Course Content (Under Construction)

Our goal is a careful study of the notion of limit in calculus, particularly convergence of sequences and series. Further subjects will include the Taylor series and applications to series solutions of differential equations. These basic concepts are fundamental in all math and particularly so in all areas of advanced analysis. We will pay particular attention to methods of proof and will even touch upon some fundamentals of math logic. Our goal is to use the important study of limits, series, sequences, convergence as an entry step into mathematical reasoning. Both of these objectives are fndamental for this course.

Basic Course Information

Math 2283 is a 3 credit course. Here is the official mathematics department description of this course:

MATH 2283 - Sequences, Series, and Foundations Introduction to mathematical reasoning used in advanced mathematics. Elements of logic. Mathematical induction. Real number system. General, monotone, recursively defined sequences. Convergence of infinite series/sequences. Taylor's series. Power series with applications to differential equations. Newton's method.

Lecture: 010 LEC , 11:15 A.M. - 12:05 P.M. , M,W Keller Hall 3-111 , TCEASTBANK .
Instructor: Professor S.Sperber.
Email: sperber@math.umn.edu
Office: 457 Vincent Hall
Phone: 612-625-4374
Office Hours: Monday, Wednesday 2:30-3:30PM
Textbook: Sequences, Series and Foundations Mathematics 2283, Wayne Richter,

Recitation:

-011-DIS, 10:10AM - 11:00AM, Tu,Th Vincent Hall 113, TCEASTBANK, Kimberly Logan,
-012-DIS, 11:15AM - 12:05PM, Tu,Th Akerman Hall 313, TCEASTBANK, Kimberly Logan

Organization of Course and Grading Policy

There will be two mid-semester in-class exams of 50 minute duration. Exams will be held in recitation on Tuesday, October 13 and Tuesday, November 17.

There is also a (cumulative) final examination which will be given (date and location to be filled in) According to one-stop "courses meeting 11:00 a.m.-11:59 a.m. MWF will have final exam at 10:30 a.m.-12:30 p.m., Monday, December 21 (check this)

Homework will be collected weekly, every Tuesday in recitation and some, if not all, of your work on each problem set will be graded.This grade will be a component of the 20% of the "homework and class participation" portion of your Final grade

The final grade will be determined according to the following guide:

mid-semester exams 40%

final 40%

homework and class participation 20%

Some Solutions to Sample Final Exam

Some solutions to sample final exam

Problem 1. For n=1, d/dx(sin x) = cos x = sin x cos ({\pi}/2) + cos x sin ({\pi}/2).

Assume d^n/dx^n (sin x) = sin (x + n({\pi}/2)).

Then d^{n+1}/dx^{n+1} sin x = cos (x + n({\pi}/2)) and

sin(x + (n+1)({\pi}/2)) = sin (x + n({\pi}/2 + {\pi}/2) = sin(x + n({\pi}/2)) cos {\pi}/2) + cos (x + n({\pi}/2)) sin {\pi}/2

2. Let a_n = (-1)^n (n+1)^2 2^n x^n.

Then |a_{n+1}/a_n| = {(n+2)/(n+1)}2|x| which goes to 2|x| as n goes to infinity.

So power series converges absolutely if |x| < 1/2 and diverges for |x| > 1/2.

When x = 1/2 series diverges (general term does not go to zero). Same for x=-(1/2).

3.a) n< (n+{1/n}) < n+1 so series diverges by comparison test with /sum (1/(n+1))

b) Diverges. General term does not go to infinity.

4. (n^2 + n + 1)/(2n^2 + 3) - (1/2) = (2n-1)/2(2n^2 +3).

Given \epsilon, we need N such that for n>N,

|(2n-1)/2(2n^2 +3)| < \epsilon

For all n, 2n-1<2n and 2n^2 + 3 > 2n^2

So for all n, |(2n-1)/2(2n^2 +3)| < 1/2n.

Choose N > 1 / 2(\epsilon).

5. Since \sum |a_n| is convergent and |a_n|/n < |a_n| the comparison test gives us that \sum |a_n|/n converges

therefore \sum a_n/n converges. If convergent + divergent is convergent then we get the divergent series is the

difference of two convergent series. Contradiction. So convergent + divergent must be divergent.

6. a) f(x) = 1/x ln(x) > 0 and decreasing for x \geq 2 so can use integral test.

Integral of f(t) dt from 2 to T is ln(ln T) - ln( ln 2) and lim = infinity as T goes to infinity.

So divergent.

b). Convergent by alternating series test.

7. e^(-1) gives an alternating series with general term (-1)^n/n!. 5! = 120 > 100 so

1 - 1 + 1/2 -1/6 + 1/24 gives e^(-1) to within 1/100.

8 lub A =1, glb A =-1, no max, no min.