[Math] Correspondence between ODE and difference equation

ordinary differential equationsrecurrence-relations

In Wikipedia about difference equations, there is some description about correspondence between ODE and difference equation:

If you consider the Taylor series of
the solution to a linear differential
equation:$$\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n$$ you
see that the coefficients of the
series are given by the nth derivative
of $f(x)$ evaluated at the point $a$. The
differential equation provides a
linear difference equation relating
these coefficients.

The rule of thumb (for equations in
which the polynomial multiplying the
first term is non-zero at zero) is
that:$$ y^{[k]} \to f[n+k]$$and more
generally $$x^m*y^{[k]} \to n(n-1)(n-m+1)f[n+k-m]$$

Example: The recurrence relationship
for the Taylor series coefficients of
the equation: $$(x^2 + 3x -4)y^{[3]} -(3x+1)y^{[2]} + 2y = 0\,$$ is given by$$n(n-1)f[n+1] + 3nf[n+2] -4f[n+3] -3nf[n+1] -f[n+2]+ 2f[n] = 0\, $$ or $$-4f[n+3] +2nf[n+2] + n(n-4)f[n+1] +2f[n] = 0.\,$$

My questions are:

  1. I was wondering what the rationale
    behind this conversion from an ODE to
    a difference equation is? Although
    having tried to read it several
    times, I was not able to understand
    it.
  2. In reverse direction, can a
    difference equation be converted to
    an ODE using this correspondence?
    How to?
  3. Is the conversion of an ODE into a
    difference equation in numerical
    methods for solving an ODE related
    to the correspondence between the
    two mentioned above?

  4. This equivalence can be used to
    quickly solve for the recurrence
    relationship for the coefficients in
    the power series solution of a linear
    differential equation.

    Problems generally solved using the
    power series solution method taught in
    normal differential equation classes
    can be solved in a much easier way.

    I was wondering how exactly the
    correspondence can make solving an
    ODE or a difference equation easier?

Thanks and regards!

Best Answer

All that's going on here is that when you differentiate a Taylor series expansion

$$f(x) = f^{(0)}(a) + \frac{f^{(1)}(a)}{1!} (x - a)^1 + \frac{f^{(2)}(a)}{2!} (x - a)^2 + ...$$

you get the Taylor series expansion

$$f'(x) = f^{(1)}(a) + \frac{f^{(2)}(a)}{1!} (x - a)^1 + \frac{f^{(3)}(a)}{2!} (x - a)^2 + ....$$

In other words, what you've done is precisely shifted the Taylor coefficients one to the left. That's all the Wikipedia article is saying.

I disagree that this makes ODEs any easier to solve, though. ODEs were already this easy to solve, you just weren't taught the right language for seeing this.

Related Question