Algorithms for approximating $\sqrt{2}$

Question

Well, "Solving" is the wrong term since I am speaking about irrational numbers. I just don't know which word is the correct word... So that can be part $1$ of my question... what is the correct word since you obviously can't "solve" an irrational number because it goes forever.

Part $2$ (my real question) are there algorithms for figuring out the answer to a problem like the square root of $2$ other than guess-and-checking your way to infinity? Again, I'm obviously not asking for an algorithm to give me the never ending answer because that's crazy... but for example if I wanted to know what the $15^{th}$ decimal place of the square root of $2$ was, is there an algorithm for that?

Thank you! (I'm new here and know nothing about how to format math questions so any help or links would be appreciated as well, thanks!)

Answer 1

You can use newton's method to compute the digits of $\sqrt{(2)}$:
Let: $$ f(x) = x^2 -2 $$ Define the iteration: $$ x_0 = 1\\ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} $$ This will converge to $\sqrt{2}$ quadratically.

If you want to compute other square roots:

Consider:
$$g(x) = x^2 - a$$


Which has the iterants: $$ x_{n+1}=\frac{1}{2}\left(x_n+\frac{a}{x_n}\right) $$ As mentioned below.

There's also what's called the continued fraction expansion of an algebraic number. You can use a finite continued fraction expansion.

As an example: $$ x_0 = 1 \\ x_1 = \frac{1}{2}\left(x_0 + \frac{2}{x_0}\right) =\frac{1}{2}\left( \large \textbf{1} + \frac{2}{ \large \mathbf{1}}\right) = \frac{3}{2}\\ x_2 = \frac{1}{2}\left(x_1 + \frac{2}{x_1}\right) = \frac{1}{2}\left( \large \mathbf{\frac{3}{2}} + \frac{2}{ \large \mathbf{\frac{3}{2}}}\right), \text{ etc. } $$

Added

Since we are using Newton's method, and you are wondering why it converges to the root of $f(x)$,

Note the following:
$\textbf{Theorem} $: Suppose that the function $f$ has a zero at $\alpha$, i.e., $f(\alpha) = 0$

If $f$ is continuously differentiable and its derivative is nonzero at $\alpha$, then there exists a neighborhood of $\alpha$ such that for all starting values $x_0$ in that neighborhood, the sequence ${x_n}$ will converge to $\alpha$.

So if we choose our starting guess appropriately, Newton's method always converges to the root of the equation if $f$ has these properties .

Answer 2

A related problem. Another way to go is the Taylor series. Derive the Taylor series of the function $\sqrt{x}$ at the point $x=1$

$$ \sqrt{x} = 1+{\frac {1}{2}} \left( x-1 \right) -{\frac {1}{8}} \left( x-1 \right) ^{2}+{\frac {1}{16}} \left( x-1 \right)^{3}-{\frac{5}{128} } \left( x-1 \right)^{4}+O\left( \left( x-1 \right) ^{5} \right). $$

If you plug in $x=2$, you get an approximate value for the $\sqrt{2}\sim 1.398437500$. Increasing the number of terms in the series improves the approximation.

Added: We can write the Taylor series of $\sqrt{x}$ explicitly by finding the $n$th derivative of $\sqrt{x}$ as

$$ \sqrt{x} = \sum _{n=0}^{\infty }\frac{\sqrt {\pi }}{2}\,{\frac {{a}^{\frac{1}{2}-n} \left( x-a\right)^{n}}{\Gamma\left( \frac{3}{2}-n \right)n! }}.$$

Substituting $a=1$ in the above formula gives the Taylor series at the point $a=1$:

$$\sqrt{x} = \sum _{n=0}^{\infty }\frac{\sqrt {\pi }}{2}\,{\frac { \left( x-1\right)^{n}}{\Gamma\left( \frac{3}{2} - n \right)n! }}.$$

Putting $x=2$ in the above equation, we have:

$$\sqrt{2} = \sum _{n=0}^{\infty }\,{\frac {\sqrt{\pi}}{2\,\Gamma\left( \frac{3}{2} - n \right)n! }}. $$

Answer 3

You can also compute square roots using continued fractions. For example for $\sqrt{2}$ you have $$ \sqrt{2}=1+(\sqrt{2}-1)=1+\frac{(\sqrt{2}-1)(\sqrt{2}+1)}{\sqrt{2}+1}=1+\frac{1}{\sqrt{2}+1} $$ where $1$ is the integer part of $\sqrt{2}$. Then repeat the process for $\sqrt{2}+1$ whose integer part is $2$: $$ \sqrt{2}+1=2+(\sqrt{2}-1)=2+\frac{(\sqrt{2}-1)(\sqrt{2}+1)}{\sqrt{2}+1}=2+\frac{1}{\sqrt{2}+1} $$ therefore by repeating the process we have $$ \sqrt{2}=1+\frac{1}{2+\frac{1}{\sqrt{2}+1}}=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\cdots}}}} $$

Answer 4

Following Rystyn's answer: it is nice to write down the decimals to understand how good the convergence is in Newton's method:

1.000000000000000000000000000000000000000000000000000000000000000000000  
1.500000000000000000000000000000000000000000000000000000000000000000000  
1.416666666666666666666666666666666666666666666666666666666666666666666  
1.414215686274509803921568627450980392156862745098039215686274509803921   
1.414213562374689910626295578890134910116559622115744044584905019200054  
1.414213562373095048801689623502530243614981925776197428498289498623195  
1.414213562373095048801688724209698078569671875377234001561013133113265  
1.414213562373095048801688724209698078569671875376948073176679737990732  
1.414213562373095048801688724209698078569671875376948073176679737990732 
1.414213562373095048801688724209698078569671875376948073176679737990732