MAT01A1. Appendix E: Sigma Notation

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Transcription:

MAT01A1 Appendix E: Sigma Notation Dr Craig 5 February 2019

Introduction Who: Dr Craig What: Lecturer & course coordinator for MAT01A1 Where: C-Ring 508 acraig@uj.ac.za Web: http://andrewcraigmaths.wordpress.com

Important information Course code: MAT01A1 NOT: MAT1A1E, MAT1A3E, MATE0A1, MAEB0A1, MAA00A1, MAT00A1, MAFT0A1 Learning Guide: available on Blackboard. Please check Blackboard twice a week. Student email: check this email account twice per week or set up forwarding to an address that you check frequently.

Important information Lecture times: Tuesday 08h50 10h25 Wednesdays 17h10 18h45 Lecture venues: C-LES 102, C-LES 103 Tutorials: Tuesday afternoons (only!) 13h50 15h25: D-LES 104 or B-LES 102 OR 15h30 17h05: C-LES 203

Important information Textbook: the textbook for this module is Calculus: Early Transcendentals (International Metric Edition) James Stewart 8th edition An e-book version of the textbook is provided to every MAT01A1 student. Click on miebooks on ulink. If you want to buy a hard copy, older editions are fine.

Other announcements 2nd year IT students who have a clash with the tutorial on a Tuesday afternoon must email Dr Craig today. No MAT01A1 tuts on Wednesdays. If you see this on your timetable, it is an error. Students with Chem. prac clash: email Mr Kgatshe ckgatshe@uj.ac.za Need help? Visit the Maths Learning Centre in C-Ring 512.

Lecturers Consultation Hours Monday: 14h40 15h25 Dr Craig (C-508) Tuesday: 11h20 13h45 Dr Robinson (C-514) Wednesday: 15h30 17h05 Dr Robinson (C-514) Thursday: 11h20 12h55 Dr Craig (C-508) Friday: 11h20 12h55 Dr Craig (C-508)

Sigma notation If a m, a m+1,..., a n are real numbers and m, n Z such that m n, then n a i = a m + a m+1 + a m+2 +... + a n 1 + a n The letter i is called the index of summation. Other letters can also be used as the index of summation. The number of terms in the sum is n m + 1.

Examples: (a) 4 i 2 = 1 2 + 2 2 + 3 2 + 4 2 = 30 (b) n i = 3 + 4 + 5 +... + (n 1) + n (c) (d) i=3 3 2 j = 2 0 + 2 1 + 2 2 + 2 3 = 15 j=0 n k=1 1 k = 1 + 1 2 + 1 3 +... + 1 n

More examples: (e) (f) 3 i 1 i 2 + 3 = 1 1 1 2 + 3 + 2 1 2 2 + 3 + 3 1 3 2 + 3 = 0 + 1 7 + 1 6 = 13 42 4 2 = 2 + 2 + 2 + 2 = 8

Exercise: Write the sum 2 3 + 3 3 +... + n 3 in sigma notation. Solution(s): Sigma notation for a particular sum is not unique. Some possible solutions: 2 3 + 3 3 +... + n 3 = n i 3 i=2 2 3 + 3 3 +... + n 3 = n 1 (j + 1) 3 j=1 2 3 + 3 3 +... + n 3 = n 2 (k + 2) 3 k=0

Which of the options below are correct sigma notation for the sum: 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 (a) 8 (3 + 2i) (b) (c) (d) 8 2(1 + j) j=1 5 k= 2 [2 + (k + 3)] 7 (3 + 2m) m=0

Theorem 2: Let c be any constant. Then n (a) ca i = c n (b) (c) a i n (a i + b i ) = n a i + n (a i b i ) = n a i n b i n b i

Theorem 2: Let c be any constant. Then n (a) ca i = c n (b) (c) a i n (a i + b i ) = n a i + n (a i b i ) = n a i n b i n b i Proof: (a) follows from the distributive law: ca m + ca m+1 +... + ca n = c(a m + a m+1 +... a n ). (b) follows from the commutativity and associativity of addition. (c) combine the last two results.

Note that in general ( n n ) ( n ) (a i b i ) a i b i For example: 2 2 1 = 2 and 2 = 4 but 2 (1 2) = 4 8.

Example: Find n 1. Solution: n 1 = 1 + 1 +... + 1 }{{} n terms = n Example: Prove the formula for the sum of the first n positive integers: n n(n + 1) i = 2

Telescoping sums: n Consider j=0 ( 1 j + 1 1 j + 2 ). This is equal to ( 1 1 1 ) ( 1 + 2 2 1 ) ( 1 + 3 3 1 ) +... 4 ( ) 1... + (n 1) + 1 1 (n 1) + 2 ( 1 + n + 1 1 ) = 1 1 n + 2 n + 2

Example: Prove the formula for the sum of the squares of the first n positive integers: n i 2 = 1 2 +2 2 +...+n 2 n(n + 1)(2n + 1) = 6 Proof: Let S = n i 2. Consider the telescoping sum: n [(1 + i) 3 i 3 ] =...

Theorem 3: Let c be a constant and n a positive integer.then (a) n 1 = n (b) n c = nc (c) n (d) n (e) n i = n(n+1) 2 i 2 = n(n+1)(2n+1) 6 i 3 = [ n(n+1) ] 2 2

The formulas in the previous theorem can also be proved by Mathematical Induction. We will teach the method of Mathematical Induction in Week 5. To see a method for proving formula (e) with a telescoping sum, see Exercise 39 in the textbook.

Which of the options below are correct sigma notation for the sum: 2 + 7 + 12 + 17 + 22 (a) n (2 + 5i) (b) (c) 5 [12 + 5(j 2)] j=1 5 (2 + 5m) k=1 (d) 4 [7 + 5(l 1)] l=0

Examples: Evaluate Find lim n n i(4i 2 3) n 3 i ) 2 ] + 1 n[( n

An important example of sigma notation: Therefore 1 + x1 1! + x2 2! + x3 3! + x4 4! +... = 1 + x + x2 2 + x3 6 + x4 24 +... x n = n! = ex n=0 e = e 1 = 1 + 1 + 1 2 + 1 6 + 1 24 +... 2.718

Extra exercises (solutions on next slide): Write in expanded form: n+3 j=n Write in sigma notation: j 2 1 3 + 9 27 + 81 243 + 729

Solutions to extra exercises: Write in expanded form: n+3 j=n j 2 = n 2 + (n + 1) 2 + (n + 2) 2 + (n + 3) 2 Write in sigma notation: 1 3 + 9 27 + 81 243 + 729 6 7 ( 1) i (3 i ) OR ( 1) k+1 (3 k 1 ) OR... i=0 k=1