Class Summary Page

for

MATH 162

Calculus with Applications, II

November, 1997

Go back to the second class summary page.

  • Monday, November 3. We discussed numerical techniques for evaluating definite integrals; techniques like the ones that we covered today are necessary in order to evaluate definite integrals such as Int(exp(-x^2),x=a..b) because the function exp(-x^2) does not have an elementary antiderivative. We began by reviewing a pair of approximation methods that were already covered in MATH 160; left-endpoint and right end-point approximations. We placed these approximations in the context of our general integration application philosophy: in order to calculate Int(f(x),x=a..b), i.e., the (signed) area between the curve y=f(x), the x-axis and the lines x=a and x=b we should chop the interval [a,b] up into n small subintervals, and approximate the areas corresponding to each of these little subintervals by some appropriately chosen rectangle. The two simplest ways to do this are to take the height of the rectangle to be either the value of the integrand f at the left-hand endpoint or the value at the right-hand endpoint. The first of these corresponds top the left-endpoint approximation (L_n) and the second leads to the right-endpoint approximation (R_n) -- see page 478 in the course text. We noticed that if f is an increasing function, then the rectangles whose areas are being used in L_n are all inscribed, while the rectangles in R_n are circumscribed. So if f is an increasing function, we have that L_n <= Int(f(x),x=a..b) <= R_n; if f is decreasing, then interchange the words inscribed and circumscribed in the preceding sentence and reverse the direction of the inequalities in the first part of this sentence. This suggested that we should try to sonehow "average" the left- and right-endpoint approximations. The first way we did this was by averaging the endpoints of each subinterval [x_(i-1),x_i] to get the midpoint X=(x_(i-1)+x_i)/2, and then taking the rectangle heights to be the values of the f at these midpoints. This gave us the midpoint approximation M_n (see page 478 in the text, and also Section 4.3). A different way to do the averaging is along the y-axis instead of the x-axis: instead of making the height of the rectangle the value of f at the average of the endpoints, we could define it to be the average of the values of f at the two endpoints. We saw that it was possible to view this last approximation as arising from the use of a trapezoid instead of a rectangle (you can see this in the upper diagram in Figure 5 on page 480 in the text, even though Stewart uses that diagram to illustrate something else). We next rewrote these four approximations using the compact summation notation in the special case whe all subintervals have the same length, (b-a)/n. Then we went to the computers and worked on a Maple worksheet, Approximate Integration. This notebook creates tables of approximations by the four different schemes for different choices of n that are similar to those found on page 481 of the course text, except that in the Maple worksheet we can estimate any definite integral that we wish. We worked through the worksheet with the improper integral Int(1/sqrt(1-x^2),x=0..1), which we knew from last class should equal Pi/2 (because this integral gives the arclength of one quarter of the unit circle). We saw that the right endpoint and trapezoidal approximations did not exist in this case because they each required the evaluation of 1/sqrt(1-x^2) at x=1. We also noted that (as expected) the midpoint approximations were converging to Pi/4 faster the left-endpoint approximations. The class was given some time to experiment with various other definite integrals. At the end of this part of the class we summarized some of our findings. We observed that the midpoint and trapezoidal approximations seemed to always converge faster than the midpoint and trapezoidal approximations. Also, the midpoint approximation converges a little faster than the trapezoidal approximation; Stewart explains why on page 480. We briefly discussed how software like Maple knows how to choose the subinterval size in order to be guaranteed that the value it is returning is accurate to a certain number of decimal places (it uses error bounds like those in box 5 on page 480), and more sophisticated approximation schemes such as Simpson's Rule were mentioned. It should be pointed out that we have only scratched the surface here; several weeks of MATH 464 (Numerical Analysis) are spent studying numerical integration methods. At the very end of the class we started a discussion of how to compute surfaces of revolution. After spinning a curve described by y=f(x) about the y-axis, we chopped it into a bunch of concentric ribbons. If we approximate each ribbon as being generated by rotating a small straight line segment about the y-axis, then calculating this approximate surface area boils down to computing the surface area of the frustum of a cone.

