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Why can linear approximation equal quadratic approximation

  1. Nov 21, 2015 #1
    Hi

    I'm having trouble visualizing why in a function such as 1/(1-x2)
    linear approximation of 1/(1-u) where u = x2 is the same as quadratic approximation of 1/(1-x2)

    The linear approximation is 1+u or 1+x2
    Quadratic approximation is the same, 1+x2

    Can someone explain to me why this happens?

    Thanks!
     
  2. jcsd
  3. Nov 21, 2015 #2

    Mark44

    Staff: Mentor

    If you carry out the long division for ##\frac 1 {1 - x^2}## you get ##\frac 1 {1 - x^2} = 1 + x^2 + x^4 + \dots##. The quadratic approximation (using terms of degree 2 or less) of ##\frac 1 {1 - x^2}## is just ##1 + x^2##.
    Similarly, if you carry out the long division for ##\frac 1 {1 - u}## you get ##\frac 1 {1 - u} = 1 + u + u^2 + \dots##. The linearization (which uses terms of degree 1 or less) is 1 + u. If ##u = x^2##, the quadratic approximation of ##\frac 1 {1 - x^2}## is the same as the linear approximation of ##\frac 1 {1 - u}##.
     
  4. Nov 22, 2015 #3
    It comes from composition laws for polynomial approximations:
    if two functions ##f## and ##g## have a polynomial approximation in 0 at order ##n##, say ## P## and ## Q## respectively, then if ##f(0) =0##, ## g \circ f ## has a polynomial approximation at order ## n##, which is the truncature at degree ##n## of ## Q \circ P##.

    In your case, ##f(x) = x^2## admits a polynomial approximation at order 2, which is ## P(x) = x^2 ##,
    as well as ##g(x) = \frac{1}{1-u}## which has polynomial approximation ##Q(x) = 1 + x + x^2 ## at order 2. Furthermore ## f(0) = 0##, so the polynomial approximation of ## g\circ f## is the truncature of ##Q\circ P = 1 + x^2 + x^4## at degree 2, which is ## 1+x^2##.
     
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