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Why is this expression bounded by 1?

  1. Aug 22, 2013 #1
    How exactly might one go about showing that

    [tex]
    \left| \frac{1 - e^{-iy}}{-iy} \right|
    [/tex]

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

    [tex]
    \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).
    [/tex]

    But this doesn't seem to be the right way to do it. For example, if you look at [itex]g(8,N)[/itex] for the first several values of [itex]N\in \mathbb N[/itex], it eventually settles down to something less than one, but it takes it awhile! Indeed, one needs to take [itex]N=14[/itex] to make [itex]g(8,N) < 1[/itex]. 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?
     
  2. jcsd
  3. Aug 22, 2013 #2

    mathwonk

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    Science Advisor
    Homework Helper

    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?
     
  4. Aug 22, 2013 #3

    phyzguy

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    Science Advisor

    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.
     
  5. Aug 23, 2013 #4
    Ok. Both of you guys seem to have a pretty decent suggestion. It looks like, for [itex]y\in \mathbb R[/itex],

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

    We first observe that it is sufficient to show [itex]f(y) \leq 1[/itex] on [itex](-2,2)[/itex]; on the complement of this interval, the obvious estimate [itex]f(y) \leq \frac{2}{|y|} \leq 1[/itex] 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, [itex]f(y)[/itex] is an even function, so it will suffice to look at [itex]0 < y < 2[/itex]. If we let [itex]x = y/2[/itex], this is equivalent to looking at [itex]g(x) = \frac{\sin x}{x}[/itex] for [itex]0 < x < 1[/itex]. But the alternating series estimation theorem shows us that [itex]\frac{\sin x}{x} < 1[/itex] for [itex]x\in (0,1)[/itex], so we're done. Does this sound right?
     
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