Are there expansions of the expression $(a+b)^{1/n}$?

Question

Is there an expansion of the expression in the bracket such as

$$ \sqrt{a + b} = (a + b)^{1/2}$$

If not do you know of a method that lets us solve such expression and ones with higher roots?

Answer 1

never underestimate the binomial theorem and the coefficients associated with it! they have played a pivotal role in the development of mathematical analysis.

whilst Pascal's triangle well deserves the attention it receives, the development for indices other than a non-negative integer opens new doors.

it is worth remembering that the expression $(1+x)^y$ is a function of both $x$ and $y$, although we usually focus on $x$ as the variable of interest.

here is one way of writing the main assertion of the binomial theorem which treats the variables on a more equal footing. as we can see, this leads immediately to a moderately interesting combinatorial result concerning the reciprocals of certain sets of integers:

for any real $a$ and complex number $z$ with $|z| \lt 1$ we have: $$ (1-z)^a = \sum_{n=0}^{\infty}\sum_{k=1}^n \frac{(-1)^k s(n,k)}{n!}a^kz^n \tag{1} $$ where the symbol $s(n,k)$ is the unsigned Stirling number of the first kind

since the left-hand side can be developed as: $$ e^{a\log(1-z)} = \sum_{k=0}^{\infty} \frac{(a\log(1-z))^k}{k!} \tag{2} $$ we may, by comparing coefficients of $a^k$ in (1) and (2), deduce that: $$ (\log(1-z))^k = (-1)^k k! \sum_{n=k}^{\infty} s(n,k)\frac{z^n}{n!} \tag{3} $$ now, since we know that: $$ \log(1-z) = -\sum_{m=1}^{\infty} \frac{z^m}m $$ the term involving $z^n$ in $(\log(1-z))^k$ may also be evaluated as the sum: $$ \sum_{m_1+\dots+m_k=n} \frac1{m_1m_2\dots m_k} $$ where each $m_i \gt 0$

as a worked example, take $n=7$ and $k=3$ so that the above sum is: $$ 3\frac1{1 \cdot 1 \cdot5}+6\frac1{1 \cdot 2 \cdot 4}+3\frac1{1 \cdot 3 \cdot 3}+3\frac1{2 \cdot 2 \cdot 3} = \frac{29}{15} $$ which is given by the term from (3) as: $$ \frac{3!}{7!}s(7,3) = \frac6{5040}\cdot 1624 = \frac{29}{15} $$