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Calculus help

Started by Wayshow, May 20, 2012, 02:38:11 PM

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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


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Wayshow

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



Taibz

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Mojo

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

Quote from: Keith on May 20, 2012, 06:33:57 PM
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:
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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.

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

JC

#10
lol you maths nerds ##
To strive.
To seek.
To find.
And not to yield.

Wayshow

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.
Ty for trying though :3



Wayshow

And.... I finished #10. Yay



Raging Mage2

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

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Keith

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