Trying to verify function as a powerseries

[Susanne217]

Homework Statement

Let \frac{1}{1-z+z^2} = \sum_{n=0}^{\infty} a_n \cdot z^n

where a_{1} = a_0 = 1 \Leftrightarrow a_{1}-a_{0} = 0 and a_2=0 and a_{n+3} + a_n = 0

The Attempt at a Solution

By the theorem of Identity

1 = 1 - z + z^2 \sum a_n \cdot z^n = a_0 + (a_{1}-a_{0})z + a_{2} \cdot z^2 + \sum_{n=3}^{\infty} a_n \cdot z^{n} - \sum_{n=1}^{\infty} a_{n} \cdot z^{n+1} + \sum_{n=0}^{\infty} a_n z^{n+2}

If I write out the first sum I get \sum_{n=3}^{\infty} a_n \cdot z^{n} = a_3 z^3 + a_{4}z^4 + \cdots

I write out a couple terms of the second sum - \sum_{n=1}^{\infty} a_{n} \cdot z^{n+1} = - a_1 \cdot z^{2} - a_2 \cdot z^{3} - \cdots

If I make the commen index n = 0

it looks like the index increase at n+3 and at a power of n+2 such that

a_0 + (a_{1}-a_{0})z + a_{2} \cdot z^2 +\sum_{n=3}^{\infty} + a_n \cdot z^{n} - \sum_{n=1}^{\infty} a_{n} \cdot z^{n+1} + \sum_{n=0}^{\infty} a_n z^{n+2} = \sum_{n=0}^{\infty} (a_{n+3} +a_n)z^{n+2}

How is this?

Sincerely
Susanne

 

[tiny-tim]

Hi Susanne!

Sorry, but this is really messy.

Write out the total coefficient of zⁿ⁺² for n≥0. to get a general equation for a_n a_n-1 and a_n-2, and then deal with z⁰ and z¹ separately.

 

[Susanne217]

Hi Susanne!

Sorry, but this is really messy.

Write out the total coefficient of zⁿ⁺² for n≥0. to get a general equation for a_n a_n-1 and a_n-2, and then deal with z⁰ and z¹ separately.

Just be clear tiny-tim,

Is my formulation correct until here

Let \frac{1}{1-z+z^2} = \sum_{n=0}^{\infty} a_n \cdot z^n

where a_{1} = a_0 = 1 \Leftrightarrow a_{1}-a_{0} = 0 and a_2=0 and a_{n+3} + a_n = 0

The Attempt at a Solution

1 = 1 - z + z^2 \sum a_n \cdot z^n = a_0 + (a_{1}-a_{0})z + a_{2} \cdot z^2 + \sum_{n=3}^{\infty} a_n \cdot z^{n} - \sum_{n=1}^{\infty} a_{n} \cdot z^{n+1} + \sum_{n=0}^{\infty} a_n z^{n+2}

My attempt to expand the three sums (using the original n = 0)
1) \sum_{n=0}^{\infty} a_n \cdot z^{n} =a_{0} + a_{1}z + a_{2}z^2+ a_{3}z^3 + a_4 z^4 + a_5 z^5 + \cdots +

2) \sum_{n=0}^{\infty} a_{n} \cdot z^{n+1} = a_0 z + a_1 z^{2} + a_2 z^3 + \cdots

3) \sum_{n=0}^{\infty} a_n z^{n+2} = a_{0}z^2 + a_{1}z^3 + a_2 z^4

I can see that if I subtract the second sum from the first I get

\sum a_{n-1}z^{n+1}

But what's my next step? to add

\sum_{n=1}^{\infty} a_{n-1}z^{n+1}+ \sum_{n=0}^{\infty} a_{n}z^{n+2}

Can that be right tiny-tim?? Cause I keep getting something else that a_{n+3}+a_n inside the sum??

If I do I end up with \sum_{n=0}^{\infty} a_{n}z^{n+1} as the joined sum of the three. But can't be right?

 

[tiny-tim]

Hi Susanne!

My attempt to expand the three sums (using the original n = 0)
1) \sum_{n=0}^{\infty} a_n \cdot z^{n} =a_{0} + a_{1}z + a_{2}z^2+ a_{3}z^3 + a_4 z^4 + a_5 z^5 + \cdots +

2) \sum_{n=0}^{\infty} a_{n} \cdot z^{n+1} = a_0 z + a_1 z^{2} + a_2 z^3 + \cdots

3) \sum_{n=0}^{\infty} a_n z^{n+2} = a_{0}z^2 + a_{1}z^3 + a_2 z^4

This is a bad idea …

it might turn out ok, but it's so easy to make a mistake if you do it that way.

Instead, keep all the ∑s so far as you can, but rewrite them so that they all say zⁿ, and they all start from n = 2 (this makes it really easy to add the ∑s )

For example, with ∑_n=0 a_nzⁿ⁺¹,

rewrite it as ∑_n=2 a_n-1zⁿ + a₀z.

Try again. ​