User:Mortice/Maths

My formula:

$$

(x+y)^n = \sum_{r=0}^n{ \frac{d^r}{dx^r} {x^n} \int^r1 dy }

...where $\backslash int\{^r\}$ is the r'th integral

The traditional formula:

$$

(x+y)^n = \sum_{r=0}^n{ P_r^n x^r y^{n-r} }

...where $P\_r^n$ is a Permutation

For example:

$$

\displaystyle (x+y)^4 = x^4+4x^3y+6x^2y^2+4xy^3+y^4

Which is the same as:

$$

(x+y)^4 = x^4*1+4x^3*y+12x^2*\frac12y^2+24x*\frac16y^3+24*\frac1{24}y^4

If you consider:

$$

P_r^n = \frac{n!}{r!(n-r)!}

and:

$$

\frac{d^r}{dx^r} {x^n} = \frac{n!}{(n-r)!}x^{(n-r)}

and:

$$

\int 1 dy = \frac1{r!}y^r

...then it all logically falls out.

Lets use $\backslash displaystyle\; n[a^r]$ meaning the expression

$\backslash displaystyle\; a^r$ occurs n times using different

variables for a, all summed together.

So $\backslash displaystyle\; x^2+y^2+z^2+7\; =\; 3[a^2]+7$.

For instance:

$$

\displaystyle (x+y)^4 = x^4+4x^3y+6x^2y^2+4xy^3+y^4

Could be written as:

$$

\displaystyle (x+y)^4 = 2[a^4]+2[4a^3b]+6a^2y^2

$$

\displaystyle (x+y+z)^3 = 3[a^3]+6[3a^2b]+6xyz

These numbers can be gleaned from the following Pascal's tetrahedron layer:

1

3 3

3 6 3

1 3 3 1

And so...

$$

\displaystyle (x+y+z)^4 = 3[a^4]+6[4a^3b]+3[6a^2b^2]+3[a^2bc]

1

4 4

6 10 6

4 10 10 4

1 4 6 4 1