The function g ( x ) = ∑ n = 0 ∞ n c n x n − 1 converges for − R < x < R .

Question

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Answer 1

Question

Chapter 10.7, Problem 64E

(a)

To determine

To show: The function g(x)=∑n=0∞ncnxn−1 converges for −R<x<R.

(b)

To determine

To show: The value (x+h)n−xnh=ncnn−1 using Mean Value Theorem.

(c)

To determine

To show: The value of |g(x)−f(x+h)−f(x)h|=|∑n=0∞ncn(xn−1−cnn−1)|.

(d)

To determine

To show: The value of xn−1−cnn−1x−cn=(n−1)dn−1n−2 using Mean Value Theorem.

(e)

To determine

The reason for |x−cn|<h and |dn−1|≤α=max{|x|,|x+h|}.

(f)

To determine

To show: The value of |g(x)−f(x+h)−f(x)h|≤|h||∑n=2∞n(n−1)cnαn−2|.

(g)

To determine

To show: The series ∑n=2∞n(n−1)αn−2 converges for −R<x<R.

(h)

To determine

To show: The limit of limh→0f(x+h)−f(x)h=g(x).

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Answer 2

Question

Chapter 10.7, Problem 64E

(a)

To determine

To show: The function g(x)=∑n=0∞ncnxn−1 converges for −R<x<R.

(b)

To determine

To show: The value (x+h)n−xnh=ncnn−1 using Mean Value Theorem.

(c)

To determine

To show: The value of |g(x)−f(x+h)−f(x)h|=|∑n=0∞ncn(xn−1−cnn−1)|.

(d)

To determine

To show: The value of xn−1−cnn−1x−cn=(n−1)dn−1n−2 using Mean Value Theorem.

(e)

To determine

The reason for |x−cn|<h and |dn−1|≤α=max{|x|,|x+h|}.

(f)

To determine

To show: The value of |g(x)−f(x+h)−f(x)h|≤|h||∑n=2∞n(n−1)cnαn−2|.

(g)

To determine

To show: The series ∑n=2∞n(n−1)αn−2 converges for −R<x<R.

(h)

To determine

To show: The limit of limh→0f(x+h)−f(x)h=g(x).

Expert Solution & Answer

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Answer 3

Question

Chapter 10.7, Problem 64E

(a)

To determine

To show: The function g(x)=∑n=0∞ncnxn−1 converges for −R<x<R.

(b)

To determine

To show: The value (x+h)n−xnh=ncnn−1 using Mean Value Theorem.

(c)

To determine

To show: The value of |g(x)−f(x+h)−f(x)h|=|∑n=0∞ncn(xn−1−cnn−1)|.

(d)

To determine

To show: The value of xn−1−cnn−1x−cn=(n−1)dn−1n−2 using Mean Value Theorem.

(e)

To determine

The reason for |x−cn|<h and |dn−1|≤α=max{|x|,|x+h|}.

(f)

To determine

To show: The value of |g(x)−f(x+h)−f(x)h|≤|h||∑n=2∞n(n−1)cnαn−2|.

(g)

To determine

To show: The series ∑n=2∞n(n−1)αn−2 converges for −R<x<R.

(h)

To determine

To show: The limit of limh→0f(x+h)−f(x)h=g(x).

Expert Solution & Answer

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See solution

Answer 4

Question

Chapter 10.7, Problem 64E

(a)

To determine

To show: The function g(x)=∑n=0∞ncnxn−1 converges for −R<x<R.

(b)

To determine

To show: The value (x+h)n−xnh=ncnn−1 using Mean Value Theorem.

(c)

To determine

To show: The value of |g(x)−f(x+h)−f(x)h|=|∑n=0∞ncn(xn−1−cnn−1)|.

(d)

To determine

To show: The value of xn−1−cnn−1x−cn=(n−1)dn−1n−2 using Mean Value Theorem.

(e)

To determine

The reason for |x−cn|<h and |dn−1|≤α=max{|x|,|x+h|}.

(f)

To determine

To show: The value of |g(x)−f(x+h)−f(x)h|≤|h||∑n=2∞n(n−1)cnαn−2|.

(g)

To determine

To show: The series ∑n=2∞n(n−1)αn−2 converges for −R<x<R.

(h)

To determine

To show: The limit of limh→0f(x+h)−f(x)h=g(x).

Expert Solution & Answer

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See solution

Answer 5

Chapter 10.7, Problem 64E

(a)

To determine

To show: The function g(x)=∑n=0∞ncnxn−1 converges for −R<x<R.

(b)

To determine

To show: The value (x+h)n−xnh=ncnn−1 using Mean Value Theorem.

(c)

To determine

To show: The value of |g(x)−f(x+h)−f(x)h|=|∑n=0∞ncn(xn−1−cnn−1)|.

(d)

To determine

To show: The value of xn−1−cnn−1x−cn=(n−1)dn−1n−2 using Mean Value Theorem.

