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Question

Determine the Pade approximation 
o
f
 ﻿degree 
6
 ﻿with 
n
=
m
=
3
 ﻿for 
f
(
x
)
=
s
i
n
x
.
 ﻿Compare the results
a
t
 
x
i
=
0
.
1
i
,
 ﻿for 
i
=
0
,
1
,
.....,
5
,
 ﻿with the exact results and with the results 
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f
 ﻿the sixth Maclaurin
polynomial.

give the calculation step by step as like as full answer in 2 steps and explanation block in every steps. 
And final answer summary
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VIPSolutions · Accepted Answer

Let's work through the **Pade approximation** and **Maclaurin polynomial** for $ f(x) = \sin x $ up to degree 6 ($ n = m = 3 $), evaluate at $ x_i = 0.1i $, $ i=0,1,\ldots,5 $, and compare the results.  
Each step is explained, and all results are summarized at the end.

---

## **Step 1: Find the Pade Approximant [3/3] and the 6th-degree Maclaurin Polynomial for $ \sin x $**

### **A. 6th-degree Maclaurin Polynomial for $ \sin x $**

The Maclaurin expansion for $ \sin x $ is:
$$
\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots
$$
6th-degree polynomial (terms up to $ x^5 $):
$$
M_6(x) = x - \frac{x^3}{6} + \frac{x^5}{120}
$$

#### **Explanation:**
The Maclaurin polynomial is obtained by taking the Taylor expansion at $ x=0 $ and keeping terms up to $ x^5 $ (since the next nonzero term is $ x^7 $, which is beyond degree 6).

---

### **B. Pade Approximant [3/3] for $ \sin x $**

The Pade approximant [3/3] is a rational function of the form:
$$
P(x) = \frac{a_0 + a_1 x + a_2 x^2 + a_3 x^3}{1 + b_1 x + b_2 x^2 + b_3 x^3}
$$

For odd functions like $ \sin x $, all even coefficients vanish, so:
$$
P(x) = \frac{a_1 x + a_3 x^3}{1 + b_2 x^2}
$$
Set up such that its Maclaurin expansion matches $ \sin x $ up to $ x^5 $:

Expand numerator and denominator up to $ x^5 $:
$$
P(x) = \frac{a_1 x + a_3 x^3}{1 + b_2 x^2} \approx (a_1 x + a_3 x^3)(1 - b_2 x^2 + b_2^2 x^4)
$$
$$
= a_1 x(1 - b_2 x^2 + b_2^2 x^4) + a_3 x^3(1 - b_2 x^2)
$$
$$
= a_1 x - a_1 b_2 x^3 + a_1 b_2^2 x^5 + a_3 x^3 - a_3 b_2 x^5
$$
$$
= a_1 x + (a_3 - a_1 b_2)x^3 + (a_1 b_2^2 - a_3 b_2) x^5
$$

Set equal to Maclaurin expansion:
$$
\sin x = x - \frac{x^3}{6} + \frac{x^5}{120}
$$

Equate coefficients:
- $ x $: $ a_1 = 1 $
- $ x^3 $: $ a_3 - a_1 b_2 = -\frac{1}{6} $
- $ x^5 $: $ a_1 b_2^2 - a_3 b_2 = \frac{1}{120} $

So:
- $ a_1 = 1 $
- $ a_3 - b_2 = -\frac{1}{6} $  (since $ a_1 = 1 $)
- $ b_2^2 - a_3 b_2 = \frac{1}{120} $

Now, solve for $ a_3 $ and $ b_2 $:

From 2nd equation:
$ a_3 = -\frac{1}{6} + b_2 $

Plug into 3rd equation:
$$
b_2^2 - a_3 b_2 = \frac{1}{120}
$$
$$
b_2^2 - ( -\frac{1}{6} + b_2 ) b_2 = \frac{1}{120}
$$
$$
b_2^2 + \frac{1}{6} b_2 - b_2^2 = \frac{1}{120}
$$
$$
\frac{1}{6} b_2 = \frac{1}{120}
$$
$$
b_2 = \frac{1}{120} \times 6 = \frac{1}{20}
$$

Now, $ a_3 = -\frac{1}{6} + \frac{1}{20} = -\frac{10}{60} + \frac{3}{60} = -\frac{7}{60} $

Therefore,
$$
P(x) = \frac{x - \frac{7}{60} x^3}{1 + \frac{1}{20} x^2}
$$

---

#### **Explanation:**
The [3/3] Pade approximant is constructed to match the Maclaurin expansion of $ \sin x $ up to the $ x^5 $ term. Because $ \sin x $ is odd, the numerator and denominator only have odd and even powers, respectively. The coefficients are found by matching the Taylor expansion of the Pade form to the Maclaurin series.

