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Greetings /π/ by Charles Murdworth - Sun, 20 Jul 2014 05:35:27 EST ID:OWPvoCXf No.14222 Ignore Report Reply Quick Reply
1405848927891.jpg -(79370 B, 911x838) Thumbnail displayed, click image for full size. 79370
I'm here with a burning question, and I'm a dipshit and this shit flies right above my head.
Let's say what if this guy actually figured out a valid formula for how frequencies interact.

https://www.youtube.com/watch?v=jVATlX4XKMk&list=PL0838D67BB8E0396C


So I will paraphrase my story. Basically, I have a jar of mushroom vodka currently fermenting in the corner of my room.
To keep the jar cool I have a small fan running.
The fan despite being cheap is a fast and efficient beast with a powerful motor.
I have verified that the motor's vibration is vibrating precisely at 459 HZ G#

>Question
So because the alleged effects of frequencies on our brains I have to assume that the binaural effect created by my room fan in conjunction with remarkably wide room resonance
is going to be doing something to my brain, and Maybe this is an opportunity to test the so called theory in analog manner.
I'll post a diagram to show exactly how this room effect is achieved. I tried to get an audio recording of it, but now luck.. maybe if I had a set of sensitive binaural mics.
Comment too long. Click here to view the full text.
1 posts omitted. Click Reply to view.
>>
William Paddlenutch - Tue, 22 Jul 2014 14:47:48 EST ID:Gw2IN3ba No.14228 Ignore Report Quick Reply
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I seriously hope you're trolling OP. Also, wtf is mushroom vodka?

This is now a timecube thread.

http://www.timecube.com/

http://youtu.be/Tn2UCqL5qyo
>>
Nigel Murdville - Tue, 22 Jul 2014 16:02:10 EST ID:gU9kqsUV No.14229 Ignore Report Quick Reply
1406059330710.jpg -(62107 B, 640x480) Thumbnail displayed, click image for full size. 62107
>>14228
what am i reading here
>>
William Paddlenutch - Tue, 22 Jul 2014 19:09:07 EST ID:Gw2IN3ba No.14231 Ignore Report Quick Reply
1406070547200.jpg -(128255 B, 1280x716) Thumbnail displayed, click image for full size. 128255
>>14229
>what am i reading here

A paradigm shift that will rock your world to its cubic core. You are beginning to understand that everything you were ever taught before the realization of cubic time amounts to nothing!
>>
Molly Worthingbury - Wed, 23 Jul 2014 02:31:55 EST ID:v2qInvsx No.14235 Ignore Report Quick Reply
>>14228
YOU ARE NOT THE WISEST HUMAN GENE RAY, YOU FILTHY WORD ANIMAL
>>
Hugh Turveyhall - Thu, 24 Jul 2014 03:02:04 EST ID:lx6v5XWF No.14244 Ignore Report Quick Reply
"a set of sensitive binaural mics."

Do you mean "two microphones"?
Or, if the motor's vibration at "precisely" 459hz doesn't change, "one microphone"? You do realize that "how frequencies interact" is addition, right?


is math real by Ernest Nopperbury - Tue, 03 Jun 2014 16:57:49 EST ID:Uq/TU4om No.14060 Ignore Report Reply Quick Reply
1401829069682.gif -(1375 B, 60x60) Thumbnail displayed, click image for full size. 1375
do discrete systems even exist outside of conceptual renderings?
I guess our number system came out of counting things by orders of ten, using fingers to represent whatever was being counted, but the map is not the territory and information doesn't translate into number of fingers so easily.. In a digital world, things stick to the program, but I can't really tell if reality is quantifiable, if you know what I mean.
2 posts omitted. Click Reply to view.
>>
Matilda Wankinnure - Sat, 05 Jul 2014 22:10:50 EST ID:kFCSHfPa No.14174 Ignore Report Quick Reply
>do discrete systems even exist outside of conceptual renderings?

/pss/ fag here. I think we can take a few different tacks with this.

First: how is it that we can conceive of discrete systems if none exist in nature? For that matter, what, exactly, is going on when we conceive of a discrete system? Does said system exist in our brains somehow?

