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A Summer of Maths

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Original post by Brit_Miller
Ah, I don't know enough about it to know what to do then. :biggrin:


It just requires a little manipulation. :smile:
Original post by Lord of the Flies
It just requires a little manipulation. :smile:


I'll leave it to someone better, I just can't see it. :eek:
Reply 82
Avatar for james22
james22
16
Original post by Lord of the Flies
Really? I don't see how... When 1<x<e1/e,  f(x)1<x<e^{1/e},\;f(x) 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.
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.
Reply 84
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james22
16
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.
Original post by james22
...


Also, there is a major difference between f(x)=xxx...f(x)=x^{x^{x^{...}}} 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



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!
(edited 11 years ago)
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:

y=2y⇒y=2  or  4y=\sqrt{2}^y\Rightarrow y=2\;\text{or}\;4
(edited 11 years ago)
Reply 87
Avatar for Blutooth
Blutooth
16
Original post by Lord of the Flies
Hm yes indeed...

y=2y⇒y=2y=\sqrt{2}^y\Rightarrow y=2

I don't understand why my reasoning is wrong though...


or y=4. Your reasoning is correct.
Reply 88
Avatar for james22
james22
16
Original post by Lord of the Flies
Also, there is a major difference between f(x)=xxx...f(x)=x^{x^{x^{...}}} 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



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/Tetration#Extension_to_infinite_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.
Reply 89
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james22
16
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.
Original post by james22
I know what I am saying is true because Euler showed it:

http://en.wikipedia.org/wiki/Tetration#Extension_to_infinite_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?
Reply 91
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james22
16
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
Reply 92
Avatar for Zuzuzu
Zuzuzu
12
Haven't thought about it in depth, but I think it should be defined for the soln set of x^y = y^x.
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 :wink:
Original post by Lord of the Flies
Here's an easy one!

Question

Evaluate lim⁡x→0sin⁡xnsin⁡nx(n>0)\displaystyle\lim_{x\to0}\frac{ \sin x^n}{\sin^n x}\quad (n>0)

Required

Spoiler




Solution:

Spoiler

(edited 11 years ago)
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? :tongue:


The way I approached it.

Spoiler

Original post by jack.hadamard
...


Nice! There are two far quicker ways of doing it though... :tongue:

Spoiler

(edited 11 years ago)
Original post by Lord of the Flies
Nice! There are two far quicker ways of doing it though... :tongue:


Spoiler

Original post by Lord of the Flies
Nice! There are two far quicker ways of doing it though... :tongue:

Spoiler



Ahhh, I was close to doing the second method, but I wasn't sure if you were allowed to split it up like that. Wouldn't have recognised the 1st method. Nice question and answer! :smile:
Reply 99
Avatar for Blutooth
Blutooth
16
Original post by jack.hadamard
Solution:

Spoiler



Didn't want to use L'Hopital?

Edit: nvm read post above. I did it by the second method, though I prefer the first method of Lord's.
(edited 11 years ago)

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