Math 0230 Calculus 2 Lectures

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1 Math 00 Calculus Lectures Chapter 8 Series Numeration of sections corresponds to the text James Stewart, Essential Calculus, Early Transcendentals, Second edition. Section 8. Sequences A sequence is a list of numbers written in a definite order. It also can be treated as a function defined at the integer numbers only. A sequence can be finite or infinite. Notation: a, a, a,..., a n,... {a, a, a,...} {a n } {a n } {a 4, a 5, a 6,...} {a n } n=4 { } Example. a) n { b) sin πn { ) n n c) n } } n= a n = sin πn, a n = n, {,,,..., n,... } a n = )n n n, { 0,, } πn, 0,..., sin,... {, 9,, 5 8, 9 6,..., )n n,... n d) The Fibonacci sequence {f n } cannot be represented by a formula and is defined recursively: f =, f =, f n = f n + f n for n, {,,,, 5, 8,,,...} { Example. Find a formula for a general term of the sequence, 4 9, 5 7, } 7, 7 4,.... Solution: a n = n + n As we can see the terms of this sequence are getting smaller and smaller closser to 0. Definition A sequence {a n } has the it L: a n = L if the terms a n can be made as close to L as we like by taking n sufficiently large. If the it exists, then the sequence is called convergent, otherwise it s called divergent. If fx) = L and fn) = a n when n is an integer, then a n = L }

2 Definition a n = means that for any positive number M there is a positive integer N such that if n > N, then a n > M. Limit Laws If a n and b n are convergent sequences and c is a constant, then ap n = a n + b n ) = a n + b n a n b n ) = a n b n ca n = c a n c = c a n b n ) = a n b n a n = b n [ a n b n if b n 0 ] p a n if p > 0 and a n > 0 Squeeze Theorem If a n b n c n for n N and a n = c n = L then b n = L. such that if n > N, then a n > M. If a n = 0, then a n = 0 Example. Find n + n. Solution: Divide numerator and denominator by the highest power of n that occurs in the denominator and then apply the Limit Laws n + n = + /n /n = Example 4. Solution: Find n +. ) n n + = 0 < n + < n ) n + /n /n = = n + when n. 0 = 0, Then by Squeeze theorem n + n = 0. ) n = 0 and n + = 0.

3 Continuity and Convergence Theorem If a n = L and the function f is continuous at L, then fa n ) = fl). The sequence {r n } is convergent if < r and divergent for all other values of r. rn = 0 if < r < Definition A sequence {a n } is called increasing if a n < a n+ for all n. It is called decreasing if a n > a n+ for all n. A sequence is called monotonic if it is either increasing or decreasing. Definition A sequence {a n } is bounded above if there is a number M such that a n M for all n. A sequence {a n } is bounded below if there is a number m such that a n m for all n. If it is bounded above and below, then {a n } is a bounded sequence. Monotonic Sequence Theorem Every bounded monotonic sequence is convergent.

4 Example. = = It is an infinite sum series) of numbers. Section 8. Series In general, if we have a sequence {a n } then we may want to add its terms: a + a + a + + a i + = This expression is called an infinite series or just a series). a n To define a value of a series consider its partial sums: s = a, s = a + a, s = a + a + a,..., s n = a + a + a n = n a i, i= where s n is called nth partial sum. The partial sums form a new sequence {s n } If the sequence {s n } is convergent and s n = s, then the series and a n = s. The number s is called the sum of the series. If {s n } is divergent, then the series a n is called divergent. n Special notation: a n = a i i= Geometric series: ar n = ar n, a 0. It is convergent when r <. In this case its sum is If r, then the geometric series is divergent. ar n = Example. Represent the number.46 as a ratio of integers. a n is called convergent a r Solution:.46 = = = ) ) n = r <.

