MATH 120 Theorem List

Transcription

1 December 11, 2016 Disclaimer: Many of the theorems covere in class were not name, so most of the names on this sheet are not efinitive (they are escriptive names rather than given names). Lecture Theorems & Useful Definitions Least Upper Boun Property Let S R. If S has at least one upper boun, then S has a least upper boun. We efine c = sup S to be the least upper boun of S. Unique Upper Boun Theorem Let S R. If S has a least upper boun, then it has exactly one upper boun (this least upper boun is unique). Domain Convention If the omain of a function is not specifie, assume that it is the largest subset of the real numbers for which the function makes sense. One-To-One Functions A function f is calle one-to-one (or injective) if whenever x y, f(x) f(y), or, equivalently, if whenever f(x) = f(y), x = y. Arithmatic of Functions (f + g)(x) = f(x) + g(x) (f g)(x) = f(x) g(x) (f g)(x) = f(x) g(x) (f/g)(x) = f(x)/g(x) Composition of Functions f g(x) = f(g(x)) 1

2 Limits Let a R. Let L R. We say lim x a f(x) = L if: There exists a real number t > 0 such that {x R : 0 < x a < t} D(f). For all real numbers ɛ > 0, there exists a real number δ > 0 such that: For all x R, if 0 < x a < δ, then f(x) L < ɛ. Limit Properties If f(x) = k, then for every a R, lim x a f(x) = k. If f(x) = x, then for every a R, lim x a f(x) = f(a). If lim x a f(x) exists an is a real number, an lim x a g(x) exists an is a real number, then lim x a (f + g)(x) = lim x a f(x) + lim x a g(x). If lim x a f(x) exists an is a real number, an lim x a g(x) exists an is a real number, then lim x a (f g)(x) = lim x a f(x) lim x a g(x). If lim x a f(x) exists an is a real number, an lim x a g(x) exists an is a real number, then lim x a (f g)(x) = (lim x a f(x))(lim x a g(x)). If lim x a f(x) exists an is a real number, an lim x a g(x) exists an is a real number not equal to zero, then lim x a (f/g)(x) = limx a f(x) lim x a g(x)). Limits are a Local Property Law Let f, g be functions. Suppose a, L R. Suppose there exists some t > 0 such that for all x with 0 < x a < t, we have g(x) = f(x). Then lim x a g(x) = L. Squeeze Theorem Let f, g, h be functions. Let a R. Suppose there exists t > 0 such that for all x satisfying 0 < x a < t, we have f(x) g(x) h(x). Suppose also that lim x a f(x) = lim x a h(x) = L. Then, lim x a g(x) = L. One-Sie Limits Let f be a function. Let a, L R. We say lim x a + f(x) = L if: There exists t > 0 such that (a, a + t) = x R : 0 < x a < t D(f). For all ɛ > 0, there exists δ > 0 such that if 0 < x a < δ, then f(x) L < ɛ. 2

3 We say lim x a f(x) = L if: There exists t > 0 such that (a t, a) = x R : 0 < a x < t D(f). For all ɛ > 0, there exists δ > 0 such that if 0 < a x < δ, then f(x) L < ɛ. One-Sie Limit Theorem If f is a function, an a, L R, then lim x a f(x) = L if an only if lim x a + f(x) = L an lim x a f(x) = L. Limts at Infinity Let f be a function. Let L R. We say that lim x f(x) = L if: There exists a number t R such that (t, ) = x R : x > t D(f). For all ɛ > 0, there exists a real number R such that if x > R, then f(x) L < ɛ. We say that lim x f(x) = L if: There exists a number t R such that (, t) = x R : x < t D(f). For all ɛ > 0, there exists a real number R such that if x < R, then f(x) L < ɛ. Infinite Limite of Sine For every L R, the statement lim x f(x) = L is false. Rational Function Degree Theorems be a rational function. f(x) = n k=0 a kx k g(x) = n k=0 b kx k f(x) If eg(g) > eg(f), then lim x R(x) = lim x g(x) = 0. f(x) If eg(g) < eg(f), then the statement lim x R(x) = lim x g(x) = L is false for all L R. Let R(x) = f(x) g(x) If eg(g) = eg(f), then lim x R(x) = lim x f(x) g(x) = an b n. Triangle Inequalities n k=0 c k n k=0 c k n k=0 c k c 0 + n k=1 c k 3

