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    (Original post by gaffer dean)
    that makes sense now, cheers!
    Don't mention it. I'm not sure why, but if you're curious and google yields nothing some of the other regulars on this thread will be able to answer that question!
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    (Original post by reubenkinara)
    Don't mention it. I'm not sure why, but if you're curious and google yields nothing some of the other regulars on this thread will be able to answer that question!
    I just checked another website they used 30 degrees and ended up with the same answer.

    Here's a screenshot of what they did:
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    (Original post by gaffer dean)
    I just checked another website they used 30 and ended up with the same answer.
    It will only give the same answer for the integral you gave if your calculator is in degrees mode. Stick to radians for trig calculus
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    (Original post by AmirHabeeb)
    It will only give the same answer for the integral you gave if your calculator is in degrees mode. Stick to radians for trig calculus
    alright thanks
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    (Original post by reubenkinara)
    Don't mention it. I'm not sure why, but if you're curious and google yields nothing some of the other regulars on this thread will be able to answer that question!
     \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}}}}

    Am I allowed to cancel out  \sqrt{1+\sqrt{1+1}}? so that it simplifies to:

     \sqrt{1+4}< \sqrt{1+2+\sqrt{1+3+ \cdots }}< \sqrt{1+3+ \sqrt{1+4}}
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    (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'm trying to avoid these problems because of chem revision but they're too tempting
    Spoiler:
    Show
    I think it would be sufficient to show that:

    \sqrt{5} < \sqrt{3 + \sqrt{4 + \sqrt{5 + ...}}} < \sqrt{4 + \sqrt{5}}

    Ok, first we show \sqrt{5} < \sqrt{4 + \sqrt{5}} \Leftightarrow 5 < 4 + \sqrt{5} which is true as \sqrt{5} > 1

    Now we show \sqrt{5} < \sqrt{3 + \sqrt{4 + ... }}

    Assume \sqrt{5} < \sqrt{3 + \sqrt{4 + ... }} \Leftrightarrow 5 < 3 + \sqrt{4 + \sqrt{5 + ...}}

    \Leftrightarrow 2 < \sqrt{4 + \sqrt{5 + ...}} which is true as \sqrt{5 + \sqrt{6 + ... }} > 1

    Finally, we show \sqrt{3 + \sqrt{4 + \sqrt{5 + ...}}} < \sqrt{4 + \sqrt{5}}

    Assume the above to be true \Leftrightarrow 3 + \sqrt{4 + \sqrt{5 + ...}} < 4 + \sqrt{5}

    \Leftrightarrow \sqrt{4 + \sqrt{5 + ...}} < 1 + \sqrt{5}

    \Leftrightarrow \frac{1}{2} \sqrt{4 + \sqrt{5 + ... }} < \phi

    \Leftrightarrow 1 + \frac{1}{4} \sqrt{5 + \sqrt{6 + ... }} < \phi + 1

    \Leftrightarrow \frac{1}{4} \sqrt{5 + \sqrt{6 + ... }} < \phi

    = \frac{1}{4} \sqrt{5 + \sqrt{6 + ...}} < \sqrt{1 + \sqrt{1 + \sqrt{1 + ... }}}

     = \sqrt{ \frac{5}{16} + \sqrt{ \frac{6}{16^{2}} + ... }} < \sqrt{1 + \sqrt{1 + ...}} which is true as each nested radical on the LHS is less than 1, and we're done

    I fear there may be a flaw in my logic, seems to simple :holmes:
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    (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've been scratching my head about a fox and a rabbit for a while, so I figured I'd give this a go.
    Given its a show that rather than a prove that:
    Spoiler:
    Show
    Consider the left inequality first, squaring and rearranging acouple of times.
    \sqrt{1+\sqrt{2+\sqrt{5}}}<\sqrt  {1+\sqrt{2+\sqrt{3+\sqrt{4+ \cdots }}}} 

\Leftrightarrow 2< \sqrt{4 + \sqrt{5 + \sqrt{6. . .}}}

\Leftrightarrow 0< \sqrt{5 + \sqrt{6 + \sqrt{7}. . .}}}
    Which it clearly is.

