Taylor Series of Function

1. Nov 24, 2008

roeb

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. Nov 24, 2008

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:

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