Adam

Alright, so I have a cat and my girlfriend has a smaller dog. (It should be noted that we don't live together) The dog is barely larger than the cat and is also very hyper. The cat just wants to chill, especially when the dog is around.

When she first brought the dog around, the cat was just curious and maybe a bit cautious of the dog, but if it could be on higher ground and silently watch the dog, it was ok. The dog would eventually spot the cat and chase it. The cat would escape, but eventually come out and let more of its curiosity get the better of it.

Then, my parents visited with their dog. Their dog, also a smaller breed, is much more "chill" than my girlfriend's dog. He and the cat were found napping on the couch side by side a few times. No issue co-existing. But, like any puppy, this dog likes to play and really had a ball with my girlfriend's dog. So, one day my girlfriend's dog, while playing with my parent's dog, noticed my cat and gave chase. My parent's dog gave chase to the girlfriend's dog (based on their interaction prior to this moment, I'm assuming he was not chasing the cat), but of course the cat didn't see it like that.

I found my cat on a window sill with an arched back, hair on end, clearly very shaken. He hid in my room all night after that. Ever since then, and it has been a while, whenever my girlfriend brings her dog over, my cat has no curiousity, he is in hiding, period. And when cornered, there is hissing. It should also be noted that my cat can still coexist with my parent's dog, despite its "involvement" in the incident.

With the plan of eventually living together, we are trying to get our pets used to each other. Part of it is getting the dog to settle down as that is one reason I think my cat is on edge. But the other part is getting the cat to "forgive and forget" that incident, as he seems to remember it.

Any ideas?

Thanks

Let A be the set of all sequences {s_n} (n=1 to infinity) in L^2 (L^2, being the typical definition, = {{s_n} (n=1 to infinity) : the sum n=1 to infinity of s_n^2 is finite} such that the sum of s_n^2 is less than 1. Prove that A is an open subset of L^2.

I figure that I am using this definition to prove this....

Let M be a metric space. We say the subset A of M is open of M if for all x in A there exists r>0 such that the entire open ball B(x;r) is contained in A.

B(x,r) = {a in A: rho(x,a)<r}

Since we are told we are in L^2, I'm trying to figure out how to use that....

I had this question answered earlier, but today my teacher told us that we should be using a particular theorem (shown below) to prove this..somehow.

The norm for sequences in L^2 has the following properties:

1) ||s||_2 => 0, s is in L^2

2) ||s||_2 =0 iff s={0}(n=1 to infinity)

3) ||cs||_2 = |c|||s||_2 where c is a constant

4) ||s+t||_2 <= ||s||_2 + ||t||_2

Any ideas?

abs(1/(x^2+1) - 1/(y^2+1) )< abs(x - y) (x and y in [0, infinity)

Working with the right hand side, using the some LCD's, factoring, and properties of the absolute value, etc, etc...

I work my way down to this.

|x-y||x+y|/(|x^2+1||y^2+1|)<|x-y| So, obviously, I have the |x-y| I'm looking for on the left, so all I need to do is show that |x+y|/(|x^2+1||y^2+1|) < 1, which by inspection sure looks like it will always be less than one on the interval given, but I can't seem to prove it.

Obesity is defined as a body mass index (BMI) of 3 kg/m2 or more. A 95% confidence

interval for the percentage of California adults aged 20 years and over who were obese was

found to be 21% to 24.9%. What was the sample size?

If I'm right, then we are talking about a proportion here. I'm having trouble figuring out what p (or q) is so that I can work backwards and find n.

Let A be the set of all sequences {s_n} (n=1 to infinity) in L^2 (L^2, being the typical definition, = {{s_n} (n=1 to infinity) : the sum n=1 to infinity of s_n^2 is finite} such that the sum of s_n^2 is less than 1. Prove that A is an open subset of L^2.

I figure that I am using this definition to prove this....

Let M be a metric space. We say the subset A of M is open of M if for all x in A there exists r>0 such that the entire open ball B(x;r) is contained in A.

B(x,r) = {x in A: rho(x,a)<r}

Since we are told we are in L^2, I'm trying to figure out how to use that....

Let {a_n} (n=1 to infinity) and {b_n} (n=1 to infinity)

be Cauchy sequences in a metric

space (M; rho): Prove that

{rho(a_n; b_n)} (n=1 to infinity) is Cauchy in R

:

I understand what it means to be cauchy in this instance. Since each sequence is cauchy, it means that the distance to any other point in that particular sequence is less than epsilon. And I need to prove that the distance from any point in one sequence within epsilon of any point of the other sequence.

So, I think, I just need to define the given sequences to be less than epsilon/something (probably 2). take that rho(a_n,b_n) a and expand say |a_n - b_n| = |a_n +/- something - b_n|, then use the triangle inequality, get epsilon, and we are done.

Just not sure what I'm not seeing in order to finish this up.

Exercise 5. Prove that if ρ and σ are both metrics for a set M , the

ρ + σ is also a metric for M .

I figure that like most other proofs of this variety, I am going to equate my metrics to epsilon/2, then find some way to add them together, use the triangle inequality, and then be done...

I'm just not sure which definition I'm supposed to be using in my notes.

The weekly repair cost for a machine (in hundreds of dollars) has a beta distribution with

Alpha = Beta = 2. If we randomly selected 52 weeks, what is the probability that the mean repair

cost is less than $60?

If a_n is a sequence greater than 0, and the sum of a_n converges, then the sum (n=1 to infinity) of sqrt(a_n)/n converges.

