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.
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
A) To express f(x) in partial fractions, we need to decompose the rational function into simpler fractions. The general form of the partial fraction decomposition is:
f(x) = A/(1+2x) + B/(3-x)
To find the values of A and B, we need to perform the following steps:
1. Clear the denominators by multiplying through by the common denominator:
(1+2x)(3-x) * f(x) = A(3-x) + B(1+2x)
2. Expand the equation:
(3 - x + 6x - 2x^2) * f(x) = 3A - Ax + B + 2Bx
3. Collect like terms:
(3 - x + 6x - 2x^2) * f(x) = (3A + B) + (-A + 2B)x + (-2A)x^2
4. Equate the coefficients of each power of x to the corresponding term on the right:
Coefficient of x^2: -2A = 0 (implies A = 0)
Coefficient of x: -A + 2B = 0 (implies B = A/2 = 0)
Coefficient of x^0: 3A + B = 1 (implies 3A + B = 1)
5. Solve the system of equations to find the values of A and B:
Substituting A = 0 into the third equation:
0 + B = 1
B = 1
Substituting B = 1 into the second equation:
-A + 2(1) = 0
A - 2 = 0
A = 2
So, the partial fraction decomposition of f(x) is:
f(x) = 0/(1 + 2x) + 1/(3 - x)
Simplifying further:
f(x) = 1/(3 - x)
B) To find the first four terms in the series expansion of f(x) in ascending powers of x, we can use the Taylor series expansion. For f(x) = 1/(3 - x), the Taylor series expansion around x = 0 is:
f(x) = f(0) + f'(0)x + f''(0)(x^2/2!) + f'''(0)(x^3/3!) + ...
To find the coefficients, we need to calculate the derivatives of f(x) and evaluate them at x = 0.
f(x) = 1/(3 - x)
Differentiating with respect to x:
f'(x) = 1/(3 - x)^2
Differentiating again:
f''(x) = 2/(3 - x)^3
Differentiating once more:
f'''(x) = 6/(3 - x)^4
Evaluating at x = 0:
f(0) = 1/(3 - 0) = 1/3
f'(0) = 1/(3 - 0)^2 = 1/9
f''(0) = 2/(3 - 0)^3 = 2/27
f'''(0) = 6/(3 - 0)^4 = 2/81
Thus, the first four terms in the series expansion of f(x) are:
f(x) ≈ 1/3 + (1/9)x + (2/27)(x^2/2!) + (2/81)(x^3/3!)
C) To find the coefficient of x^n in the series expansion of f(x) in ascending powers of x, we use the formula for the nth derivative of a function evaluated at x = 0:
f^n(x) = n!/(3 - x)^(n+1)
The coefficient of x^n in the series expansion is given by the nth derivative evaluated at x = 0 divided by n!:
Coefficient of x^n = f^n(0)/n!
Substituting x = 0 into the nth derivative:
f^n(0) = n!/(3 - 0)^(n+1) = n!/3^(n+1)
The coefficient of x^n in the series expansion of f(x) is n!/3^(n+1).
A) To express f(x) in partial fractions, we need to factor the denominator.
The denominator of f(x) is (1+2x)(3-x). We can start by finding the partial fraction decomposition of each factor individually.
For the factor (1+2x), we can write it as:
1+2x = A
For the factor (3-x), we can write it as:
3-x = B
Now we can multiply both sides by the denominators to get rid of the fractions:
1 = A((3-x)) + B((1+2x))
Simplifying this equation gives:
1 = (3A - B)x + (A + 2B)
Now we can equate the coefficients of the powers of x on both sides.
Matching the constant term (the coefficient of x^0), we have:
1 = A + 2B
Matching the coefficient of x^1 term, we have:
0 = 3A - B
Solving these two equations will give us the values of A and B.
From the first equation, we can solve for A in terms of B:
A = 1 - 2B
Substituting this into the second equation:
0 = 3(1 - 2B) - B
= 3 - 6B - B
= 3 - 7B
Solving for B:
7B = 3
B = 3/7
Substituting B back into the equation A = 1 - 2B:
A = 1 - 2(3/7)
= 1 - 6/7
= 1/7
Therefore, the expression f(x) in partial fractions is:
f(x) = 1/((1+2x)(3-x))
= 1/((3/7)(1+2x) + (1/7)(3-x))
Note: To get the final expression, you might need to simplify further if necessary.
B) To find the first four terms in the series expansion of f(x) in ascending powers of x, we can use the method of Taylor series expansion.
The general formula for the Taylor series expansion of a function f(x) around a point a is:
f(x) = f(a) + f'(a)(x-a)/1! + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...
In this case, we want the series expansion of f(x) in ascending powers of x, so we can choose a = 0.
To find the first term, we substitute x = 0 into f(x):
f(0) = 1/((1+2(0))(3-(0)))
= 1/((1)(3))
= 1/3
So, the first term in the series expansion is 1/3.
To find the second term, we differentiate f(x) with respect to x, and then substitute x = 0:
f'(x) = -5/(3-2x)^2
f'(0) = -5/(3-2(0))^2
= -5/9
So, the second term in the series expansion is (-5/9)x.
To find the third term, we differentiate f'(x) with respect to x, and then substitute x = 0:
f''(x) = 20/(3-2x)^3
f''(0) = 20/(3-2(0))^3
= 20/27
So, the third term in the series expansion is (20/27)x^2.
To find the fourth term, we differentiate f''(x) with respect to x, and then substitute x = 0:
f'''(x) = -120/(3-2x)^4
f'''(0) = -120/(3-2(0))^4
= -120/81
= -40/27
So, the fourth term in the series expansion is (-40/27)x^3.
Putting it all together, the first four terms in the series expansion of f(x) in ascending powers of x are:
f(x) = (1/3) - (5/9)x + (20/27)x^2 - (40/27)x^3
C) To find the coefficient of x^n in the series expansion of f(x) in ascending powers of x, we need to look at the x^n term and find its coefficient.
From the previous part, we have the series expansion:
f(x) = (1/3) - (5/9)x + (20/27)x^2 - (40/27)x^3 + ...
The coefficient of x^n will be the coefficient of the x^n term. Looking at the series expansion, we can see that the coefficient of x^n will be zero for n > 3, as there are no terms with powers of x greater than x^3.
Therefore, for n > 3, the coefficient of x^n in the series expansion of f(x) in ascending powers of x will be zero.