finding a recursive formula and its limit

[@meltossmedia]

I passed my test thanks to your 100 series video. Thank you, sensei.

[@blackpenredpen]

Meltoss You must also put in a good amount of work to study!! So great job to you as well!!

[@MrRyanroberson1]

the implicit formula also allows you to generalize to all indexes, a_0 = 2, a_-1 = -1/2. The recursive formula agrees: a_0 = (-4-1)/(-9/2+2) = -10/(-9+4) = 10/5 = 2. a_1 = 15/20 = 3/4

[@ffggddss]

Nice! BTW, I did the recursive formula slightly differently, just because it was the first thing that occurred to me. In order to get a whole number I could pull out of the fraction, I went: 3a[n] = (3n+6)/(3n+1) = 1 + 5/(3n+1) Then pick that apart one operation at a time, to this level: 3a[n] - 1 = 5/(3n+1) 5/(3a[n] - 1) = 3n + 1 Then subs. n+1 for n in the first line, and use what we just found: 3a[n+1] = 1 + 5/(3(n+1) + 1) = 1 + 5/(3n+1 + 3) = 1 + 5/([5/(3a[n] - 1)] + 3) Multiply top & bottom of the "big" fraction by (3a[n] - 1): ... = 1 + (15a[n] - 5)/(5 + 9a[n] - 3) = 1 + (15a[n] - 5)/(9a[n] + 2) = (24a[n] - 3)/(9a[n] + 2) Finally, divide by 3: a[n+1] = (8a[n] - 1)/(9a[n] + 2) Essentially the same; just different turns onto different roads, to wind up at the same hotel. ;-) Fred

[@blackpenredpen]

Ah I see!! Now my challenge is to go from recursive to explicit.

[@elvinsamedov547]

