15. Two functions f, g : R → R are equal up to n-th order at a if
lim
h→0
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) = a0+a1(x−a)
such that f and g are equal up to first order at a.
(b) If f
0
(a), . . . , f(n)
(a) exist, show that f and the function g defined by
g(x) = Xn
i=0
f
(i)
(a)
i!
(x − a)
i
,
are equal up to nth order at a.
Hint: The limit
limx→a
f(x) −
Pn−1
i=0
f
(i)
(a)
i!
(x − a)
i
(x − a)
n
may be evaluated by L’Hospital’s rule.

Let's go through the problem step by step.

### Part (a): Differentiability and First-Order Approximation

We need to show that a function \( f \) is differentiable at \( a \) if and only if there is a function \( g(x) = a_0 + a_1(x - a) \) such that \( f \) and \( g \) are equal up to first order at \( a \).

#### Step 1: Understanding the statement
To say that \( f \) and \( g \) are equal up to first order at \( a \) means that the following limit holds:
\[
\lim_{h \to 0} \frac{f(a + h) - g(a + h)}{h} = 0.
\]

Let's break it down:

- \( f(a + h) - g(a + h) \) represents the difference between \( f \) and \( g \) as \( x \) approaches \( a \).
- Dividing by \( h \) and taking the limit as \( h \) approaches 0, we obtain the condition that the functions are equal up to first order.

#### Step 2: \( f \) is differentiable implies \( g \) exists

Suppose \( f \) is differentiable at \( a \). Then we can write:
\[
f(x) = f(a) + f'(a)(x - a) + o(x - a),
\]
where \( o(x - a) \) represents a term that vanishes faster than \( (x - a) \) as \( x \) approaches \( a \).

Define the function \( g(x) = f(a) + f'(a)(x - a) \). Clearly, \( g(x) = a_0 + a_1(x - a) \) with \( a_0 = f(a) \) and \( a_1 = f'(a) \).

Now, we compute the limit:
\[
\lim_{h \to 0} \frac{f(a + h) - g(a + h)}{h} = \lim_{h \to 0} \frac{o(h)}{h}.
\]
Since \( o(h) \) vanishes faster than \( h \), the limit is 0. Hence, \( f \) and \( g \) are equal up to first order at \( a \).

#### Step 3: \( f \) and \( g \) equal up to first order implies \( f \) is differentiable

Now suppose \( f \) and \( g(x) = a_0 + a_1(x - a) \) are equal up to first order at \( a \). Then:
\[
\lim_{h \to 0} \frac{f(a + h) - g(a + h)}{h} = 0.
\]
This implies that:
\[
f(a + h) - g(a + h) = o(h),
\]
where \( o(h) \) again denotes a function that vanishes faster than \( h \).

Therefore,
\[
f(a + h) = g(a + h) + o(h) = a_0 + a_1h + o(h).
\]
Since \( g(x) = a_0 + a_1(x - a) \), we can differentiate \( g \) with respect to \( x \) and find \( a_1 \). This implies that \( f \) is differentiable at \( a \) with \( f'(a) = a_1 \).

Thus, \( f \) is differentiable at \( a \).

### Part (b): Higher-Order Approximation

Now, we need to show that if \( f'(a), \dots, f^{(n)}(a) \) exist, then \( f \) and the function
\[
g(x) = \sum_{i=0}^{n} \frac{f^{(i)}(a)}{i!} (x - a)^i
\]
are equal up to the \( n \)-th order at \( a \).

#### Step 1: Define the function \( h(x) \)
Let:
\[
h(x) = f(x) - \sum_{i=0}^{n-1} \frac{f^{(i)}(a)}{i!} (x - a)^i.
\]
We need to evaluate the limit:
\[
\lim_{x \to a} \frac{h(x)}{(x - a)^n}.
\]

Using the definition of \( h(x) \), we have:
\[
\frac{h(x)}{(x - a)^n} = \frac{f(x) - \sum_{i=0}^{n-1} \frac{f^{(i)}(a)}{i!} (x - a)^i}{(x - a)^n}.
\]
Apply L'Hôpital's rule \( n \) times (since the limit is an indeterminate form \( 0/0 \)):
\[
\lim_{x \to a} \frac{h(x)}{(x - a)^n} = \lim_{x \to a} \frac{f^{(n)}(x)}{n!}.
\]
Since \( f^{(n)}(a) \) exists and is continuous at \( a \), this limit is simply:
\[
\frac{f^{(n)}(a)}{n!}.
\]

Therefore, the limit is 0 if we subtract \( \frac{f^{(n)}(a)}{n!} \) from \( h(x) \), showing that \( f \) and \( g(x) \) are equal up to \( n \)-th order at \( a \).

### Summary:

- In part (a), we showed that differentiability at \( a \) is equivalent to the existence of a linear function \( g(x) \) such that \( f \) and \( g \) are equal up to first order at \( a \).
- In part (b), we showed that if \( f'(a), \dots, f^{(n)}(a) \) exist, then \( f \) and the polynomial \( g(x) \) are equal up to the \( n \)-th order at \( a \).

Would you like more details or have any specific questions on this? Here are some follow-up questions:

1. What are higher-order approximations of functions, and how do they relate to Taylor series?
2. How does L'Hôpital's rule help in evaluating limits in the context of higher-order derivatives?
3. What does the \( o(x-a) \) notation signify in the context of function approximations?
4. How does the concept of differentiability extend to higher-order derivatives?
5. Can you provide an example where a function is differentiable but not smooth?

**Tip:** When dealing with higher-order approximations, Taylor series provides a powerful tool for approximating functions near a specific point.

Explore the concept of differentiability and its relation to Taylor series approximations. Learn how L'Hôpital's rule is applied to evaluate limits involving higher-order derivatives. Suitable for advanced undergraduate students and anyone interested in calculus and mathematical analysis.

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