x^5 = (x-1)^5 => ((x-1)/x)^5 = 1 or u^5 =1 where u = (x-1)/x u^5 = e^(i*2pi*n) => u = e^(i*2pi*n/5), n = 0, 1, 2, 3, 4 u=(x-1)/x = 1 - 1/x => x = 1/(1-u) = 1/(1 - e^(i*2pi*n/5)), n = 1, 2, 3, 4. Expanding the denominator and putting in standard form gives x = 1/2 + i * 1/2 * sin(2pi*n/5) / (1 - cos(2pi*n/5)) You can substitute the exact values of these sines and cosines and simplify but I just left it like this.
For example when you add 1 (say for the graph (x+1)^2), you can essentially think of the entire x-axis moving one to the left as when you plug in -1 it becomes 0^2, when you plug in 1 it becomes 2^2, and so on. So now x = -1 takes the place of what x = 0 would be in the unchanged x^2 graph. And it works the other way, when we take one from the graph (e.g. (x-1)^2). You can think of it as the whole x-axis moving to the right by one unit as if you plug in 0, it becomes (-1)^2, plug in 1, it becomes 0^2. So now x = 1 takes the place of what x = 0 would be in the x^2 graph. Hope this helps!
x⁵ = (x-1)⁵ I would like to let (x-1) = y x - y = 1 x⁵ - y⁵ = 0 x² + y² = 1 + 2xy x³ - y³ = (1 + 3xy) (x³-y³)(x²+y²) = (1+2xy)(1+3xy) x⁵ + x³y² - y³x² - y⁵ = 1 + 3xy + 2xy + 6(xy)² x⁵ - y⁵ + x²y²(x-y) = 1+5xy + (6xy)² Let xy = a 5a² + 5a + 1 = 0 a = -5 ± √(25 - 20)/10 a = -5 ± √5/10 x - y = 1 xy = (-5 ± √5)/10 Solving the system u get the value of x
It's a decapitated quintic, with a = 0 because the x^5 cancels out. So, if a = 0, it's not technically a quintic, it just looks like one till you mess it up.
problem x⁵ = (x-1)⁵ x⁵ - (x-1)⁵ = 0 [x-(x-1)][x⁴+(x-1)x³+(x-1)²x²+(x-1)³x+(x-1)⁴]=0 x⁴+(x-1)x³+(x-1)²x²+(x-1)³x+(x-1)⁴=0 This expands to 5x⁴-10x³ +10x²-5x+1 = 0 Let x= y+1/2 5y⁴ +(5/2) y² +(1/16) = 0 80 y⁴ +40y² +1=0 y² = (-5± 2√5)/20 Two solutions for y² y² =-(√5/20) ( √5 -2 ) y² =-(√5/20) ( √5 + 2 ) Solutions for y y = -i (√5/2) √(√5-2) y = i (√5/2) √(√5-2) y = -i (√5/2) √(√5+2) y = i (√5/2) √(√5+2) Back substitution x = y + 1/2 x solutions x = (1/2)-i (√5/2) √(√5-2) x = (1/2)+i (√5/2) √(√5-2) x = (1/2)-i (√5/2) √(√5+2) x = (1/2)+i (√5/2) √(√5+2) answer x ∈ { (1/2) - i (√5/2) √(√5+2), (1/2) - i (√5/2) √(√5-2), (1/2) + i (√5/2) √(√5-2), (1/2) + i (√5/2) √(√5+2) }
I did this way (step by step, a long way) ...
