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Question: How to factor 1+x in a radical

Question:How to factor 1+x in a radical

C_R 3832 Maple
factor
20 hours ago
0

This is a follow-up question from here concerned with rearraging (not simplifying) an expression for an easier visual comparision. 
 

restart

In a Maple document I want to demonstrate (easier to follow from the output without the knowledge of Maple commands) that the following expression

expr := sqrt(2)/(sqrt(-beta^2+1)*sqrt(2+(-x-1)*beta^2))

2^(1/2)/((-beta^2+1)^(1/2)*(2+(-x-1)*beta^2)^(1/2))

 (1)

visually matches the integrand of this ellpitic integral

FunctionAdvisor(definition, EllipticF)

[EllipticF(z, k) = Int(1/((-_alpha1^2+1)^(1/2)*(-_alpha1^2*k^2+1)^(1/2)), _alpha1 = 0 .. z), MathematicalFunctions:-`with no restrictions on `(z, k)]

 (2)

This can be done with following substitution

x = 2*k^2-1

x = 2*k^2-1

 (3)

subs(x = 2*k^2-1, 2^(1/2)/((-beta^2+1)^(1/2)*(2+(-x-1)*beta^2)^(1/2)))

2^(1/2)/((-beta^2+1)^(1/2)*(-2*beta^2*k^2+2)^(1/2))

 (4)

simplify(2^(1/2)/((-beta^2+1)^(1/2)*(-2*beta^2*k^2+2)^(1/2)))

1/((-beta^2+1)^(1/2)*(-beta^2*k^2+1)^(1/2))

 (5)

Since the substitution comes without explanation, I though about rearranging expression (1) by Maple commands in a way that facilitates visual pattern matching.
How to collect the factor 1+x in (1) leading to  sqrt(2)/(sqrt(-beta^2+1)*sqrt(2-(1+x)*beta^2))

It looks as one would write by hand.

 

At a glance, this much easier suggest that k^2 = (1+x)*(1/2).
 


Download Collect_one_plus_x_-_02.mw

 

(Maple 2025.2 only for easier editing. Results are the same with Maple 2026)

 

 

Original Post - Collect_one_plus_x.mw

 
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