How do I combine series sums into one representation?

[Susanne217]

Homework Statement

I have these three series here

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

what I would like to do is to add the together as one sum representation

The Attempt at a Solution

From what I can see the commen sum of three is

\sum_{n=0}^{\infty} (a_{n+2} - a_{n+1} +a_{n})z^n is that right?

What I don't remember do I add the index's together or subtract them?

If not please direct to a webpage with rules on howto do the above right :) Cause I have a whole in my knowledge then regarding to above :(

Susanne

 

[Gib Z]

In Mathematics there's so much stuff to learn, you don't want to fill your head with rules for every thing.

Why don't you try writing each sum out in full and collect like powers of z, and see what kind of new sum that gives you.

 

[HallsofIvy]

Homework Statement

I have these three series here

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

In the second sum, let k= n+1 so that n= k-1. When n= 1, k= 0 so you will have a sum in which k runs from 0 to infinity:
\sum_{n=1}^\infty a_n a^{n+ 1}= \sum_{k=0}^\infty a_{k-1}z^k

And since the "k" or "n" in each sum is a dummy index, you can just replace "k" with "n":
\sum_{n=0}^\infty a_{n-1}z^n


Now you can add "like powers" in the two sums- although, since the first sum starts at n= 2, you have to treat n=0 and n= 1 separately.

what I would like to do is to add the together as one sum representation

The Attempt at a Solution

From what I can see the commen sum of three is

\sum_{n=0}^{\infty} (a_{n+2} - a_{n+1} +a_{n})z^n is that right?

What I don't remember do I add the index's together or subtract them?

If not please direct to a webpage with rules on howto do the above right :) Cause I have a whole in my knowledge then regarding to above :(

Susanne[/QUOTE]

 

[Susanne217]

In the second sum, let k= n+1 so that n= k-1. When n= 1, k= 0 so you will have a sum in which k runs from 0 to infinity:
\sum_{n=1}^\infty a_n a^{n+ 1}= \sum_{k=0}^\infty a_{k-1}z^k

And since the "k" or "n" in each sum is a dummy index, you can just replace "k" with "n":
\sum_{n=0}^\infty a_{n-1}z^n


Now you can add "like powers" in the two sums- although, since the first sum starts at n= 2, you have to treat n=0 and n= 1 separately.

what I would like to do is to add the together as one sum representation

The Attempt at a Solution

From what I can see the commen sum of three is

\sum_{n=0}^{\infty} (a_{n+2} - a_{n+1} +a_{n})z^n is that right?

What I don't remember do I add the index's together or subtract them?

If not please direct to a webpage with rules on howto do the above right :) Cause I have a whole in my knowledge then regarding to above :(

Susanne

[/QUOTE]

I tried I again and I always end up with \sum_{n=0}^{\infty} (a_{n+2} - a_{n+1} +a_{n})z^{n+1} ??

 

[Count Iblis]

HallsofIvy made a rare mistake above. Let me explain it without intorducing another summation variable. Let's focus on the first summation from n = 2 to infinity. You'll get the same terms in that summation if you let n run from zero to infinity and at the same time you replace n by n + 2. What you are then doing is saying that:


f(2) + f(3) + f(4) + ...=


g(0) + g(1) + g(2)+...


if we define g(n) as f(n+2), so that g(0) = f(2), g(1) = f(3) etc.

 

[Susanne217]

HallsofIvy made a rare mistake above. Let me explain it without intorducing another summation variable. Let's focus on the first summation from n = 2 to infinity. You'll get the same terms in that summation if you let n run from zero to infinity and at the same time you replace n by n + 2. What you are then doing is saying that:


f(2) + f(3) + f(4) + ...=


g(0) + g(1) + g(2)+...


if we define g(n) as f(n+2), so that g(0) = f(2), g(1) = f(3) etc.

As I understand you a_{n+2}z^{n+2}

 

[Count Iblis]

As I understand you a_{n+2}z^{n+2}

Yes, that's correct. You can write out the first few terms and convince yourself that you get the same terms.

 

[Susanne217]

HallsofIvy made a rare mistake above. Let me explain it without intorducing another summation variable. Let's focus on the first summation from n = 2 to infinity. You'll get the same terms in that summation if you let n run from zero to infinity and at the same time you replace n by n + 2. What you are then doing is saying that:


f(2) + f(3) + f(4) + ...=


g(0) + g(1) + g(2)+...


if we define g(n) as f(n+2), so that g(0) = f(2), g(1) = f(3) etc.

Any chance you would write the other two sums like the above. Then I could use that to check the rest of my result :)

Susanne

 

[Count Iblis]

Well, the second summation is of the form:

f(1) + f(2) + f(3) + ...

where f(n) = a_n z^(n+1)

If you want to write

f(1) + f(2) + f(3) + ... as

g(0) + g(1) + g(2)+...

how do yo have to define g(n) in terms of f(n)?

