Use zero- through fourth-order Taylor series expansions to predict f ( 2.5 ) for f ( x ) =In x using a base point at x = 1 . Compute the true percent relative error ε t for each approximation. Discuss the meaning of the results.

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Answer 1

Textbook Question

Chapter 4, Problem 6P

Use zero- through fourth-order Taylor series expansions to predict f ( 2.5 ) for f ( x ) =In x using a base point at x = 1 . Compute the true percent relative error ε t for each approximation. Discuss the meaning of the results.

Expert Solution & Answer

To determine

To calculate: The approximate value of f(x)=lnx at the point x=2.5 with stating iteration at x=1 using the Taylor series with the simultaneous computation of the true percentage relative error at the zero-, first-, second-, third- and fourth order approximates.

Answer to Problem 6P

Solution:

The zero first, second, third and fourth order Taylor series approximations for the function f(x)=lnx at the point x=2.5 are 0, 1.5, 0.375, 1.5 and 0.2344 with the true relative percentage error at the respective stages as 100 %, 63.7 %, 59.07 %, 63.7%, and 74.42 % respectively.

The result shows that theTaylor series approximation could be generalised as,

f(3)=0+1.51−(1.5)22+(1.5)33−(1.5)44+⋯

Explanation of Solution

Given Information:

The function f(x)=lnx with points x=2.5 and the starting iteration at x=1.

Formula Used:

The Taylor series approximation of nth order is:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2!(xi+1−xi)2+⋯+f(n)(xi)n!(xi+1−xi)n.

Calculation:

Consider the zero-order approximation for the provided function,

f(xi+1)=f(xi)

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(2.5)=f(1)=ln(1)=0

The zero order approximation gives 0.

The exact value of the function at 2.5 would be:

f(2.5)=ln(2.5)=0.9163

Thus, the true relative percentage error would be:

εt=(|true−approximate|true×100)%=(|0.9163−0|0.9163×100)%=100%

The relative percentage error at this stage is 100%.

The first-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)=0+1(1)(1.5)=1.5

The first order approximation gives 1.5.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−1.5|0.9163×100)%=63.7%

The relative percentage error at this stage is 38.91%.

The second-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2=0+1(1)(1.5)−12(1)2(1.5)2=0.375

The second order approximation gives 0.375.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−0.375|0.9163×100)%=59.07%

The relative percentage error at this stage is 59.07%.

The third-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2+f(3)(xi)3!(xi+1−xi)3

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2+f(3)(1)3!(2.5−1)3=0+1(1)(1.5)−12(1)2(1.5)2+26(1)3(1.5)3=1.5

The third order approximation gives 1.5.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−1.5|0.9163×100)%=63.7%

The relative percentage error at this stage is 63.7%

The fourth-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2+f(3)(xi)3!(xi+1−xi)3+f(4)(xi)4!(xi+1−xi)4

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2+f(3)(1)3!(2.5−1)3+f(4)(1)3!(2.5−1)4=0+1(1)(1.5)−12(1)2(1.5)2+26(1)3(1.5)3−624(1)4(1.5)4=0.2344

The third order approximation gives 0.2344.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−0.2344|0.9163×100)%=74.42%

The relative percentage error at this stage is 74.42%.

This implies that the Taylor series approximation could be generalised as,

f(3)=0+1.51−(1.5)22+(1.5)33−(1.5)44+⋯

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Answer 2

Textbook Question

Chapter 4, Problem 6P

Use zero- through fourth-order Taylor series expansions to predict f ( 2.5 ) for f ( x ) =In x using a base point at x = 1 . Compute the true percent relative error ε t for each approximation. Discuss the meaning of the results.

Expert Solution & Answer

To determine

To calculate: The approximate value of f(x)=lnx at the point x=2.5 with stating iteration at x=1 using the Taylor series with the simultaneous computation of the true percentage relative error at the zero-, first-, second-, third- and fourth order approximates.

Answer to Problem 6P

Solution:

The zero first, second, third and fourth order Taylor series approximations for the function f(x)=lnx at the point x=2.5 are 0, 1.5, 0.375, 1.5 and 0.2344 with the true relative percentage error at the respective stages as 100 %, 63.7 %, 59.07 %, 63.7%, and 74.42 % respectively.

The result shows that theTaylor series approximation could be generalised as,

f(3)=0+1.51−(1.5)22+(1.5)33−(1.5)44+⋯

Explanation of Solution

Given Information:

The function f(x)=lnx with points x=2.5 and the starting iteration at x=1.

