Compute forward and backward difference approximations of O ( h ) and O ( h 2 ) , and central difference approximations of O ( h 4 ) for the first derivative of y = cos x at x = π / 4 using a value of h = π / 12 . Estimate the true percent relative error ε t for each approximation.

Question

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Answer 1

Textbook Question

Chapter 23, Problem 1P

Compute forward and backward difference approximations of $O ( h )$ and $O ( h 2 )$ , and central difference approximations of $O ( h 4 )$ for the first derivative of $y = cos x$ at $x = π / 4$ using a value of $h = π / 12$ . Estimate the true percent relative error $ε t$ for each approximation.

Expert Solution & Answer

To determine

To calculate: The forward and backward difference approximations of $O(h)$ and $O(h2)$ , and central difference approximations of $O(h2)$ and $O(h4)$ for the first derivative of $y=cosx$ at $x=π4$ using a value of $h=π12$ . Also find the true percent error $ε$ , for each approximation.

Answer to Problem 1P

Solution:

Forward difference approximation of $O(h)$ is $−0.79108963$ with true percent error $ε=−11.877%$

Forward difference approximation of $O(h2)$ is $−0.72601275$ with true percent error $ε=−2.674%$

Backward difference approximation of $O(h)$ is $−0.60702442$ with true percent error $ε=14.154%$

Backward difference approximation of $O(h2)$ is $−0.71974088$ with true percent error $ε=−1.787%$

Central difference approximation of $O(h2)$ is $−0.69905703$ with true percent error $ε=1.138%$

Central difference approximation of $O(h4)$ is $−0.70699696$ with true percent error $ε=0.016%$

Explanation of Solution

Given information:

Function, $y=cosx$

Step size, $h=π12$

The initial value of x, $x=π4$

Formula used:

Forward difference approximation of $O(h)$ is,

$f′(xi)=f(xi+1)−f(xi)h$

Forward difference approximation of $O(h2)$ is,

$f′(xi)=−f(xi+2)+4f(xi+1)−3f(xi)2h$

Backward difference approximation of $O(h)$ is,

$f′(xi)=f(xi)−f(xi−1)h$

Backward difference approximation of $O(h2)$ is,

$f′(xi)=3f(xi)−4f(xi−1)+f(xi−2)2h$

Central difference approximation of $O(h2)$ is,

$f′(xi)=f(xi+1)−f(xi−1)2h$

Central difference approximation of $O(h4)$ is,

$f′(xi)=−f(xi+2)+8f(xi+1)−8f(xi−1)+f(xi−2)12h$

$xi+1=xi+h$ Here $i=0,1,2,...$

True percent relative error is,

$ε=(exact value − numerical valueexact value)×100$

Calculation:

Consider the function,

$f(x)=cosx$

First derivation of the function is,

$f′(x)=−sinx$

Thus, the true value of the first derivative of $y=cosx$ at $x=π4$ is,

$f′(π4)=−sin(π4)=−0.70710678$

The value of $xi−2$ is,

$xi−2=xi−2h=π4−2(π12)=π12=0.261799388$

The value of the function at $xi−2=0.261799388$ is,

$f(0.261799388)=cos(0.261799388)=0.965925826$

The value of $xi−1$ is,

$xi−1=xi−h=π4−(π12)=0.523598776$

The value of the function at $xi−1=0.523598776$ is,

$f(0.523598776)=cos(0.523598776)=0.866025404$

The value of $xi$ is,

$xi=π4=0.785398163$

The value of the function at $xi=0.785398163$ is,

$f(0.785398163)=cos(0.785398163)=0.707106781$

The value of $xi+1$ is,

$xi+1=xi+h=π4+π12=1.047197551$

The value of the function at $xi+1=1.047197551$ is,

$f(1.047197551)=cos(1.047197551)=0.5$

The value of $xi+2$ is,

$xi+2=xi+2h=π4+2(π12)=1.308996936$

The value of the function at $xi+2=1.308996936$ is,

$f(1.308996936)=cos(1.308996936)=0.258819045$

Forward difference approximation of $O(h)$ is,

$f′(xi)=f(xi+1)−f(xi)h$

For $y=cosx$ forward difference approximation is,

$f′(xi)=0.5−0.707106781(π12)=−0.2071067810.261799388=−0.79108963$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.79108963)(−0.70710678))×100=−11.877%$

Forward difference approximation of $O(h2)$ is,

$f′(xi)=−f(xi+2)+4f(xi+1)−3f(xi)2h$

For $y=cosx$ forward difference approximation is,

$f′(xi)=−0.258819045+4(0.5)−3(0.707106781)2(π12)=−0.380139388(π6)=−0.72601275$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.72601275)(−0.70710678))×100=−2.674%$

Backward difference approximation of $O(h)$ is,

$f′(xi)=f(xi)−f(xi−1)h$

For $y=cosx$ backward difference approximation is,

$f′(xi)=0.707106781−0.866025404(π12)=−0.158916230.261799388=−0.60702442$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.60702442)(−0.70710678))×100=14.154%$

Backward difference approximation of $O(h2)$ is,

$f′(xi)=3f(xi)−4f(xi−1)+f(xi−2)2h$

For $y=cosx$ backward difference approximation is,

$f′(xi)=3(0.707106781)−4(0.866025404)+0.9659258262(π12)=−0.376855447(π6)=−0.71974088$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.71974088)(−0.70710678))×100=−1.787%$

Central difference approximation of $O(h2)$ is,

$f′(xi)=f(xi+1)−f(xi−1)2h$

For $y=cosx$ central difference approximation is,

$f′(xi)=0.5−0.8660254042(π12)=−0.366025404(π6)=−0.69905703$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.69905703)(−0.70710678))×100=1.138%$

