Lecture One type of mathematical proof that goes everywhere is mathematical induction (tb 147). Induction is essentially used to show something is true for all iterations, i, of a sequence, where i N. The principle of mathematical induction is simple, and is given in principle 5.1.1. Suppose a statement P i is given. If it can be shown that 1. P is true for i = 1, and that 2. if P i is true, then P i+1 is also true, then P i is true for all i N. Here is an example. Ex. Use induction to prove that 8 n 3 n is evenly divisible by 5 for n N. First we'll note that if n = 1, => 8 1 3 1 = 5, which is, of course, evenly divisible by 5. Next assume that for some i, k N, 8 i 3 i = 5k. So 8 i + 1 3 i + 1 = 8 i 8 3 i 3 = (8 i 3 i )8 5(3 i ) = (5k)8 5(3 i ) = 5(8k 3 i ) (distributive property). Since 8, k and 3 i are all members of J, and J is closed under multiplication, then (8k 3 i ) J. So 8 i + 1 3 i + 1 is evenly divisible by 5. => 8 n 3 n is evenly divisible by 5 for n N.(p 1). //. [] Ex. Here's another one. We want to prove that the sum of the cubes of the first n natural numbers is n = n 2 (n+1) 2 /4. First note that if n = 1, => we have 1 = n 2 (n+1) 2 /4 = 1(4)/4 = 1. Next assume that for some k N, 1 + 2 3 + 3 3 + + k 3 = k 2 (k+1) 2 /4 = k. So k+1 = k 2 (k+1) 2 /4 + (k + 1) 3 = (k + 1) 2 (k 2 /4 + k + 1) = ((k+1) 2 /4)(k 2 + 4k + 4) = ((k+1) 2 /4)(k + 2) 2 = (k+1) 2 ((k + 1) +1) 2 /4, which is precisely what we desired. So, the sum of the cubes of the first n natural numbers is n = n 2 (n+1) 2 /4 (p 1). //. [] Ex Now here's a tricky dicky! We want to show that for n N, Π r=1n (1 + x 2r ) = (1 x 2^(n+1) )/(1 x 2 ), for x ±1. Let n = 1. Then Π r=1n (1 + x 2r ) = (1 + x 2 ) = (1 x 4 )/(1 x 2 ) which is true (why?). 1S
Ex (cont) Now assume that k N, and that Π r=1k (1 + x 2r ) = (1 x 2^(k+1) )/(1 x 2 ), for x ±1. Then Π r=1 (k+1) (1 + x 2r ) = (1 + x 2k+2 )(1 x 2^(k+1) )/(1 x 2 ), for x ±1 (why?) = (1 + x 2k x 2 )(1 x 4k )/(1 x 2 ), for x ±1 (why?) = (1 + x 2(k + 1) )(1 - x 2k(2) )/(1 x 2 ), for x ±1 = (1 + x 2(k + 1) )(1 - x (2k) )(1 + x (2k) )/(1 x 2 ), for x ±1 = (1 x 2^(k+2) )/(1 x 2 ), for x ±1? [] Lecture A strong form of mathematical induction is given on tb 153. It is similar in nature to what we had earlier and more generalized. In this case, if we show for statement P that 1. P is true for some j J, and that 2. if P i is true for some i J, where i > j, then P i+1 is also true, then P i is true for all i J Here is an example. Ex. It is sometimes convenient in mathematics to have some idea of how large n! is for large n ( N ). James Stirling has shown that L n!e n /( (2π)n n.5 ) = 1. Let's use the strong form to attempt to prove that n! > 2 n for all n N, where n > 3. For n = 4, 4! = 24 > 16 = 2 4. Next suppose that i N, and that i > 3, and that i! > 2 i. So based on this last truth, we know that i + 1 > 0, and (i + 1)! = i!(i + 1) > 2 i (i + 1) (Algebra rule). Since i + 1 > 4 > 2, then (i + 1)2 i > (2) 2 i = 2 (i + 1)i. => (i + 1)! > 2 (i + 1)i. So n! > 2 n for all n N, where n > 3 (p 2). //. [] Lecture Now try these problems. Student problem set H (tb 156-159) 4, 6, 8, 12, 14, 22, 36, 38 Student Solution set H H4. e) For n = 1, 25 32 = -7, which is divisible by 7. Next assume 7 divides 5 2k 2 5k for k N. Note that 5 2(k+1) 2 5(k+1) = 5 2k (25) 2 5k (4) = 5 2k (25) 2 5k (25) + 2 5k (21) = 25(5 2k 2 5k ) + 2 5k (21). Note that each of the last two terms is divisible by 7 (why?). => 7 divides evenly into 5 2k 2 5k for k N L(p 1). //. 2S
