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I What methods other than Taylor Series to solve this eq?

  1. Feb 15, 2017 #1
    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
     
  2. jcsd
  3. Feb 15, 2017 #2

    BvU

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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:
     
  4. Feb 15, 2017 #3
    Ahh, thank you, I will have to go away and plug that in :)
     
  5. Feb 15, 2017 #4

    mfb

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    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.
     
  6. Feb 15, 2017 #5
    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
     
  7. Feb 15, 2017 #6

    Charles Link

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    I
    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.
     
  8. Feb 15, 2017 #7
    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
     
  9. Feb 20, 2017 #8

    Stephen Tashi

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    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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