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Studying for finals I have come across a result that I understand how the system is derived but I cannot solve the system. I feel like this should be trivial, but I do not know where to go. Through Gaussian Quadrature I need to find constants $c_1, c_2, x_1, x_2$

Skipping all the integration and getting right to the system I derived I have:

$$c_1 + c_2 = 2$$ $$c_1x_1 + c_2x_2 = 0$$ $$c_1x_1^2 + c_2x_2^2 = \frac{2}{3}$$ $$c_1x_1^3 + c_2x_2^3 = 0$$

I never learned how to solve such systems with quadratic and cubic members.

The book says "simple algebra" can be used to solve this, and the book provides the solutions

$$c_1 = 1, c_2 = 1, x_1 = \frac{-\sqrt{3}}{3}, x_2 = \frac{\sqrt{3}}{3}$$

How are these derived?

Thank you!

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Assume that neither $x_1$ nor $x_2$ is zero (you can rule this out by simple testing).

Then the 2nd and 4th equations imply that $c_1x_1=-c_2x_2$ and $c_1x_1^3=-c_2x_2^3$, respectively. Then we can divide the corresponding sides of each of these equations to get that $$x_1^2=x_2^2$$ so the 3rd equation becomes $$(c_1+c_2)x_1^2=2/3$$ But then the first equation tells us that $$2x_1^2=2/3$$ so $$x_1=\pm\sqrt{3}/3$$ and thus also $$x_2=\pm\sqrt{3}/3.$$

Note that we can't have that $x_1=x_2$, since then the 2nd equation would imply that $c_1+c_2=0$. Therefore, $x_2=-x_1$, and the 2nd equation then implies that $c_1 = c_2$. The first equation then tells us that $$c_1=c_2=1.$$

So we have that $$\boxed{c_1=c_2=1}$$ and $$\boxed{x_1=\pm\sqrt{3}/3}$$ and $$\boxed{x_2=\mp\sqrt{3}/3}.$$

The $x_i$ can be interchanged, as long as they have opposite signs. You can see this in the original equations--swapping them doesn't change the equations.

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  • $\begingroup$ Brilliant answer. Thank you. $\endgroup$ – John Dec 5 '14 at 4:36

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