### a to the b power again...

...but a much deeper result. I've known this since 1965, but have no idea if it is well known--I've never checked.

And if it doesn't boggle your mind, either you knew it already or you're not paying attention.

We start by defining the relevant syntactic operator:

In our new notation, the constraint in "a to the b power I" is:

(

We'll be applying

Without carefully crafted syntax rules, we might construe "

We define an evaluation operator

x=0: 1 - 4 + 6 - 4 + 1 = 0

x=1: 0 - 4 + 12 - 12 + 4 = 0

x=2: 0 - 4 + 24 - 36 + 16 = 0

x=3: 0 - 4 + 48 - 108 + 64 = 0

x=4: 0 - 4 + 96 - 324 + 256 = 24

x=5: 0 - 4 + 192 - 972 + 1024 = 240

x=6: 0 - 4 + 384 - 2916 + 4096 = 1560

1 1^x - 4 2^x + 6 3^x - 4 4^x + 1 5^x =

=

=

because x (1-x)^4 = (1 - (1-x)) (1-x)^4 = (1-x)^4 - (1-x)^5.

Each term has at least 4 initial zeros, and so

(1-x)^4 - (1-x)^6 = (2 x - x^2) (1-x)^4,

therefore x^2 (1-x)^4 = 2 x (1-x)^4 - ((1-x)^4 - (1-x)^6).

1 2^x - 4 3^x + 6 4^x - 4 5^x + 1 6^x =

=

= 2

Again, each term has at least 4 initial zeros, and so

So it looks like...

These forms are very like multigrades, which are only allowed coefficients of +1 and -1. Example:

1^x + 2^x + 4^x + 7^x = 3^x + 5^x + 6^x, x = 1,2.

1^x + 2^x + 4^x + 7^x = 0^x + 3^x + 5^x + 6^x, x = 0,1,2.

Note: setting 0^0 = 1 is inherited from the binomial expansion of (1-x)^1.

*I*know it, gimme that old time religion, that's good enough for me. Nevertheless, if you, dear reader, can cite it elsewhere or have good reason to think it's new, please let me know.And if it doesn't boggle your mind, either you knew it already or you're not paying attention.

We start by defining the relevant syntactic operator:

*[x^n] = n^x.***Z***operates on abstract algebraic polynomials.***Z***distributes through terms and ignores coefficients:***Z***[a w^m + b x^n] = a m^w + b n^x.***Z**In our new notation, the constraint in "a to the b power I" is:

(

*- 1)[a^b] = 0.***Z**We'll be applying

*to expansions of powers of 1-x yielded by the binomial theorem. For clarity, everything will be illustrated for the fourth power; the obvious generalizations are all true.***Z**Without carefully crafted syntax rules, we might construe "

*[(1-x)^4] = 4^(1-x)"--not what we want.***Z**We define an evaluation operator

*yielding an abstract algebraic polynomial in one or more distinct atomic variables ("x" in what follows). In the absence of an explicit specific substitution for the abstract variable(s),***E***[...] yields a form not subject to further evaluation, thus suitable for the application of***E***.***Z***[(1-x)^4] = 1 x^0 - 4 x^1 + 6 x^2 - 4 x^3 + 1 x^4***E***[(1-x)^4] = 1 0^x - 4 1^x + 6 2^x - 4 3^x + 1 4^x***ZE**x=0: 1 - 4 + 6 - 4 + 1 = 0

x=1: 0 - 4 + 12 - 12 + 4 = 0

x=2: 0 - 4 + 24 - 36 + 16 = 0

x=3: 0 - 4 + 48 - 108 + 64 = 0

x=4: 0 - 4 + 96 - 324 + 256 = 24

x=5: 0 - 4 + 192 - 972 + 1024 = 240

x=6: 0 - 4 + 384 - 2916 + 4096 = 1560

*[(1-x)^4] has 4 initial zeros. Similarly,***ZE***[(1-x)^5] has 5 initial zeros. Therefore:***ZE**1 1^x - 4 2^x + 6 3^x - 4 4^x + 1 5^x =

*[x (1-x)^4]***ZE**=

*[(1-x)^4 - (1-x)^5]***ZE**=

*[(1-x)^4] -***ZE***[(1-x)^5]***ZE**because x (1-x)^4 = (1 - (1-x)) (1-x)^4 = (1-x)^4 - (1-x)^5.

Each term has at least 4 initial zeros, and so

*[x (1-x)^4] likewise has 4 initial zeros.***ZE***[x (1-x)^4](4) = 24.***ZE***[x (1-x)^4](5) = 360.***ZE**(1-x)^4 - (1-x)^6 = (2 x - x^2) (1-x)^4,

therefore x^2 (1-x)^4 = 2 x (1-x)^4 - ((1-x)^4 - (1-x)^6).

1 2^x - 4 3^x + 6 4^x - 4 5^x + 1 6^x =

*[x^2 (1-x)^4]***ZE**=

*[2 x (1-x)^4 - ((1-x)^4 - (1-x)^6)]***ZE**= 2

*[x (1-x)^4] -***ZE***[((1-x)^4] -***ZE***[(1-x)^6)].***ZE**Again, each term has at least 4 initial zeros, and so

*[x^2 (1-x)^4] likewise has 4 initial zeros.***ZE***[x^2 (1-x)^4](4) = 24.***ZE***[x^2 (1-x)^4](5) = 480.***ZE**So it looks like...

*[x^m (1-x)^n](k) = 0, 0 <= m, 0 <= k < n.***ZE***[x^m (1-x)^n](k) = n!, 0 <= m, 0 <= k = n.***ZE***[x^m (1-x)^n](k) = ??, 0 <= m, 0 <= n < k.***ZE**These forms are very like multigrades, which are only allowed coefficients of +1 and -1. Example:

1^x + 2^x + 4^x + 7^x = 3^x + 5^x + 6^x, x = 1,2.

1^x + 2^x + 4^x + 7^x = 0^x + 3^x + 5^x + 6^x, x = 0,1,2.

Note: setting 0^0 = 1 is inherited from the binomial expansion of (1-x)^1.

ZE[(1-x)^n](k) = k! S(k,n), where S(k,n) is a Stirling number of the second kind (see the "Explicit Formula" section).

ReplyDeleteWhat you're actually counting here is the number of surjective functions from a k element set to a n element set (which is why you get 0 when k<n and n! when k=n).

Hey, thanks. Good to know. Must check out those Stirling numbers.But there's lots more: see next "a to the b power" post.

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