EDIT: I do not think that the step in the proof at 6:28 makes sense. Sure, one can easily prove z^(2^k)+1/z^(2^k)=-1 by induction. But then it is assumed that: z^2024+1/z^2024 = z^1024.z^512.z^256...z^8 = (-1)^7 ...when in fact: (-1)^7 = (z^1024+1/z^1024)(z^512+1/z^512)...(z^8+1/z^8) ...and this right hand side does not equal z^2024+1/z^2024 - unless of course there is a lot of cancellation due to z+1/z=1 but this is certainly not obvious or likely in my opinion. Instead here is another way to approach the problem which also acts as a generalisation: Let a[n] = z^n + 1/z^n for n>=1 with: a[1] = 1 (given) a[2] = -1 (square z+1/z=1) => a[n](z + 1/z) = (z^n + 1/z^n)(z + 1/z) => a[n].a[1] = z^(n+1) + z^(n-1) + 1/z^(n-1) + 1/z^(n+1) => a[n].1 = a[n+1] + a[n-1] => a[n+1] = a[n] - a[n-1] for n>=2 Using this recurrence relation we get: a[3] = -2 a[4] = -1 a[5] = 1 a[6] = 2 a[7] = 1 a[8] = -1 The recurrence relation implies that any a[n] depends only on the previous 2 values in the sequence. Since a[7]=a[1] and a[8]=a[2], the sequence has a period of 6 ie a[6m+k]=a[k]. One can then use the values above to look up a[n] for any n=6m+k, for example: a[2024]=a[6.337+2]=a[2]=-1.
@@mikasa_sucasa Oh yes, the solution in the video seems to make an incorrect assumption at 6:28. I have edited my comment to reflect this. Well spotted!
@@PrimeNewtons if you Let u=z+1/z and want to write polynomial in u you probably will get Chebyshev polynomial I tried to find coefficients of Chebyshev polynomials by deriving and solving recurrence relation using exponential generating function or by deriving and solving ordinary differential equation using power series but I have got two sums which I cannot evaluate Sum from recurrence relation \sum\limits_{k=m}^{\lfloor\frac{n}{2} floor}{n \choose 2k} \cdot {k \choose m} Sum from ordinary differential equation 1+\sum\limits_{k=1}^{\lfloor\frac{n}{2} floor}\frac{(-1)^k}{4^k}\cdot\frac{n}{n-k}\cdot {n-k \choose k}
Yeah it's very easy to generalize, first you turn the equation into z²-z+1=0 and solve for the roots to get cos(π/3)±sin(π/3)i and it's polar form will be e^iπ/3 and now we can turn z^n+1/z^n=e^inπ/3+e^-inπ/3 which if we take the complex definition of cos which is 1/2(e^ix+e^-ix) then this will equal 2cos(nπ/3) and plugging in 2024 for n for this question will give us -1.
You can in fact generalise z^n + z^-n, just let z = cos∅ + isin∅, and by de Moivre's theorem, which states that z^n = r^n(cosn∅ + isinn∅), and we can calculate r by letting n=1, and finding the magnitude of z. We can see that r=1, |z| = √(cos²∅+sin²∅) = 1. 1^n, as long as n is real, will always be 1, so we can proceed accordingly. Now by the theory, z^n = cosn∅ + isinn∅ z^-n = cos(-n∅) + isin(-n∅), and since cosine is an even function and sine is an odd function z^-n = cosn∅ - isinn∅, so z^n + z^-n = 2cosn∅ When n=1, given in the problem, z + z^-1 = 1 2cos∅ = 1 cos∅ = 1/2, which is a very popular angle, ∅ = π/3, and we can now substitute n=2024. z^2024 + z^-2024 = 2cos(2024(π/3)), and evaluate it with the calculator (for my convenience lol), z^2024 + z^-2024 = -1
Another, more general way. Let's call f(n) = z^n + z^(-n). Then f(0) = 2 and f(1) = 1 (the last equality is from the problem statement). We have the following property that can be easily checked: f(n)f(m) = f(n+m) + f(n-m). Substituting m=1 in this last equation, we get f(n)f(1) = f(n+1) + f(n-1). Remembering that f(1) = 1, we get f(n+1) = f(n) - f(n-1), which is a recurrence relation for a sequence such that any term is the difference between the previous two terms. The first two terms of this sequence are 2 and 1, therefore the sequence is: f(0) = 2, f(1) = 1, f(2) = -1, f(3) = -2, f(4) = -1, f(5) = 1, f(6) = 2, f(7) = 1, ... The last two terms f(6) and f(7) are equal to the first two terms f(0) and f(1), therefore the sequence has period 6. Therefore, f(n) = f(n mod 6). This allows the calculation of f(n) for any n, as the first 6 values f(0) through f(5) are known. So, for example, f(2024) = f(2024 mod 6) = f(2) = -1, f(2025) = f(2025 mod 6) = f(3) = -2, f(7404) = f(7404 mod 6) = f(0) = 2, etc.
Wait what? What/how do you factorize z²⁰²⁴+z⁻²⁰²⁴ as (z¹⁰²⁴+z⁻¹⁰²⁴)(z⁵¹²+z⁻⁵¹²)(z¹²⁸+z⁻¹²⁸)(...) ? Surprisingly it works here (I'm not sure why), but it would fail with 2022 or 2025 instead of 2024.
This can be done much easier, even in your head without ever touching pen or paper. The equation z + 1/z = 1 implies z² − z + 1 = 0 so z is one of the complex cube roots of −1 since z³ + 1 = (z + 1)(z² − z + 1) This means that we have z⁶ = (z³)² = (−1)² = 1 and therefore z⁶ⁿ = (z⁶)ⁿ = 1ⁿ = 1 for any integer n. And since 2022 is evidently an even multiple of 3 and therefore an integer multiple of 6 we have 2024 ≡ 2 (mod 6) so z²⁰²⁴ + 1/z²⁰²⁴ = z² + 1/z² = (z + 1/z)² − 2 = 1 − 2 = −1
My friend, i made it with Euler's formula and the answer it's right, but i think the demonstration whene you multiply all the factors is incomplete, it really is a long multiplication.
I agree with the last comment: the final product is a bit more complicated. There are many other pesky terms. If you set a_n = z^n + 1/z^n it is easy to check that there are only 6 different values of a_n. In increasing order of n: 1, -1, -2, -1, 1, 2. the remainder of the division of 2024 by 6 is 2 so a_2024=a_2=-1. A complex approach works very well too and is pretty fast. But I do like your approach and the fact that, even though z must be a complex number (the sum of a real number and its reciprocal has a minimum of 2), you don't need to use complex number to find the answer. Great question! As always, I learned a lot watching your videos.
