Question

Use the definition of taylor series to find the taylor series (centered at c) for the function. f (x)=6/x^1 c=1

[infinity]
f(x) =Σ
n=0

Answer 1

`f(x)=\displaystyle (6)/(x)=\sum^\infty_(n=0)6(-1)^n(x-1)^n`

Explanation:

Recall the formula for Taylor Series:

`\displaystyle f(x)=f(c)+f'(c)(x-c)+(f''(c)(x-c)^2)/(2!)+(f'''(c)(x-c)^3)/(3!)+...+(f^n(c)(x-c)^n)/(n!)=\sum^\infty_(n=0)(f^n(c))/(n!)(x-c)^n`

Determine the derivative function fⁿ(c):

`\displaystyle f(c)=(6)/(c)=(6)/(1)=6\\ \\f'(c)=-(6)/(c^2)=-(6)/(1^2)=-6\\ \\f''(c)=(12)/(c^3)=(12)/(1^3)=12\\\\f'''(c)=-(36)/(x^4)=-(36)/(1^4)=-36\\\\....\\\\f^n(c)=6(-1)^(n)n!`

Therefore, the infinite series can be written as:

`\displaystyle f(x)=(6)/(x)=\sum^\infty_(n=0)(6(-1)^nn!)/(n!)(x-1)^n=\sum^\infty_(n=0)6(-1)^n(x-1)^n`