There's a mistake. The lower bound on the summation is greater than 1 for all (finite) n>1. This means that for arbitrarily large finite n, the first term in the summation is the m=2 term. Therefore, in the limit as n goes to infinity, the summation starts at m=2, not at m=1 as in the video. Said differently, passing a limit into the bound of a discrete sum is not always justified. This should make the summation evaluate to 1/(2n) rather than 1/(3n), so that the value calculated for the "first half" is 1, not 2/3, so the solution to this "evil question" is unfortunately just 1000.
Wow, so this the mother load. First of all, thank you for the comment always glad for the input. I have been going over this for quite some time over the past few days and to be honest can't really find a direct resolve on the interwebs. The short answer is that this seems to work fine with wolfram. The way I'm interpreting the situation is that using (n to root of n... or n^1/n) when used as the index of the sum (or lower bound) has to be greater than one, therefore automatically becoming 2, when we want the limit as n approaches infinity. So I have understood limits to actually flatten the result, 1/n as n->∞ is just zero not 0.0000000000...0001. But clearly some issue is at hand. I want to be clear that I'm not dismissing this comment at all. Just the opposite, I'm really curious. Could you please maybe elaborate or point to some internet link specifically addressing this issue. Do I even understand the concern correctly? Thank you very much for the engagement, I'm super curious regarding this. Glad you watched my video and got this involved. Looks like this indeed is an "evil equation" 😉
@@Roman_CKThe limit of 1/n as n→infinity is indeed 0 because it's the number it approaches. In our case, n^(1/n) does indeed approach 1, however it will never quite get to 1 for any value of n, making it so the lower bound of the sum is never quite 1. Overall, this can just be fixed by adding another floor function around it. There's never too many floors!
Pretty cool equation! I played a little bit with the evil equation on python to observe how this function approaches this limit. Regarding the lower bound of the summation, I considered two cases: floor and ceiling of the number (so that a sum from 1.5 to 5 becomes a sum from 1 to 5 if using floor or from 2 to 5 if using ceiling). If we use the floor, the functions seems to be approaching 3000 as n tends to infinity, while using celing (my prefered choice) gives us a limit of a thousand. It also looks like second case approaches the limit quicker (since it is scaled down compared to the first function). Anyway, although these observations were trivial, I really liked this problem! It made me have fun with this equation and python. I also had some concerns with the floor and ceiling function (since they are not continuous and I dont think that floor lim = lim floor).
Well I have to say. This video seems to be creating some controversy regarding some technical aspects of the math. Really fascinating, glad the people are getting this involved. Seems like I may have to tweek this limit equation, possibly make "evil equation 2". Sounds interesting.
Oh, this comment section really turned out quite critical 😅 Don't take it too harshly, it's the nature of matematicians to argue about rigor :) If you want and need any help with your next "insane equation" idea, you can ask me! I am majoring in applied maths right now, so I might know some technicalities :D
Yes, @darkking571 just below had a similar concern. I agree limit being inside is a different thing than the limit being outside. In this instance I was hoping it wouldn't make a difference but I don't actually know for sure. And I'm hoping to remove the floor functions all together. But thanks for the comment, I'm happy to see people get involved like this. 🙂
I liked the video,it truly encaptures that an equation looking intimidating at first sight could just be a simpler one,aside from that there were minor problems as pointed out by others which in my opinion don't steal away the concept of the video, anyways continue with the good work.
@@spookyleo2589 it's an expression. You could argue that the end result = initial expression is an equation, but that's not what the title and the comments mean.
Extremely cool, but you need to justify the step of lim(f(g(n))) = f(lim(g(n))) (Thats whats happening, when you "look at some part of the equation") And, as discrete sum with changing boundaries is not continuous, we cant do that at all, sadly Also, it's not clear how would summation work with fractional boundaries Aside from that, your video ideas are pretty cool! Keep it up! I don't wanna discourage you from making them ❤
Thanks for the comment. I agree, moving the limit in and out of the floor function isn't really the thing to do. It isn't a step, I was merely contemplating in which position it would look more interesting. To be honest I'm hoping to get rid of the floors and maybe ceilings all together since they seem to be rather problematic. Yes, I hope to keep up my video as well, thanks for engaging. Always welcome. 😉
Fractional boundaries don't confine to the definition of the summation operator, hence its necessary to include a floor operator alongside the lower bound for the summation to even have some meaning to it. Otherwise it just an illegal operation.
Exactly, the video shows steps which are not true in general, i.e. exchanging limits and operators/functions. A large part of analysis is devoted to prove in which situations this can be done. Unfortunately a very bad video except to get the ahs and ohs of the innocents.
