Implicit function theorem part 2

Lambda96
Messages
233
Reaction score
77
Homework Statement
see Post
Relevant Equations
Implicit function theorem
Hi,

I'm not sure if I've solved the problem correctly

Bildschirmfoto 2024-07-06 um 16.56.27.png


In order for the Implicit function theorem to be applied, the following two properties must hold ##F(x_0,z_0)=0## and ##\frac{\partial F(x_0,z_0)}{\partial z} \neq 0##. ##(x_0,z_0)=(1,2)## is a zero and ##\frac{\partial F(x_0,z_0)}{\partial z} =-x^2=-1## so both properties are fulfilled


According to the Implicit function theorem, there now exists a function ##g## for which the following relation ##x=g(z_1)## is valid in the neighborhood ##(z_0 - \epsilon, z_0 + \epsilon)##.


Then the following applies ##F(x,z_1)=F(g(z_1),z_1)##
 
Physics news on Phys.org
You're doing good! keep going. You could end up with a value of some ##\epsilon>0## in the end, because that's the the problem is asking for. Then, you could show that the statement in the problem is true using the value of ##\epsilon## you found and the implicit function theorem.
 
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...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top