# Recurrence Relation.

I was searching the internet when I came a across a question, and just couldn't solve it. I kept rearranging and substituting but kept going around in loops.

"For $n:= 1,2,3,.....,$ Let $$I_n = \int_0^1 \frac{x^{n-1}}{2-x}.dx$$ Writing $x^n =x^{n-1}(2-(2-x))$, show that this sequence of numbers satisfies the recurrence relation $$I_{n+1} = 2I_n - \frac{1}{n}$$

Extension

The value taken for $I_1 = \ln{2}$ is $0.6931$ if the recurrence relation is now used to calculate successive values for $I_n$ we find $I_{12}=-0.0189$ (you are not required to confirm the calculation). Explain carefully both why this cannot be correct and the source of the error, given that all the intermediate arithmetical operations have been performed exactly.

• It can't be right because the integrand is positive. Can you solve the recurrence? – Ross Millikan Oct 26 '13 at 14:14
• i've solved the recurrence with the help of @cameron , but just stuck on the extension part, surely its not always positive because when n is large enough it'll be negative however for the above scenario it is incorrect due to round off error. would that be correct or would the explanation need to be better? – jhon Oct 26 '13 at 14:17
• No, both the numerator and denominator are positive over all of $[0,1]$. Note that the correct value of $\log 2$ is closer to $0.69314718$, so the one you are using is low. As $n$ gets large, the numerator gets small (except very near $1$) so the integral is going toward zero. Once you cross zero, the $\frac 1n$ term pulls you farther from zero, as does the doubling of the $I_n$ You should make this more specific with the solution of the recurrence. – Ross Millikan Oct 26 '13 at 14:26

Well, \begin{align}I_{n+1} &= \int_0^1\frac{x^n}{2-x}\,dx\\ &= \int_0^1\frac{x^{n-1}\bigl(2-(2-x)\bigr)}{2-x}\,dx\\ &= \int_0^1\frac{2x^{n-1}-(2-x)x^{n-1}}{2-x}\,dx\\ &= \int_0^1\left(\frac{2x^{n-1}}{2-x}-\frac{(2-x)x^{n-1}}{2-x}\right)\,dx.\end{align} Can you take it from there?
• Try to keep the end goal in mind. We need for $I_n$ to show up, somehow. See if you can make that happen, based on what has been done so far. – Cameron Buie Oct 26 '13 at 13:24
• i would separate it into 2 integrals, take out a factor of 2 from the first part giving me $2I_n$ ?? however how would i get $1/n$ ?? – jhon Oct 26 '13 at 13:26
• actually i don't sorry, how would i get $1/n$? – jhon Oct 26 '13 at 13:28