1

The function is given that $$f(x)=e^x$$ Prove that the divided difference is positive for any distinct numbers $x_0,...x_n$ for $n\geq0$. First divided difference is $${ f[x_1]-f[x_0]\over x_1-x_0}$$ Now if $x_0>x_1$ $${ e^{x_1}-e^{x_0}\over x_1-x_0}>0$$ If $x_0<x_1$ $${ e^{x_1}-e^{x_0}\over x_1-x_0}>0$$ Hence, it is positive for any case for the first divided difference. Note that $x_0\neq x_1$ as all points are distinct. How can I prove that is it valid up to n th divided difference a.k.a $$f[x_0,x_1,...,x_n]>0$$ The divided difference here is defined as $$f[x_0,x_1,...,x_n]={f[x_1,...,x_{n}]-f[x_0,..x_{n-1}]\over x_n-x_0}$$

Don
  • 636
  • 2
    Maybe just induction will do the trick. I am not familiar with finite differences that much. Could you give the recursive definition? – Cornman Oct 03 '19 at 23:04
  • https://math.stackexchange.com/questions/95420/show-fx-0-x-1-ldots-x-n-1-for-divided-differences-of-xn?rq=1 How about using the equality given here. As the derivative of f(x) is always positive, the divided difference is positive. – Don Oct 03 '19 at 23:20
  • Induction should fail. Have you tried to calculate $f[x_0,x_1,x_2]$ and show that this is positive? Maybe one can guess a formula for $f[x_0,x_1,\dotso,x_n]$ that way and prove that formula inductively, but it should get messy – Cornman Oct 03 '19 at 23:37
  • There is a mean value theorem for divided differences. Can I apply that? If yes, then it is straight forward proof. As $f^n(x)>0$ for all x – Don Oct 03 '19 at 23:41
  • I mean this : https://en.wikipedia.org/wiki/Mean_value_theorem_(divided_differences) – Don Oct 03 '19 at 23:44
  • If you are allowed to use this result, this solves your problem, because $f^{(n)}(x)>0$ – Cornman Oct 04 '19 at 00:01

0 Answers0