Welp, just got raped by my dynamical systems exam.
Is there an easier way to approximate centre manifolds in multi-parameter space than multiplying out the taylor expansion term-by-term and equating coefficients? That shit took way too long.
I want to learn Python so today for the first time I got on Project Euler and I went to the first problem and I thought, "Oh, that's easy. You just write a loop goes through every number up to 999 and... wait, you don't need programming to do this, you can use triangular numbers." So I solved it by hand instead and that's the story of how once again I didn't learn Python. The world is safe.
I have the impression that Project Euler is more something for programmers who want to improve theirs maths, not the other way around. Never enjoyed it that much from a programming perspective.
I'd rather program something that has a real application or something that makes my life easier, instead of doing difficult pointless stuff with programming. That's what maths is for anyway :v:
Well being someone who doesn't like programming I have no idea what I should be doing to improve my programming because I usually try to actively avoid it, as my anecdote should demonstrate.
I'm just trying to get better at it because it's so damn employable.
[editline]25th June 2015[/editline]
Problem 5 is also really easy by hand. Actually, it would probably take me longer to do via coding than by hand. It took me less than a minute by hand. I guess figuring out how to articulate how to solve it into a program could actually be good practice.
[QUOTE=JohnnyMo1;48051969]Well being someone who doesn't like programming I have no idea what I should be doing to improve my programming because I usually try to actively avoid it, as my anecdote should demonstrate.
I'm just trying to get better at it because it's so damn employable.
[editline]25th June 2015[/editline]
Problem 5 is also really easy by hand. Actually, it would probably take me longer to do via coding than by hand. It took me less than a minute by hand. I guess figuring out how to articulate how to solve it into a program could actually be good practice.[/QUOTE]
As much as I prefer back of the envelope physics and maths, I can't deny that numerical simulations are becoming an ever more important area. Coding is dull at first, but after not too long at all it becomes rather straightforward and easy and it does wonders at further developing your problem solving skills.
Plus it's nice the first time you create a piece of code which does the same job as a prebuilt function but hella quicker.
A quick real life combinatorial problem I need verification for. Imagine the "Bob buys 40 watermelons for <an important situation involving watermelons>" type situation where Bob [I]actually[/I] needs to get the amount of watermelons right because if he doesn't it's a shitty situation :v: .
My dad has some amount of wooden figurines, there are four types: a house, a cylinder, a tower and a cube. He also has four colours. He wants to make sets of four figurines (one of each) but in such a way that each set is unique (i.e. each set differs from the other in the sense that at least one figurine has a different colour than the other). Also each set may only have one occurrence of each colour.
First, how many of such sets (actually lists) are there? I look at it as just having four bins where you can put in resp. 4, 3, 2 and 1 colour so the amount of unique sets is 4 x 3 x 2 x 1=24.
Now for each figurine, how many does he have to paint in some colour? My approach was taking some figurine and then looking how much sets correspond with that specific colour for a figurine. That's 6 for each figurine. Fix the first bin, that's 3 x 2 x 1 possible permutations left. So for each colour and each figurine, there should be six of them. He has to paint each type of figurine six times in the same colour (it is critical that this is correct)
Am I correct? It's kinda funny how he as an art teacher came up with such a textbook exercise :v: He's helping someone out with some project I guess and they need shitloads of unique sets of figurines.
Also, he actually needs 35 sets. Since it's hard to make a new figurine, I suggested adding another colour. My guess is that he then needs 35 copies of each figurine then (but to exhaust each permutation he'll need 5 x 24=120)?
[QUOTE=JohnnyMo1;48051969]Well being someone who doesn't like programming I have no idea what I should be doing to improve my programming because I usually try to actively avoid it, as my anecdote should demonstrate.
I'm just trying to get better at it because it's so damn employable.
[editline]25th June 2015[/editline]
Problem 5 is also really easy by hand. Actually, it would probably take me longer to do via coding than by hand. It took me less than a minute by hand. I guess figuring out how to articulate how to solve it into a program could actually be good practice.[/QUOTE]
Holy shit I can do something JohnnyMo1 can't (which soon won't be true).
By the way, if you want to get good at programming, just do something everybody does:
A calculator!
If you find making a calculator easy, then do something fucked up, like mini-max algorithm or networking (multiplayer) between two Python clients. (you can run both programs on same PC, no need for two PCs).
Or make game of life (automata).
Or make pathfinding algorithm (A-star).
[QUOTE=Fourier;48082205]Holy shit I can do something JohnnyMo1 can't (which soon won't be true).[/QUOTE]
You could do it all you like and I'd never do it if I could get away with it. Coding is boooring.
I like to compare it to making clocks (even though I never made one), you design like a whole mechanical thing and everything works together and it's so pleasing when everything interlocks and it works as expected and it is somewhat autonomous. You get to see past the syntax really quickly.
