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Author Topic: Mathematics Help Thread  (Read 95790 times)

Arx

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Re: Mathematics Help Thread
« Reply #2640 on: June 19, 2017, 02:55:27 am »

I'm completely stumped. I'm reading over a nonlinear dynamical systems paper, and at one point they go from



to



And I can't for the life of me figure out how that's correct. The best I can manage is



But I can't see how that extra factor in the denominator reduces the numerator to the form they have. I feel like I'm just being dense, anyone see what my mistake is?

All the constants are nonzero and positive.
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Re: Mathematics Help Thread
« Reply #2641 on: June 19, 2017, 03:44:46 am »


But I can't see how that extra factor in the denominator reduces the numerator to the form they have. I feel like I'm just being dense, anyone see what my mistake is?

All the constants are nonzero and positive.

Do we know anything about z?
If it's a constant like a-f, then I can't see how the we can keep a linear dependence in the numerator for the derivative but somehow reduce the resulting quadratic dependence in the denominator.
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Arx

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Re: Mathematics Help Thread
« Reply #2642 on: June 19, 2017, 04:08:44 am »

z is a constant, yes. e is also an arbitrary constant, not Euler's number. The only relationship between the constants is

b > d + z/f

I'm starting to wonder if I've missed something from context or something strange like that.

At this point I'd settle for proof that psi'(0) is negative iff z/(f^2) > e, which is the conclusion the authors draw, and as long as that's correct I think everything else hangs together.

I'm just thoroughly confused, because it seems highly unlikely that the authors would be wrong, but I can't see how they could be right either...
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Re: Mathematics Help Thread
« Reply #2643 on: June 19, 2017, 07:39:19 am »

z is a constant, yes. e is also an arbitrary constant, not Euler's number. The only relationship between the constants is

b > d + z/f

I'm starting to wonder if I've missed something from context or something strange like that.

At this point I'd settle for proof that psi'(0) is negative iff z/(f^2) > e, which is the conclusion the authors draw, and as long as that's correct I think everything else hangs together.

I'm just thoroughly confused, because it seems highly unlikely that the authors would be wrong, but I can't see how they could be right either...

Cool, then I think we can get it  :)
Inserting y=0 in the denominator makes it: f(b-d)-z = f( b - d - z/f), but b > d+z/f  =>  b - d - z/f > 0, i.e. it can't be zero and (especially squared) it will be positive and therefore not affect the sign of the derivative.

So we just have to look at the nominator of the fraction derivative: n'(y) * d(y) - d'(y)*n(y)
I get
n'(y) = (d+ey) + (f+y)*e
d'(y) = (b-d-ey) +(f+y)*(-e)

Inserting 0 as y in these and slowly lifting the (negative) parentheses ends me with a lot of cancelling parts, so only bef2 - bz  is left

I.e.  Phi'(0) = a*(bef2 - bz) / f2* 'something positive' = -ab*(z/f2-e) / 'something positive'.

I.e. The sign of the derivative at y=0 is indeed negative iff z/f2-e > 0   <=>  z/f2 > e

 :)  Let me know if you want to compare how I lift the parentheses.


(It would seem that there is an typo in the book, as the equation for phi'(0) would be correct if the denominator was squared.)
« Last Edit: June 19, 2017, 09:38:07 am by askovdk »
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Re: Mathematics Help Thread
« Reply #2644 on: June 19, 2017, 08:36:51 am »

That's what I got as well, yes. Which is good, because it means the rest of the results should be fine. If anyone has an explanation as to the derivatives, though, I'd love to see it - I'm still baffled.
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Re: Mathematics Help Thread
« Reply #2645 on: June 19, 2017, 08:46:25 am »

...  If anyone has an explanation as to the derivatives, though, I'd love to see it - I'm still baffled.

? I don't understand, - i.e. would you like to see how I get the long list of mostly cancelling terms?
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Re: Mathematics Help Thread
« Reply #2646 on: June 19, 2017, 09:05:11 am »

No, that's easy. What I'd like to know is exactly how the authors of the paper got



As the derivative of psi at 0.
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Re: Mathematics Help Thread
« Reply #2647 on: June 19, 2017, 09:37:42 am »

No, that's easy. What I'd like to know is exactly how the authors of the paper got



As the derivative of psi at 0.

Hmm, I'm sure I see the problem, - beside that I truly believe there is a typo in the book.

