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cse547, math547 DISCRETE MATHEMATICS Professor Anita Wasilewska

LECTURE 9a

CHAPTER 2 SUMS Part 1: Introduction - Lecture 5 Part 2: Sums and Recurrences (1) - Lecture 5 Part 2: Sums and Recurrences (2) - Lecture 6 Part 3: Multiple Sums (1) - Lecture 7 Part 3: Multiple Sums (2) - Lecture 8 Part 3: Multiple Sums (3) General Methods - Lecture 8a Part 4: Finite and Infinite Calculus (1) - Lecture 9a Part 4: Finite and Infinite Calculus (2) - Lecture 9b

CHAPTER 2 SUMS Part 4: Finite and Infinite Calculus (1) - Lecture 9a

Finite and Infinite Calculus Infinite Calculus review We define a derivative OPERATOR D as Df (x) = lim h 0 f (x + h) f (x) h Derivative operator D is defined for some functions on real numbers R, called differentiable functions. We denote Df(x) = f (x) and call the result a derivative f of a differentiable function f

Derivative Operator D is called an operator because it is a function that transforms some functions into different functions D is a PARTIAL function on the set R R over R, i.e. D : R R R R where R R = {f : R R} of all functions D is a partial function because the domain of D consists of the differentiable functions only, i.e. functions f for f (x + h) f (x) which lim exists h 0 h

FINITE CALCULUS Difference operator Let f be any function on real numbers R (may be partial) f : R R We define f (x) = f (x + 1) f (x) transforms ANY function f into another function g(x) = f (x + 1) f (x), so we have that : R R R R

Difference Operator Example Let f : R R be given by a formula f (x) = x m We evaluate: Df (x) = mx m 1 Reminder: D(x m ) = mx m 1 What about??? Evaluate (x 3 ) = (x + 1) 3 x 3 = 3x 2 + 3x + 1, Dx 3 = 3x 2 D

Difference Operator Question Q: Is there a function f for which f = Df Yes there is a new power of x, which transforms as nicely under, as x m does under D

Falling Factorial Power Definition of Falling Factorial Power Let f : R R be given by a formula f (x) = x m for x m = x(x 1)(x 2) (x m + 1) and m > 0 Wa also define in a similar way a notion of a rising factorial power

Rising Factorial Power Definition of Rising Factorial Power Let f : R R be given by a formula f (x) = x m for x m = x(x + 1) (x + m 1) and m > 0 Let now see what happens when the domain of f is restricted to the set of natural numbers N

Factorial Powers Let now f : N N, f (x) = x m We evaluate n n = n(n 1)(n 2) (n n + 1) = n! We evaluate 1 n = 1 2 (1 + n 1) = n! We got n n = n! and 1 n = n!

Factorial Powers We define case m = 0 x 0 = x 0 = 1 PRODUCT OF NO FACTORS 0! = 1 x R We have already proved: 1 0 = 0! = 1 0 0 = 0! = 1 n! = n n = 1 n for any n 0

Factorial Powers Let s now evaluate (x m ) = (x + 1) m x m in order to PROVE the formula: (x m ) = mx m 1 It means that on x m behaves like D on x m : D(x m ) = mx m 1

Factorial Powers Evaluate (x + 1) m = (x + 1)x(x 1) (x + 1 m + 1) = (x + 1)x(x 1) (x m + 2) Evaluate x m = x(x 1) (x m + 2)(x m + 1)

Factorial Powers Evaluate (x m ) = (x + 1) m x m = (x + 1)x(x 1) (x m + 2) x(x 1) (x m + 2)(x m + 1) = x(x 1) (x m + 2)((x + 1) (x m + 1)) = x(x 1) (x m + 2) m = mx m 1 We proved: (x m ) = mx m 1 Hwk Problem 7 is about x m

Infinite Calculus: Integration Reminder: differentiation operator D is D : R R R R Df (x) = g(x) = f (x) D is partial function Domain D = all differentiable functions D is not 1-1; D(c) = 0 all c R So inverse function to D does not exist BUT we define a reverse process to DIFFERENTIATION that is called INTEGRATION (1) We define a notion of a primitive function (2) We use it to give a general definition of indefinite integral

Infinite Calculus: Integration Definition A function F(x) = F such that DF = DF (x) = F (x) = f (x) is called a primitive function of f (x), or simply a primitive of f Shortly, F is a primitive of f iff DF = f F is a primitive of f differentiation iff f is obtained from F by The process of finding primitive of f is called integration

Fundamental Theorem Problem: given function f, find all primitive function of f (if exist) Fundamental Theorem of differential and integral calculus The difference of two primitives F 1 (x), F 2 (x) function f (x) is a constant C, i.e. of the same F 1 (x) F 2 (x) = C for any F 1,F 2 such that DF 1 (x) = f (x), DF 2 (x) = f (x)

Fundamental Theorem The Fundamental Theorem says that (1) From any primitive function F(x) we obtain all the other in the form F(x) + C (suitable C) (2) For every value of the constant C, the function F 1 (x) = F(x) + C represents a PRIMITIVE of f

