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Calculus help
Keith:
screw math youll never use it
Wayshow:
--- Quote from: Keith on May 20, 2012, 06:33:57 pm ---screw math youll never use it
--- End quote ---
i need it to graduate :@
Taibz:
I love how clean your paper is, my papers are always full of drawings and shit lol :nyan:
Elyxiatic:
This isn't technically right, but I know where it converges.
There is probably a way to "show" it converges, I'm just assuming it does.
Q11a)
a(n) = sqrt (3 + a(n-1))
a(n+1) = sqrt( 3 + sqrt (a(n))
a(n+1) = sqrt (3 + sqrt (3 + a(n-1)))
a(n+1) - a(n) = sqrt (3 + sqrt(3 + a(n-1))) - sqrt (3 + a(n-1))
If we assume series converges, a(n+1) - a(n) = 0
sqrt(3 + a(n-1)) = sqrt(3 + sqrt(3 + a(n-1)))
Square both sides
3 + a(n-1) = 3 + sqrt (3 + a(n-1)
a(n-1) = sqrt (3 + a(n-1))
Square again
a(n-1)^2 = 3 + a(n-1)
a(n-1)^2 - a(n-1) -3 = 0
a(n-1) = 2.302775637
Thats where your series converges.
What I've done is incorrect though, I find a(n-1) instead of a(n).
Unless we assume a(n-1) = a(n). But then I just assumed convergence instead of proving it.
I know it involves subtracting something like a(n+1) away from a(n) and showing it equals something.
On a side note, nobody uses this stuff in the real world anyway.
Keith:
--- Quote from: Elyxiatic on May 21, 2012, 11:58:19 pm ---This isn't technically right, but I know where it converges.
There is probably a way to "show" it converges, I'm just assuming it does.
Q11a)
a(n) = sqrt (3 + a(n-1))
a(n+1) = sqrt( 3 + sqrt (a(n))
a(n+1) = sqrt (3 + sqrt (3 + a(n-1)))
a(n+1) - a(n) = sqrt (3 + sqrt(3 + a(n-1))) - sqrt (3 + a(n-1))
If we assume series converges, a(n+1) - a(n) = 0
sqrt(3 + a(n-1)) = sqrt(3 + sqrt(3 + a(n-1)))
Square both sides
3 + a(n-1) = 3 + sqrt (3 + a(n-1)
a(n-1) = sqrt (3 + a(n-1))
Square again
a(n-1)^2 = 3 + a(n-1)
a(n-1)^2 - a(n-1) -3 = 0
a(n-1) = 2.302775637
Thats where your series converges.
What I've done is incorrect though, I find a(n-1) instead of a(n).
Unless we assume a(n-1) = a(n). But then I just assumed convergence instead of proving it.
I know it involves subtracting something like a(n+1) away from a(n) and showing it equals something.
On a side note, nobody uses this stuff in the real world anyway.
--- End quote ---
If you got a(n-1)
and need a(n)
add 1 -.^ -.^ -.^ -.^ -.^ -.^ -.^ -.^ -.^ -.^ -.^ -.^ -.^
lol i smart cuz its not wrong at all