can you help me please even if you do not answer all of the questions
math Jeff Edmonds York University Assignment 4 MATH1090 Logic Intermediate Value Theorem f(x) r a b 0 I really wanted you to follow the parse tree and prove something cool. But this was tooooo hard. …. Aaaaaah! Q112 Q236 Q325 Q420 Q57 100 1 Epsilon Delta Proofs: Key to calculus, analysis, and set theory. Continuous: Let the function f(x) be from reals to reals. f is said to be continuous everywhere iff "x, f is continuous at x iff "x, for an arbitrary definition of closeness, as x' approaches x f(x') stays close to f(x). iff "x, " infinitesimal >0, within a sufficiently small range near x, f does not vary by more than iff "x, ">0, $>0, "x'[x-,x+], |f(x')-f(x)|≤ x x' f(x) f(x) f(x') Continuous Continuous: "x, ">0, $>0, "x´, [x´[x-,x+] |f(x')-f(x)|≤] Question 1: Parse this using the context free grammar and follow the game to prove that f(x) = x3 is continuous at x=0. First you need to express the math problem as a first order logic statement. Parse it with the following Contest Free Grammar S "x S(x) | $x S(x) | SiS | SbothS | SoneS | R(T,T) | … S "x S(x) | $x S(x) | S2S | S1bothS2 | ScasesS | R(T,T) | … Mechanically traverse each player’s parse tree following the dictated script. 4. Merge in the scripts of the oracle when stuck. …. Aaaaaah! Continuous C(u) ≠ C(v) $ colouring C # colours in C ≤ 6 " planar graph G " edge u,v # nodes in G = i S(i) ≡ Planar Colouring(i) C(u) ≠ C(v) $ colouring C # colours in C ≤ 6 " planar graph G " edge u,v # nodes in G = i-1 S(i-1) ≡ Planar Colouring(i-1) I will assure you S(i-1). I need to prove S(i)]. …. Aaaaaah! Let G be an arbitrary graph. Question 1: Parse this using the context free grammar and follow the game to prove that f(x) = x3 is continuous at x=0. Continuous Continuous: "x, ">0, $>0, "x´, [x´[x-,x+] |f(x')-f(x)|≤] Continuous & Converging Converge: Consider the infinite sequence. It is said to converge to the value a iff for an arbitrary definition of closeness, the sequence eventually gets and stays close to a iff ">0, $i, "i'i, |ai'-a|≤. i i' a a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, …. 6 Continuous & Converging Try to prove the follow: Assume that f(x) is continuous everywhere. As 1, 2, 3, 4, 5, … goes to infinity, 1/1, 1/2, 1/3, 1/4, 1/5, … converges to f(1/1), f(1/2), f(1/3), f(1/4), f(1/5), … converges to f(0) 0 i i' a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, …. a f(0) 7 Continuous & Converging i i' a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, …. Question 2: "f, [Continuous: "x, ">0, $>0, "x´, [x´[x-,x+] |f(x')-f(x)|≤]] [Converging: ">0, $i, "i´, [ i´i |f(1/i´)-f(0)|≤]] Parse this using the context free grammar and follow the game to prove it. Hint: Use the fact that f is continuous at x=0. f(0) 8 Continuous: "x, ">0, $>0, "x´, [x´[x-,x+] |f(x')-f(x)|≤] Reals: eg x = 0.2904857920495839506803…. Most reals do not have a finite description Rationales/Fractions: eg x = 27/99 = 0.27272727272… Each has a finite description. Axiom Dense: "reals x,y, $rational c(x,y) Proof: Let x and y be arbitrary reals. x = 0.2904857932495839506803…. y = 0.2904853948839579020918…. Though these have an infinite number of digits, the index i of each digit is finite. There must be a first index i at which their digits differ. Define the rational c = 0.290485794 (x,y) Continuous Round brackets means that c is not x or y. Continuous: "x, ">0, $>0, "x´, [x´[x-,x+] |f(x')-f(x)|≤] Reals: eg x = 0.2904857920495839506803…. Most reals do not have a finite description Rationales/Fractions: eg x = 27/99 = 0.27272727272… Each has a finite description. Axiom Dense: "reals x,y, $rational c(x,y) Despite this, in EECS2001, we learn that there are more reals than rationales. Crazy! I can’t wait to learn that! Continuous Continuous: "x, ">0, $>0, "x´, [x´[x-,x+] |f(x')-f(x)|≤] Dense: "reals x,y, $rational c(x,y) Rationales/Fractions: Suppose instead of the reals, our universe of objects are the rationales. Let f(x) = −1 for x<√2 1="" for="" x="">√2 True. But for every rational x, the output f(x) is a well a define rational. That’s not fair √2 is not rational. We do not have to define it, because √2 is not a rational. What is f(√2)? Question 3: Follow the game to prove that f(x) is continuous at every rational x. (By symmetry assume x>√2.) Clearly this function is not continuous! …. Aaaaaah! f(x) √2 1 0 -1 Continuous Intermediate Value Theorem: "f where f(x) is a continuous function from the reals to the reals "a,b with f(a)<0 and="" f(b)="">0, $r[a,b] for which Binary search gets you closer and closer but not there. f(r)=0. f(x) a b 0 r Intermediate Value Theorem More generally, if g(x) is a continuous function and u ∈ [g(a), g(b)], then there exists a real value r∈[a,b] for which g(r) = u. This is proved from the previous by setting f(x) = g(x) − u. a b r Intermediate Value Theorem: "f where f(x) is a continuous function from the reals to the reals "a,b with f(a)<0 and="" f(b)="">0, $r[a,b] for which f(r)=0. u g(x) Intermediate Value Theorem a b r Intermediate Value Theorem: "f where f(x) is a continuous function from the reals to the reals "a,b with f(a)<0 and="" f(b)="">0, $r[a,b] for which f(r)=0. f(x) Intermediate Value Theorem 0 Clearly this is true! It is not true over the rationales! Hence to prove it, you need to differentiate between reals and rationales. f(x) √2 1 0 -1 Supremum upper bound max supremum upper bound Dense: "reals x,y, $rational c(x,y) Supremum: ["r´
r, x´S] Question 4: Prove r=5 is the supremum of S = {x | x<5} by="" following="" this="" definition’s="" parse="" tree.="" question="" 5:="" follow="" this="" axiom’s="" parse="" tree="" to="" prove="" that="" it="" is="" not="" true="" over="" the="" rationales.="" supremum="" upper="" bound="" supremum="" 5="" s="{x" |=""><5} defining="" zero="" continuous:="" "x,="" "="">0, $>0, "x´, [x´[x-,x+] |f(x')-f(x)|≤] Supremum: "S⸦R, [S is non-empty and bounded] $r [["r´r, x´S]] Defn zero: "q, [[">0, |q|≤] [q=0]] Intermediate Value Theorem: "a,b, [f(a)<0 &="" f(b)="">0] [$r[a,b], f(r)=0] Intermediate Value Theorem Question 6: Prove this. Hint: Let S = {x≤b | f(x)<0}. let r be the supremum of s. prove f(r)=0 f(x) a b 0 r don’t do it. it is too hard. let="" r="" be="" the="" supremum="" of="" s.="" prove="" f(r)="0" f(x)="" a="" b="" 0="" r="" don’t="" do="" it.="" it="" is="" too="">