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Solving Recurrence Relations by Wesley Tillingbury - Fri, 16 Oct 2015 21:23:55 EST ID:pCKCgraP No.14939 Ignore Report Reply Quick Reply
File: 1445045035508.png -(14284B / 13.95KB, 269x235) Thumbnail displayed, click image for full size. 14284
I did discrete math last year but I've forgotten a lot about how to solve recurrence relations. In the solutions to one of my assigned homeworks, I don't understand the process the professor used to find the explicit formula. For instance, in the pic related example:

C(n) = 3C(n-1) + 4; C(0) = 4

Why aren't the "+ 4"s being multiplied by the 3 as the relation is backwards substituted? Why is it exempt?
>>
Cyril Blavingstutch - Sat, 17 Oct 2015 01:50:54 EST ID:i84x+n57 No.14940 Ignore Report Quick Reply
1445061054900.jpg -(2828662B / 2.70MB, 3504x2336) Thumbnail displayed, click image for full size.
>Why aren't the "+ 4"s being multiplied by the 3 as the relation is backwards substituted? Why is it exempt?

You're right; they should be multiplied by the 3s. Whoever made your pic fucked up hardcore and has no business teaching math. I hope for your sake that it wasn't your professor.

Line 3 of the red text is where they first fucked up. Should be:

C_n = 3*[3*[3C_(n-3) + 4] + 4] +4 = 3*3*3*C_(n-3) + 4*(1 + 3 + 3*3)

This means:

C_n = 3^k*C_(n-k) + 4*(summation from i = 0 to k-1 of 3^i)

The summation is just a geometric series (https://en.wikipedia.org/wiki/Geometric_series#Formula):

C_n = 3^k*C_(n-k) + 4*(1 - 3^k)/(1 - 3) = 3^k*C_(n-k) + 2*3^k - 2
Comment too long. Click here to view the full text.
>>
Fucking Clurrykire - Sat, 17 Oct 2015 20:27:30 EST ID:i84x+n57 No.14942 Ignore Report Quick Reply
>>14940
>and verifying with the original recurrence relation
and checking against the original recurrence relation

I'll fix the link too: https://en.wikipedia.org/wiki/Geometric_series#Formula

nb
>>
Phyllis Pemmerhut - Mon, 19 Oct 2015 18:39:21 EST ID:xJK33YUo No.14944 Ignore Report Quick Reply
>>14940
Thanks a lot for your help, I'm glad that I wasn't missing something huge. It was my professor so I'll talk to her today. I actually need some more help if you're willing, with this problem:

"In class we showed that the recurrence relation for divide and conquer algorithms such as the quick and merge sorts is C_n = 2*C_n/2 + n, where C_1 = 0. Using this recurrence relation, the analysis of these algorithms is Theta(n*lg(n)). Solve the recurrence relation, using a different base case: C_1 = 15, to demonstrate whether changing the base case results in the same (or different) Big Theta result."

Now I'm not sure if you're familiar with computer science, but I don't think you need to be to solve this problem. I'm just unsure of how to approach 'solving' the relation, and how/if a different base case would change the analysis (which we perform using the Master Theorem https://en.wikipedia.org/wiki/Master_theorem#Generic_form ). Any guidance?
>>
Shit Bloblingwater - Tue, 20 Oct 2015 02:32:41 EST ID:i84x+n57 No.14945 Ignore Report Quick Reply
>>14944
I've only ever taken one CS course. But if I understand correctly, the base case should be irrelevant with regard to using the master theorem. The recurrence relation in question falls under case 2 mentioned the the Wikipedia article, and accordingly is theta(n log n). This result is independent of the base so should be true of C_1 = 15.

For solving the relation, just use the same backward substitution argument as before. Doing that gives you:

C_n = 2^k*C_(n/2^k) + k*n

Set 2^k = n:

C_n = n*C_1 + n*log_2(n)

For C_1 = 15:

C_n = 15n + n*log_2(n) = theta(n log n)

So it checks out.


Trig troubles by Molly Bingergold - Mon, 12 Oct 2015 00:12:03 EST ID:P4gPE0kz No.14933 Ignore Report Reply Quick Reply
File: 1444623123652.png -(4785B / 4.67KB, 455x54) Thumbnail displayed, click image for full size. 4785
I'm told I have to expand the left side using double angle identities but I'm not sure where to start. Can someone help me get started with this?
>>
Doris Sommerdock - Mon, 12 Oct 2015 22:13:31 EST ID:i84x+n57 No.14935 Ignore Report Quick Reply
Start with cos(2x) = 1 - 2*sin^2(x). Solve this for sin^2(x). You should be able to figure out the rest from here.
>>
Frederick Banningsire - Wed, 14 Oct 2015 19:26:09 EST ID:i84x+n57 No.14937 Ignore Report Quick Reply
>>14935
Meh, I looked at the problem again, and it's a little more involved than I initially thought. Okay, I'll just work it out fully. Doing what I previously said gives:

