Sigma otatio. Itroductio We use sigma otatio to idicate the summatio process whe we have several (or ifiitely may) terms to add up. You may have see sigma otatio i earlier courses. It is used to idicate the summatio of a umber of terms that follow some patter. Σ tells us to sum This tells us to ed with i = a i i=k This tells us to start with i = k a i tells us what to sum. Usually a i is a fuctio of i. Figure.: The schematics of sigma otatio. EXAMPLE... Here are several simple examples. Note the the letter for the idex eed ot be. i = () + () + (3) + () + () = 30. = 7 (i + ) = 0 + 7 + + 37 + 0 = 0. i=3 j = + + 3 + =. j= (d) I this example represets some fixed but ukow iteger value. Notice that k chages but does ot. (k + 3) = () + (7) + + ( + 3). k= (e) Note that the etire summatio may be symbolic: f (x i ) x = f (x ) x + f (x ) x + + f (x ) x. (f ) Here s aother symbol-lade summatio: f + 3i 3 = f + 3 3 + f + 3 + f + 9 3 + f + 3 3. If f (x) = x, the the previous sum would become + 3i 3 = + 3 3 + + 3 + + 9 3 + () 3.
math 3, day sigma otatio (g) You should be able to reverse this process ad write a sum i compact summatio otatio by recogizig appropriate patters. Or you might thik + + + 8 + + 0 = 0 k. k= 0 + + + 8 + + 0 = ( + + 3 + + + 0) = k. k= (h) This reverse process might take place symbolically: or as f (.)(0.) + f (.)(0.) + + f (.0)(0.) = (0.)[ f (.) + f (.) + + f (.0) = 0. YOU TRY IT. (Sigma Notatio). Traslate each of the followig: = cos(π) = 3 + =0 Now write each of the followig sums usig sigma otatio. (d) + 9 + + + 8 (e) 3 + 9 + 7 + 8 + 3 (f ) + + + (g) 0 + + + 3 + + 0 f ( + 0.i)(0.) f ( + 0.i) + + 9 + = 30 + = 0 0 + 3 + + 33 + 7 = 0 (d) 3 = aswer to you try it.. (e) = 3 (f ) = ( ) (g) 0 =0 YOU TRY IT. (Challege). Determie the values for each of these sums for =, 0, ad 0. k k= k k= k=0 k (d) Now determie a geeral formula for ay value of for the sum k=0 k. Notice i some of the sums we factored out a costat term. I fact applyig basic associativity ad distributivity laws for additio, we have ad ca i = ca + ca + + ca = c(a + a + + a ) = c (a i + b i ) = (a + b ) + (a + b ) + + (a + b ) I other words, we have = (a + a + + a ) + (b + b + + b ) = a i a i + b i.
math 3, day sigma otatio 3 THEOREM.. (Basic Summatio properties). For ay costat c, ca i = c (a i + b i ) = a i a i + b i.. Four Key Summatio Formulæ As we begi to tackle the area problem we will fid that we eed to use a few basic summatio formulæ repeatedly. The first is quite easy to see. Suppose that c is a costat. The times {}}{ c = c + c + + c = c. Aother simple formula that you ca figure out is the sum of the first itegers. Let S = + + 3 + + ( ) +. For example, S = + + 3 + = 0 (as ay bowler would kow). There a formula for S that C. F. Gauss (a very famous 9th cetury mathematicia) figured out whe he was. Here s how: Write the summads forwards ad backwards like so: S = + + 3 + + ( ) + + S = + ( ) + ( ) + + + S =. Now add each colum. What do you get as the total for each? How may times?. So what is the formula for S? Now solve for S. 3. Use your formula to check that S = 0. Now use it to determie S 0 ad S 00.. Suppose you wated to sum the eve itegers oly. Let T be the sum of the first eve itegers: T = + + +. What is the formula for T? Ok, your formula for the sum of the first itegers should have bee S = (+). There are two more formulæ that you should memorize: the sum of the first itegers squared ad cubed. They are listed below. Their proofs are a little harder ad we will skip them here. THEOREM.. (Summatio Formulæ). For ay positive iteger c = c i = i = (d) ( + ) ( + )( + ) i 3 = ( + ) Iterestig! The sum the first cubes is the square of the sum of the first itegers.
