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I love how these raw mathematicians consider something proved when they can understand, meanwhile the computer can prove it easily just by counting a finite num
by screenoridesaga 3y ago
I love how these raw mathematicians consider something proved when they can understand, meanwhile the computer can prove it easily just by counting a finite number of bits. What exactly would be considered proof in this case? Any explanation only mathematicians can understand?
- red_trumpet 3y agoWell, how does your proposed computer proof look like? A computer can easily calculate both sides of the equation up to say, float precision. But that's not a proof; it only tells you that both numbers are near each other!
- earlymodernlvr 3y agoThe computer can only check a finite number of cases computationally, a proof can prove the results for all N.
- feoren 3y ago> the computer can prove it easily just by counting a finite number of bits Did you miss the infinite sum there? How would you prove an infinite sum equals a transcendental number by counting finite bits? You'd have to count infinite bits.
- hyperhello 3y agoWouldn’t you be able to see the difference converging on zero at least? Unless it oscillates all over but seemed to average. I don’t know if the squeeze theorem applies.
- tsimionescu 3y agoThere are many series that converge to an extraordinarily small but non 0 number.
- damiankennedy 3y agoThe first thing you need to go with a series like this is prove that it converges. Then you can take a series that is already known to converge to pi. The choice of series will mean the difference between the proof being very hard and practically impossible. Then change that series to give 32/pi^3 instead of pi. Then deduct a mapping between groups of n in one series and groups of n in the other series.
- hatthew 3y agoIf you ask a computer to compute (1/x) + 10^(-googolplex), it'll look an awful lot like the limit is 0.