Can the Triangular Number Identity Be Simplified Further for Different Series?

[ramsey2879]

Let $$T_{n} = n*(n+1)/2$$ and n and m are integers. I discovered that

$$2*n+1 = \frac{(T_{(n-1)} -m)*(T_{(n+2)}-m) - (T_{(n-2)}-m)*(T_{(n+1)}-m)}{(T_{n} - m - 1)}$$ except for the case where the denominator is zero.

Is there a simple way to prove this identity?

 

[mathman]

Have you tried mathematical induction - it looks like it should work, but I haven't tried it. I am surprised that the expression is independent of m.

 

[CRGreathouse]

You can do a ton of simplification in the numerator -- I filled a sheet of paper doing so, though I'm loathe to post it in case of arithmetic mistakes. It's a fourth degree polynomial in two variables before simplification, and most drops out. Only a linear factor of m remains, for example, and you can collect the rest.

Of course numerical results can help convince you as well if that's your cup of tea instead.

 

[dodo]

I was lazy to try myself, but Mathematica (in the form of quickmath.com, menu Algebra / simplify) simplifies it to 2m+1. I supposed substitution and cancellation would do. Actually I thought it was a disguised homework. (Maybe I should try myself and stop talking.)

 

[ramsey2879]

I was lazy to try myself, but Mathematica (in the form of quickmath.com, menu Algebra / simplify) simplifies it to 2m+1. I supposed substitution and cancellation would do. Actually I thought it was a disguised homework. (Maybe I should try myself and stop talking.)

You must mean 2n+1, right? Thanks your word is all I need. I am too cheap to buy the program. I too was surprise that it is independent of m which is why I posted an otherwise trivia identity.

 

[ramsey2879]

You can do a ton of simplification in the numerator -- I filled a sheet of paper doing so, though I'm loathe to post it in case of arithmetic mistakes. It's a fourth degree polynomial in two variables before simplification, and most drops out. Only a linear factor of m remains, for example, and you can collect the rest.

Of course numerical results can help convince you as well if that's your cup of tea instead.

OK i will now give this a try since I have more free time> I must confess that I got a hint from another site where I orginally posted a similar problem but with 1 substituted for m. That was before I noticed that "1" could be any integer.
T(n) = A
T(n-1) = A - n
T(n-2) = A -2n + 1
T(n+1) = A +n+1
T(n+2) = A +2n+3

T(n)-m-1 = A-1-m
[T(n-1)-m]*[T(n+2)-m] = [A-n-m]*[A+2n+3-m]
[T(n-2)-m]*[T(n+1)-m] = [A-2n+1-m]*[A+n+1-m]

So the problem reduces to proving that

(A-n-m)(A+2n+3-m)-(A-2n+1-m)(A+n+1-m)=(A-1-m)(2n+1)

opps I have an errand to run Bye for now

 

[ramsey2879]

OK i will now give this a try since I have more free time> I must confess that I got a hint from another site where I orginally posted a similar problem but with 1 substituted for m. That was before I noticed that "1" could be any integer.
T(n) = A
T(n-1) = A - n
T(n-2) = A -2n + 1
T(n+1) = A +n+1
T(n+2) = A +2n+3

T(n)-m-1 = A-1-m
[T(n-1)-m]*[T(n+2)-m] = [A-n-m]*[A+2n+3-m]
[T(n-2)-m]*[T(n+1)-m] = [A-2n+1-m]*[A+n+1-m]

So the problem reduces to proving that

(A-n-m)(A+2n+3-m)-(A-2n+1-m)(A+n+1-m)=(A-1-m)(2n+1)

opps I have an errand to run Bye for now

Continuing
(A-n-m)(A+2n+3-m) -(A+n+1-m)(A-2n+1-m)
=((A-1-m)+1-n)*(A+2n+3-m) -((A-1-m)+2+n)*(A-2n+1-m)
=(A-1-m)*(4n+2) +(1-n)*(A+2n+3-m) -(2+n)*(A-2n+1-m)
=(A-1-m)*(4n+2) +(1-n)*(A-1-m +2n+4) -(2+n)*(A-1-m-2n+2)
=(A-1-m)*(2n+1) +(1-n)*(2n+4) - (2+n)(-2n+2)
=(A-1-m)*(2n+1) QED

 

[Hurkyl]

If you cross multiply, and then bring everything over to one side, you are left with the equation f = 0 for a polynomial f that is quadratic in m and quartic in n,

You can plug in 3 values of m to get three quartic polynomials in n. By interpolation, your identity holds if and only if these three quartics are all zero. (since f is quadratic in m, and we have 3 samples)

But any quartic in n is zero if and only if, whenever you select 5 values of n and plug them in, you get zero. (again, by interpolation)

So, choose 3 values of m and 5 values of n and plug all 15 combinations into your identity -- they will all work if and only if your identity works in general.



