Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer
In fact, when the Taylor expansion exists and is convergent (so the function is locally real analytic), this also gives easy solutions to partial differential equations. This basic machinery in the PDE case is called the Cauchy-Kovalevsky theorem. – Willie Wong Mar 7 '11 at 13:11
This is the Principle of Conservation of Difficulty at work, against which we are in general defenseless! :) – Mariano Suárez-Alvarez Mar 7 '11 at 18:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.