Why does this not diverge when x = inf?

  • Thread starter Thread starter denjay
  • Start date Start date
AI Thread Summary
The equation under discussion, (x^3/3) ln(1-e^{-βx}) = 0, does not diverge as x approaches infinity due to the behavior of the logarithm and the exponential function. As x increases, e^{-βx} approaches zero, allowing the use of a Taylor expansion for ln(1-t) which simplifies the expression to zero. The limit can also be analyzed using L'Hôpital's Rule, transforming it into a 0/0 indeterminate form that confirms the limit value is zero. The context of evaluating x from infinity to zero is clarified as a limit process rather than a direct substitution. Overall, the equation's behavior at infinity is consistent with the principles of calculus and asymptotic analysis.
denjay
Messages
77
Reaction score
0
Here is the equation I came across during a proof in my Statistical Mechanics textbook.

\frac{x^3}{3} ln(1-e^{-βx}) = 0

where β = \frac{1}{k_{B}T} and x is evaluated from \infty to 0.

Why does this not diverge? Is it an estimation?
 
Mathematics news on Phys.org
denjay said:
Here is the equation I came across during a proof in my Statistical Mechanics textbook.

\frac{x^3}{3} ln(1-e^{-βx}) = 0

where β = \frac{1}{k_{B}T} and x is evaluated from \infty to 0.

Why does this not diverge? Is it an estimation?

What do you mean by "x is evaluated from \infty to 0"? Just that that is the range of x, i.e., ##x \in [0,\infty)##?

At any rate, as you take ##x\rightarrow \infty##, ##e^{-\beta x}## becomes small. You can then use the taylor expansion of the logarithm, ##\ln(1+t) \approx t## as ##t\rightarrow 0## to find the leading order behavior as x gets large, and see that the expression will go to zero.

You could also use L'Hopital's rule to demonstrate this, as the limit is of the indeterminate form ##\infty \cdot 0##. Both approaches will give you the limit, but the first approach also tells you how the function behaves asymptotically as x gets large.
 
denjay said:
Here is the equation I came across during a proof in my Statistical Mechanics textbook.

\frac{x^3}{3} ln(1-e^{-βx}) = 0
Was this in the form of a limit? That context would make sense with what you said below.

$$ \lim_{x \to \infty} \frac{x^3}{3} ln(1-e^{-βx}) $$

The above is the indeterminate form [∞ * 0]. Being indeterminate, you can't tell how it will turn out.

The thing to do is to write it in a form so that L'Hopital's Rule can be used; namely,
$$ \frac{1}{3}\lim_{x \to \infty} \frac{ln(1-e^{-βx}) }{x^{-3}} $$

Now, it's the indeterminate form [0/0], so L'Hopital's can be used. The numerator is approaching 0, as is the denominator.

Applying L'H, you arrive pretty quickly at a limit value of 0.
denjay said:
where β = \frac{1}{k_{B}T} and x is evaluated from \infty to 0.
I'm assuming that β > 0. Also, we don't normally evaluate something from ∞
to 0. We can take the limit, though, as x → ∞.
denjay said:
Why does this not diverge? Is it an estimation?
 
Sorry I probably should have provided more information. That equation is the result of an integral being from 0 to ∞. So, with that equation, setting x = 0 makes the equation zero along with setting x = ∞.

And the only restriction on β is that β > 0.
 
If you are working with integrals, you should know by now that you can not "set x= \infty". You have to use limits.
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Back
Top