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Charles Matthews's user avatar
Charles Matthews's user avatar
Charles Matthews
  • Member for 14 years, 7 months
  • Last seen more than 9 years ago
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To what extent is it true that "number theory = mathematics"?
Quite hard to do justice to all that: see "Why are topological ideas so important in arithmetic?" and comments.
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Linear algebra inequality
Ah, Robinned again.
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Why are topological ideas so important in arithmetic?
See en.wikipedia.org/wiki/Diophantine_geometry for more accurate and neutral perspective (I hope). Gauss is alleged to have said about FLT that he thought it should be developed in a broader context. Kummer initiated a theory, and, as you say, this special equation had a broad influence. General methods for Diophantine equations really started with Thue, and I wouldn't call Diophantine approximation "topological"; post-Vojta there is a new view? FLT's solution I don't understand, but deformation of Galois representations are bounding Selmer groups? There is topology on one side there.
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Evaluation of the following series... $S = 1/(2\times3) + 1/(5\times6) + 1/(7\times8) + 1/(10\times11) + ... $
You realise that Alvarez adds 1/5x6, 1/6x7, 1/7x8, 1/10x11, 1/11x12 etc.?
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Evaluation of the following series... $S = 1/(2\times3) + 1/(5\times6) + 1/(7\times8) + 1/(10\times11) + ... $
I have to say that I'm not going to type more mathematical notation here, given your reaction above. It is a bit ironic that you ask for an explanation in mathematical language. It is not clear to me whether this is a challenge problem to which you already know the answer, but I'm not highly motivated yet (whether it is or not). You have ignored the request I made to formulate the series in sigma notation: "..." can waste a lot of time, as the discussion shows.
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Evaluation of the following series... $S = 1/(2\times3) + 1/(5\times6) + 1/(7\times8) + 1/(10\times11) + ... $
OK, in this forum, please don't use "compsci" notation for multiplication, and please do use sigma notation. We can all write 1/6 = 1/2 - 1/3 here, so could you verify that my a(k) is what you want?
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Evaluation of the following series... $S = 1/(2\times3) + 1/(5\times6) + 1/(7\times8) + 1/(10\times11) + ... $
It's just irritating to conceal the partial fraction decomposition, though.
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Evaluation of the following series... $S = 1/(2\times3) + 1/(5\times6) + 1/(7\times8) + 1/(10\times11) + ... $
I’m now considering Ʃ a(k)/k where a(k) = 0 when k is a square, and otherwise a(k) = -1<sup>b(k)</sup> with b(k) = k + [square root(k)], [ ] being integral part. This is meant to be equivalent to the OP's intended series, up to some factor of plus or minus 2 or something.
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Evaluation of the following series... $S = 1/(2\times3) + 1/(5\times6) + 1/(7\times8) + 1/(10\times11) + ... $
Please reformulate with a -1 to the power of floor function of square root of n, applied to alternating reciprocals, and omitting the squares. Then I think we might understand. "Classical" does not mean calculus textbook, necessarily, though.