# Find the Taylor series (check my work)?

## See the answers for work... **can someone help me work through the derivatives part so I can construct an ${f}^{n} \left(a\right)$ rule? a. $f \left(x\right) = \frac{1}{x} ^ 2$, $a = 4$ b. $f \left(x\right) = \setminus \sqrt{x}$, $a = 2$

May 3, 2018

For part A
Finished -- can someone check the steps overall?

#### Explanation:

$f ' = - 2 {x}^{-} 3$
$\setminus \Rightarrow f ' \left(4\right) = - 2 {\left(4\right)}^{-} 3$

$f ' ' = \left(- 2\right) \left(- 3\right) {x}^{-} 4$
$\setminus \Rightarrow f ' ' \left(4\right) = \left(- 2\right) \left(- 3\right) {\left(4\right)}^{-} 4$

$f ' ' ' = \left(- 2\right) \left(- 3\right) \left(- 4\right) {x}^{-} 5$
$\setminus \Rightarrow f ' ' ' \left(4\right) = \left(- 2\right) \left(- 3\right) {\left(4\right)}^{-} 5$

f^n(4)=((-1)^n(n-1)!)/(4^n)

C_n=1/(n!)*f^n(a)=((-1)^n(n-1)!)/4^n*1/(n!)=((-1)^n(n-1))/(4^n)
$\setminus \therefore f \left(x\right) = \setminus {\sum}_{n = 0}^{\setminus} \infty {C}_{n} {\left(x - a\right)}^{n} = \setminus {\sum}_{n = 0}^{\setminus} \infty {\left(- 1\right)}^{n + 1} \frac{1}{{4}^{n}} {\left(x - 4\right)}^{n}$, which is $\frac{1}{x} ^ 2$ centered at $a = 4$

May 3, 2018

For part B
Literally confused on step one. Someone help me lol

#### Explanation:

$f ' = \frac{1}{2 \setminus \sqrt{x}} = \frac{1}{2} {\left(x\right)}^{- \frac{1}{2}}$
$f ' ' = - \frac{1}{4 {x}^{\frac{3}{2}}} = \frac{1}{4} {\left(x\right)}^{- \frac{3}{2}} = \frac{1}{2} \left(- \frac{1}{2}\right) {x}^{- \frac{3}{2}}$
$f ' ' ' = \frac{3}{8 {x}^{\frac{5}{2}}} = \frac{3}{8} {\left(x\right)}^{- \frac{5}{2}} = \frac{1}{2} \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right) {x}^{- \frac{5}{2}}$

...
${f}^{4} = \frac{1}{2} \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right) \left(- \frac{5}{2}\right) {x}^{- \frac{7}{2}}$
${f}^{5} = \frac{1}{2} \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right) \left(- \frac{5}{2}\right) \left(- \frac{7}{2}\right) {x}^{- \frac{9}{2}}$

exponent seems to increase by $- \frac{1}{2} - n$?

May 4, 2018

A) $f \left(x\right) = \frac{1}{x} ^ 2$ pivoted about $x = 4$

$f \left(x\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \frac{\left(n + 1\right)}{{4}^{n + 2}} {\left(x - 4\right)}^{n}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{16} - \frac{x - 4}{32} + \frac{3 {\left(x - 4\right)}^{2}}{256} - {\left(x - 4\right)}^{3} / 256 + \ldots$

B) $f \left(x\right) = \sqrt{x}$ pivoted about $x = 2$

$f \left(x\right) = \sqrt{2} + \frac{x - 2}{2 \sqrt{2}} - {\left(x - 2\right)}^{2} / \left(16 \sqrt{2}\right) + {\left(x - 2\right)}^{3} / \left(64 \sqrt{2}\right) - \frac{5 {\left(x - 2\right)}^{4}}{4096 \sqrt{2}} + \ldots$

#### Explanation:

We seek:

A) TS of $f \left(x\right) = \frac{1}{x} ^ 2$ pivoted about $x = 4$
B) TS of $f \left(x\right) = \sqrt{x}$ pivoted about $x = 2$

We use the general definition of a TS pivot about $x = a$:

 f(x) = f(a) + (f^((1))(a))/(1!)(x-a) + (f^((2))(a))/(2!)(x-a)^2 +
 \ \ \ \ \ \ \ \ \ \ \ + (f^((3))(a))/(3!)(x-a)^3 + ... +
 \ \ \ \ \ \ \ \ \ \ \ + (f^((n))(a))/(n!)(x-a)^n + ...

