I think this problem has not been properly solved in the video. As others have also pointed out, there are two main problems with the demonstration. The "first method" has the problem of just guessing one polynomial solution (P = X^4 + 1, and showing it satisfies the original equation, which is ok) but not showing that it is the ONLY solution. In fact, it isn't if the polynomials are understood as elements of the polynomial ring over the field of complex numbers. In this case, there are 5 distinct solutions (and answers). The "second method" has the problem of not showing that the coefficients b,c,d of the X^3, X^2, and X monomials must be zero indeed (see around 9:28). Though being true, I don't think this is obvious at all, and it must be derived step by step with a lot of additional work by comparing the coefficients of the X^16, X^15, X^14, X^13, and X^12 terms of the result of the polynomial composition on the LHS. This involves some combinatorics. When doing this work, we find that coefficient a must be a 5th root of unity, coefficients b, c, d must all be 0, and coefficient e must be 1. Over the real numbers this results in the unique solution P = X^4 + 1, while over the complex numbers there are 5 solutions: P_k = alpha^k * X^4 + 1 with alpha being any primitive 5th root of unity and k an integer from the set { 0, 1, ..., 4}.
There are holes in the presentations. (1) That x^4 + 1 is a solution does not mean it is THE solution, you must discard the other possibilities. (If complex coefficients are allowed, ax^4 + 1 is such a possibility for a = every fifth root of unity.) (2) "don't waste time dealing with b c d" = "don't bother fully proving the result"... As it turns out, it's not hard to prove that b = c = d = 0 by looking at the higher power terms of P(P(x)-1): those terms are much shorter than what you might fear.
When a question asks for P(2) instead of P(x) it is often the case that there's an answer that does not involve fully working out what P(x) is. I am not seeing it in this case, but that is a common question-setter's trick.
Actually, it’s not more difficult with P (2) = ? Than it is for P (X) = ? If we define Q (X) = P(P (X) - 1), then we have Q (Q (X)) = X^16, and Q (Q (2)) = 2^16. Obviously, 2^4 is a solution, and P (2) = 17.
I knew how to solve this almost immediately. HOWEVER, i always look at the length of the video to judge how hard it is. This one is longer than most, so i thought i needed a complex method, derivatives of nested functions etc. Finally went back to my original idea and checked the video. What a mug. BTW you have the neatest blackboard writing I've ever seen.
Thank you so much for these videos, theyre helping alot. Please can u create 3 playlists for the other topics in math olympiad; algebra, combinatorics and number theory
If P(x) is just a rational function, we can remember that the square root is multivalued, so we could try both values of √16=±4. P(x)=x⁴+1 and P(x)=1/x⁴+1 both satisfy the relation, and give two options, P(2)=17 and P(2)=17/16
@@williamperez-hernandez3968 notation is just notation. If I do something using P(x)=sin(x²), the world doesn't end. Hence why I specified rational functions at the start.
My first gut was to try a change of variable by letting q(x) = p(x) - 1, then substituting out all the P's. This yields a very fast path to the solution. When we have p(q(x)) = 1 +x^16, we subtract 1 from each side and have q(q(x)) = x^16, and can see from here that x^4 satisfies q(x).
After taking b=c=d=0 you did not have to write the +....+ , however after raising the inside to the 4th power you ignore all the other terms and you should have written +...+. I'm guessing you did it right in preperation of the video and I can relate being a teacher. Thanks for another nice video
I think You might have expanded the polynomial equation (ax^4+e-1)^4 , I can feel there is much more math before your next step, though it is a small suggestion in order to avoid the gap
@@PrimeNewtons It is beautiful. The universe is differential geometry and tensor fields. Calculus on manifolds holds the key to physics; it describes Einstein's general theory of relativity and the quantum field theories of the standard model of particle physics. Together, these two theories (GR and the QFTs of the SM) describe nature in its totality; though, we have room, yet, for learning! It is my dream to contribute to humanity's understanding of what math must exist that governs reality----specifically, regarding gravitation at the scale of quantum mechanics.
I am always interested in answers that we just 'know' are impossible or should be ignored. There is something interesting to ask about what that really means!
