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? 
 A: You should be aware of the fact that, what is known in Italy as “Tartaglia triangle” is known in the rest of the world as “Pascal triangle”.
Concerning your question, what you're after is the binomial series:$$(1+x)^\alpha=\sum_{n=0}^\infty\binom\alpha nx^n,$$with$$\binom\alpha n=\frac{\alpha(\alpha-1)\cdots\bigl(\alpha-(n-1)\bigr)}{n!}.$$
A: 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*}

