| |
| |
| |
|
Page: 1 2 3 4 5 6 7 8
Comments:
<0> actually, the start variable does half of the bookkeeping for finding the subsequence.
<0> that's 3n 'steps'. 3n-3 can be achieved by modifying the initialization slightly.
<1> gotcha
<1> thx
<0> and s and sum are aliases, apparently :)
<0> Manyfold: would you do anything different?
<2> revising stuff is good :)
<3> i'd like to try this again
<3> ax + by + cz = d ?
<3> I have three points in 3D space forming a triangle, I have an x and z position on this triangle, how can I find out what the y is?
<3> How can I use this ax + by + cz = d to figure out the y?
<4> int[Sin(2t) Sqrt[Sin^4(t) + Cos^4(t)] dt. Havent solved this one for a week. Any suggestions will be appriciated :P
<1> is that [sin(x)]^4 or sin(sin(sin(..(x)))))
<1> ?
<1> t*
<4> Sin(t)*sin(t) ....
<5> yeah, that's a bizarre question on mnvl's part
<1> what's your problem HiLander ?
<5> no problem; your question was strange
<5> i've never seen anyone use sin^n(x) to denote sin(sin(...(x)...))
<1> you've never seen f^n(x) meaning f(f(f(..f(x)))) ?
<1> don't worry, you will
<5> i didn't say that
<5> i said, specifically, for the trig functions, i've never seen that definition applied
<6> if i have an equivalence relation ~ on a compact metric space X and the natural projection p:X --> X/~, what is a decent sufficient condition to be able to 'lift' a continuous map f:X/~ -> X/~ to a continuous map f':X->X?
<1> HiLander, i am not as stupid as you seem to think i am
<0> yes, luckily sin(sin(x)) and friends seem to be practically useless.
<5> mnvl: i didn't say you were stupid, i said your question was bizarre given the context
<5> frankly, i've yet to give any thought as to your intelligence
<5> and i see no reason to
<7> In the unusual case that someone did want to talk about the iterated cosine function or something, they would probably use cos^{(n)}(x), because the convention of cos^n(x) meaning {cos(x)}^n is so well-established.
<1> well Copter can you not factorise sin^4(x) + cos^4(x) ?
<4> I tried EVERYTHING I know to get rid of that square root
<0> Sin^4(t) + Cos^4(t) = (Sin^2(t) + Cos^2(t))^2 - 2 Sin^2(t)Cos^2(t) = (2 - Sin^2(2t))/2 = Cos(4t)/2. That ought to help.
<4> usually it comes in the forms where it ends up being 1 like sin^2(x)+cos^2(x) or similar things
<1> HiLander, if i ask any more questions which you may find bizarre given the context, it will not be necessary for you to comment
<5> oh, that's very nice of you
<5> but since your permission isn't really something i care about
<1> sin(2t) * sqrt(cos(4t)) ??
<5> i'll kindly decline
<0> mnvl: it's probably better to stop one step earlier then.
<8> what's off topic? is there something that can *not* be math related?
<1> no int-e i was just thinking that the last one looks like something that could be substituted
<0> hmm, the last step was wrong anyway
<9> anyone's good with algebraic geometry?
<0> (2 - Sin^2(2t))/2 = (1 + Cos^2(2t))/2, now Sin(2t) * sqrt((1+Cos^2(2t))/2) looks good actually.
<0> because it suggests a substitution.
<4> int-e: thx, finnally1
<4> !
<8> xahlee: you here too?
<1> well done int-e
<10> hi myrkraverk_
<4> Sin(2t) * sqrt((1+Cos^2(2t))/2) looks good actually., what would be the next move from hjere?
<10> myrkraverk_: do i know you from #emacs? or?
<10> myrkraverk_: your name sounds familiar but i forgot... sorry.
<8> xahlee: #emacs, I think, mostly ;)
<0> Copter: multiply by -1
<10> myrkraverk_: ah. :) I think i'm still banned there.
<8> xahlee: banned, why?
<0> (No I'm not kidding. I'm just trying to make the obvious substitution even more obvious.)
<0> no you are not.
<10> myrkraverk_: long story. http://emacswiki.org/cgi-bin/wiki/KickbanXahLeeFromEmacsChannel
<0> oh
<0> #emacs, sorry. wrong context.
<4> int-e: the expression in the root?
<10> myrkraverk_: ah, you are the one in Iceland, right?
<4> its 1+cos^2[2t]/2 goes to -1 - cos^2[2t]/2 , how does it help?
<8> xahlee: ah - I think I remember some line noise about that (for your info, I'm personally rather neutral towards any of that stuff)
<8> xahlee: yes
<0> Copter: try smaller steps
<10> myrkraverk_: nice to see you here anyhow. :D
<0> Copter: you don't need to solve the whole integral in one step, just simplifying it is enough
<8> xahlee: ;)
<0> Copter: and you can eliminate the trigonometric function now which helps a lot.
<0> *functions
<4> I cant see...
