# How do I find if $\frac{e^x}{x^3} = 2x + 1$ has an algebraic solution?

Is there some way of solving $$\frac{e^x}{x^3} = 2x + 1$$ non-numerically? How would I go about proving if there exists a closed form solution? Similarly how would I go about proving if there exists an analytic solution?

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Sounds like Lambert W? –  user2468 Mar 22 '12 at 18:47
@AméricoTavares I think you mean $e^x - 2 x^{\color{red}{4}} - x^3 = 0.$ –  user2468 Mar 22 '12 at 18:56
@J.D. Thanks. Yes, it was a mistake. If you type e^x-2x^4-x^3=0 in WolframAlpha it gives you three roots x≈-0.84922, x≈0.963883 and x≈9.92089, none with exact form. –  Américo Tavares Mar 22 '12 at 19:01
No, Lambert W won't help for this one. It would if the $+1$ wasn't there. –  Robert Israel Mar 22 '12 at 19:37
The simple continued fraction of the root near $1$ begins $[0; 1, 26, 1, 2, 4, 1, 29, 5, 10, 42, 3, 1, 3, 96, 1, 1, 1, 4, 4,\ldots]$, i.e. $$0+\dfrac{1}{1+\dfrac{1}{26+\dfrac{1}{1+\ldots}}}$$ I doubt that there is a closed form for the elements. –  Robert Israel Mar 22 '12 at 23:09

GEdgar answers the question of proving that there's no algebraic closed form solution, i.e. a solution that is an algebraic number (a root of some finite degree polynomial with rational coefficients). However, as he mentions, if there were such a solution $a$, then $a^3(2a+1)$ would also be algebraic, and couldn't equal $e^a$ which would be transcendental by Lindemann's theorem.

The best that you can do is to solve it numerically, e.g. with Newton's method. You could use the Lagrange inversion formula (as Robert Israel does), or perhaps some variation of Lambert's W function or an infinite continued fraction expansion for each of the three real solutions.

The best general technique to find real solutions is to first locate the solutions, then use an appropriate iterative method to get the required accuracy. In this case, an excellent method that you could even do by hand if you had to would be to graph $e^x$ (in blue below) versus the polynomial $x^3(2x+1)$ (in red). Because of the contrast in scale near the origin, where there are two roots, and away from it in the first quadrant, where the third root lies, it is not possible to depict these on a linear scale (without ellipsis). So I show the important views below. The first graph, of the polynomial RHS near the origin, you should be able to sketch purely from looking at the factored form. Knowing the shape of an exponential $e^x$, and knowing that it always eventually outgrows any polynomial, you can then deduce that there must be three roots.

As @RobertIsrael pointed out, there are actually infinitely many complex numbers satisfying the equation. If we rewrite it with the customary variables $z=x+iy=re^{i\theta}$ (i.e., with $z$ in stead of $x$), the equation becomes $$e^x(\cos y+i\sin y) = e^z = w = 2z^3 \left( z+\frac12 \right) = 2z^4+z^3.$$ Now we can treat the right and left sides as two functions $w=f(z)=e^z$ and $w=g(z)=2z^4+z^3$ in the complex plane. The exponential function $f(z)$ takes each vertical line to a circle (a covering map). The imaginary axis gets mapped to the unit circle, with $0$ sent to $1$ and preserving length. Lines with fixed $x$ are first scaled by $r=e^x$ and then wrapped around the circle of radius $r$ centered at the origin. Each vertical line is mapped surjectively (and periodically) onto the corresponding circle. Each horizontal strip $y\in\left(2\pi t-\pi,2\pi t+\pi\right]$ (for each $t\in\mathbb{R}$) is consequently mapped smoothly and injectively onto $\mathbb{C}\setminus\{0\}$ by $z\rightarrow e^z$. So much for the left hand side. The right hand side is a degree $4$ polynomial map. As a function of its real and imaginary components, it is $$g(z) = 2y^4 - i(8x+1)y^3 - 3x(4x+1)y^2 + ix^2(8x+3)y + x^3(2x+1) =$$ $$\left( 2y^4 - 3x(4x+1)y^2 + x^3(2x+1) \right) + \left( - (8x+1)y^3 + x^2(8x+3)y \right)\,i$$ and as a function of its polar variables, it is $$g(z) = 2z^4+z^3 = r^3 e^{3i\theta} \left( 2 \, r e^{i\theta} + 1 \right) = 2 \, r^4 e^{4i\theta} + r^3 e^{3i\theta}$$ from which we can see that circles centered at the origin get mapped to closed curves with some fixed total curvature. Here are plots of the image of $r=1$, $r=2$, and $r=\frac12$ (identifiable by their scale), which each have rotation index $4$ and total curvature $8\pi$. In contrast, the winding number is not constant over the curve's interior because the curve intersects itself multiple times and hence is also not simple.

