Question Number 100624 by Dwaipayan Shikari last updated on 27/Jun/20

Find the value of (√(2+(√(2+(√(2+(√2)))))))...∞ using cos function

Commented byMJS last updated on 27/Jun/20

why cos?  x=(√(2(√(2(√(2(√(2....))))))))  x^2 =2x  x_1 =0 wrong  x_2 =2 is the solution

Commented byDwaipayan Shikari last updated on 27/Jun/20

There was a mistake on my question. Now can you prove it  sir using cos function????  I have edited my question

Commented byDwaipayan Shikari last updated on 27/Jun/20

Sir i have found something  1+cos(π/4)=2cos^2 (π/8)  (√((1/2)(1+(1/(√2)))))=cos(π/8)  (1/2)(√(2+(√2)))=cos(π/8)=cos(π/2^3 )  1+cos(π/8)=2cos^2 (π/(16))  (1/2)(√(2+(√(2+(√2)))))=cos(π/(16))=cos(π/2^4 )  So  (1/2)(√(2+(√(2+(√(2+(√(2+(√(2+(√(2+(√(2...))))))))))))))n=cos(π/2^(n+1) )  lim_(n→∞)   (1/2)(√(2+(√(2+(√(2+(√(2+(√(2+(√2)))))))))))..n=cos(0)=1  so(√(2+(√(2+(√(2+(√(2+....∞))))))))=2  is it right???????

Answered by maths mind last updated on 27/Jun/20

u_(n+1) ^2 =u_n +2...E  u_0 =(√2)  we can proof this  withe  logic  0≤u_n ≤2...easy too see  u_(n+1) −u_n =((u_n +2−u_n ^2 )/((√(u_n +2))+u_n ))=((−(u_n −2)(u_n +1))/((√(u_n +2))+u_n ))≥0  u_n ↾ increasing   ⇒  since0 ≤U_n ≤2  ⇒∃ (w_n ) such u_n =2cos(w_n ) proof  w_n =cos^(−1) ((u_n /2)) well defind since (u_n /2)∈[0,1]  since u_n ∈[0,2] and incrasing ⇒w_n ∈[0,(π/2)]⇒  w_n  is deacrising  E⇔4cos^2 (w_(n+1) )=2+2cos(w_n )⇒2cos^2 (w_(n+1) )−1=cos(w_n )  cos(2x)=2cos^2 (x)−1⇒  E⇔cos(2w_(n+1) )=cos(w_n )  ⇒2w_(n+1) =w_n  because w_n ∈[0,(π/2)]  ⇒w_(n+1) =(w_n /2)⇒w_n is geometric sequences  ⇒w_n =w_0 ((1/2))^n   w_0 =(π/4)  u_n =2cos((π/(4.2^n )))=2cos((π/2^(n+2) ))

Commented byghiniboss last updated on 28/Jun/20

what aspect of maths is this?

Commented byprakash jain last updated on 28/Jun/20

Two topics  Sequences and series  recurrence relation