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Question is : Set $f(x)$ to be a polynomial of degree $m$. Denote $f[x_0,x_1,…,x_n]$ as the Newton Divided Difference. Given that $f[x_0,x_1,…,x_n,x]$ is a polynomial of degree $m-(n+1)$. Use this result to show that $f[x_0,x_1,…,x_n,x]=0$ for any $x_0,x_1...,x_n$, if $\deg(f)≤n$.

My thought: $\deg(f)≤n$ implies $\deg(f[x_0,x_1,…,x_n,x])＜0$, so $f[x_0,x_1,…,x_n,x]$ is the zero polynomial. This means $f[x_1,…,x_n,x]-f[x_0,x_1,…,x_n]=0$ by the definition. Also since $f[x_0,x_1,…,x_n]$ is a constant term while $f[x_1,…,x_n,x]$ is a polynomial with x in each term, so both of them should be equal to 0. Thus $f[x_0,x_1,…,x_n]=0$ for all $x_i$, $i=0,1...,n$.

I don't know my argument is correct or not, does anyone have any idea about this question?

Gary
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user1228735
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  • Hello. Could you please use mathjax and possibly also modify the text so that your question becomes more clear? Also, your thoughts about how ro solve the question and what you have done so far, should go in the question not in a separate answer. Thanks! – lcv Feb 08 '24 at 09:03
  • Are there any other hints, I am also confused now... – user1228735 Feb 08 '24 at 09:30
  • You should tell us about the notation. In particular, I guess, that your are looking at divided differences. This could help. – Gary Feb 08 '24 at 09:33
  • Yes, it's actually Newton divided difference, thank you. – user1228735 Feb 08 '24 at 09:53
  • But I am also wondering is there any other method different from using MVT to prove it – user1228735 Feb 08 '24 at 10:02

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