Can the graph of $tan^{-1}{left(frac{sin x}{x}right)}$ be expressed as $Ce^{-kx}cos(omega x + phi)$?
After graphing $sin x$, I thought of trying something interesting. I wanted to plot the angle $theta$ that a point $(x,sin x )$ makes with the origin on the $y$-axis, against $x$ on the $x$-axis.
$$tantheta = frac{sin x}{x}Rightarrow theta=tan^{-1}{left(frac{sin x}{x}right)}$$
Graphing $y = 20timestheta$ (multiply by 20 for graphical purposes):
Part of it reminded me of the graph for the damped oscillator (specifically, the $x>0$ part).
That made me wonder if it was possible to find constants $C,k,omega,$ and $phi$ such that $$theta = Ce^{-kx}cos(omega x + phi)$$
However, after toying with Grapher for a while, $y = theta$ didn't seem to decrease exponentially.
That led me to this question: is there any analytical way to find real constants $C,k,omega,$ and $phi$ such that $theta = Ce^{-kx}cos(omega x + phi)$?
Furthermore, are there any complex constants $C,k,omega, text{and } phi$?
real-analysis complex-analysis trigonometry exponential-function graphing-functions
edited Dec 9 at 4:36
user1101010
7551630
asked Apr 5 '17 at 8:46
Ujkan Sulejmani
1586
add a comment |
After graphing $sin x$, I thought of trying something interesting. I wanted to plot the angle $theta$ that a point $(x,sin x )$ makes with the origin on the $y$-axis, against $x$ on the $x$-axis.
$$tantheta = frac{sin x}{x}Rightarrow theta=tan^{-1}{left(frac{sin x}{x}right)}$$
Graphing $y = 20timestheta$ (multiply by 20 for graphical purposes):
Part of it reminded me of the graph for the damped oscillator (specifically, the $x>0$ part).
That made me wonder if it was possible to find constants $C,k,omega,$ and $phi$ such that $$theta = Ce^{-kx}cos(omega x + phi)$$
However, after toying with Grapher for a while, $y = theta$ didn't seem to decrease exponentially.
That led me to this question: is there any analytical way to find real constants $C,k,omega,$ and $phi$ such that $theta = Ce^{-kx}cos(omega x + phi)$?
Furthermore, are there any complex constants $C,k,omega, text{and } phi$?
real-analysis complex-analysis trigonometry exponential-function graphing-functions
edited Dec 9 at 4:36
user1101010
7551630
asked Apr 5 '17 at 8:46
Ujkan Sulejmani
1586
No for real constants: the derivatives are quite different.
– Bernard
Apr 5 '17 at 8:54
Certainly not, there is a serious mismatch for negative $x$.
– Yves Daoust
Apr 5 '17 at 14:47
add a comment |
After graphing $sin x$, I thought of trying something interesting. I wanted to plot the angle $theta$ that a point $(x,sin x )$ makes with the origin on the $y$-axis, against $x$ on the $x$-axis.
$$tantheta = frac{sin x}{x}Rightarrow theta=tan^{-1}{left(frac{sin x}{x}right)}$$
Graphing $y = 20timestheta$ (multiply by 20 for graphical purposes):
Part of it reminded me of the graph for the damped oscillator (specifically, the $x>0$ part).
That made me wonder if it was possible to find constants $C,k,omega,$ and $phi$ such that $$theta = Ce^{-kx}cos(omega x + phi)$$
However, after toying with Grapher for a while, $y = theta$ didn't seem to decrease exponentially.
That led me to this question: is there any analytical way to find real constants $C,k,omega,$ and $phi$ such that $theta = Ce^{-kx}cos(omega x + phi)$?
Furthermore, are there any complex constants $C,k,omega, text{and } phi$?
real-analysis complex-analysis trigonometry exponential-function graphing-functions
edited Dec 9 at 4:36
user1101010
7551630
asked Apr 5 '17 at 8:46
Ujkan Sulejmani
1586
After graphing $sin x$, I thought of trying something interesting. I wanted to plot the angle $theta$ that a point $(x,sin x )$ makes with the origin on the $y$-axis, against $x$ on the $x$-axis.
$$tantheta = frac{sin x}{x}Rightarrow theta=tan^{-1}{left(frac{sin x}{x}right)}$$
Graphing $y = 20timestheta$ (multiply by 20 for graphical purposes):
Part of it reminded me of the graph for the damped oscillator (specifically, the $x>0$ part).
