Show that for the implicit midpoint rule?

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In summary, the conversation is about a question regarding a problem statement that involves an equation with symbols representing unknown values. The person asking for help needs to provide more information, such as the complete problem statement, relevant equations, and what they have tried so far. Without this information, it is difficult for others to provide assistance.
  • #1
shehpar
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||e_i+1|| <= ||e_i||+h||f( (t_i+t_i+1)/2, y_i+y_i+1)/2)-f((t_i+t_i+1)/2, y(t_i+t_i+1)/2))||+O(h^3).

I need help about this question.if anybody able to guide me , I be thankful .
 
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  • #2
shehpar said:
||e_i+1|| <= ||e_i||+h||f( (t_i+t_i+1)/2, y_i+y_i+1)/2)-f((t_i+t_i+1)/2, y(t_i+t_i+1)/2))||+O(h^3).

I need help about this question.if anybody able to guide me , I be thankful .
You need to provide more information than this. There is a template with three parts:

  1. The complete problem statement
  2. Relevant or useful equations
  3. What you have tried

You have provided an equation with no explanation of what the symbols mean, or any other information, so don't expect much help.
 

1. What is the implicit midpoint rule?

The implicit midpoint rule is a numerical integration method used for solving ordinary differential equations. It is a second-order method that uses the derivative at the midpoint of the interval to approximate the value of the function at that point.

2. How is the implicit midpoint rule derived?

The implicit midpoint rule is derived by approximating the integral of the derivative using a midpoint approximation. This results in a two-stage method where the function value at the midpoint is used to estimate the slope at the midpoint, and then this slope is used to estimate the function value at the end of the interval.

3. What are the advantages of using the implicit midpoint rule?

The implicit midpoint rule is unconditionally stable, meaning that it can be used for a wide range of problems without having to adjust the step size. It is also a second-order method, which means that it has a higher accuracy compared to first-order methods.

4. Are there any limitations to using the implicit midpoint rule?

One limitation of the implicit midpoint rule is that it requires the solution of a nonlinear equation at each time step, which can be computationally expensive. It is also not suitable for stiff equations, as it may produce inaccurate results in these cases.

5. How can the accuracy of the implicit midpoint rule be improved?

The accuracy of the implicit midpoint rule can be improved by decreasing the step size. However, this will also increase the computational cost. Alternatively, a higher-order method can be used, such as the implicit trapezoidal rule, which has a higher accuracy but also requires solving a nonlinear equation at each time step.

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