I’d do exactly what you did for alpha^2 + beta^2 + gamma^2 but there is a slicker approach for alpha^3 + beta^3 + gamma^3 using the original polynomial. alpha^3 = -2 alpha -1 and similarly for beta^3 and gamma^3 so the sum of those cubes is therefore -2(alpha + beta + gamma) -3 = -3.
You can define the recursive sequence relation : u(n+3)+2u(n+1)+u(n) = 0 Sequences that verify this relation have a closed form of the form : u(n) = A a^n+B b^n + C c^n where a,b and c are the roots of x^3+2x+1 (inject u(n) = x^n into the relation to see why) The specific sequence we are interested in is : u(n) = a^n + b^n + c^n (A=B=C=1) To compute u(n) using the recursive relation we need a set of three values : u(0) = 3 u(1) = a+b+c = 0 (as you have shown) u(2) = a²+b²+c² = (a+b+c)²-2(ab+bc+ca) From there use u(n+3) = -2u(n+1)-u(n) and you can get all values of u(n) easily.
To find the sum of the n-powers of the roots, just divide the polynomial x^n by x^3+2x+1. The remainder will be a quadratic so it will express α^n in terms of 1, α and α^2. Hence you'll only have to compute α+β+γ and α^2+β^2+γ^2 for any power of n.
If you are masochistic try the multinomial expansion on (alpha + beta + gamma)^5 The given cubic implies alpha + beta + gamma=0, (alpha*beta + alpha*gamma + beta*gamma)=2, alpha*beta*gamma=-1 (These are called elementary symmetric polynomials in the roots that are expressible as coefficients(Vieta) Then use (alpha + beta + gamma)^5 = 0 and some tedious algebra/factoring on the expanded form right hand side. I did not work it all out, but it is probably about the same amount of effort as you presented in video.
Everyone has given really slick solutions for sum of cubes, here is my way😅 begin by noticing alpha+beta+gamma=0 thus the sum of the cubes of the roots will be 3(alpha) (beta) (gamma) = 3(-1) = -3
My favorite is using polynomial long division. Given a polynomial p(x) consider p‘(x)/p(x) You get Sum from i=0 to inf R(i) x^(-i-1) where R(i) is the sum of the roots to the power i of p(x). here we get 3/x + 0/x^2 + -4/x^3 + -3/x^4 + 8/x^5 + 10/x^6 + … So R(0) = 3 R(1) = 0 R(2) = -4 R(3) = -3 R(4) = 8 and finally R(5) = 10 :)
a+b+c=0 ab+ac+bc=2 a^2+b^2+c^2=0^2-2*2=-4 The sum of cubes is obvious from the equation. a^3+b^3+c^3=-2(a+b+c)-3=-3 Then, multiplying the equation with x^2 we have a^5+b^5+c^5=-2*(a^3+b^3+c^3)-(a^2+b^2+c^2)=-2*(-3)-(-4)=6+4=10 In fact, for any S_n=a^n+b^n+c^n we don't need the expansion of (a+b+c)^n. Every S_n can be obtained by recuration from the equation and Viette.
Rewrite the eqn as x^3 = -2x-1 to easily solve α^3+β^3+γ^3=-2(0)-3=-3 since α+β+γ=0 Rewrite the eqn as x^2 = -2-(1/x) to easily solve α^2+β^2+γ^2=-6-(2/-1)=-4 since αβ+αγ+βγ=2 and αβγ=-1
Let's note a, b, c the 3 roots of the equation (I have no greek letters) As a^3 +2.a +1 = 0, we have a^3 = -2.a -1 and a^5 = -2.a^3 -a^2 = -2.(-2.a -1) -a^2 = -a^2 +4.a +2 Idem with b and C So, a^5 + b^5 + c^5 = -(a^2 + b^2 + c^2) +4.(a + b + c) +6 Now a + b + c = 0 and a.b + a.c + b.c = 2 (symetric functions of the roots of the equation) So (a^2 + b^2 + c^2) =(a + b + c)^2 - 2.(a.b +a.c + b.c) = 0^2 - 2.2 - -4 Finally: (a^5 + b^5 + c^5) = - (-4) +4.0 + 6 = 10
an easier solution: x^3+2x+1=0 by vieta's formula Σα = -b/a = 0 S1 = 0 (where S1 is the sum of roots, S2 is the sum of the squares of the roots S3 is the sum of the cubes of the roots and so on) by considering each root and summing the 3 results together α^3+2α+1 = 0 β^3+2β+1 = 0 γ^3+2γ+1 = 0 you get S3 + 2S1 + 3 = 0 S3 + 2*0 + 3 = 0 S3 = -3 dividing the original equation by x results in x^2 + 2 + 1/x = 0 rewriting for S S2 + 6 + S(-1) = 0 {note that S(-1) is the sum of the reciprocals of the roots} by vieta's again S(-1) = -c/d = -2 solving for S2 S2 + 6 - 2 = 0 S2 = -4 multiplying the original equation by x^2 results in x^5 + 2x^3 + x^2 = 0 rewriting for S S5 + 2S3 + S2 = 0 S5 +2*-3 + -4 = 0 S5 = 6 + 4 = 10 therefore α^5 + β^5 + γ^5 = 10
ChatGPT 4o solved this problem from the statement in the video description (slightly edited): "Given that α, β, γ are roots of the equation x^3 + 2x + 1 = 0, calculate α^5 + β^5 + γ^5."