@@deoradh That's not what he says. If you apply a polynomial to a polynomial, you get a polynomial, but it's possible to get a polynomial by applying two functions, neither of which are polynomial. Take P(x)=f(x)+1 where f is any self-inverse function (there are plenty of self inverse functions which are not polynomials). Then P(P(x)-1)=f(f(x))+1=x+1, which is a polynomial. So it's not guaranteed P(x) is a polynomial at first glance just because we have a polynomial on the RHS.
It was badly explained and hand waved, but the basic idea is that by setting the coefficients equal on both sides after multiplying everything out one can construct a system of equations from the resultant coefficients of the 15th to 1st powers of x on the left and right hand sides whose only solution is that b=c=d=0. It would have taken a lot of space on the blackboard, so xe only did it explicitly for the 16th and 0th powers of x. (-:
@@JdeBPComparison of coefficients of the resulting 16th degree polynomial on both sides is the correct way to go, but it suffices to compare the coefficients of X^16, X^15, X^14, X^13, and X^12 (in that order!), and you don't really have to solve a whole system of equations, because they fall out one after the other if you do it in that order. It's a bit elaborate and involves some combinatorial thinking, but it's absolutely doable. The right way to think for the X^15 equation, for example, is to ask which terms on the expanded LHS can possibly contribute to the X^15 coefficient; there are not so many. X^15 can only come from the X^4 term of the outer polynomial. And because 15 = 4+4+4+3 is the only viable 4-partition of 15 and there are 4 ways it can be achieved (order of summands is not important!), we get the relation 4*a_4*(a_4*X^4)^3*(a_3*X^3) = 0*X^15, which translates to the equation 4*a_4^4*a_3 = 0. This gives a_3 = 0 since a_4 is not 0 (we already know from the X^16 equation that a_4 = 1 if considered a real number). And so on. I don't want to spell it all out but I hope I could convey the idea of how to rigorously do the comparison of coefficients.
set y = 1/x. You get f(1/y) +2f(y) = 3/y. But it works for x a well : f(1/x) +2f(x) = 3/x . So you get a system of 2 equations with 2 unknowns (f(x) and f(1/x). As a result, you get : f(x) = 2/x - x.
@@roi_legodu It's a polynomial solution as well. Nothing in the question says that we work with real numbers. If you work with complex numbers, 1 has 5 roots for the equation a^5 = 1. For example, [ (cos(2 pi / 5) + i * sin( 2pi /5) ] * X^4 +1 is a polynomial solution.
@@Machin008Bidule00 When working with questions like this one, you assume only working with real numbers, *unless explicitly mentioned.* Or unless you want to be quirky like you are being right now. I mean, no one here even mentioned complex numbers expect you, because it is obvious the question did not mean to involve complex numbers. We know a polynomial of degree N has N solutions in the complex plane smh
P(P(x)−1) = 1+x¹⁶ ⇒ P(P(x)−1)−1 = x¹⁶; substitute Q(x) = P(x)−1
Q(Q(x)) = x¹⁶ ⇒ Q(x) = x⁴ ⇒ P(x) = Q(x)+1 = x⁴+1.
Great solution
I also did in the exact same way.
Sorry I hadn't seen your comment before posting mine... TH-cam is sometimes lazy when refreshing the comments...
This is what I did. Thanks.
Yep that is how I did it! Nice
I think this problem has not been properly solved in the video. As others have also pointed out, there are two main problems with the demonstration.
The "first method" has the problem of just guessing one polynomial solution (P = X^4 + 1, and showing it satisfies the original equation, which is ok) but not showing that it is the ONLY solution. In fact, it isn't if the polynomials are understood as elements of the polynomial ring over the field of complex numbers. In this case, there are 5 distinct solutions (and answers).
The "second method" has the problem of not showing that the coefficients b,c,d of the X^3, X^2, and X monomials must be zero indeed (see around 9:28). Though being true, I don't think this is obvious at all, and it must be derived step by step with a lot of additional work by comparing the coefficients of the X^16, X^15, X^14, X^13, and X^12 terms of the result of the polynomial composition on the LHS. This involves some combinatorics.
When doing this work, we find that coefficient a must be a 5th root of unity, coefficients b, c, d must all be 0, and coefficient e must be 1. Over the real numbers this results in the unique solution P = X^4 + 1, while over the complex numbers there are 5 solutions: P_k = alpha^k * X^4 + 1 with alpha being any primitive 5th root of unity and k an integer from the set { 0, 1, ..., 4}.
