Current Year 12 Thread Mark VI

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  1. Etoile's Avatar
    221 01-05-2012  21:43
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    Re: Current Year 12 Thread Mark VI
    (Original post by Maths_Lover)
    :woo: Well done! :party2:



    No wonder I was having troubles... I haven't covered hyperbolic functions yet! :shakecane:

    :lol:
    (Original post by chickenonsteroids)
    Well done

    (Original post by Llewellyn)
    I'm no where near finishing ...

    But then, neither is anyone else in my class :rolleyes:

    Congrats! 95% is excellent
    Thank you everyone!! This is 40% of my grade as well, so I'd only need 2/60 marks in the exam for a D

    Llewellyn, when is your deadline? :lolwut:
  2. bananarama2's Avatar
    222 01-05-2012  21:48
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    Re: Current Year 12 Thread Mark VI
    (Original post by Etoile)
    Thank you everyone!! This is 40% of my grade as well, so I'd only need 2/60 marks in the exam for a D

    Llewellyn, when is your deadline? :lolwut:
    Setting you sights high I see
  3. Llewellyn's Avatar
    223 01-05-2012  21:55
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    Re: Current Year 12 Thread Mark VI
    (Original post by Etoile)
    Thank you everyone!! This is 40% of my grade as well, so I'd only need 2/60 marks in the exam for a D

    Llewellyn, when is your deadline? :lolwut:
    13th of May for the final draft :holmes:

    I still have faith...
  4. Maths_Lover's Avatar
    224 01-05-2012  21:59
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    Re: Current Year 12 Thread Mark VI
    (Original post by Llewellyn)
    Prove by contradiction that e is irrational given:

    \displaystyle e = \displaystyle\sum\limits_{n=0}^{ \infty} \frac{1}{n!}
    YES! Learned this over the Easter holidays.

    Assume e is rational. Then it can be expressed as a ratio of coprime integers, a and b:

    \displaystyle e = \frac{b}{a} = \displaystyle\sum\limits_{n=0}^{ \infty} \frac{1}{n!}

    \displaystyle = (a-1)!b = \displaystyle\sum\limits_{n=0}^{ \infty} \frac{a!}{n!} - upon multiplying through by a factorial.

    Now, as a and b are both positive integers, (a-1)!b must also be a positive integer, as are all of the terms in the series \displaystyle\sum\limits_{n=0}^{ \infty} \frac{a!}{n!} up until and including the term \dfrac{a!}{a!} .

    Therefore the terms beyond \dfrac{a!}{a!} must sum to a positive integer to make this possible.

    Considering the terms after \dfrac{a!}{a!} :

    \displaystyle \frac{a!}{(a+1)!} + \frac{a!}{(a+2)!} + \frac{a!}{(a+3)!} + \cdots = a! \displaystyle\sum\limits_{n=1}^{ \infty} \frac{1}{(a+n)!}

    One can say that \displaystyle\sum\limits_{n=1}^{ \infty} \frac{1}{(a+n)!} is less than \displaystyle\sum\limits_{n=1}^{ \infty} \frac{1}{(a+1)^n} = \frac{1}{(a+1)} \displaystyle\sum\limits_{n=0}^{ \infty} \frac{1}{(a+1)^n}

    Summing the resulting geometric progression to infinity:

    \displaystyle S_{\infty } \displaystyle\sum\limits_{n=0}^{ \infty} \frac{1}{(a+1)^n} = \frac{1}{1- \frac{1}{(a+1)} }

    \displaystyle = \frac{(a+1)}{a} - upon simplification.

    \displaystyle \therefore \frac{1}{(a+1)} \displaystyle\sum\limits_{n=0}^{ \infty} \frac{1}{(a+1)^n} = \frac{1}{(a+1)} \left(\frac{(a+1)}{a} \right) = \frac{1}{a}

    Now, the maximum value that \dfrac{1}{a} can take is 1, so this means that:

    \displaystyle 0 < \frac{a!}{(a+1)!} + \frac{a!}{(a+2)!} + \frac{a!}{(a+3)!} + \cdots < 1.

