The very last step divides by 4 instead of 2. It should be 52. Which means that every natural number gets hit by a some a_n (as he argues earlier in the video) except for 1 (the latter for obvious reasons).
Always confusing when he makes basic errors like this towards the end of his videos. Quite often he'll make a sign error, but then do another one later which cancels it out. But this one is just wrong and the video ends right after lol
if we set the nested root expression to x, then we get x = sqrt(x + n) and this can be written as x^2-x = n. For every natural number > 1, x^2-x is some positive integer, implying that we hit every natural number x > 1 at a_(x^2-x). So the 51st natural number we reach is 52, when n = 52^2-52 = (52*51) = 2652.
It's hard to see how the 51st member of a monotonically increasing sequence of integers is 26. If you divide by the given 2 instead of 4, you get 52, which looks more reasonable. In fact, it HAS to be 52 because we are counting the natural numbers in the bigger sequence and 1 doesn't appear there¹. ¹ in fact, aₙ = 1 gives n = 0 which, in the context of the formula for aₙ, is ... interesting. Edit: "natural numbers", not "integers"
@@SpeedcoreDancecore True: but, given we are making a sequence of roots of increasingly large numbers, it must be strictly increasing (I'd have thought - happy to be wrong about that).
About your note, the reason we infer n=0 after assuming a[n]=1 is simply because we are solving the equation a=sqrt(n+a), and thus 1=sqrt(n+1). This is only possible if n=0, which would imply a[0]=1... but that's only because we assumed that a[n]=1 was possible at all. If we go at it from the other direction, and try to compute a[0] without any assumptions, we would need to solve a=sqrt(a). There are two possible solutions there, either a=0 or a=1. Because the a[n] values aren't well defined in the first place (infinitely nested functions are inherently vague), it's not clear from the problem statement which of these is the "correct" value. If you tried to define a[0] through a limit procedure, you would get a different answer based on how you initialized the computation. For example, say we recursively defined c[k+1]=sqrt(n+c[k]), and then defined a[n] as the limit of these c[k] values. If the limit exists, it would always solve a=sqrt(n+a), and surprisingly the solution to that equation is unique for all n>0. For n=0 however, the solution is no longer unique, and the limit of our c[k] sequence will depend on how we initialize c[0]. If we initialized with c[0]=0 then you would get a[0]=0, but initialize with any c[0]>0 and you'll get a[0]=1. Infinitely nested functions intentionally obfuscate the initialization of such procedures, so they are inherently vague. Slightly related, if for each real x>0 you defined a[x] as the unique solution to a=sqrt(x+a), then taking the limit as x approaches 0 would suggest a[0]=1.
Quadratic equation x^2-x-n=0 with integer solutions suggests that n should be factor as m(m+1). a_n=m+1. For integer a_n, n must be: 1*2, 2*3,.... The 51st integer is when n=51*52, and a_n=52.
Nice video! I had kind of similar approach, but a bit of slightly different path: Like you, I arrived to the conclusion that a_n = (1+sqrt(1+4n))/2. In order for that to be a natural number, sqrt(1+4n) needs to be an odd natural number. Since n>1, the sqrt(1+4n) > 2. From this we get an equation: 1+4n = (2k + 1)^2. [where k is a positive integer] This simplifies to n = k*(k+1). [and we see that n will be a natural number for all positive integers k] To get the 51st natural number, we just now need plug in k=51. So to sum it up: k = 51 -> n = 51*52 -> a_n = (1+sqrt(1+4*51*52))/2 = 52
I miss the proof that these sequences converge. If a(1,n) = √(n) and a(k+1,n) = √(n+a(k,n)), then a(k+1,n) > a(k,n) for all k and n in N. Therefore, a(k,n) is strictly monotonly increasing for fixed n in N. Thus, the limits for k to infinity exist, if a(k,n) is bounded above. By induction: a(1,n) = √(n) < 2√(n) and if a(k,n) < 2√(n) for fixed k in N, then a(k+1,n) = √(n+a(k,n)) < √(n+2√(n)) < √(3n) < 2√(n) for all n in N. This shows: a(k,n) < 2√(n) for all k in N und the limits for k to infinity exist.
Strictly speaking, the infinite nesting notation seems to rather suggest that the actual recurrence system is f₀(t,n) = t fₖ₊₁(t,n) = fₖ(√(n+t), n) L(t,n) = lim_[k→ ∞] fₖ(t,n) aₙ = L(t₀ , n) where t₀ is some non-mentioned "initializing"/"evaluation" constant. Because in the notation, the infinitely many repeats extend into the "rightward" direction, not the leftward direction. I don't know how we can prove convergence for this definition though.
