Problem 1
Question
Compute the nth Taylor Polynomial at 0 for the exponential function.
Step-by-Step Solution
Verified Answer
The nth Taylor Polynomial for \( e^x \) at 0 is \( T_n(x) = \sum_{k=0}^{n} \frac{1}{k!} x^k \).
1Step 1: Understand the Function
We want to find the Taylor polynomial of the exponential function, which is given by \( f(x) = e^x \). This function is special because the exponential function is equal to its own derivative.
2Step 2: Taylor Series Formula
The Taylor series of a function \( f(x) \) at \( a = 0 \) is given by the formula:\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \]where \( f^{(n)}(0) \) represents the \( n \)-th derivative of \( f(x) \) evaluated at \( x=0 \).
3Step 3: Derivative of the Exponential Function
For \( f(x) = e^x \), each derivative is the same as the original function. This means:\[ f^{(n)}(x) = e^x \]Evaluating this at \( x = 0 \) gives \( f^{(n)}(0) = e^0 = 1 \) for all \( n \).
4Step 4: Construct the nth Taylor Polynomial
Substitute the derivatives into the Taylor series formula:\[ T_n(x) = \sum_{k=0}^{n} \frac{1}{k!} x^k \]This gives us the nth Taylor polynomial of the exponential function centered at 0.
Key Concepts
Exponential FunctionTaylor SeriesDerivative
Exponential Function
The exponential function, commonly represented as \( f(x) = e^x \), is a unique and fundamental function in mathematics. Its key property is that its rate of growth (derivative) is always proportional to its current value. This means
- The exponential function is equal to its own derivative.
- Therefore, the function itself is unchanged by differentiation.
Taylor Series
A Taylor series is an infinite sum of terms that are derived from the values of a function's derivatives at a single point. It's used to approximate functions with polynomials, making complex functions easier to work with. For any function \( f(x) \), the Taylor series centered at 0 is represented as
- The formula: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \]
- Each term of the series is determined by the nth derivative of \( f(x) \) evaluated at 0, divided by \( n! \), with \( x^n \) as a multiplying factor.
Derivative
The derivative of a function is a core concept in calculus that describes the rate at which a function is changing at any given point. For the exponential function \( f(x) = e^x \), the derivative holds the incredible property of being the same as the original function. This can be stated mathematically as
- \( f'(x) = e^x \)
- Higher order derivatives also yield the same expression, \( f^{(n)}(x) = e^x \), for all orders \( n \).
- Each coefficient in the Taylor polynomial \( T_n(x) = \sum_{k=0}^{n} \frac{1}{k!} x^k \) for the exponential function is simply \( \frac{1}{k!} \).
Other exercises in this chapter
Problem 1
Suppose \(f: \mathbb{R} \rightarrow \mathbb{R}\) is continuously differentiable such that \(f^{\prime}(x)>0\) for all \(x .\) Show that \(f\) is invertible on t
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Prove the product rule. Hint: Use \(f(x) g(x)-f(c) g(c)=f(x)(g(x)-g(c))+(f(x)-\) \(f(c)) g(c)\).
View solution Problem 2
Suppose I, \(J\) are intervals and a monotone onto \(f: I \rightarrow J\) has an inverse \(g: J \rightarrow I\). Suppose you already know that both \(f\) and \(
View solution Problem 2
Suppose p is a polynomial of degree d. Given any \(x_{0} \in \mathbb{R}\), show that the \((d+1)\) th Taylor polynomial for \(p\) at \(x_{0}\) is equal to \(p\)
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