Secondly: mathematics isn't all discrete. There are mathematical continua. Couldn't those be real?
>>
Martha Drummleman - Wed, 16 Jul 2014 02:37:20 EST ID:jEbtLayo No.14216 Ignore Report Quick Reply
Do our thoughts exist physically? Surely our thoughts are some kind of representation of the physical world. But are the two the same thing?
>>
Charlotte Grimridge - Thu, 17 Jul 2014 20:23:26 EST ID:Dn4zhXKX No.14218 Ignore Report Quick Reply
>>14174
There are quite a few Pythagoreans who talk about pi as an algorithm, rather than a number, and who think discrete math is true to reality and continuous math is conceited. Doron Zeilberger, for one.

Seems to go hand in hand with digital physics and universe/brain-as-computer doctrine. Personally, I don't care about whatever's the most empirical metaphysics; thought and experience are continuous.
>>
Archie Wazzletere - Sat, 19 Jul 2014 04:01:08 EST ID:d7wWJNnk No.14219 Ignore Report Quick Reply
In mathematics there are structures and theories. Structures are things like the natural numbers or real numbers, and theories are the collection of statements deduced from axioms using deduction rules. Mathematics attempts to describe these structures using theories, but due to the incompleteness theorems we know that we will never be able to know all the true facts about, say, the natural numbers. But it's clear that natural numbers exist (as counting objects) and the continuum exists as well (as distances between objects). The theories on the other hand, that attempt to describe these structures, are purely human constructs that optimistically try to approach the truth about the structures.

I like to think of mathematics sort of like a biography about a real person. The subject really exists as a corporeal thing (structures), but the biography is a human construct that tries to tell the true facts about the subject. It should be clear that a biography could never tell every little fact about a person, but if it's a really great person they're worth writing about even if you can't tell it all! What makes the biography of mathematics special is that it is a biography of the most fundamental concepts of existence.

In short, both discrete and continuous things are real. Our descriptions of them in mathematics cannot encompass all the truth about them, but what they are about is fundamentally real.

>>14218

I've heard of these people being called constructivists rather than pythagoreans. Or finitists, like zeilberger. They don't deny the possibility of continuity, the rationals are continuous and none of these guys would deny the existence of fractions. What they stand against is the notion of a "completed" infinity, the idea of being able to produce any infinite ordinal is a good point but I think is an over-reaction to a misperception of infinity in the public.
>>
Shit Smallman - Thu, 24 Jul 2014 02:50:21 EST ID:VbY4tW0M No.14243 Ignore Report Quick Reply
>>14219
Math, at least pure math, consists entirely of induction and rationalism. Any math proof you read is a priori which is fucking insane.


Simple Algebra Question by Barnaby Hadgeville - Tue, 15 Jul 2014 15:01:32 EST ID:qCDKBx4v No.14213 Ignore Report Reply Quick Reply
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Can someone explain in a very verbose way how the first term = the second term in this image?
>>
Shit Nicklewater - Tue, 15 Jul 2014 15:26:08 EST ID:Fj/YvlCk No.14214 Ignore Report Quick Reply
Yesh. Remember that a root is the same as a (fractional) power. That is:
(a+b)^2 =/= a^2+b^2 but
(ab)^2 = a^2b^2

You can factor out 4 below: sqrt( 4 * (-x^2-x) ) which is sqrt(4)*sqrt(-x^2-x)
And sqrt(4) = 2 which cancels out the 2 above.

Goddamn I wish this website supported latex.
>>
Jarvis Munningman - Tue, 15 Jul 2014 20:33:17 EST ID:qCDKBx4v No.14215 Ignore Report Quick Reply
Thank you sir. I think I'll just keep this 1 thread for all my basic questions.
>>
Hugh Turveyhall - Thu, 24 Jul 2014 02:48:19 EST ID:lx6v5XWF No.14242 Ignore Report Quick Reply
(a^m * b)^1/m = (a^m)^1/m * b^1/m = a * b^1/m


Business Math by ConfusedSoul - Wed, 23 Jul 2014 00:12:37 EST ID:BfGCwHN9 No.14233 Ignore Report Reply Quick Reply
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I, for the life of my cannot figure this solution out. This is the last question on my study guide and I am totally stumped:
On April 1st, the unpaid balance in an account was $174. A payment of $70 was made on April 8th. On April 26 a $26.00 purchase was made. The interest rate per month was 1.75% per month of the average daily balance. Find the finance charge and new balance at the end of April.
2 posts and 1 images omitted. Click Reply to view.
>>
Whitey Billingshaw - Wed, 23 Jul 2014 10:19:31 EST ID:2aEkBtv5 No.14237 Ignore Report Quick Reply
((7(174)+18(104)+5(130))/30)(0.0175)+130=132.18

Yes? No?
>>
Fucking Ponderseg - Wed, 23 Jul 2014 10:29:38 EST ID:GT+uOqbr No.14238 Ignore Report Quick Reply
"Business Math" is just story problems from 3rd grade. Math with a narrative should not be considered a separate type of math. Whoever thought of this needs to die.
>>
Molly Worthingbury - Wed, 23 Jul 2014 17:23:12 EST ID:v2qInvsx No.14239 Ignore Report Quick Reply
>>14236
Sorry I was a dick. I'm assuming the "finance charge" is the interest charged at the end of the month.

