In this video I showed how to formulate and solve basic infinitely nested radicals using examples. This is the first in a series of planned videos on this topic.
One of the best explanations ever. Mathologer channel showed a weird expression that Ramanujam came up with. It didn’t show how to solve it, instead it showed how it was constructed. Thank you!
anytime I search a problem and I see this page pop up on RUclips , I'm always grateful to God because I know that I will understand the topic...Thank you so much!!!!
Pretty Simple Problem x = the radical given it gives rise to x^2 + 6 - x -> x^2 + x - 6 = 0 -> (x - 2)(x + 3) = 0 That gives us x = 2 / x = -3 As the given expression can never give rise to a negative answer hence the answer is 2.
@@m.h.6470 Here is a very simple explanation to that. Just consider two terms and neglect all the terms for now. We get a = Root (6 - root(6)) which is less than 2 now we subtract this number a from 6 then again root it so it becomed a bit more than 2 but the deviation from 2 is lesser than before. This process keeps on going and whenever we get odd number of 6 , the outcome become more than 2 and whenever we have even number of 6 we get less than 2. But as the number of 6 keepn on increasing the deviation keeps on decreasing so finally as number of 6 approaches infinity , the deviation approaches zero and answer is 2.
@@Pramit1156 yes, I get that it is simple. I did it myself, but you need to mention it. Otherwise you may end up with bogus solutions, as the term may diverge and then the variable is 'infinity' and calculations become nonsensical...
when i was on the math team in HS (decades ago), there was a nested radical problem. I remember it because I was the only one on my team that got it. I tend not to remember when it went the other way. It is interesting when you see a type of problem for the first time and are able to solve it, you get a rush of chemical in your brain that makes you feel so satisfied. And the rush seems stronger when others didn't get it. Maybe it is just pride.
I remember how I at the age of 14 figured out how to calculate the square root by hand, i.e. I used a calculator but only basic algebra functions. Then I figured the idea worked for any integer root. Actually, it even worked for fractional roots as long as the reciprocal of the decimals is an integer. I felt so proud and that rush you described
It might even help to look at it in terms of the exponent of n being a sum of an infinite series, so it’s n^(1 + 1/2 + 1/4 + 1/8 + 1/16 + ...), which is n^2.
Properly, convergence should be established. Define partial sums Sn and then take limit as n--> infinity and it readily converges to 2. S1 = sqrt(6) S2 = sqrt(6-sqrt(6)) S3 = sqrt(6-sqrt(6-sqrt(6))) Sn = sqrt(6-S(n-1)) where the (n-1) is a subscript of course. If the limit is assumed 2, define delta: delta = abs[2-Sn] < p where p > 0 is an arbitrarily small real (epsilon) for all n > N. delta converges monotonically to 0 to establish the existence of sufficiently large N for all possible p > 0.
You're absolutely right and moreover, the sequence S(n) converges to 2 for any arbitrary positive value chosen for S(1) . The only problem is that not every nested radical may be easily expressed (if at all) as an infinite sequence, which brings the question, whether nested radicals are actually mathematical objects.
Hey I saw this really cool quirk of nested radicals, if you do Sqrt(x + sqrt(x +sqrt(x+...))) etc, and evaluate it for X = 0, you get 1, or atleast limit as x -> 0 So an infinite sum of square roots of 0 is equivalent to 1
Numbers seem pretty simple to understand, until you start thinking about infinity. Like how if you add, for example, ...666666 +...333334 your would just get ...000000. Is the sum equal to infinity because it is an infinite number or is this infinity equal to 0 because it is just infinite 0s. Zeros have an absolute value of 0 and add no value to a number unless there is a number with place value is eventually put infront of it, which never happens in this case.
I only saw the thumbnail and solved it in what I think is a simpler way, let's give an example with 6, sqrt(6-sqrt(6-.... = x, sqrt(6-x) = x and then it becomes surprisingly easy, just square both sides or even just plug in numbers because these are small numbers
For the root 6 nested radical don't we need to prove that k≠0 and how can we say that k doesn't diverge Root(6*root6) is a greater number than root 6 so it's not equal to zero but how can we prove that it doesn't diverge to infinity
At 12:25 you make an unproven assumption, that k converges. Of course, WE know, that it converges, as you constructed it, but if we didn't know that, you can't just decide it to be k. You are only allowed to calculate with infinite series, if you are sure, that they converge, otherwise your variable might be 'infinity' and all calculations will result in non-sense.
