Calculate

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February 4th 2006, 03:55 PM
DenMac21

Calculate

Here is one problem.

If
$a+b+c=0$
and
$a^2+b^2+c^2=1$

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

Help anyone?
February 4th 2006, 06:05 PM
ThePerfectHacker

Quote:

Originally Posted by DenMac21

Here is one problem.

If
$a+b+c=0$
and
$a^2+b^2+c^2=1$

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

Help anyone?

Perhaps you may do this,

$a^2+b^2+c^2=1$
Non of the $a,b,c$ could be more than 1, thus, $|a,b,c|<1$

Do this in terms of a function of one, $a=x$ where $x$ is a changing variable with $|x|<1$ . Thus,
$x+b+c=0$
$x^2+b^2+c^2=1$
Now solve these two for $b,c$ it does not matter because we can consider them to be the same here. Thus, by substitution,
$x^2+(-x-c)^2+c^2=1$ thus,
$x^2+x^2+2cx+c^2+c^2=1$ thus,
$x^2+cx+c^2=1/2$ thus,
$c^2+xc+(x^2-1/2)=0$ thus,
$c=\frac{-x\pm\sqrt{2-3x^2}}{2}$ by Quadradic Formula, thus,
$b=\frac{-x\mp\sqrt{2-3x^2}}{2}$
Thus, as a function of $x$ for $|x|<1$
$(a,b,c)=(x,\frac{-x\pm\sqrt{2-3x^2}}{2},\frac{-x\mp\sqrt{2-3x^2}}{2})$
You can check that,
$a+b+c=0$
$a^2+b^2+c^2=1$ .
Know all is left is to find what,
$a^4+b^4+c^4$ as see whether it is invaraint of $x$ .

Here it gets messy, notice that
$(n+m)^4+(n-m)^4=2n^4+12n^2m^2+m^4$
This is confirmed by simple expansion.
Thus,
$a^4+b^4+c^4$ becomes,
$x^4+\left(\frac{-x+\sqrt{2-3x^2}}{2}\right)^4+\left(\frac{-x-\sqrt{2-3x^2}}{2}\right)^4$
Now this is a variation of the identity before show, with
$n=-x,m=\frac{-x+\sqrt{2-3x^2}}{2}$
Thus,
$a^4+b^4+c^4=\frac{1}{2}$
February 4th 2006, 06:07 PM
ThePerfectHacker

I was thinking maybe you can do something else. Perhaps, show that the derivative of the function is zero, thus it must be constant. After you know that select any value of $x$ and know its value.
February 4th 2006, 07:24 PM
ticbol

Quote:

Originally Posted by DenMac21

Here is one problem.

If
$a+b+c=0$
and
$a^2+b^2+c^2=1$

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

Help anyone?

Here is one way.

a +b +c = 0 ------------(1)
a^2 +b^2 +c^2 = 1 -----------(2)

a^2 +b^2 +c^2 = 1 ------------------------------------(2)
Square both sides,
(a^4 +b^4 +c^4) +2[(ab)^2 +(ac)^2 +(bc)^2] = 1 -------(2a)
So,
a^4 +b^4 +c^4 = 1 -2[(ab)^2 +(ac)^2 +(bc)^2] ----------(2b)

a +b +c = 0 -----------------------------(1)
Square both sides,
(a^2 +b^2 +c^2) +2[ab +ac +bc] = 0 ------(1a)
Substitution, from (2),
1 +2[ab +ac +bc] = 0
2[ab +ac +bc] = -1
ab +ac +bc = -1/2
Square both sides,
[(ab)^2 +(ac)^2 +(bc)^2] +2[(a^2)bc +a(b^2)c +ab(c^2)] = 1/4 -----(3)
[(ab)^2 +(ac)^2 +(bc)^2] +2abc[a +b +c] = 1/4
Since [a+b+c] = 0, then,
[(ab)^2 +(ac)^2 +(bc)^2] = 1/4 -----------***

Substitute that into (2b),
a^4 +b^4 +c^4 = 1 -2[1/4]
Therefore,
a^4 +b^4 +c^4 = 1/2 -------------answer.

-----------------------------------------------------
I did the long multiplications on paper here. I did not show them here because, well, they are long.
You can do them on paper there to verify how I got the Eq.(2a), (1a), and (3).
February 5th 2006, 03:25 AM
DenMac21

Thanks for help.

Solutions are both correct, but ticbo's solution is "cleanly" for me.
February 5th 2006, 09:51 AM
ThePerfectHacker

I agree ticbol's solution is more elegant. Nice job ticbol!