I.e. Phi(y) = a* n(y)/d(y)
Phi'(y)=a*(n'(y)*d(y) - d'(y)*n(y) ) / d2(y)
= a*( (1*(d+ey)+(f+y)*e)*((f+y)*(b-d-ey)-z) - (((b-d-ey)+(f+y)*(-e))*((f+y)*(d+ey)+z)) ) / ((f+y)*(b-d-ey)-z)2

Inserting 0 as y

= a*( (1*(d+ey)+(f+y)*e)*((f+y)*(b-d-ey)-z) - (((b-d-ey)+(f+y)*(-e))*((f+y)*(d+ey)+z)) ) / ((f+y)*(b-d-ey)-z)2
= a*( (d+f*e)*(f*(b-d)-z) - (((b-d)-f*e))*((f*d+z)) ) / (f*(b-d)-z)2

... reducing the numerator ...

= a* (bef2 - bz) / (f*(b-d)-z)2

pulling a b out of the numerator, and an f2 out of numerator and denominator

= a*b* ( f2*(e-z/f2) ) / ( f*((b-d) - z/f ) )2
= a*b* ( f2*(e-z/f2) ) /  f2*(b-d - z/f )2

and for some reason wanting a '-' in front

= (-1)*(-1) a*b* ((e-z/f2) ) /  (b-d - z/f )2
= -ab* (  -1*(e-z/f2) ) /  (b-d - z/f )2
= -ab* (  -e +z/f2) ) /  (b-d - z/f )2
= -ab* (   z/f2 -e ) /  (b-d - z/f )2

Q.E.D.   :)   (Still assuming a typo in the book.)

Or are we talking past each others?
« Last Edit: June 19, 2017, 09:44:12 am by askovdk »
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Re: Mathematics Help Thread
« Reply #2648 on: June 19, 2017, 10:01:58 am »

Definitely looking like a typo, if my own algebra is anything to go by.
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Re: Mathematics Help Thread
« Reply #2649 on: June 19, 2017, 10:30:36 am »

The assumption that there's a typo in the paper is kinda suboptimal, but it looks correct. Oddly, WolframAlpha thinks that's not right, but there are now three different people contradicting it so \_(ツ)_/.

Thank you guys.
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Re: Mathematics Help Thread
« Reply #2650 on: June 19, 2017, 11:12:37 am »

I mean, it makes sense that there is a typo.

Let u be the numerator and v be the denominator. Both u and v can be seen as first-order polynomials in terms of z: p(z, y, parameters) = (some sign value)z + (some function)(y, parameters).

Calculating d(u/v)/dy = (u'v - uv')/v2 gives us at most a first-order polynomial in the resulting numerator (as p' = (some function)' with respect to y) and exactly second-order denominator. [Pardon my programmer-ish notation]

That cannot be achieved with the author's suggested solution.
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Re: Mathematics Help Thread
« Reply #2651 on: July 03, 2017, 08:34:09 pm »

Why is this expression true? (That "Pi" represents the pi symbol. I thought the alt code for it looked too much like an "n".)
Quote
sin(2*Pi/3) = √(3)/2

I don't understand how that conversion works.

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Re: Mathematics Help Thread
« Reply #2652 on: July 03, 2017, 09:21:09 pm »

Geometric construction using a hexagon, if I recall properly. It's not a conversion, though- the sine function (and trig functions in general) are concerning the ratio of lengths of sides of the unit triangle (hypotenuse of length 1) given the internal angle.

That sin(pi/4)=√(2)/2 simply comes from splitting a square in half along the diagonal, and using the Pythagorean theorem to calculate the length of the diagonal.
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Re: Mathematics Help Thread
« Reply #2653 on: July 04, 2017, 05:28:29 am »

Geometric construction using a hexagon, if I recall properly.

Easier than that, even. Sin(2pi/3) is just sin(pi/3) in the 2nd quadrant (if you prefer, you can think of it as sin(pi-pi/3)), which is (√3)/2. (Since the sign of sin (best phrase) doesn't change in the 2nd quadrant because it's a function only of y and r, which are both still positive.)
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Re: Mathematics Help Thread
« Reply #2654 on: July 04, 2017, 07:30:39 pm »

Yes, but then it begs the question of how we know that sin(pi/3) = (√3)/2.

There's a geometric construction one way or another, here. Though once you have the usual intervals (pi/6, pi/4, pi/3) you can use them to find the exact value everything from pi/360 to 359pi/360.

Using that sin(2x)=2sin(x)cos(x), you can find all of those divided by a power of 2.

Using that sin(3x)=sin(x+2x)=sin(x)cos(2x)+sin(2x)cos(x)=sin(x)(1-2sin2(x))+2sin(x)cos2(x)=sin(x)-2sin3(x) + 2sin(x)(1-sin2(x)) = 3sin(x)-sin3(x), you can find all of those divided by a power of 3.

Naturally sin(4x) trivially follows from sin(2x). We can keep going, but then we run into issues with the polynomials that we produce because of the Abel-Ruffini theorem: any polynomial degree 5 or above has no general solution. We can still get solutions, but they stop looking particularly coherent.
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