A Polynomial Equation | Math Competitions

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[@SyberMath]

Can @WolframAlpha solve this problem? What do you think? Check the results here: postimg.cc/HjYv7Hzy

[@user-wt1ul7ki6p]

Strictly speaking, the question is ill-formed. When we go back to the original equation, and plug in 2 as x, we can see p(2)/p(1) must equal to 2/(2-4)=-1 But you polynomial answer does not satisfy the equation. Since there are no other possibilities, the final answer should be no answer, i.e. no such polynomial could satisfy the original equation. A better form of this question would be your 2nd equation, i.e. (x-4)p(x) = x p(x).

[@YouTube_username_not_found]

It is not clear whether Q(X) = Q(x-1) implies that Q must be constant. I think one should prove this.

[@Manluigi]

You can see the geometric point of view. Q(X - 1) is the traslated polinomial of Q(X) . You dont can traslated Y=k (polinomial constant )

[@YouTube_username_not_found]

@@Manluigi That still doesn't prove anything. What if there is another polynomial that's not constant with the same property?

[@fetch7312]

Assuming that Q(x)=Q(x-1), then Q(x) must be a periodic function with a period of 1. Assuming that P(x) is a finite polynomial, this makes Q(x) a finite polynomial as well as Q(x) is a factor of P(x). If Q(x) has an infinite amount of x-values where the output is point k, then the function Q(x)-k must have an infinite amount of zeros. The Fundamental theorem of algebra states that there are n solutions in the complex plane for an nth degree polynomial. Since the vertical shift of a polynomial is also a polynomial, this implies that Q(x) must have infinitely many zeros, which means it can be factored into infinitely many roots, meaning it is an infinite polynomial. Since it is a factor of P(x), this means that P(x) must be an infinite polynomial. However, if Q(x) just equals a constant, then it satisfies the requirements while also allowing P(x) to be a finite polynomial.

[@Manluigi]

Q(X-1)is always different Q(X) and the zero are different.

[@YouTube_username_not_found]

@@fetch7312 >> "Q(x)-k must have an infinite amount of zeros. The Fundamental theorem of algebra states that there are n solutions in the complex plane for an nth degree polynomial." At this point, you could have just said that Q(x)-k is the zero polynomial because it is is the only polynomial that have infinitely many zeroes, making Q(x) = k , concluding the proof. As for the rest of your reply, it has an error which is "this implies that Q(x) must have infinitely many zeros"

[@NemanjaSo2005]

If P(0)=0, then the original expression is not defined for x=1 as P(1)/P(0) would be division by 0.

[@cm5754]

This is a valid point. If P(0) and P(1) were equal to 0 the left side of the original equation gives 0/0 = 1/(-3) which is false. The problem solves a different equation, (x-4)P(x) = xP(x-1). The original equation is only correct in the same sense that x/x=1 is correct, that is only when both sides are defined.

[@YouTube_username_not_found]

If we are working with rational fractions, then P/Q = R/S is equivalent to PS = RQ , provided that Q and S are different from the zero polynomials, even when for some values, Q and S are zero. This is because abstract polynomials are different from polynomial functions. So the reasoning in the video is acceptable for abstract polynomials. If however the P in the original equation is a polynomial function, then we cannot plug in the illegal values, but we could instead solve using limits.

[@cm5754]

@@KZbin_username_not_found In the world of rational functions, x/x is not the same expression, and not the same rational function, as (x(x-1))/(x(x-1)), although they are equal whenever both are defined, and x^2(x-1)=x^2(x-1) for all x. The fundamental question is whether the original equation is just supposed to mean the values are equal when both are defined, or is supposed to mean that whenever either side is defined, the other is also defined with the same value. The solution takes the first approach: that they are equal whenever both are defined. That is what justifies the cross multiplication.

[@cm5754]

One place this comes into this problem is the special case P(x) = 0. This is a valid solution if we are looking for solutions that are valid whenever both sides of the original equation are defined. The video does claim it as a solution. However, P(x) = 0 is not a valid solution if we want the left side to be defined whenever the right side is defined. So the collection of correct solutions does differ in this specific case depending on how we read the equation.

[@YouTube_username_not_found]

@@cm5754 Rational fractions are not to be confused with rational functions, they are like abstract abstract polynomials.

