# Taylor series expansion

• Jun 26th 2010, 10:53 AM
usagi_killer
Taylor series expansion
Find the Taylor series for http://stuff.daniel15.com/cgi-bin/ma...%5Csqrt%7Bx%7D centered around http://stuff.daniel15.com/cgi-bin/mathtex.cgi?x=1.

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

http://stuff.daniel15.com/cgi-bin/ma...D%28x-1%29%5En

http://stuff.daniel15.com/cgi-bin/ma...%7B1%7D%7B2%7D

http://stuff.daniel15.com/cgi-bin/ma...%7D%5Cright%29

http://stuff.daniel15.com/cgi-bin/ma...%7D%5Cright%29

http://stuff.daniel15.com/cgi-bin/ma...%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 http://stuff.daniel15.com/cgi-bin/ma...D%28x-1%29%5En but I can't find the pattern for http://stuff.daniel15.com/cgi-bin/ma...8-3%29%28-5%29

• Jun 26th 2010, 11:43 AM
Also sprach Zarathustra
$something=\frac{(2n)!}{n!2^n}$
• Jun 26th 2010, 12:54 PM
usagi_killer
Quote:

Originally Posted by Also sprach Zarathustra
$something=\frac{(2n)!}{n!2^n}$

Hi how is it (2n)! ? Isn't that even?
• Jun 26th 2010, 01:09 PM
Also sprach Zarathustra
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:

${(2n)!}=1\cdot 2 \cdot 3 \cdot 4 \cdot ... \cdot (2n)$
• Jun 27th 2010, 12:17 AM
usagi_killer
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:

http://stuff.daniel15.com/cgi-bin/ma...29%5E%7Bn+1%7D
• Jun 27th 2010, 04:46 AM
HallsofIvy
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 $2^n n!$.