What's equal $(x+y)^{\sqrt{2}} $ to and what's its geometric interpretation?

Question

I'm confused how I can evaluate $(x+y)^{\sqrt{2}}$ using the Newton binomial.

The Newton binomial Newten has positive integer exponents not irrational numbers, so I'm curious to check evaluation of $(x+y)^{\sqrt{2}}$ where $x$ and $y$ are real and what's its geometric interpretation.

Answer 1

$$(1+x)^{p}=\sum_0^{\infty}\binom{p}{k}x^k$$ here $p\in C$

$$(x+y)^{p}=y^p\left(1+\frac{x}{y}\right)^p=y^p\sum_{0}^{\infty}\binom{p}{k}(\frac{x}{y})^k$$

here is the expansion of $(1+x)^{1/2}$ to give an idea of how to expand. $$\displaystyle (1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3\dots$$

$$\displaystyle \boxed{(1+x)^{\frac{1}{2}}=1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3-\frac{5x^4}{128}+O(x^5)} $$

Answer 2

Let's look at $(1+t)^{\sqrt{2}}$ instead.

The series $$ \sum_{k=0}^{\infty}\binom{\sqrt{2}}{k}t^k $$ converges for $|t|<1$, which can be checked with the ratio test. The definition of the binomial coefficient for arbitrary $a$ and nonnegative integer $k$ is $$ \binom{a}{k}=\frac{a(a-1)\dotsm(a-k+1)}{k!} $$ If you need to compute $(x+y)^{\sqrt{2}}$, with $0<x<y$, you can consider $t=x/y$ and so $$ (x+y)^{\sqrt{2}}=y^{\sqrt{2}}(1+t)^{\sqrt{2}} $$

Not every analytic object needs to have a geometric interpretation: the definition of $y^{\sqrt{2}}$ is $$ y^{\sqrt{2}}=\exp(\sqrt{2}\log y) $$ (exponential and logarithm with respect to the base you prefer).