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    (Original post by joostan)
    I'm gonna go eat, I'll sort that in a bit
    Have fun eating some :sexface:
    How's your maclaurin series/ do you know the Basel problem?
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    (Original post by Felix Felicis)
    Have fun eating some :sexface:
    How's your maclaurin series/ do you know the Basel problem?
    LOL, ooo I was gonna use one of them but I don't know what they are so didn't know if posting a maclaurin series problem would of been appropriate hahaha
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    (Original post by Felix Felicis)
    Have fun eating some :sexface:
    How's your maclaurin series/ do you know the Basel problem?
    (Original post by tigerz)
    Okays, I look forward to seeing what method you decide use
    I was thinking of going polar/parametrised, but upon closer inspection this does look rather complicated.

    My maclaurin series is ok, Basel's \dfrac{\pi^2}{6} isn't it?
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    (Original post by joostan)
    Anyone got a nice question? I'm in the mood for some maths.
    To keep with the trend... Show that:

    \sqrt{1+\sqrt{2+\sqrt{5}}}<\sqrt  {1+\sqrt{2+\sqrt{3+\sqrt{4+ \cdots }}}}<\sqrt{1+\sqrt{2+\sqrt{4+ \sqrt{5}}}}
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    (Original post by joostan)
    I was thinking of going polar/parametrised, but upon closer inspection this does look rather complicated.

    My maclaurin series is ok, Basel's \dfrac{\pi^2}{6} isn't it?
    Yup, For this problem it is advisable to introduce polar coordinates (r and ϕ)
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    (Original post by tigerz)
    Yup, For this problem it is advisable to introduce polar coordinates (r and ϕ)
    What modules are you doing this yr again?
    I thought you were doing C1, C2 and an applied. But some of the questions you pose (assuming you can solve them) are surprising for an AS maths student!
    But then, you wish to do maths @ Uni.
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    Is there anyway to simplify expressions like this

     \dfrac{a}{bcos \dfrac{b}{a}}

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    (Original post by reubenkinara)
    What modules are you doing this yr again?
    I thought you were doing C1, C2 and an applied. But some of the questions you pose (assuming you can solve them) are surprising for an AS maths student!
    But then, you wish to do maths @ Uni.
    C1, C2 & S1, I can only solve 'em with the use of the internet Although there's a really simple way of solving the rabbit one, he chose to do it the mathematical way
    Some of the other questions came because I tried to get extra practice with weird papers which ended up including c3 and c4 syllabus which is why I had to get some help from here!

    (Original post by MathsNerd1)
    I see, that's fair enough and I don't think I've ever actually rolled on the floor laughing before

    Also I would try it out but I'm hunting out a few STEP questions to try instead


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    Really? we just end up losing balance, ain't really got a choice (and we can't even remember why most of the time)
    Ahh good luck

    (Original post by Lord of the Flies)
    To keep with the trend... Show that:

    \sqrt{1+\sqrt{2+\sqrt{5}}}<\sqrt  {1+\sqrt{2+\sqrt{3+\sqrt{4+ \cdots }}}}<\sqrt{1+\sqrt{2+\sqrt{4+ \sqrt{5}}}}
    I may attempt this, am I heading in the right direction?
    Spoiler:
    Show
    \sqrt{1+\sqrt{2+\sqrt{5}}} \rightarrow\ \sqrt{1+\sqrt{1+1+\sqrt{1+4}}}
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    (Original post by tigerz)
    C1, C2 & S1, I can only solve 'em with the use of the internet Although there's a really simple way of solving the rabbit one, he chose to do it the mathematical way
    Some of the other questions came because I tried to get extra practice with weird papers which ended up including c3 and c4 syllabus which is why I had to get some help from here!