Actually, I think it's easy to see that f is no worse than cubic in n and linear in m, in which case you only need to try 2 values of m and 4 values of n, for 8 combinations in all.

 

[ramsey2879]

If you cross multiply, and then bring everything over to one side, you are left with the equation f = 0 for a polynomial f that is quadratic in m and quartic in n,

You can plug in 3 values of m to get three quartic polynomials in n. By interpolation, your identity holds if and only if these three quartics are all zero. (since f is quadratic in m, and we have 3 samples)

But any quartic in n is zero if and only if, whenever you select 5 values of n and plug them in, you get zero. (again, by interpolation)

So, choose 3 values of m and 5 values of n and plug all 15 combinations into your identity -- they will all work if and only if your identity works in general.



Actually, I think it's easy to see that f is no worse than cubic in n and linear in m, in which case you only need to try 2 values of m and 4 values of n, for 8 combinations in all.

Now I have a new generalization and need to know how many instances of the new variable $$b$$ must be checked for the following equation:
Let n,b,m be integers and
Let $$T_{n} = n(n+1)/2$$ and

$$D_{(b,m,n)} = (T_{(n-b)}-m)*(T_{(n+2b)}-m)-(T_{(n+b)}-m)*(T_{(n-2b)}-m)$$

then the following equation appears to hold:

$$b*(2n+1)*(T_{n}-m-b^{2}) = D_{(b,m,n)}$$

So far I checked more than the necessary eight instances (2 values of m, for each of four values of n) for b = 1,2,3, and 4

Does this qualify as a determination of the validity of the equation or do I have to check more values of b? Again, after dividing both sides by (T(n) - m - b^2), the left hand side is independent of the variable m!

 

[Hurkyl]

Well, the general principle at work here is that a polynomial of degree d is uniquely determined by its values at d+1 points.

Equivalently, if f(x) is a polynomial of degree at most d, and you select d+1 a's, then f(x) is the zero polynomial if and only if f(a) = 0 for each a.

This application follows by iteration; if f(x, y) is a polynomial of degree at most d in x, then if you select d+1 distinct a's, then f(x, y) is the zero polynomial if and only if f(a, y) is the zero polynomial for each a. And since f(a, y) is a polynomial in y...

So basically, if your polynomial is at most degree d in some variable, you just have to make sure to use d+1 values for that variable.

This latest identity looks like it's degree 4 in b. (The left hand side has degree 3, and each term on the right hand side has two factors of degree 2) So you need to use 5 values for b. Unless it's obvious that the highest degree b's cancel, just like the highest degree n's, in which case you know the R.H.S. is only degree 3 in b.

 

[ramsey2879]

Well, the general principle at work here is that a polynomial of degree d is uniquely determined by its values at d+1 points.

Equivalently, if f(x) is a polynomial of degree at most d, and you select d+1 a's, then f(x) is the zero polynomial if and only if f(a) = 0 for each a.

This application follows by iteration; if f(x, y) is a polynomial of degree at most d in x, then if you select d+1 distinct a's, then f(x, y) is the zero polynomial if and only if f(a, y) is the zero polynomial for each a. And since f(a, y) is a polynomial in y...

So basically, if your polynomial is at most degree d in some variable, you just have to make sure to use d+1 values for that variable.

This latest identity looks like it's degree 4 in b. (The left hand side has degree 3, and each term on the right hand side has two factors of degree 2) So you need to use 5 values for b. Unless it's obvious that the highest degree b's cancel, just like the highest degree n's, in which case you know the R.H.S. is only degree 3 in b.

