Riassunto Libro “Essential Mathematics for Economics”, Sintesi del corso di Matematica Generale

Scarica Riassunto Libro “Essential Mathematics for Economics” e più Sintesi del corso in PDF di Matematica Generale solo su Docsity! AII mathematical “arguments” consists of “statements” or “propositions”. These are declarative sentences that can be either true and false and not both. This is called Principle of Excluded Middle or Tertium non datur. The propositions can be grouped in a collection P, denoted by small letters. A proposition is a statement which takes one of the two logical values, true (T) or false (F). The logical value of any proposition must be known at least in an ideal state of knowledge. Trough the table of the truth is possible to specify the logic value of a proposition in terms of its components. NEGATION The negation —p (“not p”) of a proposition p is true if p is false, and false if p is true. UNION The union p v q (“p or q”) of two propositions p and q is true if at least one ofthe propositions is true, and false otherwise. INTERSECTION | The intersection p A q (“p and q”) of two propositions p and q is false if at least one ofthe propositions is false, and true otherwise IMPLICATION Given two propositions p and q, the implication p > q (“p implies q” or “if p then q”) is the proposition which is false if p is true and q is false, in the other cases the implication is true. In the proposition p = q, p is referred to as the hypothesis and q as the thesis. BIIMPLICATION | Given two propositions p and q, the biimplication p @ q (“p biimplies q” or “p is equivalent to q”, or “p if and only if q”) is true if p and q have the same logical value, and false otherwise. TAUTOLOGIES A tautology is a composition of some prepositions which is true regardless the logical value of the propositions. Therefore, a tautology is always true. There are many examples of tautologies: Modus ponens s=(pA(p>q)=>q Reduction to the absurd s=(p>g)© (-q>-p) De Morgan's laws St=(pva) e (pA-g) Se := (pAg) (pv -g) A set is a collection of elements (or members). Generally, we denote a set by capital letters and the elements of the set by small letters. To express the relation between the element of a set and its corresponding set, we use the symbol “€”, which means “belongs to”. aeA It means that a is an element of A ag It means that a is not an element of A A = {a,b,c, ..} | A setis denoted by brackets Da Singleton of x, which only consists of the element x d Empty set, which has no elements “such that” {x:P(x)} The set of all elements x which satisfy the condition P(x). So, P(x) is a proposition depending on x. INCLUSION RELATION AcB If for two sets A and B, it happens that all the elements of A are also elements of B, then we can say that “A is a subset of B”. But, if itis not true that A c B, we write A £ B, which means that there exists an element of A that does not belong to B. EQUALITY RELATION A=B Two sets A and B are equal if they contain the same elements. So, they are equal if A © Band —BCA. But if for two sets A and B we have that A c B and B # A, then we say that “A is a proper subset of B”. This means that if every element of A is an element of B, then there is some element of B not belonging to A. & This is the symbol that denotes the proper subset. UNION AND INTERSECTION Given two sets A and B, we define the union AUB = { x : (x € A) or (x € B)}. While the intersection A N B is defined asA N B={x: (x € A) and (x € B)}. Disjoint sets: Two set A and B are said to be disjoint if A N B = @, more explicitly when the intersection is empty. PROPERTIES OF UNION AND INTERSECTION Idempotency AUA = A; ANA = A; Commutativity AUB = BUA; ANB = BNA Associativity (AUB)UC = AU(BUC); — (ANB)NC = AN(BINC Distributivity of N w.r.t. U | (AUB)NC = (ANC) U (BNC) Distributivity of U w.r.t.M | (ANB)UC = (AUC) N (BUC) DIFFERENCE AND COMPLEMENT If A and B are any two sets then we define the difference “A minus B" A\B = {x€ A :x € B}. In addition to this, if allthe sets referred to are subsets of a unique set X, called the universe or the master set, then we define the complement of a set A (with respect to X) as the set CA=X\A={xEX:x € A}. v a e As regards the complements, for any two sets A and B the following equalities hold: 1. C(AUB)= CANCB — Therefore the complement of the union is the intersection of the complements. 2. C(ANB)= CAUCB Therefore the complement of the intersection is the union of the complemenis. CARTESIAN PRODUCT The Cartesian product of two sets A and B, denoted A x B, is the set of all possible ordered pairs where the elements of A are first and the elements of B are second. Therefore, if A consists of m elements and B consists of n elements, then the Cartesian product A x B consists of mx nelements. So, in total we have m x n ordered pairs (a,b). AxB={(a, b):a € A andb e B} For example, Consider two non-empty sets A = {a:, az, a3} and B = {b1, be, bg} Cartesian product AxB = {(a1,b1), (a1,b2), (a1,bs), ( a2,b1), (a2,b2).(a2.b3), (as3.b1), (as.b2), (as,b3)}. -A*B QUANTIFIERS a The existential quantifier. It means “exists” w The universal quantifier. It means “for all” or “for every” al It means “there exists and it is unique” INDEX FAMILIES OF SETS The set of all the natural numbers N= {0, 1,2,...} The set of all positive integers N+*=N\{0} ={1,2,...} The set of all even numbers A= {0, 2, 4,...} A=fneN:3k€Nsuchthatn=2k}={2k:k e N} The set of all odd numbers B={1,3,5,...} B={en+1:neN} The elements of a sets may themselves be a set. The phrase “a set of sets” could sound confusing, so we often use the terms collection and family when we wish to emphasize that the elements of a given set are themselves sets. Suppose / is family of a set, more precisely the family of all index, and with each a € /we associate a set Aa Therefore, {Aaaa is defined as an indexed family of sets. It is possible to define the union and intersection of a family of sets, in particular of {Aa}ael, using the universal and existential quantifiers. - Union of a family of sets: Ung Aa = {x : 3a € I such that x E Aa} - Intersection of a family of sets: Naga Aa = {x : x E Aa Va € I} PARTITION OF A SET A family {Aa}aa of subsets of a set A is a partition of A if: e Aadoesnotcontain the empty set Aa# 0 e Theunion of the subsets must equal to the entire original set A Ucel Aa= A e The intersection of any two distinct sets is empty AaNAg=@ A={1,2,3} The partitions of A are: 1. {{1,2,3}:= {A} I. {{1}, {2}. {3} II.{{1}, {2,3} IV.