Let us say x=y*cos(y), what does y equals in terms of x?

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DarnCat
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Let us say x=y*cos(y), what does y equals in terms of x?

Postby DarnCat » Wed Dec 24, 2014 2:57 pm UTC

Just a problem I've been stuck on recently... sadly wolfram alpha exceeds the standard computation time so sadly I can't simply use that as I normally do. So how can you go about solving it? I'm currently at a loss, so I hopes someone here could help me solve it; any ideas?

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Sizik
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Re: Let us say x=y*cos(y), what does y equals in terms of x?

Postby Sizik » Wed Dec 24, 2014 7:42 pm UTC

So in other words, you're trying to find the inverse of f(x) = x*cos(x). Since it's not injective, you'll have to restrict the domain in order to get a partial inverse.
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Lopsidation
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Re: Let us say x=y*cos(y), what does y equals in terms of x?

Postby Lopsidation » Wed Dec 24, 2014 7:57 pm UTC

There's no inverse in terms of elementary functions. Either get lucky with your value of x, or use a numerical approximation.

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Forest Goose
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Re: Let us say x=y*cos(y), what does y equals in terms of x?

Postby Forest Goose » Wed Dec 24, 2014 9:13 pm UTC

If you like horrible things and questionable assumptions:

Assume y(x) is analytic, then

z2y4 = x2z2(y2 - 1) + y(y - 2xz)

,z being the derivative of y.

Then, substitute in a power series for y and use the rule for powers to get a nasty nasty recurrence relation.

I tried doing the above, then assuming y was quadratic, but it just gives an equation that doesn't work (which is obvious, in reflection, and could've saved a lot of time if I thought about it). At any rate, numerical methods are probably the best bet.
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Derek
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Re: Let us say x=y*cos(y), what does y equals in terms of x?

Postby Derek » Thu Dec 25, 2014 1:32 am UTC

Could the Lambert W function help solve this? The Lambert W function is the function W(z) such that z = W(z)*exp(W(z)), in other words it's the inverse of z*exp(z). And cos(z) = Re(exp(i*z)) = (exp(i*z) + exp(-i*z))/2, so it seems like we might be able to get something from this.

x = y*cos(y)
x = y/2*(exp(i*y) + exp(-i*y))
2i*x = i*y*exp(i*y) + i*y*exp(-i*y)
???

Elmach
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Re: Let us say x=y*cos(y), what does y equals in terms of x?

Postby Elmach » Thu Dec 25, 2014 2:52 am UTC

Forest Goose wrote:If you like horrible things and questionable assumptions:

Assume y(x) is analytic, then

z2y4 = x2z2(y2 - 1) + y(y - 2xz)

,z being the derivative of y.

Then, substitute in a power series for y and use the rule for powers to get a nasty nasty recurrence relation.

I tried doing the above, then assuming y was quadratic, but it just gives an equation that doesn't work (which is obvious, in reflection, and could've saved a lot of time if I thought about it). At any rate, numerical methods are probably the best bet.

I suspect you are on the right track, but I would like to know how you found that equation.

Actually... I was about to suggest a Fourier Expansion, but since that is y cos y, not y cos x, that won't work.

edit: wait x = y cos x is trivial. Whoops

Editx2
X is y cos y
Dx is dy cos y minus y dy sin y
Dy dx is one over (cos y minus y sine y)
This is an equation that might be easier to solve.

Easier in some sense.

Editx3: no it is not easier this way; the standard solution to doing odes will give you x equals y cos y.

Editx4: actually, you can reduce(?) that ode to
Dy/dx is y/(x-y sqrt(yy-xx)).
not sure if this is better. No trig functions, yes, but...???

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Forest Goose
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Re: Let us say x=y*cos(y), what does y equals in terms of x?

Postby Forest Goose » Fri Dec 26, 2014 4:55 am UTC

I made a sign error with the equation (that's what you get for doing things in notepad...) Anyways:

Put C and S for cos y and sin y, respectively.

x = yC => 1 = zC - yzS => yzS = zC - 1 => y2S2 = (x / y - 1 / z)2 = x2 / y2 - 2x / yz + 1 / z2.

Then,

x2 + y2S2= y2(C2 + S2) = y2, so

y2 = x2 + x2 / y2 - 2x / yz + 1 / z2 =>

y4z2 = x2z2(y2 + 1) + y(y - 2xz)

*While drunk on...let's say holiday cheer...I was thinking about this, and I'm pretty sure there is a nicer polynomial in x, y, and z than the one above - but I forget how I got to it; it didn't have 6th degree (combined) terms, but I think 4th was the highest...that might be more tractable if anyone can figure it out (it was almost the same method, which is why I'm struggling to remember the difference...I'll post if it comes back to mind. I'm not sure polynomials of power series is, really, going to yield a nice answer in any case - unless it is really really needed, the recurrence relation on coeffs. is just horrid - more suited to Halloween...being scary and all...I'll stop now).

--If exactness isn't required, you can always expand C to a power series, truncate at the cubic term, have a 4th degree poly in y, then solve using the quintic formula with x as a coeff; you get 4 choices per x, and I'm not sure how good the over all solution is - I haven't tried it.
Forest Goose: A rare, but wily, form of goose; best known for dropping on unsuspecting hikers, from trees, to steal sweets.


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