I had a similar problem to this one a few days ago. If we can show that both sqrt(a_n) and 1/n are both in L^2 (L^2 is the typical definition, defined as {s_n}(n=1 to infinity) : sum (n=1 to infinity) s_n^2 < infinity)

Since we know that a_n converges, and that it is obviously sqrt(a_n)^2, so sqrt(a_n) is in L^2., and (1/n)^2 = 1/n^2, and we know that that sums to pi^2/6, so 1/n is also in L^2. So, by Schwarz inequality, the product of them is absolutely convergent, and thus convergent as well.

1) GIve an example of a sequence {s_n} (n=1 to infinity) in L^2 s.t. sum from n=1 to infinity of abs(s_n) = infinity

2) Let {a_n} be in L^2. Prove that sum (n=1 to infinity) a_n/n is absolutely convergent.

For 1), i'm still trying to formulate something

For 2) I'm pretty sure this is just an application of schwarz's inequality. We let a_n just be itself and we let 1/n be the other sequence, t_n. All we have to show is that 1/n is in L^2. And we are done. Which is to say that the sum from n=1 to infinity of 1/n^2 is less than infinity, which is true, so we are done.

So, therefore the product of those sequences is absolutely convergent. That right?

Suppose Y has a probability density function f(y) = e^(2-y) for y >2 and 0 otherwise.

Find m(t) and then use m(t) to find E(Y^2)

A manufacturer wasnts to advertise a mileage interval that excludes no more than 10% of the milage on a tire. Mean mileage is 25,000, standard deviation is 4000. What interval should I suggest?

The weekly repair cost Y for a machine has a probability density function given by f(y) = 3(1-y)^2 for 0<y<1, and 0 elsewhere.

with measurements in hundreds of dollars. how much money should be budgeted each week so that the actual repair cost will exceed the budgeted amount only 10% of the time.

I know I need to integrate, but I simply am not sure on the interval I need to integrate over.

Help?

Okay, so I have searched high and low for a software that can edit in a lightning effect to a video I am making.

Can anybody point me in the direction of a free software that definitely has this effect? Or someone that has the software themselves that wouldn't mind make a quick few second clip for me?

1) Show that 1/(1+t^2) = summation(n=0 to infinity) (-1)^n*t^2n

So, for this I show that R_n (the remainder) is 0. If I did it correctly, skipping a bunch of steps here, lim(n->infinity) 1/(t*(n+1)!) which goes to 0. So, the remainder is 0.

2) Use 1) to show that arctan(x) = summation(n=0 to infinity) (-1)^n*(x^2n+1)/(2n+1) for |x|<1

I would presume that I am integrating both sides, and using the fact that I can take the integral inside the summation. Then I show that the remainder is 0. Am I on the right track?

3) Use 2) to show that summation (n=1 to infinity) (-1)^n/(3^n*(2n+1)) = pi/(2*sqrt(3)).

That simply has me lost.

Prove that if a is a real constant, then (1+x)^a = summation(n=1 to infinity) aCn (a choose n) * x^n.

Realized I need to show that the remainder of these two functions needs to be zero...

So, if I take n+1 derivatives of f, I get (a-(n+1))! (1+x)^(a-(n+1)) Since |x|<1, obviously this is < (a-(n+1))!

From here, I'm not sure where to go...or if I'm even moving in the right direction.

Consider a random variable with a geometric distribution p(y) = q^y-1 * p (y= 1, 2, 3, .....)

a) Show that Y has a distribution function F(y) such that F(i) = 1- q^i, i = 0, 1, 2, 3.....

In general, F(y) = 0 for y<0, or 1-q^i for i<= y <=i+1

b) Show that the preceding function has the properties:

1)F(- infinity) = lim (y->-infinity) F(y) = 0

2) F(infinity) = lim (y->infinity) F(y) = 1

F(y) is a nondecreasing fucntion.

I don't see what I need to do for a)

For b), they all seem obvious, but I just don't see how to write it formally.

Convergence and uniform convergence?

Let (n=1 -> infinity) x(1-x)^n.

a) Show that this series is convergent on [0,1] to f, where f(0) = 0 and f(x) = 1 for 0<x<=1.

By using the ratio test,...

x(1-x)^n+1/x(1-x)^n = 1-x

So, i find that |1-x|<1, so x is convergent on 0 to 2, so it is obviously convergent on [0,1].

What I don't understand is how to find the proper function where f(0) = 0 and f(x) = 1 on 0<x<=1.

b) Show that this series is not uniformly convergent on [0,1].

I know that there is a theorem for sequences that states that if f isn't continuous, then the sequence isn't uniformly convergent. Does the same hold here?

If the sum (n=1 -> infinity) a_n < infinity, then show that sum (n=1->infinity) a_n*x^n converges uniformly on 0<=x<=1.

Since 0<=x<=1, obviously the second series < infinity as well. I also know, by a theorem, that if you can show that a series converges at 0 and 1, means that the entire interval is also uniformly convergent.

R = lim (n-> infinity) |a_n+1/a_n| = 1, if i did that right. And |x|<=1 as well, then it converges at 0 and at 1, so there it is uniformly convergent on [0,1].

Am I close?

Let (n=1 -> infinity) x(1-x)^n.

a) Show that this series is convergent on [0,1] to f, where f(0) = 0 and f(x) = 1 for 0<x<=1.

By using the ratio test,...

x(1-x)^n+1/x(1-x)^n = 1-x

So, i find that |1-x|<1, so x is convergent on 0 to 2, so it is obviously convergent on [0,1].

What I don't understand is how to find the proper function where f(0) = 0 and f(x) = 1 on 0<x<=1.

b) Show that this series is not uniformly convergent on [0,1].

I know that there is a theorem for sequences that states that if f isn't continuous, then the sequence isn't uniformly convergent. Does the same hold here?