Domain and Range of the Infinite Power Tower Function
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- Опубліковано 26 чер 2024
- This is the part 3 of the Infinite Power Tower!
start: 0:00
Idea/example of a fixed point: 2:45
endscreen: 9:20
The graphs of x^x, x^x^x, x^x^x^x, ... / 1
*The Satisfyingly Strange Journey to the Infinite Power Tower*
Part1: Solving x^x^3=2 vs. x^x^3=3 • they don’t teach these...
Part2: Solving x^x^...=2 vs. x^x^...=3 • Infinite Power Tower E...
Part3: Domain and Range of y=x^x^... • Domain and Range of th...
Part4: Why (cbrt(3))^(cbrt(3))^... converges to 2.4 and NOT 3? • Why it doesn't converg...
Reference:
1. mathworld.wolfram.com/PowerTo...
2. arxiv.org/pdf/1908.05559
If you enjoy my videos, then you can click here to subscribe ua-cam.com/users/blackpenredpe...
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blackpenredpen,
math for fun
The range being from e raised to its additive inverse, to e raised to its multiplicative inverse, is really aesthetic.
I LOVE YOU!
I love how honest you were about not knowing the = sign part! That's great. You still had something to teach me and I appreciate it when people don't try to cover up parts they don't know. First video I've seen of yours and I just subscribed!
That blue pen is crying about not being in the channel name. It did so much work for no recognition. We salute you, blue pen.
I really wish that i could give you a hug seriously your the most amazing and kind teacher ive ever seen ....... im about to cry cause you really helped me a lot ....
A question still remains, why did you not already factored pen in blackpenredpen ? sorry dont judge me ... love ya
blackpen + redpen = pen(black + red)
Tecnically, everything in blackpenredpen is multiplying, so it would be
pen²blackred
@@tonaxysam You are assuming abelian though
Black * pen * red * pen = pen^2 * maroon
@@donutman4020 lol
2 of my fav channel
1st one mind your decision and 2 one black pen red pen
🔥🔥
You are my favorite youtuber I never miss your videos
Another mind boggling video!
You really are an amazing teacher! Thank you
Thanks! And you found my secret video!
How the comment 2 months ago...even though video was uploaded 7 hrs ago
@@siddharthsoni2101 patreon, may be.
How did you find the video?
@@puneetmishra4726 what's the meaning of patreon?
I am not sure but I remember 3b1b asked a question about this as a homework in his lockdown math series, so this video answered the question,great!!
Very good presentation!
Nice chain chomp
Can't wait to try this out on my calculator! Thanks for the Interesting video.
Cool beard! I love your chnnel, youre amazing
About what happens at the equalities: the upper limit does not generally belong to the domain (or at least it depends on your starting value), but the lower one does.
Motivation:
We can look at how the function y_n+1 = x ^ y_n behaves to justify that. We look at how a small perturbation eps around the solution behaves (with eps arbitrarily small):
- First, for the upper limit where x = e^(1/e) and y = e. You may check that if y_n = e, then y_n+1 = (e^(1/e)) ^ e = e^1 = e as well, so it is indeed a fixed point, consistent with the video. Let's now look at a perturbed value y_n = e+eps. Then y_n+1 = (e^(1/e))^(e+eps) = (e^(1/e))^e * (e^(1/e))^eps = e * e^(eps/e) = e * (1 + eps/e + 1/2*(eps/e)^2 + ...) = e + eps + 1/2*eps^2/e + ... . So we see that a perturbed y_n by an amount eps is mapped onto a perturbed y_n+1 by an amount eps+1/2*eps^2/e (plus higher-order terms that are negligible for sufficiently small eps). That extra quadratic term in eps is strictly positive. Therefore, negative perturbations eps become less perturbed, but positive perturbations become more perturbed. Therefore the solution is stable from one side but unstable from the other. This is called semi-stable, but that *is not* generally speaking a stable fixed point. (Although it does converge for the case where it is specified that y_1 = x, like in the video, since that approaches from below, albeit slowly.)
