A Summer of Maths

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  1. Lord of the Flies's Avatar
    81 03-07-2012  11:21
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    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. Brit_Miller's Avatar
    82 03-07-2012  11:26
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    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. Lord of the Flies's Avatar
    83 03-07-2012  11:29
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    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. Brit_Miller's Avatar
    84 03-07-2012  11:46
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    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. :eek:
  5. james22's Avatar
    85 03-07-2012  12:03
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    Re: A Summer of Maths
    (Original post by Lord of the Flies)
    Really? I don't see how... When 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.
  6. Lord of the Flies's Avatar
    86 03-07-2012  12:14
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    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. james22's Avatar
    87 03-07-2012  12:21
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    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. Lord of the Flies's Avatar
    88 03-07-2012  12:40
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    Re: A Summer of Maths
    (Original post by james22)
    ...
    Also, there is a major difference between 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:
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    y=x^y\Rightarrow x=y^{\frac{1}{y}}\Rightarrow x'=\dfrac{1-\ln y}{y}

    Therefore:

    \bullet\;\;x strictly increases for 1<y<e and 1<x<e^{1/e}

    \bullet\;\;x strictly decreases for y>e and 1<x<e^{1/e}\quad(\displaystyle \lim_{+\infty}y^{\frac{1}{y}}=1).

    These two statements imply that there are exactly two values y (one satisfies 1<y<e the other satisfies y>e) for each value of x (when 1<x<e^{1/e})


    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. Lord of the Flies's Avatar
    89 03-07-2012  12:49
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    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:

    y=\sqrt{2}^y\Rightarrow y=2\;\text{or}\;4
    Last edited by Lord of the Flies; 03-07-2012 at 13:27.
  10. Blutooth's Avatar
    90 03-07-2012  13:12
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    Re: A Summer of Maths
    (Original post by Lord of the Flies)
    Hm yes indeed...

    y=\sqrt{2}^y\Rightarrow y=2

    I don't understand why my reasoning is wrong though...
    or y=4. Your reasoning is correct.
  11. Lord of the Flies's Avatar
    91 03-07-2012  13:16
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    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. james22's Avatar
    92 03-07-2012  13:26
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    Re: A Summer of Maths
    (Original post by Lord of the Flies)
    Also, there is a major difference between 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:
    Show

    y=x^y\Rightarrow x=y^{\frac{1}{y}}\Rightarrow x'=\dfrac{1-\ln y}{y}

    Therefore:

    \bullet\;\;x strictly increases for 1<y<e and 1<x<e^{1/e}

    \bullet\;\;x strictly decreases for y>e and 1<x<e^{1/e}\quad(\displaystyle \lim_{+\infty}y^{\frac{1}{y}}=1).

    These two statements imply that there are exactly two values y (one satisfies 1<y<e the other satisfies y>e) for each value of x (when 1<x<e^{1/e})


    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. james22's Avatar
    93 03-07-2012  13:27
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    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. Lord of the Flies's Avatar
    94 03-07-2012  13:31
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    • Location: Paris, France
    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. james22's Avatar
    95 03-07-2012  13:42
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    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. Zuzuzu's Avatar
    96 03-07-2012  14:01
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    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. Lord of the Flies's Avatar
    97 03-07-2012  15:00
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    • Location: Paris, France
    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. jack.hadamard's Avatar
    98 03-07-2012  16:54
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    Re: A Summer of Maths
    (Original post by Lord of the Flies)
    Here's an easy one!

    Question

    Evaluate \displaystyle\lim_{x\to0}\frac{ \sin x^n}{\sin^n x}\quad (n&gt;0)

    Required

    Spoiler:
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    L'Hôpital's rule

    Solution:

    Spoiler:
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    \displaystyle \lim_{x \to 0}\ \frac{\sin(x^n)}{\sin^n(x)} = 1 for all n \in \mathbb{N}.


    Lemma 0. For every x \in \mathbb{R} with x \geq 0 the inequality \sin(x) \leq x holds.

    Spoiler:
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    Observe that 1 - \cos(x) \geq 0 for all x, and hence, we get

    \displaystyle \int_{0}^{x} 1 - \cos(x)\ dx \geq 0 for a real number x \geq 0

    \implies x - \sin(x) \geq 0

    as desired. Equality iff x = 0.


    Lemma 1. For every x \in [0, 1] we have \sin(x^n) \geq \sin^n(x) for all n \in \mathbb{N}.

    Spoiler:
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    Define \phi(x) = \sin(x^n) - \sin^n(x).