 

[D H]

HallsofIvy made a rare mistake above.

You should have offered a correction then -- using summation symbols.

What you need to do in general, Susanne217, is to first modify the series so that the power terms are of the same form for each of the series. Do not worry about the limits of the series at this stage. So, let's do that, once again picking on the second series.

\sum_{n=1}^\infty a_n z^{n+ 1}

The series on the left is simply the second series from the original post. From above, the goal here is to express the power term of each of the series in the same form. The canonical form is z^n, but here we have terms involving z^{n+1}. So, substitute k=n+1, or n=k-1. This substitution may affect other factors in a summand will affect and the sum limits as well. In this case, the factor of a_n becomes a_{k-1} since n=k-1 and the lower sum limit changes from n=1 to k=2[/tex] since k=n+1. Thus<br /> <br /> \sum_{n=1}^\infty a_n z^{n+ 1}= \sum_{k=2}^\infty a_{k-1}z^k<br /> <br /> As Halls mentioned, the index is just a dummy variable. There is nothing wrong with changing the <i>k</i> in the series on the right hand side with an <i>n</i>, yielding<br /> <br /> \sum_{n=1}^\infty a_n z^{n+ 1}= \sum_{n=2}^\infty a_{n-1}z^n<br /> The next stage is to form a single series. Now you do have to worry about the summation limits. Different example from above, suppose you are asked to compute<br /> <br /> S = \sum_{n=0}^{\infty} b_n z^n + \sum_{n=0}^{\infty} c_n z^{n+1}<br /> <br /> After the first step this becomes<br /> <br /> S = \sum_{n=0}^{\infty} b_n z^n + \sum_{n=1}^{\infty} c_{n-1} z^n<br /> <br /> Now we have a problem: The two series don&#039;t have the same limits. What you need to do is expand some of the series so that do have series with the same limits. Thus<br /> <br /> S = b_0 + \sum_{n=1}^{\infty} b_n z^n + \sum_{n=1}^{\infty} c_{n-1} z^n<br /> <br /> Finally, the series can be combined. Note that this might well mean a few stray terms that are not a part of the series.<br /> <br /> S = b_0 + \sum_{n=1}^{\infty} (b_n +c_{n-1})z^n

 

[Susanne217]

See next post...

 

[Susanne217]

Thats so funny cause I get the following result myself

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

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 that? Better I hope :)

Sincerely
Susanne

 

[Count Iblis]

In the equation after the sentence "By the theorem of Identity"

you take out some terms of the summations with seems to be motivated in order to let the summations start with a z^2 term, but then you also take out a2 z^2 from the first summation on theright hand side of that equation. That defeats the purpose of your effort.

Instead, make sure that the three summations all start with z^2. So, don't take out the a2 z^2 term and instead let the first sumatrion of the r.h.s. be from n = 2 to infinity. Then you can shift the n for all the summations so that they all run from, say, zero to infinity (or from n = 2 to infinity, it doesn't matter what you choose). Then all the summations limits are the same and thus the power of z as a function of n will now be the same, so you can combine the three summations into one and take out the z^(n + something) in the summand.

 

[Susanne217]

In the equation after the sentence "By the theorem of Identity"

you take out some terms of the summations with seems to be motivated in order to let the summations start with a z^2 term, but then you also take out a2 z^2 from the first summation on theright hand side of that equation. That defeats the purpose of your effort.

Instead, make sure that the three summations all start with z^2. So, don't take out the a2 z^2 term and instead let the first sumatrion of the r.h.s. be from n = 2 to infinity. Then you can shift the n for all the summations so that they all run from, say, zero to infinity (or from n = 2 to infinity, it doesn't matter what you choose). Then all the summations limits are the same and thus the power of z as a function of n will now be the same, so you can combine the three summations into one and take out the z^(n + something) in the summand.

Just understand you correctly Count Iblis,

I am suppose to end up with a sum that looks someward like this?

\sum( a_{n+3} + a_n ) z^{n+2}

because I need it to satisfy the condition a_{n+3} + a_n = 0 because a_{n+3} = -a_{n}

 

[Count Iblis]

Yes, that's right. But note that if you rewrite the series in a more standard way, along the lines D.H and I have been suggesting, you find a recurrence relation the expresses a_n in terms of a(n-1} and a_{n-2} and from that you can then derive that
a_{n+3} = -a_{n}.

 

[Susanne217]

Yes, that's right. But note that if you rewrite the series in a more standard way, along the lines D.H and I have been suggesting, you find a recurrence relation the expresses a_n in terms of a(n-1} and a_{n-2} and from that you can then derive that
a_{n+3} = -a_{n}.