- Next, for the lower limit where x = e^(-e) and y = 1/e. You may check that if y_n = 1/e, then y_n+1 = (e^(-e)) ^ (1/e) = e^(-1) = 1/e as well, so it is again a fixed point, as derived in the video. For the perturbed value y_n = 1/e+eps however, y_n+1 = (e^(-e))^(1/e+eps) = (e^(-e))^(1/e) * (e^(-e))^eps = 1/e * e^(-eps*e) = 1/e * (1 - eps*e + 1/2*(eps*e)^2 + ...) = 1/e - eps + 1/2*eps^2*e + ... . So we see that a perturbed y_n by an amount eps is mapped onto a perturbed y_n+1 by an amount -eps+1/2*eps^2*e (plus higher-order terms). At first sight, this seems similar to the previous case. However, interestingly, the sign of the perturbation flips, from eps towards -eps. So also the perturbations flip back and forth between the stable and unstable side and become alternately smaller and larger in magnitude! To resolve the total effect of that, we look at an additional term of the power series to find that y_n = 1/e+eps is mapped onto y_n+1 = 1/e - eps + 1/2*eps^2*e - 1/6*eps^3/e^2 + ... , and repeating the same idea for another iteration this is mapped onto y_n+2 = (e^(-e))^(1/e - eps + 1/2*eps^2*e - 1/6*eps^3*e^2 + ...) = (e^(-e))^(1/e) * (e^(-e))^(-eps) * (e^(-e))^(1/2*eps^2*e) * (e^(-e))^(-1/6*eps^3*e^2) * ... . Keeping only up to third-order terms in eps, this amounts to 1/e * (e^(eps*e)) * (e^(-1/2*eps^2*e^2)) * (e^(1/6*eps^3*e^3)) = 1/e * (1+eps*e+1/2*eps^2*e^2+1/6*eps^3*e^3) * (1-1/2*eps^2*e^2) * (1+1/6*eps^3*e^3) = 1/e + eps - 1/6*eps^3*e^2. Note that the quadratic terms cancel, but when including third-order terms a positive perturbation is mapped onto a slightly smaller perturbation, and a negative one to a slightly bigger one (since the third-order term -1/6*eps^3*e^2 always has the opposite sign compared to eps). So on both sides the perturbation tends towards zero now. Therefore, this *is* a stable fixed point.
Together, the lower limit of the domain converges, but the upper limit only "semi-converges" (which is therefore not generally convergent, and depends on the starting value for y_1). The domain includes the lower limit but not necessarily the upper.
#QED
Just to be sure, I checked this with some numeric computations, and it seems to work out as derived.
Might be a nice followup video in itself?
With apologies for the cumbersome plain text notation... ;-)
Man this is too thorough...
It holds at the upper bound too, though.
@@angelmendez-rivera351 Not if you start the iteration y_n+1 = x ^ y_n from a value y_0 > e. So the upper boundaries convergence depends on your starting point. Try it. That is called semi-convergent.
Dave Langers I don't think that matters at all. Mathematically speaking, we still say it converges at the upper bound, and this is precisely what you will find everywhere in the literature in the subject
By looking at thumbnail I was like..I know this😀..after watching full video I realised I knew only 40% of this..Thank You!🙏🏻👍🏻
Love the Mario Chomper Mic!
4:22 The equality is missing because think of g(x) = f(x) - x. f(x) is the recurrence function.
If g(x_0) = 0 and dg/dx > 0, in physics it would be an unstable equilibrium as if you got away from x_0, the repeated application of the recursion will take you further from x_0, hence it is not a convergent value.
If dg/dx < 0, in physics it would be an stable equilibrium because if you got away from x_0, the repeated application of the recursion will take you towards from x_0, hence it is a convergent value.
But if dg/dx is Zero, it is only converging from one side, but diverging from the other
Edit : 3b1b's video explaining ne it much better, obviously
ua-cam.com/video/CfW845LNObM/v-deo.html
Click here for video ua-cam.com/video/chgnCwcd8oA/v-deo.html
Click here for video ua-cam.com/video/hbj6APyN3Ro/v-deo.html
But the recurrence function is defined at d(g)/dx = 0 . Do we really have to require that the convergence be the same from left and right?
@@valeriobertoncello1809 it's only trying to give an analogy, I have edited in the post a video which explains it properly
@@VaradMahashabde What's the timestamp?
I really do appreciate the fact that you are willing to say that you don't know how to get the the equalities
So many people just pretend they know, and in my opinion that's just bad teaching
Steve sir(blackpenredpen) pls reply to this
Thank u sir for such tricky questions
I am in 8th standard
U made my interest in calculus
I learnt calculus because of u
I am from india
Keep challenging us with such questions
My name is prittish
Yes he indeed taught calculus very nicely, he's been of great help to me as well
You are an amazing inspiration. How about doing some multivariate analysis?
