Maclaurin series is a special case of Taylor series that is centered at c=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))/(3!)x^3+(f^4(0))/(4!)x^4 +...
To determine the Maclaurin polynomial of degree n=4 for the given function f(x)=x^2e^(-x) , we may apply the formula for Maclaurin series.
To list f^n(x) up to n=4 , we may apply the following formula:
Product rule for differentiation: d/(dx) (u*v) = u' *v +u*v'
Derivative property: d/(dx) (f+-g+-h) = d/(dx) f +-d/(dx) g+-d/(dx) h
Power rule for differentiation: d/(dx) x^n =n*x^(n-1)
Derivative formula for exponential function: d/(dx) e^u = e^u * (du)/(dx)
f(x)=x^2e^(-x)
Let u =x^2 then u' = 2x
v = e^(-x) then v' = e^x*(-1) =-e^(-x)
d/(dx) (x^2e^(-x)) = 2x*e^(-x) + x^2*(-e^(-x))
= 2xe^(-x) -x^2e^(-x)
Let: u =x then u' =1
v = e^(-x) then v' =-e^(-x)
Note: c = constant value.
d/(dx) c*xe^(-x) = c*d/(dx) xe^(-x)
= c*[1*e^x +x * (-e^(-x))]
= c*[e^x-xe^(-x)]
= ce^x-cxe^(-x)
d/(dx) c*e^(-x) = c*d/(dx) e^(-x)
=c*(-e^(-x))
=-ce^(-x)
f'(x) =d/(dx) (x^2e^(-x))
= 2xe^(-x) -x^2e^(-x)
f^2(x) = d/(dx) (2xe^(-x) -x^2e^(-x))
=d/(dx) 2xe^(-x) -d/(dx) x^2e^(-x)
= [2e^(-x)-2xe^(-x)] - [2xe^(-x) -x^2e^(-x)]
=2e^(-x)-2xe^(-x) - 2xe^(-x) +x^2e^(-x)
=2e^(-x)-4xe^(-x) +x^2e^(-x)
f^3(x) = d/(dx) (2e^(-x)-4xe^(-x) +x^2e^(-x))
=d/(dx) 2e^(-x) -d/(dx) 4xe^(-x)+ d/(dx) x^2e^(-x)
=[-2e^(-x)] -[4e^(-x)-4xe^(-x)]+ [2xe^(-x) -x^2e^(-x)]
=- 2e^(-x) -4e^(-x)+4xe^(-x)+ 2xe^(-x) -x^2e^(-x)
=- 6e^(-x)+6xe^(-x) -x^2e^(-x)
f^4(x) = d/(dx) ( - 6e^(-x)+6xe^(-x) -x^2e^(-x))
=d/(dx) (-6e^(-x)) + d/(dx) 6xe^(-x)- d/(dx) x^2e^(-x)
=[ 6e^(-x)]+[6e^(-x)-6xe^(-x)] -[2xe^(-x) -x^2e^(-x)]
=6e^(-x)+6e^(-x)-6xe^(-x) -2xe^(-x) +x^2e^(-x)
=12e^(-x)-8xe^(-x) +x^2e^(-x)
Plug-in x=0 for each f^n(x) , we get:
f(0)=0^2e^(-0)
=0*1
=0
f'(0)=2*0*e^(-0) -0^2e^(-0)
=2*0*1 +0*1
=0
f^2(0)=2e^(-0)-4*0*e^(-0) +0^2e^(-0)
=2*1 -4*0*1 +0*1
=2
f^3(0)=- 6e^(-0)+6*0*e^(-0) -0^2e^(-0)
=-6*1 +6*0*1 +0*1
=-6
f^4(0)=12e^(-0)-8*0*e^(-0) +0^2e^(-0)
=12*1 -8*0*1 +0*1
=12
Note: e^(-0) = e^0 =1 .
Plug-in the values on the formula for Maclaurin series, we get:
sum_(n=0)^4 (f^n(0))/(n!) x^n
= f(0)+(f'(0))/(1!)x+(f^2(0))/(2!)x^2+(f^3(0))/(3!)x^3+(f^4(0))/(4!)x^4
= 0+0/(1!)x+2/(2!)x^2+(-6)/(3!)x^3+12/(4!)x^4
= 0+0/1x+2/2x^2-6/6x^3+12/24x^4
= 0+0+x^2-x^3+1/2x^4
= x^2-x^3+1/2x^4
The Maclaurin polynomial of degree n=4 for the given function f(x)=x^2e^(-x) will be:
P(x)=x^2-x^3+1/2x^4
No comments:
Post a Comment