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Thread: Fact check

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Fact check

    Just to make sure, I am pretty sure that I am right but $\displaystyle \text{ As }x\to{0}\csc(u(x))\sim\frac{1}{u(x)}$

    Provided $\displaystyle u(0)=0$

    Here is how I decided upon that

    $\displaystyle \lim_{x\to{0}}\frac{\csc(u(x))}{\frac{1}{u(x)}}=\l im_{u(x)\to{0}}\frac{u(x)}{\sin(u(x)}$

    Now let $\displaystyle \psi=u(x)$

    and as $\displaystyle x\to{0}\Rightarrow\psi\to{0}$

    $\displaystyle \therefore\lim_{x\to{0}}\frac{u(x)}{\sin(u(x))}=\l im_{\psi\to{0}}\frac{\psi}{\sin(\psi)}=L$

    so $\displaystyle \frac{1}{L}=\lim_{\psi\to{0}}\frac{\sin(\psi)}{\ps i}=\lim_{\psi\to{0}}\frac{\sin(\psi)-\sin(0)}{\psi-0}=\cos(0)=1$

    So $\displaystyle \frac{1}{L}=1\Rightarrow{L=1}$

    $\displaystyle \therefore\csc(u(x))\sim\frac{1}{u(x)}$

    Something seems wrong

    Could someone verify this? This is for something I don't want to make a factual error on. Thanks in advance
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  2. #2
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    Quote Originally Posted by Mathstud28 View Post
    Just to make sure, I am pretty sure that I am right but $\displaystyle \text{ As }x\to{0}\csc(u(x))\sim\frac{1}{u(x)}$

    Provided $\displaystyle u(0)=0$

    Here is how I decided upon that

    $\displaystyle \lim_{x\to{0}}\frac{\csc(u(x))}{\frac{1}{u(x)}}=\l im_{u(x)\to{0}}\frac{u(x)}{\sin(u(x)}$

    Now let $\displaystyle \psi=u(x)$

    and as $\displaystyle x\to{0}\Rightarrow\psi\to{0}$

    $\displaystyle \therefore\lim_{x\to{0}}\frac{u(x)}{\sin(u(x))}=\l im_{\psi\to{0}}\frac{\psi}{\sin(\psi)}=L$

    so $\displaystyle \frac{1}{L}=\lim_{\psi\to{0}}\frac{\sin(\psi)}{\ps i}=\lim_{\psi\to{0}}\frac{\sin(\psi)-\sin(0)}{\psi-0}=\cos(0)=1$

    So $\displaystyle \frac{1}{L}=1\Rightarrow{L=1}$

    $\displaystyle \therefore\csc(u(x))\sim\frac{1}{u(x)}$

    Something seems wrong

    Could someone verify this? This is for something I don't want to make a factual error on. Thanks in advance
    The Laurent Series of cosec x is $\displaystyle \frac{1}{x} + \frac{x}{6} + \frac{7 x^3}{360} + ..... $ so

    $\displaystyle \text{cosec} \, u = \frac{1}{u} + \frac{u}{6} + \frac{7 u^3}{360} + ..... $

    As to what happens when x --> 0, you obviously need to know something about the behaviour of u = u(x) .......

    What you've got is valid I think when u ---> 0 as x --> 0 ......
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mr fantastic View Post
    The Laurent Series of cosec x is $\displaystyle \frac{1}{x} + \frac{x}{6} + \frac{7 x^3}{360} + ..... $ so

    $\displaystyle \text{cosec} \, u = \frac{1}{u} + \frac{u}{6} + \frac{7 u^3}{360} + ..... $

    As to what happens when x --> 0, you obviously need to know something about the behaviour of u = u(x) .......

    What you've got is valid I think when u ---> 0 as x --> 0 ......
    Thanks a lot Mr. Fantastic, I am sure that if you think its right, it will be right enough for students of my age.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Two more things, is it safe to say the following, are therea any caveats?