I tried to add Pi*n (why not) instead of just n and surprisingly got two possible answers, but how can it be? generation formula look like sqrt(n(n-Pi)+Pi*sqrt (...) etc. For n=3 I get the following nested radical sqrt(9-3*Pi+Pi*(sqrt(...))), solving which I obtain the equation k^2-Pi*k+3*Pi-9=0, which gives two solutions k=3 - integer, the intended one and Pi-3 - the irrational, but positive, I can't get what's wrong with it? there cannot be two values of the same real-value radical
Solution: Given, that √ of something is always positive and that 6 > √6, we can assume, that the term converges somewhere between 0 and (6 - √6). As it converges, we are allowed to replace and calculate with infinite series: x = √(6 - √(6 - √(6 - ...))) → x = √(6 - x) And this can simply be solved: x = √(6 - x) |² x² = 6 - x |+x -6 x² + x - 6 = 0 x = -1/2 ± √((1/2)² - (-6)) x = -1/2 ± √(1/4 + 24/4) x = -1/2 ± 5/2 Negative solution (-1/2 - 5/2 = -6/2 = -3) doesn't make sense → extraneous solution x = -1/2 + 5/2 = 4/2 = 2 Therefore the original term converges on 2
You're correct. That's what I always say when solving equations. Unless division by the variable is the only way, don't even consider it. And you must first show k can not be 0. Which is the case here.
One of the best explanations ever. Mathologer channel showed a weird expression that Ramanujam came up with. It didn’t show how to solve it, instead it showed how it was constructed. Thank you!
anytime I search a problem and I see this page pop up on RUclips , I'm always grateful to God because I know that I will understand the topic...Thank you so much!!!!
Pretty Simple Problem
x = the radical given
it gives rise to x^2 + 6 - x
-> x^2 + x - 6 = 0
-> (x - 2)(x + 3) = 0
That gives us x = 2 / x = -3
As the given expression can never give rise to a negative answer hence the answer is 2.
You can only do that, once you have proven, that it converges however. Otherwise it is not allowed to calculate with infinite series.
@@m.h.6470 Here is a very simple explanation to that.
Just consider two terms and neglect all the terms for now. We get a = Root (6 - root(6)) which is less than 2 now we subtract this number a from 6 then again root it so it becomed a bit more than 2 but the deviation from 2 is lesser than before. This process keeps on going and whenever we get odd number of 6 , the outcome become more than 2 and whenever we have even number of 6 we get less than 2. But as the number of 6 keepn on increasing the deviation keeps on decreasing so finally as number of 6 approaches infinity , the deviation approaches zero and answer is 2.
@@Pramit1156 yes, I get that it is simple. I did it myself, but you need to mention it. Otherwise you may end up with bogus solutions, as the term may diverge and then the variable is 'infinity' and calculations become nonsensical...
@@m.h.6470 But this proves that if this limit exists, it must be equal to 2.
@@boguslawszostak1784 IF the limit exists, yes. But ONLY if. If the limit diverges, the result may still be the same, but would be wrong.
Newton you are a joy to watch. Love to see a maths teacher with a constant bounce in his step.
I'm a high school math teacher, and this is something I never learned in school
Thank you very much for this, sir.
when i was on the math team in HS (decades ago), there was a nested radical problem. I remember it because I was the only one on my team that got it. I tend not to remember when it went the other way. It is interesting when you see a type of problem for the first time and are able to solve it, you get a rush of chemical in your brain that makes you feel so satisfied. And the rush seems stronger when others didn't get it. Maybe it is just pride.
Frrr, that's the reason why I love math so much😂
Gotta love endorphins.
I remember how I at the age of 14 figured out how to calculate the square root by hand, i.e. I used a calculator but only basic algebra functions. Then I figured the idea worked for any integer root. Actually, it even worked for fractional roots as long as the reciprocal of the decimals is an integer. I felt so proud and that rush you described
It might even help to look at it in terms of the exponent of n being a sum of an infinite series, so it’s n^(1 + 1/2 + 1/4 + 1/8 + 1/16 + ...), which is n^2.
This is very better to learn compared to comercial teaching classes without any matter
This is pretty cool! I've seen these many times and i rarely 'got' them. You've just made them palatable! ❤
Great video. Congratulations from Barcelona-Catalonia.
Thanks
Another brilliant explanation!
intriguing and very usual, as always. Keep up this awesome work please.
Once you know the trick, you can solve these in your head in no time.
For the product one, it can be further be seen by taking it to be x^1/2 x^1/2² x^1/2³ ...
=x^(1/2+1/2²+1/2³+...)
=x
Properly, convergence should be established. Define partial sums Sn and then take limit as n--> infinity and it readily converges to 2.
S1 = sqrt(6)
S2 = sqrt(6-sqrt(6))
S3 = sqrt(6-sqrt(6-sqrt(6)))
Sn = sqrt(6-S(n-1)) where the (n-1) is a subscript of course.
If the limit is assumed 2, define delta:
delta = abs[2-Sn] < p where p > 0 is an arbitrarily small real (epsilon) for all n > N. delta converges monotonically to 0 to establish the existence of sufficiently large N for all possible p > 0.
