MHB Find Max Function g: $\sqrt{10^{2016}}$

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evinda
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Hello! (Wave)

We are given the function $g: \mathbb{R}^{2016} \to \mathbb{R}$ with $g(y)=\sqrt{10 y_1^2 + 10^2 y_2^2+ \dots+ 10^{2016} y_{2016}^2} where y=(y_1, \dots, y_{2016})$.

I have found that the maximum of this function over the set $\{ y=(y_1, \dots, y_{2016}) \in \mathbb{R}^{2016}: \sum_{i=1}^{2016} y_i^2=1\}$ is $\sqrt{10^{2016}}$. Am I right? (Thinking)
 
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evinda said:
Hello! (Wave)

We are given the function $g: \mathbb{R}^{2016} \to \mathbb{R}$ with $g(y)=\sqrt{10 y_1^2 + 10^2 y_2^2+ \dots+ 10^{2016} y_{2016}^2} where y=(y_1, \dots, y_{2016})$.

I have found that the maximum of this function over the set $\{ y=(y_1, \dots, y_{2016}) \in \mathbb{R}^{2016}: \sum_{i=1}^{2016} y_i^2=1\}$ is $\sqrt{10^{2016}}$. Am I right? (Thinking)

Yep. You're right! (Nod)
 
I like Serena said:
Yep. You're right! (Nod)

Great... Thank you! (Clapping)
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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