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Assume the following situation. I want to evaluate the derivative of a function for which I have a power series. In principle this is well known: just insert the derivatives at each coefficient: $$ S(x) = \sum_{k=0}^\infty a_k \cdot x^k \to S(x)' = \sum_{k=0}^\infty (k+1)\cdot a_{k+1} \cdot x^k $$ and evaluate. So far, so good.

The convergence-radius of the power series is small, but fortunately I can reexpress it as $$ S(x) = x_0-x_1+x_2-\ldots - x_{m-1}+\sum_{k=0}^\infty a_k \cdot x_m^k $$ and I do not know, how I reflect the leading $x_k$ into the derivative. It is with a transfer-function $f(x)=b^x-1$ that $$x_1=b^x-1,x_2=b^{x_1}-1,\ldots x_m=b^{x_{m-1}}-1$$ such that $x_m$ is in the radius of the power series for $S(x)$. So my question is now how to include that leading terms in the formula for the derivative?

Is it simply to write the derivative $$ S(x)' = f'(t)_{|t=x} - f'(t)_{|t=x_1} + \ldots - f'(t)_{|t=x_{m-1}} + \sum_{k=0}^\infty (k+1)\cdot a_{k+1} \cdot x_m^k \qquad \text{???}$$ but this is just a guess...

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Note that you need to employ the chain rule, so with $x_k=\underbrace{f(\cdots(f}_k(x))\cdots)=f(x_{k-1})$ you have $$\frac d{dx} x_k = f'(x_{k-1})\cdot\frac d{dx} x_{k-1} $$ so by induction $$\frac d{dx} x_k = f'(x_{k-1})f'(x_{k-2})\cdot\ldots\cdot f'(x).$$ And of course $f'(t)=\ln b \cdot b^t$, so ultimately $$\frac d{dx} x_k = (\ln b)^k\cdot b^{x_{k-1}+x_{k-2}+\ldots+ x}.$$

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Yes, the chain-rule... and the first term in $S(x)'$ is $f'(t)_{|t=x}=1$ ? This looks so unfamiliar, but must obviously be true... <grrrm> –  Gottfried Helms Dec 14 '12 at 23:07
    
Actually, I don't know. You define $x_1=f(x)$, but didn't define $x_0$. –  Hagen von Eitzen Dec 15 '12 at 9:03
    
I've got it now, I needed to make a big table to see the pattern; perhaps I'll put my result in an extra answer. Thanks anyway, Hagen! –  Gottfried Helms Dec 28 '12 at 20:45

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