Of course, the exponential map $f(z)$ also maps circles to closed curves, and in particular, it maps central circles $z=re^{i\theta}$ with fixed $r$ to curves with reciprocal positive real intercepts and unit winding number about $w=1$. To investigate these images, let us note that for $r > 0$ fixed and constant, \left. \eqalign{ & x+iy=z=re^{i\theta} \\\\ & \frac{dz}{d\theta}=iz=-y+ix } \right\} \qquad \implies \qquad \eqalign{ \frac{dx}{d\theta}&=-&y=-&r\sin\theta \\\\ \frac{dy}{d\theta}&= &x= &r\cos\theta } so that for $w=f(z)$, \left. \eqalign{ & u+iv = w=e^z = \frac{dw}{dz} \\\\ & \eqalign{ \frac{dw}{d\theta} &= \frac{dw}{dz}\,\frac{dz}{d\theta} = w \cdot iz = ire^{x+i(y+\theta)} \\\\& = izw = i(x+iy)(u+iv) \\\\& = -(xv+yu)+(xu-yv)i } } \right\} \qquad \implies \qquad \eqalign{ \frac{du}{d\theta}&=-&re^x\sin(\theta+y) \\\\ \frac{dv}{d\theta}&= &re^x\cos(\theta+y) \\\\ \frac{dv}{du} &=-&\cot(\theta+r\sin\theta) } Identifiying $f$ and $g$ with real functions $F,G:\mathbb{R}^2\rightarrow\mathbb{R}^2$, \eqalign{ F(r,\theta) &=f(z) =e^{r\cos\theta} \left( \cos\left(r\sin\theta\right),~ \sin\left(r\sin\theta\right) \right) \\\\ G(r,\theta) &=g(z) =r^3 \left( 2r\cos{4\theta} +\cos{3\theta},~ 2r\sin{4\theta} +\sin{3\theta} \right) } our complex equation becomes a system of two simultaneous real equations: \eqalign{ e^{r\cos\theta} \cos\left(r\sin\theta\right) &= r^3 \left( 2r\cos{4\theta} +\cos{3\theta} \right) = r^3 \left( 2r\,\href{http://en.wikipedia.org/wiki/Chebyshev_polynomials#Examples}{T_4}\left(\cos\theta)\right) +T_3\left(\cos\theta)\right) \right) \\\\ e^{r\cos\theta} \sin\left(r\sin\theta\right) &= r^3 \left( 2r\sin{4\theta} +\sin{3\theta} \right) = r^3 \left( 2r\,\href{http://en.wikipedia.org/wiki/Chebyshev_polynomials#Examples}{U_4}\left(\cos\theta)\right) +U_3\left(\cos\theta)\right) \right) \sin\theta }

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Why would a closed form solution have to be algebraic? Perhaps there is a solution that is a rational multiple of $\pi$ (I doubt it, but I see no way of disproving it). –  Robert Israel Mar 22 '12 at 19:56
By the way, there are infinitely many solutions, of which three are real. –  Robert Israel Mar 22 '12 at 21:06
@Robert, those are both great points, thanks. I will try to update my post and do the question justice. –  bgins Mar 22 '12 at 22:16

If you write your equation as, say, $f(x)=t$ where $f(x) = e^x - 2 x^4 - x^3$, you can use the Lagrange inversion formula to get power series solutions. A convenient starting point is $f(1)=e-3$. Taking $a=1$ and $b=f(a) = e-3$, the Lagrange inversion formula says that for $t$ near $b$, one solution of $f(x)=t$ is $$g(t) = a + \frac{t-b}{f'(a)} + \sum_{n=2}^\infty \lim_{w \to a} \frac{d^{n-1}}{dw^{n-1}} \left(\frac{w-a}{f(w)-b}\right)^n \frac{(t-b)^n}{n!}$$ which in this case is $$g(t) = 1 + \frac{t-b}{e-11} + \frac{-e+30}{(e-11)^3} \frac{(t-b)^2}{2} + \frac{2 e^2 - 115 e +2106}{(e-11)^5} \frac{(t-b)^3}{6} + \frac{-6e^3+470 e^2 - 16237 e + 232608}{(e-11)^7} \frac{(t-b)^4}{24} + \ldots$$

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Lindemann's Theorem: If $a$ is nonzero and algebraic, then $e^a$ is transcendental. http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem
From this we see your equation has no algebraic solutions.

Note: your title does use the word "algebraic" ... even though your body says "analytic".

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I think I mean to say analytic here :) –  shuttle87 Mar 22 '12 at 20:58
But there is no such thing as an "analytic number". There are analytic functions... –  Robert Israel Mar 22 '12 at 21:01
I think he said "analytic solution" when he means "closed form solution"... –  GEdgar Mar 22 '12 at 21:15
Yes, I suspect so. –  Robert Israel Mar 22 '12 at 22:58