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I'm trying to calculate the value of $f(x_1)$ with Newton Raphson's method. The following information is given:

  • $x_0=0$
  • $x_1=1$
  • $f(x_0)=2$
  • $f'(x_0)=0$
  • $f'(x_1)=0$
  • $f''(x_0)=0$
  • $f''(x_1)=0$
  • $f[x_0,x_0,x_1,x_1] = 1$ (third degree differentiation)

It is given that we don't know that the function $f(x)$ is a polynomial or not. The differentiation has something to do with a derivative. This is the hint that is given:

$\lim \limits_{x \to y} f[x,y] = \lim \limits_{x \to y} \frac{f(y)-f(x)}{y-x} = f'(y)$

This is the hint where I am stuck:

$f[x_0,x_0,x_1,x_1]$ = $\lim \limits_{x_2 \to x_0, x_3 \to x_1}f[x_2,x_0,x_1,x_3]$

Can someone explain the last hint to me?

This is the answer but I don't understand it: answer

Close up:

closeup

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  • $\begingroup$ What is initial point? $\endgroup$ – Michael Galuza Aug 21 '15 at 7:35
  • $\begingroup$ Does f have one independent variable or two? $\endgroup$ – Narasimham Aug 21 '15 at 7:41
  • $\begingroup$ @MichaelGaluza I think you may take $x_0$ as initial point. $\endgroup$ – Stanko Aug 21 '15 at 7:48
  • $\begingroup$ @Dongo, no, $f'(x_0)=0$ $\endgroup$ – Michael Galuza Aug 21 '15 at 7:49
  • $\begingroup$ @Narasimham I don't know. $\endgroup$ – Stanko Aug 21 '15 at 7:59
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I assummed that $$ f[x_1, x_2] = \frac{f(x_1) - f(x_2)}{x_1 - x_2} $$ $$ f[x_1, x_2, x_3] = \frac{f[x_1,x_2] - f[x_2,x_3]}{x_1 - x_3} $$ $$ f[x_1, x_2, x_3, x_4] = \frac{f[x_1,x_2,x_3] - f[x_2,x_3,x_4]}{x_1 - x_4} $$

Let $x_2\to x_1$, $x_4\to x_3$. Then $$ f[x_1, x_1, x_3, x_3] = \frac{f[x_1,x_1,x_3] - f[x_1,x_3,x_3]}{x_1 - x_3}. $$ Now $$ f[x_1, x_1, x_3] = \frac{f[x_1, x_1] - f[x_1, x_3]}{x_1-x_3} = \frac{f'(x_1) - f[x_1, x_3]}{x_1-x_3},\\ f[x_1, x_3, x_3] = \frac{f[x_1, x_3] - f[x_3, x_3]}{x_1-x_3} = \frac{f[x_1, x_3]-f'(x_3)}{x_1-x_3}, $$ and $$ f[x_1, x_1, x_3, x_3] = \frac{f'(x_1) - 2f[x_1, x_3]+f'(x_3)}{(x_1 - x_3)^2}. $$

Let's substitute $x_1=0$, $x_3=1$: $$ f[0, 0, 1, 1] = f'(0) - 2f[0, 1]+f'(1)=-2f[0,1]=-2(f(1)-f(0))=1\\\Longrightarrow f(1)=f(0)-\frac12=\frac32 $$

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