Find the first three nonzero terms of the Taylor series for the function ?

Attachment image

1 Answer

Relevance
  • 3 months ago

    Taylor series of f(x) around a is:

    f(a) + f'(a)(x - a) / 1! + f"(a)(x - a)^2/ 2! +  f'"(a)(x - a)^3/ 3! +  f''"(a)(x - a)^4/ 4! + 

    In our case, assuming positive x

    f(x) = sqrt(2x - x^2)

    f'(x) = (1 - x)/sqrt(2x - x^2)

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

    f'''(x) = (3 - 3 x)/(2x-x^2)^(5/2)

    f''''(x) = -(3 (4 x^2 - 8 x + 5))/(2x-x^2)^(7/2)

    and when we plug in x=a=1

    f(1) = 1

    f'(1) = 0

    f"(1) = -1

    f'''(1) = 0

    f''''(1) = -3

    this will cause 2nd and 4th term od the series to be 0.

    First 3 non-zero terms

    f(a) = f(1) = 1

    f"(a)(x - a)^2/ 2! = f"(1)(x - 1)^2/ 2= 

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

    f''"(a)(x - a)^4/ 4! = f''"(1)(x - 1)^4/ 24 =

    = - (1/8) (x - 1)^4

Still have questions? Get answers by asking now.