Expanding Maclaurin Series

**wilcofan3** · March 11th 2009, 09:40 AM

Expand $\frac {1} {(2+x)^3}$ as a Maclaurin Series. Determine the interval of convergence.

I took some derivatives, as I have been doing with Maclaurin Series so far.

$f(x)= \frac {1} {(2+x)^3}$

$\frac {dy} {dx}= \frac {-3} {(2+x)^4}$

$\frac {d^2 y} {dx^2}= \frac {(-3)(-4)} {(2+x)^5}$

$\frac {d^3 y} {dx^3}= \frac {(-3)(-4)(-5)} {(2+x)^6}$

$\frac {d^4 y} {dx^4}= \frac {(-3)(-4)(-5)(-6)} {(2+x)^7}$

I then plugged 0 into the function and its derivatives, and began to put it in the general expansion of Maclaurin Series, but I'm not sure what the general expression/sigma notation is.

**Scott H** · March 11th 2009, 03:39 PM

Your calculations for the derivatives are correct. The Maclaurin expansion is

$f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}{2!}+f'''(0)\frac {x^3}{3!}+\cdots$

Here, we have

$\begin{aligned} f(x)&=\frac{1}{2^3}+\frac{(-3)x}{2^4}+\frac{(-3)(-4)x^2}{2^5\cdot2!}+\frac{(-3)(-4)(-5)x^3}{2^6\cdot3!}+\cdots\\ &=\frac{1}{2^3}-\frac{3\cdot x}{2^4}+\frac{3\cdot 4\cdot x^2}{2^5\cdot 2!}-\frac{3\cdot 4\cdot 5\cdot x^3}{2^6\cdot 3!}+\cdots \end{aligned} $

Now we look closely at the terms and think of a formula for the $n$ th term in terms of $n$ (in our case, starting at $n=0$ will be better).

First, we notice that this is an alternating series. We can start with the factor $(-1)^n$ and then add the factors

$x^n,\,\frac{1}{2^{n+3}},\,\frac{1}{n!},\,3\cdot 4\cdot 5\cdots \cdot (n+1)\cdot(n+2)=\frac{(n+2)!}{2},$

giving us

$\sum_{n=0}^\infty\,(-1)^n\cdot\frac{(n+2)!}{2}\cdot\frac{1}{2^{n+3}}\cd ot x^n\cdot\frac{1}{n!}=\sum_{n=0}^\infty\,(-1)^n\frac{(n+2)!x^n}{2^{n+4}\cdot n!}.$

**wilcofan3** · March 11th 2009, 04:35 PM

Originally Posted by Scott H

Your calculations for the derivatives are correct. The Maclaurin expansion is

$f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}{2!}+f'''(0)\frac {x^3}{3!}+\cdots$

Here, we have

$\begin{aligned} f(x)&=\frac{1}{2^3}+\frac{(-3)x}{2^4}+\frac{(-3)(-4)x^2}{2^5\cdot2!}+\frac{(-3)(-4)(-5)x^3}{2^6\cdot3!}+\cdots\\ &=\frac{1}{2^3}-\frac{3\cdot x}{2^4}+\frac{3\cdot 4\cdot x^2}{2^5\cdot 2!}-\frac{3\cdot 4\cdot 5\cdot x^3}{2^6\cdot 3!}+\cdots \end{aligned} $

Now we look closely at the terms and think of a formula for the $n$ th term in terms of $n$ (in our case, starting at $n=0$ will be better).

First, we notice that this is an alternating series. We can start with the factor $(-1)^n$ and then add the factors

$x^n,\,\frac{1}{2^{n+3}},\,\frac{1}{n!},\,3\cdot 4\cdot 5\cdots \cdot (n+1)\cdot(n+2)=\frac{(n+2)!}{2},$

giving us

$\sum_{n=0}^\infty\,(-1)^n\cdot\frac{(n+2)!}{2}\cdot\frac{1}{2^{n+3}}\cd ot x^n\cdot\frac{1}{n!}=\sum_{n=0}^\infty\,(-1)^n\frac{(n+2)!x^n}{2^{n+4}\cdot n!}.$

Thank you so much! I just had a lot of trouble putting it together yet I knew what to do, but that helped so much!

One quick question: The interval of convergence would be from $(-\infty,\infty)$ , right?

**Scott H** · March 11th 2009, 06:36 PM

To show that $x$ belongs to the interval of convergence, we must show that the remainder of the Maclaurin series,

$R_n(x)=\frac{f^{(n)}(c_n)}{n!}x^n\,\,\,\,\,\,\,\,\ ,\,0\le c_n\le x$

tends to $0$ as $n\rightarrow\infty$ .

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