As f and g are differentiable functions such that. f (x) = 20e 2x, g (x) = 4e 2x + 4xe 2x,

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1 srinivasan (rs7) Sample Midterm srinivasan (690) This print-out should have 0 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Determine if points ( x 0 ) e x exists, and if it does, find its value.. it 0. it correct. it 0. it does not exist As f and g are differentiable functions such that f (x) 0, g (x) 0 we have to apply L Hospital s rule again. But f (x) 0e x, g (x) e x + xe x, from which it follows that f (x) 0, g (x). Consequently, the it exists and it points Find the n th term,, of an infinite series n when the n th partial sum, S n, of. it 6. it 0 Now x 0 ( e x e x x ) x(e x ) f(x) g(x) the series is given by. S n n n +. n(n + ) correct where f, g are everywhere differentiable functions such that f(x) 0, g(x) 0. Thus L Hospital s rule can be applied. But while so f (x) 0e x 0, g (x) (e x ) + xe x, f(x) g(x) f (x) g (x) ( 0(e x ) ) e x + xe x.. n.. n. n 6. n n(n + ) Since S n a + a + +, we see that But a S, S n S n (n > ). S n n n + n +.

2 srinivasan (rs7) Sample Midterm srinivasan (690) Thus a, while Consequently, for all n. n, (n > ). n + n n + n(n + ) points Determine whether the series ( ) n is convergent or divergent, and if convergent, find its sum.. convergent, sum correct points Let h be a continuous, positive, decreasing function on [, ). Compare the values of the series and the integral. A > B. A B A B 0 n 0. A < B correct In the figure h(n) h(z) dz.. convergent, sum 7. convergent, sum 6. convergent, sum 6. divergent a a a a 6... The given series is an infinite geometric series a r n with a and r. But the sum of such a series is (i) convergent with sum a when r <, r (ii) divergent when r. Consequently, the given series is convergent, sum. 6 the bold line is the graph of h on [, ) and the areas of the rectangles the terms in the series, h(n). n Clearly from this figure we see that a h() < a h() < h(z) dz, h(z) dz,

3 srinivasan (rs7) Sample Midterm srinivasan (690) while a h() < a 6 h(6) < and so on. Consequently, 6 h(z) dz, h(z) dz, as. Thus ( )n b n+ ( ) n b n <. Consequently, by the Ratio test, the given series is absolutely convergent. keywords: Szyszko A < B points Which one of the following properties does the series have?. divergent n ( ) n n + n. conditionally convergent. absolutely convergent correct The given series has the form n ( ) n b n, b n n + n of an alternating series. But the denominator is increasing very fast, so first let s check if the series is absolutely convergent rather than simply conditionally convergent. We use the Ratio test, for then ( )n b n+ b n+ ( ) n b n b n But (n + ) + n + n + 0n + 6 n + (n + ) + n points Determine whether the series n 0 n + cos nπ is conditionally convergent, absolutely convergent or divergent.. absolutely convergent. divergent. conditionally convergent correct Since cos nπ ( ) n, the given series can be rewritten as the alternating series with Now f(n) for all n, while ( ) n n + f(x) n + > f(n) x +. ( ) n f(n) n + 6 f(n + ) n + 0. Consequently, by the Alternating Series Test, the given series converges. On the other hand,

4 srinivasan (rs7) Sample Midterm srinivasan (690) by the Limit Comparison Test and the p-series test with p /, we see that the series n 0 f(n) is divergent. Consequently, the given series is conditionally convergent points Determine which, if any, of the following series diverge. (n) n (A) n! (B) n n. both of them. B only n (n + ) n. neither of them. A only correct To check for divergence we shall use either the Ratio test or the Root test which means computing one or other of +, /n for each of the given series. (A) The ratio test is the better one to use because Now + ( n! (n + )! n! (n + )! (n + ) n+ ). n n n +, while (n + ) n+ ( ) n n + n n (n + ). n Thus + ( ) n n + e > as, so series (A) diverges. n (B) The root test is the better one to apply because /n n + 0 as, so series (B) converges. Consequently, of the given infinite series, only A diverges points Determine the interval of convergence of the power series n x n n n.. interval of cgce [, ) correct. interval of cgce (, ). interval of cgce (, ). interval of cgce [, ]. interval of cgce (0, ) ( 6. interval of cgce ], 0 7. converges only at origin 8. interval of cgce (, ) When x n n n,

5 srinivasan (rs7) Sample Midterm srinivasan (690) then + n n x n+ (n + ) n+ x n nx (n + ). In this case + x. So by the Ratio Test, the given series (i) converges when x < and (ii) diverges when x >. It remains to check for convergence at x ±. Now when x, the series reduces to n ( ) n n which is the alternating harmonic series, hence convergent by the Alternating Series Test. On the other hand when x, the series reduces to n which is the harmonic series, hence divergent. Consequently n interval of convergence [, ) points Find a power series representation for the function. f(x). f(x). f(x) f(x) tan (x). ( ) n n + xn+ n + xn ( ) n n+ n + xn+ correct. f(x). f(x) 6. f(x) n+ n + xn+ ( ) n n n + xn We know that tan x On the other hand, n + xn+ x 0 + t dt. x + x + x +... on the interval (, ). Replacing x by x we thus see that + x x + x x on (, ). But then f(x) x 0 { n 0 ( ) n x n ( ) n t n} dt {( ) n x on (, ). Consequently, tan (x) 0 } t n dt ( ) n n + xn+ ( ) n n+ n + xn+ is a power series representation for tan (x) on (, ) points

6 srinivasan (rs7) Sample Midterm srinivasan (690) 6 Find the degree Taylor polynomial of f centered at x when f(x) x lnx ln(x ) + (x ). 0 + ln(x ) + (x ). 0ln + (ln + )(x ) + (x ). 0ln + (ln + )(x ) + (x ) correct. 0 + (ln + )(x ) + (x ) 6. 0ln + ln(x ) + (x ) The degree Taylor polynomial of f centered at x is given by T (x) f() + f ()(x ) When f(x) x lnx, therefore, +! f ()(x ). f (x) lnx +, f (x) x. But when f() 0 ln, f () (ln + ), f (). Consequently, the degree Taylor polynomial centered at x of f is 0ln + (ln + )(x ) + (x ).