What Does \left[ z^{n} \right] (ln(1-z))^{2} / (1-z)^{m+1} Represent?

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I need help with solving this weird equation...

What is:
\left[ z^{n} \right] (ln(1-z))^{2} / (1-z)^{m+1}?
 
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Can't have an equation without an equals sign. Ahyup.
 
Can you do these two Taylor series (at 0): (ln(1-z))^2 and 1/(1-z)^{m+1} ?
Then take the product of the two series.
 
Meekah said:
I need help with solving this weird equation...

What is:
\left[ z^{n} \right] (ln(1-z))^{2} / (1-z)^{m+1}?
Please restate your question. This is not an equation, it is an expression. What are you trying to do with it?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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