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How to expand $ \sqrt{1 + x}$. $$ \sum_{n = 0}^\infty {{\left ( 1 \over 2\right )!}x^n \over n! \left({1 \over 2 }- n\right )!} = 1 + \sum_{n = 1}^\infty {{\left ( 1 \over 2\right )!}x^n \over n! \left({1 \over 2 }- n\right )!}$$ How can I simplify $ \left({1 \over 2 }- n\right )! $?

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A related thread. –  J. M. Aug 8 '12 at 13:50
    
yeah ... i know that, but can I manipulate those things around to prove that they are equivalent? –  hasExams Aug 8 '12 at 13:51
    
thanks ... i think i should try it. –  hasExams Aug 8 '12 at 13:53
    
Well, you're not supposed to expand out $\binom{1/2}{n}$ as factorials in this case, if you're not willing to use the formulae in my answer in the question I linked to. See Byron's approach there. –  J. M. Aug 8 '12 at 13:54
    
no .... i'll stick with yours. –  hasExams Aug 8 '12 at 13:56
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2 Answers

up vote 3 down vote accepted

Per OP's request:

One way to deal with the binomial coefficients in your series would be to use the duplication formula for the factorial:

$$\binom{2n}{n}=\frac{4^n}{\sqrt \pi}\frac{\left(n-\frac12\right)!}{n!}$$

and the reflection formula

$$\left(-n-\frac12\right)!\left(n-\frac12\right)!=(-1)^n\pi$$

The other way is to use the following definition, valid for any complex $x$:

$$\binom{x}{n}=\frac1{n!}\prod_{k=0}^{n-1}(x-k)$$

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is it this relation –  hasExams Aug 8 '12 at 14:32
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Just to motivate a more elementary solution.

Let $$f(x)=\sqrt{x+1}=(x+1)^{1/2}$$

Now, we expand around $x=0$.

$$f'(x)=\frac 1 2 (x+1)^{1/2-1}$$

$$f''(x)=\frac 1 2 \left(\frac 1 2-1 \right)(x+1)^{1/2-2}$$

$$f'''(x)=\frac 1 2 \left(\frac 1 2-1 \right)\left(\frac 1 2-2 \right)(x+1)^{1/2-3}$$

By induction, we get

$$f^{(n)}(x)=\prod_{k=0}^{n-1} \left(\frac 1 2 -k\right)(x+1)^{1/2-n}$$

Thus, we get

$$f^{(n)}(0)=\prod_{k=0}^{n-1} \left(\frac 1 2 -k\right)$$

$$f^{(n)}(0)=\prod_{k=0}^{n-1} \left(\frac {1 -2k}{2}\right)$$

$$f^{(n)}(0)=(-1)^{n}\frac{1}{2^{n}}\prod_{k=0}^{n-1} \left({2k-1}\right)$$

Now, note the product is exclusively of odd factors up to $(2n-3)$. What we do is fill in the missing even numbers, and divide to keep things the same. Note that $(2n-2)\cdots 4\cdot 2=2^{n-1} (n-1)!$

$${f^{(n)}}(0) = {( - 1)^n}\frac{1}{{{2^n}{2^{n - 1}}(n - 1)!}}\prod\limits_{k = 1}^{n - 1} {\left( {2k} \right)} \prod\limits_{k = 0}^{n - 1} {\left( {2k - 1} \right)} $$

$${f^{(n)}}(0) = \frac{{{{( - 1)}^n}}}{{{2^n}{2^{n - 1}}(n - 1)!}}\prod\limits_{k = 1}^{2n - 2} k = {( - 1)^n}\frac{1}{{{2^{2n - 1}}}}\frac{{\left( {2n - 2} \right)!}}{{\left( {n - 1} \right)!}}$$

Thus, we have that

$$\sqrt {1 + x} = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{{2^{2n - 1}}}}} \frac{{\left( {2n - 2} \right)!}}{{n!\left( {n - 1} \right)!}}{x^n}$$

$$\sqrt {1 + x} = 1 + \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}}}{{{2^{2n - 2}}}}} \frac{{\left( {2n - 2} \right)!}}{{\left( {n - 1} \right)!\left( {n - 1} \right)!}}\frac{{{x^n}}}{n}$$

$$\sqrt {1 + x} = 1 + \frac{1}{2}\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{{2^{2n}}}}} \frac{{\left( {2n} \right)!}}{{n!n!}}\frac{{{x^{n + 1}}}}{{n + 1}}$$

$$\sqrt {1 + x} = 1 + \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^{n + 1}}}}{{{2^{2n + 1}}}}} {2n\choose n}\frac{{{x^{n + 1}}}}{{n + 1}}$$

Note this gives that

$$\frac{1}{{\sqrt {1 + x} }} = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^{n + 1}}}}{{{2^{2n}}}}} {2n\choose n}{x^n}$$

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thanks ... looks like i've been making mistake!! –  hasExams Aug 8 '12 at 15:08
    
You derived the one for the reciprocal... –  J. M. Aug 8 '12 at 15:10
    
strange ... $ 1 \over \sqrt 2 $ link nevertheless helpful. Could you modify a bit ... so that it's helpful to remember. –  hasExams Aug 8 '12 at 15:12
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