Calc 4 Student, Please help me understand Existance and Uniqueness

danerape
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Ok, after going thru the proof, the only think that still eludes me is the region of definition given in the theorem itself. The rectangle where f and the partial of f with respect to y are known to be continuous.

I am thinking of this three dimensionally, and I do not know if this is the correct way to think about it? In other words, I am imagining that f and its partial are continuous "ON" the rectangle, not "IN" the rectangle. I have seen the theorem written with both words.

My 3-d take on this is that f and its partial are known continuous at every point, (x,y), within the confines of the rectangle. I am imagining f and its partial to be graphed in the z direction in other words.

I think I am confused about this because of the nature of picard iteration, where y is a function of x. I am considering y to be independent of x in my thinking, this is why I am not sure it is right.

So, is it true that continuity of f and its partial exist at every interior point in the rectangle?

Is it correct to think of this three dimensionally, as if f and the partial are being graphed ON the rectangle in the z direction.

ALSO, I HAVE A ROUGH DRAFT OF A PAPER I AM WRITING FOR STUDENTS WORKING AHEAD LIKE MYSELF, I THINK YOU CAN GET A BETTER JIST OF MY UNDERSTANDING THERE, ON PAGE 2.

Thanks

Dane
 

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Could this in some ways be analogous to thinking of a direction field? Even though we know y to be a function of x while graphing the direction field of y'=f(x,y), we still graph lineal elements, which seems analogous to graphing in the z direction. Does f being continuous in, or on the rectangle insinuate that the direction field exist in that rectangle?
 
Also, any critique of the paper is certainly welcome, before I submit it I will have it reviewed to make sure all is well.

Thanks

Dane

PS, pretty hard to understand for a mining engineering major, lol
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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