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.
systems-of-equations matlab condition-number
add a comment |
up vote
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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.
systems-of-equations matlab condition-number
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
add a comment |
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.
systems-of-equations matlab condition-number
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
systems-of-equations matlab condition-number
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
add a comment |
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
add a comment |
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 $a$ !
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $a$ !
add a comment |
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 $a$ !
add a comment |
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 $a$ !
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 $a$ !
answered Nov 17 at 3:58
Claude Leibovici
116k1156131
116k1156131
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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