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[@arynbhar]

```WE LOVE MATHEMATICS```

[@OtiumAbscondita]

@@arynbhar *COMMUNISM INTENSIFIES*

[@elvinsamedov547]

Where is my 💗 😟

[@gustavoalexanderma8587]

@@elvinsamedov547 ❤

[@OtiumAbscondita]

@@elvinsamedov547 nowhere

[@cameronspalding9792]

I’ve just realised: if we have a recursive relationship in the form a_n = f(a_n-1), f differential at L where L is a limit for a_n and |a_n-L| is of order 1/n^m where m>0 then |f’(L)|=1

[@duggydo]

This is so cool! This is something we never covered in calculus classes. I feel like the teachers must have skipped over it.

[@virendrasule3258]

6:42 but how does this method work if the recurrence of the sequence is not first order and also when an explicit expression for a_n is not available?

[@junba1810]

Finding the limit doesn't require you to know the explicit formula. So you don't need it at all.

[@matheussales4861]

If I have the recursive formula A(n)=sqrt(A(n-1)+1), with A0=1, how do I deduce it's explicit form? Fun fact, the limit of this sequence when n goes to infinit is the golden ratio ^-^.

[@angelmendez-rivera351]

Matheus Sales I may be wrong, but I believe you cannot deduce the explicit form using elementary functions, or even standard special functions.

[@archithtelukunta4599]

Matheus Sales A(n) = sqrt[1+sqrt[1+sqrt[1+...........sqrt [1+sqrt(2)]]]] .There would be (n-1) “sqrt”s

[@angelmendez-rivera351]

GaMeR 123 Correct, but this would not be expressible using elementary functions.

[@enricodibenedetto6959]

You don't have to deduce the explicit form to calculate this limit. Firstly, notice that your succession, as it's defined, is increasing (you can prove it with induction or you can simply notice that because the function sqrt(x+1) is an increasing one and a1=sqrt(2)>a0=1 so it follows that an is constantly increasing). Now, for a theorem on monotone succession, if the succession is bounded it converges, if it's not it diverges. In this case, you can prove by induction that 2 is bigger of all the terms of the succession. Now, finally, you can say that an converges to one of the solution of the fixed point equation x=sqrt(x+1). Solving it you obtain two solution a positive one and a negative one. Obviously, you have to pick the positive one and it happens to be the golden ratio. Hope to be useful. Have a good night :)

[@archithtelukunta4599]

Angel Mendez-Rivera Induction FTW!:)

[@andresxj1]

So, for recursive formula, if the sequence converges, the limit doesn't depend on the initial value?

[@blackpenredpen]

Andy Arteaga It depends! Sometimes you will get two answers (more maybe more) when you do what I did in the video. You may need to use the initial condition to determine which L it is!

[@ambigramm]

You need to learn about the mandelbrot aNd julie set. This fractal is Based around the Initial value for recursive formulars

[@juniormontalvan4062]

You deserve more subscribers man.

[@NoNameAtAll2]

He will converge to more subs

his subscribers will diverge

[@blackpenredpen]

Thank you!!

[@Qoow8e1deDgikQ9m3ZG]

why don't use a_n - a_n-1 or a_n / a_n-1 to get the recursive function?

[@mrandersonpw53]

What if, proving properly the sequence if bounded and monotone of course, the cuadratic formula gives us 2 positive solutions bounded by our sup of inf. Which one Should I choose? The closest one to the bound? Is there a bifurcation?

[@adam_salha]

Hey! I love your videos, you inspired me to study mathematics in college! I recently had an idea, if we have two fixed points A and B, and wanna see where C is such that angle ACB = 90° we can simply draw the circle of diameter [AB]. But what if the angle was different than 90°, how would the shape behave? If it's 30° for example or 45°!

[@scarbotheblacksheep9520]

Circle around the plus? That's news to me.

[@blackpenredpen]

check out Ford circle, by Numberphile. They have a nice video on it.

[@pierreabbat6157]

To check the recursive formula: a=2; (8a-1)/(9a+2)=15/20=3/4. a=5/10=1/2; (8a-1)/(9a+2)=3/6.5=6/13.

[@beegdigit9811]

2:51 What does he mean by putting a circle around the plus?

[@keescanalfp5143]

