Math 3: Quiz 6 Solutions, Fall 05 Chapter 9 Keep in mind that more than one test will wor for a given problem. I chose one that wored. In addition, the statement lim a LR b means that L Hôpital s rule was used to transform from lim a to lim b. Also, exp x) means e x.. Determine whether the series are absolutely convergent, conditionally convergent, or divergent. a) )! Ratio Test Define a )!. Then a!. Apply the Ratio Test to a to show that a is absolutely convergent. So a + ρ a + + )! )! + 0 <. b) So, by the ratio test, a converges which means that a converges absolutely. ) tan + Integral Test Define a ) tan +. Using the integral test, define fx) tan x x. The derivative of + f is f x) x tan x x + ). Since f x) is less than zero for x > 0.76538, then f is decreasing from [, ). So we need to assess the convergence of the following integral b fx) dx b tan x x + u tan x du dx x + dx b tan b π [ u u du b ] tan b π tan b) π ) b π ) π 3 π 8 π 3 3 3 π Since the integral converges, then a converges. So a converges absolutely. Limit Comparison Test To show absolute convergence, compare a to b. So a tan c b + + lim tan + π ) π. Since c is positive and finite and since b converges by the P-series Integral Test), p > ) then a converges absolutely.
Quiz 6, page c) + ln ) Comparison Test and Alternating Series Test Define a ) ln. To show absolute divergence of a, pic b ln. Since >, then a > b. Since the smaller series b diverges, then the bigger series a diverges by the comparison test. So a does not converge absolutely. Next, we need to apply the Alternating Series Test to find if the series diverges or conditionally converges. Define b ln ln x and fx) x. Since f x) ln x x < 0 for x > e, then b is eventually decreasing. Finally, since lim b ln ln x LR x x lim x x 0, then a converges by the Alternating Series Test. Since a converges, but does not converge absolutely, then a converges conditionally. d) e) ) )! 3 Divergence Test Define a ) )! 3. Since lim a ) )! 3 and lim a + )+ + )! 3 +, then a oscillates between ±. Thus, a diverges. Therefore, a diverges by the Divergence Test. ) ln ) Integral Test Define a ) ln ) and b a ln ). Applying the integral test to a will show if a converges absolutely. Define fx). Note that fx) is continuous for x >. xln x) Since f x) + ln x x < 0 for x >, then fx) is a decreasing function for x >, as ln x) 3 required by the integral test. Further, since fx) is continuous, then we can use the Integral Test. So fx) dx dx b xln x) b [ ] ln b b u ln b dx xln x) u ln x du dx x ln ln b ln b du b ln u ) ln Since fx) dx converges, then a converges. Hence, a converges absolutely.
Quiz 6, page 3 f) ) 5 P-series Integral Test) and Alternating Series Test Define a ) 5. Since a diverges It is a p-series with p 5 ), then a does not converge absolutely. Next, we need to apply the Alternating Series Test to find if it diverges or conditionally converges. Define b 5 and fx) 5. Since f x) x 5 5 x < 0, then b 6 is a decreasing sequence. Finally, since lim b 0, then a converges by the Alternating Series Test. Since a converges, but does not coverge absolutely, then a converges conditionally.. Find the radius and interval of convergence for the following power series. The first step is to find where the power series converges absolutely. To do this, use either the Ratio Test or the Root Test They ll do most of the wor). Where the series converges absolutely will give the interval of convergence. Because ρ on the endpoints of the interval of convergence, the ratio or root test is inconclusive; hence another test Comparison Test, Limit Comparison Test, Integral Test, P-series Integral Test), Alternating Series Test, etc.) should be used to chec the endpoints. a) x 5 Absolute Convergence using the Ratio Test The power series will converge when ρ <. So a + ρ a x + 5 + ) 5 x lim x + ) 5 x. So the power series converges when ρ x <. The interval of convergence is So the radius of convergence is R. Chec the endpoints x < < x <. Let x Let x x 5 x 5 5 5 ) ) 5 ) Since ) converges by the p-series test p > ), then ) converges absolutely, and hence converges. After testing the endpoints, the interval of convergence is x [, ].
Quiz 6, page b) x Absolute Convergence using the Root Test The power series will converge when ρ <. So ρ a x x ) x, since lim. So the power series converges when ρ x <. The interval of convergence is x < x < < x <. The radius of convergence is R. Chec the endpoints Let x x 3) Let x x ) ) ) Since 3) converges by the p-series test p > ), then ) converges absolutely, and hence converges. After testing the endpoints the interval of convergence is x [, ]. c) ) 3 x )! Absolute Convergence using the Ratio Test The power series will converge when ρ <. So a + 3 + x + )! ρ a [ + )]! 3 x 3 x )! + ) + ))! 3 x + ) + ) 0 Since ρ doesn t depend on x and ρ <, then the power series converges everywhere R ).
