Pretty cool function (and its inverse) Watch

01rorlin
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#21
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Wow, Tom, this is amazing stuff.

x = 45 is the word on the street.
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Zii
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#22
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#22
(Original post by pyrolol)
It's the maximum of the inverse.

(Btw there's a good interview question based on that, find the maximum of that inverse, and then say which of e^pi and pi^e is bigger.)


As is quite clear, x = e^(1/e) is the only asymptote of the originally described function (every single point thereafter can be crudely described as being at infinity) and y = 1 is the only asymptote of the inverse function, whose maximum is at x = e.
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Ed.
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(Original post by Audrey Hepburn)
:ridinghor

I shalt ride you down and take you to the gallows!
But fair maiden, your trusty stead is really only a child's toy . You couldn't ride down a badger, let alone my mathematical ways.
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Audrey Hepburn
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#24
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(Original post by Ed.)
But fair maiden, your trusty stead is really only a child's toy . You couldn't ride down a badger, let alone my mathematical ways.
You don't even do maths! And you and I both know that I have successfully ridden down many a badger! :cool:

:badger:
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Ed.
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(Original post by Audrey Hepburn)
You don't even do maths! And you and I both know that I have successfully ridden down many a badger! :cool:

:badger:
So the maths forum is only for those who do maths now-a-days is it ? Ridiculous.

Sorry I doubted that you could run down badgers :badger: Did you see this the other day ? It made me sad.
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Mcjazz
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Why have you guys hijacked this poor guy's thread? They were just trying to talk a bit of maths, have you no shame?
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Strictly Business
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#27
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#27
That is quite cool. :yy:
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Audrey Hepburn
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#28
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(Original post by Ed.)
So the maths forum is only for those who do maths now-a-days is it ? Ridiculous.

Sorry I doubted that you could run down badgers :badger: Did you see this the other day ? It made me sad.
:eek:

Ed, I think it's time that we went to work :ts:
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Audrey Hepburn
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(Original post by Mcjazz)
Why have you guys hijacked this poor guy's thread? They were just trying to talk a bit of maths, have you no shame?
:eek:











...sowwy :puppyeyes:
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Zii
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So to summarise, the biggest finite number that can be generated by x^{x^{x^{x^{\cdot^{\cdot^{\cdot}  }}}}} is e, when x = e^{\frac{1}{e}}. When x > e^{\frac{1}{e}}, the result is basically \infty.

I don't have a clue what goes on when x<1.

When x = \sqrt{2}, the result is 2.

In fact, if x^{x^{x^{x^{\cdot^{\cdot^{\cdot}  }}}}} converges to some finite y, x = y^{\frac{1}{y}}.

The inverse function of y = x^{x^{x^{x^{\cdot^{\cdot^{\cdot}  }}}}} is y = x^{\frac{1}{x}}.

The original function y has no asymptote. (e^{\frac{1}{e}}, e) is a point on the curve and all points thereafter can be generalised as (x, \infty). The inverse function has an asymptote y = 1, and a maximum y = e^{\frac{1}{e}}, at x = e.

I don't really know how to investigate what happens when x<1.
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Ed.
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(Original post by Mcjazz)
Why have you guys hijacked this poor guy's thread? They were just trying to talk a bit of maths, have you no shame?
Sorry
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Mcjazz
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#32
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(Original post by Audrey Hepburn)
:eek:











...sowwy :puppyeyes:
Too bloody right! Now, carry on fellazzz :love:
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Ed.
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(Original post by Audrey Hepburn)
:eek:

Ed, I think it's time that we went to work :ts:
It will be done, :ts:
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Mcjazz
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(Original post by Ed.)
Sorry
It's not me you need to apologise to.



(:p:)
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Ed.
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#35
(Original post by Mcjazz)
It's not me you need to apologise to.



(:p:)
I'm so back at school right now.

Sorry guys, please carry on..
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Mcjazz
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#36
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#36
:rofl:
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Dystopia
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(Original post by Zii)
In fact, if x^{x^{x^{x^{\cdot^{\cdot^{\cdot}  }}}}} converges to some finite y, x = y^{\frac{1}{y}}.
Yes. The condition x = y^(1/y) is necessary but not sufficient.

The original function y has an asymptote x = e^{\frac{1}{e}}
No. The point (e^(1/e), e) is on the graph of the original function; it isn't an asymptote.
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Zii
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(Original post by Dystopia)
No. The point (e^(1/e), e) is on the graph of the original function; it isn't an asymptote.
Oh right yeah of course! So does every point after that, ie (2, y) , (3, y) (4.9, y) etc etc have y value \infty ? And thus the graph has no asymptote?
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Swayum
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Yeah, I think that's right (that it has no asymptotes).

(Original post by Zii)
I don't have a clue what goes on when x<1.
Well here's a thought:

Pick, arbitrarily, x = 0.7

We'd first have to work out 0.7^0.7. Taking a root of a number < 1 gives a number greater than the original one (i.e. greater than 0.7). Let's call this 0.7 + dx (with dx > 0).

So our next calculation would be 0.7^(0.7 + dx). We're now taking a "bigger" root than before so our new number would have to be less than 0.7 + dx. Let's call it 0.7 + dx - dx_2.

Our next calculation becomes 0.7^(0.7 + dx - dx_2). We're now taking a "smaller" root than before so our new number would have to be greater than 0.7 + dx - dx_2. Let's call it 0.7 + dx - dx_2 + dx_3.

But as we go on and on, dx_n gets smaller and smaller. So eventually we converge to some value between 0.7 and 0.7 + dx.

0.7 was chosen "randomly", the argument above works for 0 < x < 1. What it converges EXACTLY to I don't know.
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pyrolol
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(Original post by Swayum)
Yeah, I think that's right (that it has no asymptotes).



Well here's a thought:

Pick, arbitrarily, x = 0.7

We'd first have to work out 0.7^0.7. Taking a root of a number < 1 gives a number greater than the original one (i.e. greater than 0.7). Let's call this 0.7 + dx (with dx > 0).

So our next calculation would be 0.7^(0.7 + dx). We're now taking a "bigger" root than before so our new number would have to be less than 0.7 + dx. Let's call it 0.7 + dx - dx_2.

Our next calculation becomes 0.7^(0.7 + dx - dx_2). We're now taking a "smaller" root than before so our new number would have to be greater than 0.7 + dx - dx_2. Let's call it 0.7 + dx - dx_2 + dx_3.

But as we go on and on, dx_n gets smaller and smaller. So eventually we converge to some value between 0.7 and 0.7 + dx.

0.7 was chosen "randomly", the argument above works for 0 < x < 1. What it converges EXACTLY to I don't know.
Try something like 0.05. It oscillates between 2 values. (According to wikipedia, this happens when x < e^-e.
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