To prove: That the nth Maclaurin polynomial of the given function for odd n is given as: T n ( x ) = x + 1 2 x 3 3 + 1 ⋅ 3 2 ⋅ 4 x 5 5 + .... + 1 ⋅ 3 ⋅ 5... ( n − 2 ) 2 ⋅ 4 ⋅ 6... ( n − 1 ) x n n

Question

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Answer 1

Question

Chapter 8.4, Problem 61E

To determine

To prove:

That the nth Maclaurin polynomial of the given function for odd n is given as:

$T_{n} (x) = x + \frac{1}{2} \frac{x^{3}}{3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{x^{5}}{5} + .... + \frac{1 \cdot 3 \cdot 5... (n - 2)}{2 \cdot 4 \cdot 6... (n - 1)} \frac{x^{n}}{n}$

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Answer 2

Question

Chapter 8.4, Problem 61E

To determine

To prove:

That the nth Maclaurin polynomial of the given function for odd n is given as:

$T_{n} (x) = x + \frac{1}{2} \frac{x^{3}}{3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{x^{5}}{5} + .... + \frac{1 \cdot 3 \cdot 5... (n - 2)}{2 \cdot 4 \cdot 6... (n - 1)} \frac{x^{n}}{n}$

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Answer 3

Question

Chapter 8.4, Problem 61E

To determine

To prove:

That the nth Maclaurin polynomial of the given function for odd n is given as:

$T_{n} (x) = x + \frac{1}{2} \frac{x^{3}}{3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{x^{5}}{5} + .... + \frac{1 \cdot 3 \cdot 5... (n - 2)}{2 \cdot 4 \cdot 6... (n - 1)} \frac{x^{n}}{n}$

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Answer 4

Question

Chapter 8.4, Problem 61E

To determine

To prove:

That the nth Maclaurin polynomial of the given function for odd n is given as:

$T_{n} (x) = x + \frac{1}{2} \frac{x^{3}}{3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{x^{5}}{5} + .... + \frac{1 \cdot 3 \cdot 5... (n - 2)}{2 \cdot 4 \cdot 6... (n - 1)} \frac{x^{n}}{n}$

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Answer 5

Chapter 8.4, Problem 61E

To determine

To prove:

That the nth Maclaurin polynomial of the given function for odd n is given as:

$T_{n} (x) = x + \frac{1}{2} \frac{x^{3}}{3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{x^{5}}{5} + .... + \frac{1 \cdot 3 \cdot 5... (n - 2)}{2 \cdot 4 \cdot 6... (n - 1)} \frac{x^{n}}{n}$

Answer 6

Chapter 8.4, Problem 61E

To determine

To prove:

That the nth Maclaurin polynomial of the given function for odd n is given as:

$T_{n} (x) = x + \frac{1}{2} \frac{x^{3}}{3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{x^{5}}{5} + .... + \frac{1 \cdot 3 \cdot 5... (n - 2)}{2 \cdot 4 \cdot 6... (n - 1)} \frac{x^{n}}{n}$

Answer 7

Chapter 8.4, Problem 61E

To determine

To prove:

That the nth Maclaurin polynomial of the given function for odd n is given as:

$T_{n} (x) = x + \frac{1}{2} \frac{x^{3}}{3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{x^{5}}{5} + .... + \frac{1 \cdot 3 \cdot 5... (n - 2)}{2 \cdot 4 \cdot 6... (n - 1)} \frac{x^{n}}{n}$

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 8.1 - Prob. 1PQ Ch. 8.1 - Prob. 2PQ Ch. 8.1 - Prob. 3PQ Ch. 8.1 - Prob. 4PQ Ch. 8.1 - Prob. 5PQ Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E

To prove: That the nth Maclaurin polynomial of the given function for odd n is given as: T n ( x ) = x + 1 2 x 3 3 + 1 ⋅ 3 2 ⋅ 4 x 5 5 + .... + 1 ⋅ 3 ⋅ 5... ( n − 2 ) 2 ⋅ 4 ⋅ 6... ( n − 1 ) x n n

To prove: That the nth Maclaurin polynomial of the given function for odd n is given as: Tn(x)=x+12x33+1⋅32⋅4x55+....+1⋅3⋅5...(n−2)2⋅4⋅6...(n−1)xnn

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Chapter 8 Solutions

To prove:

That the nth Maclaurin polynomial of the given function for odd n is given as:

$T_{n} (x) = x + \frac{1}{2} \frac{x^{3}}{3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{x^{5}}{5} + .... + \frac{1 \cdot 3 \cdot 5... (n - 2)}{2 \cdot 4 \cdot 6... (n - 1)} \frac{x^{n}}{n}$