I was wondering how I might go about finding the roots of this equation? Can it only be done using root approximation methods like newton's method?
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There are a lot of solutions – Henry Oct 05 '16 at 15:29
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@Henry: Can you tell anyone solution please...... – Sahil Kumar Oct 05 '16 at 16:04
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@SahilKumar One solution is about $15.643997$. Another is about $100.540911$ and there are many more – Henry Oct 05 '16 at 16:11
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@Henry: No, actually i am asking for the analytical solution procedure ( process of getting answer, analytically ) as it was asked in the question – Sahil Kumar Oct 05 '16 at 16:18
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@SahilKumar: there may be a series. For example for large integer $n$ you could try $n \pi + \dfrac{(-1)^n}{n \pi} - O\left(\dfrac 1{n^2}\right) $ – Henry Oct 05 '16 at 16:33
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@Henry. Very good idea, indeed. Just for fun, I played with it (have a look to my answer ... if you have time to waste). Cheers. – Claude Leibovici Oct 06 '16 at 09:25
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This is a transcendental equation which does not show any closed form solution and, then, numerical methods are required.
Considering the problem of the first and second positive roots, I cannot resist the pleasure to show once more the beautiful approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ whcih was proposed, more than 1400 years ago, by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (see here).
If we use it, the problem reduces to $$16 x^3-4(4 \pi -1) x^2-4 \pi x+5 \pi ^2=0$$ Using Cardano's method, it is easy to show that there are three real roots and that one of them is negative. For sure, the analytical expressions are quite ugly but perfectly workable. The numerical values are $$x_1=1.11495\qquad x_2=2.77389$$ while the exact solutions, obtained using Newton method, are$$x_1=1.11416\qquad x_2=2.77260$$
As Henry commented, the next roots appear to be closer and closer to $n\pi$. As he proposed to do it, use Taylor expansions around $x=n\pi$. This will give $$\sin (x)-\frac{1}{x}=-\frac{1}{\pi n}+(x-\pi n) \left(\frac{1}{\pi ^2 n^2}+(-1)^n\right)+O\left((x-\pi n)^2\right)$$ from which $$x_n\approx\frac{\pi n \left(\pi ^2 (-1)^n n^2+2\right)}{\pi ^2 (-1)^n n^2+1}\tag 1$$ Just for illustration purposes, I wrote down the values generated using $(1)$ and the exact solutions obtained using Newton method. $$\left( \begin{array}{cc} n & \text{from } (1) & \text{Newton} \\ 1 & 2.78740 & 2.77260\\ 2 & 6.43841 & 6.43912\\ 3 & 9.31747 & 9.31724\\ 4 & 12.6454 & 12.6455\\ 5 & 15.6440 & 15.6440\\ 6 & 18.9025 & 18.9025\\ 7 & 21.9456 & 21.9456\\ 8 & 25.1725 & 25.1725\\ 9 & 28.2389 & 28.2389 \end{array} \right)$$
Just for the fun, using $(1)$ $$x_{1000}=3141.5929718996\color{red}{472} $$ while Newton would lead to $$x_{1000}=3141.5929718996\color{red}{525}$$
Claude Leibovici
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The answer is no. You need some sort of numerical method (fixed point iteration, perhaps).
Sean Roberson
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3Would you mind elaborating on how you know the answer is no? I believe you, but there are some transcendental equations we can solve, like $\sin(x) = x$. – Jason DeVito - on hiatus Oct 05 '16 at 15:36
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I'll be honest, I really don't quite know why. I'm a grad student, yes, but I'm really still learning this stuff. I know this is one equation that can't quite be solved with pen and paper. – Sean Roberson Oct 05 '16 at 15:38
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1"I know this is one equation that can't quite be solved with pen and paper." But that's exactly it. How can you "know" this equation can't be solved analytically if you don't know of a proof? In mathematics there is no knowledge without proofs. – Oct 06 '16 at 09:28
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@Yves: Perhaps I'm using the word "solve" too loosely: I mean that the full (real) solution set is expressible in terms of "nice" values of elementary functions. (Of course, it's just ${0}$). So, what I guess I am really saying is two things. One, it is easy to verify that $x = 0$ is a solution, and two, it's not too hard to show that it's the only real solution. For example, one can argue that $\sin(x) < x$ for positive $x$ by noting that $\sin(x)/x$ measures a ratio of line segment length to arc length in a unit circle, so is automatically smaller than $1$ when $x$ is positive. – Jason DeVito - on hiatus Oct 06 '16 at 20:23
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@Yves: (continued). On the other hand, I would be surprised to discover solutions to the original equation $\sin(x) - \frac{1}{x} = 0$ can be expressed in terms of "nice" values of elementary functions. – Jason DeVito - on hiatus Oct 06 '16 at 20:27
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@JasonDeVito: In that sense I agree with you. I was also thinking of complex solutions, for which I don't think there are closed formulas. – Oct 06 '16 at 21:20