Let f(x) = 1/(1-x). Find a cubic polynomial p(x) so that f(0) = p(0), f'(0)=p'(0), f"(0)=p"(0), f'''(0)=p'''(0)

Question

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Consider if f were a more complicated function. How would this be more efficient to finding values of f?

Thank you!!

Answer 1

f(0) = 1

f '(x) = (u'v - uv')/v² where u = 1, u' = 0, v = 1 - x, v' = -1, v² = (1 - x)²

f '(x) = (0 - -1)/((1 - x)² = 1/(1 - x)² so f '(0) = 1

f ''(x) = (u'v - uv')/v² where u = 1, u' = 0, v = (1 - x)², v' = 2(1 - x)(-1) = 2x - 2, v² = (1 - x)⁴

f ''(x) = (0 - (2x - 2))/(1 - x)⁴ = (2 - 2x)/(1 - x)⁴ = 2/(1 - x)³ so f ''(0) = 2

f '''(x) = (u'v - uv')/v² where u = 2, u' = 0, v = (1 - x)³, v' = 3(1 - x)²(-1) = -3(1 - x)², v² = (1 - x)⁶

f '''(x) = (0 - 2((-3)(1 - x)²)/(1 - x)⁶ = 6/(1 - x)⁴ so f '''(0) = 6

A cubic can be written as P(x) = ax³ + bx² + cx + d

P(0) = d and putting this together with f(0) above we get d = 1

P'(x) = 3ax² + 2bx + c so P'(0) = c and putting this together with f '(0) above we get c = 1

P''(x) = 6ax + 2b and putting this together with f ''(0) above we get 2b = 2 or b = 1

P'''(x) = 6a and putting this together with f '''(0) above we get 6a = 6 so a = 1

P(x) = 1x³ + 1x² + 1x + 1 or

P(x) = x³ + x² + x + 1

Answer 2

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

f'(x)=-1(1-x)^-2(-1)=1/(1-x)²

f''(x)=-2(1-x)^-3(-1) = 2/(1-x)³

f'''(x)=-6(1-x)^-4(-1)=6/(1-x)⁴

f(0)=1, f'(0)=1,f''(0)=2, f'''(0)=6

Let p(x)=ax³+bx²+cx+d

p'(x)=3ax²+2bx+c

p''(x)=6ax+2b

p'''(x) = 6a

p(0)=f(0)=1, so d=1

p'(0)=f'(0)=1, so c =1

p''(0)= f''(0)=2, so 2b=2, b=1

p'''(0)=f'''(0)=6, so 6a=6, a =1

p(x) = x³+x²+x+1