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420chan is Getting Overhauled - Changelog/Bug Report/Request Thread (Updated July 26)

ti-83 window ranges

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- Sat, 21 Nov 2015 16:32:15 EST ONOBjqOA No.14985
File: 1448141535597.jpg -(7581B / 7.40KB, 260x194) Thumbnail displayed, click image for full size. ti-83 window ranges
Hey got a question for /math/ i'm doing an assignment where I have to graph numbers in the double digit thousands (1600, 1500 etc) what are the best ranges to set on my window so I can be able to view the graph efficiently?

GRE prep help

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- Fri, 22 May 2015 18:34:24 EST Kjt/my9S No.14740
File: 1432334064046.png -(257289B / 251.26KB, 1600x900) Thumbnail displayed, click image for full size. GRE prep help
Hey quick question for you fine folks here at /math/

So I'm studying for the GRE using Gruber's complete guide 2015. There was a math question that confused me. Basically to get the answer you must divide two seperate equations by each other.

The question is: If A is 250 percent of B, what percent of A is B?

Eventually, there comes a point where we have A = 250/100B and x/100A = B.

After that the author says we divide the first equation by the second, so we get A/(x/100A)=(250/100B)/B.

After this it's fairly simple, I'm just having trouble grasping how we can divide one separate equation by another. Could one of you clarify?

Pic Related, it's a screenshot of the part I was talking about.
7 posts omitted. Click View Thread to read.
>>
Thomas Billingdale - Thu, 28 May 2015 23:49:32 EST 1XOUKwXv No.14759 Reply
>>14758
I just gotta vent this out, whoever wrote the solution to this in your homework up there is a fucking moron. This is a 1 step problem
A = 2.5 * B
(1/2.5) * A = B

Go slap whoever the fuck gave that solution to you
>>
James Gurrychetch - Mon, 01 Jun 2015 18:05:14 EST nQdck93t No.14769 Reply
>>14759
Hey I just wanted to reply and say thanks for the help to you and everyone else in this thread. When you write it like that it is very simple, I don't know why he gave such a complicated solution. NB
>>
Simon Geffingfedging - Sat, 21 Nov 2015 04:11:03 EST SYdGx2TL No.14984 Reply
dumb jolly african-american confirmed

Is this legal?

View Thread Reply
- Wed, 11 Nov 2015 17:34:08 EST 6QpDX4yc No.14970
File: 1447281248240.jpg -(835996B / 816.40KB, 1920x1080) Thumbnail displayed, click image for full size. Is this legal?
I think it can't be done... but how else can I simplify this? I was years ago I had to do that...
3 posts omitted. Click View Thread to read.
>>
Phineas Bungold - Wed, 18 Nov 2015 11:45:09 EST i84x+n57 No.14981 Reply
>>14980
>As long as x+y is nonzero, this should work.
Pfft, never mind. If that were the case, the original expression would be undefined as well.
>>
Eliza Sudgefidge - Thu, 19 Nov 2015 00:05:39 EST aYPLZoUV No.14982 Reply
1447909539540.gif -(237904B / 232.33KB, 640x480) Thumbnail displayed, click image for full size.
>>14970

>Is this legal?

No, that will get you 5 to 7 in state prison. But you'll be out in 3 1/2 with good behavior.

double integrals are wrong

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- Mon, 26 Oct 2015 20:23:29 EST REpT3xaI No.14956
File: 1445905409356.jpg -(286545B / 279.83KB, 1725x930) Thumbnail displayed, click image for full size. double integrals are wrong
basic geometry says calculus 3 is incorrect

Is there a third dimension I'm not seeing?
1 posts omitted. Click View Thread to read.
>>
James Brannerman - Tue, 27 Oct 2015 14:09:02 EST i84x+n57 No.14958 Reply
Double integrals are for volume, bro.
>>
James Brannerman - Tue, 27 Oct 2015 14:54:03 EST i84x+n57 No.14959 Reply
>>14958
To elaborate, the double integral in the OP represents the volume of a pyramid with a square base at x = 3 and its apex at the origin. This means the base has area 3^2 and height 3. The volume of a pyramid is V = 1/3*b*h = 1/3(9)(3) = 9.
>>
Jack Sellykut - Tue, 27 Oct 2015 16:11:58 EST REpT3xaI No.14960 Reply
1445976718356.jpg -(109848B / 107.27KB, 563x750) Thumbnail displayed, click image for full size.
That makes good sense, thank you James

