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#1 Help Me ! » Linear Fractional Transformation » 2010-06-02 04:55:34
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C is a continuously differentiable curve, and f(z) is a linear fractional transformation. Define a length function
and show that this length is preserved by f. Also show that f preserves hyperbolic distance
.Sooo...
f is of the form (az+b)/(cz+d) for some complex constants a,b,c,d. And the first thing I need to show is asking that L( f(C) ) = L(C), right? This seems like it should be a relatively straightforward calculation, but I cannot make it work so I think I fudged in setting up the equality. Can someone set-up this equality explicitly, please? Or tell me this approach won't work and suggest another perspective?
Also, in my personal history I've ignored hyperbolic trig functions so I have no idea whatsoever to do with the second part.
Thank you!
#2 Help Me ! » Holomorphic self-map on D(0,1) » 2010-05-28 05:02:50
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I need to show that a holomorphic map f: D(0,1)---> D(0,1) exists such that f(1/2)=3/4 and such that f'(1/2)=2/3. No clue 
#3 Help Me ! » complex power series » 2010-04-23 03:38:06
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f is holomorphic on the complex plane. Assume that for each complex w there is at least one n such that b_n (w)=0, where b_n (w) is the coefficient of (z-w)^n in the power series expansion of f at w. Show that f is a polynomial.
I do not remember power series expansion. How is the coefficient b_n (w) found? And even though I do not remember power series expansion, wouldn't the power series expansion of a polynomial be the original polynomial?
#4 Help Me ! » algebra with complex numbers » 2010-04-05 05:30:45
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I have complex function f(z)= (z-w)/(1-w' z). Here, w is a fixed complex number on the unit circle and w' is the conjugate of w (so if w=a+bi, then w'=a-bi).
I have to show that if z is in the unit disc, then so it f(z).
This means I have to show that |(z-w)/(1-w'z)| is less than or equal to 1 whenever |z| <= 1, right?
So I went ahead with the statement |(z-w)/(1-w'z)| <= 1 and said that this holds whenever |(z-w)| <= |(1-w'z)| holds because I can "distribute" the norm over any product and, because the norm is always positive, I can multiply both sides of the inequality by the normed denominator and obtain the second form in that way.
But, when I substitute z=x+yi and w= a+bi and compute both sides of the inequality, I get at the end that the original statement holds whenever x^2 + y^2 <= x^2 + y^2, which is always true for any complex number. This seems to say that the function f takes EVERY complex number to the unit disc. Also, nothing seems to go wrong if I replace the inequality by strict equality, which says that f(z)=1 whenever x^2+y^2 = x^2 + y^2. Does this say that f takes every complex number to the unit circle because this is true of every complex number (|z|^2 = |z|^2 always)?
I went over the computation a couple of times, but found no errors. Still, I find the result very odd. Can someone tell me whether I screwed something up in my process? I feel like I'm committing some grievous algebraic sin or something, but cannot figure out what it is.
#5 Help Me ! » random vector, probability » 2010-03-16 05:53:58
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I am having trouble understanding these notations:
.I am to show that P(this set) is zero, but am not making sense of how to picture this set. (The X_i are from a random vector X, and the X_i are independent and identically distributed random variables)
#6 Help Me ! » probability: seq of r.v.'s » 2010-03-09 03:40:37
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For a sequence of arbitrary random variables, find
such that converges weakly to .I don't even know where to begin this one. It seems to say that for any random variable I can define a positive real constant that makes the values of the r.v. arbitrarily small (depending on how the constant is defined), but this does not make sense with my intuition. In essence, a random variable is just a real-valued function but I don't remember ever learning that any random sequence of real functions has this property...
#7 Re: Help Me ! » Topology: Covers of figure-eight space » 2010-02-17 05:28:16
Thanks--your explanation confirms what I inferred as I was trying to finagle a way to draw some of the covering spaces, and also bolsters my confidence in what covering spaces I did come up with. Do you happen to know how many 3-fold covering spaces of the figure-eight space there are?
I'm sure that we'll get to it soon, but out of curiosity what is the correspondence you speak of? I know I could look it up, but me and course books have rarely agreed, and the book for this course is no exception. I mean, its nice mathematically, but its got its front cover too far up its index to be pedagogically effective (as most do).
