The Mean Value Theorem Math 120 Calculus I
... It’s easier to prove Rolle’s theorem because it’s a special case, but the whole MVT can then, in turn, be proved from Rolle’s theorem. Rolle’s theorem says if a function is continuous on a closed interval and differentiable on its interior, and it takes the same value at both endpoints of the interv ...
... It’s easier to prove Rolle’s theorem because it’s a special case, but the whole MVT can then, in turn, be proved from Rolle’s theorem. Rolle’s theorem says if a function is continuous on a closed interval and differentiable on its interior, and it takes the same value at both endpoints of the interv ...
LINEAR DYNAMICS 1. Topological Transitivity and Hypercyclicity
... Ta . Let U, V be two non-empty open subsets of O(C). One may find > 0, and f, g ∈ O(C) such that {h ∈ O(C) : d(f, h) < 3} ⊆ U, {h ∈ O(C) : d(g, h) < 3} ⊆ V, where d is as given in (1.1). Now it is easy to see that there exists a closed disc K ⊂ C such that {h ∈ O(C) : kh − f k∞,K < } ⊆ U, {h ∈ ...
... Ta . Let U, V be two non-empty open subsets of O(C). One may find > 0, and f, g ∈ O(C) such that {h ∈ O(C) : d(f, h) < 3} ⊆ U, {h ∈ O(C) : d(g, h) < 3} ⊆ V, where d is as given in (1.1). Now it is easy to see that there exists a closed disc K ⊂ C such that {h ∈ O(C) : kh − f k∞,K < } ⊆ U, {h ∈ ...
Lecture #33, 34: The Characteristic Function for a Diffusion
... (Af )(t, x) = 0. The Feynman-Kac representation theorem extends this idea by providing an explicit formula for the solution of this partial di↵erential equation subject to certain boundary conditions. Theorem 21.3 (Feynman-Kac Representation Theorem). Suppose that u 2 C 2 (R), and let {Xt , t 0} be ...
... (Af )(t, x) = 0. The Feynman-Kac representation theorem extends this idea by providing an explicit formula for the solution of this partial di↵erential equation subject to certain boundary conditions. Theorem 21.3 (Feynman-Kac Representation Theorem). Suppose that u 2 C 2 (R), and let {Xt , t 0} be ...
a review sheet for test #02
... The derivative of the function f(x) is defined as f x h f x f ' x lim ...
... The derivative of the function f(x) is defined as f x h f x f ' x lim ...
TUTORIAL 6 SOLUTIONS (1) Consider the surface S given by the
... on [−1, 1]. The critical points of g occur when g 0 (y) = cos y = 0, which doesn’t happen on the interval [−1, 1]. Thus, the extrema on this boundary piece occur at the endpoints. Hence, we compute g(−1) = sin(−1) = − sin(1) and g(1) = sin(1), and see that on B1 , the minimum value of f is − sin(1) ...
... on [−1, 1]. The critical points of g occur when g 0 (y) = cos y = 0, which doesn’t happen on the interval [−1, 1]. Thus, the extrema on this boundary piece occur at the endpoints. Hence, we compute g(−1) = sin(−1) = − sin(1) and g(1) = sin(1), and see that on B1 , the minimum value of f is − sin(1) ...
ECO4112F Section 4 Integration
... Property 3 tells us that a constant (but only a constant) can be taken in front of an integral sign. This also follows from the properties of limits and sums. Property 4 follows from property 2 and 3, using c=-1. Property 5 tells us we can find the area under the graph between a and c, by splitting ...
... Property 3 tells us that a constant (but only a constant) can be taken in front of an integral sign. This also follows from the properties of limits and sums. Property 4 follows from property 2 and 3, using c=-1. Property 5 tells us we can find the area under the graph between a and c, by splitting ...
Lecture notes 2.26.14
... dividing by λ > 0 and letting λ → 0+ , we obtain 0 ≤ ∇f (x̄)T d. Similarly, dividing by λ < 0 and letting λ → 0− , we obtain 0 ≥ ∇f (x̄)T d. As a consequence, f (x̄)T d = 0 for all d ∈ Rn . This implies that ∇f (x̄) = 0. The case where x̄ is a local maximum can be dealt with similarly. Remark 4. A p ...