I think there is something missing from that proof. Why do you assume you can factorize z^2024 + 1/z^2024 as ( z^1024 + 1/z^1024)( z^512 + 1/z^512)( z^256 + 1/z^256)(....)? Using this logic, you could argue that z^n+1/z^n is always -1 or 1 for all natural numbers n, as you can generate any natural number using the sum of powers of 2. But this is not the case. For example z^3 +1/z^3 = -2 (see below for proof of that). Am I missing something? Proof for z^3+1/z^3 = -2: (z+1/z)^3 = z^3 + 3z + 3/z + 1/z^3 1^3 = z^3 + 1/z^3 + 3(z + 1/z) 1 = z^3 + 1/z^3 + 3(1) -2 = z^3 + 1/z^3
if you work out the odd exponents, you'll notice the value of the expression z^n + 1/z^n cycles over the integers {2,1,-1,-2,-1,1}. 2024 = 2 (mod 6) which corresponds to the 3rd element in cycle i.e. -1.
you can do something similar using de moivre's theorem! If we let z = e^i(theta) then z + 1/z = 2cos(theta) = 1, so cos(theta) = 1/2 giving theta = pi/3. z^2024 + 1/z^2024 = 2cos(2024 theta) = 2cos(2024pi/3). 2024pi/3 is equivalent to 2pi/3 (the modulo when dividing by 2pi) which gives that z^2024 + 1/z^2024 = 2 cos(2pi/3) = 2 * -1/2 = -1. Your pattern recognition way is very nice though!
z + 1/z = 1 => z^2 - z + 1 = 0 => z = [1 +- sqrt(1 - 4)]/2 = 1/2 +- i * sqrt(3)/2 = e^(ix) where x = 60 degrees or x = -60 degrees, we can choose either as these are conjugates and thus they satisfy the same equations Since z = e^(ix) where x is 60 degrees, we can mod its exponents in z^2024 + 1/z^2024 by 6 to get that this is equal to z^2 + 1/z^2, which we can easily discern is equal to -1 by squaring the originally equality z + 1/z = 1.
If you solve z+1/z=1, you will find the equation z^2-z+1=0, witch give us the complex cubic roots of the unity. Then, if you calculate z^2024 and apply the ideia of congruent arcs, you wiil find (for both roots): z^2024+1/z^2024 = e^(2*pi*i/3)+ e^(4*pi*i/3) =-1
Could use recursion Z(n+1)=Z(n)Z(1)-Z(n-1). (Z(n) := (x^n)+1/(x^n )) Here we are given Z(1) = 1 in which case sequence is length 6. i.e. Z (1..6)= {1, -1, -2, -1, 1, 2} (Z(6) = Z(0) =2). Hence Z(2024)=Z(rem(2024,6)) = Z(2) = -1.
You can also think of complex numbers in the one but how you think there is a general form for z^n+1/z^n=2cos(nx) this works if z have modulus 1 we can prove that z in this question have modulus 1 z+1/z=1 z²-z+1=0 If you find the roots by your own you gonna notice that the modulus of z is always one So in the expression we can rewrite as z+1/z=2cos(60) So the argument of z is 60 z^{2024}+1/z^{2024}=2cos(60.2024) if you divide 60.2024 by 360 the rest gonna be 120 so : z^{2024}+1/z{2024}=2cos(120)=-1
The transition from a sum to a product at the end was not stated clearly. Another approach: Define u(n)=z^n+1/z^n Given u(1)=1 Evaluate u(n) for n=1,2,… [u(1)^n=u(n-1)*u(1)] ==> u(n+1)=u(n)-u(n-1) Resulting in the series: u(n)=1,-1,-2,-1,1,2, (repeat) 1,-1,-2,… For n=1,2.. General solution: u(n)=u(6k+m), k=0,1… , m=1,2,..6 for any n=1,2,… u(2024)=u(6*337+2) Since m=2, u(2024)=-1
The way I did it was to solve the quadratic, gives z = the complex number of 1/2 +/- root(3)/2j (j because I'm an engineer). The modulus of z is 1, so raising z to any power the modulus remains 1. Since the argument of z is +/- pi/3 the argument of z^n is +/- (n+1 mod 6)pi/3. 2024 mod 6 is 2. So z^2024 is -1. Either value of z^2024 gives -1. I find that, more often than not,working with complex numbers using modulus and argument is far simpler than real and imaginary.
Your factorization is not true. otherwise, you could rewrite 6 as binaries, 6=4+2 and you would obtain (-1)*(-1)=1. However, as franolich3 above correctly states, z^6+1/z^6=2, so your formula is not true in general. Also, please consider using brackets after multiplication, before a negative number.
alternatively, given z+1/z=1 (clearly z is not=0),multiply b/s by z => z^2+1=z z^2=z-1 -----(1) again multiply by z => z^3=z^2-z =z-1-z=-1 {by (1)} => z^3=-1 -----(2) square b/s =>z^6=1 -----(3) now z^2024={(z^6)^337*(z^2) }= z^2 {by (3)} To evaluate E=z^2024+1/z^2024 =z^2+1/z^2 =(z+1/z)^2-2 =(1)^2-2 =1-2=-1 as found by Prime Newtons
By solving the equation you find two complex conjugate roots z= (1+i√3)/2 = 1•e^(iπ/3) = 1∠(π/3) = cos(π/3)+i•sin(π/3) and ¯z= (1-i√3)/2 = 1•e^(-iπ/3) = 1∠(-π/3) = cos(π/3)-i•sin(π/3) which have two surprising properties: (1) They are the reciprocal of each other : 1/z = 2/(1+i√3) = (2(1-i√3))/((1+i√3)(1-i√3)) = (2(1-i√ 3))/(1²-(-3)) = (1-i√3)/2 =¯z . (2) They are unity : |z| = | ¯z | = 1. Which means that z^n+1/z^n = z^n + (1/z)^n = z^n + (¯z)^n = (cos(π/3)+i•sin(π/3))^n + (cos(π/3)-i•sin(π/3))^n = cos(n*π/3)+i•sin(n*π/3) + cos(n*π/3)-i•sin(n*π/3) = 2•cos(n*π/3) in accordance with the Moivre's formula. And so z^2024+1/z^2024 = 2•cos(2024*π/3) = 2•cos(2π/3) = 2•(-½) = -1
z^2-z+1=0-->z=1/2±i*sqrt(3)/2 It's a sixth or five sixths of the way around the complex unit circle. Raising a number on the complex unit circle by a power simply multiplies its angle around the circle by that power. Because 2024 is even, but not divisible by 3, z^2024 will either be a third or two thirds of the way around the circle. Because two thirds of the way around the circle is also negative one third of the way around the circle and vice versa, z^-2024 will be the other one z^2024 will either equal -1/2+i*sqrt(3)/2 or -1/2-i*sqrt(3)/2 With z^-2024 being the other. The i*sqrt(3)/2s cancel, leaving us with -1 as the answer. An odd power not divisible by 3 would give an answer of 1, an odd power divisible by 3 would give -2, and a power divisible by 6 would give 2.