Time

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- Sat, 01 Aug 2015 17:37:45 EST eAsqVIz1 No.14841
File: 1438465065277.jpg -(32969B / 32.20KB, 342x392) Thumbnail displayed, click image for full size. Time
I have just stumbled upon a question while in the shower, what is the speed of time?
23 posts and 4 images omitted. Click View Thread to read.
>>
Fanny Wivingdale - Wed, 21 Oct 2015 15:16:21 EST bgR+wrEg No.14946 Reply
>>14919

the poster just rounded 5.39e-44 to 1e-43. i wouldnt use that in a calculation though
>>
Cornelius Gorryban - Thu, 22 Oct 2015 22:46:02 EST GKhdNJGT No.14948 Reply
>>14946
I know, but it's kind of like me rounding my height up to 10 feet tall when I'm really only about 5 and a half feet tall. It wouldn't have bugged me so much if he had at least said "approximately 1e-43", even though it's barely even half that amount. I mean, If you're going to round to zero digits of precision, why not make it 5e-44? Plus in the next sentence, he used EIGHT digits of precision!

It's cool, though. I'm over it now.
>>
Barnaby Blommerstudge - Mon, 26 Oct 2015 16:15:26 EST sT1kFNdT No.14955 Reply
>>14842

thought you do have a point in terms of the psychologcal there is an actual determinator which directly shows the value of what OP speaks, but though it seems that rather because of the psychologi involved there is beleived for a long time to be a fluxuation in this, and tough it is true in certain " frames of visioN" there is not a trace of

i want the fommula to a shitload vs an assload

View Thread Reply
- Sat, 03 Oct 2015 02:04:23 EST tiSYuoE5 No.14925
File: 1443852263874.jpg -(9330B / 9.11KB, 480x360) Thumbnail displayed, click image for full size. i want the fommula to a shitload vs an assload
can any of you meet my challenge that i am challenging here today

it is monday monday gotta get down on monday!
>>
Lillian Shittingwater - Sat, 03 Oct 2015 02:46:17 EST i84x+n57 No.14926 Reply
According to circlejerk:

buttload * 10 = 1 butt ton
butt ton * 10 = 1 assload
assload * 10 = 1 asston
asston * 10 = 1 shitload
shitload * 10 = 1 shitton
shitton * 10 = 1 fuckload
fuckload * 10 = 1 fuckton
>>
Reuben Gigglepot - Sun, 25 Oct 2015 19:34:19 EST JsLKELFF No.14953 Reply
take a pen

take a pencil

grab a pen

grab a pencil

paper or write on you will do

eventually

discover what you seek

foremost you have done

"Good at math"

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- Thu, 01 Oct 2015 01:08:46 EST M30RiRC5 No.14920
File: 1443676126033.jpg -(109223B / 106.66KB, 640x465) Thumbnail displayed, click image for full size. "Good at math"
What does it mean to be "good at math?"
5 posts omitted. Click View Thread to read.
>>
Barnaby Clullerfoot - Sun, 25 Oct 2015 04:15:05 EST AuBkeUkZ No.14951 Reply
When you can sign Brian McKnight - Back At One

Physics question about elasticity

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- Sat, 17 Oct 2015 18:55:11 EST gpkCKW1u No.14941
File: 1445122511224.jpg -(81568B / 79.66KB, 800x800) Thumbnail displayed, click image for full size. Physics question about elasticity
Yo, so I should preface this by saying this is a physics question, but it is quite mathematical.
My issue right now with computational physics research is that I am trying to prove a 2d simplification of a 3d model of a vesicle has an elastic response to applied forces. My 2d simplification of a 3d vesicle, a sphere of lipids containing a fixed volume of water, is a loop containing a fixed area. It is a discrete model in which I move the vertices representing the loop in order to reduce line-tension, but however never does so in a manner which changes area (to linear order). The issue is that when I apply an outward gaussian force to this bitch, for small magnitude forces, it converges to a new length. My best fits of this regime look like newlength=k(fapplied)^2+lengthoriginal. While this is interesting, and I am glad that my loop isn't diverging in these cases, I am not sure whether that relation actually describes an elastic surface/the elastic modulus of such a surface. For high forces it doesn't converge at all.

