For n = 1 we don't have to care about the limit existence. That's because even if it does exist, our resulting relation x³-x-1=0 doesn't have rational (let alone integer) roots. So, either the limit doesn't exist or it does but it's not an integer one. Either way it's not satisfying our condition.
For the homework, we can still use a similar argument... our sequence is still monotonically increasing, that part of the argument didn't depend on n>1. So we just need to show it's bounded. Claim: a_k < 1.5 for all k Base case: a_1 = 1 < 1.5 Inductive case: suppose a_k < 1.5, then a_k+1 = cbrt(1 + a_k) < cbrt(1 + 1.5) = cbrt(2.5) < cbrt(3.375) = 1.5 So the sequence a_k is increasing and bounded from above by 1.5... so it converges, and the limit is at most 1.5. But the sequence started at 1 and is increasing, so the limit is strictly larger than 1. Therefore the limit exists but is not a natural number.
Speaking of re-using Michael's results, we could also go for x³ = 1 + x , assuming that it was converging after all and easily see, that there is no such natural x. So it does not even matter, if it converges for n=1, we know it won't work anyway.
9:39 Well it certainly *does* make sense and of course makes the nested cube root equal to 0 --- not equal to x=1 ! Does that mean the derivation is wrong? No! What we found is that x is *one* solution of x³ = n+x. For n=0, there are three solutions: x=-1, x=0, x=1, and the actual limit is only one of them, namely x=0. But this suggests that there is a *gap* in the argument and that we need to be more careful for the "main" part: If we pick a natural number x and then let n = (x-1)x(x+1), does that really imply that x is the limit of the nested cube roots? As just seen, this is not the case if we start with x=0. So let's be more careful with x=2: As presented, we must have n = 6, and then we know that the limit is *some* root of X³ = n+X, but a priori the limit/this root could either be the 2 we started with or any of the other two roots! However, it turns out that the other two roots are complex and hence cannot be the limit of the nested cube root. Is this always the case? Well, if we start with x, let n=(x-1)x(x+1) and want to find the roots other than x of the polynomial X^3-X-n, we have to divide out the known linear factor X-x, which leads us to the quadratic X² + x X + x²-1 which has discriminant D= x² -4(x²-1) = 4 - 3x². Thus for x \ge 2, the discriminant is negative, i.e., the x we started from is the only real solution - as desired.
This raises the interesting "bonus" question: Let a be a real number and let b be the nested cube root b = sqrt[3]{ a + sqrt[3]{ a + ... }}. Convergence is readily shown for all real a, not just for natural numbers and also in this generalization it is the case that b is one of the real roots of X³-X-a. But which of the roots is b when there are several real roots? As seen above, this happens when 4 - 3a² is positive, i.e., for |a| < 2/sqrt 3.
It doesn't make sense because 0 isn't considered a natural number to begin with... that's why the argument for 2 < n² is always true just by splitting the n=1 case If you do consider 0 as a natural number then just split that case as we did with n=1 and don't forget that for the case of the limit being equal to x that is only true for n>1 because of the 2 < n² argument made which still limits the values of x that you can put into the last equation...
@@omarabdulazeez4298 The infinite nested radical cbrt(0 + cbrt(0 + cbrt(...))) certainly _does_ make sense, and is nice to include because it is the easiest case (a_k=0 for all k so immediately x=0) so it can be dealt with very quickly. Indeed there is no reason the problem shouldn't be stated over the integers, since we have cbrt(-n + cbrt(-n + cbrt(...))) = -cbrt(n + cbrt(n + cbrt(...))) - since *Z* has a bit more structure than *N* , I see no reason not to solve the problem over *Z* .
You are right, there is a gap in the argument. Everything is fine when n is a non-zero natural number, but he didn't prove it. However, you did a mistake or a typo in your comment, since 2/sqrt(3) is actually the threshold for |b|, and not |a|, for the polynomial X^3-X-a not having any other real roots than b. The threshold for a is then (2/sqrt(3))^3-2/sqrt(3)=2/(3*sqrt(3)), which is smaller than 1, unlike 2/sqrt(3). Another way of finding the number of real roots of X^3-X-a is to consider the function P(x)=x^3-x. We want the number of solutions of P(x)=a. P is a continuous function, its derivative is 3x^2-1, which is positive when |x|>1/sqrt(3) and negative when |x|
Solution to the n = 1 case: We can use the Monotone Sequence Theorem to prove it converges. We know that the sequence is increasing using the exact same method shown in the video. (2:55) Then, we can show that it is bounded above by 2. a_k < 2 --> a_k + 1 < 3 --> cuberoot(1 + a_k) < cuberoot(3) --> a_k+1 < cuberoot(3) < 2 Since the sequence is increasing, we know that its limit will be greater than 1 because a_2 > 1, also we know that it is less than 2, therefore 1 < a_k < 2, meaning that the limit cannot be a natural number since its value lies between two consecutive natural numbers QED.
