. (a) n = a n converges, so we know that a n =. Therefore, for n large enough we know that a n <. It follows that (a n ) < a n for n large enough. Thus the series converges by direct comparison to n= a n. (b) We ll show n= (a n) converges. This will mean that n= (a n) converges absolutely and therefore converges. As argued in part (a), for n large enough we know that a n <. It follows that (a n ) < a n for n large enough.therefore n= (a n) converges by direct comparison to n= a n. (c) This diverges by the Nth term test for Divergence. We know that (a n ) =. (You can argue this in several ways. One of these is using part (a).) Therefore (a n) =. (d) We ll show n= ( )n (a n ) converges. This will mean that n= ( )n (a n ) converges absolutely and therefore converges. ( )n <, so ( )n (a n ) < a n. Therefore n= ( )n (a n ) converges by direct comparison to n= a n.. (a) Insufficient information. If for instance a n = n then the series will converge but if a n = series will diverge. n then the (b) Insufficient information. If for instance a n = n then the series will converge. On the other hand, if a n = ln n then the series n= n ln n will diverge. You can show this will diverge using the integral test. (Note: this problem was harder than I anticipated when I wrote it. We would not expect you to come up with this on your own.) (c) Converges. Factor out of the sum. (d) Insufficient information. f for instance a n = n then the series will converge but if a n = n then the series will diverge.. Let S n = b + b + b +... + b n. It means that S n =. 4. (i) The series a n n converges. This can be seen as follows. a n and a n n are both series with positive terms, a n converges, and for all n, an a n n. By the comparison test for convergence, we conclude that a n n also converges. (ii) The series ( ) n a n a n diverges. Since a n converges, we must have a n =, therefore =, taking into account the fact that all a n s are positive. Thus the sequence ( )n diverges, and by the divergence test the series ( ) n a n diverges. (iii) The series (sin n)a n converges, in fact it converges absolutely. Indeed, since sin n, we have that for all n, sin n a n a n, and sin n a n converges by comparison with a n. This means that (sin n)a n also converges, because an absolutely convergent series is also convergent. (iv) It cannot be determined whether ( ) nan 4 converges or diverges. If a n = ( 4 ) n (a n = ( 4 n) ( ) for instance, then ) nan 4 = is divergent. If a n = ( ) n (an = ( n), ) then ( ) nan 4 = ( n ) is convergent. a n. First we verify the statements in the hint: (i) The derivative of an even function is odd. If f is even then f(x) = f( x). Take the derivative of each side. d dx f(x) = d dx f( x). This gives f (x) = f ( x). f is an odd function. (ii) The derivative of an odd function is even. If f is even then f(x) = f( x). Take the derivative of each side. d dx f(x) = d dx f( x). This gives f (x) = f ( x). f is an even function. Note that if g is odd then g() =.
The Taylor series generated by f about x = is given by f (n) () x n. (a) If f is even then f is odd and f is even, f is odd, etc. In other words, f (n) is even if n is even and odd if n is odd. But then if n is odd f (n) is odd and f (n) () =, so the Taylor series centered at x = has only even powers of x. (b) If f is odd then f is even and f is odd, f is even, etc. In other words, f (n) is even if n is odd and odd if n is even. But then if n is even f (n) is odd and f (n) () =, so the Taylor series centered at x = has only odd powers of x. 6. sin u = ( ) n un+ for all u. (n + )! Let u = x. Then sin x = ( ) n (x ) n+ (n + )! = ( ) n x4n+ (n + )! Divide by x. Then sin x x = ( ) n x4n+ (n + )! 7. First let s do 7b (which should have been its own problem but got stuck in the middle of this in a TeX error.) (7b) The Taylor series generated by f about x = π 4 is given by f (n) ( π 4 ) (x π 4 )n. Take the derivatives of sin x and evaluate them at π/4. We know that sin π 4 = and cos π 4 =. We get + (x π 4 )! (x π 4 )! (x π 4 ) + 4! (x π 4 )4 +! (x π 4 ) ++ (x π 4 )6 + 7! (x π 4 )7 7. (a) We know cos u = ( ) n un for all u. (n)! Let u = x. cos x = ( ) n (x ) n = ( ) n x6n (n)! (n)! x cos x = ( ) n x6n+ (n)! (c) I ll write this in summation notation, but it s actually easier to just write out the first 4 terms to answer this part of the problem.. [. ] [. ] x cos x dx = ( ) n x6n+ dx = ( ) n x6n+ (n)! (n)! dx = ( ) n (n)!. x cos x dx = ( ) n x 6n+ (n)! 6n +. = ( ) n (.) 6n+ (n)!(6n + ). x cos x dx = (.) (.)8 8 + (.)4 4! 4 (.) +... (You could simplify this more, but I won t.) (d) We can use the alternating series error estimate because the series is alternating in sign, the terms are decreasing in magnitude, and the terms tend towards zero. The third term is less than 6, so only the first two non-zero terms of the series are necessary.. x 6n+ dx
