anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Say 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
  • anonymous
Got 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
  • nowhereman
It'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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.