Ballu Equation

Ballu equation provides the relation between the exponents and the factorials. This equation is obtained from nth order backward finite difference as well as through the numerical analysis of the power factorial table. The uniqueness of the equation is that it remains unity for any value variable existing in the equation within its domain. This equation extends its horizon into infinite values and negative factorials. And this equation is one of the few known infinite summation series that forms an indeterminate form and results in a specific unique real value solution other than Riemann's Paradox.

Power factorial table (PFT)

A numerical analysis with 1st column filled with a series of numbers raised with the difference of one and raised by power $n$ yields the factorial of $n$. The analysis follows the finite difference in backward order for all the columns until the column converges to value zero.

Example

For $n=0,1,2,3$

1			1										1			1			1										1			1			1			2									1			1			1			0			6
2			1										2			2			1										2			4			3			2									2			8			7			6			6
3			1										3			3			1										3			9			5			2									3			27			19			12			6
4			1										4			4			1										4			16			7			2									4			64			37			18			6
5			1										5			5			1										5			25			9			2									5			125			61			24			6

Equation example

Summation series for $n=3$ is obtained by the following order

The first column is mathematically expressed as$$x^3:(x-1{)}^3:(x-2{)}^3:(x-3{)}^3$$The second column is mathematically expressed as$$x^3-(x-1{)}^3:(x-1{)}^3-(x-2{)}^3:(x-2{)}^3-(x-3{)}^3$$The third column is mathematically expressed as$$x^3-(x-1{)}^3-[(x-1{)}^3-(x-2{)}^3]:(x-1{)}^3-(x-2{)}^3-[(x-2{)}^3-(x-3{)}^3]$$And the last column is expressed as$$x^3-(x-1{)}^3-(x-1{)}^3+(x-2{)}^3-[(x-1{)}^3-(x-2{)}^3-(x-2{)}^3+(x-3{)}^3]$$Any row value of the last column is 6 and so, above equation can be reduced to$$x^3-3(x-1{)}^3+3(x-2{)}^3-(x-3{)}^3=6$$This equation can be simplified as$$(\genfrac{}{}{0ex}{}{3}{0})x^3-(\genfrac{}{}{0ex}{}{3}{1})(x-1{)}^3+(\genfrac{}{}{0ex}{}{3}{2})(x-2{)}^3-(\genfrac{}{}{0ex}{}{3}{3})(x-3{)}^3=6$$So, above equation is simplified in the form of summation series as$$\sum _{k=0}^3(-1{)}^k\frac{3!(x-k{)}^3}{(3-k)!k!}=3!$$or$$\sum _{k=0}^3(-1{)}^k\frac{(x-k{)}^3}{(3-k)!k!}=1$$

Generalized form

The general form of the PFT is expressed mathematically as$$\sum _{k=0}^n(-1{)}^k\frac{n!(x-k{)}^n}{(n-k)!k!}=n!$$or$$\sum _{k=0}^n(-1{)}^k\frac{(x-k{)}^n}{(n-k)!k!}=1$$

Special case

For $n=0$, the first column and the last column are the same. This implies the question equation itself is the solution equation. And $0^0$ is an indeterminate form which implies it has multiple solutions and in this context, the value of $0^0=1$.

n^th order binomial transform

Ballu equation is also obtained by binomial transform. The binomial transform, T, of a sequence, {a_n}, is the sequence {s_n} defined by

$$s_n=\sum _{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})a_k$$

Using the finite differences and nth order backward difference of binomial transform result in the Ballu equation given by$$\sum _{k=0}^n(-1{)}^k\frac{(x-k{)}^n}{(n-k)!k!}=1:\{\begin{array}{ll}x\in \mathbb{ℝ}& \\ n\in \mathbb{𝕎}\end{array}$$The Ballu equation always remains unity. Other forms of Ballu equation are$$\sum _{k=0}^n(-1{)}^k\frac{(x-k{)}^{n-m}}{(n-k)!k!}=1$$and$$\sum _{k=0}^{n-m}(-1{)}^k\frac{(x-k{)}^{n-m}}{(n-k)!k!}=1$$where $m\in \mathbb{𝕎}$ and $m\le n$.

Application

Difference between Infinity and infinity

Isolating the positive and negative terms, Ballu equation is given by$$\sum _{k=0}^n\frac{{(x-2k)}^n}{(n-2k)!(2k)!}-\sum _{k=0}^n\frac{{\left(x-(2k+1)\right)}^n}{\left(n-(2k+1)\right)!(2k+1)!}=1$$and substituting $n=\mathrm{\infty }$ or $x=\mathrm{\infty }$ results in an indeterminate form with a unique solution as unity.$$\mathrm{\infty }-\mathrm{\infty }=1$$This equation is one of the few known infinite summation series that forms an indeterminate form of $\mathrm{\infty }-\mathrm{\infty }$ and results in a specific unique real value solution.

Zero raised to zero

Using power factorial table when $n=0$, entire row yields unity which implies $0^0=1$

Also, $0^0=1$ is the only solution that satisfies the Ballu equation for any given value of $n$.

Negative factorials

When $n=1$ and $m=1$ in the Ballu equation $$\sum _{k=0}^n(-1{)}^k\frac{(x-k{)}^{n-m}}{(n-k)!k!}=1$$The denominator ends up with negative factorial and it is found that $$(-1)!=\pm \frac{1}{0}$$ This is also interpreted as$$(-1)!=\pm \mathrm{\infty }$$Similarly, every negative factorial value ends up as $$(-x)!=\pm \frac{k}{0}$$where $x\in \mathbb{\mathbb{N}}$ and $k\in \mathbb{ℝ}$ and this implies

$$(-x)!=\pm \mathrm{\infty }$$The same results are obtained by gamma function.

References

1. Kolosov Petro, On the link between finite difference and derivative of polynomials, HAL.fr articles, 6 May 2017.
2. Balram A. Shah, 2020, Special equation from Binomial transform and n^th order finite difference, DOI:10.35543/osf.io/xze4u
3. Comparison of power of sequence of all natural number and the factorial
4. Relation between power and factorial

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