    Now, we firstly observe that

    \displaystyle \frac{d\phi}{dx}\ =\ nx^{n-1}\cos(x^n) - n\sin^{n-1}(x)\cos(x)

    \ \ \ \ \ \ \geq\ nx^{n-1}\cos(x^n) - nx^{n-1}\cos(x) by applying Lemma 0.

    \ \ \ \ \ \ \geq\ nx^{n-1}\left[\cos(x^n) - \cos(x) \right]


    and secondly, since x^n \leq x in the interval, we see that

    \cos(x^n) \geq \cos(x)

    which implies that \displaystyle \frac{d\phi}{dx} \geq 0 with equality at x = 0.

    Hence, \phi(x) is increasing on (0,1] with a minimum \phi(0) = 0.


    Therefore, we see that \displaystyle \frac{\sin(x^n)}{\sin^n(x)} \geq 1 for x \in (0,1] by Lemma 1.


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

    \displaystyle 1\ \leq\ \frac{\sin(x^n)}{\sin^n(x)}\ \leq\ \frac{x^n}{\sin^n(x)}

    in the interval denoted above.


    Hence, to compute \displaystyle \lim_{x \to 0} \left(\frac{x}{\sin x}\right)^n we notice that the limit of the product is the product of the limits.

    Now, by Lemma 0, we see \displaystyle 1 \leq \frac{x}{\sin x} and similar argument shows that \tan(x) \geq x in the interval, so that \displaystyle \frac{x}{\sin x} \leq \frac{1}{\cos x}

    Therefore, by the Squeeze theorem, we find that \displaystyle \lim_{x \to 0} \left(\frac{x}{\sin x}\right)^n = 1, but then by the same theorem

    \displaystyle \lim_{x \to 0}\ \frac{\sin(x^n)}{\sin^n(x)} = 1

    for all n \in \mathbb{N}.
    Last edited by jack.hadamard; 03-07-2012 at 17:00.
  19. jack.hadamard's Avatar
    99 03-07-2012  17:30
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    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.
    Spoiler:
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    I started from this question..

    (Original post by jack.hadamard)
    ..
    Let a_1, a_2, a_3, ..., a_n be an arithmetic sequence with non-zero terms.

    Obtain an expression for \displaystyle \sum_{k=1}^{n-1} \frac{1}{a_k a_{k+1}}.
    ...
    and then derived a formula for \displaystyle \sum_{k=1}^{n-i} \frac{1}{a_k a_{k+1} a_{k+2} \cdots a_{k + (i-1)} a_{k + i}} and it turned out to be an interesting application of it.
  20. Lord of the Flies's Avatar
    100 03-07-2012  17:33
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    • Location: Paris, France
    Re: A Summer of Maths
    (Original post by jack.hadamard)
    ...
    Nice! There are two far quicker ways of doing it though...

    Spoiler:
    Show


    Lemma: \displaystyle\lim_{x\to0} \frac{ \sin x}{x}=\frac{x}{\sin x}=1
    Spoiler:
    Show

    Let f(x)=\dfrac{\sin x}{x}. Conditions for applying l'Hôpital's rule are met, therefore \displaystyle\lim_{x\to0} f(x)=\lim_{x\to0}\cos x =1

    Additionally, \displaystyle\lim_{x\to0} f(x)^{-1}=(1)^{-1}=1



    First method


    \displaystyle\frac{\sin x^n}{ \sin^n x}=\frac{x^n \sin x^n}{x^n \sin^n x}=\left(\frac{\sin x^n}{x^n}\right)\left(\frac{x}{ \sin x}\right)^n

    The limit of both brackets when x\to0 is 1 by the above lemma, hence \displaystyle \lim_{x \to 0}\ \frac{\sin x^n}{\sin^n x} = 1

    Second method

    Applying l'Hôpital's directly to the initial limit gives (differentiating for -a&lt;x&lt;a where a is real, positive, and minimal):


    \displaystyle\lim_{x \to 0}\frac{ \sin x^n}{ \sin^n x}= \lim_{x \to 0}\frac{nx^{n-1}\cos x^n}{n\cos x \sin^{n-1}x}=\lim_{x \to 0}\left(\frac{\cos x^n}{\cos x}\right)\left(\frac{x}{\sin x}\right)^{n-1}=\frac{1}{1}\cdot(1)^{n-1}=1
    Last edited by Lord of the Flies; 03-07-2012 at 18:03. Reason: Forgot an 'n'
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