I really liked the rigor of using the monotone sequence theorem to justify how the limit of the sequence was calculated. So many times these kinds of details get skipped over in math proofs leaving an uneasy feeling of not quite knowing why certain moves are justified beyond handwaving.
Where in God's name does the 8 :07 equation come from..there isNO basis for thisnguess..whybsont everyone else asking this..surely everyone else is just a flummoxed by it..kts the kind of thing thst makes math stupid and annoying or infuriating and I think should stop altogether..
@@angelmendez-rivera351 The comment at the top to which I was replying stated “these kinds of details get skipped over in math proofs”. No they don’t, or they are not proof. I’m not sure how it is possible to misinterpret my comment, but apparently this happened twice.
11:00 Thank you for clearly explaining the motivation for making the guess. I think that intuition benefits from learning the motivation behind different guesses.
The 2nd problem can be done rigorously. The sequence of functions that you constructed are, around x=1, monotonous, increasing, bounded above and analytic. Therefore it converges to an analytic function around x=1. Therefore it has a Taylor series expansion. Square the Taylor series and compare the coefficients with the right hand side, you get the linear function as the only solution.
The second problem must be solved much more rigorously than it has been. In particular, nothing proves that the limit F(x) is a polynomial function. First consider a sequence F_n(x) = sqrt(f(x) + ...+ sqrt(f(x + n - 1))), with f(x) equal to x^2 + x - 1. Then, prove by induction that the sequence is increasing, and bounded above by x + 1. Finally, let F(x) be the limit. Prove by induction that for every positive integer k, F(x) >= a(k), where a(k) is the following sequence : a(1) = 1/2 and a(k+1) = sqrt (a(k)). Not so simple as that, if we want to stay at a reasonable mathematical level. Sorry for my broken English, I'm French. Hi.
Re 2nd nested root. The lesson here is: problems that are not solvable in a given context become solvable in a more general context. "Generalise to solve": how true that often is in Mathematics
For the first sequence, wouldn't replacing the "last" 6 with an 8 just collapse the whole sequence like a house of cards to the number 2, proving this is its upper bound?
I’m taking my first rigorous proofs based class next semester and one of things in the syllabus completely foreign to me was working with sequences, i.e finding their limit, proving they converge, etc. so it is very helpful to see videos like this to get a feel for some of the tools used. These problems were also just quite fun and novel to me as well.
Michael: constructs a very nice real analysis proof that the first sequence converges, then computes the limit in a nice rigorous fashion. Also Michael: Well let’s just let this infinite radical be a function, probably well-defined, who knows. Let’s get a polynomial equation, I dunno, probably a linear function. Hey! This linear function satisfies the recurrence, we have no other choice but to declare this the answer!
@@dnuma5852 If he said that, it doesn’t make any sense, because unlike the first one, the second infinite radical is not the limit of a sequence with a simple recurrence. Instead, it’s the limit of one chain in some doubly-indexed sequence a_{m,n}, with a_{m,m}=sqrt(m^2+3m+1) and a_{m,n}=sqrt(m^2+3m+1+a_{m+1,n}), namely the limit of a_{1,n} as n goes to infinity. This is a totally different type of problem, and can’t be handled with the exact ideas as the first one. It would be useful and instructive for Michael to have covered this one rigorously too.
Noahtaul(know it all): you are among many who appear to be quite knowledgeable/proficient mathematically who have “empty math channel” Noahtaul: So why don’t you produce your own math videos and publish them on your channel?
I really liked the Penn Fact box containing a formal statement of the Monotone Convergence Theorem. This kind of thing should be very helpful to someone like me who doesn't use math too often but enjoys your videos.
For the second one, if you pick f(x)=x^2-x-1=x^2-(x+1) you get sqrt(x^2-(x+1)+sqrt((x+1)^2-(x+2)+sqrt((x+2)^2-(x+3)...)))=F(x) and the value we're looking for is F(3) but in this case, F(x)=x
Here's how I did the second one: I noticed that the numbers can be written as 2^2+1, 3^2+2, 4^2+3, ..., so I write a1=(the original formula), a2=(formula starting from 11), a3=(formula starting from 19),... and observe that a2=a1^2-2^2-1, a3=a2^2-3^2-2, ... so that a_n ^2 - a_{n+1} ^2 = (n+1)^2 + n = (n+2)^2 - (n+3) and just guessed a_n = n+2... so a_1 = 3. Of course it's not rigorous, but it works.