{{2}. {1,3} V. {{3}, {1,9} POWER SET If A is a set then the power set P(A) of A is the set whose elements are all the subsets of A. More explicitly, P(A) is a set which consists of 2” elements ot A, ex. if we consider the set A = {a,b,c} then we get P(A) = {9, A, {a}, {b}, {c}. {a,b}, {a,c}, {b,c}} The elements of the set A are 3, so the P(A) consists of 2° = 8 elements. a#gbe(a<b)or(b #a) Thestrict order relation associated to any order relation on a set A is irreflexive and transitive. : The maximum and supremum are subset of an ordered set. Let consider an order relation on a set A and let A' be a subset of A. then an element maximum(A’) (or minimum(A”?)) of A is said to be the maximum (or minimum) of A' if it satisfies the following conditions: 1. Max(A”) € A' (or min(A") € A’) 2. a <max(A”)(or min(A’) <a ) for every ac A” It is important to highlight that the minimum (or the maximum) of any ordered set must belong to the set. Therefore, the maximum of a set is an element of that set, which goes after any other element of the set. The same is for the minimum. : Let consider an order relation in a set A and let A' be a subset of A. Then an elementk E A (or h € A) is said to be an upper (or lower) bound of A’ ifa < k (or h < a) for every a € A. If there exists some upper (lower) bound of the subset A’ of A, then A’ is said to be upper (or lower) bounded. If A’ is both upper and lower bounded then it is said to be bounded. The lower (or upper) bound do not necessarily belong to A’. Therefore, if an upper (or lower) bound of a set belongs to that set, then it also is the maximum (or the minimum) of that set. Supremum and Infimum: Let A' C A be an upper bounded set. If the set K of all the upper bounds of A’ has minimum, then it is said to be the supremum of A. if A’ is lower bounded and the set H of all the lower bounds has a maximum, then it is said to be the infimum of A’. ifthe maximum or the minimum exists, then it is also the supremum or the infimum of that set. However, they do not necessarily belong to that set. Let A be a set of n elements. A totally ordered subset of A of k elements is referred to a disposition without repetitions of k elements out of n elements. In order to know how many totally ordered subset of A of k elements can be considered, we use this formula: Dak=n © (n-1) - (n-2) ... - (n-k+1) the number of the operations depends on the value of k. We have n opportunities to choose the 15° element, then, once we have chosen it, we have (n-1) opportunities to choose opportunities to choose the second element and so on, until we reach the last element (which is also called the k-?" element) that can be chosen in n-k+1 different ways. Permutations: If k = n, then the disposition without repetitions would be equal to the P, (the number of permutations of n elements). lt is computed in the following way: Pn =1:2-3-...-n The number 1 - 2- 3- ...- nis also said n factorial and it is denoted by n!. The factorial number just means to multiply a series of descending natural numbers. it is generally agreed that 01 =1. The recursive formula : (n + 1)! = (n + 1) - n! Let A be a set of n elements. Since in this case (with respect to the disposition without repetitions), we do not have any order subset imposed, if we want to know how many subset of A of elements can be considered we use this formula: Therefore, a set of k elements can be permuted in K! different ways, therefore Cnxis obtained by dividing the formula of the disposition without repetitions by k!. The second expression is obtained by dividing the ratio by (n —k)!. (£) — “‘ over k" is said to be a binomial coefficient. It holds true 2 properties: 1 Ma (2) 0sksn 2. (= + (I) sksn-1 According to the second property, the binomial coefficient can be represented by means of the Tartaglia's Triangle, which consists of a series of numbers in the triangle, in which every number is the sum of the numbers laying above it). It is possible to compute it thanks to the Newton’s binomial formula, which states that: (a+ b)" = Lito(m) at bE According to this formula, the power (a + b)" is the product of n factors equal to (a + b). This product can be expressed as the sum of all the products obtained by extracting from each of the parenthesis (a + b) either a or b. Therefore, the product a”-* b* appears exactly CD) times. Given a set A of n element, and we want to determine the number of k-tuples (a1,a2... ax) that can be formed by suing the elements of A that might also be repeated, we use the formula An = n. According to it, the first element can be chosen in n way, then we have n ways of choosing the second element and so on. N is the set of all the natural numbers. There exists some relevant properties about the natural numbers: 1. There exists a first element 0 (minimum) 2. Every natural number has an immediate successor (n+1) 3. The set of the natural numbers is unbounded form above 4. Every nonempty subset of N has a minimum ( this is because N is a well ordered set) 5 : Let S be a subset of N. Then S is equal to N if two properties are satisfied: 1. 0€ES 2. IfineSthenn+1eSforallneS The principle of induction is extremely important because it represents a way of providing that the preposition P(n) depending on n (which belongs to the set of all the natural numbers) is true for every n € N. Principle of induction : Let P(n) be a proposition depending on n. Then P(n) is true for every n 2 ©î (fi is a fixed natural number) if : a) P(î)istrue b) P(n)=P(n+1)forever n27i(nEN) If a proposition is true for the number n = 1 and if it can be shown that if the proposition is true for n, it will be also true for n + 1. Then the preposition is true for all the real numbers. The set of all the integers is denoted by a Z, which is composed by both negative and positive numbers and the zero. It is necessary an extension from natural numbers to the integers because it is not true that the equation a + x = b(a,b € N) has a solution in x € N. Therefore, we need to consider the set of the integers. In this case the equation a + x = b(a, b € Z) has a unique solution x=2-a€ Z. In addition, it is necessary to act an additional extension from the integers to the rational numbers. The set of all the rational numbers is denoted by an Q, which is defined as Q = f ih eZ,kEZ \t0}} We consider this set because it is not true that a'x=b(a,b € Z,a # 0) has a solution x € Z. Therefore, we need to consider the set of the rational numbers. In this case the equation a+ x = b(a, b € Q,a # 0) has a unique solution x = 2 EQ. However, the rational number line is not continuous because it presents some holes. For this reason, in order to guarantee that the number line is continuous, we need to introduce the set of the Real Numbers, denoted by an R. > Closed downwards set: If (X,<) is any ordered set, then a subset A of X is said to be closed downwards if, for every x,y E X (x <y)and (ye A)>x EA > Closed upwards set: If (X,<) is any ordered set, then a subset A of X is said to be closed upwards if, for every x,y E X (x < y) and (x € A)>y € A Real Numbers: A real number is defined as a subset A of the set Q of the rational numbers, which satisfies these properties: - As non empty - A is properly contained inQ - A is closed downwards - Ahas no maximum There are 2 types of real numbers: 1. Rational numbers > if r= min (Q\A) exists; 2. Irrational numbers + if min (Q\A) does not exist. Order relation: The order relation < and its associated strict order relation < on R are defined for any two real numbers a and B identified by the closed downwards sets A and B. The order relation is defined as @ < 8 © A c B. While, the strict order relation is defined as a < BP © A&B Theorem — Density of Q in R: For all numbers a, f € R such that a < B, there exists a (real) rational number # such that a<î<B Properties of the Real Numbers: + is associative, + is commutative, - is distributive with respect to +, * is associative, - is commutative, Existence of the neutral element 0 for the addition, Existence of the opposite, Existence of neutral element 1 for the multiplication, Existence of the reciprocal of every number different form zero, No zero divisors. Infimumiand supremumioniri Before defining the infimum and supremum on R, it is necessary to demonstrate that the real line is continuous. Therefore, we let A be a subset of R, which satisfies the following conditions: - Ais nonempty - A is properly containedinR - A is closed downwards - Ahas no maximum Then min(R\A) exists. At this point we can define the existence of the infimum and the supremum. Therefore we let A be any nonempty set of R which is bounded form above. Then there exists the supremum of A. While, if A is a nonempty subset of R which is bounded from below. Then, there exists the infimum of A. However, if A is any subset of the real line which is unbounded from above (below), then we write that supA=+co (infA=-00). INTERVALS'INIRI Let /be a nonempty subset of R. Then /is said to be an interval for every x, < x2 (x1,% € Dif < x < x thenx € I. In other words, an interval in R is any subset of R which together with any two points x, < x, also contains every other point x between x, and x,. ABSOLUTE VALUE: The absolute value of a real number x is defined as x= {%, FA In addition, the absolute value can also be considered as a function (defined on R) that takes all the nonnegative real values. Properties: 1. |x|l <kifandonlyif-k <x<k 2. le+yl <lx|+]p] . 3. ey <lxl- Iv I Proof of property 2: Consider that —|x| < x < |x|] and —|y| < y < |y]. If we sum term by 4. E _ let term, we obtain —(|x|] + |y]) £ x +y < |xl +]y] which is equivalent to |x + y] < 1x1 + |y1 ly ll by the 1° property. CANTOR'SINESTED'INTERVALS'ITHEOREM: Consider a sequence {I,},ewof closed and bounded intervals on the real line which is decreasing by set inclusion. Then there exists at least one real number belonging to every interval sequence. Proof: For every n € N, /n=[an, bo] and let A= {an:n € N} be the set of all the left extremes and B= {b,:n € N} the set of all the right extremes of the intervals. Lets prove that a, < b,forallm,neN - Case 1:in the first case we have that m=n because we deal with the extremes of the same interval. - Case 2: in the second case, if m<n, then we have that am < an < bn because Im contains In. Consequenitly, this also implies that am < bn. - Case 3: in the third case, if m>n, then we have that a,m < bm < bn because In contains Im. Consequently, this also implies that a, < bn. Therefore, if we define a = sup(A) and f = inf(A), we have that a < R. So for every real number y such thata <B <y, we have that y € NS In If in addition the length of the intervals tends to zero, we have that there exist one number x, belonging to every interval. Distance in R: The distance between two real number x and y is defined as d(x,y) = |x—yl Properties of the distance: 1. d(x,y)=0 Proof of Triangle inequality: In order to prove the 4°" property of the distance, we have to use 2. d(x,y)=00x=y the property of the absolute value, according to which the absolute value of the sum is less or 3. d(x,y)= d0y,2) equal to the sum of the absolute values 3|]x + y| < |x| + |y]. 4. d(x,y)+d(y,z)= d(x,2) d@,zi=|x-z|=l@-)+0-2l<lx-yl+ly=z1= d@y)+d0,2) An equation of 15° degree is of the form ax +b=0 4 ax=—b where x is the unknown. Two cases might occur: 1. a#+0— Thereis a unique solution to the equation, which is x = 2 2. a=0 Here, two other subsets may occur: a. b=0 + Then, every number x € R is a solution of the equation; b. b #0 Thereis no solution. A quadratic (or second-degree) equation is of the form ax? + bx + c = 0 where x is the unknow. The number of solutions depends on the real number A= b? — 4ac, which is said to be discriminant. Three cases might occur: - A<04+ There is no real solution of the equation; - A=0+ There is a unique solution € = i of the equation; - A>Q0—+ There are two distinct solutions x, = oli sal of the equation. Quadratic inequalities: A quadratic inequality (or inequality of second- degree) consist of 4 different types: 1. ax°+bx+c>0 2. ax?+bx+c>0 3. ax? +bx+c<0 4. ax°+bx+c<0 and x, = Open interval centred at a point: Let x, be any point of R and consider any real positive number e. Then the open interval centred at x,with radius e is defined as_ 1%, = {x € R:1x — xol < €} = ]xo — €,x0 + €[. Therefore, we can say that open interval centred at x,with radius e > 0 contains all the points (which are real numbers) whose distance from xpis smaller than e. Properties of the Trigonometric Functions: Properties of COSX and SINX Properties of TANX 1. cosx and sinx are defined on R 1. tanx is defined on R\{E+ kr ik ez} 2. |cosx| £ [sinx| £ 1 2 8. cosx and sinx are both periodic with period 7 = 2. 2. tanx takes all the real values as x varies in the open interval 5A [ 4. Fundamental identity: cos?x + sin?x =1 3. tanx is periodic with period 7 = 7 5. cosx is an even function and sinx is an odd function | 4. tanx is an odd function Value at relevant arcs: The angles oraresx Trigono- C_—-— metrico 0° | 20°) 45° 60°| 90°|120° 135° 150° 180° 210° 225° 240°|270° 300°) 315° 330° | 360% functions 1a ilaaQa © a GA hh ih > ®|3|3|3|3|j3/#«|6|*|#|#|3|a|3|a|6]|® 1/56 GI 1_A_b E_B_I anvo |9|a03 a|0!