You need to know these tricks: Substitute x=y+1/2 into the given equation and rearrange to (y+1/2)^6-(y-1/2)^6=0. Pascal's triangle 1 6 15 20 15 6 1 2[6(1/2)y^5+20(1/2)^3*y^3+6(1/2)^5*y]=0. Expand: 6y^5+5y^3+(3/8)y=0. Divide by y and multiply by 8: 48y^4+40y^2+3=0. Apply quadratic formula: y^2=(-40±32)/96=-1/12 or -3/4. Take square roots: y=±i√3/6 or ±i√3/2. Also note that y=0. Thus, x=y+1/2=(3±i√3)/6 or (1±i√3)/2 or 1/2.
0 is not a root, then divide by x⁶ 1 = (1 - 1/x)⁶ t := 1 - 1/x, t ≠ 1 t⁶ = 1 has 5 roots: {e^(⅓πni), n from 1 to 5} = = {-1, ½ ± ½sqrt(3) i, -½ ± ½sqrt(3) i} then 1 - t = 1/x can be equal: {2, ½ ± ½sqrt(3) i, 3/2 ± ½sqrt(3) i} = = {2, e^(±⅓πi), sqrt(3) e^(±⅙πi)} then x can be equal: {½, e^(±⅓πi), 1/sqrt(3) • e^(±⅙πi)} = = {½, ½ ± ½sqrt(3) i, ½ ± ⅙sqrt(3) i} it is the final answer
Treat x^6-(x-1)^6 = (x^2)^3 - ((x-1)^2)^3 instead and then apply the identities a^2-b^2=(a-b)(a+b) and a^3-b^3=(a-b)(a^2+ab_b^2) to get x^6-(x-1)^6 = (2x-1) [ x^4 + x^2(x-1)^2 + (x-1)^4]. Then note that x^4 + x^2(x-1)^2 + (x-1)^4 = [ x^2+(x-1)^2]^2-[x(x-1)]^2. Apply the identity a^2-b^2 = (a-b)(a+b) again and get x^4 + x^2(x-1)^2 + (x-1)^4 = [ x^2+(x-1)^2-x(x-1) ] [ x^2+(x-1)^2+x(x-1) ] Polynomial in each squared bracket is a quadratic one after simplification.
I always use b/2 = arithmetic mean of roots, A, and c = geometric mean squared of roots, G, with (x)2 + b*x + c for x = A +- sqrt((A)2 - G). It is a much quicker route to quadratic roots.
much simpler solution if you use complex numbers in polar form. x^6 = (x-1)^6 --> (x-1)/x = 1^(1/6) = r, where r is one of the 6 roots of 1 on the complex plane [two lie on the real axis at 1 and -1] --> x = 1/(1-r), which rules out r = 1. therefore, the 5 solutions correspond to r = exp(i*Pi*n/3), where n = 1, 2, 3, 4, and 5 (noting that n=0 is ruled out).
That's the easiest way to solve it. Showing a geometric interpretation of this equation, with the solutions describing a circle around 0 would help generalize solutions for (x-1)^n=x^n starting with the obvious solution x=1/2.
If x^6 = (x - 1)^6, then x - 1 = a x, where a is 6th root of unity, i.e. 1, j, j^2, -1, - j and - j^2. The solutions are x = 1 / (1 - a), with a ≠ 1 : { 1/2, - j, -j^2, (1 - j)/3, (1 - j^2) / 3 }. Thanks for your video.
@@jpl569 In your description of the solution, why is j the principal complex (non-real) _cube root_ of unity, rather than the principal complex (non-real) _6th root_ of unity? (More concretely, why is j = -½ + i*½√3 (with negative real part) instead of j = ½ + i*½√3 (with positive real part)? )
@@yurenchu Good question ! The 6th roots of unity are given by : z^6 = 1. Then z^3 = 1 or z^3 = -1. If z^3 = 1, we find z = 1 or j or j^2. And if z^3 = - 1, we find z = - 1 or - j or - j^2. So j is the first 3rd root of unity (j = - 1/2 + i √3 / 2), and - j^2 is the first 6th root (- j^2 = - 1/2 + i √3 / 2). An important property of j is : 1 + j + j^2 = 0, so that 1 + j = - j^2, and 1 + j^2 = - j, and so on… So that the 6th roots are also {1, 1 + j, j, -1, j^2, 1 + j^2}. Have a nice time with math !
