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[edit] January 4

[edit] Involute curves

Hello everyone,

Could anybody tell me, informally, how to plot the construction of an involute curve without involving arclengths (or just give me an efficient way of drawing an involute curve for "any" given parametric curve).

I'm trying to draw the involute of the curve described parametrically by \bigg(-2.4\cos(t)+0.6\cos(4t),-2.4\sin(t)+0.6\sin(4t)\bigg), but then I have to integrate \int_0^{a}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt which is just unwieldy and slows down the calculations to no end (as I have to calculate it at each step) (the Wolfram integrator returns a function with about 50 sines/cosines).

The method I'm using now uses the equation on the involute page : r(t)-tr^\prime(t) describes the equation of the involute when using arclength parametrization, but here I can't use arclength parametrization, nor can I use the other equations which need the evaluation of the arclength.

I'm not too willing to start coding implementations of algorithms for numeric calculation of arclengths eiter, I haven't really got the time.

Anyway, thanks to everyone that can help out. -- Xedi (talk) 04:51, 4 January 2008 (UTC)

I know this may be of little help, but the integral
\int_0^{a}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt
you are searching for, could be expressed (and easily computated by a CAS) as an elliptic integral of the second kind:
\frac{24}{5} E(\frac{3 a}{2}|-\frac{16}{9}). Pallida  Mors 05:20, 4 January 2008 (UTC)
Are you sure ? The Wolfram integrator gives an expression just in term of sines and cosines. -- Xedi (talk) 12:53, 4 January 2008 (UTC)

This is the mess I get :

Integrate[ (((2.4*Cos[4*x] - 2.4*Cos[x])^2 + (2.4*Sin[4*x] - 2.4*Sin[x]))^2)^ (1/2), x] ==

(0.*Sqrt[((2.4*Cos[x] - 2.4*Cos[4*x])^ 2 - 2.4*Sin[x] + 2.4*Sin[4*x])^ 2])/((2.4*Cos[x] - 2.4*Cos[4*x])^ 2 - 2.4*Sin[x] + 2.4*Sin[4*x]) + (5.76x* Sqrt[((2.4*Cos[x] - 2.4*Cos[4*x])^ 2 - 2.4*Sin[x] + 2.4*Sin[4*x])^ 2])/((2.4*Cos[x] - 2.4*Cos[4*x])^ 2 - 2.4*Sin[x] + 2.4*Sin[4*x]) + (2.4*Cos[x]*Sqrt[ ((2.4*Cos[x] - 2.4*Cos[4*x])^2 - 2.4*Sin[x] + 2.4*Sin[4*x])^2])/ ((2.4*Cos[x] - 2.4*Cos[4*x])^2 - 2.4*Sin[x] + 2.4*Sin[4*x]) - (0.6*Cos[4.*x]* Sqrt[((2.4*Cos[x] - 2.4*Cos[4*x])^ 2 - 2.4*Sin[x] + 2.4*Sin[4*x])^ 2])/((2.4*Cos[x] - 2.4*Cos[4*x])^ 2 - 2.4*Sin[x] + 2.4*Sin[4*x]) + (1.44*Sin[2.*x]* Sqrt[((2.4*Cos[x] - 2.4*Cos[4*x])^ 2 - 2.4*Sin[x] + 2.4*Sin[4*x])^ 2])/((2.4*Cos[x] - 2.4*Cos[4*x])^ 2 - 2.4*Sin[x] + 2.4*Sin[4*x]) - (1.92*Sin[3.*x]* Sqrt[((2.4*Cos[x] - 2.4*Cos[4*x])^ 2 - 2.4*Sin[x] + 2.4*Sin[4*x])^ 2])/((2.4*Cos[x] - 2.4*Cos[4*x])^ 2 - 2.4*Sin[x] + 2.4*Sin[4*x]) - (1.152*Sqrt[ ((2.4*Cos[x] - 2.4*Cos[4*x])^2 - 2.4*Sin[x] + 2.4*Sin[4*x])^2]* Sin[5.*x])/ ((2.4*Cos[x] - 2.4*Cos[4*x])^2 - 2.4*Sin[x] + 2.4*Sin[4*x]) + (0.36* Sqrt[((2.4*Cos[x] - 2.4*Cos[4*x])^ 2 - 2.4*Sin[x] + 2.4*Sin[4*x])^ 2]*Sin[8.*x])/ ((2.4*Cos[x] - 2.4*Cos[4*x])^2 - 2.4*Sin[x] + 2.4*Sin[4*x])

