MHB A recurrence formula for an integral

AI Thread Summary
The discussion revolves around solving a recurrence formula for an integral, specifically the equation A(p,q) = A(p-1, q) + A(p, q-1). The participants clarify that the values of p and q are positive integers and explore how different base cases affect the solution. If A(p,1) and A(1,q) are both set to 1, A(p,q) corresponds to binomial coefficients due to Pascal's rule. The complexity of the base cases can significantly influence the approach to finding a solution, with suggestions to use generating functions to derive A(p,q). Ultimately, the method of solving the recurrence depends heavily on the chosen base cases.
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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