Find the number of real solutions of the equation $2^x+x^2=1$

Question

My attempt is as follows:-

$$2^x+x^2=1$$ $$\left(1+x\cdot log(2)+\frac{x^2\cdot (log(2))^2}{2!}+\frac{x^3\cdot (log(2))^3}{3!}+\dots\right)+x^2=1$$ $$x\cdot log(2)\left(1+x^2+\frac{x\cdot log(2)}{2!}+\frac{x^2\cdot (log(2))^2}{3!}\right)=0$$

$$x=0, \left(1+x^2+\frac{x\cdot log(2)}{2!}+\frac{x^2\cdot (log(2))^2}{3!}\right)=0$$

$$x^2+\frac{x\cdot log(2)}{2!}+\frac{x^2\cdot (log(2))^2}{3!}+\dots=-1$$

Now I started to think when this can be negative, it can only be when x is negative.

I also thought about $$S=x^2+\sum_{i=1}^{\infty}\left(\frac{x^i\cdot (log(2))^i}{i+1}\right)$$. But I was not finding the way to transform it into telescopic series.

I am stuck here and not able to proceed further.

Answer 1

This should help:

Clearly $x=0$ is a solution. The other can be seen from the graph. (Numerical solution: $x = -0.572195.$)

In short: exactly two real solutions.

One function is always concave up, the other always concave down, so the maximum number of solutions is two, which is what we have.

Answer 2

Wow, you really are overcomplicating things. There's no need to use the series expansion. First, just note that both $2^x$ and $x^2$ are increasing functions on $[0,\infty)$, which means that the equation $2^x+x^2=1$ can have at most one solution on this interval. Indeed it does have exactly one solution, given by $x=0$. And then note that the $2^x$ term becomes insignificant when $x\to-\infty$, so the function is decreasing on $(-\infty,a]$ for some $a<0$, and has exactly one solution here too. So the answer is $2$.

Answer 3

Consider the function $$f(x)=2^x+x^2-1.$$ Its first and second derivatives are $$f'(x)=\ln2\cdot2^x+2x$$ and $$f''(x)=(\ln2)^2\cdot2^x+2.$$

Observe:

• Second derivative is strictly positive
• First derivative has one root at $x_{min}=\dfrac1{\ln2}W\bigg(\dfrac{(\ln 2)^2}2\bigg)\approx-0.28454$
• $f(x_{min})\approx-0.098034<0$

Therefore, $f(x)$ has two zeroes, and there are two real solutions to your equation.

Answer 4

Too long for a comment.

Starting from Andrew Chin's analysis, we could obtain a quite good estimate of the solution.

As said, the first derivative cancels at $$x_*=-\frac{1}{\log (2)}W\left(\frac{\log ^2(2)}{2}\right)$$ Now, develop the function as a Taylor series around this point $$f(x)=2^x+x^2-1=f(x_*)+\frac 12 f''(x_*)(x-x_*)^2+ O\big((x-x_*)^3\big)$$ Ignoring the higher order terms, the approximation is then $$x_\pm=x_* \pm \sqrt{-2\frac {f(x_*)}{f''(x_*)}}$$ Computing, this would give as estimates $x_-=-0.5707$ and $x_+=0.0016$ while the exact solutions are $x_-=-0.5722$ and $x_+=0$.

Answer 5

Let $x=\log_2 t$ and recast $2^x+x^2=1$ as,

$$\log_2 t +\sqrt{1-t}=0$$

Apply a perturbation approximation around the base solution $t=\frac23$ to obtain an accurate algebraic solution for the root,

$$t_r=\frac{\ln16+\sqrt3 (\ln\frac49 -2)}{\ln8-3\sqrt3}$$

The root for $x_r=\log_2 t_r$ is

$$x_r=-0.57233\> (-0.57220)$$

with the exact value in parenthesis for comparison.