Polynomials (2 Viewers)

[Fus Ro Dah]

Here's a fun question I did a couple weeks ago.

 

[seanieg89]

Here's a fun question I did a couple weeks ago.

I assume you mean no nontrivial zeroes.

 

[Carrotsticks]

I think OP's problem can be fixed by either changing the polynomial to 1 + sum(etc) or by redefining:



Because 0 most certainly lies in that circle.

 

[Carrotsticks]

Ok since nobody replied may as well give a start for anybody who wants to try.

Firstly when I saw the question the first thing I thought of was the Golden Ratio because it's the radius of the circle. But it's a ~little~ different from the usual quadratic t^2=t+1 so with a little algebra we see that it's just the positive solution for the quadratic 1-t-t^2. For simplicity sake, I'll let r = 1/2 (root 5 - 1) for 'radius' such that if |z| < r, then 1 - |z| - |z|^2 > 0

Suppose . Using the Reverse Triangle inequality for |z| < r < 1, we have:



Do you recognise the last expression perhaps? (look at the condition for |z|). Give it a try now!

 

[barbernator]

<a href="http://www.codecogs.com/eqnedit.php?latex=for~|z|<r<1\\ \\ |z|<1-|z|^2\\ \\ \therefore 1-\sum_{i=1}^{\infty }|z|^{i}> 1-\sum_{i=1}^{\infty }(1-|z|^2)^{i}\\ \\ taking~the~limiting~sum.\\ \\ 1> \frac{1-|z|^2}{|z|^2}> 0\\ \\ \therefore 1-\sum_{i=1}^{\infty }(1-|z|^2)^{i}> 0\\ \\ \therefore ~all~inequalities~are~positive\\ \\\therefore there~are~no~solutions~when~|z|< \frac{\sqrt{5}-1}{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?for~|z|<r<1\\ \\ |z|<1-|z|^2\\ \\ \therefore 1-\sum_{i=1}^{\infty }|z|^{i}> 1-\sum_{i=1}^{\infty }(1-|z|^2)^{i}\\ \\ taking~the~limiting~sum.\\ \\ 1> \frac{1-|z|^2}{|z|^2}> 0\\ \\ \therefore 1-\sum_{i=1}^{\infty }(1-|z|^2)^{i}> 0\\ \\ \therefore ~all~inequalities~are~positive\\ \\\therefore there~are~no~solutions~when~|z|< \frac{\sqrt{5}-1}{2}" title="for~|z|<r<1\\ \\ |z|<1-|z|^2\\ \\ \therefore 1-\sum_{i=1}^{\infty }|z|^{i}> 1-\sum_{i=1}^{\infty }(1-|z|^2)^{i}\\ \\ taking~the~limiting~sum.\\ \\ 1> \frac{1-|z|^2}{|z|^2}> 0\\ \\ \therefore 1-\sum_{i=1}^{\infty }(1-|z|^2)^{i}> 0\\ \\ \therefore ~all~inequalities~are~positive\\ \\\therefore there~are~no~solutions~when~|z|< \frac{\sqrt{5}-1}{2}" /></a>
finally..

 

[Carrotsticks]

Much better! However, what about the case when n_1 = 1...

 

[barbernator]

why did we have to put the restriction n1 > 1 in the first place?

 

[Carrotsticks]

Because if n_k is negative for any integer k, then it's not a polynomial anymore..

 

[barbernator]

Because if n_k is negative for any integer k, then it's not a polynomial anymore..

i mean, why was it not for n1 > 0 instead? like why did you have to set the restriction n1 = > 2 ?