It can be done really short once you find z^2+1/z^2=-1 z^2 can be replaced by complex number omega and by applying properties of cube root of unity you can get -1 in 1 step
Let f(n) = z^n+1/z^n. Then, f(1)=1 and f(2)=-1 (straightforward) but also the identity f(n+1)=f(n)-f(n-1) holds (verify!). So we obtain a recurrence relation and it turns out the pattern repeats every 6 integers (verify!), starting from 1. Then you only need to know what 2024 mod 6 is (it is 2) so f(2024) = f(2)=-1
Even though using this trick for this question is faster for these type of questions it's better and easier to turn it into a quadratic equation and solve for roots then if complex numbers come up use the polar form or just plug the roots in the second equation to get the answer. So rather than 2024 for if we had something like 2371 or something then turning the roots of quadratic equation into polar form and using the definition of the complex cosine we can generalize this equation to z^n+1/z^n=2cos(nπ/3) and plugging in 2371 for n we get 1.
Can you please explain why it's -1 though? If you had used two 4's, instead of one 8, as the last factors to reach 2024, it would have been an even number of -1's multiplied together, and hence would equal 1, not -1. I assume it's something periodic, but it would be nice to understand.
Wait what ??? I couldn't understand the last part. How can you just multiply like that ??? It is far better to just use the quadratic and make out the solution to be z = -w / -w^2 [w=imaginary cube root of one] Hence given expression gives rise to the sum of two imaginary cube root of one which is -1. This is a valid logic. But what did you do there ???
So when you added the powers of 2 together to reach 2024, when expanding that idea to the equation you were solving were you multiplying? ((z+1/z)^1024 * (z+1/z)^512 * …) or using a power tower? ((((z+1/z)^1024)^512)^…) or are both methods just the same thing represented with different notation?
Both methods you described are different. Taking the first one, all the powers of 2 are added while in the second all the powers are multiplied. For this case, 2024 is achieved by adding powers of two. Taking all the additive powers out causes it to get multiplied.
General Case: z + 1/z is real, so equating real parts: rcos(t) + 1/r cos(t) = 1, rsin(t) = 1/r sin(t) for all t ==> r = 1 , cos(t) = 1/2 (WLOG t = pi/3 in principal branch) ==> z = exp(i*pi/3) ==> z^n + z^-n (for integer n) = exp(i*npi/3) + exp(-i*npi/3) = 2cos(n*pi/3) = 2cos(k*pi/3), where n == k mod 6 n = 4,1024 ==> k = 4 ==> sum is 2cos(4pi/3) = -1
We start with the given equation: z + 1/z = 1 We rearrange this equation to solve for z: z^2 - z + 1 = 0 Solving this quadratic equation, we get: z = (1 ± √(-3))/2 Since z is a real number, the only valid solution is: z = (1 + √3i)/2 Now, we need to evaluate the expression z^2024 + 1/z^2024. Using the formula for raising a complex number to a power: (a + bi)^n = a^n - (n * b^2 * a^(n-2))/2 + i(n * a * b^(n-1)) Substituting z = (1 + √3i)/2, we get: z^2024 = ((1 + √3i)/2)^2024 = (1 + 2024√3i - 2024*3i)/4^2024 = (1 + 2024√3i - 6072i) / 16^2024 Similarly, we can calculate 1/z^2024: 1/z^2024 = 1 / ((1 + √3i)/2)^2024 = 4^2024 / (1 + 2024√3i + 2024*3i) = 16^2024 / (1 + 2024√3i + 6072i) The final expression is the sum of these two terms: z^2024 + 1/z^2024 = (1 + 2024√3i - 6072i) / 16^2024 + 16^2024 / (1 + 2024√3i + 6072i) This is the formula to calculate z^2024 + 1/z^2024 given that z + 1/z = 1.
how do we know that (z^1024+1/z^2024)(z^512+1/z^512)...(z^8+1/z^8)=(z^2024+1/z^2024)? arent there going to be many more terms than just those two for example z^1024/(z^512 times z^256...)=z^1024/z^1000=z^24?
Let z=e^it. Then z+1/z= 2cos t = 1 > cos t = 1/2 > t=pi/3 + 2pi n, n being an integer. Thus, z^2024 = e^2024(pi/3+2pi n)i = e^(674 pi i) e^(2 pi i/3) = e^(2pi i/3) = -e^(-i pi/3). So, z^2024 + z^(-2024) = -2 cos pi/3 = -1.
To get 1000 from binary, you could have started with the sum of all binaries below 1024, starting with 1, which would have resulted in 1023. Now you know you need to subtract 23. 23 in binaries is 16+4+2+1, so you need to omit these, but take all the others
I think you made a leap too far. 2024=1024+512+256+64+32+8 (so far so good) (z+1/z)^2=z^2+1/z^2 + 2 = 1^2 z^2+1/z^2 = -1 similarly for every time the power is doubled but ... (z^1024 + 1/z^1024)(z^512 + 1/z^512)= (-1)(-1) = z^1536 + 1/z^1536 + (z^512 + 1/z^512) = z^1536 + 1/z^1536 - 1 so z^1536 + 1/z^1536 = 1 + 1 = 2 so multiplying z to the power of two expression by by another, to a different power of two, is NOT the same multiplying the -1s together (note that it can work out that way e.g. using the last two powers, 32 and 8). same for 256 and 128 and for 64 and 32 The easier is solve the quadratic (z+1/z=1 => z² -z +1 = 0 => z = (1±√3)/2=e^iπ/3 or e^i5π/3 Raise to the 2024th power (2024=2 mod 6) gives z^2024 = e^i2π/3 or e^i4π/3 or (-1 ±√3/2 both give -1 when subbed in to z^1024 + 1/z^1024
For typing purposes, please let me represent "z to the n plus one over z to n" as Z(n). It looks to me like you used Z(a)Z(b) = Z(a+b). Which has not been established.
You could have done the breakdown of 2,024 into a sum of powers of 2 easily, assuming that the calculator app on your table has a programmer mode: Just convert 2,024 from decimal to binary...
Maybe I'm missing something but I don't believe this solution is complete. I haven't finished working out the answer yet, but multiplying the power of 1024 and the power of 512 is not the same as squaring; you get a power of 1536 plus a power of 512 (which is -1), equaling 1, so isolating 1536 is then 2. (I believe)
Now that I've worked it out further, there is a pattern to the powers, every time you multiply z^n + 1/z^n by z + 1/z (which is 1) , you get z^n+1 + 1/z^n+1 + z^n-1 + 1/z^n-1. What this yields is a pattern of 1,-1,-2,-1,1,2. You do have the right result, but I don't think I agree with the assumptions made to get there.
