3 + Rad(3 + Rad(x)) = x (Infinitely nested radical)
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- Опубліковано 17 тра 2024
- In this video I used the fact of infinitely nested radicals to solve a radical equation. It was intuitive to find a proper substition to make a polynomial but while tinkering with the problem, I realized it was partially an infinitely nested radical. I therefore chose to solve it that way in the light of recent videos on the same topic.
*3 + √(3 + √x) = x* has the same solution(s) as just *3 + √x = x* (for the reason you mentioned), which is *x = (x − 3)²* and we solve the quadratic *x² − 7x = 9 = 0* to obtain the solution(s) of the original equation.
The other solution *x = (7 − √13)/2* is a solution to *3 − √x = x,* so it also makes sense if we consider the other, non-principal branch of the square root.
"positivity vibe check" was so good 😄
amazing video as always
Cant x=3+√x?
Sure. Didn't notice your comment before I posted mine.
Substitute z = x - 3. Then you get a quartic in z, which has two easy solutions (z = 1 and z = -2 ). Get two further real solutions by long division. Resubstituting you see that only one of the solutions leads to a solution x of the original equation.
Ahhh!!!! Rewriting this equation as an infinetley nested sum makes it easu to solve. How genius!
just a question ... Rad() or sqrt() ? ^^
PS: it blown my mind...never let this reccurent proof be calculated by a computer.
Got it! The value for x must not only be positive as sir Newtons said, but it must be at least 3+sqrt(3). There are in fact 4 possible solutions for this problem. I am not sure how Newtons got that strange root which he rejected. 3 of the 4 possible roots were request because they were all less than 3+sqrt(3).
I thought I was settled with this problem but I need to figure out how Newtons go that 2nd root.
Please explain the phenomena of infinity in math (algebra)...I can use this very example of indefinitely nested radical but compare the case with SQRT(2....) and with SQRT(3....). if both of these 'nestations' are equal to infinity, can one say that 2=3. Of course, you will say NO. However, I am troubled by any insinuation that infinity is the same for nested roots of 3 and 2. Why?...if we conveniently accept that is true and some day calculate something in quantum mechanics and/or astrophysics we might get stuck, because it was convenient to assume at some point that ZERO factorial equals ONE, for example.
You have gotten to the point where you discuss magnitudes of infinity, which is a number theory conundrum. For the sake of working with infinity, you cannot use infinity to equate two expressions. Infinity in basic algebra is just a way to describe end behavior
@@BlazeStormCloud I agree of course. However, @4:35 in your video you are equalizing the two Ks which are different (one radical (the main one) did not matter???). According to your replied answer, that should not be allowed?
Very good
Wondering if it be worth your time to do a proof. (B^n)-1 is always divisible by B-1
(B^n)-1=b^n - 1^n and using the general expansion or whatever (b-1)(b^(n-1)+b^(n-2)...+b+1)
You can also prove this result using factor theorem, by putting f(B) = (B^n)-1 , and finding the conditions for its roots, you'll see that f(1)=0 , and by factor theorem we know for any polynomial f(x) if f(a)=0 then (x-a) is a factor of f(x) , so in our case (B-1) must be a factor of f(B) and if it's a factor then it obviously divides f(B) , and ofcourse factor theorem only works for polynomials so 'n' must belong to positive integers.
This is beautiful 😂😂😂
If you expand this into a polynomial you get "x^4-12x^3+48x^2-73x+36". So can you find an answer to polynomials using infinite radicals, if so what types?
Or would all of those types be easily solved different ways (like factoring), even if very big?
They wouldn't all (probably) be easily factorable for instance the expanded form in your comment has 2 nice factors (x-1) and (x-4) but interesting thing to note is that the roots actually change if you continue expanding using more nested terms, except for the two roots found for 3+√x=x , so I guess there probably does exist certain types of polynomials that stem from this relation which share only certain common roots (and factors) but this is only really useful for the first expanded form as by the second expansion you would be at a 8th degree polynomial and knowing 2 roots I guess you could reduce it to 6th degree but good luck getting any further than that.
Une solution unique entre 5 et6.
The equation
3 + √(3 + √x) = x
obviously does not involve an infinitely nested square root, and there really is no need to resort to infinitely nested square roots to solve this equation.