  • Wednesday, November 5. Continuing from where we left off last class, we computed the surface area of the frustum of a cone. We started with a large paper cone, and cut off its top to form a frustum. For future reference, we drew the circles at the base and top of this frustum on the board. Then we cut straight down the side of the frustum and flattened it out to see that this was just a sector of an annulus (or washer) between two concentric circles. We could use simple geometry as Stewart does in section 8.3 to calculate the area, but it is more in keeping with what we have been doing so far in the course to use calculus instead. Here's how we did it. We looked at a small arc of the upper circle; if the radius of that circle is r_1 and the angle subtended by the arc is Delta theta, then the length of this arc is (Delta theta)*r_1 -- we found this relationship by observing that when Delta theta is 2*Pi, then the arclength is the entire circumference, 2*Pi*r_1. We then looked at the corresponding arc in the bottom circle, which had length (Delta theta)*r_2, where r_2 was the radius of that circle. We then cut out the portion of the frustum between these two arcs, and upon flattening it out observed that it was nearly a trapezoid of height l (the slant length measured along the frustum), base length (Delta theta)*r_2 and top length (Delta theta)*r_1. The area of this trapezoid is (Delta theta)*((r_1+r_2)/2)*l. Summing up the approximating trapezoids all the way around the circles that determine the frustum and then taking the limit as Delta theta tends to zero gives us the following integral for the surface area of the frustum: Int(((r_1+r_2)/2)*l,theta=0..2*Pi) = 2*Pi*((r_1+r_2)/2)*l. It is convenient in thinking about this expression to regard ((r_1+r_2)/2) as an "average radius." Now we could go back and try to write an integral for a surface of revolution. If the surface was generated by rotating the curve y=f(x) (with a < x < b) about the y-axis, then the surface area can be approximated by a sum each of whose terms has the form 2*Pi*x*sqrt((Delta x)^2+(Delta y)^2); here x is the average radius of a little approximating frustum and sqrt((Delta x)^2+(Delta y)^2) is the slant length of that frustum. Just as we did when we discussed arclength calculations, we factored Delta x out of the sqrt and took the limit as Delta x went to zero which replaces the sum by an integral: SA = Int(2*Pi*x*sqrt(1+(dy/dx)^2),x=a..b). We ended with a classic example: the calculation of the surface area of a sphere. Actually, we calculated the surface area of a(n upper) hemisphere. In this example y=sqrt(r^2-x^2), so we have after doing some algebra to simplify the integrand SA = 2*Pi*r*Int(x*(r^2-x^2)^(-1/2)),x=0..r), which actually is an improper integral of type 2 -- because (r^2-x^2)^(-1/2) is undefined at x=r. We compensated for this by writing Int(x*(r^2-x^2)^(-1/2)),x=0..r) = lim_(t-->r-) Int(x*(r^2-x^2)^(-1/2)),x=0..t), and then making the substitution u=r^2-x^2. After a little more work, this gave us that the surface area of the hemishere was 2*Pi*r^2, so the surface area of the full sphere was 4*Pi*r^2. We ended by commenting on why the formula for the surface area of a sphere (4*Pi*r^2) is the derivative with respect to r of the volume of the sphere ((4/3)*Pi*r^3), in just the same way that the formula for the circumference of the circle (2*Pi*r) is the derivative with respect to r of the area of a circle (Pi*r^2). Question: Why do we get a slightly different relation for cubes and squares? [The formula for the surface area of a cube of side length x (6*x^2) is twice the derivative with respect to x of the volume of the cube ((4/3)*Pi*r^3), and the formula for the perimeter of a square (4*x) is twice the derivative with respect to x of the area of the square (x^2).]

