English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

Please find all real roots of these set of equations:

http://m1.freeshare.us/173fs730110.gif

2007-09-06 07:30:14 · 3 answers · asked by Amir 1 in Science & Mathematics Mathematics

3 answers

lets denote φi = arctan(xi)
and rewrite the equations as

-ctan(2φ1) = tan(φ2)
-ctan(2φ2) = tan(φ3)
-ctan(2φ3) = tan(φ4)
-ctan(2φ4) = tan(φ1)

their solutions are
π/2 + 2φ1+ nπ = φ2
π/2 + 2φ2+ mπ = φ3
π/2 + 2φ3+ kπ = φ4
π/2 + 2φ4+ lπ = φ1

finally
φ1 = (π/30 + nπ/15)


Answer:
there are 14 real solutions
x1 = +/-tan(6,18,30,42,54,66,78)

2007-09-06 08:32:03 · answer #1 · answered by Alexander 6 · 1 0

{x[3] = -.5773502693, x[4] = .5773502693, x[1] = -.5773502693, x[2] = .5773502693}, {x[3] = 1.*I, x[4] = 1.*I, x[1] = 1.*I, x[2] = 1.*I}, {x[1] = -1.376381920, x[2] = -.3249196962, x[4] = .3249196962, x[3] = 1.376381920}, {x[3] = -.9004040446, x[2] = -2.246036774, x[4] = .1051042353, x[1] = -4.704630111}

2007-09-06 08:08:34 · answer #2 · answered by Anonymous · 0 0

I could only find two solutions.

Solution 1:
x1 = 1/sqrt(3)
x2 = -1/sqrt(3)
x3 = 1/sqrt(3)
x4 = -1/sqrt(3)

Solution 2:
x1 = -1/sqrt(3)
x2 = 1/sqrt(3)
x3 = -1/sqrt(3)
x4 = 1/sqrt(3)

There may be more solutions.

2007-09-06 07:43:12 · answer #3 · answered by lithiumdeuteride 7 · 0 0

fedest.com, questions and answers