Another regulated functions question

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1. suppose f:[a,b]->R is a weakly increasing function, i.e. u<v gives f(u)<=f(v). Show that f is the limit of a sequence of step functions.

3. I have no idea how to start this. Please help.
 
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If you were to draw the graph of f on a computer monitor, what would the picture look like?

Now buy successively better computer monitors :)
 
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Surely different choices of f would look different wouldn't they? How do this help?
 
Of course the sequence of approximating step functions will depend on the particular function f. My hint is that you should approximate the graph of f roughly as a computer monitor does.
 
Sorry, but I don't know how a computer monitor would aproximate the graph of f, can you please elaborate?
 
Think of square pixels -- approximate the graph of f by something you might draw using the edges of a square grid.
 
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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