Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Taylor polynomials

  1. Dec 14, 2011 #1

    georg gill

    User Avatar
    Gold Member

    http://bildr.no/view/1030479

    The link above, it is my own and it is a bit disorderly, I think should explain taylor polynomials. In one assignent one had an assignment to derive taylor polynomials for

    [tex]cost^2[/tex]

    If one use the derivation rules with chain one get 2t for first derivative and [tex]4t^4[/tex]
    for second and so on. If t=0 and we are looking at a maclaurin series then every term except the first cos(0) becomes zero. I can understand that one could say [tex]u=t^2[/tex] and derieve taylor series for cosx and just use afterwards [tex]u=t^2[/tex].

    But what is even more confusing is that in one other assignment one were too find taylor polynomial for [tex]y=ln(x^2)[/tex]. Here one uses chain rule and it does work:

    [tex]y'=\frac{2}{x}[/tex] [tex]y''=\frac{-2}{x^2}[/tex] [tex]y''=\frac{4}{x^4}[/tex]


    What is the difference?
     
  2. jcsd
  3. Dec 14, 2011 #2
    Try to calculate the derivatives more carefully, because what you stated is false. The first term is not the only non-zero term (there are infinitely many non-zero terms). The chain rules gives a correct result. (perhaps you are forgetting about the product rule)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Taylor polynomials
  1. Taylor Polynomials (Replies: 2)

  2. Taylor Polynomials (Replies: 2)

Loading...