Count,

If I subtract the second sum from the first I get (Which is hopefully correct)

\sum_{n=1}^{\infty} a_{n-1}z^{n+1} if this is correct my next logical step is to add this together with the third sum which produces

\sum_{n=0}^{\infty} (??)

But there is a problem with my calculations count :(

If I add the third sum to the difference between the first two I get

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

and it sums into on only one term which looks nothing like the intended target... What am I doing wrong? Why isn't it obvious :(

 

[Count Iblis]

You are doing something wrong. Let me start from this point in your posting#12:

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

Where I put the a_2 z^2 term back inside the summation relative to what you did. What you now have are summations that all start with z^2. If you now simply replace n by n-1 in the second summation, then you need to compensate for that by increasing the lower limit from n= 1 to n = 2. So, the summand in the second summation then contains z^n as in the first summation and it starts at the same lower limit for n, but, of course, you now have a_{n-1} in there.

You can then rewrite the last summation in the same way, i.e. change n to n-2 and simultaneously increase the lower limit from n=0 to n = 2.

 

[Susanne217]

You are doing something wrong. Let me start from this point in your posting#12:

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

Where I put the a_2 z^2 term back inside the summation relative to what you did. What you now have are summations that all start with z^2. If you now simply replace n by n-1 in the second summation, then you need to compensate for that by increasing the lower limit from n= 1 to n = 2. So, the summand in the second summation then contains z^n as in the first summation and it starts at the same lower limit for n, but, of course, you now have a_{n-1} in there.

You can then rewrite the last summation in the same way, i.e. change n to n-2 and simultaneously increase the lower limit from n=0 to n = 2.

Okay then by this I simply transformer this into

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

But I can follow this being a recurcive relation half way and its suppose to be n+3 because it has three terms?

 

[D H]

Okay then by this I simply transformer this into

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

You didn't do that right.

But I can follow this being a recurcive relation half way and its suppose to be n+3 because it has three terms?

Don't worry about that yet. You will get a different recursive relationship here, and the you will derive the desired one.

 

[Susanne217]

You didn't do that right.

Good then I'm only semi stupid...

Anyway please my first step is to squize the two sums 1) and 2) together :)

If I change the index to n = 0 in the third sum I get

\sum_{n=0}^{\infty} a_n z^{n+2}

But the difference sum if I do something simular to series 1) and series 2)

their difference sum is still:

\sum_{n=0}^{\infty} a_{n+1} z^{n+2} and that's a long way off from

a_n+3 + a_n ??

 

[D H]

Let's change the index in the third sum,

\sum_{n=0}^{\infty} a_n z^{n+2}

The goal here is to get a sum that involves z^n rather than z^{n+2}. One way to avoid confusion is to invent a new dummy index variable. So, define m=n+2. With this the sum becomes

\sum_{m=2}^{\infty} a_{m-2} z^m

Now rename that index variable back to n:

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

 

[D H]

Now do the same thing to the second sum. When you are done you should have something like

1 = \left(1 - z + z^2\right) \sum a_n \,z^n = a_0 + (a_{1}-a_{0})z + \sum_{n=2}^{\infty} a_n \,z^n - \sum_{n=2}^{\infty} b_{n} z^n + \sum_{n=2}^{\infty} c_n z^n

where the b_n and c_n are expressed in terms of some appropriate element from the set \{a_n\}.

 

[Susanne217]

Now do the same thing to the second sum. When you are done you should have something like

1 = \left(1 - z + z^2\right) \sum a_n \,z^n = a_0 + (a_{1}-a_{0})z + \sum_{n=2}^{\infty} a_n \,z^n - \sum_{n=2}^{\infty} b_{n} z^n + \sum_{n=2}^{\infty} c_n z^n

where the b_n and c_n are expressed in terms of some appropriate element from the set \{a_n\}.

I'm one to something now

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

and for the second sum

Let's change the index in the third sum,

\sum_{n=0}^{\infty} a_n z^{n+1}

The goal here is to get a sum that involves z^n rather than z^{n+1}. One way to avoid confusion is to invent a new dummy index variable. So, define m=n+1. With this the sum becomes

\sum_{m=2}^{\infty} a_{m-1} z^m

Now rename that index variable back to n:

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

How is that ?? and I'm supose to combine sum 1) and 2) into a_n+3 now?

 

[D H]

Don't worry about that yet! First get the sum right. Then worry about that. You are putting the cart before the horse.

 

[Susanne217]

Don't worry about that yet! First get the sum right. Then worry about that. You are putting the cart before the horse.

Sorry, but what's my next step then? Cause the whole thing doesn't look obvious to me..

 

[D H]

The next step is to combine the series into one. You are hung up on this a_n+3 thing. You will not get the result a_n+a_n+3=0 when you form the series, so best to forget about that for now. You will get a different recursive relationship. From that you should be able to derive the result you are expecting to see.