Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Requiem

  • 2 years ago

Power series problem. (x-2)^n/(n^n) Find the radius and interval of convergence.

  • This Question is Closed
  1. aykayyy
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i actually just learned this today lol..... try the ratio test!

  2. Requiem
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I did, but am not sure if I am doing ti correctly

  3. sarahusher
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    I would do the nth root test?

  4. aykayyy
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\lim_{n \rightarrow \infty} \left| \frac{ a _{n+1} }{ a _{n} } \right|\]

  5. Requiem
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1375316023379:dw|

  6. aykayyy
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh yes @sarahusher is probably right! i didnt see the ^n.... the root test would work better

  7. Requiem
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok ill try that

  8. Requiem
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok i did the root test and got (x-2)/n

  9. aykayyy
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    then take the limit of that

  10. sarahusher
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Yep!

  11. Requiem
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i pull the x-2 out, and the limit of (1/n) would be 0

  12. sarahusher
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    exactly!

  13. Requiem
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but then i multiply the 0 by the (x-2) right?

  14. Requiem
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    which would give me 0 overall?

  15. aykayyy
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yup

  16. sarahusher
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    As limit < 1, the series will converge for every 'x'

  17. Requiem
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Im not sure how they are getting the interval of convergence which they are saying is from negative infinity to positive infinity

  18. sarahusher
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Okay, I'll explain: So when finding the radius of convergence: We know that as the limit=0<1, the series is convergent for every 'x' So for any x for any 'x' you get (ie from -infinity to +infinity) the series will converge Using the limit that you have, you get to ROC = infinity which directly gives you the interval as -infinity<x<infinity

  19. sarahusher
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Does that make sense?

  20. Requiem
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Everything else makes sense, but the ROC (x-2) Lim as 'n'-------> infinity of (1/n) = 0 0*(x-2)= 0 Im just not sure how the ROC is infiinty

  21. sarahusher
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    'The radius of convergence' r is a 'nonnegative real number' or '∞' such that the series converges if [x+L] < r , here L is your limit So our limit is 0 so we want an 'r' such that all of our 'x'<r but 'x' always converges for x<r And we worked out before that x converges for all values (ie for all R - real numbers) So it's not so much a calculation to work this out, as opposed to seeing that no matter what your value the series will converge no matter where it lies along R

  22. sarahusher
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    |dw:1375317501046:dw| Think of it like a disc, no matter where your value of 'X' lies in the disc, for this example the series will always converge

  23. Requiem
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok that makes more sense now...thanks sara :)

  24. sarahusher
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Okay, If anything isn't clear let me know :)

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy