# Thread: Question on interval of convergence

1. ## Question on interval of convergence

Question 37 from 11.8 of Stewart's Early Transcendentals -

A function is defined by
f(x) = 1 + 2x + x^2 + 2x^3 + x^4.....

that is, its coefficients are C (2n) = 1 and C (2n+1) = 2 for all n >= 0.
Find the interval of convergence of the series and find an
explicit formula for f(x)

The answer is given at the end of the book as
Interval of convergence = (-1, 1)
f(x) = (1 + 2x)/(1 - x^2)

Anyone know how to calculate f(x) because there are no examples like this in the book. I can't even see how that value of f(x) is correct, how does f(x) relate to the given power series, how does f(x) = (1 + 2x)/(1 - x^2) = 1 + 2x + x^2 + 2x^3 + x^4.....

2. ## Re: Question on interval of convergence

Originally Posted by nukenuts
Question 37 from 11.8 of Stewart's Early Transcendentals -

A function is defined by
f(x) = 1 + 2x + x^2 + 2x^3 + x^4.....

that is, its coefficients are C (2n) = 1 and C (2n+1) = 2 for all n >= 0.
Find the interval of convergence of the series and find an
explicit formula for f(x)

The answer is given at the end of the book as
Interval of convergence = (-1, 1)
f(x) = (1 + 2x)/(1 - x^2)

Anyone know how to calculate f(x) because there are no examples like this in the book. I can't even see how that value of f(x) is correct, how does f(x) relate to the given power series, how does f(x) = (1 + 2x)/(1 - x^2) = 1 + 2x + x^2 + 2x^3 + x^4.....
Here is a hint to get you started. You will need to justify why this is allowed.

$1+2x+x^2+2x^3+x^4+2x^5+...=(1+x+x^2+x^3+x^4+...)+( x+x^3+x^5+...)$

We can write each of these in summation notation

$\sum_{n=0}^{\infty}x^n+x\sum_{n=0}^{\infty}\left( x^2\right)^n$

Post back if you get stuck

3. ## Re: Question on interval of convergence

You can, very generally, use the "ratio test" or "root test" to find the radius of convergence. Most people find the ratio test easier.
The positive series $\sum a_n$ converges if $lim_{n\to\infty}\frac{a_{n+1}}{a_n}$ is less than 1 and diverges if that limit is greater than 1. For general series, you can take the absolute value to test for "absolute convergence".

Here, if n is even, n+1 is odd so we would have $a_n= x^n$ and $a_{n+1}= 2x^{n+1}$.
$\left|\frac{a_{n+1}}{a_n}\right|= \left|\frac{2x^{n+1}}{x^n}\right|= 2|x|$
If no is odd, n+1 is even and we would have $a_n= 2x^n$ and $a_{n+1}= x^{n+1}$.
$\left|\frac{a_{n+1}}{a_n}\right|= \left|\frac{x^{n+1}}{2x^n}\right|= \frac{1}{2}|x|$.

Neither of those ratios depends on n so, for convergence we must have both 2|x|< 1 and (1/2)|x|< 1. The first gives |x|< 1/2 and the second |x|< 2. In order to have convergence, they must both be true so we must have |x|< 1/2.

Do you know about "geometric series"? The sum of $\sum_{n=0}^\infty x^n = 1+ x+ x^2+ \cdot\cdot\cdot= \frac{1}{1- x}$.

To find the limit break the sum into two series:
$\sum_{n=0}^\infty C_n x^n= \sum_{\text{n even}} x^n+ \sum_{\text{n odd}} 2x^n$

Since the first sum is over the even numbers only, let n= 2j so that becomes
$\sum_{j= 0}^\infty x^{2j}= \sum_{j= 0}^\infty (x^2)^n$
That is a geometric series with "common ratio" $x^2$. Use the formula for sum of a geometric series.

Since the second sum is over the odd numbers only, let n= 2j+ 1 and that becomes
$\sum_{j= 0}^\infty 2x^{2j+1}= 2x\sum_{j= 0}^\infty (x^2)^j$
That is 2x times the sum of a geometric series.

Each geometric series will give a sum in the form of a fraction. Add the two fractions to get the form you have.

4. ## Re: Question on interval of convergence

Hello, nukenuts!

A function is defined by:. $f(x) \:=\: 1 + 2x + x^2 + 2x^3 + x^4 + \hdots$

where its coefficients are:. $C(2n) = 1$ and $C(2n+1) = 2$ for all $n \ge 0.$

(a) Find the interval of convergence of the series.
(b) Find an explicit formula for $f(x).$

Book answers:
. . Interval $(\text{-}1,1) \qquad f(x) \:=\:\frac{1 + 2x}{1 - x^2}$

$f(x) \;=\;1 + 2x + x^2 + 2x^3 + x^4 + 2x^5 + x^6 + 2x^7\hdots$

. . . . $=\;(1+2x) + x^2(1+2x) + x^4(1+2x) + x^6(1+2x) +\hdots$

. . . . $=\;(1+2x)\underbrace{\big(1 + x^2 + x^4 + x^6 + \hdots\big)}_{\text{Infinite series}}$

The series has first term $a = 1$ and common ratio $r = x^2.$

Its sum is:. $\frac{1}{1-x^2}$

. . which converges for $x^2\,<\,1 \quad\Rightarrow\quad |x| \,<\,1 \quad\Rightarrow\quad x \in (\text{-}1,1)$

Therefore: . $f(x) \;=\;(1+2x)\cdot\frac{1}{1-x^2} \;=\;\frac{1+2x}{1-x^2}$