If you look at the given equation you can see that there is some kind of repetition at the left hand side. Working from the inside out we take a number x, take the square root, add 3, and then _again_ take the square root and add 3. This means that if we define a function
f(x) = 3 + √x
on the domain [0, ∞) then the equation can be written as
f(f(x)) = x
Now let
f(x) = y
and substitute that in f(f(x)) = x and we have
f(y) = x
The function f(x) = 3 + √x is _strictly increasing_ on its domain [0, ∞) so, for any two real values x and y in this domain, if y > x then f(y) > f(x) but since f(y) = x and f(x) = y this would imply x > y which is a contradiction. Similarly, if y < x then f(y) < f(x) would imply x < y which is again a contradiction. Since y > x and y < x are excluded we must have y = x.
Thus, it follows from f(f(x)) = x that f(x) = x and, conversely, f(x) = x evidently implies f(f(x)) = x. Therefore, the equations f(f(x)) = x and f(x) = x have _the same set of solutions_ on the domain [0, ∞) which means that we can solve the equation f(x) = x to obtain the solutions of f(f(x)) = x. So, we only need to solve
3 + √x = x
Subtracting 3 from both sides and then squaring both sides gives
x = (x − 3)²
x² − 7x + 9 = 0
(x − ⁷⁄₂)² = ¹³⁄₄
x = ⁷⁄₂ + ¹⁄₂√13 ⋁ x = ⁷⁄₂ − ¹⁄₂√13
Since 3 + √x = x implies x > 3 and ⁷⁄₂ − ¹⁄₂√13 < 3 only x = ⁷⁄₂ + ¹⁄₂√13 is a solution of the original equation.
Of course you can express the sole solution of the equation 3 + √(3 + √x) = x as an infinitely nested square root, but then you need to prove that your infinitely nested square root actually converges, which is not really trivial.
Let us define an infinite sequence of finitely nested square roots by means of the recurrence relation
u₀ = 3, uₙ₊₁ = 3 + √uₙ
then proving that your inifinitely nested square root converges is equivalent to proving that the limit of uₙ for n → ∞ exists. We have proved this if we can prove that (1) the sequence (uₙ) is (strictly) increasing and (2) the sequence (uₙ) has an upper bound.
To prove (1) we need to prove that uₙ₊₁ > uₙ for any n ∈ ℤ₀⁺. This statement is evidently true for n = 0 since u₁ = 3 + √3 > 3 = u₀. Now suppose the statement is true for some nonnegative integer n = k. Then we have uₖ₊₂ − uₖ₊₁ = (3 + √uₖ₊₁) − (3 + √uₖ) = √uₖ₊₁ − √uₖ > 0 since uₖ₊₁ − uₖ > 0. So, the statement uₙ₊₁ > uₙ is true for n = 0 and also true for n = k + 1 if it is true for n = k, which implies that uₙ₊₁ > uₙ for any n ∈ ℤ₀⁺.
To prove (2) we note that since uₙ₊₁ > uₙ for any n ∈ ℤ₀⁺ we have 3 + √uₙ > uₙ for any n ∈ ℤ₀⁺ which implies √uₙ > uₙ − 3 and therefore uₙ > (uₙ − 3)² since uₙ − 3 ≥ 0 and therefore uₙ² − 7uₙ + 9 < 0 which implies ⁷⁄₂ − ¹⁄₂√13 < uₙ < ⁷⁄₂ + ¹⁄₂√13 for any n ∈ ℤ₀⁺. Of course, since ⁷⁄₂ − ¹⁄₂√13 < 3 and 3 ≤ uₙ this also means that 3 ≤ uₙ < ⁷⁄₂ + ¹⁄₂√13 for any n ∈ ℤ₀⁺.
Since the sequence (uₙ) is (strictly) increasing with an upper bound it follows that lim uₙ for n → ∞ exists. If L is this limit, then it follows from uₙ₊₁ = 3 + √uₙ that L = 3 + √L and therefore L = ⁷⁄₂ + ¹⁄₂√13 since uₙ ≥ 3 for any n ∈ ℤ₀⁺. And, as has already been proved algebraically, x = ⁷⁄₂ + ¹⁄₂√13 is the sole solution of the equation 3 + √(3 + √x) = x which can therefore indeed be represented as the infinitely nested square root defined by the sequence (uₙ).
I ain't reading allat 💀
Let y = 3 + sqrt(x)
Then 3 + sqrt(y) = x
So y + sqrt(y) = x + sqrt(x)
f(y) = f(x), where f(t) = t + sqrt(t)
f(t) is increasing, then f(x) = f(y) only when x = y
x = 3 + sqrt(x)
x**2 - 6*x + 9 = x
x**2 - 7*x + 9 = 0
x = (7 + sqrt(13))/2
I watched your video three times and did not see where you made a sloppy mistake. Very strange! Where did the 2nd root go!!!!!
I too got two solutions. I think that there might be 4 real solutions. Strange that you missed the 2nd root and possibly two others.
where are you from?
