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Advanced Calculus by Ian Hibberkit - Mon, 10 Aug 2015 18:35:46 EST ID:AuFFnCvz No.14855 Ignore Report Reply Quick Reply
File: 1439246146147.jpg -(35151B / 34.33KB, 640x480) Thumbnail displayed, click image for full size. 35151
Hey math,

I'm taking the first Advanced Calculus (intro to proof writing) course this fall. I've done some of the work in advance and am not too worried. What were your experiences in your homeland's analogous class?
4 posts omitted. Click Reply to view.
>>
Graham Nickledale - Sat, 10 Oct 2015 19:55:35 EST ID:AuFFnCvz No.14928 Ignore Report Quick Reply
OP reporting back in. Thanks for the resources. Material is super cool, but instructor has modeled it like some kind of competition. The person who has the most points (earned by presenting proofs) sets the level for an A in the class. It's a mathematical deathmatch. Only the strongest will survive.
>>
Simon Bocklebanks - Sun, 25 Oct 2015 13:17:30 EST ID:Dk8yywxc No.14952 Ignore Report Quick Reply
>>14928

That's pretty terrible. What school is that?
>>
Caroline Fubberforth - Mon, 26 Oct 2015 13:56:36 EST ID:nDKmqEtd No.14954 Ignore Report Quick Reply
>>14928
I'd like to see a game theory course graded this way.
>>
Martin Hippershaw - Fri, 30 Oct 2015 22:31:17 EST ID:AuFFnCvz No.14961 Ignore Report Quick Reply
>>14952
I go to a very science-y state university, so the math major is pretty competitive -- a positive thing, but then we have instructors who do this kind of stuff to weed people out. Pretty stressful, man.
>>
Walter Nenderbury - Mon, 14 Dec 2015 02:09:04 EST ID:VQWIWcoe No.14999 Ignore Report Quick Reply
>>14885
Do you have a copy of Galois' Dream? It's a nice text, you might like it.


I wanna be a master by Basil Fongerhood - Wed, 02 Dec 2015 06:52:11 EST ID:tkxf2V39 No.14988 Ignore Report Reply Quick Reply
File: 1449057131350.png -(8745B / 8.54KB, 790x595) Thumbnail displayed, click image for full size. 8745
Can somebody here give me a beginning, to end list of how I should study complete field of math?
3 posts omitted. Click Reply to view.
>>
Wesley Gonnerlock - Mon, 07 Dec 2015 19:17:02 EST ID:tkxf2V39 No.14993 Ignore Report Quick Reply
In what order should I study math, you know what I mean by end.
>>
Cedric Niggerfoot - Thu, 10 Dec 2015 09:32:46 EST ID:Dk8yywxc No.14994 Ignore Report Quick Reply
>>14993

Start by reading "How to prove it", the book. Tell use what parts of that you enjoy, and what you're interested in, and I'll give you a list of books to go through. The way to read a math book and understand it is:
#1: Read through the chapter once
#2: Read and memorize all the definitions
#3: Read and memorize statements of all theorems in order
#4: Read and make sure you deeply understand ALL the proofs
#5: Attempt exercises. Reread the chapter multiple times.

I want to emphasize that you can't just fly through a chapter and get anything out of it.
>>
Caroline Brookcocke - Sat, 12 Dec 2015 16:11:59 EST ID:k9kkb1hx No.14996 Ignore Report Quick Reply
>>14988
http://us.metamath.org/index.html

This site tries to enumerate all proofs and how they are connected. You can't really do anything with it though and trying to learn Mathematics using it would be like trying to learn driving by watching assembly line robots build cars.
>>
Hannah Fandale - Sun, 13 Dec 2015 12:10:00 EST ID:Dk8yywxc No.14997 Ignore Report Quick Reply
>>14995

Yep that's the one.
>>
Walter Nenderbury - Mon, 14 Dec 2015 02:01:28 EST ID:VQWIWcoe No.14998 Ignore Report Quick Reply
The correct answer is "it's impossible". There's already more math than a person could digest in a lifetime, and more is made/found every day.