  • Friday, November 7. We spent the day calculating the volume and the surface area of the solid of revolution generated by rotating about the x-axis the region bounded by the curve y=1/x, the line x=1 and the x-axis. (This problem is an extension of Exercise 21 The volume of this object was first calculated by Torricelli. We fisrt calculated the volume by slicing the volume into disks perpendiclar to the x-axis. This led us to write the volume as the improper integral Int(Pi/x^2,x=1..Infinity). We next needed to decide how to make sense of this integral; this kid of improper integral is what Stewart calls a Type 1 improper integral. Just as we already did in the case of Type 2 improper integrals, we need to approximate the improper integral by proper integrals, and then take a limit as the proper integeals tend to the improper one. Since the problem here was that the interval of integration stretched all of the way out to infinity in the positive x-direction, all that we need to do is stop the integration at some point t, and then let t tend to infinitySo we wrote the improper integral as the limit of integrals which had the same integrand but finite intervals of integration: Int(Pi/x^2,x=1..Infinity) = lim_(t-->Infinity) Int(Pi/x^2,x=1..t) = lim_(t-->Infinity) [-Pi/t--Pi/1] = Pi. Note that Torricelli found the fact that an infinitely long object could have a finite volume to be "incredible." (By the way, Torricelli used a different technique than us to computet he volume. Not only was he working, quite literally, in the days of pre-calculus, but he used a technique which is similar to what we would call the method of shells, while we integrated by disks.) After you think about this for a while, it is not so surprising, as long as the object gets thin fast enough in the direction of infinite extent. We then went on to do a calculation that we thought had a truly incredible answer: the surface area of the same solidof revolution. Thinking of slicing up the sufrace into a bunch of pieces each of which is nearly the frustum of a cone led us to the formula SA = Int(2*Pi*(1/x)*sqrt(1+(dy/dx)^2),x=1..Infinty) = ... = 2*Pi*Int(sqrt((x^4+1)/x^3,x=1..Infinity). The problem here is that it is not very easy to evaluate the integral. [I've thought about this since the class, and it is possible to do, it's just not pleasant. First make the substitution u=x^2, and follow this by the trigonometric substitution tan(theta)=u, re-express the integral entirely in terms of sines and cosines, and follow the procedure outlined in my handout "A Note on Trigonometric Integrals."] Instead of evaluating the integral directly, we just estimated (and bounded) it. We reasoned that for large values of x, we might be able to ignore the "+1" inside in the square root. Furthermore, since x^4+1 > x^4, it follows that Int(sqrt((x^4+1)/x^3,x=1..t) > Int(1/x,x=1..t) = ln(t). Thus, since SA = 2*Pi*lim_(t-->Infinity) Int(sqrt((x^4+1)/x^3,x=1..t) and ln(t) --> Infinity as t-->Infinity, we have that the surface area must be infinite -- even though the total volume was finite! On Monday, we will come back to discuss in a more general setting the "comparison" technique we used to conclude that Int(sqrt((x^4+1)/x^3,x=1..Infinity) was divergent by comparing it to Int(1/x,x=1..Infinity), i.e., to lim_(t-->Infinity) Int(1/x,x=1..t).

