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<math>\int{dti} \int{f(x_i,t_i)g(\Delta x, \Delta t) dx_i dy_i}</math>  
 
<math>\int{dti} \int{f(x_i,t_i)g(\Delta x, \Delta t) dx_i dy_i}</math>  
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f(xf, tf)
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We know that
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We know that f(stuff) is nothing more than the
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<math>f(x_i,t_i)\,</math>
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product of the phases of the incoming and outgoing  
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is nothing more than the product of the phases of the incoming and outgoing waves and the amplitude function. Because this equation only takes into account one incoming and one outgoing wave, we will need to recalculate the later equations using different f-functions for all outgoing waves and sum these results. Fortunately, as we discovered above, only the first and second outgoing waves are relevant.
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waves and the amplitude function.
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Calculation of g proves more challenging. We know the wave equation
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Calculation of g proves more challenging. We know
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<math>\frac{d^2y}{dt^2} - \frac{c^2d^2y}{dx^2} = 0</math>
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the wave equation
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However, this is for a uniform, sourceless wave. Like most generalizations, this is an unrealistic situation in the real world. What we need is a function that can generate a brief pulse. This sounds like a delta function.
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d2y/dt2 - c2d2y/dx2 = 0
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<math>\frac{d^2g}{dt^2} - \frac{c^2d^2g}{dx^2} = \delta(\Delta x)\delta(\Delta t)</math>
 
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However, this is for a uniform, sourceless wave.
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Like most generalizations, this is an unrealistic
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situation. What we need is a function that can
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generate a brief pulse. This sounds like a delta  
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function, so we will use this.
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d2g/dt2 - c2d2g/dx2 = (d)(x-xi)(t-ti)
      
This is not an easy equation to solve without  
 
This is not an easy equation to solve without  
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