Q43P f(x)=cotx,a=蟺2... [FREE SOLUTION]

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Chapter 1: Q43P (page 32)

$f (x) = c o t x, a = \frac{蟺}{2}$

Short Answer

Expert verified

The first few terms of Taylor鈥檚 series expansion are:

$\cot (x) = (x - \frac{蟺}{2}) - \frac{{(x - \frac{蟺}{2})}^{3}}{3} - \frac{2 {(x - \frac{蟺}{2})}^{5}}{15} - 鈥�$

Step by step solution

Given Information

The given function

$f (x) = c o t 鈥� x$

Definition of Taylor’s series

The Taylor鈥檚 series is a power series that yields the expansion of a function in the vicinity of a point if the function is continuous, all of its derivatives exist, and the series converges.

Rewrite cotxin required form

Rewrite and use $\cot (\frac{蟺}{2} + x) = - \tan (x)$

$\cot (x) = \cot (\frac{蟺}{2} + (x - \frac{蟺}{2}))$

$\cot (\frac{蟺}{2} + (x - \frac{蟺}{2})) = - \tan (x - \frac{蟺}{2})$

Find the Taylor's series forcotx .

Use the Maclaurin series of $\tan x$ and replace $x$ by $(x - \frac{蟺}{2})$ ,

$\tan (x) = x + \frac{x^{3}}{3} + \frac{2 x^{5}}{15} + 鈥�$

$- \tan (x) = - x - \frac{x^{3}}{3} - \frac{2 x^{5}}{15} - 鈥�$

$- \tan (x - \frac{蟺}{2}) = - (x - \frac{蟺}{2}) - \frac{{(x - \frac{蟺}{2})}^{3}}{3} - \frac{2 {(x - \frac{蟺}{2})}^{5}}{15} - 鈥�$

$\cot (x) = (x - \frac{蟺}{2}) - \frac{{(x - \frac{蟺}{2})}^{3}}{3} - \frac{2 {(x - \frac{蟺}{2})}^{5}}{15} - 鈥�$

Write first few terms of Taylor鈥檚 series about $a = \frac{蟺}{2}$ ,

role="math" localid="1657419575245" $\cot (x) = (x - \frac{蟺}{2}) - \frac{{(x - \frac{蟺}{2})}^{3}}{3} - \frac{2 {(x - \frac{蟺}{2})}^{5}}{15} - 鈥�$

Hence, first few terms of Taylor鈥檚 series expansion are:

$\cot (x) = (x - \frac{蟺}{2}) - \frac{{(x - \frac{蟺}{2})}^{3}}{3} - \frac{2 {(x - \frac{蟺}{2})}^{5}}{15} - 鈥�$

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91影视

$f (x) = c o t x, a = \frac{蟺}{2}$

Short Answer

Step by step solution

Given Information

Definition of Taylor’s series

Rewrite cotxin required form

Find the Taylor's series forcotx .

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