An explicit formula, probably has \( \\[15pt]
{x^k} \) as factor, because it looks like the Taylor series of \( \\[15pt]
{^xb} \), which is higly probable to have \( \\[15pt]
{ln(b)^k} \) as factor on the k derivative
Since the Taylor series is
\( ^xb=\sum_{k=0}^{\infty} \frac {1}{k!}*\frac{d^k (^xb)}{dx^k}|_0 *x^{k} \)
T(x,k) has to be
\( T(x,k)=\frac{d^k (^xb)}{dx^k}|_0 *\frac{x^{k}}{ln(b)^k} \)
{x^k} \) as factor, because it looks like the Taylor series of \( \\[15pt]
{^xb} \), which is higly probable to have \( \\[15pt]
{ln(b)^k} \) as factor on the k derivative
Since the Taylor series is
\( ^xb=\sum_{k=0}^{\infty} \frac {1}{k!}*\frac{d^k (^xb)}{dx^k}|_0 *x^{k} \)
T(x,k) has to be
\( T(x,k)=\frac{d^k (^xb)}{dx^k}|_0 *\frac{x^{k}}{ln(b)^k} \)
I have the result, but I do not yet know how to get it.
