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I am approximate $v = x^2 \in L^2(-1, 1)$ by orthonormal set $\{1/\sqrt 2$, $\cos \pi x$, $\sin \pi x\}$. Thus, I am computing Fourier coefficients of $x^2$ on the subspace of $X$ spaneed by $\{1/\sqrt 2$, $\cos \pi x$, $\sin \pi x\}$. A textbook says

$(v, 1/\sqrt 2) = \sqrt 2/3$,

$(v, \cos \pi x) = -4/\pi^2$,

$(v, \sin \pi x) = 0$.

So far OK. Then, $v \approx \sqrt 2/3 - 4/\pi^2 \cos \pi x$. I am confused. This should be $v \approx 1/3 - 4/\pi^2 \cos \pi x$?

Giorgos Giapitzakis
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T. B.
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1 Answers1

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Yes, you are correct, and the book has a typo. $\sqrt{2}/3$ is just the coefficient on the vector $1/\sqrt{2},$ and they forgot to multiply.

  • Thank you! The textbook is "Introductory funcitonal analysis with applications to boundary value problems and finite elements". The book is very well written, so I strongly recommend this except for this typo! – T. B. Jul 15 '21 at 15:58