This is the only valid proof. Thanks.
You're absolutely right and moreover, the sequence S(n) converges to 2 for any arbitrary positive value chosen for S(1) . The only problem is that not every nested radical may be easily expressed (if at all) as an infinite sequence, which brings the question, whether nested radicals are actually mathematical objects.
Crazy !!
Thanks for your videos.
Nested radicals sounds like a band name.
Hey I saw this really cool quirk of nested radicals, if you do
Sqrt(x + sqrt(x +sqrt(x+...))) etc, and evaluate it for X = 0, you get 1, or atleast limit as x -> 0
So an infinite sum of square roots of 0 is equivalent to 1
Wow, radicals always made me nervous. Not anymore. Thanks!!
What about convergence? Such things are OK if the "series" converges.
Very good. Thanks 🙏
Love your videos!
Please make a video on Linear Transformations. I really need your help for that topic 😭
Wow genial la explicación
The radical of the beast!
y=sqrt(6-y)
y^2=6-y
y^2+y-6=0
(y+3)(y-2)=0
y=-3 or 2, but the original problem suggests that y is positive, so y=2
Numbers seem pretty simple to understand, until you start thinking about infinity. Like how if you add, for example, ...666666 +...333334 your would just get ...000000. Is the sum equal to infinity because it is an infinite number or is this infinity equal to 0 because it is just infinite 0s. Zeros have an absolute value of 0 and add no value to a number unless there is a number with place value is eventually put infront of it, which never happens in this case.
It was interesting, good job!
The answer is 2..We can apply -1 +√4(6)+1 over 2.
I only saw the thumbnail and solved it in what I think is a simpler way, let's give an example with 6, sqrt(6-sqrt(6-.... = x, sqrt(6-x) = x and then it becomes surprisingly easy, just square both sides or even just plug in numbers because these are small numbers
For the root 6 nested radical don't we need to prove that k≠0 and how can we say that k doesn't diverge
Root(6*root6) is a greater number than root 6 so it's not equal to zero but how can we prove that it doesn't diverge to infinity
The laugh at the end though💀💀☠️.
Commenting so you get recommended by RUclips.
Reply to do the same.
In the introductory example, why would it be 'a disaster' to divide both sides by k?
Just don't want students to do it. It's a bad habit. But okay in this video, since kis not 0
At 12:25 you make an unproven assumption, that k converges. Of course, WE know, that it converges, as you constructed it, but if we didn't know that, you can't just decide it to be k. You are only allowed to calculate with infinite series, if you are sure, that they converge, otherwise your variable might be 'infinity' and all calculations will result in non-sense.
"those who stop learning stop living" whoa
8:41 Well that was unexpected!
I tried to add Pi*n (why not) instead of just n and surprisingly got two possible answers, but how can it be?
generation formula look like sqrt(n(n-Pi)+Pi*sqrt (...) etc.
For n=3 I get the following nested radical sqrt(9-3*Pi+Pi*(sqrt(...))), solving which I obtain the equation k^2-Pi*k+3*Pi-9=0, which gives two solutions k=3 - integer, the intended one and Pi-3 - the irrational, but positive, I can't get what's wrong with it? there cannot be two values of the same real-value radical
Ok, now I see , there's always negative under the root, I wonder how it still gives the correct answer next to the wrong one))
I like it😊😊😊
Very beautiful!?
Solution:
Given, that √ of something is always positive and that 6 > √6, we can assume, that the term converges somewhere between 0 and (6 - √6).
As it converges, we are allowed to replace and calculate with infinite series:
x = √(6 - √(6 - √(6 - ...)))
→
x = √(6 - x)
And this can simply be solved:
x = √(6 - x) |²
x² = 6 - x |+x -6
x² + x - 6 = 0
x = -1/2 ± √((1/2)² - (-6))
x = -1/2 ± √(1/4 + 24/4)
x = -1/2 ± 5/2
Negative solution (-1/2 - 5/2 = -6/2 = -3) doesn't make sense → extraneous solution
x = -1/2 + 5/2 = 4/2 = 2
Therefore the original term converges on 2
Easy !!!!
w2oooooooooooooo a COW!!!!!!
why 6?
👍👍👍😁🤪👋
sqrt(6 x sqrt(6 x sqrt(6 x ....
The Number of the Nested Beast !!!!
😨😨😨
7:42 Dividing by k wouldn't have been a disaster huh 😂
You're correct. That's what I always say when solving equations. Unless division by the variable is the only way, don't even consider it. And you must first show k can not be 0. Which is the case here.
@@PrimeNewtons yeah absolutely very bad idea most of the time
I solved it in like 15 sec n the video was 17 min
Nested and naked radicals ?🤣🤣🤣🤣🤣🤣
👍👍👍😁🤪👋