How can I expand $ (a+b)^{q} = a^q + b^q + \cdots $? I am wondering that I can expand $$
(a+b)^{q} = a^q + b^q + \cdots
$$
for $a,b\in\mathbb R$ and $q\ge 1$.
I know there is a binomial expansion for $q \in \mathbb Z^+$, however, I'm wondering if there exists something for real positive $q$.
 A: Suppose that $|a| > |b|$. We have the binomial series 
$$ (1 + x)^q = \sum_{n=0}^\infty \binom qn x^n, \qquad |x| < 1$$
where 
$$ \binom qn = \frac{q \cdot (q-1) \cdots (q-n+1)}{n!} $$
Hence
$$ (a+b)^q = a^q \left(1 + \frac ba\right)^q 
= a^q \sum_{n=0}^\infty \binom qn \frac{b^n}{a^n} = \sum_{n=0}^\infty \binom qn a^{q-n}b^n $$
A: Suppose that  We try to find taylor serier expansion of
$$ f(x)=(1 + x)^q =a_0+a_1x+a_2x^2+a_3x^3+..... , \qquad |x| < 1$$
$$ f(0)=a_0$$
$$ f(0)=(1 + 0)^q=1  $$
Thus $a_0=1$
Now take derivatives both sides
$$ f'(x)=q(1 + x)^{q-1} =a_1+2a_2x+3a_3x^2+..... $$
$$ f'(0)=q(1 + 0)^{q-1} =a_1$$
Thus  $a_1=q$
Now take derivatives both sides again
$$ f''(x)=q(q-1)(1 + x)^{q-2} =2a_2+3.2a_3x+4.3a_4x^2+..... $$
$$ f''(0)=q(q-1)(1 + 0)^{q-2} =2a_2$$
Thus  $a_2=\frac{q(q-1)}{2}$
If you continue in that way you will get 
Thus  $a_n=\frac{q(q-1)(q-2)....(q-(n-1))}{n!}=\binom qn$
So you can write 
$$ f(x)=(1 + x)^q =\sum_{n=0}^\infty \binom qn x^n, \qquad |x| < 1  \tag{ 1} $$
Now last step to find for 
$$ (a+b)^{q} $$
We will use the fact above thus Suppose that $|a| > |b|$. 
you can write $$ a^q(1+\frac{b}{a})^{q} $$
and suppose we call $x=\frac{b}{a}$
Because  $|a| > |b|$ , so $|x|=|\frac{b}{a}|<1$
$$ (a+b)^{q}= a^q(1+x)^{q}$$ , where $$|x|<1$$
we can use the formula in $(1)$
Finally we can write
$$ (a+b)^{q}= a^q(1+x)^{q}=a^q\sum_{n=0}^\infty \binom qn x^n, $$ , where $$|x|<1$$
If you put $x=\frac{b}{a}$
$$ (a+b)^{q}= a^q\sum_{n=0}^\infty \binom qn x^n=a^q\sum_{n=0}^\infty \binom qn (\frac{b}{a})^n, $$ , where $$|x|<1$$
$$ (a+b)^{q}= a^q\sum_{n=0}^\infty \binom qn (\frac{b}{a})^n, $$ , where $$ |a| > |b|$$ 