@@jpl569 Thanks for your reply. Ah, so it's just the consequence of how you preferred to solve the equation then (namely via z^3 = 1 first). Usually, the solutions to z^6 = 1 are given in polar form as z = e^(i*k*pi/6) , with k = 0 , 1 , 2 , 3 , 4 , 5 . So the "first" sixth root of unity (let's call it _u_ ) would naturally correspond to k=1 , which is z = u = e^(i*pi/6) = cos(pi/6) + i*sin(pi/6) = ½ + i*½√3 . The six 6th roots of unity can then be given as u^0 , u^1 , u^2 , u^3 , u^4 , and u^5 . (k=0 would correspond to the "zero'th" root, which is the positive real root z = u^0 = 1 .) I was wondering what j stood for (it wasn't clarified in your opening comment, and it's an unusual use of the letter j because in math it's often used as the imaginary unit, like i), and when I figured it out by matching your solution description to the explicit solution set, I just found it remarkable and somewhat puzzling that j apparently corresponds to the k=2 solution (or possibly even the k=4 solution), instead of the k=1 solution.
@@yurenchu Thanks for your comment. I admit that "j" may look a strange definition, but I used it in all my math studies as e^(2i π/3), because in my country everybody uses it. We also learned the calculations with j and j^2, which lead efficiently to nice results. By the way, the 6th roots of unity are e^(2i k π/6) with k = 0 , 1 , 2 , 3 , 4 , 5, not z = e^(i*k*pi/6) as you wrote... these are 12th roots... and cos (π/6) = √3 / 2 and not 1/2... 😉
By visual inspection, 0.5 is clearly a solution. Also note that if we expand (x-1)^6 it will lead off with x^6, which will cancel. Therefore we only expect five roots. The four roots other than 0.5 will be complex. To find them we proceed as follows. Divide both sides by x^6 to get (x-1)^6 / x^6 = 1 [(x-1)/x]^6 = 1 (1 - 1/x)^6 = 1 To satisfy this, 1 - 1/x is one of the sixth roots of unity. These of course are at magnitude 1, angle a multiple of 60 degrees: Also, given the symmetry of the problem negating the right hand side doesn't change the solution set, and makes further work slightly cleaner. We have -1 + 1/x = 1 -1 + 1/x = -1 -1 + 1/x = cos(60) + j*sin(60) -1 + 1/x = cos(60) - j*sin(60) -1 + 1/x = -cos(60) + j*sin(60) -1 + 1/x = -cos(60) - j*sin(60) Now add 1 to both sides of every equation: 1/x = 2 1/x = 0
Given the symmetry, the only real root has to be 0.5. obvious. Using your transformation the other four roots are the four corners on the unit circle. Excellent.
@@KipIngram The expressions of the solutions for x can be "cleaned up" (i.e. removing complex expressions from the denominator) by multiplying both numerator and denominator of a solution by the _complex conjugate_ of the denominator. (If z = a + j*b is a complex number (with real numbers a and b ), then zᶜ = a - j*b is the complex conjugate of z .) You'll find that all five solutions are of the form x = 0.5 + j*b , which is kinda surprising.
@@yurenchu Yes, I was lazy. 🙂 Thank you. I just wanted them far enough along to check the results using my calculator, which handles complex numbers natively (Swiss Micros DM42).
@@KipIngram LOL! Fair enough... I too am lazy. Namely too lazy to get out my calculator (Casio fx-82 ; I don't know if it also does complex numbers), that's why I always calculate things like this by hand, wherever/whenever I can. :-) Never heard of the Swiss Micros DM 42 before, I had to look it up: it looks quite cool, with a nice screen to render graphs and (I'm guessing) bitmap images.
Proprietà delle potenze: stesso esponente, quindi devono avere la stessa base; essendo l'esponente pari, le basi possono essere opposte, questo conduce a due equazioni: 1) x=x-1 basi con lo stesso segno, e questa è chiaramente un'equazione impossibile x-x=1 quindi 0=1. 2) x=-x+1 basi opposte questa equazione si risolve subito x+x=1 quindi 2x=1 infine x=1/2
