I believe it will continue to work, but it will not be as simple as the previous one. x = sqrt(-20+9sqrt(-20+9sqrt(-20+9sqrt(-20+9sqrt(-20+9sqrt(-20+9sqrt(ta da da da)))))
(x-5)(x-4)=0 x^2-9x+20=0 x^2=9x-20 x=sqrt(9x-20) NOTE: x=sqrt(9x-20) So x=sqrt(9(sqrt(...)-20)-20) Looks very tricky! Please finish this for me smart RUclipsrs
Let x=sqrt(20+sqrt(20+sqrt(20+...))). Since x=sqrt(20+*sqrt(20+sqrt(20+...))*), then we can write x as sqrt(20+x) So x² = +-(x+20). But x is positive because sqrt(positive stuff) is always positive, so we can write x² = x+20 --> x²-x-20 =0. We can both solve by factoring (x-5)(x+4)=0 and considering only the positive answer x=5 or by the quadratic formula x1,2 = [-(-1)+-sqrt((1)²-4(1)(-20))]/2(1) = [1+-sqrt(81)]/2, but we already know x>0, so we only consider (1+9)/2=5. In both cases we end up with x=5. But we defined x=sqrt(20+sqrt(20+sqrt(20+...))), so sqrt(20+sqrt(20+sqrt(20+...)))=5
I am 1 year late but for me this video is new. There are 2 roots for the expression (x-5)(x-4)=0, they can express as x=sqrt(-20+9*x). in the previous case we have a notable difference x= +-sqrt(-20+9*x) there are 2 different algorithms for 2 different cases therefore if we pick + or - we find the 5 or -4, but in this case we have one algorithm with 2 different solutions (Notice that if you pick the - root in the first case you change the x to -x in the second case is the almost the same but in the first case we have 20+5 and 20-4 both roots are at different sides of 20 in the real number line. Now we have 20+45 and 20+36 both at the same side from 20 in the real line). Also is obvious that we cannot do the infinite nested root we need to chop the root at some point, and we have something light this x=sqrt(-20+9* sqrt(-20+9* sqrt(-20+9* sqrt(-20+9*----- sqrt(-20+9*m))))), and this is very, very important. This means that in order to calculate x we need an initial value of m. in the first case the algorithm converges for any real value of m at 5 or -4. (even if some roots are complex, the imaginary part goes to 0 and the real part to 5 or -4). But in the second case for a real m that is not 4 the algorithm converge at 5, the interesting case is that 5 is an stable root, any number and the system converge at 5, but 4 is the unstable root is like an inverse cone in equilibrium the minimal disturbance and instated of going back to 4 the algorithm converges at 5
The process of generating this "nested square roots" helped me to get the answer (may be I'm not very sure it will work for all real nos.) to a long unanswered question that "for all real nos. there is a sequence of rationals and a sequence of irrationals that converges to that real no." THANKS
Proof that all numbers are equal: X= Sqrt(x^2)= Sqrt(1+x^2-1)= Sqrt(1+sqrt((x^2-1)^2)) Apply the sequence to the inner sqrt Sqrt(1+sqrt(1+sqrt(((x^2-1)^2-1)^2))) ... Sqrt(1+sqrt(1+...))=x It doesn't matter what x is
We can not put x= sqrt( 9x - 20 ) into infinite series, because when we start to calculate it and see what is the value that x approaches to, it will be undefined.
What about alternating between adding and subtracting? sqrt(20 + sqrt(20 + sqrt(20...))) is 5, and sqrt(20 - sqrt(20 - sqrt(20...))) is 4 I thought sqrt(20 - sqrt(20 + sqrt(20 - sqrt(20...)))) would be somewhere between 4 and 5, but it's actually ~3.887
It's much more intuitive from the other direction: We want to represent any real number x as an infinite nested square root. x = sqrt(a + sqrt(a + ...)) = sqrt(a + x) x^2 = a + x a = x^2 - x Otherwise it is entirely unclear why you would start off by multiplying with (x+4), even if it makes sense in the end.
This is absolutely true, but be careful. If x < 0 then we have to subtract the roots. For example, (-4)^2 - (-4) = 20, but -4 = -sqrt(20 - sqrt(20 - sqrt(20 - ...)))
x = a, (x-a)(x+a-1) = 0, x^2- x - a(a-1) = 0, x = sqrt(a(a-1) + x), let a(a-1) = b, a^2 - a - b = 0, (a - 1/2)^2 = b + 1/4, a = 1/2 + sqrt(b + 1/4), sqrt(b + sqrt(b + sqrt(b + ...))) = 1/2 + sqrt(b + 1/4) ----> the answer to the nested Michael Jordan problem! taking b = 1, we get 1/2 + sqrt(5)/2 = the golden ratio.
