High School Math : Understanding Taylor Series

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #1 : Understanding Taylor Series

Give the \displaystyle x^{2}  term of the Maclaurin series of the function \displaystyle f \left ( x\right ) = \frac{1}{x-2}

Possible Answers:

\displaystyle -\frac{1}{6} x^{2}

\displaystyle -\frac{1}{2} x^{2}

\displaystyle -\frac{1}{8} x^{2}

\displaystyle -\frac{1}{12} x^{2}

\displaystyle -\frac{1}{4} x^{2}

Correct answer:

\displaystyle -\frac{1}{8} x^{2}

Explanation:

The \displaystyle x^{2}  term of the Maclaurin series of a function \displaystyle f has coefficient \displaystyle \frac{f '' (0) }{2!} = \frac{f '' (0) }{2}

The second derivative of \displaystyle f can be found as follows:

\displaystyle f \left ( x\right ) = \frac{1}{x-2} = (x-2) ^{-1}

\displaystyle f' \left ( x\right ) = -1\cdot (x-2)^{-2} = -1(x-2)^{-2}

\displaystyle f'' \left ( x\right ) = (-2) ( -1(x-2)^{-3}) = 2(x-2)^{-3} = \frac{2}{(x-2)^{3}}

\displaystyle f'' \left ( 0 \right ) =\frac{2}{(0-2)^{3}} = -\frac{1}{4}

The coeficient of \displaystyle x^{2} in the Maclaurin series is therefore

\displaystyle \frac{f '' (0) }{2} = \frac{-\frac{1}{4}}{2} = - \frac{1}{8}

Example Question #2 : Understanding Taylor Series

Give the \displaystyle (x-1)^2  term of the Taylor series expansion of the function \displaystyle f (x) = \log_5 x about \displaystyle x = 1.

Possible Answers:

\displaystyle \frac{2}{ \ln 5 } (x-1)^{2}

\displaystyle \frac{1}{ \ln 25 } (x-1)^{2}

\displaystyle \frac{\ln 5}{ 2 } (x-1)^{2}

\displaystyle \ln 25\cdot (x-1)^{2}

\displaystyle - \frac{1}{ \ln 25 } (x-1)^{2}

Correct answer:

\displaystyle - \frac{1}{ \ln 25 } (x-1)^{2}

Explanation:

The \displaystyle (x-1)^{2} term of a Taylor series expansion about \displaystyle x = 1 is

\displaystyle \frac{f ' ' (1)}{2!} (x-1)^{2}= \frac{f ' ' (1)}{2}(x-1)^{2}.

We can find \displaystyle f''(x) by differentiating twice in succession:

\displaystyle f (x) = \log_5 x

 

\displaystyle f'(x) = \frac{1}{x \ln 5}

\displaystyle f'(x) =\frac{1}{ \ln 5} \cdot x^{-1}

 

\displaystyle f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left (\frac{1}{ \ln 5} \cdot x^{-1} \right )

\displaystyle f''(x) =\frac{1}{ \ln 5}\cdot \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{-1} \right )

\displaystyle f''(x) =\frac{1}{ \ln 5} \cdot \left ( -1 \cdot x^{-2} \right )

\displaystyle f''(x) =- \frac{1}{ \ln 5 \cdot x^{2}}

\displaystyle f''(1) =- \frac{1}{ \ln 5 \cdot 1^{2}} = - \frac{1}{ \ln 5 }

so the \displaystyle (x-1)^2 term is \displaystyle \frac{f''(1) }{2} \; (x-1)^{2} = - \frac{1}{2 \ln 5 } (x-1)^{2} = - \frac{1}{ \ln 25 } (x-1)^{2}

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