27, 2021 at 21:12
  • $\begingroup$ Nice, that makes sense. Thank you for the help :-) $\endgroup$
    – Lifeni
    Dec 27, 2021 at 21:27

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