Thanks for writing the actual sequence. One thing I really hate about "..."-expressions is that the actual sequence usually is not obvious (to me), in particular the starting value (which is important for the limit in the general case). I guess convention is "..." starts at zero in sums and at one in products? There are other places it gets put though, for instance roots, what then?
The first example is a typical case of an iteration of the form f[f[f[ .....].If this iteration converges , then the limit is a solution of x = f[x]. In our case f[x]= (x+6)^(1/3). This is easy to see if one looks a the graphs of x ,and f[x] , where one can see a sort of spider web which arises if one looks at the points f[f[x] ,f[f[f[x]]] etc. around the point of intersection of the two graphs.The cubic equation is not hard to solve since one can easily guess a solution.
Neat fact about the second sequence: If you pick a value of x to terminate it at, i.e. if you take some n to calculate √(f(1)+√(f(2)+√...+√f(n))) but then you replace the final f(n) with (n+2)², your sequence lands at exactly 3. Idea of proof: You'd have f(n-1)+√(n+2)² in the second last root, so that's (n-1)²+3(n-1)+1+(n+2) = n²+2n+1 = (n+1)², so that gives f(n-2)+√(n+1)² in the third last root which reduces to just n², the next one (n-1)², and so on all the way back to f(1)+√4² = f(1)+4 = 5+4 = 9, so the topmost root becomes √9 = 3.
This gives another, more "backwards" way to prove the second sequence: Start with √9, clearly 3. Extract the number 4: √9 = √(5+4) = √(5+√16) Extract the number 5: √(5+√16) = √(5+√(11+5)) = √(5+√(11+√25)) Extract the number 6: √(5+√(11+√25)) = √(5+√(11+√(19+6))) = √(5+√(11+√(19+√(36))) Keep extracting the integers one by one, and you end up at the sequence above. Since every intermediate step is equal to 3, so is also the limit. (Of course, this is not mathematically rigorous as I have written it here.)
The second one is very cool~ But I have a question (10:53) about how to explain that F is a polynomial. If F is not a polynomial then you can't suppose it as F(x) = ax + b.
It is more like: "Wouldn't it be nice if F were a linear function? Let's check if that could work." If it doesn't work, then try another approach/guess. If it does work, then there you are. You found a solution.
The second one's definition is a bit ambiguous. For me it looked like the second element is the first one plus 6, the third one is the second plus 8, and so on, so a_n = a_n-1 + 2 * (n + 1), which gives the same numbers for the first few n but then diverges from the f(x) in the video.
Yeah this, the only question I had after seeing this video was where the hell did he fart out the explicit form for f from, but the sequence of numbers in the radical being a quadratic one makes it clearer. The calculations themselves feel like they are rightfully omitted as trivial for the purposes of this video but the motivation behind finding f and the general logic of getting there would've been a good addition.
I would've liked this too. I know it's related to the fact that adding successive odd numbers gets you the squares. There's a video by Mathologer about these kinds of sequences and how to derive a polynomial from them, but it escapes me
First one is easy, basically say x = cbrt(6+cbrt(6+...), then notice that x = cbrt(6+x) since it's nested with itself, then just cube both sides and then arrenge the equation like: x^3-x = 6... plugging in small values you can notice that x is equal to 2. Am I wrong? I am being really honest, I'm not that experient in maths... is there any error or detail that I should consider before assume that "nested with itself statement"?
How would one construct a rigorous proof around 11:35 ? That F should be linear. Physicists would forge happily ahead with this, but I wonder if a true proof is hard.
Well, engineer me and not a mathematician. Let the value of the first sequence be x. Then x^3-6=x, and then x^3-x-6=0, and x=2 by inspection. But what do I know.
It seems kind of out of nowhere, the appearance of the formula in the second one. If this was a comp. problem, I suppose the hardest part would be to figure it out, after which it would be quite straightforward.
