[QUOTE=Krinkels;42183820]What is the result of Problem 4?[/QUOTE]
Star convex sets are simply connected.
This is pretty simple math, so I apologize, but my brother's been hounding me about this very simple question that his teacher accidentally messed up.
You roll a six sided die and flip a coin. What's the probability of rolling a number less than five [B]or[/B] flipping a heads?
Unlike most independent event probability questions, the question does not read "and." Had it read "and," the answer would be 1/3, since (2/3)(1/2) is 1/3. However, it read "or." This is a topic that is so insignificant it does not even merit discussion. The higher probability would trump the lower probability, since the question asks "or." So he just wants to know if the answer 2/3 would be right. Given my reasoning, yes. I'm pretty poop at math, so I'm just looking for affirmation here.
I believe it's 5/6. The probability of rolling a number less than five or flipping heads would be the complement of getting a number five or above and flipping tails. (since A or B is logically equivalent to not "not A and not B") So it would be 1 - (1/3)(1/2) = 1 - 1/6 = 5/6.
[editline]14th September 2013[/editline]
Your reasoning doesn't really work if you think about it: say you roll the die and get a 5 or a 6. Well damn. But then you flip the coin, and it comes up heads. Your chance to satisfy "less than 5 or heads" must have been higher than 2/3, since you got that extra "saving throw."
Does anyone know a formula or pattern for this sequence: [url]http://oeis.org/A114729?[/url]
Well I could construct a polynomial that would return the sequence given at those values but something tells me that's not what you're after. :v:
Haha no, not really.
There's this Mathematica code on the page:
c[n_] := Flatten[ Table[{Range[3 Floor[(k - 1)/2] + 2],
Table[{i, i}, {i, Floor[k/2] + k, 1, -1}]}, {k, n}]];
uppertrim[list_] := Fold[DeleteCases[#1, #2, 1, 1] &, list, Range[Max[list]]];
lowertrim[list_] := DeleteCases[list - 1, 0];
numbertimes[list_] := Table[Length@Position[Take[list, k], list[[k]]], {k, Length[list]}];
a[n_] := uppertrim[c[n]];
b[n_] := uppertrim[a[n]];
d[n_] := numbertimes[a[n]];
e[n_] := numbertimes[b[n]];
f[n_] := numbertimes[c[n]];
a[6]
I'm not sure what it means, though.
[QUOTE=JohnnyMo1;42173866]Algebraic topology. Grad level[/QUOTE]
What counts as grad level in the US? I nearly took a course this semester in algebraic topology which would be the equivalent to masters in the US I think, but I'm in my 3rd year, thought I should just take some easy shit instead!
[QUOTE=ThisIsTheOne;42200572]What counts as grad level in the US? I nearly took a course this semester in algebraic topology which would be the equivalent to masters in the US I think, but I'm in my 3rd year, thought I should just take some easy shit instead![/QUOTE]
Well usually a grad level class is something you'd take in your 5th year or later. I've been told by one of my professors who is a Brit that in UK universities you essentially do nothing but what you're there to study from as soon as you start, whereas here you take a bunch of general education classes. Not sure if it works like that in Straya. Presumably that lets you take higher-level classes a bit sooner but you don't get any higher education on other topics.
It really depends on the school and the course, too. I know some schools do an undergrad algebraic topology class. I could see that: do a year-long course on Munkres with a semester on the general topology section and a semester on the algebraic topology section, for instance. I think if you really want to get into stuff like cohomology theory though you kind of have to wait for grad school.
[QUOTE=JohnnyMo1;42201216]Well usually a grad level class is something you'd take in your 5th year or later. I've been told by one of my professors who is a Brit that in UK universities you essentially do nothing but what you're there to study from as soon as you start, whereas here you take a bunch of general education classes. Not sure if it works like that in Straya. Presumably that lets you take higher-level classes a bit sooner but you don't get any higher education on other topics.
It really depends on the school and the course, too. I know some schools do an undergrad algebraic topology class. I could see that: do a year-long course on Munkres with a semester on the general topology section and a semester on the algebraic topology section, for instance. I think if you really want to get into stuff like cohomology theory though you kind of have to wait for grad school.[/QUOTE]
Yeah, except in a few places we just focus exclusively on our choice of subject - we don't do majors/minors or anything and even for A-levels (last 2 years of school) you only tend to do 3-5 subjects.
