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How does one find the best polynomial approximation of order 2 of $f(t)=e^t$ in $(C[0,2\pi], ||.||) $

What I have tried:

If one orthonormalizes $1,t,t^2...$ $C[0,2\pi]$ one will get the legendre polynomials (or legendre functions?) which can be used to find the Gauss approximation: $$p(x) = \sum _{i=0}^n \frac{1}{||L_i||^2_{2, L^2[0,2\pi]}}(L_i,f)L_i(x)$$

So from this question: Orthnormalization of the system $(1,t,t^2) $ in $C[0,2\pi]$

One gets: $$L_0= \frac{1}{\sqrt{2 \pi}} , L_1 = \frac{t-\pi}{\sqrt{\frac{2}{3}}\pi}, L_{2}= \frac{t^2-(6\pi)(t-\pi)-\frac{4}{3}\pi^2}{\frac{82}{3}\sqrt{\frac{2}{5}}\pi^{5/2}}$$

this seems way too ugly to be true if one puts $L_0,L_1,L_2$ into p(x) ... what is the right way to do this?

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Your $L_1$ and $L_2$ are not quite right. –  Robert Israel Feb 1 '13 at 17:59
    
isnt: $||f||_{L2}= (\int_{0}^{2\pi} f^2)^{1/2}$ –  bakabakabaka Feb 1 '13 at 22:46
    
Yes, that's the $L^2$ norm. But you didn't specify which norm to use in $C[0,2\pi]$. In any case, that's not the problem with your $L_1$ and $L_2$. –  Robert Israel Feb 1 '13 at 23:09
    
I am a bit confused, where did i go wrong with my $L_1$, $L_2$? –  bakabakabaka Feb 2 '13 at 11:11
    
Your $\int_0^{2\pi} L_1^2 \ dt = \pi $, $\int_0^{2\pi} L_1 L_2\ dt = -\sqrt{60 \pi}/41$, and $\int_0^{2\pi} L_2^2\ dt = 61/41^2$. –  Robert Israel Feb 3 '13 at 3:20
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1 Answer

up vote 2 down vote accepted

Is $\| \cdot \|$ the supremum norm or the $L^2$ norm? For supremum norm (best uniform approximation), you can use the Remez algorithm. In Maple,

 > expand(numapprox[minimax](exp(t),t=0..2*Pi,2));

$$61.6583225- 114.6215740\,t+ 28.70843538\,{t}^{2}$$

For $L^2$ norm, you can use those orthogonal polynomials. I get

$$ {\frac {15}{4}}\,{\frac { \left( -{\pi }^{2}-3-3\,\pi +{{\rm e}^{2\, \pi }}{\pi }^{2}+3\,{{\rm e}^{2\,\pi }}-3\,{{\rm e}^{2\,\pi }}\pi \right) {t}^{2}}{{\pi }^{5}}}\\ -\frac{3}{2}\,{\frac { \left( 4\,{{\rm e}^{2\, \pi }}{\pi }^{2}+15\,{{\rm e}^{2\,\pi }}-6\,{\pi }^{2}-15-14\,{{\rm e} ^{2\,\pi }}\pi -16\,\pi \right) t}{{\pi }^{4}}}\\ +\frac{3}{2}\,{\frac {5\,{ {\rm e}^{2\,\pi }}-5-4\,{{\rm e}^{2\,\pi }}\pi -6\,\pi +{{\rm e}^{2\, \pi }}{\pi }^{2}-3\,{\pi }^{2}}{{\pi }^{3}}} $$

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