Placing $n$ linear functions so that it is best fit to another function in integral norm sense?

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Say we want to build a function which is piecewise linear $$f(x) = sum_forall k (H(x-x_k)-H(x-x_k+1))l_k(x)\l_k(x) = c_k1x+c_k2$$



And also so that it fits best possibly some function $xto g(x)$: $$l_k,x_k=min_l_k,x_kleftf(x)-g(x)$$



Please note that the line end point coordinates $x_k$ we can decide for ourselves.




I've made some numerical approaches which seem promising on this, but how can one approach it algebraically/analytically?







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  • 1




    If $H(x_k)$ represents the unit step function translated to $x_k$ the notation would be better $f(x) = sum_forall kleft(H(x-x_k)-H(x-x_k+1)right)l_k(x)$
    – Cesareo
    Jul 24 at 20:26











  • Yes it is intended to be "active between $x_k$ and $x_k+1$". Feel free to edit if you want.
    – mathreadler
    Jul 24 at 21:20











  • This kind of problem can be successfully handled with evolutionary programming -Evolution Strategies. See papers with those authors: Thomas Back, Gunter Rudolph, Hans-Paul Schwefel,
    – Cesareo
    Jul 25 at 12:59










  • @Cesareo sounds interesting. Are they related to genetic algorithms?
    – mathreadler
    Jul 25 at 18:27










  • Evolution Strategies have an Evolutionary paradigm but they use for each parameter real numbers instead a fractionary representation (chromosome). There are very efficient algorithms (CMA-ES) which converge in difficult non-convex situations.
    – Cesareo
    Jul 25 at 18:58















up vote
-1
down vote

favorite












Say we want to build a function which is piecewise linear $$f(x) = sum_forall k (H(x-x_k)-H(x-x_k+1))l_k(x)\l_k(x) = c_k1x+c_k2$$



And also so that it fits best possibly some function $xto g(x)$: $$l_k,x_k=min_l_k,x_kleftf(x)-g(x)$$



Please note that the line end point coordinates $x_k$ we can decide for ourselves.




I've made some numerical approaches which seem promising on this, but how can one approach it algebraically/analytically?







share|cite|improve this question

















  • 1




    If $H(x_k)$ represents the unit step function translated to $x_k$ the notation would be better $f(x) = sum_forall kleft(H(x-x_k)-H(x-x_k+1)right)l_k(x)$
    – Cesareo
    Jul 24 at 20:26











  • Yes it is intended to be "active between $x_k$ and $x_k+1$". Feel free to edit if you want.
    – mathreadler
    Jul 24 at 21:20











  • This kind of problem can be successfully handled with evolutionary programming -Evolution Strategies. See papers with those authors: Thomas Back, Gunter Rudolph, Hans-Paul Schwefel,
    – Cesareo
    Jul 25 at 12:59










  • @Cesareo sounds interesting. Are they related to genetic algorithms?
    – mathreadler
    Jul 25 at 18:27










  • Evolution Strategies have an Evolutionary paradigm but they use for each parameter real numbers instead a fractionary representation (chromosome). There are very efficient algorithms (CMA-ES) which converge in difficult non-convex situations.
    – Cesareo
    Jul 25 at 18:58













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











Say we want to build a function which is piecewise linear $$f(x) = sum_forall k (H(x-x_k)-H(x-x_k+1))l_k(x)\l_k(x) = c_k1x+c_k2$$



And also so that it fits best possibly some function $xto g(x)$: $$l_k,x_k=min_l_k,x_kleftf(x)-g(x)$$



Please note that the line end point coordinates $x_k$ we can decide for ourselves.




I've made some numerical approaches which seem promising on this, but how can one approach it algebraically/analytically?







share|cite|improve this question













Say we want to build a function which is piecewise linear $$f(x) = sum_forall k (H(x-x_k)-H(x-x_k+1))l_k(x)\l_k(x) = c_k1x+c_k2$$



And also so that it fits best possibly some function $xto g(x)$: $$l_k,x_k=min_l_k,x_kleftf(x)-g(x)$$



Please note that the line end point coordinates $x_k$ we can decide for ourselves.




I've made some numerical approaches which seem promising on this, but how can one approach it algebraically/analytically?









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 24 at 21:22
























asked Jul 24 at 18:44









mathreadler

13.6k71857




13.6k71857







  • 1




    If $H(x_k)$ represents the unit step function translated to $x_k$ the notation would be better $f(x) = sum_forall kleft(H(x-x_k)-H(x-x_k+1)right)l_k(x)$
    – Cesareo
    Jul 24 at 20:26











  • Yes it is intended to be "active between $x_k$ and $x_k+1$". Feel free to edit if you want.
    – mathreadler
    Jul 24 at 21:20











  • This kind of problem can be successfully handled with evolutionary programming -Evolution Strategies. See papers with those authors: Thomas Back, Gunter Rudolph, Hans-Paul Schwefel,
    – Cesareo
    Jul 25 at 12:59










  • @Cesareo sounds interesting. Are they related to genetic algorithms?
    – mathreadler
    Jul 25 at 18:27










  • Evolution Strategies have an Evolutionary paradigm but they use for each parameter real numbers instead a fractionary representation (chromosome). There are very efficient algorithms (CMA-ES) which converge in difficult non-convex situations.
    – Cesareo
    Jul 25 at 18:58













  • 1




    If $H(x_k)$ represents the unit step function translated to $x_k$ the notation would be better $f(x) = sum_forall kleft(H(x-x_k)-H(x-x_k+1)right)l_k(x)$
    – Cesareo
    Jul 24 at 20:26











  • Yes it is intended to be "active between $x_k$ and $x_k+1$". Feel free to edit if you want.
    – mathreadler
    Jul 24 at 21:20











  • This kind of problem can be successfully handled with evolutionary programming -Evolution Strategies. See papers with those authors: Thomas Back, Gunter Rudolph, Hans-Paul Schwefel,
    – Cesareo
    Jul 25 at 12:59










  • @Cesareo sounds interesting. Are they related to genetic algorithms?
    – mathreadler
    Jul 25 at 18:27










  • Evolution Strategies have an Evolutionary paradigm but they use for each parameter real numbers instead a fractionary representation (chromosome). There are very efficient algorithms (CMA-ES) which converge in difficult non-convex situations.
    – Cesareo
    Jul 25 at 18:58








1




1




If $H(x_k)$ represents the unit step function translated to $x_k$ the notation would be better $f(x) = sum_forall kleft(H(x-x_k)-H(x-x_k+1)right)l_k(x)$
– Cesareo
Jul 24 at 20:26





If $H(x_k)$ represents the unit step function translated to $x_k$ the notation would be better $f(x) = sum_forall kleft(H(x-x_k)-H(x-x_k+1)right)l_k(x)$
– Cesareo
Jul 24 at 20:26













Yes it is intended to be "active between $x_k$ and $x_k+1$". Feel free to edit if you want.
– mathreadler
Jul 24 at 21:20





Yes it is intended to be "active between $x_k$ and $x_k+1$". Feel free to edit if you want.
– mathreadler
Jul 24 at 21:20













This kind of problem can be successfully handled with evolutionary programming -Evolution Strategies. See papers with those authors: Thomas Back, Gunter Rudolph, Hans-Paul Schwefel,
– Cesareo
Jul 25 at 12:59




This kind of problem can be successfully handled with evolutionary programming -Evolution Strategies. See papers with those authors: Thomas Back, Gunter Rudolph, Hans-Paul Schwefel,
– Cesareo
Jul 25 at 12:59












@Cesareo sounds interesting. Are they related to genetic algorithms?
– mathreadler
Jul 25 at 18:27




@Cesareo sounds interesting. Are they related to genetic algorithms?
– mathreadler
Jul 25 at 18:27












Evolution Strategies have an Evolutionary paradigm but they use for each parameter real numbers instead a fractionary representation (chromosome). There are very efficient algorithms (CMA-ES) which converge in difficult non-convex situations.
– Cesareo
Jul 25 at 18:58





Evolution Strategies have an Evolutionary paradigm but they use for each parameter real numbers instead a fractionary representation (chromosome). There are very efficient algorithms (CMA-ES) which converge in difficult non-convex situations.
– Cesareo
Jul 25 at 18:58
















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