ECN115 Mathematical Methods Lecture 4 (2023-2024) PDF

ECN115 Mathematical Methods in Economics and Finance Evgenii Safonov 2023-2024. LECTURE 4 1 Outline ◁ Limits and infinite sums ◁ Continuity of a function ◁ Derivative of a function ◁ Composition of functions (next lecture’s materi...

ECN115 Mathematical Methods in Economics and Finance Evgenii Safonov 2023-2024. LECTURE 4 1 Outline ◁ Limits and infinite sums ◁ Continuity of a function ◁ Derivative of a function ◁ Composition of functions (next lecture’s material) 2 Limits of sequences R Definition. Number x ∈ is a limit of sequence (an ) if for any real number 𝜖 > 0, there exists a natural number N such that for all n > N, |an − x| < 𝜖. In this case, we say that sequence (an ) has a finite limit, or that it converges Thus, x is a limit of an if ∀𝜖 > 0, ∃N ∈ N: ∀n > N, |an − x| < 𝜖 3 For reference: infinite limits Definition. We say that a sequence (an ) tends to infinity if for any number M there is N ∈ N such that for all n > N, an > M. In this case, we write lim an = ∞ an −→ ∞ n→∞ Definition. We say that a sequence (an ) tends to minus infinity if for any number M there is N ∈ N such that for all n > N, an < M. In this case, we write lim an = −∞ an −→ −∞ n→∞ ◁ an = n, then lim an = ∞ n→∞ 2 ◁ an = −n , then lim an = −∞ n→∞ ◁ (an ) = (1, 0, 2, 0, 3, 0, 4, 0,...) does not have either finite or infinite limit ◁ an = ln(n), then lim an = ∞ n→∞ 4 There cannot be many limits of a sequence Theorem L3.1. A sequence of real numbers has at most one finite limit. Notations: x = lim an or an −→ x n→∞ Remark. The theorem generalizes to infinite limits as well; i.e. a sequence of real numbers has at most one (finite or infinite) limit 5 Arithmetic properties of limits Theorem L3.2. If sequences (an ) and (bn ) have finite limits, then the sequence (an + bn ) also has a finite limit; moreover, lim (an + bn ) = lim an + lim bn n→∞ n→∞ n→∞ Theorem L3.3. If sequences (an ) and (bn ) have finite limits, then the sequence (an · bn ) also has a finite limit; moreover, lim (an · bn ) = lim an · lim bn n→∞ n→∞ n→∞ Theorem L3.4. If sequences (an ) and (bn ) have finite limits, and lim bn ̸= 0, n→∞ an then the sequence ( ) also has a finite limit; moreover, bn (︂ an )︂ lim an n→∞ lim = n→∞ bn lim bn n→∞ 6 Limit of a sum of sequences (Proof of Theorem L3.2) Theorem L3.2. If sequences (an ) and (bn ) have finite limits, then the sequence (an + bn ) also has a finite limit; moreover, lim (an + bn ) = lim an + lim bn n→∞ n→∞ n→∞ Proof. Let a = lim an , b = lim bn n→∞ n→∞ 1) Consider arbitrary 𝜖 > 0 2) Since an → a, there is N1 such that for all n > N1 , |an − a| < 𝜖/2 3) Since bn → b, there is N2 such that for all n > N2 , |bn − b| < 𝜖/2 4) Take N = max{N1 , N2 }. Then N ≥ N1 , N2. Hence, for all n > N, |an − a| < 𝜖/2 and |bn − a| < 𝜖/2. Therefore, for all n > N 𝜖 𝜖 |an + bn − (a + b)| = |(an − a) + (bn − b)| ≤ |an − a| + |bn − b|< + = 𝜖 ⏟ ⏞ 2 2 by triangle inequality Thus, we’ve proved that a + b is the limit of an + bn 7 Infinite sums Definition. Given a sequence a, an infinite (from one side) sum of its elements is defined as follows: ∞ (︃ n )︃ ∑︁ ∑︁ aj = lim aj n→∞ j=1 j=1 the infinite sum in the left hand