Let's start with (1): Define $\mu (m)=(2m-1)!/[22m-1(m-1)!]$ and $\mu (0)=1$

Show (by induction) that this solves $\mu (m)=\mu (m-1)(1+2m)/2$ Hence, show that this gives the magnitude of the 'coefficient' of the nth derivative of $x^{-1/2}$ , and hence also $(x+1{)}^{-1/2}$. So, can you see why we have:

${\displaystyle [(1+x{)}^{-1/2}{]}^{(n)}=(-1{)}^n\mu (n)(1+x{)}^{-(1+2n)/2}}$

where $f(n)(x)=\frac{d^{n}f}{{\mathrm{dx}}^n}$?

Hence,

${\displaystyle [(1+x{)}^{-1/2}{]}^{(n)}=[(-1{)}^n(2n-1)!]/[2^{2m-1}(m-1)!(x+1{)}^{(1+2n)/2}]}$

If this is the sort of thing you were looking for, then try and see if you can get expressions for the other three. Ask if you need any more help...

Marcos

P.S. By concise I mean an expression in which we're not using things like 1x3x5x...xN (only the factorial function)



So, can you see why we have:
[(1+x)^-1/2 ]⁽ⁿ⁾ = (-1)ⁿ m (n)(1+x)^-(1+2n)/2
where f⁽ⁿ⁾ (x) = dⁿ f/dxⁿ

Hence,
[(1+x)^-1/2 ]⁽ⁿ⁾ = [(-1)ⁿ (2n-1)!]/[2^2m-1 (m-1)!(x+1)^(1+2n)/2 ]

If this is the sort of thing you were looking for, then try and see if you can get expressions for the other three. Ask if you need any more help...

Marcos

P.S. By concise I mean an expression in which we're not using things like 1x3x5x...xN (only the factorial function)