How to solve $\tan x=x^3$

Question

I was playing with functions I just learned, like sin, cos and tan, then I saw that the graph of $ y = \tan x$ and $y = x^3$ are pretty similar, that's how I thought of the equation $\tan x = x^3$. No matter how hard I try, I can't find a way to even start this problem. (Please consider the fact that I am a highschool student, if it isn't possible to solve this problem with my current knowledge please let me know.)

Answer 1

In order to remove the discontinuities, consider instead that you look for the zeros of function $$f(x)=\sin(x)-x^3\cos(x)$$ They will be closer and closer to $(2n+1)\frac \pi 2$ because of the cosine.

Using one single iteration of Newton method $$x_0=(2n+1)\frac \pi 2 \quad \implies \quad x_1=(2n+1)\frac \pi 2 -\frac 8 {\left((2n+1)\pi\right)^3}$$

Just trying

$$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution} \\ 1 & 4.702832970 & 4.702774539 \\ 2 & 7.851917536 & 7.851915909 \\ 3 & 10.99482207 & 10.99482191 \\ 4 & 14.13681301 & 14.13681299 \\ 5 & 17.27856575 & 17.27856574 \\ 6 & 20.42023481 & 20.42023481 \\ \end{array} \right)$$

and we could do much better using a single iteration of Halley or Householder method.

For example, with Halley method, for $n=1$ we would obtain $$x=\frac{3 \pi \left(32-576 \pi ^2+729 \pi ^6\right)}{2 \left(32-432 \pi ^2+729 \pi ^6\right)}=4.702774922$$

Edit

Let $k=(2n+1)\frac \pi 2$; the first iterate of Householder method is $$x=k-\frac{3 \left(2 k^6-6 k^2+1\right)}{k \left(6 k^8-36 k^4+5 k^2+18\right)}$$

For $n=1$, the estimate if off by $1.70\times 10^{-9}$.

The next method (no name) gives $$x=k-\frac{4 k \left(6 k^8-36 k^4+5 k^2+18\right)}{24 k^{12}-216 k^8+28 k^6+360 k^4-60 k^2-19}$$

For $n=1$, the estimate if off by $7.63\times 10^{-12}$.

And we could continue forever, having each time a closed form expression.

Answer 2

You are considering your functions over the real numbers.

You already know from high school that the tangent function ($\tan$) is periodic and the cubic power function ($\text{^}^3$) is not periodic.

As already stated in a comment, both functions seem to be quite the same in the interval $(−\frac{\pi}{2},\frac{\pi}{2})$ at first glance.

Let's look at the graphs of your functions over the real numbers (e.g. on Wolfram Alpha):
$\tan(x)$
$x^3$
$\tan(x)\text{ and }x^3$
$\tan(x)-x^3$

You already know from high school that $x=0$ is a solution of your equation $\tan(x)=x^3$.

Equations can be solved i. a. by numeric approximations, series approximations or algebraic manipulations.

Let's look if we can solve the equation by algebraic manipulations, or in closed form.

1.) The two functions cannot be identical over an interval.

An analytic function is a function that is infinitely differentiable (means arbitrarily often differentiable everywhere in its domain). The domain of $\tan$ and of $\text{^}^3$ is the whole real axis. Both functions are analytic.

Because both functions are analytic, there is, according to the Identity theorem for analytic functions, no interval where $\tan(x)=x^3$ for all $x$ in this interval.
Therefore there are only single $x$ with $\tan(x)=x^3$.

2.) No solutions in terms of elementary functions

$$\tan(x)=x^3$$

With the imaginary unit $i$ and the complex numbers, the function term of every trigonometric function can be represented as rational term of $e^{ix}$ and $e^{-ix}$.

So we have: $$\tan(x)=-i\frac{e^{ix}-e^{-ix}}{e^{ix}+e^{-ix}}$$ $$-i\frac{e^{ix}-e^{-ix}}{e^{ix}+e^{-ix}}=x^3$$ $$-i(e^{ix}-e^{-ix})=x^3(e^{ix}+e^{-ix})$$ $$-i(e^{ix}-e^{-ix})-x^3(e^{ix}+e^{-ix})=0$$ $$-ie^{ix}+ie^{-ix}-x^3e^{ix}-x^3e^{-ix}=0\ \ \ |\cdot e^{ix}$$ $$-i(e^{ix})^2+i-x^3(e^{ix})^2-x^3=0$$ $$-x^3(e^{ix})^2-i(e^{ix})^2-x^3+i=0$$ $$-x^3e^{2ix}-ie^{2ix}-x^3+i=0$$ $x\to\frac{t}{2i}$: $$-\frac{1}{8}it^3e^t-ie^t-\frac{1}{8}it^3+i=0$$ $$t^3e^t+8e^t+t^3-8=0$$

We see, this equation is an algebraic equation (means polynomial equation) with rational coefficients, that depends on $t$ and $e^t$. Such kind of equations don't have solutions except $0$ in terms of elementary functions.

