Find a way of computing $x_{20}$ accurately, perhaps by replacing the integrand by a truncated Taylor series. After computing $x_{20}$ to full machine precision, use the recurrence backward to get $x_{19}, x_{18}, ..., x_0$. Is $x_0$ correct? What about the other $x_n$? Does the recurrence relation behave differently when used backward, and if so, why?
This problem is the continuation of this other problem: Define $x_n=\int_{0}^{1}t^n(t+5)^{-1}dt$. Show that $x_0=\ln 1,2$ and that $x_n=n^{-1}-5x_{n-1}$ for $n=1,2,...$. Compute $x_0, x_1, ..., x_{10}$
The truth is that I do not understand the statement very well, what does a truncated Taylor series mean? How can I do this? I consider a Taylor series of $t^n(t+5)^{-1}$? This series why does it exist? I also do not understand what is going on and I am unable to do it, any help would be useful, thank you very much.
$x_{20} = \int_0^1 \frac {t^{20}}{t+5} dt$
The Taylor series:
$\frac {t^{20}}{t+5} = \frac {t^{20}}{5}\sum_\limits{n=0}^{\infty} \frac {(-1)^nt^n}{5^n}\\ x_{20} = \int_0^1 \sum_\limits{n=0}^{\infty} \frac {(-1)^nt^{n+20}}{5^{n+1}} dt = \sum_\limits{n=0}^{\infty} \frac {(-1)^nt^{n+21}}{(n+21)5^{n+1}}|_0^1 = \sum_\limits{n=0}^{\infty} \frac {1}{(n+21)5^{n+1}}$
That series will converge pretty quickly. So, you can get a decent numerical estimate based on a relatively small number of terms.
then use:
$x_{n} = \frac {1}{n} - 5x_{n-1}\\ x_{n-1} = \frac {1}{5n} - \frac {x_{n}}{5}$
However this recurrence relation is stable. If you had a large error in your estimate for $x_{20}$ and you worked it backward to $x_0$ you get very nearly the same number. Dividing the previous number in the sequence by 5 will divide the error by 5 at each iteration.