01/31/2016, 04:11 AM
If we take for granted that the red terms on the series do not change when n grows, even for real values of n, then the difference \( \\[20pt]
{^{n+\Delta}(x+1)-^n(x+1)} \) eliminates all red terms of \( \\[20pt]
{^n(x+1)} \).
So, the partial derivative \( \\[25pt]
{\frac{\partial\; ^n(x+1)}{\partial n} \,=\, d_{n+1}\,x^{n+1} \,+\, d_{n+2}\,x^{n+2} \,+\, ... \,=\, x^{n+1}\sum d_i\,x^i} \) has a taylor series that starts from the n+1 term, and \( \\[20pt]
{x^{n+1}} \) can be taken common factor.
This hints that the derivative of \( \\[15pt]
{^xa} \)may have \( \\[20pt]
{(a-1)^{x+1}} \) as a factor.
{^{n+\Delta}(x+1)-^n(x+1)} \) eliminates all red terms of \( \\[20pt]
{^n(x+1)} \).
So, the partial derivative \( \\[25pt]
{\frac{\partial\; ^n(x+1)}{\partial n} \,=\, d_{n+1}\,x^{n+1} \,+\, d_{n+2}\,x^{n+2} \,+\, ... \,=\, x^{n+1}\sum d_i\,x^i} \) has a taylor series that starts from the n+1 term, and \( \\[20pt]
{x^{n+1}} \) can be taken common factor.
This hints that the derivative of \( \\[15pt]
{^xa} \)may have \( \\[20pt]
{(a-1)^{x+1}} \) as a factor.
I have the result, but I do not yet know how to get it.
