A recurrence formula for an integral

In summary, the conversation discusses a recurrence formula for a function $A(p, q)$, where the base case $A(p, 1)$ and $A(1, q)$ are known. The value of $A(p, q)$ depends on the base case, and there is a procedure to solve the recurrence, but it also depends on the base case chosen.
  • #1
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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  • #2
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?
 
  • #3
Euge said:
Hi Zaid,

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

Of course. I should have noted that.
 
  • #4
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.
 
  • #5
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 ?
 
  • #6
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 $A(p, q) = \binom{p+q-1}{q-1}$.
 
  • #7
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 ?
 
  • #8
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 $A(p, q)$.
 

FAQ: A recurrence formula for an integral

What is a recurrence formula for an integral?

A recurrence formula for an integral is a mathematical equation that relates the value of an integral to the values of other integrals. It can be used to simplify the calculation of integrals and to find patterns in their values.

How is a recurrence formula for an integral derived?

A recurrence formula for an integral is often derived from the integration by parts formula, which relates the integral of a product to the integral of its derivative. By applying this formula repeatedly, a recurrence formula can be derived.

What are the benefits of using a recurrence formula for an integral?

Using a recurrence formula for an integral can make the calculation of integrals more efficient and can reveal patterns that can be used to simplify the calculation. It can also be used to prove the convergence of integrals.

Are there any limitations to using a recurrence formula for an integral?

One limitation of using a recurrence formula for an integral is that it only applies to certain types of integrals and may not work for more complex integrals. It also requires some knowledge of mathematical manipulation and may not be suitable for beginners.

Can a recurrence formula for an integral be used to solve real-world problems?

Yes, recurrence formulas for integrals can be applied to solve real-world problems in various fields such as physics, engineering, and economics. They can be used to model and analyze systems that involve continuous change, such as population growth or motion of objects.

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