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I have an exercise which asks me to find polynomials $P$ and $Q$ with a degree $2$ that satisfy $$\exp(z)= \dfrac{P(z)}{Q(z)} + O(z^5)\ \text{for} \ z\to 0$$ My question is: Are they actually unique ($\rightarrow $ can something be actually unique if it is expressed with Landau?) or can I just take 2 polynomials of degree 2 for example in type of our known RK-stability-function which is the $$ R(z)=\sum\limits_{k=0}^s \dfrac{z^k}{k!}$$

Thanks for your help!

Xi Tong

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  • $\begingroup$ You may want to take a look at this answer. These are known as Padé approximants (learned this bit from J.M.). Use that buzzword to search for more information. $\endgroup$ – Jyrki Lahtonen Nov 19 '13 at 8:17
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They are not unique if you allow multiplying both $P$ and $Q$ with a constant. I would make an ansatz $Q(z)=1+az+bz^2$, $P(z)=c+dz+ez^2$ and compare $\frac PQ$ with the Taylor expansion of $\exp(z)$. Note that we have five conditions for five unknowns $a,b,c,d,e$, so everything should work out nicely.

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  • $\begingroup$ Thanks! How do I get unique a,b,d and e? Because if I just set P/Q = Taylor expansion of exp(z) I only get some expression for d and e defined over a and b. So... it doesn't give me a unique solution :/ $\endgroup$ – Xi Tong Nov 19 '13 at 8:46
  • $\begingroup$ @XiTong.Develop Exp[z] as a Taylor series up to fourth order (this is siomple). Multiply it by Q; substract P from the result. Cancel the first terms. You end with five simple equations for five unknowns (a,b,c,d,e). This is just rephrasing what Hagen von Eitzen wrote. $\endgroup$ – Claude Leibovici Nov 19 '13 at 8:51
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Since Hagen von Eitzen provided the solution, just let me add this comment : what they ask you is to find the Pade[2,2] approximant of Exp[z]. This kind of expansion is always much better than Taylor series for the same number of coefficients.

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