A recurrence formula for an integral

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SUMMARY

The discussion centers on solving the recurrence relation $$A(p,q) = A(p-1 , q ) + A(p,q-1)$$ with known base cases $A(p,1)$ and $A(1,q)$. It is established that the solution depends on the values of the base cases. Specifically, if $A(p,1) = A(1,q) = 1$, then $A(p,q)$ corresponds to binomial coefficients as per Pascal's rule. Alternative base cases yield different forms of $A(p,q)$, such as $A(p,q) = \binom{p+q}{q}$ or $A(p,q) = \binom{p+q-1}{q-1}$. The discussion also highlights the importance of generating functions in solving such recurrences.

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alyafey22
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I was solving an integral and on the process of integration by parts it seems that I have a certain recurrence formula that I have to solve. Before I post it I want to understand how to solve a simpler case which is

$$A(p,q) = A(p-1 , q ) +A(p,q-1) $$

where the base case $A(p,1)$ and $A(1,q)$ is known to me. Is there a general formula to solve it ? or does that depend on the base case ?

I am completely clueless!
 
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ZaidAlyafey said:
I was solving an integral and on the process of integration by parts it seems that I have a certain recurrence formula that I have to solve. Before I post it I want to understand how to solve a simpler case which is

$$A(p,q) = A(p-1 , q ) +A(p,q-1) $$

where the base case $A(p,1)$ and $A(1,q)$ is known to me. Is there a general formula to solve it ? or does that depend on the base case ?

I am completely clueless!

Hi Zaid,

What are the conditions on $p$ and $q$? Are they positive integers?
 
Euge said:
Hi Zaid,

What are the conditions on $p$ and $q$? Are they positive integers?

Of course. I should have noted that.
 
ZaidAlyafey said:
$$A(p,q) = A(p-1 , q ) +A(p,q-1) $$

where the base case $A(p,1)$ and $A(1,q)$ is known to me.
If $A(p,1)=A(1,q)=1$, then $A(p,q)$ are binomial coefficients due to Pascal's rule.
 
Evgeny.Makarov said:
If $A(p,1)=A(1,q)=1$, then $A(p,q)$ are binomial coefficients due to Pascal's rule.

So that depends on the value of the base case ?
 
It will depend on the value of the base case. If you had $A(p,1) = p + 1$ and $A(1, q) = q + 1$, then $A(p, q) = \binom{p+q}{q}$. However, if $A(p, 1) = 1$ and $A(1, q) = q$, then $A(p, q) = \binom{p+q-1}{q-1}$.
 
Oh. My approach seems very complicated because the base case is very complex but I think it is worth the trail. I wanted to ask one more question : Is there any procedure or algorithm to solve the recurrence ? or that depends on the base case chosen ?
 
It will depend on the base cases, but you use the recurrence above to find a functional equation for the generating function of $A(p, q)$, which may help determine $A(p, q)$.
 

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