I haven't left out anything. I tried the problem using m = 45/8 and got the answer that's in the back of the book. Thank you both for your help!

Yes, there is no ambiguity. Weight is 180 = mg, and when g = 32 ft/sec^2, then m = 180/32 slugs (or 45/8 as you said). This is unfamiliar to some because we rarely work in English units. When force is in pounds, the mass is in slugs. When I said slug ft/sec^2, that is just a pound, not an alternative unit of force, so the fact that v was in ft/sec told us that the force had to be in pounds and we needed to divide by the mass in slugs to get acceleration.

I want to derive the covariance of an AR(2) process but I struggle at:

AR(2): Yt=c0+c1Yt-1+c2Yt-2+et

Cov(Yt,Yt-1)=Cov(c0+c1Yt-1+c2Yt-2+et , Yt-1)

->c1 Cov(Yt-1 Yt-1) +c2 Cov(Yt-2 Yt-1)

the first term is c1 Var(Yt-1) but how do I proceed with c2 Cov(Yt-2 Yt-1)?

[QUOTE=Acemanhattan;37157727]I have a project that I am working on and since I think I'll be asking a lot of questions I'll just post the question here so the context is clear.



Fairly straight forward, our data set (if that's what people call it) and our plot is the following:





Then it seems like an appropriate function would be v(x)=-.04x+.56, 0≥ x >6



From the discussion earlier I've settled on D(x)=-(1/72)x^2+.5 (since my velocity function is in feet I decided to make this function in feet as opposed to inches, which was what we found earlier.



Which I have looks like the following



So then if we are interested in finding total stream flow how does the following sound:

(1) Find that the average value of the function D(x) is 1/3 so that the average depth of our creek is 1/3 foot. (2) solve for when D(x)=1/3 foot to find that our average depth occurs when x=2*sqrt(2) (3) plug this x value into v(x) to find our average velocity at our average depth is .42(ish)ft/sec (4) integrate D(x) to find that there is 4ft^2 of area (and i'm assuming this implies water) in the cross section (5) multiply 4ft^2 by our average velocity of .42ft/sec to find that our total stream flow is 1.6(ish) ft^3/sec.

One way to do this is to repeat this out to infinity. The Cov(Y(t-2),Y(t-1)) = c1*Var(Y(t-2))+c2*Cov(Y(t-2),Y(t-3)) and so on.

Another way is, since I'm assuming the process is stationary, Cov(Y(t),Y(t-1)) = Cov(Y(t-2),Y(t-1)), lets call this x.

Plug x into the equation you wrote, and solving for x should be easy from there. The repeating to infinity way requires a similar observation, that the variance of the terms should remain constant.

http://en.wikipedia.org/wiki/Autoreg...lker_equations

Their phi's correspond to your c's.

All right I'm taking a modern physics class and I want people's opinions on how reasonable this is. My ta gives us a quiz, the problem is he gives us wavelength of light and work function and asks for photon energy, stopping potential and ke of electron. The photon energy is 10 ev, the work function 5 ev. So I just write down that the ke of the electron must be 5 ev and the stopping potential five V. He gives me no credit on the second or third part because I didn't show work. This is the first he mentioned that he was going to be a stickler for showing work.

Hint: Use the Orbit-Counting Lemma.

Question about limits of integration after I make my substitution also about integrating the integrand sec(x)



Starting with the lower limit of integration x=3, I think I do the following:



and then for x=infinity, following the same logic as above, this would be the case when tan is undefined at pi/2 and -pi/2 (I'm not sure why we call it -pi/2 instead of 3pi/2)

But how do I know whether to use the positive value or the negative value?

Additionally, after I got the problem simplified to 2 times the integral from a to b of sec(x), I had to go to a table to integrate it. I just realized I didn't try integration by parts, so I'm guessing that is the answer to the question "how do I integrate this without a table"?

You can see the indefinite integral isn't right because you get infinity when you evaluate the limits. The problem is the 5th line:

tan^2(theta) + 1 = sec^2(theta)

not sec(theta). That comes from

cos^2(theta) + sin^2(theta) = 1

Dividing by cos^2(theta) gives

1 + tan^2(theta) = sec^2(theta).

So you end up just integrating dtheta. Since x goes from 3 to infinity and

theta = arctan(sqrt(x))

theta goes from pi/3 to pi/2, and we get a final answer of

2(pi/2 - pi/3) = pi/3.


You could also have said from the beginning that

theta = arctan(sqrt(x))

dtheta = 1/2 * 1/[sqrt(x)*(1+x)] * dx

since the derivative of arctan(x) = 1/(1 + x^2). Then you get the integral of dtheta again.


