Given
equation 1: $$E = \sum_{k=1}^N \tau x_k g(\frac{n_k}{\tau}) + \sum_{k=1}^N n_kh(\frac{n_k}{\tau})$$
equation 2: $$E = \frac{1}{2}\gamma X^2 + \epsilon \sum_{k=1}^N |n_k| +\frac{\eta*}{\tau}\sum_{k=1}^N n_k^2 $$ where $$\eta* = \eta - \frac{1}{2}\gamma\tau$$
I am told that $n_k = \frac{X}{N}$ and $x_k = (N-k)\frac{X}{N}$ and $k = 1,...,N$
and that from equation 1 and 2 and the above definitions
point 1: $$E = \frac{1}{2}XTg(\frac{X}{T})(1-\frac{1}{N}) + Xh(\frac{X}{T})$$
point 2: $$E = \frac{1}{2}\gamma X^2+\epsilon X + (\eta - \frac{1}{2}\gamma\tau) \frac{X^2}{T }$$
note here $\tau$ is a time variablce
Can you show how to get to point 1 and point 2 from equations 1 and 2?
Thank you
I'm assuming that $T:=\tau N$.
Since $n_k$ is free of $k$, you are summing $N$ copies of the same thing in the second summation of equation 1, yielding $$N\cdot{\frac XN}\cdot h\left(\frac X{\tau N}\right)=X h\left(\frac XT\right).$$
As for the first summation in equation 1, we can substitute $x_k:=(N-k)\frac XN$ and $n_k:=\frac XN$, then pull out everything that is constant, obtaining $$ \tau\cdot\frac XN\cdot g\left(\frac X{\tau N}\right)\sum_{k=1}^N(N-k).$$ What remains in the summation can be written out (in reversed order) as $$0+1+2+\cdots+(N-1)={(N-1)N\over 2},$$ using the formula for the sum of the first $n$ natural numbers. Putting everything together and simplifying, you'll get point 1.
For equation 2, keep in mind that $n_k$ is free of $k$ so you're summing $N$ copies of the same thing in each of the summations.