The Ultimate Math Expression Thread.

Pre-College Topics High School Life
1

<p>Okay.
Let us create the ultimate CC mathematical expression.
Add/Subtract/Divide/Multiply any number, derivative, series, limit, integral, system, vector, etc.
Differentiate/integrate/limit any part of expression all you want.</p>

<p>Add as many terms as you want. Anything and everything.</p>

<p>However, please simplify things if possible. If there is something that is, for example, d(constant)=derivative of a constant=0. please simplify.</p>

<p>Note to use int() for integral, d() for derivative, sum() etc.</p>

<p>Copy+paste and add on to post above.</p>

<p>I'll start.</p>

<p>1</p>

2

<p>1*74</p>

<p>Tsenchar.</p>

3

<p>1*74/pi</p>

<p>10char</p>

4

<p>1*74/pi = 0</p>

<p>Simplify that :P</p>

5

<p>1*74/pi = 0 + 5000000000000000000000000000000000</p>

6

<p>^ Good work DarkDays. This thread can’t survive much longer… :P</p>

<p>1*74/pi = 0 + 5000000000000000000000000000000000i</p>

7

<p>1*74/pi = 0 + 5000000000000000000000000000000000i * -i + 5 * 10^32 + 100/pi - 26/pi</p>

8

<p>1*74/pi = 0 + 5000000000000000000000000000000000i * -i + 5 * 10^32 + 100/pi - 26/pi + infinity</p>

9

<p>1*74/pi = 0 + 5000000000000000000000000000000000i * -i + 5 * 10^32 + 100/pi - 26/pi + infinity - 1/2 infinity (try and wrap your brain around that)</p>

10

<p>1<em>74/pi - e^{pi</em>i}*ln(epsilon) = 0 + 5000000000000000000000000000000000i * -i + 5 * 10^32 + 100/pi - 26/pi + infinity - 1/2 infinity</p>

<p>quod erat demonstrandum</p>

11

<p>1<em>74/pi - e^{pi</em>i}*ln(epsilon) = 0 + 5000000000000000000000000000000000i * -i + 5 * 10^32 + 100/pi - 26/pi + infinity - 1/2 infinity -0.0000000000000000000000001</p>

12

<p>1<em>74/pi - e^{pi</em>i}*ln(epsilon) = 0 + 5000000000000000000000000000000000i * -i + 5 * 10^32 + 100/pi - 26/pi + infinity - 1/2 infinity -0.0000000000000000000000001 = 2^(aleph_0)</p>

<p>Trivial.</p>

13

<p>1<em>74/pi - e^{pi</em>i}*ln(epsilon) = 0 + 5000000000000000000000000000000000i * -i + 5 * 10^32 + 100/pi - 26/pi + infinity - 1/2 infinity -0.0000000000000000000000001 = 2^(aleph_0)> 2^i</p>

14

<p>(1<em>74/pi - e^{pi</em>i}*ln(epsilon) = 0 + 5000000000000000000000000000000000i * -i + 5 * 10^32 + 100/pi - 26/pi + infinity - 1/2 infinity -0.0000000000000000000000001) / 0= 2^(aleph_0)> 2^i</p>

<p>Game over.</p>

15

<p>(1<em>74/pi - e^{pi</em>i}*ln(epsilon) = 0 + 5000000000000000000000000000000000i * -i + 5 * 10^32 + 100/pi - 26/pi + infinity - 1/2 infinity -0.0000000000000000000000001) / 0^0 = 2^(aleph_0) > 2^i</p>

<p>The proof is left to the reader.</p>

16

<p>If anybody does this in the next hour I’ll fast from CC for a week.</p>

17

<p>^what do you mean by “does”
(1<em>74/pi - e^{pi</em>i}*ln(epsilon) +infinity = 0 + 5000000000000000000000000000000000i * -i + 5 * 10^32 + 100/pi - 26/pi + infinity - 1/2 infinity -0.0000000000000000000000001) / 0^0 = 2^(aleph_0) > 2^i</p>

18

<p>as in completely solve it…</p>

19

<p>i feel a little dumber after reading this thread. quite depressing.</p>

20

<p><em>Removed</em></p>

<p>just look below my post -_-</p>

<p>thanks for correcting Ilikeyou</p>

<p>Ilikeyou2</p>