3:40 More formally.: Extending the sequence to left with $a_0 = 0$, we have $a_1 > a_0$ and as $f(x):=\sqrt[3]{n+x}$ is increasing, we can do induction $a_{k+1} > a_k$ implies $a_{k+2}=f(a_{k+1} > f(a_k)=a_{k+1}$
For the case of n = 1, you can show that the sequence converges by comparing it with the sequence for n = 2. The latter is clearly larger than the former (term by term), and we know it's increasing (because the proof in the video did not need the condition n > 1), so the former converges.
The statement at 8:35 is false in one direction. Take 3=x^3-x for example. This polynomial has a real solution, but not one in the integers. It's not a problem though, because we only need it in the other direction :)
It seems somewhat irrelevant. It's true that if x is natural then n will be, but we're only considering natural n in the first place so it doesn't matter. It WOULD make a difference if we were allowing arbitrary real n in the problem, and then we could use this to show that x can only be natural for natural n.
Suppose the limit for 1 exists and equals some natural number x We know that x^3 = 1 + x, or 1 = x^3 - x = (x-1)x(x+1) > (x-1)^3 If x was greater than or equal to 2, then we’d have 1 > (x-1)^3 > (2-1)^3 = 1, so the only other option is x = 1. 1 = (1-1)1(1+1) = 0 * 1 * 2 = 0, but that doesn’t work, so x has no natural number solution.
6:29 "Using tricks from calculus you can show (2¹/³)(n¹/³) < n" I don't think you need calculus here since n is already a natural number greater than or equal to 2. Note that since 2
8:35 i feel like the statement that "x is a natural number iff n is a natural number" isn't really supported by the closure as you said it is, since for the case where n=1 the solution to that equation isn't even rational (i know Michael probably just said it without thinking as it would imply all natural numbers are a solution to this problem...). Therefore, your conclusion that n should be written as the product of 3 consecutive natural numbers is flawed; a more accurate assessment would be to observe that this is only a sufficient condition, this is true because if n=(t-1)t(t+1) for some natural number t then it is a solution to that equation, and in fact the only one since it's easy to prove that the solution is unique.
0:35 Homework First let simplify the infinitely cube root function Cbrt(n + Cbrt(n + Cbrt(n +...))) = P Cbrt(n + P) = P P³ = n + P P³ - P = n Now if n = 1 we have: P³ - P = 1 Using the Newtwo-Raphson Method we can see that: P ≈ 1.3247, which is NOT a Natural Number
@@reeeeeplease1178 Then again, the part that did not work for n=1 in the video is boundedness, whereas increasing is still clear. But the sequence for n=1 is below the sequence for n=2, hence also bounded.
@@bjornfeuerbacher5514 not necessarily. He showed that if it converges, then it doesn't converge to a natural number. In other words, there are only two possibilities. Either it converges, in which case it isn't a natural number, or it doesn't converge, in which case it isn't a natural number. Either way it isn't a natural number, so it doesn't matter whether it converges.
I've found a nice formula. The value of n such that sqrt{3}(n^3+sqrt{3}(n^3+sqrt{3}(n^3+...) is equal to x, for a natural x, is n_(x+1) = ((n_x)/x + 2x +1)(x+1), with n_0=n_1=0 and n_2=6
I'm coming from a functional programming background. There, you usually use fixpoints to represent objects with infinite nesting using only a finite amount of stuff. For this problem, I will be using Literate Haskell syntax and use "..." as an name like any other. So like the other comments, no big news, at least some justifications why you are allowed to use that shortcut. 1. Define the object "a" of the problem statement. In the video you had the free variables "n" and "...", this just makes them explicit. > a n ... = (n + ...) ** (1/3) Notice, that this definition is neither recursive, nor infinite. You are free to apply "a" twice to any object you like. After all, this is what it means for an object to have two free variables. 2. Define our search problem as a predicate over the arguments "n" and "x": > is_solution n x = (is_natural n) && (is_natural x) && (x `is_fixpoint_of` (a n)) The last term that contains "is_fixpoint_of" is the most interesting for our solution. Here is a usefull implementation of that function: > x `is_fixpoint_of` f = x - f x == 0 This is how that last term looks after some functional substitutions (and finally some math substitutions): xfa x n == x `is_fixpoint_of` (a n) == (x - (a n x) == 0) -- This is the same equation you got out of the limits == (x - (n + x) ** (1/3) == 0) -- "i - j == 0" can be substituded with "i**3 - j**3 == 0" using "x" for i and "(n + x) ** (1/3)" for j == (x ** 3 - ((n + x) ** (1/3)) ** 3 == 0) -- "(i ** (1/3)) ** 3" can be substituted with "i" == (x ** 3 - x - n == 0) This means, that the predicate "xfa" is equivalent to checking the roots of: poly x n = x ** 3 - x - n Like shown in the video, this multi-variable polynomial has two parts, and reduces to zero when "n == x ** 3 - x": poly x (x ** 3 - x) == x ** 3 - x - (x ** 3 - x) == 0
