Let A be a subset of the integers. (a) Write a careful definition (using quantifiers) for the term smallest element of A. (b) Let E be the set of even integers; that is E = {x ∈ Z : 2|x}. Prove by contradiction that E has no smallest element. (c) Prove that if A ⊆ Z has a smallest element, then it must be unique.

I need to solve math problem by using freemat or mathlab program. However, I can't understand how to make a code it.
This is one example for code ( freemat )

format long

f = @(x) (x^2) % define f(x) = x^2
a=0
b=1
N=23
dx = (b-a)/N % dx = delta_x

1. Methods:

(a) Left Rectangular Rule also known as Lower sum.

(b) Right Rectangular Rule also know as upper sum.

(c) Midpoint Rule

(d) Trapezoid Rule

(e) Simpson Rule

2. Evaluate

Integral {4/(x^2 + 1)} (Upper bound 1 Lower bound 0)

Calculate the indicated similarity or distance measures of the vectors show below (you can do this by hand or by writing code, but please show the computation either way):

x = (1,1,1,1) and y = (3,3,3,3) cosine, correlation, Euclidean

x = (0,1,0,1,0,1) and y = (1,0,1,0,1,0) cosine, correlation, Euclidean, Jaccard

x = (1,1,0,1,0,1) and y = (1,1,1,0,0,1) cosine, correlation, Lmax, Jaccard

1. Why guessing and checking is alright in solving differential equations In lecture (and possibly in other courses), you have seen differential equations solved by looking at the equation, moving parts around, reasoning about it using an analogy with eigenvalue/eigenspaces, and then seeing that the solution that we proposed actually works — i.e. satisfies all the conditions of the differential equation problem. This process should have felt a bit different than how you have seen how systems of linear equations are solved (by doing Gaussian Elimination) where it was clear that every step was valid. Indeed, it is different. Although the eigenvalue/eigenspace analogy to differential equations can be made precise and rigorous, doing that carefully is beyond the scope of this course. In effect, all of that reasoning in between seeing the problem and checking the solution can be considered a kind of inspired guessing. This should lead you to a natural question — how can we be sure that we have found all of the solutions? We’ve checked to see that the solution we found solves the equations, but maybe there are more solutions that are different. How can we be sure? After all, we are using the solution of the differential equation for its predictive power — for example, we are using the fact of RC time constants to argue that this limits the speed of digital computation. Making such inferences is only proper if we have indeed found the only solution to the differential equation. In the mathematical literature, this is sometimes referred to as the problem of establishing the “uniqueness” of solutions. The concept is also very important for us in engineering contexts. You have already seen in EE16A’s touchscreen module that node voltages need not be unique, and that is why you need to specify a ground in your circuit. You also saw this concept in EE16A’s localization module where you learned how to approach inconsistent linear equations by the method of least squares: you started with no solutions, allowed some error and then got infinitely many potential solutions with error. To make the solution unique, you had to specify that you wanted to minimize the size of the hypothesized error. This problem walks you through an elementary proof of the uniqueness of solutions to a simple scalar differential equation of the form

d/dt x(t) = αx(t) (1) with initial condition x(0) = x0  (2)

Being able to do simple proofs is an important skill, not only in its own right, but also for the systematic logical thinking that it exercises. This problem has multiple parts, but the goal is simply to help you see how you could have come up with this proof entirely on your own.

(a) Please verify that the guessed solution xd(t) = x0e αt satisfies (1) and (2).

