f (x) = x 2 (x) = (1 + x 2 ) 2 (x) = (1 + x 2 ) 2 + 8x 2 (1 + x 2 ) 3 8x (1 + x 2 ) x (1 + x 2 ) 3 48x3 f(0) = tan 1 (0) = 0 f (0) = 2
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1 Math 5 Exam # Practice Problem Solutions. Find the Maclaurin series for tan (x (feel free just to write out the first few terms. Answer: Let f(x = tan (x. Then the first few derivatives of f are: f (x = + x f x (x = ( + x f (x = ( + x + 8x ( + x f (4 (x = 8x ( + x + 6x ( + x 48x ( + x 4 f( = tan ( = f ( = f ( = f ( = f (4 ( = so the Maclaurin series for tan (x starts out as x! x = x x Replacing x with x, we get that the Maclaurin series for tan (x is. Use the first two non-zero terms of an appropriate series to give an approximation of sin(x dx. Give (with explanation an estimate of the error (the difference between your approximation and the actual value of the integral. Answer: We know that the Maclaurin series for sin x is so the Maclaurin series for sin(x is x x! + x5, sin(x dx =! + x. [ x = (! + x dx x7 7! + x = 4 + ]
2 Thus, approximating by the first two terms, we see that sin(x dx 4 = 4. By the Alternating Series Approximation Theorem, the error of this estimate is no bigger than the next term in the series, which is =, so the error is certainly less than =... Find the it without using L Hôpital s Rule. Answer: From Problem we know that x sin x cos x The Maclaurin series for cos x is so the series for cos x is sin x =! + x x! + x4, 4x! + 6x4. x sin x cos x = x! + x Dividing numerator and denominator by x yields x ( 4x! + 6x4 = x x4! + x8 4! 6x = 4! = 4 =.! + x. 4x! 6x4 4. Find the Taylor series for e x centered at. What is the interval of convergence for this series? Answer: The Maclaurin series for e x is + x + x! + x! = replacing x with x, the Maclaurin series for e x is ( x n = ( n xn n! n!. x n n!. To find the interval of convergence, we use the Ratio Test: n+ xn+ ( (n+! n ( n x n n! = x n n + = x n n + =, which is certainly less than, so this series converges absolutely for all x. the interval of convergence is (,.
3 5. Find the Maclaurin series for x cos t dt. Answer: Since the Maclaurin series for cos x is x! + x4 = ( n xn (n!, we can replace x with t to get the Maclaurin series for cos t : t6! + t = ( n t6n (n!. x cos t dt = x [ ( n t6n (n! dt = ( n t 6n+ (6n + (n! where we evaluate the integral term-by-term. The first few terms of this series are x x7 4 + x Write out the first five terms of the Taylor series for x centered at x =. Answer: Let f(x = x and evaluate the first few derivatives of f: f (x = x f (x = 4x / f (x = 8x 5/ f (4 (x = 5 6x 7/ f( = f ( = ] x = ( n x 6n+ (6n + (n! f ( = 4 Hence, the Taylor series for x centered at is or, simplifying slightly, f ( = 8 f (4 ( = (x 4! (x + 8! (x 5 6 (x 4 + (x (x 8 + (x 48 5(x 4 76
4 7. Find the Maclaurin series for f(x = +x. What is its interval of convergence? Answer: Writing +x as we can use the geometric series to see that ( x, + x = ( x n = ( n n x n. Since the equality r = r n is only valid when r <, we see that this series converges for x <, meaning that x < / (we could also have seen this by using the Ratio Test, so the interval of convergence is (,. If it s not clear that the series doesn t converge at the endpoints, it s easy to check the endpoints: when x = /, the series is ( n ( n n = ( n, which diverges. When x = /, the series is ( n ( n n =, which also diverges. the series diverges at both endpoints and the interval of convergence is as stated above. 8. Plugging in x = to the Maclaurin series for e x, we can write e as e = k= How many terms are necessary to approximate e to within 8? You may take it as known that e. Answer: Let f(x = e x. Using the Taylor Approximation Theorem, the partial sum has an error no larger than n k= k!. k! = n! M (n +! n+, where M is an upper bound on f (n+ (x for all x such that x. Since f (n+ (x = e x for any n and since e x is an increasing function, f (n+ (x = e x e = e for any x such that x. e is what we re trying to approximate, so it s not a good choice for M, but in the problem we re told we may assume e <, so is a good choice for M. the error is no larger than M (n +! n+ = (n +!. 4
5 Hence, if we choose n so that the above quantity is 8, we ll be done. Of course, if n =, then (n +! = = 4 and 4 = 8, so we can approximate e to within 8 by 9. Find all the sixth roots of. k= k! = = 8. Answer: Notice that we can write in polar form as e iπ. Hence, if re iθ is a sixth root of, then it must be the case that e iπ = ( re iθ 6 = r 6 e i(6θ, so r =. Also, we could choose θ such that 6θ = π, so θ = π/6. Of course, so we could also choose (Notice that e i π 6 = e i π 6, so we can stop here. e iπ = e i(π = e i(5π = e i(7π = e i(9π = e i(π =..., θ = π 6, 5π 6, 7π 6, 9π 6, π Since π 6 = π and 9π 6 = π, the sixth roots of are. Write i in polar form. Answer: We first compute the modulus: so i = e iθ for some θ. In particular e i π 6, e i π, e i 5π 6, e i 7π 6, e i π, e i π i = + ( = + =, θ = tan ( = 5π i = e i 5π 5
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