Question

Find the maclaurin series for sinx and cosx and show that they converge for all x to their respective functions.

Answer 1

Let's consider a particular case of Taylor Series, in the region near $x = 0$ . Such a polynomial is called the Maclaurin Series. The infinite series expansion for $f(x)$ about $x = 0$ becomes:

$$f (x) \approx f (0) + f ^ {\prime} (0) x + \frac {f ^ {\prime \prime} (0)}{2} x ^ {2} + \frac {f ^ {\prime \prime \prime} (0)}{3 !} x ^ {3} + \frac {f ^ {\mathrm {i v}} (0)}{4 !} x ^ {4} + \dots$$

Let's find the Maclaurin Series expansion for $f(x) = \sin x$ :

$$\sin x = \sin 0 + x \cos 0 + \frac {x ^ {2} \sin 0}{2} + \frac {x ^ {3} \cos 0}{3 !} + \dots = x - \frac {1}{6} x ^ {3} + \frac {1}{1 2 0} x ^ {5} - \frac {1}{5 0 4 0} x ^ {7} + \dots$$

and for $f(x) = \cos x$ :

$$\cos x = \cos 0 + x \sin 0 + \frac {x ^ {2} \cos 0}{2} + \frac {x ^ {3} \sin 0}{3 !} + \dots = 1 - \frac {1}{2} x ^ {3} + \frac {1}{2 4} x ^ {4} - \frac {1}{7 2 0} x ^ {6} + \dots$$

We plot our answer to see if the polynomial is a good approximation to $f(x) = \sin x$ .

We observe that our polynomial (in red) is a good approximation to $f(x) = \sin x$ (in blue) near $x = 0$ . In fact, it is quite good between $-3 \leq x \leq 3$ . So, the statement that the Maclaurin series converge to their respective function for all $x$ is wrong.