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    Since this one hasn't been answered yet:

    Solution 16

    Note, applying x+\sqrt{x^2+1} \to x to the former and x-\sqrt{x^2+1} \to x to the latter:

    \displaystyle\int _{-\infty}^{\infty} \displaystyle \mathfrak{I}(f)(x) dx = \displaystyle\int _{-\infty}^{\infty} \dfrac{d}{dx}[x+\sqrt{x^2+1}]f(x+\sqrt{x^2+1}) dx +\displaystyle\int _{-\infty}^{\infty} \dfrac{d}{dx}[x-\sqrt{x^2+1}]f(x-\sqrt{x^2+1}) dx

    =\displaystyle\int _{0}^{\infty} f(x)dx + \displaystyle\int _{-\infty}^{0}f(x)dx

    =\displaystyle\int _{-\infty}^{\infty} f(x) dx _{\square}

    Observe now that \displaystyle\int_{-\infty}^{\infty} \dfrac{\cos x \sin  (\sqrt{x^2+1})}{\sqrt{x^2+1}}dx = \dfrac{1}{2} \displaystyle\int_{-\infty}^{\infty}\dfrac{\sin (x+ \sqrt{x^2+1} ) - \sin (x-\sqrt{x^2+1})}{\sqrt{x^2+1}} dx

    =\dfrac{1}{2} \displaystyle\int_{-\infty}^{\infty} \left(\dfrac{x+ \sqrt {x^2+1}}{\sqrt{x^2+1}}\right)  \dfrac{\sin (x +\sqrt{x^2+1})}{x+\sqrt{x^2+1}} + \left( \dfrac{\sqrt{x^2+1} - x}{\sqrt{x^2+1}} \right) \dfrac{\sin(x-\sqrt{x^2+1})}{x-\sqrt{x^2+1}} dx

    =\dfrac{1}{2} \displaystyle\int_{-\infty}^{\infty}\left(\dfrac{\sin (x +\sqrt{x^2+1})}{x+\sqrt{x^2+1}} + \dfrac{\sin (x -\sqrt{x^2+1})}{x-\sqrt{x^2+1}}\right) + \dfrac{x}{\sqrt{x^2+1}}\left( \dfrac{ \sin (x +\sqrt{x^2+1})}{x+\sqrt{x^2+1}} - \dfrac{\sin (x -\sqrt{x^2+1})}{x-\sqrt{x^2+1}}\right) dx

    =\dfrac{1}{2} \displaystyle\int_{-\infty}^{\infty} \dfrac{\sin x}{x} dx

    =\dfrac{\pi}{2}, by applying the above result in the case f(x) =\dfrac{\sin x}{x}.
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    Problem 24

    Evaluate \displaystyle \prod_{n=1}^{1006} \sin \left(\frac{n\pi}{2013}\right)
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    (Original post by Star-girl)
    Problem 14**

    Prove that 2^{60}-1 is divisible by 61.
    2 to the power of 60 - 1 = 1152921504606846795
    1152921504606846795 / 61 = 18900352534538475
    Therefore it's divisible by 61. Problem solved
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    (Original post by Indeterminate)
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    Solution 20
    Integral
    \displaystyle\int \sqrt{r^{2} - x^{2}} dx Let x = r \sin \theta \Rightarrow dx = r \cos \theta d \theta

    \Rightarrow \displaystyle\int \sqrt{r^{2} - x^{2}} dx = r^{2} \displaystyle\int \cos^{2} \theta d \theta = \frac{r^{2}}{2} \displaystyle\int (1 + \cos 2 \theta) d \theta

    = \frac{r^{2}}{2} \left( \theta + \sin \theta \cos \theta \right) We have \theta = \arcsin \frac{x}{r} and

    \sin \theta \cdot \cos \theta = \frac{x}{r} \cdot \frac{\sqrt{r^{2} - x^{2}}}{r}} = \frac{x \sqrt{r^{2} - x^{2}}}{r^{2}}

    = \dfrac{1}{2} \left( r^{2} \arcsin \dfrac{x}{r} + x \sqrt{r^{2} - x^{2}}\right) + \mathcal{C}

    Let \arcsin \dfrac{x}{r} = \theta \Rightarrow \dfrac{x}{r} = \sin \theta and \cos \theta = \dfrac{\sqrt{r^{2} - x^{2}}}{r} \Rightarrow \tan \theta = \dfrac{x}{\sqrt{r^{2} - x^{2}}} \Rightarrow \arcsin \dfrac{x}{r} = \arctan \left( \dfrac{x} {\sqrt{r^{2} - x^{2}}} \right)

