Instructor Hentzel Office Phone: 515-294-8141 E-mail: hentzel@iastate.edu Math Department Fax: 515-294-5454 http://www.math.iastate.edu/hentzel/class.166.08 Textbook: Calculus by Varberg, Purcell, Rigdon, ninth edition. ####################################################################### # # # No Class all of next week # # # # Next Class Monday, March 24, 2008 # # # ####################################################################### Friday, Friday, March 14 9.3 Integral Test p467:16,18,22,26 Main Idea: When is "enough", enough? If you can integrate f(x), there is a very good way to see if f(1)+f(2)+f(3) ... converges. It also provides an error bound. Key Words: p-series, Integral Test, Error Bound, Positive Term Series Goal: Learn that for f DECREASING Infinity Infinity INT f(x) converges <===> SUM f(n) converges x=1 n=1 -------------------------------------------------------- In Class work: Page 460 Problem 26 Three people, A, B, and C, divide an apple as follows. First they divide it into fourths, each taking a quarter. Then they divide the left over quarter into fourths, each taking a quarter, and so on. Show that each gets a third of the apple. Logically, each gets the same amount. Since the amount remaining n is (1/4), the whole apple disappears so each must get 1/3. Math should give us the same answer. 2 3 1/4 1/4 Person A gets 1/4 + (1/4) + (1/4) + ... = --------- = ----- = 1/3. 1-1/4 3/4 -------------------------------------------------------------------------- Page 460 Problem 28 Suppose that the government pumps an extra one billion into the economy. Assume that each business and individual saves 10% of its income and spends the rest, so of the initial one billion, 90 % is respent by individuals and businesses. Of that amount, 90 % is spent, and so forth. What is the total increase in spending due to the government action? This is called the multiplier effect in economics. x = the fraction respent. 2 3 4 1 M = 1 + x + x + x + x + .... = ----------- 1-x When x = 0.9 the multiplier effect is 10 times. During times of inflation, when everybody spends their pay check instantly, before it shrinks in value, the multiplier effect goes to infinity because x is 1.00 (nothing is saved). -------------------------------------------------------------- Page 461 Problem 49 Many drugs are eliminated from the body in an exponential manner. Thus, if a drug is given in dosages of size C at time intervals of length t, the amount A of the drug n st in the body just after the (n+1) dose is _ _ 1 _ _ 2 _ _ 3 _ _ n | -kt | | -kt | | -kt | | -kt | A = C + C|e | +C| e | +C| e | + .... + C| e | n |_ _| |_ _| |_ _| |_ _| where k is a positive constant that depends on the type of drug. (a) Derive a formula for A, the amount of drug in the body just after a dose if a person has been on the drug for a very long time (assume an infinitely long time). C ------ -kt 1-e (b) Evaluate A if it is known that 1/2 of a dose is eliminated from the body in 6 hours and doses of size 2 milligrams are given every 12 hours. 2 2 2 2 + --- + --- + --- + ..... 4 2 3 4 4 2 2 8 ------------- = ------- = ------ milligrams. 1-1/4 3/4 3 6 hr 6 hr 0 hr Checking it out, 8/3 -----> 4/3 -----> 2/3 -----> 2 + 2/3 = 8/3 in the twelve hour cycle. First three doses. 5 21 85 2, -, --, -- 2 8 32 --------------------------------------------------------------- Get["font.math"]; f[n_] := Sum[2 (1/4)^i,{i,0,n}]; A = Table[{n,f[n]},{n,0,20}]; a = ListPlot[A]; b = Plot[8/3,{x,1,20},PlotStyle->{RGBColor[1,0,0]}]; c = ParametricPlot[{0,y},{y,0,1}]; d = Show[c,b,a, PlotLabel->"P463 p49; 2mg each 12 hours: half life is six hours", PlotRange->All]; Display["p49x.ps",d]; g[x_] := 2(1/4)^x; e = Plot[0,{x,0,20}]; h = ParametricPlot[{0,y},{y,0,2}]; A = Table[ Plot[ Sum[ g[x-i], {i,0,n}], {x,n,n+1}], {n,0,20}]; k = Show[e,h,A, PlotLabel->"P463 p49 2mg each 12 hours; half life is six hours"]; Display["p49y.ps",k]; (* Suppose we took the medicine every 6 hours instead *) s[x_] := 1(1/2)^x; e = Plot[0,{x,0,20}]; h = ParametricPlot[{0,y},{y,0,2}]; B = Table[ ParametricPlot[ {x/2,Sum[ s[ x-i ], {i,0,n}]}, {x,n,n+1}, PlotStyle->RGBColor[1,0,0]], {n,0,40}]; k = Show[e,h,A,B, PlotLabel->"P463 p49; red=1 mg each 6 hours; half life is six hours"]; Display["p49z.ps",k]; --------------------------------------------------------------- New Material: We now are going to start a series of techniques to determine "ball park" results about series. We simply want to know if a series converges or diverges. If a series converges, then we can next try to find the limit. If we cannot find the limit, we can give an estimate of the limit along with an error bound that tells how far we might be off. For a series a + a + a + a + .... + a + ... 