This is not the same; it is an example of the extended binomial formule given for INTEGER n over real exponents (here equal 1/2). it a corollary to the Taylor formula for analytic functions.
I want to develop (x+y)^1/2 without knowing the values of x and y. Now, if we know the values of x and y, so we can calculate the square root of the given expression obviously.
The link suggested by @Ahmad was very helpful. Thank you very much.
There is a general formula to expand expressions of that kind (x+y)^n, with "n" is an integer. for the case of an n=1/2, we can use the pascal's triangle to expand any kind of expressions.
The square root is a real analytic function from [0, infinity) to [0, infinity)
which means it has a power series expansion. That can be calculated (expanded ) in a Taylor series expansion around x. If x or x+y is negative then one has to consider complex roots and it becomes somewhat messy.
The binomial expansion is only valid for integers. The square root does not have a finite expansion - but it does have a Taylor series expansion.
This is not the same; it is an example of the extended binomial formule given for INTEGER n over real exponents (here equal 1/2). it a corollary to the Taylor formula for analytic functions.
Before answering this question, i have applied the Taylor expansion (a+b)^n on the available expansion result of (x+y)^(1/2) which is derived in the video link that @ Ahmad Takash provided, and same results were obtained.
Ok, n must be an integer, so, the expansion of (x+y)^1/n where n is an integer, will be the same. i.e.,
(x+y)^n = x^n + n/1!*x^(n-1)*y +
n*(n-1)/2!*x^(n-2)*y^2 + ...
and
(x+y)^(1/n)=x^(1/n) + (1/n)/1!*x^(1/n−1)*y+
(1/n)(1/n−1)/2!*x^(1/n−2)*y^2...
Comparing them, ... they are the same and can be applied on (x+y)^1/2 when the condition mentioned by @ Octav Olteanu is staisfied.
So, i think that the video which @ Abdelmounaim Chetoui found that was very helpful for him, doesn't contradict with the result that can be obtained when applying one of these formulas.
I made a small search and find the following site:
There is no closed form solution if n is not a positive integer. However, consider the function f(x,y)= (x+y)^u, u>0.
Then calculate the two dimensional power series expansion around the point (0,0), If n is an positive integer - then one gets the normal polynomial expansion theorem. If u is not a positive integer you get a Taylor series that does not end. I u>0 you get a real analytic function (convergent power series). One has to use care here since we might have to define this on sheets of a Reimann surface, e.g., u =1/2 and x+y 0 you get a real analytic function in two variables.