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
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!
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
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.
Beauty!!
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!
how is it easy
@@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
n*n-1=(n-1)*(n+1)
-----------------------------------------
Recursive formula:
sqrt(n*n)=sqrt(1+(n-1)*sqrt((n+1)*(n+1))
Nice🙂
Thanks
Nice!
Thank you, sir! 😁
Jesus doesn’t have any siblings.
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?
right,but how to show that,may be by induction
@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
...
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