Calculer la somme suivante de symboles de Kronecker :∑i=15δi,3.\sum_{i=1}^{5} \delta_{i,3}.i=1∑5δi,3. I solved itMark it doneAdd to my listKeep it in your listI like this problem0 likesSolutions1Reveal solutionsAre you sure? Give it a try first.Solution by darktoasterDiscussions1 useful vote∑i=15δi,3=δ1,3+δ2,3+δ3,3+δ4,3+δ5,3=0+0+1+0+0=1\sum\limits_{i=1}^5 \delta_{i,3} = \delta_{1,3} + \delta_{2,3}+\delta_{3,3}+\delta_{4,3}+\delta_{5,3} = 0 + 0 + 1 + 0 + 0 = 1i=1∑5δi,3=δ1,3+δ2,3+δ3,3+δ4,3+δ5,3=0+0+1+0+0=1\newlineEn effet par définition du symbole de Kronecker : δi,j={1, si i=j0, si i≠j\delta_{i,j} = \begin{cases} 1 \text{, si } i = j \\ 0 \text{, si } i \ne j \end{cases}δi,j={1, si i=j0, si i=jEn particulier ∀i∈[ [1,5] ],δi,3={1, si i=30, si i≠3\forall i \in [\![1,5]\!], \delta_{i,3} = \begin{cases} 1 \text{, si } i = 3 \\ 0 \text{, si } i \ne 3 \end{cases}∀i∈[[1,5]],δi,3={1, si i=30, si i=3 ReportFor an unclear, ambiguous, or possibly incorrect statement, please use the Discussion tab on the right. Report content that needs moderator intervention, such as dangerous, clearly non-mathematical, or plagiarized content.Submit
Solution by darktoasterDiscussions1 useful vote∑i=15δi,3=δ1,3+δ2,3+δ3,3+δ4,3+δ5,3=0+0+1+0+0=1\sum\limits_{i=1}^5 \delta_{i,3} = \delta_{1,3} + \delta_{2,3}+\delta_{3,3}+\delta_{4,3}+\delta_{5,3} = 0 + 0 + 1 + 0 + 0 = 1i=1∑5δi,3=δ1,3+δ2,3+δ3,3+δ4,3+δ5,3=0+0+1+0+0=1\newlineEn effet par définition du symbole de Kronecker : δi,j={1, si i=j0, si i≠j\delta_{i,j} = \begin{cases} 1 \text{, si } i = j \\ 0 \text{, si } i \ne j \end{cases}δi,j={1, si i=j0, si i=jEn particulier ∀i∈[ [1,5] ],δi,3={1, si i=30, si i≠3\forall i \in [\![1,5]\!], \delta_{i,3} = \begin{cases} 1 \text{, si } i = 3 \\ 0 \text{, si } i \ne 3 \end{cases}∀i∈[[1,5]],δi,3={1, si i=30, si i=3