#8 Re: Help Me ! » Topology: Covers of figure-eight space » 2010-02-16 13:40:16
Nope, afraid not. I'd be interested in knowing how that could help, but as that concept hasn't entered in lecture yet, my professor must have solutions in mind that don't use a universal cover.
I'm still working on this, but so far I've managed to doodle what I believe to be some 3-fold covers of the figure-eight space. I'm supposed to find all 3-fold covers of the space though. How can I find out whether I have them all? Or maybe some of mine are redundant. I think they are all distinct, but its all based on intuition. So I guess I'd also like to know how to tell whether two of my doodles represent the same cover space. Advice?
#9 Help Me ! » Topology: Covers of figure-eight space » 2010-02-16 10:09:00
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I have to find all covers of the figure-eight space where the preimage of each point has exactly 3 distinct elements in the covering space.
I am really confused on this. I know that one such covering space is 8 X {1, 2, 3}, where "8" denotes the figure-eight space. This is just 3 copies of the figure-eight. But what other ones are there? How do I "find" them? I am not even sure how to begin constructing the others, whatever they may be. Any advice would be priceless!
Thank you in advance.
#10 Re: Help Me ! » osculating circles and spherical curves » 2010-01-26 17:05:14
I think I got one direction (that c(s) spherical implies the center is in the normal plane for every normal plane), but haven't figured the other direction out.
Let's say I wanted to show that if there is some fixed point in every normal plane of a curve, then the curve must be spherical (lie on the surface of a sphere).
I did not mention this before, but the curve is in R^3.
I know the normal plane is the span of the normal and binormal vectors. So in general to say that some point a_0 is in the normal plane to some curve, c, at some point s_0 would mean that there exist m_1 and m_2 such that
But by assumption a_0 is in the normal plane for ALL s (is in every normal plane of the curve c(s) ).
Before I continue, I'd like to get someone to put a check on my intuition please:
Does a_0 being in every normal plane mean that for each s there exist m_1 and m_2 as above, or do the same m_1 and m_2 work for every point on the curve? I think that the same m_1 and m_2 should work but am uncertain.
Thanks in advance.
#11 Help Me ! » osculating circles and spherical curves » 2010-01-26 07:24:21
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I have to show:
A curve, c(s), is spherical if and only if there is some fixed point, a, located in every normal plane of c(s).
Of course, this fixed point must be the center of the sphere, but I am having a hard time showing it. I want to say that for all s, the normal vector, N(s), points directly at the center and since the curve is parametrized by arclength, the center is actually at the terminal point of N(s), but I cannot formalize this. The only things I can think to work with are the Frenet equations and the fact that the center of the osculating sphere for any point on a spherical curve is constant.
Please help! I know this is probably not a very difficult problem, but I am very uncomfortable in my differential geometry course and feel way over my head. Thanks in advance.
#12 Re: Help Me ! » Equation » 2009-12-03 07:16:56
You are supposed to simplify this, correct? Apply the Distributive Property (this should be in your book, look it up in the index if you forgot it).
I have A Equation Problem 5x-2x+8?
#13 Re: Help Me ! » expectation of standard normal random variable » 2009-12-03 07:11:24
Ok, I feel foolish.
I meant what I said when I said "double exponential", buuuuuut I had wrongly interpreted the problem to be asking to evaluate the integral of the double exponential
when, in fact, the problem statement indicated I ought evaluate the integral , as you pointed out. Thank you, Avon.With a few solid nights' rest and a bit of hindsight, I can't imagine how I did not at some point realize that I wasn't even applying the definition of expectation correctly...what a goon.
If has the standard normal distribution then
which doesn't contain what I would call a double exponential (i.e. something like ).
Is this the integral that you are having problems with?
#14 Re: Help Me ! » mental maths » 2009-11-25 07:17:44
You did not mention a course you were taking or a specific source of these exercises, so I do not know what mathematical tools you have available to work with but I think I can offer some suggestions based on the nature of the exercises and that you refer to these as "mental maths" problems.
Yes, you can solve these exactly like equations. In fact, that's exactly what they are--equations with one unknown/variable. If you want to improve on time, you will be faster the more familiar you are with simple calculations. Look online or in basic arithmetic texts for some practice problem sets with single-operation exercises. Practice with exercises of addition and subtraction of 2-digit and 3-digit numbers, and memorization of multiplication tables will help you become very quick at these mental arithmetic problems. At this link (http://www.math.com/students/practice.html) there is a list of math topics for practice. If you click on "Basic Math", you can pick operations and numbers from 0-12 and it will make problems for you and time you to see how many you can do in one minute. Search for other sites too, there are a lot of them out there.
