The following is Mathematica code. I apologize but I don't know how to write it more clearly.
B1 = ReplaceAll[A1, x1 -> 1/A2]; simply means to take A1 and within A1 replace all x1 with one divided by A2 which is written 1/A2 where A2 = xc/x1 - xc/x2; The semicolon means that the output is not displayed/printed. If the semicolon is missing the output is displayed/printed. ReplaceAll is the same as substitute.
A0 = 1/x0
A1 = xc/x0 - xc/x1
A2 = xc/x1 - xc/x2;
A3 = xc/x2 - xc/x3;
A4 = xc/x3 - xc/x4;
A5 = xc/x4 - xc/x5;
B1 = ReplaceAll[A1, x1 -> 1/A2];
B2 = ReplaceAll[B1, x0 -> 1/A1]
C1 = ReplaceAll[B2, x2 -> 1/A3];
C2 = ReplaceAll[C1, x1 -> 1/A2];
C3 = ReplaceAll[C2, x0 -> 1/A1]
D1 = ReplaceAll[C3, x3 -> 1/A4];
D2 = ReplaceAll[D1, x2 -> 1/A3];
D3 = ReplaceAll[D2, x1 -> 1/A2];
D4 = ReplaceAll[D3, x0 -> 1/A1]
E1 = ReplaceAll[D4, x4 -> 1/A5];
E2 = ReplaceAll[E1, x3 -> 1/A4];
E3 = ReplaceAll[E2, x2 -> 1/A3];
E4 = ReplaceAll[E3, x1 -> 1/A2];
E5 = ReplaceAll[E4, x0 -> 1/A1]
FullSimplify[A0]
FullSimplify[A1]
FullSimplify[B2]
FullSimplify[C3]
FullSimplify[D4]
FullSimplify[E5]
When I apply FullSimplify this is what I get:
FullSimplify[A0]
$$\frac{1}{x_0}$$
FullSimplify[A1]
$$x_c \left(\frac{1}{x_0}-\frac{1}{x_1}\right)$$
FullSimplify[A2]
$$x_c^2 \left(\frac{1}{x_0}-\frac{2}{x_1}+\frac{1}{x_2}\right)$$
FullSimplify[A3]
$$x_c^3 \left(\frac{1}{x_0}-\frac{3}{x_1}+\frac{3}{x_2}-\frac{1}{x_3}\right)$$
FullSimplify[A4]
$$x_c^4 \left(\frac{1}{x_0}-\frac{4}{x_1}+\frac{6}{x_2}-\frac{4}{x_3}+\frac{1}{x_4}\right)$$
FullSimplify[A5]
$$x_c^5 \left(\frac{1}{x_0}-\frac{5}{x_1}+\frac{10}{x_2}-\frac{10}{x_3}+\frac{5}{x_4}-\frac{1}{x_5}\right)$$
I need to show that as I continue the replacement pattern within the code the result from FullSimplifyfor the $n$-th row is:
$$x_{c}^{n-1} \sum\limits_{k=1}^{n} \frac{(-1)^{k-1} \binom{n-1}{k-1}}{x_{k-1}}$$
A0,A1,A2,A3,A4, andA5before I appliedFullSimplifyto them? – Mats Granvik Aug 13 '20 at 18:59