But doing the standard progression arithmetic->algebra/geometry/precalc -> calc, then some proof fundamentals, then hit intro level of diff eq, linear algebra, combinatorics and probability, group theory, and you'll be in a good place to head off in whatever direction you like.

ie read off a programme listing for most undergrad math programs.


ti-83 window ranges by Samuel mcshoebat - Sat, 21 Nov 2015 16:32:15 EST ID:ONOBjqOA No.14985 Ignore Report Reply Quick Reply
File: 1448141535597.jpg -(7581B / 7.40KB, 260x194) Thumbnail displayed, click image for full size. 7581
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?
>>
Lydia Maddleworth - Tue, 24 Nov 2015 13:59:54 EST ID:J4OUpAxW No.14986 Ignore Report Quick Reply
>>14985
i report yu


GRE prep help by Samuel Subblestone - Fri, 22 May 2015 18:34:24 EST ID:Kjt/my9S No.14740 Ignore Report Reply Quick Reply
File: 1432334064046.png -(257289B / 251.26KB, 1600x900) Thumbnail displayed, click image for full size. 257289
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.
5 posts omitted. Click Reply to view.
>>
Thomas Billingdale - Thu, 28 May 2015 23:41:17 EST ID:1XOUKwXv No.14757 Ignore Report Quick Reply
>>14756
Ill try my best. Take equations

1: 5 = 5
2: 10 = 10

Say I want to find equation 2 divided by equation 1, that is

We get
10/ 5 = 10 / 5
Reducing...
2 = 2

I can divide equation 2 by 1 for the same reason I can do the following (which may be a bit simpler)
Comment too long. Click here to view the full text.
>>
Thomas Billingdale - Thu, 28 May 2015 23:46:46 EST ID:1XOUKwXv No.14758 Ignore Report Quick Reply
>>14756
Holy fuck, ok i just actually read the question you are working on. What a fucking stupid way of solving this.

Wow, no words. Here is the problem

A is 250% of B, that is A = 2.5 * B
(Can think of it like, A is 100% of B, or A = B, if A is 90% of B then A = .9 B)
Gee, I wonder what % of B A is?
Divide both sides by 2.5...
B = (1/2.5)*A = .4*A
Presto, B = 40% of A
no need for all that useless shit
>>
Thomas Billingdale - Thu, 28 May 2015 23:49:32 EST ID:1XOUKwXv No.14759 Ignore Report Quick 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 ID:nQdck93t No.14769 Ignore Report Quick 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 ID:SYdGx2TL No.14984 Ignore Report Quick Reply
dumb jolly african-american confirmed


Is this legal? by Jarvis Blathershit - Wed, 11 Nov 2015 17:34:08 EST ID:6QpDX4yc No.14970 Ignore Report Reply Quick Reply
File: 1447281248240.jpg -(835996B / 816.40KB, 1920x1080) Thumbnail displayed, click image for full size. 835996
I think it can't be done... but how else can I simplify this? I was years ago I had to do that...
1 posts omitted. Click Reply to view.
>>
Nathaniel Nonkinwed - Wed, 18 Nov 2015 06:40:51 EST ID:g17i/tIt No.14979 Ignore Report Quick Reply
>>14971
That's wrong.
>>
Phineas Bungold - Wed, 18 Nov 2015 11:39:40 EST ID:i84x+n57 No.14980 Ignore Report Quick Reply
>>14979
No, he's right. Multiplying both the top and bottom by xy gives you (x^2-y^2)/(x^2+y^2+2xy) = (x^2-y^2)/(x+y)^2 = (x+y)(x-y)/(x+y)^2 = (x-y)/(x+y). As long as x+y is nonzero, this should work.
>>
Phineas Bungold - Wed, 18 Nov 2015 11:45:09 EST ID:i84x+n57 No.14981 Ignore Report Quick 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 ID:aYPLZoUV No.14982 Ignore Report Quick 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.
>>
Albert Bidgedock - Thu, 19 Nov 2015 01:46:52 EST ID:GKhdNJGT No.14983 Ignore Report Quick Reply
>>14979

Try again.