The relation to fixpoints should become obvious when considering that the "infinite" object is constructed like: > obj n = x where x = a n x This uses the variable x and the rule "a n x" to 'tie the knot'. While this will usually lead to non-termination or error-messages in most of todays compiler, you could think of a sufficiently smart evaluator, that performs the logic we have outlined above. -> Given n, it chooses an x that is a fixpoint of "a n" Note, that depending on the allowed range of types (e.g. nat, int, complex, ...) and values for n and x, this fixpoint might not be unique. (e.g. for n=0 -> x={-1,0,1})
I genuinely didn't understand why we had to use calculus here at all [I did find it interesting though]. I just set the expression to be equal to x, and then after substituting the majority of the expression with x, I got (n+x)^(1/3) = x. After cubing both sides and bringing x to the other side, we get x^3 - x = n. On factoring x, we get x(x^2-1) = n, and on using the difference of squares formula, we get x(x-1)(x-2) = n [I arrived at the answer shown in the video here but I just wanted to simply it a bit more]. This expression can be shown by x!/(x-3)!. Thus for all positive integer values of x, we can calculate n for which all values of the nested cube root are positive integers.
"I just set the expression to be equal to x" By doing so you make an unjustified assumption: that the limit exists at all. It's like saying y = 1 + 2 + 2^2 + 2^3 + ... y - 1 = 2 + 2^2 + 2^3 + ... = 2y y = -1 That's why the calculus part is important. Once we're sure the limit exists your reasoning is correct (except you factored x^2-1 as (x-1)(x-2)).
One day, two teachers, of physics and maths, sat down to solve a problem. M: I've found that the problem has a solution! P: I wouldn't start solving this if I didn't think so.
michael made a mistake - it isn't true as another commenter pointed out, giving the counterexample that (x-1)x(x+1) = 3 has a real, non-integer, solution.
Let m = ³√(n + ³√(n + ...)) Then, if this nested expression converges, m = ³√(n + m) m³ = n + m n = m³ - m Note that reversing the signs of both m and n, preserves the equality. So we need only investigate non-negative values. That is, since we seek natural numbers m, and no such m's will give n < 0, no negative n's need be checked. (This also follows from looking at the successive terms in the nested expression. It's just interesting that any positive solutions we find will yield negative counterparts.) So any integer m ≥ 0 will give an integer value of n, which must then be checked for convergence & equality. E.g., setting m = 0 gives n = 0, obviously correct; but m = 1 also gives n = 0, obviously incorrect. Other (integer) values for m give: n m 6 2 24 3 60 4 120 5 . . . .. 999,900 100 etc., ad infinitum. Do all (or any) of these converge? Of interest, a calculator iteration on the n = 1 case appears to converge on m = 1.324717957..., the sole real root of m³ - m - 1 = 0 I believe it is possible to show that the sequence of terms of the nested expression all lie between ³√n and 1 + ³√n for n ≥ 1. I haven't yet worked that out. But calculator iteration on larger and larger n values, appears to converge more and more quickly, and to the expected m values. Let's check out the professor's attack on the problem. Fred NB. This is a re-post without certain characters that I suspect are causing my original post to disappear every time I post it.
Sure enough, replacing the cube-root character with superscript-3 (³) and radical (√) worked! So what does YT have against the cube-root character? Fred
okay, i also do n+x=x³; x(x-1)(x+1) = n, means n must be a product of 3 consecutive natural numbers, x > 1 but i m not sure, is this the proof that this are ONLY solutions for n, or there can be another kinds of constructing the n?
Very good. I was sure the there was no limit until 8 minutes in. Feeling slow today. I think for n=1, the limit is cube root of 2. So n=1 is not in the set of bounded limits that converge to a natural number. But n=2 converges to 2, so it is in the solution set. But not confident of my answers here. This reinforces the correctness of my decision to matriculate, but not in math.
the limit for n=1 is not cube root of 2 btw - you have a_1 = 1, a_2 = cbrt(1 + 1) = cbrt(2), and a_3 = cbrt(1 + cbrt(2)), a_4 = cbrt(1 + cbrt(1 + cbrt(2))), and it gets more and more complicated (as expected).
n=1 case: exactly the same - just use 2n as an upper bound for the sequence. Does not converge to a natural number. n=0 case: It is effectively an empty product at it's core, so x=1 does make some amount of sense. However, x=0 is the answer that'd be derived directly from the limit that gave us the cubic. So x=1 and x=-1 are more comparable to the complex solutions you get for other values of n.
