# Thread:

1. ## Taylor series expansion

Find the Taylor series for $%5Csqrt%7Bx%7D$ centered around $x=1$.

I've gotten pretty close but I don't know how to put it back into summation format. This is my working:

$f%28x%29%20=%20%5Csum_%7Bn=0%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7Bf%5En%281%29%7D%7Bn%21%7D%28x-1%29%5En$

$f%5E1%281%29%20=%20%5Cfrac%7B1%7D%7B2%7D$

$f%5E2%281%29%20=%20%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Cleft%28-%5Cfrac%7B1%7D%7B2%7D%5Cright%29$

$f%5E3%281%29%20=%20%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Cleft%28-%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Cleft%28-%5Cfrac%7B3%7D%7B2%7D%5Cright%29$

$f%5E4%281%29%20=%20%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Cleft%28-%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Cleft%28-%5Cfrac%7B3%7D%7B2%7D%5Cright%29%5Cleft%28-%5Cfrac%7B5%7D%7B2%7D%5Cright%29$

So we have:

$\sqrt{x} = 1 + \frac{\frac{1}{2}}{1!}(x-1) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}(x-1)^2+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}(x-1)^3 + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{4!}(x-1)^4 + \cdots$

So I know the summation form must have $1+%5Csum_%7Bn=1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%28-1%29%5E%7Bn+1%7D%5Cfrac%7B1%7D%7B2%5En%7D%5Ctext%7Bsomething%7D%7D%7Bn%21%7D%28x-1%29%5En$ but I can't find the pattern for $%281%29,%20%281%29%28-1%29,%20%281%29%28-1%29%28-3%29,%20%281%29%281%29%28-3%29%28-5%29$

2. $something=\frac{(2n)!}{n!2^n}$

3. Originally Posted by Also sprach Zarathustra
$something=\frac{(2n)!}{n!2^n}$
Hi how is it (2n)! ? Isn't that even?

4. Hello!

You need to find some kind of formula to:

$1\cdot 3 \cdot 5 \cdot 7 \cdot ... \cdot (2n-1)$

but, the above equals to: $\frac{(2n)!}{n!2^n}$

Check this formula for $n=4$.
Try to prove it! It is easy!

By the way:

${(2n)!}=1\cdot 2 \cdot 3 \cdot 4 \cdot ... \cdot (2n)$

5. yup that was exactly the formula for the multiplication of odd numbers however how do you prove it? in other words how did you get that 'formula'?

Also:

can't I have something like this:

$1+%5Cfrac%7B1%7D%7B2%7D%28x-1%29%20+%20%5Cfrac%7B1%7D%7B2%7D%5Csum_%7Bn=1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%28-1%29%5En%5Cfrac%7B%281%29%283%29%5Ccdots%282n-1%29%7D%7B2%5En%7D%7D%7B%28n+1%29%21%7D%28x-1%29%5E%7Bn+1%7D$

6. A product of even numbers, 2*4*6*8*...*(2n) can be written as (2*1)(2*2)(2*3)(2*4)*...(2*n). Now factor out those "2"s: (2*2*2*2...2)(1*2*3*4...*n)= $2^n n!$.

To simplify a product of odd numbers, 1*3*5*7*...*(2n-1), multiply and divide by the "missing" even numbers:
$\frac{1*2*3*4*5*6*7*8...*(2n-1)(2n)}{2*4*6*8*...(2n)}$

The numerator is now (2n)! and the denominator is $2^n n!$.
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