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    (Original post by joostan)
    You have a to do list? O.o
    Yeah, a fair bit of it is at http://www.thestudentroom.co.uk/show...1#post43707571

    Perhaps I should just post your homework up here as a problem and you might just do it
    If only… the trouble is that it's hard enough that I can't easily get started on it
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    (Original post by Smaug123)
    Yeah, a fair bit of it is at http://www.thestudentroom.co.uk/show...1#post43707571
    If only… the trouble is that it's hard enough that I can't easily get started on it
    Ambitious
    Learn to draw? That would take me forever :lol:
    Ah, that would indeed be most problematic I shudder to think what foul devilry is in that then . . .
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    (Original post by joostan)
    Ambitious
    Learn to draw? That would take me forever :lol:
    I keep thinking "That can't be too hard!" - after all, it's just lines on paper. I'm getting a little bit better without any practice at all, too
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    (Original post by Smaug123)
    I keep thinking "That can't be too hard!" - after all, it's just lines on paper. I'm getting a little bit better without any practice at all, too
    Haha. I guess then, unlike me, you're not a malco.
    Graph sketches? Not for me.
    My art teacher in year 8 told me to drop art :laugh:
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    (Original post by joostan)
    I have no idea what just happened . . . :eek:
    I'm pretty sure your mind just got Fourier transformed...
    (Original post by Smaug123)
    If only… the trouble is that it's hard enough that I can't easily get started on it
    Motivation: the final theorem on you N+S worksheet would imply Fermat's Last Theorem should that not be the case... (not that I would even know where to start on it! :lol:)
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    (Original post by joostan)
    Haha. I guess then, unlike me, you're not a malco.
    Graph sketches? Not for me.
    My art teacher in year 8 told me to drop art :laugh:
    Don't worry, graph sketches don't have to be accurate, at any level. If the basic features are all there (cuts the axes/has turning points/asymptotes (if you don't know what any of those words mean, they aren't relevant to you yet - these rules will keep working right through to doing a maths degree) the right number of times in the right order, does vaguely like what it's supposed to do as it goes to +/- infinity, , and that's about it), you're fine. It doesn't have to be pretty. If you have to make the gap between "0" and "1" be twice the size of the gap between "1" and "2" on some axis, that's fine. Just start off with the basic bits - draw the x-axis, the points where the graph cuts that axis, ditto for the y-axis, what it does at the ends, where it goes flat, anything else interesting that happens, then just join them all up and you're done.
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    (Original post by BlueSam3)
    Don't worry, graph sketches don't have to be accurate, at any level. If the basic features are all there (cuts the axes/has turning points/asymptotes (if you don't know what any of those words mean, they aren't relevant to you yet - these rules will keep working right through to doing a maths degree) the right number of times in the right order, does vaguely like what it's supposed to do as it goes to +/- infinity, , and that's about it), you're fine. It doesn't have to be pretty. If you have to make the gap between "0" and "1" be twice the size of the gap between "1" and "2" on some axis, that's fine. Just start off with the basic bits - draw the x-axis, the points where the graph cuts that axis, ditto for the y-axis, what it does at the ends, where it goes flat, anything else interesting that happens, then just join them all up and you're done.
    Yeah, but as the curve approaches the asymptote, it's not meant to ever quite have the same gradient or curve back up or anything nasty like that but I have shaky hands and tend to have to spend a while sketching my graphs
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    (Original post by Smaug123)
    I keep thinking "That can't be too hard!" - after all, it's just lines on paper. I'm getting a little bit better without any practice at all, too
    Drawing technique is not hard, it can be developed quite rapidly. It is being able to see and portray whatever you are drawing or painting that takes years. Similarly, a lot of mathematical technique (using this term loosely) can be learned quite quickly. However, it doesn't necessarily make you 'see'. Take many A-level mathematics students who can routinely apply methods but struggle where thinking is involved, the situation is similar. So just like for mathematics, the curriculum for art also fails to equip students with the ability to 'see', that ability takes practice and is more rewarding.
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    (Original post by Ateo)
    Drawing technique is not hard, it can be developed quite rapidly. It is being able to see and portray whatever you are drawing or painting that takes years. Similarly, a lot of mathematical technique (using this term loosely) can be learned quite quickly. However, it doesn't necessarily make you 'see'. Take many A-level mathematics students who can routinely apply methods but struggle where thinking is involved, the situation is similar. So just like for mathematics, the curriculum for art also fails to equip students with the ability to 'see', that ability takes practice and is more rewarding.
    Yeah, I think I know what you mean by "see" - I've slowly been gaining this ability through my attempts to become a better introspector and rationalist. I can, with some effort of will, "just see the lines" that are in front of me - as a corollary to my learning to see what is really there in a metaphorical sense (as in, see what the evidence presents you, not what you want to get out of the evidence, in rationality training).
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    (Original post by Smaug123)
    Yeah, I think I know what you mean by "see" - I've slowly been gaining this ability through my attempts to become a better introspector and rationalist. I can, with some effort of will, "just see the lines" that are in front of me - as a corollary to my learning to see what is really there in a metaphorical sense (as in, see what the evidence presents you, not what you want to get out of the evidence, in rationality training).
    And the wonderful thing about it is that it empowers you to present your own contribution to the work in a meaningful way, whatever it may be, you decide what you create but you are able to create it in an honest and aesthetically pleasing way. Dismissing this results in terrible art, be it abstract or not. I don't see how people can appreciate an over-simplified abstract painting that is completely out of shape over a painting of say Bouguereau or Rembrandt - the beauty of their work is apparent. Main difference of art and mathematics is that the product of art is somewhat visible to all, mathematics you have to study to understand, that adds to the appreciation, not all see it.
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    I'm so confused . . .
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    (Original post by Ferrari Lexus)
    I'm so confused . . .
    Yes, the thread's been derailed somewhat. . .

