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