I think that in this video ua-cam.com/video/elQVZLLiod4/v-deo.html 3b1b does a really nice job explaining a nice (visual) solution to the puzzle (especially around 19:55). If you watch it, you immidiately understand why the equalities hold, and not only the inequalities. His approach is based on cobweb diagrams: the infinite power tower converges whenever the cobweb diagram between the functions y=a^x and y=x, starting from the input value x, converges to an intersection point of the two functions. The upper bound of this interval of convergence is found when and where y=a^x is tangent to y=x, i.e. there's only one point of intersection between the two functions in which both have slope 1.
the application of the fixed point theorem is clear for the case z>1; in this case, for zz>e^(-e), and for the bipolar oscillation for z
Amazing
There is a very interesting research paper called “exponential reiterated” that answers your questions among others.
My attempt to show that the infinite exponentiation tower converges for x = exp(1/e). It is based on showing that the sequence is monotonically increasing and bounded, hence it converges. This argument does not work for x = exp(-e), because the sequence is not monotonic. I tend to be verbose (maybe tedious!) so this comment is quite long!
Let define recursively the tetration sequence as: y(1) = x, y(n) = x^[y(n-1)] for n > 1. Note that it's a parametric family of sequences, and we limit to the case 1 ≤ x ≤ e^(1/e).
1a) Let's first show that the sequence y(n) is monotonically increasing, that is y(n+1) ≥ y(n). For the base case (n = 1) we need to prove that y(2) ≥ y(1), which reduces to x^x ≥ x, or x^x - x ≥ 0. This is a calculus exercise, and one can relatively easily show that a) x^x - x = 0 for x = 1 (by direct substitution) and that b) the derivative of x^x - x is given by x^x·[ln(x)+1] - 1 ≥ ln(x) + 1 - 1 ≥ 0 for x ≥ 1 (since x^x ≥ 1 and ln(x) ≥ 0 for x ≥ 1). This implies that x^x - x is non-negative for x ≥ 1, that is x^x ≥ x, which proves the induction base case. One can also double-check graphically using any plotting tool.
1b) Moving on to the general induction step, let's assume that y(n) ≥ y(n-1), and prove that y(n+1) ≥ y(n). We have y(n+1) = x^y(n) ≥ x^y(n-1) = y(n), which is what we want. I have used the induction hypothesis and the fact that the function x^t is monotonically increasing (as a function of t) for a fixed x ≥ 1.
2) Let's now prove that y(n) is bounded; remember that 1 ≤ x ≤ e^(1/e) ≈ 1.445. First, y(n) is obviously always positive, that is bounded from below. We now show, again by induction, that y(n) ≤ e. The base case is trivial, as y(1) = x ≤ e^(1/e) < e. For the general induction step, let's assume that y(n) ≤ e, and prove that y(n+1) ≤ e. We have y(n+1) = x^y(n) ≤ x^e ≤ [e^(1/e)]^e = e^(e/e) = e. This last inequality sheds some light on why the upper bound of the x convergence range is e^(1/e): it is the largest value of x such that x^e ≤ e.
Points 1) and 2) show that y(n) is a bounded, monotonically increasing sequence. Since ℝ is a complete metric space, y(n) converges to a limit L, which concludes the proof. Furthermore, if L = lim y(n) for n → +∞ , by taking the limits for n → +∞ of both sides of the recursive definition of y(n), we have L = lim[y(n)] = lim[x^y(n-1)] = x^lim[y(n-1)] = x^L, which shows that L is the solution to the equation L = x^L. Consistently, this equation has solutions only if x ≤ e^(1/e).
PS: step 1a) would become simpler by defining y(0) = 1 (carrying exponentiation "0 times" yields 1) and y(n) = x^[y(n)] for n > 0. This still recovers y(1) = x as it should be. But the base induction step becomes y(1) ≥ y(0), that is x ≥ 1, which is trivially true. The general induction step does not change because it still holds y(n) = x^[y(n-1)] for n ≥ 1. In point 2), the base induction step is similarly trivial: y(0) = 1 < e^(1/e) < e.
wooo very good
Is this function analytic in this range? If so, can we use analytic continuation to extend it further?
Want daily doses like michael penn
blackredpen Did you know you don't need the chain rule to calculate d/dx(log(9x)? You can just rewrite it as a sum as d/dx(log(9)+ log(x) because of the log property log(ab) = log(a)+ log(b). Log(9) becomes 0 because its a constant but log(x)
becomes 1/ln(10)x. Other than that case and a couple of other cases, I recommend using the chain rule but that is also an optimized strategy.
