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You have that $$f'_n(x) = f_{n-1}(x)$$ Now let's prove by recurence that $$|f_n(x)| \leq M \frac{x^n}{n!}$$ Indeed, $|f_0(x)| \leq M$ as it is continuous on $[0,1]$ Now if $|f_n(x)| \leq M \frac{x^n}{n!}$, then $$|f_{n+1}(x)| = \left| \int_0^x f_n(t)dt \right| \leq \int_0^x \left| f_n(t)\right| dt $$ $$\leq M \int_0^x \frac{t^n}{n!} dt = M ...


2

It's not clear whether "all the isolated singularities" is meant to include $\infty$ (i.e. in the Riemann sphere), or just those in $\mathbb C$. However, it turns out not to matter. For any rational function, the sum of the residues at all poles (including $\infty$) is $0$. In this case, the residue at $\infty$ is $0$, since $$\dfrac{z^n}{1+z+\ldots + ...


2

$$z^2 -iz=\left|z-i\right|$$ $$(a+bi)^2-i(a+bi)=\sqrt{a^2+(b-1)^2}\\a^2-b^2+2abi-ia+b=\sqrt{a^2+(b-1)^2}\\(a^2-b^2+b)+i(2ab-a)=\sqrt{a^2+(b-1)^2}\\ \to \left\{\begin{matrix} a(2b-1)=0 & \to & \\ a=0 ,or,b=\frac {1}{2} a^2-b^2+b=\sqrt{a^2+(b-1)^2} & & \end{matrix}\right.$$ now check for $a=0$ ,$b=\frac{1}{2}$ $$a=0 \to ...


2

You can take another approach, too.. Take a modulus in both side. If $z\neq i$, so $z=e^{i\theta}$. Since $im(z^2-iz)=0$ we have $\sin(2\theta)=\cos \theta$ and we obtain the solutions.


1

You won't get any farther than regulated functions. In fact, assuming that you mean $\mathcal{P}$ to be an operator that takes a class of functions to functions that are piecewise the restriction of elements of that class. $$\mathcal{P(U(P(}C))) = \mathcal{U(P(}C))$$ That is, a piecewise regulated function is regulated.This is fairly immediate if you take ...


1

No reference here, but I can at least answer your direct question of why the two conditions are equivalent. In particular, we need to figure out what the "first partial derivatives of the total derivative" look like. Assume, for the time being, that we're talking about a function $f:\Bbb R^n \to \Bbb R$. The total derivative is given by $$ Df = ...


1

It’s false as written. Take $b=10$, $r=2$, and $u=111$: $\operatorname{2-string}^{-1}(u)=14$, $\operatorname{10-string}(14)=15$, and $|\operatorname{trans}_{10,2}(111)|=2$, but $\lceil|u|\cdot\log_{10}2\rceil=\lceil3\log_{10}2\rceil=\lceil\log_{10}8\rceil=1$.


1

Here's what I found: let's pick the case where $m\quad =\quad 2k\quad /\quad k\in Z\\ $ As you know: $\sum _{ n=1 }^{ \infty }{ \frac { { t }^{ n } }{ { n }^{ m } } } \quad ={ Li }_{ m }(t)$ Let's substitute $t={ e }^{ ix }$: $\sum _{ n=1 }^{ \infty }{ \frac { { { e }^{ inx } } }{ { n }^{ m } } } \quad ={ Li }_{ m }({ e }^{ ix })$ We will use the ...


1

Let $$ F(z)=\frac{z^n}{1+z+z^2+\ldots+z^{n-1}}=\frac{P(z)}{Q(z)}. $$ Since $Q(1)=n\ne 0$, then, for every $z\ne 1$ we have $$ Q(z)=\frac{1-z^n}{1-z}, $$ and $F$ can be redefined as $$ F(z)=\begin{cases} \frac{(z-1)z^n}{z^n-1} &\mbox{ for } z\ne 1\\ \frac1n &\mbox{ for } z=1 \end{cases} $$ Therefore, the set of isolated singularities of $F_n$ is given ...



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