sin^2(x) = 1/2 - 1/2*cos(2x)

square both sides: sin^4(x) = 1/4 -1/2*cos(2x) + 1/4*cos^2(2x) (call this equation 1)

start with another double angle identity: cos(2x) = 2*cos^2(x) - 1

solve for cos^2(x): cos^2(x) = 1/2*cos(2x) + 1/2

sub 2x for x: cos^2(2x) = 1/2*cos(4x) + 1/2 (equation 2)

plug eq. 2 into eq. 1: sin^4(x) = 1/4 - 1/2*cos(2x) + 1/4(1/2*cos(4x) + 1/2)
Comment too long. Click here to view the full text.


Help me to understand by Fuck Sottinghet - Tue, 13 Oct 2015 17:55:25 EST ID:43kF2tms No.14936 Ignore Report Reply Quick Reply
File: 1444773325462.gif -(688596B / 672.46KB, 205x160) Thumbnail displayed, click image for full size. 688596
I'm doing an assignment for college, can you explain the math by providing an example please ?
This assignment involves developing a primitive plagiarism detection application
We will use the following simple comparison metric. Let

A
= (
a
0
;a
1
;

;a
n
>>
Ebenezer Bazzleway - Fri, 16 Oct 2015 21:17:28 EST ID:2a+jHHTF No.14938 Ignore Report Quick Reply
You need to tell us more. All you've told us is:
  • you need to develop a primitive plagiarism detection application
Is this to detect when a given text is plagiarism from another?

  • simple comparison metric
This is the metric used to determine similarity?

A=(a0; a1; an
Is this the entire metric? Did you forget to close parentheses or is there more? What are the variables involved?


Show sets are equinumerous by finding a specific bijection between the sets. by Phoebe Hennerlock - Tue, 22 Sep 2015 13:24:29 EST ID:SKWHYv68 No.14909 Ignore Report Reply Quick Reply
File: 1442942669344.png -(18538B / 18.10KB, 462x143) Thumbnail displayed, click image for full size. 18538
Hi /math/, looking for some help. Several problems for my advanced calculus homework follow this format, and I just don't understand how to describe the bijection. I understand bijections, and that finding one proves two sets are equinumerous, but again I don't know how. For instance, this first problem:
a) S = [0,1] and T = [1,4]
What does the answer to this look like, and how can I apply it to other problems of a similar format?
>>
James Cledgewig - Tue, 22 Sep 2015 16:57:33 EST ID:i84x+n57 No.14910 Ignore Report Quick Reply
t = 3s + 1 where s is an element of S and t is an element of T
>>
Emma Gipperhood - Tue, 22 Sep 2015 18:09:49 EST ID:RsLdN3g2 No.14911 Ignore Report Quick Reply
you want to find a function, f, from one set to the other which is injective (no two elements get sent to the same element) and surjective (each element in the "receiving set" has an associated element in the first set). More formally, injectivity is illustrated by: if f(a) = f(b) then a = b, surjectivity: For any b in the receiving set, there exists an element a in the first set such that f(a) = b.

So in the question all it wants you to do is find a function from the first set to the second set that satisfies the above two properties.

So for the first question, we could by inspection try the function f(x) = 3x + 1 like James gave. We now need to show that it's injective and surjective (hence bijective). For injection, suppose we have f(a) = f(b).
Then 3a + 1 = 3b + 1 which implies that a = b.

For surjectivity, if we take any element in T (say b), then we need to show that there exists an element in S such that f(a) = b. So a such that 3a + 1 = b. Because of the sets in question we can just take cases. If b = 4 then 1 works, if b = 1 then 0 works. Done.

Note that functions don't have to be so neat. So we could have taken the function f, defined by f(x) = 1 if x = 0 and f(x) = 4 if x is 1. Then proving it's a bijection is easy.

So the basic strategy is to first "guess" a function based on trial and error, intuition or calculation and then try to show that your guess is a bijection.
>>
Shit Hinkinshaw - Tue, 22 Sep 2015 20:56:00 EST ID:BfM+rTDn No.14912 Ignore Report Quick Reply
>>14911
So wait, maybe I'm just confused about the presentation. [1,4] doesn't mean {1,2,3,4}? Just {1,4}?
>>
Priscilla Serringpit - Wed, 23 Sep 2015 04:07:47 EST ID:i84x+n57 No.14913 Ignore Report Quick Reply
>>14912
It's a continuous interval - the segment of the real number line between 1 and 4, end points inclusive as indicated by the square brackets. Parentheses, on the other hand, indicate that an endpoint is not included in the set.
>>
Priscilla Bruzzlesit - Wed, 23 Sep 2015 14:48:51 EST ID:crJsE1sF No.14915 Ignore Report Quick Reply
>>14912
>>14913

oh yeah, my bad (in my defense I was high - and still am). The first function still works.


