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Comments:
<0> Catfive: i mean, im left with lim (x-->oo) ( 0 )
<1> yes, but if every token of a given linear type is equivalent, its safe. its how linear folk make linear arrays.
<1> without them linear types can be way too restrictive, and can only build tree like structures.
<2> edwardk : i.e. first alias a structure, then swap a linear (or affine) part, through one of the aliases
<3> % Limit[Limit[(x^2 + y^2) / (x^2 + y^4), y -> Infinity], x -> Infinity]
<4> Catfive: 0
<0> Is that...normal, to write lim(x-->oo) (0) ?
<1> ski: sure. the trick is you maintain the property of linearity which is what i'm seeking to control, not the immutability of the structure.
<1> cyclone et. al allow that form of linear containment as a sop to practicality ;)
<3> Jax - you can say that it's continuous in a 'nice' way and that if it's also differentiable, then its derivative is bounded. as noted, the Lipschitz property turns out to be very important in the more advanced theory of the calculus.
<1> anyways a good example of where that sort of thing is useful is you can use it to swap in the 'whitehole' or 'blackhole' that is used in traditional call-by-need graph reduction, to notify any other thread that attempts to access the current node that its being worked on and by what thread. the end result is still referentially transparent, and the affinity of the reduction is preserved.
<1> there are other cases where the order of operation doesn't matter, and it lets linear capabilities be used as test and set semaphores, etc.
<5> Catfive and what advantages do i have if the derivative is bounded?
<3> Jax - also, if T: V -> W is a linear transformation, then the fact that T is Lipschitz means || T(y) - T(x) || <= c || y - x ||; but since T is linear, this can be written || T(y - x) || <= c || y - x || and setting z = y - x this becomes || T(z) || <= c || z || for all z. A linear T that satisfies such a property is called a bounded linear operator, a concept of fundamental importance in functional analysis. The smallest bound of T, sup {|| ...
<3> ... T(z) || / || z ||}, is called the *norm* of T and written || T ||.
<5> ah yes
<5> we are dealing with norms in calculus 2
<3> (in fact, for a linear T the following are equivalent: (i) T is continuous at one point; (ii) T is continuous; (iii) T is bounded.)
<5> can you define linear in this context, not sure i'm thinking what you are
<1> T(ax + b) = aT(x) + T(b)
<0> % [Limit[(1/xy) * Tan[xy/(1+xy), y -> Infinity]
<3> A linear transformation T: V -> W between vector spaces V and W is a map that satisfies T(x + y) = T(x) + T(y) for all x,y in V, and T(cx) = c T(x) for all scalars c and all vectors x in V.
<4> Copter2: $Failed
<0> % Limit[(1/xy) * Tan[xy/(1+xy), y -> Infinity]
<2> edwardk : mhm
<4> Copter2: $Failed
<1> the above version is the shorthand way of sticking both conditions together ;)
<5> ah ok
<2> (hehe, two kinds of 'linear' being discussed :)
<0> % Limit[(1/xy) * Tan[xy/(1+xy)], y -> Infinity]
<1> ski: so are over to the affinity side of the lattice yet?
<4> Copter2: Tan[xy/(1 + xy)]/xy
<1> ski: or did we get through destructable ok?
<2> edward : i'm not sure
<1> anyways once you know you won't forget or destroy the object you know it will necessarily be used to it has to be evaluated and should be updated strictly from then out, hence strict at the top of the affinity lattice.
<2> edwardk : maybe we should move, since it's somewhat offtangent, and it's harder to keep track with interlacing conversations
<1> that way translating from unrestricted to relevant entails an evaluation in place.
<1> yeah
<1> hrmm
<2> #haskell-overflow ?
<1> kk
<3> (exercises for the interested: (1) Prove: (a) || T(x) || <= || T || || x || for all x; (b) if || T(x) || <= b || x || for all x, then || T || <= b. (2) Show that with the one-norm || x ||_1 = |x_1| + ... + |x_n| on R^n, then the norm of the linear functional L_a(x) = sum(i=1..n, a_i x_i) is || a ||_oo (this is the uniform norm). (3) Show similarly that if || x || = || x ||_oo, then || L_a || = || a ||_1.)
<0> ehm ahhh
<0> How do I calculate limit[(1/xy) * Tan[xy/(1+xy), x -> 0] ?
