# Roots of series of exponential raised to power of x?

1. Feb 19, 2015

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. Feb 19, 2015

### Svein

No. Rewrite the equation: $a_{1}e^{-k_{1}x}=-a_{2}e^{-k_{2}x}$ and you see that for your formula to work, a1 and a2 must have opposite signs. Otherwise you need to introduce somewhere.

3. Feb 19, 2015

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?

4. Feb 19, 2015

### mathman

Let y = e-x. If the k's are integers you have a polynomial in y, which may or may not be readily solvable.

5. Feb 20, 2015

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!

6. Feb 20, 2015

### Svein

$e^{-kx}=(e^{-x})^{k}$

7. Feb 20, 2015

### mathman

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.

8. Feb 23, 2015

### why Fenix

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.

go here
http://en.wikipedia.org/wiki/Galois_theory
to section
"A non-solvable quintic example"

9. Feb 24, 2015

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?

10. Feb 24, 2015

### why Fenix

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?

11. Feb 24, 2015

### mathman

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.

12. Feb 25, 2015

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

13. Feb 25, 2015

### mathman

It looks like you would need to use numerical methods.

14. Feb 26, 2015

### why Fenix

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...