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Differential squared

  1. Jul 26, 2012 #1
    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. jcsd
  3. Jul 26, 2012 #2
    In a Taylor Expansion, we expand [itex]f[/itex] about a point [itex]x_0[/itex]. Then we take successive derivatives at [itex]x_0[/itex]. So when you say "y=(dx)^2" do you mean that the first term in the Taylor expansion is [itex]f''(x_0)[/itex]?

    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 [itex]1/n![/itex] which is getting very small, very fast as [itex]n \to \infty[/itex].

    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)
     
  4. Jul 26, 2012 #3
    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.
     
  5. Jul 26, 2012 #4
    You mean as [itex]dx \to 0 [/itex] 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.
     
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