$|e^{-ina}-e^{-inb}|=2\sin(n\theta_0)$

Question

How can I show that $|e^{-ina}-e^{-inb}|=2\sin(n\theta_0)$ for some $\theta_0 \in[a,b]$?

I am trying this post Exercise 9, Chapter 2 of Stein's Fourier Analysis. Showing that a fourier series does not converge absolutely but converges conditionally. and was very quickly lost by this fact among others.

Is this some application of the mean value theorem that I am not spotting? How can I see this?

Answer 1

You have \begin{align} |e^{-ina}-e^{-inb}|^2 &=(\cos (-na) - \cos (-nb))^2+(\sin (-nb)-\sin (-na))^2\\ \ \\ &=2-2\cos(na)\cos(nb)-2\sin(na)\sin(nb)\\ \ \\ &=2-2\cos n(a-b)\\ \ \\ &=2-2\cos 2\tfrac{n(a-b)}2\\ \ \\ &=2-2\left(\cos^2\tfrac{n(a-b)}2-\sin^2\tfrac{n(a-b)}2\right)\\ \ \\ &=2-2\left(1-2\sin^2\tfrac{n(a-b)}2 \right)\\ \ \\ &=4\sin^2\tfrac{n(a-b)}2. \end{align} So $$ |e^{-ina}-e^{-inb}|=2\left|\sin\tfrac{n(a-b)}2\right| $$

Answer 2

This is similar to the answer from Martin Argerami, but uses a different, possibly simpler, way to compute the result. However, at the end, I show the requirement of $\theta_0 \in \left[a,b\right]$ cannot be met in general, although I also show what I consider to be a likely typo and what the answer would be then.

Using Euler's formula, then gathering the real & imaginary parts together, and using $\cos\left(-x\right) = \cos\left(x\right)$ and $\sin\left(-x\right) = -\sin\left(x\right)$, we have

$$e^{-ina} - e^{-inb} = \left(\cos\left(na\right) - \cos\left(nb\right)\right) + i\left(\sin\left(nb\right) - \sin\left(na\right)\right) \tag{1}\label{eq1}$$

Using the Sum and Difference Formulas for $\sin$ and $\cos$ to combine their differences into products, we have that

$$\cos\left(na\right) - \cos\left(nb\right) = 2\sin\left(\frac{n}{2}\left(a + b\right)\right)\sin\left(\frac{n}{2}\left(b - a\right)\right) \tag{2}\label{eq2}$$

and

$$\sin\left(nb\right) - \sin\left(na\right) = 2\sin\left(\frac{n}{2}\left(b - a\right)\right)\cos\left(\frac{n}{2}\left(a + b\right)\right) \tag{3}\label{eq3}$$

Thus, for the magnitude, note there is a common factor of $2\sin\left(\frac{n}{2}\left(b - a\right)\right)$ in both \eqref{eq2} and \eqref{eq3}, so it can be pulled out. Also, there is a $\sin$ and $\cos$ factor with the same argument of $\frac{n}{2}\left(a + b\right)$, so the square root of the sum of their squares would be $1$. Thus,

$$\left\lvert e^{-ina} - e^{-inb}\right\rvert = 2\left\lvert \sin\left(n\left(\frac{b - a}{2}\right)\right)\right\rvert \tag{4}\label{eq4}$$

Note you need to use absolute value signs due to the magnitude being non-negative. However, there may not necessarily be any $\theta_0 \in \left[a,b\right]$ which works. For example, consider that $a = 2$ and $b = 3$. The result gives a $\theta_0 = \pm \frac{b - a}{2} = \pm \frac{1}{2}$. As such, you cannot prove the question is true in general.

However, if you change $e^{-inb}$ to $e^{inb}$ instead, then it'll work, so it's likely a typo there. In this case,

$$e^{-ina} - e^{inb} = \left(\cos\left(na\right) - \cos\left(nb\right)\right) - i\left(\sin\left(na\right) + \sin\left(nb\right)\right) \tag{5}\label{eq5}$$

This changes \eqref{eq3} to

$$\sin\left(na\right) + \sin\left(nb\right) = 2\sin\left(\frac{n}{2}\left(a + b\right)\right)\cos\left(\frac{n}{2}\left(b - a\right)\right) \tag{6}\label{eq6}$$

Using the similar arguments as before, along with using \eqref{eq2}, we will now get

$$\left\lvert e^{-ina} - e^{inb}\right\rvert = 2\left\lvert \sin\left(n\left(\frac{a + b}{2}\right)\right)\right\rvert \tag{7}\label{eq7}$$

As such, in this case, $\theta_0 = \frac{a + b}{2}$ works as $\theta_0 \in \left[a,b\right]$.