I'm not sure you're using the term "empty product" correctly here. Maybe you meant empty sum, but cbrt(0 + cbrt(0 + cbrt(...))) isn't an empty sum (and neither is 0+0+0+...; an empty sum/product is a sum/product with zero terms/factors, not a sum/product in which the terms/factors are zero)
@@SlipperyTeeth I see what you're saying, but there is no empty product (indeed there is no multiplication in the statement of the problem). An empty product equals the number 1, so "an infinite stack of cube roots of an empty product" would be cbrt(cbrt(...cbrt(1))) = 1, whereas what we have is cbrt(cbrt(...cbrt(0))) = 0. I think your confusion is that the empty sum and empty product both look like " " (because they are both empty), but they are different because the operation that is being applied zero times is different - when you apply addition zero times you get the empty sum 0, whereas when you apply multiplication zero times you get the empty product 1. In fact now that I think about it, one could use the fact that the empty sum and empty product both look the same to give one of those trendy false proofs that 0=1: 0 = " " = " " = 1, QED xD
@@schweinmachtbree1013 I'm not sure where you think you got that 0. Really, it's more like cbrt(cbrt(...))... There's nothing at the core of it, because every 0 that was there vanished by addition. I was arguing that this "nothing" should be formally considered a product, inherited from the properties of cbrt (a multiplicative function). If the empty product is throwing you off, consider this different line of reasoning: 0+x=x=1*x If these are properties you would like to preserve, then it would otherwise be fair to substitute 0+ for 1* in the expression. Then by the usual limit process, it is clear that we have a sequence of 1s for the partial computations. So the string evaluates to 1.
@@SlipperyTeeth When n=0 the limit is clearly 0, not 1. We are defining cbrt(n + cbrt(n + cbrt(n + ...))) to be the limit of the sequence a_1 = cbrt(n), a_2 = cbrt(n + cbrt(n)), a_3 = cbrt(n + cbrt(n + cbrt(n))) - in general (a_k) is recursively defined by a_{k+1} = cbrt(n + a_k). When n=0 we have a_1 = cbrt(0) = 0, a_2 = cbrt(cbrt(0)) = 0, a_3 = cbrt(cbrt(cbrt(0))) = 0, so we have a_k = 0 for all k, i.e. (a_k) is the constant sequence (0) which of course converges to 0.
Not sure why this is interesting or worth 10 minutes of math complexity. If the sequence converges, then x^3 = x + n. Cardano and others showed that if n = 0, then there are 3 real roots, but if n > 0 then there is one real root, with a well-defined solution, with x*(x-1)*(x+1) = n. Therefore, for any n, there is a value of n for which this converges. If there is an a such that (a - 1) * a * (a+1) -= n, then there is a solution. Why did he take 10 minutes to reach this conclusion?
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For n = 1 we don't have to care about the limit existence. That's because even if it does exist, our resulting relation x³-x-1=0 doesn't have rational (let alone integer) roots. So, either the limit doesn't exist or it does but it's not an integer one. Either way it's not satisfying our condition.
True, but I want the limit if it exists.
@@edwardlulofs444 it exists, it's around 1.3247. The closed for is a mess and you can see it if you solve x³-x-1=0 on wolframalpha for example.
@@Czeckie Thanks
@@Czeckie the solution is not rational as it incorporates square roots
For the homework, we can still use a similar argument... our sequence is still monotonically increasing, that part of the argument didn't depend on n>1. So we just need to show it's bounded.
Claim: a_k < 1.5 for all k
Base case: a_1 = 1 < 1.5
Inductive case: suppose a_k < 1.5, then a_k+1 = cbrt(1 + a_k) < cbrt(1 + 1.5) = cbrt(2.5) < cbrt(3.375) = 1.5
So the sequence a_k is increasing and bounded from above by 1.5... so it converges, and the limit is at most 1.5. But the sequence started at 1 and is increasing, so the limit is strictly larger than 1. Therefore the limit exists but is not a natural number.
Speaking of re-using Michael's results, we could also go for x³ = 1 + x , assuming that it was converging after all and easily see, that there is no such natural x. So it does not even matter, if it converges for n=1, we know it won't work anyway.
And the limit turns out to be the plastic ratio
9:39 Well it certainly *does* make sense and of course makes the nested cube root equal to 0 --- not equal to x=1 ! Does that mean the derivation is wrong? No! What we found is that x is *one* solution of x³ = n+x. For n=0, there are three solutions: x=-1, x=0, x=1, and the actual limit is only one of them, namely x=0.
But this suggests that there is a *gap* in the argument and that we need to be more careful for the "main" part: If we pick a natural number x and then let n = (x-1)x(x+1), does that really imply that x is the limit of the nested cube roots? As just seen, this is not the case if we start with x=0. So let's be more careful with x=2: As presented, we must have n = 6, and then we know that the limit is *some* root of X³ = n+X, but a priori the limit/this root could either be the 2 we started with or any of the other two roots! However, it turns out that the other two roots are complex and hence cannot be the limit of the nested cube root.
Is this always the case? Well, if we start with x, let n=(x-1)x(x+1) and want to find the roots other than x of the polynomial X^3-X-n, we have to divide out the known linear factor X-x, which leads us to the quadratic X² + x X + x²-1 which has discriminant D= x² -4(x²-1) = 4 - 3x². Thus for x \ge 2, the discriminant is negative, i.e., the x we started from is the only real solution - as desired.
This raises the interesting "bonus" question: Let a be a real number and let b be the nested cube root b = sqrt[3]{ a + sqrt[3]{ a + ... }}. Convergence is readily shown for all real a, not just for natural numbers and also in this generalization it is the case that b is one of the real roots of X³-X-a. But which of the roots is b when there are several real roots? As seen above, this happens when 4 - 3a² is positive, i.e., for |a| < 2/sqrt 3.
It doesn't make sense because 0 isn't considered a natural number to begin with... that's why the argument for 2 < n² is always true just by splitting the n=1 case
If you do consider 0 as a natural number then just split that case as we did with n=1 and don't forget that for the case of the limit being equal to x that is only true for n>1 because of the 2 < n² argument made which still limits the values of x that you can put into the last equation...
Nice!
@@omarabdulazeez4298 The infinite nested radical cbrt(0 + cbrt(0 + cbrt(...))) certainly _does_ make sense, and is nice to include because it is the easiest case (a_k=0 for all k so immediately x=0) so it can be dealt with very quickly. Indeed there is no reason the problem shouldn't be stated over the integers, since we have cbrt(-n + cbrt(-n + cbrt(...))) = -cbrt(n + cbrt(n + cbrt(...))) - since *Z* has a bit more structure than *N* , I see no reason not to solve the problem over *Z* .
You are right, there is a gap in the argument. Everything is fine when n is a non-zero natural number, but he didn't prove it. However, you did a mistake or a typo in your comment, since 2/sqrt(3) is actually the threshold for |b|, and not |a|, for the polynomial X^3-X-a not having any other real roots than b.
The threshold for a is then (2/sqrt(3))^3-2/sqrt(3)=2/(3*sqrt(3)), which is smaller than 1, unlike 2/sqrt(3).
Another way of finding the number of real roots of X^3-X-a is to consider the function P(x)=x^3-x.
We want the number of solutions of P(x)=a. P is a continuous function, its derivative is 3x^2-1, which is positive when |x|>1/sqrt(3) and negative when |x|
Solution to the n = 1 case:
We can use the Monotone Sequence Theorem to prove it converges.
We know that the sequence is increasing using the exact same method shown in the video. (2:55)
Then, we can show that it is bounded above by 2.
a_k < 2 --> a_k + 1 < 3 --> cuberoot(1 + a_k) < cuberoot(3) --> a_k+1 < cuberoot(3) < 2
Since the sequence is increasing, we know that its limit will be greater than 1 because a_2 > 1, also we know that it is less than 2, therefore
1 < a_k < 2, meaning that the limit cannot be a natural number since its value lies between two consecutive natural numbers QED.
0:35 Homework
10:50 Good Place To Stop
3:40 More formally.: Extending the sequence to left with $a_0 = 0$, we have $a_1 > a_0$ and as $f(x):=\sqrt[3]{n+x}$ is increasing, we can do induction $a_{k+1} > a_k$ implies $a_{k+2}=f(a_{k+1} > f(a_k)=a_{k+1}$
3:50 accidental pirate 👀
For the case of n = 1, you can show that the sequence converges by comparing it with the sequence for n = 2.
The latter is clearly larger than the former (term by term), and we know it's increasing (because the proof in the video did not need the condition n > 1), so the former converges.
How is cbrt(n+am)
i also dont see how cbrt(2n) can lead to cbrt(2)*cbrt(am)
That is a typo, cbrt(n) is correct.
The statement at 8:35 is false in one direction. Take 3=x^3-x for example. This polynomial has a real solution, but not one in the integers. It's not a problem though, because we only need it in the other direction :)
It seems somewhat irrelevant. It's true that if x is natural then n will be, but we're only considering natural n in the first place so it doesn't matter. It WOULD make a difference if we were allowing arbitrary real n in the problem, and then we could use this to show that x can only be natural for natural n.