Anyhoo, how would you go about calculating the elastic relationship between the length of an elastic loop and the force applied to it?


Any help would be much appreciated.
Best,
A human.
>>
John Hennerkut - Mon, 19 Oct 2015 10:09:39 EST i84x+n57 No.14943 Reply
>Anyhoo, how would you go about calculating the elastic relationship between the length of an elastic loop and the force applied to it?

That depends on how the force is applied. Is it two parallel lines squeezing the loop from opposite sides?
>>
Cedric Guckleworth - Fri, 23 Oct 2015 18:58:36 EST gpkCKW1u No.14949 Reply
1445641116924.gif -(190352B / 185.89KB, 2000x2000) Thumbnail displayed, click image for full size.
>>14943
it's an elastic loop, think rubber band.

I found a hint at the answer here:http://www.wired.com/2012/08/do-rubber-bands-act-like-springs/
I also found an APS journal article about it that is more rigorous American Journal of Physics 31, 938 (1963); doi: 10.1119/1.1969212

Solving Recurrence Relations

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- Fri, 16 Oct 2015 21:23:55 EST pCKCgraP No.14939
File: 1445045035508.png -(14284B / 13.95KB, 269x235) Thumbnail displayed, click image for full size. Solving Recurrence Relations
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?
1 posts and 1 images omitted. Click View Thread to read.
>>
Phyllis Pemmerhut - Mon, 19 Oct 2015 18:39:21 EST xJK33YUo No.14944 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 i84x+n57 No.14945 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

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- Mon, 12 Oct 2015 00:12:03 EST P4gPE0kz No.14933
File: 1444623123652.png -(4785B / 4.67KB, 455x54) Thumbnail displayed, click image for full size. Trig troubles
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 i84x+n57 No.14935 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 i84x+n57 No.14937 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)

simplify: sin^4(x) = 3/8 - 1/2*cos(2x) +1/8*cos(4x)

sub 3x for x: sin^4(3x) = 3/8 - 1/2*cos(6x) +1/8*cos(12x)

Help me to understand

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- Tue, 13 Oct 2015 17:55:25 EST 43kF2tms No.14936
File: 1444773325462.gif -(688596B / 672.46KB, 205x160) Thumbnail displayed, click image for full size. Help me to understand
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 2a+jHHTF No.14938 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.

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- Tue, 22 Sep 2015 13:24:29 EST SKWHYv68 No.14909
File: 1442942669344.png -(18538B / 18.10KB, 462x143) Thumbnail displayed, click image for full size. Show sets are equinumerous by finding a specific bijection between the sets.
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?
2 posts omitted. Click View Thread to read.
>>
Shit Hinkinshaw - Tue, 22 Sep 2015 20:56:00 EST BfM+rTDn No.14912 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 i84x+n57 No.14913 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 crJsE1sF No.14915 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

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- Mon, 21 Sep 2015 17:45:39 EST 6m2TPZ+N No.14906
File: 1442871939125.jpg -(66247B / 64.69KB, 640x640) Thumbnail displayed, click image for full size. 2 easy and funny mathematics tricks
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 2JYFlB0V No.14907 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

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- Tue, 04 Aug 2015 15:10:07 EST 84oIcStQ No.14849
File: 1438715407888.png -(53664B / 52.41KB, 250x249) Thumbnail displayed, click image for full size. Advice on interesting maths topics
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/
5 posts omitted. Click View Thread to read.
>>
Cornelius Hammerway - Thu, 03 Sep 2015 20:35:11 EST Ayxv8mCh No.14876 Reply
>>14866
mathlogic? really? enjoy your topos theory and unemployment.
>>
Frederick Pickford - Wed, 09 Sep 2015 17:36:03 EST LzNXN+l+ No.14887 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 AdYmD9Mq No.14905 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.

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