MITES Physics III Summer Introduction 1. 3 Π = Product 2. 4 Proofs by Induction 3. 5 Problems 5


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1 MITES Physics III Summer 010 Sums Products and Proofs Contents 1 Introduction 1 Sum 1 3 Π Product 4 Proofs by Induction 3 5 Problems 5 1 Introduction These notes will introduce two topics: A notation which helps simplify long mathematical expressions and proofs by induction. A proof by induction is essentially a simple way to prove claims about positive integers. In terms of the usefulness of these topics in calculus, the information on notation is the most relevant while the introduction to proofs by induction is a crucial stepping stone in the general study of rigorous (proofbased) mathematics. Sum The Greek capital letter Σ ( Sigma ) will be the focus of this section. The essential idea is that whenever you see the symbol Σ, it means you are adding some set of quantities. Alternatively, whenever we compute the sum of some quantity in which each term in the sum obeys a fixed rule we can write the sum using Σ. This idea is best seen with examples. Consider the sum of the numbers from 1 to 10. We can write this in long form as (1) Or, using the Σ notation we can write this as 10 i 55 () which is much more concise than Eq (1). Let s decompose this notation to see how each part contributes to the definition of Σ as a sum. There are three main elements besides the Σ itself. 1
2 last number index first number expression (3) Below the Sigma we wrote index first number. The index is the number we choose to document where we are in the summation. In the first example i, is the index. The first number is the first value the index assumes in the sum. In the first example this value is 0. To the right of the Sigma is an expression which defines how terms in the summation depend on i. The expression is always some function of i where i is an integer. In the case of the first example, this function of i is just simply i. At the top of the Sigma is the last number the index i assumes. This number determines the ending point of the summation. In our first example this number is 10. To compute the sum we add all the terms from the i first number all the way to i last number with i increasing by integer steps with each successive term. Here s another example N (4) can be written as i (5) It is important to realize that the sum is independent of the index we choose to define its elements. This means we can replace i with any other variable we wish and the value of the sum will be unchanged. This independence of parameter is a general property of all types of sums and even has an analog in calculus in which the value of a definite integral is independent of the variable we use to integrate it. As a final comment, it is not difficult to introduce the notion of infinite sums into our notation. An infinite sum has no final number and instead continues on and on without end. In this way the last number can be treated as or equivalently we can take the last number to be N and simply take the limit of the sum as N goes to infinity. Here s an example lim N i i (6) 3 Π Product After we ve gone through Sigmas and Sums, Pis and Products are easy. All the information about index, first number, and expression transfer over except now we re computing products instead of sums. For example consider the product of the numbers 1 to , 68, 800 (7)
3 is written as 10 i 3, 68, 800 (8) It should be clear that as with sums the value of a product is independent of the index. Product notation gives us a different way of expressing the factorial N i N! (9) Here s another illustrative example. We will use our sigma and pi notation to express a fundamental property of exponentials. If we take the product of two exponentials e a1 e a we know we can write the result as e a1+a. Generalizing to N different exponentials e a1 e a e a N e a1+a+ +an (10) Or using our notation and the convenient notation exp(x) e x we can write ( N N ) exp(a i ) exp a i (11) 4 Proofs by Induction Now for a completely unrelated topic. Suppose we believe we understand a general property of a certain number of integers, but we don t know if the property applies to all integers 1. How can we prove the result for all integers? Well we can first show that the property is true for the number 1. Then we can prove that if the property is true for a general integer k then it must be true for an integer k + 1. In this way we see that if the rule is true for 1 then it is true for 1+1 and it is true for +13 and so on. Here s a classic example Let s add the first N integers. Starting off easy let s add 1 and (1) and another one 1 When I say integers I really mean only positive integers 3
4 (13) (14) (15) Noticing the pattern we can claim N N(N + 1) (16) Now to prove it with induction. We first must show that the result is true for N 1. This is easy; we just have 1 1(1 + 1) 1 (17) Now we must assume that the rule is true for a general integer k and then show that the rule must be true for k + 1. So we must prove that if k is true then k + 1 is true. If the formula applies to k we have k (18) And we must prove that the above formula implies a similar formula for k k + 1 (k + 1)(k + ) We begin by adding k + 1 to both sides of Eq (18) k + k k + 1 (19) Now we will isolate the rightside and perform some algebra + (k + 1) + k + 1 k + k + k + (k + 1)(k + ) 4
5 So we have k + 1 (k + 1)(k + ) (0) and the formula is therefore true for all positive integers. Here is a summary of the general steps in a Proof by Induction Proof by Induction 1)Show that the formula is true for N 1 )Assume the formula is true for k 3)Show using the formula is step ) that the formula is true for k Problems 1.) Use the properties of logarithms to show that the following formula is true ln [ N α i β i ] (ln α i ln β i ) α is the lower case Greek letter alpha and β is the lower case Greek letter beta.) Use proof by induction to prove that the following results are satisfied by all integers n a) (i 1) n b) c) i 3 n(n + 1)(n + 1) 6 j 1 n 1 j1 3.) Let ( ) n be defined as k Use induction to prove that for all positive integers n ( ) n n! k k!(n k)! (a + b) n k1 ( ) n a n k b k k 5
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