math 3, day sigma otatio EXAMPLE... Let s use these formulæ to calculate a few sums. 8 i ( + )( + ) = = 8(9)(7) = 0. This ext problem is more typical: We have a ukow upper limit for the sum. The goal is to express the aswer as simply as possible. 3i 3 = 3 3 i = 3 3 [ ( + )( + ) = ( + )( + ) = + 3 + = + 3 + = + 3 +. Now use limits to determie the value of the previous sum as 3i lim 3 = lim + 3 + = + 0 + 0 =. (d) The ability to determie limits like the oe above will be critical for solvig the area problem. Here s aother. Determie lim i 3. First we work o the sum usig Theorem.. ad.. So (e) Let S = Solutio. i 3 = [ ( i This time S = lim ) 3 + 3 i 3 i 3 [ i 3 + 3 = = [ ( + ) = + = =. 3 = lim = 0 =.. Determie lim S. i 3 + 3 () 3 = i 3 + (3) = [ ( + ) = ( + ) + = + + + = + + + = 3 + +. + So lim S = lim 3 + + = 3.
math 3, day sigma otatio YOU TRY IT.3. Use summatio properties ad formulæ to fid the followig geeral sums. Your aswer will be i terms of. Be sure to simplify. ( i ) i 3 Now use your aswers to ad to determie i (d) lim i (e) lim 3 ( + i ) 7 3 + 3 + (d) (e) 3 YOU TRY IT.. Here are two more to practice. [ i Let S =. Determie lim S. [ i Let R = i. Determie lim R. aswer to you try it.. 8 3 aswer to you try it.3. 8 A Look Ahead This ext example is much more typical of how we will actually use summatio otatio. EXAMPLE..3. Let f (x) = x o the closed iterval [0,. Determie the value of where is a arbitrary positive iteger. Solutio. First we must evaluate f f i = f i. i, i = ( i ) = i. (.) Now turig to the sum f i = [ i = [ i = i = 3 i Usig (.) is costat, use Theorem.. Split the sum, use Theorem.. Theorem.. ad Theorem..
math 3, day sigma otatio Now use the summatio formulæ. i f = 3 i = ( + )( + ) 3 ( + )( + ) = 3 = + + 3 = 3 3 = 8 3 3. Theorem.. ad Theorem.. Theorem.. Simplify Simplify Simplify ad distribute the mius sig Notice that we ca use this geeral solutio to evaluate sums for various values of without havig to redo the sum. For example if = 0, the 0 f i 0 0 = 8 3 0 3(0) =.. More geerally, as, it is easy to see that lim webwork: Click to try Problems through. f i ( 8 = lim 3 ) 3 = 8 3. (.) Why do this? What do sums such as those i ) Example..3 represet? Why should we be iterested i them? Well, f represets the of a fuctio at equally spaced poits. The poits are uits apart. We ca thik of the product f i as "height width" or the area of a rectagle. Thus the sum represets the area of rectagles that approximate the area uder the curve of f. As the figures below idicate, as the umber of rectagles gets larger, we get a more precise approximatio to the actual area uder the curve. ( i 0.0 0. 0.8...0. 0.0 0. 0.8...0. Figure.: The left-had graph shows for the fuctio f (x) = x o [0, with the iterval divided i to = 0 subitervals of width = 0. The heights of the rectagles come from the fuctio values f i = f i 0 as i chages from to 0. he sum of the areas of the 0 rectagles is. ad is a approximatio of the area uder the curve o the iterval [0,. I the right-had graph, = 0, ad the same process is repeated. The sum of the areas of all the rectagles is a better approximatio of the true area uder the curve o the iterval [0,. By lettig, we may obtai the exact area uder the curve. We foud that this area was 3 8 i (.).