Although I have a basic polynominal type equation, there is something of interest about it that applys to modulus operations, e.g.
$$(T_{(n-b)}-m)*(T_{(n+2b)}-m) == (T_{(n+b)}-m)*(T_{(n-2b)}-m) \mod 2n+1| n = index$$
Now triangular nymbers have a basic recursive formula as follows
$$T(n) = 2*T_{(n-1)}-T_{(n-2)} +1$$
Could it be possible to substitute a different integer a for $$T_{0}$$ or $$T_{1}$$ or different integers for both and get the same identity equation for the index operator?
Could there be another recursive series that would have the same identity equation for the index operator?
I am of the opinion that the answer is no to both questions but need too consider it further.

 

[ramsey2879]

I checked my last post and realized that I made a fundamental error.
Since m can be any integer then substituting $$A_{0} = T_{0} + c, A_{1} = T_{1} + c |c = integer$$ as the new starting numbers of the recursive series gives the same modulus equation since subtracting or adding c to each term of the series does not change the recurrence relation.

i.e. The T series is 0,1,3,6 ... T(n) = 2T(n-1) - T(n-2) + 1
The A series would be c+0,c+1,c+3,c+6, ... since 2c - c = c.

If on the otherhand you chose A(0) = c and A(1) = c+d+1 you essentially sifted the index by d and will get a new basic polynomial equation

$$b(2n+2d+1)(A_{n}-m-b^2) = D_{b,m,n}|A_{i}\ is\ substituted\ for\ T_{i}$$
So the mod for the new identity is 2n+2d+1

 

[ramsey2879]

I checked my last post and realized that I made a fundamental error.
Since m can be any integer then substituting $$A_{0} = T_{0} + c, A_{1} = T_{1} + c |c = integer$$ as the new starting numbers of the recursive series gives the same modulus equation since subtracting or adding c to each term of the series does not change the recurrence relation.

i.e. The T series is 0,1,3,6 ... T(n) = 2T(n-1) - T(n-2) + 1
The A series would be c+0,c+1,c+3,c+6, ... since 2c - c = c.

If on the otherhand you chose A(0) = c and A(1) = c+d+1 you essentially sifted the index by d and will get a new basic polynomial equation

$$b(2n+2d+1)(A_{n}-m-b^2) = D_{b,m,n}|A_{i}\ is\ substituted\ for\ T_{i}$$
So the mod for the new identity is 2n+2d+1

Still another variable "w" to add to the mix.
Let A(0) = 0, A(1) = 1, A(n) = 2*A(n-1)-A(n-2) + w
If w = 1 then you have the triangular series
If w = 2 then you have the series of squares
In general $$A_n = w*(n^2 -n)/2 + n$$ is a direct formula for $$A_n$$

redefine $$D_{(b,n,m)$$ as:

$$(A_{(n-b)}-m)*(A_{(n+2b)}-m) - (A_{(n+b)}-m)*(A_{(n-2b)}-m)$$

Then
$$D_{(b,n,m)} = b(w*(2n-1)+2)*(A_{n} -wb^{2} -m)$$

 

[ramsey2879]

Still another variable "w" to add to the mix.
Let A(0) = 0, A(1) = 1, A(n) = 2*A(n-1)-A(n-2) + w
If w = 1 then you have the triangular series
If w = 2 then you have the series of squares
In general $$A_n = w*(n^2 -n)/2 + n$$ is a direct formula for $$A_n$$

redefine $$D_{(b,n,m)$$ as:

$$(A_{(n-b)}-m)*(A_{(n+2b)}-m) - (A_{(n+b)}-m)*(A_{(n-2b)}-m)$$

Then
$$D_{(b,n,m)} = b(w*(2n-1)+2)*(A_{n} -wb^{2} -m)$$

It could be interesting to select different n,b,w and m to make the determinant D(b,n,m) zero. I might add that b and n are counters and must be integers while w and m are not and can be complex. It is easy to see that a complex number m can not be selected to make the determinant D(b,n,m) zero unless w is also complex, but could it be posible to have w be a complex number and yet make the determinant zero with a integer m?

