Find x^4+y^4+z^4
Thread Rating:
Poll
| No votes (0%) | |||
| No votes (0%) | |||
 | 1 vote (100%) | ||
| No votes (0%) | |||
| No votes (0%) | |||
| | 1 vote (100%) | ||
| | 1 vote (100%) | ||
| | 1 vote (100%) | ||
| | 1 vote (100%) | ||
| | 1 vote (100%) |
1 member has voted
March 30th, 2020 at 11:47:33 AM
permalink
This puzzle is definitely beer worthy. If you're looking for something a little less challenging, I recommend the easy math puzzles thread. I will admit I was stuck on this one and needed some help on a method I didn't know to solve it. Now, I have what I believe is the answer, but it would be nice to have it confirmed. That said, here is the problem:
Given:
x + y + z = 1
x^2 + y^2 + z^2 = 4
x^3 + y^3 + z^3 = 9
What is x^4 + y^4 + z^4 ?
Usual rules:
I will provide hints if everybody seems to be stuck.
Given:
x + y + z = 1
x^2 + y^2 + z^2 = 4
x^3 + y^3 + z^3 = 9
What is x^4 + y^4 + z^4 ?
Usual rules:
- Please don't just plop a URL to a solution elsewhere until a winner here has been declared.
- All those who have won a beer previously are asked to not post answers or solutions for 24 after this posting. Past winners who must chime in early, may PM me.
- Beer to the first satisfactory answer and solution, subject to rule 2.
- Please put answers and solutions in spoiler tags.
I will provide hints if everybody seems to be stuck.
Last edited by: Wizard on Mar 30, 2020
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 30th, 2020 at 2:39:33 PM
permalink
I got a PM from someone who was under the impression I didn't know the answer to my own question. As always, there is a chance I could be wrong, but I do have what I think is a correct answer. I am not crystal clear why the method I followed works, which I hope can be a topic of discussion once we at least come to a majority agreement what the answer is.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 30th, 2020 at 5:50:00 PM
permalink
I received a correct answer and solution by PM from an existing Beer Club member. The clock is ticking for non-members to get in the club before the 24-hour clock expires.
Last edited by: Wizard on Mar 30, 2020
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 30th, 2020 at 7:00:05 PM
permalink
x^4 + y^4 + z^4 = 97/6 = 16.16666 ?
I will show the calculations if correct.
March 30th, 2020 at 7:38:40 PM
permalink
Quote: ssho88
x^4 + y^4 + z^4 = 97/6 = 16.16666 ?
I will show the calculations if correct.
Yes, I agree. May I see the calculations?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 30th, 2020 at 7:42:18 PM
permalink
Quote: WizardQuote: ssho88
x^4 + y^4 + z^4 = 97/6 = 16.16666 ?
I will show the calculations if correct.Yes, I agree. May I see the calculations?
x + y + z = 1-----------------------Eq1
x^2 + y^2 + z^2 = 4-----------------Eq2
x^3 + y^3 + z^3 = 9-----------------Eq3
Eq1 * Eq2
(x + y + z)*(x^2 + y^2 + z^2) = 1 * 4 = 4
(x^3 + y^3 + z^3) + yx^2 + xy^2 + xz^2 + zx^2 + zy^2 + yz^2 = 4
Substitute Eq3 into above equation,
9 + yx^2 + xy^2 + xz^2 + zx^2 + zy^2 + yz^2 = 4
yx^2 + xy^2 + xz^2 + zx^2 + zy^2 + yz^2 = -5 -----------------Eq4
Eq1 * Eq1
(x + y + z)*(x + y + z) = 1
(x^2 + y^2 + z^2) + 2(xy + xz + yz) = 1
Substitute Eq2 into above equation,
4 + 2(xy + xz + yz) = 1
(xy + xz + yz) = -3/2 -----------------Eq5
Eq1 * Eq5
(x + y + z)*(xy + xz + yz) = 1 * -3/2 = -3/2
(yx^2 + xy^2 + xz^2 + zx^2 + zy^2 + yz^2) + 3(xyz) = -3/2
Substitute Eq4 into above equation,
So, -5 + 3(xyz) = -3/2
(xyz) = 7/6 -----------------Eq6
Eq2 * Eq5
(x^2 + y^2 + z^2)*(xy + xz + yz) = 4*-3/2 = -6
(yx^3 + zx^3 + xy^3 + zy^3 +xz^3 + yz^3) + xyz^2 + xzy^2 + yzx^2 = -6
(yx^3 + zx^3 + xy^3 + zy^3 +xz^3 + yz^3) + xyz(x + y + z) = -6
Substitute Eq6 and Eq1 into above equation,
(yx^3 + zx^3 + xy^3 + zy^3 +xz^3 + yz^3) + 7/6(1) = -6
(yx^3 + zx^3 + xy^3 + zy^3 +xz^3 + yz^3) = -6 - 7/6 = -43/6 -----------------Eq7
Eq1 * Eq3
(x + y + z)*(x^3 + y^3 + z^3) = 9
x^4 + y^4 + z^4 + (yx^3 + zx^3 + xy^3+ zy^3 + xz^3+ yz^3) = 9
Substitute Eq7 into above equation,
x^4 + y^4 + z^4 + (-43/6) = 9
x^4 + y^4 + z^4 = 9 + 43/6 = 97/6 = 16.166666
Last edited by: ssho88 on Mar 30, 2020
April 4th, 2020 at 7:04:15 AM
permalink
Wizard, how did you solve it?
