- #1
sami23
- 76
- 1
Use Binomial Series to approximate sqrt(35) with an accuracy of 10^(-7)
Formulas for binomial series: (1-x)^r and sum{from 0 to n}(r k)(x)^k
sqrt(35) = sqrt(35*36/36) = 6*sqrt(35/36) = 6*sqrt(1-(1/36))
Now it looks more like the binomial series formula:
let r = 1/2 because of the radical and x = 1/36
6*[1-(1/36)]^(1/2) and use the second formula to expand:
6* sum{from 0 to n}(1/2 k)(1/36)^k
I did from k=0 to k=5:
= 6[1-(1/2)*(1/36)+(1/8)*(1/36)^2-(1/16)*(1/36)^3+(1/32)*(1/36)^4-(1/64)*(1/36)^5+...]
But I didn't get close enough with only 5 terms. First I used 4 terms because (1/32)*(1/36)^4 gave me 1.9*10^(-8) which I thought would be enough terms needed for an accuracy of 10^(-7) but wasn't.
I eventually got 6[0.98620624] = 5.917237444
and the actual sqrt(35) = 5.916079783
+- 0.001 and I need +- 0.0000001
My professor didn't teach us to use the alternating series estimation theorem which I though would help me find how many n terms I would need to use but I don't understand. Please help. I think I'm on the right track I'm just not quite there.
Thanks,
Sami
Formulas for binomial series: (1-x)^r and sum{from 0 to n}(r k)(x)^k
sqrt(35) = sqrt(35*36/36) = 6*sqrt(35/36) = 6*sqrt(1-(1/36))
Now it looks more like the binomial series formula:
let r = 1/2 because of the radical and x = 1/36
6*[1-(1/36)]^(1/2) and use the second formula to expand:
6* sum{from 0 to n}(1/2 k)(1/36)^k
I did from k=0 to k=5:
= 6[1-(1/2)*(1/36)+(1/8)*(1/36)^2-(1/16)*(1/36)^3+(1/32)*(1/36)^4-(1/64)*(1/36)^5+...]
But I didn't get close enough with only 5 terms. First I used 4 terms because (1/32)*(1/36)^4 gave me 1.9*10^(-8) which I thought would be enough terms needed for an accuracy of 10^(-7) but wasn't.
I eventually got 6[0.98620624] = 5.917237444
and the actual sqrt(35) = 5.916079783
+- 0.001 and I need +- 0.0000001
My professor didn't teach us to use the alternating series estimation theorem which I though would help me find how many n terms I would need to use but I don't understand. Please help. I think I'm on the right track I'm just not quite there.
Thanks,
Sami