- #1

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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?

- Thread starter skarthikselvan
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- #1

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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

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[tex]e^{ln(1+e^{x})} = e^{x}[/tex]

[tex]1 + e^{x} = e^{x}[/tex]

[tex]1 + e^{x} = e^{x}[/tex]

- #3

Rach3

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

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The right conditions for the desired asymptotic behavior of [tex]\ln (1 + e^x)[/tex] areRach3 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].

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

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

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- #5

HallsofIvy

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Ouch! ln(1+ eRach3 said:Is this in a physics context or math context? It makes a difference; the expression is true in physics, and false in mathematics.

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

No, it isn't. What you wrote above would beIn pure mathematics, this is translated as

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

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

What you have is a stronger condition.

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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.HallsofIvy said:Is it really true that physicists say "equal" when they mean "approximately equal"?

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shmoe

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matt grime

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- #9

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 [tex]\approx[/tex]. It's ubiquitous in statistical physics.HallsofIvy said:Ouch! ln(1+ e^{x})= x is amathematicsexpression, not physics, and isnevertrue.

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

Rach3

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.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.

- #11

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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

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[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.

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

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