x⁶ = (x-1)⁶ Method 1: Note that x=0 is not a solution to this equation. So we can safely divide both sides by x⁶ : 1 = (x-1)⁶/x⁶ 1 = [(x-1)/x]⁶ 1 = (1 - 1/x)⁶ ... substitute u = (1 - 1/x) ... 1 = u⁶ This equation has six solutions in the complex plane: u = 1 , u = (1 ± i√3)/2 , u = (-1 ± i√3)/2 , u = -1 . If u = 1 : (1 - 1/x) = 1 -1/x = 0 x = -1/0 = undefined (could be +infinity or -infinity, or infinity*z where z is any unit direction in the complex plane) If u = (1 ± i√3)/2 : (1 - 1/x) = (1 ± i√3)/2 -1/x = (-1 ± i√3)/2 1/x = -(-1 ± i√3)/2 x = -2/(-1 ± i√3) ... multiply both numerator and denominator of RHS by (1 ± i√3) ... x = -2(1 ± i√3)/[(-1 ± i√3)*(1 ± i√3)] x = -2(1 ± i√3)/[(±i√3 - 1)*(±i√3 + 1)] x = -2(1 ± i√3)/[ (±i√3)² - 1² ] x = -2(1 ± i√3)/[ (i)²(±√3)² - 1² ] x = -2(1 ± i√3)/[ (-1)(3) - 1 ] x = -2(1 ± i√3)/[ -4 ] x = (1 ± i√3)/2 x = ½ ± i*½√3 If u = (-1 ± i√3)/2 : (1 - 1/x) = (-1 ± i√3)/2 -1/x = (-3 ± i√3)/2 1/x = -(-3 ± i√3)/2 x = -2/(-3 ± i√3) ... multiply both numerator and denominator of RHS by (3 ± i√3) ... x = -2(3 ± i√3)/[(-3 ± i√3)*(3 ± i√3)] x = -2(3 ± i√3)/[(±i√3 - 3)*(±i√3 + 3)] x = -2(3 ± i√3)/[ (±i√3)² - 3² ] x = -2(3 ± i√3)/[ (i)²(±√3)² - 3² ] x = -2(3 ± i√3)/[ (-1)(3) - 9 ] x = -2(3 ± i√3)/[ -12 ] x = (3 ± i√3)/6 x = ½ ± i*⅙√3 If u = -1 : (1 - 1/x) = -1 2 = 1/x x = ½ So there are five solutions, and they are: x = ½ ± i*½√3 , x = ½ ± i*⅙√3 , x = ½ . - - - Method 2: see reply to this comment, below.
Method 2: x⁶ = (x-1)⁶ ux = x-1 ... where u is any of the six (complex-valued) solutions to the equation u⁶ = 1 ; namely u = 1 , u = -1 , u = (½ ± i*½√3) , u = (-½ ± i*½√3) ... ux - x = -1 (u-1)x = -1 For u=1 this leads to 0x = -1 ==> x = undefined. For the other five values of u, we have the solution x = -1/(u-1) ... let uᶜ = the complex conjugate of u , and multiply both numerator and denominator of RHS by (uᶜ-1) ... x = -(uᶜ-1)/[(u-1)(uᶜ-1)] x = -(uᶜ-1)/[ uuᶜ - uᶜ - u +1 ] ... since all values of u are on the unit circle, uᶜ = 1/u and hence uuᶜ = |u|² = 1 ... x = -(uᶜ-1)/[ 1 - uᶜ - u +1 ] x = -(uᶜ-1)/[ 2 - (u + uᶜ) ] x = -(uᶜ-1)/[ 2 - 2*Re(u) ] ... where Re(u) is the real component of u ... x = -½(uᶜ-1)/[1 - Re(u)] x = -½(Re(u) - i*Im(u) - 1)/[1 - Re(u)] ... where Im(u) is the imaginary component of u ; u = Re(u) + i*Im(u) ... x = -½(Re(u) - 1 - i*Im(u))/[1 - Re(u)] x = ( -½(Re(u) - 1) + ½i*Im(u) )/[1 - Re(u)] x = -½(Re(u) - 1)/[1 - Re(u)] + ½i*Im(u)/[1 - Re(u)] x = ½ + i*½Im(u)/[1 - Re(u)] For u = -1 , Re(u) = -1 and Im(u) = 0 , so x = ½ + i*½(0)/[1 - (-1)] = ½ + i*0/2 = ½ For u = ½ ± i*½√3 , Re(u) = ½ and Im(u) = ±½√3 , so x = ½ + i*½(±½√3)/[1 - (½)] = ½ + i*½(±½√3)/[½] = ½ ± i*(½√3) For u = -½ ± i*½√3 , Re(u) = -½ and Im(u) = ±½√3 , so x = ½ + i*½(±½√3)/[1 - (-½)] = ½ ± i*½(½√3)/[1½] = ½ ± i*(½√3)/[3] = ½ ± i*(⅙√3) So there are five solutions, and they are: x = ½ , x = ½ ± i*½√3 , x = ½ ± i*⅙√3 .
Ok, going for the representation of the roots, we get that there are 5 roots, you showed 4 wich are on the complex plane, how would we find the one on the real plane? I can guess it's +1/2, but how do we prove it?
x^6=(x-1)^6 positive for all x . Then ((x-1)/x)^6 = 1. (x-1)/x is 6th root of unity. Then, (x-1)/x = e^(i2kpi/6)=> x=1/(1- e^(i2kpi/6)) for k=0,1..,5. For k=0, there's no solution. For k=3, x=1/2. Substitute k=1, 2, 4,5 and obtain two pairs of complex conjugates for x.
@@SpencersAcademy Okay I nail your Olympiad math problems. Let's see if you can nail our 9th grade regular maths problems. Solve sin(9x)=sin(5x)+sin(3x). You may record a video if you wish.
