Number of Real solutions of $e^{x^2}=ex$

Question

Find a number of real solutions of the following equation. $$e^{x^2}=ex$$

Need a pure calculus approach..

Answer 1

If you take the natural log of your equation you can arrange it to read

$$x^2-1 = \ln x.$$

The two sides are easy to graph and intersect at $x=1$. Noting that the left side is always concave up and the right is always concave down shows there is a second solution (around 0.45.)

Answer 2

Maybe help you .

take $$f(x)=e^{x^2}-ex$$so $$f'(x)=2xe^{x^2}-e=0\\x=0.761\\$$

one of roots is $x=1 $ but smaller root is between $$(0,0.761)$$ . You can find it numerically .

Answer 3

$$(e^{x^2})''=2(2x^2+1)e^{x^2}>0,$$ which says that $f(x)=e^{x^2}$ is a convex function. Thus, our equation has at most two real roots.

But $1$ is a root and it's obvious that there is a root on $(0,1)$,

which says that the answer is $2$.

Answer 4

I believe the following is a short proof sketch that the answer is 2. Clearly an integer solution is $x=1$. You can argue based on the geometry of the curves $f(x)=e^{x^{2}}$ and $g(x)=ex$ that the intersection between the two contains either zero, one, or two points.

Well you already know it is either one or two, since $x=1$ is a solution. Therefore, where the calculus comes in, is showing that $ex$ is not the tangent line to $f(x)=e^{x^{2}}$ at $x=1$. There then must be two real roots.

EDIT: Michael Rozenberg's argument above about convexity should be used to replace my handwaving about the 'shape of the functions.'

Answer 5

Disclaimer: Related but does not answer the question.

(It does however, tell you how to find the roots in terms of the Lambert W function)

Start by squaring both sides to get

$$e^{2x^2}=e^2x^2$$

Divide both sides by $-e^{2x^2-2}/2$ to get

$$-2e^{-2}=-2x^2e^{-2x^2}$$

Apply the Lambert W function to both sides to get

$$-2x^2=W_k(-2e^{-2})$$

Or,

$$x=\pm\sqrt{-\frac12W_k(-2e^{-2})}$$

Since $-e^{-1}<-2e^{-2}<0$, there are two real valued branches of the Lambert W function, so we have 4 possible solutions:

$$x=\pm\sqrt{-\frac12W_{k}(-2e^{-2})},\quad k=-1,0$$

Trivially, the solution can't be negative, since $e^{x^2}>0$, so this reduces down to

$$x=\sqrt{-\frac12W_{k}(-2e^{-2})},\quad k=-1,0$$

For $k=-1$, we find that

$$W_{-1}(-2e^{-2})=-2$$

And for $k=0$, we find that

$$W_0(-2e^{-2})=-\sum_{n=1}^\infty\frac{n^{n-1}}{n!}(2e^{-2})^n\approx-0.4064$$

Which gives respective solutions

$$x=1,0.4508$$

One may we wish to exploit different representations of our solution, such as

$$\sqrt{-\frac12W_0(-2e^{-2})}=\exp\left(-\frac12W_0(-2e^{-2})-2\right)=\frac1{2e^2}\sum_{n=0}^\infty\frac{(n+0.5)^{n-1}}{n!}(2e^{-2})^n$$

Or, perhaps if you prefer numerical iteration methods,

$$a_0=0.5,~~a_{n+1}=e^{a_n^2-1}$$

Upon which I obtain (with my unperfect calculator)

$$a_{20}=0.450763653$$