# AIME: Polynomial Functional Equation (Reupload) 2016 A Problem 11

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• Published on Nov 5, 2018
• P(7/2) seems pretty arbitrary, especially considering our given information. How do we proceed?
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## Comments • 47

• LetsSolveMathProblems  9 months ago +21

*I refilmed this video starting at 6:43 because the previous version did not explain why P cannot have imaginary roots or roots with multiplicity greater than 1. I thank the comment section for graciously pointing out that my solution is incomplete. As an additional note, I remark that it is possible to go straight into writing P(x) as Q(x)x(x-1)(x+1) at 3:36. However, I did not find this path to the solution particularly intuitive or instructive, although it is certainly shorter. It really isn't until 6:43 in the video that we have learned enough about P to (somewhat) confidently predict the factored form of P. At 3:36, this intuition is almost absent, and I could not find a good way of explaining why the next logical step was to write P(x) as Q(x) times linear factors instead of exploring the constraints further.

• Abe 9 months ago

Thanks for doing so, it's definitely very clear now.

• Sitanshu Chaudhary 8 days ago +1

This is very cool way to done a question but i have doubt of we find some more value of p(x) further we try to find out the p(x) with the help of some value however i know this is very tough
Conclusion: even i try much time but i did not conjectured to this question

• Nellvin Cervantes Month ago

If P(x) = Ax(x - 1)(x + 1), how would x = (-infinite , -2] integers get P(x) = 0? Sorry for this question.

• Agamdeep Singh 2 months ago

brilliant explanation

• Gabriel Porto 3 months ago +1

Great video and explanation. Thank you!

• LetsSolveMathProblems  3 months ago +2

My pleasure. Glad you enjoyed it. =)

• Ay Lgrs 3 months ago

Your answer was very nice, however, I still wanted to show a totally different way to solve it and which may even be shorter.
I said that if a polynomial P was solution of (x-1)×P(x+1)=(x+2)×P(x) and different from 0, then first he could be written as : a(n)x^n+a(n-1)x^n+1+...+a(0) with a(i) the coefficients and a(n) would be different of 0. Using the equation you could use the binomial theorem (on (x+1)^n only!) to simplify the result and by indentifying with x^n you would find : na(n)+a(n-1)-a(n)=2a(n)+a(n-1) so knowing that a(n) is different from 0 you guess that n=3.
After that it is easy (but long) to develop the equation with a polynomial of the third degree and so you have with an identification : a(3) is unknown, a(2)=0, a(1)=-a(3) and a(0)=0. The polynomial can therefore also be written as : a(3)(x-1)(x+1)×x (it is not necessary to notice it)
Using the fact that (P(2))²=P(3) you have 36a(3)²=a(3)×24 knowing that a(3) is different from 0, a(3)=2/3.
Finally you just plug 7/2 into your polynomial and you get 105/4

• Quantumphase 4 months ago

at 5:45 how do you get p(2) = 0?

• ramdam dam 5 months ago

Nice

• Sa B 7 months ago +3

This problem is very interesting and your solution is quite clean! :D

• Math solver point 7 months ago +1

Please solve fast
Find all polynomials P(x) such that (p(x))^2=1+ xp(x+1) for all real numbers x.

• maths genius 4 months ago

deg( p(x))^2 = deg( 1+ x p(x+1))) , deg dgree of p let n=deg( p) ====> dep( p^2) =2deg( p)=2n and deg( 1+xp(x+1) =n+1 ,===> 2n=n+1===> n=1 ====> p(x) = 1+ax , , cause for x=0 p(0)^2 = 1+0p( 1)=1 ===> ( ax+1)^2 =1+x ( 1+a( x+1)) ===> a2 x2 +2ax +1=1 +x +ax2 +ax ====> a2=a and , a+1 =2a ====> a=1 ===>p(x) =1+x

• Darren Ngo 7 months ago

Why are P(1) P(0) and P(-1) the only real zeros of P why do they count and not all the others?

