26 views

Skip to first unread message

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.

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.

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)
> Can we give the reciprocal of this equation in a

> simplified way?

>

>

> v = ?

>

> We can start by passing g to the other side.

>

> v*(1 - v^2/c^2)^(3/2) = g*t
> We can start by passing g to the other side.

>

>

> 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.

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)
>

> Richard Hachel schrieb:

>>

>> Can we give the reciprocal of this equation in a

>> simplified way?

>>

>>

>> v = ?

> The equation is suggestive of relativistic mechanics

> Martin.

R.H.

Apr 10, 2022, 4:25:59 PM4/10/22

to

"clicl...@freenet.de" schrieb:

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.

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",
>

> "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.

> >

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

Apr 11, 2022, 12:38:19 AM4/11/22

to

anti...@math.uni.wroc.pl schrieb:

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.

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
>

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

change old subscrition no longer works and new

procedure at Google is unacceptable to me.

--

Waldek Hebisch

Reply all

Reply to author

Forward

0 new messages

Search

Clear search

Close search

Google apps

Main menu