8. (a) To find the Taylor series of f(x) at x = one can compute the following: f() + f ()(x ) + f ()! (x ) + f ()! f(x) = x f() = ( 9) f (x) = ( x) f () = ( 9) f (x) = ( x) f () =! ( 9) f (x) = ()()( x) 4 f () =! ( 9) 4 f (4) (x) = 4!( x) f(4)() = 4! ( 9)... f n (x) = ( x) (n+) f (n) () = ( 9) (n+) This gives us (x ) +... + f (n) () (x ) n +... or 9 +! (x ) 9 9! (x ) +! 9 4! (x ) +... + ( ) n+ 9 (n+) (x )n +... 9 (x ) (x ) (x ) n+ (x )n + 9 9 + 9 4 +... + ( ) +... 9 (n+) An alternative method is the following: To find the Taylor series of f(x) at x = a is the same as to find the Maclaurin series of f(a + t) at t = and then plug x = t a. In our case, f(x) = x and a = so that Here t = x. (b) The geometric series /9 x 9 9. f( + t) = /( ( + t)) = /(t + 9) = ( /9)(/( + t/9)) = /9( t/9 + (t/9) (t/9) + ) = ( ) n+ t n /9 n+. n (x )n ( ) 9 n (with a = /9 and r = (x ) 9 converges to /9 9 + x < Therefore, for x < 9 the series converges to f(x). The radius of convergence is thus (c) Since 7 < 9, the series converges to f(7). (d) Since / > 9, the series diverges. 9. One can use repeated differentiation to compute the values of f and of its derivatives at, or one can simply observe that the Taylor series about x = for a polynomial in (x ) is the polynomial itself. Therefore by identifying the coefficients of the Taylor series to that of the given polynomial one gets: for f() = f () = f ()! = f (6) () = 7.. The first few derivatives of f(x) = cos x and their values at π/ are: f(x) = cos x f( π/) = f (x) = sin x f ( π/) = f (x) = cos x f ( π/) = f (x) = sin x f ( π/) = f (4) (x) = cos x f (4) ( π/) =
After that the pattern repeats itself. Using the formula giving the Taylor series in terms of the values of the derivatives we obtain: cos(x) = (x + π/) (x + π/) +! = n (x + π/)n+ ( ). (n + )! (x + π/)! (x + π/)7 7!. (a) The function cosh x is even, and it goes to infinity exponentially fast as x ±. So the graph looks like a parabola going to infinity exponentially. The function sinh x is odd, and it goes to exponentially fast as x, and to as x. Thus the graph looks like the graph of x, except that it goes much faster to infinity. (b) d cosh x dx = sinh x d sinh x dx = cosh x (c) We know the Taylor series for e x, and also for e x (by substituting x for x in the Taylor series for e x ): e x = + x + x! + x! + x4 4! + e x = x + x! x! + x4 4! Subtracting the second series from the first term by term (this is allowed since both series converge absolutely for all x) and dividing by two we get: sinh x = ex e x = x + x! + x! + = (d) Taking the derivative of the series above term by term, we obtain: cosh x = d sinh x dx x n+ (n + )! = + x! + x4 4! + = x n (n)! i. ln( +x +x x ) = ln( + x) ln( x). So in order to find Taylor series for ln x we have to find Taylor series for ln( + x) and ln( x). ln( x) = dx x and x = xn. So ln( x) = x n dx = x n x n+ dx = n + + C. Since for x = we have ln( x) = ln =, C =. Using substitution we can find Taylor series for ln( + x): ln( + x) = ln( ( x)) = Finaly, for ln +x x = ln( + x) ln( x) we have ln + x x = ( x) n+ n + + ( x) n+ n + x n+ n + = + = x x + x x4 4 +... ( x) n+ + x n+. n + If n is odd ( x) n+ + x n+ =, if n = k is even ( x) n+ + x n+ = x n+. So in the Taylor series for ln +x x we ll have only odd powers of x and the power series will be ln + x x = x k+ k + k= = x + x + x + x7 7 +.... 4
