sum(k=1..60) 1.2^k + 1 ≈ ∫[1..60] (1.2^x + 1) dx
= (ln1.2 * 1.2^x + x)[1..60]
= 309,108 seconds
A certain computer algorithm used to solve very complicated differential equations uses an iterative method. That is, the algorithm solves the problem the first time very approximately, and then uses that first solution to help it solve the problem a second time just a little bit better, and then uses that second solution to help it solve the problem a third time just a little bit better, and so on. Unfortunately, each iteration (each new problem solved by using the previous solution) takes a progressively longer amount of time. In fact, the amount of time it takes to process the k-th iteration is given by T(k) = 1.2^k + 1 seconds.
A. Use a definite integral to approximate the time (in hours) it will take the computer algorithm to run through 60 iterations. (Note that T(k) is the amount of time it takes to process just the k-th iteration.)
B. The maximum error in the computer's solution after k iterations is given by Error = 2k^-2. Approximately how long (in hours) will it take the computer to process enough iterations to reduce the maximum error to below 0. 0001?
** Please include explanations if possible !
6 answers
What is the answer to part B?
b. First, find the number of iterations to get an error of less than .0001 by finding the number to get .0001 (and knowing that any number greater will "reduce error below .0001"). So, 2k^-2=.0001, or 2k^2=1/.001, and k=sqrt(5000) or approx. 70.7107. Since the number of iterations is always whole, that means it's at .0001 by about 70, so set k=71.
Plug that into the original equation given by the problem to get 1.2^71+1=418667.7483. Divide by 3600 (to convert to hours) and your final answer is approximately 116.2966 hours!
Plug that into the original equation given by the problem to get 1.2^71+1=418667.7483. Divide by 3600 (to convert to hours) and your final answer is approximately 116.2966 hours!
^ I may have made an error here, so feel free to check my work
Above, you went from 2k^-2=0.0001, to 2k^2 = 1/.001. I think you forgot a zero. Otherwise, the answer I got was k=142, then plugging it in, I got 48689401.708 hours. Might be too big, but idk.
don't you have to plug it into the definite integral not the original equation because the question is asking for the total time for all the iterations up to 142 too not just the time it takes for the 142th iteration.
2 answers