Questions: Calculate the first 3 terms of the Taylor series about a = 0 of: tan x x - x^3/3! + x^5/5! x + x^3/3 + 2x^5/12 Does not exist x + x^2/6 + x^3/120

Transcript text: Calculate the first 3 terms of the Taylor series about a = 0 of: $\tan x$ $x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}$ $x+\frac{x^{3}}{3}+\frac{2 x^{5}}{12}$ Does not exist $x+\frac{x^{2}}{6}+\frac{x^{3}}{120}$

Solution

Solution Steps

Step 1: Define the Function

We start by defining the function $ f(x) = \tan(x) $.

Step 2: Calculate the Derivatives

Next, we compute the first three derivatives of $ f(x) $:

The first derivative is $ f'(x) = \sec^2(x) = \tan^2(x) + 1 $.
The second derivative is $ f''(x) = 2\tan^2(x) + 2 $.
The third derivative is $ f'''(x) = (2\tan^2(x) + 2)(\tan^2(x) + 1) + 2(2\tan^2(x) + 2)\tan^2(x) $.

Step 3: Evaluate Derivatives at $ x = 0 $

We evaluate the function and its derivatives at $ x = 0 $:

$ f(0) = \tan(0) = 0 $.
$ f'(0) = \sec^2(0) = 1 $.
$ f''(0) = 2\tan^2(0) + 2 = 0 + 2 = 0 $.
$ f'''(0) = 2 $.

Step 4: Construct the Taylor Series

Using the values obtained, we construct the Taylor series about $ a = 0 $: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 \] Substituting the evaluated values: \[ f(x) = 0 + 1 \cdot x + \frac{0}{2}x^2 + \frac{2}{6}x^3 = x + \frac{1}{3}x^3 \]

Step 5: Write the First Three Terms

The first three terms of the Taylor series for $ \tan(x) $ about $ a = 0 $ are: \[ f(x) = x + \frac{1}{3}x^3 \]