(e)

To determine

The reason for |x−cn|<h and |dn−1|≤α=max{|x|,|x+h|}.

(f)

To determine

To show: The value of |g(x)−f(x+h)−f(x)h|≤|h||∑n=2∞n(n−1)cnαn−2|.

(g)

To determine

To show: The series ∑n=2∞n(n−1)αn−2 converges for −R<x<R.

(h)

To determine

To show: The limit of limh→0f(x+h)−f(x)h=g(x).

Answer 6

Chapter 10.7, Problem 64E

(a)

To determine

To show: The function g(x)=∑n=0∞ncnxn−1 converges for −R<x<R.

Answer 7

Chapter 10.7, Problem 64E

(a)

To determine

To show: The function g(x)=∑n=0∞ncnxn−1 converges for −R<x<R.

Answer 8

(b)

To determine

To show: The value (x+h)n−xnh=ncnn−1 using Mean Value Theorem.

Answer 9

(b)

To determine

To show: The value (x+h)n−xnh=ncnn−1 using Mean Value Theorem.

Answer 10

(c)

To determine

To show: The value of |g(x)−f(x+h)−f(x)h|=|∑n=0∞ncn(xn−1−cnn−1)|.

Answer 11

(c)

To determine

To show: The value of |g(x)−f(x+h)−f(x)h|=|∑n=0∞ncn(xn−1−cnn−1)|.

Answer 12

(d)

To determine

To show: The value of xn−1−cnn−1x−cn=(n−1)dn−1n−2 using Mean Value Theorem.

Answer 13

(d)

To determine

To show: The value of xn−1−cnn−1x−cn=(n−1)dn−1n−2 using Mean Value Theorem.

Answer 14

(e)

To determine

The reason for |x−cn|<h and |dn−1|≤α=max{|x|,|x+h|}.

Answer 15

(e)

To determine

The reason for |x−cn|<h and |dn−1|≤α=max{|x|,|x+h|}.

Answer 16

(f)

To determine

To show: The value of |g(x)−f(x+h)−f(x)h|≤|h||∑n=2∞n(n−1)cnαn−2|.

Answer 17

(f)

To determine

To show: The value of |g(x)−f(x+h)−f(x)h|≤|h||∑n=2∞n(n−1)cnαn−2|.

Answer 18

(g)

To determine

To show: The series ∑n=2∞n(n−1)αn−2 converges for −R<x<R.

Answer 19

(g)

To determine

To show: The series ∑n=2∞n(n−1)αn−2 converges for −R<x<R.

Answer 20

(h)

To determine

To show: The limit of limh→0f(x+h)−f(x)h=g(x).

Answer 21

(h)

To determine

To show: The limit of limh→0f(x+h)−f(x)h=g(x).

Answer 22

Expert Solution & Answer

Answer 23

Expert Solution & Answer

Answer 24

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See solution

The function g ( x ) = ∑ n = 0 ∞ n c n x n − 1 converges for − R < x < R .

Concept explainers

Videos

To show: The function g(x)=∑n=0∞ncnxn−1 converges for −R<x<R.

To show: The value (x+h)n−xnh=ncnn−1 using Mean Value Theorem.

To show: The value of |g(x)−f(x+h)−f(x)h|=|∑n=0∞ncn(xn−1−cnn−1)|.

To show: The value of xn−1−cnn−1x−cn=(n−1)dn−1n−2 using Mean Value Theorem.

The reason for |x−cn|<h and |dn−1|≤α=max{|x|,|x+h|}.

To show: The value of |g(x)−f(x+h)−f(x)h|≤|h||∑n=2∞n(n−1)cnαn−2|.

To show: The series ∑n=2∞n(n−1)αn−2 converges for −R<x<R.

To show: The limit of limh→0f(x+h)−f(x)h=g(x).

Want to see the full answer?

The function g ( x ) = ∑ n = 0 ∞ n c n x n − 1 converges for − R &lt; x &lt; R .

Concept explainers

Videos

To show: The function g(x)=∑n=0∞ncnxn−1 converges for −R<x<R.

To show: The value (x+h)n−xnh=ncnn−1 using Mean Value Theorem.

To show: The value of |g(x)−f(x+h)−f(x)h|=|∑n=0∞ncn(xn−1−cnn−1)|.

To show: The value of xn−1−cnn−1x−cn=(n−1)dn−1n−2 using Mean Value Theorem.

The reason for |x−cn|<h and |dn−1|≤α=max{|x|,|x+h|}.

To show: The value of |g(x)−f(x+h)−f(x)h|≤|h||∑n=2∞n(n−1)cnαn−2|.

To show: The series ∑n=2∞n(n−1)αn−2 converges for −R<x<R.

To show: The limit of limh→0f(x+h)−f(x)h=g(x).

Want to see the full answer?

The function g ( x ) = ∑ n = 0 ∞ n c n x n − 1 converges for − R < x < R .