MHB Finding Maclaurin series of a natural log function

tmt1
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I need to find the Maclaurin series of this function:

$$f(x) = ln(1 - x^2)$$

I know that $ln(1 + x)$ equals

$$\sum_{n = 1}^{\infty}\frac{(-1)^{n - 1} x^n}{n}$$

Or, $x - \frac{x^2}{2} + \frac{x^3}{3} ...$

If I swap in $-x^2$ for x, I get:

$$-x^2 + \frac{x^4}{2} - \frac{x^5}{3} + \frac{x^6}{4} ...$$

Is this a valid method?

However, the answer in the text says

$$- (x^2 + \frac{x^4}{2} + \frac{x^6}{3} ...)$$
 
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tmt said:
I need to find the Maclaurin series of this function:

$$f(x) = ln(1 - x^2)$$

I know that $ln(1 + x)$ equals

$$\sum_{n = 1}^{\infty}\frac{(-1)^{n - 1} x^n}{n}$$

Or, $x - \frac{x^2}{2} + \frac{x^3}{3} ...$

If I swap in $-x^2$ for x, I get:

$$-x^2 + \frac{x^4}{2} - \frac{x^5}{3} + \frac{x^6}{4} ...$$

Is this a valid method?

However, the answer in the text says

$$- (x^2 + \frac{x^4}{2} + \frac{x^6}{3} ...)$$

$\displaystyle \begin{align*} \ln{ \left( 1 - x^2 \right) } &= \ln{ \left[ \left( 1 - x \right) \left( 1 + x \right) \right] } \\ &= \ln{ \left( 1 - x \right) } + \ln{ \left( 1 + x \right) } \end{align*}$

so that means

$\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \,\left[ \ln{ \left( 1 - x^2 \right) } \right] &= \frac{\mathrm{d}}{\mathrm{d}x}\,\left[ \ln{ \left( 1 - x \right) } + \ln{\left( 1 + x \right) } \right] \\ &= -\frac{1}{1 - x} + \frac{1}{1 + x} \end{align*}$

Can you relate each of these to a geometric series?
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...

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