bloodhound

With advent of new multimedia, this kind of storage will be necessary. Hey we used to think VCD's were good enough before DVD's came along.

My main area, is classical guitar, but i branch out into jazz,blues,rock. etc. But recently my technique has gone downhill due to lack of time to practice. I am currently developing my sweep picking technique. To sepultallica: How many notes can u play in a second ?

If thy did not fix something that is not broken, no progress is made

Beauty is in the eye Of the Beholder. This is new form of Haiku. A Bit like Modern Art may I say. River of Creativity.

HAH, with freedon comes responsibility...... or so they say ehemm. I quite enjoyed school actually, our school was rubbish, and one of the roughest in our part of london. so it was all good.

but issnt W an empty set??? i dont think there are any integer pair (n,m) satisfying n^2=m(m+1)/2 i tried using maple as well. [edit]OOOPS I AM SO STUPID,,, (1,1) obviously does the job[/edit]

can u show us how you derived the first limit? that result is interesting

its not suprising that a sequence of rationals gives u a irrational as a limit a famous example is the limit of ratio of consecutive fibonacci numbers for those who dont know , they are defined as [math]F_{0}=1[/math] [math]F_{1}=1[/math] [math]F_{n+1}=F_{n}+F_{n-1}[/math] for [math]n\ge 2[/math] and [math]\lim_{n\to\infty}\frac{F_{n+1}}{F_{n}}=\phi[/math] where phi is generally known as the golden ration/number and its value is [math]\frac{1+\sqrt{5}}{2}[/math]

OHHHHH,,, how stupid of me, i should have thought of each point , rather than just the endpoints. I get it now. Couple more Affine Spaces question to do for tmr. Cheers

still dont get it I thought the statement in my last post WAS what we wanted to prove... thats a bit like saying its true because it is. Well ok. it sends st. lines to st.lines. and the endpoints are in F©, but i still dont get how the line between F(x) and F(y) HAS to be in F©. Can't it be that some portions are in A'\F©

so since the line segment from x to y is in C then the line segment from F(x) to F(y) is in F©. is that all the proof?

huh? whats Spec(k[x,y,...,z])?? Where did we use the fact that C is a convex subset of A in the proff above?

I meant since F is an affine map, its takes points to points. so F can't act on vector. but you said at the beginning " F(x) = L(x)+v for some linear map L and some fixed vector v" which sort of doesnt make sense as F is defined from A->A' and not from the respective vector space V->V'. although i do understand what you are saying.

ok pick two points in F©,say F(X)=X' F(Y)=Y' then since F is affine, the line [math](1-t)\vect{x}+t\vect{y}[/math] is mapped into [math](1-t)\vect{x'}+t\vect{y'}[/math] ARGH... then is that line a subset of F©?

Let [math]F\colon \mathcal{A}\to \mathcal{A}'[/math] be an affine linear map, and let [math]\mathcal{C}[/math] be a convex subset of [math]\mathcal{A}[/math]. Show that its image [math]\mathcal{C}'=F(\mathcal{C})[/math] is a convex subset of [math]\mathcal{A}'[/math] Any Hints on how to proceed?

well the general way is to pick an element, then see what the permutation does to that element . in cyclic notation [math](a_{1}a_{2}\dotsb a_{k})[/math] is a k cycle and read as [math]a_{1}\rightarrow a_{2},a_{2}\rightarrow a_{3}\dotsb a_{k-1}\rightarrow a_{k},a_{k}\rightarrow a_{1}[/math] an example would be a rotation of a square. Label each vertex, i.e 1,2,3,4. (starting from top left, and going clockwise) rotate the square by 90 degrees anti clockwise round the origin. that will take vertex 1 to 2, 2 to 3, 3 to4, and 4 to 1 written in a cycle it gives (1234) for a cycle its order is just its lenght. take the same square and reflect on the horizontal line through its centre. that takes 1 to 2 . 2 to 1. 3 to 4 , 4 to 3 that can be written as (12)(34) when you have a product of disjoint cycles then its order is the least common multiple. ill just try to show you how it works. suppose you have sigma = (abcd)(efg)(hiklm) then sigma^n = (abcd)^n (efg)^n (hiklm)^n since they are disjoint and each of those cycles have order of its lenght. so to get sigma^k=sigma you need all those three to be themselves as well. when n=4 . the first cycle is same . the rest arent. when n=3 the second one is same, the rest arent, when n=5, the third one is same the rest arent. so you keep on going. now the first cycle is same when n is a multiple of 4, second when n is multiple of 3. and third when n is multiple of 5 so the least number k that satisfies being a multiple of 4,3,5 is lcm(4,3,5)=60

well. i uninstalled it. and it really messed up my windows hehe.

woo. Got windowblinds and Iconpackager for Stardock. All i need it DesktopX for its Objects and Widgets... someday still have to see if it hampers my performance

whats this "LARGER" you pple are on about. I think you mean LAGER.

there are various software to create fractals. http://www.ultrafractal.com/showcase.html this is a "arty" software to create fractals. here are tons of exampls of fractals http://mathworld.wolfram.com/topics/Fractals.html and definition of fractal http://mathworld.wolfram.com/Fractal.html

yes , they do it all the time, on ships anyway

oh man. u should see some of the parodies on ebay. just go to ebay.com and serch for virgin mary cheese th9is painting is quite good what about a butterfly grilled cheese

thanks a lot mandrake, i will have a stab at it sometime, i really need to brush on my l337 analysis skillz.

yes please. some hints would be good. could u give me some cauchy sequence which does not converge. [edit] oops x_n=a for all n is a cauchy sequence that doesnt converge rite[/edit] where d is the usual metric in R [edit] dOH.,,,, that ovbisouly converges it should be x_n = n