Practice factoring numbers (which will become easier the more multiplication tables you know) will also help. Also, try doing a lot of your mental maths problems with pencil and paper. Doing this kind of practice will train you to be able to do the problems in your head.
Good luck!
Hi,
I needed some help trying to solve basic mental maths problems. They are usually in multiple choice answers and I tend to work backwards going through all the answers to see which is the right one.. where I really want to improve on time. So how would I solve some questions like:
160 - ? + 170 = 280
120 / ? + 7 = 11
can you solve these like equations? whats the best way to tackle problems like these? more examples whould be helpful..Thanks
Thank you
#15 Re: Help Me ! » expectation of standard normal random variable » 2009-11-24 22:11:51
Thanks for your reply 
I understand using the fact that
is odd to compute the mean, but my original question and what I still cannot figure out, is how to compute . As we have not approached moment generating functions in this course yet, I do not think I should use that route to find a solution. Any suggestions for my original problem?The function
is an odd integratable function
Therefore,
the expectation of a standard normal random variable
must equal 0.Another way is to look at the moment generating function of a stardard normal random variable,
which is
and
#16 Help Me ! » expectation of standard normal random variable » 2009-11-24 05:22:46
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How do I calculate expectations with the standard normal random variable? The integrals generally look very intimidating if even doable sometimes. I am curious how to approach these types of problems in general, so any advice is greatly appreciated. The problem in particular I am staring at right now is to find
, where is the standard normal random variable. The primary issue I'm having is how to calculate things like this--for instance, and referring to the exercise I mentioned, I write out the integral necessary to calculate this expectation, but it is the integral of the product of a double-exponential with a single-exponential.#17 Re: Help Me ! » integrating with e » 2009-11-23 23:29:45
Thank you very much for your confirmation--I really didn't see a viable point-of-entry on this one, but its been so long since I've done calculus that I thought maybe I was missing something.
Interestingly enough, almost this exact integral appears on a homework assignment for my probability course. I had only posed this integral initially to check my understanding of the calculus relevant to the problem. I will post the original problem as a new topic (since it apparently is).
#18 Help Me ! » integrating with e » 2009-11-23 09:06:36
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I'm just not sure how to come at this. Usually with things that look this way I use a substitution, but any way I try a substitution, I have a factor of "x" in the du-expression that is supposed to be substituted for "dx" (which, to my understanding, ought to be "x-free"). I need to be able to handle integrals like this one for homework, but have an exam coming soon as well and there will be lots of similar integrals. What is a good approach?
#19 Re: Help Me ! » integration by substitution » 2009-11-16 09:57:29
Yea, it made intuitive sense, and I figured (after my post) that my method was correct when I evaluated the integral. Thank you for your confirmation of that as well as for your inclusion of the evaluation of the integral afterward (it was great to compare my work to someone else's)!
Yeah that substitution and expression for dx are correct, but you have to do a bit of algebra before you can get everything in terms of y.
After you sub in x = tan(y), apply a trig identity so that your expression turns into something more usable
#20 Help Me ! » integration by substitution » 2009-11-16 08:58:00
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I have not done calculus in years but am in a course that requires me to try and recall my calculus...I'm not doing so hot.
Integrate
using the substitution .I remember that I cannot simply stick "tan(y)" in for "x", but I also have to find an expression to substitute-in for "dx".
How do I find this expression that will take the place of "dx"? Do I do
? If not, what is the appropriate method? And if so, please let me know--a little confirmation never hurt.Thanks in advance.
#21 Help Me ! » connected components in uniform topology » 2009-11-11 07:52:45
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I can show that
is not connected by producing a separation into bounded and unbounded elements.There is a hint in the book (Munkres), but again I do not see how it helps. Munkres says "it suffices to consider the case y=0". If I consider this case, then I see that if x is in the zero component, then x-0=x must be bounded (because [0], the zero component, is bounded and I have shown the separation above). However, I am unclear about the other direction. If x is bounded, why does this necessarily imply that x must be in the 0 component, [0]? And I still don't see how this special case is sufficient to prove the general result originally stated.
Thanks in advance.
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