  • Monday, November 10. We began by deriving a pair of Comparison Tests for Improper Integrals of Type 1, that is, integrals of the form Int(f(x),x=a..Infinity) -- these comparison tests are not only important in their own right, but they are closely related to various convergence tests for infinite series that we will discuss in the last part of the course. We looked at the situation where there are a pair of functions f and g, both continuous on [a,Infinity) with f(x) >= g(x) >= 0. We observed that Int(f(x),x=a..Infinity) (respectively, Int(g(x),x=a..Infinity)) could be interpreted as the area between the graph of y=f(x) (resp. y=g(x)) and the x-axis that lies to the right of the line x=a. We write each of these improper integrals as a limit of proper integrals in the usual way: that Int(f(x),x=a..Infinity) = lim_(t->Infinity) Int(f(x),x=a..t) and Int(g(x),x=a..Infinity) = lim_(t->Infinity) Int(g(x),x=a..t). These limits can either converge or diverge; that's true for any limit. What is special here is that because the integrands f and g are nonnegative, the proper integrals Int(f(x),x=a..t) and Int(g(x),x=a..t) are nondecreasing functions of t. When we take the t->Infinity limit of a nondecreasing function, there are only two possibilities: either the limit converges, or else it diverges to inifinity. Going back to the area interpretation of the improper integrals, we see that either the areas under the curves are finite, or else they are infinite. Since there are the two possibilities for each improper integral, there are 4 (=2*2) total possibilities. We examined them, and found that one of them is impossible, but two of them can be turned into Comparison Tests.
  • It is impossible to have Int(f(x),x=a..Infinity) < Infinity and Int(g(x),x=a..Infinity) = Infinity.
  • It is possible to have Int(f(x),x=a..Infinity) = Infinity and Int(g(x),x=a..Infinity) = Infinity. In this case we observe that as soon as we know that the "smaller" area Int(g(x),x=a..Infinity) is infinite, then the "larger" area must be infinite as well. This gives us a comparison test. If we're given an improper integral Int(f(x),x=a..Infinity), and we are able to somehow come up with a suitable reference function g that satisfies f(x) >= g(x) >= 0, and if we determine that Int(g(x),x=a..Infinity) diverges, then we can conclude that Int(f(x),x=a..Infinity) was a divergent improper integral.
  • It is possible to have Int(f(x),x=a..Infinity) < Infinity and Int(g(x),x=a..Infinity) < Infinity. Here we see that if the "large" area is finite, then the "smaller" area must be as well. This gives us a second comparison test. If we're given an improper integral Int(g(x),x=a..Infinity), and we are able to somehow come up with a suitable reference function f that satisfies f(x) >= g(x) >= 0, and if we determine that Int(f(x),x=a..Infinity) converges, then we can conclude that Int(g(x),x=a..Infinity) was a convergent improper integral. Furthermore, we have the inequality that Int(g(x),x=a..Infinity) <= Int(f(x),x=a..Infinity).
  • It is possible to have Int(f(x),x=a..Infinity) = Infinity and Int(g(x),x=a..Infinity) < Infinity. This case does not lead to a comparison test since if the "larger" area is infinite, we can't make any conclusion about the "smaller" area, nor does knowing that the "smaller" area is finite tell us whether the "larger" area is finite or infinite.
    We worked through an example, Int((12*x^23+16)/(5*x^8-17),x=2..Infinity). Here we compared the integrand (12*x^23+16)/(5*x^8-17) to (12/5)*x^15, our reasons being that for large values of x we should be able to ignore the 16 in the numerator, and the 17 in the denominator. It turns out that (12*x^23+16)/(5*x^8-17) > (12/5)*x^15 for x >= 2, and since it is not hard to see that Int((12/5)*x^15,x=2..Infinity) diverges, it follows that Int((12*x^23+16)/(5*x^8-17),x=2..Infinity) is divergent as well. Looking back at this example (and ahead to the material on infinite series) we decided that it would be a good idea to begin collecting a set of 'reference' improper integrals. We examined the p-integrals Int(1/x^p,x=1..Infinity). We showed that these diverged for p <= 1, and that they converged for p > 1. Having completed our treatment in the course of integration techniques, we began to talk about our last integration application: centers of mass. We began by looking at the law of the lever (or the principle behind the see-saw) that says that in order to balance two point masses along a rod of negligible weight, we should put the fulcrum at the location so that the products of the masses with their distances from the fulcrum are equal: m_1*d_1 = m_2*d_2. This principle was illustrated using a ruler as the rod, and some Play-Doh for the two point masses. We then introduced a coordinate axis along the rod, and in terms of the locations x_1 and x_2 of the two masses, found that the center of mass (i.e., the position for the fulcrum to balance the masses) was given by X = M_y/M, where M_y = m_1*x_1+ m_2*x_2 [the moment of the system about the y-axis (or about x=0) is the sum of the individual moments of the two masses about the y-axis], and M = m_1+m_2 [the mass of the system is just the sum of the individual masses]. We ended by introducing a third mass m_3 at position x_3, and we found that in the case of three masses, the center of mass was now given by by X = M_y/M, where M_y = m_1*x_1+m_2*x_2+m_3*x_3 [the moment of the system about the y-axis (or about x=0) is the sum of the individual moments of the three masses about the y-axis], and M = m_1+m_2+m_3 [the mass of the system is just the sum of the individual masses]. We'll generalize this to the case of n masses on Wednesday. At the end of class, a review sheet to aid in preparation for the third exam was distributed.

