Find the power series

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  • #1
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z is a complex number.

[tex]f(z) = \frac{4 + 3z}{(z + 1)(z + 2)^2}[/tex]

[tex]\frac{1}{1 + z} = \frac{1}{1 - (-z)} = 1 + (-z) + (-z)^2 + \cdots = \sum_{n = 0}^{\infty}(-z)^n[/tex]

[tex]\frac{1}{(z + 2)^2} = -\frac{d}{dz} \ \frac{1}{1 - (-z - 1)} = -\frac{d}{dz}\sum_{n = 0}^{\infty}(-z - 1)^n = \sum_{n = 0}^{\infty}-n(-z-1)^{n-1} [/tex]

[tex]f(z) = \sum_{n = 0}^{\infty}-(4 + 3z) n (-z)^n (-z - 1)^{n - 1}[/tex]

Radius of convergence by the ratio test:

[tex]\lim_{n\to\infty}\left|\frac{-(4+3z)(n+1)(-z)^{n+1}(-z-1)^n}{-(4+3z)n(-z)^n(-z-1)^{n-1}}\right|=\lim_{n\to\infty}\left|\frac{(n+1)(z)(z+1)}{n}\right|[/tex]

[tex]|z(z+1)|<1[/tex]

Is the above correct?

If so, how do I find the Radius of Convergence from that?
 

Answers and Replies

  • #2
Char. Limit
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No, that doesn't look right. For one thing, you can't just combine summands like that. The product of (z+1)(z+2)^2 is going to be the product of the two sums, and you're probably going to end up with a double sum or something.
 
  • #3
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No, that doesn't look right. For one thing, you can't just combine summands like that. The product of (z+1)(z+2)^2 is going to be the product of the two sums, and you're probably going to end up with a double sum or something.
Then how can I find the power series of that expression?
 
  • #4
Char. Limit
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I'd split the function up using Partial Fractions. Then you'll get three functions that are easy to devise a power series for.
 
  • #5
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I'd split the function up using Partial Fractions. Then you'll get three functions that are easy to devise a power series for.
Could I instead use the rule [itex]R\geq\min{R,S} = 1[/itex]?

Isn't there a way to do this without the use of partial fractions?
 
  • #6
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By partial fractions, we can re-write [itex]f(z)[/itex] as
[tex]
\frac{A}{z + 1} + \frac{B}{z + 2} + \frac{C}{(z + 2)^2} = \frac{4 + 3z}{(z + 1)(z + 2)^2}.
[/tex]
By direct computations, [itex]A(z + 2)^2 + B(z + 2)(z + 1) + C(z + 1)[/itex] [itex]= Az^2 + 4Az + 4A + Bz^2 + 3Bz + 2B + Cz + C.[/itex]
To find the coefficients, we solve the system of linear equations.
[tex]
\begin{pmatrix}1 & 1 & 0 & 0\\
4 & 3 & 1 & 3\\
4 & 2 & 1 & 4\end{pmatrix}\Rightarrow
\begin{pmatrix}1 & 0 & 0 & 1\\
0 & 1 & 0 & -1\\
0 & 0 & 1 & 2\end{pmatrix}
[/tex]
So the coefficients are [itex]A = 1[/itex], [itex]B = -1[/itex], and [itex]C = 2[/itex].
Therefore, [itex]\displaystyle\frac{4 + 3z}{(z + 1)(z + 2)^2} = \frac{1}{z + 1} - \frac{1}{z + 2} + \frac{2}{(z + 2)^2}[/itex] [itex]\displaystyle\Rightarrow\frac{1}{1 - (-z)} - \frac{1}{2}\cdot\frac{1}{1 - \left(-\frac{z}{2}\right)} - \frac{d}{dz}\frac{1}{1 - \left(-\frac{z}{2}\right)}[/itex].
The rational expressions can be represented as the following power series
[tex]
\sum_{n = 0}^{\infty}(-z)^n - \frac{1}{2}\sum_{n = 0}^{\infty}\left(-\frac{z}{2}\right)^n - \sum_{n = 0}^{\infty}n\left(-\frac{z}{2}\right)^{n - 1}.
[/tex]

Now how do I find the radius of convergence
 
  • #7
Char. Limit
Gold Member
1,204
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First express it as a single sum. Due to associativeness, you can easily rewrite it like this:

[tex]\sum_{n=0}^{\infty} (-1)^n z^n - (-1)^n z^n 2^{-n-1} - (-1)^n n z^n 2^{1-n}[/tex]

Which, in a simpler form, is:

[tex]\sum_{n=0}^{\infty} \left( (-1)^n \left(1 - \frac{1}{2^{n+1}} - \frac{n}{2^{n-1}} \right) \right) z^n[/tex]

Now how would you find the radius of convergence of a single sum?
 
  • #8
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First express it as a single sum. Due to associativeness, you can easily rewrite it like this:

[tex]\sum_{n=0}^{\infty} (-1)^n z^n - (-1)^n z^n 2^{-n-1} - (-1)^n n z^n 2^{1-n}[/tex]

Which, in a simpler form, is:

[tex]\sum_{n=0}^{\infty} \left( (-1)^n \left(1 - \frac{1}{2^{n+1}} - \frac{n}{2^{n-1}} \right) \right) z^n[/tex]

Now how would you find the radius of convergence of a single sum?
Ratio test

So we end up getting

[tex]|z| < 1[/tex]

and R = 1

Correct?
 
Last edited:

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