<10> the question about algebraic geometry i have is that, you know, the problem of cl***ifying curves (polynomials in 2 vars) of higher degrees seems to have stopped in the 18th century...
<4> still stuck in the root
<8> xahlee: btw, I've resently (this week) started math at reykjavik university (in case that interests you)
<10> i mean, basically, algebraic geometry took a abstract development and went on into more and more abstract hteorems...
<0> I would deal with the root in the next step.
<10> myrkraverk_: as in, majoring in math? or started college?
<8> xahlee: taking BS in math, whatever that is of the 2 you mentioned ;)
<10> Newton is the one who did the cl***ification of 3rd degree curves
<10> but afaik, there's no general theory about what curves looks like in 4 or 5, 6 etc degrees
<8> xahlee: you mean 4, 5, 6 dimensions?
<10> afaik, even with all the powerful abstract algebraic geometry machinery, there's no answer to the basic question of what curves looks like.
<10> myrkraverk_: the graph of polynomials of degree n
<8> xahlee: oh, ok - I don't really know anything about polynomials ;P
<10> i would really like to know, from someone who had algebraic geometry, to comment about what curves look like of higher degree polynomials.
<8> xahlee: we've been doing induction (I thing the english term is) this week
<4> int-e: On a definite integral of limits 0..2pi , im unable to make that substitution coz it results with an integral with bounds 1 and 1
<10> myrkraverk_: is this your 3rd year in college?
<0> Copter: are you serious?
<0> Copter: the integral is from 0 to 2pi?
<8> xahlee: no, 1st (of math, that is) -- and we don't really have "college" here, only what pretty much amounts to high shcool and then universities
<4> maybe not ....Im guessing its 0 to 2pi, but knowing myself, im probably wrong
<10> myrkraverk_: in American speak, college is just a general term for university.
<8> xahlee: oh, k
<11> nah
<11> its different
<0> Copter: if it is from 0 to 2pi, you can just show that the result is 0 by splitting it into two parts.
<11> college and university are different
<11> the only difference is the population of students
<8> SmoothOp: different how?
<11> and maybe something else
<5> wrong
<0> Copter: or four as it may be.
<8> SmoothOp: hehe ;P
<11> well
<5> a school is a university if it has a graduate program
<11> its something
<5> a college otherwise
<11> thanks
<11> thats what i ment
<4> The original problem was to find arch length of x=asin^4t , y= acos^4t... is that 0 to 2pi? :p
<5> but in the states, we say "go to college" and not "go to university"
<11> b/c we is american
<5> regardless as to whether it's a college or university
<10> HiLander: yeah, i think HiLander summed it up.
<11> and us americans dont care
<0> that's from 0 to pi I think.
<8> HiLander: ah, k; so for undergrad things (like BS in math degrees) it doesn't matter?
<12> Weird, we still call my school a university, and it doesn't have a grad program
<12> Oh, wait. Yeah, it does. Sorry, I forgot.
<8> well, ru (my uni) obvously has grad program(s), we're so far sharing the cl***room with masters students
<11> guess not
<10> myrkraverk_: another way of what HiLander said is that: the term college and univ have a somewhat technical definitions (though not very exact), but in general, we say "go to college" and we don't say "go to university".
<0> Copter: But you better check for yourself.
<8> xahlee: k - I think I have it now ;)
<4> int-e: do I place t=0, t=pi/2 t=pi , etc..to get an idea how it looks like, just like i am doing with a polar form?
<10> nice site http://windowseat.ca/circles/index.php
<0> Copter: you should know what they mean. the curve has some anomalies.
<8> now, the reason I joined in the first place, was that I'm taking a look at homework: for p(n) = 2^n f(n), where f(n) is fibonacci, and d(n) = p(n+1) - p(n) ---> prove that d(n) always ends in a 6
<0> Copter: btw, you also seem to have pulled a factor out of a square root without taking its absolute value. that's dangerous.
<4> it was a^2
<4> I pulled it as sqrt(a^2)
<0> sqrt(a^2) = |a|
<0> not a
<0> unless a>=0
<8> if anyone cares to lend a hand feel free (it's not a return ***ignment)
<4> a>=0 wasnt stated, thx for that remark
<10> myrkraverk_: tell us where you are stuck
<3> I have three points in 3D space forming a triangle, I have an x and z position on this triangle, how can I find out what the y is? (someone is telling me this is impossible, is this true?)
<8> xahlee: I'm just starting now to look at the first few d(n) ;-D
<10> Scottc: what is your x and y? points?
<8> xahlee: I'm just kinda used to work in a group of 2 - 4 ppl
<3> the x and y are part of a point
<3> er x and z
<3> the y is the height of the point, which I need to find out.
<0> Copter: anyway, to make the substitution work, just cut the integral into small enough pieces.
Return to
#math
or
Go to some related
logs:
#ubuntu
full install of debian
paralel dhcpcd
#gentoo
#debian
beth buu
marvell 88e8001 fedora
php windows disable-odbc
#gentoo
#sdl
|
|