The Moomin wrote:And now the Iron Maiden song 'Two minutes to midnight' makes sense.

The nuclear explosions are like lightbulbs going on over my head.

The new mind blown emoji just happens to retroactively encompass this perfectly.

- Fri Dec 15, 2017 11:46 pm UTC
- Forum: Individual XKCD Comic Threads
- Topic: 1655: "Doomsday Clock"
- Replies:
**64** - Views:
**12341**

The Moomin wrote:And now the Iron Maiden song 'Two minutes to midnight' makes sense.

The nuclear explosions are like lightbulbs going on over my head.

The new mind blown emoji just happens to retroactively encompass this perfectly.

- Thu Jun 29, 2017 8:00 am UTC
- Forum: Logic Puzzles
- Topic: Basketball Player riddle
- Replies:
**1** - Views:
**3300**

A famous basketball player is lying on an operating table in the hospital. He is there for routine surgery and he's not nervous at all. They place the gas mask over his face and he hears a nurse say, "Just relax. Everything is going to be fine." This statement makes him extremely nervous. ...

- Mon Jan 30, 2017 10:49 pm UTC
- Forum: What If?
- Topic: What-if 0153: "Hide the Atmosphere"
- Replies:
**46** - Views:
**15710**

It's just an engineering problem, right? Surely this specific what-if is feasable, if economically impractical.

- Tue Dec 06, 2016 12:53 pm UTC
- Forum: Mathematics
- Topic: What's your favourite irrational number?
- Replies:
**78** - Views:
**20431**

Brjuno number(s)

- Sun Jul 10, 2016 8:49 am UTC
- Forum: Mathematics
- Topic: Trickier problem than it seems. (goat in a circular paddock)
- Replies:
**19** - Views:
**4280**

If I'm understanding the Mathworld formula correctly and haven't made any input errors, then setting d = r for their equation we get the following formula for the area, A, in terms of the radius of the paddock (the smaller circle), r, and the length of the rope (the radius of the larger circle) R: ...

- Sat Jul 09, 2016 11:46 am UTC
- Forum: Mathematics
- Topic: Trickier problem than it seems. (goat in a circular paddock)
- Replies:
**19** - Views:
**4280**

The first approximate solution I have managed to come up with thus far is:

The length of the rope is approximately 2/√3 times the radius of the circular paddock fence.

The length of the rope is approximately 2/√3 times the radius of the circular paddock fence.

- Sat May 21, 2016 9:44 am UTC
- Forum: Mathematics
- Topic: Meta-Probability Problem
- Replies:
**16** - Views:
**3859**

Regarding Normals, just apply the argument \tan{\frac{\pi}{2}x} It has both real infinities and is completely continuous. Of course, there's always \tanh^{-1}{x} As well as all the other functions that can be artificially constructed to be absolutely less than unity in domain but cover the entire re...

- Sun May 15, 2016 1:24 am UTC
- Forum: Mathematics
- Topic: Meta-Probability Problem
- Replies:
**16** - Views:
**3859**

I read the question as "All you know about the distribution is that it's biased towards heads.". If you want to bring other "knowledge" into this about intuitions regarding coin manufacturing and whatnot, then that opens a can of worms that can't be sealed again. How do we know ...

- Sat May 14, 2016 1:48 pm UTC
- Forum: Mathematics
- Topic: Meta-Probability Problem
- Replies:
**16** - Views:
**3859**

This comes to the fuzzy edge of what we mean by "probability". Ostensibly, coin flips are still independent and identically distributed, we just know that the probability of heads p is larger than 0.5. The chance of a head and a tail is 2*p*(1-p), so we can immediately conclude that the c...

- Sat May 14, 2016 4:33 am UTC
- Forum: Mathematics
- Topic: Meta-Probability Problem
- Replies:
**16** - Views:
**3859**

You have a coin, which has an unknown probability distribution biased towards heads. (or tails, but that doesn't change the theory behind this question) What is the probability you will, over two tosses, in no particular order, get a head and tail? This is not a homework question, it arose from a di...

- Tue Apr 05, 2016 10:26 pm UTC
- Forum: Mathematics
- Topic: Need help with non-linear Diophantine
- Replies:
**10** - Views:
**2699**

Long and boring case bashing (mod 8) seems to indicate that the only non-square d for which there might be solutions are d = 3-8k, for positive integers k. You probably missed something when bashing mod 8. (-10,-1000,0,100) is a solution to the equation. And? That was already included earlier with ...

