One way to speed up P5 is to just use the generating function of the fibonacci numbers, which is pretty well known. It is 1/(1-x-x^2) and subbing x=1/4 gives 1/(1-(1/4)-(1/16))=16/11, and multiplying by 1/4 gives 4/11 (as we have (1/4)^(n+1) in the sum, not (1/4)^n). I would guess this is what the contestants used to evaluate the sum
I was unaware of the feynman trick for the second problem, Is there any other way to solve it, I have tried subtitutions and stuff but cannot get to an answer
The best resource is simply doing past problems, available through the website. Also, youtube.com/@Silver-cu5up?si=WJde_QfYq1xD1L64 this channel has a whole series on different methods needed for integration bees
Oooh I had a slightly different solution for Finals question 5. Let f_n(x)=(floor((2^n)x)-floor((2^n)x-1/4))/2^n and suppose that x has binary representation given by x=0.a_1a_2...a_na_(n+1)a_(n+1).... Then f_n(x)=1/2^n if a_(n+1)=a_(n+2)=0 or f_n(x)=0 otherwise. Suppose that the binary representation of x contains it's it first 2 consecutive 0's at the (N+1)th and (N+2)th decimal place, so that max f_n(x)=1/2^N. Then the integral can be expressed as sum_{N>=1} int_(A_N)1/2^N dx =sum_{N>=1}|A_N|/2^N where A_N ={x in (0,1): binary expansion of x has the first consecutive 0's occur in the (N+1)th and (N+2)th place}. Probabilistically, |A_N|=P_N=P(in a random sequence of 0's and 1's, the first 2 consecutive 0's occur in the (N+1)th and (N+2)th position). This sum can be evaluated by identifying that P_N satisfies the linear recurrence P_N=P_(N-1)/2+P_(N-2)/4, with initial conditions, P_0=1/4 and P_1=1/8.
Some other matholy people told me about this binary representation. I love it!! There are so many different ways to look at problems like this (literally)
One way to speed up P5 is to just use the generating function of the fibonacci numbers, which is pretty well known. It is 1/(1-x-x^2) and subbing x=1/4 gives 1/(1-(1/4)-(1/16))=16/11, and multiplying by 1/4 gives 4/11 (as we have (1/4)^(n+1) in the sum, not (1/4)^n). I would guess this is what the contestants used to evaluate the sum
Some of these looked approchable while others just made me look and cry
Hello, are you using Obsidian to write?
I use Microsoft whiteboard along with a drawing pad to write
@@danielrosado3213 Thanks!
I was unaware of the feynman trick for the second problem, Is there any other way to solve it, I have tried subtitutions and stuff but cannot get to an answer
I have not found another method yet, but that doesn’t mean that none exist. Possibly there may be a series solution?
@@danielrosado3213 No, I have no clue anymore...
Thanks!!!
Bro make a good thumbnail. Only then your video will blast because content is super.
Can any one tell me the best resources to prepare for this exam?
The best resource is simply doing past problems, available through the website. Also, youtube.com/@Silver-cu5up?si=WJde_QfYq1xD1L64 this channel has a whole series on different methods needed for integration bees
@@danielrosado3213 thanks! that was very helpful
Oooh I had a slightly different solution for Finals question 5. Let f_n(x)=(floor((2^n)x)-floor((2^n)x-1/4))/2^n and suppose that x has binary representation given by x=0.a_1a_2...a_na_(n+1)a_(n+1).... Then f_n(x)=1/2^n if a_(n+1)=a_(n+2)=0 or f_n(x)=0 otherwise. Suppose that the binary representation of x contains it's it first 2 consecutive 0's at the (N+1)th and (N+2)th decimal place, so that max f_n(x)=1/2^N. Then the integral can be expressed as sum_{N>=1} int_(A_N)1/2^N dx =sum_{N>=1}|A_N|/2^N where A_N ={x in (0,1): binary expansion of x has the first consecutive 0's occur in the (N+1)th and (N+2)th place}. Probabilistically, |A_N|=P_N=P(in a random sequence of 0's and 1's, the first 2 consecutive 0's occur in the (N+1)th and (N+2)th position). This sum can be evaluated by identifying that P_N satisfies the linear recurrence P_N=P_(N-1)/2+P_(N-2)/4, with initial conditions, P_0=1/4 and P_1=1/8.
Some other matholy people told me about this binary representation. I love it!! There are so many different ways to look at problems like this (literally)
Very epic bro
Doing this before JEE feels like I'm battling a final boss where I learn new stuff constantly and patterns and just stuff....I LOVE IT
hii