Formula Used:

The Taylor series approximation of nth order is:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2!(xi+1−xi)2+⋯+f(n)(xi)n!(xi+1−xi)n.

Calculation:

Consider the zero-order approximation for the provided function,

f(xi+1)=f(xi)

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(2.5)=f(1)=ln(1)=0

The zero order approximation gives 0.

The exact value of the function at 2.5 would be:

f(2.5)=ln(2.5)=0.9163

Thus, the true relative percentage error would be:

εt=(|true−approximate|true×100)%=(|0.9163−0|0.9163×100)%=100%

The relative percentage error at this stage is 100%.

The first-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)=0+1(1)(1.5)=1.5

The first order approximation gives 1.5.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−1.5|0.9163×100)%=63.7%

The relative percentage error at this stage is 38.91%.

The second-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2=0+1(1)(1.5)−12(1)2(1.5)2=0.375

The second order approximation gives 0.375.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−0.375|0.9163×100)%=59.07%

The relative percentage error at this stage is 59.07%.

The third-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2+f(3)(xi)3!(xi+1−xi)3

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2+f(3)(1)3!(2.5−1)3=0+1(1)(1.5)−12(1)2(1.5)2+26(1)3(1.5)3=1.5

The third order approximation gives 1.5.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−1.5|0.9163×100)%=63.7%

The relative percentage error at this stage is 63.7%

The fourth-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2+f(3)(xi)3!(xi+1−xi)3+f(4)(xi)4!(xi+1−xi)4

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2+f(3)(1)3!(2.5−1)3+f(4)(1)3!(2.5−1)4=0+1(1)(1.5)−12(1)2(1.5)2+26(1)3(1.5)3−624(1)4(1.5)4=0.2344

The third order approximation gives 0.2344.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−0.2344|0.9163×100)%=74.42%

The relative percentage error at this stage is 74.42%.

This implies that the Taylor series approximation could be generalised as,

f(3)=0+1.51−(1.5)22+(1.5)33−(1.5)44+⋯

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Answer 3

Textbook Question

Chapter 4, Problem 6P

Use zero- through fourth-order Taylor series expansions to predict f ( 2.5 ) for f ( x ) =In x using a base point at x = 1 . Compute the true percent relative error ε t for each approximation. Discuss the meaning of the results.

Expert Solution & Answer

To determine

To calculate: The approximate value of f(x)=lnx at the point x=2.5 with stating iteration at x=1 using the Taylor series with the simultaneous computation of the true percentage relative error at the zero-, first-, second-, third- and fourth order approximates.

Answer to Problem 6P

Solution:

The zero first, second, third and fourth order Taylor series approximations for the function f(x)=lnx at the point x=2.5 are 0, 1.5, 0.375, 1.5 and 0.2344 with the true relative percentage error at the respective stages as 100 %, 63.7 %, 59.07 %, 63.7%, and 74.42 % respectively.

The result shows that theTaylor series approximation could be generalised as,

f(3)=0+1.51−(1.5)22+(1.5)33−(1.5)44+⋯

Explanation of Solution

Given Information:

The function f(x)=lnx with points x=2.5 and the starting iteration at x=1.

Formula Used:

The Taylor series approximation of nth order is:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2!(xi+1−xi)2+⋯+f(n)(xi)n!(xi+1−xi)n.

Calculation:

Consider the zero-order approximation for the provided function,

f(xi+1)=f(xi)

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(2.5)=f(1)=ln(1)=0

The zero order approximation gives 0.

The exact value of the function at 2.5 would be:

f(2.5)=ln(2.5)=0.9163

Thus, the true relative percentage error would be:

εt=(|true−approximate|true×100)%=(|0.9163−0|0.9163×100)%=100%

The relative percentage error at this stage is 100%.

The first-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)=0+1(1)(1.5)=1.5

The first order approximation gives 1.5.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−1.5|0.9163×100)%=63.7%

The relative percentage error at this stage is 38.91%.

The second-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2=0+1(1)(1.5)−12(1)2(1.5)2=0.375

The second order approximation gives 0.375.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−0.375|0.9163×100)%=59.07%

The relative percentage error at this stage is 59.07%.

