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#1
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Summation Operator
Hi,
I'm trying to work out a way of getting Excel 2003 to answer y=2i+1 when i=1 initially with a final value of 10. y=(2x1+1)+(2x2+1)+...+(2x10+1) I could type it by hand however my next question requires the upper limit n to be infinity for the eqn y=1/i(i-1) when i=1 initially and the final value for i=infinity y=(1/(1(1+1))+(1/2(2+1)+...(1/infinity(infinity+1) and I eed to find the approximate answer for y when the addition of the successive terms causes a lass than 0.0001% change in y, as well as the last value for i when the program stops. Am utterly stumped! Please help, Thank you. |
#2
Posted to microsoft.public.excel.worksheet.functions
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Summation Operator
In A1 enter:
..5 In A2 enter: =A1+1/(ROW()*(ROW()+1)) and copy down In B2 enter: =(A2-A1)/A1 and copy down For the first 10 rows: 0.50000000 0.66666667 33.3333333333% 0.75000000 12.5000000000% 0.80000000 6.6666666667% 0.83333333 4.1666666667% 0.85714286 2.8571428571% 0.87500000 2.0833333333% 0.88888889 1.5873015873% 0.90000000 1.2500000000% 0.90909091 1.0101010101% and for rows 999 thru 1002: 0.99900000 0.0001002004% 0.99900100 0.0001000001% 0.99900200 0.0000998004% 0.99900299 0.0000996013% so for i1000 the relative change term-to-term will be under your limit. -- Gary''s Student - gsnu200797 |
#3
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Summation Operator
y=2i+1 when i=1 initially with a final value of 10.
y=(2x1+1)+(2x2+1)+...+(2x10+1) Hi. For this first question... n=10 ? n + n*(1 + n) 120 For Second question: Chg= 0.000001 ? (Sqr(Chg^2 + Chg) - Chg)/Chg 999.000499999875 Hence, 999 is the last value. 1,000 will put you over. and I need to find the approximate answer for y when the addition of the successive terms causes a less than 0.0001% change in y n=999 ? n/(1 + n) 0.999 -- HTH :) Dana DeLouis "Sapphy" wrote in message ... Hi, I'm trying to work out a way of getting Excel 2003 to answer y=2i+1 when i=1 initially with a final value of 10. y=(2x1+1)+(2x2+1)+...+(2x10+1) I could type it by hand however my next question requires the upper limit n to be infinity for the eqn y=1/i(i-1) when i=1 initially and the final value for i=infinity y=(1/(1(1+1))+(1/2(2+1)+...(1/infinity(infinity+1) and I eed to find the approximate answer for y when the addition of the successive terms causes a lass than 0.0001% change in y, as well as the last value for i when the program stops. Am utterly stumped! Please help, Thank you. |
#4
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Summation Operator
Hi. I posted a correction a while ago, but I still don't see it. Let me try again. I hope this isn't posted twice.
I made a mistake for #2. When we round down, then the percentage difference between (n+1) and n has not reached our desired percentage change. Hence, we need to Round Up. If Chg equals 0.000001 Then =CEILING((SQRT(Chg^2 + Chg) - Chg)/Chg,1) Returns 1000. With n = 1000, then this last value is =n/(n+1) =0.999000999000999 For the first question, when n=10, then: =n*(2+n) Returns 120. -- HTH :) Dana DeLouis "Sapphy" wrote in message ... Hi, I'm trying to work out a way of getting Excel 2003 to answer y=2i+1 when i=1 initially with a final value of 10. y=(2x1+1)+(2x2+1)+...+(2x10+1) I could type it by hand however my next question requires the upper limit n to be infinity for the eqn y=1/i(i-1) when i=1 initially and the final value for i=infinity y=(1/(1(1+1))+(1/2(2+1)+...(1/infinity(infinity+1) and I eed to find the approximate answer for y when the addition of the successive terms causes a lass than 0.0001% change in y, as well as the last value for i when the program stops. Am utterly stumped! Please help, Thank you. |
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