Implicit function theorem part 2

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For the Implicit Function Theorem to apply, two conditions must be satisfied: F(x_0, z_0) = 0 and the partial derivative ∂F/∂z at (x_0, z_0) must not equal zero. In this case, with (x_0, z_0) = (1, 2), both conditions are met, as F(1, 2) = 0 and ∂F/∂z = -1. Consequently, a function g exists such that x = g(z_1) in the neighborhood of z_0. The relation F(x, z_1) = F(g(z_1), z_1) holds true, leading to the potential determination of a positive epsilon value. This confirms the validity of the problem's statement using the Implicit Function Theorem.
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Homework Statement
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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)##
 
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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.
 
First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...