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    (Original post by Jkn)
    That each definite integral takes a value that is in that set! Don't know why everyone's always outraged with use of what is probably bad notation :lol: At A-Level we obviously aren't taught about set theory but we are told we can use \in to say something is part of a set of numbers. What's so wrong about what I wrote that means you cannot understand it?
    I guessed what your intended meaning was. However, if somebody communicated to you in English and kept using "an propar grammer", then you would probably feel the same. The most important part of studying independently is to always make sure you correctly understand the things that you have learnt. For if you don't do so, then you risk to become confused for a long time.

    The elements in \mathbb{R} do not have the form (x,y). If you meant for x and y to be real numbers, then you should have omitted the brackets and instead said x,y \in \mathbb{R}. If you wanted the brackets, then you could have used the notation (x,y) \in \mathbb{R}^2, referring to a tuple. What you wrote could also be interpreted as an open interval, in which case (x,y) \subset \mathbb{R}; a subset of the real numbers and not a member.

    (Original post by Jkn)
    Why?
    Ideally, you would want to focus entirely within the boundaries of the exam. If you keep doing questions that require different methods (i.e. Laplace transforms) until the last day, then you will be inclined to think about those first. To quote Abraham Maslow, to a man who possesses a hammer, every problem starts to look like a nail.
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    (Original post by Mladenov)
    Actually, apart from Poe, Ayn Rand is the only writer whom books I can enjoy.
    Dostoevsky all the way! Tolstoy or Hugo too, for that matter.

    Lol, I used to worship Ayn Rand in Grade 11. I used to read Atlas Shrugged in class during Revision.
    But then, man, well, now, I disagree strongly with practically everything she promulgates.
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    (Original post by jack.hadamard)
    I guessed what your intended meaning was. However, if somebody communicated to you in English and kept using "an propar grammer", then you would probably feel the same. The most important part of studying independently is to always make sure you correctly understand the things that you have learnt. For if you don't do so, then you risk to become confused for a long time.

    The elements in \mathbb{R} do not have the form (x,y). If you meant for x and y to be real numbers, then you should have omitted the brackets and instead said x,y \in \mathbb{R}. If you wanted the brackets, then you could have used the notation (x,y) \in \mathbb{R}^2, referring to a tuple. What you wrote could also be interpreted as an open interval, in which case (x,y) \subset \mathbb{R}; a subset of the real numbers and not a member.
    He could have meant a Dedekind cut, and so aptly the pair is in \mathbb R (Although somehow the integrals have to represent sets of rationals. Lebesgue integration anyone?)
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    I think this might be quite a nice one (hopefully :awesome:)

    Problem 213 **/***

    By considering an appropriate contour integral of f(z)=\dfrac{\sec z}{z^5} and its residues, show that  \displaystyle \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n+1)^5} = \dfrac{5 \pi^5}{1536}
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    (Original post by Mladenov)
    Actually, apart from Poe, Ayn Rand is the only writer whom books I can enjoy.

    By the way, Jkn, do you have an elegant, and not arduous, solution to problem 198, for I have one which is a hell of an arithmetic and I am not quite enthusiastic to type it in latex.
    Oh right, I've never heard of her! Perhaps try some F. Scott Fitzgerald. Especially if you're a fan of short stories!

    Unfortunately I do not, and I do not believe such a solution exists :')
    (Original post by jack.hadamard)
    I guessed what your intended meaning was. However, if somebody communicated to you in English and kept using "an propar grammer", then you would probably feel the same. The most important part of studying independently is to always make sure you correctly understand the things that you have learnt. For if you don't do so, then you risk to become confused for a long time.

    The elements in \mathbb{R} do not have the form (x,y). If you meant for x and y to be real numbers, then you should have omitted the brackets and instead said x,y \in \mathbb{R}. If you wanted the brackets, then you could have used the notation (x,y) \in \mathbb{R}^2, referring to a tuple. What you wrote could also be interpreted as an open interval, in which case (x,y) \subset \mathbb{R}; a subset of the real numbers and not a member.