|703 3]. | aa aaa) cos |1| BE 1|0| 16 2 3 20302 za 33 3 tan [0 (E 01 Sao LL SL cor [roi LL RE LE By the definition of trigonometric function we have that: sin(r—@)=sina | cos(m—a)=-cosa | sin(m+a)=-sina cos( + a) = — cosa Trigonometric Formulae (addition and subtraction) sin(x + y) = sinx* cosy + cosx* siny sin(x — y) = sinx' cosy — cosx ‘siny cos(x +y) = cosx * cosy — sinx * siny cos(x — y) = cosx * cosy + sinx * siny Trigonometric Inverse Functions: Since the trigonometric functions sinx, cosx and tanx are periodic, they cannot be injective. However, if we consider some conventional restriction of those functions, then we can compute the inverse function, because those restrictions are strictly monotone and therefore invertible. > ARCSINE: The function arcsine of x (arcsinx) is the inverse of sinx. If we consider the restriction of sinx to the closed and bounded interval [-5.5] is increasing and takes all the values in [-1, 1]. Therefore, the arcosinx is defined on [-1, 1] and takes all the values in LE 1] > ARCCOSINE: The function arccosine of x (arccosx) is the inverse of cosx. The restriction of cosx to the closed and bounded interval [0, 7] is decreasing and takes all the values in [-1, 1]. Therefore, the arccosx is defined on [-1, 1] and takes all the values in [0, 7]. > ARCTANGENT: The function arctangent of x (arctanx) is the inverse of tanx. If we consider the restriction of tanx to the open and bounded interval H=s[is increasing and takes all the values in R. Therefore, the arctanx is defined on R and takes all the values in È Graph of cosx Graph of sinx Graph of tanx A function f:(R >) A Ris continuous ata point x, € A if the values f(x) keep themselves together near the value f (xo) when x is near xy. However in order to be more precise, it is necessary to provide a better mathematical definition of the continuity at a point. Therefore, a function f :(R >) A+ Ris said to be continuous at a point x, € A if one of the following statemenits is true: 1. For every neighbourhood V(.,) of f(xo) there exists a neighbourhood U,.,, of x, such that the following implication is true: 2. For every positive real number e there exists a positive real number é such that the following implication is true: (for every x E A) 3. For every positive real number e there exists a neighbourhood U,., of x, such that the following implication is true: We consider the basic neighbourhood, that is to say the interval centred at f(x) with radius e > 0 (and respectively, the interval centred at x, with radius é > 0). Therefore, fixing an interval centred at f(x0) is equivalent to fixing a real number e > 0 since the centre f(x) of the interval is given, as well as fixing an interval centred at x) is equivalent to fixing a real number 6 > 0 since the centre x, of the interval is given (and this corresponds to the definition of the points 1. and 2.) When we want to verify if a function is continuous at a point x,, we must start from the inequality | f@) — f(@)I< e. Continuity at an isolated point: A function f : (R >) A + R is continuous at any isolated point x, € A. As a matter of fact, if x, is an isolated point of A then there exists a neighbourhood U,., of the point x, such that U,, N A = {xo} However, it is possible to have that a function f is not continuous at a point x, of its domain, but it is only continuous to the right or to the left at the point xo. Left and right continuity at a point: A function f : (R >) A + Rissaid to be: - Left continuous ata point xy € A, if for every positive real number e there exists a positive real number é such that: xo S<x Sx > |f@)- fo) <e - Left continuous ata point x, € A, if for every positive real number e there exists a positive real number é such that: xy £x<%9+8 > ff) <e. A function is continuous at an interior point x, if and only if it is both left and right continuous at x. Continuity on a domain: A function is said to be continuous if it is continuous at every point of its domain. Continuity of the restriction: If f is a continuous function on some domain, then every restriction of f is also continuous. Sign in a neighbourhood (theorem): This theorem is important because it guarantees that the sign of a function which is continuous and different form 0 at some point is maintained on a neighbourhood of this point. Let a function be continuous at x, € A. If f(xo) > 0 (orif f(xo) < 0) then there exists a neighbourhood U, of x, such that f(®) > 0 (orif f(x) < 0) for every x € U,., N A. In addition, there exists a neighbourhood U,. of x, and a positive real number k such that f(x) > k (orif f(x) < —k) for every x EU. NA. Proof: Assume that f(x) > 0 and fix any real number (e) greater then 0 such that e < f(x). Since f is continuous at x, we have that there exists a neighbourhood U,,of x, such that f(xo) — e < f(x) < f(x) +e forall points x € U,, n A. In particular, f(x) > f(x) -e=kK>0 forall points x E U., NA. A continuous function is locally bounded: If a function is continuous at a point x, then f is bounded in a neighbourhood of xo. > Continuity of the compound function: If the functions f and g are such that the compound function g ° f exists, f is continuous at a point xy, and g is continuous at u = f(x), then the compound function g ° f is continuous at xo. Proof: Let's consider the neighbourhood V of the point z° = (92 f)(x0) = I(f(0)) = g(uo). Since g is continuous at the point u, such that u € W,, implies that the function g(u) € V,(x,)- From the continuity of f at x, there exists a neighbourhood U., of the point x, such that x € U,., implies that u = f(x) € W.,,. Therefore, x € U., implies that (9g Dl F(X) E Vyenko) > Continuity of the sum, the product and the ratio of continuous functions: If f and g are two functions which are both continuous at the point x, then the sum g+f and the product g-f are continuous at x. Moreover, if the function g(x0) # 0, then the ratio Lis continuous at xo. Proof: In order to show that the sum g+f is continuous at xo, it is necessary to fix a real number e > 0. Then we must consider the property of the absolute value, through which we get that |(f + 9)@) — (f +9 (0) = If) +9) — f(x) — go) < If) — fo) +19) — (xo): Since f is continuous at x, there exists a neighbourhood U', such that x € U',, > If@)- f@)l < 5 Since g is continuous at x, there exists a neighbourhood U”, such that x € U”., > 19) — g@)l < L Moreover, if we define U,, = U',,NU",,, then U,,is a neighbourhood of x, such that |(f + 9)() — (f+@@)l<î+î=e. : Since every constant f(x)=k is continuous, as the identity function g(x)=x, we have that every rational function is a continuous function. : If a function f : (R >) A + R is monotone, and A is an interval and the image set f(A) is an interval, then the function is continuous. Continuity of trigonometric functions: The trigonometric functions are continuous, since they are monotone on subsequent intervals, and the image of those restrictions are intervals. : If a function f : (R >)A + Ris strictly monotone, and A is an interval, then the inverse function is continuous. Continuity of the trigonometric inverse functions and the n-th root: The trigonometric inverse functions and the n-th root are continuous. INTERMEDIATE VALUE THEOREM If fis a real-valued continuous function on the interval [a, b], and n is any number between fa) and f(b), then there is a point