Question

1. [20 points] Let f(x) = e^{2x} - 2x - 1 (1). [5 points] Show that f has a zero of multiplicity 2 at x* = 0. 1 (2). [8 points] Express f(x) as a Taylor series around x* = 0 of the form f(x) = x^2h(x). Give the explicit form of h(x) and evaluate h(0). (Hint: use the Taylor series e^x = \sum_{n=0}^{\infty} \frac{1}{n!}x^n) (Answer: h(x) = \sum_{n=0}^{\infty} \frac{2n+2}{(n+2)!}x^n = 2 + \frac{4}{3}x + \frac{2}{3}x^2 + ..., h(0) = 2) (3). [7 points] Use the modified Newton method and do iteration three times to find the approximated root, starting with x_0 = 1 (Answer: x_3 \approx 0.000351)

          1. [20 points] Let
f(x) = e^{2x} - 2x - 1
(1). [5 points] Show that f has a zero of multiplicity 2 at x* = 0.
1
(2). [8 points] Express f(x) as a Taylor series around x* = 0 of the
form f(x) = x^2h(x). Give the explicit form of h(x) and evaluate
h(0).
(Hint: use the Taylor series e^x = \sum_{n=0}^{\infty} \frac{1}{n!}x^n)
(Answer: h(x) = \sum_{n=0}^{\infty} \frac{2n+2}{(n+2)!}x^n = 2 + \frac{4}{3}x + \frac{2}{3}x^2 + ..., h(0) = 2)
(3). [7 points] Use the modified Newton method and do iteration
three times to find the approximated root, starting with x_0 = 1
(Answer: x_3 \approx 0.000351)

1. [20 points] Let
f(x) = e^2x - 2x - 1
(1). [5 points] Show that f has a zero of multiplicity 2 at x* = 0.
1
(2). [8 points] Express f(x) as a Taylor series around x* = 0 of the
form f(x) = x^2h(x). Give the explicit form of h(x) and evaluate
h(0).
(Hint: use the Taylor series e^x = ∑n=0^∞ (1)/(n!)x^n)
(Answer: h(x) = ∑n=0^∞ (2n+2)/((n+2)!)x^n = 2 + (4)/(3)x + (2)/(3)x^2 + ..., h(0) = 2)
(3). [7 points] Use the modified Newton method and do iteration
three times to find the approximated root, starting with x0 = 1
(Answer: x3 ≈0.000351)

Added by Carlos R.

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