    Considering the right inequality, squaring and rearranging.
    \sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+ \cdots }}}}<\sqrt{1+\sqrt{2+\sqrt{4+ \sqrt{5}}}}

\Leftrightarrow 1+\sqrt{5} > \sqrt{4+ \sqrt{5 + \sqrt{6 \cdots}}}

\Leftrightarrow 2\phi > \sqrt{4+ \sqrt{5 + \sqrt{6 \cdots}}}

\Leftrightarrow 2\sqrt{1+\sqrt{1 + \sqrt{1 + . . . }}} > \sqrt{4+ \sqrt{5 + \sqrt{6 \cdots}}}

\Leftrightarrow \sqrt{4 + \sqrt{16 + \sqrt{256 + \sqrt{256^2 + . . . }}}}> \sqrt{4+ \sqrt{5 + \sqrt{6 \cdots}}}
    Which again, it clearly is.
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    (Original post by Felix Felicis)
    I'm trying to avoid these problems because of chem revision but they're too tempting xD
    Spoiler:
    Show
    I think it would be sufficient to prove that:

    \sqrt{5} < \sqrt{3 + \sqrt{4 + \sqrt{5 + ...}}} < \sqrt{4 + \sqrt{5}}

    Ok, first we prove \sqrt{5} < \sqrt{4 + \sqrt{5}} \Rightarrow 5 < 4 + \sqrt{5} which is true as \sqrt{5} > 1

    Now we prove \sqrt{5} < \sqrt{3 + \sqrt{4 + ... }}

    Assume \sqrt{5} < \sqrt{3 + \sqrt{4 + ... }} \Rightarrow 5 < 3 + \sqrt{4 + \sqrt{5 + ...}}

    \Rightarrow 2 < \sqrt{4 + \sqrt{5 + ...}} which is true as \sqrt{5 + \sqrt{6 + ... }} > 1

    Finally, we prove \sqrt{5} < \sqrt{3 + \sqrt{4 + \sqrt{5 + ...}}} < \sqrt{4 + \sqrt{5}}

    Assume the above to be true \Rightarrow 5 < 3 + \sqrt{4 + \sqrt{5 + ...}} < 4 + \sqrt{5}

    \Rightarrow 2 < \sqrt{4 + \sqrt{5 + ...}} < 1 + \sqrt{5}

    \Rightarrow 1 < \frac{1}{2} \sqrt{4 + \sqrt{5 + ... }} < \phi

    \Rightarrow 1 < 1 + \frac{1}{4} \sqrt{5 + \sqrt{6 + ... }} < \phi + 1

    \Rightarrow 0 < \frac{1}{4} \sqrt{5 + \sqrt{6 + ... }} < \phi

    = \frac{1}{4} \sqrt{5 + \sqrt{6 + ...}} < \sqrt{1 + \sqrt{1 + \sqrt{1 + ... }}}

     = \sqrt{ \frac{5}{16} + \sqrt{ \frac{6}{16^{2}} + ... }} < \sqrt{1 + \sqrt{1 + ...}} which is true as each nested radical on the LHS is less than 1, and we're done

    I fear there may be a flaw in my logic, seems to simple :holmes:
    Very similar to mine - damn you Felix - you beat me to it
    I've been spending a while on the rabbit and fox and nowt's happening :cry2:
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    (Original post by joostan)
    Very similar to mine - damn you Felix - you beat me to it
    I've been spending a while on the rabbit and fox and nowt's happening :cry2:
    Nice
    What problem is this? :O I'll go snoop :holmes:
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    (Original post by Felix Felicis)
    I'm trying to avoid these problems because of chem revision but they're too tempting xD
    Spoiler:
    Show
    I think it would be sufficient to prove that:

    \sqrt{5} < \sqrt{3 + \sqrt{4 + \sqrt{5 + ...}}} < \sqrt{4 + \sqrt{5}}

    Ok, first we prove \sqrt{5} < \sqrt{4 + \sqrt{5}} \Rightarrow 5 < 4 + \sqrt{5} which is true as \sqrt{5} > 1

    Now we prove \sqrt{5} < \sqrt{3 + \sqrt{4 + ... }}

    Assume \sqrt{5} < \sqrt{3 + \sqrt{4 + ... }} \Rightarrow 5 < 3 + \sqrt{4 + \sqrt{5 + ...}}

    \Rightarrow 2 < \sqrt{4 + \sqrt{5 + ...}} which is true as \sqrt{5 + \sqrt{6 + ... }} > 1

