This is a problem from the 2001 Stanford Math Tournament Algebra section. $$$$Given that $$x+y+z=3$$ $$x^2 + y^2 + z^2 = 5$$$$x^3+y^3+z^3=7$$Find $x^4+y^4+z^4$. $$$$My friend claimed that he was able to solve this question using geometric series, i.e. adding the above three equations together and forming geometric series with three terms, but I can't see how he solved it using that method. I solved it using the method described in the solutions document, but I haven't seen a solution using geometric series. Can anyone find a way to solve the problem using geometric series? Any help with this problem would be great. Thanks for the help.
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7$\begingroup$ is your friend prone to lying? $\endgroup$– obatakuAug 15, 2013 at 19:39
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$\begingroup$ Maybe he used some other method, but didn't want you to win, so he gave you a false hint to lead you in the wrong direction. $\endgroup$– rurouniwallaceAug 15, 2013 at 19:43
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1$\begingroup$ Since the only possible answer is $\frac 16 a^4 - b a^2 + \frac 43 ac + \frac 12 b^2$ with $a=x+y+z$, $b=x^2+y^2+z^2$ and $c=x^3+y^3+z^3$, there is little room for alternative theories. maybe he used not a geometric series, but an arithmetic sequence to qǵuess the next value? In that case he was lucky ... $\endgroup$– Hagen von EitzenAug 15, 2013 at 19:45
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$\begingroup$ Good thing these aren't integer solutions. cough fermat cough $\endgroup$– Ali CaglayanAug 15, 2013 at 20:09
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$\begingroup$ You should ask him to substantiate his claim. For example, it is not true that "the sum must follow an arithmetic progression". As a side note, $x^0 + y^0 + z^0 = 3 $. $\endgroup$– Calvin LinAug 15, 2013 at 20:10
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1 Answer
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This is a rather encumbered solution for a not-as-hard problem, and it is the only way I see to make sense of what your friend is saying. On the other hand, it is cool that it let us know about Newton's identities and their proof using generating functions.
Let us form the formal series \begin{align}F(x,y,z,t)&:=(x+y+z)t+(z^2+y^2+z^2)t^2+(x^3+y^3+z^3)t^3+\ldots\\&=3t+5t^2+7t^3+at^4+\ldots,\end{align} where $a:=x^4+y^4+z^4$ is the number we want to find.
Forming this power series is kind of adding the given equations (as your friend suggested), but we multiply them first by $t,t^2,t^3,...$, respectively, before adding.
Consider the polynomial \begin{align}G(x,y,z,t)&:=-1+(x+y+z)t+(xy+xz+yz)t^2-(xyz)t^3\\&=-1+3t+(xy+xz+yz)t^2-(xyz)t^3\end{align}
We multiply $F$ times $G$. It happens that we get a polynomial:
\begin{align} FG&=\left(\sum_{k=1}^{\infty}(xt)^k+\sum_{k=1}^{\infty}(yt)^k+\sum_{k=1}^{\infty}(zt)^k\right)G\\ &=-\left(\frac{xt}{1-xt}+\frac{yt}{1-yt}+\frac{yt}{1-xt}\right)(1-xt)(1-yt)(1-zt)\\ &=t\left((-x)(1-yt)(1-zt)+(-y)(1-xt)(1-zt)+(-z)(1-xt)(1-yt)\right)\\ &=-(x+y+z)t+(xy+xz+yz)t^2-(xyz)t^3\\ &=-3t+(xy+xz+yz)t^2-(xyz)t^3 \end{align}
In the second equality we used the sum of the geometric series more or less like your friend said.
Comparing coefficients of the multiplication $FG$ with the result in the last line we get (the Newton's identities for the values in this problem)
\begin{align} 2(xy+xz+yz)&=9-5\\ 3xyz&=3(xy+xz+yz)-15+7\\ 0&=3(xyz)-5(xy+xz+yz)+21-(x^4+y^4+z^4) \end{align}
From these, substituting from top to bottom, we find the value of $x^4+y^4+z^4$. Notice that the method allows us to continue using the equalities between coefficients of higher degree in $t$ to compute $x^5+y^5+z^5$, $x^6+y^6+z^6$, and so on ...
