Mean value theorem


ANSWER:

Can anyone pl explain me the solution? :sweat_smile:

2 Likes

1 Like

Thanks bro! But how do you know cos(an+1) = cos(an-1) =1 ??

1 Like

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

1 Like

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!!

3 Likes

Bro, I’m still not getting how cos(an+1) = cos(an-1) =1 ?? (Sorry if I sound irritating) :grin:

1 Like

i am trying bro. i’d be able to help then.

1 Like

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

2 Likes

Sure Lavesh!! Quite possible Vibhu!!

2 Likes

Bro sorry I did some bs earlier …check whether this clears your doubt (I did according to the solution)…

1 Like

Thanks…I’ll go through itnow(make take time😅)

1 Like

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.
So,
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… :rofl: :rofl:

1 Like

Was reading your explanation…thanks bro!(but you had to type so much :pray:)

2 Likes

i am free rn so i don’t mind bro. welcome.

1 Like

Summing-up all 3 of your’s solution…I think I understood the answer is 0. Thanks guys!!!

1 Like

i still didn’t understand how they wrote after the sin k part…

1 Like

I’m uploading a pic😂

2 Likes

Bro I think differentiation of cos(a n) gives sin (a n) which is some sink …I think

2 Likes

it might be bro…the solution is very confusing… the above post is just my interpretation of the solution.

1 Like


Guys!!! :joy: :joy:
Thanks a lot!!

4 Likes