    Ideally, you would want to focus entirely within the boundaries of the exam. If you keep doing questions that require different methods (i.e. Laplace transforms) until the last day, then you will be inclined to think about those first. To quote Abraham Maslow, to a man who possesses a hammer, every problem starts to look like a nail.
    Hmm, good point, I hadn't considered that it could be interpreted as an interval! I just put the brackets so that it was easier to read :lol: Perhaps braces?

    Very nice quote, I agree! I need to return to my step preparation :')
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    (Original post by Jkn)
    Perhaps braces?
    Not unless you write \mathcal{P}\left(\mathbb R\right) instead
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    (Original post by Mladenov)
    Problem 212**

    Evaluate \displaystyle \int_{0}^{\infty} \frac{2-2\cos x-x\sin x}{x^{4}}\, dx
    Another fun integral to do whilst avoiding exam preparation

    (I've decided to be rigorous for the sake of it)

    Problem 212

    \displaystyle I(x)= \int_{0}^{\infty} \frac{2-2\cos x-x\sin x}{x^{4}}\, dx = \int_0^{\infty} \frac{4 \sin^2 (\frac{x}{2})- x \sin(x)}{x^4} \ dx

\displaystyle = \left[-\frac{4 \sin^2 (\frac{x}{2})- x \sin(x)}{3x^3} \right]_0^{\infty} +\frac{1}{3} \int_0^{\infty} \frac{\sin(x)-x \cos(x)}{x^3} \ dx

    Noting that \displaystyle \lim_{x \to 0} \frac{4 \sin^2 (\frac{x}{2})- x \sin(x)}{x^3} = \lim_{x \to 0} \frac{x^2-x^2}{x^3} = 0,

    \displaystyle \begin{aligned} \Rightarrow I(x) = \frac{1}{3} \int_0^{\infty} \frac{\sin(x)-x \cos(x)}{x^3} \ dx = \left[-\frac{\sin(x)-x \cos(x)}{6x^2} \right]_0^{\infty} +\frac{1}{6} \int_0^{\infty} \frac{\sin(x)}{x} \ dx \end{aligned}

    Noting that \displaystyle \lim_{x \to 0} \frac{\sin(x)- x \cos(x)}{6x^2} = \lim_{x \to 0} \frac{x-x}{6x^2} = 0,

    \displaystyle \Rightarrow I(x) = \frac{1}{6} \int_0^{\infty} \frac{\sin(x)}{x} \ dx

    Let f(t) = \displaystyle \int_0^{\infty} \frac{\sin(tx)}{x} \ dx,

    \displaystyle \Rightarrow \mathcal{L} \{ f(t) \} = \int_0^{\infty} e^{-st} \int_0^{\infty} \frac{\sin(x)}{x} \ dx \ dt = \int_0^{\infty} \frac{1}{x} \mathcal{L} \{sin(tx) \} \ dx, by Fubini's Theorem.

    \displaystyle \Rightarrow \mathcal{L} \{ f(t) \} = \int_0^{\infty} \frac{1}{s^2+x^2} \ dx = \frac{\pi}{2s}

    \displaystyle \Rightarrow f(t) = \mathcal{L}^{-1} \{ \frac{\pi}{2s} \} = \frac{\pi}{2}

    Therefore \displaystyle I(x) = \frac{1}{6} f(1) =\frac{\pi}{12} \ \square
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    Quick break from STEP. :ninja:

    Solution 211


    \begin{aligned} \cos \dfrac{z}{2}=\displaystyle \prod_{k\geq 0}\left(1-\frac{z^2}{\pi^2(2k+1)^2}\right)  \Rightarrow \tan \frac{z}{2}=-2\frac{d}{dz}\ln \cos \frac{z}{2}=\sum_{k\geq 0}\frac{4z}{\pi^2(2k+1)^2-z^2}

    \displaystyle \begin{aligned}\int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh bx} \, dx=\int_0^{\infty} \frac{1}{x}\left(\frac{x^a-x^{-a}}{x^b-x^{-b}}\right) \,dx &=\int_0^{1} \frac{2}{x}\left(\frac{x^{a}-x^{-a}}{x^{b}-x^{-b}}\right) \,dx\\&=\int_0^{1} 2x^{b-a-1}\left(\frac{1-x^{2a}}{1-x^{2b}}\right) \,dx\\&=\sum_{k\geq 0}\int_0^{1} 2x^{(2k+1)b-a-1}(1-x^{2a}) \,dx \\& =\sum_{k\geq 0} \frac{4a}{b^2(2k+1)^2-a^2}\\&  = \frac{\pi}{b} \tan \frac{\pi a}{2b}\end{aligned}

