Tartaglia triangle for real field? Let's say i need to find the coefficient of the 2nd term in a binomia, we can use pascal's triangle. This works only if the power is a natural number. 
What if we want to know it for a real power binomial? We have something for that? 

We  can  derive a Pascal triangle as follows:

We expand 
  \begin{align*}
(1+x)^n=\sum_{k=0}^\infty\binom{n}{k}x^k\qquad\qquad n=0,1,2,\ldots
\end{align*}
  and obtain the coefficients 
  \begin{array}{lcccccccccccccccc}
(1+x)^0\qquad\qquad&&&&&1&&\color{lightgrey}{0}&&\color{lightgrey}{0}&&\color{lightgrey}{0}&&\color{lightgrey}{0}&\ldots\\
(1+x)^1\qquad\qquad&&&&1&&1&&\color{lightgrey}{0}&&\color{lightgrey}{0}&&\color{lightgrey}{0}&\ldots\\
(1+x)^2\qquad\qquad&&&1&&2&&1&&\color{lightgrey}{0}&&\color{lightgrey}{0}&\ldots\\
(1+x)^3\qquad\qquad&&1&&3&&3&&1&&\color{lightgrey}{0}&\ldots\\
(1+x)^4\qquad\qquad&1&&4&&6&&4&&1&\ldots\\
\end{array}

In the series above we set the upper limit to $\infty$ and note that $\binom{n}{k}=0$ if $k>n$. This way we can better see the analogy with the generalised version which we consider now.
We can now derive a generalised Pascal triangle in a similar way:

We expand
  \begin{align*}
(1+x)^{n+\frac{1}{2}}=\sum_{k=0}^\infty\binom{n+\frac{1}{2}}{k}x^k\qquad\qquad n=0,1,2,\ldots
\end{align*}
  and obtain the coefficients 
  \begin{array}{lcccccccccccccccc}
(1+x)^{\frac{1}{2}}\qquad&&&&&1&&\color{lightgrey}{\frac{1}{2}}&&\color{lightgrey}{-\frac{1}{8}}&&\color{lightgrey}{\frac{1}{16}}&&\color{lightgrey}{-\frac{5}{128}}&\ldots\\
(1+x)^{\frac{3}{2}}\qquad&&&&1&&\frac{3}{2}&&\color{lightgrey}{\frac{3}{8}}&&\color{lightgrey}{-\frac{1}{16}}&&\color{lightgrey}{\frac{3}{128}}&\ldots\\
(1+x)^{\frac{5}{2}}\qquad&&&1&&\frac{5}{2}&&\frac{15}{8}&&\color{lightgrey}{\frac{5}{16}}&&\color{lightgrey}{-\frac{5}{128}}&\ldots\\
(1+x)^{\frac{7}{2}}\qquad&&1&&\frac{7}{2}&&\color{blue}{\frac{35}{8}}&&\color{blue}{\frac{35}{16}}&&\color{lightgrey}{\frac{35}{128}}&\ldots\\
(1+x)^{\frac{9}{2}}\qquad&1&&\frac{9}{2}&&\frac{63}{8}&&\color{blue}{\frac{105}{16}}&&\frac{315}{128}&\ldots\\
\end{array}

Observe  the sum of two adjacent  entries in a row is the corresponding entry in the next row. They fulfill   the  binomial identity
\begin{align*}
\binom{\frac{9}{2}}{3}=\binom{\frac{7}{2}}{2}+\binom{\frac{7}{2}}{3}\qquad\qquad\qquad
\color{blue}{\frac{105}{16}=\frac{35}{8}+\frac{35}{16}}\\
\end{align*}
which  is quite obvious   when  we  consider
\begin{align*}
[x^k](1+x)^{\frac{9}{2}}
&=[x^k](1+x)(1+x)^{\frac{7}{2}}\\
&=[x^k](1+x)^{\frac{7}{2}}+[x^{k-1}](1+x)^{\frac{7}{2}}
\end{align*}

In the same way we can derive a generalised Pascal triangle for $\alpha\in\mathbb{C}$. 
  \begin{align*}
(1+x)^{n+\alpha}=\sum_{k=0}^\infty\binom{n+\alpha}{k}x^k\qquad\qquad n=0,1,2,\ldots
\end{align*}

