# I believe this is a Taylor series. How do I approach it, and what formulas can I use to solve this type of problem?

Suppose that $|x| < 1$. Find the sum of the series

$$2x - 4x^3 + 6x^5 - 8x^7 + \cdots$$

I'm not looking for an answer. I want to know how to appropriately solve such a question though.

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Hint: Take the integral of the series – Brad Jul 19 '14 at 19:36
How would I take the integral of this series? – DavidN Jul 19 '14 at 19:38
Integrate each term. Find an expression for that series and then differentiate. – Brad Jul 19 '14 at 19:39
Another way to put it: the terms of this expression look a lot like the power rule for derivatives. Is there a Taylor series that you could differentiate and obtain this series? – Semiclassical Jul 19 '14 at 19:40
each term is going up by x^2 and I think with a little bit more plug and chug I can find out the constant integers that the coefficients are multiplied by with respect to b-a – DavidN Jul 19 '14 at 19:43

Occasionally it can be easier to find a formula for the integral or derivative of a series. $$S(x) = 2x - 4x^3 + 6x^5 + \cdots$$

$$\int \!S(x)\, \mathrm{d}x = x^2 - x^4 + x^6 + \cdots$$

$$\int \!S(x)\, \mathrm{d}x = \frac{x^2}{1+x^2}$$

$$S = \left(\int \!S(x)\, \mathrm{d}x\right)' = \left(\frac{x^2}{1+x^2}\right)' = \frac{2 x}{\left(x^2+1\right)^2}$$

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Let $S(x)$ be the sum. Then $$x^2S(x)=2x^3-4x^5+6x^7-\cdots.$$ Thus $$S(x)+x^2S(x)=2x+2x^3+2x^5+\cdots.\tag{1}$$ On the right we recognize a geometric series, first term $2x$, common ratio $x^2$, so sum $\frac{2x}{1+x^2}$ if $|x|\lt 1$.

Finally, using (1), we find that $S(x)=\dfrac{2x}{(1+x^2)^2}$.

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If the function is a rational function, then this method is guaranteed to find the function, provided you know enough terms of the series. This method can be generalized to the so-called differential approximants method, and is then capable of finding the function if it satisfies a differential equation of some finite order with polynomial coefficients. It requires a bit more work than simply trying some ad hoc manipulations, but the advantage is that it is an automatic method that is guaranteed to find any rational function, however complicated, provided enough terms are known.

Since the function f(x) is odd we can divide by x to obtain:

$$g(x) = \frac{f(x)}{x} = 2 - 4 x^2 + 6 x^4 - 8 x^6+10 x^8\cdots$$

Since this is function of $x^2$, we can consider $h(x)=g\left(\sqrt{x}\right)$:

$$h(x) = 2 - 4 x + 6 x^2 - 8 x^3 +10 x^4\cdots$$

Then let's multiply h(x) by a polynomial $q(x) = 1 + a_1 x + a_2 x^2 +a_3 x^3+\cdots a_n x^n$ and then we choose the coefficients $a_r$ such that the highest n-1 orders of the known terms of the series of the product $q(x) h(x)$ become zero. If we find that this requires us to choose all the $a_r$ for r larger than some number equal to zero, and all the coefficients of the series of $q(x) h(x)$ except the first few end up equal to zero, then we've hit the jackpot. Let's see what we get by taking $q(x)$ to be a 3rd degree polynomial:

$$\begin{split} q(x)h(x) = 2 &+ (-4 + 2 a_1) x + (6 - 4 a_1+2 a_2) x^2 +(-8 + 6 a_1-4 a_2+2 a_3) x^3\\&+(10-8 a_1 + 6 a_2 - 4a_3)x^4 \end{split}$$

Equating the coefficients of the 3 highest powers to zero yields:

$$\begin{split}a_1 &= 2\\ a_2 &=1\\ a_3 &=0 \end{split}$$

So, we can then guess that a higher order polynomial would not change the result, all the coefficients of powers higher than 2 will be zero. Also the product simplfies to $q(x)h(x) = 2$. This should not change if one includes more terms which is easily verified to be the case. We have thus found that:

$$(1+2 x + x^2) h(x) = 2$$

Therefore:

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

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