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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the series is \[\sum_{0}^{\infty} x / (1x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im sorry that is completely wrong. 1/(1x) = \[\sum_{0}^{\infty} x^n\] im sorry that is completely wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where you are sending 0 > infinity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and you sum allll of those terms

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/(1x) 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right? you can find that series in any caluculus text book

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that series has to be memorized

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so to get to your equation you must multiply everyyy term by x because x * 1/(1x) = \[x *\sum_{n=0}^{\infty} x^n\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0multiplying by x your exponents are n and 1. when multiplying like bases you add exponents you get n + 1 so. \[x/(1x) = \sum_{n=0}^{\infty} x^{n+1}\] = \[x + x^2 + x^3 + x^4 + x^5 + .... + x^{n+1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok can i ask you one more question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so to find the interval of convergence how would i go about that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0have you learned the nth root test? or ratio test?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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) 

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you simplify that? dont froget the ABS and show me what you get

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can look at it as multiples in the numerator and denominator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you have x^(n+2) = x^n * x * x right? its all just boring exponent work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dont 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 }

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah. real quick work it out on a piece of paper while i type up the next part

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the ratio test states Xn+1 / Xn < \[\left x_{n+1}/x _{n} \right< \rho =1\] Xn+1 / Xn = x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can plug solve for x in the inequality  x  < p = 1  x  < 1 1 > x > 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can test the end points now An = X^(n+1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so to check for the end points you just plug into x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and for 1 it diverges because it doesnt converge and 1 it diverges also right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but into the series. its kind of something a little technical x^n doesnt equal series x^n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah because it is a divergent geometric series where r must be <1 r^n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the ratio test is just difficult with the exponents
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