- #1
sami23
- 76
- 1
Use binomial series to approximate sqrt (35) with an accuracy of 10^(-7)
(35) = sqrt(35*36/36) = 6*sqrt(35/36)
Formula: (1+x)^n where x=(-1/36) and n=(1/2):
6*sqrt(35/36) = 6[(1 + (- 1/36))^(1/2)] =
The coefficients of the binomial series are:
'1/2 choose 0' is 1.
'1/2 choose 1' is 1(1/2)/1 = 1/2
'1/2 choose 2' is 1/2(1/2-1)/2 = -1/8
'1/2 choose 3' is -1/8(1/2-2)/3 = 1/16
'1/2 choose 4' is 1/16(1/2-3)/4 = -5/128
from k=0 to k=4:
= 6[(1 + (1/2)*(-1/36) - (1/8)*(-1/36)^2 + (-1/16)*(1/36)^3 -
(5/128)*(-1/36)^4]
= 6[(1 - (1/2)*(1/36) + (1/8)*(1/36)^2 - (1/16)*(1/36)^3 +
(5/128)*(1/36)^4]
= 5.917237472 but it has to be more accurate
I don't know where the mistake is in the series. I used the 4 terms because (5/128)(1/36) = 2.30*10^(-8) Please help, thanks again.
(35) = sqrt(35*36/36) = 6*sqrt(35/36)
Formula: (1+x)^n where x=(-1/36) and n=(1/2):
6*sqrt(35/36) = 6[(1 + (- 1/36))^(1/2)] =
The coefficients of the binomial series are:
'1/2 choose 0' is 1.
'1/2 choose 1' is 1(1/2)/1 = 1/2
'1/2 choose 2' is 1/2(1/2-1)/2 = -1/8
'1/2 choose 3' is -1/8(1/2-2)/3 = 1/16
'1/2 choose 4' is 1/16(1/2-3)/4 = -5/128
from k=0 to k=4:
= 6[(1 + (1/2)*(-1/36) - (1/8)*(-1/36)^2 + (-1/16)*(1/36)^3 -
(5/128)*(-1/36)^4]
= 6[(1 - (1/2)*(1/36) + (1/8)*(1/36)^2 - (1/16)*(1/36)^3 +
(5/128)*(1/36)^4]
= 5.917237472 but it has to be more accurate
I don't know where the mistake is in the series. I used the 4 terms because (5/128)(1/36) = 2.30*10^(-8) Please help, thanks again.