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Comments:
<0> oh, were you asking in my Solve[f(t)==-30,t]?
<1> well maybe stupid to say sound it an object, but every sound wave i mean
<0> That means solve it for t.
<0> If you meant something in the answer, I have no idea what the answer is saying.
<2> hello
<3> I suppose that if it travels over the speed of sound, it will be reversed.
<4> since it is a polynomial of degree 5 (which can't in general be solved explicitely using square roots only) you get the answer in terms of roots to some equations
<4> % NSolve[1/(2t^3)-5t^2+3t+6==-30,t]
<5> lillpelle: {{t -> 3.000685314260021}, {t -> -2.39865782451885}, {t -> -0.24269497373454305}, {t -> 0.12033374199648428 + 0.2068006087339667*I}, {t -> 0.12033374199648428 - 0.2068006087339667*I}}
<4> there are the numerical values...
<6> It's impossible to enumarate through all the cardinals, right?
<3> What cardinals are there ?
<6> Um...
<6> How big is the set of all sets?
<7> something has to exist
<7> before it can be "big"
<3> Well, not how many are there.
<3> What are they ?
<6> What are the cardinals?
<7> the princes of the church
<3> What are the instances of cardinals ?
<6> Oh boy...
<6> For every set, there is a cardinal number. Two cardinal numbers are the same if and only if there is a one-to-one mapping between their corresponding sets.
<3> Right, but I'm not asking what cardinals are.
<8> how many sets are there
<3> I'm asking you to list the cardinals.
<6> Oh.
<6> Well, there's an infinite number of them.
<6> But is this infinity countable?
<3> You can use shortcuts, I suppose.
<3> For instance, the naturals are cardinals, so you can say the naturals, and ...
<6> But then we have natural numbers which are fixed points of thngs like the successor function, and the power-of-two function.
<6> The former is smaller than the latter, isn't it?
<9> http://wikipedia.org/wiki/List_of_large_cardinal_properties
<6> "Remarkable cardinals", too...
<10> hello
<6> They're like surreal numbers. Their names don't actually mean anything.
<10> im reading about linear operators and it says "For example, the conjugation of complex numbers is an R-linear map C->C but it is not C-linear"
<10> im a little confused about what that means.
<3> Conjugation means to change the sign of the imaginary part.
<3> So, it's linear, just looking at the R part, since nothing changes.
<10> ok yeah i got that im not sure about the C-linear, linear in the complex?
<6> Oh, things like 0, 1, 2, etc. don't actually have to be cardinal numbers.
<10> er not linear
<10> oh ok thanks Olathe cleared that up
<3> No problem.
<11> Hi.
<10> it also says, If A is an m n matrix, then A defines a linear transformation from R^n to R^m by sending the column vector x an element of R^n to the column vector Ax an element R^m. Every linear transformation between finite-dimensional vector spaces arises in this fashion" why just finite?
<10> im reading wikipedia by the way. can you not have a matrix A of infinite size. is that bogus
<3> Well, it doesn't say just finite.
<3> You can extend the idea to infinite transformations, I suppose.
<10> oh ok
<4> you can have matrices of infinite size, however there are also other linear operators
<4> ex: the shift operator on the set of infinite n-tuples: (x1,x2,x3,...) is mapped to (0,x1,x2,x3,...). Try to write down the matrix for that one.
<3> Can't you just add a row or column to an infinite identity matrix ?
<4> ex of a mapping that cannot be represented by a matrix: Let T be the mapping that takes f(x) (defined on, say [0,1]) to (Tf)(x)=integral of f(y) as y tends from 0 to x.
<4> Olathe: yes, of course :)
<11> lillpelle, in which basis?
<4> say, the standard basis
<10> lillpelle, what is the standard basis?
<11> :)
<4> the i:th basis vector is full of zeroes, except on place i, where it has a 1.
<10> ok thanks
<11> lillpelle, it's a basis?
<10> lillpelle, is that different then the identity matrix
<11> can you explain why it's a basis, plz?
<3> delta : Because you can get any vector by multiplying each basis item by the value in the vector.
<11> Olathe, sorry?
<3> delta : (1 25 11) = 1(1 0 0) + 25(0 1 0) + 11(0 0 1)
<11> Olathe, infinite n-tuples: (x1,x2,x3,...)
<3> delta : That extends to infinitely long things.
<11> Olathe, how?
<7> the problem is finiteness
<3> delta : You just have x1(1 0 0 ...) + x2(0 1 0 ...) + x3(0 0 1 ...)
<11> yes :)
<11> no, Olathe, you can't do that.
<3> Why not ?
<7> the all 1 infinite vector, for example, has no finite representation in the form Olathe is giving
<3> Can't you just sum the basis vectors ?
<7> sums are finite
<11> finitely
<3> Oh, finite representation.
<11> That's the definition of a linear combination: it must be finite.
<11> So, it can't be a basis of this vector space.
<3> Then there is no basis.
<11> oh? :)
<10> no basis?
<7> i thought you can AoC one into existence
<3> You need infinite elements in the basis.
<7> zornify, if you will
<11> everybody thought that :)
<10> cant you prove by induction?
<11> transfinite?
<3> yond : For things in the naturals.
<9> shoezorn
<10> oh ok
<12> What's the problem? Every vector space has a basis, so if you have T:V->W, you just pick your favourite bases on V and W (let's say <v_\alpha : \alpha \in \Lambda> and <w_\alpha : \alpha \in \Lambda'>) Just represent Tv_\alpha in terms of the basis on W, dump that on your matrix and you've got your matrix
<12> (although it might have uncountable number of rows and/or columns, but who cares)
<11> so simple :)
<3> I think I see how it would work.
<3> Have a set of (1 x ...), (0 0 1 x ...) (0 0 0 0 1 x ...) ... for all x in the field.
<3> That wouldn't work, either, but you could extend it a bit.
<3> I suppose you could choose all vectors with 1 as the first element as the basis.
<3> Is that a proper basis ?
<3> I suppose that and the vectors with 0 as the first element.
<3> 0 1, 0 0 1, 0 0 0 1...
<9> um, no
<9> and you're making some strange ***umptions about what vectors look like
<6> Could one say that a property of even numbers is true for all odd numbers?
<3> ihope : It depends on the property.
<4> delta: since I'm working with Hilbert spaces, I thought of a "Hilbert basis" and of course had l^2 in mind... should have been clearer.
<10> another question if i have two matrices with the same column space i can say they have the same rank and therefore the same dimensions, but not the same nullspace. but if i know one is invertible is the other invertible also? im thinking so, but im not sure why
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