One of The Hardest International Maths Olympiad Problem
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- čas přidán 27. 12. 2023
- IMO Question - 2019. Can You Solve?
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A great exponential equation! What do you think about this problem?
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It hurts me even to see _ln 1_ written down as a divisor.
ln1=0😂
Extremely simple : no solution. End point.
Every candidate solution you find under the form (i ln3) / (2k pi) should be checked in the original equation and should then be rejected.
Complex exponentiation is multivalued
So it works a single branch
@@iqwit Not really. If you consider it as a calculus operator, it can only return one value. Otherwise it's impossible to calculate anything.
I keep trying to visualize how 1^(-1.09861...i/2π) somehow makes 1 a 3. I'm failing.
La solución es negativa. (-i ln3) / (2k pi)
The minus sign is unnecessary since k can be any integer
Also reported for misinformation
the description says "IMO Question - 2019" however, there is no such question in the 2019 IMO
They usually don't ask nonsense questions. But the buzzwords "Math" and "Olympiad" seem to generate clicks on YT.
@@user-gd9vc3wq2hyes, this is not an interesting problem of the type put into Olympiads.
Ex falso sequitur quod libet 😂
According to this logic, 1^[(ln 3)/(2kπi)] = 3^(n/k), n=0,±1,±2,… (k is fixed here). This multi-valued object is not the same as 3. If you take x multi-valued itself, you get 1^[(ln 3 + 2πiℤ)/(2πiℤ*)] = 3^ℚ, which again while it contains 3 I wouldn't call that equal to 3. Do mathematicians really write that with an equality sign?
Notation : ℤ* = ℤ\{0}, ln denotes the real-valued logarithm.
Hate it as much as you like he is right... just pick the right complex logarithm. When you end up with n/k you are using two definitions of the complex log so wrong result. One definitions should be used and only one. The complex log to be unique is defined on the exponential function with a restriction... Google it. It's tge branch concept
In Russia , we would just say that x = log1(3) , which equals to 0 ,without using natural logarithms.
You might say so, but that doesn't make it correct. There is no function u |--> log1(u) because this would be the inverse of z |--> 1^z which is the constant function 1 and hence has no inverse.
Given that Russia is well known for producing great mathematicians, I doubt that's true. If x=0 as you say then the expression would say 1⁰=3 which means 1=3.
I especially liked the part in the end where you stated "don't feel bad if you got this wrong"
Between the fact that ln(1) = 0 and the zero product property, x has no solution.
no **real** solution. There's an infinite amount of complex solutions to this problem via usage of the series representation of the exponential function.
@@omarfish8940 Even complex numbers, when multiplied by 0 (or ln(1)), equal 0. And 1 to the power of any number, real or complex, equals 1.
Either 1 = 3, or there are no solutions.
someone correct me if im wrong, but doesn't 1^(-ln3*i/2kπ) equal 1??
You're absolutely right. This video's author is insane or just trolling
Sorry, but no. Anytime you put the imaginary i in an exponent, the rules all change. Instead of a simple exponentiation (say for example 1^6 = 1*1*1*1*1*1 which of course is just 1), you get 'rotations' in the complex plane. For example, 1^(i) is more like 'twisting' a vector around. The most famous is e^(i) = -1. What can I say, complex math is, well..... 'complex'.
@@mikefochtman7164 You are very nice for trying to explain it, but i'm very familiar with complex analysis. I know that, all results in the complex plane must be looked at differently than the real plane. I'm saying that in this case, where the base of the exponent is 1, the result is still 1. I looked it up and it is indeed the case.
@@user-uy8yt7ku4w I think you are too harsh on your critisism. He probably felt too confident in his answer and decided to not check the result. Honestly, if the exponential's base was anything other than 1, i wouldn't have noticed either.
It all comes down to the interpretation of 1^x in the question, which is never explained properly in the video. The solution relies on interpreting 1^x for complex x as exp(x log 1), which gives infinitely many choices if we take log 1= 2ik pi for integer k. (Equivalently, replace 1 by
exp(2ik pi).) So the video is not really solving one equation, but infinitely many: exp(2ik pi x)=3 for each k. There are no solutions when k=0 (principal branch), but for each fixed non-zero k there are in fact infinitely many (essentially because there are infinitely many choices of log 3).