  • Wednesday, November 12. We began by generalizing the expression that we had found for the center of mass of for systems of two and three particles along a line to a system of n particles along a line. We found that the center of mass, X, could be written as X = M_y/M, where M_y = m_1*x_1+ m_2*x_2 + ... + m_n*x_n [the moment of the system about the y-axis (or about x=0) is the sum of the individual moments of the n masses about the y-axis], and M = m_1+m_2 + ... + m_n [the mass of the system is just the sum of the individual masses]. In words, X is a weighted average of the locations x_1, x_2, ...,x_n of the locations of the n masses with the weights being the masses themselves. Observe that up to this point the mathematicsthat we have been using to define the center of mass is just algebra. Calculus enters when we try to find the center of mass of a linear object (a rod). Specifically, we supposed that the rod lay along the x-axis with endpoints at x=a and x=b (with a < b), and that the density of the rod was given by the (density) function r(x). I called this density function, rho, in class, but in typing up these notes, I've switched to its Roman equivalent, r, to avoid confusion with the two-dimensional case to be studied later. Note that r(x) is measured in units of mass/length.) We use the same integration application machinery we've used over and over again throughout the semester. We broke the rod into many (say, n) tiny pieces, each of which we approximated as being a point mass. The segment of the rod between x_i and x_i+Delta x_i has a mass that is approximately r(x_i)*Delta x_i (here we assume that r(x) is a continuous function), so M_y is approximately Sum(x_i*r(x_i)*Delta x_i,i=0..n-1) and M is approximately Sum(r(x_i)*Delta x_i,i=0..n-1). Taking the n->Infinity limit (with all of the Delta x_i's tending to zero as well), we get that X = M_y/M where M_y = Int(x*r(x),x=a..b) and M = Int(r(x),x=a..b). Having determined how to find the center of mass of a (one-dimensional) rod, we turned to the same problem for a two-dimensional plate (or lamina). For simplicity, we assumed here that the density rho was constant (note that the units of this rho are mass/length^2), and we assumed that our region was bounded by the curves y=f(x), y=g(x) and the lines x=a and x=b with a < b and g(x) < f(x). In order to not lose our investment in having solved the center of mass problem for rods, we decided to compute the x-coordinate of the plate by squashing it down onto the x-axis. In doing this we had to figure out how to relate the density rho and the geometry of the plate to an effective density r(x) for the rod. The principle that we used to make this correspondence was that the total mass between any x and x+Delta x would have to be the same. For the rod, we have already calculated this mass to be r(x)*Delta x, while for the plate it is rho*area, which in turn is approximately (since the area is nearly a rectangle) rho*height*base = rho*(f(x)-g(x))*Delta x. Equating r(x)*Delta x with rho*(f(x)-g(x))*Delta x, and solving for r(x) gives r(x) = rho*(f(x)-g(x)). Now proceeding just as before for the rod, we have that that X = M_y/M where M_y = Int(x*r(x),x=a..b) = Int(x* rho*(f(x)-g(x)),x=a..b) and M = Int(r(x),x=a..b) = Int(rho*(f(x)-g(x)),x=a..b). One could compute the other coordinate Y = M_x/M of the centroid (X,Y) by setting up analogous integrals with respect to y, but after the exam we will see that it is also possible to write M_x as an intergral with respect to x. A practice version of the third exam (and a solution set) was distributed at the end of class.

  • Thursday, November 13. A review session was held in ACD 302 from 7-9pm.

  • Friday, November 14. The third exam was given today.

