# Thread: Using a power series expansion to approx. Square root 3.

1. ## Using a power series expansion to approx. Square root 3.

Following up from my previous post (Power series expansion of (1-x)^-1/2. Approximating the value of square root 2.), this is the real deal. The question is; "Use a power series expansion to approximate Square Root 3. Comment on the reasonableness of your result." Disclaimer: we have been taught nothing on Power series, Taylor series or binomial series. After doing 2 hours of research online, I find myself still confused. I understand how they expand now because of these articles https://en.wikipedia.org/wiki/Taylor...eral_variables , https://en.wikipedia.org/wiki/Binomial_theorem.

I can't understand how they work to approximate square root values. Look at this and open the answer for Example 5b: 4. The Binomial Theorem .It's a procedure for approximating the value of a cube root and it looks like it worked (ofc, they used an expansion for cube root). I thought I understood it then but I tried it for the expansion of square root 3. Applying the same method; so after working out the expansion and then substituting x for 2, it just didn't work. No where near the answer.

Really stuck here. Any help is appreciated, even if you can improve my understanding of these power series expansions being used for approximation.

2. ## Re: Using a power series expansion to approx. Square root 3.

1) $\displaystyle x=2$ gives $\displaystyle \frac1{\sqrt{1-2}}=\sqrt{-1}$. You need $\displaystyle x$ such that $\displaystyle \sqrt3=\sqrt{\frac{1}{\frac13}}=\frac1{\sqrt{1-x}}$
3) How quickly does the series converge (if it does).

3. ## Re: Using a power series expansion to approx. Square root 3.

Originally Posted by Archie
1) $\displaystyle x=2$ gives $\displaystyle \frac1{\sqrt{1-2}}=\sqrt{-1}$. You need $\displaystyle x$ such that $\displaystyle \sqrt3=\sqrt{\frac{1}{\frac13}}=\frac1{\sqrt{1-x}}$
3) How quickly does the series converge (if it does).
Is it the same if I use (1+x)^-1/2. I don't really understand the differences between the different expansions for square roots and which change causes what. Is doing (1-x)^-1/2 instead just to avoid having a negative x value? I think it might be the same thing actually

4. ## Re: Using a power series expansion to approx. Square root 3.

Well I found out it does converge but really slowly. I did it up to x^9 and it only got 1.72356... which is only correct to 1 decimal places to the actual approx. of sqrt(3) of 1.732050... Since the terms are getting smaller, I guess ill just say its reasonable when i get it correct to 2 decimal places. I think that should be ok right?

5. ## Re: Using a power series expansion to approx. Square root 3.

I would suggest that it works, but it's not very reasonable. The sequence

$\displaystyle a_{n+1}=\frac12 \left( a_n + \frac3{a_n} \right)$

converges much more quickly. ($\displaystyle a_9$ will be more accurate than most calculators can display for any reasonable value of $\displaystyle a_0$).

6. ## Re: Using a power series expansion to approx. Square root 3.

Originally Posted by Bumdwarf
Is it the same if I use (1+x)^-1/2
Yes. The difference is that $\displaystyle x^{2k+1}=\left(-\tfrac23\right)^{2k+1} < 0$. The using the series for $\displaystyle \tfrac{1}{\sqrt{1-x}}$ gives the same result, but the negatives are part of the series with each $\displaystyle x^{2k+1}=\left(\tfrac23\right)^{2k+1} > 0$.

7. ## Re: Using a power series expansion to approx. Square root 3.

Originally Posted by Bumdwarf
Is it the same if I use (1+x)^-1/2.
Originally Posted by Bumdwarf
Well I found out it does converge but really slowly. I did it up to x^9 and it only got 1.72356...
which is only correct to 1 decimal places to the actual approx. of sqrt(3) of 1.732050...
With it being $\displaystyle \bigg(1 - \dfrac{2}{3}\bigg)^{-1/2},$ |2/3| is a relatively large percentage of 1.

For a faster convergence (without using the strong recursive formula in post # 5,) you could use $\displaystyle (4 - 1)^{-1/2}$.

|-1| is a relatively smaller percentage of 4. Six terms gives an approximation of about 1.732088.