    \displaystyle \frac{a!}{(a+1)!} + \frac{a!}{(a+2)!} + \frac{a!}{(a+3)!} + \cdots is now defined to be an integer between zero and one.

    Consider an integer k, which lies between zero and one. Then 0 < k < 1 . On multiplying through by k, we obtain 0 < k^2 < k , which is nonsense as the square of an integer is never less than the integer so we reach a contradiction and so there exist no such integers k.

    This means that we reach a contradiction with \displaystyle 0 < \frac{a!}{(a+1)!} + \frac{a!}{(a+2)!} + \frac{a!}{(a+3)!} + \cdots < 1 and so our original assumption must have been false.

    It follows that e is irrational.
    Last edited by Maths_Lover; 03-06-2012 at 22:06.
    Post rating:
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  5. chickenonsteroids's Avatar
    225 01-05-2012  22:02
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    Re: Current Year 12 Thread Mark VI
    (Original post by Maths_Lover)
    x
    How do you prove 2+2 will always equal 4?
  6. Llewellyn's Avatar
    226 01-05-2012  22:06
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    Re: Current Year 12 Thread Mark VI
    (Original post by Maths_Lover)
    x
    Very good +1 for the latex effort

    (Original post by chickenonsteroids)
    How do you prove 2+2 will always equal 4?
    Don't open that can of worms.
  7. Maths_Lover's Avatar
    227 01-05-2012  22:10
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    Re: Current Year 12 Thread Mark VI
    So many LaTex fails in that last post. :facepalm:
  8. bananarama2's Avatar
    228 01-05-2012  22:11
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    Re: Current Year 12 Thread Mark VI
    (Original post by chickenonsteroids)
    How do you prove 2+2 will always equal 4?
    I'd argue is axiomatic and true for physical parallels.
  9. Maths_Lover's Avatar
    229 01-05-2012  22:14
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    Re: Current Year 12 Thread Mark VI
    (Original post by chickenonsteroids)
    How do you prove 2+2 will always equal 4?
    Dear goodness no.

    (Original post by Llewellyn)
    Very good +1 for the latex effort

    Don't open that can of worms.
    Thanks. :lol: It serves as good practice in LaTex. :ahee:
  10. Maths_Lover's Avatar
    230 01-05-2012  22:16
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    Re: Current Year 12 Thread Mark VI
    (Original post by Emissionspectra)
    We had a maths test recently and I was the only one who could intergrate \int sin^3xcos^2x\ dx
    Go you. :five:

    (Original post by Etoile)
    Thank you everyone!! This is 40% of my grade as well, so I'd only need 2/60 marks in the exam for a D

    Llewellyn, when is your deadline? :lolwut:
    You're welcome. :woo:
  11. chickenonsteroids's Avatar
    231 01-05-2012  22:21
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    Re: Current Year 12 Thread Mark VI
    (Original post by wcp100)
    I'd argue is axiomatic and true for physical parallels.

    (Original post by Maths_Lover)
    Dear goodness no.
    axiomatic truths annoy me for some reason.

    Silly rationalism
  12. Maths_Lover's Avatar
    232 01-05-2012  22:22
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    Re: Current Year 12 Thread Mark VI
    Here's the next one:

    \displaystyle \int \sqrt{25 - x^2} \ dx

    A little trickier...
    Last edited by Maths_Lover; 01-05-2012 at 22:23.
  13. Llewellyn's Avatar
    233 01-05-2012  22:23
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    Re: Current Year 12 Thread Mark VI
    To expand, we need to define what 2 and 4 are. If by 2, you mean a real number that rounds to 2 and by 4 you mean a real number that rounds to four then 2.3 + 2.3 = 4.6, and by extension, 2+2 =5, not 4. You can really mess with maths through various philosophies of language.

    There is a perceived truth, or axiom, that 2+2 = 4 because it seems logical. But that is just a belief; it is not knowledge.