4 Continuity Let f be a function. Let a R. We say that f is continuous at a if: a D(f) lim x a f(x) = f(a) Intermeiate Value Theorem Let f be a function, an suppose that [a, b] D(f). Suppose that f is continuous on [a, b]. If f(a) < 0 an f(b) > 0, then there exists c (a, b) such that f(c) = 0. Extreme Value Theorem Let f be continuous on the finite close interval [a, b]. Then there exists c, [a, b] such that: f(c) f(x) for all x [a, b]. f() f(x) for all x [a, b]. Differentiability Let f be a function. Let (a, b) be an interval such that (a, b) D(f). Let c (a, b). f(x) f(c) We say that f is ifferentiable at c if lim x c x c exists an is a real number, an we call this number f (c). f(c+h) f(c) Alternatively, we say that f is ifferentiable at c if lim h 0 h exists an is a real number, an we call this number f (c). Linearization Let f be a function which is ifferentiable at c. Define T (x) = f(c) + (x c)f (c) to be the linearization of f at c. Linearization Approximation Theorem Let f be a function which is ifferentiable at c. Let T (x) be the linearization of f at c. Then lim x c f(x) T (x) x c = 0. One-Sie Derivatives If lim f(x+h) f(x) h 0 + h If lim f(x+h) f(x) h 0 h exists, we say that f has a right erivative at x, an we write it as f +(x). exists, we say that f has a left erivative at x, an we write it as f (x). 4

5 Differentiability Rules Let f an g be functions that are ifferentiable at a point x. If f(y) = k for all y R, then f (x) = 0. If f(y) = y for all y R, then f (x) = 1. (f + g)(x) = f (x) + g (x) (f g)(x) = f (x) g (x) (f g)(x) = f (x)g(x) + f(x)g (x) ( 1 f (x) = f (x) (f(x)) 2 If n Z, an f(y) = y n, then f (y) = ny n 1. If g(y) 0, then f g (x) = g(x)f (x) f(x)g (x). (g(x)) 2 Chain Rule Let f an g be functions. If g is ifferentiable at c, an if f is ifferentiable at g(c), then f g is ifferentiable at c, an (f g) (c) = f (g(c))g (c) Twinkle Twinkle Little f Theorem Let (a, b) be an interval. Let c (a, b). Let f be a function such that f is ifferentiable at c. (a, b) D(f). Then there exists a function f such that: f is continuous at c f (c) = f (c) f(x) f(c) = (x c)f (x) for all x (a, b) Positive Derivitive Implies Increasing Let (a, b) be an interval. Let c (a, b). Let f be a function such that (a, b) D(f). Suppose f is ifferentiable at c, an f > 0, then there exists a t > 0 such that: f(x) > f(c) if c < x < c + t f(x) < f(c) if c t < x < c Negative Derivative Implies Decreasing Let (a, b) be an interval. Let c (a, b). Let f be a function such that (a, b) D(f). Suppose f is ifferentiable at c, an f < 0, then there exists a t > 0 such that: f(x) < f(c) if c < x < c + t f(x) > f(c) if c t < x < c 5

6 Local Mininimums an Maximums Let f be a function. Let c D(f). We say that c is a local maximum of f if there exists some t > 0 such that for all x D(f), if x c < t, then f(x) f(c). We say that c is a local minimum of f if there exists some t > 0 such that for all x D(f), if x c < t, then f(x) f(c). Global an Local Maximums an Minimums Let f be a function. Let c D(f). If c is a global minimum for f, then c is a local minimum for f. If c is a global maximum for f, then c is a local maximum for f. Derivatives of Local Maximums an Local Minimums Let (a, b) be an interval. Let c (a, b). Let f be a function such that f is ifferentiable at c. Suppose (a, b) D(f). If c is a local maximum or local minimum of f, then f (c) = 0. Newton s Metho Newton s Metho is an applie technique to approximate a solution, x, to satisfy f(x) = 0, where f is function which is ifferentiable for some interval. Start by choosing an x 1, so n initial = 1. Then, for any value n, compute x n+1 : x n+1 = x n f(xn) f (x n) If the initial chosen value x 1 is appropriate, then as n increases, x n will approach the root of f(x). Rolle s Theorem Let [a, b] be an interval. Let f be a function continuous on [a, b] an ifferentiable on (a, b). If f(a) = f(b), then there exists c (a, b) such that f (c) = 0 Mean Value Theorem Let [a, b] be an interval. Let f be a function continuous on [a, b] an ifferentiable on (a, b). Then there exists c (a, b) such that f(b) f(a) = f (c)(b a) Generalize Mean Value Theorem Let [a, b] be an interval. Let f an g be functions continuous on [a, b] an ifferentiable on (a, b). Then there exists c (a, b) such that f (c)(g(b) g(a) = g (c)(f(b) f(a). 6

