# Powersums and ODE

1. May 15, 2006

### Mathman23

Hi

Given a power-series

$$\sum_{n=0} ^{\infty} \frac{2^n}{2n+1} z^{2n+1}$$

if f(x), for $$z = x \in \mathbb{R}$$ is the sum of the above power series. Then show that f is solution for the differential equation

$$\frac{dx}{dy} - 2xy = 1$$

My Solution

The generic form of this power series is $$f(x = z) = \sum_{n=0} ^{\infty} \frac{2^n}{2n+1} z^{2n+1}$$

Therefore $$\frac{dx}{dy} = \frac{d}{dx}(z) = \sum_{n=0} ^{\infty} 2^n \cdot z^{2n}$$

If I insert this above information into the ODE I get:

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

If this is correct do I try to put it all into one sum?

Sincerley
Fred

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

### AKG

$$\frac{dx}{dy} = \frac{d}{dx}(z) = \sum_{n=0} ^{\infty} 2^n \cdot z^{2n}$$

The above is wrong. Especially where you write $\frac{d}{dx}(z)$. Look, don't confuse yourself with x and z. You have a function:

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

To see if it is a solution to

$$\frac{dx}{dy} - 2xy = 1$$

you need to check that:

$$\frac{dx}{df} - dxf = 1$$

Note:

$$\frac{dx}{df} = \frac{1}{\frac{df}{dx}}$$

You want to verify:

$$\frac{1}{\sum_{n=0}^{\infty}2^nx^{2n}} - 2x\sum_{n=0}^{\infty}\frac{2^n}{2n+1}x^{2n+1} = 1$$

for all x.

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

### Mathman23

Okay thank You,

Just to clarify then the next is write this down a have one sum sign and then factor out a commen factor ?

/Fred

If yes then I choose $${\sum_{n=0} ^{\infty} 2^n x^{2n}}$$ as the commen denominator.

Then I get $$\sum_{n=0} ^{\infty} \frac{1- 2x^2 \sum_{n=0} ^{\infty} (2^n * x^{2n}) \sum_{n=0} ^{\infty} \frac{2^n x^{2n}}{2n+1}}{{\sum_{n=0} ^{\infty} 2^n x^{2n}}} - 1 = 0$$

If this is correct, do I need to get rid of the sum in the denominator?

Sincerely
Fred

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

### AKG

No, you get:

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

however I can't see why you'd want to look at it this way. Something's wrong with the question, since the series:

$$\sum _{n=0}^{\infty}\frac{2^n}{2n+1}x^{2n+1}$$

doesn't even converge for all $x \in \mathbb{R}$. Maybe you're only supposed to look at these series as formal power series, but then I don't see why they would say $x \in \mathbb{R}$. Actually, even looking at it that way the thing you're asked to prove appears false. I think you copied out the question incorrectly.

5. May 15, 2006

### Mathman23

Hello AKG,

Looking at the question again:

Given a power series:

$$\sum_{n=0} ^{\infty} \frac{2^n}{1 \cdot 3 \cdot 5 \cdots (2n+1)} z^{2n+1} = z + 2/(1 \cdot 3) z^3 + 2^2/(1 \cdot 3 \cdot 5) z^5 + \cdots$$

converge for all $$z \in \mathbb{C}$$.

Let f(x) for $$z = x \in \mathbb{R}$$ describe the sum of the above series.

Then show that f is a solution for

$$\frac{dx}{dy} - 2xy = 1$$

thats the exact problem. Any idears?

Sincerely

Fred.

p.s. Isn't
$$\sum_{n=0} ^{\infty} \frac{2^n}{1 \cdot 3 \cdot 5 \cdots (2n+1)} z^{2n+1} = \sum_{n=0} ^{\infty} \frac{2^n}{(2n+1)} z^{2n+1}$$

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

### HallsofIvy

Staff Emeritus
No, it isn't! The only difference I can see between the two is than in one the denominator is 1*3*...*(2n+1) and in the other the denominator is 2n+1 and those are certainly NOT the same!

Your notation is still abominable! If $z = x \in \mathbb{R}$,
then x is NOT a function of y and so $\frac{dx}{dy} - 2xy = 1$ makes no sense. Perhaps you meant
$$y(z)= \sum_{n=0} ^{\infty} \frac{2^n}{1 \cdot 3 \cdot 5 \cdots (2n+1)} z^{2n+1} = z + 2/(1 \cdot 3) z^3 + 2^2/(1 \cdot 3 \cdot 5) z^5 + \cdots$$
satisfies the equation
$\frac{dy}{dz} - 2zy = 1$

Yes,
$$\frac{dy}{dz}= \sum_{n=0} ^{\infty} \frac{2^n}{1 \cdot 3 \cdot 5 \cdots (2n-1)} z^{2n}$$
so the left side of the equation becomes
$$\sum_{n=0} ^{\infty} \frac{2^n}{1 \cdot 3 \cdot 5 \cdots (2n)} z^{2n}+\sum_{n=0} ^{\infty} \frac{2^n}{1 \cdot 3 \cdot 5 \cdots (2n+1)} z^{2(n+1)}$$

Now, In the first sum, let j= n so it becomes
$$\sum_{j=0} ^{\infty} \frac{2^j}{1 \cdot 3 \cdot 5 \cdots (2j)} z^{2j}$$
And, in the second sum, let j= n+1 so n= j-1. Of course, when n= 0, j= 1 so the second sum becomes
$$-2\sum_{j=1} ^{\infty} \frac{2^j}{1 \cdot 3 \cdot 5 \cdots (2j-1)} z^{2j}$$
except for j= 0, those are exactly the same but of different sign and so every term except the z0 term cancels. What is $\frac{2^j}{1 \cdot 3 \cdot 5 \cdots (2j)} z^{2j}$ when j= 0?

