What methods other than Taylor Series to solve this eq?

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Discussion Overview

The discussion centers around methods for solving equations of the form A1ek1t + A2ek2t = C, with a focus on alternatives to Taylor Series expansion. Participants explore various mathematical approaches and the limitations of certain methods.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Archie introduces the problem and expresses interest in methods beyond Taylor Series expansion.
  • One participant suggests substituting e^x with y, transforming the equation into a polynomial form, and then solving for y before reverting to x.
  • Another participant notes that analytic solutions in closed form are generally not available for all cases, specifically mentioning integer values of B between -3 and 4.
  • A participant explains that polynomial equations have analytic solutions only for certain orders (n=1 to n=4), while for n=5 and higher, closed form solutions do not exist.
  • Discussion includes a reference to the Abel–Ruffini theorem, which states the impossibility of solving general quintic equations by radicals, while acknowledging that specific quintic equations may still be solvable.
  • There is a mention of a trigonometric solution for quintic equations, indicating that some special cases exist.

Areas of Agreement / Disagreement

Participants express varying views on the solvability of polynomial equations, particularly regarding the existence of analytic solutions for different orders. There is no consensus on the applicability of the discussed methods to all forms of the original equation.

Contextual Notes

Limitations include the dependence on the specific form of the equations and the conditions under which certain methods may or may not yield solutions. The discussion does not resolve the applicability of the proposed methods to all cases.

Archie Medes
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Hi

If I have a problem of the form:

A1ek1t + A2ek2t = C​

where A1,A2,k1,k2,C are real and known

Or simplified:

ex + AeBx = C​

I can turn it into an nth degree polynomial by Taylor Series expansion, but I'd like to know what other methods I can study

Thanks,

Archie
 
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Hello Archie, :welcome:

You can replace ##e^x## by ##y## and then you have an equation in ##y## of the form $$y + Ay^B = C$$

Once you solved that you have ##x = \log y\ ## :smile:
 
Ahh, thank you, I will have to go away and plug that in :)
 
Note that - in general - you won't find an analytic solution in closed form. It works for integer B between -3 and 4, for all others there are just a few special cases where it works.
 
Are you talking about the substitution provided by BvU?

What is the name of the theorem/proof (sorry if poor terminology) that explains it, so I can read up?

Thanks
 
I
Archie Medes said:
Are you talking about the substitution provided by BvU?

What is the name of the theorem/proof (sorry if poor terminology) that explains it, so I can read up?

Thanks
I believe I can answer your question. Basically, mfb is referring to the form that BvU provided. It is well-known from algebra that polynomial equations (and that is the form that it has after BvU's substitution) only have analytic solutions for orders n= 4, 3,2, and 1. Order n=2 is the quadratic equation. For that one, ## x=(-b \pm \sqrt{b^2-4ac})/(2a) ##. For the cubic and quartic polynomial equations (n=3 and n=4), analytic solutions also exist, but they are quite detailed (more complicated than the quadratic form). For n=5 and larger, closed form analytic solutions do not exist.
 
Ahh thanks

A quick search for solving 5th degree polynomials brought me to the Quintic function page on wiki, where it says the impossibility was proven by the Abel–Ruffini theorem at the beginning of the 19th century
 
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Archie Medes said:
where it says the impossibility was proven by the Abel–Ruffini theorem at the beginning of the 19th century

You should keep in mind that "impossibility" in that theorem refers to solving "the general" quintic and it refers to "solution by radicals". Particular quintic equations may be solvable by radicals. There is a so-called "trigonometric" solution to quintic equations.
 

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