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Math 231 Exam 3 Review Sheet by Ruth Luo1 Alternating Series • A

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Math 231 Exam 3 Review Sheet by Ruth Luo1 Alternating Series • A Math 231 Exam 3 Review Sheet by Ruth Luo1 Concepts Covered: Alternating series, Convergence tests, Power series, Taylor/Maclaurin Series, Taylor poly- nomials and error Alternating Series P P n P P n+1 • Consider a series an = (−1) bn or an = (−1) bn. If 1. bn > 0 for all n, 2. bn is decreasing (i.e., bn > bn+1) for all n, and 3. limn!1 bn = 0, P then an converges. P n P n+1 • If S = (−1) bn or S = (−1) bn is an alternating series that satises the three conditions above, then the error RN obtained when approximating S with partial sum SN is upper bounded by bN+1, that is, jRN j < bN+1: P P P • A series an converges absolutely if janj converges (this also imples that an converges). P P P • A series an converges conditionally if janj diverges but an converges. • Exercises 1. Show that P1 n n+1 converges conditionally. n=0(−1) n2+1 2. What is the smallest value of such that the partial sum approximates the sum P1 (−1)n N SN S = n=1 n2 within 1 ? 200 Convergence tests • The following are the new convergence tests we have learned. Note that it is still important to remember the old convergence tests for series (limit comparison test, comparison test, integral test). P an+1 • Ratio test - for a series an, let L = limn!1 j j. Then an P if L < 1, then an absolutely converges, P if L > 1, then an diverges, if L = 1, then the test is inconclusive. P 1=n • Root test - for a series an , let L = limn!1 (janj) . Then the same conditions hold as in ratio test. Ratio test is useful when contains factorials. Note that (n+1)! (n+1)·(n)·(n−1)·:::·2·1 . • an n! = (n)·(n−1)·:::·2·1 = n + 1 n 1=n • Root test is useful when an contains nth powers. E.g., (jc j) = c. • Exercises - determine if the following series converge absolustely, converge conditionally, or diverge. State explicitly which test you are using. 1. P1 (−3)nn2 n=1 n! 2. P1 (−2)2n n=1 nn 1 3. P (−1)n p n n=1 n3+5 4. P1 n 5nn n=1(−1) n! Power Series [email protected]; there may be typos. 1 A power series is a function of the form P1 n for some coecients . We say that the power series • n=0 cn(x − a) cn is centered at a. There are three options for the convergence of P1 n: • n=0 cn(x − a) P1 n converges for all values of , n=0 cn(x − a) x P1 n converges only when , n=0 cn(x − a) x = a P1 n has a radius of convergence , i.e., if , then P1 n converges. n=0 cn(x − a) R jx − aj < R n=0 cn(x − a) If , then P1 n diverges. jx − aj > R n=0 cn(x − a) • Use ratio test or root test to nd the radius of convergence. When R < 1, it is necessary to check the convergence on the endpoints, a + R and a − R individually. • The most basic example of a power series is 1 1 X = xn; which converges if jxj < 1(R = 1): 1 − x n=0 From this, we can get the power series of many other functions like 1 (by integrating the power • 1+x2 ; arctan x series of 1 , 1 (by dierentiating the power series of 1 , (by integrating the power series of 1+x2 (1−x)2 1−x ) ln(1 + x) 1 ), etc. 1+x • Dierentiating and integrating power series do not change the radius of convergence. • Exercises 1. Find the radius of convergence and the interval of convergence of the power series P1 4n(x−1)n . n=0 n+1 Find the power series centered at 0 of the following functions, then state the radius of convergence: 2. 1 1+x2 3. arctan x 4. x3 arctan(4x2) 5. R x3 arctan(4x2)dx 6. x2 (5−x)3 7. x 1+5x3 8. R 1=10 x 0 1+5x3 dx 9. Write down a number which approximates R 1=10 x to within −8. 0 1+5x3 dx 10 Taylor/Maclaurin series • A Taylor series is a power series representation of a function centered at some a. Given an innitely dier- entiable function f(x), the general form of the Taylor series of f is 1 X f (n)(a) f(x) = (x − a)n: n! n=0 (n) A Maclaurin series is a Taylor series where , i.e., P1 f (0) n • a = 0 f(x) = n=0 n! x : • You should know the series of the following functions: sin x = cos x = 2 ex = arctan x = (binomial series) (1 + x)k = • Exercises Find the Taylor series of the given functions centered at a. 1. f(x) = x3 cos(x2), a = 0 2. f(x) = cos x; a = π=2 3. f(x) = ln x, a = 2 p 4. f(x) = 4 1 + x, a = 0 5. Let f(x) = x3 cos(x2). Using the power series obtained in (1), nd f (11)(0). 6. We know that the power series centered at of 1 is P1 n. Show that this is the same a = 0 f(x) = 1−x n=0 x (n) series we get when we use (a) P1 f (0) n, and (b) binomial series, to nd the power series of . n=0 n! x f(x) 7. Find the sum of π2 π4 π6 1 − 222! + 244! − 266! + ::: 8. Find the sum of 52 53 54 −5 + 2! − 3! + 4! − ::: 9. Evaluate 1−cos(2x) . limx!0 1+2x−e2x 2 (Hint: write out a couple terms of the numerator and the denominator. Note that a0+a1x+a2x +::: limx!0 2 = b0+b1x+b2x +::: a0+0+0+::: = a0 .) b0+0+0+::: b0 10. Evaluate sin(x2)−x2 cos x . limx!0 x4 Taylor polynomials and error (n) Given a Taylor series for P1 f (a) n, we have that • f(x) = n=0 n! (x − a) f(x) = TN (x) + Rn(x) (n) where PN f (a) n is called the degree Taylor polynomial (initial chunk of the Taylor TN (x) = n=0 n! (x − a) N series that has degree at most N) and f (N+1)(z) R (x) = (x − a)N+1 N (N + 1)! for some z between a and x. In general, we cannot solve for z. Instead, we can nd an upper bound (a constant) for RN (x) for all x in some interval. ∗ ∗ ∗ • Given an interval, say [c; d], to maximize RN (x), we pick a values of z and x in [c; d] such that z maximizes (N+1) ∗ (N+1) and ∗ maximizes N+1. And so f (z ) ∗ N+1 . f (z) x (x − a) jRN (x)j ≤ j (N+1)! (x − a) j • Exercises x 1. Find the degree 4 Taylor polynomial, T4(x) of e centered at a = 0. 2. Find the degree 4 Taylor polynomial, T4(x) of sin x centered at a = 0. 3. Let f(x) = sin x. Use Taylor's Theorem to nd the maximum error in the approximation f(x) ≈ T4(x) on the interval [0; 1=2] (where a = 0). x 4. Let f(x) = e . Use Taylor's Theorem to nd the maximum error in the approximation f(x) ≈ T999(x) on the interval [0; 1=10] (where a = 0). 3.
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