Binomial series #2 stuck!

• Apr 16th 2009, 08:05 PM
mollymcf2009
Binomial series #2 stuck!
$\displaystyle f(x) = \frac{5}{(1+\frac{x}{10})^4}$

$\displaystyle = 5\left[1 + \frac{x}{10})^-4\right]$

$\displaystyle = 5 \left[1 + (-4)(\frac{x}{10}) + \frac{(-4)(-5)}{2!}(\frac{x}{10})^2 + \frac{(-4)(-5)(-6)}{3!}(\frac{x}{10})^3 + ... \right]$

But I don't know if this is right, or what my nth term would be, much less the power series...Thanks!!
• Apr 16th 2009, 08:15 PM
Chris L T521
Quote:

Originally Posted by mollymcf2009
$\displaystyle f(x) = \frac{5}{(1+\frac{x}{10})^4}$

$\displaystyle = 5\left[1 + \frac{x}{10})^-4\right]$

$\displaystyle = 5 \left[1 + (-4)(\frac{x}{10}) + \frac{(-4)(-5)}{2!}(\frac{x}{10})^2 + \frac{(-4)(-5)(-6)}{3!}(\frac{x}{10})^3 + ... \right]$

But I don't know if this is right, or what my nth term would be, much less the power series...Thanks!!

Before I start, are we dealing with a Maclaurin or Taylor Series? About what point are we constructing the series?
• Apr 16th 2009, 08:33 PM
mollymcf2009
Quote:

Originally Posted by Chris L T521
Before I start, are we dealing with a Maclaurin or Taylor Series? About what point are we constructing the series?

Sorry...

Here is the whole question:

Use the binomial series to expand the function as a power series.

$\displaystyle f(x) = \frac{5}{(1+\frac{x}{10})^4}$
• Apr 16th 2009, 09:28 PM
o_O
Binomial series are given by: $\displaystyle (1+y)^k = 1 + ky + \frac{k(k-1)}{2!}y^2$$\displaystyle {\color{white}.} \ + \ \frac{k(k-1)(k-2)}{3!}y^3 + \cdots + \frac{k(k-1)(k-2)\cdots(k - n+1)}{n!}y^n + \cdots for any \displaystyle k \in \mathbb{R} and if \displaystyle |y| < 1. Here, \displaystyle y = \tfrac{x}{10} and \displaystyle k = -4 . So: \displaystyle g(x) = \left( 1 + \frac{x}{10}\right)^{-4} \displaystyle g(x) = 1 + (-4)\left(\frac{x}{10}\right) + \frac{(-4)(-5)}{2!}\left(\frac{x}{10}\right)^2 + \frac{(-4)(-5)(-6)}{3!}\left(\frac{x}{10}\right)^3 + \cdots \displaystyle + \ \frac{(-4)(-5)(-6)\cdots(-4-n+1)}{n!}\left(\frac{x}{10}\right)^n + \cdots _______________ Let's look at the general term: \displaystyle \frac{(-4)(-5)(-6)\cdots(-3-n)}{n!}\frac{x^n}{10^n} Just looking at the expanded series, we see that with even \displaystyle n, the coefficient is positive and with odd \displaystyle n, the coefficient is negative. So, if we factor out the negative signs: \displaystyle \frac{(-1)^{n} (4)(5)(6)\cdots(n+3)}{10^n \cdot n!} x^n \cdot {\color{red}\frac{3!}{3!}} = \frac{(-1)^{n} (n+3)(n+2)(n+1)n!}{3! \cdot 10^n \cdot n!}x^n = \cdots etc. • Apr 16th 2009, 09:37 PM mollymcf2009 Quote: Originally Posted by o_O Binomial series are given by: \displaystyle (1+y)^k = 1 + ky + \frac{k(k-1)}{2!}y^2$$\displaystyle {\color{white}.} \ + \ \frac{k(k-1)(k-2)}{3!}y^3 + \cdots + \frac{k(k-1)(k-2)\cdots(k - n+1)}{n!}y^n + \cdots$

for any $\displaystyle k \in \mathbb{R}$ and if $\displaystyle |y| < 1$. Here, $\displaystyle y = \tfrac{x}{10}$ and $\displaystyle k = -4$ .

So: $\displaystyle g(x) = \left( 1 + \frac{x}{10}\right)^{-4}$

$\displaystyle g(x) = 1 + (-4)\left(\frac{x}{10}\right) + \frac{(-4)(-5)}{2!}\left(\frac{x}{10}\right)^2 + \frac{(-4)(-5)(-6)}{3!}\left(\frac{x}{10}\right)^3 + \cdots$ $\displaystyle + \ \frac{(-4)(-5)(-6)\cdots(-4-n+1)}{n!}\left(\frac{x}{10}\right)^n + \cdots$

_______________

Let's look at the general term: $\displaystyle \frac{(-4)(-5)(-6)\cdots(-3-n)}{n!}\frac{x^n}{10^n}$

Just looking at the expanded series, we see that with even $\displaystyle n$, the coefficient is positive and with odd $\displaystyle n$, the coefficient is negative.

So, if we factor out the negative signs: $\displaystyle \frac{(-1)^{n} (4)(5)(6)\cdots(n+3)}{10^n \cdot n!} x^n \cdot {\color{red}\frac{3!}{3!}} = \frac{(-1)^{n} (n+3)(n+2)(n+1)n!}{3! \cdot 10^n \cdot n!}x^n = \cdots$

etc.

Ok, I understand everything except where the 3!/3! got there.

Thanks!!!!
• Apr 16th 2009, 10:09 PM
o_O
Focus on the expression: $\displaystyle \frac{4 \cdot 5 \cdot 6 \cdots (n+3)}{n!}$

We can simplify it by multiplying by "1": $\displaystyle \frac{{\color{red} 1 \cdot 2 \cdot 3} \cdot 5 \cdot 6 \cdots (n+3)}{n! \cdot {\color{red}3!}} = \frac{(n+3)!}{n! \cdot 3!}$ $\displaystyle = \frac{(n+3)(n+2)(n+1)n!}{n! \cdot 3!} = \frac{(n+3)(n+2)(n+1)}{ 6}$

So you can see why I multiplied both top and bottom by $\displaystyle 3!$ so that I can simplify that long product into a simple factorial.