You have to be real careful when interchanging limits and floors. The floor of the limit to a series that approaches an integer n from below is n. The limit of the floor of the same series is n-1. So this is totally not trivial
I was close. I read the -n/n as just being applied to the numinator and assumed that n/the nth root of n! Approached inf very slowly so I got the floor of the sum from m= 1 to inf of 2,000/m which is inf
When you say technically it's not illegal, how does eliminating the parentheses stil apply the exponent to the whole term and not only to the m that it's directly adjacent to?
NOOO! 3:19 you can pull the n out but on the screen you put it outside!!! You saved it by reputting it in right after, but for that step it was actually illegal.
3:12 is a complete joke. You cannot apply the limit just for the denominator, wtf If you rewrite you have n^(1-1/n)/m and it tends to infinity... 3:14 you cannot pull out the -1 in the exponent out of the sum... I found 3000n^(1 - 1/n) at the end, so it tends to infinity
So..... you omitted the parentheses in your equation, which completely changes the summation and where the exponent is applied (originally it looks like it's only applying to the 'm'), and yet you ask us to attempt to solve it? Wtf dude.
There's a mistake. The lower bound on the summation is greater than 1 for all (finite) n>1. This means that for arbitrarily large finite n, the first term in the summation is the m=2 term. Therefore, in the limit as n goes to infinity, the summation starts at m=2, not at m=1 as in the video. Said differently, passing a limit into the bound of a discrete sum is not always justified. This should make the summation evaluate to 1/(2n) rather than 1/(3n), so that the value calculated for the "first half" is 1, not 2/3, so the solution to this "evil question" is unfortunately just 1000.
you beat me to it
agreed
Wow, so this the mother load. First of all, thank you for the comment always glad for the input. I have been going over this for quite some time over the past few days and to be honest can't really find a direct resolve on the interwebs. The short answer is that this seems to work fine with wolfram.
The way I'm interpreting the situation is that using (n to root of n... or n^1/n) when used as the index of the sum (or lower bound) has to be greater than one, therefore automatically becoming 2, when we want the limit as n approaches infinity.
So I have understood limits to actually flatten the result, 1/n as n->∞ is just zero not 0.0000000000...0001. But clearly some issue is at hand. I want to be clear that I'm not dismissing this comment at all. Just the opposite, I'm really curious. Could you please maybe elaborate or point to some internet link specifically addressing this issue. Do I even understand the concern correctly?
Thank you very much for the engagement, I'm super curious regarding this. Glad you watched my video and got this involved.
Looks like this indeed is an "evil equation" 😉
@@Roman_CKThe limit of 1/n as n→infinity is indeed 0 because it's the number it approaches. In our case, n^(1/n) does indeed approach 1, however it will never quite get to 1 for any value of n, making it so the lower bound of the sum is never quite 1. Overall, this can just be fixed by adding another floor function around it. There's never too many floors!
I have trouble comprehending non-integer sum bound at all...
Pretty cool equation! I played a little bit with the evil equation on python to observe how this function approaches this limit.
Regarding the lower bound of the summation, I considered two cases: floor and ceiling of the number (so that a sum from 1.5 to 5 becomes a sum from 1 to 5 if using floor or from 2 to 5 if using ceiling).
If we use the floor, the functions seems to be approaching 3000 as n tends to infinity, while using celing (my prefered choice) gives us a limit of a thousand. It also looks like second case approaches the limit quicker (since it is scaled down compared to the first function).
Anyway, although these observations were trivial, I really liked this problem! It made me have fun with this equation and python.
I also had some concerns with the floor and ceiling function (since they are not continuous and I dont think that floor lim = lim floor).
Well I have to say. This video seems to be creating some controversy regarding some technical aspects of the math. Really fascinating, glad the people are getting this involved. Seems like I may have to tweek this limit equation, possibly make "evil equation 2". Sounds interesting.
Oh, this comment section really turned out quite critical 😅
Don't take it too harshly, it's the nature of matematicians to argue about rigor :)
If you want and need any help with your next "insane equation" idea, you can ask me! I am majoring in applied maths right now, so I might know some technicalities :D
bro summoned the devil
hell yess
the floor function is not continuous so you cannot take the limit in and out
Yes, @darkking571 just below had a similar concern. I agree limit being inside is a different thing than the limit being outside. In this instance I was hoping it wouldn't make a difference but I don't actually know for sure. And I'm hoping to remove the floor functions all together. But thanks for the comment, I'm happy to see people get involved like this. 🙂
Give bro the evil equation 💀
just wanted to be a litte evil
I’m very skeptical of taking limits of the individual pieces. Seems very non rigorous.
nothing wrong with you
2:33 I agree you are evil but,it is indeed illegal. If a notation is ambigious it is illegal to not clarify it. You can't omit those sorry.
I liked the video,it truly encaptures that an equation looking intimidating at first sight could just be a simpler one,aside from that there were minor problems as pointed out by others which in my opinion don't steal away the concept of the video, anyways continue with the good work.
Exchanging limits with upper limit of summation is quite the thing - it works until it doesn't.
