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Author Topic: Random Stuff  (Read 1171110 times)

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sudgy

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Re: Random Stuff
« Reply #1625 on: January 14, 2014, 03:31:50 pm »
0

If I'm not mistaken, you can also factor (1 - tan^2 x) as (1 + tan x) * (1 - tan x)

Yeah, but that didn't help me in my situation.
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   Quote from: sudgy on June 31, 2011, 11:47:46 pm

Witherweaver

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Re: Random Stuff
« Reply #1626 on: January 14, 2014, 03:36:24 pm »
0

Though in this case I'd just replace tan x by sin x/cos x and clear out all the fractions.  Seems like the most direct way.
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theory

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Re: Random Stuff
« Reply #1627 on: January 14, 2014, 03:36:28 pm »
0

Ah, right.  I had misread your post.

Math epiphanies are the best epiphanies :)
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Ozle

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Re: Random Stuff
« Reply #1628 on: January 14, 2014, 03:37:44 pm »
+4

If I'm not mistaken, you can also factor (1 - tan^2 x) as (1 + tan x) * (1 - tan x) right?

Yes


*Edit*
Damn, looks like I was wrong, darn 50/50 chance to look clever fails AGAIN!
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SirPeebles

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Re: Random Stuff
« Reply #1629 on: January 14, 2014, 03:49:10 pm »
+1

(2 tan x)/(1 - tan2x) = (2 sin x cos x)/(cos2x - sin2x) = sin(2x) / cos(2x) = tan(2x)
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sudgy

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Re: Random Stuff
« Reply #1630 on: January 14, 2014, 04:16:02 pm »
0

(2 tan x)/(1 - tan2x) = (2 sin x cos x)/(cos2x - sin2x) = sin(2x) / cos(2x) = tan(2x)

Actually, the original problem was to turn tan(2x) into 2/(cot x - tan x)
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   Quote from: sudgy on June 31, 2011, 11:47:46 pm

Witherweaver

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Re: Random Stuff
« Reply #1631 on: January 14, 2014, 04:35:07 pm »
0

(2 tan x)/(1 - tan2x) = (2 sin x cos x)/(cos2x - sin2x) = sin(2x) / cos(2x) = tan(2x)

Actually, the original problem was to turn tan(2x) into 2/(cot x - tan x)

You originally said simplify, and we can all agree that tan(2x) is the simplest form :P
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Witherweaver

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Re: Random Stuff
« Reply #1632 on: January 14, 2014, 04:36:47 pm »
0

But even 2tanx/(1-tan^2x) seems like an unnecessary middle step..

tan(2x) = 2sinxcosx/(cos^2x-sin^2x) = 2/(cosx/sinx - sinx/cosx) = 2/(cotx - tan x)
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sudgy

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Re: Random Stuff
« Reply #1633 on: January 14, 2014, 04:40:35 pm »
0

I used the identity for tan (2x) that they had used earlier.
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   Quote from: sudgy on June 31, 2011, 11:47:46 pm

Tables

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Re: Random Stuff
« Reply #1634 on: January 14, 2014, 04:44:01 pm »
0

Trig conversions, ugh, I remember doing those at A level, and then almost never once I hit first year Maths.
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Re: Random Stuff
« Reply #1635 on: January 14, 2014, 05:08:32 pm »
0

http://htwins.net/scale2/


The scale of everything
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Re: Random Stuff
« Reply #1636 on: January 15, 2014, 11:22:35 pm »
+1

This is one of the best games I've ever played:

http://www.kongregate.com/games/StuStutheBloo/no-one-has-to-die

Well, maybe it is more of an interactive story than a game, but it is great nonetheless. I can't really describe what it's about without giving spoilers, so I recommend to check it out for yourself.
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Re: Random Stuff
« Reply #1637 on: January 16, 2014, 12:04:12 am »
0

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Kirian

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Re: Random Stuff
« Reply #1638 on: January 16, 2014, 11:30:05 am »
+3

Not mine, found on a different forum:

Code: [Select]
  Such crawl                                                             
                  How Yendor
          Dogue
                      Wow
      Many ascii                         
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WalrusMcFishSr

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Re: Random Stuff
« Reply #1639 on: January 16, 2014, 01:42:38 pm »
0

Have you ever heard of "industrial musicals"?

