Lab 4. Recall, from last lab the commands Table[], ListPlot[], DiscretePlot[], and Limit[]. Go ahead and review them now you'll be using them soon.

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1 Calc II Page 1 Lab 4 Wednesday, February 19, :01 PM Recall, from last lab the commands Table[], ListPlot[], DiscretePlot[], and Limit[]. Go ahead and review them now you'll be using them soon. Use the material from the last lab to generate the recursive series: Generate the first 30 terms using the Table[] command Generate the first 30 terms as decimal approximations. Create a DiscretePlot[] of the absolute value of the difference between the terms and looks like this:. My graph Obviously it converges fairly quickly. Generate the Discrete Plot of the Log[10,<expression>] (we are making the log base 10 for ease of interpretation). Mine looks like this: Wow! That's a nice looking graph. The graph stops around 21-- but that's just because we have reached the limit of what Mathematica can determine as a difference... so the absolute value of the difference evaluates to 0 in Mathematica and the Log[] of 0 is undefined, so the graph quits. Let's approximate the slope of this graph: (that's change in y from x=5 to x=6 divided by a change in x of 1). Other sample points will give us similar values. If we wanted to do this correctly we would perform a linear regression-- but that's a

2 Calc II Page 2 topic for another class. So, if we ball park things... every 4 terms improves out accuracy by around 3 decimal places. Not too shaby! The 15th term in the sequence is: And according to our graph above, that's accurate to around +/- 1 in the 11 decimal place. So we should definitely get accuracy up to 10 places: let's check: No on to series Okay.. Moving along. In the last lab you dealt primarily with sequences (and a little bit of arc length). Now we are going to start to do some work with series. Recall that there are THREE components to keep in mind when dealing with series: The sequence of terms ( ) The series The sequence of partial sums ( ) Let's consider the geometric series. The key things to notice are the POWER is n-1 the constant term is 5 the base is 2/3 Use Table[] to generate the first 10 terms in the sequence (NOTE: this is the first 10 terms-- not the first 10 partial sums): You know that as n goes to infinity the expression goes to 0, and so goes to 0 as well. If we want to have any hope of the sequence of partial sums converging (and hence the series) then the sequence of terms MUST go to 0. FIRST TEST OF DIVERGENCE OF A SERIES: If the sequence of terms does NOT go to 0 then the series diverges. It's not immediately obvious from the output that the sequence is decreasing. Let's turn those fractions into decimal approximations with 3 decimal places: Test your understanding of the material from the last lab and produce a graph that shows the first 10 terms. My graph looks like this:

3 Calc II Page 3 It is important to remember that the convergence (or divergence) of a series is determined by the behavior of its sequence of partial sums, so let's see what we can do about looking at the terms in the sequence of partial sums. The main Mathematica function you need for partial sums is the aptly named Sum[] function. Here's how to interpret them: Sum[5(2/3)^(n-1),{n,4,10}] Notice the how each piece aligns (I'll color code): Sum[5(2/3)^(n-1),{n,4,10}] By and large we care about the series from to, but the option for any sort of partial sum is available. Let's see if Mathematica is smart enough to calculate the partial sum from 1 to n. To make this easy to follow I need to use a different variable: Sum[5(2/3)^(i-1),{i,1,n}] I get the rather unenlightening: However, if we force Mathematica to make it look nicer: then we find something that's a bit easier to understand. Is that what we should expect? According to the formula we learned in class for the nth partial sum we get: In our example above,, and. So we should get: So... does this equal what we found above? And if you look at that carefully you will see it is equivalent to the value suggested by Mathematica... so

4 Calc II Page 4 yay! The next function we want to use (again) is Limit[]. From class we know that the limit as n goes to infinity of a geometric series is. Since and, then we should get. Can Mathematica do as well? Well... Wasn't that handy? Let's try a different series. To begin with, let's define the sequence of terms and store them in a function called. Let's represent the terms in the terms sequence: this way: f[x_]:=sqrt[x+2]-sqrt[x]. Go ahead and do that (you might want to clear the definition of f first just to make sure there aren't any lingering 'special cases' floating around: Clear[f] f[x_]:=sqrt[x+2]-sqrt[x] Use Table[] to generate the 100 through the 110th term. Your output should look something like this: UseDiscretePlot[] to graph this subsequence: It certainly looks as if it is decreasing. Now let's consider the series that is formed from these terms: Using the terminology we developed in class, we define the nth partial sum for this series: We can represent this in Mathematica as follows: S[n_] := Sum[f[x], {x, 1, n}] Now let's use DiscretePlot[] to look at the first 100 partial sums as a graph:

5 Calc II Page 5 Looks as if this series is going to diverge. But can we be certain? Poke around a little for example, try looking at the graph of to (That shouldn't take too long to calculate). Doesn't look promising does it? There are several approaches we could take to convince ourselves this series diverges, but we won't see them for awhile, so we will have to be a little more imaginative. See what happens as you list out several of these terms. Let's figure out : Notice that most of these terms cancel (only 4 will remain... no matter how far we go):. = This series is an example of something called a telescoping sum. A bit of perseverance will lead us notice that when n is sufficiently large the only terms that survive the cancelling process are two from near the beginning of the partial sum. These two terms a 1 and a 2,show up as negative. Most of the intermediate terms are cancelled except for the two positive values in the last two terms. So in full generality we would say: Let's check our intuition by using Mathematica to compare our two values: SClosed[n_] := Sqrt[n + 2] + Sqrt[n + 1] - Sqrt[2] - 1 Table[SClosed[x] - S[x], {x, 1, 100}] You should see a lot of zeros. This suggests that we did things correctly. Clearly, corresponding sum diverges too. Let's examine the same problem in another fashion. Graph the function are the TERMS in the series). Look to see if it's decreasing: diverges so the (these That looks pretty convincing to me but how do I prove it carefully? Obviously, you answer in a somewhat bored tone of voice, you take the derivative of f(x) and prove that it is negative on the

6 Calc II Page 6 interval [1, ): This looks a *bit* intimidating, but the denominator is always positive on [1, ), and since the square root function preserves ORDER the numerator must always be negative (think about it). Hence the entire expression is always negative on [1, ). Surprisingly, we can get Mathematica to do this for us: Reduce[{D[f[x], x] < 0}] Give it a try. Or even easier: Reduce[f'[x]<0] Mathematica will return the interval on which this condition is satisfied. Notice, we again took advantage of Mathematica's inability to tell the difference between a sequence and a function. (Go back and review the last lab if what I just said confused you) OK so we've ascertained that the function f(x) is decreasing on [1, ). We can figure out the limit easily enough (as x goes to infinity). (Do this). This sequence is also clearly positive and continuous on this interval too... so check this out: Here's what I want you to realize... the HEIGHT of each dot is the value of a term in the sequence associated to the series. The other thing to notice is that the distance between the vertical lines is exactly one. Let me emphasize the points and enhance the graphics a bit (this will also change the x- and y-scales a tad): The area of reach rectangle is the height of the red dot in the rectangles upper, left corner (remember that these heights come from the sequence). Because these heights are decreasing, the rectangle is

7 Calc II Page 7 completely contained under the curve. The total area in grey in the picture above is the exactly value of. But... that's LESS than the area under the curve. Since the curve is always positive the AREA of the curve over the rectangles should be MORE than the area under the rectangles... But we could also draw the rectangles like this: Here the (because the function is decreasing and always positive) the area of the rectangles is MORE than the area under the curve from to If this integral DIVERGES, then series does too. So try to find the improper integral using mathematica (recall that : See what I did there? I used the infinity sign in the boundary. Handy isn't it? Clearly, Mathematica believes the sequence diverges but let's do the steps ourselves a bit more carefully just to be certain. We're going to start by figuring out the value of We can do this, but in order to keep Mathematica from confusing itself with some unwarranted assumptions, we have to add a little bit of legalese: Integrate[f[x], {x, 1, t}, Assumptions -> {Element[t, Reals],t>1}] That last bit is new to us. It's an argument to Integrate[], that tells the command that t is greater than one and that t is a real number. It's a bit of a hassle, but on the plus side we get a nice expression for the definite integral: It's not immediately clear whether or not this expression will diverge as t goes to infinity but we can ask Mathematica (and you should try to do this one): So we see from the steps that the integral diverges and hence the series diverges too. Before we move on let's look at the first 10 partial sums: Table[S[n],{n,1,10}] That's useful but well it's sort of hard to see what's happening. If we hadn't defined the function S[n]

8 Calc II Page 8 we could have generated the terms in the partial sum sequence directly: Table[Sum[f[x], {x, 1, n}],{n,1,10}] You'll get the same result of course, but now if we change Sum[] to NSum[], we get an expression which is occasionally more useful: Table[NSum[f[x], {x, 1, n}], {n, 1, 10}] Here's another cool thing (but not about that problem): Mathematica can dispense with a lot of the intermediate steps figure out what we really mean. Try the following (remember, you can get Mathematica to give you infinity by typing <esc>inf<esc>): Spiffy non? That should make all sorts of problems easier to solve! As always, it is important for you to know what's really going on. Mathematica can give the wrong results. Consider this expression: NSum[1/n^2+10^(-8)/n,{n,1, }] Give that a try see.. Mathematica says it converges but Mathematica is WRONG! Each term in the sequence is greater than. That's a multiple of the harmonic series (which you will learn about on Friday), and that diverges, so the entire expression diverges. The symbolic version of the command is, however, smarter: That should be good enough for today. Be sure to send me your Mathematica file (along with the names of your group members). If you're done early you can either leave, or ask me homework questions.