2 easy and funny mathematics tricks by No_name - Mon, 21 Sep 2015 17:45:39 EST ID:6m2TPZ+N No.14906 Ignore Report Reply Quick Reply
File: 1442871939125.jpg -(66247B / 64.69KB, 640x640) Thumbnail displayed, click image for full size. 66247
2 easy and funny mathematics tricks

https://youtu.be/CyOZVaWEIr4?list=PLsOOtD1hOoQOSZMGQ-9CHceCPTJ8Wwzzs

and

https://youtu.be/hg3noXFcjko?list=PLsOOtD1hOoQOSZMGQ-9CHceCPTJ8Wwzzs
>>
Nell Clonningdale - Mon, 21 Sep 2015 18:25:45 EST ID:2JYFlB0V No.14907 Ignore Report Quick Reply
113 (a-b concatenated with a+b).

customer pays 0.9*(1.1*price_item_1 + 1.1*price_item_2 + ... + 1.1*price_item_n) = 0.99*(price_item_1 + price_item_2 + ... + price_item_n) < (price_item_1 + price_item_2 + ... + price_item_n) which is the price he would have normally paid.

these aren't mental tricks though, they're puzzles.


Advice on interesting maths topics by Nicholas Blackcocke - Tue, 04 Aug 2015 15:10:07 EST ID:84oIcStQ No.14849 Ignore Report Reply Quick Reply
File: 1438715407888.png -(53664B / 52.41KB, 250x249) Thumbnail displayed, click image for full size. 53664
I'm a maths student going into my third year in September. I've already started looking at the possible modules I can take next year and I'm feeling a little overwhelmed; the number of modules available have increased dramatically and they all seem pretty interesting.

Measure theory, Introduction to Topology, Galois theory, Structures of Complex Systems, Commutative Algebra, Algebraic Topology, Brownian Motion, Geometric Group Theory, Algebraic Geometry, Set Theory, Markov Processes and Stochastic Analysis are the modules I'm considering but there are possibly many interesting modules that I passed up.

I'd be grateful if anyone who has studied any of these topics could let me know how they found them and if anyone's really bored they could have a quick scan of the modules available and let me know if I missed anything interesting:

http://www2.warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/year4/
http://www2.warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/year3/
3 posts omitted. Click Reply to view.
>>
Fuck Worthingway - Wed, 26 Aug 2015 22:48:36 EST ID:Dk8yywxc No.14866 Ignore Report Quick Reply
>>14849

What classes did you end up going for? I'm taking intro to modern algebra, modern analysis, descriptive set theory, and mathematical logic
>>
Cornelius Hammerway - Thu, 03 Sep 2015 20:34:25 EST ID:Ayxv8mCh No.14875 Ignore Report Quick Reply
I am a ph.d student studying lie groups and conservation laws. If I were you, I would take topology, measure theory, markov processes and stochastic analysis, and geometric group theory. My reasoning is because I had similar choices to make in undergrad and that is essentially what I picked. Do what interests you, anon, and be good at that.
>>
Cornelius Hammerway - Thu, 03 Sep 2015 20:35:11 EST ID:Ayxv8mCh No.14876 Ignore Report Quick Reply
>>14866
mathlogic? really? enjoy your topos theory and unemployment.
>>
Frederick Pickford - Wed, 09 Sep 2015 17:36:03 EST ID:LzNXN+l+ No.14887 Ignore Report Quick Reply
>>14876

I think what you're interested in is shit you elitist prick. You seriously think that your particular flavor of group theory or physics is going to put you leagues ahead of other people? Take your own advice, shut the fuck up, and go multiply some matrices rude boy.
>>
Ernest Grimridge - Mon, 21 Sep 2015 17:44:25 EST ID:AdYmD9Mq No.14905 Ignore Report Quick Reply
>>14866
I'm taking Measure Theory, Introduction to Topology, Functional Analysis I, Structure of Complex Systems, Galois Theory, General Relativity and Dynamical Systems.

I need to drop two or three of them though.

>>14854
I'm taking Algebraic Topology for sure next term. Set Theory doesn't really lead on to anything and apparently is kind of easy so I'll do it next year when I'm doing my master's thesis.

