Can Bessel Functions and Cosine be Expressed as Infinite Series?

[athrun200]

Homework Statement

Show that
\cos x=J_{0}+2\sum(-1)^{n}J_{2n}
where the summation range from n=1 to +inf

Homework Equations

Taylor series for cosine?
series expression for bessel function?

The Attempt at a Solution

My approach is to start from R.H.S.
I would like to express all bessel functions in the series form, then compare it to the taylor series of cosine.

I notice that the summation can be written as
-J_{2}+J_{4}-J_{6}+J_{8}+...
Using the recurrence relation, we have - 2J'_{3}-2J_{7}-2J_{11}-...

Therefore, R.H.S can be written asJ_{0}+ 4J'_{3}-4J_{7}-4J_{11}-...

But it seems it will be extremely difficult to deal with it. Since each term itself is a series. We are now summing up infinity many series.

I wonder if we have a better way to finish this question

 

[jackmell]

Homework Statement

Show that
\cos x=J_{0}+2\sum(-1)^{n}J_{2n}
where the summation range from n=1 to +inf

Homework Equations

Taylor series for cosine?
series expression for bessel function?

The Attempt at a Solution

My approach is to start from R.H.S.
I would like to express all bessel functions in the series form, then compare it to the taylor series of cosine.

I notice that the summation can be written as
-J_{2}+J_{4}-J_{6}+J_{8}+...
Using the recurrence relation, we have - 2J'_{3}-2J_{7}-2J_{11}-...

Therefore, R.H.S can be written asJ_{0}+ 4J'_{3}-4J_{7}-4J_{11}-...

But it seems it will be extremely difficult to deal with it. Since each term itself is a series. We are now summing up infinity many series.

I wonder if we have a better way to finish this question

How about just a few? Can we even get that far with it? I don't know. Say the first three terms:

1-x^2/2+x^4/24

Ok, just that much. Can you start writing out say 3 or 4 or five of those series and see what the first term of them looks like and then notice a trend and then use induction to equate coefficients on each side of the expression and conclude the general expression for each constant? Each will be a sum of course so that we'd have:

1=\sum_1^{\infty} a_n

-1/2=\sum_1^{\infty}b_n

1/24=\sum_1^{\infty}c_n

and so forth.

 

[athrun200]

How about just a few? Can we even get that far with it? I don't know. Say the first three terms:

1-x^2/2+x^4/24

Ok, just that much. Can you start writing out say 3 or 4 or five of those series and see what the first term of them looks like and then notice a trend and then use induction to equate coefficients on each side of the expression and conclude the general expression for each constant? Each will be a sum of course so that we'd have:

1=\sum_1^{\infty} a_n

-1/2=\sum_1^{\infty}b_n

1/24=\sum_1^{\infty}c_n

and so forth.

I manage to match the first two terms.
But when doing the n=k step in MI, I have no idea now to express it.
There are so many terms on the right contain x^(2k)

 

[athrun200]

After some works, I discovered the coefficient of x^2n on RHS is like this (shown in the picture below).

But it seems it is impossible for them to be the same.

 

[jackmell]

After some works, I discovered the coefficient of x^2n on RHS is like this (shown in the picture below).

But it seems it is impossible for them to be the same.

Ok. But that's not easy to read. Also, I'm not sure that's the best approach ok but it's essential in order to become good at math: to try things and if they don't work, well, you try something else so I admire you're willingness to at least try it and even if someone comes with a better way, you're on your way to becommng better at math. :)

 

[athrun200]

Ok. But that's not easy to read. Also, I'm not sure that's the best approach ok but it's essential in order to become good at math: to try things and if they don't work, well, you try something else so I admire you're willingness to at least try it and even if someone comes with a better way, you're on your way to becommng better at math. :)

I discovered a formula very similar to my question.
I don't know if I can work on it to get the answer. I am still trying.
Comments are very welcome!

L.H.S. is e^{i x sinθ}

 

[athrun200]

By putting θ=Pi/2, then done!