Thanks bro! But how do you know cos(an+1) = cos(an-1) =1 ??
Bro what I thought is a is neither determined to be constant or variable so I assumed a to be zero itself …I don’t know whether it is correct or wrong…as for the solution given by them …they used Cauchy Mean value theorem I think
Okay…bro…if we assume a=0 then a1/a will be equal to infinity instead of 1, am I correct??
I honestly didn’t understand the solution in the book!!
Bro, I’m still not getting how cos(an+1) = cos(an-1) =1 ?? (Sorry if I sound irritating)
i am trying bro. i’d be able to help then.
Bro according to me infinity is very large value so at infinity It does not matter u take an or an+1 or an-1 all values Will be same only
Sure Lavesh!! Quite possible Vibhu!!
Bro sorry I did some bs earlier …check whether this clears your doubt (I did according to the solution)…
Thanks…I’ll go through itnow(make take time😅)
a(n+2) = cos (a(n+1))
a(n) = cos (a(n-1))
Subracting and taking limit -
lim n infinity a(n+2) - a(n) = lim n infinity [cos (a(n+1)) - cos (a(n-1))]
Since infinity is large: a(n+1), a(n), a(n+2) will have close values.
a(n+2) - a(n) = \lambda times [a(n+1) - a(n)]
here lambda is taken as sin K, because a(n+1), a(n), a(n+2) lies b/w 0 and 1. so, a(n+2) - a(n) and [a(n+1) - a(n)] will also lie between 0 and 1.
After this i didn’t understand what they did…
Was reading your explanation…thanks bro!(but you had to type so much )
i am free rn so i don’t mind bro. welcome.
Summing-up all 3 of your’s solution…I think I understood the answer is 0. Thanks guys!!!
i still didn’t understand how they wrote after the sin k part…
I’m uploading a pic😂
Bro I think differentiation of cos(a n) gives sin (a n) which is some sink …I think
it might be bro…the solution is very confusing… the above post is just my interpretation of the solution.