These functions are equal. But I don't understand why.
$a \leftrightarrow f(x) =|\cos(2\pi x)|^2$
$b \leftrightarrow f(x) = \dfrac{\cos(4\pi x)}{2} + 0.5$
Which results both in this plot:
These functions are equal. But I don't understand why.
$a \leftrightarrow f(x) =|\cos(2\pi x)|^2$
$b \leftrightarrow f(x) = \dfrac{\cos(4\pi x)}{2} + 0.5$
Which results both in this plot:
They are the same because $|\cos(x)|^2=\cos^2(x)=\frac{1+\cos(2x)}{2}=\frac{\cos(2x)}{2}+\frac{1}{2}.$ (Note that the absolute value doesn't do anything, because of the square; i.e., $x^2=|x|^2$ for any $x$.) See here. This can be derived as follows:
For any $x$ and $y$, we have $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y).$ Thus, $\cos(2x)=\cos^2(x)-\sin^2(x)$ $\cos(2x)=\cos^2(x)-(1-\cos^2(x))$ $\cos(2x)=2\cos^2(x)-1$ and therefore $\cos^2(x)=\frac{1+\cos(2x)}{2}$
Here is another, more direct derivation, using complex exponential definition of cosine: $\cos^2(x)=\left(\frac{e^{ix}+e^{-ix}}{2}\right)^2=\frac{(e^{ix})^2+2(e^{ix})(e^{-ix})+(e^{-ix})^2}{4}=$ $\frac{e^{i(2x)}+e^{-i(2x)}+2}{4}=\frac{\cos(2x)+1}{2}$