<6> hmm working with arithmetic series, what's the best way to solve "Find the sum of all positive integers less than 3000 which are divisible by 3 or 5."? The neatest way I can think of is find the sum of the threes, add to the sum of the fives, then subtract the sum of the fifteens to cancel the doubles
<7> that seems sensible
<6> ouch
<6> <4>arctanx: thread killed
<6> quite possibly it hates me
<8> [mbot]: Yes, I hate arctanx
<9> @bo
<9> @bot
<9> % hello
<9> heh, so much lag :)
<6> all I did was % Sum[n+1,{n,0,3}] :P
<8> Cale: he killed you bot
<9> yeah, probably
<8> Cale: KILL HIM!
<6> JohnFlux: well after it failed I realised I had mathematica on my own machine, and it behaved :P
<8> eye for an eye
<9> I didn't see any messages on the bot's end, but it's not capturing this stuff now
<9> It did catch the 'ouch' there though
<4> :)
<4> Cale: "Good day to you!"
<6> hah
<9> aha
<8> mbot: do you hate arctanx now?
<8> silly bot
<5> for 2 converging sequences {a_n} and {b_n} can i say that because for every n element of N if a_n < b_n, that a must be < b too?
<5> {a_n} and {b_n} are two sequences of real numbers with limit a, b.
<9> a <= b
<5> if i know that a_n < b_n for every n in N, can i show that the limits a < b holds?
<3> Jax - let a_n = -1/n and b_n = 0 for all n.
<9> not strictly
<9> they can be equal
<5> ok but i can say, if a_n <= b_n for all n of N, that a <= b, right?
<9> You can say that too
<8> i don't see why that's true
<5> but the same thing with < doesn't hold, correct?
<8> what if one converged slower than the other?
<9> If a_n < b_n for all n, then a <= b
<10> JohnFlux: so what?
<8> hmm
<3> (since in particular if a_n < b_n, then a_n <= b_n)
<8> int-e: i guess if a was bigger than b then at some point they'd have to actually cross
<8> int-e: at a finite point
<5> can i prove that if a_n < b_n, that a < b doesn't hold?
<10> JohnFlux: ***ume a>b, set epsilon = (a-b)/3, apply definition of limes.
<9> right
<10> JohnFlux: obtain a contradiction.
<9> limes :)
<8> i don't like limes
<7> limes : small green citrus fruit
<8> my explanation was simpler
<9> Jax: You can find an example
<9> Jax: Catfive gave one already
<10> The Latin noun limes had a number of different meanings [etc.]
<9> a_n = 0 for all n, and b_n = 1/n
<5> do i need such examples at exams to prove this?
<9> Yeah
<5> evil
<9> but it's not hard to come up with them
<9> Then a_n < b_n for all n, but the limit of both sequences is 0
<5> ah yes, indeed
<8> or just 1/n and 2/n
<8> right?
<9> sure
<7> sure they both converge to zero but a_n = 1/2 b_n always
<8> if just a_n = n
<8> do we say it converges on infinite, or that it doesn't converge?
<7> the limit is n
<8> that's not a limit
<7> why?
<8> hmmm
<7> is not each term of the sequence arbitarily close to n? :)
<10> JohnFlux: depends. If we say that it converges to infinity, we bear in mind that this has its own special definition.
<7> generally
<10> JohnFlux: in particular it does not make infinity a number.
<7> in england (at least) we would say it diverges
<7> sometimes we say "diverges to positive infinity"
<7> convergence implies a real valued limit point, infinity isn't a real number
<5> hm for what kind of sequence is the limit equals to the limes sup and limes inf ?
<9> ekiM: Well, you can compactify the real line by adding a point to each end, called infinity and -infinity, which all such divergent sequences would converge to.
<9> (divergent in R)
<7> you can, but I imagine this is for a first year analysis course.. :)
<9> :)
<11> In my year of real analysis, that was covered.
<7> Jax : a constant sequence?
<9> Jax: a convergent one.
<5> a converging one
<7> oh
<7> indeed
<5> yeah but how do i show that its true ;)
<7> take the definitions and jiggle them around
<9> Suppose that the lim sup is equal to the lim inf and show that the sequence converges, and then take a convergent sequence and show that its lim sup and lim inf are equal.
<9> Hint: the limit in each case will be equal to the common value of the lim sup and lim inf
<8> btw, anyone know how to find the uncertainties for a best fit line? :/
<8> I have some data points each with an uncertainty
<8> and I have a best fit line
<8> but I don't know what the uncertainty of the gradient of the line is
<12> a million iterations of (points)+random_value_up_to_uncertainty then line fitting is my best suggestion :\
<7> how are you quantifying the uncertainty?
<8> i don't really care
<8> 1 standard deviation is fine
<8> but anything rough
<8> I have 6 points each with an error bar, then fitting a gradient to it
<11> chi square test?
<8> i want to know how many significant figures I can give the gradient to
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