If you look at the sequence 5, 11, 19, 29, 41, ... and look at successive differences, you get 6, 8, 10, ... and if you look at successive differences again, you get 2, 2, 2, 2, ... all 2's. You then work back from this. 6,8,10,... = 4+2n, so if we let the sequence 5, 11, ... be a{n} with a{1} = 5, then a{2} = a{1} + 4 + 2*1, a{3} = a{1} + 4 + 2*1 + 4 + 2*2. So a{n} = a{1} + 4(n-1) + 2(1+...+(n-1)) and since 1+...+(n-1) = (1/2)n(n-1), and a{1} = 5, we get a{n} = 5 + 4n -4 + n(n-1) = 5 + 4n -4 + n^2 -n = n^2 + 3n +1
@@HershO. Or you notice that the first number is 2² + 1, the second number is 3² + 2, the third number is 4² + 3 and so on. Hence the nth number is (n+1)² + n = n² + 3n + 1. If you don't notice that, yet another approach is to realize that since the differences between the numbers increase linearly, the numbers themselves have to increase quadratically. So you try a_n = a n² + b n + c and determine the coefficients a, b, c by inserting e. g. the numbers n = 1, n = 2 and n = 3.
The second differences in the second sequence are identically 2 [ (f(x+2) - f(x+1)) - (f(x+1) - f(x)) = 2], and you ended up with F(x) = x+2. I wonder if that's a coincidence. What sequences produce other formulas for F, such as x+3?
You can cook that up pretty easily. Choose a function f with exactly one fixed point and a derivative outside the range [-1,1]. The fixed point is a solution to the recursive equation but the sequence a, f(a), f(f(a)), f(f(f(a))), ... will drift away from the fixed point unless you start exactly at the fixed point (then it is stationary). For concreteness, say f(x) = 12x - 1. We have f(1/11) = 1/11. However the sequence 0, f(0), f(f(0)), ... = 0, -1, -13, -157 diverges. So the expression "-1 + 12(-1 + 12(-1 + 12(-1 + ...)))" has no value.
@@randomguyontheinternet_69 Yes, and you can see that it's quadratic by noticing that the size of the gaps between the numbers increase linearly! (6, 8, 10, 12, ...) It works a little bit like derivations, since the rate of change is of first order, the function itself is of second
We only have proven that: 1. *if* F is a first-degree polynomial, than F(x) = x + 2 2. F(x) = x + 2 is *a* solution of F²(x) - F(x+1) = x² + 3x + 1 However we have *not* proven, or even answered the question: 3. Is F(x) = x + 2 the *only* solution to F²(x) - F(x+1) = x² + 3x + 1?
Great question! The main problem here is -- the notation with "infinite radicals" is just not precise enough to answer the question to begin with. You need a recursive definition for that. If you have multiple possible limits, the one you get (if any) may depend on two things - how you defined the recursion representing the radical -- infinitely many possibilities! - which initial value you chose -- again infinitely many possibilities!
As others pointed out, as long as the sequence is well defined, you only have a single solution/limit point (at most). If we get multiple sol's from our cubic, we have to check each of them (rule out negative or complex sol's, rule out sol's bigger than 3 because we have seen that the limit is bounded by 3 etc.)
It would have a single "true" convergent limit (if you plug in any initial guess for x1 and iterate the formula, it would converge to this number) and the other solutions would be "valid" in the sense that if you plugged in EXACTLY that value for x1 it would work, but even a small deviation of your first guess from that value would either diverge under iteration or would converge to the "true" solution. This type of recurrence equation falls under the Fixed Point Theorem of which there's a lot of interesting convergence results to learn about. And they are connected with fractals and other interesting phenomena.
Yes and no. Then the limit would depend on where you start the sequence. So yes, there would be multiple values for the limit of the "..." expression (it would be ambiguous). But no, there can only be one limit of the (explicit) sequence. That's why I don't like "..." expressions without reference to the exact sequence.
you can probably easily find bounds for the sequence considering that under the radicals we have a polynomial function while iterating the square root is essentially an exponential process (sqrt(sqrt(sqrt(x)))=x^(1/2³)).
@@AlinaKlein953 It does follow from that but you do have to do a slight bit extra since the sequence 5, 11, 19, ... isn't bounded. (i.e. you have to prove (f(n)¹/²)ⁿ is bounded. Which it is but proving it takes a couple of steps.)
Correction: L³ = L + 6 implies L³ - L - 6 = 0. THe rest is correct in the sense that L³ - L - 6 = (L - 2)(L² + 2L + 3) so 2 is an actual root.
I really liked the rigor of using the monotone sequence theorem to justify how the limit of the sequence was calculated. So many times these kinds of details get skipped over in math proofs leaving an uneasy feeling of not quite knowing why certain moves are justified beyond handwaving.
Lmao your complaints describe the second example perfectly
These kinds of details are never left out of proofs. Never.
