Timeline for Rudin PMA 4.31 - does the elements of $E$ have to be ordered (smallest to the biggest)?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Apr 12, 2023 at 21:06 | history | edited | Dunham | CC BY-SA 4.0 |
reference to comments that add explanation
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| Apr 12, 2023 at 17:56 | comment | added | krm2233 | $f(x)$ is the sum of the $c_n$ for which $x_n<x$. If $x<1/2$ then both of $x_1$ and $x_2$ are greater than $x$ so $f(x)=0$. If $1/2<x\leq 3/4$ then $x_1$ is not less than $x$ but $x_2$ is, so $f(x)=c_2=2$. If $x>3/4$ then it is bigger than both $x_1$ and $x_2$ so that $f(x) =c_1+c_2=3$. Hopefully saying it this way also helps show that the order the $x_n$s is irrelevant to Rudin's claim. | |
| Apr 12, 2023 at 17:38 | vote | accept | niobium | ||
| Apr 12, 2023 at 19:21 | |||||
| Apr 12, 2023 at 16:52 | comment | added | niobium | And $f(x)= c_1 = 1$ for $\frac{3}{4} < x$ | |
| Apr 12, 2023 at 16:51 | comment | added | niobium | Did you mean $f(x) = 3$ for $1/2< x < 3/4$ ? It seems that you should add all the terms of the series up to $n=2$ (i.e. $c_1 + c_2$) | |
| Apr 12, 2023 at 16:35 | history | answered | Dunham | CC BY-SA 4.0 |