" x^6=(x-1)^6 positive for all x. " That's not true. As the video shows, one of the correct solutions is x = (½ + i*⅙√3) ; and for that solution, x⁶ = (½ + i*⅙√3)⁶ = [ (½(-i√3) + ½)*(i*⅓√3) ]⁶ = (½ - i*½√3)⁶ *(i⁶)*(⅓√3)⁶ = (1)*(-1)*(1/27) = -1/27 (x-1)⁶ = (½ + i*⅙√3 - 1)⁶ = -1/27 which is a negative real. (The same goes for the solution x = (½ - i*⅙√3) .) However, it doesn't affect the rest of the calculation in your comment.
Easy, the solution is x = ω. Any limit ordinal could work here really. Because any limit ordinal minus 1 is just that limit ordinal itself since it doesn't have any predecessor. As a result, there's more than 1 solution because we can't put a value of the number of limit ordinals there are. Keep in mind that limit ordinals are multiples of ω.
Why not x^6 = (x - 1)^6 => (x^3)^2 = ((x-1)^3)^2 Taking square roots of both sides, => (x^3) = +/- (x-1)^3 => x = x -1 (no solution) OR x = - (x - 1) So, x = - (x - 1) x = 1 - x 2 x = 1 x = 1/2
@@SpencersAcademy An Olympiad aspirant is very much aware of the basic identities. So, no need to reiterate those. You can directly jump redundant steps. Thank you.
In a question like this, there exist only five roots. Let me explain it this way: Let's say you expand the RHS using binomial expansion. You'll have something like this: X^6 = x^6 - 6x^5 +15x^4 - 20x^3 + 15x^2 -6x +1. Subtract x^6 from both sides makes the equation 5th-degree polynomial. And that's why we've got five roots.
No. That's a wrong approach to any mathematics question. In a case like this, it's either you follow the steps outlined in the video or you expand the right-hand side using binomial expansion.
Of course it is possible; one just has to do it properly. As outlined in several other comments, one obtais (x-1)=wx with w one of the six 6th roots of unity. For all of them apart from w=1, this is solvable: x=1/(1-w).
You need to know these tricks: Substitute x=y+1/2 into the given equation and rearrange to (y+1/2)^6-(y-1/2)^6=0. Pascal's triangle 1 6 15 20 15 6 1
2[6(1/2)y^5+20(1/2)^3*y^3+6(1/2)^5*y]=0. Expand: 6y^5+5y^3+(3/8)y=0. Divide by y and multiply by 8: 48y^4+40y^2+3=0.
Apply quadratic formula: y^2=(-40±32)/96=-1/12 or -3/4. Take square roots: y=±i√3/6 or ±i√3/2. Also note that y=0.
Thus, x=y+1/2=(3±i√3)/6 or (1±i√3)/2 or 1/2.
extremely neat! thanks!
Also thought about Pascals triangle
0 is not a root, then divide by x⁶
1 = (1 - 1/x)⁶
t := 1 - 1/x, t ≠ 1
t⁶ = 1 has 5 roots:
{e^(⅓πni), n from 1 to 5} =
= {-1, ½ ± ½sqrt(3) i,
-½ ± ½sqrt(3) i}
then 1 - t = 1/x can be equal:
{2, ½ ± ½sqrt(3) i,
3/2 ± ½sqrt(3) i} =
= {2, e^(±⅓πi), sqrt(3) e^(±⅙πi)}
then x can be equal:
{½, e^(±⅓πi), 1/sqrt(3) • e^(±⅙πi)} =
= {½, ½ ± ½sqrt(3) i,
½ ± ⅙sqrt(3) i}
it is the final answer
Excellent
By inspection x=½ --> x-1=-½ is one of the root, as raising both of them to an even number yield the same positive number.
x=0,5 0,5 - 1= - 0,5
@@habeebalbarghothy6320what
Treat x^6-(x-1)^6 = (x^2)^3 - ((x-1)^2)^3 instead and then apply the identities a^2-b^2=(a-b)(a+b) and a^3-b^3=(a-b)(a^2+ab_b^2) to get
x^6-(x-1)^6 = (2x-1) [ x^4 + x^2(x-1)^2 + (x-1)^4].
Then note that x^4 + x^2(x-1)^2 + (x-1)^4 = [ x^2+(x-1)^2]^2-[x(x-1)]^2.
Apply the identity a^2-b^2 = (a-b)(a+b) again and get
x^4 + x^2(x-1)^2 + (x-1)^4 = [ x^2+(x-1)^2-x(x-1) ] [ x^2+(x-1)^2+x(x-1) ]
Polynomial in each squared bracket is a quadratic one after simplification.
