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# A Summer of Maths Tweet

Maths and statistics discussion, revision, exam and homework help.

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1. Re: A Summer of Maths
(Original post by Brit_Miller)
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Is it undefined? I make lim x->0 f'(x)/g'(x) 0/0, so obviously g'(x)=0
No, no it is defined. I said L'Hopital's is required, not that by simply applying it you would get the result
2. Re: A Summer of Maths
(Original post by Lord of the Flies)
No, no it is defined. I said L'Hopital's is required, not that by simply applying it you would get the result
Ah, I don't know enough about it to know what to do then.
3. Re: A Summer of Maths
(Original post by Brit_Miller)
Ah, I don't know enough about it to know what to do then.
It just requires a little manipulation.
4. Re: A Summer of Maths
(Original post by Lord of the Flies)
It just requires a little manipulation.
I'll leave it to someone better, I just can't see it.
5. Re: A Summer of Maths
(Original post by Lord of the Flies)
Really? I don't see how... When returns two values, no?
No, if you take an number in this bound and apply the function to it, it converges on just 1 number. By your thinking it should be ossilating between 2 numbers, however you can see that for all x>1 if you keep doing x^x^x..., for every x you add it will increase in value so it cannot ossilate. I don't know how to show it formally but thats the logic behind it.

Try it with a few number, e.g. x=sqrt(2) means y converges to 2.
6. Re: A Summer of Maths
(Original post by james22)
No, if you take an number in this bound and apply the function to it, it converges on just 1 number. By your thinking it should be ossilating between 2 numbers, however you can see that for all x>1 if you keep doing x^x^x..., for every x you add it will increase in value so it cannot ossilate. I don't know how to show it formally but thats the logic behind it.

Try it with a few number, e.g. x=sqrt(2) means y converges to 2.
Exactly. So it cannot converge since for every additional x the value increases and is not bounded.
7. Re: A Summer of Maths
(Original post by Lord of the Flies)
Exactly. So it cannot converge since for every additional x the value increases and is not bounded.
But it is bounded, try it with sqrt(2) in a calculator, it never gets past 2.
8. Re: A Summer of Maths
(Original post by james22)
...
Also, there is a major difference between having a finite number of powers and having an infinite number of powers. In the case where the powers are infinite, I can't see what is wrong with the following:

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Therefore:

strictly increases for and

strictly decreases for and .

These two statements imply that there are exactly two values (one satisfies the other satisfies ) for each value of (when )

If what you are saying is true then there must be a mistake above - I can't spot one, but maybe that's because I'm tired, who knows!
Last edited by Lord of the Flies; 03-07-2012 at 13:19.
9. Re: A Summer of Maths
(Original post by james22)
But it is bounded, try it with sqrt(2) in a calculator, it never gets past 2.
As Blutooth pointed out:

Last edited by Lord of the Flies; 03-07-2012 at 13:27.
10. Re: A Summer of Maths
(Original post by Lord of the Flies)
Hm yes indeed...

I don't understand why my reasoning is wrong though...
or y=4. Your reasoning is correct.
11. Re: A Summer of Maths
(Original post by Blutooth)
or y=4. Your reasoning is correct.
Ah yes I should have thought of that! Phew! At least I'm not insane.
12. Re: A Summer of Maths
(Original post by Lord of the Flies)
Also, there is a major difference between having a finite number of powers and having an infinite number of powers. In the case where the powers are infinite, I can't see what is wrong with the following:

Spoiler:
Show

Therefore:

strictly increases for and

strictly decreases for and .

These two statements imply that there are exactly two values (one satisfies the other satisfies ) for each value of (when )

If what you are saying is true then there must be a mistake above - I can't spot one, but maybe that's because I'm tired, who knows!
I know what I am saying is true because Euler showed it:

http://en.wikipedia.org/wiki/Tetrati...finite_heights

I think the problem is that you are assuming the inverse is valid for all y, when it is only valid for y where there is a value for x which gives that y (if that makes sense). So while 4^1/4 and 2^1/2 give the same number for x (sqrt(2)), having y=4 is not valid because there is no value for x which gives y=4, however there is a value for x which gives y=2.

