• Mathematician Chat v. 3.999...
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[QUOTE=proboardslol;46766568]Idk if it is but its not how they did it. I just wanted to know if I was correct[/QUOTE] No, it's certainly a [i]necessary[/i] condition, but I don't see that it's obviously sufficient. If you had 4 blocks in a straight line, there would be the right number that a single L piece would leave one square left over, but it's obviously not possible to actually [i]place[/i] the piece. -- Having the sides be 2^n makes it really simple - it means each is made up of 4 blocks of size 2^n-1 First prove by induction that you can always arrange the statue into a [i]corner[/i]. Trivially true for n=0 Then for 2^n, arrange 4 blocks of size 2^n-1 so that the 'statue space' is in the middle for 3 of them, and in the corner for the last. Fill in the centre spaces to complete the construction: [img]https://dl.dropboxusercontent.com/u/4081470/corner.png[/img] To show you can always put the statue in the middle, just swap around the last block so the last 'statue space' is in the middle: [img]https://dl.dropboxusercontent.com/u/4081470/middle.png[/img]
[QUOTE=Joey90;46769239]No, it's certainly a [i]necessary[/i] condition, but I don't see that it's obviously sufficient.[/QUOTE] So you're saying I didnt really answer the question?
[QUOTE=proboardslol;46770026]So you're saying I didnt really answer the question?[/QUOTE] Yeah, you proved that it's not [i]im[/i]possible but you didn't prove that it [i]is[/i] possible.
[QUOTE=Joey90;46770137]Yeah, you proved that it's not [i]im[/i]possible but you didn't prove that it [i]is[/i] possible.[/QUOTE] :O
did I formulate this correctly? [url]http://facepunch.com/showthread.php?t=1439640&p=46775249&viewfull=1#post46775249[/url]
[QUOTE=proboardslol;46775439]did I formulate this correctly? [url]http://facepunch.com/showthread.php?t=1439640&p=46775249&viewfull=1#post46775249[/url][/QUOTE] I think it should be stated a lot more clearly - it's probably approximately correct, but it took a bit of thought. You should clearly demonstrate your algorithm, since it's not always obvious - the completely naive approach is about 4 times slower than what you've implied. In pseudocode: [CODE] for (a = 1..N) A = a^4 for (b = 1..N) B = b^4 for (c = 1..N) C = c^4 for (d = 1..N) D = d^4 if (A + B + C == D) return false [/CODE] (If you moved A, B, C and D all into the final loop, it would be ~4x slower) Then what does 'iterate an integer from 0 to 2^3[b]1[/b]-1' mean? Simply iterate but do nothing, or actually perform a calculation? The relevant calculation is raising to the 4th power. This is 'just' 3 multiplications but it raises a crucial problem - what are you doing about overflow errors?! (2^31)^4 = 2^124 which is too big even for a long, and will require special handling = SLOW If you just want a simple approximation we can gloss over a bit... But is your 6 seconds still appropriate? My guess is that it's far too low. Your calculation is then correct, but you should say what the parts are! Using the same variables as in the equation is needlessly confusing. We're also assuming we ignore the cost of adding together the integers (given the overflow concerns above, this is probably reasonable) -- tl;dr be clearer and it's likely a massive underestimate. -- I think a much more efficient way would be to work modulo various (large-ish) primes, but I suspect that's not what you wanted to demonstrate :v:
Help on how to solve this integral equation? [IMG]http://latex.codecogs.com/gif.latex?y%28x%29%3D2+%5Cint_0%5Ex%20e%5E%7By%28t%29-t%7Ddt[/IMG] [editline]25th December 2014[/editline] The y(t) in the exponent is what I'm not sure what to do with
[QUOTE=JohanGS;46791727]Help on how to solve this integral equation? [IMG]http://latex.codecogs.com/gif.latex?y%28x%29%3D2+%5Cint_0%5Ex%20e%5E%7By%28t%29-t%7Ddt[/IMG] [editline]25th December 2014[/editline] The y(t) in the exponent is what I'm not sure what to do with[/QUOTE] First thing I'd do is differentiate wrt x. Then I'd probably rearrange and integrate both sides simultaneously. You should be able to get the constant by substituting into the original equation. P.S. it's Christmas so maths may suffer
Rearrange how?
