Recitation 15

Example 1: Use Newton’s method to find all roots of the equation correct to six decimal places.

Hint: Sketch the graphs of y=3\cos x and y=x+1. The first intersection lies on -2\pi < x < -\pi, the second -\pi < x < -\pi / 2 and the last 0 < x < \pi / 2. Set f(x) = 3\cos x - x - 1. Newton’s iteration says x_{n+1} = x_n - \frac{3\cos x_n - x_n - 1}{-3\sin x_n - 1}. Take the initial approximations as -\pi, -\pi / 2, 0, we get x= -3.637958, -1.862365, -0.889470.

Example 2: Explain why Newton’s method doesn’t work for finding the root of the equation x^3-3x+6=0 if the initial approximation is chosen to be x_1 = 1.

Hint: Try to work out x_2.

Example 3: Find a formula for the general term a_n of the sequence, assuming that the pattern of the first few terms continues. (a) \left\{-3, 2, -4/3. 8/9, -16/27, \ldots\right\}. (b) \left\{1/2, -4/3, 9/4, -16/5, -25/6, \ldots\right\}.

Answer: (a) a_n = (-3)(-2/3)^{n-1}. (b) a_n = (-1)^{n+1}n^2/(n+1).

Example 4: Determine whether the sequence converges or diverges. If it converges, find the limit. (a) a_n = \frac{3+5n^2}{n+n^2}. (b) a_n=\tan\left(\frac{2n\pi}{1+8n}\right). (c) a_n = n^2/\sqrt{n^3+4n}. (d) a_n = \ln(n+1)-\ln n. (e) a_n = (\cos^2n)/2^n. (f) \{0,1,0,0,1,0,0,0,1,\ldots\}.

Hint: (a) The limit of a quotient of two polynomials of the same degree is equal to the quotient of their leading coefficients. (b) In side the tangent function, the limit is \pi/4. (c) The numerator is n^2 and the denominator is roughly n^{3/2}. (d) It is equal to \ln(1+1/n). (e) The absolute value is no greater than 1/2^n. (f) It does not converge.

Remark: (f) Intuitively, if the sequence converged, the limit would be 0. However, 1 appears indefinitely (though less and less frequently), the limit cannot be 0. For people who are interested in the rigorous proof using the definition of limit, here is the argument. For sake of contradiction, suppose the limit of the sequence is L. Let \delta = 1/2. There should be N so that L - 1/2 < a_n < L + 1/2 for all n > N. However, this means L -1/2 < 0 and 1 < L+1/2, hence 1/2 < L < 1/2, which yields a contradiction.

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