side exists when the limit on the right hand side exists. n ∑︁ Put it differently, consider a new sequence bn defined as bn = aj , then ∞ j=1 ∑︁ aj = lim bn n→∞ j=1 8 Infinite sums—example ∞ ∑︁ Example 1. Consider q j , where 0 < |q| < 1. Using Homework 2 and the j=0 definition of the infinite sum, we get: ∞ n 1 − q n+1 ∑︁ ∑︁ (︂ )︂ q j = lim q j = lim j=0 n→∞ j=0 n→∞ 1−q when the corresponding limit exists. Let us show that lim q n = 0 n→∞ 9 Infinite sums—example ∞ ∑︁ Example 1. Consider q j , where 0 < |q| < 1. Using Homework 2 and the j=0 definition of the infinite sum, we get: ∞ n 1 − q n+1 ∑︁ ∑︁ (︂ )︂ q j = lim q j = lim j=0 n→∞ j=0 n→∞ 1−q when the corresponding limit exists. Let us show that lim q n = 0 n→∞ ◁ Consider arbitrary 𝜖 > 0 ◁ Since 0 < |q| < 1, then 𝜒 = |q|−1 − 1 > 0 ◁ Using the Binomial formula, (︀ −1 )︀n |q| = (1 + 𝜒)n = 1 + n𝜒 +... > n𝜒 1 ◁ Take N > , then for all n > N, 𝜒·𝜖 1 1 |q n − 0| = |q n | = |q|n < < 0 ◁ Since 0 < |q| < 1, then 𝜒 = |q|−1 − 1 > 0 ◁ Using the Binomial formula, (︀ −1 )︀n |q| = (1 + 𝜒)n = 1 + n𝜒 +... > n𝜒 1 ◁ Take N > , then for all n > N, 𝜒·𝜖 1 1 |q n − 0| = |q n | = |q|n < < 0 there exists 𝜖 > 0 such that |f (z) − f (x)| < 𝛿 for all z such that |z − x| < 𝜖. Theorem L4.2. The two definition of continuity are equivalent. Proof. Omitted 13 Continuity of functions R Definition. A function f : D → is continuous on an open interval if it is continuous at all points x of this interval. Geometric interpretation. Continuity of a function on an interval (a, b) means that its graph on the (x, y ) plane can be drawn without pulling a pencil off paper. Continuity of common functions. Almost all functions that we have studied before are continuous (Whiteboard example(s)). 14 Outline ◁ Limits and infinite sums ◁ Continuity of a function ◁ Derivative of a function ◁ Composition of functions (next lecture’s material) 15 Tangent line to the graph of a function (example) Consider f (x) = x 3. Let us try to approximate f by a linear function near x = 2. For that, it is enough to set up two points. Naturally, one point is (x, y ) = (x, f (x)) = (2, 8). With another point (x, y ), the equation of a tangent line is: y −y ŷ (x̂) = · (x̂ − x) + y x −x 16 Tangent line to the graph of a function (example) Consider f (x) = x 3. Let us try to approximate f by a linear function near x = 2. For that, it is enough to set up two points. Naturally, one point is (x, y ) = (x, f (x)) = (2, 8). With another point (x, y ), the equation of a tangent line is: y −y ŷ (x̂) = · (x̂ − x) + y x −x ◁ Pick the second point as (3, 33 ) = (3, 27). Then, the equation is 27 − 8 ŷ (x̂) = · (x̂ − 2) + 8 i.e. ŷ = 19 · x̂ − 30 3−2 16 Tangent line to the graph of a function (example) Consider f (x) = x 3. Let us try to approximate f by a linear function near x = 2. For that, it is enough to set up two points. Naturally, one point is (x, y ) = (x, f (x)) = (2, 8). With another point (x, y ), the equation of a tangent line is: y −y ŷ (x̂) = · (x̂ − x) + y x −x ◁ Pick the second point as (3, 33 ) = (3, 27). Then, the equation is 27 − 8 ŷ (x̂) = · (x̂ − 2) + 8 i.e. ŷ = 19 · x̂ − 30 3−2 3 ◁ Pick the second point as (1.5, (1.5) ) = (1.5, 3.375). Then, 3.375 − 8 