How can I find these coefficients?

[kpkkpk]

x = 1x
x(x-1) = 1x^2 - 1x
x(x-1)(x-2) = 1x^3 - 3x^2 + 2x
x(x-1)(x-2)(x-3) = 1x^4 - 6x^3 + 11x^2 - 6x
x(x-1)(x-2)(x-3)(x-4) = 1x^5 - 10x^4 + 35x^3 - 50x^2 + 24x
...

So, I am looking for a short method how to find these coefficients ahead of each raisings of x:

1
1,-1
1,-3,2
1,-6,11,-6
1,-10,35,-50,24
...

Is the only method just to perform repetative multiplications, or is there a time and energy saving method available?

I do can find some predictable features:
-The leftmost coefficient is always "1"
-The next coefficient comes from triangular numbers (1,3,6,10,15,21,28...that is an integer when we sum up successive positive integers; for example: 1+2+3+4+5=15) with negative sign.
- The last term is a factorial with alternate + and - sign (for example: 3 factorial = 3! = 1*2*3 = 6, 4! = 1*2*3*4 =24 etc.)
- Positive and negative coefficients in a row, when summoned up, equals to 0 (for example in the last row example: 1-10+35-50+24 = 0)

Please, help me.

 

[mente oscura]

x = 1x
x(x-1) = 1x^2 - 1x
x(x-1)(x-2) = 1x^3 - 3x^2 + 2x
x(x-1)(x-2)(x-3) = 1x^4 - 6x^3 + 11x^2 - 6x
x(x-1)(x-2)(x-3)(x-4) = 1x^5 - 10x^4 + 35x^3 - 50x^2 + 24x
...

So, I am looking for a short method how to find these coefficients ahead of each raisings of x:

1
1,-1
1,-3,2
1,-6,11,-6
1,-10,35,-50,24
...

Is the only method just to perform repetative multiplications, or is there a time and energy saving method available?

I do can find some predictable features:
-The leftmost coefficient is always "1"
-The next coefficient comes from triangular numbers (1,3,6,10,15,21,28...that is an integer when we sum up successive positive integers; for example: 1+2+3+4+5=15) with negative sign.
- The last term is a factorial with alternate + and - sign (for example: 3 factorial = 3! = 1*2*3 = 6, 4! = 1*2*3*4 =24 etc.)
- Positive and negative coefficients in a row, when summoned up, equals to 0 (for example in the last row example: 1-10+35-50+24 = 0)

Please, help me.

Hello.

I wish this small my study to could help you.

[tex]{Let }\ a_i,\ k,\ n \epsilon N[/tex]

[tex]P_k = (n-1)(n-2)\ldots (n-k)=a_kn^k-a_{k-1}n^{k-1}\ldots +(-1)^{k-1}a_1n+(-1)^ka_0[/tex]

[tex]{Properties: }[/tex]

[tex]{1) } \sum_{i=0}^{k}a_i = 0[/tex]

[tex]{ 2) } \sum_{i=0}^{k}\left\vert{a_i} \right\vert{} = (k+1)![/tex]

[tex]{ 3) }\ a_k=1[/tex]

[tex]{ 4) }\ a_0=(-1)^kk![/tex]

[tex]{ 5) }\ a_{k-1}=-k \frac{k+1}{2}[/tex]

[tex]{ 6) }\ a_{k-2}= \frac{3(k-1)^4+14(k-1)^3+21(k-1)^2+10(k-1)}{24}[/tex]

Formula of recurrence, for [tex]\ a_i[/tex]

[tex]\displaystyle a_{1,k}= (-1)^{k+1}(k-1)!-a_{1, k-1}k[/tex][tex]\displaystyle a_{i,k}= a_{i-1, k-1}-a_{i, k-1}k[/tex].

Regards.

 

[Opalg]

x = 1x
x(x-1) = 1x^2 - 1x
x(x-1)(x-2) = 1x^3 - 3x^2 + 2x
x(x-1)(x-2)(x-3) = 1x^4 - 6x^3 + 11x^2 - 6x
x(x-1)(x-2)(x-3)(x-4) = 1x^5 - 10x^4 + 35x^3 - 50x^2 + 24x
...

So, I am looking for a short method how to find these coefficients ahead of each raisings of x:

1
1,-1
1,-3,2
1,-6,11,-6
1,-10,35,-50,24
...

Is the only method just to perform repetative multiplications, or is there a time and energy saving method available?

I do can find some predictable features:
-The leftmost coefficient is always "1"
-The next coefficient comes from triangular numbers (1,3,6,10,15,21,28...that is an integer when we sum up successive positive integers; for example: 1+2+3+4+5=15) with negative sign.
- The last term is a factorial with alternate + and - sign (for example: 3 factorial = 3! = 1*2*3 = 6, 4! = 1*2*3*4 =24 etc.)
- Positive and negative coefficients in a row, when summoned up, equals to 0 (for example in the last row example: 1-10+35-50+24 = 0)

Please, help me.

