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 !
What is the answer to part B?
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
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.
^ I may have made an error here, so feel free to check my work
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.
To approximate the time it will take for the computer algorithm to run through 60 iterations, we can calculate the definite integral of T(k) from 1 to 60. This will give us the total time taken to process all the iterations.
A. To calculate the definite integral, we can use the formula:
∫(1 to 60) 1.2^k + 1 dk
Integrating 1.2^k with respect to k gives us (1.2^k) / ln(1.2), and integrating 1 gives us k. So we have:
∫(1 to 60) (1.2^k + 1) dk = [(1.2^k) / ln(1.2)] + k | (1 to 60)
Substituting the limits of integration, we get:
[(1.2^60) / ln(1.2)] + 60 - [(1.2^1) / ln(1.2)] - 1
Approximately, the total time it will take the computer algorithm to run through 60 iterations is:
[(1.2^60) / ln(1.2)] + 59.2 seconds.
To convert this to hours, we divide by 3600 (since there are 3600 seconds in an hour):
[(1.2^60) / ln(1.2)] + 59.2 / 3600 hours.
B. The maximum error in the computer's solution after k iterations is given by Error = 2k^-2.
We want to find the number of iterations required to reduce the maximum error to below 0.0001. So we can set up the following equation:
2k^-2 < 0.0001
To solve this inequality, we can take the reciprocal of both sides:
k^2 > 1 / (2 * 0.0001)
Simplifying, we get:
k^2 > 5000
Taking the square root of both sides:
k > √5000
Approximately, k > 70.71.
So it will take at least 71 iterations for the computer to process enough iterations to reduce the maximum error to below 0.0001. We can use the T(k) formula to calculate the time taken for the 71st iteration:
T(71) = 1.2^71 + 1
Convert this to hours by dividing by 3600:
T(71) / 3600 hours.
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!