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# 336

Simpson's Rule - November 12, 2010
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Let h(x)=how much Homer likes x
Let b=beer, d=doughnut

int(h(x),x=d..b)=(b-d)/6*[h(a)+h(b)+4*h(d/2+b/2)]
h(x+y)=h(x)+h(y) // since Homer can only do one thing at a time
int(h(x),x=d..b)=(b-d)/6*[h(d)+h(b)+4*h(d/2)+4*h(b/2)]
h(x/2)=h(x)/2 // half of something gives half the satisfaction
int(h(x),x=d..b)=(b-d)/6*[3*h(d)+5*h(b)]
int(h(x),x=d..b)=1/2*(b-d)*[h(d)+h(b)]

We assume h(x)=const*x | how much Homer likes something is a constant of his general liking of things * the individual value of the thing
int(h(x),x=d..b)=const/2*(b^2-d^2)
1/2*(b-d)*[h(d)+h(b)]=1/2*(b-d)*(conts*d+const*b)=const/2*(b^2-d^2)

Also, since Homer is neither a happy nor a sad fellow, we can assume const=1;
Therefore h(x)=x

lol nice

Nice.

Another interpretation is that H(donut)=0 and H(beer)=0, since Homer consumes them instantly. Similarly, H((donut+beer)/2)=0. Thus, the RHS of the approximation is 0.

If the error formula were to be included, what should the unknown value "eta" in the interval (doughnut,beer) look like? A hybrid of the doughnut and the beer?

or maybe the beer and doughnut mixed in his stomach.

In that case, what exactly is Simpson's paradox?

If it doesn't fit, you must acquit?

good question

Here's a contribution to the "some fun weekend reading" part:Mathematical modelling of an Outbreak of Zombie Infection
(since my html link tag kinda doesn't work, here it is: mysite.science.uottawa.ca/rsmith43/Zombies.pdf )

And people say you'll never use math in the real world... the fools.

Does anyone know how to get from the last step of the proof of lemma 1 in Knuth's "paper" to the conclusion of the lemma?

Massive waving of hands.

see: Renteln and Dundes

... for some reason, I think the version with the Simpson heads, doughnuts, and beer mugs is easier to read. Probably this is because they're silly and colorful.

--
Furry cows moo and decompress.

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