Understanding Maclaurin Series - Math

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Question

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

Answer

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

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

Multiply by to get:

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

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

The $x^{4}$ 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 $f(x) = \sinh 3x$ .

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

Answer

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

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

Multiply by to get:

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

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

The $x^{4}$ 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 $f(x) = \sinh 3x$ .

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

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

Compare your answer with the correct one above

Question

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

Answer

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

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

Multiply by to get:

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

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

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

Compare your answer with the correct one above

Question

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

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

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

Compare your answer with the correct one above

Question

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

Answer

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

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

Multiply by to get:

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

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

The $x^{4}$ 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 $f(x) = \sinh 3x$ .

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

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

Compare your answer with the correct one above

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