11
$\begingroup$

Question:

Given $l\in \mathbb{N^+}$. $a_1,\cdots,a_l,b_1,\cdots,b_l$ are real numbers.$a_0=b_0=a_{l+1}=b_{l+1}=0$.Define $$g(m,l)=-\dfrac{\displaystyle\prod _{r=0} ^{l}{(a_m-a_r+b_{r+1}-b_m)}}{\displaystyle\prod _{r=1,r\ne m} ^{l}{(a_m-a_r+b_r-b_m)}}$$ Prove: $$\sum_{m=1}^{l}{g(m,l)}=\sum_{t=1}^{l}{(a_t-a_{t-1})b_t}$$

For example,for $l=1$,$g(1,1)=-\frac{a_1(-b_1)}{1}=(a_1-a_0)b_1$

Another example:for $l=2$, $$g(1,2)+g(2,2)=-\frac{a_1(b_2-b_1)(a_1-a_2-b_1)}{a_1-a_2+b_2-b_1}+\frac{b_2(a_2-a_1)(a_2+b_1-b_2)}{a_2-a_1+b_1-b_2}$$ $$=\frac{(a_2-a_1+b_1-b_2)(a_1 b_1+a_2 b_2-a_1 b_2)}{a_2-a_1+b_1-b_2}=\sum_{t=1}^{2}{(a_t-a_{t-1})b_t}$$

But How to prove general case,Maybe use Lagrange interpolation formula?

$\endgroup$
0
2
$\begingroup$

Fix $l$ (so we may then drop the $l$ from $g(m,l)$), and let $\alpha_{i,j}=a_{i}-a_j,\beta_{i,j}=b_i-b_j$, $d_{ij}=\alpha_{i,j}-\beta_{i,j}$. Denote by $S$ the main sum $S=\sum_{m=0}^l g(m)$.

Now, let us temporarily fix two indices $i$ and $j$. One can write

$$ g(i)=\frac{1}{d_{ij}}\frac{A(a_i-b_i)}{B(a_i-b_i)} \tag{1} $$

where $A(X)=\prod_{r=0}^{l} (X+b_{r+1}-a_r)$ is independent of $i$ or $j$ and

$$ B(X)=\prod_{1\leq r\leq n, r\neq i,j} (X+b_r-a_r) \tag{2} $$

is symmetrical in $i$ and $j$. It follows that

$$ g(i)+g(j)=\frac{1}{d_{ij}}\Bigg(\frac{A(a_i-b_i)}{B(a_i-b_i)}- \frac{A(a_j-b_j)}{B(a_j-b_j)}\Bigg) $$ and $$ g(i)+g(j)= \frac{\frac{A(a_i-b_i)B(a_j-b_j)-A(a_j-b_j)B(a_i-b_i)}{(a_i-b_i)-(a_j-b_j)}}{B(a_i-b_i)B(a_j-b_j)} \tag{3} $$

But the numerator in (3) above is in fact a polynomial. So, $d_{ij}$ does not appear in the denominator of the reduced fraction for $g(i)+g(j)$. Since $d_{ij}$ does not appear in the denominator of $g(m)$ when $m\neq i,j$, it follows that $d_{i,j}$ does not appear in the denominator of the reduced fraction for $S$.

Since this holds for any $i,j$,it follows that $S$ is a polynomial. Since each $g(m)$ has degree at most $1$ in each of its variables (as a rational fraction), $S$ also has degree at most $1$ in each of its variables.

So $S$ can be written

$$ S=C(a_1,a_2,\ldots,a_l)+\sum_{m=1}^{l} S_m(a_1,a_2,\ldots,a_l) b_m \tag{4} $$

By taking all the $b_k$'s equal to $0$, we see that $C=0$. If we take all the $b_k$'s equal to $0$ except $b_1$, we see that $g(1)=a_1b_1$ and $g(m)=0$ for $m>1$, whence $S_1=a_1$.

More generally, if all the $b_k$'s are equal to $0$ except $b_j$ for $j\gt 1$, then $g(m)=0$ except when $m=j$ or $j-1$, with $g(j-1)=\frac{b_ja_{j-1}(-a_{j-1}+a_j)}{a_{j-1}-a_j+b_j}$ and a similar formula for $g(j)$, so that $S_j=\frac{g(j-1)+g(j)}{b_j}=a_{j}-a_{j-1}$.

This proves your identity.

