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4

No, $f$ needn't be globally injective. A counterexample is $$f:\mathbb C\to \mathbb C:z\mapsto \int_0^ze^{t^2}dt$$ Why is that entire map $f$ surjective? Because by Picard's theorem it could at most skip one value $b\in \mathbb C$ i.e. $f(\mathbb C)=\mathbb C\setminus \{b\}$. Of course that potential $b$ is nonzero since $f(0)=0$. But since $f(-z)=-f(z)...


4

No, of course not. Any finite set with more than one point is compact and not convex.


3

Why wouldn't we be able to use the ratio test? Let $b_n=a_nx^n$ The ratio test states if we have the series $\sum_{n=0}^\infty b_n$ and we define $L=\lim_{n\rightarrow \infty} \left|\dfrac{b_{n+1}}{b_n}\right|$, then $L$ converges absolutely if $L<1$, $L$ diverges if $L>1$, and $L$ is inconclusive if $L=1$. $$L=\lim_{n\rightarrow \infty}\left|\...


3

It's not true. Take $N = 4, k= 2$, we get $$\frac{1}{4} \sum_{n=1}^2 e^{- 2 \pi i n } = \frac{1}{2}$$ In general, we can always take $k = 2N/4$ and make the sum $$\frac{1}{N} \sum_{n=N/4}^{3N/4 - 1} e^{- 2 \pi i n} = \frac{1}{2}.$$


3

It is false. Consider the matrix.$$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ It is not the square of a complex matrix, so $p(X)=X^2$ is not surjective.


1

Since $N$ is a multiple of $4$, let $N = 4m$. Then, ignoring the $\frac1{N}$, $\begin{array}\\ \sum\limits_{n=N/4}^{3N/4-1} e^{-4\pi ink/N} &=\sum\limits_{n=m}^{3m-1} e^{-\pi ink/m}\\ &=\sum\limits_{n=m}^{3m-1} \left(e^{-\pi ik/m}\right)^n\\ &=\sum\limits_{n=m}^{3m-1} r^n \qquad\text{where } r = e^{-\pi ik/m}\\ &=r^m\sum\limits_{n=m}^{3m-1} ...


1

A sketch, not a formal proof. (Also: my English may lack of strict terminology.) We can visualize $h = f(x,r)$ as a curved plane (3D plot of the function). The edge $x=0$ is at $h=0$. The edge $x=1$ is at $h=1$. On the edge $r=0$ there is at least one point with $h=\frac 1 2$. The same for $r=1$. Now your question: on our curved plane is there a ...


1

I'll try to address some of the misunderstandings I think you have based on the question and the comments. Someone else can tackle the problem given in $(3)$ of explaining everything about Taylor polynomials and Taylor series from the ground up if they want (though I doubt anyone will). First off, it sounds like you're conflating Taylor polynomials with ...


1

In answer to your specific points, 1) The inductive assumption is not being applied to the whole expression, but only to the first bracket, which runs to some smaller number than $k$. At that point, the intent is to show that the final element from the first bracket can be shifted into the start of the second bracket. 2) As above, the square bracket ...


1

1) How is it that Eq(2) contains $a_k$ but in that section of the proof the assumption is that the theorem is true for n≤k-1. Is not this assumption being overreached by the equation running to ak and therefore invalid? It's just a proof by complete induction: they assume that the result is true for $n \leq k-1$ and show it is also true for $n=k$. If the ...



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