3

Is there a way to solve the equation $\sin x = x\ln x$ numerically or analytically?

The only way I have been able to solve this is using a graphic calculator like Desmos, but is there another way to solve this?

Arctic Char
  • 16,007
Hector Lombard
  • 649
  • 3
  • 11
  • 1
    By analytically I'm assuming you mean a closed form solution that uses some famous set of functions? Because you can probably do it numerically but that won't give an exact closed form solution – user1001001 Sep 05 '20 at 12:54
  • If you are familiar with numerical methods, there are a few methods which can solve your problem... – Anton Vrdoljak Sep 05 '20 at 12:54
  • Yeah, sorry, I meant numerically rather than analytically. Thanks for that – Hector Lombard Sep 05 '20 at 13:04

4 Answers4

3

There is no closed form, so you need to calculate it numerically (by dichotomy for instance) to get $x\approx 1.752677...$

Nevertheless since $x\ln(x)$ is growing above $1$ quickly it is easy to show that there are only $2$ solutions, $x=0$ and another around $2$.

You can roughly estimate it by Taylor expansion of order $o(u)$ at $x=2-u$ with $u\ll 1$ and find $$u=\dfrac{2\ln(2)-\sin(2)}{1+\ln(2)-\cos(2)}\approx 1.77..$$

zwim
  • 28,563
  • 1
    $\ln$ is not defined in $0$ but $x\ln(x)$ has a limit, see https://math.stackexchange.com/q/303976/399263 – zwim Sep 05 '20 at 13:15
3

I think the simplest numerical solution for such problems is Newton's method

You would look at the function

$f(x)=\sin(x)-x\ln(x)$

and now try to find the roots. One root is obvious $x=0$.

Newton's method works as follows. You need a start value $x_0$, which you guess, and then start a recursion.

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$

To guess a starting point one could calculate some values of the function. Looking for a change in the sign.

For example $f(1)\approx 0.84$ and $f(2)\approx -0.47$

The intermediate value theorem guarantees a root in the intervall $(1,2)$, so one could take $x_0=1.5$ as the starting point.

Then

$x_1=1.5-\frac{\sin(1.5)-1.5\cdot\ln(1.5)}{\cos(1.5)-\ln(1.5)-1}\approx 1.7917$

$x_2\approx 1.7533$

$x_3\approx 1.7527$

$x_4\approx 1.7527$

It converges rather quickly.

With simple analytical arguments you can verify that these are the only two solutions. (There could be more of course)

Cornman
  • 11,065
  • 4
  • 30
  • 57
0

Hint: $\sin x = x \ln x \iff \sin x - x \ln x =0$.

So, basically on the left side of above equation you have some function, i.e. $f(x)=\sin x - x \ln x$. In other words your problem is to find a root(s) of that function. For that task we can use a few numerical methods: Newton's method, Secant method, Bisection method are most popular...

Anton Vrdoljak
  • 2,717
0

This is a transcendental equation; this means no hope for a closed form solution and numerical methods are reaquired. So, considering that you looks for the zero's of function $$f=\sin (x)- x\log(x)$$ for which $x=0$ is a trivial solution. By inspection, the solution is between $1$ and $2$. Being lazy (myself), make a Taylor expansion around $x=\frac \pi 2$; this would give $$f=\left(1-\frac{1}{2} \pi \log \left(\frac{\pi }{2}\right)\right)+\left(x-\frac{\pi }{2}\right) \left(\log \left(\frac{2}{\pi }\right)-1\right)-\frac{(2+\pi ) \left(x-\frac{\pi }{2}\right)^2}{2 \pi }+\frac{2 \left(x-\frac{\pi }{2}\right)^3}{3 \pi ^2}+\left(\frac{1}{24}-\frac{2}{3 \pi ^3}\right) \left(x-\frac{\pi }{2}\right)^4+O\left(\left(x-\frac{\pi }{2}\right)^5\right)$$ Now, using series reversion $$x=\frac{\pi }{2}-\frac{f-1+\frac{1}{2} \pi \log \left(\frac{\pi }{2}\right)}{1+\log \left(\frac{\pi }{2}\right)}-\frac{(2+\pi ) \left(f-1+\frac{1}{2} \pi \log \left(\frac{\pi }{2}\right)\right)^2}{2 \pi \left(1+\log \left(\frac{\pi }{2}\right)\right)^3 }+O\left(\left(f-1+\frac{1}{2} \pi \log \left(\frac{\pi }{2}\right)\right)^3\right)$$

Making $f=0$ leads to $x=1.74843$ while the "exact" solution (obtained by Newton method) is $1.75268$

Claude Leibovici
  • 260,315