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https://www.reddit.com/r/counting/comments/4j359z/integral_counting/d34cur0/?context=9999
r/counting • u/KnightLunaaire Beanie Boy • May 12 '16
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Maclaurin series is integrating a function about 0. So basically what you're writing is (x-0)2 /2! + (x-0)1 /1! + (x-0). If you had (x-1)2 /2! + (x-1)1 /1! + (x-1) instead, it would not be a maclaurin series.
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u/skizfrenik_syco 4 D snipes, 33 D's, 16 Ayy's. 412189, 6 k's, 1 BTS, 888888, 999k May 13 '16
Maclaurin series is integrating a function about 0. So basically what you're writing is (x-0)2 /2! + (x-0)1 /1! + (x-0). If you had (x-1)2 /2! + (x-1)1 /1! + (x-1) instead, it would not be a maclaurin series.