I Iove your channel! You cannot use "lim n→∞ a(n) exists iff lim n→∞ a(n) - a(n-1) = 0" because it's not true. Consider for example the harmonic series a(n) = 1 + 1/2 + ... + 1/n. What you can do instead is to take the Cassini formula you just proved and divide it through by F(m)·F(m-1) to obtain r(m+1) - r(m) = (-1)^m / (F(m)·F(m-1)), where r(k) := F(k) / F(k-1), and note that in the above equation the RHS(m) is alternating and the |RHS(m)| ...↓ 0 because F(m) ...↑ ∞. You can now conclude that lim n→∞ r(n) = x for some x in (0,∞). Then dividing F(n-1) + F(n) = F(n+1) through by F(n) yields 1/r(n) + 1 = r(n+1), which in the limit as n→∞ becomes 1/x + 1 = x, etc.
@@gchtrivs7897 Nice example - I think a_n = 1.5 + 0.5sin(ln(n)) would also work. So by Wikipedia, the Leibniz test also requires abs(a_n+1 - a_n) to be monotonically decreasing, ie a_n+2 - a_n+1 is smaller in magnitude than a_n+1 - a_n. He solved for this quantity so it would just take one more step. Seems like that should have been included. Edit: and of course it applies to alternating series so each difference of terms is opposite sign to the previous, he also proved that. Edit 2: It might be confusing applying this infinite sum rule to an infinite product. We are thinking of a_n as the sum of (a_n - a_n-1) + (a_n-1 - a_n-2) + ... + (a_1 - a_0) + a_0 and looking for the limit of a_n as n->infinity, so in that sense it's an infinite sum. At least I think that's the rationale?
I love your videos! You make advanced Mathematical concepts easier to grasp through your logic and it's fun filling in the blanks where we can. Keep it up!
Damn, I'm a freshman in physics and u made me understand this video. I understood everything( or almost haha), u are one of the best math channels I've seen, keep it up, it's great
You didn't actually justify the limit exists by proving the difference of consecutive terms goes to 0. The harmonic series also does that, but clearly does not converge. Simply saying that F_n > n-10^10 is sufficient though (quadratic growth in the denominator).
The Leibniz test he mentions is: “If a_n is a sequence with a_n-a_{n-1} alternating and strictly decreasing in absolute value, then a_n converges to a limit iff a_n-a_{n-1} converges to 0.” It definitely applies here exactly because of the alternating behavior of the numerator, so we’re all clear.
4:30 for what kind of sequences is this leibniz test true ? Because it seems to be false for a_n=n for example (if the limit of a_n can be +inf) And if the limit of a_n has to be finite, for the other implication we can take a_n=ln(n), ln(n)-ln(n-1) converges to 0 but a_n does not converge Am I missing something ?
It holds for the sequence of partial sums of any alternating series. But even under fairly strong hypotheses such as bounded or increasing the statement admits counterexamples.
@@chuckaway6580 "It holds for the sequence of partial sums of any alternating series" Not quite. Consider the partial sums of 1/1 - 1/2! + 1/3 - 1/4! +... The result holds if the successive differences are monotonic decreasing in absolute value.
Any chance you could flip your audio balance? On this and recent videos, as you move or face to the left of the screen (from my point of view) the audio gets louder out the right speaker, and vice versa.
it should be if a_n - a_{n-1} is an alternating sequence with limit 0, then lim a_n exists. This is easy to prove and it's what actually happened here.
2:00 I'd like to know why you choose this method instead of using old-school matrix-vector product. I feel like the latter is way more easy to understand such property: Since F_ == F_ + F_, Let A == | 1 1 | | 1 0 | , and X_ == | F_ | | F_ | , thus A*X_ = | F_ + F_ | | F_ | which is exactly | F_ | | F_ | == X_ So X_ = A*X_ = A^2*X_ = ... = A^k*X_ = A^(k+1)*X_ Solve the last two for the elements in A^(k) and you're good to go.
I don't think your Leibniz criteria is correct. If you take the series: a_(n+1) = an + 1/n if n ∈ [10^(2n),10^(2n+1)] a_(n+1)= a_n - 1/n if n ∈ [10^(2n+1),10(2n+2)] we clearly have lim a_(n+1)-a_n=0 yet, a_n has no limit. You can see that because during the increasing phase, a_n increases by ln(10^(2m+1))-ln(10^(2m)) = ln(10) Similarly, it decreases by ln(10) during the decreasing phase. Hence, a_n oscillates between two boundaries, hence has no limit.
Hey Michael if I have 2 functions defined on R with values in R that are monotone differentiable and their respective derivatives are also monotone does that mean that the ecuation f(x) =g(x) has at most 2 solutions?
This was very hard to follow due to the inverted audio (sounds left when you are facing right and vice versa). It messes with my ability to concentrate. In previous videos you used a mic which worked well. And if you have to use the mic on the camera, consider summing it in mono so there isn't such a wide and inverted stereo image. Thanks for the content!