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

Is there a third dimension I'm not seeing?
>>
George Drummerwell - Tue, 27 Oct 2015 10:46:31 EST ID:d7QflcOd No.14957 Ignore Report Quick Reply
>>14956
possibly most
>>
James Brannerman - Tue, 27 Oct 2015 14:09:02 EST ID:i84x+n57 No.14958 Ignore Report Quick Reply
Double integrals are for volume, bro.
>>
James Brannerman - Tue, 27 Oct 2015 14:54:03 EST ID:i84x+n57 No.14959 Ignore Report Quick 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 ID:REpT3xaI No.14960 Ignore Report Quick Reply
1445976718356.jpg -(109848B / 107.27KB, 563x750) Thumbnail displayed, click image for full size.
That makes good sense, thank you James


Time by Shower thought - Sat, 01 Aug 2015 17:37:45 EST ID:eAsqVIz1 No.14841 Ignore Report Reply Quick Reply
File: 1438465065277.jpg -(32969B / 32.20KB, 342x392) Thumbnail displayed, click image for full size. 32969
I have just stumbled upon a question while in the shower, what is the speed of time?
21 posts and 4 images omitted. Click Reply to view.
>>
Polly Pittgold - Fri, 25 Sep 2015 03:48:03 EST ID:i84x+n57 No.14918 Ignore Report Quick Reply
>>14914
>>14917
Planck time is around 5.39*10^-44 seconds. The Planck scale is where quantum gravitational effects become significant.

I swear, you guys are way overthinking OP's question.
>>
Augustus Gubblesare - Fri, 25 Sep 2015 19:08:49 EST ID:GKhdNJGT No.14919 Ignore Report Quick Reply
>>14918
>around 5.39

AH that's better.
>>
Fanny Wivingdale - Wed, 21 Oct 2015 15:16:21 EST ID:bgR+wrEg No.14946 Ignore Report Quick 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 ID:GKhdNJGT No.14948 Ignore Report Quick 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 ID:sT1kFNdT No.14955 Ignore Report Quick 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 by Phineas Bunforth - Sat, 03 Oct 2015 02:04:23 EST ID:tiSYuoE5 No.14925 Ignore Report Reply Quick Reply
File: 1443852263874.jpg -(9330B / 9.11KB, 480x360) Thumbnail displayed, click image for full size. 9330
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 ID:i84x+n57 No.14926 Ignore Report Quick 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
>>
William Firrylack - Sat, 03 Oct 2015 14:10:01 EST ID:GKhdNJGT No.14927 Ignore Report Quick Reply
>>14926

Get that metric shit out of here.
>>
Reuben Gigglepot - Sun, 25 Oct 2015 19:34:19 EST ID:JsLKELFF No.14953 Ignore Report Quick 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" by Archie Gennerlark - Thu, 01 Oct 2015 01:08:46 EST ID:M30RiRC5 No.14920 Ignore Report Reply Quick Reply
File: 1443676126033.jpg -(109223B / 106.66KB, 640x465) Thumbnail displayed, click image for full size. 109223
What does it mean to be "good at math?"
3 posts omitted. Click Reply to view.
>>
Archie Snodfield - Thu, 01 Oct 2015 20:23:53 EST ID:i84x+n57 No.14924 Ignore Report Quick Reply
To me it means someone excels at using abstract reasoning to solve problems. To most people it means someone can follow a formulaic procedure to solve a particular type of problem.
>>
Doris Goodville - Wed, 21 Oct 2015 15:43:01 EST ID:d7QflcOd No.14947 Ignore Report Quick Reply
applying the concepts derived form it, and also the feeling of knwoing that time spent in actualy organizing the several different patterns of thinking that revolves around the " eqaution " gives, which is very important because after a while what noramlly takes 9 hours to solve, becomes well, 8.9 hours and with repetition the number decreases and more insight to these new patterns of thinking and unifying the relation between different symbols becomes apperant, i think anyways
>>
Basil Cricklehune - Sat, 24 Oct 2015 22:42:32 EST ID:G+WyMdI9 No.14950 Ignore Report Quick Reply
like woah
>>
Barnaby Clullerfoot - Sun, 25 Oct 2015 04:15:05 EST ID:AuBkeUkZ No.14951 Ignore Report Quick Reply
When you can sign Brian McKnight - Back At One
>>
Jarvis Clayson - Sun, 01 Nov 2015 13:12:36 EST ID:ryc+8ox4 No.14962 Ignore Report Quick Reply
>>14951
what do you mean? nb


Physics question about elasticity by Rasclot - Sat, 17 Oct 2015 18:55:11 EST ID:gpkCKW1u No.14941 Ignore Report Reply Quick Reply
File: 1445122511224.jpg -(81568B / 79.66KB, 800x800) Thumbnail displayed, click image for full size. 81568
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 ID:i84x+n57 No.14943 Ignore Report Quick 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 ID:gpkCKW1u No.14949 Ignore Report Quick 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 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.


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