Taylor Series Expansion Confusion

In summary, "Taylor Series Expansion Confusion" discusses common misunderstandings surrounding the Taylor series, a mathematical tool used to approximate functions. The article highlights the conditions under which the series converges, the importance of understanding the radius of convergence, and the potential pitfalls in applying the series incorrectly. It emphasizes the need for clarity in definitions and proper application to avoid errors in mathematical analysis.
  • #1
laser1
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Homework Statement
description
Relevant Equations
.
Screenshot_1.png

For context, this is when deriving the Boltzmann distribution by using a canonical ensemble (thermodynamics).

omega is a function to represent number of microstates. According to wikipedia...

Screenshot_2.png

is the first order expansion around 0 (Maclaurin series).

My confusion: What are even differentiating with respect to? It seems like we are differentiating with respect to E, but then the epsilon looks like the "x" in the Maclaurin formula. I have more questions, but I'll leave it at this to avoid confusion for now. Thanks!
 
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  • #2
Let's analyze it from behind. The Taylor series of a function ##f(x)## at a point ##a## is given by
$$
T(f,x,a)=\sum_{n\ge 0} \dfrac{f^{(n)}(a)}{n!}(x-a)^n=f(a)+f'(a)(x-a)+O((x-a)^2)
$$
For ##f(a)## we have ##f(a)=\ln \Omega(E),## i.e. ##a=E.## Then
$$
f'(a)(x-a)=f'(E)(x-E)=\dfrac{d\ln \Omega(E)}{dx}(x-E)
$$
which means the variable is ##x=-\varepsilon + E## and the second term of the series becomes
$$
f'(E)\cdot((-\varepsilon+E)-E)=\dfrac{d\ln \Omega(E)}{d(E-\varepsilon)}(-\varepsilon+E-E)=-\dfrac{d\ln \Omega(E)}{d(E-\varepsilon)}\varepsilon
$$
We have ##d(E-\varepsilon)=dE - d\varepsilon## and with ##\varepsilon## being a constant we have ##d\varepsilon = 0.##
Proof:
\begin{align*}
\dfrac{df(E)}{d(E-\varepsilon)}&=\dfrac{df((E-\varepsilon)+\varepsilon)}{d(E-\varepsilon)}=\dfrac{df(y+\varepsilon)}{dy}\\[6pt]
&=\dfrac{df(y+\varepsilon)}{d(y+\varepsilon)}\cdot \underbrace{\dfrac{d(y+\varepsilon)}{dy}}_{=1}\\[6pt]
&=\dfrac{df(E)}{dE}
\end{align*}
In total we get
$$
\ln \Omega (E-\varepsilon)=\ln \Omega(E)- \dfrac{d\ln \Omega(E)}{dE}\varepsilon + O(\varepsilon^2)
$$

laser1 said:
I have more questions, ...
Which?
 
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  • #3
laser1 said:
According to wikipedia...

View attachment 351220
is the first order expansion around 0 (Maclaurin series).

It looks like they are really expanding around ##E##, not around 0. The function and derivative are evaluated at ##E##. The Maclaurin series is too specialized (centered at ##x_0=0##) to give a good match to your equation. You should compare it to the more general Taylor series. ##f(a)+\frac {f'(a)}{1!}(x-a)+...##. Then ##E=a##, ##E-\epsilon=x##, and ##-\epsilon=x-a##.
laser1 said:
My confusion: What are even differentiating with respect to? It seems like we are differentiating with respect to E,
That is correct.
 
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@fresh_42 Thank you, that makes sense. I also tried differentiating it with respect to epsilon around 0 and got this, but I am stuck at this step:
WhatsApp Image 2024-09-18 at 08.21.10.jpeg
 
  • #5
laser1 said:
@fresh_42 Thank you, that makes sense. I also tried differentiating it with respect to epsilon around 0 and got this, but I am stuck at this step:

Apply the chain rule.
 
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  • #6
pasmith said:
Apply the chain rule.
WhatsApp Image 2024-09-21 at 21.30.38.jpeg


I did it a few days ago and was happy, but upon closer examination: I suspect dE/d(epsilon) = -1 only if the total energy of the ensemble is a constant.

Any idea where I have gone wrong with my second method?
 
  • #7
laser1 said:
View attachment 351397

I did it a few days ago and was happy, but upon closer examination: I suspect dE/d(epsilon) = -1 only if the total energy of the ensemble is a constant.

Any idea where I have gone wrong with my second method?
I don't think that you made a mistake.
\begin{align*}
\left.\dfrac{d\Omega(E-\varepsilon)}{d\varepsilon}\right|_{\varepsilon=0}&=\left.\dfrac{d\Omega(E-\varepsilon)}{d(E-\varepsilon)}\right|_{E-\varepsilon=E}\cdot \left.\dfrac{d(E-\varepsilon)}{d\varepsilon}\right|_{\varepsilon=0}\\
&=\left.\dfrac{d\Omega(E-\varepsilon)}{d(E-\varepsilon)}\right|_{E-\varepsilon=E}\cdot\left\{\underbrace{\left.\dfrac{dE}{d\varepsilon}\right|_{\varepsilon=0}}_{=0}+\underbrace{\left.\dfrac{d(-\varepsilon)}{d\varepsilon}\right|_{\varepsilon=0}}_{=-1}\right\}\\
&=-\left.\dfrac{d\Omega(E-\varepsilon)}{d(E-\varepsilon)}\right|_{E-\varepsilon=E}
\end{align*}
The trick is now that it doesn't matter how you call your variables. We simply have
$$
-\left.\dfrac{d\Omega(E-\varepsilon)}{d(E-\varepsilon)}\right|_{E-\varepsilon=E}=-\left.\dfrac{d\Omega(x)}{dx}\right|_{x=E}=-\left.\dfrac{d\Omega(E)}{dE}\right|_{E}=-\dfrac{d\Omega(E)}{dE}
$$
 
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FAQ: Taylor Series Expansion Confusion

What is a Taylor series expansion?

A Taylor series expansion is a mathematical representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. The series is centered around a point 'a' and provides a way to approximate functions that may be difficult to compute directly.

How do you determine the Taylor series of a function?

To determine the Taylor series of a function, you need to calculate the derivatives of the function at the point 'a' where you want to center the series. The Taylor series is given by the formula:
f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)²/2! + f'''(a)(x-a)³/3! + ... ,
where f'(a), f''(a), etc. are the derivatives evaluated at 'a'.

What is the difference between Taylor series and Maclaurin series?

A Maclaurin series is a special case of the Taylor series where the expansion is centered at the point 'a = 0'. In other words, a Maclaurin series is simply a Taylor series evaluated at zero, allowing for easier calculations when approximating functions near that point.

When is a Taylor series convergent?

A Taylor series is convergent if the limit of the remainder term approaches zero as the number of terms increases. The radius of convergence can often be determined using the ratio test or root test. However, even if a Taylor series converges, it may not converge to the original function for all values of x.

What are some common mistakes when using Taylor series?

Common mistakes when using Taylor series include failing to check the radius of convergence, incorrectly calculating derivatives, and assuming that the Taylor series converges to the function for all x without verification. Additionally, one might overlook the importance of the point around which the series is expanded, which can affect the accuracy of the approximation.

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