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Taylor series for differential equation solution

  1. Nov 2, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the series solution for:
    [tex]
    y'=x^2-y^2,y(1)=1
    [/tex]


    2. Relevant equations



    3. The attempt at a solution
    I have correctly derived the series solution as:
    [tex]
    y(x)=1+(x-1)^2-\frac{(x-1)^3}{3}+\frac{(x-1)^4}{6}-...
    [/tex]
    But I cannot get the book solution for the INTERVAL OF CONVERGENCE which is |x-1|<0.04
    John
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 2, 2009 #2

    Mark44

    Staff: Mentor

    Your solution doesn't do you much good if you don't know what the general term looks like. Is the next term in the series
    [tex]-\frac{(x - 1)^5}{10}[/tex]?
     
  4. Nov 2, 2009 #3
    Yes.
    In fact I have derived the nth term as:
    [tex]
    \frac{2(-1)^{n+1}(x-1)^{n+1}}{n^2+n}
    [/tex]
    I then applied the RATIO TEST but was unable to agree the book answer.
    In fact I have been following the book method of deriving the interval of convergence by expanding:
    [tex]
    x^2-y^2
    [/tex]
    In powers of (x-1) and (y-1)
    [tex]
    x^2-y^2=2(x-1)-2(y-1)+(x-1)^2-(y-1)^2
    [/tex]
    with |x-1|<r and |y-1|<r
    Then
    [tex]
    |x^2-y^2|\leq|(x-1)^2|+|(y-1)^2|+2|(x-1|+2|y-1|
    [/tex]
    Thus
    [tex]
    |x^2-y^2|<r^2+r^2+2r+2r=2r^2+4r=M
    [/tex]
    Then the interval of convergence is:
    [tex]
    I:|x-1|<min(r,\frac{r}{3M})=min(r,\frac{r}{3(2r^2+4r)})
    [/tex]
    Then let
    [tex]
    u=\frac{r}{6r^2+12r}
    [/tex]
    Then
    [tex]
    \frac{du}{dr}=\frac{-6r^2}{(6r^2+12r)^2}
    [/tex]
    Let du/dr=0
    But then r=0[Book Answer =0.04?]
    I'm wondering if there is a book error, because I have done two other questions with this method and agreed the book answer in both cases.
    John
     
  5. Nov 3, 2009 #4

    Mark44

    Staff: Mentor

    If du/dr = -6r2/(6r2 + 12r)2) then setting du/dr = 0 would seem to imply that r = 0, but notice that both the numerator and denominator are zero when r = 0.

    The expression for du/dr has a limit as r approaches zero, however.
    [tex]\lim_{r \rightarrow 0} \frac{-6r^2}{(6r^2 + 12)^2}~=~\lim_{r \rightarrow 0} \frac{-6r^2}{36r^2(r + 2)^2}~=~\lim_{r \rightarrow 0} \frac{-1}{6(r + 2)^2}~=~-\frac{1}{24}~\approx~-.042[/tex]
    This value is numerically close to, but opposite in sign to the book's answer.

    I have some questions about the rest of the work. What escapes me about this process is why it is useful to write x2 - y2 in powers of x - 1 and y - 1. I understand the expansion itself, enabling you to write your original differential equation as y' = 2(x - 1) - 2(y - 1) + (x - 1)2 - (y - 1)2. What I'm missing is the part where this new differential equation is actually solved for y as a function of x.

    Also, why are both |x - 1| and |y - 1| < r? And where did the 3 come from in the inequality |x - 1| < min(r, r/(3M))?
     
  6. Nov 3, 2009 #5
    Hi Mark
    The book is using the following theorem:
    If f(x,y) is analytic at (x_0,y_0), i.e if f(x,y) has a Taylor series expansion in powers of (x-x_0) and (y-y_0), valid for |x-x_0|<r,|y-y_0|<r, and if for every (x,y) in this 2rx2r rectangle which has (x_0,y_0) at its center,
    [tex]
    |f(x,y|\leq M
    [/tex]
    where M is a positive number, then there is a unique particular solution y(x),analytic at x=x_0,satisfying the initial condition y(x_0)=y_0.
    In this case, of course x_0=y_0=1.
    The Taylor series expansion is valid in an interval about x_0.
    This interval is at least equal to:
    [tex]
    I:|x-x_0|<min(r,\frac{r}{3M})
    [/tex]
    Regards
    John
     
  7. Nov 3, 2009 #6

    Mark44

    Staff: Mentor

    Is the context of the theorem you cited y' = f(x, y)?
     
  8. Nov 3, 2009 #7
    Yes it is a series solution for y'=f(x,y)
    By the way your solution -0.42 is not consistent in terms of |x-1|, since modulus should be positive?
    John
     
    Last edited: Nov 3, 2009
  9. Nov 3, 2009 #8

    Mark44

    Staff: Mentor

    In post #3, you have |x - 1| < min(r, r/(6r^2 + 12r)). Apparently you took the derivative of r/(6r^2 + 12r) to find its minimum value. This expression can be simplified to 1/(6(r + 2)), if r != 0. There is a removable discontinuity at (0, 1/12).

    The graph of y = 1/(6(x + 2)) is asymptotic to the horizontal axis, and as such, has no minimum value. The slope of this curve is never zero, something we already saw.
     
  10. Nov 3, 2009 #9
    Hi Mark
    What are we concluding about this problem. Is the book answer for the interval of convergence =0.04 correct?
    John
     
  11. Nov 3, 2009 #10

    Mark44

    Staff: Mentor

    We haven't concluded anything, as far as I can tell. The theorem that you used doesn't seem to lead to a value of r for the radius of convergence. It's probably an existence theorem, that states the conditions for a series to converge without explicitly saying what the radius of convergence actually is.

    I would go back and take a look again at your series solution, y(x) = ... and see if you can find the conditions for which it converges. Are you sure that your series solution is correct? I.e., does what you have agree with what's in your textbook?
     
  12. Nov 3, 2009 #11
    Yes my series solution agrees with the book. I have done the previous and following examples in the book using their same method and have agreed both the series solutions and their intervals of convergence.
    John
     
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