If a proof skips over these details, then it is not a sufficient proof
Where in God's name does the 8 :07 equation come from..there isNO basis for thisnguess..whybsont everyone else asking this..surely everyone else is just a flummoxed by it..kts the kind of thing thst makes math stupid and annoying or infuriating and I think should stop altogether..
@@angelmendez-rivera351 The comment at the top to which I was replying stated “these kinds of details get skipped over in math proofs”. No they don’t, or they are not proof. I’m not sure how it is possible to misinterpret my comment, but apparently this happened twice.
11:00 Thank you for clearly explaining the motivation for making the guess. I think that intuition benefits from learning the motivation behind different guesses.
Lovely second problem. The Penn fact is just an AMAZING addition to the videos. Keep it up, prof. Penn
The 2nd problem can be done rigorously. The sequence of functions that you constructed are, around x=1, monotonous, increasing, bounded above and analytic. Therefore it converges to an analytic function around x=1. Therefore it has a Taylor series expansion. Square the Taylor series and compare the coefficients with the right hand side, you get the linear function as the only solution.
I don't get it... Can you please explain the Taylor series part?
@@apopet Analytic functions mean they can be defined by a power series
@@TheEternalVortex42 Thanks, but I didn't mean explain what a Taylor series is. I didn't get how the series expansion will be used.
The second problem must be solved much more rigorously than it has been. In particular, nothing proves that the limit F(x) is a polynomial function.
First consider a sequence F_n(x) = sqrt(f(x) + ...+ sqrt(f(x + n - 1))), with f(x) equal to x^2 + x - 1.
Then, prove by induction that the sequence is increasing, and bounded above by x + 1.
Finally, let F(x) be the limit. Prove by induction that for every positive integer k, F(x) >= a(k), where a(k) is the following sequence :
a(1) = 1/2 and a(k+1) = sqrt (a(k)).
Not so simple as that, if we want to stay at a reasonable mathematical level.
Sorry for my broken English, I'm French.
Hi.
Re 2nd nested root. The lesson here is: problems that are not solvable in a given context become solvable in a more general context. "Generalise to solve": how true that often is in Mathematics
The "Penn fact box" is a nice idea, I tried to click on it to see if there was a link to a sketch of a proof or to some other video.
For the first sequence, wouldn't replacing the "last" 6 with an 8 just collapse the whole sequence like a house of cards to the number 2, proving this is its upper bound?
6:10 Hey Michael, Is it not L^3-L-6=0 instead of L^3-L+6=0
Yes, but his work after that line was correct.
Another usual "glitch" from Michael.....
I’m taking my first rigorous proofs based class next semester and one of things in the syllabus completely foreign to me was working with sequences, i.e finding their limit, proving they converge, etc. so it is very helpful to see videos like this to get a feel for some of the tools used. These problems were also just quite fun and novel to me as well.
Nice, clean, logical & interesting problems. Thanks.
Michael: constructs a very nice real analysis proof that the first sequence converges, then computes the limit in a nice rigorous fashion.
Also Michael: Well let’s just let this infinite radical be a function, probably well-defined, who knows. Let’s get a polynomial equation, I dunno, probably a linear function. Hey! This linear function satisfies the recurrence, we have no other choice but to declare this the answer!
The math speaks for itself
@@adamnevraumont4027 but the logic connecting the math together does not.
i mean he did say he wasnt gonna be as rigorous with that one he left it as an exercise to the viewer
@@dnuma5852 If he said that, it doesn’t make any sense, because unlike the first one, the second infinite radical is not the limit of a sequence with a simple recurrence. Instead, it’s the limit of one chain in some doubly-indexed sequence a_{m,n}, with a_{m,m}=sqrt(m^2+3m+1) and a_{m,n}=sqrt(m^2+3m+1+a_{m+1,n}), namely the limit of a_{1,n} as n goes to infinity. This is a totally different type of problem, and can’t be handled with the exact ideas as the first one. It would be useful and instructive for Michael to have covered this one rigorously too.
Noahtaul(know it all): you are among many who appear to be quite knowledgeable/proficient mathematically who have “empty math channel”
Noahtaul: So why don’t you produce your own math videos and publish them on your channel?
Hope you keep the Penn Facts as a regular part of the channel, they're very helpful!
I really liked the Penn Fact box containing a formal statement of the Monotone Convergence Theorem. This kind of thing should be very helpful to someone like me who doesn't use math too often but enjoys your videos.
