User:Mortice/Maths

Math

2d
My formula:

$$

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

$$

...where $$ \int{^r} $$ 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.

New notation
Lets use $$\displaystyle n[a^r]$$ meaning the expression $$\displaystyle a^r$$ occurs n times using different variables for a, all summed together.

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

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

$$

3d
$$

\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