Indefinite Integral Definition of Indefinite Integral as a general form of a primitive function of f f (x)dx = F(x) + C C R where F(x) is any primitive of f i.e. DF(x) = f (x) F = f ( for short)

Proof of Fundamental Theorem Proof of the FUNDAMENTAL THEOREM of differential and Integral calculus has two parts (2) We prove: if F(x) is primitive to f (x), so is F(x) + C; i.e. D(F(x) + C) = f (x), where DF(x) = f (x) (1) We prove: F 1 (x) F 2 (x) = C i.e. from any primitive F(x) we obtain all others in the form F(x) + C We first prove (2)

Proof of Fundamental Theorem Proof of Fundamental Theorem part (2) (2) Let G(x) = F(x) + C (F(x + h) + C) (F(x) + C) D(F(x) + C) = lim h 0 h F(x + h) F(x) = lim = F (x) = f (x) h 0 h as F(x) is a primitive of f (x)

Proof of Fundamental Theorem (1) Consider F 1 (x) F 2 (x) = G(x) such that F 1 = f, F 2 = f We want to show that G(x) = C for all x R We use the definition of the derivative to evaluate G (x) = DG(x)

Proof of Fundamental Theorem D(G(x)) = (F 1 (x + h) F 2 (x + h)) (F 1 (x) F 2 (x)) lim h 0 h ( ) = F 1 (x + h) F 1 (x) lim F 2(x + h) F 2 (x) h 0 } h {{ h } Both limits exist, as F 1,F 2,primitive of f. = F 1 (x + h) F 1 (x) F 2 (x + h) F 2 (x) lim lim h 0 h h 0 h = f (x) f (x) = 0 for all x R

Proof of Fundamental Theorem We proved that F 1 (x) F 2 (x) = G(x) and G (x) = 0 for all x R But the function whose derivative is everywhere zero must have a graph whose tangent is everywhere parallel to x-asis; i.e. must be constant; and therefore we have G(x) = C for all x R This is an intuitive, nor a formal proof. The formal proof uses the Mean Value Theoremm

Proof of Fundamental Theorem Formal proof Apply the MEAN VALUE THEOREM to G(x), i.e. G(x 2 ) G(x 1 ) = (x 2 x 1 )G (ξ ) x 1 < ξ < x 2 but G (x) = 0 for all x, hence G (ξ ) = 0 and G(x 2 ) G(x 1 ) = 0, for any x 1,x 2 i.e. G(x 2 ) = G(x 1 ) for all x 1,x 2 i.e. G(x) = C This ((1)+(2)) justifies the following Definition: INDIFINITE INTEGRAL f (x) = F(x) + C, where DF(x) = F (x) = f (x)

FINITE CALCULUS

Finite Calculus Reminder: Difference Operator : R R R R For any f R R we define: f (x) = f (x + 1) f (x) is a total function on R R Remark : INVERSE to does not exist! because is not 1-1 function

Finite Calculus Example ; is not 1-1 function Take f 1 (x) = c 1, f 2 (x) = c 2 for c 1 c 2 We have that f 1 (x) f 2 (x) for all x, i.e. f 1 f 2 We evaluate f 1 (x) = f 1 (x + 1) f 1 (x) = c 1 c 1 = 0 f 2 (x) = f 2 (x + 1) f 2 (x) = c 2 c 2 = 0 f 1 = f 2 for f 1 f 2 We proved that is not 1-1 function

Finite Integration Question: Do we have a REVERSE operation to similar to the one we had for D? Answer YES! We proceed as the case of Infinite Calculus Definition A function F = F(x) is a finite primitive of f = f (x) iff F(x) = f (x) for all x R We write F = f The process of finding a finite primitive (FP) of a function f = f (x) is called a finite integration

Fundamental Theorem Problem: Given a function f = f(x), find all finite primitives of f= f(x) Fundamental Theorem of Finite Calculus The difference of two finite primitives F 1 (x), F 2 (x) of the same function f (x) is a function C(x), such that C(x + 1) = C(x) for all x R, i.e. F 1 (x) F 2 (x) = C(x) and C(x + 1) = C(x) for all x R

Fundamental Theorem The Fundamental Theorem says that (1) Given a finite primitive function F(x) of f (x) we obtain all others in the form F(x) + C(x), where the function C : R R fulfills a condition C(x + 1) = C(x) for all x R (2) For every function C(x), such that C(x + 1) = C(x) for all x R the function F 1 (x) = F(x) + C(x) is a finite primitive of f (x)

Proof of Fundamental Theorem Proof (1) Consider F 1 (x) F 2 (x) = C(x) such that F 1 = f, F 2 = f We want to show that C(x + 1) = C(x), i.e. Evaluate C(x) = C(x + 1) C(x) = 0 C(x) = (F 1 (x) F 2 (x)) = (F 1 (x + 1) F 2 (x + 1)) (F 1 (x) F 2 (x)) = (F 1 (x + 1) F 1 (x)) (F } {{ } 2 (x + 1) F 2 (x)) } {{ } F 1 F 2 = f (x) f (x) = 0

Proof of Fundamental Theorem (2) Let F 1 (x) = F(x) + C(x) and F(x) = f (x), C(x + 1) = C(x) We prove that F 1 (x) is a finite primitive of f F 1 (x) = (F(x + 1) + C(x + 1)) (F(x) + C(x)) ( F 1 = f ) = F(x + 1) F(x) + 0 = F(x) = f (x) yes!