7. May 15, 2006

### Mathman23

If You were going to express the dernominator for powerseries in a program like Maple, how would you express it?

(2n+1)! = 1*3*...*(2n+1)

What is $\frac{2^j}{1 \cdot 3 \cdot 5 \cdots (2j)} z^{2j}$ when j= 0?[/QUOTE]

My calculator finds the resultat to equal 1.

I guess that proves that the sum is solution for the ODE ?

Sincerely
Fred

8. May 15, 2006

### AKG

(2n+1)! = 1*3*...*(2n+1)

NO! Don't you know what "!" factorial is?

9. May 15, 2006

### AKG

There is STILL something wrong with your question. Could you PLEASE take the time to copy it out correctly. Do you really mean dx/dy?

10. May 15, 2006

### Curious3141

Just to explore the series itself a little more,

$$y = f(x) = \sum_{n=0} ^{\infty} \frac{2^n}{2n+1} x^{2n+1}$$

Differentiating term by term,

$$S(x) = f'(x) = \frac{dy}{dx} = \sum_{n=0} ^{\infty} (2^n) x^{2n} = 1 + 2x^2 + 4x^4 + 8x^6 + ...$$

Now observe that $$-2x^2S(x) = -2x^2 - 4x^4 - 8x^6 - ...$$

giving $$(1-2x^2)S(x) = 1$$

and $$S(x) = f'(x) = \frac{1}{1-2x^2}$$

From this we can get $$\frac{dx}{dy} = (1 - 2x^2)$$

We can also find an expression for f(x) by integrating S(x),

$$f(x) = \int f'(x)dx = \int \frac{1}{1-2x^2}dx = \frac{1}{2\sqrt{2}}\log{\frac{\sqrt{2} + 2x}{\sqrt{2} - 2x}}$$

Note that the constant of integration is zero because from the orig. series, f(0) = 0.

We can also get -2xy = -2xf(x) as $$-2xy = -\frac{x}{\sqrt{2}}\log{\frac{\sqrt{2} + 2x}{\sqrt{2} - 2x}}$$

Now we have explicit expressions for everything, but I still can't see how $$\frac{dx}{dy} - 2xy = 1$$

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

### AKG

If he meant dx/dy, it definitely doesn't work. If he meant dy/dx, it does:

$$f(x) = \sum_{n=0} ^{\infty} \frac{2^n}{1 \cdot 3 \cdot 5 \cdots (2n+1)} z^{2n+1}$$

$$f'(x) = 1 + \sum _{n=1} ^{\infty}\frac{2^nx^{2n}}{1 \cdot 3 \cdot 5 \cdots (2n-1)}$$

$$-2xf(x) = -\sum_{n=0} ^{\infty} \frac{2^{n+1}x^{2(n+1)}}{1 \cdot 3 \cdot 5 \cdots (2n+1)} = - \sum_{n=1} ^{\infty} \frac{2^nx^{2n}}{1 \cdot 3 \cdot 5 \cdots (2n-1)}$$

This whole problem was really, really easy, made incredibly difficult by horrible communication. Next time, write the question exactly as it's given to you.

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

### Curious3141

Why does (2n+1)!! appear in the denominator of your f(x)? That wasn't in the orig. question.

13. May 15, 2006

### AKG

See post #5 in this thread. The original poster thought (2n+1) = 1.3.5...(2n+1). By the way, is (2n+1)!! some notation for 1.3.5...(2n+1)? Is !! only defined for odd numbers? What's it called?

14. May 15, 2006

### Curious3141

It's the double factorial. It can be defined for evens as well the same way, except you end at 2.

You mean the question was wrong to begin with?:grumpy: :grumpy:

EDIT : Just looked at post 5. To the OP (Mathman) - PLEASE make the effort in future to write the question exactly as it appears, if you make a faulty assumption (like (2n+1)!! = (2n+1) ) then all our efforts are for naught.

Last edited: May 15, 2006
15. May 16, 2006

### Mathman23

I'm very sorry,

It was suppose to say

Hello AKG,

Looking at the question again:

Given a power series:

$$\sum_{n=0} ^{\infty} \frac{2^n}{1 \cdot 3 \cdot 5 \cdots (2n+1)} z^{2n+1} = z + 2/(1 \cdot 3) z^3 + 2^2/(1 \cdot 3 \cdot 5) z^5 + \cdots$$

converge for all $$z \in \mathbb{C}$$.

Let f(x) for $$z = x \in \mathbb{R}$$ describe the sum of the above series.

Then show that f is a solution for

$$\frac{dy}{dx} - 2xy = 1$$

thats the exact problem. Any idears?

Sincerely

Fred.

16. May 16, 2006

### Curious3141

AKG has already solved this problem in post 11.

17. May 16, 2006

### Mathman23

By the way I realised that 1*3*5****(2n+1) = (2n+1)^n.

Sorry for me being confused. I need to read the problems more preciously.

/Fred

18. May 16, 2006

### AKG

By the way I realised that 1*3*5****(2n+1) = (2n+1)^n

What? No, that's not true at all.

19. May 17, 2006

### Mathman23

Then

$$\sum_{n=0} ^{\infty} \frac{2^n}{1 \cdot 3 \cdot 5 \cdots (2n+1)} z^{2n+1}$$

Must equal

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

/Fred

Last edited: May 17, 2006