# Bessel Function and cosine

1. Nov 9, 2012

### athrun200

1. The problem statement, all variables and given/known data
Show that
$\cos x$=$J_{0}$+$2$$\sum(-1)^{n}$$J_{2n}$
where the summation range from n=1 to +inf

2. Relevant equations
Taylor series for cosine?
series expression for bessel function?

3. 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 as$J_{0}+$ $4J'_{3}$-$4J_{7}$-$4J_{11}$-...

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

2. Nov 9, 2012

### jackmell

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.

3. Nov 9, 2012

### athrun200

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)

4. Nov 9, 2012

### 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.

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5. Nov 9, 2012

### jackmell

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. :)

6. Nov 9, 2012

### athrun200

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.

L.H.S. is $e^{i x sinθ}$

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7. Nov 9, 2012

### athrun200

By putting θ=Pi/2, then done!