Difference between revisions of "Sven Bergmann"
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Do you know how to get ''smoothly'' from A to B? Well, you just need to minimize the functional expression | Do you know how to get ''smoothly'' from A to B? Well, you just need to minimize the functional expression | ||
| − | <math>\tilde {\cal S}[y(x)] \equiv | + | <math> \tilde {\cal S}[y(x)] \equiv |
\int_{x_i}^{x_f} \left(\kappa[y(x)]\right)^\nu \, ds(x) = | \int_{x_i}^{x_f} \left(\kappa[y(x)]\right)^\nu \, ds(x) = | ||
\int_{x_i}^{x_f} {|y''(x)|^\nu \over | \int_{x_i}^{x_f} {|y''(x)|^\nu \over | ||
Revision as of 13:41, 8 July 2009
I am the head of the Computational Biology Group (CBG) in the Department of Medical Genetics at the University of Lausanne.
- Address: Rue de Bugnon 27 - DGM 328 - CH-1005 Lausanne - Switzerland
- Phone at work: ++41-21-692-5452
- Cell phone: ++41-78-663-4980
- http://serverdgm.unil.ch/bergmann
- e-mail: Sven.Bergmann_AT_unil.ch
Do you know how to get smoothly from A to B? Well, you just need to minimize the functional expression <math> \tilde {\cal S}[y(x)] \equiv \int_{x_i}^{x_f} \left(\kappa[y(x)]\right)^\nu \, ds(x) = \int_{x_i}^{x_f} {|y(x)|^\nu \over
\left[1+y'(x)^2\right]^{3\nu-1 \over 2}} \,dx </math>
It turns out that it is possible to find analytically the most general solution <math>y(x)</math>, see this [paper] for details.