3.) Solutions in terms of Generalized Lambert W

$$t^3e^t+8e^t+t^3-8=0$$ $$(t^3+8)e^t+t^3-8=0$$ $$(t^3+8)e^t=-(t^3-8)$$ $$e^t=-\frac{t^3-8}{t^3+8}$$ $$\frac{t^3+8}{t^3-8}e^t=-1$$

We see, this equation cannot be solved in terms of the Special function Lambert W, but in terms of the relatively new Special function Generalized Lambert W:

$$\frac{(t-(-2))(t-(1+i\sqrt{3}))(t-(1-i\sqrt{3}))}{(t-2)(t-(-1+i\sqrt{3}))(t-(-1-i\sqrt{3}))}e^t=-1$$ $$t=W\left(^{-2,1+i\sqrt{3},1-i\sqrt{3}}_{2,-1+i\sqrt{3}),-1-i\sqrt{3}};-1\right)$$ $$x=\frac{1}{2i}W\left(^{-2,1+i\sqrt{3},1-i\sqrt{3}}_{2,-1+i\sqrt{3}),-1-i\sqrt{3}};-1\right)$$

So we have a closed form for $x$, and the series representations of Generalized Lambert W give some hints for calculating $x$.

[Mező 2017] Mező, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553

[Mező/Baricz 2017] Mező, I.; Baricz, Á.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)

[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018

Answer 3

$\def\F{\operatorname F}$

Even though the OP is a high schooler, it would be fun to exactly solve their equation. They likely want the real roots. Let’s apply Lagrange reversion after taking the multivalued inverse tangent:

$$\tan(x)=x^3\iff x=\left(k+\frac12\right)\pi+\frac i2\ln\left(\frac{2i}{x^3-i}+1\right)\mathop\iff^{} x_k=\left(k+\frac12\right)\pi+\sum_{n=1}^\infty \left(\frac i2\right)^n \frac{d^{{n-1}}}{dx^{n-1}}\left.\frac{\ln^n\left(\frac{2i}{x^3-i}+1\right)}{n!}\right|_{\left(k+\frac12\right)\pi}$$

for $0,-1\ne k\in\Bbb Z$ defining $x_0=0,x_{-1}=x_1$. Use a Stirling S1 expansion:

$$\frac{d^{{n-1}}}{dx^{n-1}}\frac{\ln^n\left(\frac{2i}{x^3-i}+1\right)}{n!}=\sum_{m=n}^\infty \frac{S_m^{(n)}(2 i)^m}{m!}\frac{d^{n-1}}{dx^{n-1}}(x^3-i)^{-m}$$

Binomial series on $\frac{d^{n-1}}{dx^{n-1}} x^{-3m}(1-ix^{-3})^{-m}$ uses hypergeometric $_3\F_2$which seemingly has no special case. We finally use $$\sum_{n=1}^\infty\sum_{m=n}^\infty a_{m,n}=\sum_{1\le n\le m \le \infty}a_{m,n}=\sum_{m=1}^\infty\sum_{n=1}^m a_{m,n}$$

to get:

$$\bbox[4px,border: 4px solid navy]{\tan(x)=x^3\implies x_k=\left(k+\frac12\right)\pi-\sum_{m=1}^\infty\sum_{n=1}^m\frac{S_m^{(n)}(-1)^ni^{m+n}2^{m-n} \Gamma(3m+n-1)}{\left(\left(k+\frac12\right)\pi\right)^{3m+n-1}\Gamma(3m)m!}\,_3\F_2\left(m+\frac n3-\frac13,m+\frac n3,m+\frac n3+\frac13;m+\frac13,m+\frac23; \frac i{\left(\left(k+\frac12\right)\pi\right)^3}\right);x_{-1}=-x_1,x_0=0}$$

test by changing value of $k$ and compare here. These links use the hypergeometric function’s series expansion.