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    (Original post by bananarama2)
    Yes, but the way GCSEs are taught means they wouldn't have a clue what a function was even if they do stuff related to them....
    Ah ok. See in France we study these things rather than dropping coins off the Eiffel tower. :teehee:
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    (Original post by Felix Felicis)
    Some more (*) problems for people who want them:

    Problem 172*

    Is \displaystyle\prod_{r=0}^{100} r! a perfect square? If not, could we remove one of the factorials to make it a perfect square?


    Solution 172*

    No, it is not a perfect square. How? Well, I can compute 100! in my mind. (like this guy: http://www.youtube.com/watch?v=8b2jOcKI798)

    Taking the premise of 100! is not a perfect square.
    A perfect square for factorials can be defined as below:

     a! \times a! = (a!)^2


     100! \times 99! \times 98! ... \times 1!

    Note that:  100! = 100 \times 99!

    So  100! \times 99! =  100 \times 99! \times 99!

    As  a! \times a! = (a!)^2 , that means:

     99! \times 99! = (99!)^2 which is a perfect square.

    We can continue this procedure mechanically by pairing up the rest of the factorials so we have:

     (100 \times (99!)^2) \times (98 \times (97!)^2) \times (96 \times (95!)^2)... \times (2 \times (1!)^2)

    Rearranging these (there are no minus terms or chicanery occurring here)

     (100 \times 98 \times 96 \times 94... \times 2) \times ((99!)^2 \times (97!)^2 \times (95!)^2 \times... (1!)^2)


    Using the LHS, we have:  2(1) (\times 2(2) \times 2(3) \times 2(4) ... \times 2(50))

    So LHS of the equation is simply:  2^5^0 \times 50!

    We know that \sqrt {2^5^0} is a perfect square (I think this is what you could call trivial), by removing  50! , we can actually end up with the initial result forming a perfect square, such that:


    \displaystyle\prod_{r=0}^{100} r!  - (50!) =  x^2
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    (Original post by Lord of the Flies)
    Ah ok. See in France we study these things rather than dropping coins off the Eiffel tower. :teehee:

    PRSOM. :lol:

    In all fairness though, Lagrange would be quite content with his question.
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    (Original post by Lord of the Flies)
    Ah ok. See in France we study these things rather than dropping coins off the Eiffel tower. :teehee:
    That question was off you first year dynamics and relativity course smart alec
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    (Original post by Jkn)
    Lightweight, you can get it form applying integration by parts to the other one by integrating \ln(x) (and differentiating the denominator) and then evaluating a fairly trivial other part with a tanx substitution
    Yeah but that took me like an hour and half to think of on the problem LotF set earlier :lol:
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    (Original post by DJMayes)
    Hyperbolics are more fun.

    (What answer do you get with that sub though?)
    The same
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    (Original post by DJMayes)
    Let  x^3 = sinh^2u

     3x^2 \ dx = 2sinhucoshu \ du

     \Rightarrow I = \frac{2}{3} \int sinh^3ucosh^2u \ dx

     = \frac{2}{3} \int sinhu(cosh^2u-1)cosh^2u \ dx

     = \frac{2}{3} \int sinhucosh^4u - sinhucosh^2u \ dx

     = = \frac{2}{3} [ \frac{1}{5}cosh^5u -\frac{1}{3}cosh^3u ]

    x = 0 so u = 0, coshu = 1, x = 2 so coshu = 3

    Finally plugging those in I get 1192/45, although this isn't really a nice answer...
    Hopefully now you will follow me to the algebraic promised land...

    Solution 178 (2)

    \displaystyle \int_0^2 ((1+x^3)x^2-x^2)\sqrt{1+x^3} dx = \left[\frac{2}{15} (1+x^3)^{\frac{5}{2}} -\frac{2}{9} (1+x^3)^{\frac{3}{2}} \right]_0^2 =\frac{1192}{45} \ \square
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    Give me 2-3 months, and I'll be able to do those integrals which look so palatable.
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    (Original post by bananarama2)
    That question was off you first year dynamics and relativity course smart alec
    Ooo I'm interested again

    I got an answer but I think it was wrong.. I might type it up later
    Spoiler:
    Show
    Basically doing some dodgy limit calculus and noting that I_1 \omega_1 = I_2 \omega_2 \Rightarrow v_1 (r+\delta r)=v_2 r. Is this the right sort of ****? :lol:

    (Original post by Felix Felicis)
    Yeah but that took me like an hour and half to think of on the problem LotF set earlier :lol:
    When was it set? :eek:
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    (Original post by Zakee)
    PRSOM. :lol:

    In all fairness though, Lagrange would be quite content with his question.
    I like the blase way physics questions are given. The sort of "a washing machine can be powered by a solar umbrella, calculate the rate of loss of mass of the sun." type questions
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    (Original post by Jkn)
    Ooo I'm interested again

    I got an answer but I think it was wrong.. I might type it up later
    Spoiler:
    Show
    Basically doing some dodgy limit calculus and noting that I_1 \omega_1 = I_2 \omega_2 \Rightarrow v_1 (r+\delta r)=v_2 r. Is this the right sort of ****? :lol:

    When was it set? :eek:
    Problem 132 xD Also, a few other integrals were set earlier as well but hey ho, nothing wrong with other people attempting them
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    (Original post by Zakee)
    Solution 172*

    No, it is not a perfect square. How? Well, I can compute 100! in my mind.
    I like all of your solution except this bit - even a computation of the number doesn't prove it isn't not a perfect square. I think a better way of doing this would be to count the number of a specific prime in your solution. For example, the number of 47's. From 47 to 100, we get 54 lots of 47. Also, we have another 47 included in each number between 94 and 100, giving another 7 lots of 47. This gives an odd number of 47's, and so it isn't a perfect square as you cannot divide the amounts of 47 equally amongst the two square factors.
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    (Original post by Jkn)
    Ooo I'm interested again

    I got an answer but I think it was wrong.. I might type it up later
    Spoiler:
    Show
    Basically doing some dodgy limit calculus and noting that I_1 \omega_1 = I_2 \omega_2 \Rightarrow v_1 (r+\delta r)=v_2 r. Is this the right sort of ****? :lol:

    When was it set? :eek:
    I did it differently to the lecture notes, but got the same answer. I didn't use dodgy limited calculus

    If you just want the direction you just need two lines of working. But you need the moment of inertia.
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    (Original post by bananarama2)
    I like the blase way physics questions are given. The sort of "a washing machine can be powered by a solar umbrella, calculate the rate of loss of mass of the sun." type questions

    My favorite is:

    "Jimmy is spinning around on the base of cylindrical chamber such that he is in contact with the base and well as the cylindrical wall. He is just on the point of lifting and the cylinder has a radius of 10m, and his mass is 60kg. Given an arbitrary compact gauge group, does a non-trivial quantum Yang-Mills theory with a finite mass gap exist?"
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    (Original post by DJMayes)
    I like all of your solution except this bit - even a computation of the number doesn't prove it isn't not a perfect square. I think a better way of doing this would be to count the number of a specific prime in your solution. For example, the number of 47's. From 47 to 100, we get 54 lots of 47. Also, we have another 47 included in each number between 94 and 100, giving another 7 lots of 47. This gives an odd number of 47's, and so it isn't a perfect square as you cannot divide the amounts of 47 equally amongst the two square factors.

    That is true. I was sort of joking about the computation part that I had calculated 100! x 99! x 98! x 97!... (If I did, then well, I might as well quit school now and just end up replacing the TI-84 plus).

    However, I didn't think of doing what you did as I just made an assumption (which I'm aware is scandalous). I'm actually fond of viewing it the way that you've shown Cheers man.

    I guess it's just my innate laziness which prevents me from being rigorous enough -sighs-. I should probably quit Mathematics and end up doing poetry. :nopity:
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    (Original post by Zakee)
    That is true. I was sort of joking about the computation part that I had calculated 100! x 99! x 98! x 97!... (If I did, then well, I might as well quit school now and just end up replacing the TI-84 plus).

    However, I didn't think of doing what you did as I just made an assumption (which I'm aware is scandalous). I'm actually fond of viewing it the way that you've shown Cheers man.
    You just knew it was't square given the rest of questions
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    (Original post by bananarama2)
    You just knew it was't square given the rest of questions

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    HA when und sees how much updating needs to be done he's going to be like:

    Spoiler:
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    (Original post by Lord of the Flies)
    HA when und sees how much updating needs to be done he's going to be like:

    Spoiler:
    Show

    I think that his fury will be more directed towards the endless questions which have been mislabeled.


    http://www.youtube.com/watch?v=i4w4wGJjuYg
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    (Original post by Lord of the Flies)
    HA when und sees how much updating needs to be done he's going to be like:

    Spoiler:
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    He'll probably return with a new account
 
 
 
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