Solving for C1 - C5 which make linear combination equal to zero for all x

Question

I have the following set of functions: $$f_1(x) = 5x$$ $$f_2(x) = 2x+3$$ $$f_3(x) = 2x^2 + 3$$ $$f_4(x) = x^2+ x$$

I need to find $C_1, C_2, C_3, C_4$ which make the linear combination equal to zero for all x.

I have tried many ways that I can think of. For example, I tried solving for $\mathbf x = \mathbf 0 * \mathbf Ainverse$ but couldn't do the gauss eliminatin to find $\mathbf Ainverse$. I proceeded to write out the linear combination and group the coefficients but I get 3 equations and 4 unknowns so that doesn't work either. Could someone give me a push in the right direction? Any advice is greatly appreciated. thank you.

For the second method I mentioned I have the following: $$(3*C_2 + 3*C_3) + (5*C_1 + 2*C_2 + C_4)x + (2*C_3 + C_4)x^2 = 0$$ This becomes 3 equations with 4 unkowns, which I cannot solve for.

Answer 1

There is not a unique solution to the problem. If you find a set of coefficients that make the polynomial identically $0$, then any constant multiple of the coefficients will do the same. We have $$ \begin{align} 3c_2+3c_3&=0 \tag 1 \\ 5c_1+2c_2+c_4&=0\tag 2 \\ 2c_3+c_4&=0 \tag 3 \end{align}$$ From ($1), c_2=-c_3.$ From $(3) c_4=-2c_3$ Substituting these values into $(2)$ gives $c_1=4c_3/5.$ Now we can set $c_3$ to any nonzero value we like to get an answer to the problem. For example, with $c_3=5,$ we get $(c_1, c_2, c_3, c_4)=(4,-5,5,-10)$

Answer 2

HINT: Let $g_1(x)=f_2(x)-(2/5)f_1(x)=3$, $g_2(x)=(1/2)(f_3(x)-g_1(x))=x^2$, $g_3(x)=f_4(x)-g_2(x)=x$. Can you find $E$, $F$, $G$ such that $Eg_1(x)+Fg_2(x)+Gg_3(x)\equiv0$?

And if you already have an equation for $C_i$ as in your edition, please observe that it can be multiplied by a (nonzero) constant, so you may fix one of your constants.

Answer 3

In polynomial space, solve the following system.

$$\begin{bmatrix} 0 & 0 & 2 & 1 \\ 5 & 2 & 0 & 1 \\ 0 & 3 & 3 & 0 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$

Reducing yields:

$$\begin{bmatrix} 1 & 0 & 0 & \frac{2}{5} \\ 0 & 1 & 0 & -\frac{1}{2} \\ 0 & 0 & 1 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$

Or, the equivalent system:

$$ \begin{align} c_1 &= -\frac{2}{5}c_4 \\ c_2 &= \frac{1}{2}c_4 \\ c_3 &= -\frac{1}{2}c_4 \end{align}$$

Here, $c_4$ is a free variable that can be chosen arbitrarily.