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Roots of series of exponential raised to power of x?

  1. Feb 19, 2015 #1
    How to solve: a1e-k1x+a2e-k2x+...+ane-knx =0 for x?

    For example in simple case of n=1,2.
    a1e-k1x+a2e-k2x=0
    the solution will be x=In (a1/a2) / [ k1-k2]. But for terms >2 what will be the solution?
     
    Last edited: Feb 19, 2015
  2. jcsd
  3. Feb 19, 2015 #2

    Svein

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    No. Rewrite the equation: [itex]a_{1}e^{-k_{1}x}=-a_{2}e^{-k_{2}x} [/itex] and you see that for your formula to work, a1 and a2 must have opposite signs. Otherwise you need to introduce somewhere.
     
  4. Feb 19, 2015 #3
    yes agree, it was a typing mistakes, sorry. because In(negative number) does not exist.

    So, again what will be the case if n>2?
     
  5. Feb 19, 2015 #4

    mathman

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    Let y = e-x. If the k's are integers you have a polynomial in y, which may or may not be readily solvable.
     
  6. Feb 20, 2015 #5
    So how can we solve that too? all e-k are raised to the power of x, likewise all y(s) in your definition are raised to the power of kx, so it is not a power series!
     
  7. Feb 20, 2015 #6

    Svein

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    [itex] e^{-kx}=(e^{-x})^{k}[/itex]
     
  8. Feb 20, 2015 #7

    mathman

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    You seem thoroughly confused! I said you get a polynomial, not a power series. The point of the original question is that the k's are different (integers?). Raising all e-k to the x power doesn't get you anywhere.
     
  9. Feb 23, 2015 #8
    Let y=ex then
    e-kix=y-ki
    you get
    a1y-k1+a2y-k2+...+any-kn =0
    or for z=1/y=1/e-x
    a1zk1+a2zk2+...+anzkn =0
    you got polynomial.


    now ask yourself about x5-x-1=0
    go here
    http://en.wikipedia.org/wiki/Galois_theory
    to section
    "A non-solvable quintic example"
     
  10. Feb 24, 2015 #9
    So how we can solve polynomial of 5th degree? and how can we calculate the roots of characteristics polynomial in problem of diagonalization of matrix of rank n>5 for example?
     
  11. Feb 24, 2015 #10
    Because people making examples, problems, tasks, are not monsters, and usually polynomials have integer roots like {0,1,-1,2,-2,3,-3} or √2 or it combination with irrational i, that you can guess, or get close to guessing by looking at derivatives and how function is running.

    Polynomial of 5th degree x5-x-1=0 is not looking scary. So why don't you solve it?
     
  12. Feb 24, 2015 #11

    mathman

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    The basic theorem is that there is no general solution for polynomials of fifth degree or higher. In practice, numerical methods are used when a specific equation is being solved.
     
  13. Feb 25, 2015 #12
    I have a polynomial of k where k could be integers or rationals, so how to solve a polynomial let`s say:
    a1x11/4+a2x9/5+a3x7/3=0

    in fact my problem is as follow:
    a1x(1+2f)+a2x(1+f)+a3x(2+f)+a4xf+a5x=0

    Where practically 1/2 <f<2
     
  14. Feb 25, 2015 #13

    mathman

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    It looks like you would need to use numerical methods.
     
  15. Feb 26, 2015 #14
    so at first post you wanted to solve something like this?
    a1e(1+2f)x+a2e(1+f)x+a3e(2+f)x+a4efx+a5ex=0
    by
    ##e^z=\sum_{n=0}^{\infty} \frac{z^n}{n!}##
    we get
    ##\sum_{n=0}^{\infty} \frac{a_1((1+2f)x)^n+a_2((1+f)x)^n+a_3((2+f)x)^n+a_4(fx)^n+a_5x^n}{n!}=0##
    ##\sum_{n=0}^{\infty} \frac{a_1(1+2f)^nx^n+a_2(1+f)^nx^n+a_3(2+f)^nx^n+a_4f^nx^n+a_5x^n}{n!}=0##
    ##A=\sum_{n=0}^{\infty} \frac{[a_1(1+2f)^n+a_2(1+f)^n+a_3(2+f)^n+a_4f^n+a_5]x^n}{n!}=0##
    for fixed f ##\in (0.5,2)##
    from Schwarz for ##a_n, b_n \in ℝ##
    ##|\sum^{\infty}_{n=0} a_n b_n|^2 \leq \sum^{\infty}_{n=0} |a_n|^2\sum^{\infty}_{n=0} |b_n|^2 ##
    or some other way i think you have to show that ##a_1(1+2f)^n+a_2(1+f)^n+a_3(2+f)^n+a_4f^n+a_5=0## for every n if A=0.
    I am not sure is it true.
    Then you will get equations for every n...
    ##a_1(1+2f)+a_2(1+f)+a_3(2+f)+a_4f+a_5=0##
    ##a_1(1+2f)^2+a_2(1+f)^2+a_3(2+f)^2+a_4f^2+a_5=0##
    .
    ..
    ...
    hope it helps...
     
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