Re: Hyperbolic Only for (x^3 + Px + Q = 0)

On Mar 13, 9:25 am, Gerry <Gerry...@xxxxxxxxx> wrote:
On Mar 13, 2:17 am, Hyperbolic-Novice <padraig...@xxxxxxxxxx> wrote:



On Mar 12, 3:47 am, Gerry Myerson <ge...@xxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:

In article
<db80ef82-f2f7-4d68-b83e-481e835ff...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,

 Hyperbolic-Novice<padraig...@xxxxxxxxxx> wrote:
On Mar 11, 5:00 am, Gerry Myerson <ge...@xxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
In article
<13d37399-0b2e-4ec5-8523-a372d8352...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,

 Hyperbolic-Novice<padraig...@xxxxxxxxxx> wrote:
Thank you Gerry for replying to my query and offering a suggestion.

yes I got this far several times with Cosh, and had no bother using
Sinh when appropriate (p > 0).

Alas I can get no further.

Some people have recommended Tanh or Cotanh but again the test-example
(x^3 - x + 1/8 = 0) leads to dead-end.

You may also find something in
J. Macdonnell, Using sinh and cosh to solve the cubic,
International Journal of Mathematical Education in Science and
Technology, Volume 18, Issue 1 January 1987 , pages 111 - 118,
also H. E. Fettis, On various methods of solving cubic equations,
National Mathematics Magazine, Vol. 17, No. 3 (Dec., 1942), pp. 117-130

--
Gerry Myerson (ge...@xxxxxxxxxxxxxxx) (i -> u for email)

Yes Gerry those are EXCELLENT resources, and I thank you for
mentioning them, but again I've poured over those specific sources and
many many others in the past year to no avail.

My query is pitched at a constituency that may in fact NOT exist at
all ie those who can DEMONSTRABLY solve this CUBIC (x^3 - x + 1/8 = 0)
by deploying a Hyperbolic INSTEAD of Cos and do so in a manner that
can be Verified/Tested for P = -1 and Q = 1/8

I have noted that there are two kinds of cubic, viz., those with three
real roots and those with one, and I have suggested that maybe the trig
functions work for only one of these cases and the hyperbolics only for
the other. What say you?

That possibly is the simple truth of the matter.

It is difficult however to give up on the idea that the range for
which the Trig functions work ie A = [(27q^2)/(-4p^3)]^0.5 < 1 can not
equally be solvable using Tanh. We need not worry about A=+ or - 1 as
the Cosh and Sinh options can account for them.

Thank you for your fedback

Padraig- Hide quoted text -

- Show quoted text -
You may find that M < 1 / 8, so 4 M < 1 / 2,
and then what do you make of arccosh 4 M?
u = (1 / 3)* acosh( 4*M )
Thus another dead-end
Padraig

i'm afraid not so.

Here's what i found in Pari's  user manual on acosh(x) :

 3.3.7 acosh(x): principal branch of cosh-1(x), i.e. such that Im(acosh
(x)) e [0, Pi].

 If x e R and x < 1, then acosh(x) is complex.

So for

 4*M = 4*(3/4)^(3/2)/8 = 0.3247595264191644925363961

we get with Pari :

 acosh(4*M) = 1.2400388648722578540497491*I

and:

 u = (1 / 3)* acosh(4*M) = 0.413346288290752618016583*I

and:

 y = cosh(u) = 0.9157818287870809327042452

and:

 x = y/(3/4)^(1/2) = 1.05745377073837789925780703

and - x certainly is a root of :

 x^3 - x + 1/8

Now the Cubic

 x^3 + p * x + q =0

has the root :

 x =  (-q/2 + ((q/2)^2 + (p/3)^3 )^(1/2) )^(1/3)
    + (-q/2 - ((q/2)^2 + (p/3)^3 )^(1/2) )^(1/3)

Your example cubic cosh :

  x^3 - x + 1/8

 x1 = 0.9304029265558516931929180

My example cubic cos :

  x^3 - 64*x - 64

 x2 = 8.4596301659070231940624562

Here's how they are related :

set p = -1 and q = 1/8

let d = x2 = 8.45963.. be the root of x^3 - 64*x - 64

and c be a certain root of the polynomial :

 ((x^3- (x/d))^3 + (p*x)^3)/x^3

which has degree 6 and has real coefficients due to d:

  x^6 - 0.3546254317464416611646952*x^4 +
0.04191973228045005105698239*x^2 - 1.001651755906516649647673

This polynomial has the 2 real roots :

  x3 = -1.057453770738377899257807

and

  x4 =  1.057453770738377899257807

and x3 is also a root of your example cubic

which gives the relationship for p :

  p = -( x3^3 - ( x3 / x2 ) ) / x3  = - 1

So here we have a relationship between a root
x2 of the cos cubic and x3 a root of the cosh cubic.

Regards

Gerry

First my apologies Gerry for saying that
'your next line is meaningless'.

While I never heard of 'Pari's user manual' what he says on acosh is
very interesting.

Thank you sincerely for your help.

I'm going to try and get Pari's user manual' here in Ireland.

Your own development of the equation is something that I am going to
study closely these coming weeks.

Again for that - Thank you very much Gerry

Padraig
.

Relevant Pages

  • Re: Hyperbolic Only for (x^3 + Px + Q = 0)
    ... Sinh when appropriate. ... the Cosh and Sinh options can account for them. ... and - x certainly is a root of: ... This polynomial has the 2 real roots: ...
    (sci.math)
  • Re: openGL problem
    ... > calls to sinh and cosh it is fine. ... > openGL but something is badly wrong with sinh and cosh. ... Unless your libm is totally whacked, ... > The expression with exp is fine but sinh and cosh give crap. ...
    (comp.os.linux.x)
  • Re: openGL problem
    ... >> I can substitute an expression using calls to exp and get the image I ... >> openGL but something is badly wrong with sinh and cosh. ... > Unless your libm is totally whacked, ...
    (comp.os.linux.x)
  • Re: openGL problem
    ... > getting from sinh() and/or cosh() to stderr, ... > (POVRay terminology, sorry, I don't know the OpenGL equivalents, but the ... > works more like what you want, you've got a camera positioning problem. ...
    (comp.os.linux.x)
  • Re: angular velocity
    ... >> coefficient second order ODE's than expressing it in terms of ... Yes, but if, as frequently happens, your boundary conditions are in ... calculated for the {cosh, sinh} pair. ...
    (sci.math)