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Closed form solving for parameters of exponential

Pino Savana

Posted 1 year ago

Hi, I have a system of exponential equations that I'll need to symbolically simplify. Any chance that with the correct inputs I can get something digestible? My aim is to have a symbolic equation that I can parametrise when I need to. No problems on the numerical side. eq1 = 1 == bc^m + a eq2 = 0.75 == bc^(tm) + a eq3 = 0 == bc^(0)+a from eq3 b =-a ss1 = Simplify[ Assuming[t > 0 < 1 && m > 0, Reduce[Rationalize@{eq1, eq2}, {a, c}, Reals]] /. m -> 260]

POSTED BY: Pino Savana

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Pino Savana

Posted 1 year ago

I coudl work with something more like - but I guess I run into the same problem. eq1 = 1 == bc^m1 + a eq2 = 3/4 == bc^(m) + a eq3 = 0 == b*c^(0) + a Reduce[{eq1, eq2, eq3}, {a, b, cm}] // ToRadicals which if parametrise for m and m1 gets me workable coefficients form a,b, and c. So i wonder if there's a way that if with some assumption could lead me to a symobolic closed form. with m1 = 100 and m=25 Solve[{eq1, eq2, eq3}, {a, b, c}, Reals] // ToRadicals {{a -> 1 - ((-3 - 1/(3 - 2 Sqrt[2])^(1/3) - (3 - 2 Sqrt[2])^( 1/3)) (-2 - 1/(3 - 2 Sqrt[2])^(2/3) + 3/(3 - 2 Sqrt[2])^( 1/3) + 3 (3 - 2 Sqrt[2])^(1/3) - (3 - 2 Sqrt[2])^(2/3))^4)/ 6912, b -> 3/16 (-3 - 1/(3 - 2 Sqrt[2])^(1/3) - (3 - 2 Sqrt[2])^(1/3)), c -> 1/(6/(-2 - 1/(3 - 2 Sqrt[2])^(2/3) + 3/(3 - 2 Sqrt[2])^(1/3) + 3 (3 - 2 Sqrt[2])^(1/3) - (3 - 2 Sqrt[2])^(2/3)))^(1/25)}}

I coudl work with something more like - but I guess I run into the same problem.

eq1 = 1 == b*c^m1 + a
eq2 = 3/4 == b*c^(m) + a
eq3 = 0 == b*c^(0) + a
Reduce[{eq1, eq2, eq3}, {a, b, cm}] // ToRadicals

which if parametrise for m and m1 gets me workable coefficients form a,b, and c. So i wonder if there's a way that if with some assumption could lead me to a symobolic closed form.

with m1 = 100 and m=25

Solve[{eq1, eq2, eq3}, {a, b, c}, Reals] // ToRadicals
{{a -> 1 - ((-3 - 1/(3 - 2 Sqrt[2])^(1/3) - (3 - 2 Sqrt[2])^(
       1/3)) (-2 - 1/(3 - 2 Sqrt[2])^(2/3) + 3/(3 - 2 Sqrt[2])^(
       1/3) + 3 (3 - 2 Sqrt[2])^(1/3) - (3 - 2 Sqrt[2])^(2/3))^4)/
    6912, b -> 
   3/16 (-3 - 1/(3 - 2 Sqrt[2])^(1/3) - (3 - 2 Sqrt[2])^(1/3)), 
  c -> 1/(6/(-2 - 1/(3 - 2 Sqrt[2])^(2/3) + 3/(3 - 2 Sqrt[2])^(1/3) + 
     3 (3 - 2 Sqrt[2])^(1/3) - (3 - 2 Sqrt[2])^(2/3)))^(1/25)}}

POSTED BY: Pino Savana

Jim Baldwin

Posted 1 year ago

Because `c` and `m` only occur as `c^m`, you can only deal with `c^m` and can't get separate estimates for `c` and `m`. Also if you change `0.75` to `3/4` and set `t` to specific values, a solution for `c^m`, `a`, and `b` occurs. t = 4; eq1 = 1 == bcm + a eq2 = 3/4 == bcm^(t) + a eq3 = 0 == b*c^(0) + a Reduce[{eq1, eq2, eq3}, {a, b, cm}] // ToRadicals (a == 1/13 (8 + (2 (81 + 13 Sqrt[57]))^(1/3)/3^(2/3) - (8 2^(2/3))/(3 (81 + 13 Sqrt[57]))^(1/3)) \|\| a == 8/13 - ((1 + I Sqrt[3]) (81 + 13 Sqrt[57])^(1/3))/(13 6^(2/3)) + (4 2^(2/3) (1 - I Sqrt[3]))/( 13 (3 (81 + 13 Sqrt[57]))^(1/3)) \|\| a == 8/13 - ((1 - I Sqrt[3]) (81 + 13 Sqrt[57])^(1/3))/(13 6^(2/3)) + (4 2^(2/3) (1 + I Sqrt[3]))/( 13 (3 (81 + 13 Sqrt[57]))^(1/3))) && b == -a && cm == 1/4 (-12 + 24 a - 13 a^2)

Because c and m only occur as c^m, you can only deal with c^m and can't get separate estimates for c and m. Also if you change 0.75 to 3/4 and set t to specific values, a solution for c^m, a, and b occurs.

t = 4;
eq1 = 1 == b*cm + a
eq2 = 3/4 == b*cm^(t) + a
eq3 = 0 == b*c^(0) + a
Reduce[{eq1, eq2, eq3}, {a, b, cm}] // ToRadicals

(a == 1/13 (8 + (2 (81 + 13 Sqrt[57]))^(1/3)/3^(2/3) - (8 2^(2/3))/(3 (81 + 13 Sqrt[57]))^(1/3)) || 
   a == 8/13 - ((1 + I Sqrt[3]) (81 + 13 Sqrt[57])^(1/3))/(13 6^(2/3)) + (4 2^(2/3) (1 - I Sqrt[3]))/(
     13 (3 (81 + 13 Sqrt[57]))^(1/3)) || 
   a == 8/13 - ((1 - I Sqrt[3]) (81 + 13 Sqrt[57])^(1/3))/(13 6^(2/3)) + (4 2^(2/3) (1 + I Sqrt[3]))/(
     13 (3 (81 + 13 Sqrt[57]))^(1/3))) && b == -a && 
 cm == 1/4 (-12 + 24 a - 13 a^2)

POSTED BY: Jim Baldwin

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