Here's a lovely ballad "My Bathroom is a Private Kind of Place" from the 1969 classic "The Bathrooms are Coming":



And who could forget that old standard "Silicones, Silicones" from 1973's "Got to Investigate Silicones":



And much, much more!

http://en.m.wikipedia.org/wiki/Industrial_musical
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Witherweaver

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Re: Random Stuff
« Reply #1640 on: January 17, 2014, 12:28:09 pm »
+1

Found a fun link on Facebook.  Some professors talk about Ramanujan summation without actually mentioning Ramanujan summation, and the comments section explodes:



Edit: The link to the Youtube page because reading the comments is the funny part.
« Last Edit: January 17, 2014, 03:46:44 pm by Witherweaver »
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Re: Random Stuff
« Reply #1641 on: January 17, 2014, 04:03:25 pm »
0

Edit: The link to the Youtube page because reading the comments is the funny part.
You can just click the embedded player to be linked to YouTube.
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Witherweaver

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Re: Random Stuff
« Reply #1642 on: January 17, 2014, 04:06:46 pm »
0

Edit: The link to the Youtube page because reading the comments is the funny part.
You can just click the embedded player to be linked to YouTube.

Right, but I wanted to emphasize the YouTube page itself.
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SirPeebles

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Re: Random Stuff
« Reply #1643 on: January 17, 2014, 05:38:38 pm »
+4

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WalrusMcFishSr

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Re: Random Stuff
« Reply #1644 on: January 17, 2014, 05:51:24 pm »
+2

Does anyone have an explanation for that series that doesn't make it seem like the universe is patronizing me?
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Witherweaver

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Re: Random Stuff
« Reply #1645 on: January 17, 2014, 05:56:35 pm »
0

Does anyone have an explanation for that series that doesn't make it seem like the universe is patronizing me?

It was brought up earlier in this thread, right?  Or was it a different thread?  Anyway:

http://en.wikipedia.org/wiki/Ramanujan_summation

It's a way of assigning values to a divergent series that makes sense. (It is consistent with the analytic continuation of the series that defines the Riemann Zeta function.)
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WalrusMcFishSr

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Re: Random Stuff
« Reply #1646 on: January 17, 2014, 06:05:18 pm »
+1

Still feels like I'm being trolled haha. Sure I've seen the math before, but I've never gotten a truly satisfying explanation for it.
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SirPeebles

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Re: Random Stuff
« Reply #1647 on: January 17, 2014, 06:10:55 pm »
+2

Does anyone have an explanation for that series that doesn't make it seem like the universe is patronizing me?

One way of viewing it is in terms of analytic continuation.  More precisely, define

Zeta(s) = (1/1)^s + (1/2)^s + (1/3)^s + (1/4)^s + ...

This series converges for real values of s greater than 1.  Naively, Zeta(-1) = 1 + 2 + 3 + 4 + ..., but of course the series does not converge at s = -1.

Instead, we try to extend Zeta(s) analytically.  Basically, this means we want a nice smooth function Z(s), and Z(s) ought to be an extension of Zeta(s) in the sense that Z(s) = Zeta(s) whenever s is a real number greater than 1.

Now, it turns out that there is no way to continuously extend Zeta(s) such that it is well-defined at s=1.  But what Riemann instead did was find an analytic continuation Z(s) which is defined for all complex values of s apart from 1.  Thus, by moving through the complex plane around 1, we do have a nice smooth extension of Zeta(s) which reaches s=-1, and Z(-1) = -1/12.

Here's another way of looking at it:

Let f(s) = 1 + s + s^2 + s^3 + ...