  • Monday, November 17. The third exam was returned today together with the solution set. We began by rederiving our integration formulas for the x-coordinate of the center of mass for a lamina of unform density. As usual, we began by slicing our region into small pieces: thin slices with sides perpendicular to the x-axis. We then replaced each of these regions by a point mass sitting at the center of the rectangle [which if the sides were x and x+Delta x with the top y=f(x) and the bottom y=g(x) is approximately (x,(f(x)+g(x))/2)]. As we saw in last Wednesday's class, the mass of this rectangle is approximately rho*(f(x)-g(x))*Delta x, so this will be the mass of the particle representing that rectangle. Summing the moments of all of these particles, dividing by the total mass and taking the limit in which the summations are replaced by integrations (see page 527 in the course text) we re-arrive at the integration formulas for M_y and M that we found on Wednesday. The nice thing about this calculation is that it gives us a simple way to calculate the moment M_x about the x-axis. The moment of an individual particle about the x-axis is (distance from y=0)*mass which is approximately [(f(x)+g(x))/2)]* rho*(f(x)-g(x))*Delta x = (rho/2)*((f(x))^2-(g(x))^2)). Taking the limit as the number of slices/particles goes to infinity, and the width of all slices goes to zero we find that M_x = Int((rho/2)*((f(x))^2-(g(x))^2)),x=a..b). We finished our discussion of center of mass by finding the centroid of a semicircular plate of radius r. The details are omitted here as this is Example 3 on page 528 in the course text. We just comment here that one of the coordinates could be found without any integration by using the symmetry priniple (see page 527). We found that the centroid of the region bounded by y=Sqrt(r^2-x^2) and the x-axis was (0,4*r/(3*Pi)), and we checked this empirically by a balancing a semicircular plate made of cardboard on a pen; the agreement was very good! With this example we completed our study of integration methods and applications. Recall that the underlying basic concept in calculus is that of the limit. There are three major specialized types of limit processes in calculus: differentiation, integration, and series. For the remainder of the course we'll study infinite series. Specifically we'll try to make sense of the process of adding up infinitely many numbers. As motivation, we began by recalling a basic approximation method from first semester calculus: the tangent line approximation. We saw that we could define the linearization L(x) of a function f(x) about the point x_0 as a function of the form a+b*(x-x_0) where the constants a and b are chosen in order to make L and its derivative L' have the same values as f and its derivative, respectively, at x_0. Solving the pair of equations L(x_0)=f(x_0) and L'(x_0)=f'(x_0) for the unknown constants a and b gave us a=f(x_0) and b=f'(x_0), which is the old result from first semester calculus. We decided to try to improve upon this approximation by adding a quadratic term that would allow the approximation to take the concavity of f into account. We looked for a function Q(x) = a + b*(x-x_0) + c*(x-x_0)^2, where now the constants a, b and c are chosen in order to make Q, Q' and Q'' have the same values at x_0 as f, f' and f'', respectively. We solved the three equations Q(x_0)=f(x_0), Q'(x_0)=f'(x_0) and Q''(x_0)=f''(x_0) for a, b and c, and found that a=f(x_0), b=f'(x_0) and c=f''(x_0)/2.

