Mixing
How to solve $cos(2^n x) = cos(x)$
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I am given the function $F:mathbb{R}tomathbb{R}$, $xmapsto 4x(1-x)$ and I have to find all $n$-cycles of the function. I have already reduced this to the problem of solving $cos(2^n x)=cos(x)$ for $xin[0,pi]$. Using Mathematica and $nin{1,2,3,4}$, I concluded that the solutions must be ${frac{2kpi}{2^n-1},frac{2kpi}{2^n+1}}cap[0,pi]$. Since the equation obviously has $2^n$ solutions and the solution set has $2^n$ elements, it only remains to verify that they are indeed solutions to the equation. I tried using $cos(2x)=2cos^2(x)-1$ and proving this by induction on $n$, but I was not able to. Thank you for any help with this problem!
trigonometry
asked Jan 17 at 6:51
Analysis801Analysis801
1267
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add a comment |
$begingroup$
I am given the function $F:mathbb{R}tomathbb{R}$, $xmapsto 4x(1-x)$ and I have to find all $n$-cycles of the function. I have already reduced this to the problem of solving $cos(2^n x)=cos(x)$ for $xin[0,pi]$. Using Mathematica and $nin{1,2,3,4}$, I concluded that the solutions must be ${frac{2kpi}{2^n-1},frac{2kpi}{2^n+1}}cap[0,pi]$. Since the equation obviously has $2^n$ solutions and the solution set has $2^n$ elements, it only remains to verify that they are indeed solutions to the equation. I tried using $cos(2x)=2cos^2(x)-1$ and proving this by induction on $n$, but I was not able to. Thank you for any help with this problem!
trigonometry
asked Jan 17 at 6:51
Analysis801Analysis801
1267
$endgroup$
add a comment |
$begingroup$
I am given the function $F:mathbb{R}tomathbb{R}$, $xmapsto 4x(1-x)$ and I have to find all $n$-cycles of the function. I have already reduced this to the problem of solving $cos(2^n x)=cos(x)$ for $xin[0,pi]$. Using Mathematica and $nin{1,2,3,4}$, I concluded that the solutions must be ${frac{2kpi}{2^n-1},frac{2kpi}{2^n+1}}cap[0,pi]$. Since the equation obviously has $2^n$ solutions and the solution set has $2^n$ elements, it only remains to verify that they are indeed solutions to the equation. I tried using $cos(2x)=2cos^2(x)-1$ and proving this by induction on $n$, but I was not able to. Thank you for any help with this problem!
trigonometry
asked Jan 17 at 6:51
Analysis801Analysis801
1267
$endgroup$
I am given the function $F:mathbb{R}tomathbb{R}$, $xmapsto 4x(1-x)$ and I have to find all $n$-cycles of the function. I have already reduced this to the problem of solving $cos(2^n x)=cos(x)$ for $xin[0,pi]$. Using Mathematica and $nin{1,2,3,4}$, I concluded that the solutions must be ${frac{2kpi}{2^n-1},frac{2kpi}{2^n+1}}cap[0,pi]$. Since the equation obviously has $2^n$ solutions and the solution set has $2^n$ elements, it only remains to verify that they are indeed solutions to the equation. I tried using $cos(2x)=2cos^2(x)-1$ and proving this by induction on $n$, but I was not able to. Thank you for any help with this problem!
trigonometry
trigonometry
asked Jan 17 at 6:51
Analysis801Analysis801
1267
asked Jan 17 at 6:51
Analysis801Analysis801
1267
edited Jan 17 at 7:09
Analysis801
asked Jan 17 at 6:51
Analysis801Analysis801
1267
asked Jan 17 at 6:51
Analysis801Analysis801
1267
asked Jan 17 at 6:51
Analysis801Analysis801
1267
1267
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add a comment |
2 Answers
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With only basic results on trigonometric equations:
begin{align}cos(2^nx)=cos x &iff 2^nxequiv pm xmod 2piiffbegin{cases}(2^n-1)xequiv 0mod 2pi\(2^n+1)xequiv 0mod 2piend{cases}\[1ex]
&iff xequiv 0mod frac{2pi}{2^n-1},bmod frac{2pi}{2^n+1}.
end{align}
answered Jan 17 at 9:23
BernardBernard
121k740116
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Using the identity $$cos A - cos B = 2sinleft(dfrac{B-A}{2}right)sinleft(dfrac{B+A}{2}right),$$ we have $$0 = cos(x)-cos(2^nx) = 2sinleft(dfrac{2^n-1}{2}xright)sinleft(dfrac{2^n+1}{2}xright),$$ which holds iff $$sinleft(dfrac{2^n-1}{2}xright) = 0 text{OR} sinleft(dfrac{2^n+1}{2}xright) = 0$$ $$dfrac{2^n-1}{2}x = kpi text{OR} dfrac{2^n+1}{2}x = kpi$$ $$x = dfrac{2kpi}{2^n-1} text{OR} x = dfrac{2kpi}{2^n+1}$$ for some $k in mathbb{Z}$.
answered Jan 17 at 7:16
JimmyK4542JimmyK4542
41.1k245107
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2 Answers
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2 Answers
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$begingroup$
With only basic results on trigonometric equations:
begin{align}cos(2^nx)=cos x &iff 2^nxequiv pm xmod 2piiffbegin{cases}(2^n-1)xequiv 0mod 2pi\(2^n+1)xequiv 0mod 2piend{cases}\[1ex]
&iff xequiv 0mod frac{2pi}{2^n-1},bmod frac{2pi}{2^n+1}.
