Question The Pony (ask me anything)

[Cobalt Storm]

Hey, Im Cobalt, and im willing to answer your question, no matter how personal or abstract.

Little bit about me if your to lazy to just look on my profile,

my sona is a female pony,
i am a male irl,
i like the color blue
my favorite food is a meatball sub.

so go ahead, ask me anything

[sigols]

How and when did you tell your parents you were a furry? If you have told them, how did you explain what a furry is? And the most important, how and when did you realise you were a furry?


Sorry for my not that good english, im from norway 

[Cobalt Storm]

Your english was better than a lot of people i know and i havent told my perrents im a furry, mainly because whenever the subject comes up they say bad things about it.

[Kittie]

What made you choose the colors of your ponisona

[Cobalt Storm]

It was mainly just my two favorite colors, yellow and blue, and the black just looks good as a second color ^.^ thats the honest answer, but here is the interesting,

The cobalt blue reflects my family line and my name, the yellow relates to my techno music, and the black reflects my rough past. and they red eyes are because i can never sleep and always have red eye.

[Kittie]

Cool

[anoni]

I do this for everyone

YOU'RE NO EXCEPTION

If you have a lawn that was n x n square metres large (where n is a positive integer), you had a lawn mower which took you "t" time to mow one square metre, you could also mow a square metre and TURN but that would take you "k" time, where k > t. What is the best way to mow the lawn? How long would it take you to mow it that way? Are there multiple equivalent ways? If there are, can you prove that they are equivalent?

[Cobalt Storm]

you would mow across one side, turn and mow next to your previous path, so that t=(nk)+n^2 thats the fastest way i see doing it.

[MrRazot]

Why do you have a face?

[Cobalt Storm]

cause faces are delicious

[anoni]

you would mow across one side, turn and mow next to your previous path, so that t=(nk)+n^2 thats the fastest way i see doing it.

This answer is partially correct! (But isn't totally) the question itself was rather difficult though.

The amount of corners you would take would be (2n - 2) not n, (if you had a 2x2 field you would have to cross two corners (As turning 180 degrees counts as one corner), if you had a three by three field you would need to cross 4 corners). Second, the amount of lines wouldn't be n² because you don't count the corners as a straight edge, so the answer would actually be n(n - 2) + 2. So the final equation for a line by line method would actually be...

  (2n - 2)k + (n(n - 2) + 2)t, or

Equation 1) 2k(n - 1) + t(n(n - 2) + 2)  (If you don't believe me try it out for n = 1, 2, 3, 4, etc)

   The other thing is that isn't actually the ONLY fastest method, if we assume you don't have to turn for the last corner (we assumed that in our last equation) the method of going around the perimeter of the garden and continuing to go around the same perimeter as it goes smaller until you get to the center of the lawn is actually equally as fast!

  The equation for this second method is pretty easy to get once you take in a few smaller cases

n = 1; 1t, 0k  (1² = 1 so it checks out)
n = 2; 2t [(1 + 1)t], 2k [(1 + 1)k] (2² = 4 so it checks out)
n = 3; 5t [(2 + 1 + 1 + 1)t], 4k [(1 + 1 + 1 + 1)k] (3² = 9 so it checks out)
n = 4; 10t  [(3 + 2 + 2 + 1 + 1 + 1)t], 6k [(1 + 1 + 1 + 1 + 1 + 1)k] (4² = 16 so it checks out)

It may be hard to see, but you'll find that the amount of edges is given by the formula...
(n - 1) + 2(n - 2) + 2(n -3) + ... + 2 + 1.

The formula for the amount of corners is given by

(2n - 2)

So the formula for the perimeter side is actually

Equation 2) 2k(n - 1) + t((n - 1) + 2(n - 2) + 2(n - 3) + ... + 2 + 1)

Now we can easily see through inspection that both equation 1) and equation 2) are the same, because the amount of squares in the garden is equal, and because the equations for the corners are the same (they have the same amount of corners) then they must have the same amount of straight edges and thus must take the same amount of time to do.

  However, if that proof does not satisfy you, we can prove it by induction

The assumption is that:
  2k(n - 1) + t(n(n - 1) + 2) = 2k(n - 1) + t((n - 1) + 2(n - 2) + 2(n - 3) + ... + 2 + 1)

We can simplify the expression by getting rid of the corners (as they are the same 2k(n - 1) equation) and dividing both sides by "t", this gives us the final result that
  n(n - 2) + 2 = (n - 1) + 2(n - 2) + 2(n - 3) + ... + 2 + 1

  This is the result that we have to prove.

  First of all, lets show that it's right for one case, let n = 1;

  1(1 - 2) + 2 = (1 - 1) + 1;
  1 = 1;

  Thus it is true for n = 1;

  Now, the method behind induction is that if we can prove it's true for the first case, can we prove it's true for the second case assuming the first case is true? Then prove it for the third case assuming the second case is true? Then the fourth case, then fifth, etc, etc, etc. So for this we assume that the xth case is true. So let n = x (normally you let n = k, but because we already used k we will use x, remember that x is a constant positive integer)

  x(x - 2) + 2 = (x - 1) + 2(x - 2) + 2(x - 3) + ... + 2 + 1.

  Now, we can assume the xth case is true, all we have to do is show that if the xth case is true, then the (x + 1)th case is true as well.

  Consider n = x + 1;

  (x + 1)(x - 1) + 2 = x + 2(x - 1) + 2(x - 3) + ... + 2 + 1;

  All we have to do is show that the RHS is equal to the LHS,

  RHS = x + 2(x - 1) + 2(x - 2) + ... + 2 + 1
Which could be written as...
  RHS = x + (x - 1) + (x - 1) + 2(x - 2) + ... + 2 + 1;

  Now remember we're assuming that the xth case is true, and the xth case states that...
  x(x - 2) + 2 = (x - 1) + 2(x - 2) + 2(x - 3) + ... + 2 + 1.

  So if we transfer that identity into our RHS equation we get

  RHS = x + (x - 1) + x(x - 2) + 2

  Now expand the RHS

  RHS = x + x - 1 + x² - 2x + 2
  RHS = x² + 1

  That "+1" is giving us problems, but we can get rid of it by knowing that 1 = 2 - 1;
  RHS = x² - 1 + 2

  Now we can factorize using standard factorization formula
  RHS = (x + 1)(x - 1) + 2

  This is the left hand side

  So therefore we  have shown that the two statements are equivalent.

  This is the full answer to the mathematics equation, that there are two fastest ways, the line method (the one you stated) and the perimeter method. Though you did give more thought than most people so thank you for actually considering the maths equations

 

[sigols]

.......... *brain starting to boil, making my head expand and then explode cause of the steam*..... What is this sorcery?! Its in f**king japanese!

[HankTOBA]

Ax or Chain saw?

[Toxic Konpeito]

I speak in calculus....I speak in calculus.....*bangs head against desk*
 
How many fingers do you have? D8