I recently found a nice identity about Partial fractions on Interesting identities about Partial fractions. The same could also be found, albeit on a different form, on A note on partial fractions. So this post is actually a note on 'A note on Partial fractions'.
It says, if $u+v=$, then for integers $m,n \geq 1$,
$\displaystyle \frac{1}{u^mv^n}=\sum_{k=0}^{m-1}\frac{\binom{n-1+k}{k}}{u^{m-k}}+\sum_{l=0}^{n-1}\frac{\binom{m-1+l}{l}}{v^{n-l}}$
The wordpress page proves this with L'Hospital's rule whereas the webpage proves this with induction. But for someone interested in Probability, two-things-adding-upto-1, the binomial coefficient and then it immediately hit me.
It's actually the 'Problem of Points', one of the classical problems of probability where two friends $A$ and $B$ play a series of games where $A$ needs $m$ wins and $B$ needs $n$ wins to win the match. Assume $A$ has a probability $p$ of winning a each game whereas $B$'s probability is $q=1-p$.
Obviously,
$\mathbb{P}(\text{A wins the match})+\mathbb{P}(\text{B wins the match})=1$
Let's compute the probability of $A$ winning the match. As Pascal reasoned, $A$ needs a minimum of $m$ games and a maximum of $m+n-1$ games to win the match. By the law of total probability,
$\begin{align}
\displaystyle\mathbb{P}(\text{A wins the match})&=\sum_{k=m}^{m+n-1}\mathbb{P}(\text{A wins the match on the }k^{th}\text{ game})\\
&=\sum_{k=m}^{m+n-1}\mathbb{P}(\text{A won }m-1\text{ games out of }k-1\text{ games})\cdot\mathbb{P}(\text{A wins the }k^{th}\text{ game})\\
&=\sum_{k=m}^{m+n-1}\binom{k-1}{m-1}p^{m-1}q^{k-m}\cdot p\\
&=\sum_{k=0}^{n-1}\binom{m+k-1}{k}p^mq^k
\end{align}$
This is 'textbook' Negative Binomial.
Interchanging $m$ with $n$ and $p$ with $q$ gives the probability of $B$ winning the match. Therefore, we finally have,
$\displaystyle \sum_{k=0}^{n-1}\binom{m+k-1}{k}p^mq^k+\sum_{k=0}^{m-1}\binom{n+k-1}{k}q^np^k=1$
Dividing both sides by $p^mq^n$, gives the identity quoted at the start of this post. Tada!!
Note that Fermat's reasoning for the Problem of points produces the following equation. Though correct, it is not particularly useful as a partial fraction decomposition.
$\displaystyle \sum_{k=0}^{n-1}\binom{m+n-1}{m+k}p^{m+k}q^{n-1-k}+\sum_{k=0}^{m-1}\binom{n+m-1}{n+k}q^{n+k}p^{m-1-k}=1$
Hope you enjoyed this.
Until then
Yours Aye
Me
It says, if $u+v=$, then for integers $m,n \geq 1$,
$\displaystyle \frac{1}{u^mv^n}=\sum_{k=0}^{m-1}\frac{\binom{n-1+k}{k}}{u^{m-k}}+\sum_{l=0}^{n-1}\frac{\binom{m-1+l}{l}}{v^{n-l}}$
The wordpress page proves this with L'Hospital's rule whereas the webpage proves this with induction. But for someone interested in Probability, two-things-adding-upto-1, the binomial coefficient and then it immediately hit me.
It's actually the 'Problem of Points', one of the classical problems of probability where two friends $A$ and $B$ play a series of games where $A$ needs $m$ wins and $B$ needs $n$ wins to win the match. Assume $A$ has a probability $p$ of winning a each game whereas $B$'s probability is $q=1-p$.
Obviously,
$\mathbb{P}(\text{A wins the match})+\mathbb{P}(\text{B wins the match})=1$
Let's compute the probability of $A$ winning the match. As Pascal reasoned, $A$ needs a minimum of $m$ games and a maximum of $m+n-1$ games to win the match. By the law of total probability,
$\begin{align}
\displaystyle\mathbb{P}(\text{A wins the match})&=\sum_{k=m}^{m+n-1}\mathbb{P}(\text{A wins the match on the }k^{th}\text{ game})\\
&=\sum_{k=m}^{m+n-1}\mathbb{P}(\text{A won }m-1\text{ games out of }k-1\text{ games})\cdot\mathbb{P}(\text{A wins the }k^{th}\text{ game})\\
&=\sum_{k=m}^{m+n-1}\binom{k-1}{m-1}p^{m-1}q^{k-m}\cdot p\\
&=\sum_{k=0}^{n-1}\binom{m+k-1}{k}p^mq^k
\end{align}$
This is 'textbook' Negative Binomial.
Interchanging $m$ with $n$ and $p$ with $q$ gives the probability of $B$ winning the match. Therefore, we finally have,
$\displaystyle \sum_{k=0}^{n-1}\binom{m+k-1}{k}p^mq^k+\sum_{k=0}^{m-1}\binom{n+k-1}{k}q^np^k=1$
Dividing both sides by $p^mq^n$, gives the identity quoted at the start of this post. Tada!!
Note that Fermat's reasoning for the Problem of points produces the following equation. Though correct, it is not particularly useful as a partial fraction decomposition.
$\displaystyle \sum_{k=0}^{n-1}\binom{m+n-1}{m+k}p^{m+k}q^{n-1-k}+\sum_{k=0}^{m-1}\binom{n+m-1}{n+k}q^{n+k}p^{m-1-k}=1$
Hope you enjoyed this.
Until then
Yours Aye
Me
No comments:
Post a Comment