5 It is a geometric series with a = and < r = 00 <. Therefore,.46 = + 46/00 /00 = + 46/00 99/00 = = = Example. x n = if x <. x Example 4. Show that the series is convergent and find its it. nn + ) Solution: nn + ) = n n +. The partial sum is n n nn + ) = i ) i + i= i= = n n + n n + + n n + = n + n + n 4 n n + n + = + n + n +. n Therefore nn + ) = nn + ) = + n + ) = n +. i= Harmonic series: is divergent. n Theorem If the series a n is convergent, then a n = 0. Test for divergence If a n DNE or a n 0, then the series Example 5. Show that the series Solution: n nn + ) n nn + ) is divergent. a n is divergent. = 0. By the test for divergence the series is divergent. 5

6 Section 8. The Integral and Comparison Tests The Integral Test Suppose a n = fn), n n 0 where f is a continuous, positive function on [n 0, ) decreasing on [m, ) for some m n 0. Then the series a n is convergent if and n=n 0 only if the improper integral fx) dx is convergent. In other words: n=n 0 If n=n 0 fx) dx is convergent, then n=n 0 a n is convergent. If fx) dx is divergent, then n=n 0 a n is divergent. n=n 0 Example. Using the Integral Test determine whether the series or divergent. n= n nn ) is convergent Solution: Consider the function fx) = x xx ) = x x x. Define its domain as x n 0 = 4 on which f is continuous and positive. Is it decreasing there? Take its derivative f x) = x x) x ) = x 6x 4x + x 9 = x + 6x 9 x x) x x) x x) The denominator is always positive inside the domain. The descriminant of the numerator is D = 9 8 = 9 < 0. When x = 0 the numerator is 9 < 0. Hence, it is negative for x 4. So, the function is decreasing. 4 fx) dx = 4 x x x t dx = t 4 x x x dx u-sub: u = x x,, du = x ) dx, u4) = 4, ut) = t t. = t t t 4 du [ ] t u = t ln u = t 4 So, the integral is divergent and hence the series is divergent. The series is called the p-series. It is convergent when p > and divergent when p. np Example. Show that the series p-series is convergent. n 6

7 Solution: We apply the Integral Test and consider the function fx) = with the domain x defined as x. On the domain f is continuous, positive, and decreasing. t fx) dx = t [ dx = ] t = x t x So, the integral is convergent and hence the series is convergent. The Copmarison Test Suppose that a n and b n are series with 0 a n b n for all n. Then If b n is convergent, then a n is convergent. If a n is divergent, then b n is divergent. Example. Using the Comparison Test determine whether the series or divergent. n=5 n is convergent n + ) n Solution: Consider the function 0 n + ) n n =. The last series is the p-series n n with p = and hence it is convergent. By the Comparison Test the series n + ) is convergent an so is n=5 n n + ). The Limit Copmarison Test Suppose that a n and b n are series with positive terms. a n If = c, where c > 0 is a finite number, then either both series converge or both diverge. b n 5 + n Example 4. Determine whether the series a n = is convergent or divergent. n ) n Solution: Consider the geometric series b n = which is convergent because r = <. a n 5 + n = n b n n = 5 + n =. n n By the Limit Copmarison Test the series 5 + n n is convergent. 7

8 Section 8.4 Other Convergence Tests The Alternating Series Test If the alternating series ) n b n, b n > 0 satisfies then the series is convergent. ) b n+ b n for all n ) b n = 0 π Example. Determine whether the series ) n sin is convergent or divergent. n) n= π Solution: It is an alternating series note sin > 0, when n ). sin x is increasing ) n) π π function on [0, π/]. Hence, sin < sin and the condition ) holds. n + n) π ) Also, sin = 0 and the condition ) holds. n By the alternating series test the series is convergent. Alternating Series Estimation Theorem If s = alternating series that satisfies ) n b n is the sum of the then ) 0 b n+ b n for all n ) b n = 0 R n = s s n b n+ Example. How many terms of the series π ) n sin we need to add in order to find n) n= the sum to the accuracy error < 0.00? ) π Solution: The condition error < 0.00 is equivalent b n+ = sin < Then n + π we get n + < sin.00) = We need to solve this inequality for n?: n + < π 0.00 = Hence, n and n 046. Answer: 046 terms. Absolute Convergence A series a n is called absolutely convergent if the series a n is convergent. 8

9 A series a n is called conditionally convergent if it is convergent but not absolutely convergent. If a series a n is absolutely convergent, then it is convergent. Example. Determine whether the series convergent, or divergent. ) n n is absolutely convergent, conditionally Solution: It is an alternating series. Denote b n = b n+ = n + < n = b n, and test the given series is convergent. Now consider ) n n = n. n = n / >. The series n is divrgent. n b n The given series is conditionally convergent. = n = 0. n n. Then b n > 0 if n, By the alternating series is divergent harmonic series. Hence the series The Ratio Test ) If a n+ a n = L <, then the series a n is absolutely convergent. ) If a n+ a n = L > or a n+ a n =, then the series a n is divergent. ) If a n+ a n =, then the Ratio Test is inconclusive. Example 4. Determine whether the series 5) n n )! is convergent. Solution: a n = 5)n n )! a n+ a n = 5 n+ n )! 5 = n + )! 5 n nn + )n + ) = 0 <. By The Ratio test the series is absolutely convergent, and hence convergent. 9

