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I have the following plot:

enter image description here

which is reflected by the given formula:

$$ \chi_{A_j}(\xi)=\begin{cases} 2, \ \ \ \ -2\le \xi<-1 \\ 1, \ \ \ \ -1\le \xi<0 \\ 2, \ \ \ \ 0\le \xi<1 \\ 3, \ \ \ \ 1\le \xi\le2 \end{cases} $$

However, when I try to write this formula in the form of:

\begin{equation} f(\xi):=\sum_{j=1}^n \eta_j\chi_{A_j}(\xi), \ \ \ \ \ \ \ \end{equation} where $j=1,2,3,4,5$ and $A_{j}$=$[\xi_{j-1}, \xi_{j}]$ and $\eta_j=[\eta_1,\eta_2,\eta_3]=[1,2,3]$

I get problems. $\eta_j$ in the sum formula does not correspond to $\eta_j$ in the plot.

How can I write the sum formula so that it reflects this function correctly?

Any ideas appreciated!

Thanks

Luthier415Hz
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    Your graph has four pieces but your sum of multiples of characteristic functions has five summands. Also the intervals $A_j$ should be half-open so as to avoid conflicts at join points. – coffeemath Apr 24 '23 at 12:36
  • Thanks, but as you can see in the formula $f(\xi)$, it has five end-points (red dots). Should I just add a $<2$ at the last interval instead of $\le 2$ and make it a red circle instead? – Luthier415Hz Apr 24 '23 at 12:39
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    If you use <2 and a circle above 2, you are altering the function in the graph and the explicit piecewise formula at the input x=2. – coffeemath Apr 24 '23 at 12:47
  • Ok, I don't know how to correct this, other than stating that the sequence of $\eta_j=[\eta_1, \eta_2, \eta_3, \eta_4]=[2,1,2,3]$, then all seems to fall in place. – Luthier415Hz Apr 24 '23 at 12:55
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    You can use the $\eta_i$ of your last comment, but your intervals have to be half-open [closed at left end, open at right] except the last and final which is closed. [Also the sum has 4 terms so upper limit on sum is 4.] – coffeemath Apr 24 '23 at 13:04
  • About your comment " but your intervals have to be half-open ][closed at left end, open at right] except the last and final which is closed. ", isn't that the case already? Thanks for the sum correction. – Luthier415Hz Apr 24 '23 at 13:06
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    When you wrote $A_j=[\eta_{j-1},\eta_j]$ those intervals are closed and intersect at endpoints. – coffeemath Apr 24 '23 at 13:10
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    Got it. Thanks for this conversation. – Luthier415Hz Apr 24 '23 at 13:12

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