ŷ (x̂) = · (x̂ − 2) + 8 i.e. ŷ = 9.25 · x̂ − 10.5 1.5 − 2 16 Tangent line to the graph of a function (example) Consider f (x) = x 3. Let us try to approximate f by a linear function near x = 2. For that, it is enough to set up two points. Naturally, one point is (x, y ) = (x, f (x)) = (2, 8). With another point (x, y ), the equation of a tangent line is: y −y ŷ (x̂) = · (x̂ − x) + y x −x ◁ Pick the second point as (3, 33 ) = (3, 27). Then, the equation is 27 − 8 ŷ (x̂) = · (x̂ − 2) + 8 i.e. ŷ = 19 · x̂ − 30 3−2 3 ◁ Pick the second point as (1.5, (1.5) ) = (1.5, 3.375). Then, 3.375 − 8 ŷ (x̂) = · (x̂ − 2) + 8 i.e. ŷ = 9.25 · x̂ − 10.5 1.5 − 2 ◁ Pick the second point as (2.1, (2.1)3 ) = (2.1, 9.261). Then, 9.261 − 8 ŷ (x̂) = · (x̂ − 2) + 8 i.e. ŷ = 12.61 · x̂ − 17.22 2.1 − 2 16 Tangent line to the graph of a function (example) Consider f (x) = x 3. Let us try to approximate f by a linear function near x = 2. For that, it is enough to set up two points. Naturally, one point is (x, y ) = (x, f (x)) = (2, 8). With another point (x, y ), the equation of a tangent line is: y −y ŷ (x̂) = · (x̂ − x) + y x −x ◁ Pick the second point as (3, 33 ) = (3, 27). Then, the equation is 27 − 8 ŷ (x̂) = · (x̂ − 2) + 8 i.e. ŷ = 19 · x̂ − 30 3−2 3 ◁ Pick the second point as (1.5, (1.5) ) = (1.5, 3.375). Then, 3.375 − 8 ŷ (x̂) = · (x̂ − 2) + 8 i.e. ŷ = 9.25 · x̂ − 10.5 1.5 − 2 ◁ Pick the second point as (2.1, (2.1)3 ) = (2.1, 9.261). Then, 9.261 − 8 ŷ (x̂) = · (x̂ − 2) + 8 i.e. ŷ = 12.61 · x̂ − 17.22 2.1 − 2 ◁ Pick the second point as (1.95, (1.95)3 ) = (1.95, 7.414875). Then, 7.414875 − 8 ŷ (x̂) = · (x̂ − 2) + 8 i.e. y = 11.7025 · x̂ − 15.405 1.95 − 2 16 Tangent line to the graph of a function (example) Consider f (x) = x 3. Let us try to approximate f by a linear function near x = 2. For that, it is enough to set up two points. Naturally, one point is (x, y ) = (x, f (x)) = (2, 8). With another point (x, y ), the equation of a tangent line is: y −y ŷ (x̂) = · (x̂ − x) + y x −x ◁ Pick the second point as (3, 33 ) = (3, 27). Then, the equation is 27 − 8 ŷ (x̂) = · (x̂ − 2) + 8 i.e. ŷ = 19 · x̂ − 30 3−2 3 ◁ Pick the second point as (1.5, (1.5) ) = (1.5, 3.375). Then, 3.375 − 8 ŷ (x̂) = · (x̂ − 2) + 8 i.e. ŷ = 9.25 · x̂ − 10.5 1.5 − 2 ◁ Pick the second point as (2.1, (2.1)3 ) = (2.1, 9.261). Then, 9.261 − 8 ŷ (x̂) = · (x̂ − 2) + 8 i.e. ŷ = 12.61 · x̂ − 17.22 2.1 − 2 ◁ Pick the second point as (1.95, (1.95)3 ) = (1.95, 7.414875). Then, 7.414875 − 8 ŷ (x̂) = · (x̂ − 2) + 8 i.e. y = 11.7025 · x̂ − 15.405 1.95 − 2 What happens when we consider x closer and closer to x? 16 Example of a derivative Since we can operate with limits now, let us consider a limit (if it exists) of the coefficient of the linear term in the considered example: (︂ 3 )︂ xn − 8 lim =? n→∞ xn − 2 where the sequence xn is such that it comes closer and closer to x; that is, xn −→ x (but xn ̸= x). ◁ How can we calculate such limits? Our problem is that we don’t know the explicit way how xn depends on n. This dependence can by arbitrary—so long as xn −→ x, and xn ̸= x. ◁ A way to overcome this problem: express xn = x + (xn − x) in our example, xn = 2 + (xn − 2) and substitute this expression to f (xn ) (in our example, to xn3 ). 