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Comments:
<0> % Limit[Log[.5]/Log[x],x->0]
<1> ihope: 0
<0> It seems 0^0 should actually be .5.
<2> hi, how can i use mbot to get the derivative of a function?
<3> N packets arrive every (L/R)N seconds, (n-1)(L/R) is the delay of each packet. is this the correct way to determine the average queueing delay depending on number of packets ? (sum(n,n,1,x-1)*(L/R) / x)
<4> % D[ArcTan[Sqrt[x+2]/Sin[x]]
<1> xerox:
<1> ToExpression::sntxi: Incomplete expression; more input is needed.
<1>
<1> $Failed
<4> boom
<4> % D[ArcTan[Sqrt[x+2]/Sin[x]]]
<1> xerox: ArcTan[Sqrt[2 + x]*Csc[x]]
<5> % D[xx,x]
<1> EvilRanter: 0
<5> huh.
<5> %D[x^2,x]
<6> azi` - http://mathbin.net/7318
<1> Title: MathBin.net - Product rule
<0> % D[x*x,x]
<1> ihope: 2*x
<7> % D[x x,x]
<1> delta: 2*x
<0> % D[2x,x]
<1> ihope: 2
<6> % D[FresnelS[x], x]
<1> Catfive: Sin[(Pi*x^2)/2]
<3> does this make any sense or am I completely off track ? :) http://www.mathbin.net/7319
<1> Title: MathBin.net - Queueing Delay
<6> ski - what do you mean?
<8> in the former case, "'" implicitely means "take the derivative of this expression, wrt 'x'"
<8> in the latter case "'" means "differentiate this function"
<8> two related, but different things
<6> ski - the first case should be written (fg)'(x)
<8> (btw, the "implicitely" was meant to refer to that it was wrt 'x', and not some other variable)
<8> Catfive : yes, that would be more coherent
<8> (x |- f(x)g(x))'
<8> would also do
<8> er
<8> (x |-> f(x)g(x))'
<7> :)
<6> because it makes you look like a programmer
<8> as i see it, the need existed long before programming started
<6> which specific notation are you having trouble with, again?
<8> i was having trouble with the *lack* of a notation
<6> for what?
<8> giving a formula for a function, without needing to give it a name
<8> there exists common specialization of it, in e.g. sum,product,integral,derivative,limes notation
<6> ski - that's what \mapsto is for
<8> hence my use of it above
<6> but you don't need it above
<9> What method of integration should I use to integrate x*sin(x^2) ?
<10> mike8901: well... take a look at the thing...
<10> which part seems to be the "most complicated" part of the integral?
<9> I tried by parts, but I couldn't evaluate the integral of cos(x^2)*x^3
<9> well, the sin(x^2) of coursae
<9> *course
<10> yeah
<10> now, do you think it'd be easier if it was sin(u)?
<9> uh, I guess.
<6> xerox - inclusion
<10> try letting u = x^2 and see what happens
<9> hm?
<10> du = 2x dx.... so dx = du/(2x)
<9> so t*sin(u)
<4> Catfive: field inclusion right
<9> I'm a bit confused
<10> well... what you did was let u = something, and you found dx
<6> xerox - inclusion is the generic meaning of the symbol
<10> so you replace dx with du/(2x) and you replace x^2 with u
<9> so I find t*(sin u)dx ?
<10> now... if you notice, the x's at the front cancel out, and you end up with (1/2)*sin(u) du
<10> where'd you get the t from?
<9> er
<4> Catfive - I thought the one that looks like C is inclusion.
<9> x*sin u d\x
<9> without the \
<8> Spivak, chapter 13, page 261 (3rd edition) : "..., integral _a ^b f = b^3 / 3 - a^3 / 3 if f(x) = x^2 for all x. This list already reveals that the notation 'integral _a ^b f' suffers from the lack of a convenient notation for naming functions defined by formulas. ..."
<9> how do I integrate that though?
<4> U turned -pi/2
<10> yeah... don't see a t there...
<9> the problem uses a t, so I accidentally put one in :P
<10> now, what is du (when u = x^2)?
<9> x^3/3
<10> uh... no (remember - differentiate, don't integrate)
<9> oh
<9> then it's just 2x
<10> yeah
<8> here using '\mapto' is naturally suggested, methinks .. but they just introduce the standard integral notation, instead :/
<10> now, treating du and dx as factors, what do you get for dx in terms of x and du?
<6> xerox - or sometimes injection
<9> huh?
<10> * treating du and dx as variables
<4> Catfive: yeah, I see why, but I never saw the pointy version before that example
<9> I'm really confused....
<10> what you basically do here is substitute the dx with something * du
<4> nice idea.
<10> so, from du = 2x dx.... you would get dx = du/(2x)
<9> on my paper thus far I have written integral of t*sin u dx
<6> xerox - by the inclusion map from a subset B_0 of B into B, I mean the map i: B_0 -> B given by i(b) = b
<6> ski - we could use \mapsto there, but the 'dx' notation is more brief, more cl***ical (for what that's worth), and also has important semantic meaning if we have, for example, the theory of differential forms available.
<10> mike8901: so, what is the original problem?
<9> to integrate t*sin(t^2)
<10> since the t should not be in the problem anywhere...
<10> ah... ook
<10> integrate t*sin(t^2) dt?
<11> http://mathworld.wolfram.com/ProductRule.html getting from (3) to (4) when you have f(x+h)*(g(x+h) - g(x)) = f(x)g'(x) you ***ume h = 0 ?
<1> Title: Product Rule -- from Wolfram MathWorld
<10> alright... don't use x in this case
<9> the problem doesn't use integral notation... it just gives a point on the original, and an equation for the derivative
<10> let u = t^2... du = 2t dt, dt = du/(2t)
<10> ah, ok
<6> azi` - http://mathbin.net/7318
<1> Title: MathBin.net - Product rule
<10> does it say that the derivative with respect to t is t*sin(t^2)?
<4> differential forms. i've heard of forms.
<9> wait
<9> where did you get dt = du/(2t)
<11> Catfive: yes, i checked that and you have the same step there
<8> (Catfive : mm, it's just that that sentence asks more for '\mapsto' than what they say after .. so if they don't give the natural answer, imo that sentence should be changed ..)
<11> Catfive: it just looks that you did it with less intermediary steps than on mathworld
<4> Catfive: is that something like a gradient?
<10> mike8901: you understand how I got du = 2t dt, right?
<6> ski - yeah, Spivak doesn't really attempt to address the general question there (it's arguably not the right context); but it's frequently addressed in books that provide a more comprehensive discussion of mathematical notation.
<9> right
<10> now, divide both sides by 2t
<10> (du and dt are to be treated as variables)
<9> ah ok
<10> now.... you substitute in t^2 = u (since u = t^2), and dt = du/(2t)
<10> you'll notice that the t's at the left would cancel out, you'd end up with a factor of (1/2) in front of the thingy...
<9> so right now, I have integral of t sin u du/2t ?
<10> and you'd have this: integral((sin(u)/2) du)
<10> yeah...
<9> ok
<10> and you know where to go from here, right?
<6> xerox - yeah, that's the differential one-form that ***ociates to each point the differential of a scalar-valued function at that point
<9> so it's just -cos u / 2 ?
<10> yes
<9> then I plug in t
<10> yeah
<10> and don't forget + C :)
<9> and get -cos t^2 / 2
<9> well I got a point
<10> yep
<9> so I can find the exact antiderivative
<9> btw, what's that method called?
<9> parts?
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