8:51 Since the first perfect square is 9, which is 3^2 (when n=2), I believe the mth odd should be 2m+1 10:53 It should be divided by 2 instead of 4, so the correct answer should be 104/2=52
True, but Michael is talking about the mth perfect square that exists at all, not the mth perfect square specifically arising from 4n+1, n>=1. Which is why he's using m=52, rather than m=51, and for that approach, 2m-1 is correct. (Maybe slightly confusing when in the box on top of the blackboard is addressing the (m+1)th perfect square, I guess, but that's the general case of something else.)
Another error in the end result for the viewers as practice... It should be clear, that the 51st (N = 51) natural number in that sequence can not be 26! The first (N = 1) is a_2 with n = 2 and a_n = 2. Someone wrote, it's 2652, and I believe that number more. As 2 = 2*1, 6 = 3*2, ..., 2652 = 51*52 = N*(N+1), this makes total sense! You wrote that as well. But then you mixed up the 4n+1 form of the square with the variable n (which is different!) As n = 51*52, not n = 26 (4n+1 = 103², not 4n+1 = 103!!!), a_n = [1 + sqrt(4n+1)] / 2 = [1 + sqrt(4*51*52+1)] / 2 = [1 + sqrt(10609)] / 2 = (1 + 103) / 2 = 52 = N + 1. This makes total sense, because this series hits every natural number a_n >= 2, and it's exactly the N = (a_n - 1)th natural number in that series! (That means, I could have avoided that explicit calculation of the root, but who cares!)
Always remember to think about what the actual sequence is and show convergence in "..." notations. For instance, set n = 0 in this example. Ehat is the actual sequence? If you say its √0, √√0, √√√0, ... the limit is 0. But if you choose any positive value replacing the "..." e.g. √0.001, √√0.001, √√√0.001, ... the limit is 1. Likewise, the equation 1.1^x = x does have a solution, but the power tower 1.1 ^ 1.1 ^ 1.1 ^ ... does not converge.
Amazing video because the maths concepts used are literally 10th grade but still show an amazing proof style that can be used in higher math. I could give this to one of my year 11 students as a perfect introduction to these problems because they already have the necessary skills to deal with quadratics. one small error in the end tho, where it should be 104/2 = 52
a^2=n+a a=(1+sqrt(1+4n))/2 Therefore 1+4n is an odd square of form (2k+1)^2 4n+1=4k^2+4k+1 n=k^2+k But this works for all values of k, so we can sub in k=51 to get n=51*52=2652 To check: a(2652)=(1+sqrt(1+10608))/2=(1+103)/2=52
I think there is a mistake at 1:04. The step of substituting in a for that infinite series assumes that the series in fact converges, which is unproven.
Not all too hard, writing a for a_n: a=sqrt(n+a) a^2-a-n=0 2a=1+sqrt(1+4n) so 4n+1 must be an odd square, but notice every odd square (2k+1)^2 is of the form 4n+1. For every positive integer k you then get n=k^2+k and a_n = k+1 k=1: n=2 and a_2 = 2 k=2: n=6 and a_6 = 3 ... k=51: n= 2652 and a_2652 = 52
Ooor. From a_n² = a_n+n, we have n = a_n³ - a_n = a_n * (a_n-1). Now we notice that a_n is a natural number if and only if n is the product of two consecutive natural numbers (and a_n is the öarger of the two). The first time this happens for n = 1 * 2, the second time for n = 2 * 3, and ... the 51st time for n = 51 * 52 (with a_n = 52)
I am quiet confusied If A52=26 that implies there hast to be some distingt n1 and n2 that result in the same an And the requsion feels 'striclty increasing' in n
L^2-L=n implies L = (1+ sqrt(1+4n))/2. We need 4n+1 to be a perfect square. We know it's odd, so (2k+1)^2=4n+1=4k^2+4k+1 implies n=k(k+1) and you get a n for every k, so let k=51.
I assume the proof was omitted because it's easy to check that the sequence defining each a_n is increasing so either converges or goes off to infinity.
@chuckaway6580 b_0 = sqrt(n) b_{k} = sqrt(n+b_{k-1}) To show it's bounded, let's say b_0 < 2n, then b_k < sqrt(n + 2n) = sqrt(3) sqrt(n) < 2n so we are good. To show it's increasing, take f(x) = sqrt(n+x), then f'(x) > 0 so it's increasing.
Title: "The most interesting iterated root problem I have seen!" So which problem is he referring to? The one he explains in the video, or the one from the thumbnail that he didn't explain?