17 Example of a derivative Continuing with our example, consider xn3 − 8 (2 + (xn − 2))3 − 8 = = xn − 2 xn − 2 23 + 3 · 22 · (xn − 2) + 3 · 2 · (xn − 2)2 + (xn − 2)3 −8 = = xn − 2 12 · (xn − 2) + 6 · (xn − 2)2 + (xn − 2)3 = = xn − 2 = 12 + 6 · (xn − 2) + (xn − 2)2 Therefore, (︂ 3 )︂ xn − 8 lim 12 + 6 · (xn − 2) + (xn − 2)2 = (︀ )︀ lim = n→∞ xn − 2 n→∞ = 12 + 6 · lim (xn − 2) + lim (xn − 2) · lim (xn − 2) = n→∞ n→∞ n→∞ = 12 + 6 · 0 + 0 · 0 = 12 18 Example of a linear approximation Recall that we have been approximating f (x) = x 3 by a linear function xn3 − 8 y = kn x + bn in this example, y = · (x − 2) + yo xn − 2 Examples of numerical values. ◁ (xn ) = (3, 1.5, 2.1, 1.95, 1.99, 2.002,...) −→ 2, then (kn ) = (19, 9, 25, 12.61, 11.7025, 11.9401, 12.012004,...) −→ 12. ◁ (xn ) = (3, 2.1, 2.01, 2.001, 2.0001,...) −→ 2, then (kn ) = (19, 12.61, , 12.0601 , 12.006001, 12.00060001,...) −→ 12. ◁ (xn ) = (1, 1.9, 1.99, 1.999, 1.9999,...) −→ 2, then (kn ) = (7, 11.41, 11.9401, 11.994001, 11.99940001,...) −→ 12. 19 Definition of a derivative. R R Definition. Consider function f : D → with D ⊆ and suppose that (a, b) ⊆ D for some real numbers a < b. The derivative of function f at point x ∈ (a, b) is denoted by f ′ (x) and is defined by f (xn ) − f (x) f ′ (x) = lim for all (xn ) such that xn ̸= x, xn −→ x n→∞ xn − x We say that the derivative at point x exists if this limit exists and is equal to the same number independently of the choice of sequence xn. In this case, we say that the function f is differentiable at point x. 20 Definition of a derivative. R R Definition. Consider function f : D → with D ⊆ and suppose that (a, b) ⊆ D for some real numbers a < b. The derivative of function f at point x ∈ (a, b) is denoted by f ′ (x) and is defined by f (xn ) − f (x) f ′ (x) = lim for all (xn ) such that xn ̸= x, xn −→ x n→∞ xn − x We say that the derivative at point x exists if this limit exists and is equal to the same number independently of the choice of sequence xn. In this case, we say that the function f is differentiable at point x. Remark 1. Note that f should be defined on some (maybe small) interval (a, b) around point x. 20 Definition of a derivative. R R Definition. Consider function f : D → with D ⊆ and suppose that (a, b) ⊆ D for some real numbers a < b. The derivative of function f at point x ∈ (a, b) is denoted by f ′ (x) and is defined by f (xn ) − f (x) f ′ (x) = lim for all (xn ) such that xn ̸= x, xn −→ x n→∞ xn − x We say that the derivative at point x exists if this limit exists and is equal to the same number independently of the choice of sequence xn. In this case, we say that the function f is differentiable at point x. Remark 1. Note that f should be defined on some (maybe small) interval (a, b) around point x. Remark 2. Note that the value of the considered limit should not depend on the choice of a particular sequence xn −→ x. 20 Definition of a derivative. R R Definition. Consider function f : D → with D ⊆ and suppose that (a, b) ⊆ D for some real numbers a < b. The derivative of function f at point x ∈ (a, b) is denoted by f ′ (x) and is defined by f (xn ) − f (x) f ′ (x) = lim for all (xn ) such