Given n + 1 points ( x i , f i ) with x i ≠ x j for i ≠ j, define L k ( x ) to be ( x − x 0 ) ( x − x 1 ) ⋯ ( x − x k − 1 ) ( x − x k + 1 ) ⋯ ( x − x n ) ( x k − x 0 ) ( x k − x 1 ) ⋯ ( x k − x k − 1 ) ( x k − x k + 1 ) ⋯ ( x k − x n ) ⋯ ( 1 )Show that L k ( x j ) is a polynomial of degree n, that L k ( x j ) = 0 for j ≠ k and L k ( x k ) = 1

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Given   n  +  1 points   (      x    i    ,      f    i    ) with       x    i    ≠      x    j   for   i  ≠  j, define       L    k    (  x  ) to be            (      x      −              x        0            )      (      x      −              x        1            )      ⋯      (      x      −              x                  k          −          1                    )      (      x      −              x                  k          +          1                    )      ⋯      (      x      −              x        n            )              (              x        k            −              x        0            )      (              x        k            −              x        1            )      ⋯      (              x        k            −              x                  k          −          1                    )      (              x        k            −              x                  k          +          1                    )      ⋯      (              x        k            −              x        n            )          ⋯  (  1  )Show that       L    k    (      x    j    ) is a polynomial of degree n, that       L    k    (      x    j    )  =  0 for   j  ≠  k and       L    k    (      x    k    )  =  1

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Use the product notation:      L    k    (  x  )  =                    ∏                  i          =          0          ,          i          ≠          k                n            (      x      −              x        i            )                      ∏                  i          =          0          ,          i          ≠          k                n            (              x        k            −              x        i            )        =      ∏          i      =      0      ,      i      ≠      k        n              (      x      −              x        i            )              (              x        k            −              x        i            )      The bottom is a product of numerical values, hence again a number. Instead the top is a product of n linear terms hence a polynomial of degree n. Nothing to prove just a matter of fact. The fact   i  ≠  k avoid singularity of the form       1    0        L    k   acts as a &quot;switch&quot; when evaluated on the original points just because of their definition      L    k    (      x    k    )  =                    ∏                  i          =          0          ,          i          ≠          k                n            (              x        k            −              x        i            )                      ∏                  i          =          0          ,          i          ≠          k                n            (              x        k            −              x        i            )        =      ∏          i      =      0      ,      i      ≠      k        n              (              x        k            −              x        i            )              (              x        k            −              x        i            )        =  1      L    k    (      x    j    )  =                    ∏                  i          =          0          ,          i          ≠          k                n            (              x        j            −              x        i            )                      ∏                  i          =          0          ,          i          ≠          k                n            (              x        k            −              x        i            )        =  (      x    j    −      x    j    )                    ∏                  i          =          0          ,          i          ≠          k          ≠          j                n            (              x        j            −              x        i            )                      ∏                  i          =          0          ,          i          ≠          k                n            (              x        k            −              x        i            )        =  0This proves (1),  f  (  x  )  =      ∑          i      =      0        n        f    i        L    i    (  x  ) will be just the form of the polynomial passing through all the given points             (              x        i            ,              f        i            )              i      =      0        n  . The       L    i   form a basis of the basis of polynomials.

Given n + 1 points ( x i </msub> , f i </msub> ) with

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