    (Original post by chickenonsteroids)
    axiomatic truths annoy me for some reason.
    Spoken like a true philosopher.

    If it comforts you, there are several arguments that can be put forward that suggest no one is right.

    And, to get off topic, that is what I dislike about how maths is taught at school. What maths actually is; is like learning a language of truth, through it I can make a series of statements and you have to accept that I am right. And from there, you can construct arguments that range from why probabilities cannot be set, to why black holes exist, etc. That is the beauty of maths, and that kind of reasoning and logic is not even slightly touched upon until University level... :grumble:
    Last edited by Llewellyn; 01-05-2012 at 22:30.
  14. Maths_Lover's Avatar
    234 01-05-2012  22:31
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    Re: Current Year 12 Thread Mark VI
    (Original post by chickenonsteroids)
    axiomatic truths annoy me for some reason.

    Silly rationalism
    Let it be... the world would be mental without axioms. :eek2:
  15. Maths_Lover's Avatar
    235 01-05-2012  22:36
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    Re: Current Year 12 Thread Mark VI
    You might like this one (or maybe not):

    \displaystyle \int^{\frac{\pi }{2} }_0 \frac{5}{3\sin \theta + 4\cos \theta } \ d\theta

    This is the last question from me tonight.
  16. chickenonsteroids's Avatar
    236 01-05-2012  22:36
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    Re: Current Year 12 Thread Mark VI
    (Original post by Llewellyn)
    To expand, we need to define what 2 and 4 are. If by 2, you mean a real number that rounds to 2 and by 4 you mean a real number that rounds to four then 2.3 + 2.3 = 4.6, and by extension, 2+2 =5, not 4. You can really mess with maths through various philosophies of language.

    There is a perceived truth, or axiom, that 2+2 = 4 because it seems logical. But that is just a belief; it is not knowledge.


    Spoken like a true philosopher.

    If it comforts you, there are several arguments that can be put forward that suggest no one is right.

    And, to get off topic, that is what I dislike about how maths is taught at school. What maths actually is; is like learning a language of truth, through it I can make a series of statements and you have to accept that I am right. And from there, you can construct arguments that range from why probabilities cannot be set, to why black holes exist, etc. That is the beauty of maths, and that kind of reasoning and logic is not even slightly touched upon until University level... :grumble:
    Knowledge a belief also. It's a justified belief and that's why knowledge seems to change a lot. Yay science

    :congrats: for the statement I put in bold. It's so true, maths is interesting but, for me, it's when you question why something is the case and not just saying 'x is this... use it'

    (Original post by Maths_Lover)
    Let it be... the world would be mental without axioms. :eek2:
    Imagine what we'd do without the Euclidean axioms :eek4:
  17. bananarama2's Avatar
    237 01-05-2012  22:44
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    Re: Current Year 12 Thread Mark VI
    (Original post by Maths_Lover)
    Here's the next one:

    \displaystyle \int \sqrt{25 - x^2} \ dx

    A little trickier...
    \frac{x}{2} \sqrt{25-x^2} + \arcsin \left(\frac{x}{5} \right)\frac{25}{2}
  18. Maths_Lover's Avatar
    238 01-05-2012  22:49
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    Re: Current Year 12 Thread Mark VI
    (Original post by chickenonsteroids)
    Imagine what we'd do without the Euclidean axioms :eek4:
    I can't! :yikes:
  19. Maths_Lover's Avatar
    239 01-05-2012  22:50
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    Re: Current Year 12 Thread Mark VI
    (Original post by wcp100)
    \frac{x}{2} \sqrt{25-x^2} + \arcsin \left(\frac{x}{5} \right)\frac{25}{2}
    Perfect!

    You are destroying these... I'll have to find some more difficult ones for you.

    Try the last one.
  20. bananarama2's Avatar
    240 01-05-2012  22:54
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    Re: Current Year 12 Thread Mark VI
    (Original post by Maths_Lover)
    Perfect!

    You are destroying these... I'll have to find some more difficult ones for you.

    Try the last one.
    Any hints for the last one?
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