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I graphed the curve defined by

$$y=3|e^{-(x/2-1/3)^2+ibx}+e^{-(x/2+1/3)^2-ibx}|$$

where $b=1000$. The graph is here:

https://dl.dropbox.com/u/9034084/screenshots/interference.png

Clearly when b goes to infinity one can talk about the envelope of this family. So my question is, what are the envelope curves (I'm curious on both upper bound and lower bound) ?

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1  
You need to have a negative sign infront of the second $b$. –  user17762 Nov 24 '12 at 7:25
    
fixed. thanks a lot. –  hyh Nov 24 '12 at 7:27
2  
How about $3\Big\lvert e^{-(x/2-1/3)^2}\pm e^{-(x/2+1/3)^2}\Big\rvert$? –  Rahul Nov 24 '12 at 7:29
    
@RahulNarain You are exactly right. (I only checked the graph. That is apparently the max possible amplitude.) –  hyh Nov 24 '12 at 7:31
    
@RahulNarain Do you know what the smoothing of the curve is, I mean the function whose value at $x$ is the average of the current function on $[x-c, x+c]$, and $c\ll b$. I'm trying to get a good picture of interference. –  hyh Nov 24 '12 at 7:37

2 Answers 2

up vote 2 down vote accepted

The graph will lie between the following curves $$3 \left( e^{-(x/2-1/3)^2}+e^{-(x/2+1/3)^2} \right)$$ and $$3 \left \vert e^{-(x/2-1/3)^2}-e^{-(x/2+1/3)^2} \right \vert$$ since $$\vert \vert c \vert - \vert d \vert \vert \leq \left \vert c e^{ibx} + d e^{-ibx} \right \vert \leq \vert c \vert + \vert d \vert$$

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I'm willing to accept this. But you should give Rahul Narain some credit in your answer. –  hyh Nov 24 '12 at 7:35
    
@hyh: There's no need for that. Marvis arrived at the solution independently and has also given the reason why these bounds work. –  Rahul Nov 24 '12 at 7:41
    
@RahulNarain OK then. The reason is obvious once someone pointed that out. –  hyh Nov 24 '12 at 7:43

We change some constants to make this easier to work with... $$\begin{eqnarray*}y &=& \left|e^{-(x-1)^2+ibx}+e^{-(x+1)^2-ibx} \right| \\ &=& \left|e^{-(x-1)^2}(\cos(bx)+i\sin(bx))+e^{-(x+1)^2}(\cos(bx)-i\sin(bx)) \right| \\ &=& \left|\left(e^{-(x-1)^2}+e^{-(x+1)^2}\right)\cos(bx) + i\left(e^{-(x-1)^2}-e^{-(x+1)^2}\right)\sin(bx) \right| \\ &=& \sqrt{\cos^2(bx)\left(e^{-(x-1)^2}+e^{-(x+1)^2}\right)^2 + \sin^2(bx)\left(e^{-(x-1)^2}-e^{-(x+1)^2}\right)^2} \\ &=&\sqrt{\left(\cos^2(bx)+\sin^2(bx)\right)\left(e^{-2(x-1)^2}+e^{-2(x+1)^2}\right) + 2\left(\cos^2(bx)-\sin^2(bx)\right)\left(e^{-(x-1)^2-(x+1)^2}\right)} \\ &=& \sqrt{e^{-2(x-1)^2}+e^{-2(x+1)^2}+2\cos(2bx)e^{-2(x^2+1)}}\end{eqnarray*}$$

The last term is the oscillatory term, with $\cos(bx)$ ranging between $-1$ and $1$. Hence, the envelopes are $\sqrt{e^{-2(x-1)^2} + e^{-2(x+1)^2} \pm 2e^{-2(x^2+1)}}$.

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