I created a function f(n,x)=(x^n)+(1/x^n) We know that f(1,x)=-1 We can also easily see that f(0,x)=1+1=2 If we look at f(n-1,x) we see that f(n-1,x)*f(1,x)=f(n,x)+f(n-2,x) and therefore f(n,x)=-f(n-1,x)-f(n-2,x) In other words, for each value of n, the function is the negative of the sum of the function's values for the previous 2 values of n We have 2 and --1 for the first 2 values We can quickly see that this algorithm will produce an infinitely repeating series 2,--1,-1,2,-1,-1... Therefore if n mod(3)=0, f(n,x)=2, otherwise f(n,x)=-1 Since 18 mod(3)=0, we know f(18,x)=2
I liked the first method the most :) ".. and good luck with that" :) Sometimes a good sense of humor is priceless :) Thanks for your daily multi-level work for us!
2nd and 3rd method are genius. I once saw a method for solving symmetric (reciprocal) equations similar to 2nd method. Will write something more to this soon.
If you express the root from the first method in polar form, raising to the 18th power becomes easy. The magnitude is 1 and the angle is 2pi/3. 1^18=1 and 2pi/3*18 is an integer multiple of 2pi, so the 18th power is 1, and the solution is 2. Okay, I'm writing this while watching the video, and I see you mentioned this later, but you did say you wanted comments on this, so posting anyway.
I actually think that your method (the professor’s fourth method), which employs De Moivre’s theorem, is superior to the other methods. Is there any other way to proceed if, for example, we had x to the 17th or 19th or even 233rd power?
@@chrisjuravich3398 I agree with you. 19th power would be r = 1, angle = 2pi/3. Then you can break it down into a + bi form and then add it to the reciprocal of a + bi.
The roots obtained in the first method are the cube roots of unity, w and w². We can use the properties of cube roots of unity, w + w² + 1 =0 and w³ = 1 to get the answer 2.
Hi, there! I guess, that sum of reversed numbers are greater than 2 or less than -2 whether x > 0 and x < 0 respectively. are we able to solve equation w/o complex numbers?
Very nice! Like it a lot! Actually, one curios note, coming from the 2nd methof with complex numbers... since x = exp(+-i*2pi/3) (important to note a unit modulus |x| = 1 here, since (sqrt(3)/2 )^2 + (1/2)^2) =1, you frequently forget about the modulus in the exponential from, btw :D) So, because of the x = exp(+-i*2pi/3), with any real N any x^N + (1/x)^N will be real, since in this case with the unit modulus (1/x)^N = (x*)^N and the imaginary part will be eliminated! and finallly, x^N = exp(+-i*N*2pi/3) and x^N + (1/x)^N = 2*cos(N*2pi/3). It's really interesting that even for really huge N, the result will still be within [-2; 2]. So, for N =18 we get 2. For example, if N = 2023, we get 2*cos(674*2pi + 2pi/3) = 2*cos(2pi/3) = -1.
Here is the 6th method along the same lines of your third. Difference of cubes factorization gives x^3-1={x-1)(x^2+x+1) and we know second factor =0 that makes the entire RHS=0 x^3=1 so x to the power any integer multiple of x =1 Thus the answer 1+1=2. If the original equation was x+1/x=1 the sum of square does the magic to make x^3=-1 and in this case the even multiples of 3rd power is 1 and the odd multiples would be -1.
X^3 is equal to 1 but that doesn’t mean x=1. X is a complex root that when cubed gives us one. The third method does not contradict itself, 1 is not a solution for x.
x(x+1/x)=-x implies x^2+x+1 =0, but that's not true for any real number x. So we can multiply both sides of the equation by 0 in form of x-1, that is 0=(x-1)(x^2+x+1)=x^3-1. Hence x^3=1. So, x^18+1/x^18=(x^3)^6+1/(x^3)^6=1+1=2 Q.E.D
When I saw the thumbnail I thought this couldn't be because f(x) = | x + 1/x | achieves its minimum value at 2. But then I thought that's as a real function ans if you allow complex arguments for x then it should be fine.
GENERALIZED PROBLEM: If x + 1/x = beta, what is rho if rho = X^alpha + 1/x^alpha? Solve for all real alpha and beta. SOLUTION: rho = 2 cos [ alpha Arccos (beta/2) ] for | beta | < 2 AND rho = 2 cosh [ alpha arccosh (beta/2) ] for | beta | .GE. 2 EXAMPLE: If alpha = pi = 3.142592654 and beta = Euler's constant = 0.57721566, then rho = -1.28439043
A simpler if more mundane way to get the result for the 3rd method is as follows. x + ⅟× = -1 ∴ x² + 1 = -x ∴ x² = -x - 1 ∴ x³ = -x² - x = -(-x - 1) - x = 1 ∴ x³ = 1.