I did that sort of Algebraic Topology in 3rd year, and then there was a more advanced course in the Masters (4th year). Of course this was at Cambridge, so I don't know how other places compare!
[QUOTE=Joey90;42205583]Yeah, except in a few places we just focus exclusively on our choice of subject - we don't do majors/minors or anything and even for A-levels (last 2 years of school) you only tend to do 3-5 subjects.
I did that sort of Algebraic Topology in 3rd year, and then there was a more advanced course in the Masters (4th year). Of course this was at Cambridge, so I don't know how other places compare![/QUOTE]
I mentioned offhand to my undergrad research professor that I would definitely go to Cambridge if I could. He told me that I could, but it would probably cost me about $40,000 a year for 4 years.
No thanks I'll just toil away in obscurity
[QUOTE=JohnnyMo1;42205893]I mentioned offhand to my undergrad research professor that I would definitely go to Cambridge if I could. He told me that I could, but it would probably cost me about $40,000 a year for 4 years.
No thanks I'll just toil away in obscurity[/QUOTE]
Yeah, international fees are pretty crazy (although not compared to some US universities!) I don't know how bursaries and loans work for international students, but I guess it would still be pretty expensive even if they were available. A shame really...
Of course if you become a professor you might be able to get yourself a free trip somehow :v:
I have declared myself a mathematics major. It remains to be seen whether this was a terrible call. My first quarter begins in a week -- Math 300 (Foundations of Modern Math) and Math 324 (Multivariate Calculus I).
[img]http://www.boredofstudies.org/wiki/images/2/26/Root_unity.GIF[/img]
Recently went through an intro into complex numbers...
I've never been mind-blown so many times in a single lecture.
[editline]test[/editline]
Also,
[img]http://upload.wikimedia.org/wikipedia/commons/4/41/Riemann_surface_log.jpg[/img]
Infinite solutions to a logarithm...
I'm not sure what that is - does it tell you anything about the logarithm besides being multivalued?
[QUOTE=Krinkels;42238535]I'm not sure what that is - does it tell you anything about the logarithm besides being multivalued?[/QUOTE]
Because exponential functions are periodic (rotates in a circle, see Euler's identity), the inverse of the exponential function will give you an infinite number of solutions that will satisfy the rotation to undo the exponent.
P.S. This only holds for complex numbers, it doesn't have this property in [img]http://upload.wikimedia.org/math/1/3/4/134676911181af05d24d406f16edf587.png[/img]
[QUOTE=Krinkels;42238535]I'm not sure what that is - does it tell you anything about the logarithm besides being multivalued?[/QUOTE]
I believe that's a graph of the imaginary value (z axis) of the (multivalued) logarithm, and the hue is the argument. Notice that the spiral goes up by 2pi every layer - this is because the complex logarithm of a number can be +/- 2*n*pi*i for any integer n.
It's a useful visualisation because it shows you how the logarithm works along contours:
For example suppose you want that log(1) = 0 (this would be the red part of that graph). If you then try to make the log continuous, you go round and say log(i) = (pi/2)*i (going anticlockwise up the spiral) so then log(-1) = pi*i (onto the yellow section) and log(-i) = (3pi/2)*i...
The confusing part is if you continue up the spiral, you get back to log(1) = 2pi*i (the pale green part) which is not the same as you started with!
This has a lot of implications in all sorts of things (useful statement there), but basically you cannot construct a continuous (single valued) complex logarithm on C\{0} - you always end up having to go 'round the spiral' and end up 2pi*i away from where you started.
Yeah, it's quite intuitive that such a thing would happen, as Euler's formula shows that exponential functions are a linear combination of sine and cosine, and (using the even/odd properties) vice versa, then think about how over the reals the inverse cosine and sine functions need their domains restricted.
