Find the Taylor series at x = 0 of the function a/(4b(x + b)) times exp(-4b/(x + 2b)), up to the x^4 term.

Step-by-step solution

1. 1

   Scale the parameter out first. Setting $t=x/b$ gives $x+b=b(1+t)$ and $x+2b=b(2+t)$, and the $b$ inside the exponential cancels completely:

   $f=\frac{a}{4b^2}\cdot\underbrace{\frac{1}{1+t}\exp\!\left(-\frac{4}{2+t}\right)}_{g(t)}$

   Now only one parameter-free series $g(t)$ has to be expanded - this is what keeps the algebra manageable.

2. 2

   Expand the exponent as a geometric series. Writing $-\frac{4}{2+t}=-\frac{2}{1+t/2}$ and using $\frac{1}{1+u}=1-u+u^2-\cdots$ with $u=t/2$:

   $-\frac{4}{2+t}=-2+t-\frac{t^2}{2}+\frac{t^3}{4}-\frac{t^4}{8}+\cdots$

   The leading $-2$ factors out as the constant $e^{-2}$, which is why $e^{-2}$ appears in every coefficient of the final answer.

3. 3

   Exponentiate the remainder. With $h(t)=t-\tfrac{t^2}{2}+\tfrac{t^3}{4}-\tfrac{t^4}{8}$, use $e^{h}=1+h+\tfrac{h^2}{2}+\tfrac{h^3}{6}+\tfrac{h^4}{24}$ and collect powers:

   $\exp\!\left(-\frac{4}{2+t}\right)=e^{-2}\left(1+t+0\cdot t^2-\frac{t^3}{12}+\frac{t^4}{24}+\cdots\right)$

   The $t^2$ coefficient cancels exactly: $-\tfrac12+\tfrac12=0$.

4. 4

   Multiply by the geometric factor $1/(1+t)$. With $\frac{1}{1+t}=1-t+t^2-t^3+t^4-\cdots$, the Cauchy product gives coefficients $1$, $1-1=0$, $1-1+0=0$, $-1+1+0-\tfrac1{12}=-\tfrac1{12}$, and $1-1+0+\tfrac1{12}+\tfrac1{24}=\tfrac18$:

   $g(t)=e^{-2}\left(1-\frac{t^3}{12}+\frac{t^4}{8}+\cdots\right)$

   Both the $t$ and $t^2$ terms vanish - the function is unusually flat at the origin.

5. 5

   Substitute back $t=x/b$.

   $f(x)=\frac{a\,e^{-2}}{4b^2}\left[1-\frac{x^3}{12b^3}+\frac{x^4}{8b^4}+\cdots\right]=\frac{a e^{-2}}{4b^2}-\frac{a e^{-2}}{48b^5}x^3+\frac{a e^{-2}}{32b^6}x^4+O(x^5)$

6. 6

   Verify the coefficients numerically. Least-squares fitting a degree-8 polynomial to $g(t)$ on $t\in[-0.3,0.3]$ returns coefficients $1.0000$, $3\times10^{-7}$, $-1\times10^{-6}$, $-0.08338$, $0.12509$ after dividing by $e^{-2}$ - matching $1,0,0,-\tfrac1{12}=-0.08333,\tfrac18=0.125$ ✓. Also $g(0)=e^{-2}=0.135335$ ✓.