1 2 3 4 n <----------------------------> The sum of an initial segment is called a partial sum. S = a + a + a + a + ... + a . We say that the n 1 2 3 4 n SERIES converges if the SEQUENCE of partial sums converges. Today, we are going to work with POSITIVE term series. Naturally, the partial sums will be increasing, so if the partial sums are bounded above, then the series will converge. Thus the test for convergence will simply be if the partial sums are bounded (above). Page 464 Example 1 1 1 1 1 Show that ---- + ---- + ---- + .... + ---- + ... 1! 2! 3! n! converges. n Notice that n! = n(n-1)(n-2)...3 2 1 >= (1/2) 2 n 1 Infinity 1 Thus SUM ---- <= SUM --------- = 2. i=1 i! i=1 n (1/2) 2 Since the partial sums are bounded, the series must converge. ....................................................... Get["font.math"]; A = Table[ {n,Sum[1/i!,{i,1,n}]},{n,1,20}]; a = ListPlot[A,PlotRange->All,PlotStyle->{PointSize[0.0100]}]; B = Table[ {n,Sum[(1/2)^(i-1),{i,1,n}]},{n,1,20}]; b = ListPlot[B,PlotStyle->{RGBColor[1,0,0],PointSize[0.020]}]; c = Plot[0,{x,0,20}]; d = ParametricPlot[{0,y},{y,0,2}]; e = Plot[E-1,{y,0,20},PlotStyle->{RGBColor[0,0,1]}]; f = Plot[2,{x,0,20}]; g = Show[f,e,d,b,c,a, PlotLabel->"P464 ex1: black=SUM 1/i!, red=SUM (1/2)^(n-1)", PlotRange->All]; Display["ex1x.ps",g]; ........................................................... A very important class of series are the p-series. 1 1 1 1 1 + ----- + ----- + ----- + ..... + ----- + ... p p p p 2 3 4 n When p = 1 this is the harmonic series and diverges. For 2 Pi p > 1 the series converges. For p = 2, the limit is ------. 6 ============================================================== What's this good for? So we take a slight deviation to signal processing and channel limits. One cannot send square wave signals. Once the bit is sent, the impulse does not stop. it dies off leaving a residue in the channel for a long period of time. If these residues are too strong you cannot hear the signal because of these residual echos. For a rectangular wave the impulses die off rate is 1/n and the residue becomes Sum 1/n. At any time, the residue can be larger than the signal. 1 For a raised cosine curve, the tail dies off as -----. Thus 2 n at any point, the residual error is bounded. This is one reason the p-series are important. ============================================================== It is unknown what it converges to when p = 3. Theorem: For DECREASING positive functions Infinity Infinity SUM f(k) and INT f(x) dx k=1 x=1 converge or diverge together. | * | * | *----- | | * | | |====*----| | | |====*----| | | | |====*----| | | | | |====*----| | | | | | |====*----| | | | | | | |====*----| | | | | | | | |====*----| -----------------+-------------------------------------------------- | | Notice that the area under the curve larger than the area inside the smaller rectangles, and at the same time less than the area under the larger rectangles. To be more precise Infinity Infinity Infinity SUM f(i) <= INT f(x) dx <= SUM f(x) i=2 x=1 i=1 Thus if the integral is bounded ... the left hand sum is bounded so the series also converges. If the integral is unbounded ... then the right hand sum is unbounded so the series also diverges. --------------------------------------------------------- Get["font.math"]; f[x_] := 5/Log[x+10]^6; a = Plot[f[x],{x,1,10}]; b = ParametricPlot[{0,y},{y,0,f[1]}]; V = Table[ ParametricPlot[{n,y},{y,0,f[n]}],{n,1,10}]; H = Table[ Plot[f[n],{x,n-1,n}],{n,2,10}]; Show[a,H,V,PlotRange->All] VV = Table[ParametricPlot[{n+1,y},{y,f[n],f[n+1]}, PlotStyle->{RGBColor[1,0,0]}],{n,1,10}]; HH = Table[ Plot[f[n],{x,n,n+1},PlotStyle->{RGBColor[1,0,0]}],{n,1,9}]; d = Show[a,b,H,V,HH,VV,PlotRange->All,PlotLabel->"P-Series Proof"] Display["ex1y.ps",d]; --------------------------------------------------------- Page 445 Example 2. Infinity 1 1 1 SUM ----- + ----- + ... + ----- ... k=1 p p p 1 2 k converges if p >1. diverges if p <= 1. Proof: Since the series is decreasing we can use the integral test. ########## When p > 1 ########## _ _ 1 | 1-p | x=t 1-p t ---- | x | t - 1 INT p = | ------ | = --------------- x=1 x |_ 1-p _| x=1 1-p t 1 1 Limit INT ----- = ----- when p > 1. t->Infinity x=1 p p-1 x Since the integral is bounded, the series converges. _ _ ########## | | x=t When p = 1 the integral is |ln(x) | = ln(t) ########### |_ _| x=1 Here, the integral is unbounded so the series diverges. Which as we already knew it would because it is the harmonic series. 1-p ########## t -1 0 <= p < 1, ---------- goes off to infinity so the series ########## 1-p must diverge. ------------------------------------------------------------------- Infinity 1 Page 466 Example 3. Does SUM ---------- converge. k=4 1.001 k Answer: This is a p-series, p=1.001 so it converges. ######## ####### ######### ------------------------------------------------------------------ Page 466 Example 4 Infinity 1 Does SUM -------- converge or diverge. k=2 k ln(k) 1 f(x) = -------- is decreasing on [2,Infinity] x ln(x) _ _ Infinity | | x=Infinity INT f(x) dx = | ln( ln(x) ) | = Infinity x=2 |_ _| x=2 Infinity 1 Therefore SUM -------- diverges by the integral test. k=2 k ln(k) -------------------------------------------------------------- Using the Integral test to give an error bound for truncating the sum at a certain point. Page 466 Example 5. Find an error bound for estimating Infinity n SUM -------- by using just the first five terms. n=1 _ _ | 2 | |_n _| e The tail of the series is less than the integral. Remember that the smaller rectangles are inside the area under the curve. _ 2 _ Infinity x | (-x ) | Infinity 1 Error <= INT ------ = | -1/2 e | = --------- n=5 _ _ |_ _| x=5 25 | 2 | 2 e |_ x _| e ------------------------------------------------------------- Get["font.math"]; f[x_] := x/E^(x^2); P1 = ParametricPlot[{0,y},{y,0,f[1]}]; A = Table[{n, Sum[f[i],{i,1,n}]},{n,1,5}]; P2 = ListPlot[A,PlotStyle->{RGBColor[1,0,0],PointSize[0.015]}]; A = Table[{n, Sum[f[i],{i,1,n}]},{n,6,10}]; P3 = ListPlot[A,PlotStyle->{RGBColor[0,0,1],PointSize[0.015]}]; P4 = Plot[0.404881,{x,1,10},PlotStyle->{RGBColor[0,1,0]}]; P5 = Show[P1,P2,P3,P4,PlotLabel->"Sum n/e^(n^2)",PlotRange->All]; Display["ex4.ps",P5]; ---------------------------------------------------------------- Page 466 Example 5. Find an upper bound for the error when the sum of the first 20 terms is used to approximate the sum of the convergent series Infinity 1 S = SUM ----- k=1 3/2 k _ _ | -1/2 | Infinity Infinity 1 Infinity dx | x | 2 Error = SUM ----- < INT --- = |------ | = -------- = 0.447214 k=21 3/2 k=20 3/2 |_-1/2 _| 20 Sqrt[20] k x ------------------------------------------------------------- Get["font.math"]; f[x_] := 1/x^(3/2); P1 = ParametricPlot[{0,y},{y,0,f[1]}]; A = Table[{n, Sum[f[i],{i,1,n}]},{n,1,20}]; P2 = ListPlot[A,PlotStyle->{RGBColor[1,0,0],PointSize[0.015]}]; A = Table[{n, Sum[f[i],{i,1,n}]},{n,21,40}]; P3 = ListPlot[A,PlotStyle->{RGBColor[0,0,1],PointSize[0.015]}]; P4 = Plot[2.59823,{x,1,40},PlotStyle->{RGBColor[0,1,0]}]; P5 = Show[P1,P2,P3,P4, PlotLabel->"P466 Ex 5 Sum 1/k^(3/2)",PlotRange->All]; Display["exy.ps",P5]; ---------------------------------------------------------------- Page 467 Example 6 How large must n be so that the partial sum S n approximates the sum of the series in Example 5 with an error of no more that 0.005? _ _ | -1/2 | Infinity Infinity 1 Infinity dx | x | 2 En = SUM ----- < INT ----- = | ------ | = -------- k=n+1 3/2 x=n 3/2 | -1/2 | Sqrt[n] k x |_ _| x=n So we want 2 ------- < 0.005 Sqrt[n] 2 -------- < Sqrt[n] 0.005 2000 ---- < Sqrt[n] 5 400 < Sqrt[n] 160,000 < n Thus we see that we have to sum a great many terms to have even reasonable accuracy.