What is the maximum value of this equation?

[yungman]

Find the maximum $\theta$ of
\left[\frac{\cos\left(\frac{kl}{2}\cos\theta\right)-\cos\left(\frac {kl}{2}\right)}{\sin\theta}\right]^2
So I need to find the maximum of
F(\theta)=\frac{\cos\left(\frac{kl}{2}\cos\theta\right)-\cos\left(\frac {kl}{2}\right)}{\sin\theta}
First I differentiate respect to $\theta$ and equate to 0
dF(\theta)=\frac{(\sin\theta)\left(-\sin\left(\frac{kl}{2}\right)\cos\theta\right)\left(-\frac{kl}{2}\sin\theta\right)-\left[\cos\left(\frac{kl}{2}\cos\theta\right)-\cos\left(\frac{kl}{2}\right)\right](-\cos\theta)}{\sin^2\theta}=0
So
\Rightarrow\;(\sin\theta)\left(-\sin\left(\frac{kl}{2}\right)\cos\theta\right)\left(-\frac{kl}{2}\sin\theta\right)-\left[\cos\left(\frac{kl}{2}\cos\theta\right)-\cos\left(\frac{kl}{2}\right)\right](-\cos\theta)=0

This is no easier to solve compare to the original equation before differentiation. I tried letting $u=\frac{kl}{2}\cos\theta\Rightarrow\;du=-\frac{kl}{2}\sin\theta d\theta$. But still it is going nowhere. Please help.

 

[Dick]

Find the maximum $\theta$ of
\left[\frac{\cos\left(\frac{kl}{2}\cos\theta\right)-\cos\left(\frac {kl}{2}\right)}{\sin\theta}\right]^2
So I need to find the maximum of
F(\theta)=\frac{\cos\left(\frac{kl}{2}\cos\theta\right)-\cos\left(\frac {kl}{2}\right)}{\sin\theta}
First I differentiate respect to $\theta$ and equate to 0
dF(\theta)=\frac{(\sin\theta)\left(-\sin\left(\frac{kl}{2}\right)\cos\theta\right)\left(-\frac{kl}{2}\sin\theta\right)-\left[\cos\left(\frac{kl}{2}\cos\theta\right)-\cos\left(\frac{kl}{2}\right)\right](-\cos\theta)}{\sin^2\theta}=0
So
\Rightarrow\;(\sin\theta)\left(-\sin\left(\frac{kl}{2}\right)\cos\theta\right)\left(-\frac{kl}{2}\sin\theta\right)-\left[\cos\left(\frac{kl}{2}\cos\theta\right)-\cos\left(\frac{kl}{2}\right)\right](-\cos\theta)=0

This is no easier to solve compare to the original equation before differentiation. I tried letting $u=\frac{kl}{2}\cos\theta\Rightarrow\;du=-\frac{kl}{2}\sin\theta d\theta$. But still it is going nowhere. Please help.

You won't be able to solve that exactly. Is this a diffraction problem or something? I believe you are supposed to use an approximation like (kl/2)<<1. Then expand the outside cos functions in the numerator into a power series and only keep the lowest nonvanishing term in kl/2.

 

[yungman]

You won't be able to solve that exactly. Is this a diffraction problem or something? I believe you are supposed to use an approximation like (kl/2)<<1. Then expand the outside cos functions in the numerator into a power series and only keep the lowest nonvanishing term in kl/2.

Thanks for the time, kl/2 can be over 10 as $k=\frac {2\pi}{\lambda}$ and l can be up to 3$\lambda$!.

So there is no easy way to solve this?

Thanks

 

[Dick]

Thanks for the time, kl/2 can be over 10 as $k=\frac {2\pi}{\lambda}$ and l can be up to 3$\lambda$!.

So there is no easy way to solve this?

Thanks

Not sure I'm quite following you there. Can you show your complete solution? No, there's no easy way to get an EXACT solution. There's an easy way to get an approximate solution if you take kl/2<<1.

