Timeline for Convergence as a function of error for the following function
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 22, 2022 at 17:46 | comment | added | Iosif Pinelis | @CfourPiO : (i) We bound $\sum_{q=1}^{n-1} a_q q$ by $\sum_{q=1}^\infty a_q q$ because the latter sum is easier to deal with (see what is done with it next); also, for large $n$ the sum $\sum_{q=1}^{n-1} a_q q$ will be close to $\sum_{q=1}^\infty a_q q$. (ii) $\epsilon_n$ takes only countably many values by the definition of a countable set (see e.g. "there exists a surjective function from ${\displaystyle \mathbb {N} }$ to ${\displaystyle S}$" in the section at en.wikipedia.org/wiki/Countable_set#Definition . | |
| Dec 22, 2022 at 10:13 | vote | accept | CfourPiO | ||
| Dec 22, 2022 at 10:12 | comment | added | CfourPiO | I also want to understand why $\epsilon_{n}$ can only take countably many values. I am not a mathematician, so my understanding can be really basic. From your approach, all I can understand is that probably the two terms involved in the original equation behave differently as $n$ increases so you introduced $\epsilon_{1n}$ and $\epsilon_{2n}$. Then, you dealt them separately to find convergence of each and then took a maximum? I think this is what I was looking for. When I said $\epsilon$, it is a common non mathematician way of saying convergence. I am happy I am learning new things here. :) | |
| Dec 22, 2022 at 10:04 | comment | added | CfourPiO | I want to understand why we use certain inequalities sometimes to find easier ways to show convergence. For example, when we say $\epsilon_{2n} = \frac{1}{n}\sum_{q=1}^{n-1} a_q q \leq \frac{1}{n}\sum_{q=1}^{\infty} a_q q$ . Logically it makes sense. However, I want to understand why it is used in this context. It could be any number less than $\infty$ and bigger than $n-1$. How do we acknowledge the error we are introducing by such inequalities. | |
| Dec 21, 2022 at 16:43 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 80 characters in body
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| Dec 21, 2022 at 15:25 | comment | added | Iosif Pinelis | @CfourPiO : (i) I meant $n$ instead of $N$ in terms of notation. (ii) As for "poorly stated", this is explained in the answer as follows: "Clearly, $\epsilon_n$ takes only countably many values; so, the equality $\epsilon_n=\epsilon$ can hold only for countably many values of $\epsilon$. Also, a closed-form expression for $\epsilon_n$ is not available. So, solutions of the equation $\epsilon_n=\epsilon$ for $n$ are not available in closed form, even when such solutions exist." | |
| Dec 21, 2022 at 15:21 | comment | added | CfourPiO | Thank you very much indeed for the detailed answer. I will look into it line by line. When you said $n$ will do instead of $N$, do you mean only in terms of notation, or something else? And, when you said the question is stated poorly, do you mean the definition of the error stated poorly? I want to understand. | |
| Dec 21, 2022 at 15:17 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |