|Monday, December 11||(Lecture 26)|
I also mentioned that the word mantissa is currently used in computer architecture. As used there, it refers to the fractional part of a floating point number.
Another aspect of substitution was addressed. Suppose a definite integral is given. Should the limits in x also be changed to limits in u, or should the "return to x-land" always be done and then x limits used? I tried to give a number of examples in order to illustrate the two possibilities. I usually go back to x-land and use the original limits because I make mistakes, sometimes many mistakes, and the return to x means I don't need to change the limits. Other people have other opinions.
I computed some areas between two curves. I used the general formulation of x=Lefty=RightTop(x)-Bottom(x) dx and illustrated this using examples where Top and Bottom have different signs.
I did a very few examples where the area is computed dy, so the formula above becomes y=Bottomy=TopRight(y)-Left(y) dy. This is discussed in section 6.1.
|Wednesday, December 6||(Lecture 25)|
Mantissa n. [L., an addition, makeweight; of Tuscan origin.] (Math.) The decimal part of a logarithm, as distinguished from the integral part, or characteristic.I discussed in detail how in "old days" students in highschool learned how to use high-accuracy tables of mantissas of common logarithms (logarithms base 10) in order to approximately compute multiplications. This skill may be obsolete since the introduction of simple and cheap calculators.
Does it necessarily follow that logarithms themselves are obsolete?
While at first glance this may seem to be true, I "analyzed" the
We showed that L´(x)=1/x using FTC. Then we used the Chain Rule
and FTC to see that if A is some fixed positive number, and
W(x)=L(Ax), then W´(x)=1/x also (almost a miracle in that the
effect of the A's cancel!). Then the Mean Value Theorem states that
W(x)-L(x)=a constant since the derivatives of L and W are
identical. If we put x=1, L(1)=0 and W(1)=L(A). Therefore the constant
is L(A), and we see that since W(x)=L(Ax)BB,
The "profile" function used to define L (that is, 1/t) is not a very complicated function. So if we are interested in computing definite integrals, we'd better be able to compute L, and the equation above may be used to simplify computations of L.
Indeed, if we use the following Maple instruction:
> f:=n->evalf(add((1/(1+(j/n)))*(1/n),j=1..n));Then the value of the function f(n) is an approximation of a Riemann sum for the definite integral from 1 to 2 of 1/t, with n subdivisions, with the right-hand endpoint of each subdivision used. Results look like this:
|Number of subdivisions||Approximate value|
of the sum
Of course, as I mentioned in class, these are approximations to ln(2), whose 10-digit value is 0.6931471806. By the way, even the computer gets tired computing Riemann sums and the last entry in the table above takes about three-quarters of a second on a moderately fast PC. Next semester we'll see numerical methods which give much better accuracy with much less computation. But the point I was trying to make here was that the conceptual idea of logs was useful, even though the computational need for logs may be obsolete.
There are many other integrals which must be computed, and which satisfy quite weird equations that come from the Chain Rule and FTC. So this whole development is something you will likely see again.
The remainder of the lecture was devoted to a certain method for obtaining indefinite integrals: substitution.
We began by studying (x3+5)400 x2 dx.
This is not a random integral, at all. If I omitted the x2 or changed it much, the integral would be quite difficult to do "by hand". The integrand (the function to be integrated) is "just" a polynomial, so there is no great theoretical obstacle to computing it. The practical difficulty of obtaining a formula for the antiderivative is what I wish to address.
One way of computing the indefinite integral is to "unfold" or "expand" (x3+5)400. By hand (even by computer!) this takes a bit of time and storage space! In fact, we recognized that we could do this by sort of guessing and reversing the Chain Rule. It is important to remember, though, that the derivative of a product of two functions is NOT equal to the product of the derivatives!
So we guesses. I wrote the Chain Rule and we identified the pieces of the computation. Then I introduced the Substitution Method, which is essentially a "bookkeeping" strategy to keep track of constants as we go from "x-land" to "u-land".
Here (x3+5)400 x2 dx is in x-land. If u=x3+5, then du/dx=3x2 and du=3x2dx and (1/3)du=x2dx. The entire integral, translated into u-land, is then u400(1/3)du. This is easy to compute in u-land, and the result is (1/401)u401(1/3)+C. Now we go back to x-land, and the final answer is (1/401)(x3+5)401(1/3)+C.
I computed some other indefinite integrals with this method:
We then computed the indefinite integral of tan(x). Since tan(x)=sin(x)/cos(x), the substitution u=cos(x) leads us to the answer -ln(cos(x))+C. Like any other indefinite integral, this can be checked using differentiation. And this indefinite integral again shows that logs will be needed in calculus!
Then I "discussed" two problems from section 5.5 of the text:
#58, which was 01x e-x2dx. The antiderivative can be computed with u=-x2.
#56, which was 02dx/(2x-3)2. I remarked that this was an error in the text (a misprint!). 2x-3=0 when x=3/2, and therefore a simple graph should convince you that the definite integral (the "area") must be infinite. If the upper bound of the integral is changed from 2 to 1, then the substitution u=2x-3 will allow us to compute the value of this definite integral.
|Monday, December 4||(Lecture 25)|
|Number of subdivisions||Approximate value|
of the sum
Maybe these results which I computed with Maple make the assertion that the limit of the Riemann sums exists and equals 2/3 more believable.
Numbers are numbers are numbers
I then discussed numbers. I asked what people could tell me about
43 43 43 43 43
Then I asked what people could tell me about these numbers:
35[x3]/[3+cos(x2)]dx 35[u3]/[3+cos(u2)]du 35[t3]/[3+cos(t2)]dt 35[v3]/[3+cos(v2)]dv 35[s3]/[3+cos(s2)]ds
These are all definite integrals, and they are all just numbers. Of course they are all equal. Again, the letters inside the integrand are dummy variables or bound variables. They are similar to the local variables in subroutines of a computer program, and they have no meaning outside of the integral itself.
I then tried to do some simple exercises with part of the FTC. It was something like this: if f(x)=2x cos(t7)dt, then f'(x)=cos(x7). And then if f(x)=2x cos(s7)ds, then f'(x)=cos(x7). The "dummy variable" strikes again!
How about if f(x)=x3
cos(s7)ds? Here we need to think a bit about definite
integrals. Here's a fact:
Adding areas fact
If a<b<c, and if f is continuous on the interval from a to c, then abf+ bcf= acf.
So adjoining areas can be added up (something many of us learn as infants, and it is a rather subtle fact!). But we also agree that aa=0: that is, a region with 0 width should have 0 area. Suppose we make some changes in the adding areas fact, and transform c into a. Then the right-hand side becomes 0, and we see that abf+ baf=0, which is the same as abf=-baf. Therefore if we want to be able to add areas and if we want to recognize that lines have area=0, we must also acknowledge that interchanging limits causes a - sign to prefix the integral. Therefore we can do this:
if f(x)=x3 cos(s7)ds, then f(x)=-3xcos(s7)ds so that f'(x) must be -cos(x7). Amazing! (Well, not really, but ...)
We can even do one further variation. Suppose F(x)=-2xsqrt(1+w6)dw. Then we know
that F'(x)=sqrt(1+x6) by FTC. What if I ask the following
Suppose G(x)=-2cos(x)sqrt(1+w6)dw. What is G'(x)?
How can we think about this? Since we know F'(x), and G(x)=F(cos(x)), this is just (!?) an application of the Chain Rule (!): G'(x)=F'(cos(x))·(the derivative of cos)=sqrt(1+(cos(x))6)·(-sin(x)). Whew! This takes some getting used to.
|Wednesday, November 29||(Lecture 24)|
A paradigm shift
Several cultures invented approaches similar to what we've done here to compute quantities such as area and volume and mass and moments. These cultures (which were located in areas corresponding to present day China, India, Egypt, and Greece) all used the same general approach: chop up the quantity into smaller pieces, approximate the pieces, take the sum, and then "compute" the limit. Of course the notation was different, but the ideas were fundamentally the same. Very clever algebraic and numerical techniques were used. See section 5.2 for information about this. It turns out that studying what seems to be a harder problem (!) earns both conceptual and computational advantages. So I attempted to do that for the specific case of f(x)=(3/2)x2+(1/2)x. I wanted to compute the area "under" this curve from 1 to 2. This is the area of the region bounded by x=1, x=2, the x-axis, and y=f(x). A direct approach is possible, but instead I studied the following problem:
Suppose A(r) is the area bounded by x=1, x=R, y=f(x), and the x-axis. What can we learn about A(r)?We saw that A(1)=0 and then that A´(R)=(3/2)R2+(1/2)R. This is an amazing result because, based on what we've already done (Initial Value Problems and the Mean Value Theorem) we can deduce A(R) totally.
The Fundamental Theorem of Calculus
I stated and commented on the Fundamental Theorem of Calculus (FTC). Then I applied it to about half a dozen problems in section 5.3 of the textbook.
Maple data and its construction
I also gave some data which I had computed using Maple. Maple is a program which is available on the Eden system and on many other computer systems around campus, since Rutgers has a site license. It is quite useful.
> f:=x->(3/2)*x^2+(1/2)*x; 2 f := x -> 3/2 x + 1/2 x > f(1);A 2 > f(2); 7Here I defined the function for Maple, and then computed f(1) and f(2).
> add(f(1+(j/50))*(1/50),j=1..50); 43001 ----- 10000 > evalf(%); 4.300100000The add instruction operates exactly like a capital sigma in traditional math notation. Maple always tries to do exact computation. So the first answer is the exact answer (!) and the second answer is a floating point approximation. % in Maple means the previously computed result.
|Number of subdivisions||Approximate answer||Predicted error|
We used FTC to see that the "true" value of the definite integral was 4.25 (maybe not too much of a surprise). I mentioned that the last number (which involved a half-million subdivisions!) took about 16 seconds to compute. I consider that a lot of time.
|Monday, November 27||(Lecture 23)|
Given a function f(x) defined on an interval [a,b], I wanted to compute the area "enclosed" by y=f(x), x=a, x=b, and the x-axis. I discussed these technical words and phrases:
I analyzed Riemann sums for a piecewise defined function, something like f(x)=2 if x<4 and f(x)=5 for x>=4, on the interval from 2 to 6. We were able to understand "any" Riemann sum, and we showed that if the length of the largest subinterval in the partition was small, then the Riemann sums would approach what we naively might call the area "under" the curve.
I then moved on to (3/2)x3+(1/2)x on the interval [1,2]. First we looked at a Riemann sum which divided the interval into 50 equal parts, and we showed by a rather simple (but clever!) geometric argument that we could estimate the "error" of that specific sum from the "true value" of the area. Then I turned things around, and asked if we could find a Riemann sum which was within 1/100,000 of the true value. We did this.
I then defined the definite integral. This is all in section 5.2 of the text.
Before I returned the second exam, I made some remarks which I attempt to record here. I apologize if these comments are too direct and too personal, but I believe they are truthful and may be necessary.
|Monday, November 20||(Review for exam #2)|
|Monday, November 20||(Lecture 21)|
I discussed how to "solve" the initial value problem f´(x)=sqrt(1+x3) with f(0)=2. From the qualitative aspect (using direction fields) I knew that f(1)>f(0)=2. What if we wanted more precise information? I showed how we can do "simple arithmetic" motivated by the Mean Value Theorem, and approximate f(1) as closely as we wished.
I began with
We will try to be a bit more indirect, more sneaky, and learn properties of area which will let us approximate what we want. What properties does "area" have?
With RULES A THROUGH E we can begin an organized campaign to compute or approximate areas. For example, here is how we could compute the area of a typical blob. I will use the RULES in the analysis that follows. I would first slice the blob by line segments to get something which had at most one curvy side. I will be done if I describe to you how to compute the area of each of the pieces. I will specialize to one of the pieces, and assert that similar things can be done to all of the other pieces. So:
Now I can approximate inside this piece of the blob and outside the blob by rectangles, and since I know the area of a rectangle, I can get some estimate of the area of the piece of the blob.
Of course this estimate is not so good, but we could improve it with some sort of scheme like this:
So we can get a collection of approximations which get closer and closer and closer to the true value of the area, and maybe that's good enough. (It is, in many cases.)
As I remarked, another instructor told me of a wonderful way of bringing home the idea of area: look at your hand and ask, what is its area? Giving a "good" answer to this may be difficult and involve quite a lot of approximation.
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