Back in 1975, my math teacher gave the following problem: x = sqrt(1+ sqrt(2 + sqrt(3 + … sqrt(1975)))) Find x. Nobody in class could solve it. So the teacher showed how to solve it. I didn't get it. Can anyone solve this?
Good explanation. I like teaching this stuff. Infinite nested fractions, square roots, powers, etc. and talking about both what they mean as sequences and how we get solutions with algebra.
Fun fact - you could use stick figures as actual variables during a math test if the variable refers to a human being. The teachers are probably not gonna like it, but it would make perfect algebraic sense - variables are just placeholders anyway.
Do you know by this method which you did like substituting x in the same equation, you can find the value of -1 ! and it will be wrong as then it will give pie = app. -4 so this sqrt 20.... is wrong
2:38 how is it "anything up to u" .. Cause in the LHS , square root of X squared is mod of X .... Therefore it can be only |X| = +√(20+X).....I maybe wrong
So if you have just: x = √(A + √(A + √...)), where A > 0, you can just go: x = √(A + x) x² - x - A = 0 x = ½(1 ± √[1+4A]) . . . but since x is necessarily > 0, x = ½(1 + √[1+4A]) And if you started with: x = √(A - √(A - √...)), then x = ½(-1 + √[1+4A]) Interesting observation, for the x = √(A + x) case: If 0 < A
Can it be rewritten like this as well? x²-x-20 =0 x = x² -20 x = (x²-20)²-20 x = ((x²-20)²-20)² - 20 And show that the expression obtained from "infinite substitution" converges to 5? If it doesn't converge, then is 5 not a solution? I suspect otherwise
well, proving the series is bounded from above by 5 is pretty easy by induction. To prove it is monotonically increasing we need a(n+1)-a(n)>0, you can multiply and divide by conjugate to see it implies a(n)-a(n-1)>0 also.
How do we know that the sequence is increasing? The adding 20 part of course works but you are square rooting. How do we prove that the adding 20 is more than the root subtracting?
When I was in calc I we called the theorem you mention around 9:30 "Bolzano's lemma", but I can't find it by that name on the internet, so I ended up calling it "MB criterion", where MB stands for Monotone-Bounded. Do you call it in any particular way?
The problem with your initial steps is you took a formula that defined one root for x and added a x = -4 root. This breaks the expansion as solely being an expansion of x = 5.
Yes I wanted to ask how is this true? x=5 graphed is just a line perpendicular to the x axis and crosses the x axis at 5 So how can -4 also be a root of that? Also we're multiplying here by a variable x isn't that wrong? Aren't we going to change the whole line into a parabola? I don't have the maths to prove that multiplying by a variable not a constant is wrong but I know dividing by a variable is since it loses us a root (which is always root zero) What i mean is let's assume we havex³+5x²+6x=0 I cannot divide by x because then I'll lose the x=0 when I factor it out So why isn't that the case here as well? Why is it okay to multiply by x and add another root???
could someone please tell me from what branch(calculus 1/2/3/4; linear algebra... you prabably know more than I) of math is this from. just please tell me I'mm dying to knowwwww. @blackpenredpen, anybody... is there a branch of math that's called (what blackpenredpen said on the last part of the video)"mathematical analysis"(I don't really know) and what would be the prerequisites?
Hello, blackpenredpen I'm a seventh-grade student in Sydney and for a math test, we were given a very confusing math problem. No one seemed to have any idea how to solve it. The question went as follows: Let A be a 2018-digit number which is divisible by 9. Let B be the sum of all digits of A, and C be the sum of all digits of B. Find the sum of all possible values of C. Do you happen to know a way to solve this?
I'm confused why the 20+x becomes 20-x for the series that converges to -4. I tested it out and it works, but I don't see the algebraic justification...
You can do it by induction. First, notice that x1 = √20 ≤ 5, so that's the base case. Now, assume x(n-1) ≤ 5. We want to show, based on this assumption, that xn ≤ 5. Since we have x(n-1) ≤ 5, we can add 20 to both sides to get 20+x(n-1) ≤ 25 Taking the square root of both sides, we get √(20+x(n-1)) ≤ 5 But by definition, xn = √(20+x(n-1)) So we have xn ≤ 5. So we get each term is bounded above by 5, and that's what it means for the sequence to be bounded above by 5.
