Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...

share|cite|improve this question
up vote 1 down vote accepted

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 $$\frac d{dx} x_k = (\ln b)^k\cdot b^{x_{k-1}+x_{k-2}+\ldots+ x}.$$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.