x⁵ = (x - 1)⁵
(x - 1)⁵ = x⁵
((x - 1)⁵)/x⁵ = x⁵/ x⁵
((x - 1)/x)⁵ = 1
set k = (x - 1)/x
k⁵ = 1
---
/// "roots of unity" of k⁵ = 1 (recall):
// note:
• argument: n·(π/5) (n = [2..10] by 2)
• module: 1 (for each)
• polar form: 1·e^(i·n·(π/5)) = e^(i·n·(π/5))
// computation:
• root #1 (of unity): e^(i·(2π/5)) = cos(2π/5) + i·sin(2π/5) ≈ 0.309 + 0.951·i
• root #2 (of unity): e^(i·(4π/5)) = cos(4π/5) + i·sin(4π/5) ≈ -0.809 + 0.587·i
• root #3 (of unity): e^(i·(6π/5)) = cos(6π/5) + i·sin(6π/5) ≈ -0.809 - 0.587·i
• root #4 (of unity): e^(i·(8π/5)) = cos(8π/5) + i·sin(8π/5) ≈ 0.309 - 0.951·i
• root #5 (of unity): e^(i·2π) = 1
---
/// case: (x - 1)/x = 0.309 + 0.951·i
(x - 1)/x = 0.309 + 0.951·i
x/x - 1/x = 0.309 + 0.951·i
1 - 1/x = 0.309 + 0.951·i
1/x = 1 - 0.309 + 0.951·i
recall: 1/a = b => a = 1/b
x = 1/(1 - 0.309 + 0.951·i)
x = 1/(0.691 + 0.951·i)
recall:
||
|| division between 2 complex numbers:
||
|| a + bi (a + bi)·(c - di) (ac + bd) + i·(bc - ad)
|| -------- = --------------------- = --------------------------------
|| c + di (c + di)·(c - di) c² + d²
||
x = (1 + 0·i)/(0.691 + 0.951·i)
note:
• a = 1
• b = 0
• c = 0.691
• d = 0.951
x = [(ac + bd) + i·(bc - ad)]/(c² + d²)
x = [(1·0.691 + 0·0.951) + i·(0·0.691 - 1·0.951)]/(0.691² + 0.951²)
■ x ≈ 0.5 - 0.688·i
---
/// case: (x - 1)/x = -0.809 + 0.587·i
(x - 1)/x = -0.809 + 0.587·i
x/x - 1/x = -0.809 + 0.587·i
1 - 1/x = -0.809 + 0.587·i
1/x = 1 + 0.809 + 0.587·i
recall: 1/a = b => a = 1/b
x = 1/(1 + 0.809 + 0.587·i)
x = 1/(1.809 + 0.587·i)
recall:
||
|| division between 2 complex numbers:
||
|| a + bi (a + bi)·(c - di) (ac + bd) + i·(bc - ad)
|| -------- = --------------------- = ---------------------------------
|| c + di (c + di)·(c - di) c² + d²
||
x = (1 + 0·i)/(1.809 + 0.587·i)
note:
• a = 1
• b = 0
• c = 1.809
• d = 0.587
x = [(ac + bd) + i·(bc - ad)]/(c² + d²)
x = [(1·1.809 + 0·0.587) + i·(0·1.809 - 1·0.587)]/(1.809² + 0.587²)
■ x ≈ 0.5 - 0.162·i
---
/// case: (x - 1)/x = -0.809 - 0.587·i
SAME REASONING AS PREVIOUSLY
x = [(1·1.809 - 0·0.587) + i·(0·1.809 + 1·0.587)]/(1.809² + 0.587²)
■ x ≈ 0.5 + 0.162·i
---
/// case: (x - 1)/x = 0.309 - 0.951·i
SAME REASONING AS PREVIOUSLY
x = [(1·0.691 + 0·0.951) + i·(0·0.691 + 1·0.951)]/(0.691² + 0.951²)
■ x ≈ 0.5 + 0.688·i
---
/// case: (x - 1)/x = 1
(x - 1)/x = 1
x/x - 1/x = 1
1 - 1/x = 1
1/x = 1 - 1
1/x = 0
NO SOLUTION !
