Can someone help me to expand the below equation for N=2 and N=3?
$f_i=\sum \{e_i:j\leq i \leq k\}$ for some $1\leq j\leq k\leq N$.
Here $e_i$ is the unit vector of coordinate i in an N-dimensional space.
Can someone help me to expand the below equation for N=2 and N=3?
$f_i=\sum \{e_i:j\leq i \leq k\}$ for some $1\leq j\leq k\leq N$.
Here $e_i$ is the unit vector of coordinate i in an N-dimensional space.
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Are you just asking to do us your homework?
Because the solution of your quest only relies on the observation that the LHS already settles the index $i$, so on the RHS it is fixed as well. Thus $f_i$ clearly becomes parallel to $e_i$ always, and then the only searched for proportional factor would simply count the set of those possible intervals $j\le k$, which contain the chosen $i$.
That is, for $N=2$ you get $f_1=2e_1$, as you could use the borders of your interval to be $j=k=1$ or $j=1$, $k=2$. By symmetry you'd get $f_2 = 2e_2$ likewise. While for $N=3$ you get $f_1=3e_1$ (using $j=k=1$ or $j=1$, $k=2$ or $j=1, k=3$) and by symmetry $f_3=3e_3$ again, whereas $f_2=4e_2$ because of the now possible border choices $j=1, k=2$ or $j=1, k=3$ or $j=k=2$ or $j=2, k=3$.
Or have I somehow misunderstood your quest? Esp. I don't get why you tagged it with #polytopes.
--- rk