Give step-by-step solution with explanation and final answer:5. In Examples 9.31 and 9.32, we have seen that in some cases, Lagrange’s remainder formula is.
not strong enough to show that lim R(x) =

Question

Give step-by-step solution with explanation and final answer:5. In Examples 9.31 and 9.32, we have seen that in some cases, Lagrange’s remainder formula is.
not strong enough to show that lim R(x) = 0. Let's develop another remainder formula.
(a) Let a be real number and let x > a, let n be a non-negative integer and let f be a

function such that £® is continuous on [a,x] and differentiable on (a,x).
(tet g:la,x] + R be the function
_ NAO) "
90 =f) 2h or.
Compute g'(0) for t€ (a0).
i) (Cauchy's remainder formula) By applying Mean Value Theorem to the function g,
show that there exists a number ¢ € (a,x) such that
NO@ ol SOO
f= P) Tem EEE ore - a),
Remark: 1f we assume further that /**) is Integrable on [a,x], then another way
of obtaining (a) (il is to use Q4(a) and then the MVT for integrals.
(6) Using the result rom (a) (i) (which obviously still holds if x < a), show tha for each of
the following functions, the remainder term at 0 satisfies
Jim Ry) =0 for each fixed x € (1,1).
(i) (Example 9.31) f(x) =In(1+x)
(i) (Example 8:32) (x) = (1+ x)”, where p isa real number.
Hint: In (b) (ii), itis useful to note that if lim [22] < 1, then lim a, = 0.

VIPSolutions · Accepted Answer

Let's go through the steps as requested:

---

### (a) (i) Compute $ g'(t) $ for $ t \in (a, x) $

Given:
$$
g(t) = f(x) - \sum_{k=0}^{n} \frac{f^{(k)}(t)}{k!}(x-t)^k
$$

Let's differentiate $ g(t) $ with respect to $ t $:

#### Step 1: Differentiate term by term

- The derivative of $ f(x) $ with respect to $ t $ is 0 (since $ x $ is fixed).
- The sum: For each term inside the sum, use the product rule:

$$
\frac{d}{dt} \left[ \frac{f^{(k)}(t)}{k!}(x-t)^k \right] = \frac{f^{(k+1)}(t)}{k!}(x-t)^k + \frac{f^{(k)}(t)}{k!} \frac{d}{dt}(x-t)^k
$$
But,
$$
\frac{d}{dt}(x-t)^k = -k(x-t)^{k-1}
$$
So,
$$
\frac{d}{dt} \left[ \frac{f^{(k)}(t)}{k!}(x-t)^k \right] = \frac{f^{(k+1)}(t)}{k!}(x-t)^k - \frac{k f^{(k)}(t)}{k!}(x-t)^{k-1}
= \frac{f^{(k+1)}(t)}{k!}(x-t)^k - \frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1}
$$

#### Step 2: Write the sum

So,
$$
g'(t) = - \sum_{k=0}^{n} \left( \frac{f^{(k+1)}(t)}{k!}(x-t)^k - \frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1} \right)
$$

#### Step 3: Simplify the sum

The sum telescopes. If you write out the terms, you'll see most terms cancel.

- The first sum: $\sum_{k=0}^{n} \frac{f^{(k+1)}(t)}{k!}(x-t)^k$
- The second sum: $\sum_{k=0}^{n} \frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1}$
    - Note: For $ k=0 $, this is 0 since $ (k-1)! $ is undefined and $ (x-t)^{-1} $ is not present.

So, adjust the second sum to start from $ k=1 $:

$$
g'(t) = - \sum_{k=0}^{n} \frac{f^{(k+1)}(t)}{k!}(x-t)^k + \sum_{k=1}^{n} \frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1}
$$

Let $ m = k-1 $ in the second sum:
$$
\sum_{k=1}^n \frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1} = \sum_{m=0}^{n-1} \frac{f^{(m+1)}(t)}{m!}(x-t)^m
$$

So,
$$
g'(t) = -\frac{f^{(n+1)}(t)}{n!}(x-t)^n
$$

---

### (a) (ii) **Cauchy's remainder formula**

By the Mean Value Theorem, there exists $ c \in (a, x) $ such that

$$
f(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k + \frac{f^{(n+1)}(c)}{n!}(x-c)^n(x-a)
$$

But according to the image, the formula is:
$$
f(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k + \frac{f^{(n+1)}(c)}{n!}(x-c)^n(x-a)
$$
So, this is the Cauchy's remainder formula.

---

### (b) **Using the result from (a)(ii):**

Show that for $ f(x) = \ln(1+x) $ and $ f(x) = (1+x)^p $, the remainder term at 0 satisfies:

$$
\lim_{n \to \infty} R_n(x) = 0, \quad \text{for each fixed } x \in (-1,1)
$$

---

#### **Example 9.31: $ f(x) = \ln(1+x) $ at $ a = 0 $**

- $ f^{(k)}(x) = (-1)^{k-1} (k-1)!/(1+x)^k $
- $ f^{(k)}(0) = (-1)^{k-1} (k-1)! $
- Taylor series: $ \sum_{k=1}^{n} (-1)^{k-1} \frac{1}{k} x^k $
- Remainder:
$$
R_{n}(x) = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1}
$$
Where $ c \in (0, x) $.

But $ f^{(n+1)}(c) = (-1)^n n!/(1+c)^{n+1} $, so
$$
R_{n}(x) = \frac{(-1)^n n!}{(n+1)!} \cdot \frac{x^{n+1}}{(1+c)^{n+1}}
= \frac{(-1)^n}{n+1} \left( \frac{x}{1+c} \right)^{n+1}
$$

As $ n \to \infty $, for $ |x| < 1 $ and $ 0 < 1+c < 2 $, $ \left| \frac{x}{1+c} \right| < 1 $, so $ R_n(x) \to 0 $.

---

#### **Example 9.32: $ f(x) = (1+x)^p $ at $ a = 0 $**

- $ f^{(k)}(x) = p(p-1)\dots(p-k+1)(1+x)^{p-k} $
- $ f^{(k)}(0) = p(p-1)\dots(p-k+1) $
- Taylor series: $ \sum_{k=0}^n \binom{p}{k} x^k $
- Remainder:
$$
R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1}
$$
Where $ c \in (0, x) $.

So,
$$
f^{(n+1)}(c) = p(p-1)\dots(p-n)(1+c)^{p-n-1}
$$
So
$$
R_n(x) = \frac{p(p-1)\cdots(p-n)}{(n+1)!} (1+c)^{p-n-1} x^{n+1}
$$

For fixed $ x \in (-1,1) $, as $ n \to \infty $, using the ratio test or noting that $ \frac{x}{1+c} < 1 $, the term goes to 0.
(Use the hint: If $ \lim_{n\to\infty} |a_{n+1}/a_n| < 1 $, then $ a_n \to 0 $.)

---

### **Final Answers**

#### (a) (i) 
$$
g'(t) = -\frac{f^{(n+1)}(t)}{n!}(x-t)^n
$$

#### (a) (ii) (Cauchy's remainder formula)
$$
f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!} (x-a)^k + \frac{f^{(n+1)}(c)}{n!} (x-c)^n (x-a), \quad \text{for some } c \in (a, x)
$$

#### (b)
For both $ f(x) = \ln(1+x) $ and $ f(x) = (1+x)^p $, for each fixed $ x \in (-1,1) $,
$$
\boxed{\lim_{n \to \infty} R_n(x) = 0}
$$

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Study Help, Fast Answers

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Related Answers

Study Help, Fast Answers