I can't help but mock people inwardly when they profess to be a nerd because they think Euler's identity is SO BOOTIFUL
[editline]19th September 2013[/editline]
Does that make me a bad person
I really hate when people wear shirts with maths or physics equations on them (it's always Maxwell's fucking equations)
My discrete mathematics lecturer is pretty chill and can make the audience laugh multiple times during a lecture. A couple weeks ago he spent an hour explaining logical qualifiers by using examples from a big brothers clone we have here in Norway. It got pretty funny when most of this examples implied that just about everyone loved themselves and pretty much everyone else in the show, and the conclusion was that everyone that participate in these shows are horrible people.
Lol, I'm stuck doing a remedial prealgebra class in college.
Going to take the placement test again and hopefully test out of the requirement.
[QUOTE=JohnnyMo1;42173737]
[IMG]http://i40.tinypic.com/2m29h4x.png[/IMG]
[/QUOTE]
Thanks Johnny! After reading this post, I went for a nap. And guess what? I had a dream about my dad giving me slips of paper with mathematical solution and theorems, identical to your picture. I needed to translate them into English, from Somalian, despite them already being in English. Well yeah.
Also passed my first discrete math "mini"-exam.
Can't believe I got full points on that question. It was gross. :v:
I love when homework seems hard and then I sit down and look at it and it's not: "Oh god oh god a sphere with two antipodal points identified? A disk with two antipodal boundary points identified!? I don't know how to compute fundamental groups of such spac- oh wait they both deformation retract to a circle never mind that's stupidly easy."
Anyone know anything about free products? Hatcher says that you can't have a free product including the same group twice back to back since then you have, e.g. g_1g_2 which just reduced to some g_3 but apparently Z*Z is a perfectly fine free product and I don't know how to reconcile this.
So I'm in Calc BC and I'm finding it fun and easy, while most other people in the class struggling to various degrees. Definitely pursuing a math degree when I start college
Warning: most math is nothing like your basic calc classes.
[editline]27th September 2013[/editline]
Take, for example, my girlfriend: she was good at and enjoyed calc, but she's a math major now and doesn't really like it.
[editline]27th September 2013[/editline]
That moment when you come across the theorem in the textbook that makes the proof you're stuck on obvious.
[I]justmathythings[/I]
[QUOTE=JohnnyMo1;42321481]Anyone know anything about free products? Hatcher says that you can't have a free product including the same group twice back to back since then you have, e.g. g_1g_2 which just reduced to some g_3 but apparently Z*Z is a perfectly fine free product and I don't know how to reconcile this.[/QUOTE]
Weird, let me know when you figure it out.
[QUOTE=JohnnyMo1;42321481]Anyone know anything about free products? Hatcher says that you can't have a free product including the same group twice back to back since then you have, e.g. g_1g_2 which just reduced to some g_3 but apparently Z*Z is a perfectly fine free product and I don't know how to reconcile this.[/QUOTE]
I [i]think[/i] what it must mean is that you have to distinguish the elements of each group somehow.
When you form a free product, you create 'words' of elements in either group, and reduce them according to the laws in each group (or equivalently alternating non-identity elements of the two groups)
But if you can't distinguish the elements - say both copies of Z have, in some sense, the 'same' numbers, you would just reduce each word down to a single number, so the free product would just be Z again which doesn't make sense.
It's a pretty stupid distinction, since you'd usually not try and do something like that, however it becomes even more important when you're trying to do free products with amalgamations, you need to make sure you reduce the right elements down.
[QUOTE=Joey90;42323544]I [i]think[/i] what it must mean is that you have to distinguish the elements of each group somehow.
When you form a free product, you create 'words' of elements in either group, and reduce them according to the laws in each group (or equivalently alternating non-identity elements of the two groups)
But if you can't distinguish the elements - say both copies of Z have, in some sense, the 'same' numbers, you would just reduce each word down to a single number, so the free product would just be Z again which doesn't make sense.
It's a pretty stupid distinction, since you'd usually not try and do something like that, however it becomes even more important when you're trying to do free products with amalgamations, you need to make sure you reduce the right elements down.[/QUOTE]
Hrm. How does the free product with amalgamation change things, and is that what is being used in Seifert Van-Kampen? Because Seifert-Van Kampen says that the fundamental group of the figure eight is Z*Z, and I'm just not positive what distinguishes that from Z.
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