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Post by Admin on Jun 1, 2023 8:04:16 GMT
Mondo
We have chosen our circle to have the radius $1+|A|+...+|E|$ and so if we add 1 inside the brackets the inequality still holds and then we can substitute and get $|z|^n$.
Vasco
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Post by Admin on Jun 1, 2023 8:07:45 GMT
Mondo
The whole idea in this problem is to get a radius big enough for the inequality to hold.
Vasco
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Post by mondo on Jun 1, 2023 8:15:26 GMT
Yes I see the $|z|>1$ need now. However I still can't agree with the last inequality in your document. You seem to claim the left hand side of it is $g(z)$ which it is not right? Also note that you multiply every coefficient by $z^{n-1}$ which makes it a completely different sum.
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Post by Admin on Jun 1, 2023 8:24:27 GMT
Yes I see the $|z|>1$ need now. However I still can't agree with the last inequality in your document. You seem to claim the left hand side of it is $g(z)$ which it is not right? Also note that you multiply every coefficient by $z^{n-1}$ which makes it a completely different sum. Mondo But we are dealing with inequalities here if $x<y$ then if $A>1$ it follows that $x<Ay$ or $x<Ay+B$ where $B>0$ Vasco
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Post by Admin on Jun 1, 2023 8:28:52 GMT
Mondo
I am not claiming that the left hand side is $|g(z)|$, just that it is greater than $|g(z)|$.
Vasco
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Post by Admin on Jun 1, 2023 8:35:29 GMT
Mondo
I will add some more explanation to my answer later today and repost it.
Vasco
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Post by Admin on Jun 1, 2023 11:01:03 GMT
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Post by Admin on Jun 1, 2023 18:47:35 GMT
Mondo
I have made a small change to the penultimate line in my answer to make it easier to follow. Use the link in post #21 to get access to it.
Vasco
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Post by mondo on Jun 1, 2023 21:25:56 GMT
Thank you for an update Vasco. Everything is fine until the line $|g(z)| < (|A| + |B| + .. + |E|)|z|^{n-1} < |z||z|^{n-1} = |z|^{n} = |f(z)|$. The part $|g(z)| < (|A| + |B| + .. + |E|)|z|^{n-1}$ is OK - higher powers must correspond to higher module + we do a sum of modules vs module of sum. So the right hand side must be bigger. However I can't get how can you say this must be less than $|z|^{n}$. This line really claims $(|A| + |B| + .. + |E|)|z|^{n-1} < |z|^{n}$ - even intuitively I can't grasp it we have a sum of $n$ number of $|z|^{n-1}$ and you claim it is smaller than a single power $|z|^{n}$. Using a real example: $A|(1+1i)^2| + B|(1+1i)^{2}| + C|(1+1i)^{2}| < |(1+1i)|^{3}$ and this is not true.
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Post by Admin on Jun 1, 2023 22:09:07 GMT
Mondo
Why do you say that the final line in your post above is not true? Since $|(1+1i)|^2=2$ then we can rewrite this as $|A|+|B|+|C|<\sqrt{2}$. Whether this holds or not depends on the values of $|A|,|B|,|C|$
Vasco
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Post by Admin on Jun 1, 2023 22:21:33 GMT
Mondo
The result in the exercise, $|g(z)|<|f(z)|$, only holds for values of $|z|$ on the circle $C$ not for all values of $|z|$.
Vasco
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Post by mondo on Jun 1, 2023 22:23:21 GMT
Mondo Why do you say that the final line in your post above is not true? Since $|(1+1i)|^2=2$ then we can rewrite this as $|A|+|B|+|C|<\sqrt{2}$. Whether this holds or not depends on the values of $|A|,|B|,|C|$ Vasco Vasco, why $|A|+|B|+|C|<\sqrt{2}$? I would say it can be rewritten into $(A+B+C)2<|1+1i|^3 = 2^{3/2}$
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Post by Admin on Jun 1, 2023 22:25:25 GMT
Mondo Why do you say that the final line in your post above is not true? Since $|(1+1i)|^2=2$ then we can rewrite this as $|A|+|B|+|C|<\sqrt{2}$. Whether this holds or not depends on the values of $|A|,|B|,|C|$ Vasco Vasco, why $|A|+|B|+|C|<\sqrt{2}$? I would say it can be rewritten into $(A+B+C)2<|1+1i|^3 = 2^{3/2}$ Mondo That's the same thing. Vasco
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Post by mondo on Jun 1, 2023 23:03:00 GMT
I looked into your updated document and I think I got it. Previously I haven't realized that $|z||z|^{n-1}$ in the final inequality is in fact $|1 + A + B + .. + E||z|^{n-1}$, now it makes sense. Thanks a lot vasco! However Vasco, why $|A|+|B|+|C|<\sqrt{2}$? I would say it can be rewritten into $(A+B+C)2<|1+1i|^3 = 2^{3/2}$ Mondo That's the same thing. Vasco It's not the sane right? In your post you haven't multiplied by $2$ on the left hand side so we were comparing unknown coefficients only. But it is not important as my example has a flaw, as I said I haven't realized $|z|$ is in fact a sum of coefficients. Maybe the choice of variables by author wasn't very fortunate here.
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Post by Admin on Jun 2, 2023 5:10:10 GMT
Mondo
My inequality is exactly the same as yours. I have just divided each side by 2 and since $2^{3/2}=2\sqrt{2}$ this gives my result.
Vasco
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