pippinsboss.bsky.social
Mathematician, tutor of A level maths, bridge player, father, grandfather and dog lover.
Maker of mathematical videos, mostly at the level of A levels
youtube.com/channel/UCnYszOhEIIdIMYyNx2yjwfg
334 posts
619 followers
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David Richeson has an interesting list at divisbyzero.com/2009/09/01/l...
I knew about some of, but quite a few were news to me
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Probably not, as I now read that the most recent Sylvester, Sylvester IV, is regarded as an antipope, so there probably won't be any more Sylvesters.
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log₁₀ 2 > 3/10
iff 10 log₁₀ 2 > 3
iff log₁₀ 1024 > 3
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That's what I get. I'll convert it to pdf and re-upload, but not just yet as I am about to go out.
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I can access some, but not others. In particular, not my own. When I spoke to someone on the MA desk, she was able to access it perfectly well, so I assumed some problem with my phone, but I still can't.
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I think that's better than my method, which was 3700÷4.
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I'll be there too, and I'm looking forward to meeting you all in person again.
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The animation looks 3-dimensional, and makes it clear that the result is still true if the quadrilateral isn't flat. The proof is, if anything, easier to see in 3D than in 2D
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Yes! It took me a while, but I cam now prove the obvious conjecture.
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Well one way is rigidly, but then they are not really seesaws any more, just one long plank. The other is with a hinge, but that only works if they can slide on their pivots, or only move though tiny angles.
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The other is about see-saws "joined together at their ends". Nothing about how they are joined. S assumed rigidly joined, and found the problem insoluble. To make the problem work, we need to assume they're hinged, and in this scenario, that turns out to be equivalent to not being joined at all!
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15×22 = 15 × 2 × 11 = 30 × 11 = 330
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the planets move in elliptical orbits. Actually he only showed that Mars does. Lesser mortals had to do the rest. Of course part of the reason he only did Mars was because he didn't have Newton-Raphson. They were both born a generation after Kepler died.
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Actually he only showed that Mars does. Lesser mortals had to do the rest. There is an anachronism though. Kepler could not have used Newton-Raphson. Newton and Raphson were born about a generation after Kepler died .
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Yes, it's a great activity. But, sorry to be picky, it ought to have the degree symbol throughout. Otherwise it means radians, and initially I thought you were defining p to be the cosine of 10 radians. (Maybe I'm over sensitive. Part of my background is signal processing, so I think in radians).
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... air resistance and other friction... because the work done going uphill is recovered when you go down.
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I have a book of tables that includes a table of 574 integrals. I use it to illustrate that integrating is hard - there's no science, just a huge list tricks one of which just might work for the case in hand
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whereas the actual dial would have to go negative via infinity. It doesn't - it just gives up.
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So I guess a long term average of half a KN is at least plausible. Actually a dial showing the reciprocal would be nicer, because as you go from driving through coasting to regenerative breaking, it would quietly go through 0 to negative...,
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I get 545N, or about half a kiloNewton. Is that reasonable? Of course most of that is against air resistance, and I have no intuition about the magnitude of that. However the car's mass is roughly 2000Kg, so going up a 10% slope, ignoring air resistance, would take about 2 KN....
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Found it!
Babylonian Quadratics, Robert D Macmillan, The Mathematics Teacher, Vol. 77, No. 1 (January 1984).
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I might be able to, but not now whilst I'm in bed!
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It goes back to the Babylonians, around 1800 BC.
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I'd say "5 times (pause) x+2 (pause) divided by 3."
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The nearby thing that is true, and may be what your S was thinking of, is
P(B) = P(A)P(B|A) + P(A')P(B|A')
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You missed out 0 and 1. They have easy disability rules too!
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Doing it in my head, I used gradients because, I decided, it would miminimise my use of working memory. Doing it in writing opens up other possibilities, but I think I'd probably do the same.
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I can confirm that mathematica doesn't find the (-1)^even solutions, at least not with Solve[]. It does admit that it might not find all the solutions, and recommends Reduce[], which I didn't know about and haven't ever used. I shan't be using it in the future either as it got stuck in a loop.
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"And", I'd say.
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Thank you. I didn't realise it either until I googled my age. Googling numbers can be quite interesting. Near where I grew up there is a Trotter's Bottom, but we are going off topic...
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Yes indeed, and the connection with the normed space is to a convex function, not a convex curve.
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Sorry about that. I noticed my phone beeping at me whilst sitting in a talk this evening. It must have componsing this message to you, but I don't speak phonish, so I don't known what it was saying.
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