Central difference approximation of $O(h4)$ is,

$f′(xi)=−f(xi+2)+8f(xi+1)−8f(xi−1)+f(xi−2)12h$

For $y=cosx$ central difference approximation is,

$f′(xi)=−(0.258819045)+8(0.5)−8(0.866025404)+0.96592582612(π12)=−2.221096451π=−0.70699696$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.70699696)(−0.70710678))×100=0.016%$

Therefore, Forward difference approximation of $O(h)$ is $−0.79108963$ with true percent error $ε=−11.877%$

Forward difference approximation of $O(h2)$ is $−0.72601275$ with true percent error $ε=−2.674%$

Backward difference approximation of $O(h)$ is $−0.60702442$ with true percent error $ε=14.154%$

Backward difference approximation of $O(h2)$ is $−0.71974088$ with true percent error $ε=−1.787%$

Central difference approximation of $O(h2)$ is $−0.69905703$ with true percent error $ε=1.138%$

Central difference approximation of $O(h4)$ is $−0.70699696$ with true percent error $ε=0.016%$

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Answer 2

Textbook Question

Chapter 23, Problem 1P

Compute forward and backward difference approximations of $O ( h )$ and $O ( h 2 )$ , and central difference approximations of $O ( h 4 )$ for the first derivative of $y = cos x$ at $x = π / 4$ using a value of $h = π / 12$ . Estimate the true percent relative error $ε t$ for each approximation.

Expert Solution & Answer

To determine

To calculate: The forward and backward difference approximations of $O(h)$ and $O(h2)$ , and central difference approximations of $O(h2)$ and $O(h4)$ for the first derivative of $y=cosx$ at $x=π4$ using a value of $h=π12$ . Also find the true percent error $ε$ , for each approximation.

Answer to Problem 1P

Solution:

Forward difference approximation of $O(h)$ is $−0.79108963$ with true percent error $ε=−11.877%$

Forward difference approximation of $O(h2)$ is $−0.72601275$ with true percent error $ε=−2.674%$

Backward difference approximation of $O(h)$ is $−0.60702442$ with true percent error $ε=14.154%$

Backward difference approximation of $O(h2)$ is $−0.71974088$ with true percent error $ε=−1.787%$

Central difference approximation of $O(h2)$ is $−0.69905703$ with true percent error $ε=1.138%$

Central difference approximation of $O(h4)$ is $−0.70699696$ with true percent error $ε=0.016%$

Explanation of Solution

Given information:

Function, $y=cosx$

Step size, $h=π12$

The initial value of x, $x=π4$

Formula used:

Forward difference approximation of $O(h)$ is,

$f′(xi)=f(xi+1)−f(xi)h$

Forward difference approximation of $O(h2)$ is,

$f′(xi)=−f(xi+2)+4f(xi+1)−3f(xi)2h$

Backward difference approximation of $O(h)$ is,

$f′(xi)=f(xi)−f(xi−1)h$

Backward difference approximation of $O(h2)$ is,

$f′(xi)=3f(xi)−4f(xi−1)+f(xi−2)2h$

Central difference approximation of $O(h2)$ is,

$f′(xi)=f(xi+1)−f(xi−1)2h$

Central difference approximation of $O(h4)$ is,

$f′(xi)=−f(xi+2)+8f(xi+1)−8f(xi−1)+f(xi−2)12h$

$xi+1=xi+h$ Here $i=0,1,2,...$

True percent relative error is,

$ε=(exact value − numerical valueexact value)×100$

Calculation:

Consider the function,

$f(x)=cosx$

First derivation of the function is,

$f′(x)=−sinx$

Thus, the true value of the first derivative of $y=cosx$ at $x=π4$ is,

$f′(π4)=−sin(π4)=−0.70710678$

The value of $xi−2$ is,

$xi−2=xi−2h=π4−2(π12)=π12=0.261799388$

The value of the function at $xi−2=0.261799388$ is,

$f(0.261799388)=cos(0.261799388)=0.965925826$

The value of $xi−1$ is,

$xi−1=xi−h=π4−(π12)=0.523598776$

The value of the function at $xi−1=0.523598776$ is,

$f(0.523598776)=cos(0.523598776)=0.866025404$

The value of $xi$ is,

$xi=π4=0.785398163$

The value of the function at $xi=0.785398163$ is,

$f(0.785398163)=cos(0.785398163)=0.707106781$

The value of $xi+1$ is,

$xi+1=xi+h=π4+π12=1.047197551$

The value of the function at $xi+1=1.047197551$ is,

$f(1.047197551)=cos(1.047197551)=0.5$

The value of $xi+2$ is,

$xi+2=xi+2h=π4+2(π12)=1.308996936$

The value of the function at $xi+2=1.308996936$ is,

$f(1.308996936)=cos(1.308996936)=0.258819045$

Forward difference approximation of $O(h)$ is,

$f′(xi)=f(xi+1)−f(xi)h$

For $y=cosx$ forward difference approximation is,

$f′(xi)=0.5−0.707106781(π12)=−0.2071067810.261799388=−0.79108963$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.79108963)(−0.70710678))×100=−11.877%$

Forward difference approximation of $O(h2)$ is,

$f′(xi)=−f(xi+2)+4f(xi+1)−3f(xi)2h$

For $y=cosx$ forward difference approximation is,

$f′(xi)=−0.258819045+4(0.5)−3(0.707106781)2(π12)=−0.380139388(π6)=−0.72601275$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.72601275)(−0.70710678))×100=−2.674%$

Backward difference approximation of $O(h)$ is,

$f′(xi)=f(xi)−f(xi−1)h$

For $y=cosx$ backward difference approximation is,

$f′(xi)=0.707106781−0.866025404(π12)=−0.158916230.261799388=−0.60702442$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.60702442)(−0.70710678))×100=14.154%$