Suppose the limit for 1 exists and equals some natural number x
We know that x^3 = 1 + x, or 1 = x^3 - x = (x-1)x(x+1) > (x-1)^3
If x was greater than or equal to 2, then we’d have 1 > (x-1)^3 > (2-1)^3 = 1, so the only other option is x = 1.
1 = (1-1)1(1+1) = 0 * 1 * 2 = 0, but that doesn’t work, so x has no natural number solution.
I would have answered with all n that can be expressed like (c+3)!/c! for every natural namber c (including 0) but anyway great video :)
Or any non-negative integer c.
@@bobh6728 yes, maibe it's even more clear
@@neogauss2344 no, one should stick to their guns about whether 0 is a natural number and tell those with the opposite opinion to deal with it xD
6:29 "Using tricks from calculus you can show (2¹/³)(n¹/³) < n"
I don't think you need calculus here since n is already a natural number greater than or equal to 2. Note that since 2
At 6:08: You could have just said that by cube rooting both sides, we obtain a_(m+1)=(2*n)^(1/3), which is less than n.
Thus, a_(m+1)
8:35 i feel like the statement that "x is a natural number iff n is a natural number" isn't really supported by the closure as you said it is, since for the case where n=1 the solution to that equation isn't even rational (i know Michael probably just said it without thinking as it would imply all natural numbers are a solution to this problem...). Therefore, your conclusion that n should be written as the product of 3 consecutive natural numbers is flawed; a more accurate assessment would be to observe that this is only a sufficient condition, this is true because if n=(t-1)t(t+1) for some natural number t then it is a solution to that equation, and in fact the only one since it's easy to prove that the solution is unique.
Given any real target > 1, the same seed formula works. For instance, seed 27.864684026710023 converges to 3.141592653589793.
0:35 Homework
First let simplify the infinitely cube root function
Cbrt(n + Cbrt(n + Cbrt(n +...))) = P
Cbrt(n + P) = P
P³ = n + P
P³ - P = n
Now if n = 1 we have:
P³ - P = 1
Using the Newtwo-Raphson Method we can see that:
P ≈ 1.3247, which is NOT a Natural Number
Don't you have to show first that it actually converges for n = 1?
@@bjornfeuerbacher5514 if it doesnt converge, we are done already
@@reeeeeplease1178 Then again, the part that did not work for n=1 in the video is boundedness, whereas increasing is still clear. But the sequence for n=1 is below the sequence for n=2, hence also bounded.
I think it converges to cube root of 2, 1.26 --- with low confidence.
@@bjornfeuerbacher5514 not necessarily. He showed that if it converges, then it doesn't converge to a natural number. In other words, there are only two possibilities. Either it converges, in which case it isn't a natural number, or it doesn't converge, in which case it isn't a natural number. Either way it isn't a natural number, so it doesn't matter whether it converges.
this may be useful, if F is continuous with bounded derivative less than 1 e.g |f'|
x = cbrt(n+cbrt(n+cbrt(n+...)))
x = cbrt(n+x)
n = x³ - x = (x-1) x (x+1)
When n is a product of three consecutive integers.
I've found a nice formula. The value of n such that sqrt{3}(n^3+sqrt{3}(n^3+sqrt{3}(n^3+...) is equal to x, for a natural x, is n_(x+1) = ((n_x)/x + 2x +1)(x+1), with n_0=n_1=0 and n_2=6
Awesome problem. Thank you Professor
You can use a_n is bounded by the cube root of 3n, and 1 is no longer a special case.
I'm coming from a functional programming background. There, you usually use fixpoints to represent objects with infinite nesting using only a finite amount of stuff. For this problem, I will be using Literate Haskell syntax and use "..." as an name like any other. So like the other comments, no big news, at least some justifications why you are allowed to use that shortcut.
1. Define the object "a" of the problem statement. In the video you had the free variables "n" and "...", this just makes them explicit.
> a n ... = (n + ...) ** (1/3)
Notice, that this definition is neither recursive, nor infinite. You are free to apply "a" twice to any object you like. After all, this is what it means for an object to have two free variables.
2. Define our search problem as a predicate over the arguments "n" and "x":
> is_solution n x = (is_natural n) && (is_natural x) && (x `is_fixpoint_of` (a n))
The last term that contains "is_fixpoint_of" is the most interesting for our solution. Here is a usefull implementation of that function:
> x `is_fixpoint_of` f = x - f x == 0
This is how that last term looks after some functional substitutions (and finally some math substitutions):
xfa x n
== x `is_fixpoint_of` (a n)
== (x - (a n x) == 0)
-- This is the same equation you got out of the limits
== (x - (n + x) ** (1/3) == 0)
-- "i - j == 0" can be substituded with "i**3 - j**3 == 0" using "x" for i and "(n + x) ** (1/3)" for j
== (x ** 3 - ((n + x) ** (1/3)) ** 3 == 0)
-- "(i ** (1/3)) ** 3" can be substituted with "i"
== (x ** 3 - x - n == 0)
This means, that the predicate "xfa" is equivalent to checking the roots of:
poly x n = x ** 3 - x - n
Like shown in the video, this multi-variable polynomial has two parts, and reduces to zero when "n == x ** 3 - x":
poly x (x ** 3 - x)
== x ** 3 - x - (x ** 3 - x)
== 0
The relation to fixpoints should become obvious when considering that the "infinite" object is constructed like:
> obj n = x where x = a n x
This uses the variable x and the rule "a n x" to 'tie the knot'. While this will usually lead to non-termination or error-messages in most of todays compiler, you could think of a sufficiently smart evaluator, that performs the logic we have outlined above.
-> Given n, it chooses an x that is a fixpoint of "a n"
Note, that depending on the allowed range of types (e.g. nat, int, complex, ...) and values for n and x, this fixpoint might not be unique. (e.g. for n=0 -> x={-1,0,1})
I’ve been out of school for 44 years and I’m still being told by some random teacher to do my homework. lol
I have had my terminal degree for only 31 years, my B.S. for 48 years. Why stop doing homework? I submitted a solution.
I genuinely didn't understand why we had to use calculus here at all [I did find it interesting though]. I just set the expression to be equal to x, and then after substituting the majority of the expression with x, I got (n+x)^(1/3) = x. After cubing both sides and bringing x to the other side, we get x^3 - x = n. On factoring x, we get x(x^2-1) = n, and on using the difference of squares formula, we get x(x-1)(x-2) = n [I arrived at the answer shown in the video here but I just wanted to simply it a bit more]. This expression can be shown by x!/(x-3)!. Thus for all positive integer values of x, we can calculate n for which all values of the nested cube root are positive integers.
"I just set the expression to be equal to x"
By doing so you make an unjustified assumption: that the limit exists at all. It's like saying
y = 1 + 2 + 2^2 + 2^3 + ...
y - 1 = 2 + 2^2 + 2^3 + ... = 2y
y = -1
That's why the calculus part is important. Once we're sure the limit exists your reasoning is correct (except you factored x^2-1 as (x-1)(x-2)).
x=(x+n)^(1/3), so the result n=x^3-x! I have no news...let's continue!
One day, two teachers, of physics and maths, sat down to solve a problem.
M: I've found that the problem has a solution!
P: I wouldn't start solving this if I didn't think so.
Yep, that why I didn't start solving the homework until I was convinced that there was a solution. Great piece of math today.
Why is that clear that if x is not an integer, (x-1)x(x+1) is not an integer either ?
michael made a mistake - it isn't true as another commenter pointed out, giving the counterexample that (x-1)x(x+1) = 3 has a real, non-integer, solution.
Let m = ³√(n + ³√(n + ...))
Then, if this nested expression converges,
m = ³√(n + m)
m³ = n + m
n = m³ - m
Note that reversing the signs of both m and n, preserves the equality. So we need only investigate non-negative values.
That is, since we seek natural numbers m, and no such m's will give n < 0, no negative n's need be checked.
(This also follows from looking at the successive terms in the nested expression. It's just interesting that any positive solutions we find will yield negative counterparts.)
So any integer m ≥ 0 will give an integer value of n, which must then be checked for convergence & equality.
E.g., setting m = 0 gives n = 0, obviously correct; but m = 1 also gives n = 0, obviously incorrect.
Other (integer) values for m give:
n m
6 2
24 3
60 4
120 5
. . . ..
999,900 100
etc., ad infinitum. Do all (or any) of these converge? Of interest, a calculator iteration on the n = 1 case appears to converge on m = 1.324717957..., the sole real root of
m³ - m - 1 = 0
I believe it is possible to show that the sequence of terms of the nested expression all lie between ³√n and 1 + ³√n for n ≥ 1. I haven't yet worked that out.
But calculator iteration on larger and larger n values, appears to converge more and more quickly, and to the expected m values.
Let's check out the professor's attack on the problem.
Fred
NB. This is a re-post without certain characters that I suspect are causing my original post to disappear every time I post it.
Sure enough, replacing the cube-root character with superscript-3 (³) and radical (√) worked!
So what does YT have against the cube-root character?
Fred
okay, i also do n+x=x³; x(x-1)(x+1) = n, means n must be a product of 3 consecutive natural numbers, x > 1
but i m not sure, is this the proof that this are ONLY solutions for n, or there can be another kinds of constructing the n?
Very fun.
Thank you, professor.
So they are six times the tetrahedral numbers i.e. six times the r=3 diagonal of Pascal's triangle.
