Taylor and Maclaurin Series - Math

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Question

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

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Answer

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

The second derivative of can be found as follows:

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

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

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

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

The coeficient of in the Maclaurin series is therefore

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

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Question

Give the term of the Taylor series expansion of the function $f (x) = \log_5 x$ about .

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Answer

The term of a Taylor series expansion about is

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

We can find by differentiating twice in succession:

$f (x) = \log_5 x$

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

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

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

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

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

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

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

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

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Question

Give the term of the Taylor series expansion of the function $f (x) = \log_5 x$ about .

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Answer

The term of a Taylor series expansion about is

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

We can find by differentiating twice in succession:

$f (x) = \log_5 x$

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

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

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

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

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

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

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

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

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Question

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

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Answer

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

The second derivative of can be found as follows:

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

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

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

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

The coeficient of in the Maclaurin series is therefore

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

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Question

Give the term of the Taylor series expansion of the function $f (x) = \log_5 x$ about .

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Answer

The term of a Taylor series expansion about is

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

We can find by differentiating twice in succession:

$f (x) = \log_5 x$

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

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

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

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

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

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

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

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

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Question

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

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Answer

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

The second derivative of can be found as follows:

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

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

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

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

The coeficient of in the Maclaurin series is therefore

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

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Question

Give the term of the Maclaurin series expansion of the function $f (x) = 7 ^{x}$ .

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Answer

The term of a Maclaurin series expansion has coefficient

$\frac{f ' ' (0)}{2!}$ = $\frac{f ' ' (0)}{2}$ .

We can find by differentiating twice in succession:

$f (x) = $7^{x}$$

$f ' (x) = $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( $7^{x}$ \right )= \ln 7 \cdot $7^{x}$$

$f ' '(x) = $\frac{\mathrm{d}$ }{\mathrm{d} x} \left (\ln 7 \cdot $7^{x}$ \right )$

$f ' '(x)= \ln 7\cdot $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( $7^{x}$ \right )$

$f ' '(x)= \ln 7\cdot \ln 7\cdot $7^{x}$$

$f ' '(x)=\left ( \ln 7 \right $)^{2}$ \cdot $7^{x}$$

$f ' '(0)=\left ( \ln 7 \right $)^{2}$ \cdot $7^{0}$ = \left ( \ln 7 \right $)^{2}$ \cdot 1 = \left ( \ln 7 \right $)^{2}$$

The coefficient we want is

$$\frac{f ' ' (0)}{2}$ =\frac{\left ( \ln 7 \right $)^{2}$$}{2}$ ,

so the corresponding term is $\frac{\left( \ln 7 \right $)^{2}$$}{2} $x^{2}$ .

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Question

Give the term of the Maclaurin series expansion of the function $f(x) = x $\tan^{-1}$ x$ .

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Answer

This can most easily be answered by recalling that the Maclaurin series for is

Multiply by to get:

$x $\tan^{-1}$ $x=x^{2 }$ - $\frac{1}{3}$ $x^{4 }$ + $\frac{1}{5}$ $x^{6 }$...$

The term is therefore $- $\frac{1}{3}$ $x^{4 }$$ .

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Question

Give the $x^ {3}$ term of the Maclaurin series of the function $f(x) = \sinh 3x$ .

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Answer

The $x^ {3}$ term of a Maclaurin series expansion has coefficient

$\frac{f ' '' (0)}{3!}$ = $\frac{f ' '' (0)}{6}$ .

We can find by differentiating three times in succession:

$f(x) = \sinh 3x$

$f'(x) = 3 \cosh 3x$

$f''(x) = 3\cdot 3\sinh 3x = 9\sinh 3x$

$f'''(x) = 9 \cdot 3\cosh 3x = 27 \cosh 3x$

$f'''(0) = 27 \cosh \left (3 \cdot 0 \right ) = 27 \cosh 0 = 27\cdot 1 = 27$

The term we want is therefore

$\frac{f ' '' (0)}{6}$ \cdot $x^{3}$ = $\frac{27}{6}$ \cdot $x^{3}$ = $\frac{9}{2}$ $x^{3}$

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Question

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

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Answer

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

The second derivative of can be found as follows:

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

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

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

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

The coeficient of in the Maclaurin series is therefore

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

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Question

Give the term of the Taylor series expansion of the function $f (x) = \log_5 x$ about .

