Answer to Question #238222 in Calculus for moe

Question #238222

Consider the function 𝑦 = tan (𝑥)


a) Show that the first two non-zero terms in the Maclaurin series of 𝑦 are 𝑥+1/3𝑥3𝑥 + 1/3 𝑥 ^3


b) Use the first two terms of the Maclaurin series of 𝑦 to estimate tan(1/3)tan ( 1/3 )


1
Expert's answer
2021-09-20T01:41:19-0400

(i)Let f(x)=tanx,then f(0)=0f(x)=sec2x=1+tan2xf(0)=1f(x)=2secx(1+tan2x)f(0)=2The Maclaurin series expansion istanx=f(0)+xf(0)+x22!f(x)+x33!f(x)+=x+x33+o(x3)(i) Let \ f(x)=\tan x, then \ f(0)=0 \\ f^{\prime}(x)=\sec ^{2} x=1+\tan ^{2} x \\ \Rightarrow f^{\prime}(0)=1 \\ f^{\prime \prime}(x)=2 \sec x\left(1+\tan ^{2} x\right)\\ \Rightarrow f^{\prime \prime}(0)=2 \\ The \ Maclaurin \ series \ expansion \ is \\ \tan x=f(0)+x f^{\prime}(0)+\frac{x^{2}}{2 !} f^{\prime \prime}(x)+\frac{x^{3}}{3 !} f^{\prime \prime}(x)+\ldots \\ =x+\frac{x^{3}}{3}+o\left(x^{3}\right)

(ii)Put x=13tan(13)=13+13(13)3+...=13+181=2881(ii) Put \ x=\frac{1}{3}\\ \therefore tan(\frac{1}{3})=\frac{1}{3}+\frac{1}{3}(\frac{1}{3})^3+...\\ =\frac{1}{3}+\frac{1}{81}\\ =\frac{28}{81}



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