Let $S(x, y, m, n) = \sum\limits_{i=0}^m \binom{n}{i}x^i y^{n-i}$, where $0 < m < n$. I want to derive the relation between $S(x, y, m, n)$ and $S(x, y, m, n-1)$.
Is there any formulas I can use?
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$\begingroup$ As a function of $y$, this $S(x,y,m,n)$ can be seen as the remainder in the Taylor expansion of order $n-m+1$ with center in the point $x$ for the function $f(t)=t^n$, so one can express it by the integral formula. $\endgroup$– Pietro MajerCommented Jan 19, 2022 at 17:30
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$\begingroup$ @PietroMajer Thanks. In that case, do I need the approximation? $\endgroup$– one userCommented Jan 19, 2022 at 17:35
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$\begingroup$ If an integral formula is ok to you, you just need to apply the integral form of the remainder. en.wikipedia.org/wiki/… $\endgroup$– Pietro MajerCommented Jan 19, 2022 at 18:20
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1 Answer
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One can construct the relation using the identity $$ \binom{n}{i} = \binom{n-1}{i} + \binom{n-1}{i-1} $$ Then, writing the $i=0$ term separately, \begin{eqnarray} S(x,y,m,n) &=& y^n + \sum_{i=1}^{m} \binom{n-1}{i} x^i y^{n-i} + \sum_{i=1}^{m} \binom{n-1}{i-1} x^i y^{n-i} \\ &=& y^n + y\sum_{i=1}^{m} \binom{n-1}{i} x^i y^{n-1-i} + \sum_{i=0}^{m-1} \binom{n-1}{i} x^{i+1} y^{n-i-1} \\ &=& y S(x,y,m,n-1) + x\left[ S(x,y,m,n-1) - \binom{n-1}{m} x^m y^{n-1-m} \right] \\ &=& (x+y) S(x,y,m,n-1) \ - \ \binom{n-1}{m} x^{m+1} y^{n-1-m} \end{eqnarray}
answered Jan 19, 2022 at 18:20