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6

All the terms in the product are less than half. Thus the infinite product is zero. (The answer by You're In My Eye is for the product $$\Pi = \prod_{n=1}^{\infty} \frac{2}{1+\pi^{{\large \frac{1}{2^n}}}}$$ The OP forgot a $2$ in the numerator of each term of the product, which is the likely cause of his confusion.)


4

Hint: You just need the fundamental theorem of calculus (the so-called "second version"). It helps to rewrite the integral with the $u$-substitution $u=t/\sqrt x $.


4

There is a general formula: $$\prod_{k=0}^\infty \frac{2}{1+x^{1/2^k}}=\frac{2}{x^2-1} \ln x$$ $$\prod_{k=1}^\infty \frac{2}{1+x^{1/2^k}}=\frac{1}{x-1} \ln x$$ $$\prod_{k=1}^\infty \frac{2}{1+\pi^{1/2^k}}=\frac{1}{\pi-1} \ln \pi=0.5345226992306749851 \dots$$ For the derivation of the above formula see the bottom of this answer Edit The original ...


3

Here are some ideas. Let $x,y \in ]a,b[$. Since $f_n \to f$ uniformly (I believe pointwise is enough though) we have $$ \lim_{n \to \infty} f_n(x) = f(x), \ \lim_{n \to \infty} f_n(y) = f(y)$$ and in particular $$ \lim_{n \to \infty} |f_n(x) - f_n(y)| = |f(x) - f(y)|$$ Since each $f_n$ is $C^1 (]a,b[)$ and $x,y \in ]a,b[$ and we know that $\{ f_n \}_{n=1}^\...


3

FYI, here is a picture illustrating Count's answer at http://math.stackexchange.com/a/1872168/241. Blue curves are the zeroth through tenth iterates of $f$; red curves are obtained by numerically inverting $F(f^{\circ N}(x)) \approx N + F(x)$ for $F(x) := \log \frac{x}{1-x} + \frac{1}{x}$. I will mention that this is an improvement on the exact fourth-order ...


3

Not quite. It's the last three terms in the second line that reduce to $-B|C|^{2}$. To see this, note that $\sum \overline{a_{j}}b_{j}=\sum\overline{a_{j}\overline{b_{j}}}=\overline{\sum a_{j}\overline{b_{j}}}=\overline{C}$. Hence, $$ \begin{aligned} &-B\overline{C}\sum a_{j}\overline{b_{j}}-BC\sum\overline{a_{j}}b_{j}+|C|^{2}\sum|b_{j}|^{2}\\ =&-B\...


2

For convenience, let $f$ be defined on $[a,b]$. For some $x' \leq b$, $f$ is monotone on $[a,x']$. Let $x$ be the supremum of these, so that for at least every $a < x' < x$, we have $f$ monotone on $[a,x']$. By continuity then, $f$ is monotone on all of $[a,x]$; for if $f(x) < f(x')$ for some $a < x'< x$, then there is an $x''$ near $x$ with $...


2

Yes, if you only have pointwise convergence. Take $(f_n)_n$ defined by $$ f_n(x) = x^2 \mathbb{1}_{[-n,n]}(x), x\in\mathbb{R}. $$ This converges pointwise to the function $f\colon x\in\mathbb{R}\mapsto x^2$, which is not bounded. But each $f_n$ is itself bounded (namely, $\lVert f_n\rVert_\infty = n^2$). Following a comment: however, if it exists, the ...


2

You could try showing that $x \neq y$ implies $x < y$ or $x > y$. Consider Cauchy sequences $(x_n)$ and $(y_n)$ representing $x$ and $y$ respectively. Remember that "$x = y$" means: for any $\delta > 0$ there exists a threshold $N > 0$ past which $\lvert x_n - y_m \rvert < \delta$ (i.e. this holds for $n,m > N$). The negation of that ...


2

Theorem. Let $x := (x_i)$ and $y := (y_i)$ be Cauchy sequences of rational numbers. Then either $\forall \epsilon > 0 : \exists N : \forall n > N : |x_n - y_n| < \epsilon$, $\exists \epsilon > 0 : \exists N : \forall n > N : x_n + \epsilon \leq y_n$, or $\exists \epsilon > 0 : \exists N : \forall n > N : x_n \geq y_n + \...


2

For any $y,z\in X$ $$\begin{align} |f_x(y)-f_x(z)|&=|d(x,y)-d(a,y)-d(x,z)+d(a,z)|\\ &\le|d(x,y)-d(x,z)|+|d(a,y)-d(a,z)|\\ &\le2\,d(y,z). \end{align} $$


1

I saw your post at "Mathimatiko ergasthri". Well, we can see that "$\frac{1}{(1+\pi^{1/2})(1+\pi^{1/4})(1+\pi^{1/8})}=\frac{(1-\pi^{1/2})(1-\pi^{1/4})(1-\pi^{1/8})}{(1+\pi^{1/2})(1+\pi^{1/4})(1+\pi^{1/8})(1-\pi^{1/2})(1-\pi^{1/4})(1-\pi^{1/8})}=\frac{(1-\pi^{1/2})(1-\pi^{1/4})(1-\pi^{1/8})}{(1-\pi^1)(1-\pi^{1/2})(1-\pi^{1/4})}=\frac{(1-\pi^{1/8})}{(1-\pi^1)}$...