At first we have x-5=0 then you multiply both sides by (x+4) I wanted to ask why this is okay to do? x=5 graphed is just a line perpendicular to the x axis and crosses the x axis at 5 So here we're multiplying by a variable x but isn't that wrong? Aren't we going to change the whole line into a parabola? I don't have the maths to prove that multiplying by a variable not a constant is wrong but I know dividing by a variable is since it loses you a root (which is always root zero) What I mean is let's say we have the equation x³+5x²+6x=0 I cannot divide by x because then I'll lose the x=0 when I factor it out So why isn't that the case here as well? Why is it okay to multiply by x and add another root???
What if we had (x-5)(x-4)=0? Can we still do it?
So that means x=5,4?😮
I mean... 5=4😮
I believe it will continue to work, but it will not be as simple as the previous one. x = sqrt(-20+9sqrt(-20+9sqrt(-20+9sqrt(-20+9sqrt(-20+9sqrt(-20+9sqrt(ta da da da)))))
The Juk. No* it doesn't.
Edit: *Spelling error: Changed "Np" to "No"
(x-5)(x-4)=0
x^2-9x+20=0
x^2=9x-20
x=sqrt(9x-20) NOTE: x=sqrt(9x-20)
So x=sqrt(9(sqrt(...)-20)-20)
Looks very tricky!
Please finish this for me smart RUclipsrs
Square root of TWAANTY
Are you saying Michael Jordan is a radical?
Steve the Cat Couch lol, yea!!
He wants to seize the means of dunk production. It explains why he had so many free throws too!
The means of produnktion
No, look again. He's saying MJ is *under* a radical.
Fred
your signature is missing an obscure quote
so we can confirm that sqrt(20+sqrt(20+sqrt(20+....))is 5
So when my little brother asks what's 2+3 I'll tell him it's infinite 20+ square roots and link him this video
you are right
dou you know the gaussian integral?
@@brianbritez2212 lol
Let x=sqrt(20+sqrt(20+sqrt(20+...))). Since x=sqrt(20+*sqrt(20+sqrt(20+...))*), then we can write x as sqrt(20+x)
So x² = +-(x+20). But x is positive because sqrt(positive stuff) is always positive, so we can write x² = x+20 --> x²-x-20 =0.
We can both solve by factoring (x-5)(x+4)=0 and considering only the positive answer x=5 or by the quadratic formula x1,2 = [-(-1)+-sqrt((1)²-4(1)(-20))]/2(1) = [1+-sqrt(81)]/2, but we already know x>0, so we only consider (1+9)/2=5.
In both cases we end up with x=5.
But we defined x=sqrt(20+sqrt(20+sqrt(20+...))), so sqrt(20+sqrt(20+sqrt(20+...)))=5
you are entertaining people with math, congratulations :)
You can do these easily with a calculator. First calculate sqrt(20). Then enter sqrt(20+Ans) and hit the = or EXE repeatedly.
point is not using the calculator and figuring it out with logic
You would think people watching maths videos would understand that
When life gives you dilemmas, make dilemmanade!
lol, why does this not have thousands of likes
sqwarut of twaanty, sqwarut of twaanty !!
So cute !!
Aviral Aryal
U cute too
You have a penguin as a profile pic?
I just recommended you to my cat. She started purring afterwards. 😍
12:34 my parrot knows you voice, the have heart you when I see your videos!
Induction:
x{1}=sqrt(20) =x{n)
I am 1 year late but for me this video is new.
There are 2 roots for the expression (x-5)(x-4)=0, they can express as x=sqrt(-20+9*x). in the previous case we have a notable difference x= +-sqrt(-20+9*x) there are 2 different algorithms for 2 different cases therefore if we pick + or - we find the 5 or -4, but in this case we have one algorithm with 2 different solutions (Notice that if you pick the - root in the first case you change the x to -x in the second case is the almost the same but in the first case we have 20+5 and 20-4 both roots are at different sides of 20 in the real number line. Now we have 20+45 and 20+36 both at the same side from 20 in the real line). Also is obvious that we cannot do the infinite nested root we need to chop the root at some point, and we have something light this
x=sqrt(-20+9* sqrt(-20+9* sqrt(-20+9* sqrt(-20+9*----- sqrt(-20+9*m))))), and this is very, very important. This means that in order to calculate x we need an initial value of m. in the first case the algorithm converges for any real value of m at 5 or -4. (even if some roots are complex, the imaginary part goes to 0 and the real part to 5 or -4). But in the second case for a real m that is not 4 the algorithm converge at 5, the interesting case is that 5 is an stable root, any number and the system converge at 5, but 4 is the unstable root is like an inverse cone in equilibrium the minimal disturbance and instated of going back to 4 the algorithm converges at 5
Hi blackpenredpen. I juste want to say that I love yours videos and I understand everything. I'm French, and I'm only 15 years old. You're the best!
