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Using Taylor's Theorem to approximate

  1. May 9, 2005 #1
    Hi Guys,

    Is there any whay I can use the following theorem to approximate the value of e^x at x=0?

    [tex] f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + ... [/tex]

    If the above function is not used for approximation, then what is it used to do?

    Thanks heaps :smile: :smile: :smile:
     
  2. jcsd
  3. May 9, 2005 #2

    arildno

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    Why bother approximating [tex]e^{0}????[/tex]
    You know what it is: [tex]e^{0}=1[/tex]

    However, since you know that, you are now in a position to approximate with only a couple of terms in the Taylor series function values like [tex]e^{0.01}, e^{0.17671234}[/tex]
     
  4. May 9, 2005 #3
    I am a n00b. Could you give me an example how Taylor approximations work?
     
  5. May 9, 2005 #4

    dextercioby

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    Sure.

    [tex] e^{0.001}\simeq e^{0}+1(0.001-0)+\frac{1}{2}(0.001-0)^{2} [/tex]

    Daniel.
     
  6. May 9, 2005 #5

    arildno

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    Let us try to approximate [tex]e^{0.01}[/tex] by forming the Taylor series about the origin:
    Then, we have:
    [tex]e^{x}=f(x)\approx{f}(0)+f'(0)(x-0)+\frac{1}{2}f''(0)(x-0)^{2}[/tex]
    where I've retained the 3 first terms in the Taylor series.
    But, now, we have:
    [tex]f(0)=e^{0}=1,f'(0)=e^{0}=1,f''(0)=e^{0}=1[/tex]
    since the derivative of the exponential function is itself
    Thus, we have:
    [tex]e^{0.01}\approx{1}+1*(0.01-0)+\frac{1}{2}*1*(0.01-0)^{2}=1+\frac{1}{100}+\frac{5}{100000}=1.01005[/tex]
     
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