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Math Help - Napier's Discovery of LN Question

  1. #1
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    Napier's Discovery of LN Question

    I have been doing some extra credit for calculus and I have done problems 1-6 using what is a variation of what Napier did to find the natural log, except to expedite the process we are using spreadsheets and such (obviously to avoid the tedious handwritten work). Anyways, here's the problem:
    Code:
    This problem is a reprise of problem 6 with 1.001 replaced by 1.000001  Compute an approximation to 
    
    which gives the area under for , using a modified Riemann sum with the (NOT equally spaced) partition
    and left hand endpoints EXCEPT neglecting the area of the last rectangle. Here denotes the largest possible power which fits in the interval . Unfortunately we can not determine by the same method we used in Problem 6, since this would require constructing a spreadsheet of such a gigantic size that it would not fit on a personal computer. Hence we will use a similar method as was devised by Napier, to bootstrap our calculations with respect to the base 1.001 so as to get a good approximation to . We begin by constructing a spreadsheet of integer powers of 1.000001 with about 1000 rows. We then search this spreadsheet for integer powers of 1.000001 bracketing the value 1.001. We find that
    where = Linearly interpolating between these two successive integer powers of 1.000001, we obtain the following estimate: (Round your answer to EXACTLY 3 decimal places after the decimal point.) Since , we obtain the following initial estimate for , using the estimate of above and the calculation of in Problem 3: This estimate of is too low. We can improve this estimate by creating a new spreadsheet which starts with the value of calculated in the spreadsheet of Problem 3 and multiplies this value by successive integer powers of 1.000001 until we get the best approximation to 7 from below. Adding this extra exponent of 1.000001 to our initial estimate of above, we obtain the improved estimate We now multiply this estimate of the number of approximating rectangles by the exact area of each rectangle, which is . This gives the following approximation to the integral : . This is very close to the actual value of the integral , which is . The ingenious idea of using these unusual nonequally spaced partitions to compute the area under thus relating it to Napier's logarithms is due to a Belgian monk, Gregory St. Vincent around 1647.
    I got the first two portions of the problem right. I am having trouble doing the third part, the "Since , we obtain the following initial estimate for , using the estimate of above and the calculation of in Problem 3:" question.

    I have tried a bunch of variations for N. 1.000001^1945, 1.000001^1946, among a ton of different decimal approximations, which is what the problem is looking for. I have tried the problem 35 times with various decimal approximations, but nothing works. If it helps: the Int(log1.001(7)) is 1946, and the estimation of log1.000001(1.001) is 999.501.
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  2. #2
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    Please see rule #6: http://www.mathhelpforum.com/math-he...ng-151418.html.

    Thread closed.

    Quote Originally Posted by anonymousperson View Post
    I have been doing some extra credit for calculus and I have done problems 1-6 using what is a variation of what Napier did to find the natural log, except to expedite the process we are using spreadsheets and such (obviously to avoid the tedious handwritten work). Anyways, here's the problem:
    Code:
    This problem is a reprise of problem 6 with 1.001 replaced by 1.000001  Compute an approximation to 
    
    which gives the area under for , using a modified Riemann sum with the (NOT equally spaced) partition
    and left hand endpoints EXCEPT neglecting the area of the last rectangle. Here denotes the largest possible power which fits in the interval . Unfortunately we can not determine by the same method we used in Problem 6, since this would require constructing a spreadsheet of such a gigantic size that it would not fit on a personal computer. Hence we will use a similar method as was devised by Napier, to bootstrap our calculations with respect to the base 1.001 so as to get a good approximation to . We begin by constructing a spreadsheet of integer powers of 1.000001 with about 1000 rows. We then search this spreadsheet for integer powers of 1.000001 bracketing the value 1.001. We find that
    where = Linearly interpolating between these two successive integer powers of 1.000001, we obtain the following estimate: (Round your answer to EXACTLY 3 decimal places after the decimal point.) Since , we obtain the following initial estimate for , using the estimate of above and the calculation of in Problem 3: This estimate of is too low. We can improve this estimate by creating a new spreadsheet which starts with the value of calculated in the spreadsheet of Problem 3 and multiplies this value by successive integer powers of 1.000001 until we get the best approximation to 7 from below. Adding this extra exponent of 1.000001 to our initial estimate of above, we obtain the improved estimate We now multiply this estimate of the number of approximating rectangles by the exact area of each rectangle, which is . This gives the following approximation to the integral : . This is very close to the actual value of the integral , which is . The ingenious idea of using these unusual nonequally spaced partitions to compute the area under thus relating it to Napier's logarithms is due to a Belgian monk, Gregory St. Vincent around 1647.
    I got the first two portions of the problem right. I am having trouble doing the third part, the "Since , we obtain the following initial estimate for , using the estimate of above and the calculation of in Problem 3:" question.

    I have tried a bunch of variations for N. 1.000001^1945, 1.000001^1946, among a ton of different decimal approximations, which is what the problem is looking for. I have tried the problem 35 times with various decimal approximations, but nothing works. If it helps: the Int(log1.001(7)) is 1946, and the estimation of log1.000001(1.001) is 999.501.
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