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Periodic function proof

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Please provide explanation to prove that f(x)=sin(x)+sin(x/sqrt(2)) is not periodic.

https://brainmass.com/math/computing-values-of-functions/periodic-function-proof-29513

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The solution is on the attached file.
I assumed that the formulas for the derivatives of sin and cos are known and also that sin(y)= sin(y+2nPI) for any n.
In other words the sin function has period 2PI.

Solution

f(x)=sin(x)+sin(x/sqrt(2))

Suppose that f(x) is periodic with period p. Then f(x+p) - f(x)=0. Now let g(x) = f(x+p) - f(x)=0 .
Since g(x)=0 then g"(x) =0 (this is the second derivative of g.
Therefore g(x)- 2g"(x)= 0.

If we do the computations(using the formulas for the derivatives of sin and cos) we'll get (from the above equation):
sin x- sin(x+p) =0 which means sin x= sin(x+p) and results that p=2n PI where n is a positive integer ( because the period of sin function is 2nPI.

But f(x+p) = f(x) so f(x+2n PI) =f(x)
So sin(x)+sin(x/sqrt(2)) = sin(x+2n PI)+sin(x/sqrt(2)+ 2n PI/sqrt2).This is equivalent to:
sin(x/sqrt(2)) = sin(x/sqrt(2)+ 2n PI/sqrt2).(because first term in the left equals first term in the right from the reason above: sin is periodical with 2nPi period).

So we have: sin(x/sqrt(2)) = sin(x/sqrt(2)+ 2n PI/sqrt2) for some n
But sin(x/sqrt(2)) = sin(x/sqrt(2)+ 2k PI)) for any k since sin is a periodical function with period multiples of 2PI

From the 2 equations above results that for some n and k we have : 2n PI/sqrt2 =2k PI
Which means n/k = sqrt 2.
This is impossible since n and k are positive integers.
So the assumption that f is periodical function is false.
Therefore f is not periodical.

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