Let's see what some old-fashioned hand-cranking can do:
x=\bigg(-2.4\cos(t)+0.6\cos(4t)\bigg),
\left(\frac{dx}{dt}\right)^2 = \bigg(2.4\sin(t)-2.4\sin(4t)\bigg)^2 = \bigg(\frac{12}{5}\bigg)^2\bigg(\sin^2(t) - 2\sin(t)\sin(4t) + \sin^2(4t)\bigg)
y=\bigg(-2.4\sin(t)+0.6\sin(4t)\bigg),
\left(\frac{dy}{dt}\right)^2 = \bigg(-2.4\cos(t)+2.4\cos(4t)\bigg)^2 = \bigg(\frac{12}{5}\bigg)^2\bigg(\cos^2(t) - 2\cos(t)\cos(4t) + \cos^2(4t)\bigg)
\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2 = \bigg(\frac{12}{5}\bigg)^2\bigg(2 - 2\sin(t)\sin(4t) - 2\cos(t)\cos(4t)\bigg)=2\bigg(\frac{12}{5}\bigg)^2\bigg(1 - \cos(3t)\bigg)
=2\bigg(\frac{12}{5}\bigg)^2\bigg(2 - 2\cos^2\bigg(\frac{3t}{2}\bigg)\bigg)=4\bigg(\frac{12}{5}\bigg)^2\bigg(\sin^2\bigg(\frac{3t}{2}\bigg)\bigg)
\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=\bigg(\frac{24}{5}\bigg)\sin\bigg(\frac{3t}{2}\bigg)
and integrating between 0 and a gives
\bigg(\frac{16}{5}\bigg)\bigg(1-\cos\bigg(\frac{3a}{2}\bigg)\bigg)
Is that right ? Gandalf61 (talk) 13:53, 4 January 2008 (UTC)
Apparently, Pallida has forgotten to differentiate. Regardless, the problem with the online Wolfram Integrator is that it doesn't simplify the results. Gandalf's derivation incorrectly assumes that \sqrt{x^2}=x and is thus valid only for 0 \le a \le \frac{2\pi}{3}. The correct general expression is \frac{16}{5}\left(1+2\left\lfloor\frac{3a}{2\pi}\right\rfloor-\cos\left(\pi\left\lfloor\frac{3a}{2\pi}\right\rfloor-\frac{3a}{2}\right)\right). -- Meni Rosenfeld (talk) 14:05, 4 January 2008 (UTC)
Ah, yes - so we are actually integrating \bigg(\frac{24}{5}\bigg)\bigg|\sin\bigg(\frac{3t}{2}\bigg)\bigg|. I stand corrected. Gandalf61 (talk) 14:14, 4 January 2008 (UTC)
I apologize for my mistake. I did differentiate, but I did it twice.
So, the integral seems to be, at last,
\int_0^a \frac{24}{5} \sqrt{\sin ^2\left(\frac{3 t}{2}\right)} \, dt
A primitive is:
-\text{sgn}\left(\sin \left(\frac{3 t}{2}\right)\right)\frac{16}{5} \cos \left(\frac{3 t}{2}\right)
Finally, the primitive evaluated at 0 (approaching from the right) is -\frac{16}{5}. Pallida  Mors 16:27, 4 January 2008 (UTC)
But this is discontinuous at some points, thus not an antiderivative everywhere and not suitable for integration. -- Meni Rosenfeld (talk) 17:45, 4 January 2008 (UTC)
Sorry Meni, what is discontinuous? The integrandum is not. It is not differentiable at points like \frac{2}{3} \pi.. Now I see your point. The primitive I gave is discontinuous, now that I see it. But it can be used to calculate the area for intervals of the form [\frac{2n\pi}{3};\frac {2(n+1)\pi}{3}] for n any non-negative integer. Pallida  Mors 18:16, 4 January 2008 (UTC)
In other words, take the primitive to be
-\text{sgn}\left(\sin \left(\frac{3 t}{2}\right)\right)\frac{16}{5} \cos \left(\frac{3 t}{2}\right)+\frac{32}{5} \left\lfloor \frac{3 t}{2 \pi }\right\rfloor.