Hi kpkkpk and welcome to MHB! You have raised an interesting question.

The product $(x)_n = x(x-1)(x-2)(x-3)\cdots(x-n+1)$ is known as a falling factorial, and its coefficients are Stirling numbers of the first kind. They crop up in various contexts in combinatorics. There is no simple formula for them. See Stirling numbers of the first kind - Wikipedia, the free encyclopedia.

 

[mente oscura]

Hi kpkkpk and welcome to MHB! You have raised an interesting question.

The product $(x)_n = x(x-1)(x-2)(x-3)\cdots(x-n+1)$ is known as a falling factorial, and its coefficients are Stirling numbers of the first kind. They crop up in various contexts in combinatorics. There is no simple formula for them. See Stirling numbers of the first kind - Wikipedia, the free encyclopedia.

Hello.(Smile)

I believe, that with my modest study, yes can calculate with facility the coefficients.

I have showed, the formulae of recurrencias for the calculation of the coefficients. I attach a file with a triangle of the above-mentioned coefficients.

Regards.

 

[chisigma]

x = 1x
x(x-1) = 1x^2 - 1x
x(x-1)(x-2) = 1x^3 - 3x^2 + 2x
x(x-1)(x-2)(x-3) = 1x^4 - 6x^3 + 11x^2 - 6x
x(x-1)(x-2)(x-3)(x-4) = 1x^5 - 10x^4 + 35x^3 - 50x^2 + 24x
...

So, I am looking for a short method how to find these coefficients ahead of each raisings of x:

1
1,-1
1,-3,2
1,-6,11,-6
1,-10,35,-50,24
...

Is the only method just to perform repetative multiplications, or is there a time and energy saving method available?

I do can find some predictable features:
-The leftmost coefficient is always "1"
-The next coefficient comes from triangular numbers (1,3,6,10,15,21,28...that is an integer when we sum up successive positive integers; for example: 1+2+3+4+5=15) with negative sign.
- The last term is a factorial with alternate + and - sign (for example: 3 factorial = 3! = 1*2*3 = 6, 4! = 1*2*3*4 =24 etc.)
- Positive and negative coefficients in a row, when summoned up, equals to 0 (for example in the last row example: 1-10+35-50+24 = 0)

Please, help me.

The polynomials You have described are known as falling polynomials and are defined as...

$\displaystyle (x)_{n} = x\ (x-1)\ ...\ (x-n+1)\ (1)$

The recursive relation is easily derivable...

$\displaystyle (x)_{n+1} = x\ (x)_{n} - n\ (x)_{n}\ (2)$

... and if You set...

$\displaystyle (x)_{n}= \sum_{k=0}^{n} c_{n,k}\ x^{k}\ (3)$

... the recursive relation (2) becomes the following relation...

$\displaystyle c_{n+1,k} = c_{n,k-1} - n\ c_{n,k},\ k=0,1,...,n+1\ (4)$

Starting from $c_{0,0}=1$ You can proceed indefinitely...

Kind regards

$\chi$ $\sigma$

 

[kpkkpk]

Hello.(Smile)

I believe, that with my modest study, yes can calculate with facility the coefficients.

I have showed, the formulae of recurrencias for the calculation of the coefficients. I attach a file with a triangle of the above-mentioned coefficients.

Regards.

Yes, indeed!
It took me some time before I understood what you meant, but now things are much clearer.
If I understood correctly, from your triangle I pick two consecutive integers (their absolute values) from the same horizontal line, then I multiply the leftmost of them with vertical row number...and then I sum them up. Every other of these numbers will be attached with + and every other with - sign.
What I then have, will be added below and between those integers I started from...in the same manner as we primitively construct Pascal´s triangle?

Thanks a lot for spending so much time for my problem!

 

[mente oscura]

Yes, indeed!
It took me some time before I understood what you meant, but now things are much clearer.
If I understood correctly, from your triangle I pick two consecutive integers (their absolute values) from the same horizontal line, then I multiply the leftmost of them with vertical row number...and then I sum them up. Every other of these numbers will be attached with + and every other with - sign.
What I then have, will be added below and between those integers I started from...in the same manner as we primitively construct Pascal´s triangle?

Thanks a lot for spending so much time for my problem!

Hello.

Excuse me you, but my use of the language Englishman is pathetic. You have understood my exposition, or have some doubt on him.

If you have some "problem" on the procedure, do not doubt in reguntarmelo, with very much pleasure, I will clarify it.

Regards.