# Math Help - Taylor Series #2

1. ## Taylor Series #2

Find the Taylor Series for f(x) centered at a=25. (assume f has a power series expansion)

$f(x) = \frac{1}{\sqrt{x}}$

I'm not going to type all these derivatives, but here are the 4th and nth:

$f^4(x) = (-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2}) x^{-\frac{9}{2}}$

So, when I calculated my nth derivative, I got:

$f^n(25) = (-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2}) (-\frac{9}{2})...-\frac{(2n+1)}{2}(25)^{2n+1}$

No idea....

EDIT: Do I need to use the binomial series here?

2. Originally Posted by mollymcf2009
Find the Taylor Series for f(x) centered at a=25. (assume f has a power series expansion)

$f(x) = \frac{1}{\sqrt{x}}$

I'm not going to type all these derivatives, but here are the 4th and nth:

$f^4(x) = (-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2}) x^{-\frac{9}{2}}$

So, when I calculated my nth derivative, I got:

$f^n(25) = (-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2}) (-\frac{9}{2})...-\frac{(2n+1)}{2}(25)^{2n+1}$

No idea....

EDIT: Do I need to use the binomial series here?
$f^{(n)}(x)=(-1)^n\frac{3 \times 5 \times \cdots \times (2n-1)}{2^n}x^{-\frac{2n+1}{2}}=(-1)^n\frac{(2n)!}{4^n n!}x^{-\frac{2n+1}{2}}.$ thus $f^{(n)}(25)=(-1)^n \frac{(2n)!}{4^n5^{2n+1}n!}.$ hence $\frac{1}{\sqrt{x}} =\sum_{n=0}^{\infty} (-1)^n \frac{\binom{2n}{n}}{4^n5^{2n+1}}(x - 25)^n.$

3. Originally Posted by NonCommAlg
$f^{(n)}(x)=(-1)^n\frac{3 \times 5 \times \cdots \times (2n-1)}{2^n}x^{-\frac{2n+1}{2}}=(-1)^n\frac{(2n)!}{4^n n!}x^{-\frac{2n+1}{2}}.$ thus $f^{(n)}(25)=(-1)^n \frac{(2n)!}{4^n5^{2n+1}n!}.$ hence $\frac{1}{\sqrt{x}} =\sum_{n=0}^{\infty} (-1)^n \frac{\binom{2n}{n}}{4^n5^{2n+1}}(x - 25)^n.$
Thanks! I am really struggling with this entire concept. Where did this $x^{-\frac{2n+1}{2}}$ come from? And where did this $4^n5^{2n+1}$ come from?
Sorry, I have been trying to understand this stuff for over a week and I can not get it...I desperately need someone to show it to me like I'm a 5 year old. Thanks so much for any help!

4. Originally Posted by mollymcf2009
Find the Taylor Series for f(x) centered at a=25. (assume f has a power series expansion)

$f(x) = \frac{1}{\sqrt{x}}$

I'm not going to type all these derivatives, but here are the 4th and nth:

$f^4(x) = (-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2}) x^{-\frac{9}{2}}$

So, when I calculated my nth derivative, I got:

$f^n(25) = (-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2}) (-\frac{9}{2})...-\frac{(2n+1)}{2}(25)^{2n+1}$

No idea....

EDIT: Do I need to use the binomial series here?
Try to spot the pattern:

$f\left(x\right)=\frac{1}{\sqrt{x}}\implies f\left(25\right)=\frac{1}{5}=\left(-\tfrac{1}{2}\right)^0\frac{1}{5^1}$
$f^{\prime}\left(x\right)=-\tfrac{1}{2}x^{-\frac{3}{2}}\implies f^{\prime}\left(25\right)=-\frac{1}{2}\frac{1}{5^3}=\left(-\tfrac{1}{2}\right)^1\cdot1\cdot\frac{1}{5^3}$
$f^{\prime\prime}\left(x\right)=-\tfrac{1}{2}\left(-\tfrac{1}{2}\right)\cdot1\cdot3\cdot x^{-\frac{5}{2}}\implies f^{\prime\prime}\left(25\right)=\left(-\tfrac{1}{2}\right)^2\cdot1\cdot3\cdot\frac{1}{5^5 }$

Thus, the nth derivative should take on the form:

$f^{(n)}\left(x\right)=\left(-\frac{1}{2}\right)^n\prod_{k=1}^{n}(2k-1)x^{-\frac{(2n+1)}{2}}\implies f^{(n)}\left(25\right)=\left(-\frac{1}{2}\right)^n\prod_{k=1}^{n}(2k-1)5^{-(2n+1)}$

Now, the taylor series should be $f\!\left(x\right)=\sum_{k=0}^{\infty}\frac{1}{k!}\ left(-\tfrac12\right)^k\left(\displaystyle\prod_{j=1}^k( 2j-1)\right)5^{-(2k+1)}\left(x-25\right)^k$

I plugged this into Maple, and it gave me this simplification (which really isn't much...)

$f(x)=\tfrac{1}{2}\sum_{k=0}^{\infty}\frac{1}{\sqrt {\pi}k!}\left(-\tfrac{1}{2}\right)^k2^{k+1}\Gamma\left(k+\tfrac{1 }{2}\right)5^{-(2k+1)}(x-25)^k$

But I see NonCommAlg beat me to this...and his way looks much, much neater...