x, € [a,b] such that f(x0) = n. Proof: In order to prove the intermediate value theorem, it is necessary to highlight that if there exist a point xo € [a,b] such that f(x) = n, then we have that f(a) < f (xo) < f(b). So, we need to consider the auxiliary function ® on [a, b] defined by (x) = f(@) — n, which is continuous since it is the difference of two continuous functions. And since f(a) < n, this means that f(a) - n < 0 = ®(a). On the other hand, since n< f(b) this means that f(b) -n>0= ®(b). Therefore we can say that at the extremes of the function (x) it has opposite values, because W(a) < 0 and (6) > 0. At this point, according to the proof, it is necessary that there exists a value xo € Ja,b[ such that ®@(x0) = 0 f(x0) — n= 0 because f(x) — n= 0 © f(x0) = n and this would complete the poof. Therefore, in order to demonstrate it, we start subdividing the interval [a, b] into two subintervals by finding the midpoint a, = ci of the interval. lf @(a,) = 0, then xo = ajand we stop the procedure because a,represents the point we are looking for. Otherwise, we keep subdividing the function in smaller and smaller intervals at whose extremes function ®(xo) takes values of opposite signs. Then, eventually we either arrive ata point x, such that the auxiliary function @ (xo) = 0 in finite number of steps or you build a sequence of closed and bounded intervals which is decreasing by set inclusion and is such that the length of the intervals tends to 0. At this point we need to prove that the auxiliary function ®(x0) is equal to 0. To do so, we use the Cantor's Theorem. Therefore, let's assume by contraposition that ® (xo) # 0 and that @ (xy) > 0. According to the “Signs in a neighbourhood”, since & is continuous on [a, b] and at x), we have that there exists a real number 5 > 0 such that @(x) > 0 for every x € ]xo — 6,x0 + S[. However this is contradictory, since the length of the intervals is smaller than 6, then we have that the interval / would be a subset of ]xo — 6,x0 + dl, and the auxiliary function @ assumes values of opposite signs at the extremes of the interval. Therefore, this completes the proof. Bolzano’s Theorem: If f is a real-valued continuous function on the interval [a, b], and f takes values of opposite sign at the extremes of [a, b], that is to say f(a)f(b) < 0, then there is a point x € Ja, b[ such that f(x) = 0. Theorem of Compaciness: lf f : (R >) A — R is a continuous function and A is a closed and bounded set (that is to say, a compact set), then the image set f(A) is also closed and bounded. Weierstrass Theorem: If f : (R >) A+ R is a continuous function and A is a closed and bounded set, then f attains both the maximum value and the minimum value, that is to say the set f(4) has both maximum element and minimum element. Proof: This theorem is connected to the Theorem of Compaciness, from which we have that f(A) is closed and bounded. Since f(A) is bounded we have that sup f(A) and inf f(A4) are both numbers. Since f(A) is closed, we must have that sup f(A) € f(A) andinff(A) € f(A). Then the theorem is proved because if the supremum of f(A) does not belong to f(4), then it would be an accumulation point. However, this represents a contradiction since f(A) is closed. Darboux Theorem: If f is a real-valued continuous function on the interval [a, b], then f takes all the values between min f ([a, b]) and max f ([a, b]). > Limit of rational functions: For any two polynomials f(x) = apx" + 1x"-1+ a2€°72 +-+ ay agx"+ax"71+ ax"? 9) = box" + bx-1+ byx92 + +-+ bm. Then we have that lim 250 box+b3xM-14 byxM2 o ifn>m xa t4+ +2) lag im = 5 ifn=m XI gm (bh + E ++ 0 "bo +5 2) 0 ifn<m > Limit of monotone functions: Let f : (R >) A + R be a monotone function, and let x) = sup(A), xo € A, then the limit sup(f(A) if f is nondecreasing inf(F(A) if f is nonincreasing © In an analogous way, this statement also lim f(@) exists, and we have that lim f(x) = { xo x>%0 holds true as regards x, = inf(A), x € A. Proof: We are only going to prove the case in which the function is nondecreasing. In order to do so, let's assume that 43 sup(f(4)) < + and fix a real number e > 0. From this, we have that there exists x € A with f(4) > 4— €, otherwise, if f@)<A4- e, then it would imply that A — e is an upper bond of f(A) smaller than A. Therefore, since f is nondecreasing we have that x > X which implies that 1= f(@) > f(@) > 4- e. Therefore, if we let 5 = x, — x, we have that |f(x) - | < e suchthat 0<|x-— x] < 8. Now let's assume that sup(f(4)) = + andfix a real number 7. Since f(A) is upper unbounded, then there exists x € A_ such that f(x) > y. Therefore, we have that x > £, and this implies that f(x) = f(&) > 7. So, we define 6 = x, — X in order to check that xo - d< x < xo which implies that f(x) > J. 1 lim = 1 Proof: The function f(x) = “2% is even due to the fact that it is the ratio of two odd functions. Then we x sinx consider x as varying strictly between 0 and È. Therefore, we have that sinx < x < tanx = 1 cosx' Dividing it by sin x, we obtain that 1 < È < At this point, we can clearly say that the function È cosa” stays between 1 and the function > both approaching 1 as x tend to 0+. Now, using the Squeeze sin Theorem we get that the limit lim 7 5 =1 . 1-cosx 1-cosx)(1+cos x 1-così x sin? x Proof: In this case we have that —T = £ X = = x2(1+cosx) = x2(1+cosx) x?(1+cosx) sin2x 1 cao sinix T—__—. Since x? (1+cosx) si "1 )2, we have that this limit tends to 1 as x approaches 0. TG 1-cosx _ 1 Therefore, we obtain that im —— =>. xo0 + 2 è nta 3. im Et = 1 Proof: In order to determine the limit lim S0E =1we put t = arcsin x. Therefore, we have that è fgetenoa aresin x approaches 0 as x approaches 0. At this point, since x = sin t, we get that lim = lim =1 Proof: In this case the proof is similar to the one of the arcsine of x. The Nepero Number + The Nepero numbers an irrational number which corresponds to e ® 2.718 .... More precisely, this n n x number e is defined as the limit of the sequence (1 + 1) as napproaches co. lim (1 + 1) =e or lim (1 + 1) =e. no now The limit of K"+ Another important limit is the limit of the sequence 4” as n approaches +00 (for every real number > 1). Proof: lf consider K>1, then we have K = 1 +8 for some positive real number 8. According to the Newton's lim K" = +00 nio binomial formula we have that k" = (1 + 6)" => 1+ né. Therefore, it is clear that 1+ né tends to +00 as n tends to +00. So because of the Theorem on the limit of comparison it is clear that lim k" = +00. . _ nico Jim k°=0 Now can also show that lim k" = 0 (for every real number 0 < k < 1)). If k is any real number such that nooo 0<k<1, then we have that lim k" = lim e = 0 since (0° tends to +co as n tends to +co because 2 > 1. nooo n.00 (1 x At this point, applying the Theorem on the limit of the reciprocal (the infinite case) ingr =0. È Powers with rational exponents > The positive basis k”, when n is a natural number, is just the product of k with itself n- times, that is to say K" = K- K- ...