What is the solution to this equation: $e^x+2x=0$

Question

I cant find any program that actually solves this type of equations and I cant find anything helpful about this type. What is the name of these equations and how do I solve this one? Thanks.

Answer 1

$$e^x=-2x$$ $$1=-2xe^{-x}$$ $$-xe^{-x}=\frac12$$ $$-x=W_k\left(\frac12\right)$$ $$\therefore x=-W_k\left(\frac12\right)$$ Where $W_k(z)$ is the $k$th branch of the Lambert-W function. The only real solution is $$x=-W_0\left(\frac12\right)=-0.35173371124919582602490930092995106517146421551711180404\dots$$ but there are infinitely many complex solutions.

Answer 2

Such an equation does not have an easy formula or a "closed form" like quadratic formula for example. If you only care about the answer, you can try to plot it as in Plot using Desmos .It turns out that it has a real root at approximately: $x=-0.352$.

Alternatively you could try a "numerical method" such as Newtons's iterative method. Such methods start at a guess and keep on improving the assumption by applying a formula. See for example: Newton's Method.

Answer 3

This is a transcendental equation. As said by others, it has no closed-form expression using the standard functions.

The derivative of the function is

$$(f(x))'=(e^x+2x)'=e^x+2.$$ As it is positive, the function is increasing and has at most one root.

Now we can search for a change of sign, by trial an error.

We first observe $$f(0)=1$$ and the next value to try must be negative. Then

$$f(-1)=e^{-1}-2<0$$ and we know that the root is in $(-1,0)$.

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From this, we draw a first approximation, by the secant method, giving

$$x\approx\frac{e}{1-3e}=-0.379921\cdots$$

Now by one iteration of Newton,

$$x\approx -0.379921-\frac{e^{-0.379921}+2(-0.379921)}{e^{-0.379921}+2}\approx-0.351631$$

and so on.

Answer 4

Newton's method is a good numerical method for solving this transcendental equation.

Graphing shows that $x_0\approx-\frac{1}{3}$ is a good initial approximation.

Newton's method provides a sequence of better approximations provided the initial guess is sufficiently close:

\begin{eqnarray} x_{n+1}&=&x_n-\frac{f(x_n)}{f^\prime(x_n)}\\ &=&x_n-\frac{e^{x_n}+2x_n}{e^{x_n}+2} \end{eqnarray}

Using $x_0=-\dfrac{1}{3}$ we get the sequence

\begin{eqnarray} x_1&=&-0.3516893316\\ x_2&=&-0.3517337110\\ x_3&=&-0.3517337112\\ x_4&=&-0.3517337112 \end{eqnarray}

So it converges rather quickly using this method.

Answer 5

To have an estimate, expand as a Taylor series $$e^x+2x=1+3 x+\frac{x^2}{2}+O\left(x^3\right)$$ Solve the quadratic to get $$x_0=\sqrt{7}-3\approx -0.354249$$ More "funny" would be the $[1,n]$ Padé approximant. Depending on $n$, you would get $$\left( \begin{array}{ccc} n & x_0^{(n)} = & x_0^{(n)} \approx \\ 0 & -\frac{1}{3} & -0.33333333 \\ 1 & -\frac{6}{17} & -0.35294118 \\ 2 & -\frac{51}{145} & -0.35172414 \\ 3 & -\frac{580}{1649} & -0.35172832 \\ 4 & -\frac{8245}{23441} & -0.35173414 \\ 5 & -\frac{140646}{399865} & -0.35173371 \\ 6 & -\frac{2799055}{7957881} & -0.35173371 \end{array} \right)$$