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Taylor polynomials

  1. Dec 14, 2011 #1

    georg gill

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    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


    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)
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