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Math Help - limit of series 1 A

  1. #16
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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
    ?
    The limit of that sequence is e^x as you should be able to recognize the MacLaurin series for e^x
    However, the derivative of y= e^x is e^x= y, not x+ y.

    If you are trying to use Picard's method to solve this equation, as Ackbeet suggested, then you are doing it wrong.

    From dy/dx= x+y we get dy= (x+y)dx so that y= \int (x+ y)dx. Of course, since we don't know y but we can approximate it. If we are given y(0)= 1 (which you don't say but is implied by "y0= 1") as a first approximation, assume y(x)= 1 for all x. Then dy/dx= x+1 so that dy= (x+1)dx and [tex]y= (1/2)x^2+ x+ C[tex] and y(0)= C= 1 gives y1= (1/2)x^2+ x+ 1, as you have.

    But then, y2= \int x+ y1 dx= \int x+ (1/2)x^2+ x+ 1 dx= \int (1/2)x^2+ 2x+ 1 dx= (1/6)x^3+ x^2+ x+ C and, again, y(0)= C= 1 gives y2= (1/6)x^2+ x^2+ x+ 1 which is NOT what you have. I expect you forgot the added x.

    Now, y3= \int x+ y2 dx= \int x+ (1/6)x^3+ x^2+ x+ 1 dx = \int (1/6)x^3+ x^2+ 2x+ 1 dx= (1/24)x^4+ (1/3)x^3+ x^2+ x+ C and y(0)= C= 1 gives y3= (1/24)x^4+ (1/3)x^3+ x^2+ x+ 1.

    One more time: y4= \int x+ y3 dy= \int x+(1/24)x^3+ (1/3)x^3+ x^2+ x+ 1 dx = \int (1/24)x^4+ (1/3)x^3+ x^2+ 2x+ 1 dx= (1/5!)x^5+ (1/12)x^4+ (1/3)x^3+ x^2+ x+ C and y(0)= C= 1 gives
    y4= (1/5!)x^5+ (2/4!)x^4+ (2/3!)x^3+ (2/2!)x^2+ x+ 1.

    I think I would conjecture now that
    yn= (1/n!)x^n+ (2/(n-1)!)x^{n-1}+ \cdot\cdot\cdot+ (2/3!)x^3+ (2/2!)x^2+ x+ 1
    and I would attempt to prove that by "induction"

    When n= 0, that says that y0= 1 which is true.

    Now assume that, for some k, yk= (1/k!)x^k+ (2/(k-1)!)x^{k-1}+\cdot\cdot\cdot+ (2/3!)x^3+ (2/2!)x^2+ x+ 1 so that
    y(k+1)= \int x+ yk dx= \int (1/k!)x^k+ (2/(k-1)!)x^{k-1}+ \cdot\cdot\cdot+ (2/3!)x^3+ (2/2!)x^2+ 2x+ 1 dx = (1/(k+1)!)x^{k+1}+ (2/k!)x^k+ \cdot\cdot\cdot+ (2/4!)x^4+ (2/3!)x^3+ x^2+ x+ C and y(k+1)(0)= C= 1 gives
    y(k+1)= (1/(k+1)!)x^{k+1}+ (2/k!)x^k+ \cdot\cdot\cdot+ (2/4!)x^4+ (2/3!)x^3+ (2/2!) x^2+ x+ 1
    just as we wish.
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  2. #17
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    Quote Originally Posted by Ackbeet View Post
    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?
    i dont have y2 of a exponent
    y2=1+ \int t+1+t+\frac{t^3}{6}dt

    i get t^2
    i should get \frac{t^2}{2}
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