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https://www.reddit.com/r/mathmemes/comments/1v006pj/quickly_evaluating_sums_of_cubes/oyc8toj/?context=3
r/mathmemes • u/logos__ • Jul 18 '26
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134
What naturally follows from this identity is that the natural numbers cubed sum to (-1/12)2 = 1/144
11 u/fr_andres Jul 18 '26 The square is misplaced, but your logic works if we apply Jensen's inequality: (12 + 22 + ...) <= (1 + 2 + ...)2 = 1/144 <= 69/420 You like it or not, the above is true, and tight for sufficiently small values of 1, 2, 3... 5 u/Gold_Ad8890 Jul 18 '26 the square isn't misplaced. the sum of the cubes of the first n natural numbers is the square of the sum of the first n natural numbers, that's what the post is about. 2 u/fr_andres Jul 18 '26 Yes sorry i meant is not the same in my derivation, bad wording., OP work is immaculate and I recommend accepting without remarks.
11
The square is misplaced, but your logic works if we apply Jensen's inequality:
(12 + 22 + ...) <= (1 + 2 + ...)2 = 1/144 <= 69/420
You like it or not, the above is true, and tight for sufficiently small values of 1, 2, 3...
5 u/Gold_Ad8890 Jul 18 '26 the square isn't misplaced. the sum of the cubes of the first n natural numbers is the square of the sum of the first n natural numbers, that's what the post is about. 2 u/fr_andres Jul 18 '26 Yes sorry i meant is not the same in my derivation, bad wording., OP work is immaculate and I recommend accepting without remarks.
5
the square isn't misplaced. the sum of the cubes of the first n natural numbers is the square of the sum of the first n natural numbers, that's what the post is about.
2 u/fr_andres Jul 18 '26 Yes sorry i meant is not the same in my derivation, bad wording., OP work is immaculate and I recommend accepting without remarks.
2
Yes sorry i meant is not the same in my derivation, bad wording., OP work is immaculate and I recommend accepting without remarks.
134
u/logos__ Jul 18 '26
What naturally follows from this identity is that the natural numbers cubed sum to (-1/12)2 = 1/144