This problem is from the Harvard-MIT Mathematics Tournament. The strategy is ti evaluate the first few applications to find the pattern for predicting future applications.
If you wonder how they come up with these problems, we're taking f = g^(-1) h g for g(x) = x^(2011) and h(x) = 1/ (1 - x), where h(h(h(x))) = x. Here g^(-1) denotes the inverse function. Then the kth iterate f^(k) is g^(-1) h^(k) g.
A fun problem! It's handy to use the notation f^n(x) to mean f composed with itself n-times. Then one has: f^3(x) = x, as proven in the video => f^(3k)(x) = x, for k≥1 One can be pedantic and prove this by induction with the induction step: Given that f^(3k)(x) = x for some k≥1 f^3(k+1)(x) = f^(3k+3)(x) = f^(3k)(f^3(x)) = f^(3k)(x) = x The notation can be extended with f^0(x) = x (f applied 0 times leaves x unchanged) in which case: f^(3k)(x) = x, for k≥0
You,....Sir Newton,....explain so well, that AMAZED me, every time. Of course, there are many math teachers like you,...as Michael Penn, Black T-shirt, Flammable, Math 505, etc. but l always prefer YOU...first. Thanks!
While questions that require you to evaluate are generally solved using general facts and removing a block by another block. This one was actually much easier than it looked, maybe it's because computing huge numbers is not generally one's strength, so the answer revolves around one's methods. Though, regardless of my rambling, I am so happy that many a times, your videos involve math that can be made easier to 'understand'; for otherwise we only 'know' rather than 'learn'. Both of which are obviously not the same 😢😅 Thank you so much! This made my day today ❤
Re: The video shake. Maybe someone has mentioned this already: It's reminiscent of interlacing done to get crisper resolution on video to be shown on old, bulky, CRT monitors. Every other frame contained every other horizontal line. When viewed on a more modern LED monitor it can show up as shake, jitter or even comb-like motion blur. Re-encoding, such as that done by TH-cam, can amplify some of the problems while smoothing others out. In short: Check camera and editing software for interlacing settings and try putting them the other way. (Short, private test videos could be the way to go here, but then, maybe you've tried all that already.)
Who is this man? He is absolutely fantastic and incredible and if he is who I think he is, he has a degree in the culinary arts and not math? Could this just be a hobby to him at this level?
@4:30 For those who didn't see why that's allowed, you can do everything with exponents: 2011'th root of blah = blah^(1/2011) So 1/(blah^(1/2011)) = (blah^(1/2011))^-1 = blah^(-1/2011) = (blah^-1)^(1/2011) = (1/blah)^(1/2011) = 2011'th root of (1/blah)
At 5:25 , instead of simplifying the denominator and canceling the exponent against the radical sign, I think it's more convenient to leave it and instead do ²⁰¹¹√( ( 1 - x²⁰¹¹ ) / ( - x²⁰¹¹ ) ) = = ²⁰¹¹√( ( -1 + x²⁰¹¹ ) / ( x²⁰¹¹ ) ) = ²⁰¹¹√( ( x²⁰¹¹ - 1 ) / ( x²⁰¹¹ ) ) = ²⁰¹¹√( 1 - 1/x²⁰¹¹ ) That way, we'll have only one occurrence of x , which helps tremendously when evaluating the third application of f : f(f(f(x))) = ²⁰¹¹√( 1 - 1/[ f(x) ]²⁰¹¹ ) = ²⁰¹¹√( 1 - 1/[ 1/(1 - x²⁰¹¹) ] ) = ²⁰¹¹√( 1 - (1 - x²⁰¹¹) ) = ²⁰¹¹√( x²⁰¹¹ ) = x EDIT: Ah, you're entering y = f(f(x)) into f(y) , instead of the other way around (entering y = f(x) into f(f(y)) ) -- Nicely done! By the way, I like how you explain things very calmly, clearly and neatly , and you have a very good voice. You are a good teacher!
not sure why you did not factor out -x^2011 from the first radical as then that simplifies greatly the answer. within the radical you would have ((1/x^2011)-1)^2011 the x^2011 become -1.
What does it mean when you have a number with another number (or symbol) in the lower right corner? For example, go to symbolabs and type in "x mod y = z", and you'll see a 3, with an x in the lower right corner.