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Answer

The term of a Taylor series expansion about is

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

We can find by differentiating twice in succession:

$f (x) = \log_5 x$

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

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

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

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

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

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

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

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

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Question

Give the term of the Maclaurin series expansion of the function $f(x) = x $\tan^{-1}$ x$ .

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Answer

This can most easily be answered by recalling that the Maclaurin series for is

Multiply by to get:

$x $\tan^{-1}$ $x=x^{2 }$ - $\frac{1}{3}$ $x^{4 }$ + $\frac{1}{5}$ $x^{6 }$...$

The term is therefore $- $\frac{1}{3}$ $x^{4 }$$ .

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Question

Give the $x^ {3}$ term of the Maclaurin series of the function $f(x) = \sinh 3x$ .

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Answer

The $x^ {3}$ term of a Maclaurin series expansion has coefficient

$\frac{f ' '' (0)}{3!}$ = $\frac{f ' '' (0)}{6}$ .

We can find by differentiating three times in succession:

$f(x) = \sinh 3x$

$f'(x) = 3 \cosh 3x$

$f''(x) = 3\cdot 3\sinh 3x = 9\sinh 3x$

$f'''(x) = 9 \cdot 3\cosh 3x = 27 \cosh 3x$

$f'''(0) = 27 \cosh \left (3 \cdot 0 \right ) = 27 \cosh 0 = 27\cdot 1 = 27$

The term we want is therefore

$\frac{f ' '' (0)}{6}$ \cdot $x^{3}$ = $\frac{27}{6}$ \cdot $x^{3}$ = $\frac{9}{2}$ $x^{3}$

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Question

Give the term of the Maclaurin series expansion of the function $f (x) = 7 ^{x}$ .

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Answer

The term of a Maclaurin series expansion has coefficient

$\frac{f ' ' (0)}{2!}$ = $\frac{f ' ' (0)}{2}$ .

We can find by differentiating twice in succession:

$f (x) = $7^{x}$$

$f ' (x) = $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( $7^{x}$ \right )= \ln 7 \cdot $7^{x}$$

$f ' '(x) = $\frac{\mathrm{d}$ }{\mathrm{d} x} \left (\ln 7 \cdot $7^{x}$ \right )$

$f ' '(x)= \ln 7\cdot $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( $7^{x}$ \right )$

$f ' '(x)= \ln 7\cdot \ln 7\cdot $7^{x}$$

$f ' '(x)=\left ( \ln 7 \right $)^{2}$ \cdot $7^{x}$$

$f ' '(0)=\left ( \ln 7 \right $)^{2}$ \cdot $7^{0}$ = \left ( \ln 7 \right $)^{2}$ \cdot 1 = \left ( \ln 7 \right $)^{2}$$

The coefficient we want is

$$\frac{f ' ' (0)}{2}$ =\frac{\left ( \ln 7 \right $)^{2}$$}{2}$ ,

so the corresponding term is $\frac{\left( \ln 7 \right $)^{2}$$}{2} $x^{2}$ .

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Question

Give the term of the Maclaurin series expansion of the function $f (x) = 7 ^{x}$ .

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Answer

The term of a Maclaurin series expansion has coefficient

$\frac{f ' ' (0)}{2!}$ = $\frac{f ' ' (0)}{2}$ .

We can find by differentiating twice in succession:

$f (x) = $7^{x}$$

$f ' (x) = $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( $7^{x}$ \right )= \ln 7 \cdot $7^{x}$$

$f ' '(x) = $\frac{\mathrm{d}$ }{\mathrm{d} x} \left (\ln 7 \cdot $7^{x}$ \right )$

$f ' '(x)= \ln 7\cdot $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( $7^{x}$ \right )$

$f ' '(x)= \ln 7\cdot \ln 7\cdot $7^{x}$$

$f ' '(x)=\left ( \ln 7 \right $)^{2}$ \cdot $7^{x}$$

$f ' '(0)=\left ( \ln 7 \right $)^{2}$ \cdot $7^{0}$ = \left ( \ln 7 \right $)^{2}$ \cdot 1 = \left ( \ln 7 \right $)^{2}$$

The coefficient we want is

$$\frac{f ' ' (0)}{2}$ =\frac{\left ( \ln 7 \right $)^{2}$$}{2}$ ,

so the corresponding term is $\frac{\left( \ln 7 \right $)^{2}$$}{2} $x^{2}$ .

