Why is this expression bounded by 1?

[AxiomOfChoice]

How exactly might one go about showing that

$$\left| \frac{1 - e^{-iy}}{-iy} \right|$$

is bounded by 1 for $y\in \mathbb R$? I thought this would be easy to show using the series expansion of $e^{-iy}$ in some way:

$$\left| \frac{1 - e^{-iy}}{-iy} \right| = \left| 1 - \frac{iy}{2} - \frac{y^2}{6} + \frac{iy^3}{24} + \ldots \right| = \lim_{N\to \infty} \left| \sum_{n=0}^N \frac{(-iy)^n}{(n+1)!} \right| = \lim_{N\to \infty} g(y,N).$$

But this doesn't seem to be the right way to do it. For example, if you look at $g(8,N)$ for the first several values of $N\in \mathbb N$, it eventually settles down to something less than one, but it takes it awhile! Indeed, one needs to take $N=14$ to make $g(8,N) < 1$. So I'm not sure a simple "inequalities are preserved under the limit" argument will work. And the series isn't alternating, so the alternating series estimation theorem won't work, either. Any ideas?

 

[mathwonk]

have you tried writing out the absolute values in top and bottom? using the basic identities e^(it) = cos(t) + i sin(t) and |a+bi|^2 = a^2 + b^2?

 

[phyzguy]

Try expanding the expression, which will give you a real function of y. You can then find the extrema of the expression inside the square root by taking the derivative and setting it equal to zero, then find the largest extremum. A little involved, but it works.

 

[AxiomOfChoice]

Ok. Both of you guys seem to have a pretty decent suggestion. It looks like, for $y\in \mathbb R$,

$$\left| \frac{1 - e^{-iy}}{-iy} \right| = \frac{2}{|y|} \left| \sin \left( \frac{y}{2} \right) \right| = f(y).$$

We first observe that it is sufficient to show $f(y) \leq 1$ on $(-2,2)$; on the complement of this interval, the obvious estimate $f(y) \leq \frac{2}{|y|} \leq 1$ does the trick. (Note that we are forcing the original function to have the value of 1 at zero; maybe I should've mentioned that!) Also, $f(y)$ is an even function, so it will suffice to look at $0 < y < 2$. If we let $x = y/2$, this is equivalent to looking at $g(x) = \frac{\sin x}{x}$ for $0 < x < 1$. But the alternating series estimation theorem shows us that $\frac{\sin x}{x} < 1$ for $x\in (0,1)$, so we're done. Does this sound right?

 

[blue_raver22]

The expression in question, \left| \frac{1 - e^{-iy}}{-iy} \right|, is bounded by 1 because it represents the absolute value of a complex number divided by its argument. In other words, it represents the magnitude of a complex number divided by its angle in the complex plane. Since the magnitude of a complex number can never exceed its angle, the expression is always bounded by 1.

To formally prove this, one could use the triangle inequality for complex numbers, which states that |a + b| \leq |a| + |b|. Applying this to the expression in question, we have:

\left| \frac{1 - e^{-iy}}{-iy} \right| = \left| \frac{1}{-iy} - \frac{e^{-iy}}{-iy} \right| \leq \left| \frac{1}{-iy} \right| + \left| \frac{e^{-iy}}{-iy} \right|

= \frac{1}{|y|} + \frac{|e^{-iy}|}{|y|}

= \frac{1}{|y|} + \frac{1}{|y|}

= \frac{2}{|y|}

Since y\in \mathbb R, we know that |y| \geq 0. Therefore, \frac{2}{|y|} \leq 2. This means that \left| \frac{1 - e^{-iy}}{-iy} \right| \leq 2 for all y\in \mathbb R.

To show that the expression is actually bounded by 1, we can use the fact that |e^{-iy}| = 1 for all y\in \mathbb R. This means that the second term in the expression, \left| \frac{e^{-iy}}{-iy} \right|, is always equal to 1. Therefore, we have:

\left| \frac{1 - e^{-iy}}{-iy} \right| = \left| \frac{1}{-iy} - \frac{e^{-iy}}{-iy} \right| \leq \left| \frac{1}{-iy} \right| + \left| \frac{e^{-iy}}{-iy} \right|

= \frac{1}{|y|}