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Powersums and ODE

  1. May 15, 2006 #1
    Hi

    Given a power-series

    [tex]\sum_{n=0} ^{\infty} \frac{2^n}{2n+1} z^{2n+1}[/tex]

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


    [tex]\frac{dx}{dy} - 2xy = 1[/tex]


    My Solution

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

    Therefore [tex]\frac{dx}{dy} = \frac{d}{dx}(z) = \sum_{n=0} ^{\infty} 2^n \cdot z^{2n}[/tex]

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

    [tex]\sum_{n=0} ^{\infty} 2^n \cdot z^{2n} - 2z \sum_{n=0} ^{\infty} \frac{2^n}{2n+1} z^{2n+1} = 1[/tex]

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

    Sincerley
    Fred
     
    Last edited: May 15, 2006
  2. jcsd
  3. May 15, 2006 #2

    AKG

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    [tex]\frac{dx}{dy} = \frac{d}{dx}(z) = \sum_{n=0} ^{\infty} 2^n \cdot z^{2n}[/tex]

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

    [tex]f(x) = \sum_{n=0}^{\infty}\frac{2^n}{2n+1}x^{2n+1}[/tex]

    To see if it is a solution to

    [tex]\frac{dx}{dy} - 2xy = 1[/tex]

    you need to check that:

    [tex]\frac{dx}{df} - dxf = 1[/tex]

    Note:

    [tex]\frac{dx}{df} = \frac{1}{\frac{df}{dx}}[/tex]

    You want to verify:

    [tex]\frac{1}{\sum_{n=0}^{\infty}2^nx^{2n}} - 2x\sum_{n=0}^{\infty}\frac{2^n}{2n+1}x^{2n+1} = 1[/tex]

    for all x.
     
    Last edited: May 15, 2006
  4. May 15, 2006 #3
    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 [tex]{\sum_{n=0} ^{\infty} 2^n x^{2n}}[/tex] as the commen denominator.

    Then I get [tex]\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[/tex]

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

    Sincerely
    Fred
     
    Last edited: May 15, 2006
  5. May 15, 2006 #4

    AKG

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    No, you get:

    [tex]\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[/tex]

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

    [tex]\sum _{n=0}^{\infty}\frac{2^n}{2n+1}x^{2n+1}[/tex]

    doesn't even converge for all [itex]x \in \mathbb{R}[/itex]. Maybe you're only supposed to look at these series as formal power series, but then I don't see why they would say [itex]x \in \mathbb{R}[/itex]. Actually, even looking at it that way the thing you're asked to prove appears false. I think you copied out the question incorrectly.
     
  6. May 15, 2006 #5
    Hello AKG,

    Looking at the question again:

    Given a power series:

    [tex]\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[/tex]

    converge for all [tex]z \in \mathbb{C}[/tex].

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

    Then show that f is a solution for

    [tex]\frac{dx}{dy} - 2xy = 1[/tex]

    thats the exact problem. Any idears?

    Sincerely

    Fred.

    p.s. Isn't
    [tex]\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}[/tex]
     
    Last edited: May 15, 2006
  7. May 15, 2006 #6

    HallsofIvy

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    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 [itex]z = x \in \mathbb{R}[/itex],
    then x is NOT a function of y and so [itex]\frac{dx}{dy} - 2xy = 1[/itex] makes no sense. Perhaps you meant
    [tex]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[/tex]
    satisfies the equation
    [itex]\frac{dy}{dz} - 2zy = 1[/itex]

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

    Now, In the first sum, let j= n so it becomes
    [tex]\sum_{j=0} ^{\infty} \frac{2^j}{1 \cdot 3 \cdot 5 \cdots (2j)} z^{2j}[/tex]
    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
    [tex]-2\sum_{j=1} ^{\infty} \frac{2^j}{1 \cdot 3 \cdot 5 \cdots (2j-1)} z^{2j}[/tex]
    except for j= 0, those are exactly the same but of different sign and so every term except the z0 term cancels. What is [itex]\frac{2^j}{1 \cdot 3 \cdot 5 \cdots (2j)} z^{2j}[/itex] when j= 0?
     
  8. May 15, 2006 #7
    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 [itex]\frac{2^j}{1 \cdot 3 \cdot 5 \cdots (2j)} z^{2j}[/itex] 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
     
  9. May 15, 2006 #8

    AKG

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

    NO! Don't you know what "!" factorial is?
     
  10. May 15, 2006 #9

    AKG

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    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?
     
  11. May 15, 2006 #10

    Curious3141

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    Just to explore the series itself a little more,

    [tex]y = f(x) = \sum_{n=0} ^{\infty} \frac{2^n}{2n+1} x^{2n+1}[/tex]

    Differentiating term by term,

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

    Now observe that [tex]-2x^2S(x) = -2x^2 - 4x^4 - 8x^6 - ...[/tex]

    giving [tex](1-2x^2)S(x) = 1[/tex]

    and [tex]S(x) = f'(x) = \frac{1}{1-2x^2}[/tex]

    From this we can get [tex]\frac{dx}{dy} = (1 - 2x^2)[/tex]

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

    [tex]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}}[/tex]

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

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

    Now we have explicit expressions for everything, but I still can't see how [tex]\frac{dx}{dy} - 2xy = 1[/tex]:confused:
     
    Last edited: May 15, 2006
  12. May 15, 2006 #11

    AKG

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

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

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

    [tex]-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)}[/tex]

    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
  13. May 15, 2006 #12

    Curious3141

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    Why does (2n+1)!! appear in the denominator of your f(x)? That wasn't in the orig. question.
     
  14. May 15, 2006 #13

    AKG

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    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?
     
  15. May 15, 2006 #14

    Curious3141

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    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
  16. May 16, 2006 #15
    I'm very sorry,

    It was suppose to say

    Hello AKG,

    Looking at the question again:

    Given a power series:

    [tex]\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[/tex]

    converge for all [tex]z \in \mathbb{C}[/tex].

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

    Then show that f is a solution for

    [tex]\frac{dy}{dx} - 2xy = 1[/tex]

    thats the exact problem. Any idears?

    Sincerely

    Fred.
     
  17. May 16, 2006 #16

    Curious3141

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    AKG has already solved this problem in post 11.
     
  18. May 16, 2006 #17
    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

     
  19. May 16, 2006 #18

    AKG

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

    What? No, that's not true at all.
     
  20. May 17, 2006 #19
    Then

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

    Must equal

    [tex]\sum_{n=0} ^{\infty} \frac{2^n}{(2n+1)!!} z^{(2n+1)}[/tex]


    /Fred

     
    Last edited: May 17, 2006
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