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Question

Give the term of the Maclaurin series expansion of the function $f(x) = x $\tan^{-1}$ x$ .

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Answer

This can most easily be answered by recalling that the Maclaurin series for is

Multiply by to get:

$x $\tan^{-1}$ $x=x^{2 }$ - $\frac{1}{3}$ $x^{4 }$ + $\frac{1}{5}$ $x^{6 }$...$

The term is therefore $- $\frac{1}{3}$ $x^{4 }$$ .

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Question

Give the $x^ {3}$ term of the Maclaurin series of the function $f(x) = \sinh 3x$ .

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Answer

The $x^ {3}$ term of a Maclaurin series expansion has coefficient

$\frac{f ' '' (0)}{3!}$ = $\frac{f ' '' (0)}{6}$ .

We can find by differentiating three times in succession:

$f(x) = \sinh 3x$

$f'(x) = 3 \cosh 3x$

$f''(x) = 3\cdot 3\sinh 3x = 9\sinh 3x$

$f'''(x) = 9 \cdot 3\cosh 3x = 27 \cosh 3x$

$f'''(0) = 27 \cosh \left (3 \cdot 0 \right ) = 27 \cosh 0 = 27\cdot 1 = 27$

The term we want is therefore

$\frac{f ' '' (0)}{6}$ \cdot $x^{3}$ = $\frac{27}{6}$ \cdot $x^{3}$ = $\frac{9}{2}$ $x^{3}$

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Question

Give the term of the Maclaurin series expansion of the function $f(x) = x $\tan^{-1}$ x$ .

Tap to reveal answer

Answer

This can most easily be answered by recalling that the Maclaurin series for is

Multiply by to get:

$x $\tan^{-1}$ $x=x^{2 }$ - $\frac{1}{3}$ $x^{4 }$ + $\frac{1}{5}$ $x^{6 }$...$

The term is therefore $- $\frac{1}{3}$ $x^{4 }$$ .

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Question

Give the $x^ {3}$ term of the Maclaurin series of the function $f(x) = \sinh 3x$ .

Tap to reveal answer

Answer

The $x^ {3}$ term of a Maclaurin series expansion has coefficient

$\frac{f ' '' (0)}{3!}$ = $\frac{f ' '' (0)}{6}$ .

We can find by differentiating three times in succession:

$f(x) = \sinh 3x$

$f'(x) = 3 \cosh 3x$

$f''(x) = 3\cdot 3\sinh 3x = 9\sinh 3x$

$f'''(x) = 9 \cdot 3\cosh 3x = 27 \cosh 3x$

$f'''(0) = 27 \cosh \left (3 \cdot 0 \right ) = 27 \cosh 0 = 27\cdot 1 = 27$

The term we want is therefore

$\frac{f ' '' (0)}{6}$ \cdot $x^{3}$ = $\frac{27}{6}$ \cdot $x^{3}$ = $\frac{9}{2}$ $x^{3}$

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Question

Give the term of the Maclaurin series expansion of the function $f (x) = 7 ^{x}$ .

Tap to reveal answer

Answer

The term of a Maclaurin series expansion has coefficient

$\frac{f ' ' (0)}{2!}$ = $\frac{f ' ' (0)}{2}$ .

We can find by differentiating twice in succession:

$f (x) = $7^{x}$$

$f ' (x) = $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( $7^{x}$ \right )= \ln 7 \cdot $7^{x}$$

$f ' '(x) = $\frac{\mathrm{d}$ }{\mathrm{d} x} \left (\ln 7 \cdot $7^{x}$ \right )$

$f ' '(x)= \ln 7\cdot $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( $7^{x}$ \right )$

$f ' '(x)= \ln 7\cdot \ln 7\cdot $7^{x}$$

$f ' '(x)=\left ( \ln 7 \right $)^{2}$ \cdot $7^{x}$$

$f ' '(0)=\left ( \ln 7 \right $)^{2}$ \cdot $7^{0}$ = \left ( \ln 7 \right $)^{2}$ \cdot 1 = \left ( \ln 7 \right $)^{2}$$

The coefficient we want is

$$\frac{f ' ' (0)}{2}$ =\frac{\left ( \ln 7 \right $)^{2}$$}{2}$ ,

so the corresponding term is $\frac{\left( \ln 7 \right $)^{2}$$}{2} $x^{2}$ .

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