How to solve $x= \sin(k- x)$?

Question

Is there a way to solve $x = \sin (k-x)$ without a computer, that is with a pocket calculator or pencil and paper?

Answer 1

Equations which mix polynomial and trigonometric terms do not show analytical solutions (this is already the case of $x=\cos(x)$) and numerical method (such as Newton) should be used.

Let us consider the case of the zero of $$f(x)=x-\sin(k-x)$$ admitting $0 <k <\pi$. We can notice that $f(0)=-\sin(k) <0$ and $f(k)=k>0$. On the other side, since $\sin(\theta)<1$, we can restrict the range between $0$ and $1$.

So, let us use Newton method starting at $x_0=1$. The iterative scheme would be $$x_{n+1}=x_n-\frac{x_n-\sin (k-x_n)}{1+\cos (k-x_n)}$$ Let us use $k=3$ as in your comment; then the successive iterates will be $$x_1=0.844648$$ $$x_2=0.820794$$ $$x_3=0.820243$$ which is the solution for six significant figures.

Obviously, you can easily do it using a simple pocket calculator.

Answer 2

As Oliver and Subhadeep Dey have suggested, drawing the graphs of $y=x$ and $y=\sin(k-x)$ and finding out their points of intersection will yield the required solutions.

However you could go for the taylor series expansion of $\sin x$ but the resulting equation would be a bit less precise (as per your demands) depending on the approximation you take.

Again, I can think of one method involving pocket calculator.Since $-1\le \sin x\le 1$,so $-1\le x\le 1$.
See, you can locate the roots by considering the function $y=x-\sin(k-x)$
So, $y_1 = f'(x) = 1 + \cos (k-x) \ge 0 \,\ \forall \,\ x\in R$
So $y=f(x)$ is an increasing function and hence it will have only one root.
$f(-1)=-[1+\sin (k+1)]\le 0$ and $f(0)=-\sin k$ and $f(1)=1-\sin(k-1) \ge 0 \,\ \text{and} \,\ \in [0,2]$
So depending on your value of $k$, you can get an estimate of which interval your root lies and almost the exact value of the root using your calculator.

Answer 3

If you need to solve the equation for several (many) values of $k$, then it can be attractive to work in reverse: you can express $k$ as a function of $x$ with

$$k=x+\arcsin(x)$$

and easily compute $(x,k)$ pairs.

When switching the values, you get a smooth monotonous curve that can be well approximated by a polynomial. (There is another branch, omitted here.) By symmetry, it is enough to consider the half with $k>0$.

Plot of $x$ as a function of $k$:

You will tabulate a number of points $(k_i,x_i)$ once for all (say $10$ of them). Then when you need to solve for a certain $k$, look for the bracketing interval $(k_i,k_{i+1})$ and use the Lagrangian interpolation with, say, the four points $i-1,i,i+1,i+2$. For efficient implementation, use the Neville computation scheme.

A more thorough analysis is required to determine the density of the points required to achieve the desired accuracy, based on the remainder formula of Lagrange. Different combinations are possible, from a single higher order polynomial to several lower order local ones, and different placement of the points.

You can even precompute the polynomial coefficients for all intervals, so that given a $k$ you find $x$ with, say, $5$ multiplies and $3$ additions, which is possible, though tedious, by hand.

For the sake of illustration, here is a plot of the third degree polynomial interpolating the whole positive range from points $x=0,x=0.5,x=0.75,x=1$.

Cubic interpolator $x = -0.0318k^3 + 0.0501k^2 + 0.4705k$:

Answer 4

if you need to use the calculator, the following method(Iteration method) very easy.

if we select $k=3$ $$\sin(3-x)\rightarrow x$$ the arrow means the cell store in calculator.

the following result when the initial $x=1$ $$0.90929742$$ $$0.86786576$$ $$0.84654367$$ and then continue

Answer 5

In case you couldn't find a calculator, or do computations but got a thread and ruler/compass/...

• Draw a circle of radius 1, and measure an arc of $k$ degrees for ACB.

• Take a portion of a long thread (Orange in figure) that is marked. and measure some length $2x$, and bisect it to $CD$ and $DE$.

• Place the first half $DC$ along the arc.

• If the endpoint $E$ of the other half touches $E'$ (so that $DE$ is perpendicular to AB) we are done. Otherwise, increment length equally at both ends $C, E$.

*We are approaching point $E'$ as the length $x$ increases from both ends, thus $D$, moving along the circle, is kept invariant as the bisection point.

Answer 6

Let $f(x) = x - \sin(k-x)$. It is trivial that there is no zero when $|x| > 1$. We also know that $f(0) = -\sin(k) \ne 0$.

The problem with the Newton's method is that it will fail when $f'(x) = 0$.

Let $x-k=y$, $g(y) = f(y+k) = y + k + sin(y)$. Let's also constrain $k \in [-pi/2, pi/2]$.

Using the Taylor's series for $sin(y)$, $g(y) \approx y + k + y - y^3/6$.

Thus: $$ g(y) = 0 \\ k + 2y - y^3/6 = 0 \\ y^3/6 - 2y - k = 0 \\ y^3 - 12y - 6k = 0 $$

I used the WolframAlpha to help solving this polinomial.

All three solutions can be seen here

You can notice that only one solution is between $[-1, 1]$. Take a look at this solution.

Let $w = 3k + \sqrt{-64+9 k^2} = 3k + i\sqrt{64-9 k^2}$. But, as $|w| = 8$, it can be rewritten as $w = 8 cis(\alpha)$, $\alpha = arg(w) = acos(3k/8)$.

Let $z = \sqrt[3]{w} = 2cis(\alpha/3)$.

Thus, our root is $y = \frac{2i(\sqrt{3}+i)}{z} - \frac{(1+i\sqrt{3})z}{2}$. Using the polar form for complex numbers, $y = 2cis\left(\frac{2pi}{3}-\frac{\alpha}{3}\right) - 2cis\left( \frac{\pi}{3} + \frac{\alpha}{3} \right)$.

But we already know that $y$ is a real number. So, let's get only the real part.

$$ y = 2cis\left(\frac{2pi}{3}-\frac{\alpha}{3}\right) - 2cis\left( \frac{\pi}{3} + \frac{\alpha}{3} \right) \\ y = 2\cos\left(\frac{2pi}{3}-\frac{\alpha}{3}\right) - 2\cos\left( \frac{\pi}{3} + \frac{\alpha}{3} \right) \\ y = -4 \sin\left( \frac{\pi}{6} - \frac{\alpha}{3} \right) \\ y = -4 \sin\left( \frac{\pi}{6} - \frac{cos^{-1}(3k/8)}{3} \right) $$

Finally, $x = k -4 \sin\left( \frac{\pi}{6} - \frac{cos^{-1}(3k/8)}{3} \right)$, if $k \in [-\pi/2, \pi/2]$.

Let's check: $$ k=\pi/4, x = 0.38744, f(x)=-0.00009688 \\ k=\pi/3, x = 0.51073, f(x)=-0.00037297 $$

Notice that any value of $k$ can be mapped in the interval $[-\pi, \pi]$.

We still have to solve for $k \in [-pi, -pi/2)$ and $k \in (pi/2, \pi]$. But, notice that, for k in these intervals, we can just map $k$ to the interval $[-\pi/2, \pi/2]$ and invert the sign of $sin(x-k)$.