What methods other than Taylor Series to solve this eq?

In summary: In the context of your original question, it is not the case that "ex + AeBx = C" is a quintic equation. It is a quadratic. You can solve for x by doing a substitution. mfb provided a substitution that would change it to y+ AyB = C. You can solve for y and then back substitute to find x.In summary, the conversation discussed solving a problem of the form ex + AeBx = C using a substitution and the limitations of finding analytic solutions to polynomial equations. The substitution provided by BvU was explained and it was mentioned that closed form analytic solutions do not exist for quintic equations and higher orders. The impossibility of solving general quint
  • #1
Archie Medes
33
5
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
 
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  • #2
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:
 
  • #3
Ahh, thank you, I will have to go away and plug that in :)
 
  • #4
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.
 
  • #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
 
  • #6
I
Archie Medes said:
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
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.
 
  • #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
 
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Likes Ravikant Rajan and Charles Link
  • #8
Archie Medes said:
where it says the impossibility was proven by the Abel–Ruffini theorem at the beginning of the 19th century

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.
 

1. What is a Taylor Series and how does it work?

A Taylor Series is a mathematical method used to approximate a function by representing it as an infinite sum of simpler functions. It works by evaluating the function and its derivatives at a single point, known as the center of the series.

2. Why would I use a method other than Taylor Series to solve an equation?

While Taylor Series can be an accurate method for approximating a function, it is not always the most efficient or practical method. Other methods, such as numerical integration or root-finding algorithms, may be better suited for solving certain equations.

3. Can you give an example of a problem where a different method is more effective than Taylor Series?

Sure, consider a function with a singularity (a point where the function is undefined) within the interval of interest. In this case, using a root-finding algorithm to locate the singularity and then using a different method to approximate the function on either side of the singularity may be more effective than using Taylor Series.

4. Are there any disadvantages to using Taylor Series?

Yes, one disadvantage is that Taylor Series can only approximate smooth functions, meaning functions with continuous derivatives. If a function is not smooth or has discontinuities, Taylor Series may not provide an accurate approximation.

5. What factors should I consider when choosing a method to solve an equation?

The choice of method depends on the specific problem and the desired level of accuracy. Some factors to consider include the type of function, the interval of interest, and any known singularities or discontinuities. It may also be helpful to compare the efficiency and accuracy of different methods for solving the same problem.

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