1. Find the Maclaurin series for f(x)=ln(1+3x) using the definition of a Maclaur
ID: 1888094 • Letter: 1
Question
1. Find the Maclaurin series for f(x)=ln(1+3x) using the definition of a Maclaurin series. [Assume that has a power series expansion]
Explanation / Answer
1) ln(1 + 3x) = S(n=1 to 8) (-1)^(n-1) 3^n x^n/n. The ratio test yields r = lim(n?8) |[(-1)^n 3^(n+1) x^(n+1)/(n+1)] / [(-1)^(n-1) 3^n x^n/n]| = 3|x| lim(n?8) n/(n+1) = 3|x|. So, this series converges (at least) when r = 3|x| < 1 ==> |x| < 1/3. Thus, the radius of convergence is 1/3. ---------------------------- 2) cos(3x)/7 = S(n=0 to 8) (-1)^n 3^(2n) x^(2n)/[7 * (2n)!] The ratio test yields r = lim(n?8) |[(-1)^(n+1) 3^(2n+2) x^(2n+2)/(7 * (2n+2)!)]| / [(-1)^n 3^(2n) x^(2n)/(7 * (2n)!)]| = 9x^2 * lim(n?8) (2n)! / (2n+2)! = 9x^2 * lim(n?8) (2n)! / [(2n+2)(2n+1) (2n)!] = 9x^2 * lim(n?8) 1 / [(2n+2)(2n+1)] = 0. Since r = 0 < 1 for all x, this series has infinite radius of convergence. ------------------------- 3) Note that (1 + t)^a = S(n=0 to 8) C(a, n) t^n, where C(a, 0) = 1 and C(a, n) = a(a-1)(a-2)...(a-n+1) / n!. Letting a = -4 and t = x/3, we have (1 + x/3)^(-4) = S(n=0 to 8) C(-4, n) (x/3)^n where for n > 0, C(-4, n) = -4 * -5 * ... * (-n-3) / n! = [(-1)^n (n + 3)! / 3!] / n! = (-1)^n (n+3)(n+2)(n+1) / 6 [note this also holds for n = 0!!]. So, we have 1/(1 + x/3)^4 = S(n=0 to 8) (-1)^n (n+3)(n+2)(n+1) x^n / (6 * 3^n) Multiply both sides by 5: 5/(1 + x/3)^4 = S(n=0 to 8) 5 (-1)^n (n+3)(n+2)(n+1) x^n / (6 * 3^n). Now, we use the Ratio Test. r = lim(n?8) |[5(-1)^(n+1) (n+4)(n+3)(n+2) x^(n+1) / (6 * 3^(n+1))] / [5 (-1)^n (n+3)(n+2)(n+1) x^n / (6 * 3^n)]| = (|x|/3) * lim(n?8) [(n+4)(n+3)(n+2)] / [(n+3)(n+2)(n+1)] = |x|/3 So, this series converges (at least) for |x|/3 < 1 ==> |x| < 3. Thus, the radius of convergence is 3. -------------------------------- 4) f(x) = 5x^(-1) f '(x) = 5 * (-1) x^(-2) f ''(x) = 5 * (-1)(-2) x^(-3) ... f^(k)(x) = 5 * (-1)^n * n! / x^(n+1) for all n = 0. ==> f^(k)(-3) = 5 * (-1)^n * n! / (-3)^(n+1) = -5 * n! / 3^(n+1) for all n = 0. So, the desired series is given by S(n=0 to 8) [-5 * n! / 3^(n+1)] (x - (-3))^n / n! = S(n=0 to 8) [-5 / 3^(n+1)] (x + 3)^n. I hope this helps!Related Questions
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