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June 15

multiple angles, converting between trig functions

Is there a closed form expression for functions like
$\sin(a\cos ^{-1}x)$
$\sin(a\tan ^{-1}x)$
$\cos(a\sin ^{-1}x)$
etc? Widener (talk) 06:32, 15 June 2012 (UTC)[reply]

Well you can always use Euler's function to get those sorts of things out in logs and powers and with the forms you've got I believe the logs should disappear, but it wouldn't be pretty. Dmcq (talk) 09:57, 15 June 2012 (UTC)[reply]

See [1]. Bo Jacoby (talk) 10:40, 15 June 2012 (UTC).[reply]

Or here, with the parentheses placed as written by the OP [2] SemanticMantis (talk) 18:40, 15 June 2012 (UTC)[reply]

As a special case, the Chebyshev polynomials of the first kind are related to

\cos

and

\cos ^{-1}

by the equation

T_{n}(x)=\cos(n\cdot \cos ^{-1}x)

(for n = 0, 1, 2, 3, ...), giving you

\sin(n\cdot \cos ^{-1}x)={\sqrt {1-(T_{n}(x))^{2}}}

.

Of course, this won't help you in the general case. — Tobias Bergemann (talk) 11:18, 15 June 2012 (UTC)[reply]

Conditions under which boundary value problems have a solution

$y''+\pi ^{2}y=f(x),y(0)=0,y(1)=0$ where f is piecewise continuous. Explain why this BVP does not in general have a solution. Widener (talk) 22:08, 15 June 2012 (UTC)[reply]

The general solution to the differential equation y"+y=0 is A cos(x)+B sin(x), so y(0)=y(2π). If you impose a boundary condition where y(0)≠y(2π) there is no solution. Bo Jacoby (talk) 07:22, 16 June 2012 (UTC).[reply]

I think Bo Jacoby means to say, "there is no homogeneous solution.” An inhomogeneous solution still exists. In fact, a simple example in physics is the solution to a the standard wave equation for a taut string whose boundary conditions are forced to some non-equilibrium value. Nimur (talk) 12:42, 16 June 2012 (UTC)[reply]

Just as an addendum, my PDE textbook, which is written from an applied mathematics perspective, mentions "solvability condition" or "compatibility condition" to describe whether a nonhomogeneous source, or indeed any boundary condition, is compatible with an implicit physical constraint, usually conservation of energy. This is very different from whether a solution exists in the terminology of pure mathematics. Existence of an "incompatible" solution might be loosely described as "no solution." I'm still hoping a mathematician can come along and explain whether the "piecewise continuous" source function, mentioned by the OP, could yield some "gotcha" that might exclude the existence of any solution, physical or otherwise. Nimur (talk) 17:15, 16 June 2012 (UTC)[reply]

"Explain why this BVP does not in general have a solution" suggests to me I'm supposed to find some f(x) for which there is no solution. Bo Jacoby showed it when f(x)=0, which I'd say is a pretty good answer. Staecker (talk) 23:14, 16 June 2012 (UTC)[reply]

He didn't, because the boundary conditions are of the form he wanted. Given we're asked for piecewise continuity, I'd be tempted to try to consider a step function, but am not well enough up on whether we need to consider weak solutions. 128.232.241.211 (talk) 08:46, 17 June 2012 (UTC)[reply]

I do not know what a homogeneous solution is. y"+y=0 is a homogeneous equation and y"+y=1 is an inhomogenous equation having the general solution y=A cos(x)+B sin(x)+1, which does not satisfy y(0)=0, y(2π)=1 for any values of A and B. Bo Jacoby (talk) 10:58, 17 June 2012 (UTC).[reply]

You're correct, I'm overthinking the problem. I have incorrectly interpreted "boundary condition" f(x) as "initial boundary condition," f(x, t=0) which is not applicable in the original question, and not a good assumption to make in general, even if we were seeking a time-evolution of the solution here. (Time for me to review ODEs and proper terminology). Nimur (talk) 13:15, 17 June 2012 (UTC)[reply]

Notice that the equation is $y''+\pi ^{2}y=f$ and not $y''+y=f$ . In this case, the complementary function is $a\sin(\pi x)+b\cos(\pi x)$ where a and b are real numbers. Imposing the boundary condition $y(0)=0$ tells us that $b=0$ . Imposing the boundary condition $y(1)=0$ again tells us that $b=0$ : we're left with a free parameter. I'd conjecture that this has something to do with the fact that, in general, there is no solution (we only really have one boundary condition). — Fly by Night (talk) 17:09, 19 June 2012 (UTC)[reply]

The linear substitution

x={u \over \pi }

transformes the equation

{d^{2}y \over dx^{2}}+\pi ^{2}y=f(x)

into

\pi ^{2}{d^{2}y \over du^{2}}+\pi ^{2}y={f({u \over \pi })}.

Divide by

\pi ^{2}

to obtain

{d^{2}y \over du^{2}}+y=g(u)

where

g(u)={f({u \over \pi }) \over \pi ^{2}}.

So the constant

\pi ^{2}

in the equation is unimportant. Bo Jacoby (talk) 07:40, 20 June 2012 (UTC).[reply]

I would imagine it would be quite important to an examiner that asked someone to solve the original ODE. — Fly by Night (talk) 22:55, 20 June 2012 (UTC)[reply]