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# Find the taylor polynomial t3(x) for the function f centered at the number:

f (x) = 7 tan^{-1}(x), a = 1, t3(x)

**Solution:**

Given:

Function f(x) = tan^{-1}(x)

Taylor polynomial up to three terms at x = a is given by

f(x) = f(a) + [(x - a) / 1!] f^{1}(a) + [ (x - a)^{2}/2! ] f^{11}(a) + [ (x - a)^{3}/3!] f^{111}(a) +....

If a = 1

Taylor polynomial up to three terms at x = 1 is given by

f(x) = f(1) + [(x - 1) / 1! ] f^{1}(1) + [(x - 1)^{2 }/ 2!] f^{11}(1) + [(x - 1)^{3}/3!] f^{111}(1) +....

f(x) = tan^{-1}(x)

f(1) = tan^{-1}(1)

f(1) = 90°

f^{1}(x) = 1 / (1 + x^{2})

⇒ f^{1}(1) = 1/(1 + 1^{2}) = 1 / 1 + 1

⇒ f^{1}(1) = 1/2

**Second order derivative f ^{11}(x)**

Differentiating f^{1} with respect to x gives

⇒ f^{11}(x) = - (2 × 2x) / (1 + x^{2})^{2}

⇒ f^{11}(x) = - 4x / (1 + x^{2})^{2}

⇒ f^{11}(1)= - 4 / (1 + 1^{2})^{2}

= -(4)/2^{2}

⇒ f^{11}(1) = - 4/4

= -1

**Third order derivative f ^{111}(x)**

Differentiating f^{11} With respect to x gives

⇒ f^{111}(x) = [(1 + x^{2})^{2} (-4) - (-4x)(2(1 + x^{2})2x)] / (1 + x^{2})^{4}

⇒ f^{111} (x) = (6x^{2 }- 2)/(1 + x^{2})^{3}

⇒ f^{111}(1) = (6(1)^{2} - 2)/(1 + 1^{2})^{3}

⇒ f^{111}(1) = (6 - 2)/(2)^{3}

⇒ f^{111}(1) = 4/(2)^{3}

⇒ f^{111}(1) = 4/8 = 1/2

Taylor polynomial up to three terms at x=a is given by

⇒f(x) = 90° + [(x - a)/1!] f^{1}(a) + [(x - a)^{2}/2!] f^{11}(a) + [(x - a)^{3}/3!] f^{111}(a) +....

⇒f(x) = 90°** **+ [(x - 1)/1!] 1/2 + [(x - 1)^{2}/2!]( -1) + 1/2 [(x - 1)^{3}/3!] +….

⇒ f(x) = 90°** **+ [(x - 1)/2] - (x - 1)^{2}/2+ (x - 1)^{3}/12+….

## Find the taylor polynomial t3(x) for the function f centered at the number:

f (x) = 7 tan - 1(x), a = 1 t3(x)

**Summary:**

The Taylor's series for the function f(x) up to the term containing of third degree is f(x) = tan^{-1}(x) = 90°** **+ [(x - 1)/2] - (x - 1)^{2}/2+ (x - 1)^{3}/12 +…

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