$\endgroup$
0
$\begingroup$

We prove the result using induction. It is already verified that the result holds for $l=2$. So assume that the following holds: \begin{equation} \sum_{m=1}^l g(m,l)=\sum_{t=1}^l (a_t-a_{t+1})b_t. \end{equation} We have \begin{align} \sum_{m=1}^{l+1} g(m,l+1) &= -\sum_{m=1}^{l+1}\frac{\prod_{r=1}^{l}a_m-a_r+b_{r+1}-b_m}{\prod_{r=1,r\neq m}^{l+1}a_m-a_r+b_{r}-b_m}(a_m+b_1-b_m)(a_m-a_{l+1}-b_m)\\ &= \sum_{m=1}^{l} g(m,l)\frac{(a_m-a_l+b_{l+1}-b_m)(a_m-a_{l+1}-b_m)}{(a_m-a_{l+1}+b_{l+1}-b_m)(a_m-a_l-b_m)}+g(l+1,l+1)\\ &= \sum_{m=1}^{l} g(m,l)\left(1-\frac{(a_{l+1}-a_l)b_{l+1}}{(a_m-a_{l+1}+b_{l+1}-b_m)(a_m-a_l-b_m)}\right)+g(l+1,l+1)\\ &= \sum_{t=1}^l(a_t-a_{t-1})b_{t}+(a_{l+1}-a_l)b_{l+1}P, \end{align} where \begin{equation} P=-\sum_{m=1}^{l} \frac{g(m,l)}{(a_m-a_{l+1}+b_{l+1}-b_m)(a_m-a_l-b_m)}+\frac{g(l+1,l+1)}{(a_{l+1}-a_l)b_{l+1}}. \end{equation} Now it suffices to prove that $P=1$.

Next, we simplify $g(l+1,l+1)$: \begin{align} \frac{g(l+1,l+1)}{(a_{l+1}-a_l)b_{l+1}}& = -\frac{1}{(a_{l+1}-a_l)b_{l+1}}\frac{\prod_{r=1}^{l}a_{l+1}-a_r+b_{r+1}-b_{l+1}}{\prod_{r=1}^{l}a_{l+1}-a_r+b_{r}-b_{l+1}}(a_{l+1}+b_1-b_{l+1})b_{l+1}\\& = -(a_{l+1}+b_1-b_{l+1})f(x)|_{x=a_{l+1}-b_{l+1}} ,, \end{align} where $f(x)=-\frac{\prod_{r=1}^{l-1}(x-a_r+b_{r+1})}{\prod_{r=1}^l x-a_r+b_{r}}$. Using partial fractions, \begin{equation} f(x)=\sum_{p=1}^l\frac{c_p}{x-a_p+b_{p}} = \frac{\sum_{p=1}^l \left(c_p \prod_{r=1,r\neq p}^lx-a_r+b_{r} \right)}{\prod_{r=1}^l x-a_r+b_{r}}. \end{equation} Thus, we get \begin{align} -\prod_{r=1}^{l-1}(x-a_r+b_{r+1})= \sum_{p=1}^l c_p \prod_{r=1,r\neq p}^l (x-a_r+b_{r})\hspace{5cm}(1) \end{align} Substituting $x=a_m-b_m$ gives the following: \begin{equation} c_m = -\frac{\prod_{r=1}^{l-1}a_m-a_r+b_{r+1}-b_m}{ \prod_{r=1,r\neq m}^l a_m-a_r+b_{r}-b_m} = g(m,l)\frac{1}{(a_m+b_1-b_m)(a_m-a_l-b_m)}. \end{equation} Substituting this back in the expression for $P$, we get \begin{align} P &= -\sum_{m=1}^{l} \frac{g(m,l)}{(a_m-a_{l+1}+b_{l+1}-b_m)(a_m-a_l-b_m)}\\ &\hspace{1cm}-(a_{l+1}+b_1-b_{l+1})\sum_{m=1}^l g(m,l)\frac{1}{(a_{l+1}-a_m+b_{m}-b_{l+1})(a_m+b_1-b_m)(a_m-a_l-b_m)}\\ &= -\sum_{m=1}^{l} \frac{g(m,l)}{(a_m-a_{l+1}+b_{l+1}-b_m)(a_m-a_l-b_m)}\left(1-\frac{a_{l+1}+b_1-b_{l+1}}{a_m+b_1-b_m}\right)\\ &= -\sum_{m=1}^{l} g(m,l)\frac{1}{(a_m-a_l-b_m)(a_m+b_1-b_m)}= -\sum_{m=1}^{l} c_m = 1. \end{align} Here, the last equation follows because $\sum_{m=1}^{l} c_m $ is the coefficient of $x$ in (1).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.