that xn ̸= x, xn −→ x n→∞ xn − x We say that the derivative at point x exists if this limit exists and is equal to the same number independently of the choice of sequence xn. In this case, we say that the function f is differentiable at point x. Remark 1. Note that f should be defined on some (maybe small) interval (a, b) around point x. Remark 2. Note that the value of the considered limit should not depend on the choice of a particular sequence xn −→ x. df Remark 3. Alternative notation: f ′ (x) = dx 20 Differentiable functions. R Definition. Suppose that a numerical function f : D → is differentiable at all points x ∈ D of its domain. It this case, we call the function f R differentiable, and we call the function g : D → defined by g (x) = f ′ (x) for all x ∈ D the derivative of the function f. 21 Derivative of a linear function R Note that function f (x) = ax + b defined for all x ∈ is linear. Therefore, it is perfectly approximated by itself and hence, we expect its derivative to be f ′ (x) = a. Let us check it. Consider arbitrary sequence xn such that xn −→ x, xn ̸= x for all n, then (axn + b) − (ax + b) a(xn − x) (ax + b)′ = lim = lim = lim a = a n→∞ xn − x n→∞ xn − x n→∞ 22 Derivative of a linear function R Note that function f (x) = ax + b defined for all x ∈ is linear. Therefore, it is perfectly approximated by itself and hence, we expect its derivative to be f ′ (x) = a. Let us check it. Consider arbitrary sequence xn such that xn −→ x, xn ̸= x for all n, then (axn + b) − (ax + b) a(xn − x) (ax + b)′ = lim = lim = lim a = a n→∞ xn − x n→∞ xn − x n→∞ What we have learned so far: ◁ If c ∈ R is a constant, then (c) ′ = 0; ◁ Next, (x)′ = 1; ◁ For f (x) = x 3 , we have calculated that f ′ (2) = 12; so, what is the formula for (x 3 )′ at other points? And what is the formula for (x k )′ where k is a natural number? 22 Derivative of x 3 Let us show that for f (x) = x 3 , we have f (x)′ = 3x 2. (︀ )︀3 )︀3 xn − x 3 (x + (xn − x) − x 3 (︀ f ′ (x) = lim = lim = n→∞ xn − x n→∞ xn − x x 3 + 3 · x 2 (xn − x) + 3 · x(xn − x)2 + (xn − x)3 −x 3 = lim = n→∞ xn − x lim 3 · x 2 + 3x · (xn − x) + (xn − x)2 = 3 · x 2 (︀ )︀ = n→∞ Substituting x = 2, we get our previous result f ′ (2) = 12 23 Derivative of a natural power of x Theorem L4.3. The function f : R → R given by f (x) = x k , where k ∈ N is R differentiable, and for all x ∈ , f ′ (x) = k · x k−1 24 Derivative of a natural power of x Theorem L4.3. The function f : R → R given by f (x) = x k , where k ∈ N is differentiable, and for all x ∈ , R f ′ (x) = k · x k−1 R Proof. Consider an arbitrary x ∈ and an arbitrary sequence xn such that xn −→ x and xn ̸= x for all n. Let us use the Binomial formula to get (︀ )︀k )︀k xn − x k (x + (xn − x) − x k (︀ f ′ (x) = lim = lim = n→∞ xn − x n→∞ xn − x k ∑︁ x k + k · x k−1 (xn − x) + Ckj · x k−j · (xn − x)j −x k j=2 = lim = n→∞ xn − x (︃ k )︃ ∑︁ = lim k ·x k−1 + Ckj · x k−j · (xn − x)j−1 = k · x k−1 n→∞ j=2 24 Additivity of the derivative R Theorem L4.4. Suppose functions f : D → and g : D → are R R differentiable at point x ∈ D, then the function (f + g ) : D → given by (f + g )(x) = f (x) + g (x) is differentiable at point x, and (f + g )′ (x) = f ′ (x) + g ′ (x) 25 Additivity of the derivative R Theorem L4.4. Suppose functions f : D → and g : D → are R differentiable at point x ∈ D, then the function (f + g ) : D → given by R (f + g )(x) = f (x) + g (x) is differentiable at point x, and (f + g )′ (x) = f ′ (x) + g ′ (x) Proof. The statement follows from f (xn ) + g (xn ) − f (x) − g (x) (f + g )′ (x) = lim = n→∞ xn − x (︂ )︂ f (xn ) − f (x) g (xn ) − g (x) = lim + = n→∞ xn − x xn − x f (xn ) − f (x) g (xn ) − g (x) = lim + lim = f ′ (x) + g ′ (x) n→∞ xn − x n→∞ xn − x Example. (x 5 + x + 1)′ = (x 5 )′ + (x)′ + (1)′ = 5x 4 + 1 + 0 = 5x 4 + 1. 25 Differentiability implies continuity Theorem L4.5. If a numerical function f : D → R is differentiable at point x ∈ D, then it is continuous at point x. 26 Differentiability implies continuity Theorem L4.5. If a numerical function f : D → R is differentiable at point x ∈ D, then it is continuous at point x. Proof. Consider arbitrary sequence (xn ) such that xn −→ x and xn ̸= x for all n. Since f is differentiable at point x, for 𝜖 = 1 > 0 there exists N such that for all n > N, ⃒ f (xn ) − f (x) ⃒ ⃒ − f ′ (x)⃒ < 1 ⃒ xn − x ⃒ Hence, for large enough n, ⃒f (xn ) − f (x) − f ′ (x) · (xn − x)⃒ ⃒ ⃒ ⃒ ⃒ < ⃒xn − x ⃒ ⃒f (xn ) − f (x)⃒ − ⃒f ′ (x) · (xn − x)⃒ ⃒ ⃒ ⃒ ⃒ ⃒ ⃒ < ⃒xn − x ⃒ since |a| ≤ |a − b| + |b| ⃒ ⃒ ⃒ ′ ⃒ ⃒ ⃒ 0 ≤ ⃒f (xn ) − f (x)⃒ < ⃒f (x) · (xn − x)⃒ + ⃒xn − x ⃒ 26 Differentiability implies continuity Theorem L4.5. If a numerical function f : D → R is differentiable at point x ∈ D, then it is continuous at point x. Proof. Consider arbitrary sequence (xn ) such that xn −→ x and xn ̸= x for all n. Since f is differentiable at point x, for 𝜖 = 1 > 0 there exists N such that for all n > N, ⃒ f (xn ) − f (x) ⃒ ⃒ − f ′ (x)⃒ < 1 ⃒ xn − x ⃒ Hence, for large enough n, ⃒f (xn ) − f (x) − f ′ (x) · (xn − x)⃒ ⃒ ⃒ ⃒ ⃒ < ⃒xn − x ⃒ ⃒f (xn ) − f (x)⃒ − ⃒f ′ (x) · (xn − x)⃒ ⃒ ⃒ ⃒ ⃒ ⃒ ⃒ < ⃒xn − x ⃒ since |a| ≤ |a − b| + |b| ⃒ ⃒ ⃒ ′ ⃒ ⃒ ⃒ 0 ≤ ⃒f (xn ) − f (x)⃒ < ⃒f (x) · (xn − x)⃒ + ⃒xn − x ⃒ Since 0 −→ 0, and lim ⃒f ′ (x) · (xn − x)⃒ + ⃒xn − x ⃒ = (|f ′ (x)| + 1) · lim ⃒xn − x ⃒ = 0 (︀⃒ ⃒ ⃒ ⃒)︀ ⃒ ⃒ n→∞ n→∞ ⃒ ⃒ then by the Squeeze Theorem (L4.1), lim ⃒f (xn ) − f (x)⃒ = 0. n→∞ 26 The product rule for differentiation Theorem L4.6 (Product Rule). Suppose numerical functions f , g : D → R R are differentiable at point x ∈ D. Then the function (f · g ) : D → given by (f · g )(y ) = f (y ) · g (y ) is differentiable at point x, and (f · g )′ (x) = f ′ (x) · g (x) + f (x) · g ′ (x) 27 The product rule for differentiation Theorem L4.6 (Product Rule). Suppose numerical functions f , g : D → R are differentiable at point x ∈ D. Then the function (f · g ) : D → given by R (f · g )(y ) = f (y ) · g (y ) is differentiable at point x, and (f · g )′ (x) = f ′ (x) · g (x) + f (x) · g ′ (x) Proof. For an arbitrary sequence xn −→ x, xn ̸= x, we have: f (xn ) · g (xn ) − f (x) · g (x) (f · g )′ (x) = lim = n→∞ xn − x (f (xn ) − f (x)) · g (x) + f (x) · (g (xn ) − g (x)) + (f (xn ) − f (x)) · (g (xn ) − g (x)) lim n→∞ xn − x f (xn ) − f (x) g (xn ) − g (x) = g (x) · lim + f (x) · lim + n→∞ xn − x n→∞ xn − x (︂ )︂ f (xn ) − f (x) + lim · lim (g (xn ) − g (x)) = using Theorem L4.5 n→∞ xn − x n→∞ = g (x) · f ′ (x) + f (x) · g ′ (x) + f ′ (x) · 0 = f ′ (x) · g (x) + f (x) · g ′ (x) 27 Corollary and Examples Corollary. For any numerical function f differentiable at point x and constant c∈ ,R (cf )′ (x) = c · f ′ (x) 28 Corollary and Examples Corollary. For any numerical function f differentiable at point x and constant c∈ ,R (cf )′ (x) = c · f ′ (x) Proof. (cf )′ (x) = (c)′ · f (x) + c · f ′ (x) = 0 + c · f ′ (x) = c · f ′ (x). 