- K. On the other hand if the exponent is negative then we have that K7" = È In addition, we mm could also have a rational exponent r = î (n>1), and then we define the power as K” = kn = km. The powers with rational exponents holds the following properties: 1. kot. kr2 = pri4r2 2 (eyz = rr 3. (AE) =" h7 4. k°9=1 5. ktT=1 The exponential function is defined as f(x) = k*, where the exponent x is said to be the independent variable or the argument. The properties of the exponential function: 1. kis defined on R andis continuous. 2. k* is increasing if K > 1 and decreasing if 0 < k < 1. In addition it is constant equal to 1 ifkK=1. 3. Ilfk>1then Jim, k*= 0* and lim = +0, if0<k<1then lim ke” =+00 and gin k* = 0*. \ Increasing function f(x) = k* withk>1 Decreasing function f(x) =k*with0<k<1 —_ a The exponential function y = f(x) = k® (for every k > 0,k # 1) admits an inverse, which is the logarithmic function x= f7!(y) = logx y. Such inverse lo gy y is said to be the /ogarithm of y to the basis k. Its domain is the set of all positive real numbers and it takes real values. Therefore, we have that y = loggx 9 x=W. The natural logarithm y = logx = log. x is the logarithm to the basis e. Therefore, when the indication of the basis of the logarithm is omitted, then we are dealing with the natural logarithm. The properties of the logarithm function (f(x) = logx x): 1. logxx is definedonthe set R** of all the positive real numbers and it is continuous. 2. logxx is increasing if 4 > 1 and decreasingif0<k<1. 3. Ifk>1then Jim logx x = —co and lim loghx = +0, if0<k<1then Jim logx x = +00 and Jim loggx = —c0. Increasing function fx) = log, x withk > 1 Decreasing function fx) = log,x with0<k<1 The logarithmic function satisfies the following properties: o loggxy=logrx +logry o loggx"=rlogrx o loggc=logab'logye The last property log, c = log, b ‘ log, c is referred to as Change of the basis. As a matter of fact, we can express log, c as the ratio of two logarithms in a fixed basis a: log, c = 282° loga b' It is important to keep in mind that, in the case of indeterminate forms (a or 0 + co), for every real number r>0 the exponential function is stronger than x”, while the logarithmic function is weaker than x”. Relevant limits with the Exponential function (x” is stronger) x > lim _—=t+00 (forallk>1) =——_______>»>» n3+%o0 N Proof of am È = +90: Since we have that k> 1, we can i RE consequently write K = 1 + è for some é > 0. Therefore, from > im x +e (forallk>1) the Newton’s binomial formula we have that: n n 1 iS n i > lim É=+% (forallk>1andforallr>0) = +9 1tns+(8? doo, x3+0 2 n n n x > lim k*-x° =0* (forall0<k<1 andforallr>0) And therefore, by the Theorem of comparison lim <= = +0 x3+%0 Relevant limits with the Logarithmic functions (x” is weaker) + . logx(1 > lim rent = 0t (for allk>1 andforallr> 0) Proof of lim ore t2 — log, e : We have that: n94%0 . po l 1 > limx"loggx=0" dorallk>1andforallr>0) li RESA lim Tiog.(1 +2) = lim log, (1 +0E ta x > lim repncta) — log e (for allk>0,k#1); lim 2242 - 1 | Atthis point, we change the variable and we put x30 x50 * 1 1 ei a OT a > tim = togk (for all K>0,kK#1); =1 _ NU x Therefore, we obtain lim (1 + 1) = e which e u>40 u Proof of im 1 — = logk :In order to prove this property, it is implies that lim Ata =e. necessary changing the variable, and we define u=k* - 10 x = logx(1 + u). Therefore, we obtain lim EI = . u tim — e 7 = logk . The result log k due to the change of the basis. The functions of the form f(x)?®: h(x) = f(x)90) = elo8f@M?® eg(logf@) > Darccosx = — 1 1 D arccosx = =-—=--===5- =- Dceosy siny +yi-costy Proof: lf we consider y = arccosx so that x = cos y (0 < y < ). Then, from the derivative of the inverse we have that 1 101 > Darctanx= — 1+x Proof: lf we consider y = arctanx so that x = tany Ci <y< 5) Then, from the derivative of the inverse we have that 1 1 1 D arctanx = == i Dtany 14tanZy 1+27 Derivative of the logarithmic and of xP: 1 xlogk Proof: lf we consider y = logxx so that x = k” (y € R). Then, from the derivative of the inverse we have that = L1-_L_=_L_ i = =_L1 Dlogrx = DIS 7 Wlogk > Flog In the particular case, when K = e, we would have D logx > Dlogrx= (in particular D log x = i xloge x 3 Dx = px! Proof: If we consider the function f(x) = xP (p € R,x > 0). Then, we can write f(x) = x? = el°8*? = ePlos*, so that dp F') = D(eP!98*) = eplosx. P_ p' L_ pxP3. x x Derivative ot f(x): We have that D {96 = D etaSCI - pat - patoet0o (5/0) log (1) + 9 EA LO) - /I0 ($@logf() + LL). fa Derivative of the absolute value f(x) = |x|: DIx] = E LOCAL PROPERTIES OF THE FIRST DEGREE Function increasing at a point: A function f : (R >) A + R is said to be increasing at a point x, € A if there exists a (<xo) and (x EU, NA) > f@) < f(x) (x > xo) and (x EU,, NA) > f(@) > f(x0) decreasing at a point x, € A if there exists a neighbourhood U,. of the point x, such that (® < xo) and (x € U,, NA) > f(®) > f(&0) at: . (x> xo) and (x € UL, NA) > f(2) < f(x0) A function which is increasing (or decreasing) at a certain point of its domain, it is not necessarily increasing (or decreasing) on every neighbourhood of that point. neighbourhood U,,of the point x, such trat . While, a function is said to be If a function is differentiable at x, and f'(x) > 0, then the function is increasing at x). While, if a function is differentiable at x, and f'(x0) < 0, then the function is decreasing at xo. Proof: If f(x) > 0, then this means that lim [OSO 3 0g, According to the theorem of the Sign in a neighbourhood, we have xo xX0 that there exists a neighbourhood U,., of x, such that x € U., \{xo} > > 0. Now, since this ratio have the same sign, which means that both f(x) — f(x) and x — x, are positive, therefore this implies that f(x) < f(xo) whenx < x; and f@) > f (xo) when x > xo. For this reason, fis increasing at xo. LE)F (0) x=X%9 Point of relative minimum and maximum: If we consider a function f :(R >) A+ R, then a point x, E A is saidtobe a point of relative minimum if there exists a neighbourhood of x, such that (x € (U., \{xo} N 4) > f@) > f(xo). While, a point xo îs said to be a point of relative maximum if there exists a neighbourhood of x, such that (x € (U,., \{x0}N 4) > f(@) > fo). Moreover, the points at which the derivative is zero, they could be either points of relative maximum or minimum. As a matter of fact, we need to assume that x, is an interior point of A and the function fis differentiable at xo. So, if xo is either a point of relative maximum or a point of relative minimum, then the derivative of x, must be equal to 0 (f'(x0) = 0). THEOREMS ON DIFFERENTIABLE FUNCTIONS ON INTERVALS ‘ROLLE'S THEOREM] The Rolle's Theorem states that if a function fis: o definedon a closed and bounded interval [a, b] Then, there exists at least a point x, that o differentiable on the open interval Ja, b[ belongs to the open interval Ja, b[, such © continuous at the point a and b that the derivative of x, is equal to 0. o (fla)=f(b) Proof: In order to prove this theorem, it is necessary noticing that the function fis continuous on the closed interval [a, b]. Therefore, according to the Weiestrass Theorem, the function fhas a minimum value m (which represents the minimum of the image set f([a,b])), and a maximum value M (which represents the maximum of the image set f.