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

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.

• $log(x+y)=log(x(1+\frac{y}{x}))=log(x)+log(1+\frac{y}{x})$, probably not what you want though – gd1035 Oct 3 '18 at 18:49

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

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

• In terms of practical usage, many programming environments offer a function log1p(), defined as $\mathrm{log1p}(x) = \log(1+x)$. This offers better accuracy if $x$ is near $0$. – njuffa Oct 3 '18 at 18:54
• Although I don't think the expansion you provided is helpful for my specific application, it does technically answer the question, and has practical merit as njuffa already said. I appreciate your answer. Best. – McMath Oct 3 '18 at 18:57
• @McMath Are you searching for approximation for large arguments $N$ or what exactly do you have in your mind? Maybe add your special context to your question to make things clear :) – mrtaurho Oct 3 '18 at 19:01

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 $$K = \log (1+\frac yx)$$.