Answer to Question #137276 in Calculus for Usman

Question

Find taylor series
F(x) = (6-x)^-7 About x=4

Answer 1

Given expression is $f(x) = \frac{1}{(6-x)^7}$

First derivative, $f'(x) = \frac{7}{(x-6)^8}$

Second derivative $f''(x) = -\frac{56}{(x-6)^9}$

Third derivative $f'''(x) = \frac{504}{(x-6)^{10}}$

Fourth derivative $f''''(x) = -\frac{5040}{(x-6)^{11}}$

and so on .........

Value of f(x) at x=4, $f(4) = \frac{1}{128}$

Value of first derivative at x=4, $f'(4) = \frac{7}{256}$

Value of second derivative at x=4, $f''(4) = \frac{7}{64}$

Value of third derivative at x=4, $f'''(4) = \frac{63}{128}$

Value of fourth derivative at x=4, $f''''(4) = \frac{315}{128}$

Taylor series of f(x)

$f(x) = f(4) + \frac{f'(4)}{1!}(x-4) + \frac{f''(4)}{2!}(x-4)^2 + \frac{f'''(4)}{3!}(x-4)^3 + \frac{f''''(4)}{4!}(x-4)^4+ .....$

$f(x) = \frac{1}{128} + \frac{7}{256}(x-4) + \frac{7}{128}(x-4)^2 + \frac{21}{259}(x-4)^3 + \frac{105}{1024}(x-4)^4 + ......$