>>14875
I don't like random variables so I think I'd find stochastic analysis hard ... sucks as if I had taken it, I could possibly do my masters thesis on it with a fields medalist as my supervisor! :/

Geometric group theory's in term two so I might do it then.

btw, how did you decide to work on lie groups? Did you find it interesting at undergraduate level?
I know it's kind of early but I'm thinking about what I should do for my master's thesis next year however everything seems so interesting that I don't know how I'm going to specialize.


Freaking math man by Hugh Nickledale - Wed, 29 Jul 2015 15:09:16 EST ID:I+l45cST No.14831 Ignore Report Reply Quick Reply
File: 1438196956144.jpg -(22700B / 22.17KB, 560x247) Thumbnail displayed, click image for full size. 22700
So pics related, why is it possible that the original formula can't be solved for x=1, but the formula after applying a series of transformation rules can be solved for x=1.
I mean, it should be the same formula but only written differently, right? Am I "not getting" something extremely basic here?
1 posts omitted. Click Reply to view.
>>
Caroline Blommleshit - Thu, 30 Jul 2015 03:51:15 EST ID:I+l45cST No.14834 Ignore Report Quick Reply
>>14832
Yes, I get how it's done with the limit and all but I'm still confused why, why does factoring allow to solve it at value 1?
What I basically don't get is: if you simplify a formula (let's take the one in the OP, but backwards) it is initally solvable for x=1, but after factoring it isn't anymore. But if you rewrite a formula it should have the same outcome for every value of the variable, right? By factoring the result of that formula is changed for one value, so it basically isn't the same formula anymore?
I'm obviously not getting something fundamental here, I hope this isn't a completely retarded question. Thanks for the reply btw.
>>
Caroline Blommleshit - Thu, 30 Jul 2015 03:57:01 EST ID:I+l45cST No.14835 Ignore Report Quick Reply
Or is the fact that the formula (x³-1)/(x²-1) isn't solvable for the value x=1 just a limitation of the 'excessive' factorization, and that's all there is to it?
>>
Albert Blingerridge - Thu, 30 Jul 2015 13:29:45 EST ID:i84x+n57 No.14836 Ignore Report Quick Reply
>>14834
>But if you rewrite a formula it should have the same outcome for every value of the variable, right? By factoring the result of that formula is changed for one value, so it basically isn't the same formula anymore?
Factoring doesn't change anything. It's the last step where the (x-1) terms are cancelled that changes things, and what you're left with is NOT equal to the original equation. Cancelling these terms is the same as division by zero (which of course is an illegal operation) UNLESS you specify x≠1. This means that the first and last functions are equal for all values of x except x=1. The last function essentially fills the "hole" in at x=1 with the limit i.e. 3/2.
>>
Angus Sommlehood - Sun, 20 Sep 2015 00:19:57 EST ID:xgdLZpNp No.14903 Ignore Report Quick Reply
>Because
>that's
>how
>squaring
>works

You get two answers when you square, because of parabolas, and that's why we have to deal with absolute values, too.
>>
John Brookstock - Sun, 20 Sep 2015 15:40:32 EST ID:BE4o4wO4 No.14904 Ignore Report Quick Reply
>>4175945
START WEARIN PURPLE WEARIN PURPLE
START WEARIN PURPLE FOR ME NOWWWWW


Probability off by factor if 10 by Faggy Blablingbanks - Mon, 14 Sep 2015 16:01:05 EST ID:gibSXrA4 No.14889 Ignore Report Reply Quick Reply
File: 1442260865295.jpg -(45730B / 44.66KB, 329x347) Thumbnail displayed, click image for full size. 45730
Using two different methods to find probability, I'm getting the same number but off by a factor of 10

>Five cards are dealt from 52-card deck; what is the probability that we draw 3 Aces & 2 Kings?
  • Method 1 (wrong):
4/52 + 3/51 + 2/50 + 4/49 + 3/48 = 9.23e-7

  • Method 2:
[(4 choose 3)*(4 choose 2)] / (52 choose 5) = 9.23e-6, which is the right answer she says.

So why is it that method 1 (which seems more intuitive to me) gives an answer that's off by 10?
2 posts omitted. Click Reply to view.
>>
Hedda Dattingmitch - Tue, 15 Sep 2015 17:55:55 EST ID:i84x+n57 No.14893 Ignore Report Quick Reply
Okay, so I was half asleep last night when I read and replied to your question. Sorry if I came off as a dick; that happens when I'm tired.

