A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
When integrating a series, is the constant, C, from the integration evaluated as part of the summation or is it as a constant "outside" of the whole summation? That's, is there a summation from 0 to infinity of the C or is it just adding the C after the main summation?
anonymous
 5 years ago
When integrating a series, is the constant, C, from the integration evaluated as part of the summation or is it as a constant "outside" of the whole summation? That's, is there a summation from 0 to infinity of the C or is it just adding the C after the main summation?

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Say I have: \[\sum_{n=0}^{\infty }x^{n}dx = \sum_{n=0}^{\infty }\frac{x^{n+1}}{n+1}+C\] Is the result this: \[\sum_{n=0}^{\infty }x^{n}dx=\sum_{n=0}^{\infty }\left (\frac{x^{n+1}}{n+1}+C \right )=\sum_{n=0}^{\infty }\frac{x^{n+1}}{n+1}+ \sum_{n=0}^{\infty }C \] OR Is the result this: \[\sum_{n=0}^{\infty }x^{n}dx=C+\sum_{n=0}^{\infty }\left (\frac{x^{n+1}}{n+1} \right )\] Thanks

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Got a missing integration in the previous post. I mean this: Say I have: \[\int\limits \sum_{n=0}^{\infty }x^{n}dx = \sum_{n=0}^{\infty }\frac{x^{n+1}}{n+1}+C\] Is the result this: \[\int\limits \sum_{n=0}^{\infty }x^{n}dx=\sum_{n=0}^{\infty }\left (\frac{x^{n+1}}{n+1}+C \right )=\sum_{n=0}^{\infty }\frac{x^{n+1}}{n+1}+ \sum_{n=0}^{\infty }C \] OR Is the result this: \[\int\limits \sum_{n=0}^{\infty }x^{n}dx=C+\sum_{n=0}^{\infty }\left (\frac{x^{n+1}}{n+1} \right )\]

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0It's actually not an 'one or the other'answer ;) Let us for the moment assume that those integrals exist and you can pull it to the inside. Then for every n there is a constant C(n) and for any choice of C(n) you get a solution. But you must notice that in order for the solution to exist the sum over the C(n) must be finite! So you are forced to choose the C(n) in such a way and then you can simply define: \[C := \sum_{n=0}^∞{C(n)}\] to get a constant for the whole integral. Thus you see that it makes no mathematical difference if you have a constant for every single element or for the whole series. Many different choices of C(n) will yield the same sum, but in the end, that is the only information you need.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.