What does the original sum mean? I can follow all the rest of this - assuming that you can indeed square such sums etc. but having learnt about infinite series I know that you have to take care that whaat you're saying makes sense.
Dr Penn it is OT but I'm following your series on differential forms, I was asking myself what does it mean to integrate a 2 form over a surface? I remember from calc2 we integrated 1forms over surfaces to obtain the flux of the vector field associated to the 1form through the surface, what is the meaning of doing an integration of an m-form? Also m-forms are integrable only on manifolds of the same dimension or is it possible to integrate a 3form on. a surface or a 2form on a 4Dsurface
![เจ็บที่ยังฮัก - ม่อน วรวิทย์ [ Official Mv ] จอนนี่มิวสิค](http://i.ytimg.com/vi/KALIGSRSjFs/mqdefault.jpg)
The very last step divides by 4 instead of 2. It should be 52. Which means that every natural number gets hit by a some a_n (as he argues earlier in the video) except for 1 (the latter for obvious reasons).
And I guess a corollary is that n = m*(m+1) yields a_n=m+1, which is the m'th natural number in the sequence.
Always confusing when he makes basic errors like this towards the end of his videos. Quite often he'll make a sign error, but then do another one later which cancels it out. But this one is just wrong and the video ends right after lol
@@hugh081 Mike is gifted mathematician but it annoys me when he makes those little mistakes.
@@hugh081 Ha it's true. Keeps us on our toes!
Yeah, it should not be the case tbat the fifty-first natural number is any increasing sequence of natural numbers is less than fifty-one.
if we set the nested root expression to x, then we get x = sqrt(x + n) and this can be written as x^2-x = n. For every natural number > 1, x^2-x is some positive integer, implying that we hit every natural number x > 1 at a_(x^2-x). So the 51st natural number we reach is 52, when n = 52^2-52 = (52*51) = 2652.
11:09 Missing « st » and + signs in the thumbnail 🤨
I thought it was gonna be easy because the thumbnail is equal to n 😅
It's hard to see how the 51st member of a monotonically increasing sequence of integers is 26. If you divide by the given 2 instead of 4, you get 52, which looks more reasonable. In fact, it HAS to be 52 because we are counting the natural numbers in the bigger sequence and 1 doesn't appear there¹.
¹ in fact, aₙ = 1 gives n = 0 which, in the context of the formula for aₙ, is ... interesting.
Edit: "natural numbers", not "integers"
If it's not strictly increasing it's possible
@@SpeedcoreDancecore True: but, given we are making a sequence of roots of increasingly large numbers, it must be strictly increasing (I'd have thought - happy to be wrong about that).
He devided by 4 instead of 2 in the last step
@@thomashoffmann8857 yes he made a mistake … this was NOT a good place to stop! 😟
About your note, the reason we infer n=0 after assuming a[n]=1 is simply because we are solving the equation a=sqrt(n+a), and thus 1=sqrt(n+1). This is only possible if n=0, which would imply a[0]=1... but that's only because we assumed that a[n]=1 was possible at all. If we go at it from the other direction, and try to compute a[0] without any assumptions, we would need to solve a=sqrt(a). There are two possible solutions there, either a=0 or a=1. Because the a[n] values aren't well defined in the first place (infinitely nested functions are inherently vague), it's not clear from the problem statement which of these is the "correct" value.
If you tried to define a[0] through a limit procedure, you would get a different answer based on how you initialized the computation. For example, say we recursively defined c[k+1]=sqrt(n+c[k]), and then defined a[n] as the limit of these c[k] values. If the limit exists, it would always solve a=sqrt(n+a), and surprisingly the solution to that equation is unique for all n>0. For n=0 however, the solution is no longer unique, and the limit of our c[k] sequence will depend on how we initialize c[0]. If we initialized with c[0]=0 then you would get a[0]=0, but initialize with any c[0]>0 and you'll get a[0]=1. Infinitely nested functions intentionally obfuscate the initialization of such procedures, so they are inherently vague.
Slightly related, if for each real x>0 you defined a[x] as the unique solution to a=sqrt(x+a), then taking the limit as x approaches 0 would suggest a[0]=1.
Quadratic equation x^2-x-n=0 with integer solutions suggests that n should be factor as m(m+1). a_n=m+1. For integer a_n, n must be: 1*2, 2*3,.... The 51st integer is when n=51*52, and a_n=52.
That is the way I solved it, too. As others have already noted, Michael's answer via a slightly more complicated argument is off by a factor of 2.
The way you manage to calculate all the (an) you are assuming that (an) exists.
