Roots of series of exponential raised to power of x?

[Adel Makram]

How to solve: a₁e^-k₁x+a₂e^-k₂x+...+a_ne^-k_nx =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?

 

[Svein]

the solution will be x=In (a1/a2) / [ k1-k2].

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 iπ somewhere.

 

[Adel Makram]

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?

 

[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.

 

[Adel Makram]

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

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!

 

[Svein]

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!

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

 

[mathman]

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!

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.

 

[why Fenix]

How to solve: a₁e^-k₁x+a₂e^-k₂x+...+a_ne^-k_nx =0 for x?

Let y=e^x then
e^-k_ix=y^-k_i
you get
a₁y^-k₁+a₂y^-k₂+...+a_ny^-k_n =0
or for z=1/y=1/e^-x
a₁z^k₁+a₂z^k₂+...+a_nz^k_n =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"

 

[Adel Makram]

Let y=e^x then
e^-k_ix=y^-k_i
you get
a₁y^-k₁+a₂y^-k₂+...+a_ny^-k_n =0
or for z=1/y=1/e^-x
a₁z^k₁+a₂z^k₂+...+a_nz^k_n =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"

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?

 

[why Fenix]

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?

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?

 

[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.

 

[Adel Makram]

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.

I have a polynomial of k where k could be integers or rationals, so how to solve a polynomial let`s say:
a₁x^11/4+a₂x^9/5+a₃x^7/3=0

in fact my problem is as follow:
a₁x^(1+2f)+a₂x^(1+f)+a₃x^(2+f)+a₄x^f+a₅x=0

Where practically 1/2 <f<2

 

[mathman]

It looks like you would need to use numerical methods.

 

[why Fenix]

in fact my problem is as follow:
a₁x^(1+2f)+a₂x^(1+f)+a₃x^(2+f)+a₄x^f+a₅x=0
Where practically 1/2 <f<2

so at first post you wanted to solve something like this?
a₁e^(1+2f)x+a₂e^(1+f)x+a₃e^(2+f)x+a₄e^fx+a₅e^x=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...