Relating Coefficients of n Length to Integers of All-Ones Digits of n Length

Today we’ll be taking a purely mathematical bent, and be exploring an interesting phenomenon concerning the coefficients of expanded form n-length polynomials and numbers consisting of all-ones digits.

For any real polynomial of the form:

z_0 + z_1 + z_2...+ z_{n-1} + z_n

Such that Z is any real number or variable defined within the real numbers; it is possible to predict the coefficients of the polynomial when it is raised to any integer power p, taking the form:

(z_0 + z_1 + z_2...+ z_{n-1} + z_n)^p

In order to predict the coefficients this procedure is followed:

First observe that Pascal’s Triangle is formed like so:

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

…and so on

Pascal’s Triangle refers to the coefficients attached to each variable in a polynomial of the form:

(z_0 + z_1)^p

Where p is the value of the specific row of Pascal’s Triangle.

Also note that 11 raised to a power p yields Pascal’s Triangle for any length, with addition errors:

1=11^0

11=11^1

121=11^2

1331=11^3

14641=11^4

15101051=11^5 with summation errors

In order to correct for the summation errors in the powers of 11, the following substitution can be made:

Let x=10

Then 11=x+1

By raising  to differing integer powers p, the powers of Pascal’s Triangle are revealed to be the coefficients of the resultant polynomial:

(x+1)^0=11^0=1

(x+1)^1=11^1=11

(x+1)^2=11^2=121

(x+1)^3=11^3=1331

And so on, again noting that the pattern becomes non-obvious at p=5 and beyond.

By noting that powers of 111 yield values that represent the powers of polynomials of length n=3 , it might be possible to attempt to employ the following substitution:

x^2=100

x=10

Hypothesizing that (x^2 + x + 1)^p would yield the appropriate coefficients. When simplified this form is non-obvious. It is necessary to define a variable  and use y instead of  in order to make the expansion yield the appropriate coefficients in a form that is obvious:

y = x^2 = 100

(y + x + 1)^p

(y + x + 1)^0 = 111^0 = 111^0=1

(y + x + 1)^1 = 111^1 = 111^1=111

(y + x + 1)^2 = 111^2 = 111^2=12321

(y + x + 1)^3 = 111^3 = 111^3=1367631

Noting that the coefficients are non-obvious starting at p=2 .

The same pattern holds true for polynomials of length ; a new variable  must be defined and substituted in order for the expansion to yield the appropriate coefficients.

I hypothesize that this pattern holds true for any polynomial for any length.

Therefore in order to predict the appropriate coefficients for any polynomial of length n, for polynomial functions that take the form:

(z_0 + z_1 + z_2...+ z_{n-1} + z_n)^p

Let x=10

Then

(x_0 + x_1 + x_2...+ x_{n-1} + x_n)^p=(\displaystyle\sum_{i=0}^{n} 10^n)^p

Such that it’s expansion yields:

(a_0x_0 + a_1x_1 + a_2x_2...+ a_{n-1}x_{n-1} + a_nx_n)^p=(\displaystyle\sum_{i=0}^{n} 10^n)^p

The expansion of:

(z_0 + z_1 + z_2...+ z_{n-1} + z_n)^p

Will yield:

(a_0z_0 + a_1z_1 + a_2z_2...+ a_{n-1}z_{n-1} + a_nz_n)^p

Such that the coefficients are appropriate prior to substitution and simplification.

Substituting x_1 and x_2 (and so on) for z and generalizing the form will yield:

(x_0 + x_1 + x_2...+ x_{n-1} + x_n)^p

Which relates powers of all-ones numbers to the coefficients of polynomial functions.

This suggests that a decomposition may be possible, which would easily predict coefficients for any polynomial of length n for any integer power p. Note that a decomposition is not implied.

What this decomposition might be, I have no idea.

~ by Ethan Mendel on June 15, 2009.

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