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

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- SlowfingerLv 63 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

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