How to get the following approximation?

[Blastrix91]

Homework Statement

I just stumbled upon an approximation I don't get where comes from

Homework Equations

F(x+dx) -F(x) = dF/dx dx

The Attempt at a Solution

My textbook just stated it out of nothing, so I have no idea where to start.

 

[Dick]

Homework Statement

I just stumbled upon an approximation I don't get where comes from

Homework Equations

F(x+dx) -F(x) = dF/dx dx

The Attempt at a Solution

My textbook just stated it out of nothing, so I have no idea where to start.

dF/dx is defined as the limit dx->0 of (F(x+dx)-F(x))/dx. So if dx is very small (F(x+dx)-F(x))/dx is approximately equal to dF/dx.

 

[Blastrix91]

So there are no approximation calculations behind? (Like a Taylor approximation or something?)

 

[Ray Vickson]

Homework Statement

I just stumbled upon an approximation I don't get where comes from

Homework Equations

F(x+dx) -F(x) = dF/dx dx

The Attempt at a Solution

My textbook just stated it out of nothing, so I have no idea where to start.

Get rid of the 'dx' thingys, which are just confusing the issue. What the approximation is saying is that for small |h| we have F(x+h) - F(x) ≈ F'(x) h, where F'(x) = derivative of F. This just follows from the definition of the derivative:
F'(x) = \lim_{h \to 0} \frac{F(x+h) - F(x)}{h}, so for small |h| (but not zero) we have
\frac{F(x+h) - F(x)}{h} \approx F'(x). Multiply through by h to get the stated result.

 

[Dick]

So there are no approximation calculations behind? (Like a Taylor approximation or something?)

You could get it by cutting off the Taylor series f(x+dx)=f(x)+f'(x)*dx+f''(x)*dx^2/2+... at the linear term too. But why complicate things?

 

[Blastrix91]

Ah now I get it. I guess what confused me was that it was called an approximation, but as you pointed out the small x wasn't going towards zero. Thank you both of you ^^

 

[Mark44]

Try it out for yourself with a simple function, f(x) = x², near the point (1, 1).

Near x₀ = 1, df = f'(1) * dx

If we look at x = 1.1 = 1 + 0.1 = x₀ + Δx,

df = f(x₀ + Δx) - f(x₀) ≈ f'(x₀) * Δx

Compare the exact change in f (df) given by the difference above, to the approximation given by the product shown above. They should be fairly close.

You can use the same idea to approximate a function value, given that you know the value of the function at some point, and the value of the derivative at the same point.

f(x₀ + Δx) ≈ f(x₀) + f'(x₀) * Δx

This is really nothing more than the first two terms of the Taylor's series for the function, around x₀.

 

[I like Serena]

The notation dx is called an infinitesimal.
It means that you should read everything as if the limit has been taken for dx to zero.
From that perspective they are not approximations, they are identities.