E-320: Teaching Math with a Historical Perspective Oliver Knill, 2011Lecture 6: CalculusThis semester, I exp erimented with a different approach to calculus. Lets see what you thinkabout it. Differences measure change, sums measure how things accumulate. The process oftaking differences is in a limit called derivative. The process of taking sums is in the limit calledintegral. These two processes are r elat ed in an intimate way. In this lecture, we want to look atthese two processes in a discrete setup where functions are evaluated o n integers.Start with the sequence of integers1, 2, 3, 4, ... .We say f (1) = 1, f(2) = 2, f(3) = 3 etc and call f a function. It assigns to a number a number.It assigns for example to the number 100 the result f (100) = 100. Now we add these numbers up.The sum of the first n numbers is calledSf (n) = f(1) + f (2) + f(3) + ... + f (n) .In our case we get1, 3, 6, 10, 15, ...It defines a new function g which satisfies g(1) = 1, g(2) = 3, g(2) = 6 etc. The new numbers areknown as the triangular numbers. From the function g we can get f back by taking difference:Dg(n) = g(n) − g(n − 1) = f(n) .For example Dg(5) = g(5) − g(4) = 15 − 10 = 5 and this is indeed f (5).Karl-Friedrich Gauss realized as a 7 year old kid when giving the task to sum up the first100 numbers that it is the same as adding up 50 times 101 which is 5050 by writing it as‘(1 + 100) + (2 + 99) + (3 + 98) . . . (100 + 1) leading to n/2 terms of (n + 1) if n is even. Takingdifferences verifies Dg(n) = (n + 1 ) n/2 − n(n − 1)/2 = n = f (n).1 3 6 10 15 21 36 45 ...n=1 n=2 n=3 n=4Lets add up the new sequence again and compute h = Sg. We get the sequence1, 4, 10, 20, 35, ...These numbers are called the tetrahedral numbers because one use h(n) marbles to build atetrahedron of side length n. For example, we need h(4) = 20 golf balls f or example to build atetrahedron of side length 4. The formula which holds for h ish(n) = n(n + 1)(n + 2)/6 .1 4 10 20 35 56 84 120 ...Also this sequence defines a function. For example, g(3) = 10. But what is g(100)? Can we finda formula fo r g(n)?n=1 n=2 n=3 n=4We can see that summing up the differences gives the function in the same way as differencingthe sum:SDf (n) = f(n) − f (0), DSf(n) = f(n)This is an arithmetic version o f the fundamental theorem of calculus. The process of addingup numbers will lead to the integralRx0f(x) dx. The process of taking differences will lead tothe derivativeddxf(x). One of the high lights of this course is t o understand the fundamentaltheorem of calculus:Rx0ddtf(t) dt = f(x) − f(0),ddxRx0f(t) dt = f(x)and see why it is such a fantastic result. You see formally that it fits the result for difference andsum. A major goal of this course will be to understand the fundamental theorem result and seeits use. But we have packed the essence of the theorem in the above version with S and D. It isa version which will lead us.1 Problem: Given the sequence 1, 1, 2, 3, 5, 8, 13, 21, . . . which satisfies the rule f(x) = f(x −1) + f(x − 2). It defines a function on the positive integers. For example, f (6) = 8. Whatis the function g = Df, if we assume f (0) = 0? Solution: We take the difference betweensuccessive numbers and get the sequence of numbers1, 0, 1, 1, 2, 3, 5, 8, ...After 2 entries, the same sequence appears again. We can also deduce directly f r om the aboverecursion that f has the property thatDf(x) = f(x − 2) . It is called the Fibonnaccisequence, a sequence of great fame.2 Problem: Take the same function f given by the sequence 1, 1, 2, 3, 5, 8, 13, 21, ... but nowcompute the function h(n) = Sf(n) obtained by summing the first n numbers up. It givesthe sequence 1, 2, 4, 7, 12, 20, 33, .... What sequence is that?Solution: Because Df(x) = f(x − 2) we have f (x) − f (0) = SDf (x) = Sf(x − 2) sothat Sf(x) = f(x + 2) − f(2). Summing the Fibonnacci sequence produces the F ibonna ccisequence shifted to the left with f(2) = 1 is subtracted. It has been relatively easy to findthe sum, because we knew what the difference operation did. This example shows:We can study differences to understand sums.The next problem illustrates this too:3 Problem: Find the next term in the sequence2 6 12 20 30 42 56 72 90 110 132 . Solution: Take differences2 6 12 20 30 42 56 72 90 110 1322 4 6 8 10 12 14 16 18 20 222 2 2 2 2 2 2 2 2 2 20 0 0 0 0 0 0 0 0 0 0.Now we can add an additional number, starting from the bottom and working us up.2 6 12 20 30 42 56 72 90 110 1321562 4 6 8 10 12 14 16 18 20 22 242 2 2 2 2 2 2 2 2 2 2 20 0 0 0 0 0 0 0 0 0 0 04 Find the next term in the sequence 3, 1 2, 33, 72, 135, 228, 357, 528, 747, 1020, 1353.... To doso, compute successive derivatives g = Df of f , then h = Dg until you see a pattern.Solution:3 12 33 72 135 228 357 528 747 1020 13533 9 21 39 63 93 129 171 219 273 3330 6 12 18 24 30 36 42 4 8 54 606 6 6 6 6 6 6 6 6 6 60 0 0 0 0 0 0 0 0 0 0We see that after taking 4 differences, we reach zero. This implies tha t the function is acubic polynomial. We can get t he next term by looking at the bottom and adding up.3 12 33 72 135 228 357 528 747 1020 135317523 9 21 39 63 93 129 171 219 273 333 3990 6 12 18 24 30 36 42 4 8 54 60 666 6 6 6 6 6 6 6 6 6 6 60 0 0 0 0 0 0 0 0 0 0