Question

Suppose \(f\) is a real function on \([a, b], n\) is a positive integer, and \(f^{(n-1)}\) exists for every \(t \in[a, b] .\) Let \(\alpha, \beta\), and \(P\) be as in Taylor's theorem \((5.15)\). Define $$ Q(t)=\frac{f(t)-f(\beta)}{t-\beta} $$ for \(t \in[a, b], t \neq \beta\), differentiate $$ f(t)-f(\beta)=(t-\beta) Q(t) $$ \(n-1\) times at \(t=\alpha\), and derive the following version of Taylor's theorem: $$ f(\beta)=P(\beta)+\frac{Q^{(n-1)}(\alpha)}{(n-1) !}(\beta-\alpha)^{*} $$

Short answer

Based on the exercise, we derived a new version of Taylor's theorem using the function \(Q(t)\) and differentiating the equation \(f(t) - f(\beta) = (t - \beta) Q(t)\) \(n - 1\) times at \(t = \alpha\). The new version of Taylor's theorem is given by: $$ f(\beta) = P(\beta) + \frac{Q^{(n-1)}(\alpha)}{(n-1)!} (\beta-\alpha)^{*} $$

Step-by-step solution

Define \(P\) and \(Q(t)\)

According to Taylor's theorem, we have the polynomial \(P\) defined as: $$ P(t) = f(\alpha) + f'(\alpha)(t-\alpha) + \dots + \frac{f^{(n-1)}(\alpha)}{(n-1)!}(t-\alpha)^{n-1} $$ Now, let's define the function \(Q(t)\) as: $$ Q(t) = \frac{f(t) - f(\beta)}{t - \beta} \text{ for } t \in [a,b], t \neq \beta $$

Differentiate using the product rule

Next, we want to differentiate the given equation, \(f(t) - f(\beta) = (t-\beta) Q(t)\), \(n-1\) times at \(t = \alpha\). First, apply the product rule on \((t-\beta)Q(t)\) to obtain the equation's first derivative with respect to \(t\). We have: $$ f'(t) = Q(t) + (t - \beta) Q'(t) $$

Differentiate the equation n-1 times

Now that we have the first derivative of the equation, we should continue differentiating it \(n-2\) more times, until we have differentiated it a total of \(n-1\) times. This way we will get an expression for \(Q^{(n-1)}(t)\). At the \(k\)-th differentiation, we will have: $$ f^{(k)}(t) = Q^{(k-1)}(t) + c_k(t-\beta)Q^{(k)}(t) $$ where \(c_k\) is a constant. Complete the \(n-1\) differentiations, and then evaluate the resulting expression at \(t = \alpha\).

Apply Taylor's theorem

Now that we have an expression for \(Q^{(n-1)}(\alpha)\), substitute it into the original version of the Taylor's theorem: $$ f(\beta) = P(\beta) + \frac{Q^{(n-1)}(\alpha)}{(n-1)!} (\beta-\alpha)^{*} $$ This is the new version of the Taylor's theorem as requested in the exercise.

Key concepts

Understanding a Real Function

When diving into mathematical analysis, one of the fundamental concepts is that of a real function. In essence, a real function is a rule that assigns to each member of the real numbers within a certain domain a unique real number. This function can be represented as a graph on a coordinate system, where the x-axis typically represents the input values while the y-axis represents the function's output. The notation \( f(x) = y \) symbolizes that function \( f \) assigns the value \( y \) to \( x \).

Specifically, in the given exercise, we're examining a function \( f \) defined on a closed interval \( [a, b] \) where the outputs are also real numbers. Understanding a real function is crucial because it sets the stage for grasping more advanced concepts like limits, continuity, and ultimately, Taylor's theorem.

Unraveling Mathematical Analysis

Mathematical analysis is the branch of mathematics dealing with limits and related theories, including differentiation and integration. It has its foundations in calculus and is concerned with the study of functions, sequences, series, and more complex mathematical entities. One of its primary goals is to provide a rigorous basis for theorems involving real functions and their behavior.

In relation to the exercise, mathematical analysis is fundamental in understanding the behavior of the function \( f \) and its derivatives over the interval \( [a, b] \). It involves applying various techniques such as the product rule and differentiation to compute derivatives which are then used in the formulation of Taylor's theorem.

Applying the Product Rule

The product rule is an essential differentiation rule used when taking the derivative of the product of two functions. It states that the derivative of a product \( u(t) \) times \( v(t) \) is given by \( u'(t)v(t) + u(t)v'(t) \). Simply put, you differentiate the first function, multiply it by the second function as is, then add the first function times the derivative of the second function.

In our exercise, we use the product rule to differentiate the equation \( f(t) - f(\beta) = (t-\beta) Q(t) \) as a product of \( (t-\beta) \) and \( Q(t) \) several times. The product rule facilitates obtaining derivatives that will ultimately lead us to the final expression of Taylor's theorem.

The Process of Differentiation

Differentiation is a cornerstone of calculus, focusing on calculating the derivative of a function, which represents the rate at which the function's value changes. To find the derivative of a function, you're essentially looking at how a tiny change in the input affects the resulting change in the output. For a real function like \( f(t) \) in our exercise, the derivative at a point \( t \) tells us the instantaneous rate of change of the function at that point.

The exercise requires us to perform this process of differentiation multiple times – specifically, \( n-1 \) times, to derive a general expression that involves \( Q^{(n-1)}(\alpha) \). This repetitive differentiation allows us to explore deeper layers of function behavior, which is particularly useful when constructing polynomial approximations of functions through techniques such as Taylor's theorem.