Symmetric polynomials In fact we have special case called power sums Newton-Girard formulas then Vieta formulas should solve the problem To use Vieta formulas you must have elementary symmetric polynomials and Newton-Girard formulas allow to express power sums in terms of elementary symmetric polynomials
Perhaps the shortest solution is to calculate a2 + b2 + c2 by using Vieta’s formula (I replaced the roots with a b c and x2 means a square) and a3 + b3 + c3 by replacing the roots in the equation. Another interesting approach (because it starts from nowhere) is the following: Let’s define X= a4 + b4 + c4 and let’s multiply this equation by a+b+c (=0 by Vieta’s formula) 0= a5 + b5 + c5 + a*(b4+c4) + b*(a4+c4) + c*(a4+b4); now by multiplying out and rearranging the summands we get (*) a5 + b5 + c5 = -ab*(a3+b3) - bc*(b3+c3) - ac*(a3+c3); Let’s transform any of the symmetric summands on the right by using the formula for the sum of cubs: ab*(a3+b3) = ab*(a+b)*(a2-ab+b2) = ab*(a+b)*((a+b)2 - 3ab)= {by using abc = -1 and a+b+c=0 we have ab= (-1)/c and (a+b)= (-c)} = c2 - 3ab Replacing these result in (*) we receive a5 + b5 + c5= -c2 + 3ab - a2 + 3bc - b2 + 3ac, by adding and subtracting 2ab + 2bc + 2ca and applying the formula for the sum of the squares we finally get a5 + b5 + c5= -(a + b + c)2 + 5(ab + bc + ca)= 0 + 5*2 = 10
Just a note for the future: standard text notation for powers is x^p, so x cubed is x^3. Or, if you have something that inserts unicode for you, superscripts are easy: x³. On Windows, the Character Map app is builtin, while on Mac, the 'keyboard and emoji viewer' menu does it.
I did it a different way. I started with something you did too: a+b+c=0 ab+ac+bc=2 abc=-1 From the first equation c=-(a+b) (let's call this A) substitute that into the other equations you get: a^2b + ab^2 = 1 (B) and ab-(a+b)^2 = 2 (C) so using (A) and the binomial theorem, c^5 =-(a+b)^5 = -(a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5) so a^5 + b^5 + c^5 = -(5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4) = -5ab(a^3 + 2a^2b + 2a^b^2 + b^3) simplify this using (B) to = -5ab(a^3 + b^3 + 2) now use the binomial theorem again, but for the cube of sum: = -5ab((a+b)^3 -3(a^2b+ab^2) + 2) use (B) again: = -5ab((a+b)^3 -3 + 2) = -5ab((a+b)^3 -1) = 5(ab - ab(a+b)^3) use (B) once more: = 5(ab - (a+b)^2) now use (C): = 5(2) = 10
A simpler approach is note that f(alpha) + f(beta) + f(gamma) = 0, which gives us the sum of cubes and then note alpha^2 f(alpha) + beta^2 f(beta) + gamma^2 f(gamma)=0 which gives us the 5th powers sum
Thank you very much for such great video. I know many great books in algebra, containing such problems. But I am looking for a better one in your opinion, could you please suggest? Thanks❤
There's also a nice side result here: the sum of the squares is negative, hence two of the roots are complex and conjugated
Well the conjugate is a given since a polynomial cannot have a complex solution without it's conplex conjugate also being a solution.
@@harry_dum7721a polynomial can have non-real solutions that are not conjugate pairs if the coefficients are non-real
@@harry_dum7721a polynomial over the reals
Since α³=-2α-1, then α³+β³+γ³=-2(α+β+γ)-3=-3
Nice, this saves us a lot of effort!
