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    (Original post by atsruser)
    Would you like to expand on that a bit? Not sure I see where the partial fractions and geometric series will arise. (Though if you can end up with a series in \cos nx that will be very useful)
    Sorry for the late reply! I've seen the method used in a similar problem before, I'll look it up and link it. :-)

    http://math.stackexchange.com/questi...osnx5-4-cosxdx
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    (Original post by Zacken)
    Sorry for the late reply! I've seen the method used in a similar problem before, I'll look it up and link it. :-)

    http://math.stackexchange.com/questi...osnx5-4-cosxdx
    That's a pretty nice approach. In fact, it relies on a similar technique to that I used in Problem 540. I'll put up a solution to that tonight.
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    (Original post by atsruser)
    That's a pretty nice approach. In fact, it relies on a similar technique to that I used in Problem 540. I'll put up a solution to that tonight.
    Yeah, I thought it was nice - personally find it better than complex residues method (especially because I haven't learnt those yet and they're out of my reach for the moment. )
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    (Original post by Zacken)
    Sorry for the late reply! I've seen the method used in a similar problem before, I'll look it up and link it. :-)

    http://math.stackexchange.com/questi...osnx5-4-cosxdx
    Damn, that's the method that I personally used to do it, though I now know that it's a special case of a more general result.

    For this the "traditional method" is messy and life's too short. Nevertheless, for the sake of completeness, I'll post that solution as an alternative once atsruser has done the honours
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    (Original post by Indeterminate)
    Damn, that's the method that I personally used to do it, though I now know that it's a special case of a more general result.

    For this the "traditional method" is messy and life's too short. Nevertheless, for the sake of completeness, I'll post that solution as an alternative once atsruser has done the honours
    I guess by doing the honours you mean post my solution to problem 544? If so, I'm happy to do that, but in fact I meant that I was going to post a solution to my problem 540, which seems to have stumped everyone. I'll do yours first though.
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    (Original post by atsruser)
    I guess by doing the honours you mean post my solution to problem 544? If so, I'm happy to do that, but in fact I meant that I was going to post a solution to my problem 540, which seems to have stumped everyone. I'll do yours first though.
    Ignore my last quote - I misread your post.

    By all means do 540 first. Meanwhile I'll get the ugly solution to 544 of the way. You can do the neat one when you have time
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    (Original post by Indeterminate)
    Ignore my last quote - I misread your post.

    By all means do 540 first. Meanwhile I'll get the ugly solution to 544 of the way. You can do the neat one when you have time
    Err, which is the ugly solution? The contour integral? If so, I've already started typing it up, but if you're going to do so, I'll save the effort.
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    (Original post by atsruser)
    Err, which is the ugly solution? The contour integral? If so, I've already started typing it up, but if you're going to do so, I'll save the effort.
    Yeah that one. I wouldn't want to impose it on anyone other than myself :lol:
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    (Original post by Indeterminate)
    Problem 544***

    Show that

    \displaystyle \int_{0}^{2\pi} \dfrac{\cos 3x}{5 - 4\cos x} \ dx = \dfrac{\pi}{12}
    Solution:
    Spoiler:
    Show

    Let \displaystyle I=\int_0^{2\pi} \frac{\cos 3x}{5-4\cos x} \ dx and let z=e^{ix} = \cos x + i\sin x

    Then \displaystyle dz =i e^{ix} dx = iz dx \Rightarrow dx = -i \frac{dz}{z}

    and

    \displaystyle 2\cos x = z+\bar{z} = z+\frac{1}{z}, 2\cos 3x =z^3+\frac{1}{z^3}

    So on the unit circle (i.e. when z=e^{ix}), we can write

    \displaystyle \frac{\cos 3x}{5-4\cos x} = \frac{1/2(z^3+1/z^3)}{5-2(z+1/z)}

    This says that on the unit circle, the expression on the RHS is purely real, with values as given by the LHS. Since x \in [0,2\pi] traverses the unit circle, we can write:

    \displaystyle I = \int_C \frac{1/2(z^3+1/z^3)}{5-2(z+1/z)} \cdot -i \frac{dz}{z} = \frac{i}{2} \int_C \frac{z^6+1}{z^3(2z^2-5z+2)} \ dz

    where C is the unit circle.

    By the Residue Theorem, \int_C f(z) \ dz = 2\pi i \sum \text{residues of f inside C}.

    Now 2z^2-5z+2=2(z-1/2)(z-2) and writing \displaystyle f(z)=\frac{z^6+1}{z^3(2z^2-5z+2)} we see that f(z) has a pole of order 3 at z=0 and a pole of order 1 at z=1/2, with the other pole being outside the unit circle, so irrelevant here.

