Hey guys, I was just wondering if any of you have any thoughts on this problem. I've exhausted all mine
Let's say we have 2 functions: f and g with and .
Then define two sequences:
1) = 3 and = for n 1
2) = 100 and = for n 1
What is the smallest positive integer m for which ?
I've tried connecting the two series with an inequality but I can't quite get one that I can prove. I've probably overlooked something blindingly obvious
Thanks in advance if you guys can help
Sorry for the late reply, i was away for a few days.
Isomorphism (see above post) was right btw, it's powers to the power of power, which makes it harder, so , but which makes it rather annoying 'cause i can't work with logs that easily. So sorry Aryth, i think you misunderstood the problem, but thanks for the reply anyway
And it also makes the terms grow waaay too fast:
(I'm not even going to type that out) etc...
I was wondering because I've got a possible inequality that could work, but I dunnno how to prove it, would induction work? (I'm a bit iffy on induction proofs of inequalities)
Would this be proveable?
Because it seems to work but I can't prove it
Thanks for your help guys.
Hi guys, I was wondering if anyone could possibly give me a few pointers as to how to prove the above inequality by induction 'cause I can't seem to work it out... But if I can prove that then I can solve the problem without too much difficulty.
Thanks for your help!
Oh, and Merry Christmas
Well, I think I have a solution, found in a last minute rush before this problem was due . Please point out any errors that I (probably) will have made
Here goes:
If I can prove the following, I can solve it:
The First Bit: (A very difficult proof)
Since (obviously )
The Second Bit: (An Induction Proof)
First, let's check and :
and
Clearly so this base case is true.
Now, suppose for some k
First, note that and that the difference between and in this case is 81 (and ), and this difference will only get bigger as the numbers get bigger.
Therefore,
Implying
What follows is basic application of the log rules:
So this must be true for (integral k of course)
The Third Bit: (An Induction Proof)
Again, we can check a base case.
and
Clearly, (If you don't believe me, check the powers).
Now, suppose for some k, that
Then since , we may write:
So this part too must be true for integral
The Fourth Bit: (The Answer (Finally))
Now, I have (hopefully) convinced you that for :
Now we need the smallest integer m such that
By the above inequality, the smallest such integer is 100 - 1 = 99.
Therefore, I proudly proclaim that the answer is 99!!!
Oh, and Merry Christmas!!!