Question

The coefficient of \(x^5\) in the Taylor series expansion of \(f(x)=\tan(x)\) about the point \(x=0\) is ____.

• A. \(2/15\)
• B. \(1/5!\)
• C. \(1/5\)
• D. \(1\)

Hint:

Multiply the series of sin x and sec x, or differentiate y' = 1 + y^2 repeatedly at x=0.

Answer 1

The Correct Option is A

Step 1: Use the known Maclaurin series of \(\sin x\) and \(\cos x\) up to \(x^5\): \[\sin x = x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots,\qquad \cos x = 1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots\]

Step 2: Find the series for \(\sec x = 1/\cos x\) up to \(x^4\): \[\sec x = 1+\frac{x^2}{2}+\frac{5x^4}{24}+\cdots\]

Step 3: Multiply \(\sin x\) by \(\sec x\) to get \(\tan x\), keeping terms up to \(x^5\): \[\tan x=\left(x-\frac{x^3}{6}+\frac{x^5}{120}\right)\left(1+\frac{x^2}{2}+\frac{5x^4}{24}\right)\]

Step 4: Collect the \(x^5\) terms from the product: \(x\cdot\frac{5x^4}{24}=\frac{5x^5}{24}\), \(-\frac{x^3}{6}\cdot\frac{x^2}{2}=-\frac{x^5}{12}\), and \(\frac{x^5}{120}\cdot 1=\frac{x^5}{120}\). Using a common denominator of 120: \[\frac{5}{24}=\frac{25}{120},\quad \frac{1}{12}=\frac{10}{120},\quad \frac{1}{120}=\frac{1}{120}\] \[\text{Coefficient}=\frac{25}{120}-\frac{10}{120}+\frac{1}{120}=\frac{16}{120}=\frac{2}{15}\]

Step 5: This matches the well known expansion \(\tan x=x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+\cdots\).

\[\boxed{\dfrac{2}{15}}\]