f(x)=x+1/x=(x²+1)/x=(x²-2x+1)/x+2=(x-1)²/x+2, giving a minimum of 2 at x=1 for positive x Similarly f(x)=(x+1)²/x-2, giving a maximum of -2 at x=-1 for negative x
Could you do like this?: if a_n were to converge, we could find the limit by solving the equation: x=x+1/x but since this has no solutions, the limit does not exist
Yes, the only extended real values that satisfy that are x=∞ and x=-∞ - since you're essentially solving 1/x=0 - and a_n>0 rules out x=-∞. Remember that, for a real sequence, an existent limit may be an extended real number, not just a real number.
@@OOKIEDOKIE Well, the focus of this video is less about whether or not it converges, but to show that it grows arbitrarily close to another divergent sequence.
A not very rigorous solution attempt that pretends a(n) is a continuous function: a(n+1)-a(n) = 1/a(n) a(n+ε)-a(n) = ε/a(n) lim ε->0 (a(n+ε)-a(n))/ε = 1/a(n) da/dn = 1/a(n) This is a separable differential equation with solutions a(n) = sqrt(2*(n+c)) From here it's easy to show that lim n -> ∞ sqrt(2*(n+c)) - sqrt(2*n) = 0 for any finite value c (simply multiply and divide the expression by sqrt(2*(n+c)) + sqrt(2*n) and you get an expression of the form 2*c/∞).
Hi Michael, I have a question: Wouldn't it be wrong to use l'hopital in the last step (min 16:16 ) since n is a natural number, which implies that the functions of n are not differentiable?? Thank you very much, I really like your content and I always see it!!
@@spkaxb1489 suppose we have a function f(x) and a sequence a_n. For every integer k, f(k)=a_k Suppose f(x) has a continuous limit at infinity. Then a_n has a discrete limit at infinity, and it equals the continuous one. As such, L'hôpital's rule would test for the continuous limit which, when we find that it exists, can be equated with the discrete limit.
I think an appropriate term for a sequence another sequence eventually comes arbitrarily close to is "attractor" ("asymptote" also works, but in my ear it's less general). There are clear analogies to the idea of a (strange) attractor in the theory of dynamical systems.
I would switch the terms - i.e. asymptotic is what I would call this type of behaviour, not least because an attractor (AFAIK) is a fixed set of points (states) in n-space, not a function with an infinite domain. Some attractors (e.g. function fixed points) manifest asymptotic behaviour. FWIW, if you set a_1 = 1, a_n ≥ √2n for n≥2, and the sequence is a pretty good approximation to √2n from very small n, which to me is even less 'attractor' (oscillating) behaviour, and more a true asymptote.
Let p and q be two strictly positive integers. (p/q)+(q/p) = (p^2/(pq)) + (q^2/(pq)) = (p^2+q^2)/(pq) . So (p/q)+(q/p) - 2 = ((p^2 - 2pq + q^2)/(pq) = ((p-q)^2)/(pq) which is greater than 0 for p≠q and equal to zero for p=q . So, for all positive rational numbers r, we can express r as (p/q) for some positive integers p and q, and so r + (1/r) - 2 is greater than or equal to zero. And, as x + (1/x) is continuous on the strictly positive real numbers, it is also everywhere at least 2. Or, an improved more direct version of that, x + (1/x) = (x/1)+(1/x) = ((x^2 + 1^2))/(1x) and so x+(1/x) - 2 = ((x-1)^2)/(x) which is non-negative for all x>0 .
Okay, I think I get it. I let Ak = ak*ak, so A(n+1) = An + 2 + 1/An, and so An = 2n + terms. Specifically, An = 2n + (A1-2+\sum 1/Ak.) Then (an-sqrt{2n}) = sqrt{An}-sqrt{2n} = (An-2n) / sqrt{An}+sqrt{2n}. This limits to zero as required as long as \sum 1/Ak grows asymptotically slower than sqrt{n} --- which it does, because the extra terms are nonnegative, and the sum is O(log n).
Looks like a typical Newton-Raphson recursion sequence. Plugging in x-y'/y=x+1/x, -dy/y=dx/x -log(y) =log(x) y=1/x So it is the recursion which takes us to the root of the equation 1/x=0, which is infinity. So the sequence diverge to infinity.
