Cosine and Something Else

Given that $k=4n$ for postive integer $n$ , we have

$a=x+\sqrt { { x }^{ 2 }-{ k }^{ 2 } } =x+i\sqrt { { k }^{ 2 }-{ x }^{ 2 } } =k\left( \dfrac { x }{ k } +i\sqrt { 1-{ \left( \dfrac { x }{ k } \right) }^{ 2 } } \right)$

$b=x-\sqrt { { x }^{ 2 }-{ k }^{ 2 } } =x-i\sqrt { { k }^{ 2 }-{ x }^{ 2 } } =k\left( \dfrac { x }{ k } -i\sqrt { 1-{ \left( \dfrac { x }{ k } \right) }^{ 2 } } \right)$

We then take advantage of this trigonometric identity

$iArcSin\left( x \right) =Log\left( ix+\sqrt { 1-{ x }^{ 2 } } \right)$

to rewrite $a,b$ as

$a=ki{ e }^{\displaystyle -i\phi }$

$b=-ki{ e }^{\displaystyle i\phi }$

where

$\phi =ArcSin\left( \dfrac { x }{ k } \right)$

Given that $ab=k^{ 2 }$ and plugging in the expressions for $a,b$ , we end up with

$\dfrac { 1 }{ 2 } \dfrac { 1 }{ { k }^{ k } } \left( { a }^{ k }+{ b }^{ k } \right) =\dfrac { 1 }{ 2 } \left( { \left( \dfrac { a }{ b } \right) }^{ \frac { 1 }{ 2 } k }+{ \left( \dfrac { b }{ a } \right) }^{ \frac { 1 }{ 2 } k } \right) =\dfrac { 1 }{ 2 } \left( { e }^{\displaystyle -ik\phi }+{ e }^{ \displaystyle ik\phi } \right)$

which, as $k\rightarrow \infty$ , becomes the exponential expression for $Cos\left( x \right)$ . Keep in mind that $k=4n$ , which eliminates pesky sign problems. Thus, as $n$ increases, it approaches the Cosine function, and from $n=4$ on, it never crosses the Cosine function more than $4$ times between $0<x<4\pi$ . For $n=3$ , it crosses the Cosine function $5$ times between $0<x<4\pi$ . Both functions are symmetric functions of $x$ .

Here's the graphic for $n=6$ , the green vertical line is where $x=4\pi$

I am able to prove that both the curves never meet and it also seems to be correct. Don't really know where I have did any mistake in the proof. Kindly go through it and comment on it....

I am new to this site and didn't get accquainted with formatting stuff.... Sorry for that

Given k= 4n

and clearly x^2 > k^2

LHS = (1/2)*((a/k)^k + (b/k)^k)

(a/k)^k * (b/k)^k = 1

and both are positive numbers as K is an even number

So by AM>=GM

we have LHS >=1

RHS = cos(x)

RHS<=1

So this condition satisfies when LHS = RHS =1

LHS = 1 => a=b=x=+/- k

so we have X = +/- K when both the curves meet

and cos(x) = 1

=> x = 2mPi or -(2m+1)Pi

Case 1 : X= 2mPi

=> X=K as K cannot be negative

=> 2mPi = K

=> Pi = K/(2m)

But we know Pi is an irrational number and cannot be expressed as m/n

Similarly for Case also

Case 2: X= -(2m+1)Pi => X=-K

=>K = (2m+1)Pi

Here also Pi condraticts the rules of irrationality

So we can say that both the curves never meet........

first check that needs to be done is to see if f(x) is real for all values of x & k. We can see that for that

$\sqrt { { x }^{ 2 }-{ k }^{ 2 } } \ge \quad 0$

Checking this condition we get

$\sqrt { { x }^{ 2 }-{ k }^{ 2 } } \ge \quad 0\quad \\ { x }^{ 2 }-{ k }^{ 2 }\ge 0\\ (x-k)(x+k)\ge 0\\ -x\le k\le x\\ x\quad varies\quad from\quad -4\pi \quad to\quad +4\pi \\ hence\quad \\ -4\pi \le k\le 4\pi \\ \\ now\quad k\quad =\quad 4n\\ so\quad -4\pi \le 4n\le 4\pi \\ \\ gives\quad -\pi \le n\le \pi \\ \\ Since\quad n\quad is\quad an\quad integer\quad <1000\quad so\quad the\quad only\quad possible\quad values\quad of\quad n\quad can\quad be\quad 1,2,3\\ \\ \\ for\quad n\quad =\quad 1\quad \\ \\ f(x)\quad =\quad (\frac { 1 }{ 2 } \frac { 1 }{ { 4 }^{ 4 } } ).((x+{ \sqrt { { x }^{ 2 }-16 } }){ ^{ 4 } }+(x-{ \sqrt { { x }^{ 2 }-16 } }){ ^{ 4 } }\\ again\quad we\quad can\quad see\quad that\quad x\quad \ge 4\quad or\quad x\quad \le -4\\ for\quad x=4\quad f(x)\quad =\quad 1\\ for\quad x>4\quad or\quad x\quad <-4\quad f(x)\quad >1\\ thus\quad f(x)\quad can\quad intersect\quad cos(x)\quad at\quad only\quad x\quad =\quad 4\quad =\quad 1.272\pi \\ cos(x)\quad lies\quad between\quad -1\quad to\quad 0\quad for\quad \pi \quad to\quad 1.5\pi .\\ \\ Hence\quad n\quad =\quad 1\quad is\quad not\quad a\quad solution\\ \\ Similarly\quad n\quad =\quad 2\quad \& \quad n\quad =\quad 3\quad is\quad checked.\\ for\quad n\quad =\quad 2\quad x\quad =\quad 2.54\pi \quad ;\quad cos(x)\quad lies\quad between\quad 0\quad \& \quad -1\quad for\quad 2.5\pi \quad to\quad 3\pi \\ \\ for\quad n\quad =\quad 3\quad x\quad =\quad 3.82\pi \quad ;\quad cos(x)\quad lies\quad between\quad 0\quad \& \quad 1\quad for\quad 3.5\pi \quad to\quad 4\pi \\ \\ Thus\quad there\quad is\quad only\quad one\quad solution\quad n\quad =\quad 3$

The solution has been found using elimination method. A detailed way wold be to plot the equations. However elimination method is quite useful to narrow down the possible solutions.