• Flavio Bernardi 5 months ago

Because if P(2) or P(-2) = 0 it would imply that all other Real values of x would return 0 for the function --> P(x) = 0 for all x, what is forbiden by the nonzero polynomial condition. Considering that, only 1, 0 and -1 can be roots of the function as they were consequences of (x-1)×P and (x+1)×P

• Wait what you do with constants? You ignore them? If you ignore them how do you do something in this form c/x+y do you cancel and division or?

• Flammable Maths 9 months ago +1

Hey! =D Can you give me your mail, or simply message me on fapablemaths@gmail.com? I've got a question :)

• LetsSolveMathProblems  9 months ago +1

Of course! Here's my email address for the channel: letssolvemathproblems7@gmail.com. =)

• Martin Epstein 9 months ago +5

Ha, you're more careful than me for taking "for every real x" so literally. I know they don't say the functional equation holds for complex x so you shouldn't assume but, like, come oooon :)

• LetsSolveMathProblems  9 months ago +3

In the original version of this video (which is deleted now), I did gloss over that fact, but I thought it may be better to explicitly verify it for the sake of rigorousness and to reduce confusion. However, if we were taking AIME--which only requires the final solution, not the explanation--it is probably a good strategy to take some things for granted. Of course, such assumptions could technically result in an erroneous solution and must be done with care. =)

• Chris Galanis 9 months ago

**attempt of solving the functional equation while f is just a continuous nonzero function**
The functional equation is
(x-1)f(x+1)=(x+2)f(x)
Substitute x with x-0.5
(x-1.5)f(x+0.5)=(x+1.5)f(x-0.5)
For all x except {±1.5}
f(x+0.5)/(x+1.5)=f(x-0.5)/(x-1.5)
For all x except {±1.5,±0.5} , devide by (x-0.5)(x+0.5) getting
f(x+0.5)/(x+0.5-1)(x+0.5)(x+0.5+1)
=
f(x-0.5)/(x-0.5-1)(x-0.5)(x-0.5+1)
Define g(x)=f(x)/(x-1)x(x+1)=f(x)/(x³-x) for all x except {-1,0,1}
So we get
g(x-0.5)=g(x+0.5)
Substitute x with x+0.5 getting
g(x)=g(x+1)
For x>1
Substitute x with lnx
g(lnx)=g(lnx+lne)
=>g(lnx)=g(lnex)
Let h(x)=g(lnx) then
h(x)=h(ex)
By substituting x with ex notice that
h(x)=h(x*e)=h(x*e*e)=...=h(x*e^k) for some integer k>2.
Thus
h(x)=h(x*e^k)
Substitute x with x/e^k
h(x)=h(x/e^k)
Take the limit of the sequence when k is close to +infty
h(x)=h(0) thus h(x)=c as constant
So g(lnx)=c is a constant function
By substituting x with e^x we get
g(x)=c thus g is constant for x>1
Therefore f(x)=c(x³-x) for x>=1
Using the property g(x)=g(x+1) we get g((0,1))=g((1,2))=c
Therefore f(x)=c(x³-x) for x>=0
Acting in a similar fashion for x

• Ky Heon 9 months ago +1

This may be a stupid question but shouldn’t M and N be prime? Because the problem expects “Prime Positive Integers”.

• Zako XY 9 months ago +3

It says "relatively prime". That just means that they can't share a common factor, not that they are both prime.

• Kyro 9 months ago

usclip.net/video/4Xp4F1h0YZM/video.html

• Hi, Can I ask a question ?

İs there any solution of this integral :
0 to inf integral (x*tanx/x^2+1)*cos((tanx)^2) dx

picture:
store.donanimhaber.com/1d/ed/79/1ded79544af148fbb3497907d4e020e5.jpg

• dave Adirondacks 9 months ago +3

im confused about nonzero- polynomial => not an infinite amount of roots, but note sin(x) has infinitely many roots and can be wirtten as an infinte product of roots , maybe i am using the wrong definition of what a polynomial is?

