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I remember a take-home physics test in high school where there was a pendulum problem and I spent a full day trying to prove that x’’= ax is generally met by a

I remember a take-home physics test in high school where there was a pendulum problem and I spent a full day trying to prove that x’’= ax is generally met by a trigonometric function. I eventually gave up, and was very surprised when I got full marks for saying to simply assume it (though the rest of the test suffered by my being stuck for so long). Besides the exponential formulation of sin and cos, I’m honestly not sure if that’s a unique answer to this day. Does anyone know?

Fronzie 6y ago
There is an exact solution for the pendulum problem: https://www.researchgate.net/publication/39575612_Exact_solution_for_the_nonlinear_pendulum https://www.researchgate.net/publication/39575612_Exact_solu... I haven't studied it well enough to say more than that about it.
- ojnabieoot 6y ago
  Thanks - I wasn’t actually aware of this but I meant to write “elementary” rather than “exact” so I’m glad you brought this up. I had supposed there was probably some incantation of special functions that solved the equation :) It’s been a long time since I’ve studied ODEs and mathematical physics but this solution seems easier than I had guessed (in the sense that it’s accessible to advanced undergrads instead of specialists).
- leephillips 6y ago
  Weird paper. Eq. 32 is supposed to be an approximation to eq. 31, but they are the same equation.
kitty_kritter 6y ago
Nope, that's pretty much all there is. The space of all differentiable linear functions is a vector space, and the derivative is a linear operator on that space (so functions and derivatives follow most of the same rules as vectors and matrices). If D is the derivative operator and x is a function of t, the equation x''=x can be written as D^2 * x = x, or 0 = (D^2 - I)x = (D + I)(D - I)x. The dimension of the kernel of (D - I) is 1, and so is the dimension of the kernel of (D + I). So the space of solutions to the original problem has dimension 0, 1, or 2. But you can find two solutions, namely cos(i * t) and sin(i * t). Those solutions are linearly independent, so they span 2 dimensions. So all solutions are going to be of the form A * cos(i * t) + B * sin(i * t). The case for x''=ax is similar.
- blt 6y ago
  I agree with the main idea of your argument, but how can it be set in the "space of all differentiable linear functions" when trigonometric functions aren't linear? I think the argument would hold on the space of real analytic functions. Maybe that is more restrictive than necessary.
  - kitty_kritter 6y ago
    Ah yes that was a slip up. You're right, it should read the "space of all differentiable functions," or maybe "all complex differentiable functions" for extra precision.

Fronzie 6y ago

There is an exact solution for the pendulum problem: https://www.researchgate.net/publication/39575612_Exact_solution_for_the_nonlinear_pendulum https://www.researchgate.net/publication/39575612_Exact_solu... I haven't studied it well enough to say more than that about it.

ojnabieoot 6y ago
Thanks - I wasn’t actually aware of this but I meant to write “elementary” rather than “exact” so I’m glad you brought this up. I had supposed there was probably some incantation of special functions that solved the equation :) It’s been a long time since I’ve studied ODEs and mathematical physics but this solution seems easier than I had guessed (in the sense that it’s accessible to advanced undergrads instead of specialists).
leephillips 6y ago
Weird paper. Eq. 32 is supposed to be an approximation to eq. 31, but they are the same equation.

kitty_kritter 6y ago

Nope, that's pretty much all there is. The space of all differentiable linear functions is a vector space, and the derivative is a linear operator on that space (so functions and derivatives follow most of the same rules as vectors and matrices). If D is the derivative operator and x is a function of t, the equation x''=x can be written as D^2 * x = x, or 0 = (D^2 - I)x = (D + I)(D - I)x. The dimension of the kernel of (D - I) is 1, and so is the dimension of the kernel of (D + I). So the space of solutions to the original problem has dimension 0, 1, or 2. But you can find two solutions, namely cos(i * t) and sin(i * t). Those solutions are linearly independent, so they span 2 dimensions. So all solutions are going to be of the form A * cos(i * t) + B * sin(i * t). The case for x''=ax is similar.

blt 6y ago
I agree with the main idea of your argument, but how can it be set in the "space of all differentiable linear functions" when trigonometric functions aren't linear? I think the argument would hold on the space of real analytic functions. Maybe that is more restrictive than necessary.
- kitty_kritter 6y ago
  Ah yes that was a slip up. You're right, it should read the "space of all differentiable functions," or maybe "all complex differentiable functions" for extra precision.