So the video is unsatisfactory: not only does it not state properly what problem is being solved, it also doesn't give a complete solution.
This is NOT an International Math Olympiad question. Certainly not from 2019 (i just checked), but the wording seems off for a genuine IMO question. It's not hard by their definition, try even A1 of that year, which is the easiest, and much harder than this imo.
Why did you call it an IMO question? Just clickbait? Did you just copycatted some reddit post? Can you reference, please?
clickbait
Didn't you do the same problem a week ago, but for 1^x = 2?
Yes
And i has the same question, is 1^x = 2 (Wolfram give no solution) equivalent to 1 = 2^(1/x) (Wolfram has solution)?
@@pokoknyaakuimut001good question. It depends on how you define 1^x for complex x.
@@andrewhone3346 It is defined as e^(x*log(1)) [natural log], and since log(1)=0 1^x=1.
@@YAWTon that is true if you take the principal branch of log, but for a non-zero complex number z there are infinitely many choices of the logarithm, given by
log z = ln |z| + i arg z,
corresponding to the infinite number of choices for the argument. So when z=1 the argument can be any integer multiple of 2 i pi, giving log 1 = 2ik pi for arbitrary integer k. The choice with k=0 (which coincides in this case with the real natural logarithm ln) is called the principal branch.
So when you define z^x = exp(x log z) for fixed complex z, as a function of x, there are infinitely many choices of this function.
28 December 😅😅😅
(Equivalent of April Fools' Day in Spain)
Your solution set is extraneous as they fail went tested in the original equation. This ignoring where you hid taking the natural log of 1 on the LHS.
Thanks budy. Great solution!
The original problem must be clarified, since the left-hand side of the equation, 1^x = 3, can have multiple values whereas the right-hand side has only one value. In fact, we seek a complex number, x, such that one of the values of 1^x is 3.
Also, there is a subtle mistake at 7:28 that inadvertently eliminates some of the solution set. When applying natural log to complex numbers, one cannot blindly proceed as if the arguments were reals. In order to get the most general solution, following (e^(i 2 k pi))^x = 3, one should proceed instead with
e^(2 k pi i x) = e^(ln3)
which implies 2 k pi i x = ln3 + (2 n pi i), for some integer n
Then the general solution is
x = - i (ln3)/(2 k pi) + (n/k), where n is any integer and k is any non-zero integer.
We can check that this works. Substituting the above into 1^x,
1^x = e^(2 m pi i x) = 3^(m/k) e^(2 pi i mn/k), where different values of m yield different values of 1^x. The value resulting when m = k is
1^x = 3 e^(2 pi i n) = 3.
For example, one of the solutions that was missed in the video is
x = 1 - i(ln3)/(2 pi). Check that this works:
1^x = e^(2 k pi i)x = e^(2 k pi i + k ln3) = 3^k
For k = 1, this gives the desired result.
'Higher' math sure looks a lot like wrong math.
How this video has more likes than dislikes? That is ridiculous. People are really bad at math
If you are allowing infinitely many choices of log 1, which is what you are doing when you interpret 1^x as exp( x log 1) , then you must also allow infinitely many choices of log 3.
You are right on target, and in fact, that mistake in the video resulted in missing an infinite subset of the solutions!
Thank you for your shareing
This video blew my mind. I love mathematics. ❤
What do You think about this approach? Much love and respect. Have a great day!❤️❤️❤️
1) x real -> No solution. OK
2) Let's suppose x complex -> x = a + ib (a, b Real numbers !)
-> 1^x = 1^(a + 1^ib) = 1^a . 1^ib
= 1 . 1^ib = 1^ib
Applying ln on complex numbers: ln(z) = ln|z| + i arg(z), we get :
ln (1^ib) = ln|1| + ib = 0 + ib = ib
Hence ln(1^x) = ib = ln(3). Pure imaginary = pure real -> Impossible. x is not a complex number
Solution: NO SOLUTION.
