# Taylor series

1. Apr 25, 2005

### retupmoc

How do i show that B_{x}(x+dx,y,z)-B_{x}(x,y,z)\approx \frac{\partial B_{x}(x,y,z)}{\partial x} dx
using a Taylor series to the first term. Using a Taylor series does B(x) = B(a) + B'(a)(x-a)? In that case what would B(x+dx) be and how can i obtain the desired result from this? Thanks in advance

2. Apr 25, 2005

### Night Owl

What's all that crazy text? Meant for the notation style thing they have here? I can't read it worth a darn like it is

3. Apr 25, 2005

### retupmoc

Ye it was a poor attempt at it. Its supposed to be Bx(x+dx,y,z)-Bx(x,y,z)=dBx/dx dxdydz

4. Apr 25, 2005

### HallsofIvy

Staff Emeritus
I take it you mean $$B_x(x+dx,y,z)- B_x(x,y,z)= \frac{\partial B_x}{/partial dx}dxdydz$$. I also assume that this is the x-component of a 3-vector.

You can't "prove" it- it's not true- except approximately which is what is intended here. The "Taylor series to the first term" is just the tangent approximation to Bx. Yes, at any given (x0, y0, z0) that is $$B_x(x_0,y_0,z_0)+ \frac{\partial B_0}{\partial x}(x_0,y_0,z_0)*(x- x_0)+ +\frac{\partial B_0}{\partial y}(x_0,y_0,z_0)*(y- y_0)+\frac{\partial B_0}{\partial z}(x_0,y_0,z_0)*(z- z_0)$$.

Now, evaluate that at (x,y,z)= (x0, y0, z0) and at (x,y,z)= (x0+ dx, y0, z0) and subtract.