it's what bprp says to it. never permitted to assume that for instance 1 2 3 - + - gets - . it's always a false step. 2 3 5 **

[@darkforcekiller]

Shouldnt the explicit formula be n+3 / 3n+4 since the first index is always 0 ( a0)

[@blackpenredpen]

Suk Mike Hawk it depends on the convention.

[@flowerwithamachinegun2692]

Suk Mike Hawk in the example above the sequence started with a_1. A sequence doesn't always have to start with a_0.

[@ranadeepkundu5357]

Some authors consider 0 as a natural number, some don't.

[@benjamimapancake6429]

I read in a book that 0 is a natural number, but whole numbers start at 1.

[@duckymomo7935]

Can’t you use Cauchy sequence Or maybe monotone convergence

[@sergioh5515]

When you solve for L, isnt that coincidentally like solving for the auxiliary equation? And cool video btw as always! Very fun of course:D

[@josephhajj1570]

Your subscribers will diverge mr Blackpenredpen

[@Archik4]

a0=2 recursive work for n=1 and give 3/4

[@damianmatma708]

Thank you for this great video :) I have a question related to your example: Suppose we have given the recursive formula: a.1 = 3/4 a.n = (8 * a_n-1 - 1) / (9 * a_n-1 + 2) Now I want to get the explicit formula (we know we should get a.n = (n+2)/(3n+1) ). Is it possible? If so, how to do that step-by-step? It would be great if you could make another video (as Part 2 or as a continuation) with my problem solved.

[@blackpenredpen]

I actually plan to make such videos in the spring. It's usually *not* easy to go from rec. to exp. For this one, I would suggest to just write out the first few terms, then look for a pattern for it. That's how I would go about it.

[@damianmatma708]

@@blackpenredpen Thank you very much :)

[@josephhajj1570]

But how can we prove it bounded (given recursive) by induction???

[@angelmendez-rivera351]

Joseph Hajj a(1) = 3/4 < 1. Now, suppose that a(n) < 1 Then a(n + 1) = (8a(n) - 1)/(9a(n) + 2) < 1 for all a(n), since, if we let m = a(n), then 8m - 1 = 9m + 2 implies -3 = m, which means that for -3 < m < 1, 9m + 2 > 8m - 1. Therefore, a(n) < 1 for all n.

[@mikeschieffer2644]

Nice video! Could you also show that asub(n-1) converges to 1/3 as n goes to infinity by using the intermediate equation you have of asub(n-1) = (n+1)/(3n-2)? Thanks for all you do!

[@angelmendez-rivera351]

Mike Schieffer Yes. Remember that a(n - 1) is just a(n) shifted by 1. Limits of sequences are shift invariant.

[@kalunlee5854]

but how can we find the explicit formula from the first term and the recursive formula?

[@prasaddash5139]

It's awesome in itself

[@SR-kd4wi]

Are there cases where recursion is simpler than explicit method?

[@sensei9767]

In most cases the recursive formular is easier to derive than the explicit one. In those cases, calculating the limit this way might be easier than finding the explicit formular.

[@angelmendez-rivera351]

Autologist Z Yes. In the vast majority of situations, one cannot express the solution of a recursive formula using elementary or known functions, in which case there is no explicit formula to begin with.

[@jhonatanandres4148]

Excelent video, do you have any video about conical sections? (Special Problems, from graphic to general formula, the opposite...?)

[@angelmendez-rivera351]

Jhonatan Andres Try checking the channel

[@jhonatanandres4148]

I think i found it

[@khbye2411]

"When n->inf, a_n goes to L. So, a_(n-1) also goes to L." I never really learnt about limits. But could anyone tell me why the a_(n-1) also goes to L? I don't understand why?

[@khbye2411]

I was wondering if a_(n-1) can tend to L, then what about a_(n-2) (third last term before the 'last' ??infinite?? term)? Could a_(n-2) also go to L?...then i thought why not generalise it to a_(n-k) to cover all the possible terms (4th last term, 5th last term...). So, I rewrote a_n in terms of a_(n-k) instead and assumed that a_(n-k) also tends to L and solved for a quadratic equation in L....and I got a repeated root where L=1/3. Like is there a reason why there's only one root? How did I even end up with L/3L from all this random stuff hahaha. Is this enough to prove that a_(n-1) will converge to L?

[@saharhaimyaccov4977]

How we yse integral to cos (tanX)

