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[120] This basic phenomenon being what more abstraction-capable post-Hegelian adults call ‘Historical Consciousness.’

[121] Eschaton’s pre- and post-procedures are convolved enough so that an actual game gets gotten up every like month or so at most, almost always on Sunday, but even then not all twelve of a year’s kids can get the hours off to play, which is why the latitude and surplus in game-personnel.

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[123] Pemulis here, dictating to Inc, who can just sit there making a steeple out of his fingers and pressing it to his lip and not take notes and wait and like inscribe [sic] it anytime in the next week and get it verbatim, the smug turd. Using the Mean-Value formula for dividing available megatonnage among Combatants whose GNP/Military // Military/Nuke ratios vary from Eschaton to Eschaton keeps you from needing to crunch out a new ratio for each Combatant each time, plus lets you multi-regress the results so Combatants get rewarded for past thermonuclear largesse [occasional verbal flourishes Hal’s — HJI]. The formula’s also provable by the Extreme Value Theorem, which the EV Theorem itself has a proof that’s just about the biggest Unit-twisting bitch in the whole of applied differentiation, but I see Hal grimacing, so we’ll keep it compact, even though this whole thing is real interesting if you’re interested and whatnot.

Say you’ve got a Combatant and a record of his past GNP/Military // Military/Nuke ratios. We want to give the Combatant the like exact average of all the past megatonnages he’s gotten in the past. The exact average is called the ‘Mean Value,’ which ought to give us a bit of a giggle, given the hostility of the context here.

So then but let A stand for the Mean Value of a Combatant’s constantly fluctuating ratio and so constantly fluctuating initial megatonnage. We want to find A and give the Combatant exactly A megatons. How to do it’s pretty elegant, and all you need for it is two pieces of data: the most his ratio’s ever been and the least it’s ever been. These two datums [sic] are called the Extreme Values of the cn-n function for which A’s the Mean Value, by the way.

So then but so let / be a continuous non-negative function (meaning the ratio) on the interval [a, b] (meaning the difference between the least the ratio’s ever been and the most it’s ever been and whatnot). Are these little explanations aggravating [sic]? Inc’s looking at me like butter would freeze. It’s hard to know what to assume v. what to explain. I’m trying to be as clear as I can be [sic]. And now he’s looking at me like I’m digressing. Why don’t you just pass that certain item back on over here, Inculator. But so we’ve got / and we’ve got [a, bj. And let r and R be the smallest and biggest values of the function /(x) on the interval [a, b]. So now check out the rectangles of height r and height R over the interval [a, b] in the diagram marked let’s go ahead and mark it say PEEMSTER:

PEEMSTER

The Mean Value we’re after, A, can now be expressed integrally as the Area of some intermediate-type rectangle whose height is taller [sic] than r but shorter [sic] than R. From here on it’s just tit. We need a constant. You always need a constant. Inc’s nodding his head sarcastically like I think I’m saying something sage. Let d be any constant, for computational reasons the closer to 1 the better, so like let d be the size of Hal’s Unit.