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anonymous

  • 5 years ago

i need to express x/1-x as a power series

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  1. anonymous
    • 5 years ago
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    1 sec

  2. anonymous
    • 5 years ago
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    ok

  3. anonymous
    • 5 years ago
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    sorry bout that mate

  4. anonymous
    • 5 years ago
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    ?

  5. anonymous
    • 5 years ago
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    so the series is \[\sum_{0}^{\infty} x / (1-x)\]

  6. anonymous
    • 5 years ago
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    yeah

  7. anonymous
    • 5 years ago
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    im sorry that is completely wrong. 1/(1-x) = \[\sum_{0}^{\infty} x^n\] im sorry that is completely wrong

  8. anonymous
    • 5 years ago
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    where you are sending 0 --> infinity

  9. anonymous
    • 5 years ago
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    what a series does is it models something else by using and infinite amount of terms. that is why you can equate what i stated above

  10. anonymous
    • 5 years ago
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    and you sum allll of those terms

  11. anonymous
    • 5 years ago
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    1/(1-x) is defined by \[\sum_{n \rightarrow0}^{\infty}x^n\] = \[1 + x^2 + x^3 + x^4 + x^5 + ... + x^n\] but this is not your equation.

  12. anonymous
    • 5 years ago
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    right? you can find that series in any caluculus text book

  13. anonymous
    • 5 years ago
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    that series has to be memorized

  14. anonymous
    • 5 years ago
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    ya

  15. anonymous
    • 5 years ago
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    so to get to your equation you must multiply everyyy term by x because x * 1/(1-x) = \[x *\sum_{n=0}^{\infty} x^n\]

  16. anonymous
    • 5 years ago
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    ok

  17. anonymous
    • 5 years ago
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    multiplying by x your exponents are n and 1. when multiplying like bases you add exponents you get n + 1 so. \[x/(1-x) = \sum_{n=0}^{\infty} x^{n+1}\] = \[x + x^2 + x^3 + x^4 + x^5 + .... + x^{n+1}\]

  18. anonymous
    • 5 years ago
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    cool right?

  19. anonymous
    • 5 years ago
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    ok can i ask you one more question

  20. anonymous
    • 5 years ago
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    yeah

  21. anonymous
    • 5 years ago
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    so to find the interval of convergence how would i go about that

  22. anonymous
    • 5 years ago
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    okay

  23. anonymous
    • 5 years ago
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    have you learned the nth root test? or ratio test?

  24. anonymous
    • 5 years ago
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    yeah

  25. anonymous
    • 5 years ago
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    so you can take the ratio test. of the series. which states you must take the absolute value of the An+1 and divide by An you should get | x^[ (n+1) +1 ] / x^(n+1) | = | x^(n+2) / x^(n+1) |

  26. anonymous
    • 5 years ago
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    can you simplify that? dont froget the ABS and show me what you get

  27. anonymous
    • 5 years ago
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    Is that just X^N?

  28. anonymous
    • 5 years ago
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    |x^N|

  29. anonymous
    • 5 years ago
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    you can look at it as multiples in the numerator and denominator

  30. anonymous
    • 5 years ago
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    you have x^(n+2) = x^n * x * x right? its all just boring exponent work

  31. anonymous
    • 5 years ago
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    you get that?

  32. anonymous
    • 5 years ago
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    in the numerator

  33. anonymous
    • 5 years ago
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    yeah

  34. anonymous
    • 5 years ago
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    you get that?

  35. anonymous
    • 5 years ago
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    So is it just X^N+1?

  36. anonymous
    • 5 years ago
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    dont forget you have a multiple of x^n in the denominator so they cancel each other x^(n+2) / x^(n+1) = { x^n * x * x } / { x^n * x }

  37. anonymous
    • 5 years ago
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    So just X then?

  38. anonymous
    • 5 years ago
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    yeah. real quick work it out on a piece of paper while i type up the next part

  39. anonymous
    • 5 years ago
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    ok

  40. anonymous
    • 5 years ago
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    the ratio test states Xn+1 / Xn < \[\left| x_{n+1}/x _{n} \right|< \rho =1\] Xn+1 / Xn = x

  41. anonymous
    • 5 years ago
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    you can plug solve for x in the inequality | x | < p = 1 | x | < 1 -1 > x > 1

  42. anonymous
    • 5 years ago
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    so that is the IOC

  43. anonymous
    • 5 years ago
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    ah ok got it

  44. anonymous
    • 5 years ago
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    you can test the end points now An = X^(n+1)

  45. anonymous
    • 5 years ago
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    so to check for the end points you just plug into x

  46. anonymous
    • 5 years ago
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    yeah

  47. anonymous
    • 5 years ago
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    perfect

  48. anonymous
    • 5 years ago
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    and for 1 it diverges because it doesnt converge and -1 it diverges also right

  49. anonymous
    • 5 years ago
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    but into the series. its kind of something a little technical x^n doesnt equal series x^n

  50. anonymous
    • 5 years ago
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    yeah because it is a divergent geometric series where r must be <1 r^n

  51. anonymous
    • 5 years ago
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    or acutally its |r| < 1 to converge and your r's are -1 and 1. any x that is a fraction, for this series, would converge.

  52. anonymous
    • 5 years ago
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    alright got it

  53. anonymous
    • 5 years ago
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    the ratio test is just difficult with the exponents

  54. anonymous
    • 5 years ago
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    ok thanks

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