## Saturday, December 9, 2006

### eek! Topic: Calculus/S&S. Level: AIME/Olympiad.

Problem: (Problem-Solving Through Problems - 5.4.15) Let $f_0(x) = e^x$ and $f_{(n+1)}(x) = xf^{\prime}(x)$ for $n = 0, 1, 2, \ldots$. Show that

$\displaystyle \sum_{n=0}^{\infty} \frac{f_n(1)}{n!} = e^e$.

Solution: Well, the LHS looks suspiciously like a Taylor series, so maybe we can find a function. A good choice is $g(x) = e^{e^x}$. Notice that $f_0(x) = g(\ln{x})$. Testing a few derivatives, we hypothesize that

$f_n(x) = g^{(n)}(\ln{x})$.

By induction, we easily have

$\frac{d}{dx}[g^{(n)}(\ln{x})] = g^{(n+1)}(\ln{x}) \cdot \frac{1}{x}$,

which exactly satisfies

$g^{(n+1)}(\ln{x}) = x \frac{d}{dx}[g^{(n)}(\ln{x})]$

so we indeed have $f_n(x) = g^{(n)}(\ln{x})$. Then $f_n(1) = g^{(n)}(0)$ and the Taylor series of $g$ centered at zero is

$\displaystyle \sum_{n=0}^{\infty} \frac{g^{(n)}(0)x^n}{n!} = \sum_{n=0}^{\infty} \frac{f_n(1)x^n}{n!}$.

Hence our desired sum is the above series evaluated at $x = 1$, which is simply

$g(1) = e^{e^{1}} = e^e$.

QED.

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Comment: A neat problem resulting from Taylor series. They are useful in all sorts of ways, especially evaluating other series. If we can reduce a given series to the Taylor series of a function like we did in this problem, evaluating it becomes plugging in a point.

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Practice Problem: (Problem-Solving Through Problems - 5.4.17) Show that the functional equation

$\displaystyle f\left(\frac{2x}{1+x^2}\right) = (1+x^2)f(x)$

is satisfied by

$f(x) = 1+\frac{1}{3}x^2+\frac{1}{5}x^4+\frac{1}{7}x^6+\cdots$ with $|x| < 1$.

1. Well, x f(ix) = sin x but plugging that in doesn't make it very nice at all :(

Can't think of a good way to do this. I tried a trig substitution x = tan t/2 but even that wasn't very effective.

2. On AoPS, someone said f(z) = arctan(iz)/iz...