Using the analytic definition of a function

JacobHempel
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Homework Statement



The best way to represent this problem is with a snapshot of the problem.

Ha.jpg


Hopefully that works.

Homework Equations



f(x) = sqrt(1^2+x^2)

The Attempt at a Solution



f'(1/2) = lim(h>0) (1-h)/B = 1-0/B = 1/B ... B = -1/(Sqrt(3)
 
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What I don't get is letter b. I don't understand the process in which started from lim(h>0) ((f(1/2+h)-(f(1/2))/h all the way to lim(h>0)1-h/b ...
 
Nope. That's definitely not the best way to do (show) it. First start with:

\lim_{h\to 0} \frac{\sqrt{1-(1/2+h)^2}-\sqrt{1-(1/2)^2}}{h}

Ok, don't even look at the printout. Now, in order to evaluate that limit (simply), we need some more h's on the bottom. We can do that by multiplying top and bottom by what, I forgot what it's called but:

\lim_{h\to 0} \frac{\sqrt{1-(1/2+h)^2}-\sqrt{1-(1/2)^2}}{h}\frac{\sqrt{1-(1/2+h)^2}+\sqrt{1-(1/2)^2}}{\sqrt{1-(1/2+h)^2}+\sqrt{1-(1/2)^2}}

Now I bet you can do all that algebra, simplify it, and then take the limit as h goes to zero. Then figure out what that paper is asking you to do with the details.
 
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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