The trick you're trying to use in method 1 doesn't work, because you're not treating getting an ace or a king as independent variables. To treat them as independent, you need to account for all possible arrangements. You did this:

(prob of 1st ace)(prob of 2nd ace)(prob of 3rd ace)(prob of 1st king)(prob of 2nd king)

But you didn't count arrangements like:

(prob of 1st ace)(prob of 1st king)(prob of 2nd ace)(prob of 2nd king)(prob of 3rd ace)

There are (5 choose 3) = (5 choose 2) = 10 possible arrangements all with the same probability, so you just need to multiply your result in method 1 by 10 to get the correct total probability.
>>
Nigel Mammlewater - Thu, 17 Sep 2015 13:54:24 EST ID:i84x+n57 No.14899 Ignore Report Quick Reply
>>14893
Actually, technically you were treating the random variables as independent simply by multiplying them together as you did. The real issue is that method 1 works on the principle that drawing a single card at a time until you have all 5 is the same as drawing all 5 at once. But since you're drawing them one at a time, you have to account for different orderings. Of course, suits don't matter in the ordering. Method 2 doesn't care about orderings; it's analogous to drawing all 5 cards at once. Method 2 just takes the number of possibilities for drawing a 5 card hand matching the conditions and divides by the number of all possible 5 card hands. This explanation should be clearer.

nb
>>
Alice Cunderchut - Thu, 17 Sep 2015 22:05:25 EST ID:qCDKBx4v No.14900 Ignore Report Quick Reply
Thank for the explanations, I didn't realize that order matters when the variables are independent (it almost seems it should be the other way around), but I'll probably stick to method 2, which is the newer one that was taught to me.

Thanks for teaching me!
>>
Alice Duckstone - Fri, 18 Sep 2015 15:32:30 EST ID:i84x+n57 No.14901 Ignore Report Quick Reply
>>14900
>Thanks for teaching me!
You're welcome.
>>
Angus Sommlehood - Sun, 20 Sep 2015 00:01:12 EST ID:xgdLZpNp No.14902 Ignore Report Quick Reply
>>14893
>>14891
>>14891
>>14891
This was the meanest thing I've seen from this site in a long time.

Bad vibes, man.
Funny bad vibes, fam.


Help Understading by Charles Drenderfuck - Sat, 29 Aug 2015 00:56:12 EST ID:08s9kFJy No.14868 Ignore Report Reply Quick Reply
File: 1440824172492.jpg -(30040B / 29.34KB, 400x346) Thumbnail displayed, click image for full size. 30040
Hey /math/. I think this is the first time I've been on this board, so I hope I'm not being terribly annoying by asking what is probably a very basic question, but I can't find an answer I understand in my textbook and google returns too many results to be useful.

I'm taking a survey of statistics class as part of a dental hygiene degree I'm pursuing and I'm having trouble understanding the procedure of a simple random sample. Specifically, this particular question:
------
You have a sampling frame of 75 people, from whom you wish to randomly select a sample of size 4. Use the portion of the random number table below to determine which people will be selected for the simple random sample, assuming the sampling frame is numbered from 01-75.

68108 89266 94730 95761 75023 48464 65544 96583 18911 16391


---

I'm having trouble understanding exactly what they're asking me to do. Am I supposed to assume that these 10 listen numbers are the first 10 (01-10) of a hypothetical list of 75 total numbers and I should thus pick 68108, 89266, 94730, and 95761 as the answer? Or is this a trick question and I can just pick any 4 of these numbers and they would be correct? Or am I just completely lost?

Thank you in advance for any help.
>>
Phoebe Gallerpitch - Sun, 30 Aug 2015 04:41:15 EST ID:i84x+n57 No.14869 Ignore Report Quick Reply
What they want you to do is to use the random numbers provided to select 4 random numbers in the range of 1-75. Because of the grouping of the random numbers provided (5 digits each), you have to use some sort of algorithm on them to get the 4 you need in the required range (two digit numbers). I came across several algorithms that work from googling; the choice is completely arbitrary. This means you must use whatever method your textbook uses (it must be in there if this question is). Otherwise, it doesn't matter if you take the last 2 digits from each random number (i.e. 08, 66, 30, 61) or the first 2 digits that fall in the range (i.e. 68, 75, 48, 65). My favorite method is putting a decimal point in front of each number, multiplying by the range (75 in this case), and then rounding down (floor function). Using this on the random number table, you get: 51, 66, 71, 71 (duplicate so throw out), and 56. If you were to get 0, it would roll over to 75.

Again, this is completely a matter of convention. You need to do what we can't do by finding out what your professor wants.
>>
Albert Pittham - Sun, 30 Aug 2015 07:29:23 EST ID:08s9kFJy No.14870 Ignore Report Quick Reply
1440934163887.jpg -(186879B / 182.50KB, 1038x1280) Thumbnail displayed, click image for full size.
>>14869

Okay, that actually makes a lot of sense. I appreciate the help, I've got a variety of dyslexia so it's always nice when someone can explain things plainly. Thank you Phoebe. I bequeath you these triangles.
>>
Doris Blytheridge - Sun, 30 Aug 2015 22:46:48 EST ID:i84x+n57 No.14871 Ignore Report Quick Reply
>>14870
No prob. Is that a pic of a crystal?
>>
Albert Blammershit - Mon, 31 Aug 2015 16:20:56 EST ID:08s9kFJy No.14872 Ignore Report Quick Reply
>>14871

Wish I knew mang.