[QUOTE=JohanGS;46792349]Rearrange how?[/QUOTE] dy/dx = e^(y-x) e^-y dy = e^-x dx e^-y = e^-x + c Then you can use the original equation to solve for c. Proof as an exercise (or perhaps I'm making it up)
Ah, I should really start using Leibniz notation more often. Things like that become so much easier to spot.
Probably way too easy question; what would be the best way of evaluating this limit? [IMG]http://latex.codecogs.com/gif.latex?%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7Bln%28x+1%29%7D%7Bln%28x%29%7D[/IMG]
Probably L'Hôpital's rule.
Gah, of course. [editline]27th December 2014[/editline] [IMG]http://latex.codecogs.com/gif.latex?%5Csum_%7Bn%3D2%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7B%5Csum_%7Bk%3D1%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Bk%5E%5Calpha%7Dn%20%5Clog%7Bn%7D%7D[/IMG] For which values of a does this converge?
[QUOTE=JohanGS;46799596]Gah, of course. [editline]27th December 2014[/editline] [IMG]http://latex.codecogs.com/gif.latex?%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7B%5Csum_%7Bk%3D1%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Bk%5E%5Calpha%7Dn%20%5Clog%7Bn%7D%7D[/IMG] For which values of a does this converge?[/QUOTE] There's at least one typo there... Mixing up your n's and k's General strategy is to use some convergence criterion, then examine the boundary case by hand. In this case of think the ratio test is probably best, and I'd guess the boundary is almost certainly 1
Yeah, I mixed it up. Fixed now. [editline]27th December 2014[/editline] It's just overwhelming to me I guess, any tips on how to divide and conquer that bitch?
If I want to prove set inclusion of something like this: (A u B ) c (A u C) Is it true that I can cancel A out from both sides and simply show that B c C ? It seems like a logical step to me.
In the following Venn diagram, B c C (the blacked-out bits have no elements). I've marked the intersection of A u B and A u C in red. The rest of A u B is empty, so B c C implies (A u B) c (A u C). [t]http://i.imgur.com/FzvgdCe.png[/t] In the following Venn diagram, A u B is a subset of A u C. Choose an element x from the green region. x is in B, but not in C, so B is not a subset of C unless the green region also happens to be empty. So, (A u B) c (A u C) does not imply B c C. [t]http://i.imgur.com/KxCS8f9.png[/t]
[QUOTE=JohanGS;46803659]Yeah, I mixed it up. Fixed now. [editline]27th December 2014[/editline] It's just overwhelming to me I guess, any tips on how to divide and conquer that bitch?[/QUOTE] Ok, the ratio test won't help much for the outer sum (I'd guessed a different typo!) The crucial thing is to analyze the inner sum - sum of 1/k^a The integral test gives us that: It converges for a > 1, this means you can bound the outer sum by a constant times 1/nlogn. You can then show that this diverges. It diverges for a < 1, and is bounded by something like n^(1-a). This means the outer sum is bounded above by something like sum 1/n^(2-a) logn, this is then bounded by 1/n^(2-a) which we know converges for a < 1. The tricky one is a=1, for this we can bound the harmonic partial sums by log, this makes the big sum like sum 1/n(logn)^2 which I think also diverges.
I have to get vectors figured out soon, so I'm doing some studying and I hit a wall with this problem: [t]http://puu.sh/dNOmV/f18550420b.jpg[/t] I need to find the scalar [img]http://puu.sh/dNOIh/839be76972.png[/img], to make [img]http://puu.sh/dNOFL/05f1d5f70e.png[/img] get in line with l (Assuming you already have W, V and U's magnitude, direction and sense)
I think there might be more information needed. Are all these vectors in the plane? Is there some relationship between U, V, l, and W? It seems like if W is parallel to l then there is no such [img]http://puu.sh/dNOIh/839be76972.png[/img].