Indefinite Sum Definition Definition of INDEFINITE SUM as a general form of a finite primitive of f = f (x) g(x)δ(x) = f (x) + C(x) g(x) = f (x) if and only if and C(x + 1) = C(x) for g : R R; f : R R, C : R R Remark : in paticular case: we can put C(x) = C for all x R as in the case of Indefinite Integral because C(x + 1) = C = C(x)

Example EXAMPLE of a CONSTANT function C = P(x) under Evaluate P(x) = sin 2πx (PERIODIC function) P(x + 1) = sin(2π(x + 1)) = sin(2πx + 2π) = P(x) We proved P(x) = P(x + 1) for all x R

Definite Integral and Definite Sum Infinite Calculus: DEFINITE INTEGRAL b a g(x)dx = f (x) b a = f (b) f (a) where f (x) = g(x) DEFINITION Finite Calculus: DEFINITE SUM where b a f (x) = g(x) g(x)δ x = f (x) b a = f (b) f (a)

Definite Sum We defined b a g(x)δ x = f (x) b = f (b) f (a) a for f (x) such that g(x) = f (x) g = f What is the MEANING of our new INTEGRAL b a g(x)δ x?

Definite Sum Reminder: b a g(x)δ x = f (x) b = f (b) f (a) for a g(x) = f (x) = f (x + 1) f (x) Let s consider a case: b = a TAKE now b = a + 1 a a g(x)δ x = f (a) f (a) = 0 a+1 a g(x)δ x = f (a + 1) f (a) = f (a) = g(a)

Definite Sum We proved that a a g(x)δ x = 0 Evaluate where a+1 a g(x)δ x = g(a) a+2 def a g(x)δ x = f (a + 2) f (a) g(x) = f (x + 1) f (x) = f (x)

Definite Sum Consider a+2 a g(x)δ x a+1 a g(x)δ x = f (a + 2) f (a) (f (a + 1) f (a)) (by definition) = f (a + 2) f (a) f (a + 1) + f (a) = f (a + 2) f (a + 1) = g(a + 1) a+2 a g(x)δ x = a+1 a g(x)δ x + g(a + 1) = g(a) + g(a + 1)

Definite Sum We proved a+1 a g(x)δ x = g(a) a+2 a g(x)δ x = g(a) + g(a + 1) Evaluate a+3 def a g(x)δ x = f (a + 3) f (a)

Definite Sum Compute a+3 a g(x)δ x a+2 a g(x)δ x = f (a + 3) f (a) (f (a + 2) f (a)) (by definition) = f (a + 3) f (a + 2) = g(a + 2) a a+3 g(x)δ x = a+2 a g(x)δ x + g(a + 2) = g(a) + g(a + 1) + g(a + 2)

Definite Sum GUESS (proof by math. induction over k) b a a+k g(x)δ x = g(a) + g(a + 1) + + g(a + k 1) a where a+k=b, and a+k-1=b-1

Definite Sum For b a b g(x)δ x a } {{ } DEFINITE SUM = g(k) a k<b } {{ } = NORMAL SUM b 1 k=a g(k) } {{ } NORMAL SUM b a g(x)δ x = f (b) f (a), where f (x) = g(x)

Definite Sum and Normal Sums Relationship between DEFINITE and NORMAL sums We defined g(x) = f (x) WE PROVED: b a g(x)δ x def = f (x) b a = f (b) f (a) b a g(x)δ x = b 1 k=a g(k) For b a

Definite and Normal SumsTheorem THEOREM For b a We write it as Reminder: b 1 k=a g(k) = b a g(x)δ x = f (x) b = f (b) f (a) a b 1 k=a g(k) } {{ } SUM Thm = b a g(x)δ x g(x) = f (x) = f (x + 1) f (x)

Definite and Normal SumsTheorem When asked of evaluating a SUM, we can evaluate a SUM INTEGRAL b g(x)δ x = f (x) b = f (b) f (a) a a } {{ } INTEGRAL where f (x) = g(x) Very easy if you know how to integrate b a g(x)δ x

Example x m dx = x m+1 m+1 INFINITE because where x m d x = x m+1 m+1 FINITE ( ) x m+1 m+1 = 1 m+1 (x m+1 ) = 1 m+1 (m + 1)x m = x m x m = x(x 1)(x 2) (x m + 1)

Example Evaluate 0 k<n k m = 0 m + 1 m + 2 m + + (n 1) m BAD n 1 k m k=0 } {{ } SUM n Thm = k m δ k 0 } {{ } INTEGRAL We know that k=0 k m δ k = k m+1 n 1 n m+1 0

Example We evaluate n k=0 k m Thm = (n+1) 0 k m δ k } {{ } Integral = n+1 0 k m δ k = k m+1 m+1 = (n + 1) m+1 0m+1 1 = (n + 1) m+1 n+1 0 Answer n k m = (n + 1) m+1 k=0

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