Difference between revisions of "2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 10"
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(b) First consider <math>x_0 > 1</math>. Let <math>x_{n+1} = 1 + \epsilon</math> and <math>x_n = 1 + \delta</math> with <math>\delta > 0</math>. The iteration gives <math>0 <\frac{\epsilon}{\delta} < \frac{2}{3}</math>. Next consider <math>\frac{1}{\sqrt{3}} < x_0 < 1</math>. As the signs of the numerator and denominator in the rational part of the iteration does not change on the interval under consideration we find that <math>x_1 > 1</math>. Finally, <math>x_0 = 1</math> produces <math>x_1 = 1</math>. | (b) First consider <math>x_0 > 1</math>. Let <math>x_{n+1} = 1 + \epsilon</math> and <math>x_n = 1 + \delta</math> with <math>\delta > 0</math>. The iteration gives <math>0 <\frac{\epsilon}{\delta} < \frac{2}{3}</math>. Next consider <math>\frac{1}{\sqrt{3}} < x_0 < 1</math>. As the signs of the numerator and denominator in the rational part of the iteration does not change on the interval under consideration we find that <math>x_1 > 1</math>. Finally, <math>x_0 = 1</math> produces <math>x_1 = 1</math>. | ||
− | To answer (c), rewrite the iteration as<math> x_{n+1} =-\frac{2(x_n)^3}{1-3(x_n)^2 }</math>, and note that for <math>0 | + | To answer (c), rewrite the iteration as<math> x_{n+1} =-\frac{2(x_n)^3}{1-3(x_n)^2 }</math>, and note that for <math>0<=x_0 < \frac{1}{\sqrt{3}}</math> the next iterate will be non-positive. Insisting that <math>-x_0 < x_1 <= 0</math>, so that <math>x_1</math> will be closer to zero than <math>x_0</math> gives the limiting case <math>x_1 =-x_0</math>, or <math>\alpha (1-3\alpha^2) =-2\alpha^3</math>, which has the solution <math>\alpha = \frac{1}{\sqrt{5}}</math>. |
Finally the implicit recurrence in (d) is obtained by running Newton backwards | Finally the implicit recurrence in (d) is obtained by running Newton backwards | ||
− | <math>a_i | + | <math>a_i-\frac{a_i^3-a_i}{3a_i^2-1}</math> = <math>-a_{i-1}, a_1 = \frac{1}{\sqrt{3}},\ldots,a_{\infty}= \frac{1}{\sqrt{5}}</math>. |
== See also == | == See also == |
Latest revision as of 04:57, 19 January 2019
Problem
Newton’s method applied to the equation takes the form of the iteration
(a) What are the roots of ?
(b) Study the behavior of the iteration when to conclude that the sequence
approaches the same root as long as you choose
. It may be helpful to start with the case
.
(c) Assume . For what number
does the sequence always approach
?
(d) For the sequence may approach either of the roots
. Can you find an (implicit) expression that can be used to determine limits
and
such that if
then the sequence approaches
. Hint:
and
approaches
when
becomes large.
Solution
(a) The roots are .
(b) First consider . Let
and
with
. The iteration gives
. Next consider
. As the signs of the numerator and denominator in the rational part of the iteration does not change on the interval under consideration we find that
. Finally,
produces
.
To answer (c), rewrite the iteration as, and note that for
the next iterate will be non-positive. Insisting that
, so that
will be closer to zero than
gives the limiting case
, or
, which has the solution
.
Finally the implicit recurrence in (d) is obtained by running Newton backwards
=
.
See also
2017 UNM-PNM Contest II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNM-PNM Problems and Solutions |