How can we prove that the expression γ m k ( λ ) = ∑ i = 1 m ( λ m i + m − 1 ) k ∏ j ≠ i ( 1 − 1 λ m i − λ m j )

Question

How can we prove that the expression      γ          m      k        (  λ  )  =      ∑          i      =      1              m                  (              λ                  m          i                    +      m      −      1      )              k            ∏          j      ≠      i            (    1    −          1                        λ                      m            i                          −                  λ                      m            j                                )

Tanner Hamilton · Accepted Answer

For a fixed     m  =  n   , let         λ    i    =      λ          n      i         and let                 σ      k        (    λ    )    =          ∑              i        =        1                    n                    λ              i                    k                    ∏              j        ≠        i                    (      1      −              1                              λ                          i                                −                      λ                          j                                          )          . Since         γ          m      K        (  λ  )   is a linear combination of         σ    k    (  λ  )                      |                          k      =      0      ,      1      ,      …      ,      K        , it is enough to prove that         σ    k    (  λ  )   is a polynomial in         λ    i                        |                          i      =      1      ,      2      ,      …      ,      n         for     k  ≥  0  . Since         σ    k    (  λ  )   is symmetric in         λ    i    , this is equivalent to proving that it is a polynomial in the elementary symmetric polynomials i.e. the coefficients of           p    (    z    )    =          ∏              i        =        1            n        (    z    −          λ      i        )      Using that:      ∏          j      ≠      i            (    1    −          1                        λ                      i                          −                  λ                      j                                )    =                    ∏                  j          ≠          i                    (              λ                  i                    −      1      −              λ                  j                    )                      ∏                  j          ≠          i                    (              λ                  i                    −              λ                  j                    )        =            −              ∏                  j                    (              λ                  i                    −      1      −              λ                  j                    )                      ∏                  j          ≠          i                    (              λ                  i                    −              λ                  j                    )        =  −            p      (              λ        i            −      1      )                      p        ′            (              λ        i            )      it follows that:      σ    k    =  −      ∑          i      =      1              n                          λ                  i                          k                          p      (              λ        i            −      1      )                      p        ′            (              λ        i            )      The sum reminisces of a Lagrange interpolation, and suggests looking at the polynomial:  f  (  z  )  =      ∑          i      =      1              n                  p      (      z      )              z      −              λ        i              ⋅                    λ                  i                          k          +          1                          p      (              λ        i            −      1      )                      p        ′            (              λ        i            )      Since                 lim              z        →                  λ          i                                    p        (        z        )                    z        −                  λ          i                      =          p      ′        (          λ      i        )       it follows that     f  (      λ    i    )  =      λ          i              k      +      1          p  (      λ    i    −  1  )  , so     f  (  z  )  =      z          k      +      1        p  (  z  −  1  )   at the n points     z  =      λ    1    ,      λ    2    ,  …  ,      λ    n    Since     p  (      λ    i    )  =  0   that means     f  (  z  )   also coincides with         r    k    (  z  )  =      z          k      +      1        p  (  z  −  1  )  mod  p  (  z  )  , and must in fact be identical to it since     deg  ⁡      r    k    &amp;lt;  deg  ⁡  p  =  n   and the two are equal at n points.Let         z          k      +      1        p  (  z  −  1  )  =      q    k    (  z  )  p  (  z  )  +      r    k    (  z  )   where         q    k     is a monic polynomial of degree     k  +  1  , then:      z          k      +      1        p  (  z  −  1  )  −      q    k    (  z  )  p  (  z  )    =        r    k    (  z  )    =    f  (  z  )    =        ∑          i      =      1              n                  p      (      z      )              z      −              λ        i              ⋅                    λ                  i                          k          +          1                          p      (              λ        i            −      1      )                      p        ′            (              λ        i            )          ⟹      −      q    k    (  0  )  p  (  0  )    =    f  (  0  )    =    −  p  (  0  )        ∑          i      =      1              n                          λ                  i                          k                          p      (              λ        i            −      1      )                      p        ′            (              λ        i            )          =    p  (  0  )        σ    k      ⟹          σ    k    =  −      q    k    (  0  )Since     p  (  z  )   is monic, the coefficients of         q    k    (  z  )   are in the same ring with the coefficients of     p  (  z  )   It then follows that         σ    k    =  −      q    k    (  0  )   is a polynomial in the coefficients of     p  (  z  )

How can we prove that the expression γ <mrow class="MJX-TeXAtom-ORD">

Answered question

Answer & Explanation

New Questions in Precalculus