User:Mortice/Maths

2d[edit]

My formula:

$(x+y)^{n}=\sum _{r=0}^{n}{{\frac {d^{r}}{dx^{r}}}{x^{n}}\int ^{r}1dy}$

...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:

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

Which is the same as:

$(x+y)^{4}=x^{4}*1+4x^{3}*y+12x^{2}*{\frac {1}{2}}y^{2}+24x*{\frac {1}{6}}y^{3}+24*{\frac {1}{24}}y^{4}$

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:

$\int 1dy={\frac {1}{r!}}y^{r}$

...then it all logically falls out.

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:

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

Could be written as:

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

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

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

And so...

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

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