Roots of series of exponential raised to power of x?

In summary, the conversation discusses the solution for the equation a1e-k1x+a2e-k2x+...+ane-knx =0 for x, where n>2. The solution involves using y=ex and the creation of a polynomial in y, with k's as integers. For polynomials of fifth degree or higher, there is no general solution, and numerical methods are used. The conversation also touches on solving a polynomial with a variable f in the exponent, and suggests using Schwarz's inequality and creating equations for each value of n.
  • #1
Adel Makram
635
15
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?
 
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  • #2
Adel Makram said:
the solution will be x=In (a1/a2) / [ k1-k2].
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.
 
  • #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?
 
  • #4
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
mathman said:
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!
 
  • #6
Adel Makram said:
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!
[itex] e^{-kx}=(e^{-x})^{k}[/itex]
 
  • #7
Adel Makram said:
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.
 
  • #8
Adel Makram said:
How to solve: a1e-k1x+a2e-k2x+...+ane-knx =0 for x?

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"
 
  • #9
why Fenix said:
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"
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
Adel Makram said:
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?
 
  • #11
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
mathman said:
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:
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
It looks like you would need to use numerical methods.
 
  • #14
Adel Makram said:
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

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...
 

FAQ: Roots of series of exponential raised to power of x?

What is a series of exponential raised to power of x?

A series of exponential raised to power of x is a mathematical expression that involves repeated multiplication of a constant (the base) by itself a certain number of times (the exponent). It can also be written in the form of a sum, where each term is a power of x.

What are the roots of a series of exponential raised to power of x?

The roots of a series of exponential raised to power of x are the values of x that make the entire expression equal to zero. In other words, they are the solutions to the equation in which the series is set equal to zero.

How do you find the roots of a series of exponential raised to power of x?

To find the roots of a series of exponential raised to power of x, you can set the expression equal to zero and then solve for x using algebraic techniques such as factoring or the quadratic formula. Additionally, you can use a graphing calculator to visually determine the roots of the series.

What is the significance of the roots in a series of exponential raised to power of x?

The roots of a series of exponential raised to power of x are important because they represent the values of x that make the expression equal to zero. They can also help determine the behavior of the series, such as whether it is increasing or decreasing, and where it crosses the x-axis.

How can understanding the roots of a series of exponential raised to power of x be useful?

Understanding the roots of a series of exponential raised to power of x can be useful in various fields of science, such as physics and biology. It can help in solving equations and modeling real-life phenomena, as well as in analyzing data and making predictions. Additionally, it can aid in understanding the behavior of exponential functions and their applications in various contexts.

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