Problem 9
Question
2-9. Two functions \(f, g: \mathbf{R} \rightarrow \mathbf{R}\) are equal up to \(\boldsymbol{n}\) th order at \(a\) if $$ \lim _{h \rightarrow 0} \frac{f(a+h)-g(a+h)}{h^{n}}=0 $$ (a) Show that \(f\) is differentiable at \(a\) if and only if there is a function \(g\) of the form \(g(x)=a_{0}+a_{1}(x-a)\) such that \(f\) and \(g\) are equal up to first order at \(a\). (b) If \(f^{\prime}(a), \ldots, f^{(n)}(a)\) exist, show that \(f\) and the function \(g\) defined by $$ g(x)=\sum_{i=0}^{n} \frac{f^{(i)}(a)}{i !}(x-a)^{i} $$ are equal up to \(n\)th order at \(a\). Hint: The limit $$ \lim _{x \rightarrow a} \frac{f(x)-\sum_{i=0}^{n-1} \frac{f^{(i)}(a)}{i !}(x-a)^{i}}{(x-a)^{n}} $$ may be evaluated by L'Hospital's rule.
Step-by-Step Solution
Verified Answer
For part (a), \(f\) is differentiable at \(a\) if it can be approximated by a linear function up to the first order. For part (b), \(f\) and its Taylor polynomial are equal up to the \(n\)-th order at \(a\).
1Step 1: Understanding the Definition
Two functions, \(f\) and \(g\), are equal up to \(n\)-th order at \(a\) if \[\lim _{h \rightarrow 0} \frac{f(a+h)-g(a+h)}{h^{n}}=0\]. This implies that the difference between the functions vanishes faster than \(h^{n}\) as \(h\) approaches 0.
2Step 1: Part (a) - Proof of Differentiability
To show that \(f\) is differentiable at \(a\), assume \(g(x) = a_0 + a_1(x - a)\). This form is a linear approximation around \(a\), where \(a_0 = f(a)\) and \(a_1\) should be \(f'(a)\) for the functions to be equal up to the first order at \(a\). The limit becomes: \[\lim_{h \rightarrow 0} \frac{f(a+h) - (f(a) + f'(a)h)}{h} = 0\], which simplifies to the definition of the derivative \(f'(a)\). Hence, \(f\) is differentiable if and only if such a \(g\) exists.
3Step 2: Part (b) - Applying the Series Expansion
Given \(f'(a), \ldots, f^{(n)}(a)\) exist, the function \(g(x)\) defined by \[g(x) = \sum_{i=0}^{n} \frac{f^{(i)}(a)}{i!}(x - a)^i\] is a Taylor polynomial of \(f\) at \(a\). To show \(f\) and \(g\) are equal up to the \(n\)-th order, consider the limit \[\lim_{x \rightarrow a} \frac{f(x) - \sum_{i=0}^{n-1} \frac{f^{(i)}(a)}{i!}(x - a)^i}{(x - a)^n}\].
4Step 3: Using L'Hospital's Rule
Apply L'Hospital's rule \(n\) times to evaluate the limit. After \(n\) differentiations, we get: \[\lim_{x \rightarrow a} \frac{\text{higher order terms of } f - g}{\text{factorial terms}} = \frac{f^{(n)}(a)}{n!} = 0\], since all lower order terms cancel out. Hence, \(f\) and \(g\) are equal up to the \(n\)-th order at \(a\).
Key Concepts
DifferentiabilityTaylor SeriesL'Hospital's RuleHigher-Order Derivatives
Differentiability
Differentiability is a fundamental concept in calculus. It describes whether a function has a well-defined tangent at a particular point.
A function \(f\) is said to be differentiable at a point \(a\) if it has a derivative at \(a\). In mathematical terms, this means the limit:
\[ \lim_{h \rightarrow 0} \frac{f(a+h) - f(a)}{h} \]
exists.
In the context of the given exercise, to show that \(f\) is differentiable at \(a\), we assume the existence of a function \(g\) in the form of \(g(x) = a_0 + a_1(x - a)\)
This can be interpreted as a linear approximation of \(f\) around the point \(a\). Here, \(a_0 = f(a)\) and \(a_1\) ideally is \(f'(a)\). If the difference between \(f\) and this linear function vanishes as \(h\) approaches 0 faster than \(h\), then \(f\) is confirmed to be differentiable at \(a\).
The limit thus becomes:
\[ \lim_{h \rightarrow 0} \frac{f(a+h) - (f(a) + f'(a) h)}{h} = 0 \]
which essentially is the definition of a derivative.
A function \(f\) is said to be differentiable at a point \(a\) if it has a derivative at \(a\). In mathematical terms, this means the limit:
\[ \lim_{h \rightarrow 0} \frac{f(a+h) - f(a)}{h} \]
exists.
In the context of the given exercise, to show that \(f\) is differentiable at \(a\), we assume the existence of a function \(g\) in the form of \(g(x) = a_0 + a_1(x - a)\)
This can be interpreted as a linear approximation of \(f\) around the point \(a\). Here, \(a_0 = f(a)\) and \(a_1\) ideally is \(f'(a)\). If the difference between \(f\) and this linear function vanishes as \(h\) approaches 0 faster than \(h\), then \(f\) is confirmed to be differentiable at \(a\).
The limit thus becomes:
\[ \lim_{h \rightarrow 0} \frac{f(a+h) - (f(a) + f'(a) h)}{h} = 0 \]
which essentially is the definition of a derivative.
Taylor Series
The Taylor series is a powerful tool in calculus to approximate functions near a specific point using polynomials. Taylor polynomials provide a way to approximate a function as a sum of its derivatives at a particular point.