[@user-iu6bz6iq8x]

This can be solved very easily by brute force which also shows the solution is unique up to a constant factor. Start with the equation in the cross multiplied form and plug in P(X) = X^n+AX^(N-1)+... (clearly any solution can be scaled linearly so no need to work with a lead coefficient). So (X-4)*P(X) = X*P(X-1) (X-4)*(X^N+A*X^(N-1) + ...) = X*( (X-1)^N +A*(X-1)^(N-1) + ...) X^(N+1) + (1-4*A)X^N + ... = X^(N+1) + (1-N*A)X^N + ... Therefore 1-4*A = 1-N*A and so N=4 which makes the rest fairly trivial. To make for less typing note that the zeroth order term is 0 on the right so must also be 0 in P (from the left side). Plug in a X^4 + A*X^3 + B*X^2 + C*X for P: (X-4)*(X^4 + A*X^3 + B*X^2 + C*X) = X( (X-1)*4 + A(X-1)^3 + B(X-1)^2 +C*(X-1) ) Multiplying out: X^5 + (A-4)X^4 + (B-4A)X^3 + (C-4B)X^2 + (0-4C)X = X * ( ( X^4 - 4X^3 + 6X^2 - 4X + 1 ) + A (X^3 - 3X^2 + 3X - 1) + B( X^2 - 2X + 1 ) + C(X - 1 ) ) Going power by power, the 5th and 4th powers are identically equal on both sides, the other 3 yield 3 equations : 3rd: B-4A = B - 3A + 6 2nd: C-4B = C -2B +3A -4 1st: -4C = -C + B - A + 1 Solve from the top down, 3rd -4A = -3A + 6 => A = -6 2nd C-4B = C-2B -18 - 4 => B = 11 1st -4C = -C + 11 + 6 + 1 => C = -6 P(X) = X^4 - 6X^3 + 11X ^2- 6X Though a but tedious, this has an advantage in that once you see N=4 you kind of know that you will get to a solution and not just going down a wormhole.

[@ImperfectKingdomSeeker]

Very slick. When I tried it, I got about halfway. But I missed the insight of thinking of 0, 1, 2, 3 as roots and expressing the polynomial as (x-1)(x-2)(x-3)*something. That is slick. And I would not have thought to plug that back into the original equation to find Q(x) with all the cancellations. Very nice.

[@leif1075]

Is that reslly slick or just random? I don't see anyone EVER thinking of that EVER..why would they..isn't it more logical to keep.pkugging other values like negstive 1, negstive 2 etc.. to look for a pattern? Agsin what he did seems random and out of nowhere to me..

[@YouTube_username_not_found]

@@leif1075 I don't find that random! It is natural to try plugging in some values and look for patterns. The first thing to notice is, whatever value we plug in, we always find a relationship between P(x) and P(x--1) , so if we could somehow find P(x--1) , we find P(x) . The thing to keep in mind is that it is easier to determine a polynomial if we deduce its roots. So now our mission is to find roots of P or at least some of them , because by finding the roots, we can factorise the polynomial into products of x -- a and another polynomial Q where a is a root of P. The second thing to notice is that in the equation P(x)(x -- 4) = P(x -- 1)x , the LHS is equal to 0 when x is 0, which is convenient because it means that we can determine that P(0) is 0 , and we can because x--4 is not 0 in this case. Now we go back to the first observation and determine in turn P(1) , P(2) and so on, and we find out that they are 0, but we should be carefull here!! The pattern stops at x = 4 because the equation gives us P(4)(4 -- 4) = P(3)•4 which reduces to P(4)•0 = 0 which is true no matter what P(4) is. So this means we can't know what P(4) is. So our sources of information ended. But it's OK , we have found some roots of P which are 0 , 1 , 2 , 3 . We can now decompose P into x(x--1)(x--2)(x--3)Q(x). Now, a natural thing to do is to plug in this expression of P into the equation, which will give us an equation with Q being the unknown (check the video) Now, what is left is to solve this equation, which was done by people in the replies to my comment. Go check them out! That would be all.😁

[@YouTube_username_not_found]

@@leif1075 Hello?? 🤨 Did you not receive my comment? please respond.

[@YouTube_username_not_found]

@@leif1075 Not sure if you have received my reply where I explained to you the thought process behind the solving method in the video so I am replying again to check this. Please like that reply to confirm.

[@TheSandkastenverbot]

When you make the general Ansatz P(x)=A*(x-a1)*...*(x-an) you get the solution by a combination of inspection and intuition ^^

[@leif1075]

Yea but that's not something I don't think anyone would ever think fl..whybwould they? Wouldn't everyone start with p(×)= x^n plus b and try tonfigure out east n is or ax^n plus b..I don't see anyone doing what he did in the video..

[@Blaqjaqshellaq]

The equation can be presented as P(x)=c*x^3 - 6*c*x^2 + 11*c*x - 6*c*x. (Thank you, Wolfram Alpha.)

[@TheSandkastenverbot]

*solution

[@manojsurya1005]

Hey nice video ,i started by plugging in values like 0,1,2.. to get the factors

[@swenji9113]

Why isn't the equation (X-4)P = XP(X-1)? It's really what the problem is about isn't it?

