1. ## Re: Simple minds...

Originally Posted by marijakopljar
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!
By definition of exponentiation, $1^{0.5}=1^{1/2}=\sqrt{1}$, which is 1.

Originally Posted by marijakopljar
How do we get the derivative f(x) = 2/3(x) to high power 3/2?
What exactly is f(x)? Do you mean that the derivative of f(x) is $\frac{2}{3}x^{2/3}$, or do you want to say that $f(x)=\frac{2}{3}x^{2/3}$ is a derivative of something else? My guess is that $f(x)=\sqrt{x}=x^{1/2}$ and its antiderivative $F(x)=\frac{2}{3}x^{3/2}$ (plus a constant). This is so because $\left(\frac{2}{3}x^{3/2}\right)'=x^{1/2}$ according to the power rule.

In plain text, you can write x^(2/3) to mean $x^{2/3}$. Note the parentheses; x^2/3 means $\frac{x^2}{3}$.

2. ## Re: linear and quadratic convergence

Sorry, antiderivative!
How do we get that antiderivative frm x2 = 1/3 x3???

3. ## Re: linear and quadratic convergence

Originally Posted by marijakopljar
How do we get that antiderivative frm x2 = 1/3 x3???
Originally Posted by emakarov
This is so because $\left(\frac{2}{3}x^{3/2}\right)'=x^{1/2}$ according to the power rule.
Edit: Sorry, this is obviously the antiderivative of a different function, but the idea is the same.

Should I also repeat my suggestion about using ^ in plain text?

4. ## Re: linear and quadratic convergence

Y, all on the net write so, but I just dont understand how do we GET this! Till now I just have to lern it by memory! I cross the whole net, but cant find the simple explanation that I miss! What would be the definition? Ok, it is area, but I cant move from it!

5. ## Re: linear and quadratic convergence

Originally Posted by marijakopljar
Y, all on the net write so, but I just dont understand how do we GET this!
By "this" do you mean the following question:
Originally Posted by marijakopljar
How do we get that antiderivative frm x2 = 1/3 x3???
By definition, an antiderivative of f is a function F such that F' = f. Since $\left(\frac{1}{3}x^3\right)'=x^2$, the function $\frac{1}{3}x^3$ is by definition an antiderivative of $x^2$.

Originally Posted by marijakopljar
What would be the definition? Ok, it is area, but I cant move from it!
Definition of what? The definition of antiderivative is not an area. It is related to area by the Fundamental theorem of calculus.

6. ## Re: linear and quadratic convergence

By definition?
But, why?

7. ## Re: linear and quadratic convergence

You ignored my two questions in post #20; why should I answer your incomplete questions?

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