Here's my solution - it's similar to ssho88's:
(x + y + z)^2 = x^2 + y^2 + z^2 + 2 (xy + xz + yz)
1 = 4 + 2 (xy + xz + yz)
xy + xz + yz = -3/2
(x + y + z)^3 = x^3 + y^3 + z^3 + 3 (x^2 y + x^2 z + y^2 x + y^2 z + z^2 x + z^2 y) + 6 xyz
1 = 9 + 3 (x^2 x + x^2 y + x^2 z + y^2 x + y^2 y + y^2 z + z^2 x + z^2 y + z^2 z - x^3 - y^3 - z^3) + 6 xyz
-8 = 3 ((x^2 + y^2 + z^2)(x + y + z) - (x^3 + y^3 + z^3)) + 6 xyz
-8 = 3 (4 * 1 - 9) + 6 xyz
xyz = (15 - 8) / 6 = 7/6
Let S represent x^4 + y^4 + z^4:
(x^2 + y^2 + z^2)^2 = x^4 + y^4 + z^4 + 2 (x^2 y^2 + x^2 z^2 + y^2 z^2)
16 = S + 2 (x^2 y^2 + x^2 z^2 + y^2 z^2)
2 (x^2 y^2 + x^2 z^2 + y^2 z^2) = 16 - S
6 (x^2 y^2 + x^2 z^2 + y^2 z^2) = 48 - 3S
(x + y + z)^4 = x^4 + y^4 + z^4 + 4 (x^3 y + x^3 z + y^3 x + y^3 z + z^3 x + z^3 y) + 6 (x^2 y^2 + x^2 z^2 + y^2 z^2)
+ 12 (x^2 yz + y^2 xz + z^2 xy)
1 = S + 4 (x^3 x + x^3 y + x^3 z + y^3 x + y^3 y + y^3 z + z^3 x + z^3 y + z^3 z - x^4 - y^4 - z^4) + 48 - 3S + 12 xyz (x + y + z)
1 = S + 4 ((x^3 + y^3 + z^3)(x + y + z) - (x^4 + y^4 + z^4)) + 48 - 3S + 12 * 7/6
1 = S + 4 (9 - S) + 48 - 3S + 14
1 = 98 - 6S
x^4 + y^4 + z^4 = S = 97/6
Here's my solution - it's similar to ssho88's:
(x + y + z)^2 = x^2 + y^2 + z^2 + 2 (xy + xz + yz)
1 = 4 + 2 (xy + xz + yz)
xy + xz + yz = -3/2
(x + y + z)^3 = x^3 + y^3 + z^3 + 3 (x^2 y + x^2 z + y^2 x + y^2 z + z^2 x + z^2 y) + 6 xyz
1 = 9 + 3 (x^2 x + x^2 y + x^2 z + y^2 x + y^2 y + y^2 z + z^2 x + z^2 y + z^2 z - x^3 - y^3 - z^3) + 6 xyz
-8 = 3 ((x^2 + y^2 + z^2)(x + y + z) - (x^3 + y^3 + z^3)) + 6 xyz
-8 = 3 (4 * 1 - 9) + 6 xyz
xyz = (15 - 8) / 6 = 7/6
Let S represent x^4 + y^4 + z^4:
(x^2 + y^2 + z^2)^2 = x^4 + y^4 + z^4 + 2 (x^2 y^2 + x^2 z^2 + y^2 z^2)
16 = S + 2 (x^2 y^2 + x^2 z^2 + y^2 z^2)
2 (x^2 y^2 + x^2 z^2 + y^2 z^2) = 16 - S
6 (x^2 y^2 + x^2 z^2 + y^2 z^2) = 48 - 3S
(x + y + z)^4 = x^4 + y^4 + z^4 + 4 (x^3 y + x^3 z + y^3 x + y^3 z + z^3 x + z^3 y) + 6 (x^2 y^2 + x^2 z^2 + y^2 z^2)
+ 12 (x^2 yz + y^2 xz + z^2 xy)
1 = S + 4 (x^3 x + x^3 y + x^3 z + y^3 x + y^3 y + y^3 z + z^3 x + z^3 y + z^3 z - x^4 - y^4 - z^4) + 48 - 3S + 12 xyz (x + y + z)
1 = S + 4 ((x^3 + y^3 + z^3)(x + y + z) - (x^4 + y^4 + z^4)) + 48 - 3S + 12 * 7/6
1 = S + 4 (9 - S) + 48 - 3S + 14
1 = 98 - 6S
x^4 + y^4 + z^4 = S = 97/6
April 4th, 2020 at 5:30:42 PM
permalink
Quote: ThatDonGuyWizard, how did you solve it?
