Sigma notation

1 Sigma notationmc-TY- Sigma -2009-1 Sigma notation is a method used to write out a long sum in a concise way. In this unit we lookat ways of using Sigma notation , and establish some useful order to master the techniques explained here it is vital that you undertake plenty of practiceexercises so that they become second reading this text, and/or viewing the video tutorial on this topic, you should be able to: expand a sum given in Sigma notation into an explicit sum; write an explicit sum in Sigma notation where there is an obvious pattern to the individualterms; use rules to manipulate sums expressed in Sigma a long sum in Sigma for use with Sigma mathcentre 20091. IntroductionSigma notation is a concise and convenient way to represent long sums. For example, we oftenwish to sum a number of terms such as1 + 2 + 3 + 4 + 5or1 + 4 + 9 + 16 + 25 + 36where there is an obvious pattern to the numbers involved.

2 The first of these is the sum ofthe first five whole numbers, and the second is the sum of the first six square numbers. Moregenerally, if we take a sequence of numbersu1, u2, u3, .. , unthen we can write the sum of thesenumbers asu1+u2+u3+..+ shorter way of writing this is to leturrepresent the general term of the sequence and putn r= , the symbol is the Greek capital letterSigmacorresponding to our letter S , and refersto the initial letter of the word Sum . So this expression means the sum of all the termsurwherertakes the values from 1 ton. We can also writeb r=aurto mean the sum of all the termsurwherertakes the values fromatob. In such a sum,aiscalled the lower limit andbthe upper PointThe sumu1+u2+u3+..+unis written in Sigma notation asn r= mathcentre 2009 Exercises1. Write out what is meant by(a)5 n=1n3(b)5 n=13n(c)4 r=1( 1)rr2(d)4 k=1( 1)k+12k+ 1(e)N i=1x2i(f)N i=1fixi2.

3 Evaluate4 k= Some examplesExampleEvaluate4 r= is the sum of all ther3terms fromr= 1tor= 4. So we take each value ofr, work outr3in each case, and add the results. Therefore4 r=1r3= 13+ 23+ 33+ 43= 1 + 8 + 27 + 64= n= this example we have used the letternto represent the variable in the sum, rather letter can be used, and we find the answer in the same way as before:5 n=2n2= 22+ 32+ 42+ 52= 4 + 9 + 16 + 25= k= mathcentre 2009 SolutionNotice that, in this example, there are 6 terms in the sum, because we havek= 0for the firstterm:5 k=02k= 20+ 21+ 22+ 23+ 24+ 25= 1 + 2 + 4 + 8 + 16 + 32= r=112r(r+ 1).SolutionYou might recognise that each number12r(r+ 1)is atriangular number, and so this exampleasks for the sum of the first six triangular numbers. We get6 r=112r(r+ 1) =(12 1 2)+(12 2 3)+(12 3 4)+(12 4 5)+(12 5 6)+(12 6 7)= 1 + 3 + 6 + 10 + 15 + 21= would we do if we were asked to evaluaten k=12k?

4 Now we know what this expression means, because it is the sum of all the terms2kwherektakes the values from 1 ton, and so it isn k=12k= 21+ 22+ 23+ 24+..+ we cannot give a numerical answer, as we do not know the value of the upper r=1( 1) , we need to remember that( 1)2= +1,( 1)3= 1, and so on. So4 r=1( 1)r= ( 1)1+ ( 1)2+ ( 1)3+ ( 1)4= ( 1) + 1 + ( 1) + 1= mathcentre 2009 ExampleEvaluate3 k=1( 1k) again, we must remember how to deal with powers of 1:3 k=1( 1k)2=( 11)2+( 12)2+( 13)2= 1 +14+19= Writing a long sum in Sigma notationSuppose that we are given a long sum and we want to express it insigma notation . How shouldwe do this?Let us take the two sums we started with. If we want to write thesum1 + 2 + 3 + 4 + 5in Sigma notation , we notice that the general term is justkand that there are 5 terms, so wewould write1 + 2 + 3 + 4 + 5 =5 k=1k.

5 To write the second sum1 + 4 + 9 + 16 + 25 + 36in Sigma notation , we notice that the general term isk2and that there are 6 terms, so we wouldwrite1 + 4 + 9 + 16 + 25 + 36 =6 k= the sum 1 +12 13+14 ..+1100in Sigma this example, the first term 1can also be written as a fraction 11. We also notice that thesigns of the terms alternate, with a minus sign for the odd-numbered terms and a plus sign forthe even-numbered terms. So we can take care of the sign by using( 1)k, which is 1whenkis odd, and+1whenkis even. We can therefore write the sum as( 1)111+ ( 1)212+ ( 1)313+ ( 1)414+..+ ( 1) mathcentre 2009We can now see thatk-th term is( 1)k1/k, and that there are 100 terms, so we would writethe sum in Sigma notation as100 k=1( 1) PointTo write a sum in Sigma notation , try to find a formula involving a variablekwhere the firstterm can be obtained by settingk= 1, the second term byk= 2, and so Express each of the following in Sigma notation :(a)11+12+13+14+15(b) 1 + 2 3 + 4 5 + 6.