I can't help but conclude that sum and functions are invariably better expressed in code. All the computer languages that tried to copy the form of maths were failures. Yet while computer languages continually evolve and improve, the syntax of maths is rigid and unchanging over hundreds of years. What we're left with is the unintuitive, awkward and clumsy first attempt of history persisting to this day.
This problem is quite interesting. I would suggest to consider this problem with 2011 replaced by n ,a positive integer. Then after some simplifications one can see that f[f[f[x]]] = x for n>1. Then the rest is easy.
@@renesperb Ah, okay. Thanks for the clarification, I thought you were referring to the argument '2011' in f(f(f(...f(f(f(2011)))...))) . By the way, I think the result applies only for odd values of n (and including n = 1).
@@yurenchu For any n which is a multiple of three the n-fold application of f[x] is just x. Now 2010 is a multiple of three , thus this simplification applies.
@@renesperb If x > 1 and n = 2 , then the denominator of f(x) is the square root of a negative term, which doesn't exist (at least not if we are operating only on the real numbers). The same goes for any other positive integer n that is even. By the way, the result applies for n=1 . (Your original comment says n>1 .)
The difference between any two-digit number and the same two digits reversed is always a multiple of 9 (I include 0 x 9 here where both digits are the same). Can you offer a simple proof?
When n = 10a+b is a two-digit integer for some digits a and b , then the reverse number is m = 10b+a , and their difference is |n - m| = = |(10a+b) - (10b+a)| = |10a - a + b - 10b| = |9a - 9b| = 9 * |a-b| which (since a and b are integers and hence (a-b) is an integer) must be a multiple of 9 .
The camera is probably shaking with excitement!!
🤣🤣🤣🤣
If you wonder how they come up with these problems, we're taking f = g^(-1) h g for g(x) = x^(2011) and h(x) = 1/ (1 - x), where h(h(h(x))) = x. Here g^(-1) denotes the inverse function. Then the kth iterate f^(k) is g^(-1) h^(k) g.
Awesome, thank you for this insight.
A fun problem! It's handy to use the notation f^n(x) to mean f composed with itself n-times. Then one has:
f^3(x) = x, as proven in the video
=> f^(3k)(x) = x, for k≥1
One can be pedantic and prove this by induction with the induction step:
Given that f^(3k)(x) = x for some k≥1
f^3(k+1)(x) = f^(3k+3)(x) = f^(3k)(f^3(x)) = f^(3k)(x) = x
The notation can be extended with f^0(x) = x (f applied 0 times leaves x unchanged) in which case:
f^(3k)(x) = x, for k≥0
The inductive proof is technically required, especially in a contest or exam situation, but it's clearly obvious.
You,....Sir Newton,....explain so well, that AMAZED me, every time. Of course, there are many math teachers like you,...as Michael Penn, Black T-shirt, Flammable, Math 505, etc. but l always prefer YOU...first. Thanks!
While questions that require you to evaluate are generally solved using general facts and removing a block by another block. This one was actually much easier than it looked, maybe it's because computing huge numbers is not generally one's strength, so the answer revolves around one's methods. Though, regardless of my rambling, I am so happy that many a times, your videos involve math that can be made easier to 'understand'; for otherwise we only 'know' rather than 'learn'. Both of which are obviously not the same 😢😅
Thank you so much! This made my day today ❤
Re: The video shake. Maybe someone has mentioned this already: It's reminiscent of interlacing done to get crisper resolution on video to be shown on old, bulky, CRT monitors. Every other frame contained every other horizontal line. When viewed on a more modern LED monitor it can show up as shake, jitter or even comb-like motion blur. Re-encoding, such as that done by TH-cam, can amplify some of the problems while smoothing others out.
In short: Check camera and editing software for interlacing settings and try putting them the other way. (Short, private test videos could be the way to go here, but then, maybe you've tried all that already.)
Thank you. I haven't tried anything yet. I will do so for the next video.
Who is this man? He is absolutely fantastic and incredible and if he is who I think he is, he has a degree in the culinary arts and not math? Could this just be a hobby to him at this level?
@4:30 For those who didn't see why that's allowed, you can do everything with exponents:
2011'th root of blah = blah^(1/2011)
So 1/(blah^(1/2011)) = (blah^(1/2011))^-1 = blah^(-1/2011) = (blah^-1)^(1/2011) = (1/blah)^(1/2011) = 2011'th root of (1/blah)
erm, what the blah?
