# Binomial Series

• March 14th 2008, 09:09 AM
BKennedy
Binomial Series
Heres my Question:

Use the binomial series to expand the function
f(x)=(3 x)\(sqrt(6+x))
as a power series

Find:
C0=
C1=
C2=
C3=
C4=
• March 14th 2008, 01:23 PM
TheEmptySet
Quote:

Originally Posted by BKennedy
Heres my Question:

Use the binomial series to expand the function
f(x)=(3 x)\(sqrt(6+x))
as a power series

Find:
C0=
C1=
C2=
C3=
C4=

$(1+x)^k=1+kx+\frac{k(k-1)}{2!}x^2+\frac{k(k-1)(k-2)}{3!}x^3+...=\sum_{n=0}^{\infty}\binom{k}{n}x^n$

where

$\binom{k}{n}=\frac{k(k-1)(k-2) \cdot\cdot\cdot (k-n+1)}{n!}$

$n \ge 1$ and $\binom{k}{0}=1$

So we need to fix up the function so it can fit the form of the theorem..

$\frac{3x}{\sqrt{6+x}}=\frac{3x}{\sqrt{6(1+\frac{x} {6})}}=$

$=\frac{3x}{\sqrt{6}} \cdot\frac{1}{\sqrt{1+\frac{x}{6}}}=\frac{3x}{\sqr t{6}}\cdot \left(1+\frac{x}{6} \right)^{-1/2}$

Now that it fits the form needed we can expand the series

$\frac{3x}{\sqrt{6}}\left(1-\frac{1}{2}\left( \frac{x}{6} \right)^1+\frac{\frac{-1}{2} \cdot \frac{-3}{2}}{2!}\left(\frac{x}{6} \right)^2 +\frac{\frac{-1}{2} \cdot \frac{-3}{2} \cdot \frac{-5}{2}}{3!}\left( \frac{x}{6} \right)^3 + \frac{\frac{-1}{2} \cdot \frac{-3}{2} \cdot \frac{-5}{2} \cdot \frac{-7}{2}}{4!}\left( \frac{x}{6} \right)^4+... \right)$

from here it is just multiplication. Good luck.(Rock)