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

Does there exist such an equation that the differences, are infinite? And if there isn't then what is the proof?

For example:

$x^2$ has a finite difference of 2.

$$ \begin{array}{c|c|c|c} n & y & \text{first differences} & \text{second differences} \\ \hline 1 & 1 & 0 & 0 \\ 2 & 4 & 3 & 2 \\ 3 & 9 & 5 & 2 \\ 4 & 16& 7 & 2 \\ 5 & 25& 9 & 2 \\ \end{array} $$

Therefore, all of the second differences are the same. Furthermore, this found by subtracting the first differences e.g. $5-3 = 2$ and $7-5=2$ etc.

share|cite|improve this question
Difference of what? Could you elaborate your example, showing why the difference of $x^3$ is $3$? – martini Sep 19 '12 at 12:56
yes, sorry about that – Jeel Shah Sep 19 '12 at 12:57
okay, I have provided an explanation as to what a difference is. – Jeel Shah Sep 19 '12 at 13:12
So $f$ has a "finite difference of $n$" if the $n$-th differences are all equal? – martini Sep 19 '12 at 13:14
in essence, yes. – Jeel Shah Sep 19 '12 at 13:23
up vote 1 down vote accepted

If $y$ is finite for every $n$, then the first differences, being differences of finite numbers, will all be finite, and then the second differences, being differences of finite numbers, will all be finite, and so on. You'll never get any infinite differences --- how could you?

Now, you can get unbounded differences, e.g., see what happens if you take $y$ to be $2^n$. But that's a different thing entirely.

share|cite|improve this answer
Can you pelase expand on "unbounded" differences? – Jeel Shah Sep 21 '12 at 0:44
Did you "see what happens if you take $y$ to be $2^n$"? Mathematics is not a spectator sport - you have to take part in it. – Gerry Myerson Sep 21 '12 at 4:01
Yes, now I have. I see what you mean. – Jeel Shah Sep 21 '12 at 4:36

What you will find is that if you start with any polynomial the $k$th differences will eventually be constant. The reason for this is that the difference of two values on a polynomial is a multiple of the value of its derivative inbetween, $$ p(x+1)-p(x)=p'(c)(x+1-x) = p'(c) $$ for some $c$ between $x$ and $x+1$. For the $k$th differences we may slightly modify this argument to show that that they are multiples of $p^{(k)}$ at some point.

However, we know that the ($n+1$) derivative of an $n$th degree polynomial will be zero, so the $n$th differences will be constant.

You will find that if you try with a function that is infinitely differentiable and all of its derivatives are positive, the $k$th differences will never be constant. A good example of such a function is $e^x$.

share|cite|improve this answer

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.