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anonymous

  • 5 years ago

can anyone help out with power series?

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  1. anonymous
    • 5 years ago
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    Yes

  2. anonymous
    • 5 years ago
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    so waht is your question?

  3. anonymous
    • 5 years ago
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    Find the radius of convergence and the interval of convergence of the power series. the sum from n=0 to infinity of 2^n x^n / n!

  4. anonymous
    • 5 years ago
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    so i used ratio test and found it converges absolutely, but im not sure how to find the Radius on convergence or interval

  5. anonymous
    • 5 years ago
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    Do you mind giving me the last few lines of your ratio test?

  6. anonymous
    • 5 years ago
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    well it boiled down to lim=|2x/n+1|

  7. anonymous
    • 5 years ago
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    COOL i got that too

  8. anonymous
    • 5 years ago
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    so limn->inifinity, 2x/(n+1) -> 0

  9. anonymous
    • 5 years ago
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    yupp, exactly

  10. anonymous
    • 5 years ago
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    that mean the interval is (-infin, infin) and radius of convergence is infinity

  11. anonymous
    • 5 years ago
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    that means*

  12. anonymous
    • 5 years ago
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    ok, just so i understand though what is the R value

  13. anonymous
    • 5 years ago
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    oh is it infinity

  14. anonymous
    • 5 years ago
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    yes

  15. anonymous
    • 5 years ago
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    so how was i able to decipher that the R value is infinity just from knowing the series converged absolutely?

  16. anonymous
    • 5 years ago
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    i guess im asking why does geting L=0 from the ratio test help me

  17. anonymous
    • 5 years ago
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    u can interpret it this way lim n->infin |x|(2/(n+1))<1 , then |x|<1/ 0 |x|<infinity=R

  18. anonymous
    • 5 years ago
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    X is a constant. So no mater what X is, the limit will go to 0, and the ratio test states that if the limit is <1, it converges. So X can be any real numbers

  19. anonymous
    • 5 years ago
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    ah ok, so becasue this series converges no matter what x-value is plugged in: my x value can range from -infinity to +infinity bc it will always converge. and then the radius for that interval is infinity also

  20. anonymous
    • 5 years ago
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    YEP that is it!!

  21. anonymous
    • 5 years ago
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    thankyou very much i appreciate it. and just for learning purposes, do i know this series is centered at 0?

  22. anonymous
    • 5 years ago
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    yes

  23. anonymous
    • 5 years ago
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    and that is bc the x-a isnt there?

  24. anonymous
    • 5 years ago
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    Becasue it is the form of c0+c1(x)+c2(x)^2 +....cn(x)^n

  25. anonymous
    • 5 years ago
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    yep!!

  26. anonymous
    • 5 years ago
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    so its in the c0 form

  27. anonymous
    • 5 years ago
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    i think i get this

  28. anonymous
    • 5 years ago
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    yea cn(x)^n then it is centered at 0, if it is cn(x-1)^n then it would be centered at x=1

  29. anonymous
    • 5 years ago
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    and in my equation here the 2^2/n! is the Cn part

  30. anonymous
    • 5 years ago
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    2^n/n!

  31. anonymous
    • 5 years ago
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    u got it?

  32. anonymous
    • 5 years ago
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    right right ok im dumb

  33. anonymous
    • 5 years ago
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    LOL no i know that is you meant.

  34. anonymous
    • 5 years ago
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    2^2 is just a typo~~

  35. anonymous
    • 5 years ago
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    yup my fault

  36. anonymous
    • 5 years ago
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    well i really do appreciate this. It may not seem it but you just helped someone who will most likely be getting early admission to UB med school ha ha Ive been getting pounded with organic and inorganic so this has been on the back burner

  37. anonymous
    • 5 years ago
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    LOL my pleasure and honor :P

  38. anonymous
    • 5 years ago
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    keep up the good work you are really patient, and a good listener. Guys like you are doing it for the love of the game if you will.

  39. anonymous
    • 5 years ago
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    LOL alright if u have any further questions, just post here :) i am gonna go now Bye :)

  40. anonymous
    • 5 years ago
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    oh can you find the radius of convergence for this power series : sum n = 0 to n =infinity ((x+3)^n)/(e^3n!)

  41. anonymous
    • 5 years ago
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    welp i think the ratio test gives you the limit to approach zero again, so i guess its infinity again?

  42. anonymous
    • 5 years ago
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    Thanks lol i was arguing with my friend about the convergence of this series.

  43. anonymous
    • 5 years ago
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    yeah like at the end of my ratio test i got x+3 ontop

  44. anonymous
    • 5 years ago
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    but the numerator is like e^3n+1 time e^3n+2 times e^3m+3 so thats infinity

  45. anonymous
    • 5 years ago
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    i just meant denominator* sorry so basicall that makes the denom infinity and the numerator x+3

  46. anonymous
    • 5 years ago
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    yea haha Thanks!~

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