Differential Squared: Do We Always Assume (dx)^2=0?

[learner928]

Quick question on high order term such as (dx)^2

in taylor expansion, higher order terms such as (dx)^2 is ignored because it's too small compared to the first order term...

if there is no first order term, can second order term still be ignored?

eg. if y=(dx)^2, do we always assume y=0 because (dx)^2 is too small?

 

[Robert1986]

In a Taylor Expansion, we expand f about a point x_0. Then we take successive derivatives at x_0. So when you say "y=(dx)^2" do you mean that the first term in the Taylor expansion is f''(x_0)?

Also, the higher order terms are not ignored because the derivatives are getting smaller. The higher order terms are ignored because the terms all have a 1/n! which is getting very small, very fast as n \to \infty.

Now, to know when to cut the terms off, you need to 1)know how precise you need to be and 2)use a remainder term to get a bound on the truncation error (here is a wikipedia article: http://en.wikipedia.org/wiki/Taylor's_theorem)

 

[learner928]

sorry, maybe I should ask my question this way:

if we have y=1+dx+(dx)^2 in the limit of x->0, we ignore the term (dx)^2 and approximate y=1+dx, is this because the term (dx)^2 is too small relative to dx or is it because the term is just absolutely too small?

ie. if we have y=1+(dx)^2 in the limit of x->0, do we still ignore (dx)^2 in this case or we can't since there is no dx term.

 

[Robert1986]

You mean as dx \to 0 right? Usually the arguments you are making are more intuitive type arguments rather than rigorous arguments. But, we ignore the dx^2 because it is much smaller than dx. In your question, the dx^2 can't be ignored.