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

1. Aug 11, 2006

### 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?

2. Aug 11, 2006

### sdekivit

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

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

3. Aug 11, 2006

### 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 $$\frac{\ln(1+e^{-x})}{x}$$.

4. Aug 11, 2006

### WigneRacah

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

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

Last edited: Aug 11, 2006
5. Aug 11, 2006

### HallsofIvy

Staff Emeritus
Ouch! ln(1+ ex)= x is a mathematics expression, not physics, and is never true.

Is it really true that physicists say "equal" when they mean "approximately equal"?

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.

Last edited: Aug 11, 2006
6. Aug 11, 2006

### Data

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. Aug 11, 2006

### 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.

8. Aug 11, 2006

### 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 highschool/freshman calculus book.

9. Aug 11, 2006

### Rach3

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.

Last edited by a moderator: Aug 11, 2006
10. Aug 11, 2006

### Rach3

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.

11. Aug 11, 2006

### 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)).

12. Aug 14, 2006

### 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.

Last edited: Aug 14, 2006