(b) To show that this solution is in fact unique, we need to consider a hypothetical y(t) that also satisfies (1) and (2)

Our goal is to show that y(t) = x(t) for all t ≥ 0. (The domain t ≥ 0 is where we have defined the conditions (1) and (2). Outside of that domain, we don’t have any constraints. ) How can we show that two things are equal? In the past, you have probably shown that two quantities or functions are equal by starting with one of them, and then manipulating the expression for it using valid substitutions and simplifications until you get the expression for the other one. However, here, we don’t have an expression for y(t) so that style of approach won’t work. In such cases, we basically have a couple of basic ways of showing that two things are the same. • Take the difference of them, and somehow argue that it is 0. • Take the ratio of them, and somehow argue that it is 1. We will follow the ratio approach in this problem. First assume that x0 6= 0. In this case, we are free to define z(t) = y(t) xd (t) since we are dividing by something other than zero. What is z(0)?

(c) Take the derivative d dtz(t) and simplify using (1) and what you know about the derivative of xd(t). (HINT: The quotient rule for differentiation might be helpful since a ratio is involved.) You should see that this derivative is always 0 and hence z(t) does not change. What does that imply for y and xd ?

(d) At this point, we have shown uniqueness in most cases. Just one special case is left: x0 = 0. Here, the division approach doesn’t seem to work because we are not permitted to divide by zero and xd(t) = 0. However, we want to show that y(t) = 0 here as well. Fundamentally, the argument we want to make is of the “it can’t possibly be otherwise” variety. Consequently, a proof by contradiction can be easier to start. In such proofs, we start by assuming the thing that we want to show is not possible. So assume that y(t) is not identically 0 everywhere for t > 0. What does this mean? This means that there is some t0 > 0 for which y(t0) = k 6= 0. (Otherwise, it would be zero everywhere.) We want to create a contradiction. It is clear that we will have no easy contradiction if we just move forward for t > t0 because we have no information given about such solutions y(t) that we can contradict. What do we know about? We have (2) which says something about y(0). This means, that we need to somehow move backward in time from t0. That way, we can hope to contradict the initial condition of 0. What do we have to work with? Well, we just did some work in the previous parts establishing uniqueness of solutions assuming nonzero initial conditions. How can we view what happens at t0 as a kind of nonzero initial condition? Apply the change of variablest = t0−τ to (1) to get a new differential equation for xe(τ) = x(t0−τ) that specifies how d dτ xe(τ) must relate to xe(τ). This should hold for −∞ < τ ≤ t0.

(e) Because the previous part resulted in a differential equation of a form for which we have already proved uniqueness for the case of nonzero initial condition, and since ye(0) = y(t0) = k 6= 0, we know what ye(τ) must be. Write the expressions for ye(τ) for τ ∈ [0,t0] and what that implies for y(t) for t ∈ [0,t0].

Consider a spacecraft that is far away from planets or other massive objects. The mass of the spacecraft is M = 1.5×105 kg. The rocket engines are shut off and the spacecraft coasts with a velocity vector v = (0, 20, 0) km/s. The space craft passes the position x = (12, 15, 0) km at which time the spacecraft fires its thruster rockets giving it a net force of F = (6 × 104 , 0, 0) N which is exerted for 3.4 s. The ejected gases have total mass that is small compared to the mass of the spacecraft. a) Where is the space craft 1 hour afterwards? b) What approximations have you made in your analysis?

Kepler’s second law is this statement: A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. We are going to prove this statement. Consider the wedge in the figure with area dA = 1 2 R 2 dθ The rate that area is swept per unit time is dA dt = 1 2 R 2 dθ dt = 1 2 R 2 ˙θ and this is true even if radius R is varying. We take the origin to be the center of the Sun and radius R is the distance between planet and Sun. The angle θ gives the position of the planet in the ecliptic plane. Kepler’s second law is equivalent to dA dt = constant or d 2A dt = 0. In class we showed that acceleration in polar coordinates can be written a = (R¨ − R ˙θ 2 )ˆr + (2R˙ ˙θ + R¨θ)θˆ Because the gravitational force is in the radial direction, the tangential component of acceleration is zero. This means that 2R˙ ˙θ + R¨θ = 0 Show that this relation is equivalent to dA/dt = constant and Kepler’s second law.

