I know it's not induction. That's why I was questioning it.(Original post by Occams Chainsaw)
Felix didn't use induction. I saw his answer and realised what my friend was trying to get me to do...
We just started FP1 so he was trying to play with it!
Not even sure if that makes sense right now. Brain = fried.
I honestly cannot see what you are trying to do....That top line isn't an identity.

bananarama2
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 14092013 23:28

User990473
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 14092013 23:32
(Original post by bananarama2)
I know it's not induction. That's why I was questioning it.
I honestly cannot see what you are trying to do....That top line isn't an identity.
I think I just proved that , didn't I? 
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 14092013 23:37
(Original post by Occams Chainsaw)
I think I'll come fix it in the morning after some sleep and save further embarrassment.
I think I just proved that , didn't I?
It's not what is written, but I think it may be have been latexed badly 
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 14092013 23:40
(Original post by bananarama2)
Aha I think that's wise.
It's not what is written, but I think it may be have been latexed badly
I guess that ones been done a few times, right? ....
Anyway, I hope to redeem myself after some kip!... It's exactly difficult lol. I'm just plugging in k+1! 
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 14092013 23:52
(Original post by Occams Chainsaw)
I've not used latex before
I guess that ones been done a few times, right? ....
Anyway, I hope to redeem myself after some kip!... It's exactly difficult lol. I'm just plugging in k+1! 
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 15092013 12:26
Any of you made up these questions yourself?
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 15092013 17:32
(Original post by bananarama2)
What I was trying to prove was what you interpreted. I now realise (after a rejuvinating 10 hours sleep and a Sunday roast) that I should have said I was trying to prove that the sum odd numbers is a square number!
A nice little pattern I noticed is this:
if I have an n*n box  or a perfect square
xxxx
xxxx
xxxx
xxxx
in this case where n = 4, x = 16
where x = a = b = c = d and we split it into (the only way I can describe it is 'mini'squares' so the sum of the xcoordinate value for a will always equal the sum for a's y coordinate etc...)
abcd
abcc
abbb
aaaa
So we have produced 7a + 5b + 3b + 1a = 16x
or, 16 (a square number) = 7 + 5 + 3 + 1...
Is that very obvious? It seems that way now I have written it down but I when I figured it out I felt pretty pleased with myself lolLast edited by User990473; 15092013 at 17:35. 
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 15092013 17:47
Using well known aproximations and ignoring higher order terms (the limits go to 0 so we can assume higher roder terms don't matter) we have
and
we can then evaluate the integrals explicitly to get the original limit to be equal to the limit of
Which is 1.
I wonder if there is a better way of doing ths though? Both L'H and series expansions feel like cheating. A bit like solving an inequality by just using calculus.Last edited by james22; 15092013 at 17:59. 
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 15092013 17:50
(Original post by james22)
An alternative solution to the one given:
Using well known aproximations and ignoring higher order terms (the limits go to 0 so we can assume higher roder terms don't matter) we have
and
we can then evaluate the integrals explicitly to get the original limit to be equal to the limit of
Which is 0.
I wonder if there is a better way of doing ths though? Both L'H and series expansions feel like cheating. A bit like solving an inequality by just using calculus. 
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 15092013 18:13
(Original post by Occams Chainsaw)
Yeah, I'm not really sure how use the latex yet. I'll look through the guide. I definitely didn't know how to display sigma. It's quite a nice tool, isn't it?!
What I was trying to prove was what you interpreted. I now realise (after a rejuvinating 10 hours sleep and a Sunday roast) that I should have said I was trying to prove that the sum odd numbers is a square number!
A nice little pattern I noticed is this:
if I have an n*n box  or a perfect square
xxxx
xxxx
xxxx
xxxx
in this case where n = 4, x = 16
where x = a = b = c = d and we split it into (the only way I can describe it is 'mini'squares' so the sum of the xcoordinate value for a will always equal the sum for a's y coordinate etc...)
abcd
abcc
abbb
aaaa
So we have produced 7a + 5b + 3b + 1a = 16x
or, 16 (a square number) = 7 + 5 + 3 + 1...
Is that very obvious? It seems that way now I have written it down but I when I figured it out I felt pretty pleased with myself lol 
Pterodactyl
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 20092013 00:04