    Problem:
    Find: (without using IBP)
    \displaystyle\int e^{ax}\cos(bx) \ dx
    and:
    \displaystyle\int e^{ax}\sin(bx) \ dx
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    (Original post by Jkn)
    Solution

    Using the integral definition of the Gamma function,

    \displaystyle \begin{aligned} \sum_{n=1}^{\infty} \frac{(-1)^n \Gamma \left(2n-\frac{4}{5} \right)}{(2n)!} = \int_0^{\infty} \left( \sum_{n=1}^{\infty} \frac{(-1)^n t^{2n}}{(2n)!} \right) t^{-\frac{9}{5}} e^{-t} \ dt = \int_0^{\infty} t^{-\frac{9}{5}} e^{-t} ( \cos t - 1) \ dt \end{aligned}

\displaystyle \begin{aligned} = \frac{1}{2} \int_0^{\infty} t^{-\frac{9}{5}} \left(e^{-t(1-i)}+e^{-t(1+i)}-2e^{-t} \right) \ dt = \frac{1}{2} \Gamma \left( - \frac{4}{5} \right) \left( \frac{1}{(1-i)^{-\frac{4}{5}}}+\frac{1}{(1+i)^{-\frac{4}{5}}}-2 \right) \end{aligned}

\displaystyle = \frac{1}{4} \Gamma \left(- \frac{4}{5} \right) (2^{\frac{2}{5}} (1+\sqrt{5})-4)

    Edit: Just realised the justification of this is 'sketchy' as we are inventing an integral representation for \Gamma(z) for Re(z)<0 but, as we define it as such, the consequences are also valid by the Bohr-Möllerup theorem and it's application in extending the definition of the Gamma function to these aloes through series representation (i.e. as the answer exists, an invalid representation of the Gamma function actually gives an answer when we use the evaluative formula with the assumption that it is valid and the representations are consistent, we must deduce that this answer is correct).

    \displaystyle \begin{aligned} \Rightarrow 2^{\frac{3}{5}} \left(1+ 4\sum_{n=0}^{\infty} \frac{(-1)^{n+1} \Gamma \left(2n-\frac{4}{5} \right)}{5 (2n)! \Gamma \left( \frac{1}{5} \right)} \right) =2^{\frac{3}{5}} \left(1-\frac{\Gamma \left( - \frac{4}{5} \right)}{5 \Gamma \left( \frac{1}{5} \right)} (2^{\frac{2}{5}} (1+\sqrt{5})-4) \right) = \frac{1+ \sqrt{5}}{2} \ \square \end{aligned}
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    (Original post by joostan)
    Yes, the thread's been derailed somewhat. . .

    Problem:
    Find: (without using IBP)
    \displaystyle\int e^{ax}\cos(bx) \ dx
    and:
    \displaystyle\int e^{ax}\sin(bx) \ dx
    dafaq is that?
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    (Original post by Ferrari Lexus)
    dafaq is that?
    Erm, what level of maths have you come across?
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    (Original post by joostan)
    Erm, what level of maths have you come across?
    just finished gcse maths!
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    (Original post by joostan)
    Yes, the thread's been derailed somewhat. . .

    Problem:
    Find: (without using IBP)
    I = \displaystyle\int e^{ax}\cos(bx) \ dx
    and:
    J = \displaystyle\int e^{ax}\sin(bx) \ dx
    I+iJ = \displaystyle \int e^{ax} (\cos(bx)+i \sin(bx)) dx = \int e^{ax + ibx} dx = \dfrac{1}{a+ib} e^{(a+ib) x} + C+iD = e^{a x} \dfrac{a - i b}{a^2 + b^2} e^{ib x}+C+iD.
    So I = \text{Re}(I+iJ) = e^{a x} \cos(bx) \dfrac{a}{a^2+b^2} + e^{ax} \dfrac{b}{a^2+b^2} \sin(bx) + C; J = \text{Im}(I+iJ) = e^{a x} \sin(bx) \dfrac{a}{a^2+b^2} - e^{a x} \cos(b x) \dfrac{b}{a^2+b^2} + D.
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    (Original post by Ferrari Lexus)
    just finished gcse maths!
    Ah, OK, therein lies the problem to solve the above integral, you need to be able to at least have a good understanding of A2 maths, without IBP you're looking at A2 Further maths.

    Integration is essentially reverse differentiation (which you may have come across GCSE)
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    (Original post by Ferrari Lexus)
    dafaq is that?
    It's an integral - it can be solved with A-level maths, or using the requested method with A-level further maths.
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    (Original post by Smaug123)
    I+iJ = \displaystyle \int e^{ax} (\cos(bx)+i \sin(bx)) dx = \int e^{ax + ibx} dx = \dfrac{1}{a+ib} e^{(a+ib) x} + C+iD = e^{a x} \dfrac{a - i b}{a^2 + b^2} e^{ib x}+C+iD.
    So I = \text{Re}(I+iJ) = e^{a x} \cos(bx) \dfrac{a}{a^2+b^2} + e^{ax} \dfrac{b}{a^2+b^2} \sin(bx) + C; J = \text{Im}(I+iJ) = e^{a x} \sin(bx) \dfrac{a}{a^2+b^2} - e^{a x} \cos(b x) \dfrac{b}{a^2+b^2} + D.
    Yup, looks good (of course), although it's less scary in factorised form. :laugh:
 
 
 
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