Calculus help

[Wayshow]

inb4lolasianaskingforhelponmaths

No but srs. I pay 1.5m/question solved. first person to solve gets it.

[Armybuilder]

Too advanced for me. Thats Tom you should be asking

[Wayshow]

I can pretty much do everything on the calculus textbook....except sequences, series. The convergence question is just wtf lol

[Taibz]

[Mojo]

I probably could have done this last year but my brain has now forgotten most of A-Level maths.

[Keith]

screw math youll never use it

[Wayshow]

screw math youll never use it

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]

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.

If you got a(n-1)
and need a(n)
add 1  -.^ -.^ -.^ -.^ -.^ -.^ -.^ -.^ -.^ -.^ -.^ -.^ -.^
lol i smart cuz its not wrong at all

[JC]

lol you maths nerds ##

[Wayshow]

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.

Ty for trying though :3

[Wayshow]

And.... I finished #10. Yay

[Raging Mage2]

school is only useful for 6 months after it ends :3

[Keith]

Screw the rules drop out while you're ahead wayshow, YOU ALREADY GOT INTO COLLEGE