The third-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2+f(3)(xi)3!(xi+1−xi)3

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2+f(3)(1)3!(2.5−1)3=0+1(1)(1.5)−12(1)2(1.5)2+26(1)3(1.5)3=1.5

The third order approximation gives 1.5.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−1.5|0.9163×100)%=63.7%

The relative percentage error at this stage is 63.7%

The fourth-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2+f(3)(xi)3!(xi+1−xi)3+f(4)(xi)4!(xi+1−xi)4

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2+f(3)(1)3!(2.5−1)3+f(4)(1)3!(2.5−1)4=0+1(1)(1.5)−12(1)2(1.5)2+26(1)3(1.5)3−624(1)4(1.5)4=0.2344

The third order approximation gives 0.2344.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−0.2344|0.9163×100)%=74.42%

The relative percentage error at this stage is 74.42%.

This implies that the Taylor series approximation could be generalised as,

f(3)=0+1.51−(1.5)22+(1.5)33−(1.5)44+⋯

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Answer 4

Textbook Question

Chapter 4, Problem 6P

Use zero- through fourth-order Taylor series expansions to predict f ( 2.5 ) for f ( x ) =In x using a base point at x = 1 . Compute the true percent relative error ε t for each approximation. Discuss the meaning of the results.

Expert Solution & Answer

To determine

To calculate: The approximate value of f(x)=lnx at the point x=2.5 with stating iteration at x=1 using the Taylor series with the simultaneous computation of the true percentage relative error at the zero-, first-, second-, third- and fourth order approximates.

Answer to Problem 6P

Solution:

The zero first, second, third and fourth order Taylor series approximations for the function f(x)=lnx at the point x=2.5 are 0, 1.5, 0.375, 1.5 and 0.2344 with the true relative percentage error at the respective stages as 100 %, 63.7 %, 59.07 %, 63.7%, and 74.42 % respectively.

The result shows that theTaylor series approximation could be generalised as,

f(3)=0+1.51−(1.5)22+(1.5)33−(1.5)44+⋯

Explanation of Solution

Given Information:

The function f(x)=lnx with points x=2.5 and the starting iteration at x=1.

Formula Used:

The Taylor series approximation of nth order is:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2!(xi+1−xi)2+⋯+f(n)(xi)n!(xi+1−xi)n.

Calculation:

Consider the zero-order approximation for the provided function,

f(xi+1)=f(xi)

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(2.5)=f(1)=ln(1)=0

The zero order approximation gives 0.

The exact value of the function at 2.5 would be:

f(2.5)=ln(2.5)=0.9163

Thus, the true relative percentage error would be:

εt=(|true−approximate|true×100)%=(|0.9163−0|0.9163×100)%=100%

The relative percentage error at this stage is 100%.

The first-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)=0+1(1)(1.5)=1.5

The first order approximation gives 1.5.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−1.5|0.9163×100)%=63.7%

The relative percentage error at this stage is 38.91%.

The second-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2=0+1(1)(1.5)−12(1)2(1.5)2=0.375

The second order approximation gives 0.375.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−0.375|0.9163×100)%=59.07%

The relative percentage error at this stage is 59.07%.

The third-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2+f(3)(xi)3!(xi+1−xi)3

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2+f(3)(1)3!(2.5−1)3=0+1(1)(1.5)−12(1)2(1.5)2+26(1)3(1.5)3=1.5

The third order approximation gives 1.5.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−1.5|0.9163×100)%=63.7%

The relative percentage error at this stage is 63.7%

The fourth-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2+f(3)(xi)3!(xi+1−xi)3+f(4)(xi)4!(xi+1−xi)4

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2+f(3)(1)3!(2.5−1)3+f(4)(1)3!(2.5−1)4=0+1(1)(1.5)−12(1)2(1.5)2+26(1)3(1.5)3−624(1)4(1.5)4=0.2344

The third order approximation gives 0.2344.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−0.2344|0.9163×100)%=74.42%

The relative percentage error at this stage is 74.42%.

This implies that the Taylor series approximation could be generalised as,

f(3)=0+1.51−(1.5)22+(1.5)33−(1.5)44+⋯

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

To calculate: The approximate value of f(x)=lnx at the point x=2.5 with stating iteration at x=1 using the Taylor series with the simultaneous computation of the true percentage relative error at the zero-, first-, second-, third- and fourth order approximates.

