Solving ODE Using Powersums: Fred's Solution

[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

 

[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.

 

[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

 

[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.

 

[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}

 

[HallsofIvy]

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}

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?

 

[Mathman23]

Hello and T.Y for Your answer,

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

 

[AKG]

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

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

 

[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?

 

[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

 

[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.

 

[Curious3141]

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.

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

 

[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?

 

[Curious3141]

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?

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?

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.

 

[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.

 

[Curious3141]

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.

AKG has already solved this problem in post 11.

 

[Mathman23]

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

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

/Fred

AKG has already solved this problem in post 11.

 

[AKG]

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

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

 

[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

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

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