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I am interested in a function which can represent the roots of the two functions which they do not share in common. The red function is y=sin((pi/2)x) and the green one is y=sin((pi/3)x) If anyone could provide help that would be awesome. The sketch is the blue function, the shape isn't necessary, only the fact that is has the roots indicated. Sketch Sin(pi/2 x) and Sin(pi/3 x)

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    Why don't you sketch a function and show us? – 2017-01-10
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    Well, I could just draw a sinusoidal function passing through those points, but its shape would be arbitrary to its purpose. It only needs to pass through those points and nothing else. – 2017-01-10
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    OK, I understand your point. Sketch it and then then we can apply Fourier. – 2017-01-10
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    Okay, sketched. – 2017-01-10

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Here's a weird solution:

$$y = \frac{\sin(\frac\pi2 x)\sin(\frac\pi3 x)}{\sin^2(\frac\pi2 x) + \sin^2(\frac\pi3 x)}$$

enter image description here

We feed in $\sin(\frac\pi2 x)$ and $\sin(\frac\pi3 x)$ into the classic analysis counterexample $f(x,y) = \frac{xy}{x^2+y^2}$.

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    Thats wonderful, maybe I can elude to what I am doing. Say you kept plotting the next f(x) with half periods consecutively increasing by 1 and you wanted to continue passing a function through the points which the set of functions don't share. I would conjecture that if you had a method for this and took the limit as n (the half period) goes to infinity, you would pass a function through all the prime numbers. – 2017-01-10
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    Just curious, am I right? – 2017-01-10
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    That's a more complicated question. How do you know that the sequence of the functions converges to a limit? – 2017-01-10
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    Thats something to find out, but do you see my idea? – 2017-01-10
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    Sure, if you could construct a function whose zeros are the numbers that appear exactly once in the sequences $(2,4,6,...), (3,6,9,...), (4,8,12,...), ...$ then its zeros would be the prime numbers. But one could probably construct such a function directly without going through this limiting process. – 2017-01-10
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    In this instance, a zero would not fall in the sequence 4n, and if it is possible, why hasn't it been done – 2017-01-10
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    @Jack I'm sure it has been done, it just isn't useful enough to be well-known. There are plenty of ways to construct real-valued functions that have zeroes exactly at the prime numbers. – 2017-01-10
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We can define a stepwise function as :

From x = 2 + 12$k$ to 4 + 12$k$ and $k\in\mathbb{Z}$: $$y = sin(2\pi(x-2)/2)$$ From x = 4 + 12$k$ to 8 + 12$k$ : $$y = sin(2\pi(x-4)/8)$$ From x = 8 + 12$k$ to 10 + 12$k$ : $$y = - sin(2\pi(x-8)/2)$$ From x = 10 + 12$k$ to 14 + 12$k$ : $$y = - sin(2\pi(x-10)/8)$$

For sure this is not the only one. We can adjust the slope to be equal at x = 2,4,8,10,... Also we can adjust amplitude.

Hope this simple one be of help.

EDIT

Here there is another one:

$$y = sin(2\pi(x-3)/12)+ 0,5 sin(2\pi(x-1)/4)$$

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    Can one function be passed through instead of a combination of 4? I would really like to elude to what I am trying to accomplish except I can't because I think someone would take it. – 2017-01-10
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If the shape isn't necessary, then how about:

$$ f(x) = \begin{cases} 0 &\textrm{ if $x$ is a root of exactly one of the two functions}\\ 1 &\textrm{ otherwise} \end{cases} $$