Question

Question :-

Find the taylor series as far as the term in x² .

i) f ( x ) = Sin^-1 x at a = 1/2

ii) f ( x ) = tan x at a = π/4 ​

Answer 1

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

Given function is

$\rm :\longmapsto\:f(x) = {sin}^{ - 1}x$

The successive differentiation of f(x) are as follow :

$\rm :\longmapsto\:f'(x) = \dfrac{1}{ \sqrt{1 - {x}^{2} } }$

can be rewritten as

$\rm :\longmapsto\:f'(x) = {\bigg(1 - {x}^{2} \bigg) }^{ - \dfrac{1}{2} }$

So,

$\rm :\longmapsto\:f''(x) = - \dfrac{1}{2} {\bigg(1 - {x}^{2} \bigg) }^{ - \dfrac{3}{2} }( - 2x)$

So,

$\rm :\longmapsto\:f''(x) = x{\bigg(1 - {x}^{2} \bigg) }^{ - \dfrac{3}{2} }$

So, the value of f(x) and its derivatives at a.

So,

$\rm :\longmapsto\:f \bigg(\dfrac{1}{2} \bigg)= {sin}^{ - 1}\dfrac{1}{2} = \dfrac{\pi}{6}$

$\rm :\longmapsto\:f'\bigg(\dfrac{1}{2} \bigg)= \dfrac{1}{ \sqrt{1 - \dfrac{1}{4} } } = \dfrac{2}{ \sqrt{3} }$

$\rm :\longmapsto\:f'' \bigg(\dfrac{1}{2} \bigg) = \bigg(\dfrac{1}{2} \bigg){\bigg(1 - \dfrac{1}{4} \bigg) }^{ - \dfrac{3}{2} }$

$\rm :\longmapsto\:f'' \bigg(\dfrac{1}{2} \bigg) = \bigg(\dfrac{1}{2} \bigg){\bigg(\dfrac{4 - 1}{4} \bigg) }^{ - \dfrac{3}{2} }$

$\rm :\longmapsto\:f'' \bigg(\dfrac{1}{2} \bigg) = \bigg(\dfrac{1}{2} \bigg){\bigg(\dfrac{3}{4} \bigg) }^{ - \dfrac{3}{2} }$

$\rm :\longmapsto\:f'' \bigg(\dfrac{1}{2} \bigg) = \bigg(\dfrac{1}{2} \bigg){\bigg(\dfrac{ \sqrt{3} }{2} \bigg) }^{ -2 \times \dfrac{3}{2} }$

$\rm :\longmapsto\:f'' \bigg(\dfrac{1}{2} \bigg) = \bigg(\dfrac{1}{2} \bigg){\bigg(\dfrac{ \sqrt{3} }{2} \bigg) }^{ -3 }$

$\rm :\longmapsto\:f'' \bigg(\dfrac{1}{2} \bigg) = \bigg(\dfrac{1}{2} \bigg){\bigg(\dfrac{2 }{ \sqrt{3} } \bigg) }^{ 3 }$

$\rm :\longmapsto\:f'' \bigg(\dfrac{1}{2} \bigg) = \bigg(\dfrac{1}{2} \bigg){\bigg(\dfrac{8}{3 \sqrt{3} } \bigg) }$

$\rm :\longmapsto\:f'' \bigg(\dfrac{1}{2} \bigg) = {\bigg(\dfrac{4}{3 \sqrt{3} } \bigg) }$

So,

By Taylor Theorem, we have as far as x²

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

$\rm :\longmapsto\: {sin}^{ - 1}x = \dfrac{\pi}{6} + \bigg(x - \dfrac{1}{2} \bigg)\dfrac{2}{ \sqrt{3} } + \dfrac{1}{2} { \bigg(x - \dfrac{1}{2} \bigg)}^{2} \dfrac{4}{3 \sqrt{3} }$

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

Given function is

$\rm :\longmapsto\:f(x) = tanx$

The successive differentiation of f(x) are as follow :

$\rm :\longmapsto\:f'(x) = {sec}^{2}x$

$\rm :\longmapsto\:f''(x) = 2{sec}x(secxtanx) = 2 {sec}^{2}x \: tanx$

So, the value of f(x) and its derivatives at a.

$\rm :\longmapsto\:f \bigg(\dfrac{\pi}{4} \bigg)= tan\dfrac{\pi}{4} = 1$

$\rm :\longmapsto\:f'\bigg(\dfrac{\pi}{4} \bigg)= {sec}^{2} \dfrac{\pi}{4} = 2$

$\rm :\longmapsto\:f''\bigg(\dfrac{\pi}{4} \bigg)=2{sec}^{2} \dfrac{\pi}{4} \: tan\dfrac{\pi}{4} = 4$

So,

By Taylor Theorem, we have as far as x²

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

$\rm :\longmapsto\: tanx = 1 + \bigg(x - \dfrac{\pi}{4} \bigg)2 + \dfrac{1}{2} { \bigg(x -\dfrac{\pi}{4}\bigg)}^{2} 4$

$\rm :\longmapsto\: tanx = 1 + 2\bigg(x - \dfrac{\pi}{4} \bigg)+ 2{ \bigg(x -\dfrac{\pi}{4}\bigg)}^{2}$

Answer 2

Answer:

$By Taylor Theorem, we have as far as x²</p><p></p><p>\rm :\longmapsto\:f(x) = f(a) + (x - a)f'(a) + \dfrac{ {(x - a)}^{2} }{2}f''(a) + - - -:⟼f(x)=f(a)+(x−a)f′(a)+2(x−a)2f′′(a)+−−−</p><p></p><p>\rm :\longmapsto\: tanx = 1 + \bigg(x - \dfrac{\pi}{4} \bigg)2 + \dfrac{1}{2} { \bigg(x -\dfrac{\pi}{4}\bigg)}^{2} 4:⟼tanx=1+(x−4π)2+21(x−4π)24</p><p></p><p>\rm :\longmapsto\: tanx = 1 + 2\bigg(x - \dfrac{\pi}{4} \bigg)+ 2{ \bigg(x -\dfrac{\pi}{4}\bigg)}^{2}:⟼tanx=1+2(x−4π)+2(x−4π)2</p><p></p><p>$