Googol by Phineas Weddlefag - Wed, 01 Apr 2015 01:56:35 EST ID:Qzq4CNd5 No.14679 Ignore Report Reply Quick Reply
File: 1427867795825.jpg -(50114B / 48.94KB, 400x392) Thumbnail displayed, click image for full size. 50114
When will it be relevant?
And the googolplex?
What use is this magnitude?
Is scale true?
8 posts and 3 images omitted. Click Reply to view.
>>
Lillian Duvingbure - Wed, 20 May 2015 19:09:40 EST ID:x6xydNWl No.14739 Ignore Report Quick Reply
>>14732
To be even more precise, i^i = e^(-pi/2)
>>
Fucking Turveydale - Sat, 23 May 2015 17:17:58 EST ID:QIXSgr8C No.14743 Ignore Report Quick Reply
>>14732
>It's weird to consider that the (multiplicative) inverse of the googolplex magnitudes are all found between 0 and 1.

...wat
>>
Edward Goodspear - Sun, 02 Aug 2015 05:16:15 EST ID:KL4t08uw No.14846 Ignore Report Quick Reply
>>14739
That real value you gave is just the principle value. There are an ifinite number of branches when raising a general complex number to a complex power like that.
>>
Nell Bindlewere - Sun, 02 Aug 2015 13:57:37 EST ID:x6xydNWl No.14847 Ignore Report Quick Reply
>>14846
This is, of course, true. I think branch cuts are the only thing I don't like about complex analysis. I like my smooth functions to be well-defined, thank you very much.
>>
Nell Bindlewere - Sun, 02 Aug 2015 13:58:36 EST ID:x6xydNWl No.14848 Ignore Report Quick Reply
>>14847
I guess by well-defined, I mean single-valued (nb)


          by Frederick Pocklock - Mon, 13 Jul 2015 23:37:34 EST ID:rBg8Fq2S No.14815 Ignore Report Reply Quick Reply
File: 1436845054145.jpg -(71439B / 69.76KB, 640x625) Thumbnail displayed, click image for full size. 71439
How do you measure intelligence?
>>
Jack Wubbernene - Tue, 14 Jul 2015 02:39:02 EST ID:i84x+n57 No.14816 Ignore Report Quick Reply
With an IQ test. /thread

In reality, intelligence is much more subtle than this. But IQ tests are the best we have and will give a rough estimate relative to the general population.
>>
Nakura !xMsGPnYjBI - Tue, 14 Jul 2015 15:06:10 EST ID:NAU1L+Of No.14817 Ignore Report Quick Reply
>>14816
Intelligence Scientists are working to improve IQ tests, remember, there's actually many different IQ tests, but only a few in wide use. Along with EQ (Emotional Intelligence) tests. IQ tests are a fantastic way to measure intelligence, although crude compared to the actual complexity of the intelligence of a human being.

I've seen people have intense chess games, losing their minds, battling to settle who loses an IQ point, dark games, for brilliant young minds. Computers are people too.
=)
>>
Molly Pedgeham - Fri, 21 Aug 2015 16:34:45 EST ID:uAPhVfTX No.14862 Ignore Report Quick Reply
>>14817

>Computers are people too.

Computers aren't people, but people are computers.
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Jack Peddlewill - Wed, 09 Sep 2015 10:23:07 EST ID:BG0DExq5 No.14884 Ignore Report Quick Reply
I have a strange view of intelligence. I think if you're a genius at making music or doing math, you might not even be that intelligent, because at that point you're a complex machine that excels at doing a job. I find myself defining intelligence in terms of how stupid you're not. What I mean exactly is this: I define intelligence as how good you are at induction. I think that's strange because solid reasoning is not inductive. On the other hand, if you're a human in a world as complex as this, you're almost never able to get all the information you need in order to acquire an accurate understanding of something. I think your ability to reach a correct understanding with incomplete information is a measure of how intelligent you are. For example, if game changing advances in solar technology happen every few months, and you have a cousin that thinks any real advance in solar technology would be followed by the government assassinating a scientist, your cousin isn't very intelligent. (True story).


Greatest Mathematical Achievements by Samuel Fanfuck - Wed, 27 May 2015 00:57:30 EST ID:So3FHAh7 No.14750 Ignore Report Reply Quick Reply
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In today's world, what's the greatest achievement a mathematician can reach?