You are my favorite youtuber
Subscribed
Subbed
Can I suggest the idea for your some future video? This idea was came to my mind at old school years. It is to find the function f(x,t) satisfied following conditions:
f(1,t) = cos(t);
f(2,t) = cos(cos(t));
f(3,t) = cos(cos(cos(t)));
and so on..
an x of course is the real number in common case, not an integer!
So, what about this ffffunction?:)
I have one question, why don't you use a simple microphone which you don't need to carry in your hand? Doesn't this become a bit uncomfortable? Btw, awesome video as always!
thanks man
I am from India. Your videos are excellent
I like watching your videos because I can’t keep up. Does that make sense?
Maybe it's better to think backwards if f (x) is defined over [a,b] it's derivative it's defined over (a,b) it loses ending points
We should get you a bigger white board. Great stuff.
Hi sir how can we suggest you or ask you intresting question's?
Please tell
Why there are comments 2 months ago even the video is 5 hours ago
Yes right...why is that ?
EDIT : now I know, he replied to one of the comments saying that this video was unlisted and he has published it now...👍
@@utsahsharma5335 exactly
The video was unlisted and somebody found the link
이름이나아무 와 bprp 보는 한국인이 있었다니 굉장히 반갑네요 엌ㅋㅋㅋㅋ
@@steve2817 저도 굉장히 반갑습니다 엌ㅋㅋ bprp 나무위키에 있는거 보셨어요? 그거 제가한거임 엌ㅋㅋ
@blackpenredpen Can you explain how the convergence of fixed point rule was derived?
How did you find this video?
but can you integrate this or find length of the curve
Given other requirement we need as viewers, it's quite bizarre that you explained fixpoints to us. interesting video tho
Cool
Wasn't this video posted earlier?
looks like i finally have a new favorite maths equation
What is the marker's brand?
Wow, you are wonderful.
One question from myside plz.
One person forgot his 4 Digit ATM pin but he remembers 4 things about his ATM pin...
1) First Digit is half of Third digit
2) sum of second and third digit is 8
3) Fourth digit is multiple of first & second digit
4) All four digits sum is 12.
So,
Generate the formula to build the ATM PIN
Let abcd be a four digit number, where a,b,c,d are elements of the set of natural numbers + {0} (since no ATM pin is negative or has decimal expansion). Also: first digit is half of the third digit (2a=c); sum of second and third digit is 8 (b+c=8); fourth digit is multiple of first and second digit (d=ab); sum of a,b,c,d is 12 (a+b+c+d=12).
We already have the value of c in terms of a (2a=c), so we can replace c in the second formula (b+2a=8). Solving for b we get b=8-2a.
Then, we replace b in the formula for d to get d=a(8-2a). By expanding the expression we get d=-2a^2+8a.
Next, the sum of all the digits is 12, namely (a)+(8-2a)+(2a)+(-2a^2+8a)=12. By arranging and cancelling terms, we get the quadratic expression -2a^2+9a-4=0
By solving for the values of a you get that a:{1/2, 4}. Since a is a natural number, the only value for a is 4.
Finally, we just plug in a for every digit formula. 2a=c -> 8=c; b+c=8 -> b=0; d=ab -> d=0.
In the end, the ATM pin is 4080
@@kevomtb6882 thanks
So basically this problem involves solving x^y = y for the domain and range where this equation is defined. If you are given y, you can the value for x, that being x = y^(1/y).
What if you are given x? Can you find the value for y?
Yes, you can also use the lambert W function for it.
Is it possible to use analytic continuation or other techniques to extend the infinite power tower function?
Yes.
The analytic continuation is given by -W[-log(z)]/log(z)
Good thank you
I couldn't understand anything but I still clicked the like button, because it's black pen red pen hence it must be good
IN ORDER TO get the inequalites, for the domain, just plug a number bigger than the bigger number.
that was cooool
Good
You are amazing 💙 sir😍😍😍👍👍❤️❤️
Mind blown
If you are oriented to a rigorous theoretical approach to the convergence behaviour, that uses no graphic/computational hints (e.g. coboweb...), a complete work is "Reiterated exponentials" by Knoebel, 1981 (free on web); divergence, convergence and alternate oscillation are rigorously explained using classic real analysis theoretical tools.
