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I have $f(x) = \sin(x)$. I thought that when I do Fourier Transform and construct epicycles, than those epicycles will draw that $\sin(x)$ function (but this is probably not case with $\sin(x)$, cause there will be just one circle? So what those epicycles actually draw? I read many times that they can draw whatever curve, so I thought it draws that function from which I did Fourier Transform.

What i think is:

$f(t) = \sin(t)$ fourier transform will be..

just $frequency = 1$ with $amplitude = 1$,

so $F(1) = 1$ and $F(x) = 0$ otherwise

$F(1) = 1$ means one unit circle when constructing epicycles, but this epicycle will not 100% draw $\sin(x)$ function, cause it is just one circle.

Krab
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  • How exactly do you construct epicycles? – Conifold Jun 03 '14 at 21:25
  • Conifold: just in my head, but sin(x) should be one circle and circle path isn't same as sin path. I don't know if i understand good that those epicycles should draw the function (sin(x) in my case). – Krab Jun 03 '14 at 21:30
  • The circle is $e^{ix}$. So $\sin x$ is the sum of two "opposite" circles: $(e^{ix}-e^{-ix})/2i$. – geodude Jun 03 '14 at 21:40
  • It might be easier in complex coordinates. Try $z(t) = e^{it} + a e^{3it}$ for a small number $a << 1$. You have to separate the real and imaginary parts. – cactus314 Jun 03 '14 at 21:42
  • In the case of the sine function, the size of every epicycle is zero. – Michael Hardy Jun 04 '14 at 00:54
  • @geodude I have a very simliar question to Krab, and I have posted it very detailed here: http://math.stackexchange.com/questions/1547004/fourier-series-and-epicycles-how-to-extract-the-radii-and-angular-velocities-f

    Edit: do you mean $(e^{ix}-e^{-ix})/2$? without the $i$ in the denominator?

    If you have any insight to share I would really appreciate it!

     – Mike Nov 26 '15 at 07:56
  • @cactus314 I have some code that does just what you mention along with a question that is very similar to Krab's here: http://math.stackexchange.com/questions/1547004/fourier-series-and-epicycles-how-to-extract-the-radii-and-angular-velocities-f

    You can actually enter your function and plot it. Also, if you have the time to share any insight with regards to my question there, I would really appreciate it!

     – Mike Nov 26 '15 at 07:59

1 Answers1

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Just $\sin t$ can't give you a circle, you need two coordinates for a point in the plane. For example, $(\cos t,\sin t)$, or in complex notation $e^{it}=\cos t+i\sin t$, draws a circle as $t$ runs over $[0,2\pi]$. If you want to approximate the graph of $y=\sin x$ with "epicycles" you need to approximate the curve $(t,\sin t)$, or $t+i\sin t$ in complex notation, with linear combinations of $\dots,e^{-it},1,e^{it},e^{2it},\dots$. Such combinations correspond to tracing rotating circles, whose centers trace other rotating circles, and their centers trace other, and so on.

Basically, you need the Fourier series of the complex valued function $t+i\sin t$. Since Fourier series converges to the function that generated it (say if the function is continuous) those epicycles will draw a piece of "whatever curve" with prescribed accuracy by cutting off the series to a finite sum. They can not approximate the whole graph at once though, because circles close and the graph of $\sin$ does not.

Conifold
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  • You are almost answering a very similar question that I've asked. If you have the time to take a look at my question, I would really appreciate if you have any insight to share.

    http://math.stackexchange.com/questions/1547004/fourier-series-and-epicycles-how-to-extract-the-radii-and-angular-velocities-f

     – Mike Nov 26 '15 at 07:53