Here is one problem.

If

$\displaystyle a+b+c=0$

and

$\displaystyle a^2+b^2+c^2=1$

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

Help anyone?

Printable View

- Feb 4th 2006, 03:55 PMDenMac21Calculate
Here is one problem.

If

$\displaystyle a+b+c=0$

and

$\displaystyle a^2+b^2+c^2=1$

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

Help anyone? - Feb 4th 2006, 06:05 PMThePerfectHackerQuote:

Originally Posted by**DenMac21**

$\displaystyle a^2+b^2+c^2=1$

Non of the $\displaystyle a,b,c$ could be more than 1, thus, $\displaystyle |a,b,c|<1$

Do this in terms of a function of one, $\displaystyle a=x$ where $\displaystyle x$ is a changing variable with $\displaystyle |x|<1$. Thus,

$\displaystyle x+b+c=0$

$\displaystyle x^2+b^2+c^2=1$

Now solve these two for $\displaystyle b,c$ it does not matter because we can consider them to be the same here. Thus, by substitution,

$\displaystyle x^2+(-x-c)^2+c^2=1$ thus,

$\displaystyle x^2+x^2+2cx+c^2+c^2=1$ thus,

$\displaystyle x^2+cx+c^2=1/2$ thus,

$\displaystyle c^2+xc+(x^2-1/2)=0$ thus,

$\displaystyle c=\frac{-x\pm\sqrt{2-3x^2}}{2}$ by Quadradic Formula, thus,

$\displaystyle b=\frac{-x\mp\sqrt{2-3x^2}}{2}$

Thus, as a function of $\displaystyle x$ for $\displaystyle |x|<1$

$\displaystyle (a,b,c)=(x,\frac{-x\pm\sqrt{2-3x^2}}{2},\frac{-x\mp\sqrt{2-3x^2}}{2})$

You can check that,

$\displaystyle a+b+c=0$

$\displaystyle a^2+b^2+c^2=1$.

Know all is left is to find what,

$\displaystyle a^4+b^4+c^4$ as see whether it is invaraint of $\displaystyle x$.

Here it gets messy, notice that

$\displaystyle (n+m)^4+(n-m)^4=2n^4+12n^2m^2+m^4$

This is confirmed by simple expansion.

Thus,

$\displaystyle a^4+b^4+c^4$ becomes,

$\displaystyle x^4+\left(\frac{-x+\sqrt{2-3x^2}}{2}\right)^4+\left(\frac{-x-\sqrt{2-3x^2}}{2}\right)^4$

Now this is a variation of the identity before show, with

$\displaystyle n=-x,m=\frac{-x+\sqrt{2-3x^2}}{2}$

Thus,

$\displaystyle a^4+b^4+c^4=\frac{1}{2}$ - Feb 4th 2006, 06:07 PMThePerfectHacker
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 $\displaystyle x$ and know its value.

- Feb 4th 2006, 07:24 PMticbolQuote:

Originally Posted by**DenMac21**

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). - Feb 5th 2006, 03:25 AMDenMac21
Thanks for help.

Solutions are both correct, but ticbo's solution is "cleanly" for me. - Feb 5th 2006, 09:51 AMThePerfectHacker
I agree ticbol's solution is more elegant. Nice job ticbol!