Notes PDF
Introductory Problems
1. Find the Taylor series for ln(1 + x) expanded about x = 0. Find the radius of convergence of this series.
Answer
1. The series is x – x2/2 + x3/3 – x4/4 + … . The radius of convergence is 1.
Solution
1. To find the taylor series, compute the first few derivatives of f(x) = ln(1 + x) and look for patterns. f'(x)= 1/(1 + x) f”(x)= -1/(1 + x)2 f”'(x)= 2/(1 + x)3 f(4)(x)= -3!/(1 + x)4 and in general f(n)(x)= (-1)n-1(n-1)!/(1 + x)n . Then f(n)(0)= (-1)n-1(n-1)! and so f(n)(0)/n! = (-1)n-1/n. Recalling f(0)= 0, the Taylor series is ln(1 + x)= x – x2/2 + x3/3 – x4/4 + … + (-1)n-1xn/n + … Apply the Ratio Test to find the radius of convergence limn → ∞|((-1)nxn+1/(n+1)) / ((-1)n-1xn/n)| = limn → ∞|(n/(n+1)) x| = |x| The series converges for |x| < 1, so the radius of convergence is 1.
2. Without computing any derivatives, find the Taylor series of cos(√(x2 + 1)) expanded about x = 0.
Answer
1. The series is 1 – (x2 + 1)/2! + (x2 + 1)2/4! – (x2 + 1)3/6! + … + (-1)n(x2 + 1)n/(2n)! + …
Solution
1. Recall the Taylor series for cos(x): cos(x) = 1 – x2/2! + x4/4! – x6/6! + … + (-1)nx2n/(2n)! + … Substitute √(x2 + 1) for x, obtaining 1 – (x2 + 1)/2! + (x2 + 1)2/4! – (x2 + 1)3/6! + … + (-1)n(x2 + 1)n/(2n)! + …
3. Evaluate the limit limx → 0(tan(x) – sin(x))/x3
Answer
1. The limit is 1/2.
Solution
1. Because the denominator is x3, we need the series for the numerator through the cubic term. The series for sin(x) is familiar: sin(x) = x – x3/3! + … . To find the terms of the Taylor series for tan(x) up through the cubic term, first compute the derivatives (tan(x))’= sec2(x) (tan(x))”= 2 sec(x) sec(x) tan(x) = 2 sec2(x) tan(x) (tan(x))”’= 2(2 sec(x) sec(x) tan(x))tan(x) + 2 sec2(x) sec2(x) = 4sec2(x)tan2(x) + 2sec4(x) Evaluating these derivatives at x = 0 gives tan'(0)= 1 tan”(0)= 0 tan”'(0)= 2 Then the first few terms of the Taylor series for tan(x) are x + x3/3 Up through cubic terms, tan(x) – sin(x) = (1/3 – (-1/3!))x3 = x3/2 Then the limit is 1/2.
4. For what value of k is this limit limx → 0(1 – kx2 – cos(3x))/x4 finite? Find the limit for that k.
Answer
1. The value of k is 9/4. For that value of k, the limit is 27/8.
Solution
1. The Taylor series for cos(3x) is 1 – (3x)2/2! + (3x)4/4! – (3x)6/6! + … . Because the denominator is x4, in order for the limit to exist the kx2 term must subtract out of the numerator, so k = 32/2! = 9/2. For k = 9/2, the value of the limit is the ratio of the coefficients of the quartic terms, so the limit is 34/4! = 81/24 = 27/8.