Estimating an Error Function from a Taylor Series

[Jewgro]

Hi,

Using Excel 2010, I am attempting to estimate the following error function using a Taylor Series when x = 1:

imgur: the simple image sharer

I am trying to find how many " n " terms I need to sum over the function to attain/estimate the value of erf(1). The only tools I can use in excel are the SUM() function and the Goal Seek/Solver add-in.

Any help would be much appreciated!

 

[Worf]

Hi and welcome to the Board
Here is an idea:
- Column A = n
- Column B = corresponding term
- Column C = error function with n terms
- Column D = difference between last two terms, when very small you can consider that the function has stabilized.
It seems that a good answer would be n=13.

Sheet2

*	A	B	C	D
1	n	term	erf	finished?
2	0	1,12838	1,12838	*
3	1	-3,38514	-2,25676	FALSE
4	2	2,82095	0,56419	FALSE
5	3	-1,316443333	-0,752253333	FALSE
6	4	0,4231425	-0,329110833	FALSE
7	5	-0,103434833	-0,432545667	FALSE
8	6	0,020373528	-0,412172139	FALSE
9	7	-0,003358274	-0,415530413	FALSE
10	8	0,000475755	-0,415054657	FALSE
11	9	-5,90807E-05	-0,415113738	FALSE
12	10	6,52998E-06	-0,415107208	FALSE
13	11	-6,50171E-07	-0,415107858	FALSE
14	12	5,88923E-08	-0,415107799	TRUE
15	13	-4,89259E-09	-0,415107804	TRUE
16	14	3,75357E-10	-0,415107804	TRUE
17	15	-2,67496E-11	-0,415107804	TRUE

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Spreadsheet Formulas

Cell Formula B2 =(1.12838*POWER(-1,A2)/FACT(A2)*(2*A2+1)) C2 =SUM(B2) B3 =(1.12838*POWER(-1,A3)/FACT(A3)*(2*A3+1)) C3 =SUM(C2,B3) D3 =IF(ABS(C3-C2)<0.0000001,"TRUE","FALSE") B4 =(1.12838*POWER(-1,A4)/FACT(A4)*(2*A4+1)) C4 =SUM(C3,B4) D4 =IF(ABS(C4-C3)<0.0000001,"TRUE","FALSE") B5 =(1.12838*POWER(-1,A5)/FACT(A5)*(2*A5+1)) C5 =SUM(C4,B5) D5 =IF(ABS(C5-C4)<0.0000001,"TRUE","FALSE") B6 =(1.12838*POWER(-1,A6)/FACT(A6)*(2*A6+1)) C6 =SUM(C5,B6) D6 =IF(ABS(C6-C5)<0.0000001,"TRUE","FALSE") B7 =(1.12838*POWER(-1,A7)/FACT(A7)*(2*A7+1)) C7 =SUM(C6,B7) D7 =IF(ABS(C7-C6)<0.0000001,"TRUE","FALSE") B8 =(1.12838*POWER(-1,A8)/FACT(A8)*(2*A8+1)) C8 =SUM(C7,B8) D8 =IF(ABS(C8-C7)<0.0000001,"TRUE","FALSE") B9 =(1.12838*POWER(-1,A9)/FACT(A9)*(2*A9+1)) C9 =SUM(C8,B9) D9 =IF(ABS(C9-C8)<0.0000001,"TRUE","FALSE") B10 =(1.12838*POWER(-1,A10)/FACT(A10)*(2*A10+1)) C10 =SUM(C9,B10) D10 =IF(ABS(C10-C9)<0.0000001,"TRUE","FALSE") B11 =(1.12838*POWER(-1,A11)/FACT(A11)*(2*A11+1)) C11 =SUM(C10,B11) D11 =IF(ABS(C11-C10)<0.0000001,"TRUE","FALSE") B12 =(1.12838*POWER(-1,A12)/FACT(A12)*(2*A12+1)) C12 =SUM(C11,B12) D12 =IF(ABS(C12-C11)<0.0000001,"TRUE","FALSE") B13 =(1.12838*POWER(-1,A13)/FACT(A13)*(2*A13+1)) C13 =SUM(C12,B13) D13 =IF(ABS(C13-C12)<0.0000001,"TRUE","FALSE") B14 =(1.12838*POWER(-1,A14)/FACT(A14)*(2*A14+1)) C14 =SUM(C13,B14) D14 =IF(ABS(C14-C13)<0.0000001,"TRUE","FALSE") B15 =(1.12838*POWER(-1,A15)/FACT(A15)*(2*A15+1)) C15 =SUM(C14,B15) D15 =IF(ABS(C15-C14)<0.0000001,"TRUE","FALSE") B16 =(1.12838*POWER(-1,A16)/FACT(A16)*(2*A16+1)) C16 =SUM(C15,B16) D16 =IF(ABS(C16-C15)<0.0000001,"TRUE","FALSE") B17 =(1.12838*POWER(-1,A17)/FACT(A17)*(2*A17+1)) C17 =SUM(C16,B17) D17 =IF(ABS(C17-C16)<0.0000001,"TRUE","FALSE") <tbody> </tbody>

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Excel tables to the web >> Excel Jeanie HTML 4

 

[Jewgro]

Thank you for the reply!

I was able to find out the answer using a similar method and it turned out that your answer and mine were very similar . I obtained a value of n = 14.

Cheers.