New Math Identity?

[cindy 2005]

Attached (vtSeries.jpg) is the math identity that I made up.

Example:

Let n = 3

it means that

[1^8+(1^8+2^8)+ (1^8+2^8+3^8)]/[1^4+(1^4+2^4)+ (1^4+2^4+3^4)] = 1/3(3^4+4*3^3+3^2-6*3+3) = 2446

 

You can try to let n = 2; result should be 130. And if n = 4, then result is 21146.

 

Is it a new one?

 

Cindy

[Turtle]

___I'm not sure idenity is the right term, rather it is an expression. When you start putting in values, it is an equation; when it is related to some structure or other, it is an identity.

___That said, it looks like it generates an interesting set. Does it relate to some structure?

___Offhand, what happens with negative numbers or fractions? ;) ;)

[Qfwfq]

A couple of years ago, instead of reading on commuter trains, I worked out a method for the summation of the first n naturals, each raised to a given exponent. What you have at numerator and denominator are these, further summed on increasing n, for exponent 8 at the numerator and 4 at the denominator. Here are the expressions for 4 and 8:

 

4: n(n + 1)(6n3 + 9n2 + n - 1)/30

 

8: n2(n + 1)2 (n4 + 2n3 - n2/3 - 4n/3 + 2/3)/8

 

sum these and take their ratio, I would expect it to give an expression equivalent to that on the rhs of your identity.

 

It is an identity, Turtle. Cindy means that the two sides are equal, for any n in N.

[cindy 2005]

A couple of years ago, instead of reading on commuter trains, I worked out a method for the summation of the first n naturals, each raised to a given exponent. What you have at numerator and denominator are these, further summed on increasing n, for exponent 8 at the numerator and 4 at the denominator. Here are the expressions for 4 and 8:

 

4: n(n + 1)(6n3 + 9n2 + n - 1)/30

 

8: n2(n + 1)2 (n4 + 2n3 - n2/3 - 4n/3 + 2/3)/8

 

sum these and take their ratio, I would expect it to give an expression equivalent to that on the rhs of your identity.

 

It is an identity, Turtle. Cindy means that the two sides are equal, for any n in N.

 

If you take a look on expression I post in jpg file (see attached) that is different on your expressions:

4: n(n + 1)(6n3 + 9n2 + n - 1)/30 means 1^4+2^4+3^4 +...+n^4. However, the statement in jpg file is expressed in form:

1^4 + (1^4+2^4) + (1^4+2^4+3^4) +(1^4+2^4+3^4+4^4)+...+(1^4+2^4^+3^4+4^4+ ...+ n^4), which is not equal to n(n + 1)(6n3 + 9n2 + n - 1)/30

 

You see the repeat number in each term. And if you take the ratio of expression 8 and 4, it cannot equal to expression that I have posted.

[Qfwfq]

I guess I was a bit too brief:

 

"further summed on increasing n"

 

meant: sum these expressions with n = 1, n = 2, n = 3 .... n = n, if you see what I mean.

 

On a math test, one would have to use two different names like n and m, but here we can get away with murder. ;)

 

I also forgot to add the LaTeX symbol ^ for raising to exponent. So the denominator is:

 

1(1 + 1)(6*1^3 + 9*1^2 + 1 - 1)/30 + 2(2 + 1)(6*2^3 + 9*2^2 + 2 - 1)/30 + ... +

+ n(n + 1)(6n^3 + 9n^2 + n - 1)/30

 

And similarly for the numerator. There will be an overall 15/4 times any other factors that might pop up. I haven't tried working out the algebra. I don't think I'll do it here at work, despite not being busy!

[Qfwfq]

I had copied the two expressions from a notebook at end lunch hour, I hope I didn't get them wrong! It's lucky I have that notebook in my case. ;)

[Qfwfq]

Offhand, what happens with negative numbers or fractions?