To find the interest, you need to know the average daily balance. There were 8 days at $174, 17 days at $104, and 4 days at $78. That's an average of $115.73. So the finance charge is .0175*115.73. And the balance is 104-the finance charge.
>>
Molly Worthingbury - Wed, 23 Jul 2014 17:30:25 EST ID:v2qInvsx No.14240 Ignore Report Quick Reply
>>14239
Oh shit, nevermind. It's a credit card.

There were 8 days at $174, 17 days at $104, and 4 days at $130. That's an average of $123. So the finance charge is .0175*123. And the balance is 130 plus the finance charge.
>>
Hugh Turveyhall - Thu, 24 Jul 2014 02:38:26 EST ID:lx6v5XWF No.14241 Ignore Report Quick Reply
Break it down into smaller problems.
First, find the average daily balance.

So the initial balance is $174, and this balance lasts for 7 days.
On the 8th, $70 is paid off, and so the new balance of $174 - $70 lasts for 18 days.
On the 26th, a $26 purchase is made, so the new balance is $174 - $70 + $26, and this lasts through the end of the month.

Thirty days hath September, April, June, and November, so that balance lasts for 5 days.

So the average daily balance is:
[ 7 * 174 + 18 * (174 - 70) + 5 * (174 - 70 + 26) ] / 30

"Cent" means 100 (century, centurion, 100 cents in a dollar), and "per" means "divide", so "per cent" means "divide this by 100"

( 1.75/100 )
Comment too long. Click here to view the full text.


Get X Y by Matilda Daggleville - Mon, 21 Jul 2014 21:33:34 EST ID:Bbuuretd No.14226 Ignore Report Reply Quick Reply
1405992814795.png -(15867 B, 439x305) Thumbnail displayed, click image for full size. 15867
If i have a angle to an object and i know the distance to it. how can i know its coordinates?
made in geogebra
>>
William Paddlenutch - Tue, 22 Jul 2014 04:30:42 EST ID:Gw2IN3ba No.14227 Ignore Report Quick Reply
(r*cos(theta), r*sin(theta)) where r is the distance and theta is the angle

http://en.wikipedia.org/wiki/Polar_coordinate_system#Converting_between_polar_and_Cartesian_coordinates
>>
Nathaniel Woshshaw - Tue, 22 Jul 2014 18:40:14 EST ID:yaOOvKGH No.14230 Ignore Report Quick Reply
>>14227
That's assuming the observer is in (0,0) of course.
>>
James Gerrynut - Tue, 22 Jul 2014 23:26:00 EST ID:Vq7cxA7H No.14232 Ignore Report Quick Reply
>>14230
thanks m8


i literally just ate my taxes by Thomas Chunninghore - Sun, 20 Jul 2014 23:50:21 EST ID:Q0uL8y2L No.14225 Ignore Report Reply Quick Reply
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DICKS EVERYWHERE


Pre-Calc question by Emma Docklegold - Sat, 19 Jul 2014 21:37:36 EST ID:sTqzkRzm No.14220 Ignore Report Reply Quick Reply
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I'm having a problem with a strange question, my textbook doesn't exactly give an explanation on how to do this one, but I'm hoping you can make a little bit more sense for me.

5.) Maximizing Revenue: The price p (in dollars) and the quantity x sold of a certain product obey the demand equation:

p= -5x +100 0 < p ≤ 20

a.) express the revenue R as a function of x

the back of the book says the answer is:

-(1/5)x^2 + 20x

but I'm not sure how to get this answer, I know I'm supposed to express it as x * p, but i don't know how the coefficients got so small
>>
Barnaby Niggerson - Sun, 20 Jul 2014 01:19:45 EST ID:xvgqavvT No.14221 Ignore Report Quick Reply
http://www.algebra.com/algebra/homework/Inequalities.faq.question.89589.html
Exact solution

Plus you can graph it on google to see if it makes sense/visualize it
Google "graph -(1/5)x^2 + 20x"
>>
Jack Pemmleham - Sun, 20 Jul 2014 21:23:55 EST ID:sTqzkRzm No.14224 Ignore Report Quick Reply
Wow, thank you so much, I'm sure to study well with this. Also I didn't know google had a graphing function, cool!