10 The Root Test ) If n a n = L <, then the series a n is absolutely convergent. ) If n a n = L > or n a n =, then the series a n is divergent. ) If n a n =, then the Root Test is inconclusive. Example 5. Determine whether the series Solution: a n = ) n n + n n ) n n + n is convergent. n n + n n an = n = <. By The Root test the series is absolutely convergent, and hence convergent. 0

11 Section 8.5 Power Series A Power Series centered at a is a series of the form c n x a) n Here we assume that x a) 0 = even when x = a). Example. For what values of x is the series n! x n convergent? Solution: We use the Ratio test: a n = n! x n, a n+ a n = n + )! x n+ n! x n = n + ) x = if x 0. Hence, the series is convergent only when x = 0. Example. For what values of x is the series Solution: The Ratio test: a n = x 5) n+ n+ n + ) n n x 5) n = x 5)n n n, x 5) n n n convergent? n x 5 x 5 = n + ) Hence, the series is convergent only when x 5 < or < x < 7. <, x 5 <. If x = or x = 7, the Ratio test gives no information. Consider these cases separately. ) n n ) n x = : x 5 = = ). The series becomes =. n n n It is an alternating decreasing series with b n = and n n = 0. By the Alternating Series test it is convergent. x = 7: x 5 =. The series becomes It is the p-series with p = > and is convergent. n n n = Therefore, the given series is convergent when x 7. For a given power series n. c n x a) n there are only three possibilieties: ) the series converges only when x = a. ) the series converges for all x. ) there is a positive number R such that the series converges if x a < R and diverges if

12 x a > R. In the last case a R, a + R) is called the interval of convergence and R is called the radius of convergence. An addition work is required to determine if the endpoints a R and a + R of the interval of convergence are included. This is how R can be found: By the Ratio test the inequality a n+ a n = c n+ c n x a < must hold. It is equivalent to the inequality x a < c n = R. c n+ Example. Find the radius of convergence and interval of convergence of the series x + ) n n x + /) n n Solution: = = 5 n 5 n 5 x + n /)n, c n = n 5, n R = c n c n+ = n 5 5n+ 5 = n n+ = 5. Hence, the series is convergent only when x + < 5 or 4 < x <. x + ) n 5 n. x = 4: x = : x + ) n 5 n = 5)n 5 n = ) n. The series becomes x + ) n = 5n 5 n 5 =. The series becomes which is divergent. n ) n which is divergent. Therefore, the radius of convergence is R = 5 and interval of convergence is 4, ).

13 Example. Section 8.6 x = x n, x <. Representing Functions as Power Series Example. Find a power series representation and the interval of convergence for Solution: = 5x + = + 5x = + 5 x) = 5 x) 5 x ) n = ) n 5n n+ xn, c n = ) n 5n. n+ The radius of convergence is R = The interval of convergence is 5, ). 5 c n c n+ = 5 n n+ n+ = 5n+ 5 = 5. Example. Find a power series representation and the interval of convergence for Solution: = x 4 x x + 4 = ) n x = 4 x 4 + x = x ) n+ x. 4 x) = x 4 4 x) It is a geometric series and it converges when x 4 <, or x < 4, or x <. The interval of convergence is, ). Differentiation and Integration of Power Series If the power series c n x a) n has radius of convergence R > 0, then [ ) d ] c n x a) n = nc n x a) n dx [ ] ) c n x a) n = C + c n x a) n+ n + The radii of convergence of the powers in ) and ) are both R. Example 4. Find a power series representation for x). Solution: x) = d [ ] [ = d ] x n = dx x dx nx n 5x +. x x + 4.