7 Higher Derivatives Definition: The n t h erivative of f is equal to f (n). ie. f = f (3) Taylor s Theorem Let [a, b] be an interval. let f be a function that is n times ifferentiable on [a, b]. Let c (a, b). Then for every x [a, b], x c, there exists x 1 between x an c such that: n 1 f(x) = f(c) + k=1 f (k) (c)(x c) k k! + f (n) (x 1 )(x c) n n! Collary of Taylor s Theorem Let [a, b] be an interval, an let f be a function that is n times ifferentiable on [a, b], with f (n) continuous on [a, b]. Let c (a, b). Suppose that for all x [a, b], f (n) (x) M. Then for all x [a, b], f(x) n 1 f (k) c k=0 We can also say that f(x) n 1 k=0 k! (x c) k M n! x c n. f (k) c k! (x c) k = O( x c n ) as x c. Lanau s Big O Notation Let f an g be functions, an let c R. We say that f(x) = O(g(x)) as x c if there exists M > 0 an δ > 0 such that if x c < δ, then f(x) M g(x). We say that f(x) = O(g(x)) as x if there exists M > 0 an R R such that if x > R, then f(x) < M g(x). Definition of Smooth A function is smooth, or infinitely ifferentiable on an interval I if an only if, for all n N, f is n times ifferentiable on I. Derivative of Inverse Functions f 1 (x) = 1 f (f 1 (x)) Inverse Function Theorem If f is ifferentiable on D(f), an D(f) is an interval, an f (x) > 0 for all x D(f) or f (x) < 0 for all x D(f), then f 1 exists an is ifferentiable on Range(f). 7

8 Exponents an Logarithms E(x) is the inverse function of log x. Useful Equalities: E(0) = 1 E(1) = e log 1 = 0 log e = 1 E(x + y) = E(x)E(y) log xy = log x + log y Stirling s Approximation to n! n! is close to (n/e) n 2πn. Hyperbolic Functions sinh x = ex e x 2 cosh x = ex +e x 2 (sinh x) = cosh x x x (cosh x) = sinh x Derivatives of Triginometric Functions x x (sin x) = cos x (cos x) = sin x x (tan x) = sec2 x 1 (arctan x) = 1+x 2 1 (arcsin x) = x x x (arccos x) = 1 1 x 2 1 x 2 Implicit Differentiation Treat y as a function of x. L Hopital s Rule for 0 0 Let (a, b) be an interval. Let f an g be functions ifferentiable on (a, b). Suppose that lim x a + f(x) = 0, lim x a + g(x) = 0, an g (x) 0 for all x (a, b). Then, if lim f (x) x a + g (x) = L, lim x a f(x) + g(x) = L. 8

9 Similarly, Suppose that lim x a f(x) = 0, lim x a g(x) = 0, an g (x) 0 for all x (a, b). Then, if lim f (x) x a g (x) = L, lim x a f(x) g(x) = L. L Hopital s Rule for 0 0 with Limits to Infinity Let f an g be functions. Suppose that lim x f(x) = 0 an lim x g(x) = 0. Suppose also that there exists R R such that f an g are ifferentiable on (R, ), an g (x) 0 for all x (R, ). f Then, for all x (R, ), if lim (x) x g (x) = L, then lim x f(x) g(x) = L. L Hopital s Rule for Let f an g be functions efine on (a, b). Suppose that f an g are ifferentiable on (a, b). Suppose that g (x) 0 for any x (a, b). Suppose that lim x a + f(x) = or lim x a + f(x) =. Suppose that lim x a + g(x) = or lim x a + g(x) =. Then, if lim f (x) x a + g (x) = L, then lim x a f(x) + g(x) = L. The two-sie version of this theorem is also true (see theorem for 0 0 ). Continous Differentiability Let (a, b) be an interval. Let c (a, b). Let f be a function that is continuous on (a, b), an ifferentiable on every points x (a, b) except for possibly at x = c. Suppose that lim x c f (x) = L for some L R. Then, f is ifferentiable at c, an f (c) = L. Anti-Derivatives Let (a, b) be an interval that may be finite or infinite. Let f be a function efine on (a, b). We call a function F an anti-erivative of f on (a, b) if for all x (a, b), F (x) = f(x). Anti-Derivatives of the Same Function on the Same Interval If F (x) is an anti-erivative of f(x) on (a, b), then so is F (x) + c for any constant c. Conversely, if F 1 (x) an F 2 (x) are both anti-erivatives of f on (a, b), then F 1 (x) F 2 (x) is a constant function. 9

