I have 5 polynomials in which a and c are real numbers.
First polynomial is
ac(1-a)[(1-a)+1][(1-a)+2] [(1-a)+3].
Please note that the coefficients associated with (1-a) are (1,1,1) also
each factor associated with (1-a) increases by 1 from [(1-a)+0] to
[(1-a)+3].
Second polynomial is
(a^2)c(1+c){(1-a)[(1-a)+1][(1-a)+2]
+(1-a)[(1-a)+1][2(1-a)+2]
+(1-a)[2(1-a)+1][2(1-a)+2]
+2(1-a)[2(1-a)+1][2(1-a)+2]}.
Please note that the coefficients associated with (1-a) are (1,1,1),
(1,1,2), (1,2,2) and (2,2,2). It seems that these coefficients can be
generated by permutations of 1 and 2.
Third polynomial is
(a^3)c(1+c)(2+c){(1-a)[(1-a)+1]
+(1-a)[2(1-a)+1]
+(1-a)[3(1-a)+1]
+2(1-a)[2(1-a)+1]
+2(1-a)[3(1-a)+1]
+3(1-a)[3(1-a)+1]}.
Please note that the coefficients associated with (1-a) are (1,1), (1,2),
(1,3), (2,2), (2,3) and (3,3).
Fourth polynomial is
(a^4)c(1+c)(2+c)(3+c)[(1-a)+2(1-a)+3(1-a)+4(1-a)].
Fifth polynomial is
(a^5)c(1+c)(2+c)(3+c)(4+c).
After I sum and simplify all these five polynomials, the summation is
equal to ac(1+ac)(2+ac)(3+ac)(4+ac) which can be expressed by the
Pochhammer symbol.
Please note that the first factor of the 5 polynomials are a, (a^2),
(a^3), (a^4) and (a^5).
The factors associated with c in these 5 polynomials are c, c(1+c),
c(1+c)(2+c), c(1+c)(2+c)(3+c) and c(1+c)(2+c)(3+c)(1+4).
The equality between these 5 polynomials and ac(1+ac)(2+ac)(3+ac)(4+ac)
can be expanded to 6 polynomials and ac(1+ac)(2+ac)(3+ac)(4+ac)(5+ac) and
so on.
I try to prove it but I just could not figure it out.
However, I do believe the proof involves Stirling numbers of the first
kind and the second kind.
Can anyone please teach me how to solve the problem?
Any reference is much appreciated.
Thanks.
Tom
.
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