Like many others I would like to know what happened to the missing 2 terms from each multiplication at the end; (z^a + z^(-a))x(z^b + z^(-b))= (z^(a+b) + z^(-(a+b)))x (z^(a-b) + z^(-(a-b)))
z + 1/z = 1 => z^2 - z + 1 = 0 Multiply by z + 1 both sides such that z not = -1 => z^3 + 1 = 0 => z = -1, -w, -w^2, where w,w^2 = complex cube roots of unity But z not = -1 => z = -w, -w^2 => z^2024 = (-w)^2024 or (-w^2)^2024 = w^2,w We know that w^3 = 1 => w^2 = 1/w z = w, 1/w => z ^2024 + 1/z^2024 = w + 1/w = w^2+1/w = -1
@@PrimeNewtons z²⁰²⁴=(z²)¹⁰¹². Any real number raised to even power is positive. Also 1/(z²⁰²⁴) is positive. And the sum of these two positive values is also positive
@@PrimeNewtons if z is real. z² is positive. 1/z² is also positive. How can the sum of these positive values be equal to -1? The same with 2024. Or am I missing something?
@hellovat_youtube You assume z is real. You can't assume it's real. In fact there is no real number z such that the sum of z and its reciprocal is equal to 1. Clearly, z is not real.
@@PrimeNewtons yes, you are right! I'm used to that usually variables are real by default and if they are not it is mentioned in the conditions. As you noticed here we have z+1/z=1, and this doesn't have real roots. A kind of implicit condition, that z is not necessary real. Thank you
I spot something wrong in your last step. To simplify, assume that what you want to calculate is: z^1536 + 1/z^1536 and note that 1536 = 1024+512 We know that z^n + 1/z^n = -1 when n= 512 or 1024. According to the logic of your last step, since you are multiplying two terms each of which has the value -1 you might expect the answer to be (-1)(-1) = +1 But: (z^1024 + 1/z^1024)*(z^512 + 1/z^512) = (-1)^2 = +1 Distributing terms gives us: z^(1024+512) + z^(512-1024) + z^(1024-512) + 1/z^(1024+512) = 1 z^1536 + 1/z^512 + z^512 + 1/z^1536 = 1 Re-arranging: z^1536 + 1/z^1536 + (z^512 + 1/z^512) = 1 z^1536 + 1/z^1536 + (-1) = 1 z^1536 + 1/z^1536 = 2
You should only like a video after watching it. That way, you don't have to comment on taking it back. Also, you should help others understand why the logic is bad. I want to learn too.
@@PrimeNewtons (z^a+z^(-a))(z^b+z(-b))=(z^(a+b) +z^(-(a+b)) + z^(a-b) +z^(b-a) two extra terms come out in the product only if prove them irrelevant could you ignore....
EDIT: I do not think that the step in the proof at 6:28 makes sense. Sure, one can easily prove z^(2^k)+1/z^(2^k)=-1 by induction. But then it is assumed that:
z^2024+1/z^2024 = z^1024.z^512.z^256...z^8 = (-1)^7
...when in fact:
(-1)^7 = (z^1024+1/z^1024)(z^512+1/z^512)...(z^8+1/z^8)
...and this right hand side does not equal z^2024+1/z^2024 - unless of course there is a lot of cancellation due to z+1/z=1 but this is certainly not obvious or likely in my opinion.
Instead here is another way to approach the problem which also acts as a generalisation:
Let a[n] = z^n + 1/z^n for n>=1 with:
a[1] = 1 (given)
a[2] = -1 (square z+1/z=1)
=> a[n](z + 1/z) = (z^n + 1/z^n)(z + 1/z)
=> a[n].a[1] = z^(n+1) + z^(n-1) + 1/z^(n-1) + 1/z^(n+1)
=> a[n].1 = a[n+1] + a[n-1]
=> a[n+1] = a[n] - a[n-1] for n>=2
Using this recurrence relation we get:
a[3] = -2
a[4] = -1
a[5] = 1
a[6] = 2
a[7] = 1
a[8] = -1
The recurrence relation implies that any a[n] depends only on the previous 2 values in the sequence. Since a[7]=a[1] and a[8]=a[2], the sequence has a period of 6 ie a[6m+k]=a[k]. One can then use the values above to look up a[n] for any n=6m+k, for example:
a[2024]=a[6.337+2]=a[2]=-1.
your solution is correct. OP’s is not
@@mikasa_sucasa Oh yes, the solution in the video seems to make an incorrect assumption at 6:28. I have edited my comment to reflect this. Well spotted!
Would be awesome to see it be generalised for z^n + 1/z^n
I'm sure there is a generalized form. I'll think about it.
@@PrimeNewtons if you Let u=z+1/z and want to write polynomial in u you probably will get Chebyshev polynomial
I tried to find coefficients of Chebyshev polynomials by deriving and solving recurrence relation using exponential generating function
or by deriving and solving ordinary differential equation using power series but I have got two sums which I cannot evaluate
Sum from recurrence relation
\sum\limits_{k=m}^{\lfloor\frac{n}{2}
floor}{n \choose 2k} \cdot {k \choose m}
Sum from ordinary differential equation
1+\sum\limits_{k=1}^{\lfloor\frac{n}{2}
floor}\frac{(-1)^k}{4^k}\cdot\frac{n}{n-k}\cdot {n-k \choose k}
You can do this quite nicely using de moivre's theorem for complex numbers :) I believe the generalised form is z^n + 1/z^n = 2 cos(n/3 pi)
Yeah it's very easy to generalize, first you turn the equation into z²-z+1=0 and solve for the roots to get cos(π/3)±sin(π/3)i and it's polar form will be e^iπ/3 and now we can turn z^n+1/z^n=e^inπ/3+e^-inπ/3 which if we take the complex definition of cos which is 1/2(e^ix+e^-ix) then this will equal 2cos(nπ/3) and plugging in 2024 for n for this question will give us -1.
You can in fact generalise z^n + z^-n, just let
z = cos∅ + isin∅, and by de Moivre's theorem, which states that
z^n = r^n(cosn∅ + isinn∅), and we can calculate r by letting n=1, and finding the magnitude of z. We can see that r=1, |z| = √(cos²∅+sin²∅) = 1. 1^n, as long as n is real, will always be 1, so we can proceed accordingly.
Now by the theory,
z^n = cosn∅ + isinn∅
z^-n = cos(-n∅) + isin(-n∅), and since cosine is an even function and sine is an odd function
z^-n = cosn∅ - isinn∅, so
z^n + z^-n = 2cosn∅
When n=1, given in the problem,
z + z^-1 = 1
2cos∅ = 1
cos∅ = 1/2, which is a very popular angle, ∅ = π/3,
and we can now substitute n=2024.
z^2024 + z^-2024 = 2cos(2024(π/3)), and evaluate it with the calculator (for my convenience lol),
z^2024 + z^-2024 = -1
Another, more general way. Let's call f(n) = z^n + z^(-n). Then f(0) = 2 and f(1) = 1 (the last equality is from the problem statement). We have the following property that can be easily checked: f(n)f(m) = f(n+m) + f(n-m). Substituting m=1 in this last equation, we get f(n)f(1) = f(n+1) + f(n-1). Remembering that f(1) = 1, we get f(n+1) = f(n) - f(n-1), which is a recurrence relation for a sequence such that any term is the difference between the previous two terms. The first two terms of this sequence are 2 and 1, therefore the sequence is:
f(0) = 2, f(1) = 1, f(2) = -1, f(3) = -2, f(4) = -1, f(5) = 1, f(6) = 2, f(7) = 1, ...
