# Thread: [SOLVED] Series

1. ## [SOLVED] Series

I am given:

And it says to find the first few coefficients, c0,c1...,to c4, which I can't find. Then it says to find the radius of convergence R of the power series. I'm not sure why they have the fraction equaling the series. I know I can use the ratio or root test to find the radius of convergence, but what exactly is the above problem stating?
Thanks,
Matt

2. Okay, I got help and figured out that I should use the form:
$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$
So then I got to:
$\frac{6x}{10}\sum_{n=0}^{\infty}\left(\frac{-x}{10}\right)^n$
and $(-1)^n\left(\frac{1}{10}\right)^n x^n$
Now, I figured out that the radius of convergence is 10, but I'm still unsure for the coefficients.

3. Hello, matt3D!

$\frac{6x}{10+x} \:=\:\sum^{\infty}_{n=0} c_nx^n$

(a) Find the first few coefficients, to $c_4$
(b) Find the radius of convergence R of the power series.
Use long division . . .

$\begin{array}{ccccccc}
& & & \frac{6}{10}x & -\frac{6}{10^2}x^2 & +\frac{6}{10^3}x^3 & - \quad\cdots \\
& & ---&---&---&--- \\
10+x & ) & 6x \\
& & 6x & +\frac{6}{10}x^2 \\
& & --- & --- \\
& & & -\frac{6}{10}x^2 \\
& & & -\frac{6}{10}x^2 & -\frac{6}{10^2}x^2 \\
& & & --- & --- \\
& & & & \frac{6}{10^2}x^3 \\
& & & & \frac{6}{10^2}x^3 & + \frac{6}{10^3}x^4 \\
& & & & ---&---\end{array}$

We see that: . $\frac{6x}{10+x} \:=\:\frac{6}{10}x - \frac{6}{10^2}x^2 + \frac{6}{10^3}x^3 - \frac{6}{10^4}x^4 + \cdots$

(a) The first five coefficients are: . $\begin{Bmatrix}c_o &=& 0 \\ \\[-4mm]c_1 &=&\frac{6}{10} \\ \\[-4mm]c_2 &=& \text{-}\frac{6}{10^2} \\ \\[-4mm]c_3 &=& \frac{6}{10^3} \\ \\[-4mm]c_4 &=& \text{-}\frac{6}{10^4}\end{Bmatrix}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The general term is: . $a_n \:=\:(\text{-}1)^{n+1}\frac{6}{10^n}x^n$ . . . for $n \geq 1$

Ratio Test: . $\frac{a_{n+1}}{a_n} \;=\; \left|\frac{6x^{n+1}}{10^{n+1}} \cdot \frac{10^n}{6x^n}\right| \;=\;\left|\frac{x}{10}\right|$

And we have: . $\left|\frac{x}{10}\right| \:<\:1 \quad\Rightarrow\quad |x| \:< \:10$

(b) Therefore, the radius of convergence is: . $R \:=\:10$

4. Originally Posted by matt3D
I am given:

And it says to find the first few coefficients, c0,c1...,to c4, which I can't find. Then it says to find the radius of convergence R of the power series. I'm not sure why they have the fraction equaling the series. I know I can use the ratio or root test to find the radius of convergence, but what exactly is the above problem stating?
Thanks,
Matt
Let's find a more general power series. Consider

$\frac{ax^m}{b+cx^p}$

Simple algebra shows that this is equivalent to

$\frac{ax^m}{b}\cdot\frac{1}{1+\frac{c}{b}x^p}$

$=\frac{ax^m}{b}\cdot\frac{1}{1+\left(\frac{c^{\tfr ac{1}{p}}x}{b^{\tfrac{1}{p}}}\right)^p}$

Now we know that

$\frac{1}{1+x^p}=\sum_{n=0}^{\infty}(-1)^nx^{np}$

$\therefore\quad\frac{1}{1+\left(\frac{c^{\tfrac{1} {p}}x}{b^{\tfrac{1}{p}}}\right)^p}=\sum_{n=0}^{\in fty}(-1)^n\left(\frac{c^{\frac{1}{p}}x}{b^{\frac{1}{p}}} \right)^{np}$

$=\sum_{n=0}^{\infty}\frac{(-1)^nc^nx^{np}}{b^n}$

$\therefore\quad\frac{ax^n}{b+cx^p}=\frac{ax^m}{b}\ cdot\frac{1}{1+\left(\frac{c^{\tfrac{1}{p}}x}{b^{\ tfrac{1}{p}}}\right)^p}$

$=\frac{ax^m}{b}\sum_{n=0}^{\infty}\frac{(-1)^nc^nx^{np}}{b^n}$

$=a\sum_{n=0}^{\infty}\frac{(-1)^nc^nx^{np+m}}{b^{n+1}}$

Using the root test we check for the ROC...to do this we must find all x such that

$\lim_{n\to\infty}\sqrt[n]{\left|\frac{(-1)^nc^nx^{np+m}}{b^{n+1}}\right|}$
$=\frac{c|x|^p}{b}<1$

Solving for x gives

$|x|<\sqrt[p]{\frac{b}{c}}$

I'll leave you the endpoints...they are fairly easy.

5. Wow, thanks Soroban, I didn't know you could use long division. Mathstud28 I'm still trying to figure your post out, but thanks for the info!