2
$\begingroup$

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...

1 Answers 1

1

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

  • 0
    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!2012-12-28