A more rigorous way to end up with the same contradiction would be to suppose the limit exists, and then take the limit of both sides of the recursion as n goes to infinity, so x + 1/x = x, thus, 1/x = 0, and therefore the limit does not exist. Which does not imply divergence to infinity, mind. Though that does follow from the lack of a limit *alongside* the monotonicity of the sequence. (I'm proposing this more rigorous way precisely because your casual disregard for rigor scares me as a more pure math guy)
@@hugh081 so the limit is the root of 1/x=0, that's simple enough, but what in Newton-Raphson guarantees that it's x=∞ instead of -∞? I assume that it's because we want ln(x) to be extended real, so x can't be negative, but just by taking the antiderivative to be ln|x|, this process itself wouldn't tell, right?
@@TC159 Yes, and no. Yes, it's not unrigorous in that it's not false, and everything (save for the comment about 1/x = 0 at the end, which is blatantly unrigorous) can (hopefully, I'm genuinely unsure) be formalized. But no, it's still unrigorous in that it's using heavy machinery in a laisse-faire way without much in the way of justification for things. At all.
13:20 This integral should have probably been from 1 to n instead of n-1 since the rectangle with height n-1 goes from x=n-1 to x=n. It makes no difference in the result but it bugged me the same.
Isn't a bigger more frustrating problem thatche doesn't justify at all why anyone would ever deduce that the limit as n goes to infinity lf asub n minus nsqrt 2.WHAT THE HELL WHERE THE HELL DOESthat clme fromwhy not asub n tikes sart 2 FOR GODS SAKE..Surely everyone is as grustrated and oizzled by thus as I an since I don't see how anyone would ever think of that no matter how smart you are right? There's just no reason to.
@@ibaijurado279 Nop. It's true that the last rectangle goes from x=n-1 to x=n but the integral he wrote goes from 1 to n-1, so that last rectangle lies off the limits of integration.
I would love to see that proof. This is as short as I could get mine: Assuming a_{n+1}=1+1/a_n converges to some value L. As n approaches infinity, a_n and a_{n+1} approach L. Substituting the terms from the original sequence: L = 1 + 1/L => L^2 = L + 1 => L^2 - L - 1 = 0 => L is either phi or -1/phi. Assuming a_0 is positive, we can prove that a_n is positive for all values of n. Therefore L is phi, thus the sequence convergences to phi.
Love this guy's videos generally but this solution seems massively over-engineered. You can easily show that a_n will eventually exceed any natural N: if a_n=1/N, so a_(n+kN+1)>N, where k=ceiling(N-a).
Would like to extend the prob with n->n+f(n,t) where t is a modular number and f a function that returns tiny positive and negative values either side of n.
Mr. Penn QUESTION PLEASE HOW or WHY would anyone deduce that the limi as n goes to infinty equals zero or even use the linit of 2 swrt of n as opposed to a limit having only a sub n terms..Surely you agree that comes out of NOWHERE and I don't see why anyone would.ever think of it..isnt that just true?? Hope you can respond please when you can.
I'm not sure i understand your question, but I'd guess there are two underlying questions. (1) Why are we looking at lim(a_n - f(n)) where f(n)=sqrt(2n) and (2) where does sqrt(2n). The answer to (1) is that for recursive defintions like a_n, we often like to find closed-form (i.e. non recursive) formula for it. That is, an expression of a_n that depends only on n and not on a_(n-1). We don't have that here, but we do have a closed form approximation, that is arbitrarily good as n gets larger. That is one way of interpreting the limit discussed in the video. For (2), where did sqrt(2n) come from? As far as the video is concerned, it was pulled from thin air. It could just be a lucky guess. But one way to make that guess would ne to plot out a bunch of values of a_n on a graph, and find a function f(n) that approximates that graph. Then, once you have a ghess, try to prove that lim(a_n-f(n)) is 0. If it is, tgen you've found a good approximation of your recursive sequence. Note that there will be infinitely many good approximations, bit generally one would pick the one with the simplest expression (e.g. sqrt(2n) instead of say sqrt(2n+1/n)+e^(-n)).
День тому+1
from Morocco thank you for your wonderful clear complete explanation....i share many of them on my facebook page.....can you please post a video to debunk the paradox π=2 is it similar to your previous dubunking 2=√2
The right question here is to find an equivalent of a_n as n goes to infty. The fact that it diverges to infty is straightforward, as a_n is increasing and it cannot converge since x+1/x = x has no real solution.
@@fahrenheit2101 to crying Kamala fans, yes... After being called the worst possible names by lefties for years, nobody is interested in being peaceful now.
a + 1/a >= 2 can be shown by AM/GM inequality. Since we get (a + 1/a)/2 >= sqrt(a/a)
Nice.
And therefore, a more elementary approach would be to show it using a quadratic, since that's the most common proof of AM/GM that I know of.
f(x)=x+1/x=(x²+1)/x=(x²-2x+1)/x+2=(x-1)²/x+2, giving a minimum of 2 at x=1 for positive x
Similarly f(x)=(x+1)²/x-2, giving a maximum of -2 at x=-1 for negative x
@@fahrenheit2101 Don't make any more posts in the forums.
Could you do like this?: if a_n were to converge, we could find the limit by solving the equation: x=x+1/x but since this has no solutions, the limit does not exist
Yes, by taking the limit in both sides of the recursion relation and finding a contradiction.
Yes, the only extended real values that satisfy that are x=∞ and x=-∞ - since you're essentially solving 1/x=0 - and a_n>0 rules out x=-∞. Remember that, for a real sequence, an existent limit may be an extended real number, not just a real number.
Yes, *but* it's important to be clear on why this works. In particular, take limits of both sides of the recursion.
Yeah thats how I was taught to do it in analysis, its much faster, I wonder why it wasnt mentioned.
@@OOKIEDOKIE Well, the focus of this video is less about whether or not it converges, but to show that it grows arbitrarily close to another divergent sequence.
For a>0, a+1/a>=2 a^2-2a+1>=0 (a-1)^2>=0.
Bit dont you agree nomone wlwuldmever..think ofnthe limit he writes here?
But after that what do you do?
@@leif1075 -- Write a correct sentence!
Thanks PROF UR the BEST
He always knows the good place to stop 😮
A not very rigorous solution attempt that pretends a(n) is a continuous function:
a(n+1)-a(n) = 1/a(n)
a(n+ε)-a(n) = ε/a(n)
lim ε->0 (a(n+ε)-a(n))/ε = 1/a(n)
da/dn = 1/a(n)
This is a separable differential equation with solutions a(n) = sqrt(2*(n+c))
From here it's easy to show that lim n -> ∞ sqrt(2*(n+c)) - sqrt(2*n) = 0 for any finite value c
(simply multiply and divide the expression by sqrt(2*(n+c)) + sqrt(2*n) and you get an expression of the form 2*c/∞).
17:32 shouldn’t the integral go from 1 to n?
(does not invalidate the proof)
Anyway, a very enjoyable video, as always!
Hi Michael, I have a question: Wouldn't it be wrong to use l'hopital in the last step (min 16:16 ) since n is a natural number, which implies that the functions of n are not differentiable??
Thank you very much, I really like your content and I always see it!!
@@spkaxb1489 suppose we have a function f(x) and a sequence a_n. For every integer k, f(k)=a_k
Suppose f(x) has a continuous limit at infinity. Then a_n has a discrete limit at infinity, and it equals the continuous one. As such, L'hôpital's rule would test for the continuous limit which, when we find that it exists, can be equated with the discrete limit.
2:06 this can be done just by using identities and not calculus!
If you expand this (sqrt(a_n) - sqrt(1/a_n))^2 > 0
you reach the result quite immediately
I think an appropriate term for a sequence another sequence eventually comes arbitrarily close to is "attractor" ("asymptote" also works, but in my ear it's less general). There are clear analogies to the idea of a (strange) attractor in the theory of dynamical systems.
I would switch the terms - i.e. asymptotic is what I would call this type of behaviour, not least because an attractor (AFAIK) is a fixed set of points (states) in n-space, not a function with an infinite domain. Some attractors (e.g. function fixed points) manifest asymptotic behaviour.
FWIW, if you set a_1 = 1, a_n ≥ √2n for n≥2, and the sequence is a pretty good approximation to √2n from very small n, which to me is even less 'attractor' (oscillating) behaviour, and more a true asymptote.