Multiply Roots by Matilda Croblingson - Wed, 02 Jul 2014 00:05:33 EST ID:qCDKBx4v No.14159 Ignore Report Reply Quick Reply
1404273933097.jpg -(109411 B, 921x523) Thumbnail displayed, click image for full size. 109411
What's the answer to #3?
>>
Cyril Sunningsted - Wed, 02 Jul 2014 11:32:24 EST ID:MTIV7/tU No.14161 Ignore Report Quick Reply
>>14159
Well x^n * x^m = (x * x * ... * x) * (x * x * ... * x) = x * x * ... x = x^(m+n), where in the first equality, the parenthesis on the left has n times x, and the one on the right has it m times.
>>
David Buzzhall - Wed, 02 Jul 2014 14:08:42 EST ID:Gw2IN3ba No.14162 Ignore Report Quick Reply
>>14161
This. I'd just like to add that this reasoning only works for integer exponents. Not that it matters in this case, since the problem is only asking for a reason and not a proof. Take a look at this:

http://math.stackexchange.com/questions/435751/proving-the-product-rule-for-exponents-with-the-same-base
>>
Nicholas Gosslegold - Wed, 02 Jul 2014 16:13:08 EST ID:e1p9nP+4 No.14163 Ignore Report Quick Reply
For non-integer exponents, the actual definition of x^a involves the exponential and logarithmic functions and the identity in the general case follows from the identity in the special case of these two functions.

The way exp and log are usually defined involves integration (one defines log first, then exp as the inverse of log) and the identity for them can be checked from the defining integral for log.

Things get fun (and much more complicated) when complex and imaginary exponents are assumed, then the additivity of the the exponents no longer true, but one instead has to deal with branching behavior. Of course, to understand that, you would have to know a little bit about Riemann surfaces and you'd be touching some extremely deep mathematics
>>
Martha Garrywire - Mon, 07 Jul 2014 16:31:05 EST ID:qCDKBx4v No.14181 Ignore Report Quick Reply
Thanks everyone. This came from some Harvard or Stanford or Princeton test from way back in the day.
I don't even remember how to do polynomial division.
>>
William Fimmermark - Mon, 14 Jul 2014 15:01:57 EST ID:K+W3fm6W No.14212 Ignore Report Quick Reply
Exponentiation is a special case of multiplication, so by multiplying exponents you increase the number of groups to be multiplied.


SICP by Jack Trotdock - Fri, 11 Jul 2014 08:14:52 EST ID:kDfCm/bw No.14204 Ignore Report Reply Quick Reply
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Hi, i enountered something weird whilst studying the answer to an exercise on SICP (computer science). The anwer declared

(1-square root of 5)/2 <1
is the same as
(square-root of 5-1)/2 <1

can someone explain? Thanks
>>
Jack Trotdock - Fri, 11 Jul 2014 09:46:10 EST ID:kDfCm/bw No.14205 Ignore Report Quick Reply
>>14204
This is the exercise
http://www.kendyck.com/math/sicp/ex1-13.xml
and you can find the point i am talking about by ctrl-f'ing "That is. let's show:"
>>
Martin Haddleberk - Fri, 11 Jul 2014 21:20:43 EST ID:Gw2IN3ba No.14209 Ignore Report Quick Reply
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>(1-square root of 5)/2 <1
>is the same as
>(square-root of 5-1)/2 <1

You forgot to include the absolute value notation. |x| = |-x| for real x is the reason why the two expressions mean the same thing.


Implicit Derivatives Explination by Simon Nodgefield - Thu, 10 Jul 2014 23:33:41 EST ID:BE8EyBvj No.14199 Ignore Report Reply Quick Reply
1405049621221.jpg -(43671 B, 771x447) Thumbnail displayed, click image for full size. 43671
So here's an excerpt from my textbook, but it's not verbose enough for me.
So they take the derivative of the y^2 = 2y(dy/dx)
My confusion comes from, why is that (dy/dx) still being multiplied to it? Should it not have disappeared because we just took the derivative?

