Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Abuse of notation wikipedia , lookup
Location arithmetic wikipedia , lookup
Mathematics of radio engineering wikipedia , lookup
Positional notation wikipedia , lookup
Large numbers wikipedia , lookup
Line (geometry) wikipedia , lookup
Big O notation wikipedia , lookup
Proofs of Fermat's little theorem wikipedia , lookup
Recurrence relation wikipedia , lookup
Arithmetic and Geometric Series Dave L. Renfro Central Michigan University November 8, 2002 I. Arithmetic Series 1. Evaluating 1 + 2 + 3 + + 100 Let S represent this sum. Here’s a neat way to evaluate S. First, add S to itself in the following way: S S = = 1 100 + + S+S = 2 99 + + + 3 98 + + + + + + + 98 3 99 2 + On the right hand side of the last equation we have times. Therefore, it follows that 2S = ( )( equation by 2 to get S: S = : 2. Evaluating 1 + 2 + 3 + + + + + 100 1 + added to itself many ). Now divide both sides of this +n Let T (n) represent this sum. Then T (1) = , T (3) = , T (7) = , 1 and T (100) = . By doing the same thing that we did in Section 1 above, we can get a closed form expression for T (n). T (n) T (n) = = T (n) + T (n) = 1 n + 2 + 3 + + (n 1) + (n 2) + + (n 2) + (n 1) + + 3 + 2 + + + + + On the right hand side of the last equation we have times. Therefore, it follows that 2T (n) = ( )( equation by 2 to get T (n): T (n) = 1 + n 1 + added to itself many ). Now divide both sides of this : The numbers T (1), T (2), etc. are called triangular numbers. For the reason why they are called this, look at some of the web pages you’ll …nd at <http://www.google.com/search?q=triangular+numbers>. 3. Numerical examples using 1 + 2 + 3 + Example 1: 3 + 6 + 9 + = 3( + 1) + 663 + + + Example 2: 4 + 7 + 10 + + ( + 3) + ( + 6) + ( = ( + + + = ( ) ( ) + ) 3T ( ) 3T ( ) = : ( + 5) + ( + 10) + ( = ( + + 3( + 663) + = + + + ) : = ( + 15) + ( + + + + ( + 1257 = ) = + 9) + + Example 3: 12 + 17 + 22 + 27 + + ) + 664 = ) ( 1 n (n 2 +n = ) ( + ( + 20) + ) ( ) T( ) 4. Algebraic examples using 1 + 2 + 3 + + + = + + ) + ) : 1 n (n 2 +n = + 1) Example 1: 1 + 2 + 3 + + (n 1) = T( ) = . Example 2: 1 + 2 + 3 + + (n + 3) + (n + 4) = T( ) = . Example 3: 1 + 2 + 3 + 2) + (n + ( + (2n 1) + 2n = T( ) = . 5. An explicit formula for the sum of an arithmetic series Suppose the …rst term of an arithmetic series is a, its common di¤erence is d, and the number of terms is n. Use a, d, and n to express the terms of this series in the manner indicated: Sn = (1’st term) + (2’nd term) + (3’rd term) + Sn = ( ) + ( ) + ( + (n’th term) ) + + [ ] If we group the a’s together, then factor a d out of the remaining terms, and …nally use an appropriate evaluation of the function T , we can …nd a closed form expression for the sum in the same way that the numerical examples 2 and 3 were done above. Sn = Sn = ( ( + ) ( + + ) + + ( ) + ) T( ( ) ( ) = + + + : + ) Now show that Sn can be written in the following two ways: Sn n 2 = [2a + (n Sn 1) d] n 2 = (a1 + an ) 6. How to recognize an arithmetic series A series is an arithmetic series if (a) consecutive terms di¤er by the same amount, or N P (b) in sigma notation the series has the form an , where an is a linear function of n. n=M II. Geometric Series 1. Prelude #1: Find a fraction for 0:37373737::: You may have seen this method in a beginning algebra class. First, set the repeating decimal equal to x and multiply both sides of this equation by 10n , where n is the period length of the repeating decimal. Then subtract the original equation from this new equation and solve for x. For the repeating decimal 0:37373737::: the period length is 2. x = 0:37373737::: 100x = 37:37373737::: 100x 99x x x = (37 + 0:37373737:::) = 37 = 37=99 (0:37373737:::) 2. Prelude #2: Find an explicit value for the in…nite sum 4 3 + 8 9 + 16 27 + Set the sum equal to x and multiply both sides of this equation by 32 , which happens to be the ratio between consecutive terms. Then subtract the original equation from this new equation and solve for x. x = 43 + 89 + 16 27 + 2 3x 2 3x 1 3x x x = = = 4 8 9 4 3 + 16 27 = 8 9 + + 32 81 + 16 27 + 32 81 + 4 3 + 8 9 + 16 27 + 3. How these two examples are similar 37 37 37 Note that 0:37373737::: = 100 + 10000 + 1000000 + . Thus, the …rst example above can be viewed as an in…nite geometric series. Moreover, the method that we used in the …rst example to …nd a fraction equivalent to 0:37373737::: is essentially same method that we used in the second example. To see this, note that in the …rst example the common ratio 1 was 100 . Although we actually multiplied the original equation in the …rst example by 100, 1 1 , multiplying by 100 would have accomplish the same thing, namely shifting and not 100 the decimals over by two spaces to allow almost everything to line up and cancel. x = 0:37373737::: = 0:00373737::: 1 100 x 1 100 x 99 100 x x x = (0:00373737:::) = 0:37 = 37=99 = (0:37 + 0:00373737:::) 37 100 4. The sum of an in…nite geometric series Consider an in…nite geometric series with common ratio r, where 1 < r < 1. [We’ll see later why we need this restriction on r.] If a is the …rst term, then we’re looking at a + ar + ar 2 + . Set the sum equal to x and multiply both sides of this equation by r. Then subtract the original equation from the new equation and solve for x. x = a + ar + ar2 + rx = ar + ar2 + ar3 + rx (r x x = ar + ar2 + ar3 + 1) x = a = a = (1 r) a + ar + ar2 + Note if a = 1 and r = 2, then this formula says that 1 + 2 + 4 + 8 + equals 1! More generally, we’ll get a negative number if a is positive and r > 1. This much alone tells us that we don’t want to use this formula when r > 1. 5. The sum of a …nite geometric series The same technique used above also works for a …nite geometric series. Suppose the …rst term of a …nite geometric sequence is a, its common ratio is r (for a …nite series no restriction on r is needed), and the number of terms is n. Use a, r, and n to express the sum of the terms of this sequence in the manner indicated: rSn Sn = (1’st term) + (2’nd term) + (3’rd term) + Sn = ( ) + ( ) + ( ) + + ( ) rSn = ( ) + ( ) + ( ) + + ( ) Sn = ( (r 1) Sn = Sn = ) + (n’th term) ( ) Now show that Sn can be written in the following way: Sn = a(1 1 rn ) r 6. Two ways to verify the formula we just found One way to show that a (rn 1) r 1 = a + ar + ar2 + + arn 1 is to simply verify by multiplication that a (rn 1) = (r + arn 1) a + ar + ar2 + 1 : Expanding the right hand side gives ar + ar2 + ar3 + a ar ar2 Combining like terms leaves us with ar3 + arn 1 arn 1 + arn a + arn = a ( 1 + rn ) = a (rn 1). Another way to verify this result is to use synthetic division.2 Since we can cancel the factors of a, it will be enough to show that rn r 1 1 1 + r + r2 + = + rn 1 + xn 1 : Note this is the same thing as showing that xn x 1 1 1 + x + x2 + = : Here’s a synthetic division table for this division: xn 1c xn 1 x xn 0 1 1 n 1 1 1 n 2 x 2 x2 x 0 1 0 1 0 1 1 1 1 1 1 0 x c o n sta nt re m a in d e r n 3 x c o n sta nt 7. Convergence of in…nite geometric series If 1 < r < 1, then rn ! 0 when n ! 1. Therefore, for these values of r we have a + ar + ar2 + + arn 1 + ! a (0 1) a a = = ; r 1 r 1 1 r which is the same formula that we got back in Section 4. However, we will not assign a numerical sum to an in…nite geometric series if jrj 1, at least when a 6= 0.3 This is because as you add more and more terms, the partial sums don’t get closer and closer to any number when r 1 or r 1. In fact, the terms you are adding will get “in…nitely large”(either positively or negatively) if r < 1 or r > 1, and you should be able to verify for yourself what happens when r = 1 or r = 1. 8. How to recognize a geometric series A series is a geometric series if (a) consecutive terms have the same ratio, or (b) in sigma N P notation the series has the form an , where an is an exponential function of n. n=M 2 Incidentally, mathematical induction can be used to prove that rn 1 is divisible by r 1 for each n = 1, 2, 3, ..., where the identity rn+1 1 = (rn 1) r + (r 1) is used for the inductive step. However, this method will only show that rn 1 is divisible by r 1. It won’t tell you what the quotient will be. 3 Obviously, if a = 0, then it is reasonable to say that the in…nite sum a + ar + ar2 + is zero.