logarithmspercentages
I'm trying to understand the formula to move from log points to percentage points. I know the same question has already been asked here: How to interpret the difference in log points
and I can follow PaulB's answer easily until the taylor expansion, is the last step that I have troubles understanding. Could anyone please help me claryfying that? It seems to me like the "-1" should be part of the log, but it's clearly not correct.
Any help would be highly appreciated
Taking logarithms can be useful because of the fact that if $x$ and $y$ are positive real numbers, then $x<y$ if and only if $\log x<\log y$, and sometimes the logarithms are easier to compare. For example, basic properties of the logarithm tell us that $\log n^2 = 2\log n$ and $\log 2^n=n\log 2$. Now look at a graph of $y=2\log x$: it rises as $x$ increases, but it also gets flatter and flatter, meaning that it’s rising more and more slowly. What about the graph of $y=(\log 2)x$? Does it get flatter as it rises, or does it keep rising at a constant slope? Which one will eventually get on top and stay there? In fact, it won’t just stay on top: it’ll keep getting further and further above the other one, so it must be growing faster.
That should help you with the comparison of $n^2$ and $2^n$. The comparison of $n^2$ with $n^2\log n$ is easier. Just look at the ratio of the two, $$\frac{n^2\log n}{n^2}=\log n\;.$$ As $n$ increases, what happens to that ratio? Does it get smaller, approach some limiting value, or get bigger? If it gets bigger, you know that the numerator, $n^2\log n$, must be growing faster than the denominator, $n^2$. If it’s roughly constant, they must be growing at about the same rate. And so on.
That leaves the comparison between $n^2\log n$ and $2^n$. Logarithms can again help here: $\log(n^2\log n)=\log n^2+\log(\log n)=2\log n+\log(\log n)$, again using basic properties of the logarithm. How does that compare with $n\log 2$, the logarithm of $2^n$, when $n$ is very large?
The way we define the Logarithm is the inverse of the exponential.
If $z= re^{i\theta}$,then $ln(z)=ln(r)+i\theta$.
Now , a problem that is there is, with this equation alone , logarithm is multi-valued i.e for every z there are infinitely many values for the logarithm . For $\theta$ differing by a value of $2\pi$ , one will get the same value. It isn't injective. This leads us restricting its domain so that each point(z) corresponds to only one value. This is referred to as a branch cut. Check the image from Wiki,
So, for log(z), simply remove any ray joining 0 and infinity.(Restricting complete rotation(0 to $2\pi$!). Also, see in the graph what happens. The principle log removes one particular ray, the negative real axis, I guess. This is only a convention, I guess.
Now, coming back to your question, you want log(iz)(PS:-The $iz$ only changes the labels on the axes) to be analytic in some given region. All you need to do is to remove a ray(not in the given region) to make it analytic in the given region.
Best Answer
We have that when x is "small" (let try with calculator)
$$\log(1+x)\approx x$$
therefore for small pecentage $\frac{\%\Delta}{100}$
$$\log\left(1+\frac{\%\Delta}{100}\right) \approx \frac{\%\Delta}{100}$$
For the last step, by exponentiation
$$\log(x)-\text{log}(y)=\log\left(\frac{\%\Delta}{100}+1\right)$$
$$e^{\log(x)-\text{log}(y)}=e^{\log\left(\frac{\%\Delta}{100}+1\right)}=\frac{\%\Delta}{100}+1$$
$$\frac{\%\Delta}{100}=e^{\log(x)-\text{log}(y)}-1$$