Question

$\\ \rm \clubsuit\ \huge{Hola\ Brainlians\ !}$
Find the Taylor series generated by the function f(x) = $\rm \dfrac{1}{x^2}$ at x = 1
$\\ \rm \spadesuit\ \orange{Best\ of\ luck\ Guys\ !}$

Answer 1

Question :

Find the Taylor series generated by the function f(x) = $\sf \tiny{\dfrac{1}{x^2}}$ at x = 1 .

Solution :

Given f(x) = $\sf \tiny{\dfrac{1}{x^2}}$ , a = 1

The Taylor series generated by f(x) at x = a is given by ,

$\bullet\ \; \displaystyle \sf \sum \limits_{k=0}^{\infty} \dfrac{f^{(k)}(a)}{k!}(x-a)^k$

We have ,

$\bullet\ \; \sf f(x) = \dfrac{1}{x^2} =(-1)^0 \dfrac{1!}{x^{0+2}}$

$\to \sf f'(x)=- \dfrac{2}{x^3} = (-1)^1 \dfrac{2!}{x^{1+2}}$

$\to \sf f''(x)=\dfrac{6}{x^4} = (-1)^2 \dfrac{3!}{x^{2+2}}$

$\to \sf f'''(x)=- \dfrac{24}{x^5} = (-1)^3 \dfrac{4!}{x^{3+2}}$

$\to \sf f^{IV}(x) =\dfrac{120}{x^6} = (-1)^4 \dfrac{5!}{x^{4+2}}$

. . .

. .

$\leadsto \sf f^{(k)}(x) = (-1)^k \dfrac{(k+1)!}{x^{(k+2)}}$

So ,

$\mapsto \sf f^{(k)}(1) = (-1)^k (k+1)!$

$\mapsto \sf \dfrac{f^{(k)}(1)}{k!}= (-1)^k (k+1)$

Therefore , the Taylor series generated by f(x) = $\sf \tiny{\dfrac{1}{x^2}}$ at x = 1 is ,

$\to \displaystyle \sf \sum \limits_{k=0}^{\infty} \dfrac{f^{(k)}(1)}{k!}(x-1)^k$

$\displaystyle \sf \to \sum \limits_{k=0}^{\infty} (-1)^k (k+1)(x-1)^k$

$\\\pink{\leadsto \sf 1-2(x-1)+3(x-1)^2-4(x-1)^3+\ .\ .\ .}$

Answer 2

Given Question :-

Find the Taylor series generated by the function

$f(x) = \rm \dfrac{1}{x^2} \: at \: x \: = \: 1$

Solution :-

$f(x) = \rm \dfrac{1}{x^2} - - (1)$

$\tt : \implies \: \boxed{ \pink{ \tt \: f(1) = 1}}$

On differentiating w. r. t. x, we get

$\tt : \implies \: f'(x) \: = \: - \dfrac{2}{ {x}^{3} } - - (2)$

$\tt : \implies \: \boxed{\pink{ \tt \: f'(1) \: = \: - 2}}$

Now, on differentiating w.r.t. x, equation (2), we get

$\tt : \implies \: f''(x) = \dfrac{( - 2)( - 3)}{ {x}^{4} } = \dfrac{6}{ {x}^{4} } - - (3)$

$\tt : \implies \: \boxed{ \pink{ \tt \: f''(1) = 6}}$

Now, on differentiating w. r. t. x, equation (3), we get

$\tt : \implies \: f'''(x) \: = \: \dfrac{(6)( - 4)}{ {x}^{5} } = - \dfrac{24}{ {x}^{5} } - - (4)$

$\tt : \implies \: \boxed{ \pink{ \tt \: f'''(1) = - 24}}$

Now, on differentiating w. r. t. x, equation (4), we get

$\tt : \implies \: f''''(x) \: = \dfrac{( - 24)( - 5)}{ {x}^{6} } = \dfrac{120}{ {x}^{6} }$

$\tt : \implies \: \boxed{ \pink{ \tt \: f''''(1) = 120}}$

Therefore,

By Taylor Series,

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

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

$\rm \: \dfrac{1}{ {x}^{2} } = 1 + ( - 2)(x - 1) + \frac{6}{2} {(x - 1)}^{2} + \frac{( - 24)}{6} {(x - 1)}^{3} + \frac{120}{24} {(x - 1) }^{4} + ..$

$\rm \: \dfrac{1}{ {x}^{2} } = 1 - 2(x - 1) + 3 {(x - 1)}^{2} - 4 {(x - 1)}^{3} + 5 {(x - 1)}^{4} + ..$