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Post by mondo on Aug 29, 2022 5:35:57 GMT
In the middle of page 65 we got power series formula for $\frac{1}{1-x^2}$ after the decomposition of partial fractions and application of above equation (4). Now, author says the interval of convergence is "|X| < |1-k| \wedge |X| < |1+k|", I am not sure if I understanding why. The sum in the equation can be simplified to $[\frac{X^J}{X^(j+1)} - \frac{X^J}{X^j+1}]$ for the case when $|X| < |j-k|$. So after the simplification we see that the sum will be $\Sigma_{j=0}^{\inf}\frac{1}{X}$ which is divergent. Is that the reason why we require $|X| < |1-k|$?
Next why we also have $X < |1+k|$? Is that to accommodate powers of negative and positive numbers in the denominator?
The fact that the radius of convergence is the distance to the nearest singularity is just an observation after we calculated $R$ and not the other way around, meaning we got $R$ from this distance right?
Thank you.
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Post by Admin on Aug 29, 2022 6:40:16 GMT
Mondo
What is that interval of convergence in your post above? Something is wrong there.
Vasco
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Post by Admin on Aug 29, 2022 6:57:23 GMT
Mondo
Needham has shown on page 65 using partial fractions that the series for $G$ can be written as the sum of two series like (3) on page 64.
The two convergence criteria for the two series on page 65 come from the convergence criteria given in (3) on page 64 for the series for $\displaystyle\frac{1}{1-x}$, which is $-1<x<1$ or $|x|<1$.
Both of these series must converge for the series for $G$ to converge and the interval is therefore:
$R=\text{min\{}|1-k|,|1+k|\}$
$|X|<|1-k|$ if the first series is to converge and $|X|<|1+k|$ if the second series is to converge.
See figure 13a at the top of page 66.
Vasco
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Post by mondo on Aug 30, 2022 6:21:06 GMT
Mondo The two convergence criteria for the two series on page 65 come from the convergence criteria given in (3) on page 64 for the series for $\displaystyle\frac{1}{1-x}$, which is $-1<x<1$ or $|x|<1$. Vasco, I think I got it although I think the convergence criteria of (3) cannot be directly applied to (4) as they are different series - in (3) $a_n = 1$ while in (4) $a_n = \frac{1}{(a-k)^{j+1}}$. However assuming $s = (1-k)$ a simple ratio test shows $ a_n = \frac{1}{s^(j+1)}$ hence $= \frac{s^(j+1)}{s^(j+2)} = \frac{1}{s}$ hence the ratio of convergence is $R = |s|$ so substituting original value it gives $R = |1-k|$, as in the book. |
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Post by Admin on Aug 30, 2022 14:04:04 GMT
Mondo The two convergence criteria for the two series on page 65 come from the convergence criteria given in (3) on page 64 for the series for $\displaystyle\frac{1}{1-x}$, which is $-1<x<1$ or $|x|<1$. Vasco, I think I got it although I think the convergence criteria of (3) cannot be directly applied to (4) as they are different series - in (3) $a_n = 1$ while in (4) $a_n = \frac{1}{(a-k)^{j+1}}$. However assuming $s = (1-k)$ a simple ratio test shows $ a_n = \frac{1}{s^(j+1)}$ hence $= \frac{s^(j+1)}{s^(j+2)} = \frac{1}{s}$ hence the ratio of convergence is $R = |s|$ so substituting original value it gives $R = |1-k|$, as in the book. Mondo The convergence test can be applied to (4) because (4) is (3) with $\bigg(\frac{X}{a-k}\bigg)$ substituted for $x$ and so instead of writing $|x|<1$ we can write $\bigg|\bigg(\frac{X}{a-k}\bigg)\bigg|<1$ or $|X|<|a-k|$ as in (4). Doing it with the ratio test as you do takes away all the geometric meaning which Needham is trying to show using figures 12, 13 and 14. The ratio test is just a recipe and doesn't give us the insight which Needham is showing us on pages 64-67. I assume you meant radius of convergence not ratio of convergence as you wrote in your post. Vasco |
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Post by mondo on Aug 31, 2022 17:18:20 GMT
Vasco, I think I got it although I think the convergence criteria of (3) cannot be directly applied to (4) as they are different series - in (3) $a_n = 1$ while in (4) $a_n = \frac{1}{(a-k)^{j+1}}$. However assuming $s = (1-k)$ a simple ratio test shows $ a_n = \frac{1}{s^(j+1)}$ hence $= \frac{s^(j+1)}{s^(j+2)} = \frac{1}{s}$ hence the ratio of convergence is $R = |s|$ so substituting original value it gives $R = |1-k|$, as in the book. Mondo The convergence test can be applied to (4) because (4) is (3) with $\bigg(\frac{X}{a-k}\bigg)$ substituted for $x$ and so instead of writing $|x|<1$ we can write $\bigg|\bigg(\frac{X}{a-k}\bigg)\bigg|<1$ or $|X|<|a-k|$ as in (4). Doing it with the ratio test as you do takes away all the geometric meaning which Needham is trying to show using figures 12, 13 and 14. The ratio test is just a recipe and doesn't give us the insight which Needham is showing us on pages 64-67. Absolutely thank you. I assume you meant radius of convergence not ratio of convergence as you wrote in your post. Yes, my mistake. I meant radius of convergence as you said. Ok, I think it is clear. One extra think I noticed is that aforementioned ratio test is also briefly discussed in this book few pages later however it is defined slightly different than in other books (especially those on real analysis or wiki page here en.wikipedia.org/wiki/Ratio_test ). $R = \lim_{n\to\infty} | \frac{c_n}{c_{n+1}}|$ in the book while often times it is defined as $R = \lim_{n\to\infty} | \frac{c_{n+1}}{c_{n}}|$. Hence we need to be careful with $R$ interpretation because depending how we calculated it the interpretation will be different! |
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Post by Admin on Aug 31, 2022 17:44:48 GMT
Mondo
The reason for the difference is that you are comparing two different things:
1. Convergence test
2. Formula for the calculation of the radius of convergence
Also notice that the book is discussing Power Series not all types of series.
Vasco
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Post by mondo on Aug 31, 2022 19:46:53 GMT
Mondo The reason for the difference is that you are comparing two different things: 1. Convergence test 2. Formula for the calculation of the radius of convergence Also notice that the book is discussing Power Series not all types of series. Vasco I don't think I compare two different things. On the bottom of page 74 we can read "The ratio test says that $R = \lim_{n\to\infty} | \frac{c_n}{c_{n+1}}|$" while in the wiki page the same ratio test is defined differently as $R = \lim_{n\to\infty} | \frac{c_{n+1}}{c_{n}}|$ (https://en.wikipedia.org/wiki/Ratio_test). However the criteria of convergence are different - in the book $R = \infty$ means coonvergence everywhere while on the wiki it means the series diverges. I am not saying there is anything wrong with it, I am only saying that we need to be careful how we interpret the results (This is mainly a note for myself)  |
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Post by Admin on Aug 31, 2022 19:57:11 GMT
Mondo
In the book on page 74 $R$ is the radius of convergence not the ratio used to test for convergence, whereas on the Wikipedia page $R$ is defined as the ratio used to test for convergence. The same letter $R$ is used to mean 2 different things.
Vasco
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Post by mondo on Aug 31, 2022 22:12:54 GMT
Ok I see. Thank you.
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