Thank you!!!! : )
I'm 12 and still understand
I've also love Blackpenredpen's videos, I'm french, and I've also 16 years old :)
Joe Makhoul so am I.
@@annonyme8529 Yees France en force 💪
The process of generating this "nested square roots" helped me to get the answer (may be I'm not very sure it will work for all real nos.) to a long unanswered question that "for all real nos. there is a sequence of rationals and a sequence of irrationals that converges to that real no."
THANKS
"Nested root, not nasty root!" :D
: )
Proof that all numbers are equal:
X=
Sqrt(x^2)=
Sqrt(1+x^2-1)=
Sqrt(1+sqrt((x^2-1)^2))
Apply the sequence to the inner sqrt
Sqrt(1+sqrt(1+sqrt(((x^2-1)^2-1)^2)))
...
Sqrt(1+sqrt(1+...))=x
It doesn't matter what x is
We can not put x= sqrt( 9x - 20 ) into infinite series, because when we start to calculate it and see what is the value that x approaches to, it will be undefined.
What about alternating between adding and subtracting?
sqrt(20 + sqrt(20 + sqrt(20...))) is 5, and sqrt(20 - sqrt(20 - sqrt(20...))) is 4
I thought sqrt(20 - sqrt(20 + sqrt(20 - sqrt(20...)))) would be somewhere between 4 and 5, but it's actually ~3.887
It's much more intuitive from the other direction: We want to represent any real number x as an infinite nested square root.
x = sqrt(a + sqrt(a + ...))
= sqrt(a + x)
x^2 = a + x
a = x^2 - x
Otherwise it is entirely unclear why you would start off by multiplying with (x+4), even if it makes sense in the end.
This is absolutely true, but be careful. If x < 0 then we have to subtract the roots.
For example, (-4)^2 - (-4) = 20, but -4 = -sqrt(20 - sqrt(20 - sqrt(20 - ...)))
x = a,
(x-a)(x+a-1) = 0,
x^2- x - a(a-1) = 0,
x = sqrt(a(a-1) + x),
let a(a-1) = b,
a^2 - a - b = 0,
(a - 1/2)^2 = b + 1/4,
a = 1/2 + sqrt(b + 1/4),
sqrt(b + sqrt(b + sqrt(b + ...))) = 1/2 + sqrt(b + 1/4) ----> the answer to the nested Michael Jordan problem!
taking b = 1, we get 1/2 + sqrt(5)/2 = the golden ratio.
When he makes a mistake, the mathematical theorems get updated
Back in 1975, my math teacher gave the following problem:
x = sqrt(1+ sqrt(2 + sqrt(3 + … sqrt(1975))))
Find x.
Nobody in class could solve it. So the teacher showed how to solve it. I didn't get it.
Can anyone solve this?
Just square both sides about 1975 times. Move everything to one side. Then find the roots of a polynomial beginning with x^(2^1975). Simple!
/s
That expression gives the Nested Radical Constant which is 1.75793275...
proof that sqrt(0+sqrt(0+sqrt(0+...))) converges to 1
Ok. I actually had a video on that but I had a mistake. It's about time to remake!
Its not increasing nor converging, sqrt(0) is not less than sqrt(0+sqrt(0)), all of that is 0
@@bluestar140 Yes. But what about the limit when e approaches 0+ of sqrt(e+sqrt(e+...)) ...? I would guess that to be 1.
x = √(0+x)
x^2=x
x(x-1)=0
(x-1)=0
x=1
Wow
Grain of salt mandatory when dealing with infinities.
"Recommend me to ... dogs, cats, fish", Oreo's ... :D
Good explanation. I like teaching this stuff. Infinite nested fractions, square roots, powers, etc. and talking about both what they mean as sequences and how we get solutions with algebra.
For An No. A it Can Be Said As
sqrt(a²-a+sqrt(a²-a+sqrt(a²-a+sqrt(a²-a+...
