Can You Solve Without Finding x & y?
Вставка
- Опубліковано 2 чер 2024
- Given x + y = -2 and x^3 + y^3 = 16, we find the value of x^4 + y^4. Instead of directly finding the values of x and y, our approach involves expressing x^4 + y^4 in terms of x + y and x^3 + y^3.
00:00 Intro
00:31 An expression for x^4 + y^4
01:44 More useful expressions
03:49 Substitution
05:08 Addendum: values of x & y
the 3×-2 written without brackets seriously tripped me up
I'm subscribed to over 1000 YT channels, but I always remember to check in every Friday morning or Thursday at Midnight if I'm awake (greetings from Montana USA). Thanks for another good one!
Thank you so much!
i used sum of cubes to (x+y)()x^2-xy+y^2) = 16 => xy = 4
then (x+y)^4 - xy( 4(x^2+y^2) + 6xy) = x^4 + y^4
found x^2 + y^2 = -4 by doing (x+y)^2 -2xy
and then plugging in to get:
x^4 + y^4 = 16- 4(-16+24) = -16
I'm a math grad, and usually I know how to and that I can solve the problems you have videos on after at most a few minutes staring at the thumbnail.
This one stumped me and I had to see how you did it!
Here is a function for the power sum for any positive integer n:
P{n} = x^n + y^n = 2^(n+1) * cos(n * pi * 2/3)
This can be determined by using the Newton-Girard identities for symmetric polynomials and symmetric power polynomials to find a recurrence relation for
P{n + 2} = -2 * P{n + 1} - 4 * P{n}
and then solving the recurrence with the given initial conditions x + y = -2 and x^3 + y^3 = 16
Alternatively, we can find a quadratic equation satisfied by x and y by noting x+y=-2 and xy=4, so x and y are the roots of t²+2t+4=0.
Noting also that t³-8=t³-2³=(t-2)(t²+2t+2²)=(t-2)(t²+2t+4), we see that x and y are the two non-real cube roots of 8, hence they are 2cis(±2π/3) (in either order).
So xⁿ+yⁿ=2ⁿcis(2nπ/3)+2ⁿcis(-2nπ/3)=2ⁿ⁺¹cos(2nπ/3)
@@MichaelRothwell1 That works, but it disobeys the instruction not to solve for x and y.
@@XJWill1 That's true!
Really nice video! I solved by doing binomial expansions of (x + y)^2, (x + y)^3 and (x + y)^4, which is a bit more work. Didn't think to try (x + y)*(x^3 + y^3) to get x^4 + y^4
x and y are solutions of x²-(-2)x+4=0.
Whenever sum and multiplication of two complex number are known, sum of their nth power can be computed. This video suggests other way is also possible.
A similar approach can be used to solve more complicated systems, like
x^2+y^2=a,
x^3+y^3=b.
If you let S=x+y and P=xy, then the equations become
S^2-2P=a,
S^3-3SP=b.
Then the first equation gives P in terms of S, and substituting that into the second gives a cubic for S. Then solve that (gross) and for each value of S, you get a value of P that you can use to solve for x and y.
Love the logical way you solve this. Showing my lack of maths knowledge. X or Y or both are complex numbers.. I can see how multiplying them will give a non complex number. Adding two complex numbers - how do you get a non complex number?
They're conjugates of each other... Like, one of them is (a+bi) and the other is (a-bi), so if you add them together, the imaginary component cancels out and you just get 2a.
@@mrphlip thanks for explaining.
In fact x and y are the roots of the quadratic equation t²+2t+4=0 (this is because x+y=-2 and xy=4).
The solutions (which can easily be found by completing the square or the quadratic formula) are t=-1±√3i, so x, y are -1±√3i in either order.
Since the non-real cube roots of 1 are (-1±√3i)/2, x and y are the non-real cube roots of 8 (which explains why x³+y³=16).
This can also be seen without solving the quadratic by factorising t³-8=t³-2³=(t-2)(t²+2t+2²).
(x + y)(x^2 + y^2) = x^3 + xy^2 + x^2 y + y^3
-2 (x^2 + y^2) = x^3 + y^3 + xy(x+y)
-2 (x^2 + y^2) = 16 - 2xy
x^2 + y^2 = xy - 8
Also, (x+y)^2 = x^2 + 2xy + y^2
4 = (x^2 + y^2) + 2xy
4 = xy - 8 + 2xy
12 = 3xy
4 = xy
x^2 + y^2 = xy - 8
x^2 + y^2 = 4 - 8 = -4
(x^2 + y^2)^2 = x^4 + y^4 + 2 x^2 y^2
16 = x^4 + y^4 + 2(4)^2
-16 = x^4 + y^4
I realize your title said "without finding x & y", but I suspected there was no solution for x and y, given the way the sign changed from negative for x + y to positive for x^3 + y^3, and then negative again for x^4 + y^4. I was able to find a pair of values for x and y such that x + y = -2, x^3 + y^3 = 16, and x^4 + y^4 has the value you obtained in the video. But the sign changes are still a bit mystifying to me. Is there a way to determine what the sign for x^n + y^n is going to be for a particular n, without finding the value of x^n + y^n?
Symmetric functions
Newton-Girard formulas
Indeed! See en.m.wikipedia.org/wiki/Newton%27s_identities
elegant
At the end: ??? y=-2-x ???
Rearrange x+y=-2 for y 👍
Solving for z^2 is easier...
x=-1+z and y=-1-z; (z-1)^3-(z+1)^3-16=-2(3z^2+1)-16=0; 6z^2+18=0; z^2=-3
x^4+y^4=(z+1)^4+(z-1)^4=2(z^4+6z^2+1)=2(9-18+1)=-16
Nice problem and solution.
We can easily find the values of x and y by noting x+y=-2 and xy=4, so x and y are the roots of t²+2t+4=0.
Noting also that t³-8=t³-2³=(t-2)(t²+2t+2²)=(t-2)(t²+2t+4), we see that x and y are the two non-real cube roots of 8, which also explains why x³+y³=16.
This is a very nice alternative solution. I like the connection to Vieta's formula for the quadratic as another nice way of solving without finding x and y explicitly!