[@НиколайШерстюк-ы7е]

Turn on captions

[@jomama3465]

Hoping to see an analytic geometry vid on your channel :))

[@nestorv7627]

How would you use this method to prove the ratio between 2 consecutive fibonacci numbers?

[@BigDBrian]

f(n) = f(n-1) + f(n-2) f(n)/f(n-1) = [ f(n-1) + f(n-2) ] / f(n-1) f(n)/f(n-1) = 1 + f(n-2) / f(n-1) note that f(n-2)/f(n-1) is the multiplicative inverse of f(n-1)/f(n-2). taking the limit as n->inf and assuming convergence, we can say that f(n)/f(n-1) = f(n-1)/f(n-2) = L substituting the earlier equation gives L = 1 + 1/L solving this familiar equation gives phi as the positive solution.

[@angelmendez-rivera351]

N S F(n) = Aφ^n + B(-φ)^(-n), so F(n + 1)/F(n) = [Aφ^(n + 1) + B(-φ)^(-n - 1)]/[Aφ^n + B(-φ)^(-n - 1)]. Now, lim F(n + 1)/F(n) = lim Aφ*φ^n/[Aφ^n] = lim φ = φ.

[@angelmendez-rivera351]

Also, F(n + 1) = F(n) + F(n - 1). Divide by F(n) to get F(n + 1)/F(n) = 1 + F(n - 1)/F(n). Now let lim F(n + 1)/F(n) = L, so L = 1 + lim F(n - 1)/F(n) = 1 + lim F(n)/F(n + 1) = 1 + 1/L, so L = 1 + 1/L. Solve for L, and the positive solution is L = φ.

[@mohittiwari8934]

Looking so smart bro 😍

[@VoteScientist]

6:14 the suspense was killing me.

[@rezamiau]

Great!!! thank you.

[@VaradMahashabde]

How to get the explicit formula from the recursive? 😕😕

[@jayapandey2541]

Black Pen Red Pen, Yay!

[@someperson188]

Given the first five terms of a sequence doesn't determine the formula for the sequence. There are infinitely many formulas for the sequence given and the limit can be anything (including +-infinity) or the sequence can have no limit. Here are some examples. (1) Let A be any real and a_n = (n+2)/(3n+1) + (A-1/3)(n-1)(n-2)(n-3)(n-4)(n-5)/n^5. Then a_1 = 3/4, a_2 = 4/7, a_3 = 5/10, a_4 = 6/13, a_5 = 7/16, and the limit of a_n as n approaches infinity is A. (2) Let a_n = (n+2)/(3n+1) +B(n-1)(n-2)(n-3)(n-4)(n-5), where B = 1 or B = -1. Then a_1 = 3/4,. . . , a_5 = 7/16, and the limit of a_n as n approaches infinity is infinity (resp., -infinity) if B =1 (resp., -1). (3) Let a_n = (n+2)/(3n+1) + ((-1)^n)(n-1)(n-2)(n-3)(n-4)(n-5)/n^5. Then a_1 = 3/4,. . . ,a_5 = 7/16, and the limit of a_n as n approaches infinity doesn't exist. Blackpenredpen is assuming that the formula for a_n is (in some sense) simplest, which I agree is the quotient of two linear functions.

[@Qoow8e1deDgikQ9m3ZG]

you missed something , he assume limit exists ....... don't be stupid..... and all theory tends to use the simplest trial solution, no one interested in your solution .... why not extend your stupid solution to complex number too .... stupid ....

[@someperson188]

@@Qoow8e1deDgikQ9m3ZG I'm replying to you in a civil manner because, unlike you, I am attempting (with great difficulty) to be civil. You obviously missed (or didn't understand) the last sentence of my comment (i.e., What does "simplest" mean? It's not easy or perhaps possible to answer, is it?). Further, in example (1) (resp., (2)), I construct a formula a_n, such that a_1, a_2 ,..., a_5 take the given values and the limit is an arbitrary real number A (resp., infinity). So assuming the limit exists doesn't help your case. My objection isn't new; it's well-known to mathematicians, but clearly not to you. I regret that my anger is showing. I won't answer any further comments from you if your comment is again impolite.

[@cristiancojocaru7546]

You are so cool !

[@BigDBrian]

explicit formula: fuck recursive formula: recursive formula

[@muhammetislambedirbeyoglu3135]

Just waow 👏👏👏

[@sheikhmansur3402]

As always ᵗʰⁱˢ ʷᵃˢ ᵖʳᵉᵗᵗʸ ᶜᵒᵒˡ 👍👍👍

[@seroujghazarian6343]

3/4+4/7+5/10+6/13+... diverges 😋

[@supriyajyoti22]

IIT PE VIDEO'S

[@maginobionrequiem9167]

Cool

[@bhuvird178]

This video u do well. Do something better in maths. U can create new formula my dear just do my dear

[@blackpenredpen]

Bhuvi Rd We have a new formula called the Steve-Peyam theorem! But this comment section is too small so I cannot share it with you.

[@bhuvird178]

U proved the formula

[@blackpenredpen]

Bhuvi Rd Yea. Do you have your own formula too?

[@bhuvird178]

Yes I have RD formula that is the formula in calculus but I don't prove that formula

[@habouzhaboux9488]

First!