28 Corollary and Examples Corollary. For any numerical function f differentiable at point x and constant c∈ ,R (cf )′ (x) = c · f ′ (x) Proof. (cf )′ (x) = (c)′ · f (x) + c · f ′ (x) = 0 + c · f ′ (x) = c · f ′ (x). Examples. We can now calculate the derivative of any polynomial function. )︀′ = (3x 4 )′ + (−2x 3 )′ + (5x 2 )′ + (−11x)′ + (15)′ (︀ 4 3x − 2x 3 + 5x 2 − 11x + 15 = 3 · 4x 3 − 2 · 3x 2 + 5 · 2x − 11 · 1 + 0 = 12x 3 − 6x 2 + 10x − 11 (−2x 9 + 3x 7 − x 3 )′ = −18x 8 + 21x 6 − 3x 2 (x 2 + 3x + 5)′ = 2x + 3 28 Slide for those who love general formulas For a general polynomial function, we have: (︃ k )︃′ k ∑︁ j ∑︁ aj · x = aj · j · x j−1 j=0 j=1 29 Another look on the derivative of a natural power of x ◁ Recall that (x)′ = 1. 30 Another look on the derivative of a natural power of x ◁ Recall that (x)′ = 1. ◁ Then (︀ 2 )︀′ x = (x · x)′ = (x)′ · x + x · (x)′ = 1 · x + x · 1 = 2x 30 Another look on the derivative of a natural power of x ◁ Recall that (x)′ = 1. ◁ Then (︀ 2 )︀′ x = (x · x)′ = (x)′ · x + x · (x)′ = 1 · x + x · 1 = 2x ◁ Similarly, (︀ 3 )︀′ x = (x 2 · x)′ = (x 2 )′ · x + x 2 · (x)′ = 2x · x + x 2 · 1 = 3x 2 30 Another look on the derivative of a natural power of x ◁ Recall that (x)′ = 1. ◁ Then (︀ 2 )︀′ x = (x · x)′ = (x)′ · x + x · (x)′ = 1 · x + x · 1 = 2x ◁ Similarly, (︀ 3 )︀′ x = (x 2 · x)′ = (x 2 )′ · x + x 2 · (x)′ = 2x · x + x 2 · 1 = 3x 2 ◁ Inductively, (︀ k+1 )︀′ x = (x k · x)′ = (x k )′ · x + x k · (x)′ = kx k−1 · x + x k · 1 = (k + 1)x k 30 Another look on the derivative of a natural power of x ◁ Recall that (x)′ = 1. ◁ Then (︀ 2 )︀′ x = (x · x)′ = (x)′ · x + x · (x)′ = 1 · x + x · 1 = 2x ◁ Similarly, (︀ 3 )︀′ x = (x 2 · x)′ = (x 2 )′ · x + x 2 · (x)′ = 2x · x + x 2 · 1 = 3x 2 ◁ Inductively, (︀ k+1 )︀′ x = (x k · x)′ = (x k )′ · x + x k · (x)′ = kx k−1 · x + x k · 1 = (k + 1)x k Thus, we can use the Product rule (Theorem L4.6) to calculate the derivative of a natural power of x (Theorem L4.3) without using the Binomial formula. 30 Derivative of 1/x R Theorem L4.7. Consider the function f : (−∞, 0) ∪ (0, ∞) → given by 1 f (x) =. The function f is differentiable at all points of its domain, and x (︂ )︂′ 1 1 = − 2 x x 31 Derivative of 1/x Theorem L4.7. Consider the function f : (−∞, 0) ∪ (0, ∞) → given by R 1 f (x) =. The function f is differentiable at all points of its domain, and x (︂ )︂′ 1 1 = − 2 x x Proof. Consider arbitrary x ̸= 0, and arbitrary xn −→ x, xn ̸= x, 0. Then ⎛ 1 1 ⎞ ⎛ x − xn ⎞ (︂ )︂′ − 1 xn x xn · x = lim ⎝ ⎠ = lim ⎝ ⎠ = ⎜ ⎟ ⎜ ⎟ x n→∞ xn − x n→∞ xn − x (︂ )︂ 1 1 1 1 1 1 = − lim = − · = − · = − 2 n→∞ xn · x x lim xn x x x n→∞ 31 Outline ◁ Limits and infinite sums ◁ Continuity of a function ◁ Derivative of a function ◁ Composition of functions (next lecture’s material) 32 Composition of functions Definition. If f : D → R and g : R → S, then the composition of f and g is the function g ∘ f : D → S given by g ∘ f (x) = g (f (x)). 