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 )! $?
what is the easiest way to represent $ \sqrt{1 + x} $ in series
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0Oops, that isn't the correct relation. – 2012-08-08
2 Answers
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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0is it [this relation](http://www.wolframalpha.com/input/?i=%281%2F2+-+n%29!+%3D+%28-1%29^%28n%2B1%29+pi%2F%28n-3%2F2%29!) – 2012-08-08
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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0strange ... $ 1 \over \sqrt 2 $ [link](http://www.wolframalpha.com/input/?i=Sum[%28-1%29^n+%282+n%29!%2F%28%28n!%29^2+4^n%29%2C+{n%2C+0%2C+Infinity}]) nevertheless helpful. Could you modify a bit ... so that it's helpful to remember. – 2012-08-08