(* CALCULUS 265 Honors Fall Semester 2000 8:00 to 8:50 MTTF Carver 0002 Instructor: Hentzel Office: 432 Carver Office Hours: 9:00-10:00 MTTF E-mail hentzel@iastate.edu phone 294-8141 Website http://www.math.iastate.edu/hentzel/honors.265 Thursday, November 9: Section 8.3 p571:7,11,12,19 p572:33,37 Main idea: Use Integration to test for convergence and Divergence. Key Words: p-series, Error bound Goal: Learn to use the integral test for DECREASING sequences and to get an error bound. Infinity 1 Does SUM -------- converge or diverge n=0 n^2 + 1 Infinity Integrate[(n^2+1)^(-1),n] = ArcTan[n] = Pi/2 - 0 0 Since ArcTan[n] -> Pi/2 as n-> Infinity, the sequence converges by the integral test. The error from truncating the series at n=11 terms is is bounded by Integrate[(n^2+1)^(-1),{n,11,Infinity}] = Pi/2 - ArcTan[11] = 0.0906599 Sum[1/(n^2+1),{n,11,Infinity}] = 0.0948812 for the actual error. Notice that the integral is only an approximation to the sum of the series. It is useful for an error bound. But the actual value of the sum of the first 11 terms is not well approximated by the integral. This is the sum n = 0 to n = 11. Sum[1/(n^2+1),{n,0,11}] = 1.98999 This is the integral. Integrate[1/(n^2+1),{n,0,11}] = ArcTan[11] = 1.48014 There are a few standard series that are generic to all we do. 1. The Geometric series. a+ar+ar^2+ar^3+....+ar^n+... 2. The harmonic series. 1+1/2+1/3+1/4+1/5+...+1/n+... p p p p p 3. The p-series. (1) + (1/2) + (1/3) + (1/4) + ... + (1/n) + ... The Integral test works very well on p-series. Does Sqrt[1] + Sqrt[1/2] + Sqrt[1/3] + Sqrt[1/4] + ... Converge or diverge. Does 1 + (1/2)^3 + (1/3)^3 + (1/4)^3 + ... + (1/n)^3 + ... Converge or diverge. Infinity 1 Does Sum ------------ Converge or diverge. n=2 n ln n For the snowflake curve. What is the length of the perimeter. (a) The number of edges is 3, 3(4), 3(4)^2, 3(4)^3, ... ,3(4)^n (b) The lengths of the edges is 1, 1/3, (1/3)^2, (1/3)^3, ... ,(1/3)^n. (c) The perimeter is 3, 3(4/3), 3(4/3)^2, 3(4/3)^3, ..., 3(4/3)^n So the perimeter goes off to infinity since (4/3)^n goes off to infinity. (d) The area of an equilateral triangle is s^2 Sqrt[3]/4. Thus the area grows by (the number of sides) (side/3)^2 Sqrt[3]/4. | n | 2 n | (1/3) | 3(4) | ------- | Sqrt[3]/4 = | 3 | | | Sqrt[3] 3 4^n Sqrt[3] 3 4^n Sqrt[3] 4^n --------- ------- = ------- ------------ = --------- --------- 4 3^(2n+2) 4 9 3^(2n ) 12 9^n Sqrt[3] -------( 1 + 4/9 + (4/9)^2 + (4/9)^3 + ... + (4/9)^n + ... ) 12 Sqrt[3] 1 Sqrt[3] 9 3 Sqrt[3] ---------- ------- = ----------- --- = ----------- 12 1-4/9 12 5 20 This is how the area grows. The total area including the original triangle of area Sqrt[3]/4 is 2 Sqrt[3] ------------- = 0.69282 5 (e) We show that the augmentations never touch. We claim that you never can get out of the circumcircle of the original triangle. At each step, the added bump is an equilateral triangle. The circumcircle of this equilateral triangle is tangent to the previous circumcircle and on the inside. Continuing in this way, all subsequent constructions take place inside a sequence of circles, each tangent to the previous and on the inside. Since all of these circles are disjoint, the triangular extensions are disjoint. Sue needs a medicine in her body. During each day her metabolism reduces the amount of medicine in her body by 60 %. That is, only 40% of what is in her body today will be there tomorrow. Each pill has 10 mg of medicine. There must be 16 mg in her body for the medicine to be effective. How many doses till this level is reached? The medicine has moderate side effects when the dose in the body reaches 16.6 mg. How many doses till this is reached? The medicine has horrible side effects when the medicine in the body reaches 16.7 mg. How many doses till this is reached?