[QUOTE=Krinkels;46816298]I think there might be more information needed. Are all these vectors in the plane? Is there some relationship between U, V, l, and W? It seems like if W is parallel to l then there is no such [img]http://puu.sh/dNOIh/839be76972.png[/img].[/QUOTE] I'll translate the problem: [QUOTE]Consider two vectors u and v that support the straight line L as shown. A third vector w is located on the same origin as u and v. Determine the scalar [IMG]http://puu.sh/dNOIh/839be76972.png[/IMG] that allows [IMG]http://puu.sh/dNOFL/05f1d5f70e.png[/IMG] to support the line L (Assume as known the magnitude, direction and sense of the 3 vectors)[/QUOTE] That's all I'm given. Maybe if W is parallel to L [IMG]http://puu.sh/dNOIh/839be76972.png[/IMG] is undefined?
There's probably a better way to do this, but for now: The tips of u and v are both points on the line l. Use the magnitude, direction and sense to derive the coordinates of these two points, and then derive the equation of the line l in terms of a parameter t. Find the pair (t,[img]http://puu.sh/dNOIh/839be76972.png[/img]) where l(t) and [img]http://puu.sh/dNOFL/05f1d5f70e.png[/img]([img]http://puu.sh/dNOIh/839be76972.png[/img]) intersect.
[QUOTE=Krinkels;46816451]There's probably a better way to do this, but for now: The tips of u and v are both points on the line l. Use the magnitude, direction and sense to derive the coordinates of these two points, and then derive the equation of the line l in terms of a parameter t. Find the pair (t,[img]http://puu.sh/dNOIh/839be76972.png[/img]) where l(t) and [img]http://puu.sh/dNOFL/05f1d5f70e.png[/img]([img]http://puu.sh/dNOIh/839be76972.png[/img]) intersect.[/QUOTE] Sounds good. But I'm in an online class, so these things are evaluated, and I think it was to be solved via vector logic. [editline]29th December 2014[/editline] Mmm, nevermind, I did what you said with the vector's components. I got an awful result but it seems to work.
Sorry about the awful result. I'm still unable to figure it out using vector logic. I guess an important thing is that you can also parametrize l as kv + (1-k)u. It seems like the kind of problem for which vector projection may be useful, but I can't see how to apply it here.
Don't worry, but it seems to be correct, since if I make w parallel to l, lambda becomes undefined.
[QUOTE=Krinkels;46816631]Sorry about the awful result. I'm still unable to figure it out using vector logic. I guess an important thing is that you can also parametrize l as kv + (1-k)u. It seems like the kind of problem for which vector projection may be useful, but I can't see how to apply it here.[/QUOTE] You basically have it then - if l is the line kv + (1-k)u, we want the intersection of this with t*w i.e. for what k and t does kv + (1-k)u = tw By looking at the rows of the vectors this is effectively 2 simultaneous equations which can be solved for k and t. The infinitely many solutions case arises when u,v,w are parallel, the zero solutions case arises when w is parallel to l but u and v are not. (I have replaced lambda with t for ease of typing)
Alright, I think I understoond how line parametrization works and how did you got to wt=kv+(1-k)u, but I got lost after that, can you explain a little more?
f(x,y)=sqrt(4-x^2-2xy-y^2) "Draw the largest possible domain" - how do you interpret that question?
f(x,y)=sqrt(4-(x+y)^2) I assume f has to have a real output, and since it consists of a square root, you'll have to investigate where the square root has a >=0 argument. Another hint: [sp]Try to figure out for which (x,y) pairs 4>=(x+y)^2 (assuming f has to be real).[/sp]
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