    So the integral is \dfrac{1}{2} \left(r^{2} \arctan \left( \dfrac{x}{\sqrt{r^{2} - x^{2}}} \right) + x \sqrt{r^{2} - x^{2}} \right) + \mathcal{C} as required.


    circle
    The area of the circle is given by 4 \displaystyle\int_{0}^{r} \sqrt{r^{2} - x^{2}} dx Let x = r \sin \theta \Rightarrow dx = r \cos \theta d \theta and r \to \frac{\pi}{2} and 0 \to 0

    \Rightarrow 4 \displaystyle\int_{0}^{r} \sqrt{r^{2} - x^{2}} dx = 2r^{2} \displaystyle\int_{0}^{\pi /2} (\cos 2 \theta + 1) d \theta = \pi r^{2}
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    (Original post by Felix Felicis)
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    Damn, I was just about to start writing this one up!
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    (Original post by Boy_wonder_95)
    2 to the power of 60 - 1 = 1152921504606846795
    1152921504606846795 / 61 = 18900352534538475
    Therefore it's divisible by 61. Problem solved
    The funny thing is... my method of using FLT also only takes 2 lines of work :lol:
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    (Original post by und)
    Damn, I was just about to start writing this one up!
    Ninja'd :ninja:
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    (Original post by Felix Felicis)
    ....
    Hmm I'm really tempted to ask problems that are well-known theorems under disguise (as I think they serve to educate interesting points) but it's hard to judge the difficulty
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    (Original post by ukdragon37)
    Hmm I'm really tempted to ask problems that are well-known theorems under disguise (as I think they serve to educate interesting points) but it's hard to judge the difficulty
    Lay it on me (Not today though, tired, sleepy time )
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    (Original post by Noble.)
    The funny thing is... my method of using FLT also only takes 2 lines of work :lol:
    Not fair, FLT isn't part of the A Level syllabus is it?
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    (Original post by Boy_wonder_95)
    2 to the power of 60 - 1 = 1152921504606846795
    1152921504606846795 / 61 = 18900352534538475
    Therefore it's divisible by 61. Problem solved
    We were having trouble proving that 2^700^70^7-1 is divisible by 101. Show us your ways.
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    (Original post by Boy_wonder_95)
    Not fair, FLT isn't part of the A Level syllabus is it?
    Spoiler:
    Show
    It is, and it's a proof you need to know. FLT, by the way, is short for Fermat's Last Theorem, there's a section on it in C4 :ninja:
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    (Original post by Felix Felicis)
    Spoiler:
    Show
    It is, and it's a proof you need to know. FLT, by the way, is short for Fermat's Last Theorem, there's a section on it in C4 :ninja:
    In the context Noble uses it, FLT means Fermat's Little Theorem here (which you don't need to know at A-Level :p: ).
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    (Original post by Farhan.Hanif93)
    In the context Noble uses it, FLT means Fermat's Little Theorem here (which you don't need to know at A-Level :p: ).
    Haha, I know, I was just trying to mess with Boy Wonder xD
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    (Original post by Felix Felicis)
    Haha, I know, I was just trying to mess with Boy Wonder xD
    You managed to mess with me in the process.

    I can see this thread becoming prime procrastination material for my revision...
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    (Original post by Farhan.Hanif93)
    You managed to mess with me in the process.

    I can see this thread becoming prime procrastination material for my revision...
    I know, I was going to say the same. It's sad when you have to class doing 'other' maths as procrastination :lol:
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    Problem 25**/***

    1. Show that for any (potentially infinite) set A, there is no surjection f : A \to \mathcal{P}(A), where \mathcal{P}(A) is the powerset of A.
    2. Hence show there is no surjection from the the set A to set of functions A \to A if |A|>1.
    3. Determine the relationship between \mathbb{R} and \mathbb{N} \to \mathbb{N}.
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    (Original post by Noble.)
    I know, I was going to say the same. It's sad when you have to class doing 'other' maths as procrastination :lol:
    If only I could understand groups by doing integrals, then I'd be on to something...

    Out of interest, are questions on things like groups etc. suitable for this thread?
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    (Original post by Farhan.Hanif93)
    If only I could understand groups by doing integrals, then I'd be on to something...

    Out of interest, are questions on things like groups etc. suitable for this thread?
    If we're going to procrastinate, I don't see the harm in combining a bit of revision :lol:
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    (Original post by Farhan.Hanif93)
    If only I could understand groups by doing integrals, then I'd be on to something...

    Out of interest, are questions on things like groups etc. suitable for this thread?
    It's fine as long as it's marked accordingly. If possible it would be good if everything could be defined so at a stretch anyone could have a go, but it's not a requirement.
 
 
 
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