You may recognize this as a geometric series, and note that f(s) = 1/(1-s).  Or at least, these functions are equal when |s| < 1. F(s) = 1/(1-s) is the analytic continuation of f(s).  So while f(-1) = 1 - 1 + 1 - 1 + ... does not converge, the analytic continuation gives us F(-1) = 1/(1-(-1)) = 1/2.

Another example, which is perhaps less controversial, is the function g(s) = (s2 - 1)/(s-1).  Now, g(1) is undefined, but clearly g(s) "simplifies" by writing g(s) = (s2 - 1)/(s-1) = (s+1)(s-1)/(s-1) = s+1.  If we want to be precise, 1 is not in the domain of g(s), but 1 is in the domain of the analytic continuation G(s) = s+1, and there G(1) = 2.
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WalrusMcFishSr

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Re: Random Stuff
« Reply #1648 on: January 17, 2014, 06:16:12 pm »
0

Thanks! That's as good an explanation as I've yet seen. It still feels bothersome that adding together only positive numbers would give you a negative. But I guess if we're dealing with infinity and the complex plane then all bets are off for ordinary intuition.
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Witherweaver

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Re: Random Stuff
« Reply #1649 on: January 17, 2014, 06:21:15 pm »
+1

Okay, so, we usually define an infinite sum as the limit of the sequence of partial sums:

Sum(f(n), n=0..infinity) = lim_{N->infinity} [Sum(f(n),n=0..N)]*

whenever that limit exists.  When that limit does not exist, we say the series diverges.  And we don't really do anything with it.  You may wonder if there is a consistent way to assign a value to Sum(f(n), n=0..infinity) in the case when the series diverges.  This would, of course, not be the same as the limit of the sequence of partial sums, as that limit does not exist.  However, the sequence of partial sums may still have some properties, even if it doesn't have a limit. 

Basically (I don't know much about this, I'm just reading the wikipedia page), the Ramanujan sum relates the partial sum to an integral (of the function f) by considering the sum to be an approximation (Trapezoidal) to the integral.  It turns out that the error of such a representation is fairly tractable.  From this representation one can find some values associated with the partial sums, and one of those values is the one in question here (the -1/12 for 1+2+3+...). 

It turns out that this isn't just random playing around, but actually has relevance.  When you look at the function

thing(s) = Sum(1/n^s, n=1..infinity),

which is defined for all complex s with Real(s) > 1,

a theory of Complex Analysis says this has a unique extension to the entire Complex plane (except the pole at s=1).  This extension is called the Riemann Zeta function Zeta(s), and we recognize Zeta(2) is a regular convergent infinite series with value Pi^2/6.  But it extends to other numbers as well, like Zeta(0) and Zeta(-1).  It turns out Zeta(-1) = -1/12.  (I don't know how this is shown in full rigor, but these things are known.) 

Well, we know Zeta(s) = Sum(1/n^s, n=1..infinity) for any s with Real(s) > 1.  This does not hold when s=-1.  However, Zeta(-1) = -1/12 is defined, and if we plug in s=-1 into the series on the right, it is exactly

Sum(1/n^{-1}, n = 1..infinity) = Sum(n,n=1..infinity) = 1+2+3+4+...

So when we say

1+2+3+ .... = -1/12,

this is an abuse of notation.  What this really means is that -1/12 is the value at a particular point of a function that is the analytic continuation of a particular series, where that series, if you were to plug in the same point, would give 1+2+3+.... . You can't "plug in" the point since it is not in the domain, but it makes perfect sense to extend the function thing(s) that I wrote above to the entire complex plane (though it has a pole when s=1).  So it's not completely bogus.



*So the sequence of partial sum for 1+2+3+... would be the sequence {1, 1+2=3, 1+2+3=6, 1+2+3+4=10, ...}, which is obviously divergent.
« Last Edit: January 17, 2014, 06:23:04 pm by Witherweaver »
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