I love these videos. TWAAANTY
You're videos are addictive :-D
it was still working and for both 4 and 5 when i used x-4 instead of x+4. The equation 9x-20 gives a perfect square for both 4 and 5. Thank you
20 plusss square root of twanty tatatat, very relaxing to hear
"If Xn goes to Hell(L), Xn-1 also goes to Hell(L)"
Square root of twaaanty
Blackpenredpen is the type of zuniga that teaches you how to solve for x using michael jordans...
Fun fact - you could use stick figures as actual variables during a math test if the variable refers to a human being.
The teachers are probably not gonna like it, but it would make perfect algebraic sense - variables are just placeholders anyway.
2:50- "ok this marker is supposed to be blue but"
lol i had to delete my comment after he said that
Do you know by this method which you did like substituting x in the same equation, you can find the value of -1 ! and it will be wrong as then it will give pie = app. -4 so this sqrt 20.... is wrong
書いてある事は平易なのに、
X=5からここまで自力で広げちゃう中学生ってほとんど居なさそう！
so basically
if x=sqrt(y+sqrt(y+sqrt(y+...)))
then x=(1+sqrt(1+4y))/2
2:38 how is it "anything up to u" .. Cause in the LHS , square root of X squared is mod of X .... Therefore it can be only |X| = +√(20+X).....I maybe wrong
Using this method, we can actually show that 1=sqrt(0+sqrt(0+sqrt(0...
So his recent video bprp brought me here... Earlier I was tensed to see some over intelligent beings fighting over this simple maths.
:0
How do we show that it is indeed bounded by five and doesn't blow up?
So if you have just: x = √(A + √(A + √...)), where A > 0, you can just go:
x = √(A + x)
x² - x - A = 0
x = ½(1 ± √[1+4A])
. . . but since x is necessarily > 0,
x = ½(1 + √[1+4A])
And if you started with: x = √(A - √(A - √...)), then
x = ½(-1 + √[1+4A])
Interesting observation, for the x = √(A + x) case:
If 0 < A
Another interesting note: if you write on a calculator sqrt, 2, sqrt, 2, sqrt, 2... and = you will get almost 2. : )
Can it be rewritten like this as well?
x²-x-20 =0
x = x² -20
x = (x²-20)²-20
x = ((x²-20)²-20)² - 20
And show that the expression obtained from "infinite substitution" converges to 5? If it doesn't converge, then is 5 not a solution? I suspect otherwise
Michael Jordan is RAD
Was that Doraemon theme song at first?
Please make more videos on Analysis. I have an exam coming up and I struggle to know how to answer these questions in the manner that they want.
well, proving the series is bounded from above by 5 is pretty easy by induction. To prove it is monotonically increasing we need a(n+1)-a(n)>0, you can multiply and divide by conjugate to see it implies a(n)-a(n-1)>0 also.
Is every Natural number expressible by this method Or this way
I couldn't see two stuffs due to sun light
How do we know that the sequence is increasing?
The adding 20 part of course works but you are square rooting.
How do we prove that the adding 20 is more than the root subtracting?
I think with induction
The sqrt doesn't subtract. a>b>0 implies √a>√b so √(20+√20)>√20
There is a shortcut for this just multiply the number by it's predecessor and you get number to be put in infinitusmal root
This was a green pen 🥸
Green marker
Red marker
... yay?
Twanty intensifies
Back in the day, when the legend was *greenpenredpen* or *bluepenredpen* .... 😁
Square root of twunty
What if the nesting is finite?
A problem from my high school: Find
Sqrt(1+sqrt(2+sqrt(3+... sqrt(1975)...))).
I still don't know the answer.
So michael Jordan is radical?
So many videos without blackpen :(
When I was in calc I we called the theorem you mention around 9:30 "Bolzano's lemma", but I can't find it by that name on the internet, so I ended up calling it "MB criterion", where MB stands for Monotone-Bounded. Do you call it in any particular way?
Also yeah the theorem is kinda obvious when you already know some math but I call it like that when I help freshmen with their homework lol
Bolzano-Weierstrass Theorem
Entertaining by Maths wowwww dudee keep up the good work :)
Love this video!
x=5, x-5=0, 5-5= 0 tada,tada
The problem with your initial steps is you took a formula that defined one root for x and added a x = -4 root. This breaks the expansion as solely being an expansion of x = 5.
This is not a problem. If a number x satisfies (x-5) = 0, then that same number x also satisfies (x-5)(x-4) = 0(x-4) = 0.
