# Question on Taylor power series

• Apr 1st 2008, 12:48 PM
emttim84
Question on Taylor power series
Ok, so the instructions state to apply Taylor's Theorum to find the power series (centered at c) for the function, and find the radius of convergence.

Problem: $f(x)=x^{1/2}$ with the center at $c=1$

I took the first through the fourth derivative and then evaluated them at the center to get an idea of the pattern, here's what I got:

1st Derivative: $f'(x)=\frac{1}{2x^{1/2}}$ Evaluation: $f'(1)=\frac{1}{2}$

2nd Derivative: $f''(x)=-\frac{1}{4x^{3/2}}$ Evaluation: $f''(1)=-\frac{1}{4}$

3rd Derivative: $f'''(x)=\frac{3}{8x^{5/2}}$ Evaluation: $f'''(1)=\frac{3}{8}$

4th Derivative: $f''''(x)=-\frac{15}{16x^{7/2}}$ Evaluation: $f''''(1)=-\frac{15}{16}$

I can use the Taylor series formula easily enough, however, I'm not sure how they came up with the answer for $f^n(1)=(-1)^{n-1}\frac{1*3*....*5*(2n-3)}{2^n}$ which holds true when n is equal to or greater than 2.

Now, I can write up the power series and perform the ratio test to find the radius of convergence on my own easily enough, however, I'm unfamiliar with this notation. I understand the $(-1)^{n-1}$ in the answer because that must be true for the signs to alternate back and forth. Due to the denominators in each evaluation, I can also see where $2^n$ comes into play since the denominator is just 2 raised to an exponent equal to n. What I don't see is where that frickin' $1*3*....*5*(2n-3)$ comes into play. (Headbang) Can anyone explain to me what that notation is, why it's part of the answer and how I might have to treat it when taking a limit? Thanks in advance guys.
• Apr 1st 2008, 01:22 PM
wingless
I'll ignore the minus signs because we can make them with $(-1)^n$ later.

$f^2 = \frac{1}{2}\cdot \frac{1}{2}$

$f^3 = \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{3}{2}$

$f^4 = \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{3}{2} \cdot\frac{5}{2}$
....

Did you see the pattern?

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

First find $\prod^{n-2}_{k=0}\left ( \frac{1}{2}+k \right )$.

$\prod^{n-2}_{k=0}\left ( \frac{1}{2}+k \right )$

$\prod^{n-2}_{k=0}\left ( \frac{2k+1}{2} \right )$

$\left ( \frac{1}{2} \right ) \cdot \left ( \frac{3}{2} \right ) \cdot \left ( \frac{5}{2} \right ) \cdot \left ( \frac{7}{2} \right ) \cdot \cdot \cdot \cdot \left ( \frac{2n-3}{2} \right ) \cdot$

And that makes,

$\frac{1 \cdot 3\cdot 5 \cdot 7 \cdot \cdot \cdot \cdot \left ( 2n-3\right ) }{2^{n-1}}$
• Apr 1st 2008, 01:22 PM
Mathstud28
Isnt that
just (2n-3)!
• Apr 1st 2008, 01:33 PM
wingless
Quote:

Originally Posted by Mathstud28
just (2n-3)!

No, (2n-3)! = 1.2.3.4.5....(2n-3)

But this is 1.3.5.7.....(2n-3)
(Without the even numbers)
• Apr 1st 2008, 01:41 PM
Mathstud28
Forgive me I am stupid
but then isnt it (2n-1)! since 2n-1 gives the series of odd numbers?