For instance, (dx/dy)(x^2) = 2x
It does not equal = (2x)(dy/dx)

You see? Why is it staying there? What's going on?
>>
Edward Gummlehall - Fri, 11 Jul 2014 03:53:38 EST ID:Gw2IN3ba No.14202 Ignore Report Quick Reply
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>For instance, (dx/dy)(x^2) = 2x

This is incorrect. You seem to be confusing the function dx/dy for the operator d/dx. The excerpt you posted is making use of a result in calculus known as the chain rule. The chain rule states that if you have a function f(y(x)), you can calculate its derivative using this formula: d/dx f(y(x)) = (df/dy)*(dy/dx). So to use the excerpt as an example, in this case f(y(x)) = y^2. So df/dy = 2y, but because you don't have an expression for y as a function of x to differentiate, you just leave the dy/dx as is. So multiplying the derivative of the outside function (df/dy = 2y) by the derivative of the inside function (dy/dx) gives you the total derivative with respect to x: d/dx (y^2) = 2y dy/dx.

http://en.wikipedia.org/wiki/Chain_rule
>>
Simon Nodgefield - Fri, 11 Jul 2014 12:45:58 EST ID:BE8EyBvj No.14206 Ignore Report Quick Reply
>So to use the excerpt as an example, in this case f(y(x)) = y^2
Well I guess this is where I'm confused. Why/how can the chain rule be applied here?
If we apply the chain rule to y^2, why don't we apply it to x^2?

and
>confusing the function dy/dx for the operator d/dx
So.. if you use d/dx on an X, then since x is on the top/bottom, they cancel out. But if you use it d/dx on a y, they don't cancel, and you're left with dy/dx.
That's how I'm understanding it, is that wrong?
>>
Martin Haddleberk - Fri, 11 Jul 2014 20:57:18 EST ID:Gw2IN3ba No.14208 Ignore Report Quick Reply
>>14206
>If we apply the chain rule to y^2, why don't we apply it to x^2?

You can, but it's trivial. So we would have f(x(x)) = x^2. Applying the chain rule you get: df/dx = (df/dx)*(dx/dx) = [d/dx (x^2)]*(dx/dx) = (2x)*(1) = 2x.

>So.. if you use d/dx on an X, then since x is on the top/bottom, they cancel out. But if you use it d/dx on a y, they don't cancel, and you're left with dy/dx.
That's how I'm understanding it, is that wrong?

Pretty much, that's how it works. d/dx (f) tells you by what factor to multiply to an infinitesimal increase in x to get an infinitesimal increase in f. So d/dx (x) yielding 1 just tells you that for an infinitesimal increase in x, you need to multiply by a factor of 1 to get an infinitesimal increase in x.


End behavior of derivatives of wave functions. by Cedric Sidgefut - Sun, 06 Jul 2014 18:17:11 EST ID:yGgK6aCs No.14175 Ignore Report Reply Quick Reply
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Hey /math/! I'm working through every single problem of Griffith's textbook on Quantum Mechanics to prepare for the course, and I've come across a question that I wanted to ask you guys. Often I'm having to evaluate expressions from infinity to negative infinity. With wave functions these terms always disappear because of their end behavior. In my mind, this must be true of their higher derivatives as well. I'm wondering if my intuition is wrong, and if it's not, is there a relatively pretty proof of it? Thanks in advance.
>>
Fuck Blimblestock - Sun, 06 Jul 2014 22:50:17 EST ID:p8Vzq3fC No.14176 Ignore Report Quick Reply
>Often I'm having to evaluate expressions from infinity to negative infinity. With wave functions these terms always disappear because of their end behavior.
Isn't it with the |w|^2 that the total integral needs to be finite?

Intuitively, I think you're right, but also can't do a proof.


The case for monotonic functions would be easy to prove,but not all wave functions are monotonic.
>>
Fuck Blimblestock - Sun, 06 Jul 2014 22:58:39 EST ID:p8Vzq3fC No.14177 Ignore Report Quick Reply
>>14175
Also, for the monotonic proof it's very easy. By definition for a monotonic function that also approaches 0 as gets larger/smaller, lim x->infinity f(x)-f(x+dx) = 0. Therefor, f'(x) approaches 0 as x increases.

