Power Series: Question about constants

In summary, the problem is to find y(x) where x is given as a function of a0, a1, and x0. The equation has two arbitrary constants, a0 and a1, and the author is looking for another way to solve it.
  • #1
Saladsamurai
3,020
7

Homework Statement



So I have attached the problem in image:

Pr and A are just numbers (constants) that are given. I solved the equation by power series solution. However, I am just confused because it is a second order DE but I only have one arbitrary constant of integration ao. I am not sure where I went wrong? Should I have defined BOTH a0 and a1 to be arbitrary? I never got the hang of the little details with series solutions; perhaps I can get some input on this.

Thanks :smile:

PowerSeries.jpg


Also, I would be interested to hear any other ideas about ways to solve this equation, but my power series question is my main concern.
 
Last edited:
Physics news on Phys.org
  • #2
I didn't look very closely, but it looks like you're on the right track. Since your equation is second order, you will have two arbitrary constants, which in your case are a0 and a3. Problems like this have two initial conditions y(x0) = y0, and y'(x0) = y1. Since your series is a Maclaurin series, x0 = 0.
 
  • #3
Look at the step where you change the index so it starts at n = -2. The other two sums start at n = 0 so they can't be grouped in the sum at that step. Furthermore, n = -1 and n = -2 don't give anything in the first sum and n= 0 doesn't give anything in the second. So write that step:

[tex]\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n+A\sum_{n=1}^{\infty}na_nx^n
-A\sum_{n=0}^\infty a_nx^n[/tex]

Now n = 0 will give a2 in terms of a0 and n = 1 will give a3 in terms of a1, with a0 and a1 arbitrary.
 
  • #4
Sweet! I went through it again and got a0 and a1 as my arbitrary constants. Out of curiosity, anyone see another way to solve this? I was thinking that if we divide through by y' we would have:

y"/y' + Ax - Ay/y' = 0

Now the first term can be written ln[y']'. If I could write the last term in a similar fashion (i.e. as a derivative), I could integrate directly.
 

Related to Power Series: Question about constants

1. What is a power series?

A power series is a mathematical series in which each term is a constant multiplied by a variable raised to a non-negative integer power. It can be represented as Σcnxn, where cn is the constant and x is the variable.

2. What is the difference between a power series and a polynomial?

A power series is an infinite sum of terms, while a polynomial is a finite sum. Additionally, the exponent of the variable in a power series can be any non-negative integer, while in a polynomial, it can only be a positive integer.

3. How do you find the radius of convergence for a power series?

The radius of convergence for a power series is the distance from the center of the series to the nearest point at which the series converges. It can be found by using the ratio test, which involves taking the limit of the absolute value of the ratio of consecutive terms.

4. Can constants be included in a power series?

Yes, constants can be included in a power series. In fact, the term c0 is often referred to as the constant term in a power series representation.

5. How are power series used in science?

Power series are used in a variety of scientific fields, including physics, engineering, and economics. They are often used to approximate functions and make calculations easier, as well as to model real-world phenomena such as the growth of populations or the behavior of electrical circuits.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
726
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
339
  • Calculus and Beyond Homework Help
Replies
1
Views
931
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
842
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
2K
Back
Top