Part III: The Power Stuff: Advanced Techniques in Java

Maker Quick Response Code in Java Part III: The Power Stuff: Advanced Techniques

To summarize, you have the following: a2 = a4 = a6 = -a0 2! a0 4! -a0 6!

So now you can relate the even coefficients in general If n = 2m (in which m stands for a positive integer or zero): a n = a 2m =

^ -1h a 0 ^ 2mh !

m = 0, 1, 2, 3

Because you set a0 based on the initial conditions for a given problem, you can now find the even coefficients of the solution

Now you can move on to the odd coefficients Turn back to the recurrence relation in the previous section for the solution: (n + 2)(n + 1)an+2 + an = 0 You can see that for n = 1 you get: (3)(2)a3 + a1 = 0 So: a3 = - a1 ^ 3 h^ 2h

Taking a clue from the even coefficients in the previous section, you can see that (3)(2) = 3!, so you get this: a3 = - a1 3!

Substituting a3 in terms of a1 into this equation gives you: a5 = or: a5 = a1 5! a1 ^ 5 h^ 4 h^ 3!h

Substituting n = 5 into the recurrence relation gives you: (7)(6)a7 + a5 = 0 or: a7 = -a5 ^7 h^ 6 h

which means that: a7 = - a1 7!

To sum up, the odd coefficients you have so far are: a3 = a5 = a7 = - a1 3! a1 5! - a1 7!

So now you can relate the odd coefficients of the solution this way in general, if n = 2m + 1: a n = a 2m + 1 =

^ -1h a 1 ^ 2m + 1h !

m = 0, 1, 2, 3

Using the two equations for even and odd coefficients from the previous sections, you get the following general solution to your differential equation in series terms:

^ -1h x 2m ^ 2mh !

^ -1h x 2m + 1 ^ 2m + 1h !

That s the solution to the differential equation in series terms In this case, the two series are recognizable as cos(x) and sin(x) Here s sin(x):

^ -1h x 2n + 1 = sin x ^ 2n + 1h !

And here s cos x:

^ -1h x 2n = cos x ^ 2nh !

So you can rewrite the solution as: y = a0 cos(x) + a1 sin(x) The terms a0 and a1 are arbitrary constants (just like c0 and c1), and they re set by matching the initial conditions

How do you use a power series to solve a differential equation for which you don t already know the solution For instance, what if you have an equation like this: d2y dy -x + 2y = 0 dx dx 2 Give it a try, using a power series like this: y=

Differentiating the power series from the previous section gives you the following equation: y l= ! na n x n - 1

n =1 3

And differentiating this gives you: y'' = ! n^ n - 1h a n x n - 2

3 n=2

Now for the fun part: substitution! Substituting these equations into the original differential equation gives you this result:

x n-2

Now it s time to equate powers of x on the left side of the equation to 0 on the right side Combining the coefficients for powers of x gives you this equation: [2a2 + 2a0] + [6a3 + a1] x + 12a4 x 2 + [20a5 a3] x 3 + [(n + 2)(n + 1) an+2 (n 2)an ] x n = 0 Note that every power of x must be 0 for the equation to work, so the last term in the previous equation gives you the recurrence relation for this solution: (n + 2)(n + 1) an+2 (n 2)an = 0 After some simplifying, you get: an+2 =