ii. If x = + x = 4 and x = +x. So x =. iii. Using the first four nonzero terms of series in (a) for x = / we get ln + + + 7 7 = 6 764.69477. Using the first four terms of the series for ln( + x) and x = we get Using calculator ln.694786. i. n= ( e π ii. Series n= n + n +n ln + 4 = 7.8 ) n is a geometric series with r = e π. Since e π <, the series converges. is divergent by N-th tern test: n + n + n = + n + n =. iii. Series ( ) n π n= n is convergent by AST: signs alternates and absolute value of terms is decreasing and goes to. iv. Series n= n 4 n converges. To see this we can apply either comparison or ratio test. A. For any n n 4 n n 4. Series n n= n 4 is geometric with r = n 4 and hence converges. By comparison test series we started with also converges. a B. n+ a n = n+ 4 n 4 n+ (n+)! = n 4(n+) Ratio test. v. Series n= =. Series is convergent by n(n 4) converges by it comparison test. Let s compare this series to n. n(n 4) n = n n(n 4) = 4 n = By it comparison test this series converges, since n= n converges. vi. Terms of the series n= cos n (.) change signs. Let s show, that this series is absolutely convergent. n For any n cos n (.) n (.). n n= (.) is geometric series with r = n. and hence converges. By comparison test series n= cos n (.) converges. This means that series we started with is n absolutely convergent and hence convergent. 4. For any real number p, the alternating p-series is the infinite series Σ ( ) n n= n p = p + p 4 p +... a) Show that the alternating p-series diverges for p. Solution: There are two cases to consider. p = : Then the general term is ( ) n, so the it of the nth term does not exist the sequence alternates between and. By the n-th term test, the series diverges. p < : If p is negative, n = n p and p is positive. The it of a positive power of n as n approaches p infinity is infinity, so the nth term goes to infinity in absolute value. Therefore the series again diverges by the nth term test. b) Show that the alternating p-series is convergent but not absolutely convergent if < p. Solution: To see that the series is convergent, notice that ) it is alternating, ) the terms are decreasing: n p < (n + ) p so n > p (n+), and ) the general term goes to zero: n p, so implies p n. p Therefore the alternating series test applies to show the convergence of the series. As for the lack of absolute convergence: when we take absolute values we get the conventional p-series, which we learned
(using the integral test) diverges for p. c) Show that the alternating p-series is absolutely convergent for p >. Solution: As above, absolute convergence means that the series must converge upon taking absolute values, but upon taking absolute values we again get the usual p-series. Since p > it converges, which gives absolute convergence of the alternating p-series..a) Use the Ratio Test: Look at a n+ a n. Using lhospitals rule, establish the following it. n(ln(n + )))x n+ (n + )(ln(n + ))x n = x then conclude using the ratio test that the radius of convergence of the power series is. b) When x = use the alternating series test to prove that the corresponding series converges. for x = proceed like this; use the substitution u = ln( + x) to show that the integral (+x) ln(+x) dx diverges. By direct comparison, conclude that x ln(+x) dx also diverges. Now by the integral test, conclude that the series with x = diverges. Finally, conclude that the interval of convergence is [, ) 6. Use the Ratio Test: Look at a n+ a n. Establish the following it. n(x ) n+ (n + )(x ) n+ = x then conclude using the ratio test that the radius of convergence of the power series is. When x = 4 conclude that the series is the harmonic series and thus diverges. When x = use the alternating series test to conclude the series converges. Thus the interval of convergence is [, 4) 7.Use the Ratio Test: Look at a n+ a n. Establish the following it. n+ (x ) n+ n+ (x ) n = x now note that this it is less than precisely when x. <. Thus the radius of convergence is. When x = 7. the series is just which clearly diverges by the nth term test for divergence. When x =. the series is just ( ) n which also diverges by the nth term test for divergence. THus the interval of convergence is (., 7.) 8. There are of course (infinitely) many power series of given radius of convergence. we will give one example in each case. For convenience we can consider a power series centered at, given by a nx n. i. R = If a n+ x n+ a n x n > for all x, then by ratio test the series diverges for all x, and the radius of convergence is. If we take a n =, so that a n+ /a n = n +, then (n + )!x n+ x n = (n + )x =, for all x. In particular the series xn has radius of convergence. 6
