# Laplace transform of a Taylor series expansion

1. Jul 6, 2009

### Mapes

I'm reading a paper on tissue cell rheology ("Viscoelasticity of the human red blood cell") that models the creep compliance of the cell (in the s-domain) as

$$J(s) = \frac{1}{As+Bs^{a+1}}$$

where $0\leq a\leq 1$. Since there's no closed-form inverse Laplace transform for this expression, they explore early-time ($t\rightarrow 0$) and late-time ($t\rightarrow \infty$) behavior by using a Taylor series expansion around $s\rightarrow \infty$ and $s\rightarrow 0$, respectively. This is said to yield

$$J(t)\approx \frac{t^a}{B\Gamma(a+1)}-\frac{At^{2a}}{B^2\Gamma(2a+1)}+\frac{A^2t^{3a}}{B^3\Gamma(3a+1)}$$

for the early-time behavior and

$$J(t)\approx \frac{1}{A}-\frac{Bt^{-a}}{A^2\Gamma(1-a)}$$

for the late-time behavior. However, I just can't see how these expressions arise. I know that the Laplace transform of $t^a$ is

$$L[t^a]=\frac{\Gamma(a+1)}{s^{a+1}}$$

and so presumably

$$L\left[\frac{t^a}{\Gamma(a+1)}\right]=\frac{1}{s^{a+1}}\mathrm{,}\quad L\left[\frac{t^{-a}}{\Gamma(1-a)}\right]=\frac{1}{s^{-a+1}}$$

but I can't figure out where these terms would appear in a Taylor series expansion. When I try to expand $J(s)$ in the manner of

$$f(x+\Delta x)\approx f(x) + f^\prime(x)\Delta x +\frac{1}{2}f^{\prime\prime}(x)(\Delta x)^2$$

I get zero or infinity for each term. Unfortunately, Mathematica is no help in investigating an expansion around $s\rightarrow\infty$ or $s\rightarrow 0$; it just returns the original expression. Perhaps I'm making a silly error, or perhaps the paper skipped an important enabling or simplifying step. Any thoughts?

Last edited: Jul 6, 2009
2. Jul 8, 2009

### benorin

Both series expansions below are geometric series: $\frac{1}{1-x}=\sum_{k=0}^{\infty}x^k\mbox{ for }|x|<1$.

For $\left| {\scriptstyle \frac{B}{A}}s^{a} \right| < 1,$ we have

$$J(s) = \frac{1}{As+Bs^{a+1}} = \frac{1}{As}\cdot\frac{1}{1+{\scriptstyle \frac{B}{A}}s^{a}} = \frac{1}{As}\sum_{k=0}^{\infty}\left(-1\right)^k \left(\frac{B}{A}}\right)^k s^{ak}=\frac{1}{A}\sum_{k=0}^{\infty}\left(-1\right)^k \left(\frac{B}{A}}\right)^k s^{ak-1}$$

$$J(s) = \frac{1}{As}-\frac{B}{A^2s^{1-a}}}+\frac{B^2}{A^3s^{1-2a}}}-\cdots$$

hence

$$J(t) = \frac{1}{A}u(t)-\frac{Bt^{-a}}{A^2\Gamma (1-a)}}+\frac{B^2t^{-2a}}{A^3\Gamma (1-2a)}}-\cdots$$​

where $u(t)$ is the unit step function...

And for $\left| {\scriptstyle \frac{A}{B}} s^{-a} \right| < 1,$ we have

$$J(s) = \frac{1}{As+Bs^{a+1}} = \frac{1}{Bs^{a+1}}\cdot\frac{1}{ {\scriptstyle \frac{A}{B}}s^{-a}}+1} =\frac{1}{Bs^{a+1}}\sum_{k=0}^{\infty}\left(-1\right)^k \left(\frac{A}{B}}\right)^k s^{-ak}=\frac{1}{B}\sum_{k=0}^{\infty}\left(-1\right)^k \left(\frac{A}{B}}\right)^k s^{-ak-a-1}$$

$$J(s) = \frac{1}{Bs^{a+1}}-\frac{A}{B^2s^{2a+1}}+\frac{A^2}{B^3s^{3a+1}}-\cdots$$​

hence

$$J(t) = \frac{t^{a}}{B\Gamma (a+1)}-\frac{At^{2a}}{B^2\Gamma (2a+1)}+\frac{A^2t^{3a}}{B^3\Gamma (3a+1)}-\cdots$$​

3. Jul 8, 2009

### Mapes

Thank you benorin, that makes things perfectly clear. I've only seen the geometric series expansion once or twice before in my field and wouldn't have thought to use it. It would have been nice if the paper had mentioned that they used this technique.

Thanks again!