A spherical hollow is made in a sphere of radius R = 11.3 cm such that its surface touches the outside surface of the sphere and passes through its center (see Figure). The mass of the sphere before hollowing was M = 57.0 kg. What is the magnitude of the gravitational force between the hollowed-out lead sphere and a small sphere of mass m = 4.2 kg, located a distance d = 0.55 m from the center of the lead sphere?

A bicycle with 19-in.-diameter wheels has its gears set so that the chain has a 6-in. radius on the front sprocket and 3-in. radius on the rear sprocket. The cyclist pedals at 190 rpm.
Round answers to 2 decimal places as needed.

The linear speed of the bicycle in inches per minute = ???

The speed of the bike in miles per hour = ???

Please show how to work on the calculator.

The typical daily energy use of a household is described by a function f : [−12, 12] → [0, ∞). The value of f(x) is the consumption rate at given time x. Both points x = −12 and x = 12 of the domain correspond to 3am so f(12) = f(−12). It is also known that 9am and 9pm (x = −6 and x = 6, respectively) are local maxima of the consumption rate. In this question we model the rate by a polynomial of degree 4 that is f(x) = a4x4 + a3x3 + a2x2 + a1x + a0and assume that the total energy consumption during the day equals 24.

1. (a) Write a linear system of equations describing the following properties: f(12) = f(−12), [4]x = −6, 6 are stationary points of f and the total energy consumption is 24.

Derive Cauchy’s Integral Theorem for a multiply connected region with one hole

suppose that T : V → V is a linear map on a finite-dimensional vector space V such that dim range T = dim range T2. Show that V = range T ⊕null T. (Hint: Show that null T = null T2, null T ∩ range T = {0}, and apply the fundamental theorem of linear maps.)

1(i) Show, if (X, d) is a metric space, then d∗ : X × X → [0,∞) defined by d∗(x, y) = d(x, y) /1 + d(x, y) is a metric on X. Feel free to use the fact: if a, b are nonnegative real numbers and a ≤ b, then a/1+a ≤ b/1+b .

1(ii) Suppose A ⊂ B ⊂ R. Show, if A is not empty and B is bounded below, then both inf(A) and inf(B) exist and moreover, inf(A) ≥ inf(B).

find the general solution of the equation using cauchy euler. show complete solution

x^2 y" - 3xy' + 3y = 2x^4 e^x

For each of (a)-(d) below, decide whether the statement is True or False. If True, explain why. If False, give a counterexample.

(a) Two non-parallel planes in R 3 will always meet in a line.

(b) Let v, x1, x2 ∈ R 3 with x1 6= x2. Define two lines L1 and L2 by

L1 : (x, y, z) = x1 + sv, s ∈ R,

L2 : (x, y, z) = x2 + tv, t ∈ R.

Then the lines L1 and L2 will never intersect.

(c) Let v1, v2, v3 be three non-zero vectors in R 3 . Then 0 (the zero vector) is a linear combination of v1, v2 and v3. (d) If A and B are two 2 × 2 matrices, and AB = 02×2 (where 02×2 is the 2 × 2 zero matrix), then at least one of A or B must be the zero matrix.

1. (a) Define, with precision and in a form suitable for using in a proof, the least upper bound of a nonempty subset S ⊂ R that is bounded above.

(b) Define, with precision and in a form suitable for using in a proof, an open set in a metric space (X, d).

(c) Give an example, if possible of a function f : X → Y and subsets A, B ⊂ X such that f(A ∩ B) is not equal to f(A) ∩ f(B).

(d) Give an example, if possible, of a subset S ⊂ R that is bounded above, but that has no least upper bound.

(e) Define f : R → R 2 by f(t) = (t, t2 ) and let E = {(x1, x2) : x1 < 1, x2 < 4}. Find f −1

(E). (f) Is there a set A and an onto function f : A → P(A)?

Homogenous problem. Change of variable

(x-y+3)dx + (x+y-1) dy = 0