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Midpoint and Trapezoidal Rule (EASY)

  • #1
I am solving a sample test for my test on monday, this is the question:

Table(1) have results from applying the midpoint and trapezoidal rule on 2, 4 and 8 panels

npanels....midpoint.......trapezoid
2...........2.19998820...2.48452890
4...........2.26805993...2.33225855
8...........2.28422520...2.30015924

1) do these number support the claim that integrand is concave down? Why?

2) Calculate the value of Simpson's rule using 8 panels.

3) what is the minimum number of function evaluations needed to calculate Simpson's rule from scratch on 8 panels?

************************

# 2 is easy: is 2/3(midpoint)+1/3(trappezoid)

Please help me with the other ones.

Thanks
 

Answers and Replies

  • #2
berkeman
Mentor
56,459
6,369
I am solving a sample test for my test on monday, this is the question:

Table(1) have results from applying the midpoint and trapezoidal rule on 2, 4 and 8 panels

npanels....midpoint.......trapezoid
2...........2.19998820...2.48452890
4...........2.26805993...2.33225855
8...........2.28422520...2.30015924

1) do these number support the claim that integrand is concave down? Why?

2) Calculate the value of Simpson's rule using 8 panels.

3) what is the minimum number of function evaluations needed to calculate Simpson's rule from scratch on 8 panels?

************************

# 2 is easy: is 2/3(midpoint)+1/3(trappezoid)

Please help me with the other ones.

Thanks
Please show us more of your work, per the PF Rules for schoolwork.
 
  • #3
Please show us more of your work, per the PF Rules for schoolwork.
there is nothing else to show
 
  • #4
berkeman
Mentor
56,459
6,369
there is nothing else to show
Have you actually read the PF rules?
 
  • #5
Have you actually read the PF rules?
I have, but I don't understand your question. There is nothing else to show, there is no math involve on the other two steps. You or someone that knows what they are doing, should be able to answer question 1 and 3 by looking at the data.

You want more work

2) 2/3(2.28422520)+1/3(2.30015924)=2.28954

Happy?

Thanks, I just need some guidance or someone that can answer this simple question.
 
  • #6
berkeman
Mentor
56,459
6,369
I have, but I don't understand your question. There is nothing else to show, there is no math involve on the other two steps. You or someone that knows what they are doing, should be able to answer question 1 and 3 by looking at the data.

You want more work

2) 2/3(2.28422520)+1/3(2.30015924)=2.28954

Happy?

Thanks, I just need some guidance or someone that can answer this simple question.
No, sorry. That doesn't help me. Part of the problem is that I don't understand what you are asking. If you could provide more details and background information, that would be a bit help. What is the "trapezoidal rule" that you are referring to? how does it apply to the numbers that you posted?

I'm not trying to be a pain here, but honestly, it would help if you posted a lot more background info, and lead us more into helping us provide hints to what you don't understand.
 
  • #7
well, that how the question is written on the sample test.

Trapezoidal rule
From Wikipedia, the free encyclopedia
Jump to: navigation, search
This article is about the quadrature rule for approximating integrals. For the Explicit trapezoidal rule for solving initial value problems, see Heun's method.
The function f(x) (in blue) is approximated by a linear function (in red).
Illustration of the composite trapezoidal rule (with a non-uniform grid)
Illustration of the composite trapezoidal rule (with a uniform grid)

In mathematics, the trapezium rule (also known as the trapezoid rule, or the trapezoidal rule in American English) is an approximate technique for calculating the definite integral

\int_{a}^{b} f(x)\,dx.

The trapezoidal rule works by approximating the region under the graph of the function f(x) as a trapezoid and calculating its area. It follows that

\int_{a}^{b} f(x)\, dx \approx (b-a)\frac{f(a) + f(b)}{2}.

To calculate this integral more accurately, one first splits the interval of integration [a,b] into n smaller subintervals, and then applies the trapezoidal rule on each of them. One obtains the composite trapezoidal rule:

\int_a^b f(x)\,dx \approx \frac{b-a}{n} \left[ {f(a) + f(b) \over 2} + \sum_{k=1}^{n-1} f \left( a+k \frac{b-a}{n} \right) \right].

This can alternatively be written as:

\int_a^b f(x)\,dx \approx \frac{b-a}{2n} \left(f(x_0) + 2f(x_1) + 2f(x_2)+\cdots+2f(x_{n-1}) + f(x_n) \right)

where

x_k=a+k \frac{b-a}{n},\text{ for }k=0, 1, \dots, n

(one can not use a non-uniform grid, as the composite trapezoidal rule assumes equally sized segments).
midpoint
http://www.youtube.com/watch?v=fKtSR0_xf5g"
 
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  • #8
740
27
Your extract from wikipedia is too algebraic for me. If you were to sketch the two cases: (1) concave down and (2) concave up, and draw on each the interpretation of the mid-point and trapezoidal rules, you might then see a way to answering the question.
 

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