10
$\begingroup$

Find all $x\in\mathbb{R}$ such that: $$ \left( \sqrt{2-\sqrt{2} }\right)^x+\left( \sqrt{2+\sqrt{2} }\right)^x=2^x\,. $$


Immediately we notice that $x=2$ satisfies the equation.

Then we see that $LHS=a^x+b^x$, where $a<1$ and $b<2$, therefore $RHS$ grows faster (for larger $x$, $LHS\approx b^x<2^x$)

Hence $x=2$ is the only real solution.


Unfortunately I don't know whether this line of reasoning is correct. Moreover, if it is indeed correct, how to write this formally?

$\endgroup$
  • 6
    $\begingroup$ Rewrite the equation in the form $a^x+b^x=1$, where $0<a<b<1$. Note that the function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(t):=a^t+b^t\text{ for all }t\in\mathbb{R}$ is strictly decreasing with $f(2)=1$. $\endgroup$ – Batominovski Sep 1 at 11:09
  • 1
    $\begingroup$ @Batominovski thank you, I've got it now. $\endgroup$ – MartinYakuza Sep 1 at 11:15
6
$\begingroup$

It is not difficult (formula for the double angle) to show that $$\sin \left(\frac{ \pi }{8} \right)= \sqrt{ \frac{2- \sqrt{2} }{4} }$$ which in combination with the trigonometric one gives $$\cos^2\left(\frac{ \pi }{8} \right)=1-\sin^2\left(\frac{ \pi }{8} \right)=\frac{2+ \sqrt{2} }{4}\Rightarrow \cos \left( \frac{ \pi }{8} \right)= \sqrt{ \frac{2+ \sqrt{2} }{4} }$$ thus our equation can be expressed equivalently in the form $$\left( \sqrt{2-\sqrt{2} }\right)^x+\left( \sqrt{2+\sqrt{2} }\right)^x=2^x$$ $$\left( \sqrt{\frac{2- \sqrt{2} }{4}}\right)^x+\left( \sqrt{\frac{2+ \sqrt{2} }{4}}\right)^x=1 $$ $$\sin^x\left( \frac{ \pi }{8} \right)+\cos^x \left( \frac{ \pi }{8} \right)=1$$ of course thanks to the trigonometric one $x=2$ is a trivial solution. Uniqueness of this solution is due to the fact $\sin \& \cos \le 1$. Formally, you can consider cases $x>2$ or $x<2$ and estimate the left side.

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ The $\sin, \cos$ relation is a fantastic spot. $\endgroup$ – Teresa Lisbon Sep 1 at 15:39
4
$\begingroup$

Another way.

Rewrite our equation in the following form: $$\left(\sqrt{\frac{2-\sqrt2}{2+\sqrt2}}\right)^x+1=\left(\frac{2}{\sqrt{2+\sqrt{2}}}\right)^x.$$ We see the the left side decreases and the right side increases,

which says that our equation has one real root maximum.

But $2$ is a root and we are done!

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Divide your original equation by $2^x$ getting

$((\sqrt{2+\sqrt2}/2)^x+((\sqrt{2-\sqrt2}/2)^x=1.$

As $\sqrt2<2$, both terms on the left side are exponential functions with base strictly between $0$ and $1$. Therefore, the left side is strictly decreasing forcing no more than one solution. Thus any solution that might be found by inspection must be the only one.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.