    We have:

    \displaystyle \text{res}(f(z),1/2) = \lim_{z \rightarrow 1/2}(z-1/2)f(z) = -\frac{65}{24}
    \displaystyle \text{res}(f(z),0) = \lim_{z \rightarrow 0} \frac{1}{2!} z^3 f''(z) = \frac{21}{8}

    as can easily be shown if you can be bothered with the tedium. I plugged the second one into Wolfram. (Sue me)

    Hence I = \frac{i}{2} \times 2\pi i(\frac{21}{8}-\frac{65}{24}) = \frac{\pi}{12}

    QED.

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    This will either be trivial or reasonably tricky.

    Problem 545***

    Take f:\;\mathbb{R}\to\mathbb{N}. Show that for some n the set of reals mapped to n has cardinality of the continuum.
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    (Original post by Lord of the Flies)
    This will either be trivial or reasonably tricky.

    Problem 545***

    Take f:\;\mathbb{R}\to\mathbb{N}. Show that for some n the set of reals mapped to n has cardinality of the continuum.
    Here's an attempt.

    Spoiler:
    Show

    Assume that for all n \in \mathbb{N}, the preimage f^{-1}(n) is countable. Then the domain D of f is the union of all preimages of its range, which is a countable union of countable sets, so countable. Hence the cardinality of D is not the cardinality of \mathbb{R}, and we have a contradiction. Hence for some n \in \mathbb{N}, f^{-1}(n) has the cardinality of \mathbb{R}.
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    (Original post by atsruser)
    Here's an attempt.
    Spoiler:
    Show

    Assume that for all n \in \mathbb{N}, the preimage f^{-1}(n) is countable. Then the domain D of f is the union of all preimages of its range, which is a countable union of countable sets, so countable. Hence the cardinality of D is not the cardinality of \mathbb{R}, and we have a contradiction. Hence for some n \in \mathbb{N}, f^{-1}(n) has the cardinality of \mathbb{R}.
    I think you're assuming the continuum hypothesis there you should replace "countable" with "of cardinality less than the continuum" throughout (except in "countable union of", which remains true).
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    (Original post by atsruser)
    ...
    As above. Getting around this apparent issue is the point of the question.
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    (Original post by atsruser)
    Problem 540 (**/***)

    a) Show that:

    \displaystyle \int_0^{2\pi} \frac{r^2\sin^2 2\theta}{1-2r^2\cos 2\theta+r^4} \ d\theta = \pi r^2

    stating any conditions that must hold for this to be valid.

    b) Find the value of \displaystyle \int_0^{2\pi} \frac{4\sin^2 2\theta}{17-8\cos 2\theta} \ d\theta

    (Hints available if necessary)
    Spoiler:
    Show

    a) First note that we can factorise the denominator of the integrand:

    1-2r^2\cos 2\theta +r^4 = (1-r^2\alpha)(1- r^2\beta) \Rightarrow \alpha+\beta = 2\cos 2\theta,\  \alpha \beta =1

    which we can solve to find:

    \alpha = \cos 2\theta +i\sin 2\theta = e^{i2\theta}, \beta = \cos 2\theta -i\sin 2\theta = e^{-i2\theta}

    So

    1-2r^2\cos 2\theta +r^4 = (1-r^2e^{i2\theta})(1-r^2e^{-i2\theta}) = (1-z^2)(1-\bar{z}^2)

    with z=r e^{i\theta}

    This leads us to consider

    \displaystyle \frac{1}{1-z^2} = \frac{1-\bar{z}^2}{(1-z^2)(1-\bar{z}^2)}=\frac{1-r^2\cos 2\theta+ir^2\sin 2\theta}{1-2r^2\cos 2\theta +r^4}

    so that \displaystyle \text{Im} \frac{1}{1-z^2} = \frac{r^2\sin 2\theta}{1-2r^2\cos 2\theta +r^4}

    Now for |z| < 1 \Rightarrow r < 1 we have:

    \displaystyle \frac{1}{1-z^2} = 1+z^2+z^4+ \cdots
    =(1+r^2\cos 2\theta +  r^4\cos 4\theta + \cdots)+i(r^2\sin 2\theta +r^4\sin 4\theta + \cdots)

    Hence \displaystyle \frac{r^2 \sin 2 \theta}{1-2r^2\cos 2\theta +r^4} = r^2\sin 2\theta +r^4\sin 4\theta + \cdots (1)

    But since \displaystyle \int_0^{2\pi} \sin n\theta \sin m\theta \ d\theta = \begin{cases} 0 & n \ne m \\ \pi & n=m \end{cases}

    then on multiplying both sides of (1) by \sin 2\theta, we have:

    \displaystyle \int_0^{2\pi} \frac{r^2 \sin^2 2 \theta}{1-2r^2\cos 2\theta +r^4} \ d\theta = \pi r^2 for 0 < r < 1

    I stole this approach from an example in "Visual Complex Analysis".

    b) The obvious choice here is r=2, but the integral is only defined for 0 < r < 1. However, take r=\frac{1}{2} and we see that:

    \displaystyle \int_0^{2\pi} \frac{\frac{1}{4} \sin^2 2 \theta}{1-2\frac{1}{4}\cos 2\theta +\frac{1}{16}} \ d\theta =  \int_0^{2\pi} \frac{4 \sin^2 2 \theta}{17-8\cos 2\theta} \ d\theta = \frac{\pi}{4}

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    (Original post by Smaug123)
    I think you're assuming the continuum hypothesis there you should replace "countable" with "of cardinality less than the continuum" throughout (except in "countable union of", which remains true).
    That's a good point. Do you think that argument works with that change? I haven't done this kind of stuff in so many years that I'm unable to tell if something subtle will go wrong.
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    (Original post by atsruser)
    Solution:
    Spoiler:
    Show

    Let \displaystyle I=\int_0^{2\pi} \frac{\cos 3x}{5-4\cos x} \ dx and let z=e^{ix} = \cos x + i\sin x

    Then \displaystyle dz =i e^{ix} dx = iz dx \Rightarrow dx = -i \frac{dz}{z}

    and

    \displaystyle 2\cos x = z+\bar{z} = z+\frac{1}{z}, 2\cos 3x =z^3+\frac{1}{z^3}

    So on the unit circle (i.e. when z=e^{ix}), we can write

    \displaystyle \frac{\cos 3x}{5-4\cos x} = \frac{1/2(z^3+1/z^3)}{5-2(z+1/z)}

    This says that on the unit circle, the expression on the RHS is purely real, with values as given by the LHS. Since x \in [0,2\pi] traverses the unit circle, we can write:

    \displaystyle I = \int_C \frac{1/2(z^3+1/z^3)}{5-2(z+1/z)} \cdot -i \frac{dz}{z} = \frac{i}{2} \int_C \frac{z^6+1}{z^3(2z^2-5z+2)} \ dz

    where C is the unit circle.

    By the Residue Theorem, \int_C f(z) \ dz = 2\pi i \sum \text{residues of f inside C}.

    Now 2z^2-5z+2=2(z-1/2)(z-2) and writing \displaystyle f(z)=\frac{z^6+1}{z^3(2z^2-5z+2)} we see that f(z) has a pole of order 3 at z=0 and a pole of order 1 at z=1/2, with the other pole being outside the unit circle, so irrelevant here.

    We have:

    \displaystyle \text{res}(f(z),1/2) = \lim_{z \rightarrow 1/2}(z-1/2)f(z) = -\frac{65}{24}
    \displaystyle \text{res}(f(z),0) = \lim_{z \rightarrow 0} \frac{1}{2!} z^3 f''(z) = \frac{21}{8}

    as can easily be shown if you can be bothered with the tedium. I plugged the second one into Wolfram. (Sue me)

    Hence I = \frac{i}{2} \times 2\pi i(\frac{21}{8}-\frac{65}{24}) = \frac{\pi}{12}

    QED.

    Nicely done, and I don't blame you for that second residue :lol:

    Besides finding the second derivative, another way of calculating it would involve considering a small circle around the origin. The residue would then be the coefficient of z^2 in the expansion of

    \dfrac{1}{(2z-1)(z-2)}

    but that's hardly mess-free
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    Problem 546 (*/**)

    Evaluate \displaystyle I_n = \int_0^\frac{\pi}{2} \frac{1}{1+\tan^n(x)} \ dx
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    (Original post by atsruser)
    Problem 546 (*/**)

    Evaluate \displaystyle I_n = \int_0^\frac{\pi}{2} \frac{1}{1+\tan^n(x)} \ dx
    A level methods possible or no? I don't want to be wasting time doing a uni question hehe.


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    (Original post by physicsmaths)
    A level methods possible or no? I don't want to be wasting time doing a uni question hehe.
    Yes. None of that fancy college kid integral stuff here - just good ol' plain home integration, just like Mama used to do.

    It's an easyish STEP level question, I would guess.
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    (Original post by atsruser)
    Yes. None of that fancy college kid integral stuff here - just good ol' plain home integration, just like Mama used to do.

    It's an easyish STEP level question, I would guess.
    Kl, I'm guessing a reduction formula then maybe evaluate it all the way to. I_0, do I really have to use latex can't I just post a pic of a solution?


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