Backward difference approximation of $O(h2)$ is,

$f′(xi)=3f(xi)−4f(xi−1)+f(xi−2)2h$

For $y=cosx$ backward difference approximation is,

$f′(xi)=3(0.707106781)−4(0.866025404)+0.9659258262(π12)=−0.376855447(π6)=−0.71974088$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.71974088)(−0.70710678))×100=−1.787%$

Central difference approximation of $O(h2)$ is,

$f′(xi)=f(xi+1)−f(xi−1)2h$

For $y=cosx$ central difference approximation is,

$f′(xi)=0.5−0.8660254042(π12)=−0.366025404(π6)=−0.69905703$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.69905703)(−0.70710678))×100=1.138%$

Central difference approximation of $O(h4)$ is,

$f′(xi)=−f(xi+2)+8f(xi+1)−8f(xi−1)+f(xi−2)12h$

For $y=cosx$ central difference approximation is,

$f′(xi)=−(0.258819045)+8(0.5)−8(0.866025404)+0.96592582612(π12)=−2.221096451π=−0.70699696$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.70699696)(−0.70710678))×100=0.016%$

Therefore, Forward difference approximation of $O(h)$ is $−0.79108963$ with true percent error $ε=−11.877%$

Forward difference approximation of $O(h2)$ is $−0.72601275$ with true percent error $ε=−2.674%$

Backward difference approximation of $O(h)$ is $−0.60702442$ with true percent error $ε=14.154%$

Backward difference approximation of $O(h2)$ is $−0.71974088$ with true percent error $ε=−1.787%$

Central difference approximation of $O(h2)$ is $−0.69905703$ with true percent error $ε=1.138%$

Central difference approximation of $O(h4)$ is $−0.70699696$ with true percent error $ε=0.016%$

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Answer 3

Textbook Question

Chapter 23, Problem 1P

Compute forward and backward difference approximations of $O ( h )$ and $O ( h 2 )$ , and central difference approximations of $O ( h 4 )$ for the first derivative of $y = cos x$ at $x = π / 4$ using a value of $h = π / 12$ . Estimate the true percent relative error $ε t$ for each approximation.

Expert Solution & Answer

To determine

To calculate: The forward and backward difference approximations of $O(h)$ and $O(h2)$ , and central difference approximations of $O(h2)$ and $O(h4)$ for the first derivative of $y=cosx$ at $x=π4$ using a value of $h=π12$ . Also find the true percent error $ε$ , for each approximation.

Answer to Problem 1P

Solution:

Forward difference approximation of $O(h)$ is $−0.79108963$ with true percent error $ε=−11.877%$

Forward difference approximation of $O(h2)$ is $−0.72601275$ with true percent error $ε=−2.674%$

Backward difference approximation of $O(h)$ is $−0.60702442$ with true percent error $ε=14.154%$

Backward difference approximation of $O(h2)$ is $−0.71974088$ with true percent error $ε=−1.787%$

Central difference approximation of $O(h2)$ is $−0.69905703$ with true percent error $ε=1.138%$

Central difference approximation of $O(h4)$ is $−0.70699696$ with true percent error $ε=0.016%$

Explanation of Solution

Given information:

Function, $y=cosx$

Step size, $h=π12$

The initial value of x, $x=π4$

Formula used:

Forward difference approximation of $O(h)$ is,

$f′(xi)=f(xi+1)−f(xi)h$

Forward difference approximation of $O(h2)$ is,

$f′(xi)=−f(xi+2)+4f(xi+1)−3f(xi)2h$

Backward difference approximation of $O(h)$ is,

$f′(xi)=f(xi)−f(xi−1)h$

Backward difference approximation of $O(h2)$ is,

$f′(xi)=3f(xi)−4f(xi−1)+f(xi−2)2h$

Central difference approximation of $O(h2)$ is,

$f′(xi)=f(xi+1)−f(xi−1)2h$

Central difference approximation of $O(h4)$ is,

$f′(xi)=−f(xi+2)+8f(xi+1)−8f(xi−1)+f(xi−2)12h$

$xi+1=xi+h$ Here $i=0,1,2,...$

True percent relative error is,

$ε=(exact value − numerical valueexact value)×100$

Calculation:

Consider the function,

$f(x)=cosx$

First derivation of the function is,

$f′(x)=−sinx$

Thus, the true value of the first derivative of $y=cosx$ at $x=π4$ is,

$f′(π4)=−sin(π4)=−0.70710678$

The value of $xi−2$ is,

$xi−2=xi−2h=π4−2(π12)=π12=0.261799388$

The value of the function at $xi−2=0.261799388$ is,

$f(0.261799388)=cos(0.261799388)=0.965925826$

The value of $xi−1$ is,

$xi−1=xi−h=π4−(π12)=0.523598776$

The value of the function at $xi−1=0.523598776$ is,

$f(0.523598776)=cos(0.523598776)=0.866025404$

The value of $xi$ is,

$xi=π4=0.785398163$

The value of the function at $xi=0.785398163$ is,

$f(0.785398163)=cos(0.785398163)=0.707106781$

The value of $xi+1$ is,

$xi+1=xi+h=π4+π12=1.047197551$

The value of the function at $xi+1=1.047197551$ is,

$f(1.047197551)=cos(1.047197551)=0.5$

The value of $xi+2$ is,

$xi+2=xi+2h=π4+2(π12)=1.308996936$

The value of the function at $xi+2=1.308996936$ is,

$f(1.308996936)=cos(1.308996936)=0.258819045$

Forward difference approximation of $O(h)$ is,

$f′(xi)=f(xi+1)−f(xi)h$

For $y=cosx$ forward difference approximation is,

$f′(xi)=0.5−0.707106781(π12)=−0.2071067810.261799388=−0.79108963$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.79108963)(−0.70710678))×100=−11.877%$