What's the greatest "honor" a mathematician can receive?
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Priscilla Blingerlin - Sat, 01 Aug 2015 00:22:09 EST ID:mJf1tiy7 No.14837 Ignore Report Quick Reply
>>14823


Section:
MATHEMATICS
Before they die, aging mathematicians are racing to save the Enormous Theorem's proof, all 15,000 pages of it, which divides existence four ways
ASEEMINGLY ENDLESS VARIETY OF FOOD WAS SPRAWLED OVER SEVERAL TABLES at the home of Judith L. Baxter and her husband, mathematician Stephen D. Smith, in Oak Park, Ill., on a cool Friday evening in September 2011. Canapes, homemade meatballs, cheese plates and grilled shrimp on skewers crowded against pastries, pates, olives, salmon with dill sprigs and feta wrapped in eggplant. Dessert choices included -- but were not limited to -- a lemon mascarpone cake and an African pumpkin cake. The sun set, and champagne flowed, as the 60 guests, about half of them mathematicians, ate and drank and ate some more.
The colossal spread was fitting for a party celebrating a mammoth achievement. Four mathematicians at the dinner- Smith, Michael Aschbacher, Richard Lyons and Ronald Solomon -- had just published a book, more than 180 years in the making, that gave a broad overview of the biggest division problem in mathematics history.
Their treatise did not land on any best-seller lists, which was understandable, given its title: The Classification of Finite Simple Groups. But for algebraists, the 350-page tome was a milestone. It was the short version, the CliffsNotes, of this universal classification. The full proof reaches some 15,000 pages- some say it is closer to 10,000 -- that are scattered across hundreds of journal articles by more than 100 authors. The assertion that it supports is known, appropriately, as the Enormous Theorem. (The theorem itself is quite simple. It is the proof that gets gigantic.) The cornucopia at Smith's house seemed an appropriate way to honor this behemoth. The proof is the largest in the history of mathematics.
And now it is in peril. The 2011 work sketches only an outline of the proof. The unmatched heft of the actual documentation places it on the teetering edge of human unmanageability. "1 don't know that anyone has read everything," says Solomon, age 66, who studied the proof his entire career. (He retired from Ohio State University two years ago.) Solomon and the other three mathe…
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Priscilla Blingerlin - Sat, 01 Aug 2015 00:23:10 EST ID:mJf1tiy7 No.14838 Ignore Report Quick Reply
>>14837



That loss would be, well, enormous. In a nutshell, the work brings order to group theory, which is the mathematical study of symmetry. Research on symmetry, in turn, is critical to scientific areas such as modern particle physics. The Standard Model -- the cornerstone theory that lays out all known particles in existence, found and yet to be found -- depends on the tools of symmetry provided by group theory. Big ideas about symmetry at the smallest scales helped physicists figure out the equations used in experiments that would reveal exotic fundamental particles, such as the quarks that combine to make the more familiar protons and neutrons.
Group theory also led physicists to the unsettling idea that mass itself -- the amount of matter in an object such as this magazine, you, everything you can hold and see -- formed because symmetry broke down at some fundamental level. Moreover, that idea pointed the way to the discovery of the most celebrated particle in recent years, the Higgs boson, which can exist only if symmetry falters at the quantum scale. The notion of the Higgs popped out of group theory in the 1960s but was not discovered until 2012, after experiments at CERN's Large Hadron Collider near Geneva.
Symmetry is the concept that something can undergo a series of transformations -- spinning, folding, reflecting, moving through time -- and, at the end of all those changes, appear unchanged. It lurks everywhere in the universe, from the configuration of quarks to the arrangement of galaxies in the cosmos.
The Enormous Theorem demonstrates with mathematical precision that any kind of symmetry can be broken down and grouped into one of four families, according to shared features. For mathematicians devoted to the rigorous study of symmetry, or group theorists, the theorem is an accomplishment no less sweeping, important or fundamental than the periodic table of the elements was for chemists. In the future, it could lead to other profound discoveries about the fabric of the universe and the nature of reality.
Except, of course, that it is a mess: the equations, corollaries and conjectures of the proof have been tossed amid more than 500 journal articles, some buried in thick vol…
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Priscilla Blingerlin - Sat, 01 Aug 2015 00:25:22 EST ID:mJf1tiy7 No.14839 Ignore Report Quick Reply
>>14838

