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Math Help - Problem with Taylor's Theorem

  1. #1
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    Problem with Taylor's Theorem

    EDIT: tHE QUESTION CAN BE SEEN HERE:
    IMG_2571.jpg - Uploader v6

    Hey guys, im having a problem with using taylor's theorem. Well actually i dont know where to start. I need to use taylors theorem to find the value of the BESSEL FUNCTION OF ORDER 1 of J(1->5)(0) where J(x) Σ k=0-->
    (-1)^k x^2k+1
    / k!(k+1)! 2^2k+1

    (scroll down for real translation)

    So i guess i need to find the 1st 5 terms of this. But the only kind of taylors we have been taught is starting with a function and deriving. This starts with an infinite series so i dont know how to approach this. Sorry its a little bad looking so let me know if you need any clarification. Thanks!

    Chris
    Last edited by chrsr345; December 9th 2007 at 07:26 AM.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by chrsr345 View Post
    Hey guys, im having a problem with using taylor's theorem. Well actually i dont know where to start. I need to use taylors theorem to find the value of J(1->5)(0) where J(x) Σ k=0-->
    (-1)^k x^2k+1/ k!(k+1)! 2^2k+1

    So i guess i need to find the 1st 5 terms of this. But the only kind of taylors we have been taught is starting with a function and deriving. This starts with an infinite series so i dont know how to approach this. Sorry its a little bad looking so let me know if you need any clarification. Thanks!

    Chris
    i won't solve it yet.. i want to clarify things..

    is it J(0) from 1 to 5, where J(x) = \sum_{k=0}^\infty {\frac{(-1)^k x^{2k+1}}{k!(k+1)!}}, i dont know where to locate 2^{2k+1}..
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  3. #3
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    Kalagota, thank you! the final term is in the denominator. This is it
    J(x) = \sum_{k=0}^\infty {\frac{(-1)^k x^{2k+1}}{k!(k+1)!2^{2k+1}}}
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  4. #4
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    Oh by the way guys, this is a Bessel function...i dont know what that is as its not even in my book.
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  5. #5
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    Quote Originally Posted by chrsr345 View Post
    Oh by the way guys, this is a Bessel function...i dont know what that is as its not even in my book.
    No it is not. That is not the Bessel function. I know what Bessel function is when it slaps me in the face.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    No it is not. That is not the Bessel function. I know what Bessel function is when it slaps me in the face.
    Actually, it is. It's J_1(x).

    -Dan
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  7. #7
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    Hmm, then i dont know what it is, this is what the sheet from the teacher says. (see link in original post). Can anyone atleast give me a start on how to evaluate this using taylors theorem?
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by chrsr345 View Post
    Oh by the way guys, this is a Bessel function...i dont know what that is as its not even in my book.
    A Bessel function of the first kind is a solution to the differential equation:
    x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} + (x^2 - n^2)y = 0
    which are nonsingular at the origin. Bessel functions are commonly seen when solving electrostatic (or magnetostatic) equations in cylindrical coordinates. (Though they are seen in many other places as well.)

    There is a recurrence relation for these Bessel functions:
    J^{\prime}_n(x) = J_{n - 1}(x) - \frac{n}{x}J_n(x)

    That will allow you to compute the derivatives for your Taylor series.

    I would look up Bessel functions on the web or in a book (I'm using Mathematical Methods for Physicists by Arfken) as it appears your instructor is expecting you to look up this information. (Note: I have not given you everything you need to find your Taylor series.)

    -Dan
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  9. #9
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    Topsquark, thanks and yea i did see that info online which ive been searching. But i still dont know how to relate any of the info to what i know of taylors theorem. Ive been taught to start with a normal function and find the derivatives and 4-5 terms or so. So i really have no clue
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  10. #10
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    Just be careful some strange stuff (which he cannot discuss here) happen when n is an integer.
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  11. #11
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    So are there sites that give a specific step process on how to handle this thing?
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  12. #12
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    Quote Originally Posted by chrsr345 View Post
    EDIT: tHE QUESTION CAN BE SEEN HERE:
    IMG_2571.jpg - Uploader v6

    Hey guys, im having a problem with using taylor's theorem. Well actually i dont know where to start. I need to use taylors theorem to find the value of the BESSEL FUNCTION OF ORDER 1 of J(1->5)(0) where J(x) Σ k=0-->
    (-1)^k x^2k+1/ k!(k+1)! 2^2k+1

    (scroll down for real translation)

    So i guess i need to find the 1st 5 terms of this. But the only kind of taylors we have been taught is starting with a function and deriving. This starts with an infinite series so i dont know how to approach this. Sorry its a little bad looking so let me know if you need any clarification. Thanks!

    Chris
    You have:

    J_1(x) = \frac{(-1)^0}{0!1!} \frac{x}{2} + \frac{(-1)^1}{1!2!} \frac{x^3}{2^3} + \frac{(-1)^2}{2!3!} \frac{x^5}{2^5} + ... <br />
=\frac{x}{2} - \frac{x^3}{16} + \frac{x^5}{384} - ...<br />

    Now by Taylors theorem expanding J_1(x) about 0, we get:

     <br />
J_1(x)=J_1(0) + J_1^{(1)}(0)x + J_1^{(2)}(0)x^2/2! + J_1^{(3)}(0)x^3/3!++ J_1^{(4)}(0) x^4/4! + J_1^{(5)}(0) x^5/5! + ..<br />

    Now equate the terms in x^5 in the two series, gives:

     <br />
J_1^{(5)}(0)/5!=\frac{1}{384}<br />

    or:

     <br />
J_1^{(5)}(0)=\frac{1}{384}5!=15/44<br />

    ZB
    Last edited by Constatine11; December 9th 2007 at 11:08 PM. Reason: correct arithmetical error
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  13. #13
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    Thank you constatine! Im gonna try this out and make sure it makes sense to me. Much appreciated!
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  14. #14
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    J_1(x)=J_1(0) + J_1^{(1)}(0)x + J_1^{(2)}(0)x^2/2! + J_1^{(3)}(0)x^3/3!++ J_1^{(4)}(0) x^4/4! + J_1^{(5)}(0) x^5/5! + ..

    Hmm, you have the (0) followed by the X's here, i thought X is whats being evaluated at 0...
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  15. #15
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    Quote Originally Posted by chrsr345 View Post
    J_1(x)=J_1(0) + J_1^{(1)}(0)x + J_1^{(2)}(0)x^2/2! + J_1^{(3)}(0)x^3/3!++ J_1^{(4)}(0) x^4/4! + J_1^{(5)}(0) x^5/5! + ..

    Hmm, you have the (0) followed by the X's here, i thought X is whats being evaluated at 0...
    In general, a Taylor expansion of a function f(x) about x = a is
    f(x) \approx f(a) + \frac{1}{1!}f^{\prime}(a)(x - a) + \frac{1}{2!}f^{\prime \prime}(a)(x - a)^2 + \frac{1}{1!}f^{\prime \prime \prime}(a)(x - a)^3 + ~...

    Just take a = 0 for this one.

    -Dan
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