# equation

26 views

### Richard Hachel

Apr 4, 2022, 8:56:32 AM4/4/22
to
Hello friends.

Can we give the reciprocal of this equation in a simplified way?

t=(v/g)(1-v²/c²)^3/2

v= ?

We can start by passing g to the other side.

v.(1-v²/c²)^3/2 = gt

A mathematician knows

R.H.

### nob...@nowhere.invalid

Apr 4, 2022, 1:33:14 PM4/4/22
to

Richard Hachel schrieb:
>
> Can we give the reciprocal of this equation in a
> simplified way?
>
> t = (v/g)*(1 - v^2/c^2)^(3/2)
>
> v = ?
>
> We can start by passing g to the other side.
>
> v*(1 - v^2/c^2)^(3/2) = g*t
>
> A mathematician knows
>
> R.H.

In the above, I have replaced the original UTF-8 "²" by an ASCII "^2",
placed the exponent "3/2" in parentheses, and supplied a few asterisks
"*" indicating multiplication.

The equation is suggestive of relativistic mechanics, it is probably
involving a function t(v), and the "reciprocal" of interest would be
v(t), whereas c and g would be constants.

What happens when a Computer Algebra System (CAS) is asked to solve
this equation? Unfortunately, Derive 6.10 takes longer than I cared to
wait, even with c,g,t,v declare positive real.

Martin.

### Richard Hachel

Apr 4, 2022, 2:05:26 PM4/4/22
to
Le 04/04/2022 à 19:33, "clicl...@freenet.de" a écrit :
>
> Richard Hachel schrieb:
>>
>> Can we give the reciprocal of this equation in a
>> simplified way?
>>
>> t = (v/g)*(1-v²/c²)^(1.5)
>>
>> v = ?

> The equation is suggestive of relativistic mechanics

Yes it is.

> Martin.

R.H.

### nob...@nowhere.invalid

Apr 10, 2022, 4:25:59 PM4/10/22
to

"clicl...@freenet.de" schrieb:
Looks like everybody alse has gone watching the run-up to WW3. Here's
how to attack this problem on Derive 6.10.

Converted to a polynomial through squaring and subsequent
multiplication by c^6*g^2, the equation for v(t) reads:

vv^4 - 3*cc*vv^3 + 3*cc^2*vv^2 - cc^3*vv + cc^3*gg*tt = 0

where cc = c^2, gg = g^2, tt = t^2, vv = v^2. The discriminant of the
quartic results to:

cc^9*gg^2*tt^2*(gg*tt - 27*cc)

Along with additional tests this indicates that no real solutions exist
for tt > 27*cc/gg while two real solutions exist for tt < 27*cc/gg.

Since c and g can be absorbed into scale factors for v and t, they may
be set to unity; with the restrictions cc = gg = 1 and either tt > 27
or 0 < tt < 27, Derive 6.10 produces the four solutions right away.
As usual, they involve trigonometric functions for the trisection of
angles instead of complex cube roots; all four are complex for tt >
27/256, and two of them are real for 0 < tt < 27/256.

I am a bit confused about finding tt < 27/256 from the solutions
instead of the tt < 27 indicated by the discriminant.

Martin.

### anti...@math.uni.wroc.pl

Apr 10, 2022, 10:36:14 PM4/10/22
to
clicl...@freenet.de <nob...@nowhere.invalid> wrote:
>
> "clicl...@freenet.de" schrieb:
> >
> > Richard Hachel schrieb:
> > >
> > > Can we give the reciprocal of this equation in a
> > > simplified way?
> > >
> > > t = (v/g)*(1 - v^2/c^2)^(3/2)
> > >
> > > v = ?
> > >
> > > We can start by passing g to the other side.
> > >
> > > v*(1 - v^2/c^2)^(3/2) = g*t
> > >
> > > A mathematician knows
> > >
> > > R.H.
> >
> > In the above, I have replaced the original UTF-8 "??" by an ASCII "^2",
AFAICS Galois group of this quartic is full symmetric group.
So answer to original question seem to be "No".

