Natural Log Proof

  • Thread starter Nusc
  • Start date
  • #1
753
2

Homework Statement


Prove the following statement:
[tex] ln|1+\sigma x | = \frac{1}{2} ln|1-x^2| + \frac{\sigma}{2} ln| \frac{ |1+x|}{|1-x|}
[/tex]

Homework Equations




The Attempt at a Solution


Starting from right to left would be easier:
[tex]
= \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}|]
[/tex]
Then I get stuck. Does anyone know what I'm missing?
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,163
6,781
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
753
2
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.
[tex] \frac{1}{2}( C + D ) + \frac{1}{2}( C - D ) \sigma
[/tex]

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

The answer is

[tex]
ln2 + \frac{1}{2}ln|4CD|+\frac{\sigma }{2} ln|\frac{C}{D} |
[/tex]

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: [tex] \frac{\sigma }{2} ln|\frac{C}{D} | [/tex]
 
Last edited:
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,365
1,033

Homework Statement


Prove the following statement:
[tex] ln|1+\sigma x | = \frac{1}{2} ln|1-x^2| + \frac{\sigma}{2} ln| \frac{ |1+x|}{|1-x|}
[/tex]

Homework Equations




The Attempt at a Solution


Starting from right to left would be easier:
[tex]
= \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}|]
[/tex]
Then I get stuck. Does anyone know what I'm missing?
You appear to be a step or two away.

Added in Edit:
Never mind! I was looking at the wrong side of the equation.
 
Last edited:
  • #6
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722

Homework Statement


Prove the following statement:
[tex] ln|1+\sigma x | = \frac{1}{2} ln|1-x^2| + \frac{\sigma}{2} ln| \frac{ |1+x|}{|1-x|}
[/tex]

Homework Equations




The Attempt at a Solution


Starting from right to left would be easier:
[tex]
= \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}|]
[/tex]
Then I get stuck. Does anyone know what I'm missing?

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
753
2
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?

[tex] \frac{1}{2}( C + D ) + \frac{1}{2}( C - D ) \sigma
[/tex]

with obtaining
[tex] \frac{\sigma }{2} ln|\frac{C}{D} | [/tex] at least in the final answer?


[tex]
ln [ \frac{1}{2} D ( \frac{C}{D} +1 ) + \frac{\sigma}{2} D ( \frac{C}{D} -1) ]
[/tex]
 
  • #8
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,163
6,781
Was there a typo?
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.)
 
  • #10
753
2
Is there a table of asymptotic expansions or reference I can refer to to get the desired result? I'm kind of stumped.
 
  • #11
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,163
6,781
Is there a table of asymptotic expansions or reference I can refer to to get the desired result? I'm kind of stumped.
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.
 

Related Threads on Natural Log Proof

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
12
Views
8K
Replies
1
Views
2K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
3
Views
6K
Top