Looked a simple question from the start but didn't hope of it going through concepts of limits.
Heretically taking n=0 does make sense!
It does make the question rather trivial, which is not the same as making a nonsense of it.
Hlo sir I couldn't find partial differential and multiple integrals and beta and gamma function in playlist pls send me the link sir
Great video thanks dr
Very good. I was sure the there was no limit until 8 minutes in. Feeling slow today. I think for n=1, the limit is cube root of 2. So n=1 is not in the set of bounded limits that converge to a natural number. But n=2 converges to 2, so it is in the solution set. But not confident of my answers here. This reinforces the correctness of my decision to matriculate, but not in math.
the limit for n=1 is not cube root of 2 btw - you have a_1 = 1, a_2 = cbrt(1 + 1) = cbrt(2), and a_3 = cbrt(1 + cbrt(2)), a_4 = cbrt(1 + cbrt(1 + cbrt(2))), and it gets more and more complicated (as expected).
@@garethma7734 You are probably right. . . .
Je pose x=racine3(n+.....).
Alors x^3=n+x
Puis x^3-x=n. Et n=x(x+1)(x-1)
Can you please solve for the Product of (N²-X²) where X goes from 1 to (N-1)?
factor each N²-X² into N-X and N+X, then express each of those resulting products in terms of factorials, and the result will simplify to
(2N+1)! / N
Well, ooda fought it, eh?
Cool !!
n=1 case: exactly the same - just use 2n as an upper bound for the sequence. Does not converge to a natural number.
n=0 case: It is effectively an empty product at it's core, so x=1 does make some amount of sense. However, x=0 is the answer that'd be derived directly from the limit that gave us the cubic. So x=1 and x=-1 are more comparable to the complex solutions you get for other values of n.
I'm not sure you're using the term "empty product" correctly here. Maybe you meant empty sum, but cbrt(0 + cbrt(0 + cbrt(...))) isn't an empty sum (and neither is 0+0+0+...; an empty sum/product is a sum/product with zero terms/factors, not a sum/product in which the terms/factors are zero)
@@schweinmachtbree1013 The zeros vanish and you're left with an infinite stack of cube roots of an empty product.
@@SlipperyTeeth I see what you're saying, but there is no empty product (indeed there is no multiplication in the statement of the problem). An empty product equals the number 1, so "an infinite stack of cube roots of an empty product" would be cbrt(cbrt(...cbrt(1))) = 1, whereas what we have is cbrt(cbrt(...cbrt(0))) = 0.
I think your confusion is that the empty sum and empty product both look like " " (because they are both empty), but they are different because the operation that is being applied zero times is different - when you apply addition zero times you get the empty sum 0, whereas when you apply multiplication zero times you get the empty product 1. In fact now that I think about it, one could use the fact that the empty sum and empty product both look the same to give one of those trendy false proofs that 0=1: 0 = " " = " " = 1, QED xD
@@schweinmachtbree1013 I'm not sure where you think you got that 0. Really, it's more like cbrt(cbrt(...))... There's nothing at the core of it, because every 0 that was there vanished by addition. I was arguing that this "nothing" should be formally considered a product, inherited from the properties of cbrt (a multiplicative function).
If the empty product is throwing you off, consider this different line of reasoning: 0+x=x=1*x If these are properties you would like to preserve, then it would otherwise be fair to substitute 0+ for 1* in the expression. Then by the usual limit process, it is clear that we have a sequence of 1s for the partial computations. So the string evaluates to 1.
@@SlipperyTeeth When n=0 the limit is clearly 0, not 1. We are defining cbrt(n + cbrt(n + cbrt(n + ...))) to be the limit of the sequence a_1 = cbrt(n), a_2 = cbrt(n + cbrt(n)), a_3 = cbrt(n + cbrt(n + cbrt(n))) - in general (a_k) is recursively defined by a_{k+1} = cbrt(n + a_k). When n=0 we have a_1 = cbrt(0) = 0, a_2 = cbrt(cbrt(0)) = 0, a_3 = cbrt(cbrt(cbrt(0))) = 0, so we have a_k = 0 for all k, i.e. (a_k) is the constant sequence (0) which of course converges to 0.
Uups.... when I started the video I guessed that there are no solutions..... A little bit wrong.... as there are infintely many solutions. 🙂
Me too.
Not sure why this is interesting or worth 10 minutes of math complexity.
If the sequence converges, then x^3 = x + n. Cardano and others showed that if n = 0, then there are 3 real roots, but if n > 0 then there is one real root, with a well-defined solution, with x*(x-1)*(x+1) = n. Therefore, for any n, there is a value of n for which this converges. If there is an a such that (a - 1) * a * (a+1) -= n, then there is a solution. Why did he take 10 minutes to reach this conclusion?
3:20
43 minut late
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