Let p and q be two strictly positive integers. (p/q)+(q/p) = (p^2/(pq)) + (q^2/(pq)) = (p^2+q^2)/(pq) .
So (p/q)+(q/p) - 2 = ((p^2 - 2pq + q^2)/(pq) = ((p-q)^2)/(pq) which is greater than 0 for p≠q and equal to zero for p=q .
So, for all positive rational numbers r, we can express r as (p/q) for some positive integers p and q, and so r + (1/r) - 2 is greater than or equal to zero.
And, as x + (1/x) is continuous on the strictly positive real numbers, it is also everywhere at least 2.
Or, an improved more direct version of that,
x + (1/x) = (x/1)+(1/x) = ((x^2 + 1^2))/(1x)
and so x+(1/x) - 2 = ((x-1)^2)/(x) which is non-negative for all x>0 .
Okay, I think I get it. I let Ak = ak*ak, so A(n+1) = An + 2 + 1/An, and so An = 2n + terms. Specifically, An = 2n + (A1-2+\sum 1/Ak.)
Then (an-sqrt{2n}) = sqrt{An}-sqrt{2n} = (An-2n) / sqrt{An}+sqrt{2n}. This limits to zero as required as long as \sum 1/Ak grows asymptotically slower than sqrt{n} --- which it does, because the extra terms are nonnegative, and the sum is O(log n).
x + 1/x = (sqrt(x) - 1/sqrt(x))^2 + 2 >=2
Done.
Why did did he write 1/8 at 10:30?
simple error, he corrects it at 12:04
Looks like a typical Newton-Raphson recursion sequence.
Plugging in x-y'/y=x+1/x,
-dy/y=dx/x
-log(y) =log(x)
y=1/x
So it is the recursion which takes us to the root of the equation 1/x=0, which is infinity. So the sequence diverge to infinity.
A more rigorous way to end up with the same contradiction would be to suppose the limit exists, and then take the limit of both sides of the recursion as n goes to infinity, so x + 1/x = x, thus, 1/x = 0, and therefore the limit does not exist. Which does not imply divergence to infinity, mind. Though that does follow from the lack of a limit *alongside* the monotonicity of the sequence.
(I'm proposing this more rigorous way precisely because your casual disregard for rigor scares me as a more pure math guy)
@@hugh081 so the limit is the root of 1/x=0, that's simple enough, but what in Newton-Raphson guarantees that it's x=∞ instead of -∞? I assume that it's because we want ln(x) to be extended real, so x can't be negative, but just by taking the antiderivative to be ln|x|, this process itself wouldn't tell, right?
@@fahrenheit2101 ? It's not unrigorous
@@TC159 Yes, and no. Yes, it's not unrigorous in that it's not false, and everything (save for the comment about 1/x = 0 at the end, which is blatantly unrigorous) can (hopefully, I'm genuinely unsure) be formalized.
But no, it's still unrigorous in that it's using heavy machinery in a laisse-faire way without much in the way of justification for things. At all.
@@xinpingdonohoe3978 f/f' having constant sign
13:20 This integral should have probably been from 1 to n instead of n-1 since the rectangle with height n-1 goes from x=n-1 to x=n. It makes no difference in the result but it bugged me the same.
No it shouldn't. The rectangle of height 1/k goes from x=k-1 to x=k.
@@ibaijurado279 You're right, I had a dumb moment, disregard.
Isn't a bigger more frustrating problem thatche doesn't justify at all why anyone would ever deduce that the limit as n goes to infinity lf asub n minus nsqrt 2.WHAT THE HELL WHERE THE HELL DOESthat clme fromwhy not asub n tikes sart 2 FOR GODS SAKE..Surely everyone is as grustrated and oizzled by thus as I an since I don't see how anyone would ever think of that no matter how smart you are right? There's just no reason to.
@@leif1075Did you watch the video? The proof was pretty clear..
@@ibaijurado279 Nop. It's true that the last rectangle goes from x=n-1 to x=n but the integral he wrote goes from 1 to n-1, so that last rectangle lies off the limits of integration.
Interestingly a_{n+1}=1+1/a_n does converge to phi!
