# Pretty cool function (and its inverse)Watch

10 years ago
#21
Wow, Tom, this is amazing stuff.

x = 45 is the word on the street.
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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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10 years ago
#23
(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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10 years ago
#24
(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!

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10 years ago
#25
(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!

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 Did you see this the other day ? It made me sad.
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10 years ago
#26
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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10 years ago
#27
That is quite cool.
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10 years ago
#28
(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 Did you see this the other day ? It made me sad.

Ed, I think it's time that we went to work
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10 years ago
#29
(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?

...sowwy
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#30
So to summarise, the biggest finite number that can be generated by is , when . When , the result is basically .

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

When , the result is .

In fact, if converges to some finite , .

The inverse function of is .

The original function has no asymptote. is a point on the curve and all points thereafter can be generalised as . The inverse function has an asymptote , and a maximum , at .

I don't really know how to investigate what happens when .
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10 years ago
#31
(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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10 years ago
#32
(Original post by Audrey Hepburn)

...sowwy
Too bloody right! Now, carry on fellazzz
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10 years ago
#33
(Original post by Audrey Hepburn)

Ed, I think it's time that we went to work
It will be done,
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10 years ago
#34
(Original post by Ed.)
Sorry
It's not me you need to apologise to.

()
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10 years ago
#35
(Original post by Mcjazz)
It's not me you need to apologise to.

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

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10 years ago
#36
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10 years ago
#37
(Original post by Zii)
In fact, if converges to some finite , .
Yes. The condition x = y^(1/y) is necessary but not sufficient.

The original function has an asymptote
No. The point (e^(1/e), e) is on the graph of the original function; it isn't an asymptote.
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#38
(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 ? And thus the graph has no asymptote?
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10 years ago
#39
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 .
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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10 years ago
#40
(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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