SpiritOfWrath

... I refuse to comment. Cole is powerful enough as is.

Tenth nodded. "At least consider, though, that these people had the entire multiverse at their disposal, and they chose these. But I see your point."

Well, like I said, I don't do too much for the high tech stuff. I'll get back to you once my Author starts and finishes his homework.

There is a cost. Though I guess if you do "stabilize" them, making them independent of me, that is irrelevant. But in order to do so, you would need effective manipulation of the Overflow, as well as a sound understanding in Andarik. One that I don't have.

"Still, we should exercise caution. You of all people should know not to underestimate the technologies or magics of another people."

OKOKOKOKOKOKOKOKOKOKOKOKOKOKOKOKOKOK It doesn't help that I cant type math characters. ORIGINAL TEXT (for reference): The limit as a approaches infinity of ((((1 - x^a)/(1 - x))^0.5)/(x^(a/2-1) THAT EQUALS x/(x-1)^2 AND THE ONLY NUMBERS THAT GIVE RATIONAL ANSWERS ARE SQUARES PLUS ONE BUT 17 is 16+1 AND 170 is 169+1 IT’S SO WONDERFUL SO! The first time I noticed a pattern from this was... 11! When you take the square root of any number with only ones as its digits, it results in one of two things: In the case that there are an even number of digits, it is equal to the same number of threes, with the decimal in the middle. Then it has a bunch of numbers after that, but those aren't important. In the other case, with odd digits, it just gives you a bunch of numbers, stating with a one and a zero. (It does this because when you add a digit, you are essentially [there is a bit more to it) multiplying by the square root of ten.) AND SO! I wanted to make and equation to model this, because why not! I was kind of stuck on how to represent the number of digits as the input, but then... I remembered sigma notation (don't kill me, that's what it's called) To convert number of digits to the actual number, I used __a__ \ > (10^n) / ______ n = 0 I didnt have the symbol, so it's weird. That essentially just equals 10^0 + 10^1 + 10^2 + 10^3 + ... + 10^a, which converts it. Now, this is a geometric series, which means I can represent the sum without sigma notation pretty easily. So that there is equal to (1 - 10^a)/(1 - 10) or (1 - 10^a)/(-9), which I'll defined as f(a) Now that I had that, I wanted to pursue the threes thing. So if I take the square root of f(a), it should give me that pattern that I saw up there. And it did, but I wasn't really satisfied yet, as I wanted the first 3 to always be in the ones place, as... i dunno. huh. So i divide it by 10 to the a/2 - 1 power, giving the effect I wanted. AND I REALIZED IT CAN BE ABRA KEDABRA-ed BY A LIMIT! So if you put int in a limit as a approaches infinity, you get 10/3, or 3.33333333333333333333333 repeating forever. But I wasn't satisfied. I thought, well, a is no longer changing, so why not change the base of the exponent (currently 10, ya know) So I got the original equation, which was kinda complex, and it was: The limit as a approaches infinity, of ((((1 - 10^a)/(1 - 10))^0.5)/(10^(a/2-1). Substituting x in for the 10s, The limit as a approaches infinity of ((((1 - x^a)/(1 - x))^0.5)/(x^(a/2-1) Next, I notice that the value of this at 5 was 2.5 exactly, or 5/2 COOL! But I WAS NOT SATISFIED I wanted to see why specifically 5 and 10 gave coherent answers, while all the other ones gave... incoherent answers. I found that 20, 100, and others didn't work, but then I stumbled upon 17, which did, giving 17/4 So the rational values I get have been: 5/2, 10/3, and 17/4 The numerator increases by a shifting value each time (more on that) but the denominator increased by a CONSTANT VALUE! Success! Still, I wanted to find a pattern for the numerator, and soon found that the amount it increased by increased by 2 each time, indicating a quadratic relation! But this didnt mean that ut was quadratic, only that the numbers that were rational happened to be a shift on perfect squares. This, in turn, indicated that there was a square root! 5 - 1 = 4! square root to 2, the denominator of 5! 10 - 1 = 9! square root to 3, the denominator of 10! And this repeated itself... I HAD FOUND MY PATTERN! That weird limit equation was equal to x / (x-1)^0.5 SUCCESS! And to top it all off, both 17 and 170 worked in the equation. So there. YES!

I dunno... hmmmm... my specialty is fantasy. Decidedly refuses to look at the Illuminated. HEY HERE! It's an AI controlled little spider bot that contains a piercing bolt - essentially, its light enough not to be noticed by the larger things, but it packs enough of a punch that it can be highly effective when it reaches an unprotected point. My author (and some of my creations) use it to slay wyverns.

A note appears to Rebus. Heyheyhey! Make sure conceal it from the sixth sense as well! (Narration sense, ya know)

THIS TIME IT WAS SPIRITOFWRATH NOT ME

Berin had been sleeping, "HEY HEY HEY! YOU SHOULD GO ATTACK THAT OTHER PERSON THE ONE WHO SHOT AT YOU I THINK YOU COULD WIN!" This came from the small girl that seemed to be Berin's sister. She didn't understand the reality of the situation.

...there is no explanation. I just have a *twitch* habit of *twitch* stttaabbiiinnggg people. (Unless u are talking about the math)

*STAB!* MOOOOOOOOOOOOOOOOOONEY!

Pull a random person on a gun. Refuse to please the NUMBERs

I finish my not so normal sandwich, then release a very special nugget. I smiteth all the heretics of the heathen world, bringing them to BOW TO THE NUGGET WRATH!

Investiture has to come from somewhere Theory: Wit = all of the Shards

Are you volunteering? yes. This ^^^^^^^ I agree. LAVA IS WET

Granted, except for the no Banes part. See previous post for Bane. I wish for WIND AND TRUTH

person who capitalizes MORE, in

*gives music* Its the very end of the book, when all the action ends and you cannot put it down.

I would grant that, but no boons can affect you. Guess I’ll make up for it in Banes. You are missing an alternating finger based on the day of the Rosharan week. I wish for WIND AND TRUTH

Breaking Benjamin Meshuggah AND LAST BUT NOT LEAST Gojira The Cure U2

2223 unfix the numbers

Granted. For unrelated reasons, you die. I wish to have complete and utter control over the dead.

GUYS GUYS IF YA GET The limit as a approaches infinity of ((((1 - x^a)/(1 - x))^0.5)/(x^(a/2-1) THAT EQUALS x/(x-1)^2 AND THE ONLY NUMBERS THAT GIVE RATIONAL ANSWERS ARE SQUARES PLUS ONE BUT 17 is 16+1 AND 170 is 169+1 IT’S SO WONDERFUL