Nice video!
I had kind of similar approach, but a bit of slightly different path:
Like you, I arrived to the conclusion that a_n = (1+sqrt(1+4n))/2.
In order for that to be a natural number, sqrt(1+4n) needs to be an odd natural number.
Since n>1, the sqrt(1+4n) > 2.
From this we get an equation: 1+4n = (2k + 1)^2. [where k is a positive integer]
This simplifies to n = k*(k+1). [and we see that n will be a natural number for all positive integers k]
To get the 51st natural number, we just now need plug in k=51. So to sum it up:
k = 51
-> n = 51*52
-> a_n = (1+sqrt(1+4*51*52))/2 = 52
Again the question on the thumbnail is different from the actual question on the video. On the thumbnail the + are missing.
I miss the proof that these sequences converge. If a(1,n) = √(n) and a(k+1,n) = √(n+a(k,n)), then a(k+1,n) > a(k,n) for all k and n in N. Therefore, a(k,n) is strictly monotonly increasing for fixed n in N. Thus, the limits for k to infinity exist, if a(k,n) is bounded above. By induction: a(1,n) = √(n) < 2√(n) and if a(k,n) < 2√(n) for fixed k in N, then a(k+1,n) = √(n+a(k,n)) < √(n+2√(n)) < √(3n) < 2√(n) for all n in N. This shows: a(k,n) < 2√(n) for all k in N und the limits for k to infinity exist.
Same, ty. (Ob sich da ein "und" in deinen Beweis geschlichen hat? xD)
Strictly speaking, the infinite nesting notation seems to rather suggest that the actual recurrence system is
f₀(t,n) = t
fₖ₊₁(t,n) = fₖ(√(n+t), n)
L(t,n) = lim_[k→ ∞] fₖ(t,n)
aₙ = L(t₀ , n)
where t₀ is some non-mentioned "initializing"/"evaluation" constant.
Because in the notation, the infinitely many repeats extend into the "rightward" direction, not the leftward direction.
I don't know how we can prove convergence for this definition though.
8:51 Since the first perfect square is 9, which is 3^2 (when n=2), I believe the mth odd should be 2m+1
10:53 It should be divided by 2 instead of 4, so the correct answer should be 104/2=52
The first perfect square is the number 1
@@thomashoffmann8857 But the minimum value of n is 1, and in order to get a 1 for 4n+1, n needs to be 0
@@張謙-n3levery square of an odd number is of the form 4n+1.
But the odd numbers are 1,3,5,7... Which are generated by 2m-1.
True, but Michael is talking about the mth perfect square that exists at all, not the mth perfect square specifically arising from 4n+1, n>=1. Which is why he's using m=52, rather than m=51, and for that approach, 2m-1 is correct.
(Maybe slightly confusing when in the box on top of the blackboard is addressing the (m+1)th perfect square, I guess, but that's the general case of something else.)
Forgot to say ODD perfect square, but you know what I mean
Another error in the end result for the viewers as practice... It should be clear, that the 51st (N = 51) natural number in that sequence can not be 26! The first (N = 1) is a_2 with n = 2 and a_n = 2. Someone wrote, it's 2652, and I believe that number more. As 2 = 2*1, 6 = 3*2, ..., 2652 = 51*52 = N*(N+1), this makes total sense! You wrote that as well. But then you mixed up the 4n+1 form of the square with the variable n (which is different!) As n = 51*52, not n = 26 (4n+1 = 103², not 4n+1 = 103!!!), a_n = [1 + sqrt(4n+1)] / 2 = [1 + sqrt(4*51*52+1)] / 2 = [1 + sqrt(10609)] / 2 = (1 + 103) / 2 = 52 = N + 1. This makes total sense, because this series hits every natural number a_n >= 2, and it's exactly the N = (a_n - 1)th natural number in that series! (That means, I could have avoided that explicit calculation of the root, but who cares!)
Always remember to think about what the actual sequence is and show convergence in "..." notations. For instance, set n = 0 in this example. Ehat is the actual sequence? If you say its √0, √√0, √√√0, ... the limit is 0. But if you choose any positive value replacing the "..." e.g. √0.001, √√0.001, √√√0.001, ... the limit is 1. Likewise, the equation 1.1^x = x does have a solution, but the power tower 1.1 ^ 1.1 ^ 1.1 ^ ... does not converge.
Amazing video because the maths concepts used are literally 10th grade but still show an amazing proof style that can be used in higher math. I could give this to one of my year 11 students as a perfect introduction to these problems because they already have the necessary skills to deal with quadratics. one small error in the end tho, where it should be 104/2 = 52
Why the division by 4 and not by 2 in the very last step? Also how can 26 be the 51st natural number if the sequence a_n is strictly increasing?
