September 15, 2008
This means i’m not a mathematician:
“When mathematicians and physicists are left alone in a room, one of the games they’ll play is called a Fermi problem, in which they try to figure out the approximate answer to an arbitrary problem,” said Rebecca Saxe, a cognitive neuroscientist at the Massachusetts Institute of Technology who is married to a physicist. “They’ll ask, how many piano tuners are there in Chicago, or what contribution to the ocean’s temperature do fish make, and they’ll try to come up with a plausible answer.”
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Posted by outinfour
February 16, 2008
Chapter 3 of ACEF: Rootfinding algorithmOne of the simplest rootfinding techniques is the Newton-Raphson Method. Ingredients:
- Differentiable Function [necessary]
- A function with a unique root [so we don't have to worry about a good starting value]
- A computer [only necessary if you don't want to waste your time]
The idea is to calculate the first derivative or Jacobian [matrix]. Start somewhere, hopefully close to the root. Use the first guess to come up with a second guess via the first derivative and the function value. Let be specific. Consider x^3-2 below:
From the graph, we can see the answer is about 1.25. Take a starting guess, say 1. At x=1, two things are true:
- f(x)<0
- f’(x)>0
(1) and (2) together tells you to increase x to find that special x to make f(x)=0. Do this again and again till you get x*.Matlab output
(t) f(x(t-1))1.3333 -1.00001.2639 0.37041.2599 0.01901.2599For the first guess of x=1, we get f(x)=-1. The second guess tell you to go higher by the amount -f(x)/f’(x). This takes us to 1.333 — too high, because f(x)>0. The method tells you to go down by -f(x)/f’(x). We get to f(x)>0… etc. Until we settle down to 1.2599.
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Posted by outinfour
January 15, 2008
Taylor series expansion of y=ln(x) around x=1.
Interesting is how, for small changes around 1, ln(x) and the linear expansion is quite close…

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Posted by outinfour