The red fractions in the denominator also converge to the square root of 2. 3/2, 7/5, 17/12, 41/29,... (√2 + 1)^n = a + b√2----> this formula also produces the coefficients of the fractions in the denominator which are a/b. Also, notice the sequence of integers in the fractions. The sum of numerator and denominator in each fraction is the denominator of the next fraction, and the sum of the numerator and twice the denominator in each fraction becomes the numerator of the following fraction. Example: 3/2--3 +2 = 5, so 5 is the denominator of the next fraction which is 7/5. 3 + 2x2 = 7, which is the numerator of the next fraction. Finally, the sequence produces all numbers that are both simultaneously square and triangular. The first fraction is technically 1/1. Now 1^2 x 1^2 = 1, which is the first number that is both square and triangular. The second fraction is 3/2, so we get 3^2 x 2^2 = 9 x 4 = 36, which is indeed the second number that is simultaneously square and triangular. Third fraction, 7/5. 7^2 x 5^2 = 49x25=1,225 which is the third one. Etc. Very cool sequence you picked.
This is similar to how I worked out the asymptotic limit of the ratios of consecutive numbers in the Fibonacci sequence, although I didn't know to prove that there WAS a limit.
If we assume that the limit exists, we can simply solve for x = 1 + 1/(x+1) => x - 1 = 1/(x+1) => (x-1)(x+1) = 1 => x^2 - 1 = 1 => x^2 = 2 Now then the question is whether there is any starting a_1 for which the limit approaches the negative branch of sqrt2
Just by the nature of this recursive definition, you can see that "infinity" is not even possible, because the left hand side would approach infinity while the right hand side will approach 1, which is a clear contradiction.
The negative solution for L²=2 is for the limit when n tends to minus infinity. It's easy to see that if you know the formula for the general term a_n: a_n = √2×((1+√2)^n+(1-√2)^n)/((1+√2)^n-(1-√2)^n) And the process to find the above formula could be a nice video as well. 😉
@@ryderpham5464 I didn't try that approach. I used more basic methods. First, I supposed a_n = p_n/q_n where p_n and q_n are integers. Then, I applied the recursive equation of a_n to find a recursive linear system for p_n and q_n, which can be written in matrix notation as R_n = AR_n-1, where R_n is a column matrices with p_n and q_n, and R_n-1 the same, but with indexes n-1 instead of n. Applying recursively the matrix equation, we get R_n = A^(n-1)R_1. To calculate A^(n-1), I diagonalized A. The rest of the process is pretty straightforward.
Something my teacher in school said is that when we try to calculate recursive sequences, we take the U(n+1) as a function f(Un+1)= f(x) and then solve for f(x)=x. I tried it on this question and it gave 2 answers, sqr(2) and -sqr(2)
When I noticed the values a(n) oscillated around the resulting limits, I tried to express the members of the sequence using an alternating series: a(n)= 1 + ( 1/2 - 1/10 + 1/60 - 1/348 + 1/2030 - . . .1/M(n-1)) = 1 + SUM_(k=1)^(n-1) (-1)^(k+1)/M(k) ...for n>1; a(1)=1 where M(k) can be expressed from denominators in the fractions: 1/1, 3/2, 7/5, 17/12, 41/29 ==> 1*2, 2*5, 5*12, 12*29.... M(k) = 1/8*((1+√2)^(2k+1) + (1-√2)^(2k+1) - 2*(-1)^k) How to prove that members in the series have to be integers reciprocal ?
Could show the convergence of odd and even subsequences first, then show two subsequences converge to the same limit, and hence the whole sequence converges.
After testing a few iterations starting with a_1 = 2, there seems to be convergence towards sqrt(2) once again. Same thing for a_1 = 11. My guess is that this will always be the case as long as the sequence is convergent, and that all that changes is how quickly it converges. Note that when he finds the limit in this video, he does so without using the initial value even once. You could probably prove that sqrt(2) is a stable fixed point of the sequence, but I’ll leave that for people who know what they’re doing.
Any chance we will get some Calc 3 on this channel? Thanks for all the content man
I was lost trying to figure this one out. Perfect explanation. Well done.
Thanks for the help. I really had no idea how to solve this problem.
The red fractions in the denominator also converge to the square root of 2. 3/2, 7/5, 17/12, 41/29,... (√2 + 1)^n = a + b√2----> this formula also produces the coefficients of the fractions in the denominator which are a/b. Also, notice the sequence of integers in the fractions. The sum of numerator and denominator in each fraction is the denominator of the next fraction, and the sum of the numerator and twice the denominator in each fraction becomes the numerator of the following fraction.
Example: 3/2--3 +2 = 5, so 5 is the denominator of the next fraction which is 7/5. 3 + 2x2 = 7, which is the numerator of the next fraction.
Finally, the sequence produces all numbers that are both simultaneously square and triangular. The first fraction is technically 1/1. Now 1^2 x 1^2 = 1, which is the first number that is both square and triangular. The second fraction is 3/2, so we get 3^2 x 2^2 = 9 x 4 = 36, which is indeed the second number that is simultaneously square and triangular. Third fraction, 7/5. 7^2 x 5^2 = 49x25=1,225 which is the third one. Etc. Very cool sequence you picked.
OMGGGGGG you're such a genius always helping me with my doubts
Thanks for the video. Very helpful 👍
Awesome video man, really helped
Math is beautiful. You proved it one more time. This proof is genius. Love it.
