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    (Original post by Zacken)
    just a few suggestions, you might want to add in what's wrong with the AM-GM inequality "proof" that you showed and how to fix it
    Will edit now.
    (Original post by Zacken)
    P.S: Nice touch adding the proof by contrapositive, I just want to add my thoughts on it for anybody who doesn't see why A \implies B \iff \neg B \implies \neg A
    I'm a big fan of the contrapositive
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    (Original post by joostan)
    Will edit now.
    Thanks!

    I'm a big fan of the contrapositive
    As am I - I've only recently started learning off the logic stuff and it's often much easier to do A \implies B then B \implies A by making one of them the contrapositive because it's rare to have both directions of the implication be easy.

    On the other hand... there's many a time I've skipped a step and done A \implies B and then \neg B \implies \neg A instead of \neg A \implies \neg Band realise that I've proven the same thing twice - massive facepalms. :laugh:
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    (Original post by Zacken)
    One recurring theme is to exploit the symmetry of trigonometric functions during integrating. Using this in conjunction with:

    \displaystyle

\begin{equation*}\int_a^b f(x) \, \mathrm{d}x = \int_a^b f(a+b-x) \, \mathrm{d}x\end{equation*}

    is particularly useful.

    Here's an example: \int_0^{\pi/4} \log (1 + \tan \theta) \, \mathrm{d}\theta. How is this relevant here? Call our integral I, then using the substitution \theta \mapsto \frac{\pi}{4} - \theta we have:

    \displaystyle 

\begin{equation*} I = \int_0^{\pi/4} \log \frac{2}{1 + \tan \theta} \, \mathrm{d}\theta = \frac{\pi}{4}\ln 2 - I \iff I = \frac{\pi}{8}\ln 2\end{equation*}

    Extension to this, using the same idea, roughly - attempt \displaystyle \int_0^{\pi/2} \log \sin \theta \, \mathrm{d}\theta
    Am I on the right track?

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    (Original post by Insight314)
    Am I on the right track?

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    You messed up your negative signs, you should get \int \ln \sin \theta \, \mathrm{d}\theta = \int \ln \cos \theta \, \mathrm{d}\theta \neq -\int \ln \cos \theta \, \mathrm{d}\theta with the limits from 0\, to \pi/2 on each integral but I can't be arsed typesetting.

    Then you should add both integrals to get 2I = \int \ln (\sin \theta \cos  \theta) \, \mathrm{d}\theta with the usual limits and then think about \sin 2\theta.
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    (Original post by Zacken)
    You messed up your negative signs, you should get \int \ln \sin \theta \, \mathrm{d}\theta = \int \ln \cos \theta \, \mathrm{d}\theta \neq -\int \ln \cos \theta \, \mathrm{d}\theta with the limits from 0\, to \pi/2 on each integral but I can't be arsed typesetting.

    Then you should add both integrals to get 2I = \int \ln (\sin \theta \cos  \theta) \, \mathrm{d}\theta with the usual limits and then think about \sin 2\theta.
    But doesn't the substitution change the limits from 0 to pi/2, to pi/2 to 0 so then I switch it by adding a negative sign.

    I actually did it that way the first time I tried the integral but then I noticed that the substitution changes the limits.

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    (Original post by Insight314)
    But doesn't the substitution change the limits from 0 to pi/2, to pi/2 to 0 so then I switch it by adding a negative sign.

    I actually did it that way the first time I tried the integral but then I noticed that the substitution changes the limits.

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    Yes - but you forgot that u = \pi/2 - x adds in a factor of -1 since \mathrm{d}u/\mathrm{d}x = -1. This is why I would caution using the notation u = f(x) instead of x \mapsto f(x) at this stage. You can use the fancy notation when you're more comfortable with everything.
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    (Original post by Zacken)
    Yes - but you forgot that u = \pi/2 - x adds in a factor of -1 since \mathrm{d}u/\mathrm{d}x = -1.
    Oh **** yes. Also, I remember that then you get an integral with ln 2 or something like that as integrand which is easy to calculate but the other integral with an integrand \sin2\theta and then you do the same addition of 1/2 integral of cos + 1/2 integral of sin but with 2\theta this time and then you continue ad infinitum. How does this simplify it? I can't find the workings out on this integral I did yesterday, so I might be wrong.
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    (Original post by Insight314)
    Oh **** yes. Also, I remember that then you get an integral with ln 2 or something like that as integrand which is easy to calculate but the other integral with an integrand \sin2\theta and then you do the same addition of 1/2 integral of cos + 1/2 integral of sin but with 2\theta this time and then you continue ad infinitum. How does this simplify it? I can't find the workings out on this integral I did yesterday, so I might be wrong.
    \displaystyle 2I = \int_0^{\pi/2} \ln \left(\frac{1}{2} \sin 2\theta \right) \, \mathrm{d}\theta = \int_0^{\pi/2} \ln \frac{1}{2} \, \mathrm{d}\theta + \int_0^{\pi/2} \ln \sin 2\theta \, \mathrm{d}\theta