The last two terms f(6) and f(7) are equal to the first two terms f(0) and f(1), therefore the sequence has period 6. Therefore, f(n) = f(n mod 6). This allows the calculation of f(n) for any n, as the first 6 values f(0) through f(5) are known. So, for example,
f(2024) = f(2024 mod 6) = f(2) = -1,
f(2025) = f(2025 mod 6) = f(3) = -2,
f(7404) = f(7404 mod 6) = f(0) = 2,
etc.
Just beautiful.......❤
Thanks so much....
It's mod 7,not 6
@@georiashang1120 But f(7) = 1 and f(7 mod 7) = f(0) = 2. So, if I calculated correctly f(0) and f(7), it cannot be mod 7.
@@jlmassir f(n(mod 7)+1)=f(n) the period is always 7 , not 6.
@@jlmassir trust me, I'm a Chinese, born and studied maths in communist China(only soviet union math defeats us,LOL).
Wait what? What/how do you factorize z²⁰²⁴+z⁻²⁰²⁴ as (z¹⁰²⁴+z⁻¹⁰²⁴)(z⁵¹²+z⁻⁵¹²)(z¹²⁸+z⁻¹²⁸)(...) ?
Surprisingly it works here (I'm not sure why), but it would fail with 2022 or 2025 instead of 2024.
It works for any even number, but NOT if it's divisble by 6. So, if you use powers of 2 (for the combination) there is no problem; it will work.
Yeah, that is some fake algebra.
What we could do is write
1 ∕ 𝑧²⁰²⁴
= 1 ∕ (𝑧¹⁰²⁴𝑧⁵¹²𝑧²⁵⁶𝑧¹²⁸𝑧⁶⁴𝑧³²𝑧⁸)
= (1 ∕ 𝑧¹⁰²⁴) ∕ (𝑧⁵¹²𝑧²⁵⁶𝑧¹²⁸𝑧⁶⁴𝑧³²𝑧⁸)
= (−1 − 𝑧¹⁰²⁴) ∕ (𝑧⁵¹²𝑧²⁵⁶𝑧¹²⁸𝑧⁶⁴𝑧³²𝑧⁸)
= (−1 ∕ 𝑧⁵¹² − 𝑧⁵¹²) ∕ (𝑧²⁵⁶𝑧¹²⁸𝑧⁶⁴𝑧³²𝑧⁸)
= 1 ∕ (𝑧²⁵⁶𝑧¹²⁸𝑧⁶⁴𝑧³²𝑧⁸)
︙
= 1 ∕ 𝑧⁸
Thereby,
𝑧²⁰²⁴ + 1 ∕ 𝑧²⁰²⁴ = 𝑧⁸ + 1 ∕ 𝑧⁸ = −1
Yeah that's also what I thought
@@dirklutz2818really? Is there a proof for that?
This keeps me awake
This can be done much easier, even in your head without ever touching pen or paper. The equation
z + 1/z = 1
implies
z² − z + 1 = 0
so z is one of the complex cube roots of −1 since
z³ + 1 = (z + 1)(z² − z + 1)
This means that we have z⁶ = (z³)² = (−1)² = 1 and therefore z⁶ⁿ = (z⁶)ⁿ = 1ⁿ = 1 for any integer n. And since 2022 is evidently an even multiple of 3 and therefore an integer multiple of 6 we have 2024 ≡ 2 (mod 6) so
z²⁰²⁴ + 1/z²⁰²⁴ = z² + 1/z² = (z + 1/z)² − 2 = 1 − 2 = −1
Did the same thing
This is a fantastic solution.
My friend, i made it with Euler's formula and the answer it's right, but i think the demonstration whene you multiply all the factors is incomplete, it really is a long multiplication.
I agree with the last comment: the final product is a bit more complicated. There are many other pesky terms.
If you set a_n = z^n + 1/z^n it is easy to check that there are only 6 different values of a_n. In increasing order of n: 1, -1, -2, -1, 1, 2. the remainder of the division of 2024 by 6 is 2 so a_2024=a_2=-1.
A complex approach works very well too and is pretty fast. But I do like your approach and the fact that, even though z must be a complex number (the sum of a real number and its reciprocal has a minimum of 2), you don't need to use complex number to find the answer.
Great question! As always, I learned a lot watching your videos.
I think there is something missing from that proof. Why do you assume you can factorize z^2024 + 1/z^2024 as ( z^1024 + 1/z^1024)( z^512 + 1/z^512)( z^256 + 1/z^256)(....)?
Using this logic, you could argue that z^n+1/z^n is always -1 or 1 for all natural numbers n, as you can generate any natural number using the sum of powers of 2. But this is not the case. For example z^3 +1/z^3 = -2 (see below for proof of that). Am I missing something?
Proof for z^3+1/z^3 = -2:
(z+1/z)^3 = z^3 + 3z + 3/z + 1/z^3
1^3 = z^3 + 1/z^3 + 3(z + 1/z)
1 = z^3 + 1/z^3 + 3(1)
-2 = z^3 + 1/z^3
The squaring computation part.
z³=-1 and 1/z³=-1 sum=-2
please elaborate on step 2024 = 1024+512+256+128......., I couldn't get it.
(z+1/z)³=z³+3z+3/z+(1/z)³, correct me if I'm wrong, but you've appeared to have mistaken (1/z)³ for 1/z³ in the last step
@@miscellaneous3280(1/z)^3 is the same as 1/z^3
At 6:48, I remember the life before the created powers😂.
All the love for you Mr and keep going.
I didn't know it's that easy... I would had probaby made a quadratic and solved it for waay to much time ... Thanks!
I was going to do the same thing.
I was about to follow the same approach.
I think its beautiful when the comments are filled with suggestions on the other methods. Thank you so much!
if you work out the odd exponents, you'll notice the value of the expression z^n + 1/z^n cycles over the integers {2,1,-1,-2,-1,1}. 2024 = 2 (mod 6) which corresponds to the 3rd element in cycle i.e. -1.
its 2024 which is 2 mod 6.
@@franciscook5819 fixed
This is the best explanation so far.
