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Post by salzburger99 on Sept 12, 2018 1:04:52 GMT
"If P(a) converges then the length |Cnan| of each term must die away to zero as n goes to infinity [why?]" Why?
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Gary
GaryVasco
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Post by Gary on Sept 12, 2018 4:56:05 GMT
"If P(a) converges then the length |C na n| of each term must die away to zero as n goes to infinity [why?]" Why? I can only offer intuition here. Suppose that $|c_n a^n|$ were to be constant or increasing. Then as $n \rightarrow \infty$, $\tilde{P(z)}$ would increase indefinitely and eventually overshoot $A$ and probably escape the disc of convergence. So it can only converge to A by ever smaller increments. |
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Post by Admin on Sept 12, 2018 11:07:14 GMT
Hi
I agree with what Gary has written above, but I must say that I think the way this is written in the book could be very confusing.
When I read it, it seems to me that it could be interpreted as saying that each separate term of the series individually tends to zero, which would mean that the whole series, as a sum of terms all equal to zero, would tend to zero. This would not make sense.
The only interpretation that makes sense is Gary's interpretation. I have found similar confusing statements in other textbooks in the sections on power series.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 12, 2018 14:49:03 GMT
Hi I agree with what Gary has written above, but I must say that I think the way this is written in the book could be very confusing. When I read it, it seems to me that it could be interpreted as saying that each separate term of the series individually tends to zero, which would mean that the whole series, as a sum of terms all equal to zero, would tend to zero. This would not make sense. The only interpretation that makes sense is Gary's interpretation. I have found similar confusing statements in other textbooks in the sections on power series. Vasco It did not occur to me that the word "each" could easily be read as "every". It would be better to write "each successive term".
Gary
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Post by Admin on Sept 13, 2018 13:58:06 GMT
Hi salzburger99
Do these replies help you to understand why the sequence of terms must tend to zero as $n$ tends to $\infty$?
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 13, 2018 16:35:26 GMT
I would like to amend the third sentence of my reply to read:
"Then as $n \rightarrow \infty$, $P(z)$ would proceed (or oscillate) indefinitely and eventually overshoot $A$ and probably escape the disc of convergence."
The important change is from $\tilde{P(z)}$ to $P(z)$.
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Post by Admin on Sept 14, 2018 21:08:13 GMT
Hi
On page 68 at the top we have the definition of convergence:
if $n>m>N$ and $P(z)$ converges then the inequality (6), on the same page, must hold. Now take the case when $n=m+1$, and we obtain, using (6)
$|c_na^n|<2\epsilon$ for $n>N$. Done.
Vasco
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