Allow me to present a rigorous argument. Let f(z):= sqrt(1+z). The function z-f(z) is injective over I:=[0,infty) because it is strictly increasing over this interval, which is equivalent to f'(z)=0 (yes, yes, everything is fine at the boundary). Note that over I the image of f, denoted Im(f), is a subset of [0,infty). Let the n-fold composition of f with itself be denoted by f_n. Note also that Im(f_n) is a subset of [0,infty) over I. z-f_n(z) is injective over I for all n since by induction: 1-(f_(n+1))'(z) =1-f'(f_n(z))•(f_n)'(z) >1-1•1 =0 Note that 0-f_3(0)=-10. Thus there is a unique number x>=0 with x-f_3(x)=0, by MVT and injectivity. Let y:=f(x)=sqrt(1+x)>=0. We have: y =sqrt(1+x) =f_4(x) =f_3(y) Thus y is a zero of z-f_3(z) over the interval I. By uniqueness, y=x, i.e. sqrt(1+x)=x. Since x>=0, it is the unique positive root of z^2-z-1. That is, x=φ. ◾️
@SyberMath
Жыл бұрын
Wow! 🤩
@leif1075
Жыл бұрын
@@SyberMath The swuare root of a number ca be negstive in the real world..did you misspell at 4:30?
@davinheagertans4275
Жыл бұрын
Thanks for the cool problems. Always thought provoking and leading to other ideas.
@saboorhalimi
Жыл бұрын
Nice problem and nice solution❤
@SyberMath
Жыл бұрын
Thanks 😊💕
@braydentaylor4639
Жыл бұрын
Everyone and their grandmother knows what that infinite nested radical is equal to
@wakelamp
Жыл бұрын
My grandmother was sadly poor at maths ::-)
@marcomandati6715
Жыл бұрын
Are there other complex solutions to this equation?
@Hexer1985
Жыл бұрын
As soon as it is about a continued square root containing only "1", you know it's about the golden ratio.
@mystychief
Жыл бұрын
Trying x's in excel gave me eventually the answer 1.618, so probably phi, the golden ratio. We know that phi*phi = phi+1. For x = phi, 1+x = phi*phi so sqrt(1+x) is phi again. Etc., etc., the y becomes phi again. Often with these questions I find the answer with excel and then it gets easier. I take a starting x and do a column with x-es: last x + 1, last x + 1, again, etc. etc. and a column with y-f(x) which should be zero at some point or we find the closest to zero. Finetune with a new starting value of x and the next x every time +0.1. Again but then + 0.01 until the answer(s) get more precise. You can do it also with 2 variables x and y in a matrix, more gets too complicated in excel. You can give zero's a different color in excel so that you spot the zero(s) more easily (conditional formatting).
@joyli9893
Жыл бұрын
I call this a “golden root”
@broytingaravsol
Жыл бұрын
x=(1+√5)/2
@赖皮球
Жыл бұрын
x=sqrt(1+x); x=-+(sqrt(5)-1)/2
@kranthi143B
Жыл бұрын
But in the given problem there are finite square roots.
@mcwulf25
Жыл бұрын
These types of problem always seem to give that golden ratio type of solution.
@SyberMath
Жыл бұрын
☺️
@joyli9893
Жыл бұрын
x = phi
@sergiogarciaperez4492
Жыл бұрын
Would you mind solving x for 4x^(3) - 3 - cosx = 0 in a future video, please ?
@tiripoulain
Жыл бұрын
What’s the motivation? It doesn’t look too interesting
@sergiogarciaperez4492
Жыл бұрын
@@tiripoulain ok then how do you solve it?
@tiripoulain
Жыл бұрын
@@sergiogarciaperez4492 I wouldn’t spend time trying to find a closed form if the problem is not motivated properly or if you cannot assure me that the answer is interesting in any way.
@SyberMath
Жыл бұрын
No analytical method found. Graphing indicates they intersect at about x=0.963. What is the source?
@sergiogarciaperez4492
Жыл бұрын
@@SyberMath it was a problem proposed in an exam of the Spanish equivalent of American 12th grade which said to proof that x^3 -3x = sen(x) -1 had only one solution. (Firstly using Bolzano theorem and then Rolle theorem, in which you have to do f'(x) = 0)
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