Advanced Search

Question: How to factor 1+x in a radical and some more questions

Question:How to factor 1+x in a radical and some more questions

C_R 3827 Maple
collect simplification factor
2 hours ago
0

This is a follow-up question from here concerned with rearraging (not simplifying) an expression for an easier comparision. Because of radicals involved programatic manipulation turned out to be surprisingly complicated.

(Edited in Maple 2025.2 only for easier typing. Results are the same in Maple 2026)

restart

In a Maple document I want to demonstrate 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)

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 command in a way that facilitates human pattern matching.
Q1: How to collect the factor 1+x in (1) leading to  sqrt(2)/(sqrt(-beta^2+1)*sqrt(2-(1+x)*beta^2))

(LeafCount is one less than in expression (1) and, in my opinion, it also looks as one would write by hand)

 

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

Q2: Is there a way to get to 1/(sqrt(-beta^2+1)*sqrt(1-((1+x)*(1/2))*beta^2)) by absorbing 1/sqrt(2)into the second square root of the denominator?
Asking more generally: How to absorb the numerator of a rational expression into the a square root of the nominator``

 

 

Assuming there is a solution for Q1

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

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

 (6)

we could think of grouping the factors of the rational expression into two factors where one of them contains the part to be absorbed and the corresponding square root

op(intermediate)

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

 (7)

factor1 := (2^(1/2), 1/(-beta^2+1)^(1/2), 1/(2-(1+x)*beta^2)^(1/2))[2]

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

 (8)

In the below expression we want to absorb sqrt(2) under the root

factor2 := intermediate/factor1

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

 (9)

Taking the inverse of the square gets us to the desired radicand

1/factor2^2

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

 (10)

Now taking the squareroot of it leads to something undesirable

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

(1/2)*(-2*beta^2*x-2*beta^2+4)^(1/2)

 (11)

Making it innert and taking the inverse

1/%sqrt(1/factor2^2)

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

 (12)

leads to the desired result

final := factor1/%sqrt(1-(1/2)*(1+x)*beta^2)

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

 (13)

This way has two weaknesses: First manual grouping into two factors. Second, taking the inverse of the inverse combined with a squaring operation and the innert root.
Q3: Any suggestions to make this way less awkward?

  NULL

Download Collect_one_plus_x.mw
Please Wait...