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:

I got the first two portions of the problem right. I am having trouble doing the third part, the "Since https://webwork.math.ohio-state.edu/...70f7251291.png, we obtain the following initial estimate for https://webwork.math.ohio-state.edu/...73e5daf3b1.png, using the estimate of https://webwork.math.ohio-state.edu/...61a09457d1.png above and the calculation of https://webwork.math.ohio-state.edu/...1438043ae1.png in Problem 3:" question.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 https://webwork.math.ohio-state.edu/...138add4e11.png for https://webwork.math.ohio-state.edu/...7a73dcc6e1.png, using a modified Riemann sum with the (NOT equally spaced) partition and left hand endpoints EXCEPT neglecting the area of the last rectangle.

Here https://webwork.math.ohio-state.edu/...73e5daf3b1.png denotes the largest possible power which fits in the interval https://webwork.math.ohio-state.edu/...7a73dcc6e1.png. Unfortunately we can not determine https://webwork.math.ohio-state.edu/...73e5daf3b1.png 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 https://webwork.math.ohio-state.edu/...73e5daf3b1.png. 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 https://webwork.math.ohio-state.edu/...70877535c1.png =

Linearly interpolating between these two successive integer powers of 1.000001, we obtain the following estimate:

https://webwork.math.ohio-state.edu/...9b7848d011.png

(Round your answer to EXACTLY 3 decimal places after the decimal point.) Since https://webwork.math.ohio-state.edu/...70f7251291.png, we obtain the following initial estimate for https://webwork.math.ohio-state.edu/...73e5daf3b1.png, using the estimate of https://webwork.math.ohio-state.edu/...61a09457d1.png above and the calculation of https://webwork.math.ohio-state.edu/...1438043ae1.png in Problem 3:

https://webwork.math.ohio-state.edu/...8e45d5fda1.png

This estimate of https://webwork.math.ohio-state.edu/...73e5daf3b1.png is too low. We can improve this estimate by creating a new spreadsheet which starts with the value of https://webwork.math.ohio-state.edu/...d0caed5a81.png 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 https://webwork.math.ohio-state.edu/...73e5daf3b1.png above, we obtain the improved estimate

https://webwork.math.ohio-state.edu/...8e45d5fda1.png

We now multiply this estimate https://webwork.math.ohio-state.edu/...73e5daf3b1.png of the number of approximating rectangles by the exact area of each rectangle, which is . This gives the following approximation to the integral https://webwork.math.ohio-state.edu/...f6b8f45931.png: .

This is very close to the actual value of the integral https://webwork.math.ohio-state.edu/...f6b8f45931.png, which is .

The ingenious idea of using these unusual nonequally spaced partitions to compute the area under https://webwork.math.ohio-state.edu/...138add4e11.png thus relating it to Napier's logarithms is due to a Belgian monk, Gregory St. Vincent around 1647.

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.