"tuning" a recursive sequence
Вставка
- Опубліковано 19 тра 2024
- 🌟Support the channel🌟
Patreon: / michaelpennmath
Channel Membership: / @michaelpennmath
Merch: teespring.com/stores/michael-...
My amazon shop: www.amazon.com/shop/michaelpenn
🟢 Discord: / discord
🌟my other channels🌟
mathmajor: / @mathmajor
pennpav podcast: / @thepennpavpodcast7878
🌟My Links🌟
Personal Website: www.michael-penn.net
Instagram: / melp2718
Twitter: / michaelpennmath
Randolph College Math: www.randolphcollege.edu/mathem...
Research Gate profile: www.researchgate.net/profile/...
Google Scholar profile: scholar.google.com/citations?...
🌟How I make Thumbnails🌟
Canva: partner.canva.com/c/3036853/6...
Color Pallet: coolors.co/?ref=61d217df7d705...
🌟Suggest a problem🌟
forms.gle/ea7Pw7HcKePGB4my5
12:47 you have a mistake here : 2^n/10^n = (1/5)^n and not 1/5
Yes, actually, in the case a1 > 1/100, the sequence grows to infinity, as he himself said already some minutes before.
Which means, the limit is zero.
He cut too much off of the polynomial expansion.
@@kajdronm.8887 Not necessarily, since it is a strict inequality, even if it tends to 0 it is already bigger than 0. Hop this helps 👍
@@stephomn ??? The only requirement is that the limit is zero...
@@stephomn But why is it a strict inequality?
The argument at the end doesn’t work because the sequence after taking the first two terms of the binomial actually goes to 0 so you haven’t shown it’s greater than 0. You actually need to look at something like the middle term to get the bound you want.
You can probably prove it most easily by taking the ratio of successive terms, which is a^(2^(n-1))/10. Since a > 1, eventually the numerator will be larger than the denominator so the ratio is >1, which shows it diverges to infinity.
Alternative method - take logs of the original recursive sequence. You then have a linear recurrence relation in L_n = log(a_n) which you can solve with standard techniques to get the required result. Taking logs is OK as we start with a_1 >= 0, and by inspection, a_n >= 0 with a_1 >= 0.
That is what I did.
You are one of the best quality math channels, really❤
What I did was actually find the recursion with a slightly intuitive way instead of induction.
Equivalent to solving P(n+1)=10ⁿ P²(n) , It would be very much helpful if it were P(n+1)=P²(n) instead which would straight away give the solution as P(n)= P(0)²^ⁿ
But how can we use this intuition? I let P(n)=Q(n)*P(0)²^ⁿ , where Q(x) is another random function, this helps out a lot because
[P(0)^ 2^n+1] Q(n+1)=10ⁿ *[P(0)^ 2^n+1]Q²(n) which is delightful because the part cancels out and we have basically Q(n+1)= 10ⁿ Q²(n) ;
Now if Q(n) isnt already obvious just take log(base10) on both sides and name the new function Q*(x)=log(Q(x))
Q*(n+1) = n + 2 Q*(n) which can be solved using generating functions
@ 8:52 if we impose that |a1|
The problem specifies that a1 be positive.
after finding the closed form for the sequence , we can also extract the required argument for a_1 by using D'alembert's rule
12:07 Isn’t that lower bound still 0?
Other method. Put alpha = sqrt(1/10 * sqrt( 1/100 * sqrt( 1 / 1000 * ... ))). It is possible to show that the quantity alpha is the only value of a_1 for which the sequence (a_n) does not "explodes very fast" to zero or infinity. Because a_n = sqrt(1/10^n * sqrt(1/10^{n+1} * sqrt(1/10^{n+2} * ... ) ) ) is somehow close to 1/10^n, and the other sequences are more like r^{2^n} where r < 1 or r > 1. It is also possible to show that alpha = 1/100.
If I'm not wrong, we could also have -1/100
Yes but the problem asks for numbers a1>=0
BRO CAN I BORROW YOUR BRAIN FOR MY MATH EXAM ?
Me too please!!
good one ! ( he's got a good personality too!)
If you watch enough of these, yeah probably a part of it
For upcoming jee adv I really want it
Stop yelling your post in all caps.
Well, you failed in the test.😂
First we can factor out `1/(1000*a1)`, which is a finite positive quantity which we can ignore for the purpose of demonstrating divergence to ∞ at the limit n->∞.
Then we're just left with `(1+x)^(2^n) / 10^n`. Taking the logarithm gives `log(1+x)*2^n - n*log(10)`.
For n>1 this is larger than `log(1+x)*2^n / n - log(10)` and can easily be shown to diverge to +∞ for n->∞ and log(1+x) > 0 using L'Hôpital's rule
Canadian Intermediate Math Contest out of Waterloo is actually for grade 9 and 10 students. I remember taking them in high school and doing terribly, and I never saw a question with limits
13:14
My first thought here is; can you just take the log10() on both sides?
bn = log10(an)
b(n+1) = n + 2*b(n)
i wonder if we can do it using generating functions/series
When trying to solve this problem, I tried to figure out what those exponents 0,1,4,11,26,57 were in closed form. They came from the recursion b_(n+1)=2b_(n)+n (at least that's one way to write the idea...0x2+1->1x2+2->4x2+3->11x2+4 etc).
This felt like a naturally occurring recursion! It didn't seem to increase a lot faster than 2^n, so 1/100 was the intuitive upper bound...but what was the closed form?
I wrote the generating function and tried to get that to help (narrator: it didn't), quickly OEIS'd it, which gave me Eulerian numbers. Turning to Wikipedia, it had a closed form for for this particular case.
I was hoping Michael would do a constructive proof of this closed form, instead of a "hey look at the pattern" induction. Looking back, I should have done the same (since it clearly was something off of a power of 2), but is there a simple way to "constructively" get this closed form?
Generating functions do work, it’s a pretty simple recursion. But it’s probably easier if you know how to solve these non homogeneous recurrences, eg see this article: www.math.hkust.edu.hk/~mabfchen/Math2343/Recurrence.pdf
I agree, I find constructive proofs are much more satisfying. I'm not really familiar with generating functions but this recursion results in whats called an "Arithmetico-geometric sequence" because its a combination of an arithmetic and geometric sequence.
In the wikipedia page of Arithmetico-geometric sequences they have a constructive proof of the closed form under "Sum of the terms". It essentially boils down to using the sum of a geometric sequence twice.
Yes it is.
Write a few beginning and a few ending terms of b_n one under the other, multiply from bottom to top by 2^1, 2^2... so that, adding, you reduce telescopically.
You will get a sum of k*2^(n-k). Then extract 2^n in the front of the sum and the sum of k*(1/2)^k remains.
You extract another 1/2 in front of the sum, so 2^(n-1)*the sum of k*(1/2)^(k-1) remains.
Now, the sum of k*x^(k-1) is the derivative of the sum of x^k. The sum of x^k is the sum of a geometric progression, write the formula, then the derivative, then replace x with 1/2. And so on.
Or, you can reindex n-k=t to work with 2^t, not (1/2)^k.
Strict inequalities are not preserved by limits !!!!! The reasoning at the end is false
Last time i was this early