This is similar to how I worked out the asymptotic limit of the ratios of consecutive numbers in the Fibonacci sequence, although I didn't know to prove that there WAS a limit.
If we assume that the limit exists, we can simply solve for x = 1 + 1/(x+1) => x - 1 = 1/(x+1) => (x-1)(x+1) = 1 => x^2 - 1 = 1 => x^2 = 2
Now then the question is whether there is any starting a_1 for which the limit approaches the negative branch of sqrt2
Coincidentally, we had the limit of a recursive sequence on my further maths test today. But it was way duckin harder
What was the question?
This is brilliant!
You're a life saver
Very cool question!
awesome explanation
Just by the nature of this recursive definition, you can see that "infinity" is not even possible, because the left hand side would approach infinity while the right hand side will approach 1, which is a clear contradiction.
Bro n is tending to infinity and not An. An is tending to root 2 from the left side
How do you prove that it conv tho? I know how to do it for a series but for this one its wierd cus i cant say wether an+1< or > an for n>n0
THANK U SO MUCH
The negative solution for L²=2 is for the limit when n tends to minus infinity.
It's easy to see that if you know the formula for the general term a_n:
a_n = √2×((1+√2)^n+(1-√2)^n)/((1+√2)^n-(1-√2)^n)
And the process to find the above formula could be a nice video as well. 😉
could you derive the formula with a generating function? I attempted doing A(x) =a_n x^n but it doesn't seem to work out nicely with the denominators
@@ryderpham5464 I didn't try that approach. I used more basic methods. First, I supposed a_n = p_n/q_n where p_n and q_n are integers. Then, I applied the recursive equation of a_n to find a recursive linear system for p_n and q_n, which can be written in matrix notation as R_n = AR_n-1, where R_n is a column matrices with p_n and q_n, and R_n-1 the same, but with indexes n-1 instead of n. Applying recursively the matrix equation, we get R_n = A^(n-1)R_1. To calculate A^(n-1), I diagonalized A. The rest of the process is pretty straightforward.
@@EngMorvan interesting!
Every a_n is rational though
@@user-ik2kd9mb5t yup. By construction, u c that the formula always provides rational numbers for any natural n.
Something my teacher in school said is that when we try to calculate recursive sequences, we take the U(n+1) as a function f(Un+1)= f(x) and then solve for f(x)=x.
I tried it on this question and it gave 2 answers, sqr(2) and -sqr(2)
Can u use more video's equation like this?
Another technique (my favorite) is to express the sequence in terms of n. Then solving like a normal function.
nice vid dawg
When I noticed the values a(n) oscillated around the resulting limits, I tried to express the members of the sequence using an alternating series:
a(n)= 1 + ( 1/2 - 1/10 + 1/60 - 1/348 + 1/2030 - . . .1/M(n-1)) = 1 + SUM_(k=1)^(n-1) (-1)^(k+1)/M(k) ...for n>1; a(1)=1
where M(k) can be expressed from denominators in the fractions: 1/1, 3/2, 7/5, 17/12, 41/29 ==> 1*2, 2*5, 5*12, 12*29....
M(k) = 1/8*((1+√2)^(2k+1) + (1-√2)^(2k+1) - 2*(-1)^k)
How to prove that members in the series have to be integers reciprocal ?
wow another nice way to calculate sqrt2
helpful
Could show the convergence of odd and even subsequences first, then show two subsequences converge to the same limit, and hence the whole sequence converges.
what if you have two roots for L?
The common term should have either sqrt(2)+1 or sqrt(2)-1 for this sequence.
❤️ from india
Nice i guessed it from the start that it was 2^(1/2)
how?
@@Justin-gk8hu Maybe because of sqrt(2)'s continued fraction?
sqrt(2) = 1 + 1/(2 + 1/(2 + 1/2 + ...)))
2nd from India
I personally think using a web plot and showing how that converges at the intersection of x = 1+1/(1+x) would have been more fun
But how can we proof that the sequence converges?
We can probably prove that the sequence is decreasing and bounded below by induction
Mathias Fjøsne but the sequence is clearly not decreasing. Perhaps we should start with showing the difference between a_n and a_(n+1) approaches 0.
Where's the beard gone? Did you take a continuous fraction off it every day?
We need to prove that it converges however. So this is not valid reasoning
a2n is decreasing and lower bounded; a2n+1 is increasing and upper bounded
a(0)=0, a(1)=1
a(n+2)=1/2 [a(n+1)+a(n)]
Now try this method 🤣 You'll get L=L so no result
b) Prove your proof :)
What if a(1) was set to be iqual to 2?
After testing a few iterations starting with a_1 = 2, there seems to be convergence towards sqrt(2) once again. Same thing for a_1 = 11. My guess is that this will always be the case as long as the sequence is convergent, and that all that changes is how quickly it converges. Note that when he finds the limit in this video, he does so without using the initial value even once. You could probably prove that sqrt(2) is a stable fixed point of the sequence, but I’ll leave that for people who know what they’re doing.
Thats how irrational numbers are made, arent they?
you don't even know if the limit exists, you cannot say a_n=L. Try to firts see that {a_n}n is monotone and bounded, then you can supose a_n=L
First 🥇😅
Look up the hp logo and rotate you’re phone 180 degrees… Thank me later!!
Umm π radiants…