Is Ln(1+exp(x)) = X When X Is A Large Number? - Physics Forums

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Is ln(1+exp(x)) = x when x is a large number?
  • Context: Undergrad
  • Thread starter Thread starter skarthikselvan
  • Start date Start date Aug 11, 2006
Click For Summary

Discussion Overview

The discussion centers around the expression ln(1+exp(x)) and its behavior as x approaches a large number. Participants explore whether this expression can be equated to x in different contexts, particularly in physics versus mathematics.

Discussion Character

  • Debate/contested
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • Some participants propose that for large x, ln(1+exp(x)) can be approximated as x due to the dominance of exp(x) in the expression.
  • Others argue that this approximation holds true in a physics context but is not valid in pure mathematics, where the expression is considered false.
  • A later reply suggests that the limit of the difference between ln(1+exp(x)) and x approaches zero as x approaches infinity, indicating a nuanced relationship rather than a strict equality.
  • Some participants highlight the use of asymptotic expressions in physics, where "=" may imply an approximation rather than strict equality.
  • Concerns are raised about the clarity of communication in textbooks regarding approximations, with examples provided of how large numbers are treated in educational contexts.
  • One participant mentions the Mean Value Theorem to analyze the difference between ln(1+exp(x)) and x, suggesting that the difference is small for large x.
  • Another participant emphasizes that while the limit approaches a certain behavior, it does not imply that ln(1+exp(x)) equals x in a strict mathematical sense.

Areas of Agreement / Disagreement

Participants do not reach a consensus; there are multiple competing views regarding the validity of the expression ln(1+exp(x)) = x in different contexts, particularly between physics and mathematics.

Contextual Notes

Participants note that the interpretation of the expression may depend on the context in which it is used, with distinctions made between mathematical rigor and physical approximations. The discussion also highlights the potential for confusion arising from the use of "=" in physics to denote approximate equality.

skarthikselvan Messages 4 Reaction score 0 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? Physics news on Phys.org
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sdekivit Messages 91 Reaction score 0 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 Messages 34 Reaction score 0
Rach3 said: ... 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. Last edited: Aug 11, 2006 HallsofIvy Science Advisor Homework Helper Messages 42,895 Reaction score 983
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 \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. Last edited by a moderator: Aug 11, 2006 Data Messages 998 Reaction score 0
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. shmoe Science Advisor Homework Helper Messages 1,971 Reaction score 2 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 Science Advisor Homework Helper Messages 9,361 Reaction score 6 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
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 \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 Rach3
HallsofIvy said: 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 Messages 1 Reaction score 0 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 Messages 218 Reaction score 0 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

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