Answer to Problem 6P

Solution:

The zero first, second, third and fourth order Taylor series approximations for the function f(x)=lnx at the point x=2.5 are 0, 1.5, 0.375, 1.5 and 0.2344 with the true relative percentage error at the respective stages as 100 %, 63.7 %, 59.07 %, 63.7%, and 74.42 % respectively.

The result shows that theTaylor series approximation could be generalised as,

f(3)=0+1.51−(1.5)22+(1.5)33−(1.5)44+⋯

Explanation of Solution

Given Information:

The function f(x)=lnx with points x=2.5 and the starting iteration at x=1.

Formula Used:

The Taylor series approximation of nth order is:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2!(xi+1−xi)2+⋯+f(n)(xi)n!(xi+1−xi)n.

Calculation:

Consider the zero-order approximation for the provided function,

f(xi+1)=f(xi)

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(2.5)=f(1)=ln(1)=0

The zero order approximation gives 0.

The exact value of the function at 2.5 would be:

f(2.5)=ln(2.5)=0.9163

Thus, the true relative percentage error would be:

εt=(|true−approximate|true×100)%=(|0.9163−0|0.9163×100)%=100%

The relative percentage error at this stage is 100%.

The first-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)=0+1(1)(1.5)=1.5

The first order approximation gives 1.5.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−1.5|0.9163×100)%=63.7%

The relative percentage error at this stage is 38.91%.

The second-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2=0+1(1)(1.5)−12(1)2(1.5)2=0.375

The second order approximation gives 0.375.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−0.375|0.9163×100)%=59.07%

The relative percentage error at this stage is 59.07%.

The third-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2+f(3)(xi)3!(xi+1−xi)3

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2+f(3)(1)3!(2.5−1)3=0+1(1)(1.5)−12(1)2(1.5)2+26(1)3(1.5)3=1.5

The third order approximation gives 1.5.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−1.5|0.9163×100)%=63.7%

The relative percentage error at this stage is 63.7%

The fourth-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2+f(3)(xi)3!(xi+1−xi)3+f(4)(xi)4!(xi+1−xi)4

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2+f(3)(1)3!(2.5−1)3+f(4)(1)3!(2.5−1)4=0+1(1)(1.5)−12(1)2(1.5)2+26(1)3(1.5)3−624(1)4(1.5)4=0.2344

The third order approximation gives 0.2344.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−0.2344|0.9163×100)%=74.42%

The relative percentage error at this stage is 74.42%.

This implies that the Taylor series approximation could be generalised as,

f(3)=0+1.51−(1.5)22+(1.5)33−(1.5)44+⋯

Answer 8

Expert Solution & Answer

To determine

To calculate: The approximate value of f(x)=lnx at the point x=2.5 with stating iteration at x=1 using the Taylor series with the simultaneous computation of the true percentage relative error at the zero-, first-, second-, third- and fourth order approximates.

Answer to Problem 6P

Solution:

The zero first, second, third and fourth order Taylor series approximations for the function f(x)=lnx at the point x=2.5 are 0, 1.5, 0.375, 1.5 and 0.2344 with the true relative percentage error at the respective stages as 100 %, 63.7 %, 59.07 %, 63.7%, and 74.42 % respectively.

The result shows that theTaylor series approximation could be generalised as,

f(3)=0+1.51−(1.5)22+(1.5)33−(1.5)44+⋯

Explanation of Solution

Given Information:

The function f(x)=lnx with points x=2.5 and the starting iteration at x=1.

Formula Used:

The Taylor series approximation of nth order is:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2!(xi+1−xi)2+⋯+f(n)(xi)n!(xi+1−xi)n.

Calculation:

Consider the zero-order approximation for the provided function,

f(xi+1)=f(xi)

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(2.5)=f(1)=ln(1)=0

The zero order approximation gives 0.

The exact value of the function at 2.5 would be:

f(2.5)=ln(2.5)=0.9163

Thus, the true relative percentage error would be:

εt=(|true−approximate|true×100)%=(|0.9163−0|0.9163×100)%=100%

The relative percentage error at this stage is 100%.

The first-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)=0+1(1)(1.5)=1.5

The first order approximation gives 1.5.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−1.5|0.9163×100)%=63.7%

The relative percentage error at this stage is 38.91%.

The second-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2=0+1(1)(1.5)−12(1)2(1.5)2=0.375

The second order approximation gives 0.375.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−0.375|0.9163×100)%=59.07%

The relative percentage error at this stage is 59.07%.