There are holes in the presentations. (1) That x^4 + 1 is a solution does not mean it is THE solution, you must discard the other possibilities. (If complex coefficients are allowed, ax^4 + 1 is such a possibility for a = every fifth root of unity.) (2) "don't waste time dealing with b c d" = "don't bother fully proving the result"... As it turns out, it's not hard to prove that b = c = d = 0 by looking at the higher power terms of P(P(x)-1): those terms are much shorter than what you might fear.
When a question asks for P(2) instead of P(x) it is often the case that there's an answer that does not involve fully working out what P(x) is. I am not seeing it in this case, but that is a common question-setter's trick.
how you do then?
Actually, it’s not more difficult with P (2) = ? Than it is for P (X) = ?
If we define Q (X) = P(P (X) - 1), then we have Q (Q (X)) = X^16, and Q (Q (2)) = 2^16.
Obviously, 2^4 is a solution, and P (2) = 17.
@jpl569 This is way easier, and i think this is the method we use in high school. Loved it, thank you for reminder ✌🏿
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I knew how to solve this almost immediately. HOWEVER, i always look at the length of the video to judge how hard it is. This one is longer than most, so i thought i needed a complex method, derivatives of nested functions etc. Finally went back to my original idea and checked the video. What a mug. BTW you have the neatest blackboard writing I've ever seen.
Glad to Find your videos, full of curiosity and good explanation, Subscribed :)
Thank you so much for these videos, theyre helping alot. Please can u create 3 playlists for the other topics in math olympiad; algebra, combinatorics and number theory
I love you Prime Newton.
You are an excellent teacher and I enjoy your videos. I would like to see you solving more problems of Mathematical Olympiad.
If P(x) is just a rational function, we can remember that the square root is multivalued, so we could try both values of √16=±4. P(x)=x⁴+1 and P(x)=1/x⁴+1 both satisfy the relation, and give two options, P(2)=17 and P(2)=17/16
Polynomials, by definition, only have non-negative integer exponents. That is why problem used P(x) instead of f(x).
@@williamperez-hernandez3968 notation is just notation. If I do something using P(x)=sin(x²), the world doesn't end. Hence why I specified rational functions at the start.
@@xinpingdonohoe3978he did say polynomial in the video though
@@FunctionallyLiteratePerson I know, which is why I specified the extension to rational functions.
My first gut was to try a change of variable by letting q(x) = p(x) - 1, then substituting out all the P's. This yields a very fast path to the solution.
When we have p(q(x)) = 1 +x^16, we subtract 1 from each side and have q(q(x)) = x^16, and can see from here that x^4 satisfies q(x).
In raising to power 4 the mixed terms according to binomial coefficients are not introduced, they shoud have. So missing 4* 6* 4* terms.
yeah i was about to say that
Thank you for being rigorous. Solving by knowing the answer and showing works always feels like it skips some steps.
Nice thought exercise!
Seeing [e = 1] breaks my brain even if it's true. 😂
Hello newton sir, thank you for replying to my previous comment. You are my favourits math youtuber!
After taking b=c=d=0 you did not have to write the +....+ , however after raising the inside to the 4th power you ignore all the other terms and you should have written +...+. I'm guessing you did it right in preperation of the video and I can relate being a teacher. Thanks for another nice video
That's what I'm wondering too
I think You might have expanded the polynomial equation (ax^4+e-1)^4 , I can feel there is much more math before your next step, though it is a small suggestion in order to avoid the gap
Thank you for your great videos i did correct question about epsilon delta because of you in calculus 1 exam in university😁
I have asked before: with a degree in culinary arts how did you get so incredibly good at math teaching and do you teach math someplace? Thanks
I love your videos. I'm curious what you know of Riemannian geometry.
I have not studied it yet.
@@PrimeNewtons It is beautiful. The universe is differential geometry and tensor fields. Calculus on manifolds holds the key to physics; it describes Einstein's general theory of relativity and the quantum field theories of the standard model of particle physics. Together, these two theories (GR and the QFTs of the SM) describe nature in its totality; though, we have room, yet, for learning! It is my dream to contribute to humanity's understanding of what math must exist that governs reality----specifically, regarding gravitation at the scale of quantum mechanics.
You can try composing x^16 + 1 with itself minus one and you'll quickly see the pattern
Nice! I really was struggling with what the question was.
