An Interesting Infinite Radical

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Күн бұрын

Пікірлер: 24

@WindyNight114 Ай бұрын

Very cool video. I wish I was creative enough to understand these concepts. I minored in mathematics but spent most of my courses on linear algebra and probability. Number theory and stuff like this has always captured my interest. Can’t wait for the next one!

@angelishify Ай бұрын

n*n-1=(n-1)*(n+1) ----------------------------------------- Recursive formula: sqrt(n*n)=sqrt(1+(n-1)*sqrt((n+1)*(n+1))

@arnavgaur3897 Ай бұрын

Beauty!!

@SyberMath Ай бұрын

😍❤️😍

@leif1075 Ай бұрын

@@SyberMathdon't you agree this method is totally contrived and a cheat and no one would think of it? Could you not solve in a real wayblike rewriting as dqrt 1 plus n times sqrt 1 plus n plus 1 sqrtv1 plus n plus 2 where n is 3 and so on?

@leif1075 Ай бұрын

@@SyberMathHope you can PLEASE RESPOND THIS IS NKT A REAL.METHODBITBISNA CHEAT

@demenion3521 Ай бұрын

the video is a bit misleading. i could also write 5=sqrt(25)=sqrt(1+24)=sqrt(1+3*sqrt(64))=sqrt(1+3*sqrt(1+63))=sqrt(1+3*sqrt(1+4*sqrt((63/4)²))) etc. just because starting with 4 yields integers at each step doesn't mean that 4 is the actual solution to the problem. you really need to show convergence for this and derive the result rather than deriving the infinite expression

@chucksucks8640 Ай бұрын

The fact that the sequence of radicals doesn't equal infiniti makes it convergent. I think that is the point he is trying to make.

@leif1075 Ай бұрын

Thank you IM SICK OF CONTRUVED CRAP.And plys tge reason to start with 4 is because you have 1 plus 3 is it not?

@stephenshefsky5201 Ай бұрын

Let f[k] = sqrt(1 + k sqrt(1 + (k + 1) sqrt(1 + . . .))). So, f[k] = sqrt(1 + k f[k+1]). Invert: f[k+1] = (f[k]^2 - 1) / k = (f[k] + 1)(f[k] - 1) / k. Convergence and uniqueness aside, it is easy to prove that this equation is solved by f[k] = k + 1. The original problem was to find f[3], so the answer is 3 + 1 = 4. Great problem!

@tunistick8044 Ай бұрын

how is it easy

@yusufdenli9363 Ай бұрын

@@tunistick8044 Let be f(k)= a.k+b f(k)^2 = a^2.k^2 + 2akb + b^2 a^2.k^2 + 2akb + b^2 -1 = k.(ak +a+b) a^2.k^2 + 2akb + b^2 -1 =a.k^2 + k(a+b) b^2 - 1 = 0 b=1 or b= -1 a^2 = a a=1 or a=0 2ab = a+b a=1 , b=1 f(k)=k+1

@leif1075 Ай бұрын

@@yusufdenli9363but there us simplyvNO intelligent, valid, logical reason anyone would ever think of that to begin withbright?? For gids sake there isn't sobwhybis this method even allowed??

@janakkathayat616 Ай бұрын

Nice🙂

@SyberMath Ай бұрын

Thanks

@scottleung9587 Ай бұрын

Nice!

@SyberMath Ай бұрын

Thank you, sir! 😁

@yusufdenli9363 Ай бұрын

5 = √25 = √1+24 =√1+4.6 =√1+4.√36 =√1+4.√1+35 =√1+4.√1+5.7 =√1+4.√1+5.√49 ... =√1+4.√1+5.√1+6.√1+7... =5 So, =√1+n.√1+(n+1).√1+(n+2).√1+(n+3).... = n+1 Is this right?

@hazalouldi7130 Ай бұрын

right,but how to show that,may be by induction

@yusufdenli9363 Ай бұрын

@hazalouldi7130 n = √n^2 =√1+(n-1).(n+1) =√1+(n-1).√(n+1)^2 =√1+(n-1).√1+n.(n+2) =√1+(n-1).√1+n.√(n+2)^2 =√1+(n-1).√1+n.√1+(n+1).(n+3) =√1+(n-1).√1+n.√1+(n+1).√(n+3)^2 ...