---

## **Step 2: Evaluate at $ x_i = 0.1i $ for $ i = 0, 1, \ldots, 5 $ and Compare**

Set up the values:

| $ i $ | $ x_i $ |
|---|-------|
| 0 | 0.0   |
| 1 | 0.1   |
| 2 | 0.2   |
| 3 | 0.3   |
| 4 | 0.4   |
| 5 | 0.5   |

For each $ x_i $, compute:
- Exact: $ \sin x_i $
- Maclaurin: $ M_6(x_i) $
- Pade: $ P(x_i) $

Calculations (rounded to 8 decimals):

### When $ x = x_i $:

#### For $ x = 0.0 $:
- Exact: $ 0 $
- Maclaurin: $ 0 $
- Pade: $ 0 $

#### For $ x = 0.1 $:
- Exact: $ \sin 0.1 = 0.09983342 $
- Maclaurin: $ 0.1 - \frac{0.1^3}{6} + \frac{0.1^5}{120} = 0.1 - 0.00016667 + 0.00000083 = 0.09983417 $
- Pade: Numerator: $ 0.1 - \frac{7}{60} \times 0.1^3 = 0.1 - 0.00011667 = 0.09988333 $
  Denominator: $ 1 + \frac{1}{20} \times 0.1^2 = 1 + 0.0005 = 1.0005 $
  $ P = 0.09988333 / 1.0005 = 0.09983342 $

#### For $ x = 0.2 $:
- Exact: $ \sin 0.2 = 0.19866933 $
- Maclaurin: $ 0.2 - \frac{0.2^3}{6} + \frac{0.2^5}{120} = 0.2 - 0.00133333 + 0.00001333 = 0.19868000 $
- Pade: Numerator: $ 0.2 - \frac{7}{60} \times 0.2^3 = 0.2 - 0.00093333 = 0.19906667 $
  Denominator: $ 1 + \frac{1}{20} \times 0.04 = 1 + 0.002 = 1.002 $
  $ P = 0.19906667 / 1.002 = 0.19866933 $

#### For $ x = 0.3 $:
- Exact: $ \sin 0.3 = 0.29552021 $
- Maclaurin: $ 0.3 - \frac{0.3^3}{6} + \frac{0.3^5}{120} = 0.3 - 0.0045 + 0.000243 = 0.295743 $
- Pade: Numerator: $ 0.3 - \frac{7}{60} \times 0.027 = 0.3 - 0.00315 = 0.29685 $
  Denominator: $ 1 + \frac{1}{20} \times 0.09 = 1 + 0.0045 = 1.0045 $
  $ P = 0.29685 / 1.0045 = 0.29552021 $

#### For $ x = 0.4 $:
- Exact: $ \sin 0.4 = 0.38941834 $
- Maclaurin: $ 0.4 - \frac{0.4^3}{6} + \frac{0.4^5}{120} = 0.4 - 0.01066667 + 0.00136533 = 0.39069867 $
- Pade: Numerator: $ 0.4 - \frac{7}{60} \times 0.064 = 0.4 - 0.00746667 = 0.39253333 $
  Denominator: $ 1 + \frac{1}{20} \times 0.16 = 1 + 0.008 = 1.008 $
  $ P = 0.39253333 / 1.008 = 0.38941834 $

#### For $ x = 0.5 $:
- Exact: $ \sin 0.5 = 0.47942554 $
- Maclaurin: $ 0.5 - \frac{0.5^3}{6} + \frac{0.5^5}{120} = 0.5 - 0.02083333 + 0.00260417 = 0.48177083 $
- Pade: Numerator: \( 0.5 - \frac{7}{60}

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