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3. Use the Leibniz test (alternating series test) to test whether the power series for arctan(x) centered at 0 converge...
3) Later in this course, we will learn that the function, arctan x, is equivalent to a power series for x on the interval -1sxs: 2n+1 (-1)" arctan x = We can use this power series to approximate the constant π . a) First, evaluate arctan1). (You do not need the series to evaluate it.) b) Use your answer from part (a) and the power series above to find a series representation for (The answer will be just a series-not...
(1 point) Find the first five non-zero terms of power series representation centered at x = 0 for the function below. f(x) = arctan(3) Answer: f(x) = + 0 1 /4 What is the radius of convergence? Answer: R= 4
(1 point) Find the first five non-zero terms of power series representation centered at x = 0 for the function below. f(x) = arctan(3) Answer: f(x) = + 0 1 /4 What is the radius of convergence? Answer: R= 4
In the following, we will tse a kmown power series to approximate 1/2 arctan(r) dr to within 0.00001 of the actual value of the definite integral (a) [2pt] Use a known power series representation to express (ctan(x) as a Maclaurin series. What is the radius of the series convergence? 1/2 (b) [4pts] Use your answer from part (a) to express(r) dr as an alternating series (c) [6pts] Your series in part (b) will converge by the Alternating Series Test. (You...
Question 6 (20 points). Determine whether the following series converge or diverge, and by which test: o arctan(k) k-0 oo ok k! k-0
Question 6 (20 points). Determine whether the following series converge or diverge, and by which test: o arctan(k) k-0 oo ok k! k-0
2. The following series are not power series, however, we can still use the ratio test to find the set of x for which it converges. Use the ratio test or the root test to find all values of x for which the following series converges. Remember to check the points where the test is inconclusive. (a) (5 points) (In(x))" Only check for x > 0
Use the power series itxË (-1)"X", Ixl < 1 -n=0 to determine a power series for the function, centered at 0, 14 02 7 f(x) (x + 1) dx2 ( x + 1 00 f(x) no Determine the interval of convergence. (Enter your answer using interval notation.) 3. [-17.69 Points] DETAILS LARCALC11 9.2.061. Find all values of x for which the series converges. (Enter your answer using interval notation.) 00 (8x)" n=1 For these values of x, write the sum...
7. (a) Use the well known Maclaurin series expansion for the cosine function: f (x ) = cos x = 1 x? 2! + 4! х 6! + (-1)" (2n)! . * 8! 0 and a substitution to obtain the Maclaurin series expansion for g(x) = cos (x²). Express your formula using sigma notation. (b) Use the Term-by-Term Integration Theorem to obtain an infinite series which converges to: cos(x) dx . y = cos(x²) (c) Use the remainder theorem associated...
3) Pol the diferential equation: (a) The point o -1 is an ordinary point. Compute the recursion formula for the coefficients of the power series solution centered at zo- -1 and use it to compute the first three nonzero terms of the power series when (-1)--s and y(-1)-0. (25 points) (b) The point 0 is a regular singular point Compute the associated Buler equation and compute the recursion formula for the coefficients of the series solution centered at o 0...
Use a power series centered about the ordinary point x0 = 0 to solve the differential equation (x − 4)y′′ − y′ + 12xy = 0 Find the recurrence relation and at least the first four nonzero terms of each of the two linearly inde- pendent solutions (unless the series terminates sooner). What is the guaranteed radius of convergence?
(2+3+1+1+1=8 points) Roughly, the Limit Comparison Test allows one to determine whether a given DO series an converges or diverges based on the computation of the limit an L = lim no ba 00 where on is another series. In this exercise, the Limit Comparison Test is used to determine whether the series shown below converges or diverges: yาง m4 +5n - 4 1. Write your choice of bn (Your answer should be in terms of n and simplified fully.)...