H36. For n = 2, Π r=2n,(2r - 1)/(2r - 3) = 3. For n = 3, Π r=2n,(2r - 1)/(2r - 3) = 3(5/3) = 5. For n = 4, Π r=2n,(2r - 1)/(2r - 3) = 5(7/5) = 7. For n = 5, Π r=2n,(2r - 1)/(2r - 3) = 7(9/7) = 9. So we can guess that the pattern is that Π r=2n,(2r - 1)/(2r - 3) = 2n 1. [] Lecture Now we wish to take up recursively defined sequences. Again, we will assume that n N. A recursive sequence is a sequence of numbers {a 1, a 2, a 3, } whereby the value of a n+1 is found using the immediately preceding value(s). Ex. A simple case of this the definition of the sequence {2k}, k N. Here 2 (k+1) = 22 k. [] Ex. Consider the following sequence {a k }, k N, defined by a 1 = 7, a k+1 = (a k /2)I k (ev) + ((a k 1)/2)I k (odd). Find the first 7 terms of the sequence are {7, 3, 1.5,.25,.125, -.4375, -.21875, }. [] Lecture Now we'll take a look at some special sequences. First (and simplest) is the arithmetic sequence, {a n }. This is one in which each term differs by the same value d R, so that (tb 162) a n = a 1 + (n 1)d. As an additional result, we have that the partial sum of such a sequence is S n = (n/2)[2a + (n 1)d]. A geometric sequence, {a n }, is one in which each term differs by a common ratio r R + \{1}, so that (tb 163) a n = ar n-1, OR = ra n-1. Again, an additional result is that the partial sum is defined by S n = a(1-r n )/((1 r). Now let's do some examples! Ex Write out the formula for the following sequences and find the sum of the first 7 terms of each. {a n } = {3, 6, 9, } Here, {a n } is an arithmetic sequence in which a n = 3 + 3(n-1), and S 7 = 3.5[6 + 18] = 84. {a n } = {3, 6, 12, } Here, {a n } is an geometric sequence in which a n = 3 2 n-1, and S 7 = 3[(1 2 7 )/(1-2)] = 381. 3S
Ex. (cont) {a n } = {5, 2, -1, -4, } Here, {a n } is an arithmetic sequence in which a n = 5-3(n-1), and S 7 = 3.5[10 + 6(-3)] = -28. {a n } = {9, 3, 1, } Here, {a n } is an geometric sequence in which a n = 9(1/3) n-1, and S 7 = 9[(1 (1/3) 7 )/(1 1/3)] =ca 13.4938. [] Lecture One of the most important sequences in mathematics is the Fibonacci sequence. It was named for Filius Bonaccii (1180 1228), a famous Italian mathematician. Fibonacci discovered a sequence defined as {f n }, where f 1 = f 2 = 1, and f n = f n-1 + f n-2, for n > 2. The actual sequence looks like this {fn} = {1, 1, 2, 3, 5, 8, 13, 21, } The sequence occurs quite often in nature and is commondly known and used by scientists and mathematicians alike. The nth term of the Fibonacci sequence can be approximated by the closest integer to (.5 + (5)/2) n / (5). Ex One interesting question is which terms are even, and which are odd in the Fibonacci sequence? It certainly appears that every 3d term would be even, and the rest odd. Let's use induction to find out. We want to show that f 3i, i N, can be evenly divided by two, and that the preceding two numbers are odd. We know that is the case for f 1 = f 2 = 1, and f 3 = 2 which is certainly divisible by two. Next, assume that f 3i is divisible by two for some i N, and that f 3i-1 and f 3i-2 are odd. So f 3i = 2k, for some k N. Then f 3i + 1 = f 3i + f 3i-1 (p 4), which is odd since that's an odd and even sum, and f 3i + 2 = f 3i + f 3i+1 (p 4), which is odd since that's an odd and even sum, and f 3i + 3 = f 3i+1 + f 3i+2 (p 4), which is even since that's an odd and odd sum. 4S
Ex (cont.) So then, by induction (p 1), we have the result f 3i is even, and f 3i-2, and f 3i-1 are odd, where i N. //. [] Student problem set I (tb 167-170) 2, 8, 14, 15, 18, 21, 32, 38, 52 Student Solution set I I21. 81, 169, 400. I52. Use the last example. [] Lecture As it turns out, recurrence relations can be solved using Characteristic polynomials. If a sequence defined by recurrence relation can be written as a n = ra n-1 + sa n-2 + f(n), where r, s R, and f is some function of n, then by rewriting the definition as a n - ra n-1 - sa n-2 = 0 (f(n) = 0), then Theorem 5.3.1 tells us that the recurrence relation can be solved by solving the characteristic polynomial equation x 2 rx s = 0. The solution is then giv'n by n n n a n = c 1 x 1 + c 2 x 2 (or nc 2 x 2 for double roots). Ex. Consider the recurrence relation {a n } in which a 1 = 