Series and Recurrences

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This is a tutorial made solely for the purpose of education and it was designed for students taking Applied Math 0340. It is primarily for students who have some experience using Mathematica. If you have never used Mathematica before and would like to learn more of the basics for this computer algebra system, it is strongly recommended looking at the APMA 0330 tutorial. As a friendly reminder, don'r forget to clear variables in use and/or the kernel.

Finally, the commands in this tutorial are all written in bold black font, while Mathematica output is in regular fonts. This means that you can copy and paste all comamnds into Mathematica, change the parameters and run them. You, as the user, are free to use the scripts to your needs for learning how to use the Mathematica program, and have the right to distribute this tutorial and refer to this tutorial as long as this tutorial is accredited appropriately.

 

Part V. Series and Recurrences

Recurrences
Sturm-Liouville problem
Heat transfer equations
Wave equations
Laplace equation
Applications

 

 

 

5.1. Recurrences

I. Case Sensitivity

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To solve the linear constant coefficient recurrence (also called the difference equation) of the second order

y_{n+2} =6y_{n+1} -y_n               subject to y_0 = 0,   y_1 = 1

you may try some options:

Here the first pair represents the coefficients of the difference equation of order 2; next pair tells Mathematica the first two initial values of the recurrence. The final number (10) is the number of terms the user wants to see as output. Another option:

If you know that its solution is represented through the Chebyshev polynomial of the second kind at the point x=3, then type:

If the generating function Sum[y[n]*z^n , {n,0,infinity}] is known, then (to see the first 10 coefficients)

Consider another recurrence that generates the Pell numbers:

y_{n+2} = 2 y_{n+1} + y_n ,         y_0 =0, y_1 =1,     n=0,1,2,.... .

Then these numbers can be generated as follows:

 

Recurrences, although a very tedious computation method by hand, is very simple to do in Mathematica. The best way to learn how to do recurrences in Mathematica are by examples, and a perfect example for this topic is the Fibonacci sequence.

 

5.2. Generating Functions

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Recurrences, although a very tedious computation method by hand, is very simple to do in Mathematica. The best way to learn how to do recurrences in Mathematica are by examples, and a perfect example for this topic is the Fibonacci sequence.

The process of shifting (or slipping) indices can be obtained by the following Mathematica code:

As an example, consider shifting indices in the sum:

 

 

 

5.3. Series Solutions for the first Order Equations

I. How to

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Consider the initial value problem for the Abel equation

y' = y^3 +x,     y(0) =0

We find its power series expansion using the following Mathematica commands:

This problem can be solved in slightly different way. First, we define the differential operator (non-linear because the given equation is non-linear)

Then we set n to be 13, the number of terms in the power series:

The next cell says to create a sum of n terms and effectively turn it into a series saying that the terms beyond n are indefinite. O[x]^(n+1) indicates that we know nothing about terms of order n+1 and beyond. So we define the (finite) series

Substitute this expression into the differential operator

Equate the coefficients to 0. Recall that the operator && means And.

Find the coefficients a[i] in the poer series of y[x] using Reverse[Table[a[i],{i,0,n}]]. The latter makes the form of the
coefficients agree with what you would find by hand: the next term is expressed through the previous terms.

Solve::svars: Equations may not give solutions for all "solve" variables. >>

To get the solution series, set a[0] equals to zero:

 

 

1.5.5. Series Solutions

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Example: solve the IVP:
(4-x^2 )y'[x]+y[x]==0, y[0]==1

When we apply the initial condition, we set a0=1 in the general solution.

Now we solve the initial value problem exactly:

Note that the exact solution is |2-x|^(1/4)/|2+x|^(1/4) with absolute values instead of parenthethis.

formula=Simplify[(2-x)^(1/4)/(2+x)^(1/4)]

var=Table[D[y[x],i]/.x->0,{i,1,11}]
????sols=Solve[sysofeqs,vars]
sersol=Series[y[x],{x,0,11}/.sols[[1]]]

Legendre's equation

We obtain a formula for a[n] by replacing each occurrence of n in a[n+2] by n-2.

 

 

1.5.5a. Bessel functions

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