Solve: $2^{\cos^{2014}x} - 2^{\sin^{2014} x} = \cos^{2013} (2x)$

Solution 1:

Initially, i tried to give a solution using only mathematical results from Romanian X class level. (Tenth class.) There was an error in the proof, see the comment of Saad, so i will (have to) use more.

Let $t=\cos 2x\in[-1,1]$ as in the comment of Troposphere. Then $\cos^2 x=\frac {1+t}2$, $\sin^2 x=\frac {1-t}2$, so with $N=2014/2=1007$ (odd) we have to study the function: $$ f:[-1,1]\to\Bbb R\ ,\qquad f(t)= 2^{\textstyle\left(\frac {1+t}2\right)^N} - 2^{\textstyle\left(\frac {1-t}2\right)^N} - t^{2N-1}\ . $$ It is the sum of two odd functions, one of them being $t\to -t^{2N-1}$, so it is an odd function. Let us show that the only points $t$ with $f(t)=0$ are the points $0$ and $\pm 1$. It is enough to study $f$ only on $[0,1]$.

From now on, $t\ge 0$.

LATER EDIT: As pointed out by Saad in the comments, there was an error in the initial solution, so here is an attempt to fix it, still using the already typed argument, which works only on a part of the interval $[0,1]$.

We will try to use inequalities to show $f\ge 0$ on $[0,1]$, and also control the points where equality may hold. First of all, consider two inequalities. (To get a control of $2^a-2^b$ in terms of $(a-b)$ we cannot use in our case for $a\ge b\ge 0$ Bernoulli's_inequality, this was the error, so a correction is needed.) For the helper function $h(x)=2^x=e^{x\ln 2}$ we find (Lagrange or Taylor) an intermediate point $c$ with $2^a-2^b=h(a)-h(b)=h'(c)(b-a)=2^c\ln 2(b-a)$, so $$ 2^a - 2^b \ge 2^b(a-b)\ge (a-b)\ . $$ We plug in for $a,b$ the values $\left(\frac {1\pm t}2\right)^N$, so let us also look closer at $a-b$, here we are doing stuff corresponding to the 10. class. Binomial expansion gives: $$ \begin{aligned} \left(\frac {1+t}2\right)^N - \left(\frac {1-t}2\right)^N & =\frac 1{2^N}\Big( \ (1+t)^N-(1-t)^N\ \Big) \\ & =\frac 1{2^N}\cdot 2\left( \ \binom N1 t +\binom N3 t^3 +\dots +\binom NN t^N\ \right) \\ & \ge\frac 1{2^N}\cdot 2\left( \ \binom N1 +\binom N3 +\dots +\binom NN\ \right)t^N \\ &=\frac 1{2^N}\Big( \ (1+1)^N-(1-1)^N\ \Big)\; t^N \\ &=t^N\ . \end{aligned} $$ The equality in the used inequality $t\ge t^N$ holds only in the points $t=0,1$, so this is necessary. (And sufficient at this last point.)

Putting all together, for $t\in[0,1]$ $$ \begin{aligned} f(t) &\ge \ln 2\left[ \left(\frac {1+t}2\right)^N - \left(\frac {1-t}2\right)^N\right] -t^{2N-1} \\ &\ge t^N\ln 2-t^{2N-1}=t^N(\ln 2-t^{N-1}) \\ &\ge 0\qquad\text { for $t$ in the interval $J=\left[0,\ (\log 2)^{1/(N-1)}\right]=[0,\ 0.9996357394\dots]$} \end{aligned} $$ and we cannot go with this idea to also cover the $1$, since in $1$ and in a "small" neighbourhood we a expect negative sign.

To see with bare hands that $0.9995$ works for the estimation above as upper bound, we have

$t^{N-1}\le t^{1000}\le\left(1-\frac 1{2000}\right)^{2000/2}\le \sqrt \lim\left(1-\frac 1n\right)^n=\frac 1{\sqrt e} %<\frac 1{\sqrt{2.56}} <\frac 1{1.6}=\frac 1{8/5}=\frac 58=0.625<\ln 2$.

One can adjust it for $0.9996$ by using instead $\left(1-\frac 1{2500}\right)^{2500/2.5}\le \left(\lim\left(1-\frac 1n\right)^n\right)^{1/2.5}=\frac 1{e^{2/5}}=\exp\left(-\frac 25\right) \approx 0.670320\dots < \ln 2$.