An accurate, deep study of the complete proof presented in this article provides all the necessary background if you are interested in a rigorous approach to further generalization of this problem, (e.g. infinite power tower with variable exponents) as is treated by more recent literature(*)
(* see e.g.: F.J. Toledo "On the convergence of infinite towers of powers and
logarithms for general initial data: applications to Lambert W
function sequences")
Lagrange resolvent...Please explain that
You can have a limit of
lim a→∞ (x↑↑a) maybe?
infinity
I mean, if x > 1, (x↑↑a) > x^a, and the limit when a -> ∞ of x^a is ∞, so lim a→∞ (x↑↑a) = ∞ if x > 1. If x
I really missing the marathon series, Lets do it again plzz
Hello blackpenredpen, I looked at math on the internet (cuz it's fun) and I found a video that showed an amazing way to prove Euler's formula, plz put it in a video
Let f(X)=e^(-ix)*(cos(X)+i*sin(X))
We'll take the derivative and after simplification (I won't write it because it will be so annoying to write in a comment) we will get that the derivative is 0 meaning that our original function is a constant c and after plugging into the function X=0 we will get c=1 meaning that:
e^(-ix)*(cos(X)+i*sin(X))=1
Meaning that:
e^ix = cos(X)+i*sin(X)
And that's it, I think that this is so amazing and I hope you put this in a video (plz)
You are just amazing 💖❤️, boss of Math
Please🙏
Click here for mathematics video ua-cam.com/video/chgnCwcd8oA/v-deo.html
Click here for proof of Green's theorem ua-cam.com/video/hbj6APyN3Ro/v-deo.html
Blackpenredpen 😎😎😎 cool
I have a doubt related to integral of √tanx. For the integration of dx/x²-a², we can also use 1/2a ln|x-a/x+a| instead of hyperbolic tangent right.
Edit: please clear my doubt. I hv an exam in my skl tomorrow
Technically speaking, the usage of ln|(x - a)/(x + a)|/(2a) is correct.
5:09 I thought this formula was only applicable if the base were a constant. Can someone explain how he used it for (x^y), even though x is dependent on y?
You're right. I realized that too when I watched it. I've been working on this function on my own for a while before seeing this video. I had different results
超可愛的麥克風
It's been a while since I've taken calculus, can someone remind me why we know the function converges if the absolute value of the function's derivative is less than one?
The magnitude of the deviation from the solution is multiplied by that derivative with each iteration. So if its magnitude is smaller than 1, it means that the deviation decreases and the solution converges (at least when sufficiently near). Similarly, if the derivative is positive, you converge from one side, and if the derivative is negative you hop around from left to right around the solution.
Very intersting arxiv paper , simple enough for first year university students 👍
How did you find this video?
@@pbj4184 Time travel??!!!
@@NikitaNair Someone I asked said they found a link to this video in the description of some other video. Watching videos before they're released is really cool!
Bprp what is the factorial of π
Sorry I'm not understand that how you tell about the fixed point, for x^y =y , you are considering the input as y ,the output is y too, then for the coordinates you just got (x,5) let say y=5 ,but x isn't fixed isn't it ? How come the point fixed ? For one more ,the x should be the input the y would be the output isn't it ? Is the y in the x^y is the input ,while the y on the right hand is output , the whats x suppose to be ?
but y=x/2 has a slope smaller than 1 and it still doesnt converge. shouldnt the slope be zero at infinity for it to converge in infinity?
but y is not infinity so at the border of our domain our range and our slope will be infinity or rather undefined
nice pokemon
Can I ask a question?
I wanted to solve an ODE: y'' + xy' + ky = 0, where k is constant.
My gut says to use Laplace or Fourier Transform method
@@2appleboy Thanks. This brings me nearer to the solution. this is the key to solve 2D wave equation.
You use infinite series too.
Actually I did find that I got the wrong 2D wave equation solution. The r part should look like xy'' + y' + kxy = 0 instead. I found that both Laplace/Fourier Transform don't help me get to the solution I saw in WolframAlpha.
I have to learn more functions before trying.
Moreover, those solutions are non-elementary...3D wave equation is much easier.
3b1b has a Lockdown math video about this infinite power tower
Did you bought the wireless mic???😂😂😂
😂
It's *Chain Chomp* from Super Mario Brothers! Love it! :D
Cool..
Why is it unlisted??
How do you find unlisted videos?
Nice Mic!!!!!
Blackpenredpen this is a challenge for you
Find the value of x when x⁹⁹=99to the power x
Note that x is not equal to 99
I would like to see this solved in few different ways. And at least with Lambert W function. There are few solutions on the internet, but I can't wrap my head around them for some reason. Like this one: math.stackexchange.com/questions/1292773/solving-xy-yx-analytically-in-terms-of-the-lambert-w-function. How does −ln(y) equals −y*ln(x)/x in second last step? What am I missing here?