For the second one, if you pick f(x)=x^2-x-1=x^2-(x+1)
you get
sqrt(x^2-(x+1)+sqrt((x+1)^2-(x+2)+sqrt((x+2)^2-(x+3)...)))=F(x)
and the value we're looking for is F(3)
but in this case, F(x)=x
or you just notice Fˆ2(x) - F(x+1) = xˆ2 + 3x + 1 = (x+2)ˆ2 - (x+3), i.e., F(x) = x+2.
How amazing and how beautiful the answers are
Here's how I did the second one: I noticed that the numbers can be written as 2^2+1, 3^2+2, 4^2+3, ..., so I write a1=(the original formula), a2=(formula starting from 11), a3=(formula starting from 19),...
and observe that a2=a1^2-2^2-1, a3=a2^2-3^2-2, ...
so that a_n ^2 - a_{n+1} ^2 = (n+1)^2 + n = (n+2)^2 - (n+3)
and just guessed a_n = n+2... so a_1 = 3.
Of course it's not rigorous, but it works.
Thanks for writing the actual sequence. One thing I really hate about "..."-expressions is that the actual sequence usually is not obvious (to me), in particular the starting value (which is important for the limit in the general case). I guess convention is "..." starts at zero in sums and at one in products? There are other places it gets put though, for instance roots, what then?
The first example is a typical case of an iteration of the form f[f[f[ .....].If this iteration converges , then the limit is a solution of x = f[x]. In our case
f[x]= (x+6)^(1/3). This is easy to see if one looks a the graphs of x ,and f[x] , where one can see a sort of spider web which arises if one looks at the points f[f[x] ,f[f[f[x]]] etc. around the point of intersection of the two graphs.The cubic equation is not hard to solve since one can easily guess a solution.
The second series of numbers is also the convergents to the continuous fraction of sqrt(2)
For the top one I figured if a(n+1) for infinity is cubert(6 + an), I could substitute it as such:
x = cuberoot(6+x) where the obvious answer is 2
The method used for the second radical was very clever
Neat fact about the second sequence: If you pick a value of x to terminate it at, i.e. if you take some n to calculate
√(f(1)+√(f(2)+√...+√f(n)))
but then you replace the final f(n) with (n+2)², your sequence lands at exactly 3.
Idea of proof:
You'd have f(n-1)+√(n+2)² in the second last root, so that's (n-1)²+3(n-1)+1+(n+2) = n²+2n+1 = (n+1)², so that gives f(n-2)+√(n+1)² in the third last root which reduces to just n², the next one (n-1)², and so on all the way back to f(1)+√4² = f(1)+4 = 5+4 = 9, so the topmost root becomes √9 = 3.
This gives another, more "backwards" way to prove the second sequence:
Start with √9, clearly 3.
Extract the number 4: √9 = √(5+4) = √(5+√16)
Extract the number 5: √(5+√16) = √(5+√(11+5)) = √(5+√(11+√25))
Extract the number 6: √(5+√(11+√25)) = √(5+√(11+√(19+6))) = √(5+√(11+√(19+√(36)))
Keep extracting the integers one by one, and you end up at the sequence above. Since every intermediate step is equal to 3, so is also the limit.
(Of course, this is not mathematically rigorous as I have written it here.)
@@DavidSavinainen Seems rigorous enough, a sequence of 3, 3, 3, 3, ... obviously has the limit 3 ;)
Awesome video. Thank you
For the second problem, if you started the x^2 + 3x + 1 function with f(0) = 1, then the solution would be 2, just like the first problem.
It is interesting to note that if the square root is used in the radical above instead of the cube root, it gives the same result as the radical below
Thank you, i've managed to figure out how to solve a problem that i was stuck on for a while! 😀
The second one is very cool~
But I have a question (10:53) about how to explain that F is a polynomial. If F is not a polynomial then you can't suppose it as F(x) = ax + b.
There isn't any reason behind F being polynomial. It's just a smart assumption.
It is more like:
"Wouldn't it be nice if F were a linear function? Let's check if that could work."
If it doesn't work, then try another approach/guess.
If it does work, then there you are. You found a solution.
The second one's definition is a bit ambiguous. For me it looked like the second element is the first one plus 6, the third one is the second plus 8, and so on, so a_n = a_n-1 + 2 * (n + 1), which gives the same numbers for the first few n but then diverges from the f(x) in the video.
while the second problem is defined ambiguously, the quadratic in the video is actually the explicit form of the sequence you described.
