Benedict Irwin edited Other.tex  over 9 years ago

Commit id: 4beddf539e7a2ae963270a4879052bae09231fb1

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\section{ Other }  Define \begin{equation} a functional\begin{equation}  \hat{O}^{S_k}_{x_1,x_2}(f,g):=\sum_{k=\lceil \hat{O}^{>S_k<}_{x_1,x_2}(g,f):=\sum_{k=\lceil  \int_{x_1}^{x_2} g(x)dx \rceil}^{\lfloor \int_{x_1}^{x_2} f(x) dx \rfloor} S_k \end{equation}  Let $S_k=k$, let $x_1=1$ and $x_2=2$. Also let $f=x^2, g=x$. Then, \begin{equation}  \hat{O}^{>k<}_{1,2}(x,x^2)=\sum_{k=\lceil \int_{1}^{2} xdx \rceil}^{\lfloor \int_{1}^{2} x^2 dx \rfloor} k  \end{equation}  So,  \begin{equation}  \hat{O}^{>k<}_{1,2}(x,x^2)=\sum_{k=\lceil 3/2 \rceil}^{\lfloor 7/3 \rfloor} k = \sum_{k=2}^2 k = 2  \end{equation}  This could be the inward sum as opposed to the outward sum \begin{equation}  \hat{O}^{}_{x_1,x_2}(g,f):=\sum_{k=\lfloor \int_{x_1}^{x_2} g(x)dx \rfloor}^{\lceil \int_{x_1}^{x_2} f(x) dx \rceil} S_k  \end{equation}  Which has a subtle difference, but here would be \begin{equation}  \hat{O}^{}_{1,2}(x,x^2)=\sum_{k=\lfloor 3/2 \rfloor}^{\lceil 7/3 \rceil} k = \sum_{k=1}^3 k = 6  \end{equation}  Let $S_k=k$.