Found it! here is a good explanantion about the whole solution: ua-cam.com/video/ZVnW6WAM_Co/v-deo.html
Are you looking for all complex solutions, or only one real solution? Because if x is not an integer, then 99^x := exp[ln(99)·(x + 2nπi)] for any integer n. If x^99 = exp[ln(99)·(x + 2nπi)], then x = exp[ln(99)/99·(x + 2nπi)], and x = exp[ln(x) + 2mπi] for any integer m. Therefore, ln(x) + 2mπi = ln(99)/99·(x + 2nπi), 99·ln(x) + 198mπi = ln(99)·x + 2·ln(99)·nπi, implying ln(99)·x - 99·ln(x) = [198m - ln(198)·n]·πi.
But why do you have a chain chomp?
Misss you bprp
wait didnt you already post this video a few months ago?
It was unlisted and I just published it now
@@blackpenredpen How to watch unlisted videos? I've asked others but I thought it would be best if I asked you 😁
AZE..Salam aleykum...Halal olsun əmoğlu😁😊
Can you name a best book of calculus (differentiation,calculus, area related to calclus ,and more .) Which can make me to zero to best?
Please any one @ *blackpenredpen*
I recommend searching for "Paul's Online Notes". It's a great resource that covers everything from Algebra to Differential equations clearly (and for free).
Best of luck!
Doesn't it converge for x=0?
Only 420 people found this secret video?
Ahh, he commented in one of the comments that this video was unlisted and he published it now.
How did you find this video?
Your mic is the best
How did you find this video?
@@shinyeontae When he published it
@@shinyeontae Oh I see. That's nice
Least popular video of BPRP ever after so much time...
Great Video, no crowds at all😄😄😃😁😁😂
This video is unlisted
So you might think that way😅😅😅
How did you find this video?
What about -1 ?
Doesn't d/dy(x^y) equals d/dy(y) which is 1, so the point is not convergent?
Иван Павлов No, because y is not a variable, y is an unknown.
0:52 that's what she said🥺😞😭😭😭
Given an acute non-isosceles triangle ABC with circumcircle Γ. M is the midpoint of segment BC and N is the midpoint of arc BC of Γ (the one that doesn't contains A). X and Y are points on Γ such that BX || CY || AM. Assume there exists point Z on segment BC such that circumcircle of triangle XYZ is tangent to BC. Let ω be the circumcircle of triangle ZMN. Line AM meets ω for the second time at P. Let K be a point on w such that KN || AM, ωb be a circle that passes through B, X and tangents to BC and be a circle that passes through B, X and tangents to BC and ωc be a circle that passes through C, Y and tangents to BC. Prove that circle with center K and radius KP is tangent to 3 circles ωb, ωc & Γ.
Can you solve this question plz...
@Leonhard Euler nope its an igo question (iranian geometry olympiad)
Since when did he have a blue pen
Check the 3b1b's "Power tower puzzle" video
Hey!
Can you please make a video about the Zeta function at irrational number Such as 'e' Or pi''
Please try it
Hey blackpenredpen please reply👍👍 or any body can
Do you mean the Riemann Zeta function ζ(s)? If so, there is nothing interesting about the function evaluated at e or π. In fact, there is nothing to be said about the function evaluated at those values, because no closed form for the function at those values exists.
A very thank you To you
Can anyone explain how did the inequality at 4:15 come from
Any proof??
I really tried a lot to find it
Please someone help
How can we prove the criteria for convergence of fixed point??
🙏🙏🙏
It's bugging me for days
Description
@@Bambani-tr7jw I don't get it brother
It is just showing the criteria
"Absolute value of first derivative of function of y should be less than 1"
@@Bambani-tr7jw how does this conclusion come from
I don't get it
@@jibiteshsaha4392 check the description of the video. He put a reference to an article which should explain it.
Edit: it can be nicely visualised with a 'cobweb' construction (see the second reference in the video description).
@@hetsmiecht1029 I looked into it and I didn't get how the fixed point convergence criteria comes
I just don't want to memorize it and consider it is true
Or else I would be interested in chemistry
I tried but didn't get the specific part of reference in which that proof is given
*You are really my true teacher,I am doing Mathmatics at in India,first year*
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Are you japanese or Russian?