"but then diverges from the f(x) in the video"
No, it doesn't, it gives the same values. Apparently you made a calculation error somewhere.
🙂 I would found it helpful to prove the functional form of f using the fact that its increment form sequence 6, 8, 10, 12, ...
Yeah this, the only question I had after seeing this video was where the hell did he fart out the explicit form for f from, but the sequence of numbers in the radical being a quadratic one makes it clearer. The calculations themselves feel like they are rightfully omitted as trivial for the purposes of this video but the motivation behind finding f and the general logic of getting there would've been a good addition.
I would've liked this too. I know it's related to the fact that adding successive odd numbers gets you the squares. There's a video by Mathologer about these kinds of sequences and how to derive a polynomial from them, but it escapes me
It has to do with discrete derivatives
F(x) is always rational. that blows my mind. no matter how deep we step into this sequence, we can start there and the tail will be rational.
First one is easy, basically say x = cbrt(6+cbrt(6+...), then notice that x = cbrt(6+x) since it's nested with itself, then just cube both sides and then arrenge the equation like: x^3-x = 6... plugging in small values you can notice that x is equal to 2.
Am I wrong? I am being really honest, I'm not that experient in maths... is there any error or detail that I should consider before assume that "nested with itself statement"?
How would one construct a rigorous proof around 11:35 ? That F should be linear. Physicists would forge happily ahead with this, but I wonder if a true proof is hard.
Well, engineer me and not a mathematician. Let the value of the first sequence be x. Then x^3-6=x, and then x^3-x-6=0, and x=2 by inspection. But what do I know.
but whoever invented the second exercise really thinks it's so easy to understand that those numbers come from that second-degree polynomial?
how did you arrive at the. x^2+3x+1 as f(x)/
from the start, we can deduce: S = (6 + S)^(1/3) thus (S - 2) * (S^2 + 2S + 3) = 0 -- > S = 2 (in R)
You can only deduce _that_ *if* S exists... which is what Michael does first.
It seems kind of out of nowhere, the appearance of the formula in the second one. If this was a comp. problem, I suppose the hardest part would be to figure it out, after which it would be quite straightforward.
If you look at the sequence 5, 11, 19, 29, 41, ... and look at successive differences, you get 6, 8, 10, ... and if you look at successive differences again, you get 2, 2, 2, 2, ... all 2's. You then work back from this. 6,8,10,... = 4+2n, so if we let the sequence 5, 11, ... be a{n} with a{1} = 5, then a{2} = a{1} + 4 + 2*1, a{3} = a{1} + 4 + 2*1 + 4 + 2*2.
So a{n} = a{1} + 4(n-1) + 2(1+...+(n-1))
and since 1+...+(n-1) = (1/2)n(n-1), and a{1} = 5, we get
a{n} = 5 + 4n -4 + n(n-1) = 5 + 4n -4 + n^2 -n = n^2 + 3n +1
@@Chalisque ahh thanks for that! I was thinking there had to have been a way to derive it naturally, this is a neat useful trick.
@@HershO. Or you notice that the first number is 2² + 1, the second number is 3² + 2, the third number is 4² + 3 and so on. Hence the nth number is (n+1)² + n = n² + 3n + 1.
If you don't notice that, yet another approach is to realize that since the differences between the numbers increase linearly, the numbers themselves have to increase quadratically. So you try a_n = a n² + b n + c and determine the coefficients a, b, c by inserting e. g. the numbers n = 1, n = 2 and n = 3.
The second differences in the second sequence are identically 2 [ (f(x+2) - f(x+1)) - (f(x+1) - f(x)) = 2], and you ended up with F(x) = x+2. I wonder if that's a coincidence. What sequences produce other formulas for F, such as x+3?
It's the Christmas Three!
Could you provide a counterexample of a nested sequence that has a solution to the equation we always solve, but does NOT converge.
You can cook that up pretty easily. Choose a function f with exactly one fixed point and a derivative outside the range [-1,1]. The fixed point is a solution to the recursive equation but the sequence a, f(a), f(f(a)), f(f(f(a))), ... will drift away from the fixed point unless you start exactly at the fixed point (then it is stationary).