    Really? we just end up losing balance, ain't really got a choice (and we can't even remember why most of the time)
    Ahh good luck



    I may attempt this, am I heading in the right direction?
    Spoiler:
    Show
    \sqrt{1+\sqrt{2+\sqrt{5}}} \rightarrow\ \sqrt{1+\sqrt{1+1+\sqrt{1+4}}}+.  ..
    Also (this probably won't help):
    Spoiler:
    Show
    \sqrt{1+\sqrt{2+\sqrt{5}}} \Rightarrow\ \sqrt{1+\sqrt{1^2+1+\sqrt{2^2+1}  +...\sqrt{n^2+1}}}}
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    (Original post by reubenkinara)
    Also (this probably won't help):
    Spoiler:
    Show
    \sqrt{1+\sqrt{2+\sqrt{5}}} \Rightarrow\ \sqrt{1+\sqrt{1^2+1+\sqrt{2^2+1}  +...\sqrt{n^2+1}}}}
    :O that just gave me an idea, lmaooo thank you
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    Could anyone tell me the solution of some M3 questions:

    Q1. Part of a hallow spherical shell, centre O and radius a, is removed to form a bowl with a plane circular rim. The bowl is fixed with the rim uppermost and horizontal. The centre of the circular rim is (4a/3) vertically above the lowest point of the bowl. A marble is projected from the lowest point of the bowl with speed u. Find the minimum value of u for which the marble will leave the bowl and not fall back to it. [Ans: (17ga/3)^(1/2) ]

    Q2. A particle P of mass m is attached to two strings AP and BP. The points A and B are on the same horizontal level and AB= (5a/4).
    The string AP is inextensible and AP = (3a/4)
    The string BP is elastic and BP= a
    The modulus of elasticity of BP is k.
    Show that the natural length of BP is (5ka)/(3mg+5k).

    Thanks
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    (Original post by reubenkinara)
    Also (this probably won't help):
    Spoiler:
    Show
    \sqrt{1+\sqrt{2+\sqrt{5}}} \Rightarrow\ \sqrt{1+\sqrt{1^2+1+\sqrt{2^2+1}  +...\sqrt{n^2+1}}}}
    Unless Lord of the Flies meant otherwise, I think \sqrt{1+\sqrt{2+\sqrt{5}}} terminates at  \sqrt{5} .
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    (Original post by Scientific Eye)
    Unless Lord of the Flies meant otherwise, I think \sqrt{1+\sqrt{2+\sqrt{5}}} terminates at  \sqrt{5} .
    The first part does, I thing the middle one goes on and then last one terminates..Not sure :/
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    (Original post by Scientific Eye)
    Unless Lord of the Flies meant otherwise, I think \sqrt{1+\sqrt{2+\sqrt{5}}} terminates at  \sqrt{5} .
    I see. Ignore me then Tigerz!
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    could someone help me with this question: I am asked to use the substitution x=sin \theta to find the exact value of: (Question 4 June 2005)

    now i know \frac{dx}{d \theta} =cos\theta and dx=cos \theta d \theta but why is the upper limit \frac{\pi}{6}?
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    (Original post by reubenkinara)
    I see. Ignore me then Tigerz!
    No worries, its the thought that counts I think I can use it in the middle of the equation as its \sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+ \cdots }}}}

    Spoiler:
    Show
    Is it mathematical legal to simplify like this:

    \sqrt{1+\sqrt{1+1+\sqrt{1+4}}}< \sqrt{1+\sqrt{1+1+\sqrt{1+2+ \sqrt{1+3+ \cdots }}}}< \sqrt{1+ \sqrt{1+1+\sqrt{1+3+ \sqrt{1+4}}}}

    \Rightarrow \sqrt{1+4} <\sqrt{1+2+\sqrt{1+3+ \cdots }}< \sqrt{1+3+ \sqrt{1+4}}
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    (Original post by gaffer dean)
    could someone help me with this question: I am asked to use the substitution x=sin \theta to find the exact value of: (Question 4 June 2005)

    now i know \frac{dx}{d \theta} =cos\theta and dx=cos \theta d \theta but why is the upper limit \frac{\pi}{6}?
    Because when you use substitution for definite integration you need to substitute the old limits to find the new ones so if x=sin\theta then 0.5=sin\theta so \theta=arcsin(0.5)=\dfrac{\pi}{6  }
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    (Original post by reubenkinara)
    Because when you use substitution for definite integration you need to substitute the old limits to find the new ones so if x=sin\theta then 0.5=sin\theta so \theta=arcsin(0.5)=\dfrac{\pi}{6  }
    but arcsin(0.5) is also 30 so why is the upper limit not 30?
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    (Original post by gaffer dean)
    but arcsin(0.5) is also 30 so why is the upper limit not 30?
    If I'm right, all trig calculus has to be done in radians.
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    (Original post by reubenkinara)
    If I'm right, all trig calculus has to be done in radians.
    that makes sense now, cheers!
 
 
 
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