f(n)=
-n9/3628800+11n8/725760-11n7/30240+121n6/24192-7513n5/172800+8591n4/34560-341693n3/362880+84095n2/36288-177133n/50400+7381/2520.
The computations are lengthy and would take up too much text space to include here.
f(11)=-1.569×10-11 approx. Note that f(0)=a1=7381/2520. f(n) for n between 1 and 10 does give 1/n. But outside that range the polynomial doesn't work to give 1/n.
If the coefficients are multiplied by 10! (=3628800), then factorised, in order we get:
-1, 11×5, -11×5!, 112×150, -11×683×21, 112×71×3×5×7, -11×31063×2×5, 112×139×22×53, -11×16103×6!, 112×61×25×32×5. Note the preponderance of 11 in the numerator.
METHOD
Best to use Excel or other comparable spreadsheet app to do the row manipulations and calculations below.
Let f(n)=∑arnr-1 for 1≤r≤10 be the polynomial of degree 9.
Since f(n)=1/n, ∑arnr-1=1/n, so ∑arnr=1. For example, ∑ar×3r=1.
Create a table with column headers a1, a2, ..., a10, and an extra column called SUM and 55 rows.
First of the 55 rows consists of all 1s, including the SUM column.
Second row is 2, 4, 8, ..., 512, 1024, 1.
Third row is 3, 9, 27, ..., 59049, 1
Continue to row 10 containing 10, 100, 1000, ..., 10000000000, 1.
That completes rows 1-10 (R1-R10).
The next 9 rows (R11-R19) consist of row operations. Include the SUM column in the operations.
R11=R2-2R1, R12=R3-3R1, ..., R19=R10-10R1.
This operation forces zeroes into the a1 column (C1).
To create R20, divide the number in C2R12 by that in C2R11 this is the multiplier m=R12/R11. Then perform the operation R20=R12-mR11. This produces 0 in C2R20 but non-zero values in the rest of the row.
Similarly, in C2, m=R13/R11, and R21=R13-mR11, so R21=0 but non-zero values in the rest of the row.
Proceed like this up to R27. R28-R34 follow the same pattern as for R20-R27. The number of rows to be processed like this decreases by 1 with each batch. So we have R35-R40, then R41-R45, R46-R49, R50-R52, R53-R54, and finally R55, which will have a zero in C9 and 3628800 in C10, and -1 in the SUM column.
The SUM value at the end of the first row of each batch should be 1 or -1, and the sign alternates between batches. Other patterns will emerge during the process, and these should help to check the calculations.
From R55 we can calculate a10=SUM/3628800=-1/3628800. Now we can use R53 or R54 to find a9. R53 (the first row of the 2-row batch) is probably the easiest to work with. R53 effectively reads:
362880a9+16329600a10=1, so a9=(1-16329600a10)/362880.
If R54 had been used it would have produced the same result. It's a good idea to check coefficients occasionally by using rows other than the first row of each batch. Every row will produce the same result if no errors have been made.
Moving upward to the next batch of 3 rows and selecting R50 we can substitute for a9 and a10 so a8=(-1-1451520a9-30240000a10)/40320, (note that 362880=9! and 40320=8!) and so on until all the coefficients have been found. There is a lot of computation involved, so a calculator is really necessary which can supply the required accuracy. It helps if the coefficients are expressed as fractions rather than decimals, and it may take a little effort to convert decimals to exact fractions if the calculator displays decimals for some coefficients. The final polynomial f(x) is shown in small print at the beginning of this solution.
More to follow...
