0

How would I expand the following function as a power series, around $\eta=0$?

$$g_0(1,\eta)=\frac{\left(\frac{PV}{NkT}\right)_0-1}{4\eta}$$

Note that:

$$\left(\frac{PV}{NkT}\right)_0=1+\frac{3\eta}{\eta_c-\eta}+\sum_{k=1}^4kA_k\left(\frac{\eta}{\eta_c}\right)^k$$

Then we have:

$$g_0(1,\eta)=\frac{\frac{3\eta}{\eta_c-\eta}+\sum_{k=1}^4kA_k\left(\frac{\eta}{\eta_c}\right)^k}{4\eta}$$

Jackson Hart
  • 1,600

2 Answers2

1

Starting from your last equation: $$g_0(1,\eta)=\frac{\frac{3\eta}{\eta_c-\eta}+\sum_{k=1}^4kA_k\left(\frac{\eta}{\eta_c}\right)^k}{4\eta}$$ you have: $$g_0(1,\eta)=\frac{3}{4(\eta_c-\eta)}+\sum_{k=1}^4\frac k{4\eta_c^k}A_k\eta^{k-1}$$ The terms after the + sign are ok, you just need to deal with the expansion of $(\eta_c-\eta)^{-1}$. You can give $\eta_c$ as a factor, then you have $$\frac3{3\eta_c}\left(1-\frac{\eta}{\eta_c}\right)^{-1}$$ For $|x|<1$ you have $$\frac 1{1-x}=x+x^2+x^3+...=\sum_{n=1}^\infty x^n$$

Andrei
  • 37,370
  • I believe it should $\frac{3}{4\eta_c}$ right? Then is the total answer: $\sum_{n=1}^\infty \left(\frac{\eta}{\eta_c}\right)^n+\sum_{k=1}^4\frac k{4\eta_c^k}A_k\eta^{k-1}$. I need everything to be in terms of $\frac{\eta}{\eta_c}$ – Jackson Hart Mar 27 '19 at 18:12
  • I need everything to be in terms of $\frac{\eta}{\eta_c}$ I need to multiply this function by another polynomial: $C_{01}r^0 \left(\frac{\eta}{\eta_c}\right)+C_{11}r^1\left(\frac{\eta}{\eta_c}\right)+C_{21}r^2\left(\frac{\eta}{\eta_c}\right)+C_{31}r^3\left(\frac{\eta}{\eta_c}\right)+C_{41}r^4\left(\frac{\eta}{\eta_c}\right)+C_{51}r^5\left(\frac{\eta}{\eta_c}\right)+C_{61}r^6\left(\frac{\eta}{\eta_c}\right)$ If I take your solution and multiply by this polynomial, how do I get an expansion about $\eta=0$? I need to get the coefficients in front of each term – Jackson Hart Mar 27 '19 at 18:19
  • I accepted this answer since it answered my question. I posted a related question on another thread if you are interested in looking at it. – Jackson Hart Mar 27 '19 at 18:30
0

Hint:

The sum is a polynomial with no constant term and its handling is not a problem.

Then

$$\frac{\eta}{\eta_c-\eta}=1-\frac1{1-\dfrac\eta{\eta_c}}=\frac\eta{\eta_c}+\frac{\eta^2}{\eta_c^2}+\frac{\eta^3}{\eta_c^3}+\cdots$$

  • I'm not sure how to get the full answer. I'm just trying to get a different representation of my equation, just checking to see if series expansion would work. – Jackson Hart Mar 26 '19 at 19:01
  • @JacksonHart: my hint explains that it would indeed work, but converge for $|\eta|<|\eta_c|$. –  Mar 27 '19 at 07:45