Solving for simultaneous equations

[Albert]

( i ) prove the following simultaneous equations (1) (2) and (3) has no real solution

$a+b+c=1-----(1)$

$a^2+b^2+c^2=2----(2)$

$a^3+b^3+c^3=3----(3)$

( ii ) using (1)(2)and(3)find the value of :

$a^4+b^4+c^4$

 

[Jester]

Re: solving for simultaneous equations

Here's my solution to part (i). Not really the most elegant but gets it done. Solving eqn (1) for $c$ and substituting into the remaining two equations gives (with appropriate scaling)

$b^2 +(a-1)b +a^2-a-\dfrac{1}{2} = 0$

$(a-1)b^2+(a-1)^2b-a^2+a+\dfrac{2}{3}=0$

Eliminating $b$ between these two gives that $a$ satisfies the cubic

$6a^3-6a^2-3a-1 = 0$

Via a variety of techniques we can show that this cubic has only one real root and it's lies in the interval $(1,2)$ so $a > 1$.

Next we will show that the first equation for $b$ has complex roots for this value of $a$. To show this is to show that

$(a-1)^2 - 4(a^2-a-\dfrac{1}{2}) < 0$ or $-3a^2+2a+3 < 0$

As $a$ satisfies this cubic then

$-3a^2+2a+\dfrac{3}{2} = - \dfrac{1}{2a} - a + \dfrac{3}{2} < 0$ if $a > 1.$

The latter part can be established as follows: Let $f(a) =- \dfrac{1}{2a} - a + \dfrac{3}{2} $. Since $f(1) = 0$ and $f'(a) <0$ if $a>1$ then $f(a) < 0$ for $a>1$.

 

[MarkFL]

Re: solving for simultaneous equations

I'll take a stab at part (ii):

From the form of the given expressions, we may use an auxiliary equation of:

\(\displaystyle (r-a)(r-b)(r-c)=r^3-(a+b+c)r^2+(ab+ac+bc)r-abc=0\)

to determine the linear homogeneous recursion:

\(\displaystyle S_{n+3}=(a+b+c)S_{n+2}-(ab+ac+bc)S_{n+1}+abcS_{n}\)

where we are given:

\(\displaystyle S_1=1,\,S_2=2,\,S_3=3\)

If we square (1), we find:

\(\displaystyle (a^2+b^2+c^2)+2(ab+ac+bc)=1\)

Using (2), this becomes:

\(\displaystyle 2+2(ab+ac+bc)=1\)

Hence:

\(\displaystyle ab+ac+bc=-\frac{1}{2}\)

If we cube (1), we find:

\(\displaystyle (a^3+b^3+c^3)+3(a+b+c)(ab+ac+bc)-3abc=1\)

Using (3) and our previous result, this becomes:

\(\displaystyle 3-\frac{3}{2}-3abc=1\)

Hence:

\(\displaystyle abc=\frac{1}{6}\)

And so our recursion becomes:

\(\displaystyle S_{n+3}=S_{n+2}+\frac{1}{2}S_{n+1}+\frac{1}{6}S_{n}\)

With $n=1$, we then find:

\(\displaystyle a^4+b^4+c^4=S_4=3+\frac{1}{2}\cdot2+\frac{1}{6} \cdot1=\frac{25}{6}\)

 

[Opalg]

For me, this is an ideal opportunity to use Newton's identities.

Let $x^3- px^2 +qx - r = 0$ be the equation whose roots are $a,\ b,\ c$, and let $S_k = a^k+b^k+c^k \ (k=1,2,3,4)$. Newton's identities say that

$\phantom{1}p = S_1 = 1$,
$2q = pS_1 - S_2 = 1-2 = -1$,
$3r = qS_1 - pS_2 + S_3 = -\frac12 -2 +3 = \frac12$,
$\phantom{1}0 = rS_1 - qS_2 + pS_3 - S_4$.​

Thus $p=1$, $q = -\frac12$, $r = \frac16$, and the cubic equation is $x^3 - x^2 - \frac12x - \frac16 = 0$. That is the same equation that Jester found for $a$, and as he pointed out it only has one real root. So $a,\ b,\ c$ cannot all be real.

Finally the last of those Newton's identities above shows that $S_4 = \frac16 + 1+ 3 = \frac{25}6$.

 

[Albert]

( i )using the result from MarkFL :

$a+b+c=1,\,\,ab+bc+ca=\dfrac {-1}{2},\,\, abc=\dfrac {1}{6}$

$(ab+bc+ca)^2=a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)=$$ \dfrac {1}{4}$

$\therefore a^2b^2+b^2c^2+c^2a^2=$$ \dfrac {1}{4} - 2abc(a+b+c)=\dfrac {1}{4} - \dfrac {1}{3}=\dfrac {-1}{12}<0$

This proves the first part :the simultaneous equationsno has no real solution

( ii )

$(2)^2:$$(a^2+b^2+c^2)^2=a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)=4$


$\therefore a^4+b^4+c^4=4-2\times(\dfrac {-1}{12})=\dfrac {25}{6}$ #