Complex Polynomial Inequality Verification

Definition

For z,ai∈Cz, a_i \in \mathbb C and n∈R+n \in \mathbb R^+

P(z)=a0+a1z+a2z2+⋯+anznP(z) = a_0 + a_1 z + a_2 z^2 + \dots + a_n z_n

with an≠0a_n \neq 0 (i.e. has a nnth term) is called a *polynomial* of degree nn.

The inequality we wish to verify is

Theorem

∣1P(z)∣<2∣an∣Rn\left\vert\frac{1}{P(z)}\right\vert \lt \frac{2}{\vert a_n\vert R^n}

whenever ∣z∣>R\vert z\vert > R for some R∈R+R \in \mathbb R^+. In other words, ∣1P(z)∣\left\vert\frac{1}{P(z)}\right\vert is bounded above by 2∣an∣Rn\frac{2}{\vert a_n \vert R^n} when zz is sufficiently large.

We proceed by abuse of algebra.

Proof

Let w=a0zn+a1zn−1+a2zn−2+⋯+an−1z\displaystyle{w = \frac{a_0}{z^n} + \frac{a_1}{z^{n-1}} + \frac{a_2}{z^{n-2}} + \dots + \frac{a_{n-1}}{z}}, then P(z)=(an+w)znP(z) = (a_n+ w)z^n. Also note that ∣P(z)∣=∣an+w∣⋅∣zn∣\vert P(z)\vert = \vert a_n + w \vert \cdot \vert z^n \vert. Since ∣zn∣=∣z∣n\vert z^n \vert = \vert z \vert ^n, we can write ∣P(z)∣=∣an+w∣⋅∣z∣n\vert P(z)\vert = \vert a_n + w \vert \cdot \vert z \vert ^n.

We see that

wzn=a0+a1z+a2z2+⋯+an−1zn−1wz^n = a_0 + a_1 z + a_2 z^2 + \dots + a_{n-1}z^{n-1}

So by the general triangle inequality

∣w∣∣z∣n≤∣a0∣+∣a1∣∣z∣+∣a2∣∣z∣2+⋯+∣an−1∣∣z∣n−1\vert w \vert \vert z \vert ^n \leq \vert a_0 \vert + \vert a_1 \vert \vert z \vert + \vert a_2 \vert \vert z \vert ^2 + \dots + \vert a_{n-1} \vert \vert z \vert ^{n-1}

or,

∣w∣≤∣a0∣∣z∣n+∣a1∣∣z∣n−1+∣a2∣∣z∣n−2+⋯+∣an−1∣∣z∣\vert w \vert \leq \frac{\vert a_0 \vert}{\vert z \vert ^ {n}} + \frac{\vert a_1 \vert}{\vert z \vert ^ {n - 1}} + \frac{\vert a_2 \vert}{\vert z \vert ^ {n - 2}} + \dots + \frac{\vert a_{n - 1}\vert}{\vert z \vert}

Certainly, if we pick a large enough zz, the following is true:

∣w∣<∣an∣\vert w \vert \lt \vert a_n \vert

For some reason, the above will not be nearly as useful to us as the following, so that will be what we will work with.

∣w∣<∣an∣2\vert w \vert \lt \frac{\vert a_n \vert}{2}

That is, the anzna_n z^n term asymptotically dominates the rest - for a large enough zz. So by the triangle inequality,

∣an+w∣≥∣∣an∣−∣w∣∣\vert a_n + w \vert \geq {\Large\vert} \vert a_n \vert - \vert w \vert {\Large\vert}

Since ∣w∣<∣an∣2\vert w \vert \lt \frac{\vert a_n \vert}{2}, then ∣∣an∣−∣w∣∣>∣an∣2{\Large\vert} \vert a_n \vert - \vert w \vert {\Large\vert}> \frac{\vert a_n \vert}{2} — that is, taking away less than half of ∣an∣\vert a_n \vert leaves more than half of ∣an∣\vert a_n \vert. So we then have

∣an+w∣≥∣∣an∣−∣w∣∣>∣an∣2\vert a_n + w \vert \geq {\Large\vert} \vert a_n \vert - \vert w \vert {\Large\vert} > \frac{\vert a_n \vert}{2}

All provided that zz is of sufficient size. Since ∣an+w∣⋅∣z∣n>∣an∣2⋅∣z∣n\vert a_n + w \vert \cdot \vert z \vert^n > \frac{\vert a_n \vert}{2} \cdot \vert z \vert ^n, we have

∣P(z)∣=∣an+w∣⋅∣z∣n>∣an∣2⋅∣z∣n\vert P(z) \vert = \vert a_n + w \vert \cdot \vert z \vert^n > \frac{\vert a_n \vert}{2} \cdot \vert z \vert^n

Then, if ∣z∣>R\vert z \vert > R, we have

∣P(z)∣>∣an∣2⋅∣z∣n>∣an∣2⋅Rn\vert P(z) \vert > \frac{\vert a_n \vert}{2} \cdot \vert z \vert ^n > \frac{\vert a_n \vert}{2} \cdot R^n

or,

∣1P(z)∣<2∣an∣Rn\left\vert \frac{1}{P(z)}\right\vert \lt \frac{2}{\vert a_n \vert R^n}

And thus the inequality is verified.