    Solution 212

    \begin{aligned}\displaystyle \int_0^{\infty}\frac{2-2\cos x-x\sin x}{x^4}\,dx=\frac{1}{3} \int_0^{\infty}\frac{\sin x-x\cos x}{x^3}\,dx=\frac{1}{6}\int_0^{ \infty} \frac{\sin x}{x}\,dx =\frac{\pi}{12}
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    (Original post by ukdragon37)
    Not unless you write \mathcal{P}\left(\mathbb R\right) instead
    What's that? :lol:
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    (Original post by ukdragon37)
    Not unless you write \mathcal{P}\left(\mathbb R\right) instead
    Read this (taken from here). If it still doesn't help, then try this.
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    (Original post by jack.hadamard)
    Read this (taken from here). If it still doesn't help, then try this.
    :rofl:
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    (Original post by Lord of the Flies)
    \begin{aligned}\displaystyle \int_0^{\infty}\frac{2-2\cos x-x\sin x}{x^4}\,dx=\frac{1}{3} \int_0^{\infty}\frac{x\cos x-\sin x}{x^3}\,dx
    Sign error :rolleyes: ...the dangers of emitting working
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    (Original post by Jkn)
    Sign error :rolleyes: ...the dangers of emitting working
    He's far from emitting working
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    (Original post by Jkn)
    Sign error :rolleyes: ...the dangers of emitting working
    Danger is my middle name, #YOLO.

    (thanks)

    (Original post by bananarama2)
    He's far from emitting working
    LOL I hadn't noticed.
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    (Original post by bananarama2)
    He's far from emitting working
    Taking emitting working to the extreme :lol:
    (Original post by Lord of the Flies)
    Danger is my middle name, #YOLO.

    (thanks)
    Living on the edge :cool:

    Is that how little working you did on paper? :eek: (the first two steps that is)
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    A bit of a joke...

    Problem 214
    **

    Evaluate \displaystyle \int_0^{\infty} \frac{3 \sin(x)-x \cos(x)-2x}{x^5} \ dx
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    (Original post by jack.hadamard)
    Read this (taken from here). If it still doesn't help, then try this.
    (Original post by bananarama2)
    :rofl:
    I'll have you know I'm normally sufficiently conscientious :mad:

    (Original post by Jkn)
    What's that? :lol:
    Funny that you know what Laplace transforms are but not power sets. My point exactly about how discrete mathematics is taught in this country.

    EDIT: Having seen the following elsewhere on TSR I felt I have to bring it to the attention of this thread. Please do not do this in your STEP papers.

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    (Original post by ukdragon37)
    Funny that you know what Laplace transforms are but not about power sets. My point exactly about how discrete mathematics is taught in this country.
    :rofl:

    I think it's good to learn all of these integration techniques early though! Integration is a topic taught at A-Level so why not get a year or two ahead? :lol: Set theory, on the other hand, I eagerly look forward to learning next year or over the summer

    Edit: And also, I'm trying to train myself to think mathematically and I believe the best way of doing that, at my age, is to do olympiad-style problems rather than learning things I haven't really been introduced to yet. I consider definite integration to be olympiad-style despite the fact it's not included in pre-university competitions (which is probably due to the breadth of knowledge needed rendering it difficult to be successful unless you are older and are more comfortable with the advanced techniques)

    Edit 2: PRSOM^ that picture :lol:
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    (Original post by ukdragon37)
    My point exactly about how discrete mathematics is taught in this country.
    And how glad of that I am
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    Solution 214

    -\dfrac{\pi}{48}

    From 212.

    (Original post by Jkn)
    Is that how little working you did on paper? :eek: (the first two steps that is)
    Yes, making a sign mistake on both IBPs, and thus getting to the correct value anyway. :lol:
 
 
 
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