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/sci/ - Science & Math

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10621806 No.10621806 [Reply] [Original]

[math]
\text{Let }P(z)=z^{n} + c_1 z^{n-1} + c_2 z^{n-2} + \cdots + c_n \text{ be a polynomial in the complex}
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\text{variable } z \text{, with real coefficients } c_k \text{. Suppose that } |P(i)|<1 \text{. Prove that there}
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\text{exist real numbers } a \text{ and } b \text{ such that } P(a+bi)=0 \text{ and } (a^2+b^2+1)^2 < 4b^2 + 1 \text{.}
[/math]

>> No.10621810

Previous Thread >>10615013

>> No.10621849

Reminder that Re P(i) is the alternating sum of the even indexes while Im P(i) is the alternating sum of the uneven indexes.

>> No.10622136

bumping for visibility

>> No.10622501

>>10621849
no it's not

>> No.10622534
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10622534

>>10621806
Daily reminder that Putnam winners get to become billionaires. It's LITERALLY OVER for brainlets :(

>> No.10622621

>>10621806
A real polinomial factors as a product of degree 2 and degree 1 real polinomials. For one of the factors |F(i)|<1. Withoul loss of generality n=1 or 2.
Assume P(x)=x^2-2ax+a^2+b^2 for the root (a+bi). Then |P(i)|^2=(a^2+b^2+1)-4b^2-1.
The case n=1 is even easier.

>> No.10623695

Nice problem, but why the pedophile cartoon?

>> No.10623703

>>10623695
Because OP’s a pedophile