For all integer values of 2^a=n. If z+1/z=1, z^n+1/z^n= -1.
you can do something similar using de moivre's theorem! If we let z = e^i(theta) then z + 1/z = 2cos(theta) = 1, so cos(theta) = 1/2 giving theta = pi/3. z^2024 + 1/z^2024 = 2cos(2024 theta) = 2cos(2024pi/3). 2024pi/3 is equivalent to 2pi/3 (the modulo when dividing by 2pi) which gives that z^2024 + 1/z^2024 = 2 cos(2pi/3) = 2 * -1/2 = -1. Your pattern recognition way is very nice though!
z + 1/z = 1 => z^2 - z + 1 = 0 => z = [1 +- sqrt(1 - 4)]/2 = 1/2 +- i * sqrt(3)/2 = e^(ix) where x = 60 degrees or x = -60 degrees, we can choose either as these are conjugates and thus they satisfy the same equations
Since z = e^(ix) where x is 60 degrees, we can mod its exponents in z^2024 + 1/z^2024 by 6 to get that this is equal to z^2 + 1/z^2, which we can easily discern is equal to -1 by squaring the originally equality z + 1/z = 1.
Loved this! Love seeing binary representation of numbers come up in problems!
If you solve z+1/z=1, you will find the equation z^2-z+1=0, witch give us the complex cubic roots of the unity. Then, if you calculate z^2024 and apply the ideia of congruent arcs, you wiil find (for both roots):
z^2024+1/z^2024 = e^(2*pi*i/3)+ e^(4*pi*i/3) =-1
Very elegant solution. Thumbs up to Prime Newtons! 🎉
Could use recursion Z(n+1)=Z(n)Z(1)-Z(n-1). (Z(n) := (x^n)+1/(x^n )) Here we are given Z(1) = 1 in which case sequence is length 6.
i.e. Z (1..6)= {1, -1, -2, -1, 1, 2} (Z(6) = Z(0) =2). Hence Z(2024)=Z(rem(2024,6)) = Z(2) = -1.
You can also think of complex numbers in the one but how you think there is a general form for z^n+1/z^n=2cos(nx) this works if z have modulus 1 we can prove that z in this question have modulus 1
z+1/z=1
z²-z+1=0
If you find the roots by your own you gonna notice that the modulus of z is always one
So in the expression we can rewrite as
z+1/z=2cos(60)
So the argument of z is 60
z^{2024}+1/z^{2024}=2cos(60.2024)
if you divide 60.2024 by 360 the rest gonna be 120 so :
z^{2024}+1/z{2024}=2cos(120)=-1
Wecan just use complex no. Z= r e^itheta and get the condition r= 1 by equating imaginary part and then use e^itheta= cos theta +i sin theta
The transition from a sum to a product at the end was not stated clearly.
Another approach:
Define u(n)=z^n+1/z^n
Given u(1)=1
Evaluate u(n) for n=1,2,…
[u(1)^n=u(n-1)*u(1)]
==> u(n+1)=u(n)-u(n-1)
Resulting in the series:
u(n)=1,-1,-2,-1,1,2, (repeat) 1,-1,-2,… For n=1,2..
General solution:
u(n)=u(6k+m), k=0,1… , m=1,2,..6
for any n=1,2,…
u(2024)=u(6*337+2)
Since m=2, u(2024)=-1
Your handwritting is so pretty and the sound of the chalk in the chalkboard sounds so good!
The way I did it was to solve the quadratic, gives z = the complex number of 1/2 +/- root(3)/2j (j because I'm an engineer). The modulus of z is 1, so raising z to any power the modulus remains 1. Since the argument of z is +/- pi/3 the argument of z^n is +/- (n+1 mod 6)pi/3. 2024 mod 6 is 2. So z^2024 is -1. Either value of z^2024 gives -1. I find that, more often than not,working with complex numbers using modulus and argument is far simpler than real and imaginary.
Your factorization is not true. otherwise, you could rewrite 6 as binaries, 6=4+2 and you would obtain (-1)*(-1)=1. However, as franolich3 above correctly states, z^6+1/z^6=2, so your formula is not true in general. Also, please consider using brackets after multiplication, before a negative number.
Please explain how did you get that series of products, that looks like ok, but not obvious how it's derived
First of all z + 1/z will be always greater than 2 so this is complex no.
In the last step I believe something is missed out.the first calculation is (X^2+1/X^)(X^4+1/X^4) this gives (X^6+1/X^6+X^2+1/X^2).
alternatively,
given z+1/z=1 (clearly z is not=0),multiply b/s by z
=> z^2+1=z
z^2=z-1 -----(1)
again multiply by z
=> z^3=z^2-z =z-1-z=-1
{by (1)}
=> z^3=-1 -----(2)
square b/s
=>z^6=1 -----(3)
now z^2024={(z^6)^337*(z^2)
}= z^2 {by (3)}
To evaluate
E=z^2024+1/z^2024
=z^2+1/z^2
=(z+1/z)^2-2
=(1)^2-2
=1-2=-1
as found by Prime Newtons
By solving the equation you find two complex conjugate roots z= (1+i√3)/2 = 1•e^(iπ/3) = 1∠(π/3) = cos(π/3)+i•sin(π/3) and ¯z= (1-i√3)/2 = 1•e^(-iπ/3) = 1∠(-π/3) = cos(π/3)-i•sin(π/3) which have two surprising properties:
(1) They are the reciprocal of each other : 1/z = 2/(1+i√3) = (2(1-i√3))/((1+i√3)(1-i√3)) = (2(1-i√ 3))/(1²-(-3)) = (1-i√3)/2 =¯z .
(2) They are unity : |z| = | ¯z | = 1.
Which means that z^n+1/z^n = z^n + (1/z)^n = z^n + (¯z)^n = (cos(π/3)+i•sin(π/3))^n + (cos(π/3)-i•sin(π/3))^n = cos(n*π/3)+i•sin(n*π/3) + cos(n*π/3)-i•sin(n*π/3) = 2•cos(n*π/3) in accordance with the Moivre's formula.
And so z^2024+1/z^2024 = 2•cos(2024*π/3) = 2•cos(2π/3) = 2•(-½) = -1
Another way is to express z in exponential complex form and work from there
z^2-z+1=0-->z=1/2±i*sqrt(3)/2
It's a sixth or five sixths of the way around the complex unit circle. Raising a number on the complex unit circle by a power simply multiplies its angle around the circle by that power.
Because 2024 is even, but not divisible by 3, z^2024 will either be a third or two thirds of the way around the circle.
Because two thirds of the way around the circle is also negative one third of the way around the circle and vice versa, z^-2024 will be the other one
z^2024 will either equal -1/2+i*sqrt(3)/2 or -1/2-i*sqrt(3)/2
With z^-2024 being the other.
The i*sqrt(3)/2s cancel, leaving us with -1 as the answer.
An odd power not divisible by 3 would give an answer of 1, an odd power divisible by 3 would give -2, and a power divisible by 6 would give 2.
