Three.OneFour
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 2d comment How can I split this into its' real and imaginary parts, and simplify? Please tell us next time what your real goal is (here: estimating $| \sum \cos(n) |$). Posting your own thoughts and results is great, but if you don't tell us what you really want to do it's hard for us to determine a) what you want, as in your other question, and b) whether you took a wrong/overcomplicated turn somewhere already. 2d comment How can I split this into its' real and imaginary parts, and simplify? Yes. You should start to think of complex numbers geometrically: If $z = x + iy$ is the result of the sum we have shown that $|z|$ is bounded, i.e. the distance between the point $(x, y)$ and the origin on the plane $\mathbb R^2$ is bounded. This also means that both $x$ and $y$ are bounded. 2d comment How can I split this into its' real and imaginary parts, and simplify? @volcanomane See my edit. 2d revised How can I split this into its' real and imaginary parts, and simplify? added 281 characters in body 2d answered How can I split this into its' real and imaginary parts, and simplify? Apr16 comment ((a ⇔ b) ⇒ c) ⇔ (a ⇔ (b ⇒ c)) tautology, contradiction, or neither? Thanks, I already changed that. Apr16 comment Lipschitz condition not satisfied Because your argument shows that the fraction goes to infinity for $u, v \to 0$. Therefore it cannot be bounded by any constant $L$ (constant in the sense that must be independent of $u$ and $v$). Apr16 answered ((a ⇔ b) ⇒ c) ⇔ (a ⇔ (b ⇒ c)) tautology, contradiction, or neither? Apr16 comment Lipschitz condition not satisfied Yes, your reasoning is sufficient. Apr16 answered Lipschitz condition not satisfied Apr14 comment $X$ is inner product space then its completion is Hilbert space? You can complete any metric space by the technique mentioned and the completion is unique up to an isometry (Wikipedia). Apr14 revised $X$ is inner product space then its completion is Hilbert space? added 194 characters in body Apr14 answered $X$ is inner product space then its completion is Hilbert space? Apr10 comment How to write the below proof rigorously for : If $p$ is a boundary point of $S$, then $p$ is a lim point of $S$ I assumed that $p \not \in S$ (as the original poster judging from his definition). How else does this not conform with the definition of a limit point? Apr10 revised How to write the below proof rigorously for : If $p$ is a boundary point of $S$, then $p$ is a lim point of $S$ added 353 characters in body Apr10 answered How to write the below proof rigorously for : If $p$ is a boundary point of $S$, then $p$ is a lim point of $S$ Apr2 comment What initial value do I have to take at the beginning? Although I couldn't find a reason in your link, you're right of course — I missed the change of behavior of your curves at $y_0 = 0$. Even though everything went well here, I still think that parametrizing the initial value boundary of the domain for the construction of characteristic curves is the "safer" approach. Apr1 answered Finiteness condition Mar28 answered is $N(f)=\int_{0}^{1} |f(t)|dt$ a norm on $E$(set of all continous real valued functions defined on [0,1])? Mar28 comment Non-equivalence of norms. I added another sentence to clarify what I meant. Boundedness, which is usually understood as @egreg defined it, doesn't require the sequence to converge (nor the sequence of norms).