- Tue Apr 05, 2016 10:21 pm UTC
- Forum: Individual XKCD Comic Threads
- Topic: 1663: "Garden"
- Replies:
**671** - Views:
**125319**

- Tue Apr 05, 2016 8:55 am UTC
- Forum: Mathematics
- Topic: Need help with non-linear Diophantine
- Replies:
**10** - Views:
**2699**

[rambling] Long and boring case bashing (mod 8) seems to indicate that the only non-square d for which there might be solutions are d = 3-8k, for positive integers k. Warning: this result is based on lots of ugly scribbling which I don't feel like typing up since I feel there ought to be a more ele...

- Sat Nov 21, 2015 3:06 am UTC
- Forum: Individual XKCD Comic Threads
- Topic: 1606: "Five-Day Forecast"
- Replies:
**115** - Views:
**20336**

Vehemence wrote:Once our solar system consumes itself, days and years will cease having any meaning. Crap, I made myself sad.

Ten seconds after the last panel I got an existential crisis. I need to work on this.

- Wed Oct 28, 2015 11:14 am UTC
- Forum: Individual XKCD Comic Threads
- Topic: 1594: "Human Subjects"
- Replies:
**33** - Views:
**6213**

Funny, the OED seems to think you can use snuck. It does say it's mainly North American, but it is still a word. http://www.oxforddictionaries.com/definition/english/snuck And to people in regions where it's dominant, "sneaked" sounds exactly as unpleasant on the ear as "weeped"...

- Fri Sep 11, 2015 12:07 pm UTC
- Forum: Mathematics
- Topic: Wanted: Elegant proof of sin(x)/x limit
- Replies:
**49** - Views:
**7793**

Qaanol wrote:I know of a simple way to show that (1-cos(x))/x → 0, but I have not yet found a correspondingly transparent illustration for sin(x)/x.

Have you posted this proof already? If not, then please do? I am bad at spotting posts.

- Wed Jun 03, 2015 11:00 pm UTC
- Forum: Mathematics
- Topic: Almost Nesbitt's inequality
- Replies:
**8** - Views:
**2379**

I think for my first post, the original question was to find the lower bound of the expression, assuming a, b, c are all strictly positive.

- Wed May 20, 2015 11:52 pm UTC
- Forum: Mathematics
- Topic: Almost Nesbitt's inequality
- Replies:
**8** - Views:
**2379**

how about a/sqrt(b^2 + c^2) and it's permutations

also, the same as above, but without the radicals

what will be on the other side of the inequalities if we allow all real values of a,b and c?

also, the same as above, but without the radicals

what will be on the other side of the inequalities if we allow all real values of a,b and c?

- Wed May 20, 2015 8:52 am UTC
- Forum: Mathematics
- Topic: Almost Nesbitt's inequality
- Replies:
**8** - Views:
**2379**

no. what if you replace it with the euclidean distance formula? i suppose that'll still be positive.

this isn't the inequality i'm after. the one i remember had a non-zero constant on the other side...

this isn't the inequality i'm after. the one i remember had a non-zero constant on the other side...

- Wed May 20, 2015 7:49 am UTC
- Forum: Mathematics
- Topic: Almost Nesbitt's inequality
- Replies:
**8** - Views:
**2379**

Is there still an inequality if we remove the bounds for a,b,c ?

- Wed May 20, 2015 7:47 am UTC
- Forum: Mathematics
- Topic: Almost Nesbitt's inequality
- Replies:
**8** - Views:
**2379**

No, looks like i just have it incorrect... I do remember some kind of twist on nesbitt's though, involving radicals... but that means this is not it.

- Wed May 20, 2015 7:37 am UTC
- Forum: Mathematics
- Topic: Almost Nesbitt's inequality
- Replies:
**8** - Views:
**2379**

Prove a/sqrt(b+c) + c/sqrt(b+a) + c/sqrt(a+b) > 0 for positive a,b,c

Not sure if this is even correct, but i saw this somewhere long ago.

Edit: I'm thinking Cauchy-Schwarz right now.

Not sure if this is even correct, but i saw this somewhere long ago.

Edit: I'm thinking Cauchy-Schwarz right now.

- Tue Apr 21, 2015 1:40 pm UTC
- Forum: Mathematics
- Topic: What is Math?
- Replies:
**6** - Views:
**1792**

I believe you mean "iteratively", not "recursively", since you're going forward. :D Having left academia and then returned to it, I define mathematics somewhat more holistically than most career academics I have met. I believe that mathematics is the study of systematic processe...