It can be done really short once you find z^2+1/z^2=-1 z^2 can be replaced by complex number omega and by applying properties of cube root of unity you can get -1 in 1 step
well, if this would have been in a multiple choice test we SHOULD use complex numbers !!
Let f(n) = z^n+1/z^n. Then, f(1)=1 and f(2)=-1 (straightforward) but also the identity f(n+1)=f(n)-f(n-1) holds (verify!). So we obtain a recurrence relation and it turns out the pattern repeats every 6 integers (verify!), starting from 1. Then you only need to know what 2024 mod 6 is (it is 2) so f(2024) = f(2)=-1
Even though using this trick for this question is faster for these type of questions it's better and easier to turn it into a quadratic equation and solve for roots then if complex numbers come up use the polar form or just plug the roots in the second equation to get the answer. So rather than 2024 for if we had something like 2371 or something then turning the roots of quadratic equation into polar form and using the definition of the complex cosine we can generalize this equation to z^n+1/z^n=2cos(nπ/3) and plugging in 2371 for n we get 1.
Can you please explain why it's -1 though? If you had used two 4's, instead of one 8, as the last factors to reach 2024, it would have been an even number of -1's multiplied together, and hence would equal 1, not -1. I assume it's something periodic, but it would be nice to understand.
2024 is not a power of 4. How would you get it?
@@PrimeNewtons Please see my comment and example above.
Wait what ??? I couldn't understand the last part. How can you just multiply like that ???
It is far better to just use the quadratic and make out the solution to be z = -w / -w^2 [w=imaginary cube root of one] Hence given expression gives rise to the sum of two imaginary cube root of one which is -1. This is a valid logic. But what did you do there ???
z=cos(pi/3) +- isin(pi/3), which equals e^(i*± pi/3) . So z^6 equals 1. z^2024 = (z^6)^(337+2) = z^2
So when you added the powers of 2 together to reach 2024, when expanding that idea to the equation you were solving were you multiplying? ((z+1/z)^1024 * (z+1/z)^512 * …) or using a power tower? ((((z+1/z)^1024)^512)^…) or are both methods just the same thing represented with different notation?
Both methods you described are different.
Taking the first one, all the powers of 2 are added while in the second all the powers are multiplied.
For this case, 2024 is achieved by adding powers of two.
Taking all the additive powers out causes it to get multiplied.
General Case:
z + 1/z is real, so equating real parts:
rcos(t) + 1/r cos(t) = 1,
rsin(t) = 1/r sin(t) for all t
==> r = 1 , cos(t) = 1/2 (WLOG t = pi/3 in principal branch)
==> z = exp(i*pi/3)
==> z^n + z^-n (for integer n)
= exp(i*npi/3) + exp(-i*npi/3)
= 2cos(n*pi/3)
= 2cos(k*pi/3), where n == k mod 6
n = 4,1024 ==> k = 4
==> sum is 2cos(4pi/3) = -1
It can be efficiently solved using complex alzebra .. just by putting 2024 in the place of n where the ans comes out to be 2cos(nπ/3)
We start with the given equation: z + 1/z = 1
We rearrange this equation to solve for z:
z^2 - z + 1 = 0
Solving this quadratic equation, we get:
z = (1 ± √(-3))/2
Since z is a real number, the only valid solution is:
z = (1 + √3i)/2
Now, we need to evaluate the expression z^2024 + 1/z^2024.
Using the formula for raising a complex number to a power:
(a + bi)^n = a^n - (n * b^2 * a^(n-2))/2 + i(n * a * b^(n-1))
Substituting z = (1 + √3i)/2, we get:
z^2024 = ((1 + √3i)/2)^2024
= (1 + 2024√3i - 2024*3i)/4^2024
= (1 + 2024√3i - 6072i) / 16^2024
Similarly, we can calculate 1/z^2024:
1/z^2024 = 1 / ((1 + √3i)/2)^2024
= 4^2024 / (1 + 2024√3i + 2024*3i)
= 16^2024 / (1 + 2024√3i + 6072i)
The final expression is the sum of these two terms:
z^2024 + 1/z^2024
= (1 + 2024√3i - 6072i) / 16^2024 + 16^2024 / (1 + 2024√3i + 6072i)
This is the formula to calculate z^2024 + 1/z^2024 given that z + 1/z = 1.
Clearly wrong in the last step you want to sum not to multiply
Please explain the last step
That was short, simple and kind of fun.
how do we know that (z^1024+1/z^2024)(z^512+1/z^512)...(z^8+1/z^8)=(z^2024+1/z^2024)? arent there going to be many more terms than just those two for example z^1024/(z^512 times z^256...)=z^1024/z^1000=z^24?
Let z=e^it. Then z+1/z= 2cos t = 1 > cos t = 1/2 > t=pi/3 + 2pi n, n being an integer. Thus, z^2024 = e^2024(pi/3+2pi n)i = e^(674 pi i) e^(2 pi i/3) = e^(2pi i/3) = -e^(-i pi/3). So, z^2024 + z^(-2024) = -2 cos pi/3 = -1.
To get 1000 from binary, you could have started with the sum of all binaries below 1024, starting with 1, which would have resulted in 1023. Now you know you need to subtract 23. 23 in binaries is 16+4+2+1, so you need to omit these, but take all the others
I think you made a leap too far.
2024=1024+512+256+64+32+8 (so far so good)
(z+1/z)^2=z^2+1/z^2 + 2 = 1^2
z^2+1/z^2 = -1 similarly for every time the power is doubled but ...
(z^1024 + 1/z^1024)(z^512 + 1/z^512)= (-1)(-1) = z^1536 + 1/z^1536 + (z^512 + 1/z^512)
= z^1536 + 1/z^1536 - 1
so z^1536 + 1/z^1536 = 1 + 1 = 2
so multiplying z to the power of two expression by by another, to a different power of two, is NOT the same multiplying the -1s together (note that it can work out that way e.g. using the last two powers, 32 and 8).
same for 256 and 128 and for 64 and 32
The easier is solve the quadratic (z+1/z=1 => z² -z +1 = 0 => z = (1±√3)/2=e^iπ/3 or e^i5π/3
Raise to the 2024th power (2024=2 mod 6)
gives z^2024 = e^i2π/3 or e^i4π/3 or (-1 ±√3/2
both give -1 when subbed in to z^1024 + 1/z^1024
Thank you sir. Great explanation. Thank you so much you are awesome. Love from India. 😊😊
We get z^3 +1=0 from where z^3=-1. z^2024=(z^3)^674.z^2=z^2.Similarly for 1/z^2024.
For typing purposes, please let me represent "z to the n plus one over z to n" as Z(n). It looks to me like you used Z(a)Z(b) = Z(a+b). Which has not been established.
You could have done the breakdown of 2,024 into a sum of powers of 2 easily, assuming that the calculator app on your table has a programmer mode: Just convert 2,024 from decimal to binary...
