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Induction (I think)

Given any a1>0, define (an)n>=1 by an+1=((an)^2 + 3)/(2an).

Prove that an >= root3 for all n >= 2

Any ideas?

assume it converges

as n tends to infinity a(n) and a(n+1) become very close together, assume they are a

so

a = (a^2+3)/2a

2a^2 = a^2 + 3

a^2 = 3

a= sqroot(3)

so if the sequence converges it will converge to root(3)

it could also converge to -sqrt(3) but this only happens when starting value is less than 0. This is prohibited in the question

This is a P2 induction question

Reply 2

flyinghorse

thanks

btw what exam board are you on? I've never done induction and I'm starting to think that I've been screwed by edexcel, since induction's proving to be ever so slightly important...

Reply 3

Jonny W

Step 1
Define f :frown:

0, infinity) -> R by f(x) = (x^2 + 3)/(2x).

Then df(x)/dx = 1/2 - 3/(2x^2). So f(x) is minimized at x = sqrt(3). Since f(sqrt(3)) = 6/(2sqrt(3)) = sqrt(3) it follows that f(x) >= sqrt(3) for all x.

--

Step 2
We can prove by induction that a(n) > 0 for any positive integer n.

--

Conclusion
For any integer n >= 2,

a(n)
= f(a(n - 1)) . . . using Step 2
>= sqrt(3) . . . using Step 1

Reply 4

amo1

is induction in edecel p6

Reply 5

Jonny W

amo1

is induction in edecel p6

Yes, but (in my opinion) flyinghorse's question is too hard to appear on an A-level exam.

Reply 6

seekaye

i meant to say it was a about iteration not induction. sorry.

the sequence is defined iterativley and the limit of the sequence being a root of the equation (the one you get by letting a(n) and a(n+1) be the same letter) is part of the P2 (OCR) syllabus