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Posts posted by Nicoco
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Ah when you simplify the function, it indeed is 2, but then you worked out the fact that the function clearly isnt defined in 1. Actually, as my analysis prof would say, it is absolutely futile to speak of any function without clearly stating on which domain you work.
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You can extend the factorial to non-natural numbers (including the real numbers), by using Stirlings Approximation I believe.
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Its' not that hard, the hypothenuse of the triangle is not a straight line, just hold a ruler against it...
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I found this definition of column space: The vector space generated by the columns of a matrix viewed as vectors.
Now looking at your solution, how can a vector with 4 components be a base for 4 vectors with only three components? So if this definition is the right one, I would guess that all the vectors except the zero-vector are the base... I'm going to look this up later to be sure.
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it's fairly simple i think. Suppose there are two solutions to the system, x and x'. Then both Ax=b and Ax'=b hold. So Ax=Ax'. Multiply by C to get CAx=CAx', which can be reduced to x=x'. So there is only one solution.
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I don't know if this is the right place to post this, but lately I cannot load ANY Latex equations, they all give the error message '[LaTeX Error: couldn't open URL] ' ... What's up with that??
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But what I am trying to figure out' date=' is where Euler got the idea to take the limit of (1+x/n)^n.
That's such an odd thing to just do for no reason at all.
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I don't think that is a relevant question at all, I mean, there are a lot of other theorems where one could ask where the hell they got that idea from (like Schinzel's Hypothesis H).
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I don't think it is really useful trying to picture something in the real world when reading mathematics. I mean, 2- and 3-dimensional cases aren't that important most of the time, and I'm not quite able to picture things in n-dimensional space for n>3 (and I don't think most people can)...
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Yes, it has to be with and instead of the and . I mixed up two definitions.
By the way, in the meantime I was able to prove it in a different (but somewhat more tricky) way, you might be interested.
First, I give the definition of the -thickening of a set :
is the collection of all points within a distance of a point of A.
It's similar to a Ball in a metric space.
Next, let and . Then it is easy to see that
Take , then is easely follows that
Combining these last two inclusions we see
You can prove, using the same arguments, that
But your approach is a tad more easy I think...
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sorry, double posted this...
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Does anyone have an idea how to prove the following:
[math] h(A\cup B, C\cup D) \leq \max\{h(A,C), h(B,D)\} [/math]
with the Hausdorff distance. I can see it on a drawing, but I'm not able to prove it correctly.
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I think that's A...
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0/0 isn't defined if I'm correct. The same for 0^0.
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Hi, I'm new here, I'm really into mathematics (geometry and number theory being my favourite subject, although analysis (especially topology) is lotsa fun too. Live in Belgium, currently in the first year of university here. Sad but true my English isn't good enough for a witty remark or a good pun, so I think I'll stop here.
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Suppose you have a base {e_1, e_2, e_3} in R_3. You don't know anything else about them. Is it possible, if you are given another base {u_1, .. , u_3} to give a change-of-base matrix from the e_i's to the u_i's???
Thanks!
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Lotto Calculations!
in Linear Algebra and Group Theory
Posted
The only way you can be 100% sure to win the 10pound price is to buy tickets with ALL the combinations. IFF you have all the combinations for 3 out of 42 (or 49 or whatever), then there are no other combinations left, and you can collect the money (regardless of your tremendous loss). The reason is that if you don't have all the combinations, there is still some chance that the outcome can be one of the combinations you didn't pick.
There is some better solution because you only need three correct numbers, so you can actually put more then one combination in one ticket (I mean, if you buy a ticket, you already have some combinations covered due to the fact that there are more then three numbers on the ticket)