Your problems in your videos are becoming very same now.Just the values change the questions remain the same.Try to diversify the problems you present.I would suggest more geometry problems ( u can use coordinate geometry to solve those problems as rarely anyone uses that method) or even ratio proportion and probability would be good and very helpful to learn from you because your videos taught me so much so much about algebra
Here is a method using only geometry. We have x + ⅟× = -1 ∴1 + x + x² = 0. In the complex plane, call the origin O and the point (1, 0) A. The vector ŌĀ represents the number 1. From A draw a vector ĀĒ at angle θ to the (increasing) real axis. ĀĒ represents the number x. From point E draw vector ĒŌ at angle 2θ to the horizontal through E. This represents the number x². The vector ends at O because ŌĀĒŌ represents 1 + x + x² = 0. From elementary geometry we see that OÂE = AÊO = 180⁰ - θ. ∴ ΔOAE is isosceles with |ĒŌ| = |ŌĀ| = 1. ∴ |x²| = 1 ∴ |x| = 1. ∴ |ĒŌ| = |ŌĀ| = |ĀĒ| = 1 i.e. ΔΟAE is equilateral. ∴ 180⁰ - θ = 60⁰ ⇒ θ = 120⁰. ∴ |⅟×| = 1 and ⅟× makes angle -120⁰ with the real axis. x¹⁸ makes angle (18)(120)⁰ with the real axis, a multiple of 360⁰, and |x¹⁸| = |x|¹⁸ = 1. ∴ x¹⁸ = 1. Similarly ⅟×¹⁸ = 1. ∴ x¹⁸ + ⅟×¹⁸ = 2.
@@SyberMath x+1/x=-1, from this equation x isn't equal to one or -1then x^2+x+1=0, multiplied by x-1 both sides We have x^3-1=0,x^3=1, 1/x^3=1,x^18+1/x^18=2
Super explanations!
I created a function f(n,x)=(x^n)+(1/x^n)
We know that f(1,x)=-1
We can also easily see that f(0,x)=1+1=2
If we look at f(n-1,x) we see that
f(n-1,x)*f(1,x)=f(n,x)+f(n-2,x) and therefore
f(n,x)=-f(n-1,x)-f(n-2,x)
In other words, for each value of n, the function is the negative of the sum of the function's values for the previous 2 values of n
We have 2 and --1 for the first 2 values
We can quickly see that this algorithm will produce an infinitely repeating series 2,--1,-1,2,-1,-1...
Therefore if n mod(3)=0, f(n,x)=2, otherwise f(n,x)=-1
Since 18 mod(3)=0, we know f(18,x)=2
The 3rd method surprised me in a pleasant way.
You could do problems on:
Probability
Permutations and combinations
Geometry
Coordinate geometry
vote up
I liked the first method the most :) ".. and good luck with that" :) Sometimes a good sense of humor is priceless :) Thanks for your daily multi-level work for us!
You are welcome!
Enjoyed all solutions very much🇨🇦
Great to hear!
I like the representation of e^(2*pi*i/3) in method one. You can quickly see x^3+1/x^3 = 2 and the sequence is modulo 3.
Third method really amazing.
That third method is amazing! I didn't see that x²+x+1 was the other part of the factorisation of x³-1, so you caught me off guard!
😲😜
2nd and 3rd method are genius. I once saw a method for solving symmetric (reciprocal) equations similar to 2nd method. Will write something more to this soon.
If you express the root from the first method in polar form, raising to the 18th power becomes easy. The magnitude is 1 and the angle is 2pi/3. 1^18=1 and 2pi/3*18 is an integer multiple of 2pi, so the 18th power is 1, and the solution is 2.
Okay, I'm writing this while watching the video, and I see you mentioned this later, but you did say you wanted comments on this, so posting anyway.
I actually think that your method (the professor’s fourth method), which employs De Moivre’s theorem, is superior to the other methods. Is there any other way to proceed if, for example, we had x to the 17th or 19th or even 233rd power?
@@chrisjuravich3398 I agree with you. 19th power would be r = 1, angle = 2pi/3. Then you can break it down into a + bi form and then add it to the reciprocal of a + bi.
The roots obtained in the first method are the cube roots of unity, w and w². We can use the properties of cube roots of unity, w + w² + 1 =0 and w³ = 1 to get the answer 2.
Exact same method!
in first method we can put x = w then, w^18 = 1 so 1 + 1/1 = 2
8:24 how can x-cubed be 1 when x cannot be 1 with this method ?
x is a complex cube root of 1
Although I solved it using the second method I loved the third one ❤❤❤❤
The second method was my second method but the Euler's identity method was my first method.
Hi, there! I guess, that sum of reversed numbers are greater than 2 or less than -2 whether x > 0 and x < 0 respectively. are we able to solve equation w/o complex numbers?
The 3rd solution is very very good!
🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷
Glad to hear that!