I had missed your query Turtle. I was also meaning to say, this stuff pertains to what has been termed "umbralitic" calculus. Nnnnnoooo! Nothing to do with an umbrella!!! ;)

 

Umbralitic from the Latin for 'shadow' because it's the "shadow" of real-valued calculus on the naturals, or integers too I guess. I was told this by a friend to whom I talked of my achievement when I had accomplished it and was only wondering about the roots of the polynomials (this turns out to be a very tricky thing). Upon this guy's info I then found stuff on the web, including a 1980 publication of more or less what I had done.

 

The reason I had got onto the endeavour was ideed because I had reasoned, knowing the well-known and easily proven summation for exponent 1 which is n(n + 1)/2, that this bears analogy to the integration of x in dx and, for arbitrary exponent, it presumably could be an expression analogous to the integration of x^e in dx: (x^(e + 1))/(n + 1)

 

It turns out this is so, I found a proof and a highly recursive recipe for finding the polynomial, which must then be divided by n+1. It also means Cindy's numerator and denominator are umbralitically the integral of the integral of x^e which, now I think of it, gives me a vague idea on how it might be worked out.

 

It took a lot of doing, but I've seen there can be many approaches, leading to different recipes. I worked out results up to e = 8 (just fitting Cindy's purpose!) and the meet results I found on the web, which I have seen for n up to 10. My recipe demands increasing sweat, for increasing exponent.

 

Anyway, to answer your question, fractional i. e. rational values of n means that you get near the set of reals, in which the rationals are topologically dense. Therefore, if by summation over n you mean integration in the measure dx restricted from R to Q, what you get is the "normal" intergral of x^e!!! ;)

[Turtle]

___I clearly went in over my head; thanks for throwing a line Qwfwq!

___A couple years ago I developed an expression for the nth s-sided polygonal/figurate number. I don't have it at hand (it's packed for moving), but if you think that it's something worthwhile I'll dig it out.

___I am fit to be tide. ;)

[Qfwfq]

Friday I posted the expression for the 7th power instead of that for the 8th, I should have at least thought to check it for being of the 9th degree.

:eek: :eek:

Oh, well... :eek: burn my house down and expose me to public derision! My old notes are a bit scribbled.

 

The necessary expressions are all at Mathworld:

 

http://mathworld.wolfram.com/PowerSum.html

 

and they are (19) through (28) or (29) through (38). Saturday I worked it all out and it matches up with Cindy's rhs, so verifying the identity.

 

Take expressions (36) and (32) and simplify by the common factors n(n + 1)(2n + 1) (I hadn't thought of that Friday!), dividing the numeric factors you get the 1/3, replace the power of n in each monomial with the approriate expression (19) through (24) and rewrite them as polynomials. Instead of performing a tedious polynomial division it is quicker to verify the known ratio, the rhs in Cindy's attachment, multiplying it by the denominator and checking it against the numerator.

[cindy 2005]

Friday I posted the expression for the 7th power instead of that for the 8th, I should have at least thought to check it for being of the 9th degree.

:) :)

Oh, well... :) burn my house down and expose me to public derision! My old notes are a bit scribbled.

 

The necessary expressions are all at Mathworld:

 

http://mathworld.wolfram.com/PowerSum.html

 

and they are (19) through (28) or (29) through (38). Saturday I worked it all out and it matches up with Cindy's rhs, so verifying the identity.

 

Take expressions (36) and (32) and simplify by the common factors n(n + 1)(2n + 1) (I hadn't thought of that Friday!), dividing the numeric factors you get the 1/3, replace the power of n in each monomial with the approriate expression (19) through (24) and rewrite them as polynomials. Instead of performing a tedious polynomial division it is quicker to verify the known ratio, the rhs in Cindy's attachment, multiplying it by the denominator and checking it against the numerator.

 

Hi,

 

I was out of town and got back. Your solution is good for this case (8)/(4) (I use your notation) only. However, other cases (such as (9)/(5), etc.) may require more works to do and it is difficult to predict the result.

Thanks,

Cindy

[Qfwfq]

However, other cases (such as (9)/(5), etc.) may require more works to do and it is difficult to predict the result.