Which can be shown with a miraculous two line proof
I would love to see that proof. This is as short as I could get mine:
Assuming a_{n+1}=1+1/a_n converges to some value L. As n approaches infinity, a_n and a_{n+1} approach L.
Substituting the terms from the original sequence:
L = 1 + 1/L
=> L^2 = L + 1
=> L^2 - L - 1 = 0
=> L is either phi or -1/phi. Assuming a_0 is positive, we can prove that a_n is positive for all values of n. Therefore L is phi, thus the sequence convergences to phi.
@@ihatesaturn69420 proving that all it's terms are positive doesn't imply convergence so nowhere did you prove the sequence is convergent
@@jAKUB-g2y Really? Well consider me lost dude. Mind posting the proof?
Looks like Euler's method to solve y' = 1/y
Good old L'H shows it's vanishing
Love this guy's videos generally but this solution seems massively over-engineered. You can easily show that a_n will eventually exceed any natural N: if a_n=1/N, so a_(n+kN+1)>N, where k=ceiling(N-a).
He is not trying to show that a_n diveges, but rather that a_n can ve approximated arbitrarily well by sqrt(2n) for large enough n.
Is it me or does that look like newtons method
It's a recursive definition for a sequence that happens to have a form similar to the one in Newton's method.
Would like to extend the prob with n->n+f(n,t) where t is a modular number and f a function that returns tiny positive and negative values either side of n.
Mr. Penn QUESTION PLEASE HOW or WHY would anyone deduce that the limi as n goes to infinty equals zero or even use the linit of 2 swrt of n as opposed to a limit having only a sub n terms..Surely you agree that comes out of NOWHERE and I don't see why anyone would.ever think of it..isnt that just true?? Hope you can respond please when you can.
Can you write your question more clearly? It seems a little unclear what you are asking
I'm not sure i understand your question, but I'd guess there are two underlying questions. (1) Why are we looking at lim(a_n - f(n)) where f(n)=sqrt(2n) and (2) where does sqrt(2n).
The answer to (1) is that for recursive defintions like a_n, we often like to find closed-form (i.e. non recursive) formula for it. That is, an expression of a_n that depends only on n and not on a_(n-1). We don't have that here, but we do have a closed form approximation, that is arbitrarily good as n gets larger. That is one way of interpreting the limit discussed in the video.
For (2), where did sqrt(2n) come from? As far as the video is concerned, it was pulled from thin air. It could just be a lucky guess. But one way to make that guess would ne to plot out a bunch of values of a_n on a graph, and find a function f(n) that approximates that graph. Then, once you have a ghess, try to prove that lim(a_n-f(n)) is 0. If it is, tgen you've found a good approximation of your recursive sequence. Note that there will be infinitely many good approximations, bit generally one would pick the one with the simplest expression (e.g. sqrt(2n) instead of say sqrt(2n+1/n)+e^(-n)).
from Morocco thank you for your wonderful clear complete explanation....i share many of them on my facebook page.....can you please post a video to debunk the paradox π=2 is it similar to your previous dubunking 2=√2
I prefer functional analysis and operator algebra. It is easier to construct things, and make it real.
The right question here is to find an equivalent of a_n as n goes to infty. The fact that it diverges to infty is straightforward, as a_n is increasing and it cannot converge since x+1/x = x has no real solution.
At least there are your videos here hitting me up with something more satisfying than the electoral result 😢...
Trump loves your tears
Did you not notice that this is a limit that does not exist
Please keep politics out of this channel. I just want to enjoy math in peace.
@@terryr9052 That's a *bad* thing
@@fahrenheit2101 to crying Kamala fans, yes... After being called the worst possible names by lefties for years, nobody is interested in being peaceful now.
The function in your thumbnail yields the harmonic series, which we all know diverges to infinity.
drawing a cobweb diagram makes this quite easy to see
This is not a time for any video in view of the horrific presidential result!!!
Not a moment of silence but a week!!!
Your tears are delicious
Shut up bruh.
The whole world does not revolve around USA
I voted for Harris too, but the world must go on. Making videos is this guy's livelihood, you know.
Please keep politics out of this channel. I just want to enjoy math in peace.
If y'all were less silent before the election, maybe you could've garnered some more votes instead of voting that clown into power *again*