14 Example 5. Find a power series representation for ln5x + ) and its radius of convergence. Solution: ln5x + ) = 5 = C + c n = ) n n n. 5x + dx = ) n 5 n+ n+ n + ) xn+ = C + 5n The radius of convergence is R = [ ] ) n 5n+ xn n+ ) n 5n n n xn. c n c n+ = 5 n n n n+ n + ) = 5 n+ 5 n + n = 5. 4

15 Section 8.7 Taylor and Maclaurin Series Taylor Series If f has a power series representation at a: fx) = c n x a) n, x a < R, then c n = f n) a). So, n! fx) = f n) a) n! x a) n = fa) + f a)x a) + f a)! x a) + f a) x a) +! Maclaurin Series The Taylor series of f centerd at 0 is called Maclaurin Series: f n) 0) fx) = x n = f0) + f 0)x + f 0) x + f 0) x + n!!! Example. Find the Maclaurin series for fx) = sin x. Solution: We need to find coefficients c n = f n) 0) : n! c 0 = f0) = sin 0 = 0, c = f 0) = cos 0 =, c = f 0)! c = f 0)! = cos 0! =!, c 4 = f 4 0) 4! = sin 0 4! = sin 0 = 0, c 5 = f 5 0) 5! Hence, sin x = x x! + x5 5! x7 7! + = ) n x n+ n + )! The nth-degree Taylor polynomial of f at a is T n x) = Then fx) = T n x). The remainder of the Taylor series is R n x) = fx) T n x). = 0, = cos 0 0 = 5!,... n c i x a) i. i=0 Taylor s Formula R n x) = f n+) z) n + )! x a)n+, for some z strictly between x and a. Example. Find the Maclaurin series for fx) = + x) k, where k is any real number. Solution: fx) = + x) k, f0) = f x) = k + x) k, f0) = k f x) = kk ) + x) k, f0) = kk ) f x) = kk )k ) + x) k, f0) = kk )k ) 5

16 ... f n) 0) = kk )k ) k n + ) We obtain The Binomial Series + x) k = where ) k = n ) k x n n kk )k ) k n + ), k is any real number and x <. n! Example. Use the Binomial series to expand the function fx) = Find its radius of convergence. Solution: fx) = 8 x) / = 8 x )) / = 8 x ) /. 8 Using the Binomial series with k = and with x replaced by x, we have 8 / ) n = 8 x x ) / = 8 = [ ) ) x ) + 8 )! ) n x 8 ) x 8 ) + ) ) 5 n + ) ] x ) n n! 8 )! ) ) 5 8 x as a power series. x ) 8 = [ + 6 x +! 6 x + 5 ] 5 n )! 6 x + + x n + n! 6 n The series is convergent when x <, that is, x < 8. So the radius of convergence is R = 8. 8 Several Special Series see page 484. e x x n = n! = + x! + x! + x! + sin x = ) n x n+ n + )! = x x! + x5 5! x7 7! + cos x = ) n xn n)! = x! + x4 4! x6 6! + 6

17 and etc. Example 4. Find the Maclaurin series for fx) = e x cos x. ] ] Solution: fx) = [ + x + [ x + x 6 + x4 4 + x + x4 4 = + x + x + x 6 + x4 4 x x x4 4 + x4 4 + = + x x x

18 Section 8.8 Applications of Taylor Polynomials Recall The nth-degree Taylor polynomial of f at a is T n x) = Then fx) = T n x). n c i x a) i. i=0 Approximating Functions by Taylor Polynomials fx) T n x). The remainder of the Taylor series is R n x) = fx) T n x). Taylor s Formula R n x) = f n+) z) n + )! x a)n+, for some z strictly between x and a. Alternating Series Estimation Theorem R n = s s n b n+ Example. a) What is the maximum error possible in using the approximation cos x x + x4 4 to find cos π 0 when 0. x 0.? Use this approximation to find cos π 0 decimal places. Notice, that π 0 < 0.. correct to five b) For what values of x is this approximation accurate to within 0.000? Solution: a) We can apply the alternating series estimation theorem if we consider the given series as an alternating series with b =, b = x, b = x4 4, b 4 = x6 70. Then R b 4. If 0. x 0., then the maximum value for b 4 is 0.) = Hence the error is smaller than and the approximation gives an estimate correct to five decimal places. π π 4 0) 0) The estimate is cos x + 4 = b) The error is smaller than if b 4 = x6 70 < We have to solve this inequality: x 6 < , x 6 < 0.07, x <

19 So the given approximation is accurate to within when x < If we used the Taylor s formula then we would get T 4 x) = T 5 x), R 5 x) = f 6) z) x 6, 6! where f 6) z), since f 6) z) is either cos z or sin z. Then the Taylor s formula gives the same estimate for the error as the alternating series does: R 5 x) = f 6) z) x 6 6! x