Approximating square roots using binomial expansion.

Question

Through the binomial expansion of $(1 - 2x)^\frac{1}{2}$, I am required to find an approximation of $\sqrt2$.

Binomial expansion

$ (1 + x)^n = 1 + \frac{n}{1}x + \frac{n(n-1)}{1*2}x^2 + ... $

Thus, the expansion of $(1 - 2x)^\frac{1}{2}$: $$ = 1 - x -\frac{1}{2}x^2 - \frac{1}{2}x^3 + ... $$

The suggested way, is to choose a value for $x$ so that $(1-2x)$ has the form $2*$'a perfect square'. This can be done by taking $x = 0.01$. Thus, $(1 - 2x)=(1-2*0.01) = 0.98 = 2*0.7^2$

And $$ (1 - 2x)^\frac{1}{2} = 0.98^\frac{1}{2} = 0.7\sqrt2$$ Which is equal to the previously established expansion, so we can now go ahead and find $\sqrt2$. The problem I am facing, is that there was no mention of how the value of $x=0.01$ was arrived at.

Is there an easy way to determine an appropriate value for $x$?

Answer 1

Personally, I wouldn't have done it that way. So here is how I would've done it:

Method 1:

$$\sqrt2=\sqrt{1+1}=1+\frac12-\frac18+\dots\approx1+\frac12-\frac18=\frac{11}8=1.375$$

which is much clearer to me, since it avoids having to take decimals raised to powers and gives you something you can easily do by hand.

$$1.375^2=1.890625$$

Obviously it approaches the correct value as you take more terms.

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Method 2:

This is called fixed-point iteration/Newton's method, and it basically goes like this:

$$x=\sqrt2\implies x^2=2$$

$$2x^2=2+x^2$$

Divide both sides by $2x$ and we get

$$x=\frac{2+x^2}{2x}$$

Now, interestingly, I'm going to call the $x$'s on the left $x_{n+1}$ and the $x$'s on the right $x_n$, so

$$x_{n+1}=\frac{2+(x_n)^2}{2x_n}$$

and with $x_0\approx\sqrt2$, we will then have $x=\lim_{n\to\infty}x_n$. For example, with $x_0=1$,

$x_0=1$

$x_1=\frac{2+1^2}{2(1)}=\frac32=1.5$

$x_2=\frac{2+(3/2)^2}{2(3/2)}=\frac{17}{12}=1.41666\dots$

$x_3=\dots=\frac{577}{408}=1.414215686$

And one can quickly check that $(x_3)^2=2.000006007\dots$, which is pretty much the square root of $2$.

Answer 2

It is often easier to begin with a slightly different problem to simplify slightly. Stay with me and you will see why.

Instead of finding the square root of 2, look at the square root of 200 and note that:

200 = 14² + 4

This is part of a general approach that says if N = a² + b, where b << a, then we may write:

$$\sqrt N = a(1 + \frac b {a^2} ) =a[1 + \frac b {2a^2} - \frac {b^2} {8a^4} + \frac {b^3} {16a^6} - ... ] = a + \frac b {2a} - \frac {b^2} {8a^3} + ...$$

Now, if we use N = 200, a = 14, b = 4, we get the first two terms give:

$$ \sqrt 200 = 14\frac 1 7 $$

That is $ \sqrt 2 = 1.4142857...$ which is correct to 5 significant figures. By including the next term, we get $\sqrt 200 = 14 + \frac 1 7 - \frac 1 {1372}$. This gives the superb result:

$ \sqrt 2 = 1.414212827...$, which is correct to 6 significant figures.

APPENDIX - Another example of $\sqrt{13}$

Using the same approach and only the first two terms, we note that 36² = 1296. Thus, $\sqrt{1300} = 36\frac 1 {18} = $. This gives:

$\sqrt{13} = 3.60555 $, correct to 6 significant figures.