1. ## Taylor series expansion

Find the Taylor series for $\sqrt{x}$ 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:

$f(x) = \sum_{n=0}^{\infty} \frac{f^n(1)}{n!}(x-1)^n$

$f^1(1) = \frac{1}{2}$

$f^2(1) = \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)$

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

$f^4(1) = \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)$

So we have:

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

So I know the summation form must have $1+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\frac{1}{2^n}\text{something}}{n!}(x-1)^n$ but I can't find the pattern for $(1), (1)(-1), (1)(-1)(-3), (1)(1)(-3)(-5)$

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

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

4. Hello!

You need to find some kind of formula to:

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

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

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

By the way:

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

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:

$1+\frac{1}{2}(x-1) + \frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^n\frac{(1)(3)\cdots(2n-1)}{2^n}}{(n+1)!}(x-1)^{n+1}$

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)= $\displaystyle 2^n n!$.

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

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