... dividing by λ > 0 and letting λ → 0+ , we obtain 0 ≤ ∇f (x̄)T d. Similarly, dividing by λ < 0 and letting λ → 0− , we obtain 0 ≥ ∇f (x̄)T d. As a consequence, f (x̄)T d = 0 for all d ∈ Rn . This implies that ∇f (x̄) = 0. The case where x̄ is a local maximum can be dealt with similarly. Remark 4. A p ...
MATH141 – Tutorial 2
... Solution. We know that −1 ≤ x ≤ 1, since this is the domain on which we integrate. Squaring both sides gives 0 ≤ x2 ≤ 1, we get 0 ≤ x2 ≤ 1 ...
... Solution. We know that −1 ≤ x ≤ 1, since this is the domain on which we integrate. Squaring both sides gives 0 ≤ x2 ≤ 1, we get 0 ≤ x2 ≤ 1 ...
aCalc02_3 CPS
... B. f(2) must be negative C. f(x)=0 must have a solution between 3 and 4 D. f(x) must be a linear equation with a slope of 6 E. None of the above ...
... B. f(2) must be negative C. f(x)=0 must have a solution between 3 and 4 D. f(x) must be a linear equation with a slope of 6 E. None of the above ...
Homework 8 - UC Davis Mathematics
... (a) Determine what well known function is obtained from sinh(x) + cosh(x). (b) Determine what well known function is obtained from cosh(x) − sinh(x). (c) Compute the derivatives of both sinh(x) and cosh(x). (d) What is the difference between the relationship of the derivatives of the hyperbolic trig ...
... (a) Determine what well known function is obtained from sinh(x) + cosh(x). (b) Determine what well known function is obtained from cosh(x) − sinh(x). (c) Compute the derivatives of both sinh(x) and cosh(x). (d) What is the difference between the relationship of the derivatives of the hyperbolic trig ...
CSE 232A Midterm Winter 2007
... Problem 2 (10 points) Given relation R(A, B, C), express the fact that A, B is a key on R using an algebraic dependency of the kind E1 ⊆ E 2 where E1 , E2 are valid RA expressions. That is, denoting with E(D) the result of applying query E on database D, find E1 and E2 such that for every database ...
... Problem 2 (10 points) Given relation R(A, B, C), express the fact that A, B is a key on R using an algebraic dependency of the kind E1 ⊆ E 2 where E1 , E2 are valid RA expressions. That is, denoting with E(D) the result of applying query E on database D, find E1 and E2 such that for every database ...
Lecture 17 - University of Chicago Math
... Statement of Second Fundamental Theorem of Calculus Theorem: Suppose f (x) is integrable on [a, b], and that F (x) is an antiderivative of f . Then Z b f (t) dt = F (b) − F (a) a ...
... Statement of Second Fundamental Theorem of Calculus Theorem: Suppose f (x) is integrable on [a, b], and that F (x) is an antiderivative of f . Then Z b f (t) dt = F (b) − F (a) a ...
Problem Set 3 Partial Solutions
... Solution: For this problem we need to recall the differentiation formula: ...
... Solution: For this problem we need to recall the differentiation formula: ...
Ken`s Cheat Sheet 2014 Version 11 by 17
... Instantaneous Rate of Change = Slope of a Tangent Line ...
... Instantaneous Rate of Change = Slope of a Tangent Line ...
Lecture 18: More continuity Let us begin with some examples
... 1. Not every continuous function on a non compact set is uniformly continuous. If E is any non-closed subset of R then there exists a continuous function on E that is both unbounded and not uniformly continuous. Take x0 to be any limit point of E that is not in E. Then f (x) = (x − x0 )−1 is continu ...
... 1. Not every continuous function on a non compact set is uniformly continuous. If E is any non-closed subset of R then there exists a continuous function on E that is both unbounded and not uniformly continuous. Take x0 to be any limit point of E that is not in E. Then f (x) = (x − x0 )−1 is continu ...
Norm and Derivatives
... of Hadamad differentiable function at x DG (x) set Frechet differentiable function x. In application to find Frechet or Hadamard derivative generally we shout try first to determine the form of derivative deducing Gateaux derivative acting on h,df(h) for a collection of directions h which span B ...
... of Hadamad differentiable function at x DG (x) set Frechet differentiable function x. In application to find Frechet or Hadamard derivative generally we shout try first to determine the form of derivative deducing Gateaux derivative acting on h,df(h) for a collection of directions h which span B ...
answers, in pdf - People @ EECS at UC Berkeley
... Note that this rule doesn’t depend on whether f (x) and/or g(x) are positive or negative. It is valid so long as the graph of f (x) lies above the graph of g(x) for all x from x = 2 to x = 3. • 6.4 #8 Find the area of the region. Solution Note that 2 ≥Rx(2 − x) for 0 ≤ x ≤ R 22. So the area is ...
