Maclaurin series of y(x) is
y(x)=y(0)+y′(0)⋅x+y′′(0)⋅x2/2+y′′′(0)⋅x3/6+o(x3)
We have
y(0)=tan(0)=0;
y'(x)=tan'(x)=cos2(x)1 ; y'(0)=cos2(0)1=11=1;
y''(x)=(cos2(x)1)′=cos4(x)1′⋅cos2(x)−1⋅cos2(x))′=
=cos4(x)2⋅cos(x)⋅sin(x)=cos3(x)2⋅sin(x);
y′′(0)=cos3(0)1⋅sin(0)=12⋅0=0;
y'''(x)=(cos3(x)2⋅sin(x))′=cos6(x)2⋅(sin(x))′⋅cos3(x)−2⋅sin(x)⋅(cos3(x))′=
=cos6(x)2⋅cos4(x)+6⋅sin2(x)⋅cos2(x)=cos2(x)2−cos2(x)6⋅tan2(x);
y'''(0)=cos2(0)2−cos2(0)6⋅tan2(0)=12−16⋅0=2
Thus we have
y(x)=tan(x)=0+1⋅x+0⋅x2/2+2⋅x3/6+o(x3)=x+3x3+o(x3)
(b) Use the first two terms of the Maclaurin series of y to estimate tan (1/3)
tan(1/3)=31+3(31)3=31+811=8128=0.345679 -approx value
Exat value with 6 digits:
Thus error rate=0.0006 ≈0,001 by order
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