[QUOTE=JohanGS;48084333][IMG]https://s-media-cache-ak0.pinimg.com/originals/e3/a4/ef/e3a4ef2bb014fce26fd754dd22b0ed05.jpg[/IMG][/QUOTE]
Wow thanks, that is actually super useful.
Passed my second calc exam! Only 2 more then the final
[QUOTE=Number-41;48081713]A quick real life combinatorial problem I need verification for. Imagine the "Bob buys 40 watermelons for <an important situation involving watermelons>" type situation where Bob [I]actually[/I] needs to get the amount of watermelons right because if he doesn't it's a shitty situation :v: .
My dad has some amount of wooden figurines, there are four types: a house, a cylinder, a tower and a cube. He also has four colours. He wants to make sets of four figurines (one of each) but in such a way that each set is unique (i.e. each set differs from the other in the sense that at least one figurine has a different colour than the other). Also each set may only have one occurrence of each colour.
First, how many of such sets (actually lists) are there? I look at it as just having four bins where you can put in resp. 4, 3, 2 and 1 colour so the amount of unique sets is 4 x 3 x 2 x 1=24.
Now for each figurine, how many does he have to paint in some colour? My approach was taking some figurine and then looking how much sets correspond with that specific colour for a figurine. That's 6 for each figurine. Fix the first bin, that's 3 x 2 x 1 possible permutations left. So for each colour and each figurine, there should be six of them. He has to paint each type of figurine six times in the same colour (it is critical that this is correct)
Am I correct? It's kinda funny how he as an art teacher came up with such a textbook exercise :v: He's helping someone out with some project I guess and they need shitloads of unique sets of figurines.
Also, he actually needs 35 sets. Since it's hard to make a new figurine, I suggested adding another colour. My guess is that he then needs 35 copies of each figurine then (but to exhaust each permutation he'll need 5 x 24=120)?[/QUOTE]
I think everything is correct. We treat the figurines as the bins and the colours as the objects.
For 4 figurines and 4 colours it's a permutation of 4 objects, which is 4! = 4x3x2x1=24.
Your reasoning for each figurine needing to be painted the same colour 6 times is solid too.
For the final bit my understanding is that there are 4 figurines and 5 colours, which will give 5!/1! = 5x4x3x2 =120 like you said.
[QUOTE=JohnnyMo1;48086240]You could do it all you like and I'd never do it if I could get away with it. Coding is boooring.[/QUOTE]
Somewhat agree, designing stuff with pencil or doing math is .. more stimulating. And free, you don't need to fuck with compiler errors and shit
Sorry for the homework tier question.
How do I find the Taylor Series for f(x) = x^4e^(-3x^2)) about x = 0?
[img]http://i.imgur.com/b2ErFLc.png[/img]
(Wouldn't that be a Maclaurin Series? That's how the problem was worded.)
I'm getting that all the derivatives of f(0) = 0.
I haven't calculated it out, but I don't think they're all zero. When you take the first derivative, the product rule splits it into two terms. One is going to have a factor x^3 and the other will have a factor x^5 from the derivative of the exponential. But it will be zero at x = 0. Take the next derivative though, and the x^3 splits into an x^2 term and an x^4 term and x^5 splits into x^4 and x^6. As you keep going, the x^2 term creates an x term, and then a constant term. That one will be nonzero at x = 0. So the first nonzero term will be the order x^4 term, and then you'll have higher nonzero terms as well.
Ah, ok. Thank you.
I was hesitant to keep differentiating because the amount of times i'd have to use the product rule was starting to look ridiculous.
Since the first derivative also created an x^5 term, two orders higher than the x^3 term, by the time that gets derivatived down to a constant, it'll be the O(x^6) term of the expansion, then the next will be O(x^8), etc.
[editline]3rd July 2015[/editline]
You just have to figure out how to determine the coefficients now. It doesn't seem to be too hard, I don't think.
[QUOTE=Neckbird;48114932]
(Wouldn't that be a Maclaurin Series? That's how the problem was worded.)[/QUOTE]
Maclaurin series are just a special case of Taylor series. Just a terminology thing.
[url]http://vixra.org/abs/1506.0012[/url]
P=NP totes solved
[editline]7th July 2015[/editline]
An outline of the proof for the uninitiated:
"Assume P =/= NP. Then P =/= NP. deal w/ it math nerdes"
[editline]7th July 2015[/editline]
Also this person is a co-author on a recently submitted paper which is clearly not definitely mathgen at all, guys.
