Problem 8 In Problems $6-10$, compute th... [FREE SOLUTION]

Chapter 7: Problem 8

In Problems $6-10$, compute the Taylor polynomial of degree $n$ about $a=0$ for the indicated functions. $$ f(x)=e^{3 x}, n=3 $$

Short Answer

Expert verified

The Taylor polynomial of degree 3 for $ f(x) = e^{3x} $ at $ a = 0 $ is $ T_3(x) = 1 + 3x + \frac{9}{2}x^2 + \frac{9}{2}x^3 $.

Step by step solution

Understand the Taylor Series Formula

The Taylor series of a function $ f(x) $ about $ a = 0 $ is given by $ T_n(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots + \frac{f^{(n)}(0)x^n}{n!} $. Here, we will compute terms up to the 3rd degree ($ n=3 $).

Calculate the Derivatives

The function is $ f(x) = e^{3x} $. Compute the derivatives needed:1. $ f(x) = e^{3x} $2. First derivative: $ f'(x) = 3e^{3x} $3. Second derivative: $ f''(x) = 9e^{3x} $4. Third derivative: $ f'''(x) = 27e^{3x} $

Evaluate the Derivatives at $ x = 0 $

Substitute $ x = 0 $ into each derivative to find their values:1. $ f(0) = e^{0} = 1 $2. $ f'(0) = 3e^{0} = 3 $3. $ f''(0) = 9e^{0} = 9 $4. $ f'''(0) = 27e^{0} = 27 $

Form the Taylor Polynomial of Degree 3

Substitute the values from Step 3 into the Taylor series formula:\[ T_3(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} \]\[ = 1 + 3x + \frac{9x^2}{2} + \frac{27x^3}{6} \]Simplifying, we get:$ T_3(x) = 1 + 3x + \frac{9}{2}x^2 + \frac{27}{6}x^3 $$ = 1 + 3x + \frac{9}{2}x^2 + \frac{9}{2}x^3 $

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivatives

In calculus, derivatives measure how a function changes as its input changes. They tell us the rate at which a function is increasing or decreasing at any point. For the function $ f(x) = e^{3x} $, the process of finding derivatives involves applying the chain rule repeatedly. Here's what happens step by step:

The original function is $ f(x) = e^{3x} $.
The first derivative, $ f'(x) = 3e^{3x} $, tells us how fast the function is changing at each point $ x $.
The second derivative, $ f''(x) = 9e^{3x} $, provides information on the curvature of the graph at any point - whether it's curving towards or away from the axis.
Finally, the third derivative, $ f'''(x) = 27e^{3x} $, can give insights about the rate of change of the curvature itself.

Understanding derivatives helps to piece together how a curve behaves as a whole, and they are essential in forming Taylor polynomials.

Exponential Function

The exponential function, often denoted as $ e^x $, is a mathematical function that has constant proportionality. It's unique in that its rate of growth is proportional to its current value. Specifically, the exponential function $ e^{3x} $ grows faster as $ x $ increases because the exponent is also increasing. Here are some important points to understand:

It's defined by a constant base, $ e $, raised to the power of $ 3x $.
The number $ e $ is approximately equal to 2.71828 and is known as Euler's number. It's an irrational number, similar to $ \pi $.
Exponential functions appear frequently in models where growth processes are multiplicative, like populations or investment returns.

The beauty of the exponential function is its simplicity in calculus, particularly the property that the derivative of $ e^{3x} $ is proportional to its original expression, cascading smoothly in more complex calculations like Taylor series expansions.

Series Expansion

Taylor series expansion is a method to represent functions as infinite sums of terms. A Taylor series approximates a given function around a specific point, which can make complex functions more digestible with simple polynomials. Here's how it connects to our problem:

We aim to create a polynomial approximation of $ f(x) = e^{3x} $ around $ a = 0 $ up to the third degree.
The formula for constructing the Taylor polynomial involves taking derivatives of the function, evaluating them at the point of expansion (here, $ x = 0 $), and inserting these into the Taylor series formula.
For degree 3, we compute up to the third derivative, plugging in the values to get terms like $ f(0), \, f'(0)x, \, \frac{f''(0)x^2}{2}, \, \frac{f'''(0)x^3}{6} $.

By adding up these terms, we form a polynomial that closely approximates the original function $ f(x) = e^{3x} $ near $ x = 0 $, leading to the expression $ T_3(x) = 1 + 3x + \frac{9}{2}x^2 + \frac{9}{2}x^3 $. This approximation is not only simpler to work with but provides a clear representation of how $ e^{3x} $ behaves locally, near the point of expansion.

91影视

In Problems \(6-10\), compute the Taylor polynomial of degree \(n\) about \(a=0\) for the indicated functions. $$ f(x)=e^{3 x}, n=3 $$

Short Answer

Step by step solution

Understand the Taylor Series Formula

Calculate the Derivatives

Evaluate the Derivatives at \( x = 0 \)

Form the Taylor Polynomial of Degree 3

Key Concepts

Derivatives

Exponential Function

Series Expansion

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