Multiply by alpha^2 and you can do the same to obtain the 5th powers
Exactly what I was about to say 👍
2:30, by the same token α^3+β^3+γ^3 = -2(α+β+γ)-3, which simplifies the calculation further.
Yeah. I also noticed that. From other hand, we got some identity as byproduct.
I’d do exactly what you did for alpha^2 + beta^2 + gamma^2 but there is a slicker approach for alpha^3 + beta^3 + gamma^3 using the original polynomial. alpha^3 = -2 alpha -1 and similarly for beta^3 and gamma^3 so the sum of those cubes is therefore -2(alpha + beta + gamma) -3 = -3.
You can define the recursive sequence relation :
u(n+3)+2u(n+1)+u(n) = 0
Sequences that verify this relation have a closed form of the form :
u(n) = A a^n+B b^n + C c^n
where a,b and c are the roots of x^3+2x+1
(inject u(n) = x^n into the relation to see why)
The specific sequence we are interested in is :
u(n) = a^n + b^n + c^n (A=B=C=1)
To compute u(n) using the recursive relation we need a set of three values :
u(0) = 3
u(1) = a+b+c = 0 (as you have shown)
u(2) = a²+b²+c² = (a+b+c)²-2(ab+bc+ca)
From there use u(n+3) = -2u(n+1)-u(n) and you can get all values of u(n) easily.
I love this approach!
To find the sum of the n-powers of the roots, just divide the polynomial x^n by x^3+2x+1. The remainder will be a quadratic so it will express α^n in terms of 1, α and α^2. Hence you'll only have to compute α+β+γ and α^2+β^2+γ^2 for any power of n.
Another way
x^3 + 2 x + 1 = 0
implies alpha + beta + gamma = 0
alpha * beta + beta * gamma
+ gamma * alpha = 2
x^5 + 2 x^3 + x^2 = 0
x^5 = 2 ( 2 x + 1) - x^2
Hereby
alpha ^5 + beta ^ 5 + gamma ^5
=- alpha ^2 -beta ^ 2 - gamma ^5
+ 4 ( alpha + beta + gamma)
+ 6
= - ( alpha + beta + gamma) ^2
+ 2 (alpha * beta + beta * gamma
+ gamma * alpha)
+6
= 2 * 2 + 6
= 10
If you are masochistic try the multinomial expansion on (alpha + beta + gamma)^5
The given cubic implies alpha + beta + gamma=0, (alpha*beta + alpha*gamma + beta*gamma)=2, alpha*beta*gamma=-1
(These are called elementary symmetric polynomials in the roots that are expressible as coefficients(Vieta)
Then use (alpha + beta + gamma)^5 = 0 and some tedious algebra/factoring on the expanded form right hand side.
I did not work it all out, but it is probably about the same amount of effort as you presented in video.
Everyone has given really slick solutions for sum of cubes, here is my way😅 begin by noticing alpha+beta+gamma=0 thus the sum of the cubes of the roots will be 3(alpha) (beta) (gamma) = 3(-1) = -3
P = a + b + c; Q = ab + bc + ac; R = abc;
a^5 + b^5 + c^5 = P^5 - 5(P^2 - Q)(PQ - R)
out of nowhere backwards derivation that shows none of the work you did, only the result...
@@HoSza1say that to Gauss and Ramanujan
@@mephist43Did you see any of them appear here? 😮
Easier? let d(x)=x^3+2*x+1=0 => x^5 = -x^2 +4*x+2 (i.e. remainder from x^5/d(x)). Given a+b+c =0 => a^5+b^5+c^5 = 6-(a^2+b^2+c^2).
Now (a+b+c)^2=a^2+b^2+c^2+2*(a*b+a*c+b*c). now (a*b+a*c+b*c)=2 > a^5+b^5+c^5 = 6-(-4) = 10.
My favorite is using polynomial long division.
Given a polynomial p(x) consider
p‘(x)/p(x)
You get
Sum from i=0 to inf
R(i) x^(-i-1)
where R(i) is the sum of the roots to the power i of p(x).
here we get
3/x + 0/x^2 + -4/x^3 + -3/x^4 + 8/x^5 + 10/x^6 + …
So
R(0) = 3
R(1) = 0
R(2) = -4
R(3) = -3
R(4) = 8
and finally
R(5) = 10
:)
a+b+c=0
ab+ac+bc=2
a^2+b^2+c^2=0^2-2*2=-4
The sum of cubes is obvious from the equation.
a^3+b^3+c^3=-2(a+b+c)-3=-3
Then, multiplying the equation with x^2 we have
a^5+b^5+c^5=-2*(a^3+b^3+c^3)-(a^2+b^2+c^2)=-2*(-3)-(-4)=6+4=10
In fact, for any S_n=a^n+b^n+c^n we don't need the expansion of (a+b+c)^n. Every S_n can be obtained by recuration from the equation and Viette.