Nice one! 👍
Interesting🎉
I always use b/2 = arithmetic mean of roots, A, and c = geometric mean squared of roots, G, with (x)2 + b*x + c for x = A +- sqrt((A)2 - G). It is a much quicker route to quadratic roots.
much simpler solution if you use complex numbers in polar form.
x^6 = (x-1)^6 --> (x-1)/x = 1^(1/6) = r, where r is one of the 6 roots of 1 on the complex plane [two lie on the real axis at 1 and -1] --> x = 1/(1-r), which rules out r = 1. therefore, the 5 solutions correspond to r = exp(i*Pi*n/3), where n = 1, 2, 3, 4, and 5 (noting that n=0 is ruled out).
Nice one 👏
x/(x -1) = e ^i(k π/3) for k = 1 2, 3, 4,5..There are 5 roots as this is a polynomial of degree 5.
Excellent
Almost exactly my thought when I saw the problem.
That's the easiest way to solve it. Showing a geometric interpretation of this equation, with the solutions describing a circle around 0 would help generalize solutions for (x-1)^n=x^n starting with the obvious solution x=1/2.
🎉🎉 great
very good
Thank you. Glad you enjoyed it.
oho ur great
I would lije ti apperciat ur methods
Thanks. I am grateful.
If x^6 = (x - 1)^6, then x - 1 = a x, where a is 6th root of unity, i.e. 1, j, j^2, -1, - j and - j^2.
The solutions are x = 1 / (1 - a), with a ≠ 1 :
{ 1/2, - j, -j^2, (1 - j)/3, (1 - j^2) / 3 }.
Thanks for your video.
You're welcome, brother
@@jpl569 In your description of the solution, why is j the principal complex (non-real) _cube root_ of unity, rather than the principal complex (non-real) _6th root_ of unity?
(More concretely, why is j = -½ + i*½√3 (with negative real part) instead of j = ½ + i*½√3 (with positive real part)? )
@@yurenchu Good question ! The 6th roots of unity are given by : z^6 = 1. Then z^3 = 1 or z^3 = -1.
If z^3 = 1, we find z = 1 or j or j^2. And if z^3 = - 1, we find z = - 1 or - j or - j^2.
So j is the first 3rd root of unity (j = - 1/2 + i √3 / 2), and - j^2 is the first 6th root (- j^2 = - 1/2 + i √3 / 2).
An important property of j is : 1 + j + j^2 = 0, so that 1 + j = - j^2, and 1 + j^2 = - j, and so on…
So that the 6th roots are also {1, 1 + j, j, -1, j^2, 1 + j^2}.
Have a nice time with math !
@@jpl569 Thanks for your reply. Ah, so it's just the consequence of how you preferred to solve the equation then (namely via z^3 = 1 first).
Usually, the solutions to z^6 = 1 are given in polar form as z = e^(i*k*pi/6) , with k = 0 , 1 , 2 , 3 , 4 , 5 . So the "first" sixth root of unity (let's call it _u_ ) would naturally correspond to k=1 , which is z = u = e^(i*pi/6) = cos(pi/6) + i*sin(pi/6) = ½ + i*½√3 . The six 6th roots of unity can then be given as u^0 , u^1 , u^2 , u^3 , u^4 , and u^5 . (k=0 would correspond to the "zero'th" root, which is the positive real root z = u^0 = 1 .)
I was wondering what j stood for (it wasn't clarified in your opening comment, and it's an unusual use of the letter j because in math it's often used as the imaginary unit, like i), and when I figured it out by matching your solution description to the explicit solution set, I just found it remarkable and somewhat puzzling that j apparently corresponds to the k=2 solution (or possibly even the k=4 solution), instead of the k=1 solution.
@@yurenchu Thanks for your comment. I admit that "j" may look a strange definition, but I used it in all my math studies as e^(2i π/3), because in my country everybody uses it.
We also learned the calculations with j and j^2, which lead efficiently to nice results.
By the way, the 6th roots of unity are e^(2i k π/6) with k = 0 , 1 , 2 , 3 , 4 , 5, not z = e^(i*k*pi/6) as you wrote... these are 12th roots... and cos (π/6) = √3 / 2 and not 1/2... 😉
By visual inspection, 0.5 is clearly a solution. Also note that if we expand (x-1)^6 it will lead off with x^6, which will cancel. Therefore we only expect five roots. The four roots other than 0.5 will be complex. To find them we proceed as follows.
