# need answer for this form

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#### Roshdy

##### Member level 3
dear all

pi/(2ΔΘ)
∑ cos(a sin nΔΘ)
n=1

I also need its limit when ΔΘ->0

I am afraid that there is missing a factor ΔΘ in your formula.

Consider the integration of function

f(Θ)=cos(a*sin(Θ))

over interval [0,pi/2].

Let's start from the definition of definite integrals, instead of the inverse of the derivative.

Choose a big N and then set

ΔΘ=(pi/2)/N,
which means
N=(pi/2)/ΔΘ (the upper limit of your sum)

Then the interval [0, pi/2] is divided into N subintervals:

[0,ΔΘ], [ΔΘ,2ΔΘ], ..., [(N-1)ΔΘ,NΔΘ] with the same length ΔΘ. Now, construct the sum:

f(ΔΘ)ΔΘ + f(2ΔΘ)ΔΘ + ... + f(NΔΘ)ΔΘ

Let N-> ∞, according to the definition of integral, this sum converges to the integral of f(Θ) over [0,p1/2], which is the 0th order Bessel function J0(a).

The difference of this one with yours is only the factor ΔΘ.

### Roshdy

Points: 2
nothing lost, please try to help me to solve the form above

Ok, let me explain why (I said something is missing) by some special cases:

1. a=0. This is an extreme case. Let's assume N=pi/(2ΔΘ). Then, N->Infinity when ΔΘ->0. In this case, since cos(a*sin nΔΘ)=cos(0)=1, your sum will be (let me denote that sum by SUM)
SUM=N, which tends to Infinity when ΔΘ->0.

2. When a>0 (similar for a<0) and N is large (or ΔΘ small), sin(nΔΘ) is very small for not big n's. Thefore, cos(a*sin nΔΘ) will be close to 1. That means your SUM will tend to big.

So, my conclusion is that whatever the SUM is, it'll tend to Infinity when ΔΘ->0. This is why I said your formula is missing a factor ΔΘ, which would have balanced your SUM down to a finite value.

Yes, whatever the SUM is, it still makes sense to sum it over to a simpler form. Sorry, this is the part that I missed in the last post. Here is the reason that we probably can't make it simpler. Assume that we can, and let me denote the simplified form by F(N). In other words,
SUM=F(N).
Multiplying the two sides by ΔΘ (=pi/(2N)) produces

SUMΔΘ=F(N)ΔΘ
or
(pi/2)SUM/N=(pi/2)F(N)/N

The left hand side will tend to the 0th order bessel function as I indicated last time, and we all know that the 0th order bessel function cannot be expressed in a close form (1/2 order can, by the way). Now, look at the right hand side. When N->Infinity (ΔΘ->0), it's just a limit and will tend to a close form (because only derivatives are possibly involved). Therefore, we end up with a conflict.

Points: 2