yes

Thanks, that clears up a lot of things.

Okay so P should multiply column 1 by 1, column 2 by 2, column 3 by 3, and column 4 by 4.

So P should be
[1 0 0 0
0 2 0 0
0 0 3 0
0 0 0 4]

But, when I do P[T]_B'P^-1 I don't get [T]_B unless one of my previous answers was wrong.

Yeah your T_B' is wrong -- your columns are written in terms of the B-basis it looks like.

OMG I want to die. Spent like 3 hours trying to figure it out. Fixed T_B' by dividing row 2 by 2, row 3 by 3, and row 4 by 4.

Sure enough, PT_B'P^-1 = T_B

Thanks a lot Wyman

I don't have solutions, so I just want to verify that this is correct.

a) We have an orthogonal basis, so this is y dot u1 / u1 dot u1 * u1= 6u1 = [6606]

b)We now have enough info to know yhat = [6 6 0 6] + [0 -2 0 2] =[6 4 0 8]

c) y=yhat + z
So y is [6 4 0 8] + [4 -2 4 -2]

And checking quickly, those are in fact orthogonal

d) This should be norm{y - yhat} = norm{z} = sqrt(40)

Does this look right?

Excellent. Exactly right. You did the first part by inspection, then you did the x*e^x by integration by parts. You can do them both by inspection, or both by integration by parts. So it's still too easy. Try these:

x*cos(x)

sqrt(x)*ln(x)

x*e^x / (x+1)^2

Yes, and hopefully you see intuitively what is going on here in terms of the geometry.

Yeah I see, yhat is the orthogonal projection onto the plane W=span{u1, u2}, so we can express y as the sum of yhat and some vector (z) in the orthogonal complement of W. This is somewhat similar to breaking up forces into horizontal and vertical components in physics problems.

I've never written a general expression in terms of n before, and I'm having a hard time figuring out how.

I have found a pattern (I think) such that for any n I can tell you the indefinite integral for (x^n)*(e^-x):


The problem is that I don't know how to terminate the problem since the pattern will repeat itself to infinity:

Yes the indefinite integral is harder. But notice this is a definite one from 0 to infinity which simplifies things. You have written the general expression as function of x with a definite integral on the left which is not right. You probably forgot the limits trying to find the more general problem. You are on the right path though, think how you would solve eg a x^4*e^-x type thing then x^3... etc

Okay, so I'm trying to teach myself some topology and am getting stuck on this question.

Let (X,T) be any topological space, and G be the set of all homeomorphisms from X to itself.

1) show that G is a group under composition of functions
2) show that if X = [0,1], then G is infinite
3) if X = [0,1], does G commute?

1 and 3 I am fine with, 2 I am stuck on. I can show it for certain topologies like discrete, trivial, and euclidean, but I can't think of a way to say that for any topology on X it will be true. For euclidean you can just take x^n, n>0 as the set of homeomorphisms, which is clearly infinite but stuck on the general case.

I got the indefinite one. It was fun. Do the definite one first as that's much easier and it will help you figure out the pattern for the indefinite one. You need to know that the e^-infinity times x^infinity will always be zero like we saw yesterday with that one problem with x*e^x. As x goes to infinity, e^-x goes to zero faster than any power of x goes to infinity.

When you're ready for the indefinite one, write the polynomial as a summation of x^k as k goes from 0 to n-1, and add the -x^n outside the sum. Then look at how the numbers increase from one line to the next as a function of both n and k. Try to cook up a coefficient of x^k that depends on n and k that will do that. If you want the coefficients to alternate sign, write -1 to a power which is even for +1 and negative for -1.

Have you done any computer programming? Ever use recursive functions? This can be done recursively. You write a function that gives the coefficients for any n, and it calls the one for n-1. Thinking of it that way helped me find the solution, but you don't need to write it as a recursive function, just a sum. Think about the sum like a loop, and how would you code the loop to give the right numbers.

The definite one can be used with a generating function for the generalized coupon collector's problem. That allows you to figure out the average number of craps rolls to get all of the numbers 2-11. Or the probability of rolling every number before a 7. That bet is offered at Wynn.

Meant the definite one for the generalized coupon collector's problem, not the indefinite one.

looking for indefinite integral of x*cos(x)
using derivative of x and integral of cos(x) for my product: 1*sin(x)
that integrates to: -cos(x)
Subtract that from the product of xsin(x)
x*sin(x)-(-cos(x) = x*sin(x)+cos(x)
indefinite integral of x*cos(x) = x*sin(x)+cos(x)

looking for indefinite integral of x^(1/2)*ln(x)
using integral of x^(1/2) and derivative of ln(X) as product = (2/3)x^(1/2)
which integrates to (4/9)x^(3/2)
subtract that from [(2/3)x^(3/2)]*[ln(x)]
[(2/3)x^(3/2)]*[ln(x)]-[(4/9)x^(3/2)]
indefinite integral of x^(1/2)*ln(x) = (2/3)x^(3/2)*(ln(x)-(2/3))

Still working on that other one.

Those are correct, except be sure to add C or you lose a point.

Right you get for the indefinite;



Which you can verify by induction etc. But that was a bit tough if not familiar with sums hence the definite easier trying to introduce Ace the recursive idea the easy way first as a means to make him imagine that learning various areas of mathematics eventually helps his imagination to widen up and start using multiple arsenals together without fear because something looks initially ugly or whatever...Eventually getting into derangements, incomplete gamma functions, generating functions for other topic problems and other funny things seemingly unrelated at surface.

Okay, so updating for help, and posterity.



I think that as our x gets really big our e^-4(-4^4....etc) term is going to go to zero. On the other hand our e^-0(-0^4....) term is going to evaluate to -24. So as the one gets closer and closer to zero, shouldn't the whole expression get closer and closer to -24?

I am still at a loss for how to generalize as my pattern is willing to extend itself longer and longer as my x grows.

Try this;

Call the expression for the integral with n exponent I(n). Then I(n-1) is the integral with 1 less exponent. Try to relate I(n) with I(n-1). Do you notice now something?

So for clarity, I understand you to mean for me to take the following equations

I(n)=x^n * e^-x
I(n-1)=x^(n-1) * e^-x

and try to see how they can be expressed generally to see if I see a relationship?

Yes but notice i called I(n) the integral as function of n not the integrated expression. So try to take the integral I(n) first and see if you can relate it with I(n-1).

Remember that you use partial integration technique to always reduce the exponent of x^n until it is 0 where the effort stops. Take a general n express I(n) using the I(n-1) expression while treating both of them yet unknown functions of n and see if you spot an easy relation between the 2 that can give you a hint what is the general expression.

You need a (-1)^(k+n) so they alternate sign, right?

I(n) is a number for each value of n. Write down that sequence of numbers for n = 1,2,3,4... and see how to get from one to the next.

I think I've been doing math for the better part of 12 hours, I'm off to bed. Will resume tomorrow. Thanks for the help.

The definite integral is a number for each value of n. You get those numbers by evaluating at infinity and 0 for each n. I(n) are those numbers. They aren't functions of x. You just want to find the pattern of the sequence. Like if they were 2,4,6,8...you would have I(n) = 2*n for n = 1,2,3,4... Find a pattern like that for your I(n).