Approximating the summation of the logarithm of a function

Question

I am working through a textbook for self-study and came across an approximation that I'm unsure how to solve $$ \sum_{n=1}^{N} \mathrm{ln}[A - (n-1)\epsilon ] = N \mathrm{ln}[A - (N-1)\epsilon/2] + \mathcal{O}(\epsilon^{2}), $$ where $A$ and $\epsilon$ are positive constants. We can assume $N\epsilon \ll A$. I tried using the Euler-Maclaurin sum formula but I can't get the approximation back. I'm starting to think there's something simple that I missed, perhaps with a geometric summation. Thoughts?

Answer 1

We can rewrite our sum as a product, using this equation from Wikipedia.

$$\prod_n{f(n)} = \exp\left( \sum_n{( \log{f(n)} )} \right) \tag{1}$$ $$\sum_n{( \log{f(n)} )} = \ln{ \left( \prod_n{f(n)} \right) } \tag{2}$$

So we get

$$\ln{ \left( \prod_n{ \ln[A-(n-1)\epsilon] } \right) } \tag{3}$$ $$\ln{ \Big( \ln[A-0\epsilon] \cdot \ln[A-1\epsilon]\cdots \ln[A-(N-1)\epsilon] \Big) } \tag{4}$$ $$\ln{ \Big( \ln \Big[(A-0\epsilon) + (A-1\epsilon) + \dots + (A-(N-1)\epsilon) \Big] \Big) } \tag{5}$$ $$\ln{ \left( \ln \left[ NA-\frac{N^2+N}{2}\epsilon - N\epsilon/2 \right] \right) } \tag{6}$$ $$\ln{ \left( \log{(N)}\epsilon/2 + \ln \left[ \frac{A}{\epsilon/2}-N \right] \right) } \tag{7}$$ $$\epsilon/2 + \ln{ \left( \log{(N)} + \ln \left[ \frac{A-N \epsilon/2}{(\epsilon/2)^2}\right] \right) } \tag{8}$$ $$\epsilon/2 + \ln{ \left( \log{(N)} + \ln \left[ A-N \epsilon/2\right] + O((\epsilon/2)^2)\right) } \tag{9}$$ $$O((\epsilon/2)^2) + \ln{ \left( \log{(N)} + \ln \left[ A-N \epsilon/2\right] \right) } \tag{9}$$ $$O((\epsilon/2)^2) + \ln{ \left( \ln \left[ (A-N \epsilon/2\right)^N] \right) } \tag{10}$$

Now we can reverse the operation we did between (2) and (3) to get

$$O((\epsilon/2)^2) + \sum_{n=1}^N{ \ln \left[ A-N \epsilon/2 \right] } \tag{11}$$ $$N \ln \left[ A-N \epsilon/2 \right] + O(\epsilon^2) \tag{12}$$

This wasn't obvious to me!

Answer 2

Here's some naive fiddling.

Writing $c$ for $\epsilon$ because lazy, using $=\ln(1-c)=-c+O(c^2)$, and assuming that $c \to 0$,

$\begin{array}\\ \sum_{n=1}^{N} \ln[A - (n-1)c ] &=\sum_{n=1}^{N} (\ln(A)+\ln[1 - (n-1)c/A ]\\ &=N\ln(A)+\sum_{n=1}^{N} \ln(1 - (n-1)c/A)\\ &=N\ln(A)-\sum_{n=1}^{N} ((n-1)c/A+O(c^2)))\\ &=N\ln(A)-N(N-1)c/(2A)+O(Nc^2) \qquad\text{assuming the big-oh is uniform}\\ &=N(\ln(A)-(N-1)c/(2A))+O(Nc^2)\\ &\approx N(\ln(A)+\ln(1-(N-1)c/(2A))+O(Nc^2)\\ &= N(\ln(A(1-(N-1)c/(2A)))+O(Nc^2)\\ &= N(\ln(A-(N-1)c/2))+O(Nc^2)\\ \end{array} $