[@SyberMath]

Yes

[@Utesfan100]

We can use induction on x=-i to show that P(x)=0 for all integrers

[@YouTube_username_not_found]

The Zero polynomial: Am I a joke to you?! By the way, it only has 4 zeroes. P(4) can't be deduced from the equation, it could be anything.

[@YouTube_username_not_found]

I apologize! 😅 Guess I didn't read your comment carefully. P(-1) can't be determined by plugging in values for x.

[@yoav613]

Very nice

[@mtaur4113]

Hindsight solution is very hindsight, lol telescope.

[@robertcotton8481]

A wave function with period of 1

[@bsmith6276]

Which could work, if not for the fact we are given P(x) is a polynomial in the introduction and title.

[@robertcotton8481]

@@bsmith6276 he asked if their was a function that not polynomial so I answered

[@user-vc6uk1eu8l]

Nice

[@SyberMath]

Thanks

[@swenji9113]

Using small x in the thumbnail makes it look like it's a functional equation about polynomials. I think the equation to solve is about rational fractions, not about the underlying fonctions. If not, you would have to explain where x lies and for which values the equation holds

[@YouTube_username_not_found]

I am curious 😃! In what language did you study math?

[@swenji9113]

@@KZbin_username_not_found In french

[@YouTube_username_not_found]

Ahhh 😃! C'est bien! SVP, donnez moi la définition d'une fonction.

[@swenji9113]

@@KZbin_username_not_found Une définition classique est un triplet (f,A,B) où A est le domaine, B l'ensemble d'arrivée, et f est un sous ensemble de A×B composé de couples (a,b) et telle que pour chaque a dans A, il existe un unique b dans B tel que (a,b) appartient à f. En réalité on n'a pas besoin d'être si rigoureux, et on ne l'est pas, lorsqu'on définit des fonctions pour la première fois ...

[@YouTube_username_not_found]

@@swenji9113 Mais la définition que je connais est celle ci: une fonction est un triple ... telle que pour tout a dans A il existe *au plus un* b dans B telle que (a,b) appartient à f , çàd, ce n'est pas nécessaire d'avoir tout élément a avoir un image par f . la définition que vous avez donné est celle d'une *application* , selon mes connaissances. D'ailleurs, Wikipedia note cette distinction entre fonction et application dans l'article nommé "application" .

[@kimsanov]

How we can divide both sides to e.g (x-3) if (x-3) could be zero?

[@justcommenting5117]

We assume it isn't

[@tixanthrope]

Once again, you neglected to give the domain in which the given relation holds :( x=4 is obviously bad quick inspection yields x = 1 is bad too. I will try to solve the problem while trying to figure out the said domain. edit: x (x-1) (x-2) (x-3) Q(x) where Q is any polynomial with Q(x) = Q(x-1) ie. any non-zero constant polynomial.

[@bsmith6276]

Point discontinuities at x=1, 2, 3, and 4 in the starting fractional arrangement. The first thing was to rewrite as (x-4)*P(x) = x*P(x-1), which removes said discontinuities and is usually the way this sort of problem is presented. Maybe sybermath wanted something that looked different and in the process accidentally added discontinuites.

[@cm5754]

@@bsmith6276 The other issue is that the solution presented includes P(x) = 0 as a solution. It isn't clear this is actually a solution to the original equation, since there would be no points at all in the domain of P(x)/P(x-4).

[@phlroams-dc2vl]

P(x) = x/4 ?

[@lawrencejelsma8118]

This is polynomial construction confusing. You didn't go from p(-1), p(-2), and x < 0 cases of all polynomial points=0 unless your domain is supposed to be x>=0 that I missed. I have no clue what (x + 1)(x + 2)(x + 3) composite multiplied into c(x-1)(x-2)(x-3)Q(x) forms since only x negative forces p(-x something) always equalled 0.

[@bsmith6276]

The first half of the video is finding P(0) through P(3) without knowing anything specific about the polynomial, just values that can be inferred by evaluating (x-4)*P(x) = x*P(x-1) at values that can actually be solved. That is why sybermath started at x=0, that makes the right side collapse to 0, which then gives us an actual value of 0 for P(0). Then with P(0) known it became possible to find the value for P(1) by evaluating at x=1, and keep on going until trying evaluating at x=4 did not give us anything new. So now we are at about the 5:00 mark and we have four known values P(0)=0, P(1)=0, P(2)=0, and P(3)=0. Then all those are roots of the polynomial, and thus the factorization of P(x) must include factors x, x-1, x-2, and x-3. But there may be more factors, and that what Q(x) represents - any factors the initial work did not find. Then P(x)=x*(x-1)*(x-2)*(x-3)*Q(x) for some polynomial Q(x). Then the rest of the video is just plugging in our partially constructed P(x) into the original equation and simplify to see what happens, in this case we get Q(x)=Q(x-1) which for polynomials can only happen when Q(x) is a constant. So then the conclusion is P(x) = C*x*(x-1)*(x-2)*(x-3) for arbitrary constant C.

