Approximating Binomial Series I'm so close

  • Thread starter Thread starter sami23
  • Start date Start date
  • Tags Tags
    Binomial Series
Click For Summary
SUMMARY

The forum discussion focuses on using the Binomial Series to approximate the square root of 35 with a precision of 10^(-7). The initial approach involved rewriting sqrt(35) as 6*sqrt(1-(1/36)) and applying the binomial series expansion. However, the user, Sami, struggled to achieve the desired accuracy with only 5 terms, resulting in an approximation of 5.917237444 compared to the actual value of 5.916079783. Feedback from other users highlighted the need for correct binomial coefficients and the importance of the alternating series estimation theorem to determine the necessary number of terms for the approximation.

PREREQUISITES
  • Understanding of Binomial Series and its formulas, particularly (1-x)^r and the summation notation.
  • Familiarity with Taylor series and their applications in approximating functions.
  • Knowledge of binomial coefficients and their calculation.
  • Basic concepts of series convergence and error estimation techniques.
NEXT STEPS
  • Study the Alternating Series Estimation Theorem to understand how to determine the number of terms needed for a specific accuracy.
  • Learn about binomial coefficients and their role in series expansions, focusing on the correct application in binomial series.
  • Explore Taylor series and their applications in function approximation, particularly for square roots.
  • Practice deriving and using the Binomial Series with different values to solidify understanding of convergence and accuracy.
USEFUL FOR

Mathematics students, educators, and anyone interested in numerical methods for function approximation, particularly in the context of series expansions and error analysis.

sami23
Messages
69
Reaction score
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
 
Physics news on Phys.org
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
\[\left(\begin{array}{c}\alpha & k \end{array}\right)=\frac{1}{r!}{\prod_{j=0}^{r-1}{(\alpha-j)}}\]
 
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).
 
Notice the (-1)^k in the formula eys_physics posted. Also notice the binomial coefficients generally alternate in sign. Most of the terms in your series should have the same sign.
 

Similar threads

Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K