 

[Fus Ro Dah]

<a href="http://www.codecogs.com/eqnedit.php?latex=for~|z|<r<1\\ \\ |z|<1-|z|^2\\ \\ \therefore 1-\sum_{i=1}^{\infty }|z|^{i}> 1-\sum_{i=1}^{\infty }(1-|z|^2)^{i}\\ \\ taking~the~limiting~sum.\\ \\ 1> \frac{1-|z|^2}{|z|^2}> 0\\ \\ \therefore 1-\sum_{i=1}^{\infty }(1-|z|^2)^{i}> 0\\ \\ \therefore ~all~inequalities~are~positive\\ \\\therefore there~are~no~solutions~when~|z|< \frac{\sqrt{5}-1}{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?for~|z|<r<1\\ \\ |z|<1-|z|^2\\ \\ \therefore 1-\sum_{i=1}^{\infty }|z|^{i}> 1-\sum_{i=1}^{\infty }(1-|z|^2)^{i}\\ \\ taking~the~limiting~sum.\\ \\ 1> \frac{1-|z|^2}{|z|^2}> 0\\ \\ \therefore 1-\sum_{i=1}^{\infty }(1-|z|^2)^{i}> 0\\ \\ \therefore ~all~inequalities~are~positive\\ \\\therefore there~are~no~solutions~when~|z|< \frac{\sqrt{5}-1}{2}" title="for~|z|<r<1\\ \\ |z|<1-|z|^2\\ \\ \therefore 1-\sum_{i=1}^{\infty }|z|^{i}> 1-\sum_{i=1}^{\infty }(1-|z|^2)^{i}\\ \\ taking~the~limiting~sum.\\ \\ 1> \frac{1-|z|^2}{|z|^2}> 0\\ \\ \therefore 1-\sum_{i=1}^{\infty }(1-|z|^2)^{i}> 0\\ \\ \therefore ~all~inequalities~are~positive\\ \\\therefore there~are~no~solutions~when~|z|< \frac{\sqrt{5}-1}{2}" /></a>
finally..

Your solution is essentially correct, but I think it's not explained very well. For example when using the Limiting Sum, you could have added that we can use it since |z|<1 rather than jumping directly into it. Also I think proofs should be such that even the average student could understand it. I see you have proven that P(z)>0 but you didn't really say why so somebody reading this having basic knowledge of Polynomials would be lost. Perhaps you could have added something like 'To prove that P(z) has no solutions with , we must prove that for all n_k where k E N and for all z E C, P(z)>0 and therefore has no solutions'.

i mean, why was it not for n1 > 0 instead?

I don't understand what you're asking?

 

[barbernator]

Your solution is essentially correct, but I think it's not explained very well. For example when using the Limiting Sum, you could have added that we can use it since |z|<1 rather than jumping directly into it. Also I think proofs should be such that even the average student could understand it. I see you have proven that P(z)>0 but you didn't really say why so somebody reading this having basic knowledge of Polynomials would be lost. Perhaps you could have added something like 'To prove that P(z) has no solutions with , we must prove that for all n_k where k E N and for all z E C, P(z)>0 and therefore has no solutions'.



I don't understand what you're asking?

yeh i would have made my solution more proper, i just cbfd latexing it lol.

well I assumed that the proof would have shown for all n1 > 0, yet carrot had said that I also need to consider the case when n1 = 1, but i don't see why we have to consider that case separately in the proof, and why he implemented the restriction n1 => 2 in the first place

 

[Fus Ro Dah]

yeh i would have made my solution more proper, i just cbfd latexing it lol.

well I assumed that the proof would have shown for all n1 > 0, yet carrot had said that I also need to consider the case when n1 = 1, but i don't see why we have to consider that case separately in the proof, and why he implemented the restriction n1 => 2 in the first place

Think about where comes from. What degree is the polynomial from which it came?

Also I'm not really sure if this disc is the maximal disc of which the solutions exist outside.

 

[barbernator]

Think about where comes from. What degree is the polynomial from which it came?

Also I'm not really sure if this disc is the maximal disc of which the solutions exist outside.

it comes from the +ve quadratic solution. So how do we prove for the case where n1=1 then?

 

[Carrotsticks]

Here's my solution for n_1 greater than or equal to 2 (without full explanation atm)



EDIT: Typo, one of my bottom summation limits was 1 instead of 2 and one of the upper limits was k instead of infinity.