# Problem with Taylor's Theorem

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Dec 8th 2007, 04:04 PM
chrsr345
Problem with Taylor's Theorem
EDIT: tHE QUESTION CAN BE SEEN HERE:

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
• Dec 8th 2007, 04:38 PM
kalagota
Quote:

Originally Posted by chrsr345
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 $\displaystyle J(0)$ from 1 to 5, where $\displaystyle J(x) = \sum_{k=0}^\infty {\frac{(-1)^k x^{2k+1}}{k!(k+1)!}}$, i dont know where to locate $\displaystyle 2^{2k+1}$..
• Dec 8th 2007, 05:18 PM
chrsr345
Kalagota, thank you! the final term is in the denominator. This is it
$\displaystyle J(x) = \sum_{k=0}^\infty {\frac{(-1)^k x^{2k+1}}{k!(k+1)!2^{2k+1}}}$
• Dec 9th 2007, 07:01 AM
chrsr345
Oh by the way guys, this is a Bessel function...i dont know what that is as its not even in my book.
• Dec 9th 2007, 07:42 AM
ThePerfectHacker
Quote:

Originally Posted by chrsr345
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.
• Dec 9th 2007, 07:52 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
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 $\displaystyle J_1(x)$.

-Dan
• Dec 9th 2007, 07:53 AM
chrsr345
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?
• Dec 9th 2007, 08:02 AM
topsquark
Quote:

Originally Posted by chrsr345
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:
$\displaystyle x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} + (x^2 - n^2)y = 0$
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:
$\displaystyle J^{\prime}_n(x) = J_{n - 1}(x) - \frac{n}{x}J_n(x)$

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
• Dec 9th 2007, 08:12 AM
chrsr345
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 :(
• Dec 9th 2007, 11:39 AM
ThePerfectHacker
Just be careful some strange stuff (which he cannot discuss here) happen when n is an integer.
• Dec 9th 2007, 11:48 AM
chrsr345
So are there sites that give a specific step process on how to handle this thing?
• Dec 9th 2007, 01:22 PM
Constatine11
Quote:

Originally Posted by chrsr345
EDIT: tHE QUESTION CAN BE SEEN HERE:

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:

$\displaystyle 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} + ...$$\displaystyle =\frac{x}{2} - \frac{x^3}{16} + \frac{x^5}{384} - ...$

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

$\displaystyle J_1(x)=J_1(0) + J_1^{(1)}(0)x + J_1^{(2)}(0)x^2/2! + J_1^{(3)}(0)x^3/3!++ J_1^{(4)}(0) x^4/4! + J_1^{(5)}(0) x^5/5! + ..$

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

$\displaystyle J_1^{(5)}(0)/5!=\frac{1}{384}$

or:

$\displaystyle J_1^{(5)}(0)=\frac{1}{384}5!=15/44$

ZB
• Dec 9th 2007, 01:29 PM
chrsr345
Thank you constatine! Im gonna try this out and make sure it makes sense to me. Much appreciated!
• Dec 9th 2007, 04:21 PM
chrsr345
$\displaystyle J_1(x)=J_1(0) + J_1^{(1)}(0)x + J_1^{(2)}(0)x^2/2! + J_1^{(3)}(0)x^3/3!++ J_1^{(4)}(0) x^4/4! + J_1^{(5)}(0) x^5/5! + ..$

Hmm, you have the (0) followed by the X's here, i thought X is whats being evaluated at 0...
• Dec 9th 2007, 05:50 PM
topsquark
Quote:

Originally Posted by chrsr345
$\displaystyle J_1(x)=J_1(0) + J_1^{(1)}(0)x + J_1^{(2)}(0)x^2/2! + J_1^{(3)}(0)x^3/3!++ J_1^{(4)}(0) x^4/4! + J_1^{(5)}(0) x^5/5! + ..$

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
$\displaystyle f(x) \approx f(a) + \frac{1}{1!}f^{\prime}(a)(x - a) + \frac{1}{2!}f^{\prime \prime}(a)(x - a)^2 + \frac{1}{1!}f^{\prime \prime \prime}(a)(x - a)^3 + ~...$

Just take a = 0 for this one.

-Dan
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last