  • Wednesday, November 19. We continued our work on polynomial approximations, this time looking for a cubic approximation C(x) = a + b*(x-x_0) + c*(x-x_0)^2 + d*(x-x_0)^3. As before we required that make C, C', C'' and C''' to have the same values at x_0 as f, f', f'' and f''', respectively. These conditions lead to a=f(x_0), b=f'(x_0), c=f''(x_0)/2 and d=f'''(x_0)/6. In looking carefully at our work we started to see a pattern: the 6 in the denominator of d is 1*2*3 (=3!), the 2 in the denominator of c is 1*2 (=2!). We can think of b as having a denominator of 1 (=1!). Indeed, we define 0!=1, so we can also think of there being a 0! in the denominator of a. [One way to understand why 0!=1 is that it makes the formulas work out nicely. Another is to observe that when n is a positive integer, n! is the number of ways to arrange n objects; so 0! should be the number of ways to arrange 0 objects --- which is 1, don't arrange anything because there's nothing to arrange.] We went ahead and first defined T_n(x) to be the n^th order polynomial that agrees with f and its first n derivatives at x_0 (so, e.g., T_1(x)=L(x), T_2(x)=Q(x) and T_3(x)=C(x)) and based on our analysis we made the guess that if we were to write this polynomial as T_n(x) = a_0 + a_1*(x-x_0) + a_2*(x-x_0)^2 + ... + a_n*(x-x_0)^n = Sum(a_i*(x-x_0)^i,i=0..n), then the k^th constant a_k is given by f^(k)(x_0)/k! [the k^th derivative of f evaluated at x_0, and divided by k!]. Two comments are in order before we go any further:
    (i) We usually call these polynomial Taylor polynomials (hence the T_n notation), but in the special case when x_0=0, it is customary to call these polynomials Maclaurin polynomials.
    (ii) In including a_0 as the zeroth term in Sum(a_i*(x-x_0)^i,i=0..n), we have the understanding that when x=x_0 we interpret (x-x_0)^0 as being 1. To see why we should do this, recall that the sigma notation is just a shorthand way of writing a_0 + a_1*(x-x_0) + a_2*(x-x_0)^2 + ... + a_n*(x-x_0)^n, which reduces to a_0 + a_1*0 + a_2*0 + ... a_n*0^n = a_0 when x=x_0. Be aware that this convention has nothing to do with the limits of the indeterminate type 0^0 that we considered earlier in the course when we were discussing L'Hospital's Rule.
    We ended the class by proving that this guess was correct. We broke the polynomial a_0 + a_1*(x-x_0) + a_2*(x-x_0)^2 + ... + a_n*(x-x_0)^n up into three parts: a part where all of the powers were less than k, a kth order monomial, and a part where all of the powers were greater than k. We then differentiated the polynomial k times and evaluated it at x_0. The first part gave us zero (since the kth derivative of any polynomial with order less than k must always be zero), and the third part also gave us zero (because after differentiating this term k times, we can still factor out at least one power of x-x_0, which becomes zero when we evaluate it a x=x_0. There's a different story with the middle term: when we take k derivatives (with respect to x) of a_k*(x-x_0)^k, we get a_k*k!. Therefore, the result of diiferentiating the polynomial a_0 + a_1*(x-x_0) + a_2*(x-x_0)^2 + ... + a_n*(x-x_0)^n k-times and then evaluating the result at x=x_0 gives us a_k*k!. Since we were looking for the polynomial that had the property that its kth derivative matched the kth derivative of f at x_0, we see that we must have f^(k)(x_0) = a_k*k!, or a_k = f^(k)(x_0)/k!.

  • Friday, November 21. Today we found a formula for the Maclaurin polynomials for the function f(x) = 1/(1-x). If we informally take a limit of these polynomials (note that we have not yet said anything precise about what such a limit should be) we end up with the geometric series, the most important of all infinite series and the prototypical power series. Specializing to the case of a particular x (we let x=r), we saw that the series of functions became an infinite series of numbers, and we decided to try to make sense of the act of summing up infinitely many numbers by first only summing the a certain number of terms (say, n), and then letting n tend to infinity. We noted that these partial sums form a sequence themselves, and that summing an infinite series means the same thing as taking the limit of its partial sums. We have now reduced the problem of trying to make sense of adding up all of the terms of one sequence to the problem of computing the limit of some other sequence (the sequence of partial sums). We ended the day with a clever multiply-and-subtarct trick that gave us the formula S_n = (1-r^(n+1))/(1-r) for the nth partial sum of the geometric sequence {r^n}.

  • Monday, November 24.

  • Wednesday, November 26.

  • Friday, November 28. No class today. (Thanksgiving Break)

    Go to the next page of class summaries.

    Go back to the main 162 page.

    This page last modified 12/27/97