 

[yungman]

Not sure I'm quite following you there. Can you show your complete solution? No, there's no easy way to get an EXACT solution. There's an easy way to get an approximate solution if you take kl/2<<1.

What I meant is I have to work with $\frac{kl}{2}$≥10 or so. So the approx of kl/2<<1 is not going to work for me.

Do I use power series representation to solve it? Still the $\cos(\frac{kl}{2}\cos\theta)$ is going to be tricky.

 

[Dick]

What I meant is I have to work with $\frac{kl}{2}$≥10 or so. So the approx of kl/2<<1 is not going to work for me.

Do I use power series representation to solve it? Still the $\cos(\frac{kl}{2}\cos\theta)$ is going to be tricky.

No, power series only useful if you expand in a small parameter. If kl/2 is large, I'd suggest just plotting the function for kl/2=10 and larger numbers and looking at the graph to try and get another idea. That's what I'm doing. You do only have to look at the plot in the range [0,2pi] or even [0,pi/2]. It is still periodic.

 

[yungman]

Thanks.

 

[Ray Vickson]

What I meant is I have to work with $\frac{kl}{2}$≥10 or so. So the approx of kl/2<<1 is not going to work for me.

Do I use power series representation to solve it? Still the $\cos(\frac{kl}{2}\cos\theta)$ is going to be tricky.

For a = kl/2, the function f(θ) on [0,2π] has 14 local optima in (0,2π) (7 local maxima and 7 local minima). In other words, there would be 14 separate solutions to f'(θ) = 0. It appears from the graph that there are two global maxima (at θ = 0.8647431283 and θ = 2.276849525), both giving f = 2.389314458 .

 

[Dick]

For a = kl/2, the function f(θ) on [0,2π] has 14 local optima in (0,2π) (7 local maxima and 7 local minima). In other words, there would be 14 separate solutions to f'(θ) = 0. It appears from the graph that there are two global maxima (at θ = 0.8647431283 and θ = 2.276849525), both giving f = 2.389314458 .

So that's for a=10, yes? Now Yungman should try to figure out a way to guess an approximation of where that first global max is. Try plotting the numerator and the denominator separately and thinking about why they look like they do. Doing that I can make a guess that the first global max should come at about the first value of θ>0 where cos(10*cosθ)=1. That you can solve for. It's about 0.891. Which is not a bad estimate. If kl/2 is very large you might be able to make even closer estimates.

 

[yungman]

So that's for a=10, yes? Now Yungman should try to figure out a way to guess an approximation of where that first global max is. Try plotting the numerator and the denominator separately and thinking about why they look like they do. Doing that I can make a guess that the first global max should come at about the first value of θ>0 where cos(10*cosθ)=1. That you can solve for. It's about 0.891. Which is not a bad estimate. If kl/2 is very large you might be able to make even closer estimates.

Sounds like writing an excel program and plot a point every 3 deg is in order! Is there any free online program I can use. I am not an expert in math, I don't keep up with what's available online.

Many thank to everyone that put in the time to help me.

Alan

 

[haruspex]

dF(\theta)=\frac{(\sin\theta)\left(-\sin\left(\frac{kl}{2}\right)\cos\theta\right)\left(-\frac{kl}{2}\sin\theta\right)-\left[\cos\left(\frac{kl}{2}\cos\theta\right)-\cos\left(\frac{kl}{2}\right)\right](-\cos\theta)}{\sin^2\theta}=0

In case it matters, that's wrong. There's a misplaced right parenthesis. Should be
dF(\theta)=\frac{(\sin\theta)\left(-\sin\left(\frac{kl}{2}\cos\theta\right)\right)\left(-\frac{kl}{2}\sin\theta\right)-\left[\cos\left(\frac{kl}{2}\cos\theta\right)-\cos\left(\frac{kl}{2}\right)\right](-\cos\theta)}{\sin^2\theta}=0

 

[Dick]

In case it matters, that's wrong. [/tex]

Nah, doesn't really matter, I think it's just a typo. That whole route is a dead end anyway.