1, a 2 = 3, and a n = 2a n-1 + 3a n-2 for n > 2. A quick look at the first terms of the sequence shows that {a n } = {1, 3, 9, 27, } The characteristic polynomial is giv'n by x 2 2x 3. The roots of this polynomial are -1, and 3. So we can rewrite a n as c 1 (-1) n + c 2 (3) n, and c 1 and c 2 are determined by initial conditions using Linear Algebra. -1 3 1 Adding row 1 to row 2 we obtain 1 9 3-1 3 1 0 12 4 Now solving for c1 and c2 we get c 1 = 0, c 2 = 1/3. So, as expected by looking at the original sequence {a n } is defined explicitly by a n = (1/3)(3 n ) = 3 n-1. [] Lecture Now what happens if f(n) is not 0 (as assumed earlier). 5S
Lecture Theorem 5.3.2 (tb 172) tells us that if {a n } is a sequence defined by a n = ra n-1 + sa n-2 + f(n) (as seen earlier), and p n is any particular solution to {a n }, and q n is the solution to the homogeneous affiliate of {a n }, then a n = p n + q n is the general solution. Ex Consider the recurrence relation {a n }, defined by a n = a n-1 + 4a n-2 + 3n, with a 1 = a 2 = 1. Using a linear guess for f(n), we'll select p n = a + bn, where a, b R. So a n = a + bn = a + b(n 1) + 4(a + b(n 2)) + 3n = 5a + 5bn 9b + 3n. => 5a 9b = a, and 5b + 3 = b. This gives us b = -.75, and 4a = 9b, so a = -27/16 = 1.6875. So a particular solution (ignoring initial conditions) is then given by p n = 1.6875 -.75n. Now the characteristic equation for the homogeneous sequence is x 2 x 4 = 0, and has solutions x = -1.56, 2.56. So the homogeneous solution is q n = c 1 (-1.56) n + c 2 (2.56) n. So the overall solution is a n = p n + q n = 1.6875 -.75n + c 1 (-1.56) n + c 2 (2.56) n. Now, reinstating the initial conditions we have 1 = 1.6875 -.75 - c 1 (1.56) + c 2 (2.56), and 1 = 1.6875 1.5 c 1 (1.56) 2 + c 2 (2.56) 2. So the resulting linear system is -1.56 2.56.0625-2.4336 6.5536.8125 Which reduces to c 1 =.4183, and c 2 =.2793. So the final explicit form for {an } is a n = 1.6875 -.75n +.4183(-1.56) n +.2793(2.56) n. [] Student problem set J (tb 174-175) 4, 8, 14, 20 Student Solution set J J4. c 1 = c 2 = 1. J14. c 1 = c 2 = -3. [] Lecture Now we want to introduce something quite useful in other areas of mathematics. If {a n } is a sequence, then the affiliated generating function is simply the expression f(x) = a 0 + a 1 x + a 2 x 2 +, for x in R. 6S
Lecture If g(x) is the generating function of a sequence {b n }, then we can define (tb 176) addition and multiplication of generating functions as follows: f(x) + g(x) = (a 0 + b 0 ) + (a 1 + b 1 )x + (a 2 + b 2 )x 2 +, and f(x)g(x) = (a 0 b 0 ) + (a 1 b 0 + a 0 b 1 )x + (a 0 b 2 + a 1 b 1 + a 2 b 0 )x 2 + Note that the nth term of the product (coefficient of x n ) is giv'n by a 0 b n + a 1 b n-1 + + a n-1 b 1 + a n b 0. Ex A good thing to see is that generating functions are similar to power series. In fact, (1 x) -1 = 1 + x + x 2 + Note that this is the generating function for the sequence {a n } = {1, 1, 1, }. [] Ex Consider the geometric sequence {a n } = {5(1/3) n-1 }. The generating function is giv'n by f(x) = 5 + (5/3)x + (5/9)x 2 + According to the previous example and our previous information (p 3) f(x) = 5(1 x/3) -1. [] Lecture Another important formula giv'n on tb 177 is that (1 x) -2 = 1 + 2x + 3x 2 + is the generating function for {n}, the sequence of N. Here is another example. Ex Consider the recurrence relation {a n } defined by a n = a n-1 /3, and a 1 = 5. Set f(x) = a 1 + a 2 x + a 3 x 2 +, and xf(x)/3 = xa 1 /3 + a 2 x 2 /3 + a 3 x 3 /3 + Then f(x) xf(x)/3 = a 1 + (a 2 a 1 /3)x + (a 3 a 2 /3)x + = 5 (why?) So f(x)(1 x/3) = 5 => f(x) = 5(1 x/3) -1, as seen earlier. [] Student problem set K (tb 181-182) 1, 2, 6, 10, 14 Student Solution set K K2. c) (1 2x) -1, e)3/(1 x). K10. Use p 178 as a guide. K14. Use partial fractions. [] Lecture Next, let's take a look at discrete random variables. 7S