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anonymous
 4 years ago
Calc II Alternating Remainder Estimate Help Needed
anonymous
 4 years ago
Calc II Alternating Remainder Estimate Help Needed

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Determine the smallest number of terms required to approximate the sum of the series with an error less than 0.0001. \[\sum_{n}^{\infty}(7^k)/(6^kk!)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I understand that you would set the sum equal to .0001 and solve for k. Right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I can simplify the equation by saying (1/k!)x(7/6)^k right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If I understood the problem correctly, there are a couple of ways to do it. Remember that the remainder estimate for alternating series is given by:\[R_n \le b_{n+1}\]So yeah, you can simplify it and put in the form:\[0,0001 \le \frac{7^{k+1}}{6^{k+1}(k+1)!} = \frac{1}{(k+1)!}*(\frac{7}{6})^{k+1}\]Or if you have a calculator, you can start expanding some terms until one of them is < 0.0001

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I did that and I got to the seventh term, it was the right answer! Thanks! But on the test I won't be able to use a calculator, can this problem be solved some way w/o one?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If they ask for the smallest number, I guess you have to 1) manually compute the values or 2) solve an equation of the form above. If they do not, just pick a reasonable number and compute it, like n = 10. Very likely the error will be really small and it would suffice the boundary for Rn

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And by the way, doing a series test without a calculator sucks a lot, huh

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I agree with @bmp and add that you must fire your teacher.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Im not sure if we will be given this same type of problem but we did go over a problem like this in class that was WAY easier that we solved with very little effort

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Why should one compute 7^k for k=1 to 7 and similar problems by hand. Those antiquated teachers should be trained to undersatand that the world has changed and they should follow the flow.
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