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# The Proof is Trivial! watch

1. (Original post by joostan)
How pretty are you expecting the answer to be?
Spoiler:
Show
At present I have the rather unpleasant expression you have, , and with the , and I don't much fancy seeing if it simplifies.
Oh effing eff - I meant abscissa - that'll teach me to use those fancy college kid words. I'll edit.
2. (Original post by atsruser)
Oh effing eff - I meant abscissa - that'll teach me to use those fancy college kid words. I'll edit. Maybe you could spoiler your reply though?
It is spoilered. . .?
3. (Original post by joostan)
It is spoilered. . .?
Ignore me. I'm tired and heading off to bed. I look forward to your complete argument in the morning.
4. (Original post by atsruser)
Ignore me. I'm tired and heading off to bed. I look forward to your complete argument in the morning.
Using the standard notation here to save TeXing all this junk.
Spoiler:
Show
Note that:

or .
Noting that all the terms of are necessarily positive, we deduce that we must have: .
Thus we have a single stationary point, which is in fact a minimum, at .
To see that this is a minimum, simply note that and so the derivative changes sign from negative to positive as x increases at .
5. (Original post by joostan)
...
Minor typo, can't be negative - but lovely solution! This notation fuzzles it up enough for me not to understand it.
6. (Original post by Zacken)
Minor typo, can't be negative - but lovely solution! This notation fuzzles it up enough for me not to understand it.
whupsy daisy -that was rather than , will edit.
It is a little confusing, but if you write it out it's simple enough, I was being lazy.
7. (Original post by joostan)
Using the standard notation here to save TeXing all this junk.
Spoiler:
Show
Note that:

or .
Noting that all the terms of are necessarily positive, we deduce that we must have: .
Thus we have a single stationary point, which is in fact a minimum, at .
To see that this is a minimum, simply note that and so the derivative changes sign from negative to positive as x increases at .
That looks fine, but I think that there's a much simpler argument:

Spoiler:
Show
Note that

Now let since for all . So have the same turning points, and so do their sum.

Thus we need only consider the first term in the series for , which trivially gives a minimum of
8. (Original post by atsruser)
That looks fine, but I think that there's a much simpler argument:
Spoiler:
Show
Note that

Now let since for all . So have the same turning points, and so do their sum.

Thus we need only consider the first term in the series for , which trivially gives a minimum of
It is a little smoother I guess, but I calculated the ordinates first.
9. Since I took a problem, here's one I made up back, it's not very hard.
Problem 572:*
Compute the exact value of:

Where this fraction continues in this recurring way ad infinitum.
Find an expression of this type (known as a continued fraction) for , and .
10. (Original post by joostan)
Since I took a problem, here's one I made up back, it's not very hard.
Problem 572:*
Yaay! Iterative stuff!

Compute the exact value of:

Where this fraction continues in this recurring way ad infinitum.

So that where we pick the positive root for obvious reasons.

Find an expression of this type (known as a continued fraction) for
We have:

So:

and .
We have:

11. Problem 401:

Determine whether

1/1^5 + 1/2^5 + 1/3^5...

is irrational.
12. (Original post by Krollo)
Problem 401:

Determine whether

1/1^5 + 1/2^5 + 1/3^5...

is irrational.
1^5 is irrational so the sum of an irrational number and some other numbers are irrational, hence the sum is irrational. QED.
13. (Original post by Zacken)
1^5 is irrational so the sum of an irrational number and some other numbers are irrational, hence the sum is irrational. QED.
^ me trying to learn analysis

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14. (Original post by Zacken)
1^5 is irrational so the sum of an irrational number and some other numbers are irrational, hence the sum is irrational. QED.
Such proof. . .
Many brilliance. . .
Spoiler:
Show
Wow.
15. (Original post by Zacken)
Yaay! Iterative stuff!

So that where we pick the positive root for obvious reasons.

We have:

So:

We have:

gah, just as I finished my solution I refresh the page and you've got there first! Atleast I got the same answers
16. (Original post by EnglishMuon)
gah, just as I finished my solution I refresh the page and you've got there first! Atleast I got the same answers
Ah, sorry - that's always frustrating!
17. (Original post by Zacken)
x
Having looked more closely I now notice you seem to have made a typo.
In both cases you're missing the preceding .
18. (Original post by joostan)
Having looked more closely I now notice you seem to have made a typo.
In both cases you're missing the preceding .
Urgh, thanks! Fixed. :-)
19. (Original post by Zacken)
Ah, sorry - that's always frustrating!
Lol don't be sorry, I'll have to be faster next time
20. (Original post by EnglishMuon)
Lol don't be sorry, I'll have to be faster next time
And I'll have to make less mistakes.

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