[QUOTE=JohnnyMo1;48144468][url]http://vixra.org/abs/1506.0012[/url]
P=NP totes solved
[editline]7th July 2015[/editline]
An outline of the proof for the uninitiated:
"Assume P =/= NP. Then P =/= NP. deal w/ it math nerdes"
[editline]7th July 2015[/editline]
Also this person is a co-author on a recently submitted paper which is clearly not definitely mathgen at all, guys.[/QUOTE]
I can't make any sense of the last "sentence" in that pdf. it just keeps going and going...
[QUOTE=PopLot;48148740]I can't make any sense of the last "sentence" in that pdf. it just keeps going and going...[/QUOTE]
I like that he repeats the same phrase like 4 times in one sentence.
[QUOTE=JohnnyMo1;48144468][url]http://vixra.org/abs/1506.0012[/url]
P=NP totes solved
[editline]7th July 2015[/editline]
An outline of the proof for the uninitiated:
"Assume P =/= NP. Then P =/= NP. deal w/ it math nerdes"
[editline]7th July 2015[/editline]
Also this person is a co-author on a recently submitted paper which is clearly not definitely mathgen at all, guys.[/QUOTE]
So we associate the P with problems already solved?
What the fuck, this is crazy. Will read tomorrow again.
Does anybody know of a general method to describe the connections of a 'flat' grid bent into an n-dimensional torus?
For example if I take a square, roll it into a cylinder (with no ends), and then bend it around so that the edges connect I can form a torus; this leads to every grid element having the exact same number of adjacent cells (diagonal elements included) as any other. It's very easy to jot down an adjacency matrix for the grid elements in such a system, but I'm trying to figure out how to generalise this to n-dimensions.
Basically I want to program an n-dimensional game of life in such a way that the space I use has no weird boundaries which fuck up the simulation.
[QUOTE=sltungle;48153560]Does anybody know of a general method to describe the connections of a 'flat' grid bent into an n-dimensional torus?
For example if I take a square, roll it into a cylinder (with no ends), and then bend it around so that the edges connect I can form a torus; this leads to every grid element having the exact same number of adjacent cells (diagonal elements included) as any other. It's very easy to jot down an adjacency matrix for the grid elements in such a system, but I'm trying to figure out how to generalise this to n-dimensions.
Basically I want to program an n-dimensional game of life in such a way that the space I use has no weird boundaries which fuck up the simulation.[/QUOTE]
Don't think of it as a 3D torus, think of it as a square where the opposite sides are identified - then it's really easy to generalise.
Just imagine your n-dimensional space being modulo N (so if any dimension is >N, you simply subtract to get a value between 0 & N).
If you're writing a program to do this, they're pretty good at doing that sort of thing!
[QUOTE=Joey90;48154885]Don't think of it as a 3D torus, think of it as a square where the opposite sides are identified - then it's really easy to generalise.
Just imagine your n-dimensional space being modulo N (so if any dimension is >N, you simply subtract to get a value between 0 & N).
If you're writing a program to do this, they're pretty good at doing that sort of thing![/QUOTE]
That's what I've currently got programmed up; for example stepping right (so moving over a 'column' and staying in the same row) is new_column = modulo(old_column+1,N), where N is the side length of the grid. It's easy to verify it makes sense in two-dimensions because I CAN just take a sheet of paper and roll it up to confirm that kind of thing makes sense, but I'm not entirely certain that applies to n-dimensions.
[editline]9th July 2015[/editline]
Although I guess it should because the second dimension only becomes 'apparent' (i.e. it's only clear you're not on a one dimensional line) on the N+1 element (first element in the second row), so the third dimension will only become apparent on element N^2+1, and I suppose the fourth dimension would become apparent on element N^3+1.
Just imagine that the N-dimensional cube in which you're running your simulation is tiled infinitely along every dimension. That way it's easy to see that the modular arithmetic will give you sensible results. I really don't see a point in visualizing it as a torus.
I feel I need to brush up on jacobians and eigenvalues, anything fun I problems I can do?
[QUOTE=pebkac;48155958]Just imagine that the N-dimensional cube in which you're running your simulation is tiled infinitely along every dimension. That way it's easy to see that the modular arithmetic will give you sensible results. I really don't see a point in visualizing it as a torus.[/QUOTE]
It's a little bit harder than it would otherwise be to generalise because I'm also considering diagonal elements, and wrapping with diagonal elements is a bit tricky. For example, on an NxN grid EVERY element, i, has a ±1, ±N and ±(N-1) neighbour; the ±(N+1) case isn't quite so simple with only N(N-2) elements having such a neighbour; the remaining 2N elements have neighbours at sites which are scattered around in a fashion which hasn't yet clicked to me.
This probably wouldn't be so difficult if I wasn't using Matlab to program, but oh well.
[QUOTE=Ithon;48162042]I feel I need to brush up on jacobians and eigenvalues, anything fun I problems I can do?[/QUOTE]
Play around with multivariate Newton-Raphson and PageRank stuffs, maybe?
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