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finding solutions with exponential series expansion

asked 2018-11-22 16:53:33 -0600

anonymous user

Anonymous

Hi, I couldnt recall how to solve the following for t

0.111t = 1- e^(-0.3t) . (The ans is t=8).

Thank you in advance !! Mich

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answered 2018-11-23 12:14:26 -0600

Emmanuel Charpentier gravatar image

updated 2018-11-23 16:38:38 -0600

The steps (and graphs !) of this solution are available in Sagecell

Let's see. We suppose for now that this equation is to solve in reals :

var("t", domain="real")
Ex=0.111*t-(1- e^(-0.3*t))

We note an "obvious" root :

Ex.subs(t=0)
0

Is there an explicit solution ?

Ex.log_expand().log().solve(t)
[t == 1000/111*(2*e^(3/10*t) - 1)*e^(-3/10*t)]

No such luck : this is an implicit solution, no more informative than the original equation. Let's try a numerical solution. What is the shape of the representative curve of this expression ? A curve with a minimum around about 3, null at 0 and about 8, decreasing below 3 and increasing above 3 (see Sagecell). What can we prove ?

The first derivative is

Ex.diff(t)
-0.300000000000000*e^(-0.300000000000000*t) + 0.111000000000000

An increasing curve, null about 3. Can we prove that ? The second derivative is

Ex.diff(t,2)
0.0900000000000000*e^(-0.300000000000000*t)

Yay ! This is strictly positive for all reals. Therefore, the zero we "see" on the first derivative at about 3 is the only one, and is

Ex.diff(t).solve(t)
[t == 10*log(1/37*100^(1/3)*37^(2/3))]
Ex.diff(t).solve(t)[0].rhs().n()
3.31417424447956

Now we have proven that Ex is decreasing before this zero and increasing after that. At the minimum, it is :

Ex.subs(t==Ex.diff(t).solve(t)[0].rhs()).n()
-0.262126658862769

Since it is negative at this point and not upward bound (exercise for the reader : prove it !), it has two roots, one (0) being already known. The graphs suggest that Ex is positive for t=10. Let's try to find a numerical solution :

Ex.find_root(0.1,10,t)
8.25101463236195

which is the sought answer.

Left to the reader : what are the complex roots (if any...) ? Fair warning : this one is much harder than the real case (bound to problems open since the XIXth century, and still not solved...).

EDIT : It turns out that Sympy can give us the non-trivial real root in symbolic form :

sage: import sympy
sage: sympy.solve(Ex, t)
[10*LambertW(-100*exp(-100/37)/37)/3 + 1000/111,
 10*LambertW(-100*exp(-100/37)/37, -1)/3 + 1000/111]

The numerical values are as expected :

sage: [s.n() for s in sympy.solve(Ex, t)]
[8.25101463236202, 0.e-123]

This function is known to Sage, but the conversion is incorrect (loses the second argument)

sage: [s._sage_() for s in sympy.solve(Ex, t)]
[10/3*lambert_w(-100/37*e^(-100/37)) + 1000/111,
 10/3*lambert_w(-100/37*e^(-100/37)) + 1000/111]

This is now Trac#26752

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Thank you for your help!! ( I m still surprised that the eqn. looks simple, but yet it can be tricky to solve. But appreciate your help !)

Mich gravatar imageMich ( 2018-11-24 17:46:10 -0600 )edit
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answered 2018-11-23 17:44:26 -0600

slelievre gravatar image

The value 8 is not a solution, but there is a solution whose approximate solution is 8.2510146...

Here are some ways to work on solving this equation.

Define t as a symbolic variable in the symbolic ring, and call eq the equation.

sage: t = SR.var('t')
sage: eq = 0.111*t == 1-exp(-0.3*t)

Try to solve with solve: unfortunately, Sage returns an equation which is equivalent to the equation we started with.

sage: solve(eq, t)
[t == 1000/111*(e^(3/10*t) - 1)*e^(-3/10*t)]

Use find_root to find an approximate solution between 5 and 10.

sage: eq.find_root(5, 10)
8.251014632362164

or

sage: find_root(eq, 5, 10)
8.251014632362164

The computation above is done using floating-point computations and it is not clear which digits are exact.

For a computation using arbitrary precision, use the mpmath library: define a function equal to the difference of the left-hand side and the right-hand side of eq and use mpmath.findroot to look for a root near 8:

sage: import mpmath
sage: mpmath.findroot(lambda t: 1 - exp(-0.3*t) - 0.111*t, 8)
mpf('8.2510146323620207')

See the answer by @Emmanuel Charpentier for how to use SymPy to get an exact solution in symbolic form. It is likely Giac or FriCAS could do it too.

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Neither Fricas nor Giac can solve this symbolically :

sage: giac("solve(111*t/1000-(1-exp(-3*t/10)),t)")
list[-6.76523232004e-14,8.25101463236]
sage: fricas("solve(111*t/1000-(1-exp(-3*t/10)),t)")
[]

Mathematica gives an answer subsuming Sympy's and asserting the existence of an infinity of (complex) solutions :

sage: mathematica("Reduce[111*t/1000==1-Exp[-3*t/10],t]")
(Element[C[1], Integers] && 
  t == 1000/111 + (10*ProductLog[C[1], -100/(37*E^(100/37))])/3) || t == 0

(Reformatted from Mathematica's help) :

ProductLog][z]
gives the principal solution for w in z == w e^w. 
ProductLog[k,z]
gives the k^th solution.

From Sage's lambert_w? :

This function satisfies the equation

  z = W_n(z)  e^{W_n(z)}

Same difference...

Emmanuel Charpentier gravatar imageEmmanuel Charpentier ( 2018-11-24 13:26:11 -0600 )edit

Thank you for your help!! ( I m still surprised that the eqn. looks simple, but yet it can be tricky to solve. But appreciate your help !)

Mich gravatar imageMich ( 2018-11-24 17:45:51 -0600 )edit

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Asked: 2018-11-22 16:53:33 -0600

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Last updated: Nov 23