# Solving an apparently “simple” system of linear equations

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.

• 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 '18 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 '18 at 9:02
• What do you mean? – EmThorns Nov 16 '18 at 10:44
• I shall be back tomorrow morning – 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 $$a$$ !