I'm studying numerical analysis and I'm having difficulty in solving this problem:
The question is : $$f(1)=2, \quad f'(1)=3 \quad f''(1)=1$$ $$f(2)=6,\quad f'(2)=7,\quad f''(2)=8$$ and I need to find f in lowest order possible with newton's divided difference method.
Well, I can just find $5^{th}$ order polynomial by setting $~f=a+bx+cx^2+\cdots~$, but I want to know how to apply newton's method.
You need an extension of Newton's method known as Hermite's method.
For that, you need to know that a divided difference with identical nodes is defined in terms of derivatives of $f$. In particular, $$f[a,a]=f'(a)$$ and $$f[a,a,a]=\frac{f''(a)}2.$$ In general, you'll have $$f[a,\ldots,a]=\frac{f^{(n)}(a)}{n!}$$ if $a$ is repeated $n+1$ times.
So, you'll have the following formula for the polynomial you're looking for: $$P(X)=f[1]+f[1,1](x-1)+f[1,1,1](x-1)^2+f[1,1,1,2](x-1)^3+f[1,1,1,2,2](x-1)^3(x-2)+f[1,1,1,2,2,2](x-1)^3(x-2)^2.$$
For a divided difference like $f[1,1,1,2]$, you have to use the inductive definition, so $$f[1,1,1,2]=\frac{f[1,1,1]-f[1,1,2]}{1-2},$$ where $f[1,1,1]=\frac{f''(1)}{2!}$ and for $f[1,1,2]$ you have to use again the inductive definition, so $$f[1,1,2]=\frac{f[1,1]-f[1,2]}{1-2}=\frac{f'(1)-\frac{f(1)-f(2)}{1-2}}{1-2},$$ and then $$f[1,1,1,2]=\frac{\frac{f''(1)}2-\frac{f'(1)-\frac{f(1)-f(2)}{1-2}}{1-2}}{1-2}.$$ So in the end all the divided differences can be expressed in terms of $f$ and its derivatives.