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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 math.stackexchange.com/questions/139119/… you just asked a few hours ago? –  Gerry Myerson May 1 '12 at 5:48
    
Gerry Myerson: I've flagged that question for deletion, as it was causing confusinon. –  hello May 1 '12 at 5:51
    
Are $X,Y$ taking real numbers? –  wxu May 1 '12 at 5:55
    
wxu: Yes they're real numbers –  hello May 1 '12 at 5:56
2  
Instead of flagging the previous version for deletion, it would have been better just to edit it. –  Gerry Myerson May 1 '12 at 7:37
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1 Answer

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

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

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

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