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Let' say you have a graph as such:

!(http://blogs.scientificamerican.com/roots-of-unity/files/2014/12/sinxsin2x.jpg) !

(Black one)

And you know that it's function is in the general form of cosx+sinx

Is there a way to find about what the exact function is using observation and calculation?

Thank you so much. I appreciate any input.

city7lights
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  • Do you know Fourier series? If $f\in L^1$ is $2p$ periodic, you can construct it's Fourier series, and infinite sum $\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n(\frac{n\pi x}{p})+b_n(\frac{n\pi x}{p})$. Of course in general, to find $a_n$ and $b_n$, you need to actually know what $f$ is, not just have a picture of it. – user153582 Nov 01 '15 at 01:24
  • Since it is 0 at $x=0$, it can only have $\sin$ components. Since there are two types of local maxima and local minima, I'd expect two $\sin$ components. Since there's no scale, I'd make one component $\sin(x)$, and the other is $\sin(2x)$, $\sin(3x)$. But since there is one primary peak, and one extra secondary peak per period, I'd expect the second component to be $\sin(2x)$. – David Nov 01 '15 at 03:36
  • @david Thank you! Okay, but would you be able to calculate exact numbers for the value before x, the horizontal shift, etc if you have the x and y values of the graph? And how did you make the assumption that the second one must be sin(2x)? I greatly appreciate your help :) – city7lights Nov 02 '15 at 20:19
  • The fact the the function is zero at $x=0$ means that there can only be $\sin$. Because there is one small and one large peak per "period", I guess that the frequency of the second component was double that of the the first. If it was 3 times the frequency, there would be 2 minor peaks and one major peak. Does that make sense? – David Nov 02 '15 at 21:38
  • @david yes, that makes sense :) However, other than that, is there no way of figuring out the period, and other shifts? Even if I have all of the x and y coordinates? – city7lights Nov 04 '15 at 13:51

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