# Thread: expansion of a function

1. ## expansion of a function

when reading a paper, the author said that
$\displaystyle y(t) = Ur(t)\int_{-\infty}^t dt^\prime \exp\left(-\frac{t-t^\prime}{\tau} - U\int_{t^\prime}^t d t^{\prime\prime} r(t^{\prime\prime})\right)$
can be expanded as expression involving $\displaystyle r(t), r^\prime(t) ...$
(if $\displaystyle r^\prime / r \ll r$)
$\displaystyle y(t) \approx \frac{r}{1+rU\tau} +r^\prime\frac{Ur\tau^2}{(1+rU\tau)^3} + \cdots$,
I don't know how he arrived at this conclusion. Can anyone enlighten me with some suggestions?

2. Originally Posted by wclayman
when reading a paper, the author said that
$\displaystyle y(t) = Ur(t)\int_{-\infty}^t dt^\prime \exp\left(-\frac{t-t^\prime}{\tau} - U\int_{t^\prime}^t d t^{\prime\prime} r(t^{\prime\prime})\right)$
can be expanded as expression involving $\displaystyle r(t), r^\prime(t) ...$
(if $\displaystyle r^\prime / r \ll r$)
$\displaystyle y(t) \approx \frac{r}{1+rU\tau} +r^\prime\frac{Ur\tau^2}{(1+rU\tau)^3} + \cdots$,
I don't know how he arrived at this conclusion. Can anyone enlighten me with some suggestions?
That is awfully hard to read. Could you maybe explain the variables? Is $\displaystyle U$ just a constant? When you write $\displaystyle U\int_t^t' dt'' r(t'')$ is that just then $\displaystyle U(t-t')$?

3. Sorry for the inconvenience, the only functions these formulas is $\displaystyle r(x)$ and $\displaystyle y(t)$, both $\displaystyle U$ and $\displaystyle \tau$ are constants.