Question

$\begin{gathered} \sf Let \: f(x) = {e}^{x + {x}^{2} } \: for \:real \: x. \: From \: among \\ \sf the \: following. \: Choose \: the \: Taylor \: series \\ \sf approximation \: of \: f(x) \: around \: x = 0, \: which \\ \sf includes \: all \: powers \: of \: x \: less \: than \: or \: equal \:to \: 3\end{gathered}$

$\sf A. \: 1+x+x²+x³$

$\sf B. \: 1 + x + \frac{3}{2} {x}^{2} + {x}^{3}$
$\sf C. \: 1 + x + \frac{3}{2} {x}^{2} + \frac{7}{6} {x}^{3}$
$\sf D. \: 1 + x + 3 {x}^{2} + 7x {}^{3}$


​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\: \: f(x) = {e}^{x + {x}^{2} }$

Let us first evaluate first three differential coefficients of f(x).

So,

$\red{\rm :\longmapsto\: \: f'(x) = {e}^{x + {x}^{2} } \dfrac{d}{dx}(x + {x}^{2})}$

$\red{\rm \:  =  \: f(x)(1 + 2x)}$

$\red{\rm\implies \:f'(x) = f(x)(1 + 2x)}$

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

$\green{\rm :\longmapsto\:f''(x) = \dfrac{d}{dx}f(x)(1 + 2x)}$

$\green{\rm \:  =  \: f(x)\dfrac{d}{dx}(1 + 2x) + (1 + 2x)\dfrac{d}{dx}f(x)}$

$\green{\rm \:  =  \: 2f(x) + (1 + 2x)f'(x)}$

$\green{\rm\implies \:\rm \: f''(x) =  \: 2f(x) + (1 + 2x)f'(x)}$

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

$\purple{\rm :\longmapsto\:f'''(x) = \dfrac{d}{dx}(2f(x) + (1 + 2x)f'(x))}$

$\purple{\rm \:  =  \: 2f'(x) + (1 + 2x)f''(x) + 2f'(x)}$

$\purple{\rm \:  =  \: 4f'(x) + (1 + 2x)f''(x)}$

$\purple{\rm\implies \:\rm \:f'''(x)  =  \: 4f'(x) + (1 + 2x)f''(x)}$

Now, Let's evaluate the value of differential coefficients at x = 0

So,

$\rm :\longmapsto\:f(0) = {e}^{0 + 0} = {e}^{0} = 1$

$\red{\rm\implies \:f'(0) = f(0) = 1}$

$\green{\rm\implies \:\rm \: f''(0) =  \: 2f(0) + f'(0) = 2 + 1 = 3}$

$\purple{\rm\implies \:\rm \:f'''(0)  =  \: 4f'(0) + f''(0) = 4 + 3 = 7}$

Now, Taylor Series up to three degree is given by at x = 0 is

$\rm :\longmapsto\:f(x) = f(0) + xf'(0) + \dfrac{ {x}^{2} }{2!}f''(0) + \dfrac{ {x}^{3} }{3!}f'''(0)$

So, on substituting the values as evaluated above, we get

$\rm :\longmapsto\:{e}^{x + {x}^{2} } = 1+ x + \dfrac{3{x}^{2} }{2} + \dfrac{7{x}^{3} }{6}$

• Hence, Option (C) is correct.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Taylor Series at x = a

$\rm :\longmapsto\:f(x) = f(a) + (x - a)f'(a) + \dfrac{ {(x - a)}^{2} }{2!}f''(a) + - - - \infty$

Mc Lauren's Series

$\rm :\longmapsto\:f(x) = f(0) + xf'(0) + \dfrac{ {x}^{2} }{2!}f''(0) + \dfrac{ {x}^{3} }{3!}f'''(0) + - - - \infty$

Answer 2

Answer:

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\: \: f(x) = {e}^{x + {x}^{2} }$

Let us first evaluate first three differential coefficients of f(x).

So,

$\purple{\rm :\longmapsto\: \: f'(x) = {e}^{x + {x}^{2} } \dfrac{d}{dx}(x + {x}^{2})}$

$\purple{\rm \:  =  \: f(x)(1 + 2x)}$

$\purple{\rm\implies \:f'(x) = f(x)(1 + 2x)}$

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

$\red{\rm :\longmapsto\:f''(x) = \dfrac{d}{dx}f(x)(1 + 2x)}$

$\red{\rm \:  =  \: f(x)\dfrac{d}{dx}(1 + 2x) + (1 + 2x)\dfrac{d}{dx}f(x)}$

$\red{\rm \:  =  \: 2f(x) + (1 + 2x)f'(x)}$

$\red{\rm\implies \:\rm \: f''(x) =  \: 2f(x) + (1 + 2x)f'(x)}$

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

$\green{\rm :\longmapsto\:f'''(x) = \dfrac{d}{dx}(2f(x) + (1 + 2x)f'(x))}$

$\green{\rm \:  =  \: 2f'(x) + (1 + 2x)f''(x) + 2f'(x)}$

$\green{\rm \:  =  \: 4f'(x) + (1 + 2x)f''(x)}$

$\green{\rm\implies \:\rm \:f'''(x)  =  \: 4f'(x) + (1 + 2x)f''(x)}$

Now, Let's evaluate the value of differential coefficients at x = 0

Let's evaluate the value of differential coefficients at x = 0So,

$\rm :\longmapsto\:f(0) = {e}^{0 + 0} = {e}^{0} = 1$

$\red{\rm\implies \:f'(0) = f(0) = 1}$

$\green{\rm\implies \:\rm \: f''(0) =  \: 2f(0) + f'(0) = 2 + 1 = 3}$

$\purple{\rm\implies \:\rm \:f'''(0)  =  \: 4f'(0) + f''(0) = 4 + 3 = 7}$

Now, Taylor Series up to three degree is given by at x = 0 is

$\rm :\longmapsto\:f(x) = f(0) + xf'(0) + \dfrac{ {x}^{2} }{2!}f''(0) + \dfrac{ {x}^{3} }{3!}f'''(0)$

So, on substituting the values as evaluated above, we get

$\rm :\longmapsto\:{e}^{x + {x}^{2} } = 1+ x + \dfrac{3{x}^{2} }{2} + \dfrac{7{x}^{3} }{6}$

• Hence, Option (C) is correct.

$\purple{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\pink{\rule{45pt}{7pt}}\blue{\rule{45pt}{7pt}}$

Taylor Series at x = a

$\rm :\longmapsto\:f(x) = f(a) + (x - a)f'(a) + \dfrac{ {(x - a)}^{2} }{2!}f''(a) + - - - \infty$

Mc Lauren's Series

$\rm :\longmapsto\:f(x) = f(0) + xf'(0) + \dfrac{ {x}^{2} }{2!}f''(0) + \dfrac{ {x}^{3} }{3!}f'''(0) + - - - \infty$