[a,b])). At this point, one of the two following cases may occur: >» m=M:Inthis case we have that the function is a constant and therefore the derivative of x, is equal to 0, for every xo € [a, b]. > m<M:|Inthis case, at least one of the points of minimum and maximum, that we call x,, belongs to the open interval Ja, b[. There it must be an interior point. Otherwise, if the f(x,) is equal to the minimum value and f(x) is equal to the maximum value, and x} = a and x, = b, then we have that maximum and the minimum value coincides with the extremes of the interval because the f(a) is equal to the minimum value and f(b) is equal to the maximum value. Then, since f(a) = f(b), this would imply that the minimum value is equal to the maximum value. However, this contradicts our initial assumption m < M. Therefore, since x, is an interior point of the interval [a, b] and it is either a point of relative maximum or relative minimum, according to the Corollary of the definition of relative maximum or minimum, we have that the derivative of x, is equal to 0 and the proof is completed. The Lagrange's Theorem states that if a function fis: Th h . — h I o defined on a closed and bounded interval [a, b] to Mae OPE InIENATA Di cucine ongs o differentiable on the open interval Ja, b[ 10) - SO -f@ © continuous at the point a and b f'(x0) = a > Graphical interpretation of the theorem: As regards the graphical interpretation, we can consider that {®/@ dI L (A is the slope of the straight line passing through the point (a, f(a)) and (b, f(b)) belonging to the graph of f "Then this theorem guarantees the existence of at least one point x, that belongs to the open interval ]a, b[ such that the tangent to the graph of fthrough the point (xo, f(x0)) is parallel to such secant, due to the fact that it has the same slope as this secant. Proof: The Lagrange' Theorem is clearly a generalization of the Rolle's Theorem. Indeed, if we consider the assumptions of the Rolle's theorem, and in particular the one that states that f(a) = f(b), applied to the formula f'(x0) = O) we would get as a result 0. > Consequences of Lagrange’s Theorem: lf fis a function which is defined on an interval /, and f'(x) = 0 for all x € I, than fis a constant. Proof: In order to prove this consequence of the Lagrange's theorem, it is necessary to consider two points x; and x, € I with x, < x. Then the Lagrange's theorem is applicable to frestricted to the closed and bounded interval [x1,x2], and therefore there exists a point xo € ]x1,x2[ such thatf'(x0) = = SODO LED, f(x) = f(x2). Therefore, only in the case in which f(x1) = f(x2) the result is 0. And since the same reasoning can be repeated for all points x, and x, € I with x, < x, we have that fis constant. Primitive function: If we consider a function f : (R >) A + R, then a real-valued function Fon Ais said to be primitive of fif F'@)=f( vx€eA. o If both F and Gare primitive functions of a function f defined on an interval /, then F and G differ by a constant k, which means that there exits some constant X € R such that F(X) — G() = K forallx E L Proof: Since F and G are both primitive of f we have that F'(x) = G'(x)= f(x) forallx€I, implying that (F — GY @)=0 for all x € I. Therefore, if we consider F — G as a function, which has derivative equal to 0 on /, then it is a constant k, or equivalently (F — G)(x) =F@)-G@)=K o If fis a function which is defined on an interval /, and f'(x) > 0 (or f'(x) < 0) for allx € I, then fis increasing (or decreasing). Proofî: First, itis necessary to assume that f'(x) > 0 for allx € / and consider any two points x, and x, € I with x, < x2. Then Lagrange's theorem is applicable to f restricted to the closed and bounded interval [x,, x], and therefore there exists a point xo €]x,, x2[ such that f'(x0) = Sense >0-f(x2)-f(x1)>0 f(x) < f(x). Since the same reasoning can be 27X1 repeated for all points x, and x, € I with x, < xz we have that fis increasing. The same arguments can be applied to the case when f'(x)< 0 forallx EI. The theorem of L’Hòpital is essential in order to solve indeterminate forms like (0) or (£) and it states that it is necessary to consider two functions fand g which are differentiable on a neighbourhood U,. \{xo} of the point xo, and to assume nat gS@+ 0 forevery x € U,,, \{xo}. In addition, we need to assume that dim f@)= dm 9(x) =0 or oo. If there exists Jim È L©. then there Fo LG) I exist dim 1 . So we have that dim TA . ) x>x990) Theorem on the limit of the derivative: Let f be a function which is differentiable on a neighbourhood U,, \{xo} and continuous at xo. lf there exists lim f(x), then there exists lim f(x,) and therefore we have that f'(x,) = lim f(@). xoxo xoxo xoxXo LOCAL APPROXIMATION BY MEANS OF POLYNOMIALS LINEAR APPROXIMATION: A function f is said to be the linear approximation to a function f at a point x, if f satisfies those 3 properties: 1. fisa linear function, that is to say f@) = mx+ g; 2. fx) =f(x0): 8. lim LOS _ o, XX *TX0 The graph of the linear approximation to a function fat a point x) is the tangent to the graph fat the point (xo, f(x0)). The linear approximation to a function f atx» is f@)= f@)+f o) 20). Theorem — Linear approximation implies differentiability: There exists a linear approximation f to a function f at a point xo if and only if f is differentiable at xo. In this case it must be f(x) = f(x) + f'(x)(x— xo). Proof: First of all, we need to prove that if fis differentiable at the point x, then the function f defined as FG) = f(x0) + f'(x0)(x — xo) is the linear approximation to a function f at a point xy. Therefore, we must verify that the 3 conditions required are satidfied. As concern the first property, clearly f is a linear function. In addition, we have that F(x0) = f (xo), so that also the second property is verified. In order to verify the third property, it is necessary to consider that tim LOLO _ tim LOSCO MI fim LOLA _ f(x) = 0. xiXo X7X0 XX x=X0 xXx X7X0 Conversely, we prove that if the linear approximation to fat the point