Very good. Thanks 🙏
where does that 2 you multiplied with come from
Just substitute z = cosx+isinx
and 1/z = cosx-isinx
So z+1/z=2cosx
2cosx=1 » x=π/3
So then :
z²⁰²⁴+1/z²⁰²⁴=2cos(2024x)
= 2cos(2024×π/3) = -1
Thank you Sir.
Du you teach math on school, are you official a teacher?
Maybe I'm missing something but I don't believe this solution is complete. I haven't finished working out the answer yet, but multiplying the power of 1024 and the power of 512 is not the same as squaring; you get a power of 1536 plus a power of 512 (which is -1), equaling 1, so isolating 1536 is then 2. (I believe)
Now that I've worked it out further, there is a pattern to the powers, every time you multiply z^n + 1/z^n by z + 1/z (which is 1) , you get z^n+1 + 1/z^n+1 + z^n-1 + 1/z^n-1. What this yields is a pattern of 1,-1,-2,-1,1,2. You do have the right result, but I don't think I agree with the assumptions made to get there.
(and please forgive my way of typing it out which may not follow the right order of operations, it's challenging to do on a phone keyboard!)
What's the name of the device are you using?
where can i get your t shirts
How electron is formed?
Another method:
z + 1/z = 1 ⟹ z^2 - z + 1 = 0 ⟹ z^3 + 1 = 0 ⟹ z^3 = - 1
⟹ z^2024 = z^2025/z = (z^3 )^675/z = (-1)^675/z = (-1)/z = - 1/z
⟹ z^2024 + 1/z^2024 = - 1/z - z = - (z + 1/z) = - 1
Cant u solve using the property of w(omega)
Complx cube root of unit?
I seem to be familiar with the scientific calculator you use
If z^2024 is -1, isn't the final answer -2 ?
Nice work,thanks
That is elegantly done, and I used De Moivre's theorem and overcomplicated it!
It could've been solved faster if z was replaced with a negative omega.
Like many others I would like to know what happened to the missing 2 terms from each multiplication at the end;
(z^a + z^(-a))x(z^b + z^(-b))=
(z^(a+b) + z^(-(a+b)))x
(z^(a-b) + z^(-(a-b)))
4ever fascinating by calculating 😮
z^3=-1,we have z^2024=z^2, so z^2024+(1/z^2024)=z^2+(1/z^2)=-1
nice. but I am not used to the notation -1 x -1 x -1 x ..... better to write -1 x (-1) x (-1) x ....
those who stop learning, stop living!
Yeah, its right because if someone is learning to stop everywhere he becomes a dead body !!
I thought all you had to do is since the power is even, then it is negative
z + 1/z = 1
=> z^2 - z + 1 = 0
Multiply by z + 1 both sides such that z not = -1
=> z^3 + 1 = 0
=> z = -1, -w, -w^2, where w,w^2 = complex cube roots of unity
But z not = -1
=> z = -w, -w^2
=> z^2024 = (-w)^2024 or (-w^2)^2024 = w^2,w
We know that w^3 = 1 => w^2 = 1/w
z = w, 1/w
=> z ^2024 + 1/z^2024 = w + 1/w = w^2+1/w = -1
But what is Z?
z*2024+z*-2024=2cos(2024.p/3)=2cos(2p/3)=-1
It's not true. The sum of two positive values can't be negative! (Added: here we have another case)
Positive? Where did you get that?
@@PrimeNewtons z²⁰²⁴=(z²)¹⁰¹². Any real number raised to even power is positive. Also 1/(z²⁰²⁴) is positive. And the sum of these two positive values is also positive
@@PrimeNewtons if z is real. z² is positive. 1/z² is also positive. How can the sum of these positive values be equal to -1? The same with 2024. Or am I missing something?
@hellovat_youtube You assume z is real. You can't assume it's real. In fact there is no real number z such that the sum of z and its reciprocal is equal to 1. Clearly, z is not real.
@@PrimeNewtons yes, you are right! I'm used to that usually variables are real by default and if they are not it is mentioned in the conditions. As you noticed here we have z+1/z=1, and this doesn't have real roots. A kind of implicit condition, that z is not necessary real. Thank you
You found z^2024 NOT z^2024+1/z^2024
Cool!
Awesome hehe
I spot something wrong in your last step. To simplify, assume that what you want to calculate is:
z^1536 + 1/z^1536
and note that 1536 = 1024+512
We know that z^n + 1/z^n = -1 when n= 512 or 1024.
According to the logic of your last step, since you are multiplying two terms each of which has the value -1 you might expect the answer to be (-1)(-1) = +1
But:
(z^1024 + 1/z^1024)*(z^512 + 1/z^512) = (-1)^2 = +1
Distributing terms gives us:
z^(1024+512) + z^(512-1024) + z^(1024-512) + 1/z^(1024+512) = 1
z^1536 + 1/z^512 + z^512 + 1/z^1536 = 1
Re-arranging:
z^1536 + 1/z^1536 + (z^512 + 1/z^512) = 1
z^1536 + 1/z^1536 + (-1) = 1
z^1536 + 1/z^1536 = 2
Done in 5 seconds😎
Cool
z+1/z=1 then z^2024+1/(z^2024)=-1
z + 1/z =1 ERROR
z+1/z=1,z^2-z+1=0,(z+1)(z^2-z+1)=0(z+1),z^3+1^3=0,
z^3+1=0,z^3=-1,z^3×z =-1×z,z^4=-z
z^4+1/z^4=-z-1/z=-(z+1/z)=-1
f(2n) = -1
No, f(0)=+2 and f(6)=+2
@@dirklutz2818 ok, sorry , i meant, f(n^2) = -1
Did it in mind
2 hoga answer
Because z is womegs
Z^3=1 이용하면 쉽게 풀수 있는데 왜 저렇게 힘들게 풀이 하는지 이해가 안되네
expn=1+1=2
I should take back like bad logic
You should only like a video after watching it. That way, you don't have to comment on taking it back. Also, you should help others understand why the logic is bad. I want to learn too.
@@PrimeNewtons (z^a+z^(-a))(z^b+z(-b))=(z^(a+b) +z^(-(a+b)) + z^(a-b) +z^(b-a) two extra terms come out in the product only if prove them irrelevant could you ignore....
Think it's +1
Reponse=-1
J’ai le même résultat avec une autre méthode, pas plus rapide!!!!
wow
-1
pls solve this question from my homework its too hard for me. If a=7-4sqrt3, find sqrt a + 1/sqrt a
sqrt(a)=sqrt(4)-sqrt(3)
I think you can solve the rest by yourself easily
@@vedwargantiwar4610 nah sqrt(a)=sqrt(7-4*sqrt(3))
@@aneeshbro my guy
sqrt(7-4sqrt(3))=sqrt4-sqrt3
@@aneeshbro the answer is 4 I think so
Somebody solved that for you on bprp.