Quiz 6, page 5 d) e) f) ) x + +!) + )! Absolute Convergence using the Ratio Test The power series will converge when ρ <. So a + x +)+ +!) + )! ρ a +)+ + )! + )! x + x! + )! + )! + )! x! + ) + )! x + ) + ) 0 Since ρ doesn t depend on x and ρ <, then the power series converges everywhere R ). π x 5) + )! Absolute Convergence using the Ratio Test The power series will converge when ρ <. So a + π + x 5) +) + )! ρ a + ) + )! π x 5) πx 5) + + )! + 3)! x 5) πx 5) + )! + 3) + ) + )! πx 5) + 3) + ) 0 Since ρ doesn t depend on x and ρ <, then the power series converges everywhere R ). ) x + ) 5 Absolute Convergence using the Ratio Test The power series will converge when ρ <. So a + + + ρ 5) x + a 5) x +. x + 5
Quiz 6, page 6 So the power series converges when ρ 5 x + <. The interval of convergence is 5 x + < x + < 5 5 < x + < 5 9 < x <. So the radius of convergence is R 5. Chec the endpoints Letx 9 Letx ) 9 ) 5 + ) ) 5 + ) 5 5 ) ) ) 5 5 ) 5) 6) Both 5) and 6) diverge by the divergence test. After testing the endpoints the interval of convergence is 9 < x < 9, ). 3. Find the Maclaurin series for the following: a) ln + x) Geometric Series Approach The Geometric series r converges to. Define r x. Then r + x x) ) x 7) Integrating both sides of 7) yields: + x dx ) x dx ln + x) ) x dx ) x + + Starting at instead of 0 yields ) x ln + x)
Quiz 6, page 7 Brute Force First, derive a general formula for the th derivative of fx) ln + x). So fx) ln + x) f x) + x f ) x) ) )! + x) f 3) x) ) )! + x) 3.. f ) x) ) )! + x), So fx) f0) + f ) 0) x! ) )! x! ) x. b) e x Notice that for fx) e x, f ) x) e x, so fx) f ) 0) x!! x x!. c) xe x The power series for e x was found in problem 3b. Since the power series for x is itself, then the power series for xe x is the product xe x x x! ) + x d) ln x ) + x Taing derivatives of gx) ln would be difficult since we would use the quotient x rule for each derivative. However, if we mae use of the identity ) + x ln ln + x) ln x), x x + then deriving the Maclaurin series would be easier. The Maclaurin series for ln + x) was derived in problem 3a. Finding the Maclaurin series for ln x) and then subtracting it!.
Quiz 6, page 8 ) +x from the Maclaurin series for ln + x) will give the Maclaurin series for ln x. Define y x. Then ) ln x) ln + y) y ) x) ) x x So ln ) + x ln + x) ln x) x ) x + x x x + x3 3 x + x5 5 x6 6 + x7 7 x8 8 + + x + x + x3 3 + x + x5 5 + x6 6 + x7 7 + x8 8 + x + x3 3 + x5 5 + x7 7 + x9 9 + x + e) x x + ) x + + x Define fx) x + ). Since the second part of fx) loos similar to, then we can + x) use the geometric series for a short cut. Since + x ) x 8) Taing the derivative of both sides of 8) and dividing both sides by yields: d dx + x d ) x dx So + x) ) d dx x + x) ) x 9) x fx) x + ) x [ )] + x x + x Combining 0) with 9) yields: note that the argument is x instead of x) fx) x ) + x x x ) ) ) x + ) 0)
Quiz 6, page 9 f) x 5 + x + 3x + Easy way: Since fx) x 5 + x + 3x + is a polynomial, then its power series is itself. Brute Force: So fx) x 5 + x + 3x + f0) f x) 5x + 8x 3 + 3 f 0) 3 f ) x) 0x 3 + x f ) 0) 0 f 3) x) 60x + 8x f 3) 0) 0 f ) x) 0x + 8 f ) 0) 8 f 5) x) 0 f 5) 0) 0 f 6) x) 0 f 6) 0) 0 and f ) x) 0 f ) 0) 0 for 6) fx) f ) 0) x! 5 f ) 0) x! f 0) 0) + f ) 0)x + f ) 0)! x + f 3) 0) 3! + 3x + 0! x + 0 3! x3 + 8! x + 0 5! x5 + 3x + x + x 5 x 3 + f ) 0)! x + f 5) 0) x 5 5! x 5 + x + 3x + Note that this means that the power series for a polynomial is the polynomial itself!
Quiz 6, page 0. Find the Taylor series for the following: a) fx) x +, about x Use binomial theorem) I ll mae a change of variable and find a Maclaurin series first, then transform bac to the original variable. Define y x so x y + ). So find the Maclaurin series of y +. fx) x + y + + y ) + y Since + y is a binomial series with and argument y, then fx) + y n0 ) y n ) n n0 ) x ) n n n b) fx), about x. x Define y x. Find the Maclaurin series for series. So. This is nothing more than a geometric + y fx) x + y y) ) y ) x ) 5. Extra credit: Do not wor on this problem unless you have all the previous wor done and you feel you now this chapter well. Find the sum of these series: a) b) Solutions will be given after class is over to those requesting it ) π 3 cos ) See me if you want a hint)