# Powerseries problem issue

1. May 2, 2006

### Hummingbird25

HELP: Powerseries problem issue

Hello People,

I have this here power series problem which is frustrating me:

Given the power series

$$(*) f(z) = \sum_{n=0} ^{\infty} (2n+1)z^n$$

for |z| < 1 and $$z \in \mathbb{C}$$

has the Radius of Convergens R = 1.

Show that the power series (*) can be written as another power series

g(z) = f(z) - z f(z)

which gives the result $$g(z) = \frac{1+z}{1-z}$$

Proof:

The power series (*) can be written as

$$f(z) = \sum_{n=0} ^{\infty} 2z(n \cdot z^{n-1}) + z^n$$

Since R = 1

then

$$(1-z) \sum_{n=0} ^{\infty} (2n+1)z^n - z \sum_{n=0} ^{\infty} (2n+1)z^n = 1$$

But how I proceed from here?

Sincerely Yours
Hummingbird25

Last edited: May 2, 2006
2. May 2, 2006

### Physics Monkey

Hi Hummingbird25,

Your first step should be to simply calculate the terms in the series for g(z) defined as f(z) - z f(z), $$g(z) = \sum_{n=0} (2n+1) z^n - \sum_{n=0} (2n+1)z^{n+1}$$. If you can't find the pattern directly, just write out a few terms in the series and things will become quite clear. You will find that the result can be easily summed into a function you know.

3. May 2, 2006

### Hummingbird25

Hello Physics Monkey and thank You for You kind and Quick answer,

If I write out the sum I get:

$$g(z) = (1-z) + (3z - 3z^2) + (5z^2-5z^3) + \ldots +$$

But what does that show ?

Sincerely Hummingbird25

Last edited: May 2, 2006
4. May 2, 2006

### Physics Monkey

Happy to help. Try combining like powers.

Edit: Also, you made a mistake in writing out the series. There should be plus signs between each of your parenthesis.

5. May 2, 2006

### Curious3141

Try rearranging the parantheses so that like powers are bracketed together (z terms together, z^2 together, etc.)

6. May 2, 2006

### Hummingbird25

Hello Curious3141, and thank You very much for Your answer,

If I arrange them together that gives

$$g(z) = (1-z-3z) - (3z - 5z)z + (-5z^3 + 7z^3) + (-7z^4 + 9z^4) + \ldots +$$

Sincerely Yours Hummingbird25

Last edited: May 2, 2006
7. May 2, 2006

### Hummingbird25

But written it out as sums you mean ?

$$g(z) = f(z) - z f(z) = \sum_{n=0} ^{\infty} -(2n+1) \cdot z + 2n +1) z^n$$

Last edited: May 2, 2006
8. May 2, 2006

### Physics Monkey

I would recommend you stick with the approach of writing out terms. Having grouped liked powers together all you need to do is subtract. Do you notice anything? You may need to do a few more terms in the series to see the pattern.

9. May 2, 2006

### Hummingbird25

Hi

Then the summation must be

$$g(z) = (1-z+3z) - (3z - 5z)z + (-5z^3 + 7z^3) + (-7z^4 + 9z^4) + \ldots +$$

Is that what You mean?

Sincerley Hummingbird25

Last edited: May 2, 2006
10. May 2, 2006

### Curious3141

No, let me illustrate (then you should do the rest yourself).

$$g(z) = (1) + (-z + 3z) + (-3z^2 + 5z^2) + (-5z^3 + 7z^3) + \ldots +$$

Do you see anything happening within each paranthesis ?

11. May 2, 2006

### Physics Monkey

Yes, that is what I mean. Now carry out the subtraction for each term in brackets. What do you notice?

12. May 2, 2006

### Hummingbird25

Yes the series act $$1+ z^n$$ ??

Sincerley
Hummingbird25

13. May 2, 2006

### Curious3141

Ah, no. There's a summation sign and a factor of 2 missing there. Try again ?

14. May 2, 2006

### Hummingbird25

Okay

I will try again,

my surgestion is

$$\sum_{n=0} ^ {z} 1^n = 1+z$$

That factor of two must I add that within the sum or outside the sum?

Sincerley Hummingbird

Last edited: May 2, 2006
15. May 2, 2006

### Curious3141

No, that's not right.

You've gotten to this point right ?

$$g(z) = 1 + 2z + 2z^2 + 2z^3 + ... = 1 + 2(z + z^2 + z^3 + ...)$$

Put that into summation notation. Also see if you can spot what sort of basic series is represented there.

Last edited: May 2, 2006
16. May 2, 2006

### Hummingbird25

That looks like the sum $$\sum_{n=0} ^{?} (1+2z^n)$$

17. May 2, 2006

### Curious3141

The question mark should be replaced by infinity. Also, when you put the '1' within the summation it means it's going to be added in each term, meaning you'll be doing it an infinite number of times. So that isn't right.

Also are you sure the lower bound for the summation is zero ?

18. May 2, 2006

### Hummingbird25

Okay then the sum must be?

$$2 \sum_{n=?} ^{\infty} z^{n+1} = \cdots = \frac{1+z}{1-z}$$

Sincerely Hummingbird25

Last edited: May 2, 2006
19. May 2, 2006

### Curious3141

You're getting there, it's still not written out quite correctly. Let me show you, I hope you can understand it (it's important to understand more than just remember).

BTW, I made a mistake in writing f(z) before, I meant g(z).

$$g(z) = 1 + 2(z + z^2 + z^3 + ...) = 1 + 2\sum_{n=1}^{\infty}z^n$$

Clear so far ? You can start the sum from n = 0 but then the expression in the summation has to become z^(n+1) - you understand ? And don't forget to add the one outside the summation.

Now, the summation is actually a geometric series with first term z and a common ratio of z. It will converge for |z| < 1. The sum of a geometric series with first term a and common ratio r is :

$$S = \frac{a}{1-r}$$

So, using that here,

$$g(z) = 1 + 2\sum_{n=1}^{\infty}z^n = 1 + 2(\frac{z}{1-z}) = \frac{1-z}{1-z} + \frac{2z}{1-z} = \frac{1+z}{1-z}$$

And now, since $$g(z) = f(z) - zf(z)$$, you can solve for $$f(z)$$. Presumably, that's what the question wants you to do at the end even if it's not stated as such.

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