anonymous
  • anonymous
Calc II Alternating Remainder Estimate Help Needed
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Determine 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
  • anonymous
I understand that you would set the sum equal to .0001 and solve for k. Right?
anonymous
  • anonymous
I can simplify the equation by saying (1/k!)x(7/6)^k right?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
If 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
  • anonymous
I 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
  • anonymous
If 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
  • anonymous
And by the way, doing a series test without a calculator sucks a lot, huh
anonymous
  • anonymous
I agree with @bmp and add that you must fire your teacher.
anonymous
  • anonymous
Im 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
  • anonymous
Why 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.