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Taylors Theorem

  1. Apr 27, 2008 #1


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    Show that the taylor series generated by [tex]f(x)=e^{x}[/tex] about x=0 converges to f(x) for ever real value of x.

    Taylors theorem states that:

    f(b) = P_{n} + \frac{e^c}{(n+1)!} b^{n+1}

    where P_n is the taylor polynomial of order 'n' and the following term is the error term and c is some value between 0 and b. If the error term approaches 0 as n approaches infinity, then the series converges to f(x).

    Does this mean that to answer the question all i need to do is show that the error term approaches zero as n gets large? If so, how would i do so?

    Thanks in advance for the help,
  2. jcsd
  3. Apr 27, 2008 #2


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    Well a non-rigorous approach would be to state that a factorial will grow faster than an exponential, especially since b is constant. The denominator of the error term will therefore outgrow the numerator and thereby kill of the whole term.
  4. Apr 27, 2008 #3


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    Yes, all you need to do is show that the error term approaches 0 as n gets large.

    Since ex is an increasing function and c< b, ec< eb so
    [tex]\frac{e^nb^{n+1}}{(n+1)!}< e^b\frac{b^{n+1}}{(n+1)!}[/tex].

    Now, to expand on exk's comment, How many factors are there in bn+1? How many factors are there in (n+1)!? What happens as soon as n> b?
  5. Apr 27, 2008 #4


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    That inequality you posted...is that assuming that b is positive? And is the e^n in the left hand side supposed to be an e^c?

    As for your second question, the bn+1 has n+2 factors? And the (n+1)! has 2n factors?

    Im still a little stuck for as how id use this information to answer the question.

    Thanks for the replies :smile:
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