Divide both sides by x^6 to get
(x-1)^6 / x^6 = 1
[(x-1)/x]^6 = 1
(1 - 1/x)^6 = 1
To satisfy this, 1 - 1/x is one of the sixth roots of unity. These of course are at magnitude 1, angle a multiple of 60 degrees: Also, given the symmetry of the problem negating the right hand side doesn't change the solution set, and makes further work slightly cleaner. We have
-1 + 1/x = 1
-1 + 1/x = -1
-1 + 1/x = cos(60) + j*sin(60)
-1 + 1/x = cos(60) - j*sin(60)
-1 + 1/x = -cos(60) + j*sin(60)
-1 + 1/x = -cos(60) - j*sin(60)
Now add 1 to both sides of every equation:
1/x = 2
1/x = 0
Excellent
Given the symmetry, the only real root has to be 0.5. obvious. Using your transformation the other four roots are the four corners on the unit circle. Excellent.
@@KipIngram The expressions of the solutions for x can be "cleaned up" (i.e. removing complex expressions from the denominator) by multiplying both numerator and denominator of a solution by the _complex conjugate_ of the denominator. (If z = a + j*b is a complex number (with real numbers a and b ), then zᶜ = a - j*b is the complex conjugate of z .) You'll find that all five solutions are of the form x = 0.5 + j*b , which is kinda surprising.
@@yurenchu Yes, I was lazy. 🙂 Thank you. I just wanted them far enough along to check the results using my calculator, which handles complex numbers natively (Swiss Micros DM42).
@@KipIngram LOL! Fair enough... I too am lazy. Namely too lazy to get out my calculator (Casio fx-82 ; I don't know if it also does complex numbers), that's why I always calculate things like this by hand, wherever/whenever I can. :-)
Never heard of the Swiss Micros DM 42 before, I had to look it up: it looks quite cool, with a nice screen to render graphs and (I'm guessing) bitmap images.
Cube root of each side gives us
x^2 = (x-1)^2 = x^2 -2x + 1
0 = -2x + 1
2x = 1
x = 1/2
Proprietà delle potenze: stesso esponente, quindi devono avere la stessa base; essendo l'esponente pari, le basi possono essere opposte, questo conduce a due equazioni: 1) x=x-1 basi con lo stesso segno, e questa è chiaramente un'equazione impossibile x-x=1 quindi 0=1. 2) x=-x+1 basi opposte questa equazione si risolve subito x+x=1 quindi 2x=1 infine x=1/2
True in Real numbers but their are others roots in Imaginary numbers 👍
Hvala❤
x⁶ = (x-1)⁶
Method 1:
Note that x=0 is not a solution to this equation. So we can safely divide both sides by x⁶ :
1 = (x-1)⁶/x⁶
1 = [(x-1)/x]⁶
1 = (1 - 1/x)⁶
... substitute u = (1 - 1/x) ...
1 = u⁶
This equation has six solutions in the complex plane:
u = 1 , u = (1 ± i√3)/2 , u = (-1 ± i√3)/2 , u = -1 .
If u = 1 :
(1 - 1/x) = 1
-1/x = 0
x = -1/0 = undefined (could be +infinity or -infinity, or infinity*z where z is any unit direction in the complex plane)
If u = (1 ± i√3)/2 :
(1 - 1/x) = (1 ± i√3)/2
-1/x = (-1 ± i√3)/2
1/x = -(-1 ± i√3)/2
x = -2/(-1 ± i√3)
... multiply both numerator and denominator of RHS by (1 ± i√3) ...
x = -2(1 ± i√3)/[(-1 ± i√3)*(1 ± i√3)]
x = -2(1 ± i√3)/[(±i√3 - 1)*(±i√3 + 1)]
x = -2(1 ± i√3)/[ (±i√3)² - 1² ]
x = -2(1 ± i√3)/[ (i)²(±√3)² - 1² ]
x = -2(1 ± i√3)/[ (-1)(3) - 1 ]
x = -2(1 ± i√3)/[ -4 ]
x = (1 ± i√3)/2
x = ½ ± i*½√3
If u = (-1 ± i√3)/2 :
(1 - 1/x) = (-1 ± i√3)/2
-1/x = (-3 ± i√3)/2
1/x = -(-3 ± i√3)/2
x = -2/(-3 ± i√3)
... multiply both numerator and denominator of RHS by (3 ± i√3) ...
x = -2(3 ± i√3)/[(-3 ± i√3)*(3 ± i√3)]
x = -2(3 ± i√3)/[(±i√3 - 3)*(±i√3 + 3)]
x = -2(3 ± i√3)/[ (±i√3)² - 3² ]
x = -2(3 ± i√3)/[ (i)²(±√3)² - 3² ]
x = -2(3 ± i√3)/[ (-1)(3) - 9 ]
x = -2(3 ± i√3)/[ -12 ]
x = (3 ± i√3)/6
x = ½ ± i*⅙√3
If u = -1 :
(1 - 1/x) = -1
2 = 1/x
x = ½
So there are five solutions, and they are:
x = ½ ± i*½√3 , x = ½ ± i*⅙√3 , x = ½ .
