Cosh of a variable

Hi. I need to solve the equation to get a variable from another variable. The equation is:

C(w)=Cosh[acosh(x1)+acosh(x2)+acosh(x3)+acosh(x4)]

x1=(w-0.641)/(1-0.641*w)

x2=(w+0.641)/(1+0.641*w)

x3=w

x4=w

Now since w ranges from -inf to inf I'm supposed to get C(w) as a function of w. Any guidance on how I can solve this?

Answer 1

[John D'Errico]

You do it by writing much like what you did, but in the proper order, and using proper syntax. For example, using symbolic tools, we might do this:

syms w
x1=(w-0.641)/(1-0.641*w);
x2=(w+0.641)/(1+0.641*w);
x3=w;
x4=w;

LEARN TO USE SEMICOLONS! As well, LEARN TO USE PARENS PROPERLY. Functions use (), NOT []. Just because you think an expression is easier to read if you use [] in some places, that does not make MATLAB understand your syntax. {} and () do different things.

C = cosh(acosh(x1)+acosh(x2)+acosh(x3)+acosh(x4))

C = 

fplot(C)

If you wanted to write this purely using function handles, it is as easy.

x1 = @(w) (w-0.641)./(1-0.641*w);
x2 = @(w) (w+0.641)./(1+0.641*w);
x3 = @(w) w;
x4 = @(w) w;

As you can see here, I used ./ to denote a division, so this will work for vectorized computations.

Cfun = @(w) cosh(acosh(x1(w))+acosh(x2(w))+acosh(x3(w))+acosh(x4(w)))

Cfun = function_handle with value:
    @(w)cosh(acosh(x1(w))+acosh(x2(w))+acosh(x3(w))+acosh(x4(w)))

And now we can evaluate the function handle in Cfun as:

Cfun([0:0.25:1]')

ans = 5×1

1.0000 0.6242 -0.3057 -0.9999 1.0000

Be careful, as for large magnitude inputs, it seems there is a small imaginary part that seems to arise when the computations are performed in double precision.

[John]

Hi John. Thank you so much for this. Is there a possibility that I could get C as a function of w without cosh or acosh function?

[John D'Errico]

Sure. All you need to do is recognize the identity...

cosh(X) = (exp(X) + exp(-X))/2

That allows you to eliminate the cosh calls. As for the acosh, no such luck.

But I think you do not mean something so trivial. The answer then is no. As it is, you should/can be happy there is a simply evaluatable expression. But there will be no simple expression in the form of a polynomial, or something as trivial to write.

Could you express this function as a polynomial approximation? Probably. Well, possibly. And without a moderate amount of analysis, I would not tell you where that polynommial approximation would have any value in terms of convergence. But you might try to approximate the result in the form of a truncated Taylor series approximation that would be valid over some interval, or you might try using a Pade approximant. I don't see the point of using an approximation here that will be highly problematic in terms of convergence, when you have a rather simple expression that can be evaluated directly.

[John]

Thank you John. That makes a lot of sense. Just one more question. Is there a reason, why I can't get a value for C less than -1?