For concreteness, say f(x) = 12x - 1. We have f(1/11) = 1/11. However the sequence 0, f(0), f(f(0)), ... = 0, -1, -13, -157 diverges. So the expression "-1 + 12(-1 + 12(-1 + 12(-1 + ...)))" has no value.
@@cmilkau Thanks, but when I worte "nested" i meant sequences of the form a(n+1)=(ka(n)+m)^(1/p) where p>=2.
My biggest problem to solve the second question is to note that 5, 11, 19, ... Come from f(x) = x² + 3x - 1
Can find parabola given the 3 pairs : (1,5) , (2,11) , (3,19)
i really like the question but may you explain how did we get the f(x) = x^2 + 3x + 1
same question want to know as well
@@randomguyontheinternet_69 Yes, and you can see that it's quadratic by noticing that the size of the gaps between the numbers increase linearly! (6, 8, 10, 12, ...)
It works a little bit like derivations, since the rate of change is of first order, the function itself is of second
th-cam.com/video/scQ51q_1nhw/w-d-xo.html
@@cmilkau thanks bro !!
Penn fact is the best
We only have proven that:
1. *if* F is a first-degree polynomial, than F(x) = x + 2
2. F(x) = x + 2 is *a* solution of F²(x) - F(x+1) = x² + 3x + 1
However we have *not* proven, or even answered the question:
3. Is F(x) = x + 2 the *only* solution to F²(x) - F(x+1) = x² + 3x + 1?
did you find the equation n*n + 3*n + 1 from trial and error ??
6 minutes to get to L^3 - L + 6 = 0... Should take 6 seconds
Very nice
La prima è facile. 2....t^3-6=t
A very nice nested radical will be if you have only 1s inside.
fun! thank you!! And why don't you give some other similar questions?
2 doesn not seem to be a root of the equation L3-L+6=0; we should read L3-L-6=0
For the first one, what if we had gotten a cubic that had more than one real root? Would that mean we have multiple values for the limit?
Then I suppose we would be able to find upper and lower bounds to the expression to rule the other values.
Great question! The main problem here is -- the notation with "infinite radicals" is just not precise enough to answer the question to begin with. You need a recursive definition for that.
If you have multiple possible limits, the one you get (if any) may depend on two things
- how you defined the recursion representing the radical -- infinitely many possibilities!
- which initial value you chose -- again infinitely many possibilities!
As others pointed out, as long as the sequence is well defined, you only have a single solution/limit point (at most).
If we get multiple sol's from our cubic, we have to check each of them (rule out negative or complex sol's, rule out sol's bigger than 3 because we have seen that the limit is bounded by 3 etc.)
It would have a single "true" convergent limit (if you plug in any initial guess for x1 and iterate the formula, it would converge to this number) and the other solutions would be "valid" in the sense that if you plugged in EXACTLY that value for x1 it would work, but even a small deviation of your first guess from that value would either diverge under iteration or would converge to the "true" solution.
This type of recurrence equation falls under the Fixed Point Theorem of which there's a lot of interesting convergence results to learn about. And they are connected with fractals and other interesting phenomena.
Yes and no. Then the limit would depend on where you start the sequence. So yes, there would be multiple values for the limit of the "..." expression (it would be ambiguous). But no, there can only be one limit of the (explicit) sequence. That's why I don't like "..." expressions without reference to the exact sequence.
"The Q*bert of 6"
How do we know the second radical converges？
you can probably easily find bounds for the sequence considering that under the radicals we have a polynomial function while iterating the square root is essentially an exponential process (sqrt(sqrt(sqrt(x)))=x^(1/2³)).
Herschfeld's convergence theorem
@@AlinaKlein953 It does follow from that but you do have to do a slight bit extra since the sequence 5, 11, 19, ... isn't bounded. (i.e. you have to prove (f(n)¹/²)ⁿ is bounded. Which it is but proving it takes a couple of steps.)
How do you even compute partial radicals? Given f(n) should be nested first and then from n down to 1? Totally unclear, I am confused...
@@cedriclorand1634 You do them from inside out. The inner most radical first, then the one that includes it, then the one that includes that, etc.
I wish everyone would take the time to properly prove convergence for these "..."-expressions. A lot of nonsense would disappear from youtube.
X=2 in just a minute mentally
I did it in 30 secs while eating. Not even lying.
I solved both of these mentally..I probably got lucky :)
is it just a coincidence that a vid about radicals of 6 has 666 likes?
Yes. Yes it is