 

[ramsey2879]

It could be interesting to select different n,b,w and m to make the determinant D(b,n,m) zero. I might add that b and n are counters and must be integers while w and m are not and can be complex. It is easy to see that a complex number m can not be selected to make the determinant D(b,n,m) zero unless w is also complex, but could it be posible to have w be a complex number and yet make the determinant zero with a integer m?

The answer to my question is yes. Does anyone care to provide suitable positive values for both n and b such that w is any complex or irrational number, m is an integer and the determinant D(b,n,m) is zero?

 

[ramsey2879]

The answer to my question is yes. Does anyone care to provide suitable positive values for both n and b such that w is any complex or irrational number, m is an integer and the determinant D(b,n,m) is zero?

Since $$D=b(w(2n-1)+2)*(A_{n}-wb^2-m)$$ the only hope to make D zero with w complex and integer b,n,m is to make the factor $$A_{n} -wb^{2}-m = 0$$

So we have m=n and
$$wT_{(n-1)} - wb^{2} = 0$$
This has the solutions (n,b):(1,0) (2,1), (9,6), (50,35), (289,204), (1682,1189) ...
b is the sequence http://www.research.att.com/~njas/sequences/A001109
n is the sequence http://www.research.att.com/~njas/sequences/A055997

In the formula A(n)-m = n + w(T(n-1))-n, the n's cancel and w^2 can be factored out of the determinant.

I determined that for zero determinants
T(n-1-b) = T(A055997(n)-1-A001109(n)) = T(A053141(n))
T(n-1-2b) =T(A055997(n)-1-2A001109(n)) = T(-A001652-1) = T(A001652(n))
T(n-1+b) =T(A055997(n)-1+A001109(n)) = T(A053141(n+1))
T(n-1+2b) =T(A055997(n)-1+2A001109(n)) =T(A001652(n+1))


A(053141) = {0,2,14,84,492,2870...}
A(001652) = {0,3,20,119,696, ...}

The notes to these sequences states that the triangular numbers for sequence A001652(n) are twice the corresponding triangular numbers for A053141(n) so obviously the determinants are each zero.

However, the determinants with just the arguments substituted for the actual triangular numbers except for the bottom right where the substitution of one more than the argument is made are all zero also. This is new.
For instance
|2,3|
|14,21|
is zero

 

[ramsey2879]

Further generalizing the last post
Let $$A_{n} = A(053141) =\{0,2,14,84,...6A_{(n-1)}-A_{(n-2)}+2\}$$
Let $$B_{n} = A(001652) =\{0,3,20,119,...6B_{(n-1)}-B_{(n-2)}+2\}$$
and
Let $$G_{n,m} = (2m+1)A_{n} + m$$
Let $$H_{n,m} = (2m+1)B_{n} + m$$

If $$T_{a} = a(a+1)/2$$ then

$$T_{H_{n,m}} + T_{m} = 2*T_{G_{n,m}}$$

Also $$T_{a} - T_{m} = b^2$$ where

$$a = \left(G_{n,m}+G_{n+1,m}\right)/2$$ and
$$b = \left(G_{n+1,m} - G_{n,m}\right)/2$$

If m = 0 then the formula gives the square triangular numbers.

 

[ramsey2879]

Still another variable "w" to add to the mix.
Let A(0) = 0, A(1) = 1, A(n) = 2*A(n-1)-A(n-2) + w
If w = 1 then you have the triangular series
If w = 2 then you have the series of squares
In general $$A_n = w*(n^2 -n)/2 + n$$ is a direct formula for $$A_n$$

redefine $$D_{(b,n,m)$$ as:

$$(A_{(n-b)}-m)*(A_{(n+2b)}-m) - (A_{(n+b)}-m)*(A_{(n-2b)}-m)$$

Then
$$D_{(b,n,m)} = b(w*(2n-1)+2)*(A_{n} -wb^{2} -m)$$

A new twist, let $$m = A_{n} - R$$ i.e a variable with n, w and b fixed and R varying then the following expression has a value which increases by 1 for every increase of R by 1.

$$\frac{D_{(b,n,m)}}{b(w*(2n-1)+2)}$$

In other words you have a determinant $$\{A_{00},A_{01},A_{10},A_{11}\}$$ which increases by b(w*(2n-1)+2) for each increase of each of the terms by 1, so the value of the determinant reduces to $$(R+K)*b(w*(2n-1)+2)$$