Concerning discriminant, the above looks wrong. I get:

factor(discriminant(univariate(pp, v2)))

9 4 4 2 2
(43) c2 g t (256 g t - 27 c2)
Type: Factored(Polynomial(Integer))

where c2 is c^2.

--
Waldek Hebisch

### nob...@nowhere.invalid

Apr 11, 2022, 12:38:19 AM4/11/22
to

anti...@math.uni.wroc.pl schrieb:
Yes, when copying the discriminant formula, I accidentally dropped the
numerical factor from the leading term 256*a^3*e^3 for a quartic
written a*x^4 + b*x^3 + c*x^2 + d*x + e = 0. For the corrected
restriction 0 < tt < 27/256, Derive 6.10 produces the real solutions:

vv = SQRT(2)*3^(3/4)*SQRT(SQRT(8*2^(1/3)*tt^(2/3)*(3*SQRT(3) -
SQRT(27 - 256*tt))^(2/3) - 2*2^(2/3)*tt^(1/3)*(2*2^(2/3)*tt^(1/3)*
(SQRT(27 - 256*tt) + 3*SQRT(3))^(1/3) + SQRT(3))*(3*SQRT(3) -
SQRT(27 - 256*tt))^(1/3) + 2*2^(2/3)*tt^(1/3)*(SQRT(27 - 256*tt) +
3*SQRT(3))^(1/3)*(2*2^(2/3)*tt^(1/3)*(SQRT(27 - 256*tt) +
3*SQRT(3))^(1/3) - SQRT(3)) + 3) - 2^(2/3)*tt^(1/3)*(3*SQRT(3) -
SQRT(27 - 256*tt))^(1/3) - 2^(2/3)*tt^(1/3)*(SQRT(27 - 256*tt) +
3*SQRT(3))^(1/3) + SQRT(3))/12 - 3^(3/4)*(SQRT(2*2^(2/3)*tt^(1/3)*
(3*SQRT(3) - SQRT(27 - 256*tt))^(1/3) + 2*2^(2/3)*tt^(1/3)*(SQRT(27 -
256*tt) + 3*SQRT(3))^(1/3) + SQRT(3)) - 3*3^(1/4))/12

vv = - SQRT(2)*3^(3/4)*SQRT(SQRT(8*2^(1/3)*tt^(2/3)*(3*SQRT(3) -
SQRT(27 - 256*tt))^(2/3) - 2*2^(2/3)*tt^(1/3)*(2*2^(2/3)*tt^(1/3)*
(SQRT(27 - 256*tt) + 3*SQRT(3))^(1/3) + SQRT(3))*(3*SQRT(3) -
SQRT(27 - 256*tt))^(1/3) + 2*2^(2/3)*tt^(1/3)*(SQRT(27 - 256*tt) +
3*SQRT(3))^(1/3)*(2*2^(2/3)*tt^(1/3)*(SQRT(27 - 256*tt) +
3*SQRT(3))^(1/3) - SQRT(3)) + 3) - 2^(2/3)*tt^(1/3)*(3*SQRT(3) -
SQRT(27 - 256*tt))^(1/3) - 2^(2/3)*tt^(1/3)*(SQRT(27 - 256*tt) +
3*SQRT(3))^(1/3) + SQRT(3))/12 - 3^(3/4)*(SQRT(2*2^(2/3)*tt^(1/3)*
(3*SQRT(3) - SQRT(27 - 256*tt))^(1/3) + 2*2^(2/3)*tt^(1/3)*(SQRT(27 -
256*tt) + 3*SQRT(3))^(1/3) + SQRT(3)) - 3*3^(1/4))/12

which now involve real cube roots instead of trigonometric functions,
where again cc = gg = 1 for simplicity.

Martin.

PS @ Waldek: I saw unanswered posts by Ralf on fricas-devel.

### anti...@math.uni.wroc.pl

Apr 11, 2022, 10:19:11 PM4/11/22
to
clicl...@freenet.de <nob...@nowhere.invalid> wrote:
>
> PS @ Waldek: I saw unanswered posts by Ralf on fricas-devel.

I a no longer subscribed to fricas-devel: due to email
change old subscrition no longer works and new
procedure at Google is unacceptable to me.

--
Waldek Hebisch