    But you want an I in there somewhere, so the sub u = 2x (x \mapsto \frac{x}{2}) gets you that and some nice thinking about the symmetry of sine gets you the home run.
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    (Original post by Zacken)
    \displaystyle 2I = \int_0^{\pi/2} \ln \left(\frac{1}{2} \sin 2\theta \right) \, \mathrm{d}\theta = \int_0^{\pi/2} \ln \frac{1}{2} \, \mathrm{d}\theta + \int_0^{\pi/2} \ln \sin 2\theta \, \mathrm{d}\theta

    But you want an I in there somewhere, so the sub u = 2x (x \mapsto \frac{x}{2}) gets you that and some nice thinking about the symmetry of sine gets you the home run.
    I got I = -\frac{\pi}{3}\ln 2
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    (Original post by Insight314)
    I got I = -\frac{\pi}{3}\ln 2
    Haha, you're the third person to give me that incorrect answer since I've posted that problem.
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    (Original post by Zacken)
    Haha, you're the third person to give me that incorrect answer since I've posted that problem.
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    The limits change after the sub u = 2x.
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    (Original post by Zacken)
    We know that A \implies B is a true statement literally every single time unless we have the case that A is false and B is true, we can rewrite this as: A \implies B is a true statement unless \neg A \wedge B.
    :no: Look at some truth tables. https://en.wikipedia.org/wiki/Truth_table

    The conditional connective, is false, if, and only if the premise is true but the conclusion is false.
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    (Original post by Zacken)
    The limits change after the sub u = 2x.
    And I did change them? Then I split the integral into limits consisting of 0 \text{ to } \pi/2 each.
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    (Original post by zetamcfc)
    :no: Look at some truth tables. https://en.wikipedia.org/wiki/Truth_table

    The conditional connective, is false, if, and only if the premise is true but the conclusion is false.
    :facepalm: That's what I meant, I'll edit that. Thanks.
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    (Original post by Insight314)
    And I did change them? Then I split the integral into limits consisting of 0 \text{ to } \pi/2 each.
    Oh, yes.

    You seem to be saying that \int_0^{\pi} \ln \sin u \, \mathrm{d}u = 2 \times \frac{1}{2} \int_0^{\pi/2} \ln \sin u \, \mathrm{d}u.

    You want to say that \int_0^{\pi} \ln \sin u \, \mathrm{d}u = 2I \neq I
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    (Original post by Zacken)
    Oh, yes.

    You seem to be saying that \int_0^{\pi} \ln \sin u \, \mathrm{d}u = 2 \times \frac{1}{2} \int_0^{\pi/2} \ln \sin u \, \mathrm{d}u.

    You want to say that \int_0^{\pi} \ln \sin u \, \mathrm{d}u = 2I \neq I
     u = 2\theta \Rightarrow \frac{du}{d\theta} = 2 \text{, so } d\theta = \frac{1}{2} du which is where this 1/2 comes from.
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    (Original post by Insight314)
     u = 2\theta \Rightarrow \frac{du}{d\theta} = 2 \text{, so } d\theta = \frac{1}{2} du which is where this 1/2 comes from.
    But you multiplied the entire thing by 2 the line before.

    We have 2I = -\frac{\pi}{2} \ln 2 + \frac{1}{2} \int_0^{\pi} \ln \sin u \ \mathrm{d}u

    You decided to multiply everything by 2 to get: 4I = -\pi \ln 2 + \int_0^{\pi} \ln \sin u \, \mathrm{d}u.

    But we know that the last term is 2I so 4I = -\pi \ln 2 + 2I \iff I = -\frac{\pi}{2}\ln 2
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    (Original post by Zacken)
    But you multiplied the entire thing by 2 the line before.

    We have 2I = -\frac{\pi}{2} \ln 2 + \frac{1}{2} \int_0^{\pi} \ln \sin u \ \mathrm{d}u

    You decided to multiply everything by 2 to get: 4I = -\pi \ln 2 + \int_0^{\pi} \ln \sin u \, \mathrm{d}u.

    But we know that the last term is 2I so 4I = -\pi \ln 2 + 2I \iff I = -\frac{\pi}{2}\ln 2
    Oh yes, I saw my mistake. I think that was more of a silly mistake to be quite honest.

    This was a really fun question, thanks a lot.
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    (Original post by Insight314)
    Oh yes, I saw my mistake. I think that was more of a silly mistake to be quite honest.

    This was a really fun question, thanks a lot.
    It was.

    No problem! I'll try and think up some more when I've got some free time on my hands. :-)
 
 
 
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