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Math Jaxed

edit approved Jan 26, 2020 at 6:17

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Here is an approach that suggests (for d$d$ smaller than j$j$, say d^2$d^2$ less than j$j$) the sum is not larger than dj(j!)$dj(j!)$. Unfortunately, we are only bounding part of the sum by d(j/2 - d)(j!)$d(j/2 - d)(j!)$. These are the parts which have an a_i$\alpha_i$ term at least as large as (j/2+d+1)$(j/2+d+1)$.

Note that there are d$d$ terms whose product is j!$j!$: these are the d$d$ tuples where all but one of the a_i$\alpha_i$ are zero. Now let's set k=0$k=0$, and look at a single tuple whose largest a_i$\alpha_i$ is (j-k)$(j-k)$. By replacing this a_i$\alpha_i$ by (j-k-1)$(j-k-1)$ and adding 1$1$ to one of the other d-1$d-1$ places, we get d-1$d-1$ distinct tuples whose largest term is (j-k-1)$(j-k-1)$ and whose sum of (the products derived from each of) these d-1$d-1$ tuples is at most (k+d-1)/(j-k)$(k+d-1)/(j-k)$ times the single tuple, meaning the sum of all (products derived from) tuples with largest element (j-k-1)$(j-k-1)$ is less than the sum of all (products derived from) tuples with largest element (j-k)$(j-k)$. So when 2k$2k$ is less than j+1-d$j+1-d$, we get the sum of tuples with largest element (j-k)$(j-k)$ is less than d(j!)$d(j!)$. So we can bound a large part of the sum by (j+1-d)d(j!)/2$(j+1-d)d(j!)/2$. If we could extend this argument down to k=j/d$k=j/d$, we would have the sum bounded above by (j-j/d)d(j!)$(j-j/d)d(j!)$.

Gerhard "Turning Multiplication Back Into Addition" Paseman, 2020.01.25.

Here is an approach that suggests (for d smaller than j, say d^2 less than j) the sum is not larger than dj(j!). Unfortunately, we are only bounding part of the sum by d(j/2 - d)(j!). These are the parts which have an a_i term at least as large as (j/2+d+1).

Note that there are d terms whose product is j!: these are the d tuples where all but one of the a_i are zero. Now let's set k=0, and look at a single tuple whose largest a_i is (j-k). By replacing this a_i by (j-k-1) and adding 1 to one of the other d-1 places, we get d-1 distinct tuples whose largest term is (j-k-1) and whose sum of (the products derived from each of) these d-1 tuples is at most (k+d-1)/(j-k) times the single tuple, meaning the sum of all (products derived from) tuples with largest element (j-k-1) is less than the sum of all (products derived from) tuples with largest element (j-k). So when 2k is less than j+1-d, we get the sum of tuples with largest element (j-k) is less than d(j!). So we can bound a large part of the sum by (j+1-d)d(j!)/2. If we could extend this argument down to k=j/d, we would have the sum bounded above by (j-j/d)d(j!).

Gerhard "Turning Multiplication Back Into Addition" Paseman, 2020.01.25.

Here is an approach that suggests (for $d$ smaller than $j$, say $d^2$ less than $j$) the sum is not larger than $dj(j!)$. Unfortunately, we are only bounding part of the sum by $d(j/2 - d)(j!)$. These are the parts which have an $\alpha_i$ term at least as large as $(j/2+d+1)$.

Note that there are $d$ terms whose product is $j!$: these are the $d$ tuples where all but one of the $\alpha_i$ are zero. Now let's set $k=0$, and look at a single tuple whose largest $\alpha_i$ is $(j-k)$. By replacing this $\alpha_i$ by $(j-k-1)$ and adding $1$ to one of the other $d-1$ places, we get $d-1$ distinct tuples whose largest term is $(j-k-1)$ and whose sum of (the products derived from each of) these $d-1$ tuples is at most $(k+d-1)/(j-k)$ times the single tuple, meaning the sum of all (products derived from) tuples with largest element $(j-k-1)$ is less than the sum of all (products derived from) tuples with largest element $(j-k)$. So when $2k$ is less than $j+1-d$, we get the sum of tuples with largest element $(j-k)$ is less than $d(j!)$. So we can bound a large part of the sum by $(j+1-d)d(j!)/2$. If we could extend this argument down to $k=j/d$, we would have the sum bounded above by $(j-j/d)d(j!)$.