10 Common/Useful Anti-Derivatives On (0, ), the anti-erivative of 1 x is log x. On R: The anti-erivative of e x is e x. The anti-erivative of e cx is 1 c ecx. The anti-erivative of sin x is cos x. The anti-erivative of cos x is sin x. The anti-erivative of P (x) = n k=0 a kx k is n 1 k=0 k+1 a kx k+1. Anti-Derivative Notation If F is the anti-erivative of f, then we write f(x)x = F (x) + c, an F (x) is also calle the inefinite integral of f. First Orer Differential Equations A first orer ifferential equation is an equation of the form y (x) = f(x, y(x)). Here, y(x) is a function of one variable, an f(x, y) is a function of two variables. We can also write this as y = f(x, y). In practice, we are given the function f, an we nee to fin y. Initial Value Problems An initial value problem is a problem in the form: y (x) = f(x, y(x) y (x 0 ) = c 0 f(x, y) is a function we are given, x 0 an c 0 are given real numbers, an y(x) is a function for which to be solve. If f is continuous an the erivative of f exists an is continuous, then we can always solve an initial value problem an there is only one solution (Recall: f is a function of two variables, so these terms have not been efine). First Orer Linear Differential Equations If P (x) an Q(x) are functions that are continuous on the interval (a, b), then a ifferential equation of the form y (x) + P (x)y(x) = Q(x) is calle a first orer linear ifferential equation. For a first orer linear ifferential equation, f(x, y(x)) is of the form f(x, y) = Q(x) P (x). First Orer Linear Initial Value Problems If P (x) an Q(x) are functions that are continuous on (a, b), then: 10

11 y (x) + P (x)y(x) = Q(x) y(x 0 ) = c 0 is calle a first orer linear initial value problem. This initial value problem has a unique solution y(x) = c 0 e A(x) + e A(x) B(x), where A(x) is the anti-erivative of P (x) with A(x 0 ) = 0, an B(x) is the anti-erivative of e A(x) Q(x), with B(x 0 ) = 0. Homogeneous First Orer Linear Differential Equations If P (x) is a continuous (a, b), then an equation of the form y (x) + P (x)y(x) = 0 is a homogeneous first orer linear ifferential equation. If the solution y(x) is never equal to zero, then we can ivie by y(x) to obtain y (x) = P (x) (i.e. x log y(x) = P (x)). Also, if P (x) is continuous, then P (x) has an anti-erivative (to be prove in MATH 121). y(x) Homogeneous First Orer Linear Initial Value Problems If P (x) is continuous on the interval (a, b), then the initial value problem: y (x) + P (x)y(x) = 0 y(x 0 ) = c 0 has a unique solution, given by y(x) = c 0 e A(x), where A(x) is the unique anti-erivative of P (x) that satisfies A(x 0 ) = 0. We write A(x) = x x 0 P (z)z. Maximum Intervals of Existence The maximum interval of existence for an initial value problem is the largest open interval containing x 0 for which there exists a solution to the above ifferential equation. Secon Orer Differential Equations A secon orer ifferential equation is an equation of the form y (x) = f(x, y(x), y (x)). A linear secon orer ifferential equation is an equation of the form y (x)+p (x)y (x)+q(x)y(x) = R(x). If R(x) = 0, then this is calle a homogeneous equation. If P (x) an Q(x) are also constants, i.e. P (x) = p an Q(x) = q, then we get y (x)+py (x)+qy(x) = 0, where p, q R. This is calle a homogeneous linear secon orer ifferential equation with constant coeffecients. 11

12 Secon Orer Initial Value Problems If f(x, y, z) is continuous, an the erivative of f with respect to y is continuous, an the erivative with respect to z is continuous, then the initial value problem y (x) = f(x, y(x), y (x)) y(x 0 ) = c 0 y (x 0 ) = c 1 has a unique solution on some maximum interval of existence that contains x 0. Multiple Solutions to Homogeneous Linear Secon Orer Differential Equation with Constant Coeffecients For a homogeneous linear secon orer ifferential equation with constant coeffecients, if y 1 (x) an y 2 (x) are solutions, then y(x) = y 1 (x) + y 2 (x) is also a solution. Also, if y(x) is a solution, then Ay(x) is a solution for all A R. Imaginary Numbers i is efine to be a solution to the equation x = 0, i.e. i 2 = 1 (an ( i) 2 = 1). Any number of the form bi, where b R, is calle an imaginary number. If a, b R, then any number of the form a + bi is calle a complex number. If z = a + bi is a complex number, we efine Re(z) = a an Im(z) = b, which are the real an imaginary parts of z. If z an w are complex numbers, then z w = z w, so i i = i 2 = 1 = 1, so it must be the case that i = 1. Euler s Formula If x R, then e ix = cos(x) + i sin(x). This is a more general form of Euler s ientity (e iπ + 1 = 0). Homework Theorems & Useful Definitions Set Theorems (Homework #1) Suppose that S T =. Then, S \ T = S an T \ S = T. Suppose that S T. Then, S \ T =. Suppose that S an T are finite sets (so S, T R). Then, S T = S + T S T. Suppose that S R. If x R is an upper boun for S, then x + 1 is also an upper boun for S. 12