Yes I wanted to ask how is this true? x=5 graphed is just a line perpendicular to the x axis and crosses the x axis at 5
So how can -4 also be a root of that?
Also we're multiplying here by a variable x isn't that wrong? Aren't we going to change the whole line into a parabola?
I don't have the maths to prove that multiplying by a variable not a constant is wrong but I know dividing by a variable is since it loses us a root (which is always root zero)
What i mean is let's assume we havex³+5x²+6x=0 I cannot divide by x because then I'll lose the x=0 when I factor it out
So why isn't that the case here as well? Why is it okay to multiply by x and add another root???
En realidad no es un error. Lo que está haciendo es crear un ejercicio a partir de la solución, no resolviendo una ecuación.
My dog loves this video
could someone please tell me from what branch(calculus 1/2/3/4; linear algebra... you prabably know more than I) of math is this from. just please tell me I'mm dying to knowwwww.
@blackpenredpen, anybody... is there a branch of math that's called (what blackpenredpen said on the last part of the video)"mathematical analysis"(I don't really know) and what would be the prerequisites?
Doreamon bgm in starting
Hello, blackpenredpen
I'm a seventh-grade student in Sydney and for a math test, we were given a very confusing math problem. No one seemed to have any idea how to solve it. The question went as follows:
Let A be a 2018-digit number which is divisible by 9. Let B be the sum of all digits of A, and C be the sum of all digits of B. Find the sum of all possible values of C.
Do you happen to know a way to solve this?
Did you solve it?
I put it into Desmos but it converged to 5.00000016769
Desmos approximates stuff
I think that sqrt(1+sqrt(1+sqrt(1+...))) is nice
Is that a MVMT watch that you’re wearing?
Off the cuff guess you have to bring in imaginary numbers into the mix.
Sometimes maths seens dark magic
TWAANTY
I hereby apologize for claining that "the series is not bounded above 5 but less"
Can't we generate those roots with just x^2 - x= ywith x as our 5 and solved for y our 20 which then goes to the nest x=(y+x)^1/2
Lol the doraemon theme song
Clickbait, there is no mention of Michael Jordan in the video
is that a green pen? what happened to the black?!
Nasty square roots 😈
You can substitute an equation within itself?
I'm confused why the 20+x becomes 20-x for the series that converges to -4. I tested it out and it works, but I don't see the algebraic justification...
Because each time you substitute x which is *negative* sqrt(20+x) :)
This is... nesty
so x converges to 5
proof:
5 = sqrt(25) = sqrt(20 + 5) = sqrt(20 + sqrt(25)) = sqrt(20 + sqrt(20 + 5)) ...
It doesnt work for x=1 and x=0
Sounds like the doraemon theme at 0:00
This video is dedicated to anty
If x = 5
x = √20+x
5 = √20+5
5 = √25 which satisfyes the problem
But whats the proof that sqrt(20 + sqrt(20+.... is bounded by 5?
it seems like it but how do we prove it?
You can do it by induction. First, notice that x1 = √20 ≤ 5, so that's the base case.
Now, assume x(n-1) ≤ 5. We want to show, based on this assumption, that xn ≤ 5.
Since we have x(n-1) ≤ 5, we can add 20 to both sides to get
20+x(n-1) ≤ 25
Taking the square root of both sides, we get
√(20+x(n-1)) ≤ 5
But by definition, xn = √(20+x(n-1))
So we have xn ≤ 5.
So we get each term is bounded above by 5, and that's what it means for the sequence to be bounded above by 5.
greenpenredpen ??
This thiory where we can use it?
No where
√(1+2√1+3√1+4√1+5√1+.....) Ramanujan
At first we have x-5=0 then you multiply both sides by (x+4)
I wanted to ask why this is okay to do?
x=5 graphed is just a line perpendicular to the x axis and crosses the x axis at 5
So here we're multiplying by a variable x but isn't that wrong? Aren't we going to change the whole line into a parabola?
I don't have the maths to prove that multiplying by a variable not a constant is wrong but I know dividing by a variable is since it loses you a root (which is always root zero)
What I mean is let's say we have the equation x³+5x²+6x=0 I cannot divide by x because then I'll lose the x=0 when I factor it out
So why isn't that the case here as well? Why is it okay to multiply by x and add another root???
michael jordan is the square root of two times michael jordan
michael jordan is two
Thanks ❤️
ω=−1±12−4√2=−1±−3√2=−1±i3√2
Is this the Omega you were talking about?
Wait... omega? The infinite ordinal?
Wasn’t MJ number 23?