Question #238222

Consider the function š‘¦ = tan (š‘„)


a) Show that the first two non-zero terms in the Maclaurin series of š‘¦ are š‘„+1/3š‘„3š‘„ + 1/3 š‘„ ^3 …


b) Use the first two terms of the Maclaurin series of š‘¦ to estimate tan(1/3)tan ( 1/3 )


Expert's answer

(i)Let f(x)=tan⁔x,then f(0)=0f′(x)=sec⁔2x=1+tan⁔2x⇒f′(0)=1f′′(x)=2sec⁔x(1+tan⁔2x)⇒f′′(0)=2The Maclaurin series expansion istan⁔x=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=13∓tan(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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