**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 $.

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

ReplyDeleteCan'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.

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

ReplyDelete