Confusion with integration of sums

Click For Summary
SUMMARY

The discussion centers on the integration of sums in multivariable calculus, specifically the equation involving binomial coefficients: \(\sum_{r=0}^{k} \binom{n}{4r+1} x^{n-4r-1} y^{4r+1} = \sum_{r=0}^{b} \binom{n}{4r+3} x^{n-4r-3} y^{4r+3}\). The user seeks to determine if limits can be applied to both sides of the equation and subsequently converted into an integral. The consensus is that while limits can be approached, proving equality for all values of \(N\) requires careful consideration, as two functions may converge without being equal.

PREREQUISITES
  • Understanding of multivariable calculus concepts
  • Familiarity with binomial coefficients and their properties
  • Knowledge of limits and convergence in mathematical analysis
  • Basic principles of mathematical induction
NEXT STEPS
  • Study the properties of binomial coefficients in depth
  • Learn about limits and convergence in multivariable calculus
  • Explore mathematical induction techniques for proving identities
  • Investigate the process of converting sums to integrals in calculus
USEFUL FOR

Students and educators in mathematics, particularly those focusing on calculus and combinatorial identities, as well as researchers exploring advanced mathematical proofs and techniques.

Nick Jackson
Messages
13
Reaction score
0
Hello guys, since I am new at sums and multivariable calculus I faced a problem when I stumbled upon this: \sum_{r=0}^{k} \binom{n}{4r+1} x^{n-4r-1} y^{4r+1} = \sum_{r=0}^{b} \binom{n}{4r+3} x^{n-4r-3} y^{4r+3} Well, the problem is that I don't know if it's possible to put a limit in every part of the equation and then convert it to an integral (I am trying to prove that \forall x , y \in \mathbb{R}, \quad \exists n \in \mathbb{N} , n \in \mathbb{N} , n \neq 0 so that the relation holds). Can I do it or does it violate any rule? And if it is possible to do it how would the multivariable integration be? (If you want the relations between n, k, b just ask)
 
Physics news on Phys.org
The usual way to approach proving such a result is by mathematical induction. Two functions of N can approach the same limit as N approaches infinity without always being equal to each other. So proving the left and right hand sides have the same limit would not prove the two sides are equal for all values of N.
 
Oh i get it! Thanks a lot!
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K