You also could have done this by u-substitution

u = sqrt(x).

Then you get

integral of 2/(u^2 + 1) = 2*arctan(u) = 2*arctan(sqrt(x))

which is the indefinite integral.


You didn't need it, but integrating sec(x) involves a trick. You multiply by

[sec(x)+tan(x)] / [sec(x) + tan(x)]

and let u = sec(x) + tan(x). My calculus book states "Only experience prompts us to multiply numerator and denominator by sec(x) + tan(x)".

Thanks Bruce. Calling (tan^2(theta)+1) sec(theta) was a silly mistake.

I still don't understand why we only evaluate the positive square root of x though?

Theta can't be negative or you wouldn't have been able to say sqrt(tan^2(theta)) = tan(theta) = sqrt(x) as you did because tan(theta) would be negative while sqrt(tan^2(theta)) and sqrt(x) are always positive. You have to decide up front when you do your change of variables what range theta is going to go through as x goes through its range. The natural choice is to let theta go from pi/3 to pi/2 as x goes from 3 to infinity. But when you say x = tan^2(theta), that would be true if theta goes from pi/3 to pi/2, but it would also be true if theta goes from -pi/3 to -pi/2, or even if theta goes from negative to positive, or from positive to negative. By itself it says nothing about the range. It is explicit if you write theta in terms of x as theta = arctan(sqrt(x)). Then as x goes from 3 to infinity, we see that theta goes from pi/3 to pi/2. Your equation comes from that equation by taking the tangent of both sides and then squaring both sides. The squaring introduces the negative solutions that we didn't want. It's OK to integrate from theta = -pi/3 to -pi/2 if you want, but then you would have to set sqrt(tan^2(theta)) = -tan(theta) = -sqrt(x). That would introduce a negative sign, and you'd get -2*(-pi/2 - -pi/3) = pi/3 as before. That would correspond to the change of variables theta = arctan(-sqrt(x)), and x = tan^2(theta) would still be true. You know you need to get a positive answer since the thing you're integrating is always positive.

Makes perfect sense. Thanks Bruce.

Numerical Analysis...

Our professor gave us some tips about this question, but I still have a few questions now that I'm working through it.

m! / k!(m-k)! describes the number of ways of choosing a subset of k objects from a set of m elements.

Suppose decimal machine numbers are of the form:

+/- 0.d1 d2 d3 d4 x 10^n, with 1<= d1 <= 9, and 0 <= d2,d3,d4 <= 9 and |n|<= 15

What is the largest value of m for which the number can be computed for all k without causing overflow?

In other words, basically the largest number is 0.9999*10^15, some 15 digit number. So, I checked for m! which do not exceed 15 digits. 17! is the largest such number. So, if k! = 1, then:

17! / 1!(17-1)! = 17...

How do I algebraically (or some other quantitative way?) find the smallest number that could be in the denominator, providing the largest evaluation of this combination fraction?

I mean, I know, n_C_0 = 1, n_C_n = 1, n_C_1 = n, n_C_(n-1) = n... so intuitively, it leads me to believe the largest combo should just be where r = n/2.

If I use 17_C_9, I get some 5 digit number. So... I'm still looking for some algebraic way or numerical method to continue. I think my start on this problem is decent, and I understand it well, but I need a little guidance or tips. I do have MATLAB. I could just create some script, but I'd like to understand the algebra a little better too!

Thanks!

53_C_26 or 53_C_27 is the largest number below 9999*10^15... but I would just like to give a good proof... I think the part my weakest point is that n_C_r is largest when r = n/2... still not sure how to show that. I'm looking at:

m! / (m/2)!(m-m/2)! = m!/[(m/2)!]^2

EDIT:

Basically, I'm trying to show that:

x*x is always greater than (x+1)*(x-1) = x^2 - 1. Then, I can just use induction to say this is also true for any (x+n)*(x-n)? But... how does this help me with factorials??

I'm really confusing myself now...

So you try to see for what k the m!/(k!/(m-k)!) is maximum?

well try this;

for a given m fixed you want to maximize

the function f(k)=m!/k!/(m-k)!

notice f(k+1)=m!/(k+1)!/(m-k-1)!

So f(k+1)/f(k)=k!*(m-k)!/(k+1)!/(m-k-1)!=(m-k)/(k+1).


Now as long as (m-k)/(k+1) is larger than 1 the function grows as k grows.
Also notice that (m-k)/(k+1) is a declining function of k anyway. So you know the maximum will occur at f(k) when the (m-k)/(k+1)<=1 for the first time as the next term f(k+1) will be immediately smaller or equal to that (and the next smaller etc) .

