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I read that the dot product of two vectors $v_1$ and $v_2$ is equivalent to the integral of the product of two like functions.

For example, the vectors $[1,0]$ and $[0,1]$ are orthogonal and so their dot product is equal to $0$.

Now, on the unit circle I treated these two vectors as special cases of the following functions which are always $90$ degrees out of phase to each other on the unit circle. \begin{align} f(x) &= e^{i\pi x}\\ g(x) &= e^{i \pi (x+\pi/2)} \end{align}

However, I am unable to get the product of these two functions and integrate them in such a ways as to get zero.

Where am I going wrong with my Math/intuition?

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First, note that you actually want $e^{ix}$ and $e^{i(x + \pi/2)}$ (without the constant of $\pi$ already there). Then, \begin{align} \int_0^{2\pi} (fg)(x) dx &= \int_0^{2\pi} e^{ix} e^{i(x + \pi/2)}\ dx\\ &= \int_0^{2\pi} e^{i(2x + \pi/2)}\ dx\\ &= \left[ \frac{1}{2i} e^{i(2x + \pi/2)} \right|_0^{2\pi}\\ &= \frac{1}{2i} \cdot \left(e^{i(9\pi /2)} - e^{i(\pi/2)} \right)\\ &= \frac{1}{2i} \cdot ((\cos(9\pi/2) + i\sin(9\pi/2)) - (\cos(\pi/2) + i\sin(\pi/2)))\\ &= 0 \end{align}

In general, we call $\int_I f(x)g(x)\ dx$ the inner product of the functions $f$ and $g$, which is usually defined on some interval $I$. Here's a link to another answer that gives even more insight on the inner product of two functions.

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    $\begingroup$ It is easier to see if you take $e^{\frac{i\pi}{2}}$ outside the integral sign. The integral is then the result of going around a circle twice. $\endgroup$ – herb steinberg Feb 23 '18 at 1:24
  • $\begingroup$ @herbsteinberg yes that’s certainly a much more intelligent way of going about it! $\endgroup$ – AlkaKadri Feb 23 '18 at 1:51

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