Consider the problem of evaluating the integrals

$y_n={\displaystyle \underset{0}{\overset{1}{}}}\frac{x^n}{x+10}dx$

for $n=1,2,$dots. Analytically we notice that

$y_n+10y_{n-1}={\displaystyle \underset{0}{\overset{1}{}}}\frac{x^n+10x^{n-1}}{x+10}dx={\displaystyle \underset{0}{\overset{1}{}}}{x}^{n-1}dx=\frac{1}{n}.$

Also,

$y_0={\displaystyle \underset{0}{\overset{1}{}}}\frac{1}{x+10}dx=ln(11)-ln(10).$

This gives us the following algorithm for computing $y_0,y_1,$dots :

Compute $y_0=ln(11)-ln(10).$

For $n=1,2,$dots, evaluate $y_n=\frac{1}{n}-10y_{n-1}.$

$($a$)$ Show that $y_n<mn\; id="EditorElement\_92717">1</mn>$ for all $n$ and $y_n$ decreases monotonically to $0$ as $n.($A picture

would suffice for this!$)$

$($b$)$ Program the algorithm in a computer and find $y_n$ for $n=1,2,$dots,$30.$ What happens?

Why does this happen?

$($c$)$ Derive an algorithm for computing the values of these integrals based on evaluating

the value of $y_{n-1}$ from the value of $y_n.$

$($d$)$ Suppose you want to calculate the values of $y_0,y_1,$dots,$y_N$ for some number $N$ with

an absolute error of less than $$ for some $>0.$ Since the naive algorithm above

produces large errors $($even though it is exact in theory$),$ you will use the algorithm

in $($c$).$ But you need a starting value. Show that there exists MinN such that if you

take $y_M=0$ and use the algorithm in $($c$),$ then, if calculations are performed with

infinite precision, the absolute errors in the calculations of $y_0,y_1,$dots,$y_N$ will be less

than $.$

$($e$)$ Explain why rounding errors in the computer do not produce excessive errors using

this algorithm.

$($f$)$ Use your algorithm $($and the computer$)$ to find the value of $y_{30}$ to an accuracy of

${10}^{-5}.$ Explain how you chose $M$ in this case.