Numerical approximation (with error estimate)
Task number: 3074
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.