The third-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2+f(3)(xi)3!(xi+1−xi)3

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2+f(3)(1)3!(2.5−1)3=0+1(1)(1.5)−12(1)2(1.5)2+26(1)3(1.5)3=1.5

The third order approximation gives 1.5.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−1.5|0.9163×100)%=63.7%

The relative percentage error at this stage is 63.7%

The fourth-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2+f(3)(xi)3!(xi+1−xi)3+f(4)(xi)4!(xi+1−xi)4

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2+f(3)(1)3!(2.5−1)3+f(4)(1)3!(2.5−1)4=0+1(1)(1.5)−12(1)2(1.5)2+26(1)3(1.5)3−624(1)4(1.5)4=0.2344

The third order approximation gives 0.2344.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−0.2344|0.9163×100)%=74.42%

The relative percentage error at this stage is 74.42%.

This implies that the Taylor series approximation could be generalised as,

f(3)=0+1.51−(1.5)22+(1.5)33−(1.5)44+⋯

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

To calculate: The approximate value of f(x)=lnx at the point x=2.5 with stating iteration at x=1 using the Taylor series with the simultaneous computation of the true percentage relative error at the zero-, first-, second-, third- and fourth order approximates.

Answer 12

Answer to Problem 6P

Solution:

The zero first, second, third and fourth order Taylor series approximations for the function f(x)=lnx at the point x=2.5 are 0, 1.5, 0.375, 1.5 and 0.2344 with the true relative percentage error at the respective stages as 100 %, 63.7 %, 59.07 %, 63.7%, and 74.42 % respectively.

The result shows that theTaylor series approximation could be generalised as,

f(3)=0+1.51−(1.5)22+(1.5)33−(1.5)44+⋯

Answer 13

Explanation of Solution

Given Information:

The function f(x)=lnx with points x=2.5 and the starting iteration at x=1.

Formula Used:

The Taylor series approximation of nth order is:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2!(xi+1−xi)2+⋯+f(n)(xi)n!(xi+1−xi)n.

Calculation:

Consider the zero-order approximation for the provided function,

f(xi+1)=f(xi)

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(2.5)=f(1)=ln(1)=0

The zero order approximation gives 0.

The exact value of the function at 2.5 would be:

f(2.5)=ln(2.5)=0.9163

Thus, the true relative percentage error would be:

εt=(|true−approximate|true×100)%=(|0.9163−0|0.9163×100)%=100%

The relative percentage error at this stage is 100%.

The first-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)=0+1(1)(1.5)=1.5

The first order approximation gives 1.5.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−1.5|0.9163×100)%=63.7%

The relative percentage error at this stage is 38.91%.

The second-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2=0+1(1)(1.5)−12(1)2(1.5)2=0.375

The second order approximation gives 0.375.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−0.375|0.9163×100)%=59.07%

The relative percentage error at this stage is 59.07%.

The third-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2+f(3)(xi)3!(xi+1−xi)3

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2+f(3)(1)3!(2.5−1)3=0+1(1)(1.5)−12(1)2(1.5)2+26(1)3(1.5)3=1.5

The third order approximation gives 1.5.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−1.5|0.9163×100)%=63.7%

The relative percentage error at this stage is 63.7%

The fourth-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1−xi)+f''(xi)2(xi+1−xi)2+f(3)(xi)3!(xi+1−xi)3+f(4)(xi)4!(xi+1−xi)4

Now replace 2.5 for xi+1 and 1 for xi to obtain,

f(3)=f(1)+f'(1)(2.5−1)+f''(1)2(2.5−1)2+f(3)(1)3!(2.5−1)3+f(4)(1)3!(2.5−1)4=0+1(1)(1.5)−12(1)2(1.5)2+26(1)3(1.5)3−624(1)4(1.5)4=0.2344

The third order approximation gives 0.2344.

Thus,

εt=(|true−approximate|true×100)%=(|0.9163−0.2344|0.9163×100)%=74.42%

The relative percentage error at this stage is 74.42%.

This implies that the Taylor series approximation could be generalised as,

f(3)=0+1.51−(1.5)22+(1.5)33−(1.5)44+⋯

Answer 14

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Use zero- through fourth-order Taylor series expansions to predict f ( 2.5 ) for f ( x ) =In x using a base point at x = 1 . Compute the true percent relative error ε t for each approximation. Discuss the meaning of the results.

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Answer to Problem 6P

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