I think its the same problem as saying that the inverse of y=sqrt(x) is x=y^2, in the latter you could have y=3 or y=-3, both giving x=9, but y cannot be -3 because sqrt(x) is always positive.

I think the whole problem boils down to us not doing it rigourously enough so we end up with an inverse which isn't quite right.
13. Re: A Summer of Maths
(Original post by Lord of the Flies)
Ah yes I should have thought of that! Phew! At least I'm not insane.
As above, y=4 does not work because there is no value for x giving y=4, if you say x=sqrt(2) then just by estimating it using excel it clearly converges to 2, and only to 2, it does not ossilate.
14. Re: A Summer of Maths
(Original post by james22)
I know what I am saying is true because Euler showed it:

http://en.wikipedia.org/wiki/Tetrati...finite_heights

I think the problem is that you are assuming the inverse is valid for all y, when it is only valid for y where there is a value for x which gives that y (if that makes sense). So while 4^1/4 and 2^1/2 give the same number for x (sqrt(2)), having y=4 is not valid because there is no value for x which gives y=4, however there is a value for x which gives y=2.

I think its the same problem as saying that the inverse of y=sqrt(x) is x=y^2, in the latter you could have y=3 or y=-3, both giving x=9, but y cannot be -3 because sqrt(x) is always positive.

I think the whole problem boils down to us not doing it rigourously enough so we end up with an inverse which isn't quite right.
Yes I see. Do you know how one would go about justifying the fact that the inverse only exists for 0<x<1 then?
15. Re: A Summer of Maths
(Original post by Lord of the Flies)
Yes I see. Do you know how one would go about justifying the fact that the inverse only exists for 0<x<1 then?
I'll have to think about that, but actually it exists for e^-1<=x<=e^1/e
16. Re: A Summer of Maths
Haven't thought about it in depth, but I think it should be defined for the soln set of x^y = y^x.
17. Re: A Summer of Maths
(Original post by james22)
I'll have to think about that, but actually it exists for e^-1<=x<=e^1/e
Sorry, typo, that's what I meant
18. Re: A Summer of Maths
(Original post by Lord of the Flies)
Here's an easy one!

Question

Evaluate

Required

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L'Hôpital's rule

Solution:

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for all .

Lemma 0. For every with the inequality holds.

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Observe that for all , and hence, we get

for a real number

as desired. Equality iff .

Lemma 1. For every we have for all .

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Define .

Now, we firstly observe that

by applying Lemma 0.

and secondly, since in the interval, we see that

which implies that with equality at .

Hence, is increasing on with a minimum .

Therefore, we see that for by Lemma 1.

By another application of Lemma 0, we can obtain the following inequality

in the interval denoted above.

Hence, to compute we notice that the limit of the product is the product of the limits.

Now, by Lemma 0, we see and similar argument shows that in the interval, so that

Therefore, by the Squeeze theorem, we find that , but then by the same theorem

for all .
Last edited by jack.hadamard; 03-07-2012 at 17:00.
19. Re: A Summer of Maths
(Original post by Farhan.Hanif93)
I'll finish it off when I get back if someone hasn't already.
Are you still interested in this question?

The way I approached it.
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I started from this question..

..
Let be an arithmetic sequence with non-zero terms.

Obtain an expression for .
...
and then derived a formula for and it turned out to be an interesting application of it.
20. Re: A Summer of Maths
...
Nice! There are two far quicker ways of doing it though...

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Lemma:
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Let . Conditions for applying l'Hôpital's rule are met, therefore

First method

The limit of both brackets when is by the above lemma, hence

Second method

Applying l'Hôpital's directly to the initial limit gives (differentiating for where is real, positive, and minimal):

Last edited by Lord of the Flies; 03-07-2012 at 18:03. Reason: Forgot an 'n'