[QUOTE=AlienCreature;45263917]A bit late, but he did say without putting cards back. Might be wrong, but won't the probability be:
1/10 + 1/9 + 1/8?
Might be horribly wrong, didn't read the link above.[/QUOTE]
1/9 is the probability that [i]given that the first card is not a 5[/i] the second card is a 5. Thus in your sum you have to multiply by the probability that it's [i]not[/i] already been picked (9/10). Doing the same for the other probability as well, you get:
1/10 + (9/10)*(1/9) + (9/10)*(8/9)*(1/8)
which conveniently (but not surprisingly!) cancels to give
1/10 + 1/10 + 1/10
Time to bump this thread with this game I've been playing for a couple of days, I think you guys will like it: [url]http://euclidthegame.com/Tutorial/[/url]
You basically start out with the ability to draw straight lines and circles and then you're given a task of constructing a certain geometric object on each level. It starts out easy, like having to construct a parallel line or an equilateral triangle, but it gets harder as you go along. The last few levels are actually really tough if you've never seen this kind of thing done before, but it feels great to figure them out all by yourself.
Well, I kind of had to cheat on the last level, but all I did was use Wolfram alpha to [sp]calculate the exact value of a certain cosine.[/sp]
I've been mingling around in my head for ages with trig functions to try to figure out how to trisect a line, still haven't got anything. My personal rule is to figure it out without writing anything down, but I'm getting desperate.
I thought I could get the square root of three by constructing the tangent of a 60 degree angle, then somehow inverting and squaring that value to get 1/3 (as soon as you have any line in thirds, you can easily transfer that to the one you have to trisect).
The problem is that I don't know how to construct a line that is the square of another line. Or if that's even possible.
[QUOTE=Number-41;45348329]I've been mingling around in my head for ages with trig functions to try to figure out how to trisect a line, still haven't got anything. My personal rule is to figure it out without writing anything down, but I'm getting desperate.
I thought I could get the square root of three by constructing the tangent of a 60 degree angle, then somehow inverting and squaring that value to get 1/3 (as soon as you have any line in thirds, you can easily transfer that to the one you have to trisect).
The problem is that I don't know how to construct a line that is the square of another line. Or if that's even possible.[/QUOTE]
At first I thought it was a trick question, then I realised I was thinking of trisecting an angle... I think it would make a good final level :v:
As for the line...
[sp]Triangles Triangles Triangles[/sp]
[QUOTE=Number-41;45348329]I've been mingling around in my head for ages with trig functions to try to figure out how to trisect a line, still haven't got anything. My personal rule is to figure it out without writing anything down, but I'm getting desperate.
I thought I could get the square root of three by constructing the tangent of a 60 degree angle, then somehow inverting and squaring that value to get 1/3 (as soon as you have any line in thirds, you can easily transfer that to the one you have to trisect).
The problem is that I don't know how to construct a line that is the square of another line. Or if that's even possible.[/QUOTE]
Multiplying a certain length by square root of 3 is definitely possible (not sure what exactly you mean by "tangent of a 60 degree angle" though), but inverting and then squaring it doesn't really sound like something that could be possible, you're probably overthinking it. There's a more general technique that allows you to split a line segment into n equal parts for any n, and it relies on [sp]similar triangles.[/sp]
So I got a Saxon Pre-Algebra and Algebra 1 textbook. Tried doing lesson one in the Algebra 1 book, but my math skills are so crusty, I reverted to the pre-algebra book. Gonna try to do one lesson everyday.
I guess math is one of those skills where if you don't use it, you lose it.
[QUOTE=cqbcat;45405224]So I got a Saxon Pre-Algebra and Algebra 1 textbook. Tried doing lesson one in the Algebra 1 book, but my math skills are so crusty, I reverted to the pre-algebra book. Gonna try to do one lesson everyday.
I guess math is one of those skills where if you don't use it, you lose it.[/QUOTE]
Shit, man, my life is like a constant struggle to stuff new math in my brain faster than old math falls out.
Isn't that pretty much how the sciences are too? You learn it once so that you can forget it, but you have the ability to quickly learn it again when necessary.
[QUOTE=Falubii;45406975]Isn't that pretty much how the sciences are too? You learn it once so that you can forget it, but you have the ability to quickly learn it again when necessary.[/QUOTE]
Yeah. I've definitely found that the value of my education has been less learning specific stuff and more learning how to think correctly and absorb information more easily.
I've been having an issue with a symbolic logic problem if anyone can lend a hand.
Find a useful denial for [;(\exists! x)A(x);]
The textbook gives me that [;(\exists! x)A(x);] is equivalent to [;(\exists x)A(x) \wedge (\forall y)(\forall z)((A(y) \wedge A(z)) \Rightarrow y = z);]
So I thought that I could do this
[;\neg [(\exists x)A(x) \wedge (\forall y)(\forall z)((A(y) \wedge A(z)) \Rightarrow y = z)];]
Then
[;(\forall x)\neg A(x) \wedge (\exists y) (\exists z) \neg ((A(y) \wedge A(z)) \Rightarrow y = z);]
Then
[;(\forall x)\neg A(x) \wedge (\exists y)(\exists z) ((A(y) \wedge A(z)) \wedge y \neq z);]
According to the textbook, I'm almost right, but I'm missing something big.
