# Differential squared

1. Jul 26, 2012

### 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?

2. Jul 26, 2012

### 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)

3. Jul 26, 2012

### 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.

4. Jul 26, 2012

### 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.