    Finally, we prove \sqrt{5} < \sqrt{3 + \sqrt{4 + \sqrt{5 + ...}}} < \sqrt{4 + \sqrt{5}}

    Assume the above to be true \Rightarrow 5 < 3 + \sqrt{4 + \sqrt{5 + ...}} < 4 + \sqrt{5}

    \Rightarrow 2 < \sqrt{4 + \sqrt{5 + ...}} < 1 + \sqrt{5}

    \Rightarrow 1 < \frac{1}{2} \sqrt{4 + \sqrt{5 + ... }} < \phi

    \Rightarrow 1 < 1 + \frac{1}{4} \sqrt{5 + \sqrt{6 + ... }} < \phi + 1

    \Rightarrow 0 < \frac{1}{4} \sqrt{5 + \sqrt{6 + ... }} < \phi

    = \frac{1}{4} \sqrt{5 + \sqrt{6 + ...}} < \sqrt{1 + \sqrt{1 + \sqrt{1 + ... }}}

     = \sqrt{ \frac{5}{16} + \sqrt{ \frac{6}{16^{2}} + ... }} < \sqrt{1 + \sqrt{1 + ...}} which is true as each nested radical on the LHS is less than 1, and we're done

    I fear there may be a flaw in my logic, seems to simple :holmes:
    Story of my life >.< I keep replacing chem with maths

    (Original post by joostan)
    Very similar to mine - damn you Felix - you beat me to it
    I've been spending a while on the rabbit and fox and nowt's happening :cry2:
    Haha both of you are amazing, want a hint? woohoo! payback
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    (Original post by Felix Felicis)
    Nice
    What problem is this? :O I'll go snoop :holmes:
    Its on p201 I think
    I'm having trouble generating a suitable function for the motion of the fox
    Looks a little bit like M2 so I was taking a crash course
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    (Original post by tigerz)
    Haha both of you are amazing, want a hint? woohoo! payback
    Never!
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    (Original post by joostan)
    Never!
    LOOL good luck I told you i'd find a good question
    Felix here:
    There is a rabbit that runs in a perfect circle of radius r with a constant speed v. A fox chases the rabbit, starting from the center of the circle and also moves with a constant speed v such that it is always between the center of the circle and the rabbit. How long will it take for the fox to catch the rabbit?
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    (Original post by DJMayes)
    ...
    Did M2 Jan 2010 for the first time and got 70 , I wouldn't say it's that bad although my method for quite a few was different to the MS. Dropped all the marks on the last question as I didn't know what to do. I mean why did they use dy/dx for projectiles?
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    On another note...it appears The Polymath has been banned again :rofl:
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    (Original post by Boy_wonder_95)
    Did M2 Jan 2010 for the first time and got 70 , I wouldn't say it's that bad although my method for quite a few was different to the MS. Dropped all the marks on the last question as I didn't know what to do. I mean why did they use dy/dx for projectiles?
    Was this for the final part of the question? That also confused me but would it because they would have the same gradient but just different signs? I'm just taking a guess really and it's most likely incorrect.


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    (Original post by MathsNerd1)
    Was this for the final part of the question? That also confused me but would it because they would have the same gradient but just different signs? I'm just taking a guess really and it's most likely incorrect.


    Posted from TSR Mobile
    Yh 8c I believe it was, I thought something similar; that the vertical velocity would be the same but in the opposite direction but I got it all wrong. Still managed a method mark for using v = u + at :lol:
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    (Original post by Boy_wonder_95)
    Yh 8c I believe it was, I thought something similar; that the vertical velocity would be the same but in the opposite direction but I got it all wrong. Still managed a method mark for using v = u + at :lol:
    That's fair enough then and that would've been the only place I dropped marks too, however I put the correct answer for the impulse question but then decided to cross it out as I actually thought it wouldn't make much sense of the direction afterwards if it got hit at that angle, so I lost a good amount of marks there but still got 65/75 in the end which wasn't all that bad considering the grade boundaries
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    Seems like lots of ppl posting on this thread have lots of warning points... Wonder why.. :confused:
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    (Original post by thers)
    Seems like lots of ppl posting on this thread have lots of warning points... Wonder why.. :confused:
    Haven't you heard how rowdy mathematicians can get? :confused:
 
 
 
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