Yes. I'm proud to say I solved it for the general case without using the Newton Girard identities. It was quite a mess and took a while.
My solution looks similar to yours, just eyeballing it.
I will post something soon, as I plan to make an Ask the Wizard question out of it.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
April 6th, 2020 at 9:18:30 PM
permalink
I was feeling a little bored so decided to solve the problem for the general case. Let me state is as follows:
Let:
a = x + y + z
b = x2 + y2 + z2
c = x3 + y3 + z3
Solve for x4 + y4 + z4 in terms of a, b, and c.
Let d = x4 + y4 + z4
Let's start with a and b to find values for expressions involving both a and b.
a2 = (x + y + z)2 =
x2 + y2 + z2 + 2*(xy + xz + yz) =
b + 2(xy + xz + yz)
(1) xy + xz + yz = (a2 - b)/2
a * b = (x + y + z)*( x2 + y2 + z2)
= x3 + y3 + z3 + xy2 + xz2 + yx2 + yz2 + zx2 + zy2
= c + xy2 + xz2 + yx2 + yz2 + zx2 + zy2
(2) xy2 + xz2 + yx2 + yz2 + zx2 + zy2 = ab - c
a3 = (x + y + z)3 =
x3 + y3 + z3 + 3*( xy2 + xz2 + yx2 + yz2 + zx2 + zy2) + 6xyz =
c + 3*(ab - c) + 6xyz
6xyz = a3 - c - 3*(ab-c)
6xyz = a3 + 2c - 3*ab
(3) xyz = (a3 + 2c - 3ab)/6
b2 = (x2 + y2 + z2)2 = x4 + y4 + z4 + 2*( x2y2 + x2z2 + y2z2)
(4) x2y2 + x2z2 + y2z2 = (b2 - d)/2
a2 * b = (x2 + y2 + z2 + 2*(xy + xz + yz)) * (x2 + y2 + z2) =
x4 + y4 + z4 + 2*(x2y2 + x2z2 + y2z2) + 2(x3y + x3z + y3x + y3z + z3x + z3y) + 2(x2yz + xy2z + xyz2) =
d + 2*(x2y2 + x2z2 + y2z2) + 2(x3y + x3z + y3x + y3z + z3x + z3y) + 2xyz*(x+y+z) =
d + 2*(x2y2 + x2z2 + y2z2) + 2(x3y + x3z + y3x + y3z + z3x + z3y) + a(a3 + 2c - 3ab)/3 =
d + (b2 - d) + 2(x3y + x3z + y3x + y3z + z3x + z3y) + a(a3 - 3ab + 2c)/3 =
b2 + 2(x3y + x3z + y3x + y3z + z3x + z3y) + a(a3 - 3ab+ 2c)/3
2(x3y + x3z + y3x + y3z + z3x + z3y) = a2 * b - b2 - a(a3 - 3ab+ 2c)/3
x3y + x3z + y3x + y3z + z3x + z3y = a2 * b/2 - b2/2 - a4/6 + a2b/2 - ac/3
(5) x3y + x3z + y3x + y3z + z3x + z3y = a2 * b - b2/2 - a4/6 - ac/3
a4 = (x + y + z)4 =
x4 + y4 + z4 + 4*(x3y + x3z + y3x + y3z + z3x + z3y) + 6*(x2y2 +x2z2 + y2z2) + 12*(x2yz + xy2z + xyz2) =
d + 4*( a2b - b2/2 - a4/6 - ac/3) + 6*((b2 - d)/2) + 12*xyz*(x+y+z) =
d + 4a2b - 2b2 - (2/3)*a4 - (4/3)ac + 3*b2 - 3d + 12*((a3 + 2c - 3ab)/6)*a =
d + 4a2b - 2b2 - (2/3)*a4 - (4/3)ac + 3b2 - 3d + 2a4 + 4ac - 6a2b =
4/3*a4 + (8/3)ac - 2a2b + b2
2d = 1/3*a4 + (8/3)ac - 2a2b
d = a4/6 + (4/3)ac - a2b + b2/2
Going back to the original problem were a = 1, b = 4, and c = 9:
d = 14/6 + (4/3)1*9 - 12*4 + 42/2 = 97/6 = 16 1/6.