... Note that this rule doesn’t depend on whether f (x) and/or g(x) are positive or negative. It is valid so long as the graph of f (x) lies above the graph of g(x) for all x from x = 2 to x = 3. • 6.4 #8 Find the area of the region. Solution Note that 2 ≥Rx(2 − x) for 0 ≤ x ≤ R 22. So the area is ...
Final Exam topics - University of Arizona Math
... Let f be a function which is defined on some open interval containing “a” except possibly at x = a. We write if f(x) grows arbitrarily large by choosing x sufficiently close to “a”. Intermediate Value Theorem: Let f be a function which is continuous on the closed interval [a, b]. Suppose that d is a ...
... Let f be a function which is defined on some open interval containing “a” except possibly at x = a. We write if f(x) grows arbitrarily large by choosing x sufficiently close to “a”. Intermediate Value Theorem: Let f be a function which is continuous on the closed interval [a, b]. Suppose that d is a ...
Math 171 Final Exam Review: Things to Know
... − Advice: Since the second derivative test is much more efficient than the first derivative test, you should always attempt the second derivative test for determining local extrema at a critical point. If the second derivative at the critical point is zero, the test is inconclusive so you must use the ...
... − Advice: Since the second derivative test is much more efficient than the first derivative test, you should always attempt the second derivative test for determining local extrema at a critical point. If the second derivative at the critical point is zero, the test is inconclusive so you must use the ...
STABILITY OF ANALYTIC OPERATOR
... In 1984, Johnson proved a bounded convergence theorem for the operator-valued function space integral [7]. This is the rst stability theorem for the integral as a bounded linear operator on L2 (Rn ) where n is any positive integer. In [10], Johnson and Skoug introduced stability theorems for the in ...
... In 1984, Johnson proved a bounded convergence theorem for the operator-valued function space integral [7]. This is the rst stability theorem for the integral as a bounded linear operator on L2 (Rn ) where n is any positive integer. In [10], Johnson and Skoug introduced stability theorems for the in ...
Operators on normed spaces
... In this chapter we investigate continuous functions from one normed space to another. The class of all such functions is so large that any attempt to understand their properties will fail, so we will focus on those continuous functions that interact with the vector space structure in a meaningful wa ...
... In this chapter we investigate continuous functions from one normed space to another. The class of all such functions is so large that any attempt to understand their properties will fail, so we will focus on those continuous functions that interact with the vector space structure in a meaningful wa ...
Direct Variation
... You can say that y varies directly as x A direct variation must also pass through the origin ie) if x=0 then y=0 y y y y k 1 and k 2 1 2 x1 x2 x1 x2 k is also called the constant of proportionality, and y is said to be directly proportional to x y1 y2 ...
... You can say that y varies directly as x A direct variation must also pass through the origin ie) if x=0 then y=0 y y y y k 1 and k 2 1 2 x1 x2 x1 x2 k is also called the constant of proportionality, and y is said to be directly proportional to x y1 y2 ...
Geodesic ray transforms and tensor tomography
... Problem: if D A is gauge equivalent to D 0 , then FA = F0 = 0. Need a generalized gauge transformation that arranges a sign for FA . This breaks the m-tensor structure of the equation, but is manageable if the gauge transform is holomorphic. Fourier analysis in θ (Guillemin-Kazhdan 1978): ...
... Problem: if D A is gauge equivalent to D 0 , then FA = F0 = 0. Need a generalized gauge transformation that arranges a sign for FA . This breaks the m-tensor structure of the equation, but is manageable if the gauge transform is holomorphic. Fourier analysis in θ (Guillemin-Kazhdan 1978): ...
4.4 - korpisworld
... The value of f (c) has a very special meaning. It is called the average value of f on the interval [a, b] . This means, if you were to add up every single y-value along the curve from a to b (which is impossible to do, since there are an infinite number of them) and divide the number of y-values you ...
... The value of f (c) has a very special meaning. It is called the average value of f on the interval [a, b] . This means, if you were to add up every single y-value along the curve from a to b (which is impossible to do, since there are an infinite number of them) and divide the number of y-values you ...