Rewrite the eqn as x^3 = -2x-1 to easily solve α^3+β^3+γ^3=-2(0)-3=-3 since α+β+γ=0
Rewrite the eqn as x^2 = -2-(1/x) to easily solve α^2+β^2+γ^2=-6-(2/-1)=-4 since αβ+αγ+βγ=2 and αβγ=-1
Let's note a, b, c the 3 roots of the equation (I have no greek letters)
As a^3 +2.a +1 = 0, we have a^3 = -2.a -1
and a^5 = -2.a^3 -a^2 = -2.(-2.a -1) -a^2 = -a^2 +4.a +2
Idem with b and C
So, a^5 + b^5 + c^5 = -(a^2 + b^2 + c^2) +4.(a + b + c) +6
Now a + b + c = 0 and a.b + a.c + b.c = 2 (symetric functions of the roots of the equation)
So (a^2 + b^2 + c^2) =(a + b + c)^2 - 2.(a.b +a.c + b.c) =
0^2 - 2.2 - -4
Finally: (a^5 + b^5 + c^5) = - (-4) +4.0 + 6 = 10
an easier solution:
x^3+2x+1=0
by vieta's formula Σα = -b/a = 0
S1 = 0 (where S1 is the sum of roots, S2 is the sum of the squares of the roots S3 is the sum of the cubes of the roots and so on)
by considering each root and summing the 3 results together
α^3+2α+1 = 0
β^3+2β+1 = 0
γ^3+2γ+1 = 0
you get
S3 + 2S1 + 3 = 0
S3 + 2*0 + 3 = 0
S3 = -3
dividing the original equation by x results in
x^2 + 2 + 1/x = 0
rewriting for S
S2 + 6 + S(-1) = 0 {note that S(-1) is the sum of the reciprocals of the roots}
by vieta's again S(-1) = -c/d = -2
solving for S2
S2 + 6 - 2 = 0
S2 = -4
multiplying the original equation by x^2 results in
x^5 + 2x^3 + x^2 = 0
rewriting for S
S5 + 2S3 + S2 = 0
S5 +2*-3 + -4 = 0
S5 = 6 + 4 = 10
therefore α^5 + β^5 + γ^5 = 10
ChatGPT 4o solved this problem from the statement in the video description (slightly edited):
"Given that α, β, γ are roots of the equation x^3 + 2x + 1 = 0, calculate α^5 + β^5 + γ^5."
Can You prove that
√((100!×99!×98!.....4!×3!×2!)50!) is a integer?
Symmetric polynomials
In fact we have special case called power sums
Newton-Girard formulas then Vieta formulas should solve the problem
To use Vieta formulas you must have elementary symmetric polynomials
and Newton-Girard formulas allow to express power sums in terms of elementary symmetric polynomials
Symmetric polynomials and logarithmic derivatives (hence Newton-Girard formulas) are very powerful tools.
Let roots be a, b, c
Noting abc = -1 , ab +bc+ ca = 2 , a + b+ c = 0,
a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab + bc + ca) = -4
And a^3 = -2a - 1
⇒ a^5 = (a^2)(-2a -1) = -2a^3 -a^2
I) a^5 = -a^2 +4a + 2
II) b^5 = -b^2 +4b + 2
III) c^5 = -c^2 +4c + 2
--------------------------------------
Adding
(a^5 + b^5 + c^5) = -(-4) + 4(0) + 6 = 10
that was kinda thrilling
Let a, b, c be roots of x³+2x+1=0 Then: • a+b+c=0
• ab+bc+ca=2
• abc=-1
Note that
• 0=(a+b+c)²
=a²+b²+c²+2(ab+bc+ca)
=a²+b²+c²+2×2 --> a²+b²+c²=-4
• 0=(a+b+c)³
=a³+b³+c³+3(ab+bc+ca)(a+b+c)
-3abc
=a³+b³+c³+3 --> a³+b³+c³=-3
• 12=(a³+b³+c³)(a²+b²+c²)
=a⁵+b⁵+c⁵+a³b²+a³c²
+b³a²+b³c²+a²c³+b²c³
=a⁵+b⁵+c⁵+a²b²(a+b+c)-a²b²c
+b²c²(b+c+a)-ab²c²
+a²c²(c+a+b)-a²bc²
=a⁵+b⁵+c⁵-a²b²c-ab²c²-a²bc²
=a⁵+b⁵+c⁵-abc(ab+bc+ac)
=a⁵+b⁵+c⁵+2
Therefore a⁵+b⁵+c⁵=10
If a+b+c=0 then a³+b³+c³=3abc or in this case alpha beta and gamma(i couldnt find them in my keyboard)
It works for all numbers btw
With luck and more power to you.
