Kronecker delta

Kronecked delta is maths object that has some formulas.

$m\,\delta _{n,m-2}+(m+5)\,\delta _{n,m+1}=\left({\frac {5}{3}}n-{\frac {2}{3}}m+{\frac {10}{3}}\right)\,\delta _{|n-m+1/2|,3/2}$

First, resolve deltas into one by taking symmetric and anti-symmetric combinations:

$n=(m-2,m+1)$ is equivalent to $n=(m-1/2)\pm 3/2$. So deltas can be combined into: $\delta _{|n-m+1/2|,3/2}$.

Then, since the coefficients are order-1 in $m$, the trick is to insert some quantity of $n$ into some order-1 expression in $m$:

$(\alpha n+\beta m+\gamma )\delta _{|n-m+1/2|,3/2}$.

Now solve for $\alpha ,\beta ,\gamma$ with $n=m-2$ and $n=m+1$, which will be 4 equations in 3 unknowns. These equations must have one degeneracy or else it will not be possible to reduce the form.

General formula

Order 1 coefficients

If

$x_{mn}=(a_{1}m+a_{0})\,\delta _{n,\alpha _{1}m+\alpha _{0}}+(b_{1}m+b_{0})\,\delta _{n,\beta _{1}m+\beta _{0}},$

then combined form will be

$x_{mn}=\left(p_{1}n+q_{1}m+q_{0}\right)\delta _{\left|n-{\frac {(\alpha _{1}+\beta _{1})m+(\alpha _{0}+\beta _{0})}{2}}\right|,\left|{\frac {(\alpha _{1}-\beta _{1})m+(\alpha _{0}-\beta _{0})}{2}}\right|}$

where the equations to solve are:

$\alpha _{1}p_{1}+q_{1}=a_{1}$

$\beta _{1}p_{1}+q_{1}=b_{1}$

$\alpha _{0}p_{1}+q_{0}=a_{0}$

$\beta _{0}p_{1}+q_{0}=b_{0}$

This will be solvable if $\alpha _{0}=\beta _{0}$ and $a_{0}=b_{0}$ and $\alpha _{1}\neq \beta _{1}$ to:

$p_{1}={\frac {a_{1}-b_{1}}{\alpha _{1}-\beta _{1}}}$

$q_{1}=a_{1}-\alpha _{1}\left({\frac {a_{1}-b_{1}}{\alpha _{1}-\beta _{1}}}\right)$

$q_{0}=a_{0}-\alpha _{0}\left({\frac {a_{1}-b_{1}}{\alpha _{1}-\beta _{1}}}\right)$

or, if $\alpha _{1}=\beta _{1}$ and $a_{1}=b_{1}$ and $\alpha _{0}\neq \beta _{0}$ to:

$p_{1}={\frac {a_{0}-b_{0}}{\alpha _{0}-\beta _{0}}}$

$q_{1}=a_{1}-\alpha _{1}\left({\frac {a_{1}-b_{1}}{\alpha _{1}-\beta _{1}}}\right)$

$q_{0}=a_{0}-\alpha _{0}\left({\frac {a_{1}-b_{1}}{\alpha _{1}-\beta _{1}}}\right)$

Thus:

$x_{mn}=\left[\left({\frac {a_{1}-b_{1}}{\alpha _{1}-\beta _{1}}}\right)n+\left(a_{1}-\alpha _{1}\left({\frac {a_{1}-b_{1}}{\alpha _{1}-\beta _{1}}}\right)\right)m+\left(a_{0}-\alpha _{0}\left({\frac {a_{1}-b_{1}}{\alpha _{1}-\beta _{1}}}\right)\right)\right]\delta _{\left|n-\left({\frac {\alpha _{1}+\beta _{1}}{2}}\right)m-\alpha _{0}\right|,\left|{\frac {\alpha _{1}-\beta _{1}}{2}}\right|m}$

${\textrm {if}}\quad \alpha _{0}=\beta _{0}\quad {\textrm {and}}\quad a_{0}=b_{0}\quad {\textrm {and}}\quad \alpha _{1}\neq \beta _{1}$

or:

$x_{mn}=\left[\left({\frac {a_{0}-b_{0}}{\alpha _{0}-\beta _{0}}}\right)n+\left(a_{1}-\alpha _{1}\left({\frac {a_{1}-b_{1}}{\alpha _{1}-\beta _{1}}}\right)\right)m+\left(a_{0}-\alpha _{0}\left({\frac {a_{1}-b_{1}}{\alpha _{1}-\beta _{1}}}\right)\right)\right]\delta _{\left|n-\alpha _{1}m-{\frac {(\alpha _{0}+\beta _{0})}{2}}\right|,\left|{\frac {(\alpha _{0}-\beta _{0})}{2}}\right|}$

${\textrm {if}}\quad \alpha _{1}=\beta _{1}\quad {\textrm {and}}\quad a_{1}=b_{1}\quad {\textrm {and}}\quad \alpha _{0}\neq \beta _{0}$

otherwise not reducible to one kronecker delta.

