Chapter 30: Problem 23
Find the first two nonzero terms of the Maclaurin expansion of the given functions. $$f(x)=\tan x$$
Short Answer
Expert verified
The first two non-zero terms of the Maclaurin expansion of \( \tan x \) are \( x \) and \( \frac{1}{3} x^3 \).
Step by step solution
01
Understand Maclaurin Series
The Maclaurin series is a special case of the Taylor series that expands a function around 0. It's given by \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \). To find the non-zero terms, we'll differentiate \( f(x) = \tan x \) and evaluate terms at \( x = 0 \).
02
Calculate First Derivative
Differentiate \( f(x) = \tan x \) to get the first derivative: \( f'(x) = \sec^2 x \). Evaluate this at \( x = 0 \): \( f'(0) = \sec^2(0) = 1 \). This gives the first non-zero term \( x \) in the expansion.
03
Calculate Second Derivative
Find the second derivative \( f''(x) = 2\sec^2x \cdot \sec x \cdot \tan x = 2\sec^2x \tan x \). Evaluate this at \( x = 0 \): \( f''(0) = 2 \cdot 1 \cdot 0 = 0 \). The second term is zero.
04
Calculate Third Derivative
Differentiate a third time to get \( f'''(x) = 2\sec^2 x \cdot \sec^2 x + 2\sec^2 x \cdot \tan^2 x \). Evaluate this at \( x = 0 \): \( f'''(0) = 2 \cdot 1 + 0 = 2 \). The third non-zero term contributes \( \frac{2}{3!} x^3 \).
05
Write the Nonzero Terms
Now combine the non-zero terms found: the first non-zero term is \( x \) and the second is \( \frac{1}{3} x^3 \). Thus, the first two non-zero terms in the expansion are \( x + \frac{1}{3} x^3 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor Series
The Taylor Series is a powerful mathematical tool used to approximate functions. It expresses a function as an infinite sum of terms calculated from the function's derivatives at a single point. The Maclaurin Series is a special type of Taylor Series that approximates functions around the point zero. It's given by the formula:\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \]This series is immensely powerful because it can simplify complex functions into something more manageable. Particularly, by taking multiple derivatives of a function at zero, we can form a polynomial that approaches the function's real behavior over small values of \( x \).
- Start with the function.
- Take its derivatives.
- Evaluate each derivative at zero.
- Add each term together for the series.
Differentiation
Differentiation is a key component in forming a Taylor or Maclaurin series. It involves finding the derivative of the function, which measures how the function changes as the input changes. For the function \( \tan x \), its derivatives are unique and form the basis for its series expansion.When we differentiate \( \tan x \), we start with:- First derivative: \( f'(x) = \sec^2x \)- Second derivative: \( f''(x) = 2\sec^2x \cdot \sec x \cdot \tan x \)With differentiation, we take these derivatives and substitute \( x = 0 \). This simplifies to finding the crucial terms of the series:- At \( x = 0 \), \( \sec^2(0) = 1 \)- Expands to terms like \( x \) and higher powers.Thus, each term in the series reflects how the function "slides" or changes as \( x \) increases from zero.
Trigonometric Functions
Trigonometric functions like \( \tan x \) behave in periodic and often complex manners. In calculus and series expansions, we deal with understanding and approximating their values using familiar mathematical techniques. Expanding \( \tan x \) into a Maclaurin series involves breaking it into manageable parts with derivatives.Key characteristics of trig functions in expansions:
- Are periodic, meaning they repeat values in regular intervals.
- Involve derivatives that are themselves trigonometric identities.
- Can be expressed via their infinite series for precise approximation.