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How can we construct all the Chebyshev rational approximations of degree $3$ for $f(x) = \cos(x)$.

So, I note that we first get the Taylor series of $\cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} +\ldots$ and try to use Pade's theorem, but I don't see how to organize this for this example.


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I think the coefficient for $x^4$ is $1/24$. – Yimin Feb 16 '13 at 7:40
And Taylor expansion is just for Pade's approximation m =3, n =0 case. You might have to try to give (m,n) = (2,1),(1,2),(0,3) cases. – Yimin Feb 16 '13 at 7:41
Thanks but I could include the other 3 cases. However, how can we organize for this example? Can you show a calculation? Thanks – Buddy Holly Feb 16 '13 at 19:50
The calculation really consumes time. You may follow this – Yimin Feb 16 '13 at 20:31
I just don't understand it. Can you please work out some here? Thanks so much – Buddy Holly Feb 17 '13 at 7:38
up vote 1 down vote accepted

For $(m,n) = (3,0)$ ,the Taylor expansion is just enough.

For $(m,n) = (2,1)$, we expect our approximation as $F(x) = \displaystyle\frac{a_0+a_1x^1+a_2x^2}{1+b_1x}$ which approximates our goal up to order 4.

You have to show that $F(0) =\cos(0) = 1$, $F'(0) = (\cos)'(0) = 0$, $F''(0) = (\cos)''(0) = -1$, $F(0)''' = (\cos)'''(0) = 0$. 4 equations solve 4 unknowns. Or you may take

$$F(x) = (a_0+a_1x^1+a_2x^2)(1-b_1x+b_1^2x^2-b_1^3x^3+\dots)$$

expand it and the coefficients should coincide with the Taylor expansion for the first several terms.

For $(m.n) = (1,2)$, take $F(x) =\dfrac{a_0+a_1x^1}{1+b_1x+b_2x^2}$.

For $(m,n) = (0,3)$, take $F(x) =\dfrac{1}{1+b_1x+b_2x^2+b_3x^3}$.

And do the same thing as the second case does. It consumes time.

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