Classic Michael Penn mistake
a^2=n+a
a=(1+sqrt(1+4n))/2
Therefore 1+4n is an odd square of form (2k+1)^2
4n+1=4k^2+4k+1
n=k^2+k
But this works for all values of k, so we can sub in k=51 to get
n=51*52=2652
To check:
a(2652)=(1+sqrt(1+10608))/2=(1+103)/2=52
I think there is a mistake at 1:04. The step of substituting in a for that infinite series assumes that the series in fact converges, which is unproven.
Not all too hard, writing a for a_n:
a=sqrt(n+a)
a^2-a-n=0
2a=1+sqrt(1+4n)
so 4n+1 must be an odd square, but notice every odd square (2k+1)^2 is of the form 4n+1.
For every positive integer k you then get n=k^2+k and a_n = k+1
k=1: n=2 and a_2 = 2
k=2: n=6 and a_6 = 3
...
k=51: n= 2652 and a_2652 = 52
Ooor. From a_n² = a_n+n, we have n = a_n³ - a_n = a_n * (a_n-1). Now we notice that a_n is a natural number if and only if n is the product of two consecutive natural numbers (and a_n is the öarger of the two). The first time this happens for n = 1 * 2, the second time for n = 2 * 3, and ... the 51st time for n = 51 * 52 (with a_n = 52)
I am quiet confusied
If A52=26 that implies there hast to be some distingt n1 and n2 that result in the same an
And the requsion feels 'striclty increasing' in n
Michael made a mistake, as usual, in the end. He divided by 4 when he should have divided by 2, so the correct answer should be 52.
The value of n for the 51st natural value is 51*52 = 2652 while the value of a sub n for that n is 104 / 2 = 52.
L^2-L=n implies L = (1+ sqrt(1+4n))/2. We need 4n+1 to be a perfect square. We know it's odd, so (2k+1)^2=4n+1=4k^2+4k+1 implies n=k(k+1) and you get a n for every k, so let k=51.
=> n^51/102= 2^51/102=2^1/2 4^51/102=2 The 51th natural number is 2
Hang on, as a_n strictly increases with n, there's not enough natural numbers below 26 to have 51 of them. Think there's a mistake near the end.
Then please detail the mistake in your comment.
@@MyOneFiftiethOfADollar Wrong last division, by 4 instead of 2.
@@MyOneFiftiethOfADollar no
IDK why i am watching this. I don't even like math that much. Except geometry maybe.
One of us. ONE OF US!
You had an error at the end: The denominator should be 2 rather than 4, so the 51st natural number in the sequence is 52.
should divide by 2 at the end not 4...
Do we need to prove {an} is converge before using the recursion ?
Right
I assume the proof was omitted because it's easy to check that the sequence defining each a_n is increasing so either converges or goes off to infinity.
@chuckaway6580
b_0 = sqrt(n)
b_{k} = sqrt(n+b_{k-1})
To show it's bounded, let's say b_0 < 2n, then b_k < sqrt(n + 2n) = sqrt(3) sqrt(n) < 2n so we are good.
To show it's increasing, take f(x) = sqrt(n+x), then f'(x) > 0 so it's increasing.
Cool video, though different problem than the thumbnail
Because a_n^2 - a_n = 2 * triangular number. So just list out 2*triangular numbers and find the 51st term.
The thumbnail needs to be fixed.
Title: "The most interesting iterated root problem I have seen!"
So which problem is he referring to? The one he explains in the video, or the one from the thumbnail that he didn't explain?
a_n = (1+sqrt(4n+1))/2, so when n=0, a_n = 1, so the nested square root we started with, when n=0, sums up to 1?!? What did I get wrong?
Nothing. Why do you think something went wrong?
What does the original sum mean? I can follow all the rest of this - assuming that you can indeed square such sums etc. but having learnt about infinite series I know that you have to take care that whaat you're saying makes sense.
cannot get why m+1th odd square is equivalent on finding mth natural number in sequence
Amazing!
Dr Penn it is OT but I'm following your series on differential forms, I was asking myself what does it mean to integrate a 2 form over a surface? I remember from calc2 we integrated 1forms over surfaces to obtain the flux of the vector field associated to the 1form through the surface, what is the meaning of doing an integration of an m-form? Also m-forms are integrable only on manifolds of the same dimension or is it possible to integrate a 3form on. a surface or a 2form on a 4Dsurface
I wish you do analytic number theory course