antispam@math.uni.wroc.pl schrieb:
clicliclic@freenet.de <nobody@nowhere.invalid> wrote:
"clicliclic@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", 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.
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.
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.
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.
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