What is the expansion of $\log(N+x) = \log(N) + [\dots\text{blank}\dots] $? ($N \in \mathbb{R}+$ and $0 \leq x \leq 1)$.

Question

I'm working on a math problem which might be solvable if I can re-express $\log(N+x)$ as $\log(N) +$ 'something.

The problem I am having with the Taylor series expansion about $x=0$ is that it carries infinitely higher powers of $N$ in the terms of the expansion, see here

Do you know how I might expand $\log(N+x)$ as $\log(N) +$ something? Any advice/comments/suggestions would really go a long way as I'm quite stuck.

Thanks for taking the time to consider this.

-McMath.

Answer 1

I do not know if this is helpful for you in your situation but you could write

$$\log(x+y)=\log(x)+\log\left(1+\frac yx\right)$$

Answer 2

Well, if you want $\log(x + y) = log x + K$ then the only way to do that (assuming $x > 0$) is

$K = \log(x+y) - \log x = \log (\frac {x+y}x) = \log (1+ \frac yx)$.

or to put it another way:

$\log(x +y) = \log(x(1 + \frac yx)) = \log x + \log(1 + \frac yx)$.

or to put it a third wy:

$\log x + K = \log x*m$ where $\log m = K$. And $x*m = x +y$. So $K = \log (1+\frac yx)$.