# Natural Log Proof

1. Oct 22, 2016

### Nusc

1. The problem statement, all variables and given/known data
Prove the following statement:
$$ln|1+\sigma x | = \frac{1}{2} ln|1-x^2| + \frac{\sigma}{2} ln| \frac{ |1+x|}{|1-x|}$$
2. Relevant equations

3. The attempt at a solution
Starting from right to left would be easier:
$$= \frac{1}{2} ln|(1+x)(1-x)| + \frac{\sigma}{2} ln| 1+x| - \frac{\sigma}{2} ln(1-x) \\ = \frac{1}{2} [ ln(1+x) + ln(1-x) + \sigma ln(1+x) - \sigma ln(1-x)] \\ =\frac{1}{2} [ln|(1+x)(1+\sigma)| + ln|(1-x)(1-\sigma)|] \\ = \frac{1}{2} [ ln |(1+x)^{1+\sigma} (1-x)^{1-\sigma} |]\\ = \frac{1}{2} ln |( 1+x ) (1+x)^\sigma \frac{1+x}{(1-x)^\sigma}|]$$
Then I get stuck. Does anyone know what I'm missing?

2. Oct 23, 2016

### haruspex

The statement to be proved is clearly not true in general. Try σ=0. Are you sure you have stated the whole question, word for word?

3. Oct 23, 2016

### Nusc

Do you know of any identities that would give me the left hand side? Any references? I cannot find anything online. Calculus textbooks just give integrals and not identities for logs.

I need to take the natural log of the following expression.
$$\frac{1}{2}( C + D ) + \frac{1}{2}( C - D ) \sigma$$

The idea was to use that identity, then rearranging above
$$\frac{1}{2}( C + D ) [ 1 + \frac{(C-D)}{(C+D)} \sigma ]$$

$$ln2 + \frac{1}{2}ln|4CD|+\frac{\sigma }{2} ln|\frac{C}{D} |$$

Without that identity, It's pretty hard to achieve that result. Do you know of anything I can use?

What's really important is that I at least get this form: $$\frac{\sigma }{2} ln|\frac{C}{D} |$$

Last edited: Oct 23, 2016
4. Oct 23, 2016

### SammyS

Staff Emeritus
You appear to be a step or two away.

Never mind! I was looking at the wrong side of the equation.

Last edited: Oct 23, 2016
5. Oct 23, 2016

### Nusc

Last edited by a moderator: Oct 23, 2016
6. Oct 23, 2016

### Ray Vickson

Your desired result is false. For small $|x|$ we can expand both sides in a series in $x$. For ease of writing, I will use the symbol $a$ instead of $\sigma$. Let
$$L(x) = \ln (1+ ax) , \;\; R(x) = \frac{1}{2} \ln(1-x^2) + \frac{a}{2} \ln \left(\frac{1+x}{1-x}\right).$$
We have
$$L(x) = ax - \frac{1}{2} a^2 x^2 + \frac{1}{3} a^3 x^3 - \frac{1}{4} a^4 x^4 + O(x^5),$$
but
$$R(x) = ax - \frac{1}{2} x^2 + \frac{1}{3} a^3 x^3 - \frac{1}{4} x^4 + O(x^5).$$
Note that in $L(x), R(x)$ I have omitted the absolute-value signs; those signs are redundant when $|x| < 1$, which is true for small $|x|$.

7. Oct 23, 2016

### Nusc

Was there a typo? Is there a way to determine if there is one?

Or is there an easier way to simplify the natural log of the expression?

$$\frac{1}{2}( C + D ) + \frac{1}{2}( C - D ) \sigma$$

with obtaining
$$\frac{\sigma }{2} ln|\frac{C}{D} |$$ at least in the final answer?

$$ln [ \frac{1}{2} D ( \frac{C}{D} +1 ) + \frac{\sigma}{2} D ( \frac{C}{D} -1) ]$$

8. Oct 23, 2016

### haruspex

For it even to be approximately true you need to change the 1/2 factor in the first expression on the right to σ/2. That makes it more symmetric, so a lot more reasonable.
It still is not going to be generally true. The roles of the σ on the left and the σ's on the right are too different. I would guess you are only supposed to show this as asymptotic behaviour for small x. (It is equivalent to (1+x)n approximating enx.)

9. Oct 23, 2016

### haruspex

a3?

10. Oct 23, 2016

### Nusc

Is there a table of asymptotic expansions or reference I can refer to to get the desired result? I'm kind of stumped.

11. Oct 23, 2016

### haruspex

First step is to correct the RHS of the statement to be proved. See the first para of post #8.
Next, as Ray points out, on the assumption that |x|<1 and |σx|<1, all those modulus signs can be thrown away.
Then you can exponentiate both sides to get rid of the logarithms. Post what you get.