Data605hw14

##Question: This week, we’ll work out some Taylor Series expansions of popular functions.

1. f(x) = 1/1−x

2. f(x) = e^x

3. f(x) = ln(1+x)

4. f(x) = x^(1/2)

For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion.

(a) f(x) = 1/1−x

Answer: f(0) = 1

f’(0) = 1/(1−0)^2=1

f’’(0) = −2/(0−1)^3=2

Using the Maclaurin Series (Taylor series centered at 0), we get ∞ ∑ f^(n)(0)/n!(xn)=f(0)+f′(0)/1!(x)+…+f^(n)(0)/n!(xn) n=0 =1+1/1!(x)+2/2!(x^2)+…=1+x+x2+…

The series representation of this function is: ∞ ∑ xn for |x| < 1 n=0

(b) f(x) = e^x

Answer: f(0) = e^0 = 1

f’(0) = e^0 = 1

f’’(0) = e^0 = 1

Using the Maclaurin series (Taylor series at zero): n=0 ∑∞f^(n)(0)/n!(xn)=f(0)+f′(0)/1!x+…+f^(n)(0)/n!xn

=1+1/1!(x)+1/2!(x^2)+1/3!(x3)+…=1+x+x^2/2+x3/6+…

The series representation of this function is: ∞ ∑ x^k/k! k=0

f(x) = ln(1+x)

Ans:

f(0) = ln(1+0) = ln(1)= 0

f’(0) = 1/1+0=1

f’’(0) = −1/(1+0)^2=−1

f’’’(0) = 2/(1+0)^3=2

f^(4)(0) = −6/(1+0)^4=−6

Using the Maclaurin series (Taylor series at zero): ∞ ∑f^(n)(0)/n!(xn)=f(0)+f′(0)/1!x+…+f^(n)(0)/n!xn n=0 =0+1/1!x+(−1)/2!x^2+2/3!x3+(−6)/4!x^4+…=x−x2/2+x^3/3−x4/4+…

The series representation of this function is: ∞ ∑(−1)^k+1xk/k k=1

f(x) = x^1/2

(x^1/2)′=1/2x−1/2 (x^{1/2)′′=−3/4x}−3/2 (x^1/2)′′′= 5/8x^−5/2 Taylor expansion for x = 1: x^1/2=(1)1/2+1/21^{−1/2(x−1)/1!−3/41}−3/2(x−1)^2/2!+ 5/8^{1−5/2(x−1)}3/3!+…

T2(x)=1+1/2(x−1)−3/4(x−1)^{2/2!+5/8(x−1)}3/3!