Express tan(x + π/4) using the formula of Taylor series.

Steps to Solve

1. Taylor Series Formula The Taylor series expansion of a function $f(x)$ about the point $a$ is given by: $$ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots $$

2. Identify the Function and Point For this problem, we want to expand $f(x) = \tan(x)$ around the point $a = \frac{\pi}{4}$.

3. Calculate Function Value at a First, compute $\tan\left(\frac{\pi}{4}\right)$: $$ f\left(\frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1 $$

4. Compute Derivatives Next, we compute the derivatives of $\tan(x)$ at $x = \frac{\pi}{4}$.

• First derivative: $$ f'(x) = \sec^2(x) $$ Calculating at $x = \frac{\pi}{4}$: $$ f'\left(\frac{\pi}{4}\right) = \sec^2\left(\frac{\pi}{4}\right) = 2 $$

• Second derivative: $$ f''(x) = 2\sec^2(x)\tan(x) $$ Calculating at $x = \frac{\pi}{4}$: $$ f''\left(\frac{\pi}{4}\right) = 2 \cdot 2 \cdot 1 = 4 $$

• Third derivative: $$ f'''(x) = 2\sec^2(x)(\tan^2(x) + 1) $$ Calculating at $x = \frac{\pi}{4}$: $$ f'''\left(\frac{\pi}{4}\right) = 2 \cdot 2 \cdot (1^2 + 1) = 8 $$

5. Construct the Taylor Series Expansion Substituting the derivatives back into the Taylor series formula: $$ \tan(x) = 1 + 2\left(x - \frac{\pi}{4}\right) + \frac{4}{2!}\left(x - \frac{\pi}{4}\right)^2 + \frac{8}{3!}\left(x - \frac{\pi}{4}\right)^3 + \ldots $$

Simplifying those terms leads to: $$ \tan(x + \frac{\pi}{4}) = 1 + 2\left(x - \frac{\pi}{4}\right) + 2\left(x - \frac{\pi}{4}\right)^2 + \frac{4}{3}\left(x - \frac{\pi}{4}\right)^3 + \ldots $$