Shit, wait. That doesn't necessarily mean the integral from -infinity to infinity is finite, does it? Fuck man. I literally just picked up my old analysis textbook to get this shit sorted.
>>
Martin Poffinghall - Sun, 06 Jul 2014 23:18:05 EST ID:yGgK6aCs No.14179 Ignore Report Quick Reply
>>14177
Well I don't really care what the integral evaluates to, as long as all higher derivatives approach zero on both sides. I'm so n00b that I haven't even done analysis and I'm tinkering with these complex functions. Touch and go, my nigga.
>>
Martin Haddleberk - Fri, 11 Jul 2014 20:09:05 EST ID:Gw2IN3ba No.14207 Ignore Report Quick Reply
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The wave functions in QM are what are known as square-integrable functions. These are functions that satisfy the condition that the function multiplied by its complex conjugate integrated from - infinity to + infinity must yield a finite value. But Griffiths mentions on page 14 (Second Edition) that in addition to (not as a consequence of) this requirement, the wave function must also go to zero at + or - infinity. So the higher derivatives of wave functions must go to zero at + or - infinity from this second requirement and the simple fact that the derivative of a constant (in this case zero) is zero.


Learn math from the very beginning by Cornelius Bappernick - Fri, 20 Jun 2014 20:32:26 EST ID:arK+ls13 No.14106 Ignore Report Reply Quick Reply
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How can I learn math from the bottom up? I want to start at the beginning, understand all the concepts, and really have an education in math. Basically I want to start over.

I know how to add, subtract, multiply, divide, and I know a little about real numbers and whole numbers, imaginary numbers, all that stuff. I'm really good at basic algebra, but I don't really "understand" it, nor can I name everything about it. I know nothing about calculus or geometry, or trigonometry.

I heard that math is the "Language of the Universe" so I want to learn it so I can understand science well. Thanks
3 posts omitted. Click Reply to view.
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Polly Honeylock - Fri, 27 Jun 2014 15:31:53 EST ID:MXyuPRIu No.14137 Ignore Report Quick Reply
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If you can do all the exercise on the first chapter, you've got a pretty good idea of what math is
>>
Shit Nocklestirk - Mon, 30 Jun 2014 03:28:52 EST ID:e1p9nP+4 No.14145 Ignore Report Quick Reply
If you are comfortable enough with your algebra, I would suggest that you read and work through (aka do every fucking exercise) in Spivak's Caluclus or Apostol's Calculus. I never used Apostol, some people swear by it. Spivak has the advantage that he starts from the very beginning and discusses number systems and the usual axioms and that will help orient your thinking in the right directions.

Don't waste time on Euclidean geometry (I study algebraic geometry and I don't know shit about that subject), and certainly don't try jumping directly into topology as is suggested above.
>>
James Goshhall - Fri, 04 Jul 2014 19:23:28 EST ID:8Js26kRu No.14167 Ignore Report Quick Reply
>>14106
What I would recommend to you is trying to push along further in your math and pay careful attention to how you are learning. Then I would go back to earlier subjects and review through the material so you can see what you missed and why you didn't understand that in the first place. This will help you strategically in more complicated mathematics. You may actually be able to pick up some new information from just studied subjects by going over it again with a better foundation and be more capable of abstract thought instead of just plugging in calculations.
>>
Matilda Wankinnure - Sat, 05 Jul 2014 22:06:32 EST ID:kFCSHfPa No.14173 Ignore Report Quick Reply
>>14137
Holy shit, I was just working on the first chapter of that exact book. Spooky.
>>
Clara Weckleman - Fri, 11 Jul 2014 00:13:27 EST ID:hd9LVxdX No.14201 Ignore Report Quick Reply
I actually did start over too.

https://www.khanacademy.org/ you can learn from the ground up. Sheldon Axler's books are also the best out there, because they assume you have little knowledge and are self contained. For example his Trigonometry stuff in his book 'Precalculus' assumes you don't remember any Trig. Google his name and buy/pirate his books.

http://www.amazon.com/Mathematics-Content-Methods-Meaning-Dover/dp/0486409163/ Is a good book as it gives a broad overview of all math without having to do rigorous proofs.

http://www.amazon.com/All-Mathematics-You-Missed-Graduate/dp/0521797071/
All the math you missed for graduate school is higher level but one of the best books out there.

If you want the ultimate understanding in Math, basically as good as you can get without starting to specialize since it's impossible to know everything, and you already have some undergraduate self studying done (ie: you know calculus, you know discrete math already, you know probability)

  1. Principles of Mathematical Analysis by Rudin
  2. Algebra by Artin (and Sheldon Axler's Linear Algebra done right)
  3. Topology by Munkres
  4. Princeton's Companion to Mathematics http://press.princeton.edu/titles/8350.html
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