Your original conditions are almost unreadable.
Are you saying
and ?
We have:
a1 = 2, a(n+1) = (an^2 +2) / 2an
Then comes the induction proof:
" Use mathematical induction to show that: an>= 0 for all n"
This should also mean that a(n+1) is greater than, or equal to 0. So that:
a(n+1) >= 0
(an^2 +2) / 2an >= 0
an^2 +2 >= 0
Where do I go from here?
Or could I just say that an has to be greater than 0, because if an=0 then the denominator for a(n+1) would be 0?
Maybe I could say:
The only way that a(n+1) < 0, is if an < 0, since the numerator is always greater than 0 because an^2 > 0.
Therefor all I have to prove is that a(n+1) always is greater than zero.
So if:
a(n+1) >= 0
Then:
(an^2 +2) / 2an >= 0
an^2 +2 >= 0
Which is always greater than zero.
Does that sound correct?
Unfortunately, no.
This is equivalent to the original problem that asks to show that a(n) >= 0 for all n. So you have not simplified anything.all I have to prove is that a(n+1) always is greater than zero.
You can't assume what you have to prove. It's like expecting that the cashier in a grocery will give you cash for buying stuff.So if:
a(n+1) >= 0
There is no way around the steps that I mentioned above. The only way to gloss over them is knowing them so well that it is boring to write every detail.
There is also a b) part to this task, which I can't seem to solve
It goes:
Show that for all values of n
This is what I'm thinking:
If we assume that: , we have to show that this is also the case for
From before we have that: and . And we also know that: for all n.
I have tried putting .
But from here I'm not able to show that
Any ideas?