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Lecture 1 Overview (MA2730,2812,2815) Lecture 1 Overview (MA2730,2812,2815) lecture 1 Overview (MA2730,2812,2815) lecture 1 Contents of the teaching and assessment blocks MA2730: Analysis I Lecture slides for MA2730 Analysis I Analysis — taming infinity Maclaurin and Taylor series. Simon Shaw Sequences. people.brunel.ac.uk/~icsrsss Improper Integrals. [email protected] Series. College of Engineering, Design and Physical Sciences Convergence. bicom & Materials and Manufacturing Research Institute Brunel University LATEX 2ε assignment in December. Question(s) in January class test. Question(s) in end of year exam. September 21, 2015 Web Page: http://people.brunel.ac.uk/~icsrsss/teaching/ma2730 Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel MA2730, Analysis I, 2015-16 MA2730, Analysis I, 2015-16 Overview (MA2730,2812,2815) lecture 1 Overview (MA2730,2812,2815) lecture 1 Lecture 1 Lecture 1 MA2730: topics for Lecture 1 Functions — Level 1 revision What is a function? How do we define a function? What ingredients do we need? A function is a triple: Lecture 1 A rule, for example f(x)= x2 + sin(√x). A domain, for example. Revision of functions and factorials (from Level 1) D = [0, ) R. ∞ ⊆ Maclaurin polynomials — some intuition A range, for example. Taylor polynomials R = [ 1, ) R − ∞ ⊆ Definitions and formality We then write f : D R with rule f(x)= x2 + sin(√x). → Examples and exercises Often the range is replaced with the co-domain. The co-domain is The introductory material covered in this lecture can be found in usually R. Usually the domain is taken to be the largest subset of The Handbook in Chapter 1 and Section 1.1. See, for example, R that is possible. equation (1.2) with a = 0. So we usually work with just the rule, but bear in mind that this is just a lazy convenience! Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel MA2730, Analysis I, 2015-16 MA2730, Analysis I, 2015-16 Overview (MA2730,2812,2815) lecture 1 Overview (MA2730,2812,2815) lecture 1 Lecture 1 Lecture 1 What is a factorial? Motivation: plot of sin(x) and a polynomial We’re going to use these a lot. x3 x5 x7 3! = 3 2 1 = 6 x + × × − 3! 5! − 7! 5! = 5 4 3 2 1 = 120 1.5 × × × × sin(x) 1! = 1 polynomial 0! = 1 — just accept it! 1 In general, 0.5 n(n 1)(n 2) (3)(2)(1) if n > 1 0 if n N 0 then n!= − − · · · ∈ ∪ { } 1 if n = 0. −0.5 We will also need the mean value theorem and Rolle’s theorem, −1 but these can wait. −1.5 Although it would be commendably pro-active of you to revise −10 −5 0 5 10 them sooner rather than later Shaw , bicom, mathematics, CEDPS, IMM, CI, Brunel Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel MA2730, Analysis I, 2015-16 MA2730, Analysis I, 2015-16 Overview (MA2730,2812,2815) lecture 1 Overview (MA2730,2812,2815) lecture 1 Lecture 1 Lecture 1 Motivation: plot of sin(x) and a special polynomial Most functions are (nearly) polynomials x3 x5 x7 x9 x11 x13 x15 x17 x19 x21 x + + + + + − 3! 5! − 7! 9! − 11! 13! − 15! 17! − 19! 21! So what? 1.5 sin(x) Question: How does your calculator compute sin(1)? polynomial 1 All computers can do is add and subtract... ...in binary 0.5 Answer: they use approximations — like polynomials 0 Maclaurin and Taylor Series −0.5 In this first part of Analysis I we are going to study Maclaurin and −1 Taylor series. −1.5 −10 −5 0 5 10 This is very important material; the results will pervade much of your subsequent mathematical studies. Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel MA2730, Analysis I, 2015-16 MA2730, Analysis I, 2015-16 Overview (MA2730,2812,2815) lecture 1 Overview (MA2730,2812,2815) lecture 1 Lecture 1 Lecture 1 Who are they? Very Important Results. Colin Maclaurin (1698 — 1746) Brook Taylor (1685 — 1731) The Maclaurin series expansion for sin(x) x3 x5 x7 ∞ ( 1)nx2n+1 sin(x)= x + + = − − 3! 5! − 7! · · · (2n + 1)! n=0 X Similarly, for cos(x) and exp(x): x2 x4 x6 ∞ ( 1)nx2n cos(x) = 1 + + = − , − 2! 4! − 6! · · · (2n)! n=0 We just met the Maclaurin series expansion for sin(x): X x2 x3 x4 ∞ xn ex = 1+ x + + + + = . 3 5 7 n 2n+1 2! 3! 4! · · · n! x x x ∞ ( 1) x n=0 sin(x)= x + + = − X − 3! 5! − 7! · · · (2n + 1)! n=0 X There are of profound importance — remember them. forever! Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel MA2730, Analysis I, 2015-16 MA2730, Analysis I, 2015-16 Overview (MA2730,2812,2815) lecture 1 Overview (MA2730,2812,2815) lecture 1 Lecture 1 Lecture 1 Let’s develop some intuition. Digging deeper. The Maclaurin series expansion for sin(x) Calculating the Maclaurin series expansion for sin(x) x3 x5 x7 ∞ ( 1)nx2n+1 We just found a = 0, so we now need a , a , a . such that, sin(x)= x + + = − 0 1 2 3 − 3! 5! − 7! · · · (2n + 1)! n=0 2 3 4 X sin(x)= a1x + a2x + a3x + a4x + · · · How can we calculate this? How could you calculate a1? HINT: differentiate. Well, start by assuming that coefficients a0, a1, a2,... exist such that, d 2 3 4 Apply to both sides: sin(x)= a0 + a1x + a2x + a3x + a4x + dx · · · 2 3 How could you calculate a0? cos(x)= a + 2a x + 3a x + 4a x + 1 2 3 4 · · · Now, how can we find a1? Take x = 0 again. We get a1 = 1. How about taking x = 0? We get a0 = 0. Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel MA2730, Analysis I, 2015-16 MA2730, Analysis I, 2015-16 Overview (MA2730,2812,2815) lecture 1 Overview (MA2730,2812,2815) lecture 1 Lecture 1 Lecture 1 Stop and look back Countdown Calculating the Maclaurin series expansion for sin(x) Derive the Maclaurin series expansion for sin(x) We started by assuming, x3 x5 x7 ∞ ( 1)nx2n+1 sin(x)= x + + = − 2 3 4 − 3! 5! − 7! · · · (2n + 1)! sin(x) = a + a x + a x + a x + a x + n=0 0 1 2 3 4 · · · X d 2 gave: cos(x) = a + 2a x + 3a x2 + 4a x3 + Start with: sin(x)= a0 + a1x + a2x + dx 1 2 3 4 · · · · · · Taking x = 0 gave a0 = 0. Differentiating and taking x = 0 gave a1 = 1. How could you find a2? Differentiate and take x = 0. How could you find a3? Differentiate and take x = 0. Off you go then. 5! seconds. Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel MA2730, Analysis I, 2015-16 MA2730, Analysis I, 2015-16 Overview (MA2730,2812,2815) lecture 1 Overview (MA2730,2812,2815) lecture 1 Lecture 1 Lecture 1 Here it is. Child’s play... The Maclaurin expansion of sin(x) Start with the assumption that, 2 3 4 sin(x)= a0 + a1x + a2x + a3x + a4x + · · · We have just discovered how to derive Take x = 0 to get a0 = 0 and differentiate, x3 x5 x7 ∞ ( 1)nx2n+1 2 3 sin(x)= x + + = − cos(x)= a1 + 2a2x + 3a3x + 4a4x + − 3! 5! − 7! · · · (2n + 1)! · · · nX=0 Take x = 0 to get a1 = 1 and differentiate, This is the Maclaurin expansion of sin(x). sin(x) = 2a + 3 2a x + 4 3a x2 + − 2 × 3 × 4 · · · The same ‘x = 0 and differentiate’ technique can be applied for Take x = 0 to get a2 = 0 and differentiate, ‘any’ function on the left hand side... cos(x) = 3 2a + 4 3 2a x + − × 3 × × 4 · · · Take x = 0 to get a = 1/3! ..., 3 − Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel MA2730, Analysis I, 2015-16 MA2730, Analysis I, 2015-16 Overview (MA2730,2812,2815) lecture 1 Overview (MA2730,2812,2815) lecture 1 Lecture 1 Lecture 1 Exercise Solution: boardwork. Examples x2 x3 x4 ∞ xn Show that ex =1+ x + + + + = 2! 3! 4! · · · n! n=0 X Start by assuming ex = a + a x + a x2 + a x3 + and then use 0 1 2 3 · · · the ‘x = 0 and differentiate’ technique. You have cosec4(π/4) minutes. Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel MA2730, Analysis I, 2015-16 MA2730, Analysis I, 2015-16 Overview (MA2730,2812,2815) lecture 1 Overview (MA2730,2812,2815) lecture 1 Lecture 1 Lecture 1 Polynomial approximation An application Remember that earlier we saw that chopping off 3 5 So sin(x) x x + x near to x = 0. A computer/calculator ≈ − 3! 5! x3 x5 x7 ∞ ( 1)nx2n+1 could use this in an algorithm. sin(x)= x + + = − − 3! 5! − 7! · · · (2n + 1)! Computers cannot look up the value of sin(x). n=0 X Computers can only add and subtract in binary. Multiplication is at the x5 term gave a good approximation to sin(x) near to x = 0. like repeated addition and division is like repeated subtraction. 1.5 3 5 sin(x) x x polynomial And that is how they can do arithmetic: + . x + looks like: 1 − ×÷ − 3! 5! 0.5 They cannot do anything more complicated than this. x3 x5 Key Point: if x is close to zero 0 But they can calculate x 3! + 5! . 2 3 4 − then x , x , x ,... all become −0.5 And that gives sin(x) accurately near to x = 0, for example.
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