Order 2 coefficients

If

$x_{mn}=(a_{2}m^{2}+a_{1}m+a_{0})\,\delta _{n,\alpha _{1}m+\alpha _{0}}+(b_{2}m^{2}+b_{1}m+b_{0})\,\delta _{n,\beta _{1}m+\beta _{0}},$

then combined form will be

$x_{mn}=\left(p_{2}n^{2}+p_{1}n+q_{2}m^{2}+q_{1}m+q_{0}\right)\delta _{\left|n-{\frac {(\alpha _{1}+\beta _{1})m+(\alpha _{0}+\beta _{0})}{2}}\right|,\left|{\frac {(\alpha _{1}-\beta _{1})m+(\alpha _{0}-\beta _{0})}{2}}\right|}$

where the equations to solve are:

$\alpha _{1}^{2}p_{2}+q_{2}=a_{2}$

$\beta _{1}^{2}p_{2}+q_{2}=b_{2}$

$\alpha _{1}(p_{1}+2\alpha _{0}p_{2})+q_{1}=a_{1}$

$\beta _{1}(p_{1}+2\beta _{0}p_{2})+q_{1}=b_{1}$

$\alpha _{0}(p_{1}+\alpha _{0}p_{2})+q_{0}=a_{0}$

$\beta _{0}(p_{1}+\beta _{0}p_{2})+q_{0}=b_{0}$

This will be solvable if $\alpha _{0}=\beta _{0}$ and $a_{0}=b_{0}$ and $|\alpha _{1}|\neq |\beta _{1}|$ to:

$p_{2}={\frac {a_{2}-b_{2}}{\alpha _{1}^{2}-\beta _{1}^{2}}}$

$p_{1}={\frac {a_{1}-b_{1}}{\alpha _{1}-\beta _{1}}}-2\alpha _{0}p_{2}$

$q_{2}=a_{2}-\alpha _{1}^{2}p_{2}$

$q_{1}=a_{1}-\alpha _{1}\left(p_{1}+2\alpha _{0}p_{2}\right)$

$q_{0}=a_{0}-\alpha _{0}\left(p_{1}+\alpha _{0}p_{2}\right)$

$x_{mn}=\left(p_{2}n^{2}+p_{1}n+q_{2}m^{2}+q_{1}m+q_{0}\right)\delta _{\left|n-\left({\frac {\alpha _{1}+\beta _{1}}{2}}\right)m-\alpha _{0}\right|,\left|{\frac {\alpha _{1}-\beta _{1}}{2}}\right|m}$

or, if $\alpha _{1}=\beta _{1}\neq 0$ and $a_{2}=b_{2}$ and $\alpha _{0}\neq \beta _{0}$ to:

$p_{2}={\frac {1}{2\alpha _{1}}}{\frac {a_{1}-b_{1}}{\alpha _{0}-\beta _{0}}}$

$p_{1}={\frac {a_{0}-b_{0}}{\alpha _{0}-\beta _{0}}}-(\alpha _{0}+\beta _{0})p_{2}$

$q_{2}=a_{2}-\alpha _{1}^{2}p_{2}$

$q_{1}=a_{1}-\alpha _{1}\left(p_{1}+2\alpha _{0}p_{2}\right)$

$q_{0}=a_{0}-\alpha _{0}\left(p_{1}+\alpha _{0}p_{2}\right)$

$x_{mn}=\left(p_{2}n^{2}+p_{1}n+q_{2}m^{2}+q_{1}m+q_{0}\right)\delta _{\left|n-\alpha _{1}m-\left({\frac {\alpha _{0}+\beta _{0}}{2}}\right)\right|,\left|{\frac {\alpha _{0}-\beta _{0}}{2}}\right|}$

otherwise not reducible to one kronecker delta.

Order 3 coefficients

If

$x_{mn}=(a_{3}m^{3}+a_{2}m^{2}+a_{1}m+a_{0})\,\delta _{n,\alpha _{1}m+\alpha _{0}}+(b_{3}m^{3}+b_{2}m^{2}+b_{1}m+b_{0})\,\delta _{n,\beta _{1}m+\beta _{0}},$

then combined form will be

$x_{mn}=\left(p_{3}n^{3}+p_{2}n^{2}+p_{1}n+q_{3}m^{3}+q_{2}m^{2}+q_{1}m+q_{0}\right)\delta _{\left|n-{\frac {(\alpha _{1}+\beta _{1})m+(\alpha _{0}+\beta _{0})}{2}}\right|,\left|{\frac {(\alpha _{1}-\beta _{1})m+(\alpha _{0}-\beta _{0})}{2}}\right|}$

You do algebra yourself.