• Nejla Akyuz 6 months ago

A polynomial has a finite degree, if it has an infinte degree it is a power series expansion of a function.

• dave Adirondacks 9 months ago

@佐藤裕也 thanks for the answer

• 佐藤裕也 9 months ago +6

sin(x) is not a polynomial because sin(x) has infinitely many terms.

• Garrett Van Cleef 9 months ago +6

Good work. Slow down a bit though. Like A LOT. Hard to follow your videos as you rocket through them. Take a breath between concepts. Let your audience absorb what you say before moving on!

• Logan Kageorge 5 months ago

@LetsSolveMathProblems If half the people say you're going too fast and half say you're too slow, then your pacing is perfect! That's what the speed up and slow down buttons are for! :D

• Joseph Martos 7 months ago

@LetsSolveMathProblems i think that you explain carefully just what you must to and you skip just what you have to; but i think that you have to slow down a bit the speed of your speaking; like a 20% lol. that would be awesome. thanks for the amazing content btw

• Hoo Dini 8 months ago

I wholly disagree garrett

• WhiterockFTP 9 months ago

Your speed is perfect!

• LetsSolveMathProblems  9 months ago +9

Ironically, two of the most common constructive criticisms I receive are that I explain too quickly and that I explain too slowly. Since my audience is very diverse in terms of mathematical maturity, it is challenging for me to decide what prerequisite concepts I may take for granted, which parts of the video should be given a particular emphasis, and which parts I can dismiss as trivial. I still struggle with the pacing of the video every day. In general, I try to provide a slower, more thorough explanations on easier problems. This particular problem is pretty challenging: it is one of the final five problems on AIME (American Invitational Mathematical Competition), which means it is only solved by top of the top of math students in the United States within limited time. Consequently, I presumed that a quick pacing would be appropriate. If you found it too quick, I do apologize. I will strive to improve my pacing in the future videos (perhaps by considering where I should insert pauses). I appreciate your constructive criticism.

• ClubstepDJ 9 months ago +6

I can't do logic as much as you do, i mean, stuffs like "if this happens then we know that..."
Also, the logic that tells you the roots stuff is kinda confusing a bit

• Nicolas Nauli 5 months ago +1

@Joseph Martos Hopefully I can try to explain to you since I'm more of a junior at this topic. :DD

1) When you set a polynomial, any polynomial let's say P(x) like for example x^3 + 3x^2 + 3x + 1 and set it to 0, we can find its roots correct? This is because we want to turn the polynomial into it's "factors" like in x^3 + 3x^2 + 3x + 1, which is (x+1)(x+1)(x+1). When we set it to zero we can solve each linear expression (x+1) by setting it also as 0. Because if the whole thing is equal to 0, any one of them can also be equal to 0, hacking the whole equation and turning everything into 0, which makes the equation true. (in other words, 0 times anything is 0)

Now I have actually asked myself, what would a polynomial to the power of infinity be like? Well, you will definitely end up with infinitely many roots just like you will end up with 3 roots for a cubic expression (they may sometimes be the same roots, like in (x+1)(x+1)(x+1), but you will still have a cubic expression since you multiply by x three times). If that is so, that means the roots will include every number you can possibly think of (it sort of doesn't make sense but it's because of the special power of infinity). Therefore, when you plug in any number, it will always become 0, because that number is also included in the infinite group of roots, no matter who or what that number is. But as that is not possible for polynomials, because polynomials by definition have finite roots, you cannot have infinite roots for this question.

