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One of my math professors gave me the following challenge. It isn't graded, it's just for fun.

Consider the function:

\begin{equation*} f_n(x)=x+3^3x^3+5^3x^5+...+(2n-1)^3x^{2n-1},~x \in (0, 1). \end{equation*}

I want to find which of the following functions $f_n$ is getting close to as $n$ gets larger:

$ \displaystyle a)\frac{x(x+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$ $\displaystyle b) \frac{x(x^2+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$

$\displaystyle c) \frac{x^2(x+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$ $\displaystyle d) \frac{x^2(x^2+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$

Based on some tests i ran in mathematica by giving $n$ and $x$ values, it looks like $b)$ is the answer, but I am not sure. Can anyone confirm or deny this, and show how one might find the right answer, either with pen and paper or by using mathematica or maple or some other software?

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up vote 6 down vote accepted

Yes looks like it must be b)

Consider the following:

There is only one $x$ factor, so this eliminates choices c) and d).

The function is odd, i.e $f(-x) = -f(-x)$, so this eliminates a).

Thus b) must be the answer.

To find the answer by hand you can do the following:

Start with

$$f(x) = x + x^3 + x^5 + \cdots = \frac{x}{1-x^2} $$

Now differentiate

$$f'(x) = 1 + 3x^2 + 5x^4 + \cdots $$

Multiply by $x$

$$f'(x)x = x + 3x^3 + 5x^5 + \cdots $$

Differentiate again

$$(f'(x)x)' = 1 + 3^2 x^2 + 5^2 x^4 + \cdots $$

Multiply by $x$, differentiate again and then multiply by $x$ to give the answer.

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This is a nice solution! To anyone checking this, you should know there's one small error: after the third differentiation, you need to multiply by x one more time. – Michael Lugo Sep 4 '10 at 16:00
@Michael: You are right. I will edit the answer. Thanks! – Aryabhata Sep 4 '10 at 16:05

Yes, b).

In Maple, if you do

 map(factor, sum((2*i-1)^3*x^(2*i-1),i=1..n));

The result will be 2 terms, one which is shape $(x^2)^{(n+1)}*r(n,x)$ with $r(n,x)$ polynomial in $n$ and rational in $x$, and the second term is (exactly) (b).

What this is is actually writing $$\sum_{i=1}^{n} (2i-1)^3x^{2i-1} = \sum_{i=1}^{\infty} (2i-1)^3x^{2i-1} - \sum_{i=N-1}^{\infty} (2i-1)^3x^{2i-1}$$ (and the term order gets flipped in Maple's output). Both the resulting sums are 'easy' to do by hand.

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Here's yet another way to verify that it's (b). Let Mathematica do partial fractions decomposition:


This gives $\frac{1}{2 (x+1)}-\frac{7}{2(x+1)^2}+\frac{6}{(x+1)^3}-\frac{3}{(x+1)^4}+\frac{1}{2 (x-1)}+\frac{7}{2(x-1)^2}+\frac{6}{(x-1)^3}+\frac{3}{(x-1)^4}$, which can be expanded into power series using the geometric series and its derivatives.

Of course this is ugly, and requires you to know the answer in advance. Moron's answer is much better. ;-)

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This one might be another way of approaching the problem:

$f_n$ is a polynomial, so it is the Taylor expansion of itself at every point; then, we can pick any point in $[0,1]$, say 0, and see what the Taylor expansions of your four possibilities are. In this case you'll see that $b)$ has a Taylor expansion at $x =0$ which is exactly the same as you $f_n$, so that's your answer.

On the other hand, since I did the calculations on my computer, I don't really know how messy it is to find the answer this way. (As Moron pointed out, also a quick check of the simmetry of the functions gives the right answer)

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