I think this reformulated function overcomes the problems you have pointed out. Pallida  Mors 18:36, 4 January 2008 (UTC)

The function defined with those strange brackets is the floor function. Pallida  Mors 18:39, 4 January 2008 (UTC)
Okay, but I have already given a primitive which is simpler and also happens to cross the origin, so it readily answers the original question of \int_0^a\|\mathbf{x}'(t)\|\ dt. -- Meni Rosenfeld (talk) 18:58, 4 January 2008 (UTC)
Yep, thanks a lot, that solves the problem quite nicely. -- Xedi (talk) 23:30, 4 January 2008 (UTC)

[edit] fuzzy logic

give some examples of fuzzy logic problems in electrical field?10:21, 4 January 2008 (UTC)

Fuzzy logic, Fuzzy electronics and this might help get you started. -- Meni Rosenfeld (talk) 14:07, 4 January 2008 (UTC)

[edit] LaTeχ Freeware

saw this [1]

and wondered if anyone knew of any good freeware for entering LaTeχ into Microsoft office. Thanks! 172.200.130.39 (talk) 15:20, 4 January 2008 (UTC)

also is it true to say that \lim_{n \to 0}\bigg(n \cdot \sum_{r=a}^{b \cdot n^{-1}} f(r \cdot n) \bigg) \equiv \int_{a}^{b} f(x)\, dx —Preceding unsigned comment added by 172.200.130.39 (talk) 16:28, 4 January 2008 (UTC)

It should be a\cdot n^{-1} in the summation. You need to take the limit from the right \left(\lim_{n\to0^+}\right)). A simple = sign might be more appropriate than the \equiv sign. That said, you'll have to specify what you mean by a summation where the bounds are not integers, but for any reasonable definition, this equality is correct. -- Meni Rosenfeld (talk) 17:41, 4 January 2008 (UTC)
For non integer bounds can you not just say \int_{a}^{b} f(x)\, dx = \int_{0}^{b-a} f(x-a)\, dx =(b-a)\int_{0}^{1} f(\frac{x-a}{b-a})\, dx, 172.200.130.39 (talk) 19:09, 4 January 2008 (UTC)
This still leaves you with non-integer bounds in the summation, provided n is general. If you restrict n to be \tfrac{1}{k} with k \in \mathbb{N}^+ this transformation could work (though it should be \int_{a}^{b} f(x)\, dx = \int_{0}^{b-a} f(x+a)\, dx =(b-a)\int_{0}^{1} f(x(b-a)+a)\, dx). -- Meni Rosenfeld (talk) 19:14, 4 January 2008 (UTC)
And for those very unreasonable definitions where the formula above might not hold, you might want to take a look at Lebesgue integral. -- The Anome (talk) 18:42, 4 January 2008 (UTC)
I forgot to mention my assumption that f is continuous, or more generally, that the Darboux\Riemann integral exists and is referred to in the formula. This doesn't have much to do with the summation. -- Meni Rosenfeld (talk) 18:53, 4 January 2008 (UTC)
f doesn't even have to be that "unreasonable" for the limit not to exist, e.g. let f be 1 on rationals and 0 on irrationals. Then by letting n \rightarrow 0 first through the irrationals and then through the rationals gives different values Silverfish70 (talk) 20:40, 4 January 2008 (UTC)
I just <sarcasm>love</sarcasm> it when people take my words out of context. "reasonable" referred to the definition of summation, not to f, to limits, or to integration. It goes without saying that for the Dirichlet function, the limit above and the Riemann integral do not exist, while the Lebesgue integral does. -- Meni Rosenfeld (talk) 20:48, 4 January 2008 (UTC)