pansano conjecture

PANSANO CONJECTURE YOWAN PRADHAN MAXKAY [email protected] ABSTRACT: By analyzing the paper of Pansano, I have solved his conjecture by using (1 − 𝑥 𝑓(𝑛 ) )/𝑥 1−𝑔(𝑛) where f(n) and g(n) are independent of x and vice versa and have given some special properties of his identity. INTRODUCTION THEOREM Let n be the non negative integer, then there exist a function f(n) and g(n) which is independent of x then 1 𝑛 𝑘 (𝑓 𝑛 𝑘+𝑔 𝑛 )

𝑛 𝑘 𝑘=0 (−1)

= (𝑓 𝑛 )𝑛 𝑛!

1 𝑛 𝑘=0 (𝑓 𝑛 𝑘+𝑔 𝑛 )

PROOF Firstly, we know that (1)

(1 − 𝑥 𝑓(𝑛) ) =

𝑛 𝑘

𝑛 𝑘 𝑘=0 (−1)

𝑥𝑓

𝑛 𝑘

Where f(n) is independent of x and vice versa and n is non-negative integer. Dividing both side of (1) by 𝑥 1−𝑔(𝑛) ,where g(n) is also independent of x and so on. (1 − 𝑥 𝑓(𝑛) )/𝑥 1−𝑔(𝑛) =

(2)

𝑛 𝑘 𝑘=0(−1)

𝑛 𝑘

𝑥𝑓

𝑛 𝑘+𝑔 𝑛 −1

Now, integrating both side we have 1 (1 0

(3)

− 𝑥 𝑓(𝑛) )

1 𝑥 1−𝑔 𝑛

𝑑𝑥 =

𝑛 𝑘 𝑘=0 (−1)

𝑛 𝑘

1 (𝑓 𝑛 𝑘+𝑔 𝑛 )

From L.H.S we have (4)

1 (1 0

− 𝑥 𝑓(𝑛) )

1 𝑥 1−𝑔 𝑛

𝑑𝑥 =

1 0

𝑥 𝑚 (1 − 𝑥)𝑛 𝑑𝑥 =𝛽(1 + 𝑚 ,1 + 𝑛)

Where m =1-g(n)/f(n) as we know both the function are independent of x then m is also independent of x. So m is function of n. Then , (5)

1 0

𝑥 𝑚 (1 − 𝑥)𝑛 𝑑𝑥 =𝛽 1 + 𝑚 ,1 + 𝑛 =

Now, we easily know that

𝛤 𝟏+𝒎 𝜞 𝟏+𝒏 𝛤(𝑛+𝑚 +2)

𝛤 𝟏+𝒎

(6)

𝛤(𝑛+𝑚 +2)

𝑛 𝑘=0 1/(1

=

+ 𝑘 + 𝑚)

By using (4) we get 1 (1 0

(7)

− 𝑥 𝑓(𝑛) )

1 𝑥 1−𝑔 𝑛

𝑛 𝑘 =0 1/(1

𝑑𝑥 =n!

+ 𝑘 + 𝑚)

Where m =1-g(n)/f(n) (8) By using (3) we get

𝑛 𝑘 𝑘=0 (−1)

(9)

𝑛 𝑘 𝑘=0 (−1)

(10)

𝑛 𝑘

𝑛 𝑘

1 (𝑓 𝑛 𝑘+𝑔 𝑛 )

1 (𝑓 𝑛 𝑘+𝑔 𝑛 )

= n!

1 𝑛 𝑘=0 𝑓 𝑛 𝑘+𝑔 𝑛 𝑓 𝑛

= (𝑓 𝑛 )𝑛 n!

1 𝑛 𝑘=0 𝑓 𝑛 𝑘+𝑔(𝑛)

Here we see that the correct form of Pansano conjecture is given by 𝑛 𝑘 𝑘=0 (−1)

𝑛 𝑘

1 (𝑓 𝑛 𝑘+𝑔 𝑛 )

= (𝑓 𝑛 )𝑛 n!

1 𝑛 𝑘=0 𝑓 𝑛 𝑘+𝑔(𝑛)

This satisfies all the identities given by Pansano. PROPERTIES OF IDENTITY: (A)

𝑛 𝑘 𝑘=0 (−1)

𝑛 𝑘

1 (𝑎𝑛𝑘 −𝑛+1)

= n!

1 𝑛 𝑘 =0 𝑎𝑛𝑘 −𝑛+1

Proof: By using expression (10) if we replace f(n) = an and g(n) = 1-n then we get result 𝑛 𝑘 𝑘=0 (−1)

(B)

𝑛 𝑘 𝑘=0(−1)

𝑛 𝑘

1 (𝑛𝑘 +𝑛−1)

𝑛 𝑘

1 (𝑎𝑛𝑘 −𝑛+1)

= (𝑛)𝑛 n!

= (𝑎𝑛)𝑛 n!

1 𝑛 𝑘=0 𝑎𝑛𝑘 −𝑛+1

1 𝑛 𝑘 =0 𝑛𝑘 +𝑛−1

Proof: By using expression (10) if we replace f(n) = n and g(n) =n-1 then we get the result 𝑛 𝑘 𝑘=0 (−1)

𝑛 𝑘

1 (𝑛𝑘 +𝑛−1)

= (𝑛)𝑛 n!

1 𝑛 𝑘 =0 𝑛𝑘 +𝑛−1

Now using expression (10) we define a unique propertie which is as :

Let 𝐻𝑛 (f , g) =

𝑛 𝑘

𝑛 𝑘 𝑘=0(−1)

1 (𝑓 𝑛 𝑘+𝑔 𝑛 )

(C) If a, b and c are function of n, then 𝐻𝑛 (a , b/c) - 𝑐 𝑛 −1 𝐻𝑛 (ac , b) = 0. Proof: From the expression (10) if we replace f = a and g =b/c then we have 𝐻𝑛 (a , b/c) =

𝑛 𝑘 𝑘=0 (−1)

𝑛 𝑘

1 (𝑎𝑘 +𝑏/𝑐)

which implies that

𝐻𝑛 (a , b/c) = 𝑐 𝑛−1 𝐻𝑛 (ac , b) Therefore, 𝐻𝑛 (a , b/c) - 𝑐 𝑛−1 𝐻𝑛 (ac , b) = 0. Proved CONCLUSION: While proving the Pansano conjecture we found actual and correct form of his conjecture and we also give some identity properties of this identity. REFERENCE (A) Pansano Brett, An interesting identities, http://arxiv.org/abs/1503.04678 (B) [I.S. Gradshteyn and I.M. Ryzhik, Table of Integrals, Series, and Products, Academic Press, INC., 1980]