I am always interested in answers that we just 'know' are impossible or should be ignored. There is something interesting to ask about what that really means!
The first method seems to be incomplete. Nowhere the claim says that (x^4 + 1) is the only possible form of P(x)
Great reminder of the existence of polynomial compositions! 😂
I would've taken a similar route. It was obviously to me that P(x) with a 4th degree polynomial.
How do we know P(x) is a polynomial in the first place?
It was stated in the original question. Also, that's like saying f(x) being a function
Composing a transcendental function with itself will not reduce to the polynomial form across all x, as seen in the setup.
@@deoradh That's not what he says. If you apply a polynomial to a polynomial, you get a polynomial, but it's possible to get a polynomial by applying two functions, neither of which are polynomial.
Take P(x)=f(x)+1 where f is any self-inverse function (there are plenty of self inverse functions which are not polynomials). Then P(P(x)-1)=f(f(x))+1=x+1, which is a polynomial. So it's not guaranteed P(x) is a polynomial at first glance just because we have a polynomial on the RHS.
I love math
0:47 the way you are holding the chalk is much similar to the pic which you set in your profile picture.😃😃😃
Is there an error around 10:47? There is no reason to put a^4 * x^16, and then to try to ^4 once again.
Noice! Quite a neat solution
a=1 is clear from the beginning but how can u proof b=c=d=0
It was badly explained and hand waved, but the basic idea is that by setting the coefficients equal on both sides after multiplying everything out one can construct a system of equations from the resultant coefficients of the 15th to 1st powers of x on the left and right hand sides whose only solution is that b=c=d=0. It would have taken a lot of space on the blackboard, so xe only did it explicitly for the 16th and 0th powers of x. (-:
@@JdeBP i tried but find it too complex to solve
@@JdeBPComparison of coefficients of the resulting 16th degree polynomial on both sides is the correct way to go, but it suffices to compare the coefficients of X^16, X^15, X^14, X^13, and X^12 (in that order!), and you don't really have to solve a whole system of equations, because they fall out one after the other if you do it in that order. It's a bit elaborate and involves some combinatorial thinking, but it's absolutely doable.
The right way to think for the X^15 equation, for example, is to ask which terms on the expanded LHS can possibly contribute to the X^15 coefficient; there are not so many. X^15 can only come from the X^4 term of the outer polynomial. And because 15 = 4+4+4+3 is the only viable 4-partition of 15 and there are 4 ways it can be achieved (order of summands is not important!), we get the relation 4*a_4*(a_4*X^4)^3*(a_3*X^3) = 0*X^15, which translates to the equation 4*a_4^4*a_3 = 0. This gives a_3 = 0 since a_4 is not 0 (we already know from the X^16 equation that a_4 = 1 if considered a real number). And so on.
I don't want to spell it all out but I hope I could convey the idea of how to rigorously do the comparison of coefficients.
f(x)+2f(1/x)=3x . Can you please try and solve it , it was in my desk for weeks and i still didn't manage to solve it yet
set y = 1/x. You get f(1/y) +2f(y) = 3/y. But it works for x a well : f(1/x) +2f(x) = 3/x . So you get a system of 2 equations with 2 unknowns (f(x) and f(1/x). As a result, you get : f(x) = 2/x - x.
not difficult to guess P(x) can be x^4+1, I just wonder is it possible to find all P(x) even if P(x) is not only can be polynomials
Some video on word problem of functions
Your video is uploaded too late
(e-1)^4+(e-1)=0 e=1
Find P(2) P(P(x)-1)=1+x^16
First😂
second
P(2)=2^4+1=17 final answer
P(X) = X^4 * e^(2i.n.pi/5) +1 is also a solution...
But it’s not a polynomial solution. The question is clear, you have to find the polynomial
@@roi_legodu It's a polynomial solution as well. Nothing in the question says that we work with real numbers. If you work with complex numbers, 1 has 5 roots for the equation a^5 = 1. For example, [ (cos(2 pi / 5) + i * sin( 2pi /5) ] * X^4 +1 is a polynomial solution.
@@Machin008Bidule00 When working with questions like this one, you assume only working with real numbers, *unless explicitly mentioned.* Or unless you want to be quirky like you are being right now.
I mean, no one here even mentioned complex numbers expect you, because it is obvious the question did not mean to involve complex numbers. We know a polynomial of degree N has N solutions in the complex plane smh