- - -
Method 2: see reply to this comment, below.
Method 2:
x⁶ = (x-1)⁶
ux = x-1
... where u is any of the six (complex-valued) solutions to the equation u⁶ = 1 ; namely u = 1 , u = -1 , u = (½ ± i*½√3) , u = (-½ ± i*½√3) ...
ux - x = -1
(u-1)x = -1
For u=1 this leads to 0x = -1 ==> x = undefined.
For the other five values of u, we have the solution
x = -1/(u-1)
... let uᶜ = the complex conjugate of u , and multiply both numerator and denominator of RHS by (uᶜ-1) ...
x = -(uᶜ-1)/[(u-1)(uᶜ-1)]
x = -(uᶜ-1)/[ uuᶜ - uᶜ - u +1 ]
... since all values of u are on the unit circle, uᶜ = 1/u and hence uuᶜ = |u|² = 1 ...
x = -(uᶜ-1)/[ 1 - uᶜ - u +1 ]
x = -(uᶜ-1)/[ 2 - (u + uᶜ) ]
x = -(uᶜ-1)/[ 2 - 2*Re(u) ]
... where Re(u) is the real component of u ...
x = -½(uᶜ-1)/[1 - Re(u)]
x = -½(Re(u) - i*Im(u) - 1)/[1 - Re(u)]
... where Im(u) is the imaginary component of u ; u = Re(u) + i*Im(u) ...
x = -½(Re(u) - 1 - i*Im(u))/[1 - Re(u)]
x = ( -½(Re(u) - 1) + ½i*Im(u) )/[1 - Re(u)]
x = -½(Re(u) - 1)/[1 - Re(u)] + ½i*Im(u)/[1 - Re(u)]
x = ½ + i*½Im(u)/[1 - Re(u)]
For u = -1 , Re(u) = -1 and Im(u) = 0 , so
x = ½ + i*½(0)/[1 - (-1)] = ½ + i*0/2 = ½
For u = ½ ± i*½√3 , Re(u) = ½ and Im(u) = ±½√3 , so
x = ½ + i*½(±½√3)/[1 - (½)] = ½ + i*½(±½√3)/[½] = ½ ± i*(½√3)
For u = -½ ± i*½√3 , Re(u) = -½ and Im(u) = ±½√3 , so
x = ½ + i*½(±½√3)/[1 - (-½)] = ½ ± i*½(½√3)/[1½] = ½ ± i*(½√3)/[3] = ½ ± i*(⅙√3)
So there are five solutions, and they are:
x = ½ , x = ½ ± i*½√3 , x = ½ ± i*⅙√3 .
Excellent job!!
the easier way is to apply square root on both sides of the equation.
Ok, going for the representation of the roots, we get that there are 5 roots, you showed 4 wich are on the complex plane, how would we find the one on the real plane? I can guess it's +1/2, but how do we prove it?
In the video, I have one real root and 4 complex roots, making it five roots altogether. You can go ahead and watch how I got 1/2 as the real root.
x^6=(x-1)^6 positive for all x . Then ((x-1)/x)^6 = 1. (x-1)/x is 6th root of unity. Then, (x-1)/x = e^(i2kpi/6)=> x=1/(1- e^(i2kpi/6)) for k=0,1..,5. For k=0, there's no solution. For k=3, x=1/2. Substitute k=1, 2, 4,5 and obtain two pairs of complex conjugates for x.
Fantastic. You nailed it.💯
@@SpencersAcademy Okay I nail your Olympiad math problems. Let's see if you can nail our 9th grade regular maths problems. Solve sin(9x)=sin(5x)+sin(3x). You may record a video if you wish.
" x^6=(x-1)^6 positive for all x. "
That's not true. As the video shows, one of the correct solutions is x = (½ + i*⅙√3) ; and for that solution,
x⁶ = (½ + i*⅙√3)⁶ = [ (½(-i√3) + ½)*(i*⅓√3) ]⁶ = (½ - i*½√3)⁶ *(i⁶)*(⅓√3)⁶ = (1)*(-1)*(1/27) = -1/27
(x-1)⁶ = (½ + i*⅙√3 - 1)⁶ = -1/27
which is a negative real. (The same goes for the solution x = (½ - i*⅙√3) .)
However, it doesn't affect the rest of the calculation in your comment.
Easy, the solution is x = ω. Any limit ordinal could work here really. Because any limit ordinal minus 1 is just that limit ordinal itself since it doesn't have any predecessor. As a result, there's more than 1 solution because we can't put a value of the number of limit ordinals there are. Keep in mind that limit ordinals are multiples of ω.
That's excellent, brother.
