Problem 22 - Calculus

Question

Find the value of \(f^{(5)}(1)\) if \(f(x)\) is approximated near \(x=1\) by the Taylor polynomial $$ p(x)=\sum_{n=0}^{10} \frac{(x-1)^{n}}{n !} $$

Step-by-Step Solution

Answer

The value is 1.

Step 1: Identify the Taylor Polynomial

The given Taylor polynomial for \(p(x)\) is centered at \(x=1\) and is expressed as:\[p(x) = \sum_{n=0}^{10} \frac{(x-1)^n}{n!}\]This is the Taylor series expansion for \(e^{x-1}\), truncated at \(n=10\). This means \(f(x) = e^{x-1}\), so we replace the series up to \(n=10\) with the function \(e^{x-1}\).

Step 2: Differentiate the Function

Since the function represented by our Taylor series is \(f(x) = e^{x-1}\), we can differentiate it to find the higher derivatives. For \(f(x) = e^{x-1}\), we have \(f'(x) = e^{x-1}\), similarly \(f''(x) = e^{x-1}\), \(f^{(3)}(x) = e^{x-1}\), and generally, \(f^{(n)}(x) = e^{x-1}\) for all \(n\).

Step 3: Evaluate the Fifth Derivative at x=1

Since every derivative of \(f(x)=e^{x-1}\) is the same function, we have \(f^{(5)}(x) = e^{x-1}\). To find \(f^{(5)}(1)\), substitute \(x = 1\) into the derivative:\[f^{(5)}(1) = e^{1-1} = e^0 = 1\]

Key Concepts

Derivative

In calculus, a derivative represents the rate at which a function is changing at any given point. It's akin to finding the slope of a function at a specific point. A derivative of a function provides a way to understand how the output value of that function changes as the input changes. In the context of Taylor series, the derivative plays a key role: it helps determine the coefficients of the terms in the polynomial. For a function like \( f(x) = e^{x-1} \), each derivative \( (f'(x), f''(x), \,...)\) is equal to the original function \( e^{x-1} \). This is because the exponential function's derivative remains consistent across all orders of differentiation. This consistency is crucial in deriving the Taylor polynomial, which provides a local approximation of a function.

Taylor Polynomial

A Taylor polynomial is a finite sum used to approximate a function near a specified point, known as the center. The polynomial is formed by using the derivatives of the function at that center point. The Taylor polynomial is especially useful because it transforms potentially complicated functions into simple-to-calculate polynomials. The Taylor polynomial of degree \( n \) is written as

• \( T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \dots + \frac{f^{(n)}(a)}{n!}(x-a)^n \)

In the original problem, the function \( p(x) \) is a Taylor polynomial centered at \( x=1 \) representing the function \( e^{x-1} \). This function is expanded using its derivatives to match the typical power series format.

Function Approximation

Function approximation involves using simpler, more manageable representations to estimate more complex functions. Taylor series and polynomials provide a powerful method for function approximation, which makes them valuable in both practical computations and theoretical analysis. By centering a Taylor series at a point, one can closely model a function's behavior near that point. This is incredibly helpful in computing functional values that might otherwise be difficult or impossible to determine directly.

• The Taylor polynomial \( p(x) \) in the problem is an approximation of \( f(x) = e^{x-1} \).
• It was used to find derivatives and thus precisely determine \( f^{(5)}(1) \).

This application highlights how Taylor series can provide effective approximations by using constrained degrees of polynomials.

Higher Derivatives

The concept of higher derivatives is pivotal in understanding the behavior of more complex functions over time. When a function is differentiated multiple times, each successive derivative is referred to as a higher derivative. For example, the second derivative tells us about the concavity of the function, while the third derivative can provide insights into changes in the rate of concavity.
In our exercise, calculating the fifth derivative \( f^{(5)}(x) \) of the function \( f(x) = e^{x-1} \) involves recognizing a key property: every derivative of this exponential function turns out to be the same due to its unique structure. Therefore, regardless of how many times you differentiate, the function remains \( e^{x-1} \). This reveals a consistency and symmetry that simplifies many calculations, making Taylor series an efficient tool for approximating functions and their derivatives.