## anonymous 5 years ago i need to express x/1-x as a power series

1. anonymous

1 sec

2. anonymous

ok

3. anonymous

sorry bout that mate

4. anonymous

?

5. anonymous

so the series is $\sum_{0}^{\infty} x / (1-x)$

6. anonymous

yeah

7. anonymous

im sorry that is completely wrong. 1/(1-x) = $\sum_{0}^{\infty} x^n$ im sorry that is completely wrong

8. anonymous

where you are sending 0 --> infinity

9. anonymous

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

and you sum allll of those terms

11. anonymous

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

right? you can find that series in any caluculus text book

13. anonymous

that series has to be memorized

14. anonymous

ya

15. anonymous

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

ok

17. anonymous

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

cool right?

19. anonymous

ok can i ask you one more question

20. anonymous

yeah

21. anonymous

so to find the interval of convergence how would i go about that

22. anonymous

okay

23. anonymous

have you learned the nth root test? or ratio test?

24. anonymous

yeah

25. anonymous

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

can you simplify that? dont froget the ABS and show me what you get

27. anonymous

Is that just X^N?

28. anonymous

|x^N|

29. anonymous

you can look at it as multiples in the numerator and denominator

30. anonymous

you have x^(n+2) = x^n * x * x right? its all just boring exponent work

31. anonymous

you get that?

32. anonymous

in the numerator

33. anonymous

yeah

34. anonymous

you get that?

35. anonymous

So is it just X^N+1?

36. anonymous

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

So just X then?

38. anonymous

yeah. real quick work it out on a piece of paper while i type up the next part

39. anonymous

ok

40. anonymous

the ratio test states Xn+1 / Xn < $\left| x_{n+1}/x _{n} \right|< \rho =1$ Xn+1 / Xn = x

41. anonymous

you can plug solve for x in the inequality | x | < p = 1 | x | < 1 -1 > x > 1

42. anonymous

so that is the IOC

43. anonymous

ah ok got it

44. anonymous

you can test the end points now An = X^(n+1)

45. anonymous

so to check for the end points you just plug into x

46. anonymous

yeah

47. anonymous

perfect

48. anonymous

and for 1 it diverges because it doesnt converge and -1 it diverges also right

49. anonymous

but into the series. its kind of something a little technical x^n doesnt equal series x^n

50. anonymous

yeah because it is a divergent geometric series where r must be <1 r^n

51. anonymous

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

alright got it

53. anonymous

the ratio test is just difficult with the exponents

54. anonymous

ok thanks