Forward difference approximation of $O(h2)$ is,

$f′(xi)=−f(xi+2)+4f(xi+1)−3f(xi)2h$

For $y=cosx$ forward difference approximation is,

$f′(xi)=−0.258819045+4(0.5)−3(0.707106781)2(π12)=−0.380139388(π6)=−0.72601275$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.72601275)(−0.70710678))×100=−2.674%$

Backward difference approximation of $O(h)$ is,

$f′(xi)=f(xi)−f(xi−1)h$

For $y=cosx$ backward difference approximation is,

$f′(xi)=0.707106781−0.866025404(π12)=−0.158916230.261799388=−0.60702442$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.60702442)(−0.70710678))×100=14.154%$

Backward difference approximation of $O(h2)$ is,

$f′(xi)=3f(xi)−4f(xi−1)+f(xi−2)2h$

For $y=cosx$ backward difference approximation is,

$f′(xi)=3(0.707106781)−4(0.866025404)+0.9659258262(π12)=−0.376855447(π6)=−0.71974088$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.71974088)(−0.70710678))×100=−1.787%$

Central difference approximation of $O(h2)$ is,

$f′(xi)=f(xi+1)−f(xi−1)2h$

For $y=cosx$ central difference approximation is,

$f′(xi)=0.5−0.8660254042(π12)=−0.366025404(π6)=−0.69905703$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.69905703)(−0.70710678))×100=1.138%$

Central difference approximation of $O(h4)$ is,

$f′(xi)=−f(xi+2)+8f(xi+1)−8f(xi−1)+f(xi−2)12h$

For $y=cosx$ central difference approximation is,

$f′(xi)=−(0.258819045)+8(0.5)−8(0.866025404)+0.96592582612(π12)=−2.221096451π=−0.70699696$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.70699696)(−0.70710678))×100=0.016%$

Therefore, Forward difference approximation of $O(h)$ is $−0.79108963$ with true percent error $ε=−11.877%$

Forward difference approximation of $O(h2)$ is $−0.72601275$ with true percent error $ε=−2.674%$

Backward difference approximation of $O(h)$ is $−0.60702442$ with true percent error $ε=14.154%$

Backward difference approximation of $O(h2)$ is $−0.71974088$ with true percent error $ε=−1.787%$

Central difference approximation of $O(h2)$ is $−0.69905703$ with true percent error $ε=1.138%$

Central difference approximation of $O(h4)$ is $−0.70699696$ with true percent error $ε=0.016%$

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Answer 4

Textbook Question

Chapter 23, Problem 1P

Compute forward and backward difference approximations of $O ( h )$ and $O ( h 2 )$ , and central difference approximations of $O ( h 4 )$ for the first derivative of $y = cos x$ at $x = π / 4$ using a value of $h = π / 12$ . Estimate the true percent relative error $ε t$ for each approximation.

Expert Solution & Answer

To determine

To calculate: The forward and backward difference approximations of $O(h)$ and $O(h2)$ , and central difference approximations of $O(h2)$ and $O(h4)$ for the first derivative of $y=cosx$ at $x=π4$ using a value of $h=π12$ . Also find the true percent error $ε$ , for each approximation.

Answer to Problem 1P

Solution:

Forward difference approximation of $O(h)$ is $−0.79108963$ with true percent error $ε=−11.877%$

Forward difference approximation of $O(h2)$ is $−0.72601275$ with true percent error $ε=−2.674%$

Backward difference approximation of $O(h)$ is $−0.60702442$ with true percent error $ε=14.154%$

Backward difference approximation of $O(h2)$ is $−0.71974088$ with true percent error $ε=−1.787%$

Central difference approximation of $O(h2)$ is $−0.69905703$ with true percent error $ε=1.138%$

Central difference approximation of $O(h4)$ is $−0.70699696$ with true percent error $ε=0.016%$

Explanation of Solution

Given information:

Function, $y=cosx$

Step size, $h=π12$

The initial value of x, $x=π4$

Formula used:

Forward difference approximation of $O(h)$ is,

$f′(xi)=f(xi+1)−f(xi)h$

Forward difference approximation of $O(h2)$ is,

$f′(xi)=−f(xi+2)+4f(xi+1)−3f(xi)2h$

Backward difference approximation of $O(h)$ is,

$f′(xi)=f(xi)−f(xi−1)h$

Backward difference approximation of $O(h2)$ is,

$f′(xi)=3f(xi)−4f(xi−1)+f(xi−2)2h$

Central difference approximation of $O(h2)$ is,

$f′(xi)=f(xi+1)−f(xi−1)2h$

Central difference approximation of $O(h4)$ is,

$f′(xi)=−f(xi+2)+8f(xi+1)−8f(xi−1)+f(xi−2)12h$

$xi+1=xi+h$ Here $i=0,1,2,...$

True percent relative error is,

$ε=(exact value − numerical valueexact value)×100$

Calculation:

Consider the function,

$f(x)=cosx$

First derivation of the function is,

$f′(x)=−sinx$

Thus, the true value of the first derivative of $y=cosx$ at $x=π4$ is,

$f′(π4)=−sin(π4)=−0.70710678$

The value of $xi−2$ is,

$xi−2=xi−2h=π4−2(π12)=π12=0.261799388$

The value of the function at $xi−2=0.261799388$ is,

$f(0.261799388)=cos(0.261799388)=0.965925826$

The value of $xi−1$ is,

$xi−1=xi−h=π4−(π12)=0.523598776$

The value of the function at $xi−1=0.523598776$ is,

$f(0.523598776)=cos(0.523598776)=0.866025404$

The value of $xi$ is,

$xi=π4=0.785398163$

The value of the function at $xi=0.785398163$ is,

$f(0.785398163)=cos(0.785398163)=0.707106781$

The value of $xi+1$ is,

$xi+1=xi+h=π4+π12=1.047197551$

The value of the function at $xi+1=1.047197551$ is,

$f(1.047197551)=cos(1.047197551)=0.5$

The value of $xi+2$ is,

$xi+2=xi+2h=π4+2(π12)=1.308996936$

The value of the function at $xi+2=1.308996936$ is,

$f(1.308996936)=cos(1.308996936)=0.258819045$

Forward difference approximation of $O(h)$ is,

$f′(xi)=f(xi+1)−f(xi)h$

For $y=cosx$ forward difference approximation is,

$f′(xi)=0.5−0.707106781(π12)=−0.2071067810.261799388=−0.79108963$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.79108963)(−0.70710678))×100=−11.877%$

Forward difference approximation of $O(h2)$ is,

$f′(xi)=−f(xi+2)+4f(xi+1)−3f(xi)2h$

For $y=cosx$ forward difference approximation is,

$f′(xi)=−0.258819045+4(0.5)−3(0.707106781)2(π12)=−0.380139388(π6)=−0.72601275$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.72601275)(−0.70710678))×100=−2.674%$

Backward difference approximation of $O(h)$ is,

$f′(xi)=f(xi)−f(xi−1)h$

For $y=cosx$ backward difference approximation is,

$f′(xi)=0.707106781−0.866025404(π12)=−0.158916230.261799388=−0.60702442$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.60702442)(−0.70710678))×100=14.154%$

Backward difference approximation of $O(h2)$ is,

$f′(xi)=3f(xi)−4f(xi−1)+f(xi−2)2h$

For $y=cosx$ backward difference approximation is,

$f′(xi)=3(0.707106781)−4(0.866025404)+0.9659258262(π12)=−0.376855447(π6)=−0.71974088$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.71974088)(−0.70710678))×100=−1.787%$

Central difference approximation of $O(h2)$ is,

$f′(xi)=f(xi+1)−f(xi−1)2h$

For $y=cosx$ central difference approximation is,

$f′(xi)=0.5−0.8660254042(π12)=−0.366025404(π6)=−0.69905703$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.69905703)(−0.70710678))×100=1.138%$

Central difference approximation of $O(h4)$ is,

$f′(xi)=−f(xi+2)+8f(xi+1)−8f(xi−1)+f(xi−2)12h$

For $y=cosx$ central difference approximation is,

$f′(xi)=−(0.258819045)+8(0.5)−8(0.866025404)+0.96592582612(π12)=−2.221096451π=−0.70699696$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.70699696)(−0.70710678))×100=0.016%$

Therefore, Forward difference approximation of $O(h)$ is $−0.79108963$ with true percent error $ε=−11.877%$

Forward difference approximation of $O(h2)$ is $−0.72601275$ with true percent error $ε=−2.674%$

Backward difference approximation of $O(h)$ is $−0.60702442$ with true percent error $ε=14.154%$

Backward difference approximation of $O(h2)$ is $−0.71974088$ with true percent error $ε=−1.787%$

Central difference approximation of $O(h2)$ is $−0.69905703$ with true percent error $ε=1.138%$

Central difference approximation of $O(h4)$ is $−0.70699696$ with true percent error $ε=0.016%$

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

To calculate: The forward and backward difference approximations of $O(h)$ and $O(h2)$ , and central difference approximations of $O(h2)$ and $O(h4)$ for the first derivative of $y=cosx$ at $x=π4$ using a value of $h=π12$ . Also find the true percent error $ε$ , for each approximation.

Answer to Problem 1P

Solution:

Forward difference approximation of $O(h)$ is $−0.79108963$ with true percent error $ε=−11.877%$

Forward difference approximation of $O(h2)$ is $−0.72601275$ with true percent error $ε=−2.674%$

Backward difference approximation of $O(h)$ is $−0.60702442$ with true percent error $ε=14.154%$

Backward difference approximation of $O(h2)$ is $−0.71974088$ with true percent error $ε=−1.787%$

Central difference approximation of $O(h2)$ is $−0.69905703$ with true percent error $ε=1.138%$

Central difference approximation of $O(h4)$ is $−0.70699696$ with true percent error $ε=0.016%$

Explanation of Solution

Given information:

Function, $y=cosx$

Step size, $h=π12$

The initial value of x, $x=π4$

Formula used:

Forward difference approximation of $O(h)$ is,

$f′(xi)=f(xi+1)−f(xi)h$

Forward difference approximation of $O(h2)$ is,

$f′(xi)=−f(xi+2)+4f(xi+1)−3f(xi)2h$

Backward difference approximation of $O(h)$ is,

$f′(xi)=f(xi)−f(xi−1)h$

Backward difference approximation of $O(h2)$ is,

$f′(xi)=3f(xi)−4f(xi−1)+f(xi−2)2h$

Central difference approximation of $O(h2)$ is,

$f′(xi)=f(xi+1)−f(xi−1)2h$

Central difference approximation of $O(h4)$ is,

$f′(xi)=−f(xi+2)+8f(xi+1)−8f(xi−1)+f(xi−2)12h$

$xi+1=xi+h$ Here $i=0,1,2,...$

True percent relative error is,

$ε=(exact value − numerical valueexact value)×100$

Calculation:

Consider the function,

$f(x)=cosx$

First derivation of the function is,

$f′(x)=−sinx$

Thus, the true value of the first derivative of $y=cosx$ at $x=π4$ is,

$f′(π4)=−sin(π4)=−0.70710678$

The value of $xi−2$ is,

$xi−2=xi−2h=π4−2(π12)=π12=0.261799388$

The value of the function at $xi−2=0.261799388$ is,

$f(0.261799388)=cos(0.261799388)=0.965925826$

The value of $xi−1$ is,

$xi−1=xi−h=π4−(π12)=0.523598776$

The value of the function at $xi−1=0.523598776$ is,

$f(0.523598776)=cos(0.523598776)=0.866025404$

The value of $xi$ is,

$xi=π4=0.785398163$

The value of the function at $xi=0.785398163$ is,

$f(0.785398163)=cos(0.785398163)=0.707106781$

The value of $xi+1$ is,

$xi+1=xi+h=π4+π12=1.047197551$

The value of the function at $xi+1=1.047197551$ is,

$f(1.047197551)=cos(1.047197551)=0.5$

The value of $xi+2$ is,

$xi+2=xi+2h=π4+2(π12)=1.308996936$

The value of the function at $xi+2=1.308996936$ is,

$f(1.308996936)=cos(1.308996936)=0.258819045$

Forward difference approximation of $O(h)$ is,

$f′(xi)=f(xi+1)−f(xi)h$

For $y=cosx$ forward difference approximation is,

$f′(xi)=0.5−0.707106781(π12)=−0.2071067810.261799388=−0.79108963$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.79108963)(−0.70710678))×100=−11.877%$

Forward difference approximation of $O(h2)$ is,

$f′(xi)=−f(xi+2)+4f(xi+1)−3f(xi)2h$

For $y=cosx$ forward difference approximation is,

$f′(xi)=−0.258819045+4(0.5)−3(0.707106781)2(π12)=−0.380139388(π6)=−0.72601275$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.72601275)(−0.70710678))×100=−2.674%$

Backward difference approximation of $O(h)$ is,

$f′(xi)=f(xi)−f(xi−1)h$

For $y=cosx$ backward difference approximation is,

$f′(xi)=0.707106781−0.866025404(π12)=−0.158916230.261799388=−0.60702442$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.60702442)(−0.70710678))×100=14.154%$

Backward difference approximation of $O(h2)$ is,

$f′(xi)=3f(xi)−4f(xi−1)+f(xi−2)2h$

For $y=cosx$ backward difference approximation is,

$f′(xi)=3(0.707106781)−4(0.866025404)+0.9659258262(π12)=−0.376855447(π6)=−0.71974088$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.71974088)(−0.70710678))×100=−1.787%$

Central difference approximation of $O(h2)$ is,

$f′(xi)=f(xi+1)−f(xi−1)2h$

For $y=cosx$ central difference approximation is,

$f′(xi)=0.5−0.8660254042(π12)=−0.366025404(π6)=−0.69905703$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.69905703)(−0.70710678))×100=1.138%$

Central difference approximation of $O(h4)$ is,

$f′(xi)=−f(xi+2)+8f(xi+1)−8f(xi−1)+f(xi−2)12h$

For $y=cosx$ central difference approximation is,

$f′(xi)=−(0.258819045)+8(0.5)−8(0.866025404)+0.96592582612(π12)=−2.221096451π=−0.70699696$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.70699696)(−0.70710678))×100=0.016%$

Therefore, Forward difference approximation of $O(h)$ is $−0.79108963$ with true percent error $ε=−11.877%$

Forward difference approximation of $O(h2)$ is $−0.72601275$ with true percent error $ε=−2.674%$

Backward difference approximation of $O(h)$ is $−0.60702442$ with true percent error $ε=14.154%$

Backward difference approximation of $O(h2)$ is $−0.71974088$ with true percent error $ε=−1.787%$

Central difference approximation of $O(h2)$ is $−0.69905703$ with true percent error $ε=1.138%$

Central difference approximation of $O(h4)$ is $−0.70699696$ with true percent error $ε=0.016%$

Answer 8

Expert Solution & Answer

To determine

To calculate: The forward and backward difference approximations of $O(h)$ and $O(h2)$ , and central difference approximations of $O(h2)$ and $O(h4)$ for the first derivative of $y=cosx$ at $x=π4$ using a value of $h=π12$ . Also find the true percent error $ε$ , for each approximation.

Answer to Problem 1P

Solution:

Forward difference approximation of $O(h)$ is $−0.79108963$ with true percent error $ε=−11.877%$

Forward difference approximation of $O(h2)$ is $−0.72601275$ with true percent error $ε=−2.674%$

Backward difference approximation of $O(h)$ is $−0.60702442$ with true percent error $ε=14.154%$

Backward difference approximation of $O(h2)$ is $−0.71974088$ with true percent error $ε=−1.787%$

Central difference approximation of $O(h2)$ is $−0.69905703$ with true percent error $ε=1.138%$

Central difference approximation of $O(h4)$ is $−0.70699696$ with true percent error $ε=0.016%$

Explanation of Solution

Given information:

Function, $y=cosx$

Step size, $h=π12$

The initial value of x, $x=π4$

Formula used:

Forward difference approximation of $O(h)$ is,

$f′(xi)=f(xi+1)−f(xi)h$

Forward difference approximation of $O(h2)$ is,

$f′(xi)=−f(xi+2)+4f(xi+1)−3f(xi)2h$

Backward difference approximation of $O(h)$ is,

$f′(xi)=f(xi)−f(xi−1)h$

Backward difference approximation of $O(h2)$ is,

$f′(xi)=3f(xi)−4f(xi−1)+f(xi−2)2h$

Central difference approximation of $O(h2)$ is,

$f′(xi)=f(xi+1)−f(xi−1)2h$

Central difference approximation of $O(h4)$ is,

$f′(xi)=−f(xi+2)+8f(xi+1)−8f(xi−1)+f(xi−2)12h$

$xi+1=xi+h$ Here $i=0,1,2,...$

True percent relative error is,

$ε=(exact value − numerical valueexact value)×100$

Calculation:

Consider the function,

$f(x)=cosx$

First derivation of the function is,

$f′(x)=−sinx$

Thus, the true value of the first derivative of $y=cosx$ at $x=π4$ is,

$f′(π4)=−sin(π4)=−0.70710678$

The value of $xi−2$ is,

$xi−2=xi−2h=π4−2(π12)=π12=0.261799388$

The value of the function at $xi−2=0.261799388$ is,

$f(0.261799388)=cos(0.261799388)=0.965925826$

The value of $xi−1$ is,

$xi−1=xi−h=π4−(π12)=0.523598776$

The value of the function at $xi−1=0.523598776$ is,

$f(0.523598776)=cos(0.523598776)=0.866025404$

The value of $xi$ is,

$xi=π4=0.785398163$

The value of the function at $xi=0.785398163$ is,

$f(0.785398163)=cos(0.785398163)=0.707106781$

The value of $xi+1$ is,

$xi+1=xi+h=π4+π12=1.047197551$

The value of the function at $xi+1=1.047197551$ is,

$f(1.047197551)=cos(1.047197551)=0.5$

The value of $xi+2$ is,

$xi+2=xi+2h=π4+2(π12)=1.308996936$

The value of the function at $xi+2=1.308996936$ is,

$f(1.308996936)=cos(1.308996936)=0.258819045$

Forward difference approximation of $O(h)$ is,

$f′(xi)=f(xi+1)−f(xi)h$

For $y=cosx$ forward difference approximation is,

$f′(xi)=0.5−0.707106781(π12)=−0.2071067810.261799388=−0.79108963$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.79108963)(−0.70710678))×100=−11.877%$

Forward difference approximation of $O(h2)$ is,

$f′(xi)=−f(xi+2)+4f(xi+1)−3f(xi)2h$

For $y=cosx$ forward difference approximation is,

$f′(xi)=−0.258819045+4(0.5)−3(0.707106781)2(π12)=−0.380139388(π6)=−0.72601275$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.72601275)(−0.70710678))×100=−2.674%$

Backward difference approximation of $O(h)$ is,

$f′(xi)=f(xi)−f(xi−1)h$

For $y=cosx$ backward difference approximation is,

$f′(xi)=0.707106781−0.866025404(π12)=−0.158916230.261799388=−0.60702442$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.60702442)(−0.70710678))×100=14.154%$

Backward difference approximation of $O(h2)$ is,

$f′(xi)=3f(xi)−4f(xi−1)+f(xi−2)2h$

For $y=cosx$ backward difference approximation is,

$f′(xi)=3(0.707106781)−4(0.866025404)+0.9659258262(π12)=−0.376855447(π6)=−0.71974088$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.71974088)(−0.70710678))×100=−1.787%$

Central difference approximation of $O(h2)$ is,

$f′(xi)=f(xi+1)−f(xi−1)2h$

For $y=cosx$ central difference approximation is,

$f′(xi)=0.5−0.8660254042(π12)=−0.366025404(π6)=−0.69905703$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.69905703)(−0.70710678))×100=1.138%$

Central difference approximation of $O(h4)$ is,

$f′(xi)=−f(xi+2)+8f(xi+1)−8f(xi−1)+f(xi−2)12h$

For $y=cosx$ central difference approximation is,

$f′(xi)=−(0.258819045)+8(0.5)−8(0.866025404)+0.96592582612(π12)=−2.221096451π=−0.70699696$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.70699696)(−0.70710678))×100=0.016%$

Therefore, Forward difference approximation of $O(h)$ is $−0.79108963$ with true percent error $ε=−11.877%$

Forward difference approximation of $O(h2)$ is $−0.72601275$ with true percent error $ε=−2.674%$

Backward difference approximation of $O(h)$ is $−0.60702442$ with true percent error $ε=14.154%$

Backward difference approximation of $O(h2)$ is $−0.71974088$ with true percent error $ε=−1.787%$

Central difference approximation of $O(h2)$ is $−0.69905703$ with true percent error $ε=1.138%$

Central difference approximation of $O(h4)$ is $−0.70699696$ with true percent error $ε=0.016%$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

To calculate: The forward and backward difference approximations of $O(h)$ and $O(h2)$ , and central difference approximations of $O(h2)$ and $O(h4)$ for the first derivative of $y=cosx$ at $x=π4$ using a value of $h=π12$ . Also find the true percent error $ε$ , for each approximation.

Answer 12

Answer to Problem 1P

Solution:

Forward difference approximation of $O(h)$ is $−0.79108963$ with true percent error $ε=−11.877%$

Forward difference approximation of $O(h2)$ is $−0.72601275$ with true percent error $ε=−2.674%$

Backward difference approximation of $O(h)$ is $−0.60702442$ with true percent error $ε=14.154%$

Backward difference approximation of $O(h2)$ is $−0.71974088$ with true percent error $ε=−1.787%$

Central difference approximation of $O(h2)$ is $−0.69905703$ with true percent error $ε=1.138%$

Central difference approximation of $O(h4)$ is $−0.70699696$ with true percent error $ε=0.016%$

Answer 13

Explanation of Solution

Given information:

Function, $y=cosx$

Step size, $h=π12$

The initial value of x, $x=π4$

Formula used:

