Asked by Hey
If f(x)=1/((1+2x)(3-x))
A)Express f(x) in partial fractions.
B)find the first four terms in the series expansion of f(x) in ascending powers of x.
C) find the coefficient of x^n in the series expansion of f x in ascending powers of x.
A)Express f(x) in partial fractions.
B)find the first four terms in the series expansion of f(x) in ascending powers of x.
C) find the coefficient of x^n in the series expansion of f x in ascending powers of x.
Answers
Answered by
Reiny
let 1/((1+2x)(3-x)) = A/(1+2x) + B(3-x) = ( A(3-x) + B(1+2x))/((1+2x)(3-x))
1 = A(3-x) + B(1+2x)
let x = 3 ---> 1 = 7B, or B = 1/7
let x= -1/2 ---> 1 = 7A/2 or A = 2/7
then 1/((1+2x)(3-x)) = 2/(7(1 + 2x)) + 1/(7(3 - x))
2/(7 + 14x) = (2/7)(1 + 2x)^-1
= (2/7)(1 + (-1)(2x) + (-1)(-2)/2!(2x)^2 + (-1)(-2)(-3)/3!(2x)^3 + .... )
= 2/7 - 4x/7 + 8x^2 /7 - 16x^3 / 7 + ....
1/(7(3 - x)) = (1/7)(3 - x)^-1
= (1/7)(3^-1 + (-1)(3^-2)(-x) + (-1)(-2)/2! (3^-3)(-x)^2 + (-1)(-2)(-3)/3! (3^-4)(-x)^3 + ...
= (1/7)(1/3 + x/9 + x^2/27 + x^3 / 81 + ...
= 1/21 + x/63 + x^2 /189 + x^3 / 567 + ...
then f(x) = 1/((1+2x)(3-x))
= 2/7 - 4x/7 + 8x^2 /7 - 16x^3 / 7 + .... + 1/21 + x/63 + x^2 /189 + x^3 / 567 + ...
= (2/7 + 1/21) + x(1/63 - 4/7) + x^2(1/189 + 8/7) + x^3(1/567 - 16/7) + ...
= 1/3 - (5/9)x + (31/27)x^2 + (-185/81)x^3 + ....
general term for (2/7)(1 + 2x)^-1
= (2/7)(-1)(-2)(-3)...(-n)/n! (1^(-1-n)(2x)^n
= +(2/7)(2^n)x^n , for even n, and -(2/7)(2^n)x^n for odd n
Ok, your turn to find the general term for (1/7)(3 - x)^-1
then add up the 2 general terms
1 = A(3-x) + B(1+2x)
let x = 3 ---> 1 = 7B, or B = 1/7
let x= -1/2 ---> 1 = 7A/2 or A = 2/7
then 1/((1+2x)(3-x)) = 2/(7(1 + 2x)) + 1/(7(3 - x))
2/(7 + 14x) = (2/7)(1 + 2x)^-1
= (2/7)(1 + (-1)(2x) + (-1)(-2)/2!(2x)^2 + (-1)(-2)(-3)/3!(2x)^3 + .... )
= 2/7 - 4x/7 + 8x^2 /7 - 16x^3 / 7 + ....
1/(7(3 - x)) = (1/7)(3 - x)^-1
= (1/7)(3^-1 + (-1)(3^-2)(-x) + (-1)(-2)/2! (3^-3)(-x)^2 + (-1)(-2)(-3)/3! (3^-4)(-x)^3 + ...
= (1/7)(1/3 + x/9 + x^2/27 + x^3 / 81 + ...
= 1/21 + x/63 + x^2 /189 + x^3 / 567 + ...
then f(x) = 1/((1+2x)(3-x))
= 2/7 - 4x/7 + 8x^2 /7 - 16x^3 / 7 + .... + 1/21 + x/63 + x^2 /189 + x^3 / 567 + ...
= (2/7 + 1/21) + x(1/63 - 4/7) + x^2(1/189 + 8/7) + x^3(1/567 - 16/7) + ...
= 1/3 - (5/9)x + (31/27)x^2 + (-185/81)x^3 + ....
general term for (2/7)(1 + 2x)^-1
= (2/7)(-1)(-2)(-3)...(-n)/n! (1^(-1-n)(2x)^n
= +(2/7)(2^n)x^n , for even n, and -(2/7)(2^n)x^n for odd n
Ok, your turn to find the general term for (1/7)(3 - x)^-1
then add up the 2 general terms
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.