Using differential nad linear approximation

[dimpledur]

Homework Statement

I just have a question regarding "estimating" the value of a function at a given point. Say we have a function $f(x, y, z, w)$ and we want to know the value of that function at $f(1.99, 1.001, 8.02, 2.01)$

Can we simply do the following:

let $f=f(2, 1, 8, 2)$ and
$df=\frac{\delta f}{\delta x}dx+\frac{\delta f}{\delta y}dy+\frac{\delta f}{\delta z}dz+\frac{\delta f}{\delta w}dw$

Then say:
$f(1.99, 1.001, 8.02, 2.01)=f(2, 1, 8, 2)(1-df)$

? or is that completely wrong?

 

[HallsofIvy]

Almost. You can, by the way get [itex]\partial[/tex], which is what I think you meant rather that $\delta$, by using "\partial" in the LaTeX.

But it is not "f(1- df)" , it is simply f- df. where df is the the partial derivatives at (2, 1, 8, 2) multiplied by dx= -0.01, dy=0.001, dz= 0.02, and dw= 0.01.

 

[sandy.bridge]

Quick question, I always thought that the dx would be in percentage form. For example, (1-(1.99/2))=dx? Is it suffice to merely say dx=2-1.99?