    $\displaystyle \lim_{x\to\infty}f(x)^{\frac{1}{x}}=\lim_{x\to\inf ty}\frac{f(x+1)}{f(x)}$

    and if $\displaystyle \lim_{x\to{c}}|f(x)|$ converges then $\displaystyle \lim_{x\to{c}}f(x)$ converges as well

    I am pretty dang sure, but never hurts to check

    Just to make sure I am not misuderstood

    I know that

    if $\displaystyle \lim_{x\to\infty}|f(x)|=0$

    then $\displaystyle \lim_{x\to\infty}f(x)=0$

    But I remember reading this assertion somewhere, I know it has pitfalls, but what are the restrictions

    the more and more I think about it, the more I think its completely incorrect except at the few exceptions
    Last edited by Mathstud28; Jun 10th 2008 at 10:14 PM.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Does it work if f(x) = 1/x, for example?
    Yes..yes it does

    $\displaystyle \lim_{x\to\infty}\bigg(\frac{1}{x}\bigg)^{\frac{1} {x}}$

    let $\displaystyle \xi=\frac{1}{x}$

    Now as $\displaystyle x\to\infty\Rightarrow\xi\to{0}$

    Giving us the commonly known limit

    $\displaystyle \lim_{\xi\to{0}}\xi^{\xi}=1$

    and from this theroem

    $\displaystyle \lim_{x\to\infty}\bigg(\frac{1}{x}\bigg)^{\frac{1} {x}}=\lim_{x\to\infty}\frac{\frac{1}{x+1}}{\frac{1 }{x}}=\lim_{x\to\infty}\frac{x}{x+1}=1$

    But for my second I thought of a counter example

    $\displaystyle f(x)=(-1)^x$

    So what are the restrictions?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Can I make this generalization?

    Let $\displaystyle f(\tau,\zeta,\theta)=\cos^{\tau}(\zeta\theta)$ where both $\displaystyle \tau$ and $\displaystyle \zeta$ are fixed constants, and $\displaystyle \tau\in\mathbb{Z}$

    that $\displaystyle -1\leq{f(\tau,\zeta,\theta)}\leq{1}$?
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  7. #7
    Behold, the power of SARDINES!
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    Quote Originally Posted by Mathstud28 View Post
    Can I make this generalization?

    Let $\displaystyle f(\tau,\zeta,\theta)=\cos^{\tau}(\zeta\theta)$ where both $\displaystyle \tau$ and $\displaystyle \zeta$ are fixed constants, and $\displaystyle \tau\in\mathbb{Z}$

    that $\displaystyle -1\leq{f(\tau,\zeta,\theta)}\leq{1}$?
    As long as $\displaystyle \zeta \in \mathbb{R}$ the cosine is bounded between -1 and 1.

    $\displaystyle \cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$

    The cosine function is not bounded in the complex plane.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    As long as $\displaystyle \zeta \in \mathbb{R}$ the cosine is bounded between -1 and 1.

    $\displaystyle \cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$

    The cosine function is not bounded in the complex plane.
    Thanks EmptySet, But I should think you would know me a little better than that...I do know the bounds of cosine and its unboundedness in the complex plane, what I am saying is that if I said this in a formal setting would it be held as true, does the exponent change anything? I know that for a tau of even parity we have that

    $\displaystyle 0\leq\cos^{\tau}(\zeta\theta)\leq{1}$

    and for odd parity

    $\displaystyle -1\leq\cos^{\tau}(\zeta\theta)\leq{1}$

    So can what I said be completely true, or is there a caveat I must include, I do not think so but I should be safe
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  9. #9
    Behold, the power of SARDINES!
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    Quote Originally Posted by Mathstud28 View Post
    Thanks EmptySet, But I should think you would know me a little better than that...I do know the bounds of cosine and its unboundedness in the complex plane, what I am saying is that if I said this in a formal setting would it be held as true, does the exponent change anything? I know that for a tau of even parity we have that

    $\displaystyle 0\leq\cos^{\tau}(\zeta\theta)\leq{1}$

    and for odd parity

    $\displaystyle -1\leq\cos^{\tau}(\zeta\theta)\leq{1}$

    So can what I said be completely true, or is there a caveat I must include, I do not think so but I should be safe
    I think that your analysis is correct. (But then again I may be crazy )

    Sorry I guess I missed the point of your question. It did seem odd for you to ask....
    Rock On...
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