Find the Taylor series (check my work)?

See the answers for work...
**can someone help me work through the derivatives part so I can construct an #f^n(a)# rule?

a. #f(x)=1/x^2#, #a=4#
b. #f(x)=\sqrt(x)#, #a=2#

3 Answers
May 3, 2018

For part A
Finished -- can someone check the steps overall?

Explanation:

#f'=-2x^-3#
#\rArrf'(4)=-2(4)^-3#

#f''=(-2)(-3)x^-4#
#\rArrf''(4)=(-2)(-3)(4)^-4#

#f'''=(-2)(-3)(-4)x^-5#
#\rArrf'''(4)=(-2)(-3)(4)^-5#

#f^n(4)=((-1)^n(n-1)!)/(4^n)#

#C_n=1/(n!)*f^n(a)=((-1)^n(n-1)!)/4^n*1/(n!)=((-1)^n(n-1))/(4^n)#
#\therefore f(x)=\sum_(n=0)^\inftyC_n(x-a)^n=\sum_(n=0)^\infty(-1)^(n+1)1/(4^n)(x-4)^n#, which is #1/x^2# centered at #a=4#

May 3, 2018

For part B
Literally confused on step one. Someone help me lol

Explanation:

#f'=1/(2\sqrtx)=1/2(x)^(-1/2)#
#f''=-1/(4x^(3/2))=1/4(x)^(-3/2)=1/2(-1/2)x^(-3/2)#
#f'''=3/(8x^(5/2))=3/8(x)^(-5/2)=1/2(-1/2)(-3/2)x^(-5/2)#

...
#f^4=1/2(-1/2)(-3/2)(-5/2)x^(-7/2)#
#f^5=1/2(-1/2)(-3/2)(-5/2)(-7/2)x^(-9/2)#

exponent seems to increase by #-1/2-n#?

May 4, 2018

A) # f(x)=1/x^2 # pivoted about #x=4#

# f(x) = sum_(n=0)^oo (-1)^n((n+1))/(4^(n+2))(x-4)^n#
# \ \ \ \ \ \ \ = 1/16 -(x-4)/32 +(3(x-4)^2)/256 - (x-4)^3/256 + ...#

B) #f(x)=sqrt(x)# pivoted about #x=2#

# f(x) = sqrt(2) + (x-2)/(2sqrt(2)) - (x-2)^2/(16sqrt(2)) +(x-2)^3/(64sqrt(2)) -(5(x-2)^4)/(4096sqrt(2)) + ...#

Explanation:

We seek:

A) TS of #f(x)=1/x^2# pivoted about #x=4#
B) TS of #f(x)=sqrt(x)# pivoted about #x=2#

We use the general definition of a TS pivot about #x=a#:

# f(x) = f(a) + (f^((1))(a))/(1!)(x-a) + (f^((2))(a))/(2!)(x-a)^2 + #
# \ \ \ \ \ \ \ \ \ \ \ + (f^((3))(a))/(3!)(x-a)^3 + ... + #
# \ \ \ \ \ \ \ \ \ \ \ + (f^((n))(a))/(n!)(x-a)^n + ... #

Part (A):

Differentiating wrt to #x# we compute the first few derivatives:

# f^((0))(x) = 1/x^2 #

# f^((1))(x) = (-2)/x^3 = (-)(2!)/(x^3) #

# f^((2))(x) = (-2)(-3)/x^4 = (+)(3!)/(x^4) #

# f^((3))(x) = (-2)(-3)(-4)/x^5 = (-)(4!)/(x^5) #

And we reasonably conclude that:

# f^((n))(x) = (-1)^n((n+1)!)/(x^(n+2)) #

Allowing us to construct a Taylor Series, pivoted about #x=4#:

# f(x) = f^((0))(4) + (f^((1))(4))/(1!)(x-4) + (f^((2))(4))/(2!)(x-4)^2 + #
# \ \ \ \ \ \ \ \ \ \ \ + (f^((3))(4))/(3!)(x-4)^3 + ... + #
# \ \ \ \ \ \ \ \ \ \ \ + (f^((n))(4))/(n!)(x-4)^n + ... #

# \ \ \ \ \ \ \ = 1/4^2 + (-(2!)/(4^3))/(1!)(x-4) + ((3!)/(4^4))/(2!)(x-4)^2 + #
# \ \ \ \ \ \ \ \ \ \ \ + (-(4!)/(4^5))/(3!)(x-4)^3 + ... + #
# \ \ \ \ \ \ \ \ \ \ \ + ((-1)^n((n+1)!)/(4^(n+2)))/(n!)(x-4)^n + ... #

# \ \ \ \ \ \ \ = 1/4^2 - ((2)/(4^3))(x-4) + ((3)/(4^4))(x-4)^2 + #
# \ \ \ \ \ \ \ \ \ \ \ - ((4)/(4^5))(x-4)^3 + ... + #
# \ \ \ \ \ \ \ \ \ \ \ + (-1)^n((n+1))/(4^(n+2))(x-4)^n + ... #

# \ \ \ \ \ \ \ = 1/16 -(x-4)/32 +(3(x-4)^2)/256 - (x-4)^3/256 + ...#

# \ \ \ \ \ \ \ = sum_(n=0)^oo (-1)^n((n+1))/(4^(n+2))(x-4)^n #

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Part (B):

Differentiating wrt to #x# we compute the first few derivatives:

# f^((0))(x) = sqrt(x) #

# f^((1))(x) = (1/2)x^(-1/2) = 1/(2x^(1/2)) = #

# f^((2))(x) = (1/2)(-1/2)x^(-3/2) = -1/(2^2x^(3/2))#

# f^((3))(x) = (1/2)(-1/2)(-3/2)x^(-5/2) = 3/(2^3x^(5/2)) #

# f^((4))(x) = (1/2)(-1/2)(-3/2)(-5/2)x^(-7/2) = (1.3.5)/(2^4x^(7/2)) #

# f^((5))(x) = (1/2)(-1/2)(-3/2)(-5/2)(-7/2)x^(-9/2) = (1.3.5.7)/(2^5x^(9/2)) #

Allowing us to construct a Taylor Series, pivoted about #x=2#:

# f(x) = f^((0))(2) + (f^((1))(2))/(1!)(x-2) + (f^((2))(2))/(2!)(x-2)^2 + #
# \ \ \ \ \ \ \ \ \ \ \ + (f^((3))(2))/(3!)(x-2)^3 + (f^((4))(2))/(4!)(x-2)^4 + ... + #

# \ \ \ \ \ \ \ = sqrt(2) + (1/(2 \ 2^(1/2)))/(1!)(x-2) + (-1/(2^2 \ 2^(3/2)))/(2!)(x-2)^2 + #
# \ \ \ \ \ \ \ \ \ \ \ + (3/(2^3 \ 2^(5/2)))/(3!)(x-2)^3 + ((1.3.5)/(2^4 \ 2^(7/2)))/(4!)(x-2)^4 + ... + #

# \ \ \ \ \ \ \ = sqrt(2) + (x-2)/(2sqrt(2)) - (x-2)^2/(16sqrt(2)) +(x-2)^3/(64sqrt(2)) -(5(x-2)^4)/(4096sqrt(2))+ ...#