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Total no. of real solution of the equation $x^2 = 2^{-x}$

My Try:: Let $f(x) = x^2-2^{-x}$

Diff. both side w.r.to $x\;,$ we get

$f^\prime(x) = 2x+2^{-x}\ln(2)$

$f^{\prime\prime}(x)=2+2^x.\ln(2)>0 \; \forall x\in \mathbb{R}$

Here $f^{\prime\prime}(x)>0\forall x\in \mathbb{R}$

So $f(x)$ is an Concave upward function that means if function $f(x)$ has real roots then it must be $2$

But using wolframalpha We get $3$ real solution.

So were is my mistake


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As I told you on your previous question: you're supposed to accept your favorite answers, if they're satisfactory. You've accepted $0$ answers on $13$ questions so far. –  Git Gud Jan 26 '13 at 11:07
Oh Sorry Git Gud –  juantheron Jan 26 '13 at 11:21
Don't apologize to me. Go to your profile and accept answers for your previous questions. –  Git Gud Jan 26 '13 at 11:27
How can i accept answer would you like to explain me .i was thinking that votes means accepting answer. –  juantheron Jan 26 '13 at 11:39
No. When you over your mousse below the downvote arrow, a 'check' sign will appear. You should click on it to accept an answer. If you do this properly, the check sign will be visible permanently and it will turn green. –  Git Gud Jan 26 '13 at 11:40

3 Answers 3

up vote 1 down vote accepted

Another approach is using the Lambert-W function.

Since $x^2=2^{-x}\Rightarrow \log(2)x/2\,e^{\log(2)x/2}=\pm\log(2)/2$, we get that $$ x=\frac2{\log(2)}\mathrm{W}\left(\pm\frac{\log(2)}{2}\right) $$ where $\mathrm{W}$ is the Lambert-W function.

For positive arguments, $\mathrm{W}$ only has one real branch, so the only positive real root of the equation is N[2/Log[2] LambertW[0, Log[2]/2], 20]: $$ x\doteq0.76666469596212309311 $$ For negative arguments, $\mathrm{W}$ has two real branches. The negative real solutions are N[2/Log[2] LambertW[0, -Log[2]/2], 20] $$ x=-2 $$ and N[2/Log[2] LambertW[-1, -Log[2]/2], 20] $$ x=-4 $$

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Because $$ f''(x)=2-2^{-x}\ln^2 2 $$ and it is not always positive. You need differentiate one more time to see that $f'''>0$ and conclude that $f$ have at most three solutions

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Thanks Norbert. $f''(x)=2-2^{-x}\ln^2(2)$ and $f'''(x)=2^{-x}\ln^2(3)>0$ means $f'(x)$ is Concave upward function So It has at most $2$ solution.So how can i prove that $f(x)$ has at most $3$ solution. –  juantheron Jan 26 '13 at 11:25
Suppose $f$ has four roots or more. Take a set of four consecutive roots in increasing order $\{a,b,c,d\}$. What does Rolle tell you about $f(a)$ & $f(b)$? What about $f(b)$ & $f(c)$? And finally, what about $f(c)$ & $f(d)$? –  Git Gud Jan 26 '13 at 11:30

Let $f(x)=x^2-2^{-x}$, then $f''(x)=2-(\log2)^22^{-x}$. Since $f''$ is increasing from $f''(-\infty)=-\infty$ to $f''(+\infty)=+\infty$, $f'$ is decreasing on $(-\infty,x]$ and increasing on $[x,+\infty)$, for some $x$. Two cases can occur.

  • Either $f'(x)\geqslant0$, then $f'$ is nonnegative everywhere and $f$ is increasing. Since $f(-\infty)=-\infty$ and $f(+\infty)=+\infty$, this implies that $f=0$ has exactly one root.

  • Or $f'(x)\lt0$, then $f'$ is positive then negative then positive.

Since $f(-3)\gt0$ and $f(0)\lt0$, the second case occurs. Thus, the function $f$ is negative then positive then negative then positive, and it has exactly three zeroes, one in each interval $(-\infty,-3)$, $(-3,0)$ and $(0,+\infty)$. The zeroes in the first two intervals are easy-to-guess integers, and the positive zero happens to lie in the interval $(\frac12,1)$ since $f(\frac12)\lt0\lt f(1)$.

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