Lấy căn bậc 6 hai vế,ta có x=x-1 hay x-x=-1 hãy x-x=-1 hay0x=-1h ấy x=-1/0.Phương trình vô nghiệm
x^6 그래프를 상상해보면, y축을 기준으로 좌우 대칭이니 그 그래프를 오른쪽으로 1만큼 이동시킨 그래프와의 교점의 x좌표가 0과 1의 1:1 내분점일꺼라 생각할래요. 그편이 훨씬 간단하고 저에겐 더 아름답게 느껴지거든요
That's fantastic, brother.
x/(x-1) =1**1/6
x/(x-1) =1
solve for x
Why not
x^6 = (x - 1)^6
=> (x^3)^2 = ((x-1)^3)^2
Taking square roots of both sides,
=> (x^3) = +/- (x-1)^3
=> x = x -1 (no solution) OR x = - (x - 1)
So,
x = - (x - 1)
x = 1 - x
2 x = 1
x = 1/2
Nice one
maybe it's 1? x pow 6 minus x pow 6 we can erase and - 1 pow 6 it's 1
{(x^3)^2-[(x-1)^3]^2}=0
[x^3-(x-1)^3][x^3+(x-1)^3]=0
{(1)[x^2+x(x-1)+(x-1)^2]}{(2x-1)[x^2-x(x-1)+(x-1)^2]}=0
7:22 "your trick" is more confusing than just applying the formula.
Otherwise an excellent video, the most challanging problem I saw for today 😉
Glad it helped!
X=1/2
why you don't take the 6th root for both side at first?
When you do that, you're gonna miss other vital solutions for x
It will become x = x+1 which has no solution
@@dante_loves_plzzaIf power is even |x|=x+1
@@cosmolbfu67 that's |x^2n| = (x+1)^2n, I'm talking about x = x+1
If you do this, some of the results will be not available
1/2
0.5
Very good. But speed up speed up. You are toooooo slow.
Okay, thanks for the tip.
@@SpencersAcademy An Olympiad aspirant is very much aware of the basic identities. So, no need to reiterate those. You can directly jump redundant steps.
Thank you.
@@tirtharajbanerjee Point noted. Thanks for the tips
Esse trem em português já é complicado imagina em grego. Fala português pra facilitar.
You must have 6 solutions. I contend in the X yi plane, you have two coincidental roots at -.5.
In a question like this, there exist only five roots.
Let me explain it this way:
Let's say you expand the RHS using binomial expansion. You'll have something like this:
X^6 = x^6 - 6x^5 +15x^4 - 20x^3 + 15x^2 -6x +1.
Subtract x^6 from both sides makes the equation 5th-degree polynomial. And that's why we've got five roots.
X=1\2
|x|=|x-1| =>
X= 1/2
?
x^6=(x-1)^6 then x=+/-(x-1) x=x-1 no solution other x=-(x-1) hence x=1/2
X:= 0.5
Меня тоже удивило почему в ответе нет 0,5
I have a doubt
On what, please?
Shud be a quintic equation
А разве нельзя сразу из обеих частей извлечь корень шестой степени?
No. That's a wrong approach to any mathematics question. In a case like this, it's either you follow the steps outlined in the video or you expand the right-hand side using binomial expansion.
Of course it is possible; one just has to do it properly. As outlined in several other comments, one obtais (x-1)=wx with w one of the six 6th roots of unity. For all of them apart from w=1, this is solvable: x=1/(1-w).
|x|=|x-1|. Dos soluciones x=1-x,x=1/2 y la solución de -x=x-1, x=1/2. O sea sólo x=1/2
Why not solve 1=(1-y)^6 where y=1/x
X = -0.5
When would this equation to solve appear in a "real-world" situation? 🤔
Probably the stuff used to make the concepts of the equasions used to make the components for the machine you are typing that question on.
Quite a lot . Work on QM , and EM.
What are the acronyms QM and EM? Please clarify.
WHY THIS COMPLICATION LET X-1=Y THEN X^6 = Y^6 i.e X NOT EQUAL X-1 NOT VALID EQUATION
x=1/2....
If u take the 6th root, u get x=plus/minus( x-1). The plus sign gives u no solution but the minus sign gives x=-x+1 so x=1/2
That's fantastic. There are other solutions for x. You just got one. Which is the only real value.
No real root. X cannot equal x minus 2
Should have worked on 6 answers.
Why? The polynomial has degree 5
X is to the power of 6!
@@mk-jl3zd yep, but there is a subtraction
Your tricks are too long and difficult.
6 is even number. that's why x=x-1 (1) or x=-(x-1) (2)
(1) has no solutions
(2) x=0.5
Too much talking in very easy places
Valore assoluto di x = valore assoluto di ( x -1) e finisce in un minuto
You would o ly be having just one value of x. There are still other values