Finding inner product and expression for norm

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?

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*}
• 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? Dec 27, 2021 at 20:40
• 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? Dec 27, 2021 at 20:55
• @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. Dec 27, 2021 at 21:12