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Get X Y by Matilda Daggleville - Mon, 21 Jul 2014 21:33:34 EST ID:Bbuuretd No.14226 Ignore Report Reply Quick Reply
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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

Nathaniel Woshshaw - Tue, 22 Jul 2014 18:40:14 EST ID:yaOOvKGH No.14230 Ignore Report Quick Reply
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
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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Pre-Calc question by Emma Docklegold - Sat, 19 Jul 2014 21:37:36 EST ID:sTqzkRzm No.14220 Ignore Report Reply Quick Reply
File: 1405820256520.gif -(1083B / 1.06KB, 26x27) Thumbnail displayed, click image for full size. 1083
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
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
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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
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
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:

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
This is the exercise
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
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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.

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?

>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
>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
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
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.
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
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
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.

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
Comment too long. Click here to view the full text.

Need help getting back into math. by Edward Druggledock - Mon, 07 Jul 2014 22:21:07 EST ID:kQQi+WA1 No.14182 Ignore Report Reply Quick Reply
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I made it through Calculus when I was in highschool. But it's been 6 years and I just started at a community college. I just need help getting started again. They put me in college algebra or MTH 111. And I'm looking at my first homework assignment. It's supposed to be review but I'm having a little trouble.
(5/2)x + (1/4)x = (7/4)x
Fractions are confusing me. I just put the x with the numerator since x = (x/1) and simplified it to 11x/4 = 7x/4.
Is that right? Not even sure what to do next, hopin you guys can help or point me to a web site that can.
Also, I forgot how much school sucks. I think math will be cool though once I really get back into it. Thanks guys!
5 posts omitted. Click Reply to view.
Edward Druggledock - Mon, 07 Jul 2014 23:30:07 EST ID:kQQi+WA1 No.14188 Ignore Report Quick Reply
Ok, sorry everybody for doing my homework here. I found a site called purplemath.com that's pretty helpful. If you know of any other websites that are helpful let me know though. Thanks!
Edward Druggledock - Tue, 08 Jul 2014 00:18:50 EST ID:kQQi+WA1 No.14189 Ignore Report Quick Reply
I do need help remembering how to factor actually. I understand how to find a common factor in some equations but I need to someone to show me how they go about factoring something like this:
x^2 + 15x + 56
I know the answer is (x+8) (x+7)
I just don't know how to get there. I mean sort of but not really. I'm about to go to work so if someone can show me their work on how to get that answer in the next couple hours I would really appreciate it!
Edward Druggledock - Tue, 08 Jul 2014 02:29:11 EST ID:kQQi+WA1 No.14190 Ignore Report Quick Reply
shit nevermind this whole thread, I am struggling so hard, but it's slowly coming back.
Jack Bagglebut - Tue, 08 Jul 2014 19:59:35 EST ID:Dn4zhXKX No.14191 Ignore Report Quick Reply
With these simple factorization problems, where the answer looks like (x+a)(x+b) and you have to figure out a and b,
think of how (x+a)(x+b) multiplies out

x^2+ax+bx+ab = x^2+(a+b)x+ab

In this case, we have x^2+15x+56
So you have to find two numbers that simultaneously multiply to 56, AND add to 15.
Start with ab=56
In the whole numbers, there are a few different pairs that multiply to 56: (1,56),(2,28),(4,14),(7,8)
But only one of these pairs adds up to 15.
Clara Weckleman - Thu, 10 Jul 2014 23:51:01 EST ID:hd9LVxdX No.14200 Ignore Report Quick Reply
https://www.khanacademy.org/ has everything from kindergarten to college math in short video presentations. Look up fractions, radicals whatever you need.

Also go buy Sheldon Axler's "Precalculus' Version 2. Either used or pirate it.
Math will actually make sense if you read it. It assumes you have zero highschool knowledge and is self contained.

Measuring how evenly spread out numbers are by Frederick Punderstet - Thu, 26 Jun 2014 18:46:12 EST ID:lH5+Lnwt No.14126 Ignore Report Reply Quick Reply
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Is there a decent formula for measuring how "diverse" a set of values are?
For instance if I have 5 values between 0 and 1...
0, 0.25, 0.5, 0.75, 1 = maximum diversity
0, 0, 0, 0, 0, = minimum diversity
1, 1, 1, 1, 1, = minimum diversity also

So I want to measure how close the differences between the values are to 0.25, ideally in a way that will return 1 in the best case and 0 in the worst case.