Very nice! Like it a lot! Actually, one curios note, coming from the 2nd methof with complex numbers... since x = exp(+-i*2pi/3) (important to note a unit modulus |x| = 1 here, since (sqrt(3)/2 )^2 + (1/2)^2) =1, you frequently forget about the modulus in the exponential from, btw :D) So, because of the x = exp(+-i*2pi/3), with any real N any x^N + (1/x)^N will be real, since in this case with the unit modulus (1/x)^N = (x*)^N and the imaginary part will be eliminated! and finallly, x^N = exp(+-i*N*2pi/3) and x^N + (1/x)^N = 2*cos(N*2pi/3). It's really interesting that even for really huge N, the result will still be within [-2; 2]. So, for N =18 we get 2. For example, if N = 2023, we get 2*cos(674*2pi + 2pi/3) = 2*cos(2pi/3) = -1.
😊
Here is the 6th method along the same lines of your third. Difference of cubes factorization gives
x^3-1={x-1)(x^2+x+1) and we know second factor =0 that makes the entire RHS=0
x^3=1 so x to the power any integer multiple of x =1
Thus the answer 1+1=2.
If the original equation was x+1/x=1 the sum of square does the magic to make x^3=-1 and in this case the even multiples of 3rd power is 1 and the odd multiples would be -1.
In the third method we exclude 1 as solution and in the next step accepts 1 as solution. Did I missed something?
X^3 is equal to 1 but that doesn’t mean x=1. X is a complex root that when cubed gives us one. The third method does not contradict itself, 1 is not a solution for x.
Taking w,w^2 roots we can get easily
x(x+1/x)=-x implies x^2+x+1 =0, but that's not true for any real number x. So we can multiply both sides of the equation by 0 in form of x-1, that is 0=(x-1)(x^2+x+1)=x^3-1. Hence x^3=1. So, x^18+1/x^18=(x^3)^6+1/(x^3)^6=1+1=2 Q.E.D
When I saw the thumbnail I thought this couldn't be because f(x) = | x + 1/x | achieves its minimum value at 2.
But then I thought that's as a real function ans if you allow complex arguments for x then it should be fine.
Same
x^3=1
Answer=2
А как быть с тем, что x≠1 и одновременно x^3=1?
X=-1/2+-i*sqrt(3)/2
2
GENERALIZED PROBLEM: If x + 1/x = beta, what is rho if rho = X^alpha + 1/x^alpha? Solve for all real alpha and beta.
SOLUTION: rho = 2 cos [ alpha Arccos (beta/2) ] for | beta | < 2 AND rho = 2 cosh [ alpha arccosh (beta/2) ] for | beta | .GE. 2
EXAMPLE: If alpha = pi = 3.142592654 and beta = Euler's constant = 0.57721566, then rho = -1.28439043
A simpler if more mundane way to get the result for the 3rd method is as follows.
x + ⅟× = -1
∴ x² + 1 = -x
∴ x² = -x - 1
∴ x³ = -x² - x = -(-x - 1) - x = 1
∴ x³ = 1.
x^18=1....2
Its de. Mwavre
Plz don't keep saying first method
In high school we learned first selotion.
Second selotion..
Your problems in your videos are becoming very same now.Just the values change the questions remain the same.Try to diversify the problems you present.I would suggest more geometry problems ( u can use coordinate geometry to solve those problems as rarely anyone uses that method) or even ratio proportion and probability would be good and very helpful to learn from you because your videos taught me so much so much about algebra
True
Maybe some Putnam problems? I really need thise 😢
Multivariable Inequality would be good too
Here is a method using only geometry.
We have x + ⅟× = -1
∴1 + x + x² = 0.
In the complex plane, call the origin O and the point (1, 0) A.
The vector ŌĀ represents the number 1.
From A draw a vector ĀĒ at angle θ to the (increasing) real axis. ĀĒ represents the number x.
From point E draw vector ĒŌ at angle 2θ to the horizontal through E. This represents the number x².
The vector ends at O because ŌĀĒŌ represents 1 + x + x² = 0.
From elementary geometry we see that OÂE = AÊO = 180⁰ - θ.
∴ ΔOAE is isosceles with |ĒŌ| = |ŌĀ| = 1.
∴ |x²| = 1
∴ |x| = 1.
∴ |ĒŌ| = |ŌĀ| = |ĀĒ| = 1
i.e. ΔΟAE is equilateral.
∴ 180⁰ - θ = 60⁰ ⇒ θ = 120⁰.
∴ |⅟×| = 1 and ⅟× makes angle -120⁰ with the real axis.
x¹⁸ makes angle (18)(120)⁰ with the real axis, a multiple of 360⁰, and |x¹⁸| = |x|¹⁸ = 1. ∴ x¹⁸ = 1.
Similarly ⅟×¹⁸ = 1.
∴ x¹⁸ + ⅟×¹⁸ = 2.
Wow!!! 😍
@@SyberMath x+1/x=-1, from this equation x isn't equal to one or -1then
x^2+x+1=0, multiplied by x-1 both sides We have x^3-1=0,x^3=1,
1/x^3=1,x^18+1/x^18=2
very Nice
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