$f(x)=f(0)+f<\mathrm{quote}/>(0).x/1!+f<\mathrm{quote}/><\mathrm{quote}/>(0).x^2/2!+...$

where $f<\mathrm{quote}/>$ denotes $\mathrm{df}/\mathrm{dx}$ etc. [you will notice that if you substitute $x=0$; of differentiate then substitute $x=0$; or differentiate twice then substitute $x=0$ etc.; the series has the correct value for $f(0)$ and all its derivatives]. Substituting:

$f(x)=1+n.x/1!+n(n-1).x^2/2!+...$

Now suppose we have $g(x,y)=(1+x+y{)}^n$

The corresponding power series expansion is

$g(x,y)=[g(0,0)]+[g_x.x+g_y.y]+[1.g_{xx}.x^2+2.g_{xy}.x.y.+g_{yy}.y^2]/2!+[g_{xxx}.x^3+3.g_{xxy}.x^2.y+3.g_{yyx}.x.y^2+1.g_{yyy}.y^3]/3!+...$

where $g_{xxy}$ denotes ${\partial }^{3}g/{\partial }^{2}x\partial y$ etc. Notice the appearance of the binomial coefficients in the square brackets. You will notice again that with this construction, all the partial derivatives of $g$ at $x=y=0$ are recovered exactly.

Therefore in our example we have

$g(x,y)=1+n.[x+y]+n.(n-1).[1.x^2+2.x.y+y^2]/2!+n.(n-1).(n-2)[x^3+3.x^2.y+3.x.y^2+1.y^3]/3!+...$

This approach can be expanded further for more variables.