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I was doing the following problem:

given $a, b, c, d \in \mathbb{R}$ and

$a+b+c+d=1$, $a^2+b^2+c^2+d^2=2$, $a^3+b^3+c^3+d^3=3$ and $a^4+b^4+c^4+d^4=4$.

Find $a^n+b^n+c^n+d^n$ (I am looking for a closed kind of formula)

Now, I can already do it in one way. We can find the elementary symmetric polynomials using newton's identity and then solve the quartic to find the values of $a, b, c, d$.

But I realise that my method is computationally heavy. And more importantly once the number of variables is increased there is no guarantee that we can solve for the roots. Also, the problem does not really need the values of $a, b, c, d$ explicitly.

The above observations motivate me to find an alternative way to find a an alternative technique to solve problems like this(may be in a more general setting).

So, can someone help me?

Martin Sleziak
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Subham Jaiswal
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    See http://math.stackexchange.com/questions/1661035/prove-fraca3b3c33-fraca7b7c77-left-fraca5b5c55 – lab bhattacharjee Mar 01 '16 at 06:49
  • @labbhattacharjee, the techniques in the links do not show any method to find an explicit formula for $a^n+b^n+c^n+d^n$. Sorr, for not explicitly mentioning that. They just use newton identities to recursively find the values. I am aware for that method. BTW, thanks for the links. – Subham Jaiswal Mar 01 '16 at 07:42
  • An intuitive guess is $n$ – Nikunj Mar 01 '16 at 09:02
  • Note that $(a(a-1))^2+(b(b-1))^2+(c(c-1))^2+(d(d-1))^2=0$. But they are not real numbers. – S.C.B. Mar 01 '16 at 09:16

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If we take $a,b,c,d$ as the roots of a monic polynomial $$ p(x)=x^4-e_1 x^3+e_2 x^2-e_3 x+e_4 $$ Newton's identities give: $$ \left\{\begin{array}{rcl}e_1 &=& 1\\ 2e_2 &=& e_1-2 \\ 3e_3 &=& e_2 - 2e_1 +3 \\ 4e_4 &=& e_3-2e_2+3e_1-4 \end{array}\right. $$ hence it follows that: $$ e_1=1,\quad e_2 = -\frac{1}{2},\quad e_3 = \frac{1}{6},\quad e_4 = \frac{1}{24}, $$ $$ p(x) = x^4-x^3-\frac{x^2}{2}-\frac{x}{6}+\frac{1}{24} $$ and:

$$ a^n+b^n+c^n+d^n = \text{Tr}\,\begin{pmatrix} 0 & 0 & 0 & -\frac{1}{24} \\ 1 & 0 & 0 & \frac{1}{6} \\ 0 & 1 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & 1\end{pmatrix}^n .$$

Since the dominant root of $p(x)$ is about $1.42016$ and the other ones are quite close to zero, $a^n+b^n+c^n+d^n$ almost behaves like $2^{n/2}$ for large $n$s.

Jack D'Aurizio
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  • Thanks. Sir, can you take a look at this question http://math.stackexchange.com/questions/1671738/explaining-newton-polygon-in-elementary-way – Subham Jaiswal Mar 01 '16 at 12:01