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04-Aug-2016 03:31:11

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04-Aug-2016 07:41:51

Ancient Drew

Ancient Drew

Posts: 5,732 Rune Posts by user Forum Profile RuneMetrics Profile
Love, Nature, Fate: Catalysts of the Universe!

The physical parts of the universe, by which I mean the stars, planets etc., are created by nature. All life comes into being due to concoctions of chemicals and the right environment to grow in. This is ensured by nature and as such this force should be considered a catalyst of the universe.

Fate may be considered one or a number of paths taken through the course of time, through which love and nature also make use of. Everything goes to balance as is ensured by fate; death stops an overabundance of life ravaging nature, and without fate the good things in life, such as love, cannot be appreciated at their fullest. This comes to show that fate is a universal catalyst as well, and it enriches both love and nature!

Love is a bit harder to explain, but I will try. Souls are a vital part of the universe, and they are enriched by love which grows over time. Life nurtured with love will tend to nature in the same way, allowing for faster catalytic growth in the universe. And while death, an agent of fate, claims loved ones and causes tragedy, it also has the potential to reveal the resolve of the strongest wills. These wills can be used for creation or destruction, and as such love can also be considered a vital catalyst of the universe, and all of these things as such bring balance. Prepare for hell on RuneScape in Naval Cataclysm!

Pokemon battle? Friend Code: 4614-0426-2439

12-Aug-2016 21:10:29

Saikie

Saikie

Posts: 29 Bronze Posts by user Forum Profile RuneMetrics Profile
e^(i*pi)+1=0, or for all you Tau enthusiasts, e^(i*tau)=1.

My favourite mathematical proof.
Claim: There exists irrationals x and y such that x^y is rational.
Proof: Consider sqrt(2)^sqrt(2). Now sqrt(2)^sqrt(2) must either be rational or irrational.
If sqrt(2)^sqrt(2) is rational we are done (in this case
x=sqrt(2)
and
y=sqrt(2)
).
If sqrt(2)^sqrt(2) is irrational, then (sqrt(2)^sqrt(2))^sqrt(2)=2 and we are done (in this case
x=sqrt(2)^sqrt(2)
and
y=sqrt(2)
).
Beauty is the first test: there is no permanent place in the world for ugly mathematics
-G.H. Hardy

18-Aug-2016 16:28:01

Saikie

Saikie

Posts: 29 Bronze Posts by user Forum Profile RuneMetrics Profile
YYinh said :
4A6F6B6B75 said :
Problem, Archimedes?

Hehe :D

That's sneaky :P

Unfortunately (or perhaps fortunately!) this does not hold. The object defined by iterations and the circle can be shown to not be the same object and so the "proof" fails to show that pi is 4. In general, two functions f and g can have little difference, that is |f(x)-g(x)|<e, e>0, for all a<=x<=b, but have their derivatives very different, that is |f'(x)-g'(x)|>m for some constant m>0. The circle and the object considered can be represented by functions p(x,y) and q(x,y) where x(t) and y(t) depend on, say, 0<=t<=1; parameterisations that can be shown to behave like f and g descirbed above.
Beauty is the first test: there is no permanent place in the world for ugly mathematics
-G.H. Hardy

18-Aug-2016 16:59:34

Saikie

Saikie

Posts: 29 Bronze Posts by user Forum Profile RuneMetrics Profile
4A6F6B6B75 said :
^It's a decent troll though, and I bet you didn't notice that

4! = 24


Saikie said :
e^(i*pi)+1=0, or for all you Tau enthusiasts, e^(i*tau)=1.
Euler's identity! I really hope this is covered in my future complex analysis courses!

I took the exclamation mark as punctuation rather than factorial, bad Rex!

Indeed, it is Euler's identity! If you are doing Complex Analysis, you most certainly will be meeting it, or at least meet e^(ix)=cos(x)+i*sin(x), where x is real.

Hope you enjoy discs and epsilon-delta proofs, you'll be seeing lots of them :D .
Beauty is the first test: there is no permanent place in the world for ugly mathematics
-G.H. Hardy

18-Aug-2016 18:31:42 - Last edited on 18-Aug-2016 18:31:53 by Saikie

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