digital root by roupam ghosh

introduction

hello, my name is roupam ghosh
in this site i have stated a theorem that i have found as an application for digital root
For my other works in digital root please click here

mail me at:bappan420@yahoo.co.in

first we take some basic things into account

the digital root of a number (say n) is the single digit obtained after adding up its digits and adding up the digits of the result,and continuing thus upto a single digit
in this site we denote the digital root of a number(say n) by [n]
in this site we also denote 'not equal to' by <>
the numbers n,m,x and k assumed are all integers and for k only,k is greater than or equal to -9and less than or equal to 9
if n=9m+k then [n]=k
[[n]+[m]]=[n+m]
[[n]*[m]]=[n*m]
[[n]^m]=[n^m]

the theorem

the statement

if the xth term of a given sequence be u(x),then if we can find at least one pair of terms u(9m+k) and u(9n+k) such that [u(9m+k)]<>[u(9n+k)] then the xth term of the sequencecannot be expressed by a polynomial in x ie.,f(x)

proof

let we have found u(9m+k)and u(9n+k) such that [u(9m+k)]<>[u(9n+k)]
let the xth term of the given sequence can be expressed by a polynomial f(x),where f(x)=a+b*x+...+r*x^p
hence u(9m+k)=f(9m+k) and u(9n+k)=f(9n+k)
now,[u(9m+k)]=[f(9m+k)]
=[a+b*(9m+k)+...+r*(9m+k)^p]
=[[a]+[b*(9m+k)]+...+[r*(9m+k)^p]]
=[[a]+[[b]*[(9m+k]]+...+[[r]*[(9m+k)^p]]]
=[[a]+[[b]*[(9m+k)]]+...+[[r]*[(9m+k)]^p]]
=[[a]+[[b]*[k]]+...+[[r]*[k]^p]]
=[[a]+[[b]*[k]]+...+[[r]*[k^p]]]
=[[a]+[b*k]+...+[r*k^p]]
=[a+b*k+...+r*k^p]

similarly we will see that [u(9n+k)]=[f(9n+k)]=[a+b*k+...+r*^p]

hence,[u(9m+k)]=[u(9n+k)]
but we found that [u(9m+k)]<>[u(9n+k)]
hence,our supposition of f(x) must be wrong
hence the xth term of the sequence cannot be expressed by a polynomial f(x)