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Does anyone know any upper bounds or known results on LOWER BOUNDS for binary forms i.e.

if you have F(X,Y)=$X^n+YX^{n-1}+Y^2X^{n-2}+...+XY^{n-1}+Y^{n}$, I need to find a lower bound

for F interms of Y for e.g. $|F(X,Y)| \geq Y^n$ (not saying this is true).

I have found upper bounds in the literature extensively but no lower bounds yet.

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    How does this differ from the question http://math.stackexchange.com/questions/139119/binary-forms-of-degree-n you just asked a few hours ago?2012-05-01
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    Gerry Myerson: I've flagged that question for deletion, as it was causing confusinon.2012-05-01
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    Are $X,Y$ taking real numbers?2012-05-01
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    wxu: Yes they're real numbers2012-05-01
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    Okay, what is the bound for $n=1$, i.e., $|X+Y|$, is zero?2012-05-01
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    Notice for $n$ odd choosing $X=-Y$ gives $F(X,Y)=0$.2012-05-01
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    wxu: good point, I'm particularly interested in the case where n=4.2012-05-01
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    wxu: I suppose it doesn't make much sense for odd n. In the case n=2 it's 3/4Y^22012-05-01
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    So for $n=4$, you only need to figure out the minimal values of $|t^4+t^3+t^2+t+1|$, where $t\in \mathbb{R}$.2012-05-01
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    wxu: I'm having trouble seeing why this is. If I found the minimal values of that function, how does that translate into a lower bound in terms of Y?2012-05-01
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    $F(X,Y)=F(X/Y,1)Y^n$, write $t=X/Y$...2012-05-01
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    anon: wow great, thank you guys.2012-05-01
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    Instead of flagging the previous version for deletion, it would have been better just to edit it.2012-05-01

1 Answers 1

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Setting $X=-Y$ gives $F=0$ for $n$ odd, so assume it's even. Observe

$$F(X,Y)=X^n+YX^{n-1}+\cdots Y^{n-1}X+Y^n =Y^nF(X/Y,1).$$

We make the change of variable $Z=X/Y$ and instead try to minimize $F(Z,1)$. Differentiating,

$$\frac{d}{dZ}\sum_{k=0}^n Z^k=\frac{d}{dZ}\frac{Z^{n+1}-1}{Z-1}=\frac{\big((n+1)Z^n\big)(Z-1)-(Z^{n+1}-1)(1)}{(Z-1)^2}.$$

We set this equal to $0$. The potential extrema occur at the roots of the polynomial in the numerator,

$$nZ^{n+1}-(n+1)Z^n+1=0. \tag{$*$}$$

For a particular $n$, we need to compute the roots above, and then plug them into $F(Z,1)$ in order to compare and see which corresponds to the global minimum. Note that $F(1,1)=n$.

There are a couple issues we swept under the rug that we must address though. For $n$ even, the polynomial $F(Z,1)$ is always nonnegative. For it is clearly nonnegative on $Z\ge0$, and by the geometric sum formula we have (remember $n$ is even!)

$$F(-Z,1)=\frac{(-Z)^{n+1}-1}{(-Z)-1}=\frac{Z^{n+1}+1}{Z+1}.$$

Since this is a ratio of two positive numbers for $Z>0$, we must have $F(-Z,1)>0$. This is why it was valid to minimize $F(Z,1)$ "instead" of $|F(Z,1)|$; they're the same! Secondly, our polynomial grows without bound in the $+\infty$ and $-\infty$ directions so there is no infimum case to worry about.


For $n=4$, WolframAlpha gives

$$Z_0=-\frac{1}{4}\left(1+\sqrt[3]{\frac{25}{3(4\sqrt{6}-9)}}-\sqrt[3]{\frac{5(4\sqrt{6}-9)}{9}}\right)\approx -0.60583;$$

$$F(Z_0,1)\approx0.673553.$$