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Math Help - limit of series 1 A

  1. #1
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    limit of series 1 A

    i have dy/dx=x+y

    y0=1
    y1=1+x+(x^2)/2
    y2=1+x+(x^2)/2+(x^3)/6
    what is the limit of the series yn
    ?
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  2. #2
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    Two comments:

    1. It looks an awful lot like the exponential series.
    2. Theoretically, assuming certain smoothness conditions are met, which it looks like they are, the series should converge to the solution of the DE. You're essentially doing Picard iterations there, if I'm not mistaken.
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  3. #3
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    yes its itterations and i need to find yn where
    n goes to infinity
    ?
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  4. #4
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    Why can't you just solve the DE exactly? It's first order linear...
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  5. #5
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    So, just to be clear, you're setting up the following iteration scheme:

    \displaystyle y_{n+1}(x)=\int(x+y_{n}(x))\,dx, right? If so, I don't get what you get. I'm seeing

    y_{0}=1,

    y_{1}=\displaystyle\int(x+1)\,dx=x+\frac{x^{2}}{2}  ,

    \displaystyle y_{2}=\int\left(x+x+\frac{x^{2}}{2}\right)\,dx=x^{  2}+\frac{x^{3}}{6},\dots

    Is there an initial condition or something I should know about?
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  6. #6
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    yes y(0)=1
    the whole question is about approximating

    so i cant solve it like first order linear
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  7. #7
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    Ok. So y(0) = 1 implies that

    y_{0}=1,

    y_{1}=\displaystyle\int(x+1)\,dx=1+x+\frac{x^{2}}{  2},

    y_{2}=\displaystyle\int\left(x+1+x+\frac{x^{2}}{2}  \right)\,dx=1+x+x^{2}+\frac{x^{3}}{6},

    y_{3}=\displaystyle\int\left(x+1+x+x^{2}+\frac{x^{  3}}{6}\right)dx=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6  }+\frac{x^{4}}{24},\dots

    That's what you got, I believe. The sequence \{y_{n}\} converges to e^{x}. To convince yourself of that fact, just expand e^{x} in a Taylor series about 0, and you'll see that it's the exact same series as you're getting here.

    How rigorous a proof does this require?
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  8. #8
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    as much rigorous as you can give me
    my orof is very strict
    and i only started with converging serieses
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  9. #9
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    Try a proof by induction.
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  10. #10
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    ok
    what expressio for n=k to take
    and what expression for n=k+1 to take
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  11. #11
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    Well, it looks like you want to show that, aside from the first term,

    \displaystyle y_{n}=\sum_{j=0}^{n+1}\frac{x^{j}}{j!}.

    What ideas does that give you?
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  12. #12
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    y_{n+1}\displaystyle =1+\int (t+ \sum_{j=0}^{n}\frac{t^{j}}{j!})dt
    how to prove that it equals \displaystyle y_{n}=\sum_{j=0}^{n+1}\frac{x^{j}}{j!}
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  13. #13
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    An induction proof has two parts: the base case, and the inductive step. What's your base case?
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  14. #14
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    aahh i understand now
    thanks
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  15. #15
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by transgalactic View Post
    i have dy/dx=x+y

    y0=1
    y1=1+x+(x^2)/2
    y2=1+x+(x^2)/2+(x^3)/6
    what is the limit of the series yn
    ?
    We have the first order linear DE...

    \displaystyle y^{'} = y + x\ \ ,\ y(0)=1 (1)

    ... and we have 'forgotten' the procedure to solve it... no problem because we can use an alternative approach!...

    Let's suppose that y(x) is analytic in x=0 so that is...

    \displaystyle y(x)= \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\ x^{n} (2)

    But the y^{(n)} (0) can be derived from (1) as follows...

    \displaystyle y^{(0)} (0) = 1

    \displaystyle y^{(1)}(x)= y+x \implies y^{(1)}(0)= 1

    \displaystyle y^{(2)}(x) = 1 + y^{(1)}(x) = 1+x+y \implies y^{(2)}(0)= 2

    \displaystyle y^{(3)}(x) =  1+x+y \implies y^{(3)}(0)= 2

    \dots

    \displaystyle y^{(n)}(x) =  1+x+y \implies y^{(n)}(0)= 2

    \dots

    ... so that is...

    \displaystyle y(x)= 1 + x + 2\ \sum_{n=2}^{\infty} \frac{x^{n}}{n!} = 2\ e^{x} -1 -x (3)

    Kind regards

    \chi \sigma
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