13 Unusual Function #1 (Homework #2) { 1 x Q f(x) = 0 x R \ Q For all a R an for all L R, the statement lim x a f(x) = L is false. Unusual Function #2 (Homework #2) { x 3 x Q g(x) = 0 x R \ Q lim x 0 g(x) = 0 an for all a R \ {0} an L R, the statement lim x a g(x) = L is false. Limit Transitivity (Homework #2) Let f an g be functions, an let a R. Suppose that lim x a g(x) = L an that lim x L f(x) = M an finally that f(l) = M. Then, lim x a f g(x) = M. One-Sie Limits an Limits at Infinity (Homework #3) Let g(x) = 1 x. Let f(x) be a function an suppose that (0, ) D(f). Then: If lim x 0 + f(x) = L, then lim x f g(x) = L. If lim x f g(x) = L, then lim x 0 + f(x) = L. Prouct Rule for one-sie limits (Homework #3) Let f an g be functions. Let a R. Suppose lim x a + f(x) = L 1 an lim x a + g(x) = L 2. Then, lim x a +(fg)(x) = L 1 L 2. Rational Continuity Equivalency (Homework #3) Let f an g be functions with D(f) = D(g) = R, an suppose that f an g are continuous. Suppose that for every x Q, f(x) = g(x). Then, for every x R, f(x) = g(x). 13

14 Unusual Function #3 (Homework #3) { 1 x = q f(x) = p 2 p in lowest form 0 x R \ Q For all a Q, f is not continuous at a. For all a R \ Q, f is continuous at a. Arithmetic of Functions, Continuity, an Differentiability (Homework #4) Let f an g be functions. If f is continuous at c, an g is not continuous at c, then f + g is not continous at c. If f is ifferentiable at c, an g is not ifferentiable at c, then f + g is not ifferentiable at c. Special Polynomials (Homework #4) Let x 1. Then for all n N, n k=0 xk = xn+1 1 x 1, an n k=1 nxn 1 = nxn 1 (n+1)x n +1 (x 1) 2. Ientity Point (Homework #5) Let f be a function that is continuous on the interval [0, 1], an suppose that for all x [0, 1], 0 f(x) 1. Then, there must be at least one point c [0, 1] with f(x) = c. Tangent/Secant Relation (Homework #5) Let f be a function that is continuous on the interval [a, b] an ifferentiable on the interval (a, b). Suppose that there exists M > 0 so that for each x (a, b), we have f (x) M. Then, f(b) f(a) M b a. Big O notation an Quotient Limits (Homework #5) Let f an g be functions such that D(f) = D(g) = R. Suppose f = O(g) as x. g(x) Then, the statement lim x f(x) = 0 is false. Big O Transitivity (Homework #6) Let f, g, an h be functions. Suppose that f(x) = O(g(x)) as x c an g(x) = O(h(x)) as x c. Then, f(x) = O(h(x)) as x c. 14

15 Polynomial Multiplication (Homework #6) If P (x) an Q(x) are polynomials, then their prouct P (x)q(x) is also a polynomial. Unusual Function #4 (Homework #7) { e 1/x x > 0 g(x) = 0 x 0 g is infinitely ifferentiable on R. Injectivity an Derivatives (Homework #7) Let f be a function with D(f) = R such that f(x) is ifferentiable on all x R. Suppose there exists a, b R such that f (a) > 0 an f (b) < 0. Then, f is not one-to-one. Hyperbolic Sine an Cosine of Sums (Homework #8) sinh(x + y) = sinh(x) cosh(y) + cosh(x) sinh(y) cosh(x + y) = cosh(x) cosh(y) = sinh(x) sinh(y) Quotients of Polynomials (Homework #9) Let f an g be polynomials with non-zero leaing coeffecients a 0 an b 0, respectively. If eg(f) = eg(g), then lim x f(x) g(x) = a b. If eg(f) < eg(g), then lim x f(x) g(x) = 0. 15