Lecture 6: Problem session

Problem of the day

This problem will be collected during class on January 29.
Pick two positive integers and generate the following sequence. The first integer picked is the first number and the second is the second. The third is the sum of 1 and the first divided by the second, add first, then divide. Then get the fourth by doing the same thing with the third and second. Continue the process, getting the fifth, sixth, etc. For example, suppose the first number is 3 and the second is 5. Then the third would be (5+1)/3=2 and the fourth would be (2+1)/5=3/5, and the fifth is ((3/5)+1)/2=4/5. Now compute the sixth number in the sequence: ((4/5)+1)/(3/5)=3 and the one following that is: (3+1)/(4/5)=5 so you can see that the sequence starts all over again. Such sequences are called periodic. This one has period 5 because the sixth term is the same as the first, etc. Repeat the process with two other initial picks. Again you get periodicity. Do you always get a periodic sequence? Explain in detail.

Assignments

These are the problems you should work before today, February 5.
Review problems, page 75, 4n+1, for n=0..22;
Section 1.1, page 92, 4n+1, for n=0...24;
Section 1.2, page 102, 1-6, and 4n, for n=3...23;
Section 1.3, page 113, 4n+1, for n=0...19.
These are the problems you should work before February 12.
Section 1.4, page 120, problems 6n+1, for n=0...15;
(skip 1.5) 1.6, page 149, problems 6n+1, for n=0...16.
These are the problems you should work before February 19.
Section 1.7, page 160, 4n+1, for n=0,...,24.
Section 1.8, page 171, problems 6n+1, for n=0...8;
These are the problems you should work before February 26.
The first test, covering Chapters P and 1 and sections 2.1 and 2.2 is on February 26.
Section 2.1, page 191, problems 6n+1, for n=0...12.
Section 2.2, page 025, problems 8n+1, for n=0...11.

Brief review of the lecture

There was a problem session today. Todays quiz was:
Find an equation for a circle with center at the point $(-1,2)$ if the point $(2,-2)$ belongs to the circle.
Click here for a solution to quiz 2 As usual, this is material left from last semester. Read it at you oun risk!
First, we talked about various ways to solve a quadratic equation; that is, ways to find the zeros of a quadratice polynomial. Of course, if the quadratic factors, then its easy to solve: if a₂x²+a₁x+a₀= a(x-r)(x-s), then the zeros of the quadratic are r and s. But if it fails to factor (with integer coefficients), the problem is harder. We must use the method of COMPLETING THE SQUARE. WE learned that we can get lots of information from the quadratic by completing the square (like the location of the vertex), and we also understand why the quadratic formula works as well.

1st degree polynomials are called linear and can be put into the form ax+b. Linear polynomials have at least one real root.

An 2nd degree polynomials are called quadratic and can be put into the form ax²+bx+c. If a quadratic polynomial can be facorted, the factors are linear. A quadratic polynomial has at most two real roots.

Using a process called completing the square , the quadratic formula for the roots of a quadratic polynomial was derived. The student is referred to page 114 of the text for another look at this important derivation. You will find that the technique of completing the square is required throughout this course. Learn it now! In this lecture we used the technique to find that x²-3x+7=0 has no real solutions.

Another important idea which completing the square brings to light is the discriminant D of a quadratic expression. It is given by D=b²-4ac, where b is the coefficient of the linear term, a is the coefficient of the square term, and c is the constant term. You can put each quadratic expression into one of the six categories: A. two real roots, opens upward; B. one (repeated) root, opens upward; C. No real roots, opens upward; D. two real roots, opens downward; E. one (repeated) root, opens downward; and F. No real roots, opens downward. You are expected to be able to carry out this classification.

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