# A question about the log of a rational function

• I
• mmzaj
In summary: This can be done by finding the roots of the denominator and then checking where the argument of the logarithm is equal to a multiple of ##2\pi##. This will give the locations of the jumps.
mmzaj
We have the rational function :
$$f(x)=\frac{(1+ix)^{n}-1}{(1-ix)^{n}-1}\left(\frac{1-ix}{1+ix}\right)^{n/2}\;\;\;,\;\;n\in \mathbb{Z}^{+}$$
It's not hard to prove that :
$$\frac{(1+ix)^{n}-1}{(1-ix)^{n}-1}=(-1)^{n}\prod_{k=1}^{n-1}\frac{x+i(\xi_{n}^{k}-1)}{x-i(\xi_{n}^{k}-1)}\;\;\;,\;\;\xi_{n}^{k}=e^{2\pi i k/n}$$
Now we want to compute $$\log f(x)$$ for x>0. The logarithm of the individual factors can be written as :

$$\log\left(\frac{x+i(\xi_{n}^{k}-1)}{x-i(\xi_{n}^{k}-1)}\right)=2i\tan^{-1}\left(\frac{x}{1-\xi_{n}^{k}}\right)+i\pi;\;\;\;\;x>0$$
So, one would expect:
$$\log f(x)=-in\tan^{-1}(x)-i\pi+2i\pi n+2i\sum_{k=1}^{n-1}\tan^{-1}\left(\frac{x}{1-\xi_{n}^{k}}\right)$$
But it looks nothing like what wolframalpha returns. What am i doing wrong here ?

Last edited:
It might be helpful to say what wolfram alpha returns.

graphically, there seems to be a difference between what I've calculated and the plot of ##\log f(x)## by multiples of ##2\pi## . but i am not able to locate the exact locations of the jumps.

You spent so much effort to type in post #1 and then you fail with some copies and pastes or links?
I hope micromass' crystal ball isn't out of order like mine currently is.

fresh_42 said:
You spent so much effort to type in post #1 and then you fail with some copies and pastes or links?
I hope micromass' crystal ball isn't out of order like mine currently is.
have you ever used WF ? it doesn't return results for general n ! and posting one example won't be of help if it doesn't say where the jumps are ! thanks for the very helpful and constructive post anyways !

If you refuse to give further information, then there is not much we can do. All I can say is that the complex logarithm is multivalued, so this can explain jumps of order ##2\pi##.

micromass said:
If you refuse to give further information, then there is not much we can do. All I can say is that the complex logarithm is multivalued, so this can explain jumps of order ##2\pi##.
where should i expect the jumps to happen ? that's where i am stuck. and we can just forget about the graphical discrepancy and correct my analytic calculation.

Why don't you show us the graphics you got? I won't reply further to this thread if you don't show us what you did in wolframalpha.

it boils down to finding the discontinuities of ##\log f(x)##

## 1. What is the definition of a rational function?

A rational function is a mathematical function that can be expressed as the quotient of two polynomial functions, where the denominator is not equal to zero.

## 2. How do I find the logarithm of a rational function?

To find the logarithm of a rational function, you can use the quotient rule for logarithms. This states that the logarithm of a quotient is equal to the difference of the logarithms of the numerator and denominator.

## 3. What is the purpose of taking the logarithm of a rational function?

Taking the logarithm of a rational function can help simplify the function or make it easier to graph. It can also be used to solve equations involving rational functions.

## 4. Can the logarithm of a rational function be negative?

Yes, the logarithm of a rational function can be negative. This can occur if the denominator is greater than the numerator, resulting in a fraction less than 1, which has a negative logarithm.

## 5. Are there any restrictions when taking the logarithm of a rational function?

Yes, when taking the logarithm of a rational function, the value of the function must be greater than zero. This is because the logarithm of a negative number or zero is undefined in the real number system.

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