Based on the exercise, the function \(g(x)\) provided by the expansion:
\[ g(x) = \sum_{i=0}^{n} \frac{f^{(i)}(a)}{i!}(x - a)^i \]
is a Taylor polynomial of \(f\) at \(a\). This represents the series expansion of \(f\) up to the \(n\)-th order term.
It says that the function \(f\) can be approximated by its derivatives at \(a\). For better accuracy, more terms (higher-order derivatives) should be included. Each term is a scaled version of the distance \((x - a)\) raised to the respective power.
If \(f\) and this polynomial \(g\) are equal up to the \(n\)-th order, it means their difference divided by \((x - a)^n\) vanishes as \(x\) approaches \(a\).
Based on the exercise, the function \(g(x)\) provided by the expansion:
\[ g(x) = \sum_{i=0}^{n} \frac{f^{(i)}(a)}{i!}(x - a)^i \]
is a Taylor polynomial of \(f\) at \(a\). This represents the series expansion of \(f\) up to the \(n\)-th order term.
It says that the function \(f\) can be approximated by its derivatives at \(a\). For better accuracy, more terms (higher-order derivatives) should be included. Each term is a scaled version of the distance \((x - a)\) raised to the respective power.
If \(f\) and this polynomial \(g\) are equal up to the \(n\)-th order, it means their difference divided by \((x - a)^n\) vanishes as \(x\) approaches \(a\).
L'Hospital's Rule
L'Hospital's rule is a method for finding limits of indeterminate forms, particularly of the type \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
L'Hospital's rule states that if: \[ \lim_{x \rightarrow c} \frac{f(x)}{g(x)} \] is an indeterminate form, we can differentiate the numerator \(f(x)\) and the denominator \(g(x)\) to evaluate the limit:
\[ \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)} \]
provided the limit on the right-hand side exists.
In the exercise context, L'Hospital's rule is applied multiple times to simplify an expression involving higher-order terms of \(f\). By differentiating both the numerator and the denominator successively, we eventually reduce the expression to a form that can be directly evaluated.
This helps in proving that the polynomial \(g(x)\) effectively approximates \(f(x)\) up to the \(n\)-th order near the point \(a\).
L'Hospital's rule states that if: \[ \lim_{x \rightarrow c} \frac{f(x)}{g(x)} \] is an indeterminate form, we can differentiate the numerator \(f(x)\) and the denominator \(g(x)\) to evaluate the limit:
\[ \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)} \]
provided the limit on the right-hand side exists.
In the exercise context, L'Hospital's rule is applied multiple times to simplify an expression involving higher-order terms of \(f\). By differentiating both the numerator and the denominator successively, we eventually reduce the expression to a form that can be directly evaluated.
This helps in proving that the polynomial \(g(x)\) effectively approximates \(f(x)\) up to the \(n\)-th order near the point \(a\).
Higher-Order Derivatives
Higher-order derivatives provide deeper insights into the behavior of functions. The first derivative \(f'(x)\) reveals the function's slope, while second and higher derivatives give information about concavity, inflection points, and more.
For instance, the second derivative \(f''(x)\) describes the curvature of the function. The third derivative \(f'''(x)\) and beyond continue to offer more refined details.
In this exercise, we use higher-order derivatives to derive the function \(g(x)\) which is the Taylor polynomial. Each derivative plays a role in constructing \(g(x)\) as a sum of terms:
\[ g(x) = \sum_{i=0}^{n} \frac{f^{(i)}(a)}{i!} (x - a)^i \]
This sum represents a polynomial which approximates \(f\) near \(a\). Each term reflects the contribution of the ith derivative at \(a\) to the overall function behavior.
By ensuring the remaining difference after this series expansion vanishes faster than \((x - a)^n\), we prove that \(f\) and \(g\) are essentially the same up to the \(n\)-th order at \(a\).
For instance, the second derivative \(f''(x)\) describes the curvature of the function. The third derivative \(f'''(x)\) and beyond continue to offer more refined details.
In this exercise, we use higher-order derivatives to derive the function \(g(x)\) which is the Taylor polynomial. Each derivative plays a role in constructing \(g(x)\) as a sum of terms:
\[ g(x) = \sum_{i=0}^{n} \frac{f^{(i)}(a)}{i!} (x - a)^i \]
This sum represents a polynomial which approximates \(f\) near \(a\). Each term reflects the contribution of the ith derivative at \(a\) to the overall function behavior.
By ensuring the remaining difference after this series expansion vanishes faster than \((x - a)^n\), we prove that \(f\) and \(g\) are essentially the same up to the \(n\)-th order at \(a\).
Other exercises in this chapter
Problem 7
2-7. Let \(f: \mathbf{R}^{n} \rightarrow \mathbf{R}\) be a function such that \(|f(x)| \leq|x|^{2}\). Show that \(f\) is differentiable at 0 .
View solution Problem 8
2-8. Let \(f: \mathbf{R} \rightarrow \mathbf{R}^{2}\). Prove that \(f\) is differentiable at \(a \in \mathbf{R}\) if and only if \(f^{1}\) and \(f^{2}\) are, an
View solution Problem 12
2-12. A function \(f: \mathbf{R}^{n} \times \mathbf{R}^{m} \rightarrow \mathbf{R}^{p}\) is bilinear if for \(x, x_{1}, x_{2} \in \mathbf{R}^{n}\), \(y, y_{1}, y
View solution Problem 13
2-13. Define \(I P: \mathbf{R}^{n} \times \mathbf{R}^{n} \rightarrow \mathbf{R}\) by \(I P(x, y)=\langle x, y\rangle\). (a) Find \(D(I P)(a, b)\) and \((I P)^{\
View solution