Math, asked by Anonymous, 11 hours ago

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 ​

Answers

Answered by mathdude500
7

\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}

Answered by sorrySoSORRY
32

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>

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