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Show that there is exactly one value of x which satisfies the equation $$2\cos^2 (x^3+x)=2^x+2^{-x} $$

I solved this using Taylor's series:

$$2^x+2^{-x}=2\{1+\frac {x^2 \{\ln2\}^2}{2!}+\frac {x^4 \{\ln2\}^4}{4!}....\} $$

$$\cos^2 (x^3+x)=1+\cos (2x^3+2x)=2-\frac{{2x^3+2x}^2}{2!}.... $$

Equating the two gives $$ x^2\{{\ln2}^2+.....\}$$ $$\implies x=0 $$

But I can't seem to be able to find an argument justifying the existence if only one root. If I said that as the other term is an infinite series hence it's roots are undefined, would that be correct?

endolith
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  • The root $x=0$ can be spotted without series. And the calculation does not even produce the series for the cosine term. And even if we had the two series, finding whether $\text{Series} 1=\text{Series} 2$ has a solution seems like a very difficult problem. – André Nicolas Apr 13 '14 at 15:11

1 Answers1

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As for real $x, 2^{\pm x}>0$

$$\frac{2^x+2^{-x}}2\ge \sqrt{2^x\cdot 2^{-x}}=1\ \ \ \ (1)$$

So, we have $$2\cos^2(x^3+x)=2^x+2^{-x}\ge2$$

$$\iff \sin^2(x^3+x)\le0$$

For real $x,\displaystyle\sin^2(x^3+x)\ge0\implies$ the relation will hold iff

$\displaystyle2^x+2^{-x}=2$ and $\displaystyle\sin^2(x^3+x)=0\iff\sin(x^3+x)=0$

But, the equality in $(1)$ will hold iff $\displaystyle2^x=2^{-x}\iff 2^{2x}=1$

Can you take it home from here?

Another way:

If $\displaystyle a+a^{-1}=2\cos\theta, a^2-2a\cos\theta+1=0$

$\displaystyle\implies a=\cos\theta\pm i\sin\theta$

If $a$ is real, $\sin\theta$ must be $\displaystyle0\implies a=\cos\theta=\pm1$

Again, if $\displaystyle a>0, a=1\implies \theta=2n\pi$ where $n$ is any integer

lab bhattacharjee
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  • very nice and easy looking answer! – user88595 Apr 13 '14 at 13:21
  • Oh no, I was able to solve the problem (coincidently, exactly the same solution!) But I was interested in justifying the particular solution that I gave. –  Apr 13 '14 at 13:22