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Comments:
<0> hi
<1> bye #math !
<2> f(x)= 1/(x-1) ==> x=1 is asimptota ?
<3> yes, it's an asymptote
<2> hm okay thanks
<2> how about f(x)=log(x-1) is x=1 is asymptote here too ?
<3> Yes, and a much more serious one
<4> does anyone know how solve (1/x)=tan(1/x)
<2> how so ?
<2> if it were ln(x-1) would it be any different ?
<5> hmm....I don't understand why it is an asymptote at all...
<5> ln(0) = 1
<5> and log(0) = 1
<3> Other way around, joerg
<3> ln(1) = 0
<2> hehe
<5> oh....sry ;)
<2> you fantesized or how ever your write it :P
<3> govner: No, the difference lies in that you can't multiply it by any number of (x - 1) factors to make it go away
<3> And so we say that it's an essential singularity
<2> in other words please :X
<3> It's probably not important :)
<3> They're both asymptotes, yes
<3> I don't mean to confuse
<2> i know you dont
<4> *does anyone know how to solve (1/x)=tan(1/x)
<3> MWG: Numerically :)
<6> MWG_Thomas: you reduce it to x=tan(x)
<4> is there no other way?
<3> If there were a closed form we would have used it in diff eq
<7> @plot (x^2-4)^2-x^2+3 xmin:-3,xmax:3
<8> csk0: http://nanographer.nanosouffle.net/?func=%28x%5E2-4%29%5E2-x%5E2%2B3%20xmin:-3%0Axmax:3%0A
<3> Instead we just said 'let b_n be the nth root of tan(x) = x'
<3> So based just on that experience, no, there's no closed form
<6> MWG_Thomas: if you want, you can make one up. define bleem(y) as x s.t. tan(x)=y*x. then it's bleem(1)
<3> :)
<5> 0 = tan(0) isn't it? :)
<3> Which is basically what we did
<3> Also, 1/tan(x) = tan(1/x) ?
<3> I didn't know that
<4> 1/x not equal to 0
<3> Oh, substitution
<3> haha, obvious
<2> thermoplyae is sqrt(x-1) x=1 asymptote too ?:X
<9> When using semantic tableaux, how might you get an infinite branch?
<9> I have a proof of correctness/completeness for semantic tableaux and that's a case it deals with, but I have no idea how it could happen
<10> (That's not me, by the way.)
<10> I'm off to teach now.
<9> hmm
<11> hi
<11> i heard i might be able to get a little math help in here?
<9> question?
<11> when you integrate cosine
<11> you would get -sin, correct?
<9> no, sin
<11> wtf, my teacher lied to me
<3> govner: No, I don't believe so
<9> if I could use the bot I could prove it
<12> Revv: the derivative of sin(x) is cos(x)
<9> but given diff of sin = cos
<9> well, i'll be here for a while, if a semantic tableaux expert happens to show up
<6> revv: integral of cos is sin, of sin is -cos, actually integral 0..x sin(y)dy is 1-cos(x)
<2> so when i look for range of defintion i need to remember that when it comes to log/ln asypmtote exists and when it comes to squareroot it doesnt ?
<3> Yes
<3> Since sqrt(x - 1) is defined for x = 1 and has value 0
<12> you should consider the 'range' and 'domain' for functions, then the same for it's derivative
<3> It can't get closer and closer to the line x = 1 because it actually has a value there
<11> ****, this physics **** is killing me, I don't even know how to start the problem
<12> go to #physics
<3> A wussy definition of asymptote, but it's probably what you have to work with
<11> thx
<2> oh right
<3> yeah, we don't want it either, go to #physics :)
<9> is there a #computation ? :P
<2> what i mean is that when i draw a function i've been told that in sqrt you odnt have asymptote where it's not defign
<12> yes
<2> i dont draw anything
<2> there's #eraction , from what i've heard :P
<9> the domain of tan includes pi/2?
<9> (***uming thats where the aymptote is, it's been a while)
<9> MY point is if the value of x is in the domain, but it's undefined, you have an asymptote, but for square root the domain is just x = 0 and up?
<3> govner: That's not always accurate for all cases
<3> (x - 1) / (x - 1) for instance
<3> But it certainly fits the bill with sqrt
<13> Is this a proper fraction? (x^2) / ( (x - 1) (x^2 + x + 1) )
<2> in the example you've just gave i know that's a hole where 1 is :X
<3> Right
<3> Puppies: Yes, I believe so
<2> how about 1/sqrt(x-1)
<2> in x=1 is that an asymptote ?
<3> It is
<6> steve: well, what value would it have there?
<9> it'd be undefined?
<6> steve: that should answer your question
<2> i've been given this function f(X)=(x^2+4)*e^(x/2) now they ask me to explain why the function doesn
<9> sqrt(-1) is undefined too? (ignoring complex numbers)
<2> 't cut the x axis
<6> steve: yup
<9> so, why is there not one big fat asymptote, or is that just not what they do?
<6> ?
<14> govner: because x is always positive
<3> govner: What can we say about the two factors in terms of whether they are positive, negative, or zero?
<14> i mean
<14> f(x)
<9> yea, I'm just being silly
<2> that the function no matter what i put into it , it will always return a postivie number?>
<6> steve: aymptote is an abused term, there are different ideas that get the same name. you need to be clear what you mean in the context.
<14> govner: yes
<2> thanks aFlag :)
<2> does this function have a vertical asymptote ?
<3> Does it?
<2> i think so :X
<2> at minus -x
<3> Why do you think that>
<3> -x?
<15> f(X)=(x^2+4)*e^(x/2)
<2> where the values of x are negative :X
<15> that the function?
<2> yes
<3> That is a confusing statement, and I don't know quite where to start
<2> if x=-100000000
<15> well
<3> Is f(x) defined there?
<15> i was taught
<15> if there is no denominator
<2> it is..
<15> you cant have a Vertical asymptote
<3> Then there can't be a vertical asymptote there
<2> wait checking what denominator means
<15> you kow
<15> a fraction
<15> numerator / denominator
<6> tz`dtop: hint, you do have a horizontal one
<2> yes just checked
<13> Does mbot have the capabilities to do long division on 2 polynomials?
<15> yes trwbw
<15> where that whole equation = 0 should be where the HA is
<15> i think :)
<2> so wait
<2> is there or isn't there ?
<15> there isnt
<3> Puppies: Yes, but please pm mbot
<2> really? so what happens in x=-1000000
<3> Nothing special, govner
<3> Gets tiny, I guess
<15> you will see
<2> :) how tiny sir ?
<15> at x=-100000
<15> its obvious
<15> there is a HA
<2> HA ?
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