bio | website | gowers.wordpress.com |
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visits | member for | 5 years, 1 month |
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Mathematics professor at Cambridge
Nov 28 |
answered | Intuitive explanation to Probability question |
Nov 26 |
answered | A random walk matrix has eigenvalue 1 with multiplicty 1 - why? |
Nov 23 |
answered | Finding the new zeros of a “perturbed” polynomial |
Nov 23 |
awarded | Nice Answer |
Nov 22 |
answered | Ramsey Theory, monochromatic subgraphs |
Nov 21 |
comment |
Can infinity shorten proofs a lot?
Not the latter as I go by my middle name. Happy to make the change but have not managed to find where I can do it. (Please excuse my utter incompetence.) |
Nov 20 |
awarded | Commentator |
Nov 20 |
comment |
Continued fractions using all natural integers
I think you're right -- and indeed I was worried that something like that would happen. So it seems that the quotients have to grow rather fast to make the number transcendental. |
Nov 20 |
answered | Continued fractions using all natural integers |
Nov 20 |
answered | Are there any good nonconstructive “existential metatheorems”? |
Nov 19 |
awarded | Nice Answer |
Nov 18 |
awarded | Nice Answer |
Nov 18 |
answered | Beamer hints and tips |
Nov 18 |
answered | What's so great about blackboards? |
Nov 18 |
comment |
Can infinity shorten proofs a lot?
I'm not sure I count induction either. It seems to me that one is not relying on infinity in a serious way to sum the first million cubes: one is saying, "If it's valid for 1 then it's valid for 2, and if it's valid for 2 then it's valid for 3, and ... and if it's valid for 999999 then it's valid for 1000000." That is a ridiculously long proof, but induction allows one to shorten it by giving a rule for when it's OK to put in the "..." (A more formal way of making the point is that the axiom of infinity isn't part of first-order Peano arithmetic.) |
Nov 17 |
awarded | Critic |
Nov 17 |
awarded | Supporter |
Nov 17 |
comment |
Can infinity shorten proofs a lot?
You can get the inequality quite easily I think. The log of n! is the sum of log m from 1 to n, which is at least the integral from 1 to n of log x, which is nlogn-n. Done. |
Nov 17 |
awarded | Nice Question |
Nov 17 |
awarded | Nice Question |