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

[skarthikselvan]

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?

 

[sdekivit]

e^{ln(1+e^{x})} = e^{x}

1 + e^{x} = e^{x}

 

[Rach3]

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

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

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

 

[WigneRacah]

...
In pure mathematics, this is translated as

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

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

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

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

and

\lim_{x \rightarrow + \infty} \ln (e^x+1) - x = 0.

 

[HallsofIvy]

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+ e^x)= 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

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

No, it isn't. What you wrote above would be
\lim_{x\rightarrow \infty}ln(e^x+1)- x= 0
What you have is a stronger condition.

 

[Data]

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.

 

[shmoe]

That's terrible. Why not just put in a squiggly equals like \approx and say they are leaving out the justification that this approximation doesn't muck up the result.

 

[matt grime]

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.

 

[Rach3]

Ouch! ln(1+ e^x)= 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 \approx. 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.

 

[Rach3]

No, it isn't. What you wrote above would be
\lim_{x\rightarrow \infty}ln(e^x+1)- x= 0
What you have is a stronger condition.

I think my condition is weaker - \lim_{x \rightarrow \infty} \frac{f(x)}{x} = 0 does not imply \lim_{x \rightarrow \infty} f(x) = 0. In this particular case, they're both true.

 

[cthulito]

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)).

 

[Robokapp]

e^{ln(1+e^{x})} = e^{x}

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 1+e^{x}=e^{x} to be proven if it's true or false. In math it's false...obviously. Reason is that you can rewrite as 1=e^{x}-e^{x} 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.