Obviously! The method is however the same and the formulae in that article suffice for many cases. The similar formulae for higher exponents would need to be worked out for other cases.

 

:)

[cindy 2005]

Obviously! The method is however the same and the formulae in that article suffice for many cases. The similar formulae for higher exponents would need to be worked out for other cases.

 

:)

 

 

Wow, I think you may give a solution for case you just have done for power (8). Let's find the expression of

1^8+(1^8+2^8) + (1^8+2^8+3^8) + ...+ 1^8+2^8+3^8+...+n^8) = ? Can you express this in term of n, which I can't get it.

 

Cindy

[Qfwfq]

Using (26) in the Mathworld page, replace each power of n with the expression for summation of that power. As (29) has the powers 1, 3, 5, 7, 8 and 9, you need expressions (19), (21), (23), (25), (26) and (27) so multiply each of these by the coefficient in (26) tidy up the polynomial and divide the whole thing by 90.

 

For a ratio such as in your initial post, it's less tedious to factor out n(n + 1) and any other common factors, looking at the factored forms of those expressions.

[cindy 2005]

Using (26) in the Mathworld page, replace each power of n with the expression for summation of that power. As (29) has the powers 1, 3, 5, 7, 8 and 9, you need expressions (19), (21), (23), (25), (26) and (27) so multiply each of these by the coefficient in (26) tidy up the polynomial and divide the whole thing by 90.

 

For a ratio such as in your initial post, it's less tedious to factor out n(n + 1) and any other common factors, looking at the factored forms of those expressions.

 

I just checked through the ways you hint but it did not work. I don't get the result as you stated. The powers I found are 1, 3, 5, 7 and 9 (did not have 8). Why do we have to start with the expression (26)? I believe that the expression

1^8+(1^8+2^8) + (1^8+2^8+3^8) + ...+ 1^8+2^8+3^8+...+n^8) = ? can be reduced in terms of n (with small number of terms).

 

The way you stated might not work well to to final statement because it involves a lot of powers to simplify, and it makes the expression is "too big" to reduce. If you already reduce the expression, please post your result of

1^8+(1^8+2^8) + (1^8+2^8+3^8) + ...+ 1^8+2^8+3^8+...+n^8) = ?

Not consider "ratio" of (8) and (4). We need to solve the power (8) only as indicated the expression above in terms of n.

[Qfwfq]

But, your expression

 

1^8+(1^8+2^8) + (1^8+2^8+3^8) + ...+ 1^8+2^8+3^8+...+n^8)

 

is the lhs of (26) repeated for increasing n (n = 1, n = 2,..., n = n). The rhs of (26) has the powers 1, 3, 5, 7, 8 and 9. With a bit of patience you get a polynomial of the 10th degree. :xx:

 

10*(27) + 45*(26) + 60*(25) - 42*(23) + 20*(21) - 3*(19) all divided by 90

[cindy 2005]

But, your expression

 

1^8+(1^8+2^8) + (1^8+2^8+3^8) + ...+ 1^8+2^8+3^8+...+n^8)

 

is the lhs of (26) repeated for increasing n (n = 1, n = 2,..., n = n). The rhs of (26) has the powers 1, 3, 5, 7, 8 and 9. With a bit of patience you get a polynomial of the 10th degree. :xx:

 

10*(27) + 45*(26) + 60*(25) - 42*(23) + 20*(21) - 3*(19) all divided by 90

 

Yes. Your result is correct, but need to reduce to simple form.

 

Thanks,

 

Cindy

[Qfwfq]

reduce to simple form.

C'mon, Cindy it only takes a bit of patience!

 

I'm not busy now so I tried it, I get:

 

(n^10)/90 + (n^9)/9 + 5(n^8)/12 + 2(n^7)/3 - 7(n^6)/45 - 7(n^5)/15 - 5(n^4)/18 +

+ 5(n^3)/18 + 2(n^2)/9 - (n^1)/15

 

You can certainly write it factored a bit if you prefer.