[;(\forall x)\neg A(x) \vee (\exists y) (\exists z) ((A(y) \wedge A(z)) \wedge y \neq z);]
How would that and turn to an or? I mean it makes sense thinking about it, but I don't know how that happens within the logic theorems.
Edit:
I'm dumb. I had to have another look at DeMorgan's laws.
Yep, every time you negate and or or it flips to the other one. Think of this statement: "p or q." This is saying, "If either p or q or both are true, this whole statement is true." Now we want to negate that, so when is the statement false? Only when both p and q are false. So the negation has to be: "~p and ~q." The connective has to switch.
[editline]17th July 2014[/editline]
Similarly, "p and q" is only true when both p and q are simultaneously true, so the negation is a statement where at least one of the statements is false: "~p or ~q."
But while math's on the mind. Has anyone here ever taken the Putnam exam? My professor and I are signing up our school to compete although we know nobody is going to do well. It's mostly for shits and giggles.
I took the last one.
I was the only one at my school. Most don't like six hour exams.
A comment on a Vi Hart video:
"Transcendence theory is frustrating to say the least. For example, we know that pi and e are transcendental numbers. [B]We do not know whether e*pi or e+pi are transcendental; however, we do know that both cannot be algebraic.[/B] There is a lot of work to be done in this field." •
Uh...
What they meant is:
We know e*pi and e+pi cannot both be algebraic.
If they were both algebraic, then the equation x^2+(e+pi)x+(e*pi)=0 would
have roots e and pi, so e and pi would be algebraic [url=http://www.math.niu.edu/~rusin/known-math/95/transcend](contradiction)[/url]
Well that makes much more sense. Gotta choose that phrasing carefully.
sqrt(x+3) = sqrt(x-2) + sqrt(x-5)
I'm doing some repetition before university. Pretty simple task, but somewhere I get it wrong, could someone show me all steps so I could see where it went wrong?
[QUOTE=JohanGS;45515756]sqrt(x+3) = sqrt(x-2) + sqrt(x-5)
I'm doing some repetition before university. Pretty simple task, but somewhere I get it wrong, could someone show me all steps so I could see where it went wrong?[/QUOTE]
By inspection, 6.
If you want to do it 'properly' (although inspection for something like this is perfectly legit). You can square both sides to get rid of the first root:
x+3 = x-2 + 2*sqrt((x-2)(x-5)) + x-5
then shuffle things so the remaining root is on its own:
10-x = 2*sqrt((x-2)(x-5))
square again to get rid of all the roots
(10-x)^2 = 4(x-2)(x-5)
Then when you expand out you get a quadratic to solve.
One root will be positive, and the correct solution, the other will be negative and is not correct*
*you could define a square root function on C that would make it correct, but that's probably not what you're looking for.
Nah, 6 is all I need. I just forgot to multiply with that extra 2 when squaring.
[editline]27th July 2014[/editline]
Thanks!
I need to come up with some way to construct an absorbing boundary for a wave function such that the total area under the modulus squared of the function remains 1 (this is a quantum mechanics problem for a project I'm working on). The particle's wave-function is allowed to move into an absorbing region from one side but can never leave again. This is bugging me.
[editline]11th August 2014[/editline]
Oh, and I intend on modelling this system computationally so I can't just say, "pretend the absorbing region is infinitely long," unfortunately. The absorbing region is of a finite size WITHIN the boundaries of the system of interest (but it's perfectly, 100% absorbing).
Wait, when you say the total area under the modulus squared of the function remains 1, what is meant by the function? The wave function, or the potential of the absorbing region? Am I interpreting this correctly: you're looking for some potential what will effectively admit and trap a particle?
Also, mightn't this be more suited for the physics thread?
Do we still have the physics thread? I couldn't find it.
And yes, that's correct. Whatever happens inside of the barrier can't effect anything outside, though (we're investigating the quantum random walk in 2-Dimensions with absorbing regions - when the walker hits the barrier in the classical walk it's absorbed which is no problem to code, but my supervisor and I are having trouble trying to construct boundary conditions at the absorber for the quantum case).
Can't wait until Thursday evening when I have some free time to dig deeper into this.
Fields medalists were announced today. Is it just me, or is Mirzakhani kind of a fox?
how easy or difficult would it be to teach yourself a semester's worth of calculus? looks like that's what i'm going to have to do, my teacher is just as clueless as i am
btw just learning limits right now, doing pretty okay with those
Imho I struggled the most with the formal limit definitions, after that it was doable. It really depends on the syllabus.
Also why is he/she your teacher if they're clueless? Is it just another student or something?
i should rephrase: he's brilliant with calculus, but clueless as how to teach students. his knowledge isn't directly transferable, unfortunately. i believe he was an engineer at nasa or something, long time ago.
Well if you can ask him questions then I think it's very doable. Especially if you also keep maths.stackexchange in mind and stuff like that.
y^2 + 3xy - 10x^2 + y + 5x = 0
Hint on how to solve for y?
It's just a quadratic equation in y? So just look at x as if it were a constant.
And your solution will depend on x.
Oh, right, you can do it like that. Thanks.
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