Any ideas? Thanks
3 posts omitted. Click Reply to view.
Walter Binnerchodge - Thu, 26 Jun 2014 23:11:10 EST ID:HbNSsmiC No.14130 Ignore Report Quick Reply
Just at first glance it would appear that this may be workable for finite sets.

Given a metric space (A, d) then we can define a diversity function

dv (AxA) = sum from j=1 to |AxA| d ((x, y)) j / |AxA|

Or the sum of all distance for all cartesian pairs divided by the set size of the cartesian product.
Walter Binnerchodge - Thu, 26 Jun 2014 23:46:48 EST ID:HbNSsmiC No.14132 Ignore Report Quick Reply
I didn't read the post after about being normalized but I think if you choose all cartesian pairs that dont have a distance of zero and divide by that set size and the size of the max distance on the interval.
Phineas Clondlefield - Mon, 30 Jun 2014 14:57:15 EST ID:d64DnQ1v No.14150 Ignore Report Quick Reply
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In an ideally diverse set the distance between consecutive elements would be (max - min)/num_elements

Couldn't you just take the average of the distance between every consecutive element and then see how much this average deviates from the ideal distance?
Reuben Pisslefoot - Mon, 30 Jun 2014 19:15:37 EST ID:jkYtMoE5 No.14153 Ignore Report Quick Reply
Standard deviation seems like it could be used here.
But putting the answers on a 0-1 scale of minimum to maximum diversity is the hard part
Rebecca Cessleford - Thu, 10 Jul 2014 16:56:48 EST ID:Vnayk5eu No.14198 Ignore Report Quick Reply

This guys correct. Given a partially ordered set A and a diversity mapping dvr: A -> [0,1] then for max diversity we should have it such that each consecutive pair should have a distance of (max(A)-min(A))/|A|.

dvr(A) = ((sum(i = 0 to |A|-1) (a_i-a_i+1))/|A|)/((max(A)-min(A))/|A|)

have I discovered this? by Isabella Surrylodge - Fri, 27 Jun 2014 02:59:35 EST ID:ySRm4RKM No.14133 Ignore Report Reply Quick Reply
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David Sibbledure - Fri, 27 Jun 2014 13:18:40 EST ID:p8Vzq3fC No.14136 Ignore Report Quick Reply
Discovered what?
Nigger Winkinpat - Sat, 05 Jul 2014 17:44:54 EST ID:Dn4zhXKX No.14171 Ignore Report Quick Reply
Probably. You can call it Surrylodge's Lemma. Although if your proof is computer assisted, it might be controversial.

OK math lets count by Beatrice Trotwater - Sat, 24 May 2014 17:18:16 EST ID:Le9Cy2Bo No.14031 Ignore Report Reply Quick Reply
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Nicholas Bablinglock - Sun, 29 Jun 2014 10:57:19 EST ID:KRSGkVa1 No.14140 Ignore Report Quick Reply
Reuben Pisslefoot - Mon, 30 Jun 2014 19:14:03 EST ID:jkYtMoE5 No.14152 Ignore Report Quick Reply
Nigger Smallford - Tue, 01 Jul 2014 19:51:57 EST ID:MTIV7/tU No.14156 Ignore Report Quick Reply
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My nigga.

This picture is called fuck numbers.
Betsy Cruzzlelat - Fri, 04 Jul 2014 13:27:06 EST ID:p8Vzq3fC No.14166 Ignore Report Quick Reply
for those that don't know https://www.youtube.com/watch?v=w-I6XTVZXww
It's a pretty fantastic video, especially if you watch the one about 1/2-1/2+1/2-1/2...
Phoebe Crarrynat - Fri, 04 Jul 2014 23:27:31 EST ID:yHx/72Ow No.14168 Ignore Report Quick Reply

not this again...

The sum of the natural numbers is NOT -1/12, this is part of something called Ramanujan summation where you *ASSIGN* values to divergent series so that you can deal with them concretely. It does not mean that if you added all the natural numbers you would get -1/12, that is troll science and the video is completely and totally misleading.

1+2+3+4+... is a divergent series and you cannot add it up to get a negative number, or any number at all for that number. 1-1+1-1+... is not 1/2. Ramanujan is used in the specialized study of infinite series and it is FALSE to say that these things really add up to what is claimed in the video. It really sickens me the way these "educated" people dumb down extremely specialized things to produce something that is completely false in the end. Ugh.

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