33 Composition of functions Definition. If f : D → R and g : R → S, then the composition of f and g is the function g ∘ f : D → S given by g ∘ f (x) = g (f (x)). Examples ◁ If f (x) = 3x + 1 and g (x) = x 2 , then g ∘ f (x) = g (3x + 1) = (3x + 1)2 = 9x 2 + 6x + 1 f ∘ g (x) = f (x 2 ) = 3(x 2 ) + 1 = 3x 2 + 1 33 Composition of functions Definition. If f : D → R and g : R → S, then the composition of f and g is the function g ∘ f : D → S given by g ∘ f (x) = g (f (x)). Examples ◁ If f (x) = 3x + 1 and g (x) = x 2 , then g ∘ f (x) = g (3x + 1) = (3x + 1)2 = 9x 2 + 6x + 1 f ∘ g (x) = f (x 2 ) = 3(x 2 ) + 1 = 3x 2 + 1 ◁ If f (x) = 4x 6 and g (x) = 7, then g ∘ f (x) = g (4x 6 ) = 7 f ∘ g (x) = f (7) = 4 · 76 = 470596 33 Composition of functions Definition. If f : D → R and g : R → S, then the composition of f and g is the function g ∘ f : D → S given by g ∘ f (x) = g (f (x)). Examples ◁ If f (x) = 3x + 1 and g (x) = x 2 , then g ∘ f (x) = g (3x + 1) = (3x + 1)2 = 9x 2 + 6x + 1 f ∘ g (x) = f (x 2 ) = 3(x 2 ) + 1 = 3x 2 + 1 ◁ If f (x) = 4x 6 and g (x) = 7, then g ∘ f (x) = g (4x 6 ) = 7 f ∘ g (x) = f (7) = 4 · 76 = 470596 ◁ If f (x) = e x and g (x) = −x 2 , then g ∘ f (x) = g (e x ) = − (e x )2 = −e 2x 2 f ∘ g (x) = f (−x 2 ) = e −x 33 Composition of functions—properties The composition operation is always associative: h ∘ (g ∘ f ) = (h ∘ g ) ∘ f for three functions f : X → Y , g : Y → Z , h : Z → W , since for any x ∈ X , the function h ∘ (g ∘ f ) : X → W , coincides with function (h ∘ g ) ∘ f : X → W h ∘ (g ∘ f )(x) = h(g (f (x))) = (h ∘ g ) ∘ f (x) Therefore, we can write the composition of several functions f , g , h simply as h∘g ∘f For example, if h(x) = |x|, g (x) = sin(x), f (x) = x 3 , then h ∘ g ∘ f (x) = h(g (f (x))) = h(g (x 3 )) = h(sin(x 3 )) = | sin(x 3 )| 34 Composition of functions—properties The composition operation is, in general, not commutative: g ∘ f ̸= f ∘ g 35 Composition of functions—properties The composition operation is, in general, not commutative: g ∘ f ̸= f ∘ g R For example, let f : → [0, ∞) is given by f (x) = e x , and let √ g : [0, ∞) → [0, ∞) is given by g (x) = x. Then the function g ∘ f : R → (0, ∞) is well-defined and given by √ 1 x g ∘ f (x) = g (f (x)) = g (e x ) = e x = (e x ) 2 = e 2 √ From the other hand, the function f ∘ g is defined only for x ≥ 0, since x is undefined for x < 0. 35 Composition of functions—properties The composition operation is, in general, not commutative: g ∘ f ̸= f ∘ g R For example, let f : → [0, ∞) is given by f (x) = e x , and let √ g : [0, ∞) → [0, ∞) is given by g (x) = x. Then the function g ∘ f : R → (0, ∞) is well-defined and given by √ 1 x g ∘ f (x) = g (f (x)) = g (e x ) = e x = (e x ) 2 = e 2 √ From the other hand, the function f ∘ g is defined only for x ≥ 0, since x is undefined for x < 0. Finally, even if the functions f ∘ g and g ∘ f are defined on the same domain, in general, f ∘ g (x) ̸= g ∘ f (x). In the example above, for x ∈ [0, ∞), we have √ √ x f (g (x)) = f ( x) = e x ̸= e 2 = g (f (x)) 35