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 ^3x+1/3x3 …
b) Use the first two terms of the Maclaurin series of 𝑦 to estimate tan(1/3)tan ( 1/3 )tan(1/3)
(i)Let f(x)=tanx,then f(0)=0f′(x)=sec2x=1+tan2x⇒f′(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)(i)Let f(x)=tanx,then f(0)=0f′(x)=sec2x=1+tan2x⇒f′(0)=1f′′(x)=2secx(1+tan2x)⇒f′′(0)=2The Maclaurin series expansion istanx=f(0)+xf′(0)+2!x2f′′(x)+3!x3f′′(x)+…=x+3x3+o(x3)
(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}(ii)Put x=31∴tan(31)=31+31(31)3+...=31+811=8128
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