Infinite Series: [2/(x^2 - 3x + 2)]?
The fraction [2/(x^2 - 3x + 2)] can be written as an infinite series. Find the sum of the first four terms of the series expansion for x = -1 and x = -2.
The fraction [2/(x^2 - 3x + 2)] can be written as an infinite series. Find the sum of the first four terms of the series expansion for x = -1 and x = -2.
Comments
By Partial Fractions,
f(x) = 2/(x^2 - 3x + 2) = 2/(x - 2) - 2/(x - 1).
Since we only need four terms, we'll do this via differentiation.
I'll do this for x = -1 (the other case is similar).
f(x) = 2/(x - 2) - 2/(x - 1) ==> f(-1) = -2/3 + 1 = 1/3
f'(x) = -2/(x - 2)^2 + 2/(x - 1)^2 ==> f'(-1) = -2/9 + 1/2 = 5/18
f''(x) = 4/(x - 2)^3 - 4/(x - 1)^3 ==> f''(-1) = -4/27 - 1/2 = -35/54
f'''(x) = -12/(x - 2)^4 + 12/(x - 1)^4 ==> f'''(-1) = -4/27 + 3/4 = 65/108.
So,
f(x) = (1/3) + (5/18)(x+1) + (-35/54) (x+1)^2 / 2! + (65/108) (x+1)^3 /3!
+ ...
I hope this helps!
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