# Approximating Binomial Series I'm so close

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 beacause (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

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I think you have used an incorrect formula
The correct one is:
$$(1-x)^{\alpha}=\sum_{k=0}^{\infty}(-1)^k$\left(\begin{array}{c}\alpha & k \end{array}\right)x^k$$$

where
$$$\left(\begin{array}{c}\alpha & k \end{array}\right)=\frac{1}{r!}{\prod_{j=0}^{r-1}{(\alpha-j)}}$$$

Dick
Homework Helper
eys_physics is correct. The series isn't alternating. And starting at the (1/36)^4 term your binomial coefficients aren't correct either. I think you just extrapolated from the form of the earlier terms. Wrongly. What you've got going there is a geometric series. If that were correct you could sum it and show sqrt(35) is rational (!?!?!?!).

(35) = sqrt(35*36/36) = 6*sqrt(35/36)
6*sqrt(35/36) = 6[(1 - 1/36)^(1/2)] =

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]

= 6.082762528 which is different from sqrt(35) = 5.916079783

I got a different series after calculating for the binomial. I calculated up to k=4 because the last term was 2.3*10^(-8) and it wanted an accuracy of 10^(-7).

Dick