Question

Question #326d7

Answer 1

`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(x) = -3.35528 - 0.4576\times(x-2) - 0.6047\times(x-2)^2`
`\qquad \qquad \qquad - 0.1845\times(x-2)^3 + \cdots`

Explanation:

Taylor expansion of a function `f(x)` 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(2) = -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(x) = -3.35528 - 0.4576\times(x-2) - 0.6047\times(x-2)^2`
`\qquad \qquad \qquad - 0.1845\times(x-2)^3 + \cdots`