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High School Math : Understanding Maclaurin Series

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Example Question #1 : Understanding Maclaurin Series

Give the $x^{4}$ $\displaystyle x^{4}$ term of the Maclaurin series expansion of the function $f(x) = x \tan^{-1} x$ $\displaystyle f(x) = x \tan^{-1} x$ .

Possible Answers:

$- \frac{3}{4} x^{4 }$ $\displaystyle - \frac{3}{4} x^{4 }$

$- \frac{1}{3} x^{4 }$ $\displaystyle - \frac{1}{3} x^{4 }$

$\displaystyle 0$

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

$- \frac{1}{12} x^{4 }$ $\displaystyle - \frac{1}{12} x^{4 }$

Correct answer:

$- \frac{1}{3} x^{4 }$ $\displaystyle - \frac{1}{3} x^{4 }$

Explanation:

This can most easily be answered by recalling that the Maclaurin series for $\tan^{-1} x$ $\displaystyle \tan^{-1} x$ is

$\tan^{-1} x = x - \frac{1}{3} x^{3 } + \frac{1}{5} x^{5 }...$ $\displaystyle \tan^{-1} x = x - \frac{1}{3} x^{3 } + \frac{1}{5} x^{5 }...$

Multiply by $\displaystyle x$ to get:

$x\tan^{-1} x = x\cdot x - x\cdot \frac{1}{3} x^{3 } + x\cdot \frac{1}{5} x^{5 }...$ $\displaystyle x\tan^{-1} x = x\cdot x - x\cdot \frac{1}{3} x^{3 } + x\cdot \frac{1}{5} x^{5 }...$

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

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

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Example Question #2 : Understanding Maclaurin Series

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

Possible Answers:

$-9x^{3}$ $\displaystyle -9x^{3}$

$9x^{3}$ $\displaystyle 9x^{3}$

$\frac{9}{2} x^{3}$ $\displaystyle \frac{9}{2} x^{3}$

$-\frac{9}{2} x^{3}$ $\displaystyle -\frac{9}{2} x^{3}$

$\displaystyle 0$

Correct answer:

$\frac{9}{2} x^{3}$ $\displaystyle \frac{9}{2} x^{3}$

Explanation:

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

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

We can find f'''(x) $\displaystyle f'''(x)$ by differentiating three times in succession:

$f(x) = \sinh 3x$ $\displaystyle f(x) = \sinh 3x$

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

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

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

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

The term we want is therefore

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

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High School Math : Understanding Maclaurin Series

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