At the beginning it felt so weird seeing you make the same mistake twice saying "F sub m plus 2" when it's clearly "F sub m plus 1", but overall nice video as always!
Maybe proof with a_n+1-a_n looks suspiciously. there is one way to proof it beatiful and clean. So, F_n=(x^n-y^n)/sqrt(5), where x is (1+sqrt5)/2 and y is -1\x. Hence, F_n+1\F_n is (x^(n+1)-y^(n+1))/(x^n-y^n). Our goal is to proof that this converges to x. In other words, x^(n+1)-y^(n+1) is almost equal to x^(n+1)-y^n*x. i.e. y^(n+1)-y^(n-1) converges to zero. y^(n-1)(y^2-1), but the absolute value of y is less than 1. qed
damn im getting confused by the audio, when you go to the right the left headphone gets louder and vice versa. Can you fix it ? GREAT VIDEO as always looking forward for more
I Iove your channel! You cannot use
"lim n→∞ a(n) exists iff lim n→∞ a(n) - a(n-1) = 0"
because it's not true. Consider for example the harmonic series a(n) = 1 + 1/2 + ... + 1/n. What you can do instead is to take the Cassini formula you just proved and divide it through by F(m)·F(m-1) to obtain
r(m+1) - r(m) = (-1)^m / (F(m)·F(m-1)),
where r(k) := F(k) / F(k-1), and note that in the above equation the RHS(m) is alternating and the |RHS(m)| ...↓ 0 because F(m) ...↑ ∞. You can now conclude that lim n→∞ r(n) = x for some x in (0,∞). Then dividing F(n-1) + F(n) = F(n+1) through by F(n) yields
1/r(n) + 1 = r(n+1),
which in the limit as n→∞ becomes 1/x + 1 = x, etc.
The limit of the harmonic series does exist, so the counter example isn't a counter example. The limit of the*sum* of the harmonic series doesn't.
@@HMWUCSB i think you're confusing "series" and "sequence". the limit of the harmonic sequence is zero. the limit of the harmonic series is +∞.
I don't think lim(a_(n+1)-a_n)=0 implies that a_n has limit.
You re right, an = ln(n) is a counter example.
@@maxime_weill We know that this ratio which limit we want to calculate is =1 for every n>1.
@@leastsignificantbit5069 That's still not enough. Consider a_n = 1.5 + 0.5sin(sqrt(n))
It sounded weird to me too, the harmonic behaves differently, isn't it?
@@gchtrivs7897 Nice example - I think a_n = 1.5 + 0.5sin(ln(n)) would also work.
So by Wikipedia, the Leibniz test also requires abs(a_n+1 - a_n) to be monotonically decreasing, ie a_n+2 - a_n+1 is smaller in magnitude than a_n+1 - a_n. He solved for this quantity so it would just take one more step. Seems like that should have been included. Edit: and of course it applies to alternating series so each difference of terms is opposite sign to the previous, he also proved that.
Edit 2: It might be confusing applying this infinite sum rule to an infinite product. We are thinking of a_n as the sum of (a_n - a_n-1) + (a_n-1 - a_n-2) + ... + (a_1 - a_0) + a_0 and looking for the limit of a_n as n->infinity, so in that sense it's an infinite sum. At least I think that's the rationale?
It is good to know exactly when to stop, thank you Michael c:
I love your videos! You make advanced Mathematical concepts easier to grasp through your logic and it's fun filling in the blanks where we can. Keep it up!
First time I've heard the word "overlining".
Damn, I'm a freshman in physics and u made me understand this video. I understood everything( or almost haha), u are one of the best math channels I've seen, keep it up, it's great
You didn't actually justify the limit exists by proving the difference of consecutive terms goes to 0. The harmonic series also does that, but clearly does not converge. Simply saying that F_n > n-10^10 is sufficient though (quadratic growth in the denominator).
The Leibniz test he mentions is: “If a_n is a sequence with a_n-a_{n-1} alternating and strictly decreasing in absolute value, then a_n converges to a limit iff a_n-a_{n-1} converges to 0.” It definitely applies here exactly because of the alternating behavior of the numerator, so we’re all clear.
Right. You'd still have to prove the alternating behavior though. I did it in my comment above.
GCH Trivs I don’t know what you mean, the difference is clearly alternating because there’s a (-1)^n in the numerator. No thought needed.
@@noahtaul You're right. Didn't notice it.
4:30 for what kind of sequences is this leibniz test true ? Because it seems to be false for a_n=n for example (if the limit of a_n can be +inf)
And if the limit of a_n has to be finite, for the other implication we can take a_n=ln(n), ln(n)-ln(n-1) converges to 0 but a_n does not converge
Am I missing something ?
It holds for the sequence of partial sums of any alternating series. But even under fairly strong hypotheses such as bounded or increasing the statement admits counterexamples.
@@chuckaway6580 "It holds for the sequence of partial sums of any alternating series"
Not quite. Consider the partial sums of 1/1 - 1/2! + 1/3 - 1/4! +...
The result holds if the successive differences are monotonic decreasing in absolute value.
You are a wonderful teacher, especially talented to present and explain hard stuff in an easy way! Where do you find these tricky problems?
Fibonacci grows without bound, this sounds convincing for this problem for sure
that's trivial if lim(f_n+1/f_n)=phi
Please make a video on differentiation of n factorial. I have searched it on internet, but cannot find a satisfying method.
I miss a F2 at the denominator, is it a 1?
Any chance you could flip your audio balance? On this and recent videos, as you move or face to the left of the screen (from my point of view) the audio gets louder out the right speaker, and vice versa.
4:39 that criterion for convergence obviously doesn't hold. Are you testing us, Michael?
it should be if a_n - a_{n-1} is an alternating sequence with limit 0, then lim a_n exists. This is easy to prove and it's what actually happened here.
That's beautiful! I love the initial proofs and how they show up in the final proof. And then we have phi. Gorgeous 😍
Great video again!! Keep up the good work!!
no matter where you go with Fibonacci, the golden ratio will always be there
You are literally my favorite person. Amazing vids.
You are the best teacher of Maths I have ever seen .Keep it up !
4:41
It's supposed to be => , not isn't it?
Why are you writing on a chalk board?
How much should I contribute to make sure you will have the same size marker board?
2:00
I'd like to know why you choose this method instead of using old-school matrix-vector product. I feel like the latter is way more easy to understand such property:
Since F_ == F_ + F_,
Let A ==
| 1 1 |
| 1 0 | ,
and
X_ ==
| F_ |
| F_ | ,
thus
A*X_ =
| F_ + F_ |
| F_ |
which is exactly
| F_ |
| F_ |
== X_
So X_ = A*X_ = A^2*X_ = ... = A^k*X_ = A^(k+1)*X_
Solve the last two for the elements in A^(k) and you're good to go.
The product kinda looks nicer with the -1 put in the numerator
I don't think your Leibniz criteria is correct.
If you take the series:
a_(n+1) = an + 1/n if n ∈ [10^(2n),10^(2n+1)]
a_(n+1)= a_n - 1/n if n ∈ [10^(2n+1),10(2n+2)]
we clearly have lim a_(n+1)-a_n=0
yet, a_n has no limit.
You can see that because during the increasing phase, a_n increases by ln(10^(2m+1))-ln(10^(2m)) = ln(10)
Similarly, it decreases by ln(10) during the decreasing phase.
Hence, a_n oscillates between two boundaries, hence has no limit.
Hey Michael if I have 2 functions defined on R with values in R that are monotone differentiable and their respective derivatives are also monotone does that mean that the ecuation f(x) =g(x) has at most 2 solutions?
This was very hard to follow due to the inverted audio (sounds left when you are facing right and vice versa). It messes with my ability to concentrate. In previous videos you used a mic which worked well.
And if you have to use the mic on the camera, consider summing it in mono so there isn't such a wide and inverted stereo image.
Thanks for the content!
Thank you a lot! Each your video is so nice! Such a beautiful math.
Give me the name of book about the Fibonacci numbers, please
At the beginning it felt so weird seeing you make the same mistake twice saying "F sub m plus 2" when it's clearly "F sub m plus 1", but overall nice video as always!
Very nice work ! I did actually learn lot from you sir. Thanks.
Why didn't you still post a vedio today
Maybe proof with a_n+1-a_n looks suspiciously. there is one way to proof it beatiful and clean. So, F_n=(x^n-y^n)/sqrt(5), where x is (1+sqrt5)/2 and y is -1\x. Hence, F_n+1\F_n is (x^(n+1)-y^(n+1))/(x^n-y^n). Our goal is to proof that this converges to x. In other words, x^(n+1)-y^(n+1) is almost equal to x^(n+1)-y^n*x. i.e. y^(n+1)-y^(n-1) converges to zero. y^(n-1)(y^2-1), but the absolute value of y is less than 1. qed
awesome video!
from Nepal
This video have a strange sound, every time you turn left and speak toward that direction I can hear your voice coming from right and vice versa.
These are for which students either college or school
kindergarten
You are doing Great Job Love from India.
you keep saying Fm+2 instead of Fm+1...kinda confusing sometimes
Nice explanation.
Spiral out!
Great job!
damn im getting confused by the audio, when you go to the right the left headphone gets louder and vice versa.
Can you fix it ?
GREAT VIDEO as always looking forward for more
checked the comments just for this :D Was also watching this while wearing headphones and it made it way harder for me to stay focussed :(
Omg very talented love from India very good keep doing math for fun
شكرا أستاذ 👍
1,68
Great😻😻👌👌👍I am in Middle school in India I have watched every one of your videos
i just cant watch it with that audio.. :(
Вау
I am third commentor of this video ,by the way sir please understand us boldly so that we can understand easy way