Minimisation of integral including absolute values











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I need to solve the following



$$min_{c,d} int_0^1 |ct+d-t^2| dt.$$



Since the integrand is of degree two, I considered to split the integral in three. My problem however, is that the bounds would then depend on $c$ as well as $d$. Is there any extension of the Leibnitz integral rule (https://en.wikipedia.org/wiki/Leibniz_integral_rule) that lets you deal with two parameters in the bound?










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  • Your polynomial factorizes as $(t-a_0)(t-a_1)$ and you know the sign on intervals $]-infty,a_0[,[a_0,a_1],]a_1,+infty[$ so you can decompose the integral in parts without absolute value. I did something similar here for $exp$ instead of $t^2$. math.stackexchange.com/questions/2344512/…. Notice that because of minimization request, $a_0$ and $a_1$ are inside $[0,1]$
    – zwim
    Nov 20 at 21:53












  • Could you elaborate om why $a_0, a_1 in [0,1]$?
    – timudk
    Nov 21 at 17:53










  • Look at the graph in the post I linked. You have to minimize the area between the curve and a line. Since $t^2$ is convex you can achieve that only when the line cuts the curve inside $[0,1]$ else this would just be a big belly contributing a lot to the result. With a cross, the area is spread on both sides off $y=t^2$ and the sum is minimized.
    – zwim
    Nov 21 at 19:47












  • I am just trying to find a formal (mathematical) argument instead of a pictorial argument.
    – timudk
    Nov 21 at 19:53















up vote
0
down vote

favorite
1












I need to solve the following



$$min_{c,d} int_0^1 |ct+d-t^2| dt.$$



Since the integrand is of degree two, I considered to split the integral in three. My problem however, is that the bounds would then depend on $c$ as well as $d$. Is there any extension of the Leibnitz integral rule (https://en.wikipedia.org/wiki/Leibniz_integral_rule) that lets you deal with two parameters in the bound?










share|cite|improve this question
























  • Your polynomial factorizes as $(t-a_0)(t-a_1)$ and you know the sign on intervals $]-infty,a_0[,[a_0,a_1],]a_1,+infty[$ so you can decompose the integral in parts without absolute value. I did something similar here for $exp$ instead of $t^2$. math.stackexchange.com/questions/2344512/…. Notice that because of minimization request, $a_0$ and $a_1$ are inside $[0,1]$
    – zwim
    Nov 20 at 21:53












  • Could you elaborate om why $a_0, a_1 in [0,1]$?
    – timudk
    Nov 21 at 17:53










  • Look at the graph in the post I linked. You have to minimize the area between the curve and a line. Since $t^2$ is convex you can achieve that only when the line cuts the curve inside $[0,1]$ else this would just be a big belly contributing a lot to the result. With a cross, the area is spread on both sides off $y=t^2$ and the sum is minimized.
    – zwim
    Nov 21 at 19:47












  • I am just trying to find a formal (mathematical) argument instead of a pictorial argument.
    – timudk
    Nov 21 at 19:53













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0
down vote

favorite
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0
down vote

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I need to solve the following



$$min_{c,d} int_0^1 |ct+d-t^2| dt.$$



Since the integrand is of degree two, I considered to split the integral in three. My problem however, is that the bounds would then depend on $c$ as well as $d$. Is there any extension of the Leibnitz integral rule (https://en.wikipedia.org/wiki/Leibniz_integral_rule) that lets you deal with two parameters in the bound?










share|cite|improve this question















I need to solve the following



$$min_{c,d} int_0^1 |ct+d-t^2| dt.$$



Since the integrand is of degree two, I considered to split the integral in three. My problem however, is that the bounds would then depend on $c$ as well as $d$. Is there any extension of the Leibnitz integral rule (https://en.wikipedia.org/wiki/Leibniz_integral_rule) that lets you deal with two parameters in the bound?







functional-analysis optimization maxima-minima






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edited Nov 20 at 22:28









Jean Marie

28.6k41849




28.6k41849










asked Nov 20 at 21:05









timudk

33




33












  • Your polynomial factorizes as $(t-a_0)(t-a_1)$ and you know the sign on intervals $]-infty,a_0[,[a_0,a_1],]a_1,+infty[$ so you can decompose the integral in parts without absolute value. I did something similar here for $exp$ instead of $t^2$. math.stackexchange.com/questions/2344512/…. Notice that because of minimization request, $a_0$ and $a_1$ are inside $[0,1]$
    – zwim
    Nov 20 at 21:53












  • Could you elaborate om why $a_0, a_1 in [0,1]$?
    – timudk
    Nov 21 at 17:53










  • Look at the graph in the post I linked. You have to minimize the area between the curve and a line. Since $t^2$ is convex you can achieve that only when the line cuts the curve inside $[0,1]$ else this would just be a big belly contributing a lot to the result. With a cross, the area is spread on both sides off $y=t^2$ and the sum is minimized.
    – zwim
    Nov 21 at 19:47












  • I am just trying to find a formal (mathematical) argument instead of a pictorial argument.
    – timudk
    Nov 21 at 19:53


















  • Your polynomial factorizes as $(t-a_0)(t-a_1)$ and you know the sign on intervals $]-infty,a_0[,[a_0,a_1],]a_1,+infty[$ so you can decompose the integral in parts without absolute value. I did something similar here for $exp$ instead of $t^2$. math.stackexchange.com/questions/2344512/…. Notice that because of minimization request, $a_0$ and $a_1$ are inside $[0,1]$
    – zwim
    Nov 20 at 21:53












  • Could you elaborate om why $a_0, a_1 in [0,1]$?
    – timudk
    Nov 21 at 17:53










  • Look at the graph in the post I linked. You have to minimize the area between the curve and a line. Since $t^2$ is convex you can achieve that only when the line cuts the curve inside $[0,1]$ else this would just be a big belly contributing a lot to the result. With a cross, the area is spread on both sides off $y=t^2$ and the sum is minimized.
    – zwim
    Nov 21 at 19:47












  • I am just trying to find a formal (mathematical) argument instead of a pictorial argument.
    – timudk
    Nov 21 at 19:53
















Your polynomial factorizes as $(t-a_0)(t-a_1)$ and you know the sign on intervals $]-infty,a_0[,[a_0,a_1],]a_1,+infty[$ so you can decompose the integral in parts without absolute value. I did something similar here for $exp$ instead of $t^2$. math.stackexchange.com/questions/2344512/…. Notice that because of minimization request, $a_0$ and $a_1$ are inside $[0,1]$
– zwim
Nov 20 at 21:53






Your polynomial factorizes as $(t-a_0)(t-a_1)$ and you know the sign on intervals $]-infty,a_0[,[a_0,a_1],]a_1,+infty[$ so you can decompose the integral in parts without absolute value. I did something similar here for $exp$ instead of $t^2$. math.stackexchange.com/questions/2344512/…. Notice that because of minimization request, $a_0$ and $a_1$ are inside $[0,1]$
– zwim
Nov 20 at 21:53














Could you elaborate om why $a_0, a_1 in [0,1]$?
– timudk
Nov 21 at 17:53




Could you elaborate om why $a_0, a_1 in [0,1]$?
– timudk
Nov 21 at 17:53












Look at the graph in the post I linked. You have to minimize the area between the curve and a line. Since $t^2$ is convex you can achieve that only when the line cuts the curve inside $[0,1]$ else this would just be a big belly contributing a lot to the result. With a cross, the area is spread on both sides off $y=t^2$ and the sum is minimized.
– zwim
Nov 21 at 19:47






Look at the graph in the post I linked. You have to minimize the area between the curve and a line. Since $t^2$ is convex you can achieve that only when the line cuts the curve inside $[0,1]$ else this would just be a big belly contributing a lot to the result. With a cross, the area is spread on both sides off $y=t^2$ and the sum is minimized.
– zwim
Nov 21 at 19:47














I am just trying to find a formal (mathematical) argument instead of a pictorial argument.
– timudk
Nov 21 at 19:53




I am just trying to find a formal (mathematical) argument instead of a pictorial argument.
– timudk
Nov 21 at 19:53















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