Let:
a = x + y + z
b = x2 + y2 + z2
c = x3 + y3 + z3
Solve for x4 + y4 + z4 in terms of a, b, and c.
Let d = x4 + y4 + z4
Let's start with a and b to find values for expressions involving both a and b.
a2 = (x + y + z)2 =
x2 + y2 + z2 + 2*(xy + xz + yz) =
b + 2(xy + xz + yz)
(1) xy + xz + yz = (a2 - b)/2
a * b = (x + y + z)*( x2 + y2 + z2)
= x3 + y3 + z3 + xy2 + xz2 + yx2 + yz2 + zx2 + zy2
= c + xy2 + xz2 + yx2 + yz2 + zx2 + zy2
(2) xy2 + xz2 + yx2 + yz2 + zx2 + zy2 = ab - c
a3 = (x + y + z)3 =
x3 + y3 + z3 + 3*( xy2 + xz2 + yx2 + yz2 + zx2 + zy2) + 6xyz =
c + 3*(ab - c) + 6xyz
6xyz = a3 - c - 3*(ab-c)
6xyz = a3 + 2c - 3*ab
(3) xyz = (a3 + 2c - 3ab)/6
b2 = (x2 + y2 + z2)2 = x4 + y4 + z4 + 2*( x2y2 + x2z2 + y2z2)
(4) x2y2 + x2z2 + y2z2 = (b2 - d)/2
a2 * b = (x2 + y2 + z2 + 2*(xy + xz + yz)) * (x2 + y2 + z2) =
x4 + y4 + z4 + 2*(x2y2 + x2z2 + y2z2) + 2(x3y + x3z + y3x + y3z + z3x + z3y) + 2(x2yz + xy2z + xyz2) =
d + 2*(x2y2 + x2z2 + y2z2) + 2(x3y + x3z + y3x + y3z + z3x + z3y) + 2xyz*(x+y+z) =
d + 2*(x2y2 + x2z2 + y2z2) + 2(x3y + x3z + y3x + y3z + z3x + z3y) + a(a3 + 2c - 3ab)/3 =
d + (b2 - d) + 2(x3y + x3z + y3x + y3z + z3x + z3y) + a(a3 - 3ab + 2c)/3 =
b2 + 2(x3y + x3z + y3x + y3z + z3x + z3y) + a(a3 - 3ab+ 2c)/3
2(x3y + x3z + y3x + y3z + z3x + z3y) = a2 * b - b2 - a(a3 - 3ab+ 2c)/3
x3y + x3z + y3x + y3z + z3x + z3y = a2 * b/2 - b2/2 - a4/6 + a2b/2 - ac/3
(5) x3y + x3z + y3x + y3z + z3x + z3y = a2 * b - b2/2 - a4/6 - ac/3
a4 = (x + y + z)4 =
x4 + y4 + z4 + 4*(x3y + x3z + y3x + y3z + z3x + z3y) + 6*(x2y2 +x2z2 + y2z2) + 12*(x2yz + xy2z + xyz2) =
d + 4*( a2b - b2/2 - a4/6 - ac/3) + 6*((b2 - d)/2) + 12*xyz*(x+y+z) =
d + 4a2b - 2b2 - (2/3)*a4 - (4/3)ac + 3*b2 - 3d + 12*((a3 + 2c - 3ab)/6)*a =
d + 4a2b - 2b2 - (2/3)*a4 - (4/3)ac + 3b2 - 3d + 2a4 + 4ac - 6a2b =
4/3*a4 + (8/3)ac - 2a2b + b2
2d = 1/3*a4 + (8/3)ac - 2a2b
d = a4/6 + (4/3)ac - a2b + b2/2
Going back to the original problem were a = 1, b = 4, and c = 9:
d = 14/6 + (4/3)1*9 - 12*4 + 42/2 = 97/6 = 16 1/6.
Last edited by: Wizard on Apr 7, 2020
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
April 7th, 2020 at 8:27:32 PM
permalink
Here is my PDF solution.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
