Answer:
c_0 = 0
c_1 = 0
c_2 = -2
c_3 = 4/3
c_4 = -5/3
Explanation:
To find the coefficients of the power series representation of the function f(x) = 4xln(1+2x), we can use the formula for the nth coefficient of a power series:
c_n = f^(n)(0) / n!
where f^(n)(0) denotes the nth derivative of f(x) evaluated at x = 0.
Let's calculate the first few coefficients:
c_0:
To find c_0, we need to evaluate f(x) at x = 0:
f(0) = 4(0)ln(1+2(0)) = 0
So, c_0 = f^(0)(0) / 0! = f(0) = 0.
c_1:
To find c_1, we need to find the first derivative of f(x) and evaluate it at x = 0:
f'(x) = 4ln(1+2x) + 4x * (1/(1+2x))(2) = 4ln(1+2x) + 8x / (1+2x)
f'(0) = 4ln(1+2(0)) + 8(0) / (1+2(0)) = 0
So, c_1 = f^(1)(0) / 1! = f'(0) = 0.
c_2:
To find c_2, we need to find the second derivative of f(x) and evaluate it at x = 0:
f''(x) = 4(1/(1+2x))(2) + 8(1+2x)(-1) = 4 / (1+2x)^2 - 8 / (1+2x)
f''(0) = 4 / (1+2(0))^2 - 8 / (1+2(0)) = 4 - 8 = -4
So, c_2 = f^(2)(0) / 2! = f''(0) / 2 = -4 / 2 = -2.
c_3:
To find c_3, we need to find the third derivative of f(x) and evaluate it at x = 0:
f'''(x) = -4(2)(1/(1+2x))^3 - 8(-1)(1/(1+2x))^2(2) = -8 / (1+2x)^3 + 16 / (1+2x)^2
f'''(0) = -8 / (1+2(0))^3 + 16 / (1+2(0))^2 = -8 + 16 = 8
So, c_3 = f^(3)(0) / 3! = f'''(0) / 6 = 8 / 6 = 4/3.
c_4:
To find c_4, we need to find the fourth derivative of f(x) and evaluate it at x = 0:
f''''(x) = 8(3)(1/(1+2x))^4 - 16(2)(1/(1+2x))^3(2) = 24 / (1+2x)^4 - 64 / (1+2x)^3
f''''(x) = 24 / (1+2x)^4 - 64 / (1+2x)^3
f''''(0) = 24 / (1+2(0))^4 - 64 / (1+2(0))^3 = 24 - 64 = -40
So, c_4 = f^(4)(0) / 4! = f''''(0) / 24 = -40 / 24 = -5/3.
Therefore, we have the following coefficients:
c_0 = 0
c_1 = 0
c_2 = -2
c_3 = 4/3
c_4 = -5/3
To find the radius of convergence (R) of the power series, we can use the ratio test. The ratio test states that if the limit of the absolute value of the ratio of consecutive coefficients is L, then the radius of convergence is given by R = 1/L.
Let's apply the ratio test:
L = lim (n→∞) |c_(n+1)/c_n|
= lim (n→∞) |[(4/(n+1))ln(1+2/(n+1))] / [(4/n)ln(1+2/n)]|
= lim (n→∞) |(n/(n+1)) * [ln(1+2/(n+1)) / ln(1+2/n)]|
By taking the limit, we can simplify the expression inside the absolute value:
L = lim (n→∞) |(n/(n+1)) * [ln(1+2/(n+1)) / ln(1+2/n)]|
= lim (n→∞) |(n/(n+1)) * [ln((n+3)/(n+1)) / ln((n+2)/n)]|
= lim (n→∞) |[(n/(n+1)) * (n+1)/(n+2)]|
= lim (n→∞) |n/(n+2)|
= 1
Since the limit of the ratio is 1, the radius of convergence (R) is given by R = 1/L = 1/1 = 1.
Therefore, the radius of convergence of the power series is R = 1.