end{align}
answered Jan 17 at 9:23
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
With only basic results on trigonometric equations:
begin{align}cos(2^nx)=cos x &iff 2^nxequiv pm xmod 2piiffbegin{cases}(2^n-1)xequiv 0mod 2pi\(2^n+1)xequiv 0mod 2piend{cases}\[1ex]
&iff xequiv 0mod frac{2pi}{2^n-1},bmod frac{2pi}{2^n+1}.
end{align}
answered Jan 17 at 9:23
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
With only basic results on trigonometric equations:
begin{align}cos(2^nx)=cos x &iff 2^nxequiv pm xmod 2piiffbegin{cases}(2^n-1)xequiv 0mod 2pi\(2^n+1)xequiv 0mod 2piend{cases}\[1ex]
&iff xequiv 0mod frac{2pi}{2^n-1},bmod frac{2pi}{2^n+1}.
end{align}
answered Jan 17 at 9:23
BernardBernard
121k740116
$endgroup$
With only basic results on trigonometric equations:
begin{align}cos(2^nx)=cos x &iff 2^nxequiv pm xmod 2piiffbegin{cases}(2^n-1)xequiv 0mod 2pi\(2^n+1)xequiv 0mod 2piend{cases}\[1ex]
&iff xequiv 0mod frac{2pi}{2^n-1},bmod frac{2pi}{2^n+1}.
end{align}
answered Jan 17 at 9:23
BernardBernard
121k740116
answered Jan 17 at 9:23
BernardBernard
121k740116
answered Jan 17 at 9:23
BernardBernard
121k740116
answered Jan 17 at 9:23
BernardBernard
121k740116
121k740116
add a comment |
add a comment |
$begingroup$
Using the identity $$cos A - cos B = 2sinleft(dfrac{B-A}{2}right)sinleft(dfrac{B+A}{2}right),$$ we have $$0 = cos(x)-cos(2^nx) = 2sinleft(dfrac{2^n-1}{2}xright)sinleft(dfrac{2^n+1}{2}xright),$$ which holds iff $$sinleft(dfrac{2^n-1}{2}xright) = 0 text{OR} sinleft(dfrac{2^n+1}{2}xright) = 0$$ $$dfrac{2^n-1}{2}x = kpi text{OR} dfrac{2^n+1}{2}x = kpi$$ $$x = dfrac{2kpi}{2^n-1} text{OR} x = dfrac{2kpi}{2^n+1}$$ for some $k in mathbb{Z}$.
answered Jan 17 at 7:16
JimmyK4542JimmyK4542
41.1k245107
$endgroup$
add a comment |
$begingroup$
Using the identity $$cos A - cos B = 2sinleft(dfrac{B-A}{2}right)sinleft(dfrac{B+A}{2}right),$$ we have $$0 = cos(x)-cos(2^nx) = 2sinleft(dfrac{2^n-1}{2}xright)sinleft(dfrac{2^n+1}{2}xright),$$ which holds iff $$sinleft(dfrac{2^n-1}{2}xright) = 0 text{OR} sinleft(dfrac{2^n+1}{2}xright) = 0$$ $$dfrac{2^n-1}{2}x = kpi text{OR} dfrac{2^n+1}{2}x = kpi$$ $$x = dfrac{2kpi}{2^n-1} text{OR} x = dfrac{2kpi}{2^n+1}$$ for some $k in mathbb{Z}$.
answered Jan 17 at 7:16
JimmyK4542JimmyK4542
41.1k245107
$endgroup$
add a comment |
$begingroup$
Using the identity $$cos A - cos B = 2sinleft(dfrac{B-A}{2}right)sinleft(dfrac{B+A}{2}right),$$ we have $$0 = cos(x)-cos(2^nx) = 2sinleft(dfrac{2^n-1}{2}xright)sinleft(dfrac{2^n+1}{2}xright),$$ which holds iff $$sinleft(dfrac{2^n-1}{2}xright) = 0 text{OR} sinleft(dfrac{2^n+1}{2}xright) = 0$$ $$dfrac{2^n-1}{2}x = kpi text{OR} dfrac{2^n+1}{2}x = kpi$$ $$x = dfrac{2kpi}{2^n-1} text{OR} x = dfrac{2kpi}{2^n+1}$$ for some $k in mathbb{Z}$.
answered Jan 17 at 7:16
JimmyK4542JimmyK4542
41.1k245107
$endgroup$
Using the identity $$cos A - cos B = 2sinleft(dfrac{B-A}{2}right)sinleft(dfrac{B+A}{2}right),$$ we have $$0 = cos(x)-cos(2^nx) = 2sinleft(dfrac{2^n-1}{2}xright)sinleft(dfrac{2^n+1}{2}xright),$$ which holds iff $$sinleft(dfrac{2^n-1}{2}xright) = 0 text{OR} sinleft(dfrac{2^n+1}{2}xright) = 0$$ $$dfrac{2^n-1}{2}x = kpi text{OR} dfrac{2^n+1}{2}x = kpi$$ $$x = dfrac{2kpi}{2^n-1} text{OR} x = dfrac{2kpi}{2^n+1}$$ for some $k in mathbb{Z}$.
answered Jan 17 at 7:16
JimmyK4542JimmyK4542
41.1k245107
answered Jan 17 at 7:16
JimmyK4542JimmyK4542
41.1k245107
answered Jan 17 at 7:16
JimmyK4542JimmyK4542
41.1k245107
answered Jan 17 at 7:16
JimmyK4542JimmyK4542
41.1k245107
41.1k245107
add a comment |
add a comment |
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