91Ó°ÊÓ

In general, the Taylor polynomial of order \(n\) is defined as: $$ P_n(x) = f(a) + f'(a)(x-a) + \frac{1}{2}f''(a)(x-a)^2 + \cdots + \frac{1}{n!}f^{(n)}(a)(x-a)^n $$ Where \(f^{(n)}(a)\) is the \(n\)th derivative of \(f(x)\) evaluated at \(x=a\). In this problem, the function is \(f(x) = \tan x\), and so we will first calculate the higher-order derivatives and then plug them into the Taylor polynomial formula. We are also given that a = 0. Thus our Taylor polynomial expression becomes: $$ P_n(x) = f(0) + f'(0)(x-0) + \frac{1}{2}f''(0)(x-0)^2 + \cdots + \frac{1}{n!}f^{(n)}(0)(x-0)^n $$

Now we compute the first few derivatives of the function \(f(x) = \tan x\). First derivative: $$ f'(x) = \frac{d}{dx}(\tan x) = \sec^2{x} $$ Second derivative: $$ f''(x) = \frac{d^2}{dx^2}(\tan x) = \frac{d}{dx}(\sec^2{x}) = 2\sec{x}\sec{x}\tan{x} = 2\sec^2{x}\tan{x} $$

Now, we evaluate these derivatives at \(a=0\): $$ f(0) = \tan{0} = 0 $$ $$ f'(0) = \sec^2{0} = 1 $$ $$ f''(0) = 2\sec^2{0}\tan{0} = 0 $$

For \(n=0,1,2\): - \(P_0(x) = f(0) = 0\) - \(P_1(x) = f(0) + f'(0)(x-0) = 0 + 1x = x\) - \(P_2(x) = f(0) + f'(0)(x-0) + \frac{1}{2}f''(0)(x-0)^2 = 0 + 1x + 0 = x\)

Now we will graph the Taylor polynomials \(P_0(x)\), \(P_1(x)\), and \(P_2(x)\), as well as the original function \(f(x)=\tan x\). Be sure to set an appropriate window to properly visualize the graphs. Since the tangent function has asymptotes at odd multiples of \(90^\circ\) or \(\pi/2\), consider focusing on a window that contains one or two periods of the graph, for example, \(-2\pi \leq x \leq 2\pi\). For the y-axis, choose a range large enough to display the graphs, for example, \(-10 \leq y \leq 10\). You will notice that the second order Taylor polynomial and the first order Taylor polynomial are equal, but despite this the Taylor polynomials only provide a reasonable approximation close to the x-axis and do not capture the vertical asymptotes present in the tangent function.