EDIT : Shortest path (thanks to Wikipedia). Knowing:
r^z = e^(z.ln r), where r is a real positive number and z is a complex number,
1^x = e^(x.ln1) = e^0 = 1
Therefore 1^x always = 1. Equations with any other value have no solutions.
That development is not correct.
Note that (e^x)^y = e^(xy) is NOT a valid identity in general with complex x and y. In your case, the wrong step is going from (e^(i2k\pi))^x=3 to e^(i2k\pi x)=3 (while applying natural log in both sides), because (e^(i2k\pi))^x is not equal to e^(i2k\pi x).
The correct identity is (e^x)^y = e^(y\log(e^x)), remembering that log is multivalued.
Please, see en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities
I don't know where the problem is, but this doesn't seem to work.
1^[-i ln3 / 2kπ] = [1 ^ (-ln3 / 2kπ)] ^ i = 1 ^ i = 1.
It’s wrong tho, you can’t write the x into the exponent as a factor because the exponent has I into it which makes this rule not work
the 'solution' is more complicated than the 'problem'
Higher Maths? How high to you have to be to think this proof is valid!
I think, normal school math, age 15 to 18 should do.
How dumb to you have to be to not think this proof is valid!
seems valid to me
What's wrong with the proof? Of course x is a complex number, anyone can say from the start. All the elementary textbooks introducing complex number has plenty such equations as exercise. I wished though he would solve for the real and imaginary parts of x.
There is no solution. The modulus of 1 to any power of x element C is always 1. The modulus of 3 is 3. So, there is no way. I only say this: De Moivre. Please no comments. I won‘t answer. It’s like jihad, to argue with proponents of this topic.
A plot of solution set, (k sequence) in complex plane, could be interesting ...
Resolving with worfram alpha ...
In solution there is an additive term ...
Instead of ln3, there is a (ln3 + i 2n pi) term (indexed by n) in numerator, over the same divisor, indexed by k
If one is multiple with any integers ,power. It is always one.
what if you use imaginary numbers? (badum, tsss)
This is not an Olympiad question, just clickbait. It would be more useful if you made a video explaining complex exponentials and logarithms.
what are you talking about ? i have all the problems from IMO 2019, including the Shorlisted Problems with solutions. This problem doesn't appear anywhere !
This is a hard problem because it requires specific knowledge that Eulers formula evaluates to 1 for every theta that’s a multiple of 2 pi. Substituting 1 for Eulers formula with the condition necessary is insanity - how would you solve this in an actual competition without knowing this fact?
Anyone who did not know this basic fact would not be participating in a math competition!
What if we now take the limit as k goes to infinity?
what if limit doesnt exist?
Doesn’t matter, it’s still true for all integers no matter how large
Nothing? If K is an integer then nothing changes since this is for all integers.
3 also complex. 3=3*exp(2n*pi*i)
Could you explain it?
It's been 30 years since I've worked with imaginary numbers. I get that i^2 = -1, so if x=i, then x^2 = -1, but what if x=-i? Is x^2 = 1 or -1? If x^2 = -1, does that mean that i = -i, or am I just imagining things?
If x=-i, then x^2 = (-i)^2. But (-i)^2 = i^2 = -1. Your mistake was when squaring -i, you get +i^2, not -i^2
@@mikefochtman7164 So does that mean that i = -i?
@@Alex_Plante no bruh its basic maths
(i)^2=(-i)^2 but it doesn't mean i=-i
@@Alex_PlanteNo sorry, let me try again. I know all the 'negative' signs can be confusing. First we have set x= -i Now we square it which is just (-i) * (-i). Now before we start thinking about the 'i' part, notice we have two negative values multiplied together. Basic rule, a times another , the results will be positive. So (-i) * (-i) = + i^2. Now that we've done that, NOW let's consider the special property of that 'i' symbol. The 'i' is defined as the sqrt(-1) (square root of negative one) (Be careful reading the 'i' and the '1' as they look similar). Remember we had +i^2. We can now replace the 'i' with sqrt(-1). So now we have +(sqrt(-1))^2. Well squaring the square root of something gives us back the original So +(sqrt(-1))^2 = +( -1). So we finally end up with just -1. (-i)^2 = -1. I hope this helps.
i just multiply by 3 on each side and get 3^x = 9, x=2
Then you would just be wrong.
