Why is Adding One Necessary in Binomial Series for 1/\sqrt{1-x^{2}}?

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When using a binomial series to expand 1/\sqrt{1-x^{2}} I come up with the correct answer except that I do not add the number one to my answer. Why do I have to add one to the series, should this not arise when calculating the sum?
 
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If x is zero the result is 1. You should definitely have a 1 in the series. What DID you do?
 
When expanding it I come up with the following result:

\sum (1*3* \ldots (2n-1)*x^{2n}) / 2^{n}n!

According to the answer key the answer is:

1 + \sum ((1*3* \ldots (2n-1)*x^{2n}) / 2^{n}n!)

Where does the one come from?
 
You didn't put limits on your summations. It looks like in the books answer the limits are 1 to infinity. In your answer they are 0 to infinity. Isn't 1 the n=0 term in your series?
 
Yes ... they separated out the n=0 term, because they thought the student would not understand the product 1\cdot3\cdots(-1)[/itex]
 
Thanks, I understand it now.
 
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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