[@lawrencejelsma8118]

@@bsmith6276 ... Thanks for the recap. But p(0)(-1 - 4) = (-1)p(-1) shows p(-1) = 0, p(-1)(-2 - 4) = (-2)p(-2) shows p(-2) = 0 etc. Where every k < 0, p(k) = 0. This wasn't the case in positive j, j > 0 where (4 - 4)p(4) = (3)p(3) at j = 4 showed p(4) didn't necessarily have to equal 0 like all x < 4 were. I didn't watch carefully and didn't understand Q(x) = C part. With p(4), p(5), p(6), ... Not known I thought that was Q(x) like electrical engineering step functions if they were constants. Since p(n + 1)A1 = p(n)A0 type of thing we might think p(n+1) - p(n) = constant of some slope where each p(j) had different Ajs not equal to 0 but can since we don't know what p(4) is since (4 - 4) was 0 being at least destroying any relationship to p(3) = 0. I did understand the x(x - 1)(x - 2)(x - 3) times a constant went from x = 0 to x = 3. I don't understand what x = -1, -2, -3, -4, ..., -k, ... also have p(k) = 0 for any k < 0 which confuses me in this polynomial construct. Then whatever p(4) turned out to be we ended at (0)p(4) = (3)p(3) = 0 for any p(4) not necessarily = 0 and being a constant vs Q(x) is not resolved by being multiplied by 0.

[@bsmith6276]

@@lawrencejelsma8118 How are you getting p(0)(-1 - 4) = (-1)p(-1)? Evaluating (x-4)*P(x) = x*P(x-1) at x=-1 yields -5*P(-1) = -1*P(-2) and evaluating it at x=0 yields -4*P(0) = 0*P(-1).

[@lawrencejelsma8118]

@@bsmith6276 ... You are correct! I was probably thinking of the step ahead. (x - 4)p(x) = x(p(x-1)) should be at x = 0 (-4)p(0) = 0(p(-1)) so p(-1) can be anything like p(4). Then at x= -1 you show, -5p(-1)=(-2)p(-2) or p(-1) undefined makes p(-2), p(-3), undefined etc. in the negative ks. p(-2), p(-3), etc. can't be defined either. Thanks! I see my mistake there. I think it is x(x - 1)(x -2)(x - 3) [ C((1/(x- 4)] rather than just x(x -1)(x - 2)(x - 3) times C he got I think because x can't equal 4. So we have to divide by (x - 4) by the given equation to maintain the ratio of the given equation. I was thinking of p(n + 1) - p(n) = (n/(n - 4))(n + 1 - n) formula myself with my negative xs error assumption wrong by not plugged in correctly.

[@lawrencejelsma8118]

Let's rewrite my money mumbo: The first equation P(x) = [x/(x - 4)] P(x - 1) should be P(x - 1) = C(x - 1)(x - 2)(x - 3)(x - 4) when P(x) = Cx(x - 1)(x - 2)(x - 3) and x not equal to 4. Then [x/(x - 4)][C(x - 1)(x - 2)(x - 3)(x - 4)] multiplies out to P(x)/(x - 4) with his derived P(x) substitution into his original ratio equation is why I keep seeing division by x - 4 and x not 4 being undefined.

[@FisicTrapella]

Notice that P(x-1) must be different of zero; and as x runs over any value, that means that P(x) can't be zero. But you set that P(0)=P(1)=P(2)=P(3)=0... 🤔🤔🤔🤔🤔🤔

[@fetch7312]

I think this could be solved simply by adding a limit statement of lim(x>a) before the P(x) and then switching the x's on the other side with a's. That way L'Hopital's rule solves the issues of running into the indeterminate form 0/0.

[@FisicTrapella]

@@fetch7312 The point is that some results are not indeterminate, such as x=4 which gives you a P(4)/0...Watching at the expression I thought that P(x) must be different of zero; but....

[@swenji9113]

I believe the problem should be formulated in terms of rational fractions. You can divide P by P(X-1) as long as P(X-1) isn't the 0 polynomial. If you see polynomials as functions, then the equation to solve has a lot of issues indeed, we don't even know where x lies. Is it for all x in R? In C? Is it for a given x?

[@swenji9113]

The confusion between P (or P(X) which is the same) and P(x) creates many issues if one isn't careful

[@FisicTrapella]

@@swenji9113 You're right, but in any case, P(x) should be different of zero. That means P(x) has no roots and this can be true for Real numbers but no (I think) in the Complex world...

[@mircoceccarelli6689]

👍👍👍😁🤪👋