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Is it possible to find exact solutions (in $\mathbb{R}$) of equations of the type

$$\alpha_1\sin(\beta_1 t)+\alpha_2\sin(\beta_2 t)+1=0$$ for $\alpha_i,\beta_i\in\mathbb{R}$?

In a comment to this question, Yulia V gave an explanation why there is probably none: "it is like a polynom, but more general, and there are no general solutions for polynoms for the degree greater than 4." But maybe in this particular case (two sines and one constant), somebody could infirm/confirm that intuition -- which is probably also right here -- with some even better arguments?

Community
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anderstood
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1 Answers1

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Depending on the four given constants we have a set of real roots and complex roots. It is a rotation of two vectors with different speeds with a constant matching y-projection.

However, the derivative can be solved in closed form by letting $$ \cos T = \dfrac {\alpha_1 \beta_1 }{\sqrt{(\alpha_1 \beta_1)^2 + (\alpha_2 \beta_2)^2} } $$ etc. which can be integrated. (Sorry no, my error !)

EDIT1: Amplitude and frequency modulation of signals are quite different this way.

Let $$x(t)= \alpha_1\sin(\beta_1 t)+\alpha_2\sin(\beta_2 t)+1$$

Differentiating $$x^{'}(t)= \alpha_1 \beta_1\cos(\beta_1 t)+\alpha_2 \beta_2\cos(\beta_2 t) $$

Letting $$ \cos T = \dfrac {\alpha_1 \beta_1 }{\sqrt{(\alpha_1 \beta_1)^2 + (\alpha_2 \beta_2)^2} },\, \sin T = \dfrac {\alpha_2 \beta_2 }{\sqrt{(\alpha_1 \beta_1)^2 + (\alpha_2 \beta_2)^2} } $$

Then $$x^{'}(t)= const \cdot [ \cos T \cos(\beta_1 t) + \sin T \cos(\beta_2 t) ] $$

which cannot be simplified due to three frequencies. Read under signal frequency modulation, you may still be lucky.

EDIT2: Second order differentiation

$$ - x^{''}(t)= \alpha_1 {\beta_1}^2 \sin(\beta_1 t)+\alpha_2 {\beta_2}^2\sin(\beta_2 t) $$

letting each signal expressible in terms of derivatives.

Narasimham
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  • I'm sorry but I'm missing the idea. If I understand correctly, you're considering $\alpha_1e^{i\beta_1 t}+\alpha_2e^{i\beta_2 t}+i=0$ and its imaginary part (the $y$-projection). But I don't get what you mean afterwards. Could you give a few more steps? – anderstood Dec 24 '14 at 00:45
  • My earlier work with $ q \sin p t + p \sin q t = 0 $ lead to a harmonic first derivative, but roots of $ x(t) $ are not closed form solvable. – Narasimham Dec 24 '14 at 08:11
  • So just to make sure, the details you give are developments to somehow illustrate that $x(t)$ has no closed-form roots, right? I don't see how to relate the expressions of $x(t)$ and $-x''(t)$. – anderstood Dec 24 '14 at 16:37
  • Yes, except in special cases. The time graph is a never-ending fluctuation within bounds of $ \pm ( \alpha_1+ \alpha_2)$ ; Also, $ \sin( \beta_1 t), \sin( \beta_2 t) $ each can be solved in terms of $ x(t), x^{''}(t) $ – Narasimham Dec 24 '14 at 17:27