Cool, the result also can be written by a tetration so
²2011
Great stuff as always.
At 5:25 , instead of simplifying the denominator and canceling the exponent against the radical sign, I think it's more convenient to leave it and instead do
²⁰¹¹√( ( 1 - x²⁰¹¹ ) / ( - x²⁰¹¹ ) ) =
= ²⁰¹¹√( ( -1 + x²⁰¹¹ ) / ( x²⁰¹¹ ) )
= ²⁰¹¹√( ( x²⁰¹¹ - 1 ) / ( x²⁰¹¹ ) )
= ²⁰¹¹√( 1 - 1/x²⁰¹¹ )
That way, we'll have only one occurrence of x , which helps tremendously when evaluating the third application of f :
f(f(f(x))) = ²⁰¹¹√( 1 - 1/[ f(x) ]²⁰¹¹ )
= ²⁰¹¹√( 1 - 1/[ 1/(1 - x²⁰¹¹) ] )
= ²⁰¹¹√( 1 - (1 - x²⁰¹¹) )
= ²⁰¹¹√( x²⁰¹¹ )
= x
EDIT: Ah, you're entering y = f(f(x)) into f(y) , instead of the other way around (entering y = f(x) into f(f(y)) ) -- Nicely done!
By the way, I like how you explain things very calmly, clearly and neatly , and you have a very good voice. You are a good teacher!
I like your teach ❤
not sure why you did not factor out -x^2011 from the first radical as then that simplifies greatly the answer. within the radical you would have ((1/x^2011)-1)^2011 the x^2011 become -1.
Shouldn't we prove it by reccurence?
What does it mean when you have a number with another number (or symbol) in the lower right corner?
For example, go to symbolabs and type in "x mod y = z", and you'll see a 3, with an x in the lower right corner.
I can't help but conclude that sum and functions are invariably better expressed in code. All the computer languages that tried to copy the form of maths were failures.
Yet while computer languages continually evolve and improve, the syntax of maths is rigid and unchanging over hundreds of years. What we're left with is the unintuitive, awkward and clumsy first attempt of history persisting to this day.
I agree
This problem is quite interesting. I would suggest to consider this problem with 2011 replaced by n ,a positive integer. Then
after some simplifications one can see that f[f[f[x]]] = x for n>1. Then the rest is easy.
The derivation f(f(f(x))) = x applies to any real number, not just to x = n where n is a positive integer.
@@yurenchu The exponent is n ,and the variable is x .
@@renesperb Ah, okay. Thanks for the clarification, I thought you were referring to the argument '2011' in f(f(f(...f(f(f(2011)))...))) .
By the way, I think the result applies only for odd values of n (and including n = 1).
@@yurenchu For any n which is a multiple of three the n-fold application of f[x] is just x. Now 2010 is a multiple of three , thus this simplification applies.
@@renesperb If x > 1 and n = 2 , then the denominator of f(x) is the square root of a negative term, which doesn't exist (at least not if we are operating only on the real numbers). The same goes for any other positive integer n that is even.
By the way, the result applies for n=1 . (Your original comment says n>1 .)
@ 15:30 Shouldn’t it be 2011^2011
Very nice video!!
Thanks Sir
Sir close to the end of your video u changed the 2010 to 2011 but 2011 isn't divisible by 3
The difference between any two-digit number and the same two digits reversed is always a multiple of 9 (I include 0 x 9 here where both digits are the same). Can you offer a simple proof?
When n = 10a+b is a two-digit integer for some digits a and b , then the reverse number is m = 10b+a , and their difference is
|n - m| =
= |(10a+b) - (10b+a)|
= |10a - a + b - 10b|
= |9a - 9b|
= 9 * |a-b|
which (since a and b are integers and hence (a-b) is an integer) must be a multiple of 9 .
@@yurenchu Thanks. I really should have thought about this before posting.
@@XiOjala You're welcome.
Isn't x only valid for the range between -Inf to 1?
No, that’s an odd root. I got confused for the same reason, but it’s actually fine
@@miegas4 Valid for x < 1 and for x > 1 .
x = 1 is a bit problematic though.
EDIT : Actually, x = 0 is also problematic.
Well done,!?
Assume f(f(x))=-1/xf(x) so you get fff(x) =x
me thinking the answer should have 2011 somewhere.
Hey
This was very interesting and very fun, the way some of this problemas are stated seems kinda confusing