Solving an apparently “simple” system of linear equations











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I have to solve this apparently simple system of linear equations:



$$ a t^5+b t^4+c t^3+d t^2+e t=0 \
a (t/2)^5+b (t/2)^4+c (t/2)^3+d (t/2)^2+e (t/2)=0 \
a/6 t^6+b/5 t^5+c/4 t^4+d/3 t^3+e t^2/2=0 \
a/42 t^7+b/30 t^6+c/20 t^5+d/12 t^4+e/6 t^3=v \
a/336 t^8+b/210 t^7+c/120 t^6+d/60 t^5+e/24 t^4=y $$



in the unknowns $a,b,c,d,e$ where I assume $t=0.8$, $y=0.1$ and $v=0.45$.
Even if the coefficient matrix is only 5x5, the system is ill-conditioned and I cannot easily find a solution in MATLAB with the methods of matrix inversion $A^{-1} b$, linsolve and lsqr.



Do you have any suggestion to solve this problem?
Thanks,



Regards,



E.










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  • It doesn't look like there will be any nice solution at all. But you can multiply some of these equations by a nice constant to get nice integer coefficients, then work from there, though your answers are still going to be ugly.
    – YiFan
    Nov 16 at 3:34










  • I made a mistake typing the equations and I do not get anymore a result !! Sorry for that.
    – Claude Leibovici
    Nov 16 at 9:02










  • What do you mean?
    – EmThorns
    Nov 16 at 10:44










  • I shall be back tomorrow morning
    – Claude Leibovici
    Nov 16 at 17:40















up vote
0
down vote

favorite












I have to solve this apparently simple system of linear equations:



$$ a t^5+b t^4+c t^3+d t^2+e t=0 \
a (t/2)^5+b (t/2)^4+c (t/2)^3+d (t/2)^2+e (t/2)=0 \
a/6 t^6+b/5 t^5+c/4 t^4+d/3 t^3+e t^2/2=0 \
a/42 t^7+b/30 t^6+c/20 t^5+d/12 t^4+e/6 t^3=v \
a/336 t^8+b/210 t^7+c/120 t^6+d/60 t^5+e/24 t^4=y $$



in the unknowns $a,b,c,d,e$ where I assume $t=0.8$, $y=0.1$ and $v=0.45$.
Even if the coefficient matrix is only 5x5, the system is ill-conditioned and I cannot easily find a solution in MATLAB with the methods of matrix inversion $A^{-1} b$, linsolve and lsqr.



Do you have any suggestion to solve this problem?
Thanks,



Regards,



E.










share|cite|improve this question






















  • It doesn't look like there will be any nice solution at all. But you can multiply some of these equations by a nice constant to get nice integer coefficients, then work from there, though your answers are still going to be ugly.
    – YiFan
    Nov 16 at 3:34










  • I made a mistake typing the equations and I do not get anymore a result !! Sorry for that.
    – Claude Leibovici
    Nov 16 at 9:02










  • What do you mean?
    – EmThorns
    Nov 16 at 10:44










  • I shall be back tomorrow morning
    – Claude Leibovici
    Nov 16 at 17:40













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have to solve this apparently simple system of linear equations:



$$ a t^5+b t^4+c t^3+d t^2+e t=0 \
a (t/2)^5+b (t/2)^4+c (t/2)^3+d (t/2)^2+e (t/2)=0 \
a/6 t^6+b/5 t^5+c/4 t^4+d/3 t^3+e t^2/2=0 \
a/42 t^7+b/30 t^6+c/20 t^5+d/12 t^4+e/6 t^3=v \
a/336 t^8+b/210 t^7+c/120 t^6+d/60 t^5+e/24 t^4=y $$



in the unknowns $a,b,c,d,e$ where I assume $t=0.8$, $y=0.1$ and $v=0.45$.
Even if the coefficient matrix is only 5x5, the system is ill-conditioned and I cannot easily find a solution in MATLAB with the methods of matrix inversion $A^{-1} b$, linsolve and lsqr.



Do you have any suggestion to solve this problem?
Thanks,



Regards,



E.










share|cite|improve this question













I have to solve this apparently simple system of linear equations:



$$ a t^5+b t^4+c t^3+d t^2+e t=0 \
a (t/2)^5+b (t/2)^4+c (t/2)^3+d (t/2)^2+e (t/2)=0 \
a/6 t^6+b/5 t^5+c/4 t^4+d/3 t^3+e t^2/2=0 \
a/42 t^7+b/30 t^6+c/20 t^5+d/12 t^4+e/6 t^3=v \
a/336 t^8+b/210 t^7+c/120 t^6+d/60 t^5+e/24 t^4=y $$



in the unknowns $a,b,c,d,e$ where I assume $t=0.8$, $y=0.1$ and $v=0.45$.
Even if the coefficient matrix is only 5x5, the system is ill-conditioned and I cannot easily find a solution in MATLAB with the methods of matrix inversion $A^{-1} b$, linsolve and lsqr.



Do you have any suggestion to solve this problem?
Thanks,



Regards,



E.







systems-of-equations matlab condition-number






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share|cite|improve this question











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asked Nov 15 at 23:47









EmThorns

6




6












  • It doesn't look like there will be any nice solution at all. But you can multiply some of these equations by a nice constant to get nice integer coefficients, then work from there, though your answers are still going to be ugly.
    – YiFan
    Nov 16 at 3:34










  • I made a mistake typing the equations and I do not get anymore a result !! Sorry for that.
    – Claude Leibovici
    Nov 16 at 9:02










  • What do you mean?
    – EmThorns
    Nov 16 at 10:44










  • I shall be back tomorrow morning
    – Claude Leibovici
    Nov 16 at 17:40


















  • It doesn't look like there will be any nice solution at all. But you can multiply some of these equations by a nice constant to get nice integer coefficients, then work from there, though your answers are still going to be ugly.
    – YiFan
    Nov 16 at 3:34










  • I made a mistake typing the equations and I do not get anymore a result !! Sorry for that.
    – Claude Leibovici
    Nov 16 at 9:02










  • What do you mean?
    – EmThorns
    Nov 16 at 10:44










  • I shall be back tomorrow morning
    – Claude Leibovici
    Nov 16 at 17:40
















It doesn't look like there will be any nice solution at all. But you can multiply some of these equations by a nice constant to get nice integer coefficients, then work from there, though your answers are still going to be ugly.
– YiFan
Nov 16 at 3:34




It doesn't look like there will be any nice solution at all. But you can multiply some of these equations by a nice constant to get nice integer coefficients, then work from there, though your answers are still going to be ugly.
– YiFan
Nov 16 at 3:34












I made a mistake typing the equations and I do not get anymore a result !! Sorry for that.
– Claude Leibovici
Nov 16 at 9:02




I made a mistake typing the equations and I do not get anymore a result !! Sorry for that.
– Claude Leibovici
Nov 16 at 9:02












What do you mean?
– EmThorns
Nov 16 at 10:44




What do you mean?
– EmThorns
Nov 16 at 10:44












I shall be back tomorrow morning
– Claude Leibovici
Nov 16 at 17:40




I shall be back tomorrow morning
– Claude Leibovici
Nov 16 at 17:40










1 Answer
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To make the problem simpler and assuming $t neq 0$, let us reawrite the equations as
$$a t^4+b t^3+c t^2+d t+e=0tag 1$$
$$frac{a t^4}{32}+frac{b t^3}{16}+frac{c t^2}{8}+frac{d t}{4}+frac{e}{2}=0tag 2$$
$$frac{a t^4}{6}+frac{b t^3}{5}+frac{c t^2}{4}+frac{d t}{3}+frac{e}{2}=0 tag3$$
$$frac{a t^4}{42}+frac{b t^3}{30}+frac{c t^2}{20}+frac{d t}{12}+frac{e}{6}=frac v {t^3} tag4$$
$$frac{a t^4}{336}+frac{b t^3}{210}+frac{c t^2}{120}+frac{d t}{60}+frac{e}{24}=frac y {t^4} tag5$$



Now, solve equations $(2)$, $(3)$, $(4)$, $(5)$ for $b,c,d,e$. This gives
This gives as solutions
$$b=-frac{5 left(a t^8+1120 t v-2240 yright)}{2 t^7}qquad c=frac{15 left(a t^8+2016 t v-3920 yright)}{7 t^6}qquad d=-frac{5 left(a t^8+2604 t v-4704 yright)}{7 t^5}qquad e=frac{a t^8+2800 t v-3920 y}{14 t^4}$$



Plug these values in $(1)$ to get
$$a t^4+b t^3+c t^2+d t+e=-frac{140 (t v-2 y)}{t^3}$$ which can be $0$ only if $tv=2y$ what ever could be the value assigned to $a$ !






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    1 Answer
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    To make the problem simpler and assuming $t neq 0$, let us reawrite the equations as
    $$a t^4+b t^3+c t^2+d t+e=0tag 1$$
    $$frac{a t^4}{32}+frac{b t^3}{16}+frac{c t^2}{8}+frac{d t}{4}+frac{e}{2}=0tag 2$$
    $$frac{a t^4}{6}+frac{b t^3}{5}+frac{c t^2}{4}+frac{d t}{3}+frac{e}{2}=0 tag3$$
    $$frac{a t^4}{42}+frac{b t^3}{30}+frac{c t^2}{20}+frac{d t}{12}+frac{e}{6}=frac v {t^3} tag4$$
    $$frac{a t^4}{336}+frac{b t^3}{210}+frac{c t^2}{120}+frac{d t}{60}+frac{e}{24}=frac y {t^4} tag5$$



    Now, solve equations $(2)$, $(3)$, $(4)$, $(5)$ for $b,c,d,e$. This gives
    This gives as solutions
    $$b=-frac{5 left(a t^8+1120 t v-2240 yright)}{2 t^7}qquad c=frac{15 left(a t^8+2016 t v-3920 yright)}{7 t^6}qquad d=-frac{5 left(a t^8+2604 t v-4704 yright)}{7 t^5}qquad e=frac{a t^8+2800 t v-3920 y}{14 t^4}$$



    Plug these values in $(1)$ to get
    $$a t^4+b t^3+c t^2+d t+e=-frac{140 (t v-2 y)}{t^3}$$ which can be $0$ only if $tv=2y$ what ever could be the value assigned to $a$ !






    share|cite|improve this answer

























      up vote
      0
      down vote













      To make the problem simpler and assuming $t neq 0$, let us reawrite the equations as
      $$a t^4+b t^3+c t^2+d t+e=0tag 1$$
      $$frac{a t^4}{32}+frac{b t^3}{16}+frac{c t^2}{8}+frac{d t}{4}+frac{e}{2}=0tag 2$$
      $$frac{a t^4}{6}+frac{b t^3}{5}+frac{c t^2}{4}+frac{d t}{3}+frac{e}{2}=0 tag3$$
      $$frac{a t^4}{42}+frac{b t^3}{30}+frac{c t^2}{20}+frac{d t}{12}+frac{e}{6}=frac v {t^3} tag4$$
      $$frac{a t^4}{336}+frac{b t^3}{210}+frac{c t^2}{120}+frac{d t}{60}+frac{e}{24}=frac y {t^4} tag5$$



      Now, solve equations $(2)$, $(3)$, $(4)$, $(5)$ for $b,c,d,e$. This gives
      This gives as solutions
      $$b=-frac{5 left(a t^8+1120 t v-2240 yright)}{2 t^7}qquad c=frac{15 left(a t^8+2016 t v-3920 yright)}{7 t^6}qquad d=-frac{5 left(a t^8+2604 t v-4704 yright)}{7 t^5}qquad e=frac{a t^8+2800 t v-3920 y}{14 t^4}$$



      Plug these values in $(1)$ to get
      $$a t^4+b t^3+c t^2+d t+e=-frac{140 (t v-2 y)}{t^3}$$ which can be $0$ only if $tv=2y$ what ever could be the value assigned to $a$ !






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        To make the problem simpler and assuming $t neq 0$, let us reawrite the equations as
        $$a t^4+b t^3+c t^2+d t+e=0tag 1$$
        $$frac{a t^4}{32}+frac{b t^3}{16}+frac{c t^2}{8}+frac{d t}{4}+frac{e}{2}=0tag 2$$
        $$frac{a t^4}{6}+frac{b t^3}{5}+frac{c t^2}{4}+frac{d t}{3}+frac{e}{2}=0 tag3$$
        $$frac{a t^4}{42}+frac{b t^3}{30}+frac{c t^2}{20}+frac{d t}{12}+frac{e}{6}=frac v {t^3} tag4$$
        $$frac{a t^4}{336}+frac{b t^3}{210}+frac{c t^2}{120}+frac{d t}{60}+frac{e}{24}=frac y {t^4} tag5$$



        Now, solve equations $(2)$, $(3)$, $(4)$, $(5)$ for $b,c,d,e$. This gives
        This gives as solutions
        $$b=-frac{5 left(a t^8+1120 t v-2240 yright)}{2 t^7}qquad c=frac{15 left(a t^8+2016 t v-3920 yright)}{7 t^6}qquad d=-frac{5 left(a t^8+2604 t v-4704 yright)}{7 t^5}qquad e=frac{a t^8+2800 t v-3920 y}{14 t^4}$$



        Plug these values in $(1)$ to get
        $$a t^4+b t^3+c t^2+d t+e=-frac{140 (t v-2 y)}{t^3}$$ which can be $0$ only if $tv=2y$ what ever could be the value assigned to $a$ !






        share|cite|improve this answer












        To make the problem simpler and assuming $t neq 0$, let us reawrite the equations as
        $$a t^4+b t^3+c t^2+d t+e=0tag 1$$
        $$frac{a t^4}{32}+frac{b t^3}{16}+frac{c t^2}{8}+frac{d t}{4}+frac{e}{2}=0tag 2$$
        $$frac{a t^4}{6}+frac{b t^3}{5}+frac{c t^2}{4}+frac{d t}{3}+frac{e}{2}=0 tag3$$
        $$frac{a t^4}{42}+frac{b t^3}{30}+frac{c t^2}{20}+frac{d t}{12}+frac{e}{6}=frac v {t^3} tag4$$
        $$frac{a t^4}{336}+frac{b t^3}{210}+frac{c t^2}{120}+frac{d t}{60}+frac{e}{24}=frac y {t^4} tag5$$



        Now, solve equations $(2)$, $(3)$, $(4)$, $(5)$ for $b,c,d,e$. This gives
        This gives as solutions
        $$b=-frac{5 left(a t^8+1120 t v-2240 yright)}{2 t^7}qquad c=frac{15 left(a t^8+2016 t v-3920 yright)}{7 t^6}qquad d=-frac{5 left(a t^8+2604 t v-4704 yright)}{7 t^5}qquad e=frac{a t^8+2800 t v-3920 y}{14 t^4}$$



        Plug these values in $(1)$ to get
        $$a t^4+b t^3+c t^2+d t+e=-frac{140 (t v-2 y)}{t^3}$$ which can be $0$ only if $tv=2y$ what ever could be the value assigned to $a$ !







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 17 at 3:58









        Claude Leibovici

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