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Taylor Series of Function

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data
    Expand V(z + dz, t).
    I have seen problems like this in both my EnM and semiconductor courses but it's bothering me because I don't understand how the Taylor series is being used in this case...

    2. Relevant equations

    3. The attempt at a solution

    Taylor series: f(a) + f'(x-a) + ....

    let f = V(z+dz, t)

    Here is where I get stuck... I don't really know what to do

    I guess the first term: V(z,t) makes sense
    but the second term: dV(z,t)/dz delta(z) doesn't make any sense to me at all...

    What the heck happens in the second term?
  2. jcsd
  3. Nov 24, 2008 #2


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    The formula for the taylor series of [itex]f(x,t)[/itex] about the point [itex]x=a[/itex] is actually:

    [tex]f(x,t)=f(a,t)+\left( \left \frac{\partial f(x,t)}{\partial x} \right| _{x=a} \right) \frac{(x-a)}{1!}+\left( \left \frac{\partial ^2f(x,t)}{\partial x^2} {\right|}_{x=a} \right) \frac{(x-a)^2}{2!}+\ldots[/tex]

    In your case, you would use this for [itex]x=z+dz[/itex] and [itex]a=z[/itex] so [itex]x-a=dz[/itex] and

    [tex]\Rightarrow f(z+dz,t)=f(z,t)+\left( \left \frac{\partial f(x,t)}{\partial x} \right| _{x=z} \right) \frac{dz}{1!}+\left( \left \frac{\partial ^2f(x,t)}{\partial x^2} {\right|}_{x=z} \right) \frac{(dz)^2}{2!}+\ldots[/tex]

    But(!) [itex]dz[/itex] is an infinitesimal, so all the terms of order [itex](dz)^2[/itex] and higher are VERY small and so they can be neglected:

    [tex]\Rightarrow f(z+dz,t) \approx f(z,t)+\left( \left \frac{\partial f(x,t)}{\partial x} \right| _{x=z} \right)dz[/tex]
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