Can someone please help me with these problems , I need to know how we get the f
ID: 2868338 • Letter: C
Question
Can someone please help me with these problems , I need to know how we get the final answer cz I have more problems like these problems, especially last one!!!
Thanks in advance for your assistance,
Find the Maclaurin series for the function f(x) = x2e-x by using a Maclaurin series that you already know. The series has the form f(x) = Sigma n = 0 to infinity cn xn. Enter the first few coefficients below. c0 = c1 = c2 = c3 = c4 = c5 = c6 = Find the Maclaurin series for the function f(x) = e-x/3 by using a Maclaurin series that you already know. The series has the form f(x) = Sigma n = 0 to infinity cn xn. Enter the first few coefficients below. c0 = c1 = c2 = c3 = c4 = Find the Maclaurin series for the function f(x) = cos(x) using the definition of Maclaurin series. The series has the form f(x) = Sigma n = 0 to infinity cn xn.Enter the first few coefficients below. c0 = c1 = c2 = c3 = c4 = Find the radius of convergence. (Enter INF for infinity.) R =Explanation / Answer
maclaurin series
f(x) = f(0) + f'(0)x + f"(0)x^2/2! + f"'(0)x^3/3! + f""(0)x^4/4!+.....
we already know that
a) f(x)= x^2 (e^-x)
f(0)= 0 = Co
f'(x) = 2x*e^(-x) - x^2*e^(-x)
= 2x*e^(-x) - f(x) => f'(0) = 0 = C1
f"(x) = 2e^(-x) - 2xe^(-x) - f'(x) = 2e^(-x) - 4x*e^(-x) + f(x) ; f"(0) = 2 => C2 = f"(0)/2! = 1
f"'(x) = -2e^(-x) - 4e^(-x) + 4xe^(-x) + 2xe^(-x) - f(x); f"'(0) = -6 => C3 = f"'(0)/3! = -1
f""(x) = 6e^(-x)+ 6e^(-x) - 6xe^(-x) - 2x*e^(-x) + x^2e^(-x) = 12 => C4 = f""(0) = 12/24 = 1/2
f""'(x) = -20e^(-x) + 8xe^(-x) + 2xe^(-x) - f(x) => f""'(0) = -20 => C5 = f""'(0)/5! = -20/120 = -1/6
f''''''(0) = 30 => C6 = f"""(0)/6! = 30/720 = 1/24
b) e^(-x/3)
we know that
e^(-x) = 1 - x+ x^2/2! - x^3/3! + x^4/4! - .....
put x= x/3 in above maclaurin expansion we get
e^(-x/3) = 1 - x/3 + x^2 / 18 - x^3/162 +x^4/ 1944 - .......
so Co = 1
C1 = -1/3
C2 = 1/18
C3 = -1/162
C4 = 1/1944
C) f(x) = cosx
f(0) =1
f'(x) = -sinx => f'(0) = 0
f"(x) = -cos x => f"(0) = -1
f"'(x) = sin x => f"'(0) = 0
f""(x)= cosx => f""(0) = 1
here the coefficients are
Co =1
C1 = 0/1!
C2 = -1/2! = -1/2
C3 = 0/3! = 0
C4 = 1/4! = 1/24
and radius of convergence = INF {since cos x is always convergent for any value of x}