3 * 1^x =/= 3^x
3^x * 1^x = (3+1)^x = 3^x
Do tell. What color is the sky in your world?
How complex!
It's a pity we can't simplify
I’m puzzled why this was considered “hard” (if it even was on the olympiad to begin with). I suspect clickbait…
I'm calling BS on this whole thing.
Should it be infinity since 1/(infinity)=0?
No, infinity can never be a part of an equation as a value because it is a term not a value. 1/0 IS NOT infinity, it's undefined
@@tophatjones6241 1/0 is not undefined, 0/0 is. However, 1/0 is not infinty aswell.
@@canr7721/0 is undefined, what are you talking about?
Division is inverse of multiplication. If c=a/b, then bc=a. I.e., in 3=6/2, then 2×3=6. If b=0, you'd get 0c=a. Since we're trying to find 'c', you can easily see that any number has the same outcome: zero for 'a', which is a fallacy. That's why we call it 'undefined', as it doesn't matter what you plug in, it's always wrong.
There are many more subtleties, Wikipedia has a good explanation, and bprp and other math channels too.
@@canr7721/0 is undefined. 0/0 is indeterminate.
Math takes us to a strange universe. If 1^x = 3 has no real solution, otherwise has infinities complex solutions...
Considering tha a complex number is not a number, but a vector (x +iy), there is no solution where x≠0, and infinities solutions where x=0.
Magic!
=> 1 = 3
=> No solution
No 'real' solution
Maybe you should plug your "solution" into the initial equation and check whether it is satisfied.
How about trying x equals to infinity, since 1/(infinity)=0?
@@jharmley6882 So you suggest that 1^infinity =3? Does 1^infinity just equal 3, or also 5, or 8, or 10, or any (positive, real) number? Can one use this to prove that all numbers are equal?
@@user-gd9vc3wq2h well ln 1 is equal to 0, and 1/(infinity) is equal to 0, so I thought that would work
Good ahead and plug your solution back into the original equation. I dare you. It will still be equal to one (1). It will work as well as your previous problem of 1^x = 2.
wait can we even say it's a solution? bcs k can be literally any natural number. so we'll just have infinite solutions.
Strictly speaking, it's a solution set.
welp, but the set contains infinite solutions@@rogerkearns8094
@@rogerkearns8094, well it is only seeming to lead to such solution set . when checking these results you'll get
1 = 3 , indicating that the equation was false, isn't it . or please show us that this conclusion is wrong .
Très belle démonstration fausse ! Pourquoi? Dans C certaines formules sont fausses.
Par exemple (a^m)^n =a^(mn) est fausse pour a=2 m=6i n=i (calculatrice hp prime).
Ah bon ?!?!!! Ce n’est pas parce que la calculatrice hp le dit…
Anybody knows that you multiply powers when you take their exponents. (3³)⁴=3¹², (because 3x4=12), try it in your calculator. But don't forget the parentheses. If you remove them, precedence is top to bottom.
@@Misteribel en.m.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities
@@Misteribel "Anybody knows". LOL.
To all French people, at least in scientific contexts, please speak fxxxxxg ENGLISH. I come from teaching in French universities, it's crazy the linguistic protectionism
1^=1
X=нет решения, потому что 1 в любой степени равно 1
Simple, there is no real solution problem solved.
Good explanation
Muy simple, no tiene solucion , profesor se complica para explicar esto, mucho mas claro es Juan
240
未看先猜有三角函數
Te faltó definir, X€C
Ay!
Nope.
Silly. Do problems that make sense if you want viewers.
..but then ∀m,n∈ℤ and k∈ℤ* we should also have
3 = (1^m)^(i·ln(3)/(2πk))·1^n,
which is equivalent to staying that ∀r∈ℚ and n∈ℤ we have
3 = 1^((r·i·ln(3)/(2π))+n).
In particular, if r=0 then ∀n∈ℤ 1^n = 3. 🤔