Earth
why you pronounce root as ''rat''?
He said rad....for "radical".
Why shouldn't we consider the other solutions to x? Aren't they technically real, but they are a biproduct of the square root function.
Let y= √x ;
3+√(3+y)=y ;
√(3+y)= (y^2)-3 ;
3+y=(y^4)-6(y^2)+9 ;
(y^4)-6(y^2)-y+6=0 ;
y=1 is a solution ;
(y-1)((y^3)+(y^2)-5y-6)=0 ;
y=-2 is a solution ;
(y-1)(y+2)((y^2)-y-3)=0 ;
for (y^2)-y-3=0 they are two solutions, the same k values you got.
Meaning that they are 4 solutions x=1, x=4, x=(7+ √13)/2, and x=(7-√13)/2.
x=(7+ √13)/2 uses both positive outputs of the sqRoot function.
x=(7-√13)/2 uses both negative outputs.
x=4, uses inner negative, outer positive.
x=1, uses inner positive, outer negative.
Sorry, i get 1 and 4 as additional solutions. Putting them into the original equation for testing shows that those are indeed valid solutions:
a) x == 1
3 + sqrt(3 + sqrt(1)) = 1
sqrt(4) = -2
± 2 = -2
=> selected: -2 = -2
b) x == 4
3 + sqrt(3 + sqrt(4)) = 4
sqrt(3 ± 2) = 1
=> selected: sqrt(3-2) = 1
sqrt(1) = 1
1 = 1
Obviously, the two solutions of the quadtratic term are not the only ones.
Simply multiplying out and reshuffling the original equation gives the equation x^4 - 12 x^3 + 48 x^2 - 73 x + 36 = 0
from where the first solution x = 1 is visible at first glance.
Dividing by that first solution (x-1) gives x^3 - 11 x^2 + 37 x -36 = 0
from where the second solution is quickly grasped by trying as x = 4.
Dividing by that second solution (x-4) gives x^2 - 7x + 9 = 0
which delivers the same results as in the video.
Since the idea with the infinitely nested converging term is nice, i will not downvote, but since two solutions are missing, i have to abstain from upvoting, sorry.
I usually don't comment when I see wrong calculations. However, I have see you comment good stuff in the past. Graph y=rad(3+rad(x)) and y=x-3. They only intersect once. Sqrt 4 is 2 . Not plus or minus 2
This guy never fails to make me feel stupid 😂💔
Why u call square root as rad? I don’t get it
radical
👍👍👍😁🤪👋
x = k^2
If you try verifying the solution it is wrong. There must be an error…
No, the solution is correct. Apparently you don't understand how to denest denestable nested square roots. Note that if
x = ⁷⁄₂ + ¹⁄₂√13
then
√x = ¹⁄₂ + ¹⁄₂√13
You can easily verify this by squaring ¹⁄₂ + ¹⁄₂√13 which gives (¹⁄₂)² + (¹⁄₂√13)² + 2(¹⁄₂)(¹⁄₂√13) = ¹⁄₄ + ¹³⁄₄ + ¹⁄₂√13 = ⁷⁄₂ + ¹⁄₂√13.
Consequently, with x = ⁷⁄₂ + ¹⁄₂√13 and therefore √x = ¹⁄₂ + ¹⁄₂√13 we get
3 + √(3 + √x) = 3 + √(3 + ¹⁄₂ + ¹⁄₂√13) = 3 + √(⁷⁄₂ + ¹⁄₂√13) = 3 + ¹⁄₂ + ¹⁄₂√13 = ⁷⁄₂ + ¹⁄₂√13 = x
so the solution checks out.
W
x-3=sqrt(3+sqrt(x))
(x-3)^2=3+sqrt(x)
x^2-6x+6=sqrt(x)
ok how about instead of solving a depressed quartic i do a funny trick
x=3+sqrt(3+sqrt(3+sqrt...))
x=3+sqrt(x)
x-sqrt(x)-3=0
sqrt(x)=(1+-sqrt(13))/2
sqrt(13)>1 so sqrt(x)=(1+sqrt(13))/2
x=(7+sqrt(13))/2
check: 3+sqrt(3+sqrt(x))=(7+sqrt(13))/2
6+2sqrt(3+sqrt(x))=7+sqrt(13)
2sqrt(3+sqrt(x))=1+sqrt(13)
3+sqrt(x)=(7+sqrt(13))/2
sqrt(x)=(1+sqrt(13))/2
x=(7+sqrt(13))/2✅
Pin?
First ❤🎉
Nah I was first
@@secsiiiinuh uh I was