How to simplify this series?

[Anwesa Roy]

I am trying to find z transform of nu

In that I encounter the following problem:

How do I find the simplification of the following series?

z^-1 + 2z^-1 + 3z^-3 + ...... to infinity

Please describe the mathematical steps.

 

[FvM]

z^-1 + 2z^-1 + 3z^-3 + ...... to infinity

Makes no sense. Specify an unequivocal construction rule.

Infinite series means that it might be described by a simple IIR expression H(z) = a(z)/b(z)

 

[KlausST]

Hi,

Shouldn't it be:

z^-1 + 2z^-2+ 3z^-3 + ...... to infinity

?

Klaus

 

[Anwesa Roy]

Makes no sense. Specify an unequivocal construction rule.

Infinite series means that it might be described by a simple IIR expression H(z) = a(z)/b(z)

Sorry I am wrong. The series should be


z^-1 + 2z^-2+ 3z^-3 + ...... to infinity

- - - Updated - - -

Yes true. Sorry for the mistake

 

[elwo06]

Sorry I am wrong. The series should be


z^-1 + 2z^-2+ 3z^-3 + ...... to infinity

- - - Updated - - -

Yes true. Sorry for the mistake

It converges for |z|>1 (root test or ratio test).
So you know for |z|>1
z/(z-1)=1/(1-1/z)=1+1/z+1/z^2+1/z^3+..
and the convergence is uniform (M test or whatever),
so differentiate both sides to obtain
D(z/(z-1))=-1/z^2-2/z^3-3/z^4-..
So your series is
-zD(z/z-1)=z/(z-1)^2

 

[albbg]

Let's call S=z^-1 + 2z^-2+ 3z^-3 +...+nz^-n
divide now both LHS and RHS by z^-1:

S/z^-1=1 + 2z^-1+ 3z^-2 +...+nz^-n+1

we can note that (2z^-1+ 3z^-2 +...+nz^-n+1) - (z^-1+ z^-2 +...+z^-n+1) = z^-1 + 2z^-2+ 3z^-3 +...+(n-1)z^-n+1

if we call now A = z^-1+ z^-2 +...+z^-n+1 then, summing it to both LHS and RHS:

S/z^-1 A=1 + z^-1+ 2z^-2 +...+(n-1)z^-n+1

but z^-1+ 2z^-2 +...+(n-1)z^-n+1 = S(n-1)
and S(n-1) = S - nz^-n thus:

S/z^-A = 1 - S - nz^-n

S[1/z^-1 +1] = 1 + A -nz^-n from which: S = (1 + A - nz^-n)/(1/z^-1 +1)

We have to find A:

A(k) = z^-1+z^-2+...+z^-k

dividing both sides by z^-1

A(k)/z^-1 = 1 + x^-1+x^-2+...+x^-n

as before x^-1+x^-2+...+x^-n = A(k-1) = A(k) - x^-k

after some passage:

A(k) = (x^-1 - x^-k-1)/(1 - x^-1)

we have to find A(n-1), that is k=n-1:

A=A(n-1) = (x^-1 - x^-n)/(1 - x^-1)

You have now to substitute this result in S then to have the sum to inf just calculate lim(n-->inf) S

even if I paid attention, please check that all my passages are corrrect.

In any case the solution found by elwo06 is faster and more elegant.

 

[elwo06]

It converges for |z|>1 (root test or ratio test).
So you know for |z|>1
z/(z-1)=1/(1-1/z)=1+1/z+1/z^2+1/z^3+..
and the convergence is uniform (M test or whatever),
so differentiate both sides to obtain
D(z/(z-1))=-1/z^2-2/z^3-3/z^4-..
So your series is
-zD(z/z-1)=z/(z-1)^2

I mean the convergence of the derived series term by term is uniform