# Question 326d7

Nov 22, 2017

f(x) = \sum_{n=0}^{n=\infty} a_n(x-x_0)^n; \qquad a_0 = f(x_0); \qquad a_n = 1/(n!)(\frac{d^nf}{dx^n})_{x_0}$f \left(x\right) = - 3.35528 - 0.4576 \setminus \times \left(x - 2\right) - 0.6047 \setminus \times {\left(x - 2\right)}^{2}$
$\setminus q \quad \setminus q \quad \setminus q \quad - 0.1845 \setminus \times {\left(x - 2\right)}^{3} + \setminus \cdots$

#### Explanation:

Taylor expansion of a function $f \left(x\right)$ around $x = {x}_{0}$ is given by

f(x) = \sum_{n=0}^{n=\infty} a_n(x-x_0)^n; \qquad a_0 = f(x_0); \qquad a_n = 1/(n!)(\frac{d^nf}{dx^n})_{x_0}

Given : \qquad f(x) = \ln(\sinx); \qquad x_0 = 2

$f \left(2\right) = - 3.35528$
(df)/(dx) = 1/(\sinx).\cosx = cotx; \qquad a_1 = 1/(1!)\cot(2) = -0.4576;
\frac{d^2f}{dx^2} = -\csc^2x; \qquad a_2 = -1/(2!)\csc(2)=-0.6047;
\frac{d^3f}{dx^3} = 2\csc^2x\cotx; \qquad a_3 = 2/(3!)\csc^2(2)\cot(2) -0.1845#

$f \left(x\right) = - 3.35528 - 0.4576 \setminus \times \left(x - 2\right) - 0.6047 \setminus \times {\left(x - 2\right)}^{2}$
$\setminus q \quad \setminus q \quad \setminus q \quad - 0.1845 \setminus \times {\left(x - 2\right)}^{3} + \setminus \cdots$