Taylor Series for Showing B_{x}(x+dx)-B_{x}(x) Approximation

[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

 

[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

 

[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

 

[HallsofIvy]

I take it you mean [tex]B_x(x+dx,y,z)- B_x(x,y,z)= \frac{\partial B_x}{/partial dx}dxdydz[/tex]. 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 B_x. Yes, at any given (x₀, y₀, z₀) that is [tex]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)[/tex].

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