That made me wonder if it was possible to find constants $C,k,omega,$ and $phi$ such that $$theta = Ce^{-kx}cos(omega x + phi)$$
However, after toying with Grapher for a while, $y = theta$ didn't seem to decrease exponentially.
That led me to this question: is there any analytical way to find real constants $C,k,omega,$ and $phi$ such that $theta = Ce^{-kx}cos(omega x + phi)$?
Furthermore, are there any complex constants $C,k,omega, text{and } phi$?
real-analysis complex-analysis trigonometry exponential-function graphing-functions
real-analysis complex-analysis trigonometry exponential-function graphing-functions
edited Dec 9 at 4:36
user1101010
7551630
asked Apr 5 '17 at 8:46
Ujkan Sulejmani
1586
edited Dec 9 at 4:36
user1101010
7551630
asked Apr 5 '17 at 8:46
Ujkan Sulejmani
1586
edited Dec 9 at 4:36
user1101010
7551630
edited Dec 9 at 4:36
user1101010
7551630
edited Dec 9 at 4:36
user1101010
7551630
7551630
asked Apr 5 '17 at 8:46
Ujkan Sulejmani
1586
asked Apr 5 '17 at 8:46
Ujkan Sulejmani
1586
asked Apr 5 '17 at 8:46
Ujkan Sulejmani
1586
1586
No for real constants: the derivatives are quite different.
– Bernard
Apr 5 '17 at 8:54
Certainly not, there is a serious mismatch for negative $x$.
– Yves Daoust
Apr 5 '17 at 14:47
add a comment |
No for real constants: the derivatives are quite different.
– Bernard
Apr 5 '17 at 8:54
Certainly not, there is a serious mismatch for negative $x$.
– Yves Daoust
Apr 5 '17 at 14:47
No for real constants: the derivatives are quite different.
– Bernard
Apr 5 '17 at 8:54
No for real constants: the derivatives are quite different.
– Bernard
Apr 5 '17 at 8:54
Certainly not, there is a serious mismatch for negative $x$.
– Yves Daoust
Apr 5 '17 at 14:47
Certainly not, there is a serious mismatch for negative $x$.
– Yves Daoust
Apr 5 '17 at 14:47
add a comment |
2 Answers
2
active
oldest
votes
The damping that you see is
$$frac{arctandfrac{sin x}x}{sin x}.$$
It is very close to the hyperbola $dfrac1x$, as you remain in the linear part of the arc tangent.
answered Apr 5 '17 at 14:51
Yves Daoust
124k671221
Is there an explanation as to why the damping/envelope is represented by that expression?
– Ujkan Sulejmani
Apr 5 '17 at 16:49
@UjkanSulejmani: look again, this is obvious.
– Yves Daoust
Apr 5 '17 at 17:40
add a comment |
The envelope of $frac{sin x}{x}$ is $pm frac{1}{x}$, and for large $x$, $arctan frac{1}{x} approx frac{1}{x}$. So the decay of the function is inverse to $x$, not exponential, and no constant $k$ exists (real or complex; an imaginary part to $k$ will just introduce a sinusoidal oscillation in the envelope, in addition to the exponential decay or growth).
answered Apr 5 '17 at 14:07
Connor Harris
4,303723
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The damping that you see is
$$frac{arctandfrac{sin x}x}{sin x}.$$
It is very close to the hyperbola $dfrac1x$, as you remain in the linear part of the arc tangent.
answered Apr 5 '17 at 14:51
Yves Daoust
124k671221
Is there an explanation as to why the damping/envelope is represented by that expression?
– Ujkan Sulejmani
Apr 5 '17 at 16:49
@UjkanSulejmani: look again, this is obvious.
– Yves Daoust
Apr 5 '17 at 17:40
add a comment |
The damping that you see is
$$frac{arctandfrac{sin x}x}{sin x}.$$
It is very close to the hyperbola $dfrac1x$, as you remain in the linear part of the arc tangent.
answered Apr 5 '17 at 14:51
Yves Daoust
124k671221
Is there an explanation as to why the damping/envelope is represented by that expression?