Forward difference approximation of $O(h)$ is,

$f′(xi)=f(xi+1)−f(xi)h$

Forward difference approximation of $O(h2)$ is,

$f′(xi)=−f(xi+2)+4f(xi+1)−3f(xi)2h$

Backward difference approximation of $O(h)$ is,

$f′(xi)=f(xi)−f(xi−1)h$

Backward difference approximation of $O(h2)$ is,

$f′(xi)=3f(xi)−4f(xi−1)+f(xi−2)2h$

Central difference approximation of $O(h2)$ is,

$f′(xi)=f(xi+1)−f(xi−1)2h$

Central difference approximation of $O(h4)$ is,

$f′(xi)=−f(xi+2)+8f(xi+1)−8f(xi−1)+f(xi−2)12h$

$xi+1=xi+h$ Here $i=0,1,2,...$

True percent relative error is,

$ε=(exact value − numerical valueexact value)×100$

Calculation:

Consider the function,

$f(x)=cosx$

First derivation of the function is,

$f′(x)=−sinx$

Thus, the true value of the first derivative of $y=cosx$ at $x=π4$ is,

$f′(π4)=−sin(π4)=−0.70710678$

The value of $xi−2$ is,

$xi−2=xi−2h=π4−2(π12)=π12=0.261799388$

The value of the function at $xi−2=0.261799388$ is,

$f(0.261799388)=cos(0.261799388)=0.965925826$

The value of $xi−1$ is,

$xi−1=xi−h=π4−(π12)=0.523598776$

The value of the function at $xi−1=0.523598776$ is,

$f(0.523598776)=cos(0.523598776)=0.866025404$

The value of $xi$ is,

$xi=π4=0.785398163$

The value of the function at $xi=0.785398163$ is,

$f(0.785398163)=cos(0.785398163)=0.707106781$

The value of $xi+1$ is,

$xi+1=xi+h=π4+π12=1.047197551$

The value of the function at $xi+1=1.047197551$ is,

$f(1.047197551)=cos(1.047197551)=0.5$

The value of $xi+2$ is,

$xi+2=xi+2h=π4+2(π12)=1.308996936$

The value of the function at $xi+2=1.308996936$ is,

$f(1.308996936)=cos(1.308996936)=0.258819045$

Forward difference approximation of $O(h)$ is,

$f′(xi)=f(xi+1)−f(xi)h$

For $y=cosx$ forward difference approximation is,

$f′(xi)=0.5−0.707106781(π12)=−0.2071067810.261799388=−0.79108963$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.79108963)(−0.70710678))×100=−11.877%$

Forward difference approximation of $O(h2)$ is,

$f′(xi)=−f(xi+2)+4f(xi+1)−3f(xi)2h$

For $y=cosx$ forward difference approximation is,

$f′(xi)=−0.258819045+4(0.5)−3(0.707106781)2(π12)=−0.380139388(π6)=−0.72601275$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.72601275)(−0.70710678))×100=−2.674%$

Backward difference approximation of $O(h)$ is,

$f′(xi)=f(xi)−f(xi−1)h$

For $y=cosx$ backward difference approximation is,

$f′(xi)=0.707106781−0.866025404(π12)=−0.158916230.261799388=−0.60702442$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.60702442)(−0.70710678))×100=14.154%$

Backward difference approximation of $O(h2)$ is,

$f′(xi)=3f(xi)−4f(xi−1)+f(xi−2)2h$

For $y=cosx$ backward difference approximation is,

$f′(xi)=3(0.707106781)−4(0.866025404)+0.9659258262(π12)=−0.376855447(π6)=−0.71974088$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.71974088)(−0.70710678))×100=−1.787%$

Central difference approximation of $O(h2)$ is,

$f′(xi)=f(xi+1)−f(xi−1)2h$

For $y=cosx$ central difference approximation is,

$f′(xi)=0.5−0.8660254042(π12)=−0.366025404(π6)=−0.69905703$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.69905703)(−0.70710678))×100=1.138%$

Central difference approximation of $O(h4)$ is,

$f′(xi)=−f(xi+2)+8f(xi+1)−8f(xi−1)+f(xi−2)12h$

For $y=cosx$ central difference approximation is,

$f′(xi)=−(0.258819045)+8(0.5)−8(0.866025404)+0.96592582612(π12)=−2.221096451π=−0.70699696$

True percent error is,

$ε=(exact value − numerical valueexact value)×100=((−0.70710678)−(−0.70699696)(−0.70710678))×100=0.016%$

Therefore, Forward difference approximation of $O(h)$ is $−0.79108963$ with true percent error $ε=−11.877%$

Forward difference approximation of $O(h2)$ is $−0.72601275$ with true percent error $ε=−2.674%$

Backward difference approximation of $O(h)$ is $−0.60702442$ with true percent error $ε=14.154%$

Backward difference approximation of $O(h2)$ is $−0.71974088$ with true percent error $ε=−1.787%$

Central difference approximation of $O(h2)$ is $−0.69905703$ with true percent error $ε=1.138%$

Central difference approximation of $O(h4)$ is $−0.70699696$ with true percent error $ε=0.016%$

Answer 14

Ch. 23 - 23.1 Compute forward and backward difference...Ch. 23 - 23.2 Repeat Prob. 23.1, but for evaluated at...Ch. 23 - 23.3 Use centered difference approximations to...Ch. 23 - Use Richardson extrapolation to estimate the first...Ch. 23 - Repeat Prob. 23.4, but for the first derivative of...Ch. 23 - 23.6 Employ Eq. (23.9) to determine the first...Ch. 23 - 23.7 Prove that for equispaced data points, Eq....Ch. 23 - Compute the first-order central difference...Ch. 23 - Prob. 9P Ch. 23 - Develop a user-friendly program to apply a Romberg...

Compute forward and backward difference approximations of O ( h ) and O ( h 2 ) , and central difference approximations of O ( h 4 ) for the first derivative of y = cos x at x = π / 4 using a value of h = π / 12 . Estimate the true percent relative error ε t for each approximation.

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