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• August 13th 2012, 01:52 AM
marijakopljar
Hallo, All!
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
• August 13th 2012, 03:42 AM
emakarov
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.
• August 13th 2012, 05:47 AM
marijakopljar
Please, why f(x) has to be x, I just can't connect...
• August 13th 2012, 05:51 AM
emakarov
Quote:

Originally Posted by marijakopljar
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.
• August 15th 2012, 03:08 AM
marijakopljar
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...."
• August 15th 2012, 03:13 AM
marijakopljar
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.
• August 15th 2012, 04:13 AM
emakarov
$\sqrt{\frac{1}{5}}$, $\sqrt{\frac{2}{5}}$, ... 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.
• August 16th 2012, 03:08 AM
marijakopljar
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?
• August 16th 2012, 03:19 AM
emakarov
Quote:

Originally Posted by marijakopljar
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
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
$\sqrt{\frac{1}{5}}$, $\sqrt{\frac{2}{5}}$, ... 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.
• August 16th 2012, 07:41 AM
marijakopljar
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!!!
• August 16th 2012, 07:50 AM
emakarov
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
Anyway, on the pic it is more than 0.4 and is greater that 0.2. What is the logic?

$\sqrt{1/5}\approx0.45$.
• August 16th 2012, 09:42 PM
marijakopljar
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!)
• August 17th 2012, 12:26 AM
emakarov
Re: Integrals (absolute beginner)
Quote:

Originally Posted by marijakopljar
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.
• August 17th 2012, 01:25 AM
marijakopljar