Timeline for A generalization of intermediate value theorem on R^k
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 17, 2013 at 8:19 | vote | accept | tckwok | ||
| Jan 17, 2013 at 8:12 | vote | accept | tckwok | ||
| Jan 17, 2013 at 8:16 | |||||
| Jan 17, 2013 at 4:36 | comment | added | domotorp | Yes, clear now, thx. Is anything known about how many dividing points are needed for the inverse integer generalization in higher dims? | |
| Jan 16, 2013 at 19:37 | comment | added | Sergei Ivanov | Everything in the proof is translation-invariant, there is no way terms like $r(b)+r(a)$ can appear. What he proves is that you can subdivide $[a,b]$ into $n$ intervals $[y_i,y_{i+1}]$ and equip them with signs $\varepsilon_i=\pm 1$ so that $\sum \varepsilon_i (r(y_{i+1})-r(y_i))=0$. This means that the sum of the terms equipped with pluses equals the sum of the terms equipped with minuses. But the total sum of all differences equals $f(b)-f(a)$. So the sum of "positive" terms equals $(f(b)-f(a))/2$. And by changing sign you may assume that there are at most $n/2$ of "positive" terms. | |
| Jan 16, 2013 at 18:51 | comment | added | domotorp | I knew the k=2 result for functions (ie. f(x)=(x,g(x)), but not this one. More sadly, I cannot even understand it. Why can it not happen that the sum of the segments is 1/2(r(b)+r(a))? If x_1 and x_{n+1} have opposite signs, this seems to be the case. | |
| Jan 16, 2013 at 17:39 | history | answered | Sergei Ivanov | CC BY-SA 3.0 |