We will use the upper bound $0.9996$ below.

We need some argument to also "cover" the point $1$ and the neighbourhood $[0.9996,\ 1]$. For this, let us write $t=1-2h$, where $h$ is between $0$ and $$H=0.0002=\frac 1{5000}\ $$ Then using Lagrange and intermediate points $c_1,c_2,\dots$ in $[0,h]\subseteq[0,H]$ we have $$ \begin{aligned} f(t) &=2^{\textstyle\left(\frac {1+t}2\right)^N} - 2^{\textstyle\left(\frac {1-t}2\right)^N} - (1-2h)^{2N-1} \\ &=2^{(1-h)^N} - 2^{h^N} - (1-2h)^{2N-1} \\ &= \left[1-(1-2h)^{2N-1}\right] -\left[ 2^1 - 2^{(1-h)^N}\right] -\left[ 2^{h^N} - 2^0\right] \\ &= \left[1-(1-2h)^{2N-1}\right] -2^{(1-c_1)^N}\ln 2\cdot \left[1-(1-h)^N\right] -2^{c_2^N}\ln 2 \cdot h^N \\ &\ge \underbrace{ \left[1-(1-2h)^{2N-1}\right] -2\ln 2\left[1-(1-h)^N\ +\ h^N\right] }_{:=g(h)} \\ &=:g(h)\ , \\[4mm] &\qquad\text{ and $g$ is increasing on $[0,H]$, because} \\[4mm] g'(h) &=2(2N-1)(1-2h)^{2(N-1)} -2N\ln 2\cdot (1-h)^{N-1} - 2N\ln 2\cdot h^{N-1} \\ &\ge 4001 (1-2h)^{2(N-1)} - 1400 (1-h)^{N-1} - 1400 h^{N-1} \\ &= 1400 (1-2h)^{2(N-1)} \underbrace{ \left[\frac{4000}{1400} -\left(\frac{1-h}{(1-2h)^2}\right)^{N-1}\right]}_{=B_1} \\ &\qquad+ 1400(1-2h)^{2(N-1)} \underbrace{ \left[\frac{1}{1400} -\left(\frac h{(1-2h)^2}\right)^{N-1}\right]}_{=B_2} \\ &\ge 0\ , \\[4mm] &\qquad\text{with the following justification...} \end{aligned} $$ The second bracket $B_2$ is positive, since $h/(1-2h)^2\le H/(1-2H)^2<\frac 1{10}$, and we raise this to the power $(N-1)$. Let us consider the first one, $B_1$. The function $h\to \frac{1-h}{(1-2h)^2}$ is increasing on $[0,H]$, so it is enough to estimate $B_2$ in $H$. This is easily done with a computer, which gives $\left(\frac {1-H}{(1-2H)^2}\right)^{N-1}\approx 1.00060032\dots ^{1006} \approx 1.8289\dots < 2.857142\dots =\frac{4000}{1400}$, so $B_2(h)\ge B_2(H)>0$.

(With bare hands i would maybe look at $1.000625=\frac 1{1600}$, further raised to the power $1600$ and/or $1601$, and $e$ is between these powers... Then some small powers of $1.000625$ like $1.000625^6$ will have a modest contribution, it is easy to estimate it.)

Putting all together, $g'\ge 0$ on $[0,H]$, so $f(t)=g(h)\ge g(0)=0$, showing the inequality $f(t)\ge 0$ on the full interval $[0,1]$.

So $f$ is zero in and only in $-1,0,1$.

(If a check is explicitly needed, $f(0)=2^{(1/2)^N}-2^{(1/2)^N}-0=0$, $f(1)=2^1-2^0-1=0$, $f(-1)=-f(1)=0$.)

We come back to the $x$-world, and solve $\cos 2x\in\{-1,0,1\}$ to obtain $2x\in\Bbb Z\cdot \frac \pi 2$, i.e. the solutions $x$ of the given equation are $$ \color{blue} {x\in\Bbb Z\cdot \frac \pi 4\ .} $$

Note: The (corrected) solution needs some analytic tools specific to the XI.th class in the Romanian school program. I will try to reduce the level, but for now i have no idea how to cover with "simple" inequalities the behaviour of $f$ near $1$...