Answer to Question #238656 in Linear Algebra for Lubi

Question

Consider the function y=tan(x)

(a) Show that the first two non-zero terms in the Maclaurin series of y are x+1/3x³...

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

Answer 1

Maclaurin series of y(x) is

$y(x)=y(0)+y'(0)\cdot x+y''(0)\cdot x^2/2+y'''(0)\cdot x^3/6+o(x^3)$

We have

y(0)=tan(0)=0;

y'(x)=tan'(x)=$\frac{1}{cos^2(x)}$ ; y'(0)=$\frac{1}{cos^2(0)}=\frac{1}{1}=1;$

y''(x)=$\left( \frac{1}{cos^2(x)}\right)'=\frac{1'\cdot cos^2(x)-1\cdot cos^2(x))'}{cos^4(x)}=\\$

=$\frac{2\cdot cos(x)\cdot sin(x) }{cos^4(x)}=\frac{2\cdot sin(x)}{cos^3(x)};$

$y''(0)=\frac{1\cdot sin(0)}{cos^3(0)}=\frac{2\cdot 0}{1}=0;$

y'''(x)=$\left (\frac{2\cdot sin(x)}{cos^3(x)} \right)'=\frac {2\cdot (sin(x))'\cdot cos^3(x)-2\cdot sin(x)\cdot(cos^3(x))'} {cos^6(x)}=\\$

$=\frac{2\cdot cos^4(x)+6\cdot sin^2(x)\cdot cos^2(x)}{cos^6(x)}= \frac{2}{cos^2(x)}-\frac{6\cdot tan^2(x)}{cos^2(x)};$

y'''(0)=$\frac{2}{cos^2(0)}-\frac{6\cdot tan^2(0)}{cos^2(0)}=\frac{2}{1}-\frac{6\cdot 0}{1}=2$

Thus we have

y(x)=tan(x)=$0+1\cdot x+0\cdot x^2/2+2\cdot x^3/6+o(x^3)=x+\frac{x^3}{3}+o(x^3)$

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

tan(1/3)=$\frac{1}{3}+\frac{\left( \frac{1}{3} \right)^3}{3}=\frac{1}{3}+\frac{1}{81}= \frac{28}{81}=0.345679$ -approx value

Exat value with 6 digits:

Thus error rate=0.0006 $\approx 0,001$ by order