Step by Step solution about How to resolve the algorithm Stirling numbers of the second kind step by step in the Mathematica / Wolfram Language programming language

Explanation:

The provided Wolfram code performs calculations related to Stirling numbers of the second kind, which are used to count the number of ways to partition a set of objects into a given number of non-empty subsets.

Line 1:

TableForm[Array[StirlingS2, {n = 12, k = 12} + 1, {0, 0}], TableHeadings -> {"n=" <> ToString[#] & /@ Range[0, n], "k=" <> ToString[#] & /@ Range[0, k]}]

• This line creates a TableForm that displays a table of Stirling numbers of the second kind.
• Array[StirlingS2, {n = 12, k = 12} + 1, {0, 0}] generates a 2D array with dimensions n + 1 x k + 1 and fills it with the Stirling numbers of the second kind.
• TableHeadings -> {"n=" <> ToString[#] & /@ Range[0, n], "k=" <> ToString[#] & /@ Range[0, k]} sets the column and row headings for the table, which are the values of n and k, respectively.

Output:

This generates a table that contains Stirling numbers of the second kind from n = 0 to n = 12 for each value of k from k = 0 to k = 12.

Line 2:

Max[Abs[StirlingS2[100, #]] & /@ Range[0, 100]]

• This line finds the maximum absolute value of the Stirling numbers of the second kind for n = 100 from k = 0 to k = 100.
• StirlingS2[100, #] evaluates the Stirling number of the second kind for n = 100 and each k in the range.
• Abs takes the absolute value of each result.
• Max finds the maximum value among the absolute values.

Output:

This produces a single value representing the maximum absolute value of the Stirling number of the second kind for n = 100.

TableForm[Array[StirlingS2, {n = 12, k = 12} + 1, {0, 0}], TableHeadings -> {"n=" <> ToString[#] & /@ Range[0, n], "k=" <> ToString[#] & /@ Range[0, k]}]
Max[Abs[StirlingS2[100, #]] & /@ Range[0, 100]]