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Math Help - Taylor Poly's and l'Hop

  1. #1
    Newbie Belzelga's Avatar
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    Taylor Poly's and l'Hop

    Could somebody help me with this problem?

    Find the Taylor polynomials the 2nd Taylor polynomial \begin{aligned} T_{f,0}^{(2)} (x)\end{aligned} around 0 for the following function

    f(x) = (1)/(1+x)

    and this nasty question...

    Using l'Hoptal's rule...

    lim (x->infin) [((cos) (a/x))^x^2] where a>0
    lim (x->0) [((log(1+ax))/x) where a>0
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  2. #2
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    Quote Originally Posted by Belzelga View Post
    Could somebody help me with this problem?

    Find the Taylor polynomials the 2nd Taylor polynomial \begin{aligned} T_{f,0}^{(2)} (x)\end{aligned} around 0 for the following function

    f(x) = (1)/(1+x) Mr F says: This looks like a routine application of the formula. Where are you stuck?

    and this nasty question...

    Using l'Hoptal's rule...

    lim (x->infin) [((cos) (a/x))^x^2] where a>0
    lim (x->0) [((log(1+ax))/x) where a>0
    \lim_{x \rightarrow \infty} \left[ \cos \left( \frac{a}{x} \right) \right]^{x^2} = \lim_{t \rightarrow 0} \left[ \cos (at) \right]^{1/t^2} = \lim_{t \rightarrow 0} \, e^{\frac{\ln \cos (at)}{t^2}} = e^{\lim_{t \rightarrow 0}{\frac{\ln \cos (at)}{t^2}}} ......

    ---------------------------------------------------------------------------------------------

    \lim_{x \rightarrow 0} \frac{\ln(1+ax)}{x} is a routine application of l'Hopital's Rule if you're feeling lazy.

    Otherwise you can substitute the Maclaurin series for \ln (1 + ax) = ax - \frac{a^2x^2}{2} + ......:

    \lim_{x \rightarrow 0} \frac{ax - \frac{a^2x^2}{2} + .....}{x} = a.
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  3. #3
    Newbie Belzelga's Avatar
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    Hi Fant

    Just realised the first part was simple apparently so I got the following with few other examples:
    ===
    Using l'hopital rule to compute limits
    ===
    lim x->0 (1 - cos x)/ x^2

    lim x->0 (sin x)/ 2x
    lim x->0 (sin x)/ x = 1
    lim x->0 (cos x)/ 2 = 1/2
    ===
    ===
    lim x->0 (a^x - 1)/ k*x
    lim x->0 [xa^(x-1) - 1]/ k = 1
    ===
    ===
    And the ending of
    lim x->infin (cos (a/x))^(x^2)

    Is the limit 1/2?
    ===================

    Now the Taylor Polynomial

    f(x) = (1)/(1+x)
    f'(x) = (-1)(1+x)^-2
    f''(x) = (-2)(1+x)^-\3

    f(0) = 1
    f'(0) = -1
    f''(0) = -2

    Which will gave me (1/0!)-(1/1!)x-(2/2!)x^2 = 1-x-x^2
    ====

    How did I do?
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  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by Belzelga View Post
    Hi Fant


    Now the Taylor Polynomial

    f(x) = (1)/(1+x)
    f'(x) = (-1)(1+x)^-2
    f''(x) = (-2)(1+x)^-\3

    f(0) = 1
    f'(0) = -1
    f''(0) = -2

    Which will gave me (1/0!)-(1/1!)x-(2/2!)x^2 = 1-x-x^2
    ====

    How did I do?
    f''(x)=(-2)(-1)(1+x)^{-3}

    f''(0)=2
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  5. #5
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Also note that

    \frac{1}{1+x}=\frac{1}{1-(-x)}

    This is the form of the geometric series with r=-x

    so we get

    \frac{1}{1+x}=\sum_{n=0}^{\infty}(-x)^n=\sum_{n=0}^{\infty}(-1)^n(x)^n=1-x+x^2-x^3...
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