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 20092013 16:52
If we change integer to positive integer:
Spoiler:Show
is not divisible by 4 (consider mod 4) and is is also not divisible by any primes of the form 4k1 (factorise as , then since these don't have integer norms* they can be divisible by (Gaussian) primes with integer norms (ie. those of the form 4k1)). However is either divisible by 4, is of the form 4k1 (in which case it must have a factor of the form 4k1) or is negative (which we have not allowed to make the statement true). Now that we have has a factor that doesn't the given result follows.
*x=0 is a simple special case.
The statement could also be made true by changing the denominator to .

Lord of the Flies
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 20092013 18:00
Solution 347
WLOG , let be s.t. and .
Define and with and . Note that and . For any
Obviously divides both initial numbers and hence 
jack.hadamard
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 21092013 17:55
Problem 349 / **
Let be such that for all . Prove that
where is the th Fibonacci.
Problem 350 / **
The Tribonacci numbers, , are defined as
and for .
Find . 
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 21092013 18:48
Solution 349
I will use the identity
There are several ways to prove this, most straight forward is to use the explicit formula for the fibonacci numbers and it is just an exercise in summing geometric series.
Using standard results from analysis we have that h is increasing and that
This is the best inequality we can have, as h can be made as close to as we like.
Using both of these we have
You may not be happy with the starred inequality, but this is true because is immediate from 
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 23092013 03:04
(Original post by Pterodactyl)
This may have been on here before so if it has please tell me and I'll delete it.
Problem 346 *
Use IBP and set x  \frac{ab}{a+b} = v and arccos(...) = u
Our uv term (arcos1arccos1) can be safely set as 0 if I am to make any geometrical sense of this.
After a bit of algebra bashing we obtain;
We now work on the RHS.
We set
Taking out the constants from the integral, we see our integral reduces down to
As y/(1y^2) is obviously 0 over the interval as it is an odd function.
This is now a well known integral;
And thus
Now (for the nice part) a sketch shows this describes a semicircle which intersects the x axis at a and b, and whereby the integrand describes the angle (for each x).
I'll take a look at the tribonacci when I have some more time at a computer. 
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 23092013 03:34
(Original post by jack.hadamard)
Problem 350 / **
The Tribonacci numbers, , are defined as
and for .
Find .
I do have a method that will certainly work, but it involves solving teh recurrance relation to get it into an explict form, squaring that and then summing all the geometric series. I don't want to do this ans I doubt it was the intention.Last edited by james22; 23092013 at 03:38. 
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 23092013 15:18
(Original post by Llewellyn)
Surprisingly tricky (I was barking down the wrong tree at 2am with inverse trig identities and far too many substitutions) but this has a very nice geometric link. (apologies if this is a repeat question, I haven't followed this thread much). Apologies also for the concise workings.
Use IBP and set x  \frac{ab}{a+b} = v and arccos(...) = u
Our uv term (arcos1arccos1) can be safely set as 0 if I am to make any geometrical sense of this.
After a bit of algebra bashing we obtain;
We now work on the RHS.
We set
Taking out the constants from the integral, we see our integral reduces down to
As y/(1y^2) is obviously 0 over the interval as it is an odd function.
This is now a well known integral;
And thus
Now (for the nice part) a sketch shows this describes a semicircle which intersects the x axis at a and b, and whereby the integrand describes the angle (for each x).
I'll take a look at the tribonacci when I have some more time at a computer.
Problem 351 **
Problem 352 */**
Last edited by Pterodactyl; 23092013 at 17:24. 
Felix Felicis
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 23092013 17:07
Problem 353*/**
For a positive integer , let be the total sum of the intervals of such that in .
Find .
Problem 354*/**
Evaluate .Last edited by Felix Felicis; 23092013 at 18:00.
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