This is called the first order approximation or the linear approximation.

[roeb]

Homework Statement

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...

Homework Equations

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?

 

[gabbagabbahey]

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

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

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

\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

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

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