We called this engineering level maths.
Should have called it devil equation to make it more cool!
nah it should have been called the "nevil equation"
It's an expression, not an equation...
@@beatn2473 it is an equation
@@spookyleo2589 it's an expression. You could argue that the end result = initial expression is an equation, but that's not what the title and the comments mean.
_D_evil
Nicee workk, keep it up
thanks
Ceil and Floor functions are used in python libraries They are mainly math functions
Those damn ridly snakes...
Weirdly enough... I now understand the floor and ceiling function because of this video. Maybe you should try adding mod next time for the funsies
Sadly enough... me too... thanks
Extremely cool, but you need to justify the step of lim(f(g(n))) = f(lim(g(n)))
(Thats whats happening, when you "look at some part of the equation")
And, as discrete sum with changing boundaries is not continuous, we cant do that at all, sadly
Also, it's not clear how would summation work with fractional boundaries
Aside from that, your video ideas are pretty cool! Keep it up! I don't wanna discourage you from making them ❤
Thanks for the comment. I agree, moving the limit in and out of the floor function isn't really the thing to do. It isn't a step, I was merely contemplating in which position it would look more interesting. To be honest I'm hoping to get rid of the floors and maybe ceilings all together since they seem to be rather problematic. Yes, I hope to keep up my video as well, thanks for engaging. Always welcome. 😉
Fractional boundaries don't confine to the definition of the summation operator, hence its necessary to include a floor operator alongside the lower bound for the summation to even have some meaning to it.
Otherwise it just an illegal operation.
Exactly, the video shows steps which are not true in general, i.e. exchanging limits and operators/functions. A large part of analysis is devoted to prove in which situations this can be done.
Unfortunately a very bad video except to get the ahs and ohs of the innocents.
The Evil's number.
You have to be real careful when interchanging limits and floors.
The floor of the limit to a series that approaches an integer n from below is n.
The limit of the floor of the same series is n-1.
So this is totally not trivial
Theres too many n 💀
Take a random problem and write it without brackets to make the problem 1000× more difficult
Got to the end and was SHOCKED at your sub count. You’re gonna be huge one day, and I’ll be able to say I was a pioneer!
Sounds great. Yeah I'm kind of surprised about the subs too, never saw it coming.
same
Should’ve used the gamma function instead of a factorial
Great work
It's very nice
Do we use Riemann sum here?
we might as well
That was a nice one!
Ah always a pleasure Nacao, glad you like it. 😄
This is a limit, not an equation
my limit is the equation...
I was close. I read the -n/n as just being applied to the numinator and assumed that n/the nth root of n! Approached inf very slowly so I got the floor of the sum from m= 1 to inf of 2,000/m which is inf
n/nnnnnnnnnnnnnnnnn
When you say technically it's not illegal, how does eliminating the parentheses stil apply the exponent to the whole term and not only to the m that it's directly adjacent to?
Lol lazy math writing, ab/c = a(b/c)
NOOO! 3:19 you can pull the n out but on the screen you put it outside!!! You saved it by reputting it in right after, but for that step it was actually illegal.
bro, n(1+2)=1n+2n. Just use the distributive property. It's extremely easy.
@@ElectricGamer_YT NOOO! you can't do that when there's an exponent!!! You need to care of the exponent.
3:12 is a complete joke. You cannot apply the limit just for the denominator, wtf
If you rewrite you have n^(1-1/n)/m and it tends to infinity...
3:14 you cannot pull out the -1 in the exponent out of the sum...
I found 3000n^(1 - 1/n) at the end, so it tends to infinity
2:22 how did you do that?
I just tried in mathjax, but if you do engulf whole sum under that power, the -n/n would've been much much higher
that's probably my only complaint with this sum - placement is too conventionally-on-m
it's inreadable in other ways
n/nnnnnn....
Woww mannnnnnn😮😮
Thanks
ervil lebaron type equation
😅😅😊
Very dubious equation
Check jee advanced math you'll definitely got quality content 😀
the math solid, questions r liquid
sqrt side is 1000
what happened at 4:39? is 1/3 a tetration? huh?
No, 1/3 is the index of the radical.
Lol that'd be crazy, but sadly fractional tetration is still not well defined 😔
In 3:14 that summation should be 1/n + 1/(2n) = 3/(2n) you made a mistake, so the answer is 3000
No way I was the 666th like!!!
Oh shit bro, just missed it (667)
@@Roman_CK Awwww...so close
So..... you omitted the parentheses in your equation, which completely changes the summation and where the exponent is applied (originally it looks like it's only applying to the 'm'), and yet you ask us to attempt to solve it? Wtf dude.
That's because I'm evil 😈
@@Roman_CK Well, I can't argue with that 😂 got me there.
this video has 666 likes!!!!
It did at some pint!
lol
Congrats on first comment.