Maclaurin series is a special case of Taylor series that is centered at a=0. The expansion of the function about 0 follows the formula:
f(x)=sum_(n=0)^oo (f^n(0))/(n!) x^n
or
f(x)= f(0)+(f'(0))/(1!)x+(f^2(0))/(2!)x^2+(f^3(0)x)/(3!)x^3+(f^4(0))/(4!)x^4 +...
To determine the Maclaurin polynomial of degree n=3 for the given function f(x)=tan(x) , we may apply the formula for Maclaurin series.
We may list f^n(x) as:
f(x) = tan(x)
f'(x)=d/(dx) tan(x) =sec^2(x)
f^2(x)=d/(dx)sec^2(x) =2tan(x)sec^2(x)
f^3(x)=d/(dx) 2tan(x)sec^2(x) =6sec^4(x)-4sec^2(x)
Plug-in x=0, we get:
f(0)=tan(0)
=0
f'(0)=sec^2(0) or (sec(0))^2
= 1^2
=1
f^2(0)=2tan(0)sec^2(0)
=2*0*1
=0
f^3(0)=6sec^4(0)-4sec^2(0)
=6(sec(0))^4-4(sec(0))^2
=6*1^4 -4*1^2
= 6-4
=2
Applying the formula for Maclaurin series, we get:
sum_(n=0)^3 (f^n(0))/(n!) x^n =0+1/(1!)x+0/(2!)x^2+2/(3!)x^3
=0+1/1x+0/(1*2)x^2+2/(1*2*3)x^3
= 0+x+0/2x^2+2/6x^3
=x+2/6x^3
=x+x^3/3
The Maclaurin polynomial of degree n=3 for the given function f(x)=tan(x) will be:
P_3(x)=x+x^3/3
Friday, August 17, 2012
Calculus of a Single Variable, Chapter 9, 9.7, Section 9.7, Problem 24
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