ii. R = This time we should look for a n s that make this it exist and a n+ x n+ a n x n <, for all x, then the series converges for all x, and so have radius of convergence. If we take a n = /, so that a n+ /a n = /(n + ), then x n+ /(n + )! x n / = x n + = <, Therefore the series xn has radius of convergence. (Does this series look familiar?) iii. R = Now if a n+ /a n = /, then a n+ x n+ a n x n = x, So by ratio test, the series converges if x x <, and diverges if >. Therefore the radius of convergence of this series is. Taking a n = would satisfy a n n+ /a n = /, and these can be the coefficients of our series. 9. The center of the power series should be the the center of its interval of convergence. So a power series with interval of convergence (, )should be centered at 6, and has radius of convergence 6. So the power series should look like a n(x 6) n. And similar to the above question, if we take a n = (x 6) n then the series 6 has radius of convergence 6. And this particular series doesn t converge n for the endpoints x =, (where we get ), and x = (where we get ( )n ).. Suppose at the end of the n-th day, there re s n bees in the beehive, out of which a n bees are newcomers. Obviously s n = a +... + a n. At the end of the first day, the beehive has only bee, which is a newcomer, i.e. a =, s =. Since every newcomer brings home new roommates the next day, we have a n+ = a n. Hence by induction a n = n. Then s n = + +... + n = n+. We need to solve n for n+ = 64, and the solution is n = 6.. Draw a picture for yourself.. e x = i. area of s = = 4 area of s = ; area of s = ; area of s = / ii. sum of the areas of all the squares = 4 + + + + + +... = 4 (This is a geometric series.) x n Therefore, e = ( ) n. = 8 6 n, This is an alternating series with the terms decreasing in magnitude and tending towards zero, so we can use the alternating series error estimate. We look for the first term whose absolute value is less than. n = 7 is the smallest n for which <. Therefore we need to use the approximation e 6 ( ) n - we go up to the term. This ends up being 7 non-zero terms. In other words, +!! + 4!! + is an approximation of e with error less than.. This is the 4th degree Taylor polynomial for cos x about x =. We know the Taylor series generated by cos x about x = converges to cos x For x <. this is an alternating series with the terms decreasing in magnitude and tending towards zero, so we can use the alternating series error estimate. The first unused term has magnitude x6. For x <. we know x6 < ( )6 = 6 = 7 = 4 7
4. As this is an alternating series, we can estimate the error by using the alternating series error estimate. This is that the size of the error is less than the first term that is not included in the approximation. In this case, we want the error to be less than than.. Thus we would like to find an odd number n whose inverse is less than.. The first one is. Thus our last term can be /99, which means that we have terms. This can be seen by noting that the last term of the partial sum with n terms will be plus or minus /( + (n )).. a. The easiest way to do this is to use the binomial series expansion with k = /. The result is + x x 8 b. When x =. this series is alternating if we add the first two terms to make a single term. Thus, we can use the alternating series error estimate. The first term not included is /6, which our upper bound for the error. c. When x =. this series is no longer alternating. Thus we have to use Taylor with remainder and Taylor s inequality i.e. f(x) = T n (x) + R n (x) where R n (x) M (n + )! x a n+. and M is a constant that is greater than or equal to f (n+) (c) for all c between x and a. Here, we have n = (because we have three terms), so we have to find an upper bound for f () between and. Taking derivatives, f () (x) = 4 + x / which obtains its maximum at x =. Thus we can take M equal to /4, and substituting our M as well as a =, we get R (x) x 4! = x 4. so R (.) (.) 4. 8