## Numerical approximation (with error estimate)

Using a Taylor polynomial, approximately compute $$\sqrt 2$$ and estimate the error.

• #### Resolution

Let $\binom{a}{n} = \frac{a(a-1)\cdots(a-(n-1))}{n!}$ for $$a \in \mathbb R$$ and $$n \in \mathbb N_0$$.

Let us write the Taylor polynomial for the function $$(1 + x)^{\frac{1}{2}}$$:

$(1 + x)^{\frac{1}{2}} = 1 + \binom{\frac12}1 x + \binom{\frac12}2 x^2 + \binom{\frac12}3 x^3 + R_4(x),$ where $$R_4(x)$$ is the remainder of the Taylor polynomial.

The remainder also has the property that $$R_4(x) = \frac{f^{(4)}(c)}{4!}\cdot(x-0)^4$$ for some $$c$$ between $$1$$ and $$x$$, where $$f = (1 + x)^{\frac12}$$. So we compute the fourth derivative $f^{(4)}(x) = -\frac{15}{16}(1+x)^{-\frac72}$ So for $$x = 1$$ the absolute value of the error will be at most $$\frac{15}{16 {\cdot} 4!} = 0.0390625$$.

So finally we can substitute into the Taylor polynomial:

$(1 + 1)^{\frac{1}{2}} \approx 1 + \binom{\frac12}1 1 + \binom{\frac12}2 1 + \binom{\frac12}3 1 = 1 + \frac12 - \frac18 + \frac3{48} = 1.4375.$

The actual value of $$\sqrt 2$$ is approximately $$1.4142$$. If we wanted to determine it more precisely, we see that we could use a higher-degree Taylor polynomial.