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The Proof is Trivial!

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Original post by joostan
How pretty are you expecting the answer to be?

Spoiler


Oh effing eff - I meant abscissa - that'll teach me to use those fancy college kid words. I'll edit.
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. . .?
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.
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. :tongue:

Spoiler

(edited 8 years ago)
Original post by joostan
...


Minor typo, xx can't be negative - but lovely solution! This notation fuzzles it up enough for me not to understand it. :laugh:
Original post by Zacken
Minor typo, xx can't be negative - but lovely solution! This notation fuzzles it up enough for me not to understand it. :laugh:


whupsy daisy -that was xln(x)x\ln(x) rather than xx, will edit.
It is a little confusing, but if you write it out it's simple enough, I was being lazy.
(edited 8 years ago)
Original post by joostan
Using the standard notation here to save TeXing all this junk. :tongue:

Spoiler



That looks fine, but I think that there's a much simpler argument:

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Original post by atsruser
That looks fine, but I think that there's a much simpler argument:

Spoiler


It is a little smoother I guess, but I calculated the ordinates first.
Since I took a problem, here's one I made up back, it's not very hard.
Problem 572:*
Compute the exact value of:
x=13+12+126+12+126+12+...x=13+\dfrac{1}{2+\dfrac{1}{26+ \dfrac{1}{2+\frac{1}{26+\frac{1}{2+...}}}}}
Where this fraction continues in this recurring way ad infinitum.
Find an expression of this type (known as a continued fraction) for x=1+52x=\dfrac{1+\sqrt{5}}{2}, and x=2x=\sqrt{2}.
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! :biggrin:


Compute the exact value of:

x=13+12+126+12+126+12+...x=13+\dfrac{1}{2+\dfrac{1}{26+ \dfrac{1}{2+\frac{1}{26+\frac{1}{2+...}}}}}

Where this fraction continues in this recurring way ad infinitum.


Unparseable latex formula:

\displaystyle [br]\begin{equation*}x -13 = \frac{1}{2 + \frac{1}{26 + x-13}} \Rightarrow \frac{1}{x-13} = 2 + \frac{1}{x+13} \Rightarrow \frac{1}{x-13} - \frac{1}{x+13} = 2 \end{equation*}



So that x2=182x=182x^2 = \sqrt{182} \Rightarrow x = \sqrt{182} where we pick the positive root for obvious reasons.

Find an expression of this type (known as a continued fraction) for x=1+52x=\dfrac{1+\sqrt{5}}{2}


We have:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}2x = 1 + \sqrt{5} \Rightarrow (2x-1)^2 = 5 \Rightarrow 4x^2 - 4x -4 = 0 \iff x^2 - x - 1 = 0\end{equation*}



So:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} x^2 = 1 + x \Rightarrow x = 1 + \frac{1}{x} = 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \cdots}}}\end{equation*}



and x=2x=\sqrt{2}.


We have:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\sqrt{2} = 1 + \sqrt{2} - 1 = 1 + \frac{1}{1 + \sqrt{2}} \Rightarrow \sqrt{2} = 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2+ \cdots}}}

(edited 8 years ago)
Problem 401:

Determine whether

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

is irrational.
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.
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

Posted from TSR Mobile
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

(edited 8 years ago)
Original post by Zacken
Yaay! Iterative stuff! :biggrin:




Unparseable latex formula:

\displaystyle [br]\begin{equation*}x -13 = \frac{1}{2 + \frac{1}{26 + x-13}} \Rightarrow \frac{1}{x-13} = 2 + \frac{1}{x+13} \Rightarrow \frac{1}{x-13} - \frac{1}{x+13} = 2 \end{equation*}



So that x2=182x=182x^2 = \sqrt{182} \Rightarrow x = \sqrt{182} where we pick the positive root for obvious reasons.



We have:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}2x = 1 + \sqrt{5} \Rightarrow (2x-1)^2 = 5 \Rightarrow 4x^2 - 4x -4 = 0 \iff x^2 - x - 1 = 0\end{equation*}



So:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} x^2 = 1 + x \Rightarrow x = 1 + \frac{1}{x} = \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \cdots}}}\end{equation*}





We have:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\sqrt{2} = 1 + \sqrt{2} - 1 = 1 + \frac{1}{1 + \sqrt{2}} \Rightarrow \sqrt{2} = \frac{1}{2 + \frac{1}{2 + \frac{1}{2+ \cdots}}}



gah, just as I finished my solution I refresh the page and you've got there first! Atleast I got the same answers :tongue:
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 :tongue:


Ah, sorry - that's always frustrating! :tongue:
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 1+1+.
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 1+1+.


Urgh, thanks! Fixed. :-)
Original post by Zacken
Ah, sorry - that's always frustrating! :tongue:

Lol don't be sorry, I'll have to be faster next time :wink:
Original post by EnglishMuon
Lol don't be sorry, I'll have to be faster next time :wink:


And I'll have to make less mistakes. :tongue:

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