x, exits, then fis differentiable at x). So, according to the first condition of the definition of the linear approximation we have that f(x0) = mxo + q = f(x0). Therefore, we get that a=f(x0) mx, sothat f(x) = f(x) +m(&— xo). At this point, we can show that condition 3 implies that fis differentiable at xo and m = f'(xo). We have that Jim te SO _ im OSOTISI _ lip OSO 4 tim OSO _ xo xoxo XX xoxo *T%O xe XT%O lim [CO+MG=20) FG _ Jim xxO x=X0 x>X9 XIX man: i = m. Therefore, the proof is now completed. The Differential: Let fbe differentiable at a point x,. Then the differentiable of fat x), denoted by df, is the change of the linear approximation of fat x,,that is df = f(@) — f( xo). From the definition of linear approximation, we have that df = f(@)— f(x0) = f(x0) + f'(xo)(x— x0) — f(x0) = f (0) + f'(xo)(x — x0) — F(X) = f'(x0) (x — xo) = f'(x0)Ax. In this formula Ax stands for the change of the independent variable, which is (x — xo). Therefore, the differential dfof fat the point xy has its expression equal to df = f'(xo)Ax. However, if fis an identity function f = î with ix) =x for allx, then we could apply the same expression since D i(x) = 1. Therefore, the differential of the identity dx would be di = dx = 1: Ax = Ax According to the Leibniz notation, we could substitute the change of the independent variable with the symbol of the differential identity dx in the expression of the differential of f, such that the expression would be df = f'(x9)dx. In addition, it is possible to get rid of the consideration of the point x, and using a more generic point x of the domain, and we would get df = f'(x)dx, which can also be written as f(x) = ta = ui The Linear Asympiote: A linear function p(x) = mx + q is said to be a linear asymptote of fas x tends to 00 (or x tends to -00) if lim |f@) — pl = lim |f@)-mx-q|=0 (or lim |f(@)- p@)|=0 ) xt x34%0 x3-% Characterization: A linear function p(x) = mx + q is said to be a linear asymptote of fas x tends to 00 (or x tends to - 0) if and only if the following conditions are satisfied: 1. lim /O=m (or lim 12 = m); xt x xi x 2. im FA) —mx)=q (or im (f@) mx) = a). Oblique asymptote: If a linear function p(x) = mx + q is an asymptote of a function f, then p is said to be: 1. An oblique asymptote or a slant asymptote if m # 0; 2. A horizontal asymptote ifm= 0. Moreover, a function f such that Jim f(x) = c0 (+00,—co) has the straight-line x = x, € R as a vertical asymptote. AX0 LT AZ VV] = ia 1 LO]z1=1V 0] 0] Aa Vo gela tAVSY: The mean value theorem states that it is necessary to let fbe a continuous function on a closed and bounded interval [a, b]. Then there exists a point è € [a,b] such that b LIOMO=fO0-a). Proof: In order to prove the Mean value theorem it is essential to consider that, since fis continuous on the closed and bounded interval [a, b], by the Weiestrass Theorem, f has a minimum (m) and a maximum (M.. In addition, we need to notice that a < b is a tagged partition of the interval [a, b] (n = 1,a = x < x; = b). Therefore, m(b — a)is a lower integral sum, and M(b — a) is an upper integral sum. This is the reason why, from the definition of the definite integral, we must have that m(b — a) < SJ} dx < M(b — a). Then we divide the inequalities by (b — a), and we SP FOdx . Sr@ar get that m< và < M. In this case we have that và minimum and the maximum of f. Therefore, by the intermediate value theorem, there must exists a value È € [a,b] such that f@)=n 0 Ta f(x)dx = f(€)(b — a) and this consideration completes the proof. Lal TZ gi 0)51= O) 7 Net: The Fundamental theorem of the integral calculus states that it is essential to let fbe a continuous function on an interval / Then we need to consider any point x € I. And we define the integral function F on the interval /as Fx) = SE fat (with x € 1). Then F'(x) = f(x) foreveryx E I. Proof: In order to prove this theorem, we need to notice that the integral function F(x) = SE fdt is defined for = n, which is a particular number between the every x € I, because if x € I is fixed, then we can say that the definite integral SE fat exists because fis continuous on /. Now, let's consider the incremental ratio of F at any point xy € I, which is : - e — ro Sf ne RE) = FF) _ RSOUR'IOd SO According to the Mean value theorem for definite integrals, there o XxX x=Xq exists, for every x € /, a point f(x) such that JE sOd — f(€(x)), where È (x) is between x — x. Therefore, we have xo x-x0 that x > xo, which implies that î(x) — xo. At the end, by the fact that f is continuous on /and by the theorem on the limit of compound functions, we have that F'(x0) = lim RE) = Jim F(EM) = dm f(€) = f(x0). This assumption xX0 xo to completes the proof because the point xo, that belongs to the interval, is arbitrary chosen. Mieli t[of = iS ae] The Torricellis Theorem represents a consequence of the Fundamental Theorem of the integral calculus. The Torricelli's theorem states that it is necessary to let f be a continuous function on an interval [a, b]. If G represents any primitive of f, then we have that SE fax = G(b)-— G(a)=G()|b. Proof: lf we fix any point x € [a,b]. Then, we have that SE fOMax = JT fax + [? fax = [? fax — [L f(x)dx = F(b) — F(a), where Fis the integral function defined in the fundamental theorem of the integral calculus. In addition, if we consider any primitive function G of f, then we have that F(x) — G(x) = k for some constant k from one of the consequences of the Lagrange’s Theorem, so that SL f@Mdx = F(b)— F(a) = G(b)+k- G(a)—k = G(b)- G(a). The quantity G(b) — G(a) = G(x)IÈ is said to be the the change of G between a and b. Chapter 9 —- Differential Calculus hapter 10 - Integral Calcul SOME ELEMENTS OF LOGIC. SET THEORY... CHAPTER 2 - MAPPINGS AND RELATION MAPPINGS.. BINARY RELATION: ORDER RELATIONS... CHAPTER 3 - COMBINATORICS.... DISPOSITIONS WITHOUT REPETIONS. COMBINATIONS. DISPOSTIONS WITH REPETITIONS. CHAPTER 4 - REAL NUMBERS... NATURAL NUMBERS INTEGERS AND RATIONAL NUMBERS. QUADRATIC EQUATIONS AND QUADRTIC INEQUALITIES. 6 6 6 7 7 CHAPTER 5 - REAL-VALUED FUNCTIONS OF ONE REAL VARIABLE... © TRIGONOMETRIC FUNCTIONS AND THEIR INVERSES.. CHAPTER 6 - CONTINUOUS FUNCTIONS CONTINUOUS FUNCTION AnD ITS PROPERTIES. CHAPTER 7 - LIMITS... FINITE LIMITS... LIMITS WITH THE TRIGONOMETRIC FUNCTIONS. CHAPTER 8 - EXPONENTIAL AND LOGARITHMIC FUNCTIONS... POWERS WITH REAL EXPONENTS. THE EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAPTER 9 - DIFFERENTIAL CALCULUS.... THE CONCEPT OF DERIVATE... IMPORTANT DERIVATIVES.. THEOREMS ON DIFFERENTIATION LOCAL PROPERTIES OF THE FIRST DEGREE... THEOREMS ON DIFFERENTIABLE FUNCTIONS ONINTERVALS LOCAL APPROXIMATION BY MEANS OF POLYNOMIALS LOCAL CONVEXITY AND LOCAL CONCAVITY..... CHAPTER 10 - INTEGRAL CALCULUS INDEFINITE INTEGRAL . DEFINITE INTEGRAL ..