3) @LetsSolveMathProblems actually already explained the logic of the possible situations when you plug in numbers and constantly getting zeros, so could ask a more specific question? In case you want to know why we don't have to worry about non-integers also being a root, it is simply because of the behaviour of the functional equation given to us. We have a relation between P(x + 1) and P(x) where the zero marathon starts when x = 1 and then the left side becoming zero, showing that P(0) is 0. You do the same thing to the next P(x), where we are constantly getting values that result in 0 when we keep decreasing by one, from x=1 to x=0, x=0 to x=-1, x=-1 to x=-2 and so on.

As you can see we have an infinite number of integer roots again, going to negative infinity, and the only way we will have infinite values of x for one P(x) is if P(x) is a constant, otherwise you will end up with x to power of infinity with infinite turning points, which is not possible. But this is not possible as P(x) is a non zero polynomial, and so if doing this forever is not allowed, what else can we do? Well, if we made (x-1) equal to 0 on the left hand side of the functional equation why not make (x+2) on the right side equal to 0? AT THIS MOMENT I WAS LIKE BOOM, THIS MATH USclipR IS A GENIUS. When we make x = -2, the right side P(-2) will get eaten by the zero from (-2 + 2), and then this is equal to the left side: (-2 - 1)P(-2+1) which is equal to -3P(-1) and that still proves that P(-1) = 0 but at the same time, it also means that P(-2), which got eaten by the (x+2) when it becomes zero, doesn't have to be 0. Also doing this again and again, will get you away from 0, so you will never meet 0 again because P(-2) cannot be 0 and is a non-zero value. You can confirm it by observing the behaviour of the equation and maybe plugging in some numbers. :)

4) After knowing that only P(-1), P(0) and P(1) are the ONLY special plug-ins that will result in a cubic polynomial, all we have to do is just join them to together. We know that this polynomial is equal to zero when x = -1, 0 and 1, so it is like solving a cubic equation except that the values of x are 'given' to you. We reverse the process and conclude that for this polynomial to be zero when I plug in x = -1 for example, I need an (x+1) to be one of the factors so that everything will be zero, and the fact holds true that P(x) = 0 when I put in -1. We do the same for 0 and 1 and we get x(x-1)(x+1) altogether and you can see that when you plug in any of the 3 numbers, you will end up with 0 in the end.

• Nani 6 months ago

@LetsSolveMathProblems also 2 please

• Joseph Martos 7 months ago

@LetsSolveMathProblems 1);3) and 4) thanks

• LetsSolveMathProblems  9 months ago +3

Mathematical logic does take some time to get used to. It personally took me a good deal of struggling with math textbooks and problems before I developed some fluency in the language of mathematics. When you state, "the logic that tells you the roots stuff is kinda confusing," what particular parts of the explanation are you alluding to? That is, are you confused on 1) why P(x) = 0 if it has infinitely many roots, 2) how the functional equation was used to find -1, 0, 1 are roots of P, 3) how the functional equation was used to find that there are no other real roots of P, and/or 4) how the functional equation was used to derive the factored form of P? I would love to elaborate more on these points if you wish me to.

• About Math 9 months ago +2

beatiful problem!

• Ionising Plum 9 months ago +1

you were mocked, by a USclipr, Known as Flammable Maths: here is the link:
usclip.net/video/4Xp4F1h0YZM/video.html
make sure to go to 32 seconds to 38

• LetsSolveMathProblems  9 months ago +1

Thank you for the compliment, but I believe Flammable has some rigorous, applicable videos, as well, along with his comedic ones. After all, in the end it's about spreading joy of mathematics (no matter the method). Hopefully both Flammy and I have achieved and continue to achieve that delightful goal. =)

• Ionising Plum 9 months ago

@LetsSolveMathProblems Your mathematics is also more applicable

• Ionising Plum 9 months ago

@LetsSolveMathProblems, No, your videos are much more elaborate, Papa flammy is just plain comedic

• LetsSolveMathProblems  9 months ago +1

Ha ha! I knew my more egregious (albeit unintentional) mispronunciations would eventually be featured in a video. Perhaps I should watch Flammy's video and learn proper mathematics. =)