So m-k<=k+1 or m-1<=2*k or k>=(m-1)/2

So if m= 2*n+1 k=n is where the maximum is.

If m=2*n then the ratio f(k+1)/f(k)=(m-k)/(k+1) becomes <1 at k=n as it goes to n/(n+1)<1 so the max is at k=n

So the max is always at k=Integerpart(m/2)

And it is then

when m=2n;

(2n)!/n!^2

and when m=2n+1

(2n+1)!/n!/(n+1)!

Using Stirling approx. for n!=n^(n+1/2)*exp(-n)*(2*Pi)^(1/2)*(1+1/(12n)+...)

You can see that

(2n)!/(n!)^2~ (2*n)^(2n+1/2)*exp(-2n)*/(n^(2n+1)*exp(-2n))/(2*Pi)^(1/2)=


=2^(2*n)/(n^(1/2))/(Pi)^(1/2)

This goes over 10^15 at n=27 so the maximum even you can put in it is m=2*26=52

so 52!/26!/26!=0.49592*10^15

At m=53 you have 53!/26!/27!=0.97347*10^15

so m=53 is the highest integer given the constraints here. (you can use Stirling for all cases to see that its an increasing function of m or n=int(m/2) )

Of course if you use eg decimal logarithms you can avoid this problem. You can then produce expressions for the first 5 digits and the exponent separately and that way manage to solve exponentially much higher m problems.

http://en.wikipedia.org/wiki/Stirling_approximation

C(n,k) is largest for k = n/2 for n even, and k = (n-1)/2 and (n+1)/2 for n odd which are always the same. You can see that from Pascal's triangle. These are just the binomial coefficients.

The nth line (starting with line 0) is C(n,k) for k going across from 0 to n. It's largest in the middle, and the largest keeps getting larger for larger n. Each number is the sum of the 2 above it. That's from the recursion

C(n,k) = C(n-1,k-1) + C(n-1,k).

Do you have to represent all the digits of the number to the one's place? If so, then you can't even do C(17,6) = 12376 which requires 5 digits. You can do C(16,8) and all the 16's because the 5 digit ones all end in 0, so you can use the exponent.

> C(16,0:16)
[1] 1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368
[13] 1820 560 120 16 1

Usually we don't worry about representing all the digits for large numbers any more than we would worry about representing all the decimal digits for small numbers. If you just need it not to overflow, then you can do C(53,k) for all k, but not C(54,k) or C(55,k) or anything higher which would have one larger than C(55,27) since Pascal's triangle for C(56,28) will add C(55,27) to C(55,28), and so on.

Also, we can compute C(m,k) for a number higher than m! since we can do

m/k/(m-k) * (m-1)/(k-1)/m-k-1) * ...

or even sum logs instead of multiplying and then exponentiate, but there would be loss of precision in these steps if you care about that.

If understand correctly, when using sigma notation, the number at the bottom "i=m" denotes the integer we start at, and the number at top "n" denotes the integer we end at, but what notation denotes the number of sub intervals we are using? Is it just implied by the way we define delta x ?

Thanks, masque de Z and BruceZ! Really interesting! I didn't think I was really opening a can of worms like this, but numerical analysis is a whole different math class (admittedly, linear algebra was quite different from other math classes I had before as well).

FWIW, this chapter is all about "Round-off Errors and Computer Arithmetic." I don't really think there are right answers nor are we intended to find the most optimal way to compute large (or small) machine numbers. I really think the objective is just to demonstrate that we know how we may run into these types of errors and the possible causes of them.

Thanks again!

There are n-m+1 terms, so if you're talking about defining the integral, this is the number of subintervals of size delta x. The number of terms is the length of the whole interval divided by delta x. In post 3956 I gave



The length of the interval is 1, so there are 1/delta_t intervals, and this is reflected in the 1/delta_t - 1 + 1 = 1/delta_t terms.

You also know that if you flip a coin m times, the most likely number of heads is m/2 for m even, and the probability of getting k heads is C(m,k)*(1/2)^m. Since this is maximum at k=m/2, C(m,k) is maximum at k=m/2.

Let's say we have a curve of an unknown function and we want to estimate the area under it using rectangles. Let's say then that we set up a table where our x values are the number of rectangles we use and our y values are the areas that correspond to each x value. As we estimate more and more rectangles the change in our estimated area should get smaller and smaller with respect to the previous estimation. Is there a way to use this sort of information to detect what limit the area is approaching?

nvm i think i didnt understand you

Well y can't be area any more than x can be. The area can only be the product of x and y

it sounds like you are talking about a using some sort of numerical method in order to approximate an integral? Am i missing something?