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Math Help - linear and quadratic convergence

  1. #1
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    linear and quadratic convergence

    Hallo, All!
    Please, could anyone explain me what are linear and quadratic convergence?
    I just dont understand following: if f(p) is from interval (0,1) then there is a linear convergence.
    Can someoe help?
    Best regards, M
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  2. #2
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    Re: linear and quadratic convergence

    Are you sure the place where you read this fact does not have the relevant definitions? It is probably talking about the rate of convergence.
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  3. #3
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    Re: linear and quadratic convergence

    Please, why f(x) has to be x, I just can't connect...
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  4. #4
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    Re: linear and quadratic convergence

    Quote Originally Posted by marijakopljar View Post
    Please, why f(x) has to be x, I just can't connect...
    It does not. Right now I am working with f(x) = sin(x). In other words, I don't understand what you mean.
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  5. #5
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    Re: linear and quadratic convergence

    If it does not , then ok. I just foudn it on net and that confused me.


    I wil cut and paste:
    Limit (mathematics) - Wikipedia, the free encyclopedia

    "Convergence and fixed pointA formal definition of convergence can be stated as follows. Suppose as goes from to is a sequence that converges to a fixed point , with for all . If positive constants and exist with


    then as goes from to converges to of order , with asymptotic error constant

    Given a function with a fixed point , there is a nice checklist for checking the convergence of p.

    1) First check that p is indeed a fixed point:

    2) Check for linear convergence. Start by finding . If...."
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  6. #6
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    Integration

    "As a first approximation, look at the unit square given by the sides x = 0 to x = 1 and y = f(0) = 0 and y = f(1) = 1. Its area is exactly 1. As it is, the true value of the integral must be somewhat less. Decreasing the width of the approximation rectangles shall give a better result; so cross the interval in five steps, using the approximation points 0, 1/5, 2/5, and so on to 1. Fit a box for each step using the right end height of each curve piece, thus √(1⁄5), √(2⁄5), and so on to √1 = 1. Summing the areas of these rectangles, we get a better approximation for the sought integral, namely..."

    PLEASE, HOW DO WE GET SQUARE FRM 1/5 ETC? I just dont get it. It shoudl be the surface under the curve.
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  7. #7
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    Re: linear and quadratic convergence

    \sqrt{\frac{1}{5}}, \sqrt{\frac{2}{5}}, ... are the values of \sqrt{x}, which is the function that is being integrated, at the approximation points \frac{1}{5},\frac{2}{5},\dots,1. These values are used to find the area of yellow rectangles in the picture. This area, in turn, is an approximation of the area under the graph of \sqrt{x}.

    It would help if you explain the transition from the rate of convergence to integrals. They are not directly related. E.g., you don't need the definition of the rate of convergence to define integrals.
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  8. #8
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    Re: linear and quadratic convergence

    Thank U, but I still dont get it. The surface under the curve is the sum of all the sums of the green surfaces. Ok. But how do we get the quadrat square?
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  9. #9
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    Re: linear and quadratic convergence

    Quote Originally Posted by marijakopljar View Post
    The surface under the curve is the sum of all the sums of the green surfaces.
    No, the area of green rectangles is an approximation to the area under the curve. Note that I did not mention green rectangles, only yellow ones. You like to change the subject without explanation: first from the rate of convergence to integrals, then from yellow to green, don't you?

    Quote Originally Posted by marijakopljar View Post
    But how do we get the quadrat square?
    By "quadrat square" do you mean square root? I already explained this:
    Quote Originally Posted by emakarov View Post
    \sqrt{\frac{1}{5}}, \sqrt{\frac{2}{5}}, ... are the values of \sqrt{x}, which is the function that is being integrated, at the approximation points \frac{1}{5},\frac{2}{5},\dots,1.
    In other words, \sqrt{1/5} is the height of the leftmost yellow rectangle. This height is multiplied by the width 1/5 to get the area of the rectangle.
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  10. #10
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    Integrals (absolute beginner)

    Thank You!
    But the problem is that I cant understand why is the height of the yellow square square roor frm 1/5. Anyway, on the pic it is more than 0.4 and is greater that 0.2. What is the logic? That is the point!!!
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  11. #11
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    Re: Integrals (absolute beginner)

    To approximate the area under the graph, the height of the rectangle is chosen to be the y-coordinate of the point on the graph above 1/5. Do you know what the graph of a function is? The graph of f(x) is a set of points with coordinates (x, f(x)) for various x. Here the function is f(x)=\sqrt{x}. Which point is located on the graph above 1/5? By definition of the graph, it is f(1/5)=\sqrt{1/5}.

    Quote Originally Posted by marijakopljar View Post
    Anyway, on the pic it is more than 0.4 and is greater that 0.2. What is the logic?
    \sqrt{1/5}\approx0.45.
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  12. #12
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    Re: Integrals (absolute beginner)

    Please (untill U still have nerves!!!) how did we get that height is square root from 1/5? That is the main problem!
    (P.S. Please, excuse me, I am in a terreble hurry, and still this, cause I am not gifted for mth at all!)
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  13. #13
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    Re: Integrals (absolute beginner)

    Quote Originally Posted by marijakopljar View Post
    Please (untill U still have nerves!!!) how did we get that height is square root from 1/5?
    I can only repeat what I said in post #11.
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  14. #14
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    Re: linear and quadratic convergence

    THANK U SO MUCH, completely clear now, it is a drastical example how (extremely) bad concentratin influences math!!!(released!!!)
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  15. #15
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    Simple minds...

    Hallo!
    Me again! Please, how do we get F(1) -F(0) = 2/3 which is right, but 1 to high power 0.5 is still 1! How do we get the derivative f(x) = 2/3(x) to high power 3/2?
    Last edited by marijakopljar; August 17th 2012 at 01:53 AM.
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