1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Taylor Series for Any (x) = Function (x) for Any (x) ?

  1. Jul 1, 2011 #1

    morrobay

    User Avatar
    Gold Member

    When a Taylor Series is generated from a functions n derivatives at a single point,
    then is that series for any value of x equal to the original function for any value x ?
    For example graph the original function (x) from x= 0 to x = 10.
    Now plug into the Taylor Expansion for x , values from 0 to 10 and graph.
    Are the two plots approximate or equal ?
    Numerical example not to be worked but just for question :
    Suppose f(x) = 4x^3 + 8x^2 - 3x +2
     
    Last edited: Jul 2, 2011
  2. jcsd
  3. Jul 2, 2011 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Hi morrobay! :smile:

    In your example, you just have a polynomial, which means that the Taylor expansion will equal the function if the expansion is taken far enough (i.e. if you take 4 terms in the expansion).

    In general, a polynomial will always equal the Taylor expansion if you take the expansion far enough.

    A more interesting case are things like sin(x). The Taylor expansion of sine will never really equal the sine but it will converge to it. That is, the more terms in the expansion you take, the better the expansion will approximate the sine.
    For example, if you take 6 terms, then you get quite a good approximation:

    [tex]sin(x)\sim x-\frac{x^3}{3!}+\frac{x^6}{6!}[/tex]

    When you take the entire Taylor series (that is: when you take all the terms), then you get equality:

    [tex]\sin(x)=\sum_{k=0}^{+\infty}{\frac{(-1)^nx^{2n+1}}{(2n+1)!}}[/tex]

    However, there are certain functions in which the Taylor expansions do not approximate the function well. Take the Taylor expansion of log(x) at 1. For points larger than 2, the Taylor expansions form very, very bad approximations of the function.
     
  4. Jul 2, 2011 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Even more interesting is the Taylor series for [itex]f(x)= e^{-1/x}[/itex] if x is not 0, 0 if x= 0. That is infinitely differentiableat x= 0 and repeated derivatives are rational functions time [itex]e^{-1/x}[/itex] if x is not 0, 0 if x is 0. That is, the Taylor series for this function exists and is identically 0 for all 0. Clearly, f(x) is not 0 except at x= 0 so this is a function whose Taylor series exist for all x but is not equal to the function value except at x= 0.

    So, no, for general functions, the Taylor's series is not necessarily equal to the function value. The "analytic functions" are specifically defined to be those for which it is true: a function is analytic, at x= a, if and only if its Taylor's series at x= a is equal to the function value for all x in some neighborhood of a.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook