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,



  • $\begingroup$ 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. $\endgroup$ – YiFan Nov 16 '18 at 3:34
  • $\begingroup$ I made a mistake typing the equations and I do not get anymore a result !! Sorry for that. $\endgroup$ – Claude Leibovici Nov 16 '18 at 9:02
  • $\begingroup$ What do you mean? $\endgroup$ – EmThorns Nov 16 '18 at 10:44
  • $\begingroup$ I shall be back tomorrow morning $\endgroup$ – Claude Leibovici Nov 16 '18 at 17:40

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

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

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 $a$ !


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.