Part (A):

Differentiating wrt to $x$ we compute the first few derivatives:

${f}^{\left(0\right)} \left(x\right) = \frac{1}{x} ^ 2$

 f^((1))(x) = (-2)/x^3 = (-)(2!)/(x^3)

 f^((2))(x) = (-2)(-3)/x^4 = (+)(3!)/(x^4)

 f^((3))(x) = (-2)(-3)(-4)/x^5 = (-)(4!)/(x^5)

And we reasonably conclude that:

 f^((n))(x) = (-1)^n((n+1)!)/(x^(n+2))

Allowing us to construct a Taylor Series, pivoted about $x = 4$:

 f(x) = f^((0))(4) + (f^((1))(4))/(1!)(x-4) + (f^((2))(4))/(2!)(x-4)^2 +
 \ \ \ \ \ \ \ \ \ \ \ + (f^((3))(4))/(3!)(x-4)^3 + ... +
 \ \ \ \ \ \ \ \ \ \ \ + (f^((n))(4))/(n!)(x-4)^n + ...

 \ \ \ \ \ \ \ = 1/4^2 + (-(2!)/(4^3))/(1!)(x-4) + ((3!)/(4^4))/(2!)(x-4)^2 +
 \ \ \ \ \ \ \ \ \ \ \ + (-(4!)/(4^5))/(3!)(x-4)^3 + ... +
 \ \ \ \ \ \ \ \ \ \ \ + ((-1)^n((n+1)!)/(4^(n+2)))/(n!)(x-4)^n + ...

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{4} ^ 2 - \left(\frac{2}{{4}^{3}}\right) \left(x - 4\right) + \left(\frac{3}{{4}^{4}}\right) {\left(x - 4\right)}^{2} +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus - \left(\frac{4}{{4}^{5}}\right) {\left(x - 4\right)}^{3} + \ldots +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus + {\left(- 1\right)}^{n} \frac{\left(n + 1\right)}{{4}^{n + 2}} {\left(x - 4\right)}^{n} + \ldots$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{16} - \frac{x - 4}{32} + \frac{3 {\left(x - 4\right)}^{2}}{256} - {\left(x - 4\right)}^{3} / 256 + \ldots$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \frac{\left(n + 1\right)}{{4}^{n + 2}} {\left(x - 4\right)}^{n}$

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Part (B):

Differentiating wrt to $x$ we compute the first few derivatives:

${f}^{\left(0\right)} \left(x\right) = \sqrt{x}$

${f}^{\left(1\right)} \left(x\right) = \left(\frac{1}{2}\right) {x}^{- \frac{1}{2}} = \frac{1}{2 {x}^{\frac{1}{2}}} =$

${f}^{\left(2\right)} \left(x\right) = \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) {x}^{- \frac{3}{2}} = - \frac{1}{{2}^{2} {x}^{\frac{3}{2}}}$

${f}^{\left(3\right)} \left(x\right) = \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right) {x}^{- \frac{5}{2}} = \frac{3}{{2}^{3} {x}^{\frac{5}{2}}}$

${f}^{\left(4\right)} \left(x\right) = \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right) \left(- \frac{5}{2}\right) {x}^{- \frac{7}{2}} = \frac{1.3 .5}{{2}^{4} {x}^{\frac{7}{2}}}$

${f}^{\left(5\right)} \left(x\right) = \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right) \left(- \frac{5}{2}\right) \left(- \frac{7}{2}\right) {x}^{- \frac{9}{2}} = \frac{1.3 .5 .7}{{2}^{5} {x}^{\frac{9}{2}}}$

Allowing us to construct a Taylor Series, pivoted about $x = 2$:

 f(x) = f^((0))(2) + (f^((1))(2))/(1!)(x-2) + (f^((2))(2))/(2!)(x-2)^2 +
 \ \ \ \ \ \ \ \ \ \ \ + (f^((3))(2))/(3!)(x-2)^3 + (f^((4))(2))/(4!)(x-2)^4 + ... +

 \ \ \ \ \ \ \ = sqrt(2) + (1/(2 \ 2^(1/2)))/(1!)(x-2) + (-1/(2^2 \ 2^(3/2)))/(2!)(x-2)^2 +
 \ \ \ \ \ \ \ \ \ \ \ + (3/(2^3 \ 2^(5/2)))/(3!)(x-2)^3 + ((1.3.5)/(2^4 \ 2^(7/2)))/(4!)(x-2)^4 + ... +

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{2} + \frac{x - 2}{2 \sqrt{2}} - {\left(x - 2\right)}^{2} / \left(16 \sqrt{2}\right) + {\left(x - 2\right)}^{3} / \left(64 \sqrt{2}\right) - \frac{5 {\left(x - 2\right)}^{4}}{4096 \sqrt{2}} + \ldots$