– Ujkan Sulejmani
Apr 5 '17 at 16:49
@UjkanSulejmani: look again, this is obvious.
– Yves Daoust
Apr 5 '17 at 17:40
add a comment |
The damping that you see is
$$frac{arctandfrac{sin x}x}{sin x}.$$
It is very close to the hyperbola $dfrac1x$, as you remain in the linear part of the arc tangent.
answered Apr 5 '17 at 14:51
Yves Daoust
124k671221
The damping that you see is
$$frac{arctandfrac{sin x}x}{sin x}.$$
It is very close to the hyperbola $dfrac1x$, as you remain in the linear part of the arc tangent.
answered Apr 5 '17 at 14:51
Yves Daoust
124k671221
answered Apr 5 '17 at 14:51
Yves Daoust
124k671221
answered Apr 5 '17 at 14:51
Yves Daoust
124k671221
answered Apr 5 '17 at 14:51
Yves Daoust
124k671221
124k671221
Is there an explanation as to why the damping/envelope is represented by that expression?
– Ujkan Sulejmani
Apr 5 '17 at 16:49
@UjkanSulejmani: look again, this is obvious.
– Yves Daoust
Apr 5 '17 at 17:40
add a comment |
Is there an explanation as to why the damping/envelope is represented by that expression?
– Ujkan Sulejmani
Apr 5 '17 at 16:49
@UjkanSulejmani: look again, this is obvious.
– Yves Daoust
Apr 5 '17 at 17:40
Is there an explanation as to why the damping/envelope is represented by that expression?
– Ujkan Sulejmani
Apr 5 '17 at 16:49
Is there an explanation as to why the damping/envelope is represented by that expression?
– Ujkan Sulejmani
Apr 5 '17 at 16:49
@UjkanSulejmani: look again, this is obvious.
– Yves Daoust
Apr 5 '17 at 17:40
@UjkanSulejmani: look again, this is obvious.
– Yves Daoust
Apr 5 '17 at 17:40
add a comment |
The envelope of $frac{sin x}{x}$ is $pm frac{1}{x}$, and for large $x$, $arctan frac{1}{x} approx frac{1}{x}$. So the decay of the function is inverse to $x$, not exponential, and no constant $k$ exists (real or complex; an imaginary part to $k$ will just introduce a sinusoidal oscillation in the envelope, in addition to the exponential decay or growth).
answered Apr 5 '17 at 14:07
Connor Harris
4,303723
add a comment |
The envelope of $frac{sin x}{x}$ is $pm frac{1}{x}$, and for large $x$, $arctan frac{1}{x} approx frac{1}{x}$. So the decay of the function is inverse to $x$, not exponential, and no constant $k$ exists (real or complex; an imaginary part to $k$ will just introduce a sinusoidal oscillation in the envelope, in addition to the exponential decay or growth).
answered Apr 5 '17 at 14:07
Connor Harris
4,303723
add a comment |
The envelope of $frac{sin x}{x}$ is $pm frac{1}{x}$, and for large $x$, $arctan frac{1}{x} approx frac{1}{x}$. So the decay of the function is inverse to $x$, not exponential, and no constant $k$ exists (real or complex; an imaginary part to $k$ will just introduce a sinusoidal oscillation in the envelope, in addition to the exponential decay or growth).
answered Apr 5 '17 at 14:07
Connor Harris
4,303723
The envelope of $frac{sin x}{x}$ is $pm frac{1}{x}$, and for large $x$, $arctan frac{1}{x} approx frac{1}{x}$. So the decay of the function is inverse to $x$, not exponential, and no constant $k$ exists (real or complex; an imaginary part to $k$ will just introduce a sinusoidal oscillation in the envelope, in addition to the exponential decay or growth).
answered Apr 5 '17 at 14:07
Connor Harris
4,303723
edited Apr 5 '17 at 14:37
answered Apr 5 '17 at 14:07
Connor Harris
4,303723
answered Apr 5 '17 at 14:07
Connor Harris
4,303723
answered Apr 5 '17 at 14:07
Connor Harris
4,303723
4,303723
add a comment |
add a comment |
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No for real constants: the derivatives are quite different.
– Bernard
Apr 5 '17 at 8:54
Certainly not, there is a serious mismatch for negative $x$.
– Yves Daoust
Apr 5 '17 at 14:47