Question #238656

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).



1
Expert's answer
2021-09-22T17:51:54-0400


Maclaurin series of y(x) is

y(x)=y(0)+y(0)x+y(0)x2/2+y(0)x3/6+o(x3)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)=1cos2(x)\frac{1}{cos^2(x)} ; y'(0)=1cos2(0)=11=1;\frac{1}{cos^2(0)}=\frac{1}{1}=1;

y''(x)=(1cos2(x))=1cos2(x)1cos2(x))cos4(x)=\left( \frac{1}{cos^2(x)}\right)'=\frac{1'\cdot cos^2(x)-1\cdot cos^2(x))'}{cos^4(x)}=\\

=2cos(x)sin(x)cos4(x)=2sin(x)cos3(x);\frac{2\cdot cos(x)\cdot sin(x) }{cos^4(x)}=\frac{2\cdot sin(x)}{cos^3(x)};

y(0)=1sin(0)cos3(0)=201=0;y''(0)=\frac{1\cdot sin(0)}{cos^3(0)}=\frac{2\cdot 0}{1}=0;

y'''(x)=(2sin(x)cos3(x))=2(sin(x))cos3(x)2sin(x)(cos3(x))cos6(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)}=\\

=2cos4(x)+6sin2(x)cos2(x)cos6(x)=2cos2(x)6tan2(x)cos2(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)=2cos2(0)6tan2(0)cos2(0)=21601=2\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+1x+0x2/2+2x3/6+o(x3)=x+x33+o(x3)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)=13+(13)33=13+181=2881=0.345679\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 0,001\approx 0,001 by order


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