6 + 20(c)(x1 )2+ (x2 )2+ (x3 )2+ (x4 ) Rules for use with Sigma notationThere are a number of useful results that we can obtain when weuse Sigma notation . Forexample, suppose we had a sum of constant terms5 k= does this mean? If we write this out in full then we get5 k=13 = 3 + 3 + 3 + 3 + 3= 5 3= general, if we sum a constantntimes then we can writen k=1c=c+c+..+c ntimes=nc . mathcentre 2009 Suppose we have the sum of a constant timesk. What does this give us? For example,4 k=13k= (3 1) + (3 2) + (3 3) + (3 4)= 3 (1 + 2 + 3 + 4)= 3 10= we can see from this calculation that the result also equals3 (1 + 2 + 3 + 4) = 34 k=1k ,so that4 k=13k= 34 k=1k .In general, we can say thatn k=1ck= (c 1) + (c 2) +..+ (c n)=c (1 +..+n)=cn k=1k .Suppose we have the sum ofkplus a constant. What does this give us? For example,4 k=1(k+ 2) = (1 + 2) + (2 + 2) + (3 + 2) + (4 + 2)= (1 + 2 + 3 + 4) + (4 2)= 10 + 8= we can see from this calculation that the result also equals(4 2) + (1 + 2 + 3 + 4) = (4 2) +4 k=1k ,so that4 k=1(k+ 2) = (4 2) +4 k=1k.

7 In general, we can say thatn k=1(k+c) = (1 +c) + (2 +c) +..+ (n+c)= (c+c+..+c) ntimes+ (1 + 2 +..+n)=nc+n k=1k . mathcentre 2009 Notice that we have written the answer with the constantncon the left, rather than asn k=1k+nc ,to make it clear that the Sigma refers just to thekand not to the constantnc. Another way ofmaking this clear would be to write(n k=1k)+nc .In fact we can generalise this result even further. If we haveany functiong(k)ofk, then we canwriten k=1(g(k) +c) =nc+n k=1g(k)by using the same type of argument, and we can also writen k=1(ag(k) +c) =nc+an k=1g(k)whereais another constant. We can also consider the sum of two different functions, such as3 k=1(k+k2) = (1 + 12) + (2 + 22) + (3 + 32)= (1 + 2 + 3) + (12+ 22+ 32)= 6 + 14= that(1 + 2 + 3) + (12+ 22+ 32) =3 k=1k+3 k=1k2,so that3 k=1(k+k2) =3 k=1k+3 k= general, we can writen k=1(f(k) +g(k)) =n k=1f(k) +n k=1g(k),and in fact we could even extend this to the sum of several functions mathcentre 2009 Key PointIfaandcare constants, and iff(k)andg(k)are functions ofk, thenn k=1c=nc ,n k=1ck=cn k=1k ,n k=1(k+c) =nc+n k=1k ,n k=1(ag(k) +c) =nc+an k=1g(k),n k=1(f(k) +g(k)) =n k=1f(k) +n k=1g(k).

8 We shall finish by taking a particular example and using sigmanotation. Suppose that we wantto find the mean of a set of examination marks. Nowmean=total sum of marksno. of valuesSo if the marks were 2, 3, 4, 5 and 6 we would havemean=2 + 3 + 4 + 5 + 65=205 But more generally, if we have a set of marksxi, whereiruns from 1 ton, we can write themean using Sigma notation . We writemean=1nn i= By writing out the terms explicitly, show that(a)5 k=13k= 35 k=1k(b)6 i=14i2= 46 i=1i2(c)4 n=15 = 4 5 = 20(d)8 k=1c= Write out what is meant by4 k=11(2k+ 1)(2k+ 3). mathcentre 2009 Answers1.(a)5 n=1n3= 13+ 23+ 33+ 43+ 53(b)5 n=13n= 31+ 32+ 33+ 34+ 35(c)4 r=1( 1)rr2= 12+ 22 32+ 42(d)4 k=1( 1)k+12k+ 1=13 15+17 19(e)N i=1x2i=x21+x22+x23+..+x2N(f)N i=1fixi=f1x1+f2x2+f3x3+..+fNxN2. (a)11+12+13+14+15=5 k=11k(b) 1 + 2 3 + 4 5 + 6 ..+ 20 =20 k=1( 1)kk(c)(x1 )2+ (x2 )2+ (x3 )2+ (x4 )2=4 k=1(xk ) k=11(2k+ 1)(2k+ 3)=1(3)(5)+1(5)(7)+1(7)(9)+1(9)(11) mathcentre 2009