Is ln(1+exp(x)) = x when x is a large number?

  • Thread starter skarthikselvan
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In summary, the person is asking if the expression ln(1+exp(x)) = x is correct and how it can be proven. This expression is not true in mathematics, but can be considered true in physics as an asymptotic approximation. The correct expression converges to the asymptotic approximation in the limit of large numbers. This approximation is often denoted by the symbol \approx in physics. However, in mathematics, this expression is false and can be rewritten as 1 = e^x - e^x, which is not true.
  • #1
skarthikselvan
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Hi all,

I have an expression like this. ln(1+exp(x)) and x is a huge number.
Can I write like this ?

ln(1+exp(x)) = x. If it is right, how can I prove this?
 
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  • #2
[tex]e^{ln(1+e^{x})} = e^{x}[/tex]

[tex]1 + e^{x} = e^{x}[/tex]
 
  • #3
Is this in a physics context or math context? It makes a difference; the expression is true in physics, and false in mathematics.

Hand-waving goes like this: for large x, e^x >> 1, so e^x + 1 = e^x, and ln(1+e^x)=ln(e^x)=x.

In pure mathematics, this is translated as

[tex]\lim_{x \rightarrow \infty} \frac{\ln(e^x+1) - x}{x}=0[/tex]

which is easily proven through the intermediate [tex]\frac{\ln(1+e^{-x})}{x}[/tex].
 
  • #4
Rach3 said:
...
In pure mathematics, this is translated as

[tex]\lim_{x \rightarrow \infty} \frac{\ln(e^x+1) - x}{x}=0[/tex]

which is easily proven through the intermediate [tex]\frac{\ln(1+e^{-x})}{x}[/tex].

The right conditions for the desired asymptotic behavior of [tex]\ln (1 + e^x)[/tex] are

[tex]\lim_{x \rightarrow + \infty} \frac{\ln (e^x+1)}{x}=1[/tex]

and

[tex]\lim_{x \rightarrow + \infty} \ln (e^x+1) - x = 0[/tex].
 
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  • #5
Rach3 said:
Is this in a physics context or math context? It makes a difference; the expression is true in physics, and false in mathematics.
Ouch! ln(1+ ex)= x is a mathematics expression, not physics, and is never true.

Hand-waving goes like this: for large x, e^x >> 1, so e^x + 1 = e^x, and ln(1+e^x)=ln(e^x)=x.
Is it really true that physicists say "equal" when they mean "approximately equal"?

In pure mathematics, this is translated as

[tex]\lim_{x \rightarrow \infty} \frac{\ln(e^x+1) - x}{x}=0[/tex]
No, it isn't. What you wrote above would be
[tex]\lim_{x\rightarrow \infty}ln(e^x+1)- x= 0[/tex]
What you have is a stronger condition.
 
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  • #6
HallsofIvy said:
Is it really true that physicists say "equal" when they mean "approximately equal"?

Quite frequently. My undergrad thermodynamics text (An Intro. to Thermal Physics by Daniel Schroeder) has a section on "large numbers" and "very large numbers" with statements like "if N is a large number and n is a small number, then n+N = N," and "if N is a large number then N^N a very large number," and "a very large number is unchanged when multiplied by a large number." It is fairly amusing to read, but approximations like this are so necessary in an introduction to the material in the text that I can see why it's there. It'd be nice if they'd say things that are actually correct though.
 
  • #7
That's terrible. Why not just put in a squiggly equals like [tex]\approx[/tex] and say they are leaving out the justification that this approximation doesn't muck up the result.
 
  • #8
It's worse than that in some places. I had a tantrum sufficient to cause me to write to the editors of a textbook because they kept on using tsuch things as sqrt(2)=1.4. that wasn't physics (nor even large numbers) but a high school/freshman calculus book.
 
  • #9
HallsofIvy said:
Ouch! ln(1+ ex)= x is a mathematics expression, not physics, and is never true.

That's why I asked if it was in a physics context - because by convention there is a weaker version of "=" used in physics which adapts to physicist's notions of approximation. Sometimes it's referred to as an "asymptotic expression", which is a short hand for "the correct expression converges to the asymptotic approximation in the limit of large N". Schroeder's text (mentioned in Data's post) outright uses "N+n=N, for large N". Other literature uses the less cavalier squiggly [tex]\approx[/tex]. It's ubiquitous in statistical physics.

This scheme of approximate "=" certainly doesn't belong in a math forum; however it seems likely that the OP's question came from a physics context, which is why I risked bringing this up.
 
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  • #10
HallsofIvy said:
No, it isn't. What you wrote above would be
[tex]\lim_{x\rightarrow \infty}ln(e^x+1)- x= 0[/tex]
What you have is a stronger condition.

I think my condition is weaker - [tex]\lim_{x \rightarrow \infty} \frac{f(x)}{x} = 0[/tex] does not imply [tex]\lim_{x \rightarrow \infty} f(x) = 0[/tex]. In this particular case, they're both true.
 
  • #11
Remember the Mean Value Theorem:

log(1+exp(x)) - x = log(1+exp(x)) - log(exp(x)) = (1+exp(x)-exp(x))*(log)'(c)

where exp(x)<=c<=exp(x)+1. Thus

log(1+exp(x)) - x = 1/c <= exp(-x).

Thus log(1+exp(x)) = x + O(exp(-x)).
 
  • #12
[tex]e^{ln(1+e^{x})} = e^{x}[/tex]

Okay...to me the person is asking to have the limit as x->oo but really what he wants is a simpler mathematical expression.

So...let's rewrite it.

e^ln(a)=a is a simple Log property...so all you're left with is [tex]1+e^{x}=e^{x}[/tex] to be proven if it's true or false. In math it's false...obviously. Reason is that you can rewrite as [tex]1=e^{x}-e^{x}[/tex] so we have 1=0 which is false.

You asked "large numbers" not "infinity" so the answer is no, but it's assumed to be in measures because of signifficant figure approximation if you're using it for phisical events.
 
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FAQ: Is ln(1+exp(x)) = x when x is a large number?

What is the value of ln(1+exp(x)) when x is a large number?

As x approaches infinity, ln(1+exp(x)) approaches x. This means that when x is a large number, ln(1+exp(x)) is approximately equal to x.

Why is it important to know the value of ln(1+exp(x)) for large values of x?

Knowing the value of ln(1+exp(x)) for large values of x is important in many mathematical and scientific calculations, such as in statistics, thermodynamics, and population modeling. It can also help in understanding the behavior of exponential functions and their relationship with logarithmic functions.

How is the value of ln(1+exp(x)) related to the natural logarithm of x?

The natural logarithm of x, ln(x), is the inverse function of the exponential function, e^x. When x is a large number, ln(1+exp(x)) is approximately equal to x, which means that ln(x) is approximately equal to 1+exp(x). This shows the relationship between the two functions.

Can you provide an example of when ln(1+exp(x)) is equal to x for a large value of x?

For example, if x=1000, ln(1+exp(1000)) is approximately equal to 1000. This can also be seen by substituting x=1000 in the equation ln(1+exp(x)) = x, which gives ln(1+exp(1000)) = 1000.

Are there any limitations to using ln(1+exp(x)) = x for large values of x?

While ln(1+exp(x)) is approximately equal to x for large values of x, there may be some error in the calculation due to the limitations of computer precision. Additionally, this equation may not hold true for extremely large values of x, as the behavior of exponential and logarithmic functions may change at very high values.

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