I have $f:\mathbb{R}\rightarrow \mathbb{R}, f(x)=\sin(4x)+\cos(x\sqrt{2})$ and I need to find the period of this function. I found that the period of $\sin(4x)=\pi/2$ and the period of $\cos(x\sqrt{2})=\pi\sqrt{2}$. Why doesn't this function have a period and how can I approach the exercise?
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WA says the function is not periodic – Dr. Sonnhard Graubner May 08 '19 at 14:12
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How to see that without a computer? – DaniVaja May 08 '19 at 14:13
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You can assume $$f(x)$$ is periodic, so $$f(x)=f(x+T)$$ and then you will get a contradiction – Dr. Sonnhard Graubner May 08 '19 at 14:19
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For $$x=0$$ you would get $$f(0)=f(T)$$ – Dr. Sonnhard Graubner May 08 '19 at 14:21
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Or a different, more general solution: https://math.stackexchange.com/questions/1356802/the-sum-of-two-continuous-periodic-functions-is-periodic-if-and-only-if-the-rati – Botond May 08 '19 at 14:25
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@Botond Not really a duplicate, as this one mixes sin and cos terms. Of course, the general solution will be applicable here as well. – Ingix May 08 '19 at 14:35
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@Ingix you are right, but Hagen's answer in the linked question totally covers it. – Botond May 08 '19 at 14:37
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1Because this "periodic" function is composed of two other periodic functions, one of period $2\pi$ and the other of period $\sqrt{2}\pi$ then if this function were periodic, it would repeat every time both of these functions reach the end of a repeat at the same point. So $2\pi n=\sqrt{2}\pi m$ where $n$ and $m$ are the number of repeats. I think if no integers $m$ and $n$ exist that satisfy this equation then it is not periodic – Henry Lee May 08 '19 at 14:52
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https://math.stackexchange.com/questions/873723/how-to-find-the-period-of-the-sum-of-two-trigonometric-functions – lab bhattacharjee May 08 '19 at 15:23