---
/// final results:
■ root #1: x ≈ 0.5 - 0.688·i
■ root #2: x ≈ 0.5 - 0.162·i
■ root #3: x ≈ 0.5 + 0.162·i
■ root #4: x ≈ 0.5 + 0.688·i
🙂
x^5 = (x-1)^5 => ((x-1)/x)^5 = 1 or u^5 =1 where u = (x-1)/x
u^5 = e^(i*2pi*n) => u = e^(i*2pi*n/5), n = 0, 1, 2, 3, 4
u=(x-1)/x = 1 - 1/x => x = 1/(1-u) = 1/(1 - e^(i*2pi*n/5)), n = 1, 2, 3, 4.
Expanding the denominator and putting in standard form gives x = 1/2 + i * 1/2 * sin(2pi*n/5) / (1 - cos(2pi*n/5))
You can substitute the exact values of these sines and cosines and simplify but I just left it like this.
Please remind me: when we move a curve in the positive direction along the x axis, why do we subtract the number of places instead of adding them?
For example when you add 1 (say for the graph (x+1)^2), you can essentially think of the entire x-axis moving one to the left as when you plug in -1 it becomes 0^2, when you plug in 1 it becomes 2^2, and so on. So now x = -1 takes the place of what x = 0 would be in the unchanged x^2 graph.
And it works the other way, when we take one from the graph (e.g. (x-1)^2). You can think of it as the whole x-axis moving to the right by one unit as if you plug in 0, it becomes (-1)^2, plug in 1, it becomes 0^2. So now x = 1 takes the place of what x = 0 would be in the x^2 graph.
Hope this helps!
x⁵ = (x-1)⁵
I would like to let
(x-1) = y
x - y = 1
x⁵ - y⁵ = 0
x² + y² = 1 + 2xy
x³ - y³ = (1 + 3xy)
(x³-y³)(x²+y²) = (1+2xy)(1+3xy)
x⁵ + x³y² - y³x² - y⁵ = 1 + 3xy + 2xy + 6(xy)²
x⁵ - y⁵ + x²y²(x-y) = 1+5xy + (6xy)²
Let xy = a
5a² + 5a + 1 = 0
a = -5 ± √(25 - 20)/10
a = -5 ± √5/10
x - y = 1
xy = (-5 ± √5)/10
Solving the system u get the value of x
No Solutions in real numbers my first guess
Isn't it a quintic? 😅
Kind of but the ^5 term cancels off immediately so you’re left with a quartic
It's a decapitated quintic, with a = 0 because the x^5 cancels out. So, if a = 0, it's not technically a quintic, it just looks like one till you mess it up.
problem
x⁵ = (x-1)⁵
x⁵ - (x-1)⁵ = 0
[x-(x-1)][x⁴+(x-1)x³+(x-1)²x²+(x-1)³x+(x-1)⁴]=0
x⁴+(x-1)x³+(x-1)²x²+(x-1)³x+(x-1)⁴=0
This expands to
5x⁴-10x³ +10x²-5x+1 = 0
Let
x= y+1/2
5y⁴ +(5/2) y² +(1/16) = 0
80 y⁴ +40y² +1=0
y² = (-5± 2√5)/20
Two solutions for y²
y² =-(√5/20) ( √5 -2 )
y² =-(√5/20) ( √5 + 2 )
Solutions for y
y = -i (√5/2) √(√5-2)
y = i (√5/2) √(√5-2)
y = -i (√5/2) √(√5+2)
y = i (√5/2) √(√5+2)
Back substitution
x = y + 1/2
x solutions
x = (1/2)-i (√5/2) √(√5-2)
x = (1/2)+i (√5/2) √(√5-2)
x = (1/2)-i (√5/2) √(√5+2)
x = (1/2)+i (√5/2) √(√5+2)
answer
x ∈ { (1/2) - i (√5/2) √(√5+2),
(1/2) - i (√5/2) √(√5-2),
(1/2) + i (√5/2) √(√5-2),
(1/2) + i (√5/2) √(√5+2) }