Analytical solution of an infinite series

In summary, this series is covergent by Ratio test. I already have the numerical solution of the above. However, I am interested in an analytical solution (approximation) of the value of infinite series (in terms of β,R, y, A).
  • #1
Matheco
4
0
How to find the value of an infinite series. for e.g.[itex]Ʃ_{n=1}^{\infty} (β^{n-1}y^{R^{n}}e^{A(1-R^{2n})}) [/itex]

where β<1, R<1, y>1, and A>0?

Note that this series is covergent by Ratio test. I already have the numerical solution of the above. However, I am interested in analytical solution (approximation) of the value of infinite series (in terms of β,R, y, A).
 
Physics news on Phys.org
  • #2
Hey Matheco and welcome to the forums.

One suggestion I have is to see if you can find a taylor series or DE series (like the Hypergeometric or Bessel functions) and see if you can get a form that is similar.

Google should give you some results for tables of series in mathematics somewhere.

Other than that, you might try looking at the structure of the differential equations and integrals and see if you can construct a DE system that matches your equation.

Depending on how badly you want to solve it, it could be an interesting research problem.
 
  • #3
Thanks chiro for your reply. At the moment I am clueless about how to deal with this series. Could you recommend a reference where series are approximated by methods suggested by you?
 
  • #4
Matheco said:
At the moment I am clueless about how to deal with this series.

Just one. That's all I'd be interested in at this point. That is, can I find just one series that I can compute it's analytic expression?

The term [itex]y^{R^n}[/itex] is indicative of a lacunary function:

http://en.wikipedia.org/wiki/Lacunary_function

So we have a start. How much can we simplify your series, keep the lacunary term, and find an analytic expression for the resulting series?

How about we let [itex]\beta=1/2, R=1/2, A=1[/itex]

Then we get:

[tex]\sum_{n=1}^{\infty} \frac{1}{2^{n-1}} y^{\frac{1}{2^n}}e^{(1-\frac{1}{4^n})}[/tex]

Can we find an analytic expression for that one? If not, then can we come up with another set of non-trivial values of the parameters and find the analytic expression for the resulting series? Suppose that was your goal: find just one non-trivial series that you can compute any kind of analytic expression for including an analytic expression containing special functions like Bessel, Erf, hypergeometric, Lambert W, or other functions.

Could you do that?
 
  • #5


I understand the importance of finding analytical solutions for complex mathematical problems. In this case, we are looking at an infinite series with certain constraints on the variables β, R, y, and A. The series is also convergent, which means that it has a finite sum.

To find the analytical solution for this infinite series, we can use the techniques of summation methods such as the Euler-Maclaurin formula or the Abel-Plana formula. These methods involve manipulating the series and approximating it with a simpler function that can be integrated or differentiated. This allows us to find a closed-form expression for the sum of the series.

In this specific case, we can use the Ratio Test to show that the series is convergent. Then, we can apply one of the summation methods mentioned above to find an analytical solution for the sum of the series. This solution will be in terms of the variables β, R, y, and A, as requested.

It is important to note that the analytical solution will be an approximation, as it is not always possible to find an exact closed-form expression for an infinite series. However, it will provide a valuable insight and understanding of the behavior of the series and can be used for further analysis and applications.

In conclusion, finding an analytical solution for an infinite series requires the use of summation methods and manipulation of the series. With the given constraints and the fact that the series is convergent, it is possible to find an analytical approximation for the sum of the series in terms of the variables β, R, y, and A.
 

Related to Analytical solution of an infinite series

1. What is an analytical solution of an infinite series?

An analytical solution of an infinite series is a method of finding the exact sum or value of an infinite series, rather than using an approximation or estimation. It involves using mathematical techniques such as algebra, calculus, and geometry to manipulate the series and determine its exact value.

2. How is an analytical solution of an infinite series different from a numerical solution?

An analytical solution involves finding the exact value of an infinite series, while a numerical solution involves using numerical methods to approximate the value. Analytical solutions are generally considered more accurate, but may be more complex to calculate.

3. What types of series can be solved analytically?

Many common types of series can be solved analytically, such as geometric series, arithmetic series, and power series. However, some more complex series may require numerical methods to find a solution.

4. What are some real-world applications of analytical solutions of infinite series?

Analytical solutions of infinite series have many practical applications, such as in finance to calculate compound interest, in physics to model natural phenomena, and in engineering to design and analyze systems. They are also used in computer science and data analysis to process and analyze large amounts of data.

5. Are there any limitations to using analytical solutions for infinite series?

While analytical solutions are generally considered more accurate, they may not always be feasible or practical to use. Some series may have complex or unknown patterns, making it difficult to find an analytical solution. In these cases, numerical methods may be a better option. Additionally, analytical solutions may be computationally intensive and time-consuming, making them less practical for real-time applications.

Similar threads

Replies
9
Views
2K
Replies
1
Views
870
Replies
5
Views
979
Replies
3
Views
1K
Replies
7
Views
2K
Replies
2
Views
1K
Replies
4
Views
843
Replies
7
Views
2K
Replies
2
Views
1K
Back
Top