- Mon Mar 30, 2015 6:19 am UTC
- Forum: General
- Topic: Need help finding a webcomic
- Replies:
**3** - Views:
**1598**

*screams in joy* YES THIS IS EXACTLY WHAT I HAVE BEEN SEARCHING FOR THANKYOU SO MUCH *gives Deva countably infinite internet cookies*

- Sun Mar 29, 2015 10:35 am UTC
- Forum: General
- Topic: Need help finding a webcomic
- Replies:
**3** - Views:
**1598**

A few months ago, I was reading this particularly entertaining nerdy webcomic. It was similar to XKCD, but focused on social issues and technology much more, with a heavy amount of sarcasm used. There was something on dystopian futures from time to time. The colour scheme of the website was blue and...

- Wed Mar 11, 2015 2:01 am UTC
- Forum: Mathematics
- Topic: Forming New Polynomials using properties of roots
- Replies:
**5** - Views:
**1581**

Strongly suspect the answer is wrong. Used Sum and Product to get the same polynomial I got earlier, and still couldn't figure out how to eliminate p^{2}x from the equation.

- Tue Mar 10, 2015 9:23 am UTC
- Forum: Mathematics
- Topic: Forming New Polynomials using properties of roots
- Replies:
**5** - Views:
**1581**

x 3 + px - 1 = 0 has three real, non-zero roots (α,β,γ) a) Find the values of α 2 + β 2 + γ 2 and α 4 + β 4 + γ 4 in terms of p, and hence show that p is negative. b) Find the monic cubic equation with co-efficients in terms of p, whose roots are α 2 , β 2 , γ 2 I've completed part a), and almost co...

- Fri Feb 27, 2015 12:03 pm UTC
- Forum: Mathematics
- Topic: Polynomials : Multiplicity of roots Homework problem
- Replies:
**11** - Views:
**3131**

Turns out the question was incorrectly stated. r = -1

- Sun Feb 22, 2015 12:08 pm UTC
- Forum: Mathematics
- Topic: Polynomials : Multiplicity of roots Homework problem
- Replies:
**11** - Views:
**3131**

Are a and b restricted to integers? Well according to the post made to correct my assumption, no. ka, kb, kr is a solution to the system of equations, and k can be any real number, so therefore, a and b do not have to be integers. Although the answer to my homework is asking to show that they are. ...

Each result contains a slightly different list each time, so it is rather difficult to search through each list exhaustively for all the oxidisers, compare them to a databse entry and eliminate repeats. Also, Chlorine Trifluoride didn't exactly turn up in any of the immediate results. From the top o...

- Sat Feb 21, 2015 8:46 pm UTC
- Forum: Mathematics
- Topic: Polynomials : Multiplicity of roots Homework problem
- Replies:
**11** - Views:
**3131**

As stated, the problem is incomplete. If (a,b,r) is a solution so is (k a, kb, kr). Maybe you need to assume that r=1? P(1) = a + b + r =0 (1) P'(1)= (n+1)a + nb = 0 (2) n+1 ≠ n |(n+1)a| = |nb| n+1|a| = n|b| n and n+1 are positive integers, although I'm not sure how to use this fact. |a| = n |b| = ...

- Sat Feb 21, 2015 2:20 pm UTC
- Forum: Mathematics
- Topic: Polynomials : Multiplicity of roots Homework problem
- Replies:
**11** - Views:
**3131**

gmalivuk wrote:You've used the root of P'(x) to get some information about a and b.

You have not yet used the root of P(x) to get any additional information.

I'm still not getting anywhere. Substitution, isolation, solving simultaneously. I've tried all of those.

What are all the known and theoretical oxidisers in existence? (Trying to acquire an exhaustive list here)

- Sat Feb 21, 2015 1:27 pm UTC
- Forum: Mathematics
- Topic: Polynomials : Multiplicity of roots Homework problem
- Replies:
**11** - Views:
**3131**

P(x) = ax n+1 + bx n + r P(x) has a double root of x = 1 show that a = n, and b = -(n+1) What I have done so far: P(x) has a double root of x = 1, so P'(x) has a root of x = 1 P'(x) = (n+1)ax n + nbx n-1 P'(1) = (n+1)a + nb = 0 (n+1)a = -nb n ≠ n+1 So it follows that a = -n and b = n+1 or a = n and ...