Way too fast! I didn't even know it was over!
Perhaps the shortest solution is to calculate a2 + b2 + c2 by using Vieta’s formula (I replaced the roots with a b c and x2 means a square) and a3 + b3 + c3 by replacing the roots in the equation.
Another interesting approach (because it starts from nowhere) is the following:
Let’s define X= a4 + b4 + c4 and let’s multiply this equation by a+b+c (=0 by Vieta’s formula)
0= a5 + b5 + c5 + a*(b4+c4) + b*(a4+c4) + c*(a4+b4); now by multiplying out and rearranging the summands we get
(*) a5 + b5 + c5 = -ab*(a3+b3) - bc*(b3+c3) - ac*(a3+c3);
Let’s transform any of the symmetric summands on the right by using the formula for the sum of cubs:
ab*(a3+b3) = ab*(a+b)*(a2-ab+b2) = ab*(a+b)*((a+b)2 - 3ab)= {by using abc = -1 and a+b+c=0 we have ab= (-1)/c and (a+b)= (-c)}
= c2 - 3ab
Replacing these result in (*) we receive
a5 + b5 + c5= -c2 + 3ab - a2 + 3bc - b2 + 3ac, by adding and subtracting 2ab + 2bc + 2ca and applying the formula for the sum of the squares we finally get
a5 + b5 + c5= -(a + b + c)2 + 5(ab + bc + ca)= 0 + 5*2 = 10
Just a note for the future: standard text notation for powers is x^p, so x cubed is x^3.
Or, if you have something that inserts unicode for you, superscripts are easy: x³. On Windows, the Character Map app is builtin, while on Mac, the 'keyboard and emoji viewer' menu does it.
This is the first youtube problem ive ever solved
I did it a different way.
I started with something you did too:
a+b+c=0
ab+ac+bc=2
abc=-1
From the first equation c=-(a+b) (let's call this A)
substitute that into the other equations you get:
a^2b + ab^2 = 1 (B)
and
ab-(a+b)^2 = 2 (C)
so using (A) and the binomial theorem, c^5 =-(a+b)^5 = -(a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5)
so a^5 + b^5 + c^5 = -(5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4)
= -5ab(a^3 + 2a^2b + 2a^b^2 + b^3)
simplify this using (B) to = -5ab(a^3 + b^3 + 2)
now use the binomial theorem again, but for the cube of sum:
= -5ab((a+b)^3 -3(a^2b+ab^2) + 2)
use (B) again: = -5ab((a+b)^3 -3 + 2) = -5ab((a+b)^3 -1)
= 5(ab - ab(a+b)^3)
use (B) once more: = 5(ab - (a+b)^2)
now use (C): = 5(2) = 10
A simpler approach is note that f(alpha) + f(beta) + f(gamma) = 0, which gives us the sum of cubes and then note alpha^2 f(alpha) + beta^2 f(beta) + gamma^2 f(gamma)=0 which gives us the 5th powers sum
Thank you very much for such great video.
I know many great books in algebra, containing such problems. But I am looking for a better one in your opinion, could you please suggest?
Thanks❤
who needs netflix
Math vids and chill
A very nice approach is to use polynomial long division involving the function and its derivative
x^3+2x+1=0 does not have 3 solution it ahs only one if you graph the function
it has 1 real solution but two complex solutions.
4:52 I want to try to solve this system of equations.. I will post the results as a reaction
It does not help, as you end up with another cubic equation. In that case you can just calculate the original roots and you are done as well
1) α³ + β³ + γ³ + 2(α + β + γ) + 3 = 0 tanks @ivanerofeev. 2) x^2 + 2 + 1/x = 0 α ^2 + β^2 + γ^2 + 6 + (αβ + αγ +βγ)/αβγ = 0 α ^2 + β^2 + γ^2 + 6 + (2)/(-1) =0 α ^2 + β^2 + γ^2 = -4 😎