Simple finite groups are analogous to atoms. They are the basic units of construction for other, larger things. Simple finite groups combine to form larger, more complicated finite groups. The Enormous Theorem organizes these groups the way the periodic table organizes the elements. It says that every simple finite group belongs to one of three families-or to a fourth family of wild outliers. The largest of these rogues, called the Monster, has more than 1053 elements and exists in 196,883 dimensions. (There is even a whole field of investigation called monsterology in which researchers search for signs of the beast in other areas of math and science.) The first finite simple groups were identified by 1830, and by the 1890s mathematicians had made new inroads into finding more of those building blocks. Theorists also began to suspect the groups could all be put together in a big list.
Mathematicians in the early 20th century laid the foundation for the Enormous Theorem, but the guts of the proof did not materialize until midcentury. Between 1950 and 1980-a period which mathematician Daniel Gorenstein of Rutgers University called the "Thirty Years' War" -- heavyweights pushed the field of group theory further than ever before, finding finite simple groups and grouping them together into families. These mathematicians wielded 200-page manuscripts like algebraic machetes, cutting away abstract weeds to reveal the deepest foundations of symmetry. (Freeman Dyson of the Institute for Advanced Study in Princeton, N.J., referred to the onslaught of discovery of strange, beautiful groups as a "magnificent zoo.")
Those were heady times: Richard Foote, then a graduate student at the University of Cambridge and now a professor at the University of Vermont, once sat in a dank office and witnessed two famous theorists -- John Thompson, now at the University of Florida, and John Conway, now at Princeton University -- hashing out the details of a particularly unwieldy group. "It was amazing, like two Titans with lightning going between their brains," Foote says. "They never seemed to be at a loss for some absolutely wonderful and totally off-the-wall techniques for doin…
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Priscilla Blingerlin - Sat, 01 Aug 2015 00:26:13 EST ID:mJf1tiy7 No.14840 Ignore Report Quick Reply
>>14839



Gorenstein envisioned a series of books that would neatly collect all the disparate pieces and streamline the logic to iron over idiosyncrasies and eliminate redundancies. In the 1980s the proof was inaccessible to all but the seasoned veterans of its forging. Mathematicians had labored on it for decades, after all, and wanted to be able to share their work with future generations. A second-generation proof would give Gorenstein a way to assuage his worries that their efforts would be lost amid heavy books in dusty libraries.
Gorenstein did not live to see the last piece put in place, much less raise a glass at the Smith and Baxter house. He died of lung cancer on Martha's Vineyard in 1992. "He never stopped working," Lyons recalls. "We had three conversations the day before he died, all about the proof. There were no good-byes or anything; it was all business."
PROVING IT AGAIN
THE FIRST VOLUME of the second-generation proof appeared in 1994. It was more expository than a standard math text and included only two of 30 proposed sections that could entirely span the Enormous Theorem. The second volume was published in 1996, and subsequent ones have continued to the present -- the sixth appeared in 2005.
Foote says the second-generation pieces fit together better than the original chunks. "The parts that have appeared are more coherently written and much better organized," he says. "From a historical perspective, it's important to have the proof in one place. Otherwise, it becomes sort of folklore, in a sense. Even if you believe it's been done, it becomes impossible to check."
Solomon and Lyons are finishing the seventh book this summer, and a small band of mathematicians have already made inroads into the eighth and ninth. Solomon estimates that the streamlined proof will eventually take up 10 or 11 volumes, which means that just more than half of the revised proof has been published.
Solomon notes that the 10 or 11 volumes still will not entirely cover the second-generation proof. Even the new, streamlined version includes references to supplementary volumes and previous theorems, proved elsewhere. In some ways, that reach speaks to the cumulative nature of mathematics: every proof is a product not only of its time but of all the thousands of years of thought that came before.
In a 2005 article in the Notices of the American Mathematical Society, mathematician E. Brian Davies of King's College London pointed out that the "proof has never been written down in its entirety, may never be written down, and as presently envisaged would not be comprehensible to any single individual." His article brought up the uncomfortable idea that some mathematical efforts may be too complex to be understood by mere mortals. Davies's words drove Smith and his three co-authors to put together the comparatively concise book that was celebrated at the party in Oak Park.
The Enormous Theorem's proof may be beyond the scope of most mathematicians -- to say nothing of curious amateurs -- but its organizing principle provides a valuable tool for the future. Mathematicians have a long-standing habit of proving abstract truths decades, if not centuries, before they become useful outside the field.
"One thing that makes the future exciting is that it is difficult to predict," Solomon observes. "Geniuses come along with ideas that nobody of our generation has had. There is this temptation, this wish and dream, that there is some deeper understanding still out there."
THE NEXT GENERATION
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James Tillingson - Sat, 01 Aug 2015 20:08:09 EST ID:i84x+n57 No.14843 Ignore Report Quick Reply
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>>14837
>>14838
>>14839
>>14840
Cool thanks. That's quite the predicament. I hope the other two guys live to see its completion.


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