1

Observe that $\max(0, 2ry - y^2)$ is $0$ except on the interval $[0, 2r]$. Therefore, assuming $r > 0$ and setting $\alpha = \min(d, 2r)$, the integral boils down to $$\begin{align*} F &= \int_0^\alpha 2eE(1 - y/d)\sqrt{2ry - y^2}\,\mathrm dy =\\ &= 2eE\int_0^\alpha\sqrt{2ry - y^2}\,\mathrm dy - \frac1d \int_0^\alpha y\sqrt{2ry - y^2}\,\mathrm dy =...


1

You don't need the continuity of $D^kf$. Note that the homogeneous polynomial $p(X):=D^kf(x_0).(X,\ldots,X)$ is continuous and $<0$ on $S^{n-1}$, hence satisfies $$p(X)\leq-\mu<0\qquad(X\in S^{n-1})\ .$$ By Taylor's theorem we therefore have $$f(x_0+X)-f(x_0)={1\over k!}\,p(X)+o(|X|^k)\leq-{|X|^k\over k!}\bigl(\mu+o(1)\bigr)\qquad(X\to0)\ .$$ Here the ...


1

Note that $k$ is necessarily even. Since you are working in finite dimensional space, the theorem of Weierstraß about maxima on compact sets applies and thus there is some $c>0$ such that $$ D^kf(x_0)[v^k]\le -c·\|v\|^k $$ where $-c$ is the maximum of the left side as a function of $v$ on the unit sphere $\{v:\|v\|=1\}$. Then by continuity of $D^kf$ ...


1

As Omnomnomnom answered, what you need is to apply the fundamental theorem of calculus. Applied to $$I=\int_0^{a(x)} f(x,t) \, dt$$ it leads to $$\frac {dI}{dx}=a'(x) f(x,a(x))+\int_0^{a(x)} f'_x(x,t) \, dt$$ Applied to your case $a(x)=x^2$, $f(x,t)=e^{x t^{-2}}$ this gives $$\frac {dI}{dx}=2 x e^{-\frac{1}{x^3}} - \int_0^{x^2}\frac{e^{-\frac{x}{t^2}}}{t^...


1

The essential property is that given a compact metric space $X$ and $r>0$ there is a finite number of balls covering $X$. Let $0<\lambda<1$ and make the simplifying assumption that $X$ may be covered by two closed balls $B_0$ and $B_1$ of size 1. Then assume (again for simplicity) that each of $\Lambda_0=B_0$ and $\Lambda_1=B_1$ (both are compact ...


1

The general outline is to write your given compact metric space $A$ as a decomposition into two sets, then four sets, then eight sets, etc., in the exact same fashion that the Cantor set is decomposed, except that whereas the decomposition elements of the Cantor set are pairwise disjoint, the decomposition elements in $A$ need not be pairwise disjoint. Be ...


1

Let $X$ be a compact metric space. The key point is that we can subdivide $X$ into smaller and smaller regions, such that at each stage there are only finitely many regions total. Specifically, we want for each $n$ a partition $$X=Y^n_1\sqcup . . . \sqcup Y^n_{k_n}$$ of $X$ such that each $Y^n_i$ has size (that is, diameter) less than (say) $2^{-n}$. If we ...


1

Let $f_n(x)=|x|$ for $|x|\le n$, and let $f(x)=n$ for $|x|\gt n$.


1

Let $x \in \mathbb{R}$. Choose $N \in \mathbb{N}$ such that $N > x$. For all $n > N$, we have $n > x$, so $x \notin [n,n+1]$, and therefore $1_{[n,n+1]}(x) = 0$. To reiterate, for all $n > N$ we have $1_{[n,n+1]}(x) = 0$. Therefore $\lim_{n \rightarrow \infty} 1_{[n,n+1]}(x) = 0$. In terms of $\epsilon$-$N$: For every $x \in \mathbb{R}$ and ...


1

All positive $a_0$ iterates to zero in finite time, see the graph of the iterating map. For negative $a_0$ there are fixed points at every $x=-m^2/(m+1)$, $m\geq 1$ (period 1 orbits). Either $a_0<-1$ eventually lands at one of those fixed points (which are unstable for $m\geq 2$), or it maps to the interval $[-1,0[$ after a finite number of iterations. ...


1

Assuming that $a_0 \gt 0$, note that for $a_n \gt 1$, $a_{n+1} \lt \lfloor a_n \rfloor$, so $\lfloor a_n+1\rfloor \le \lfloor a_n\rfloor -1$ so the sequence is bounded above by $a_n, a_n-1, a_n-2\dots 1$ and must reach $0$ and be constant there. Now assume $a_0 \lt 0$. We have $a_1=\lfloor a_0 \rfloor \{a_0\} \gt \lfloor a_0 \rfloor $ so the sequence ...


1

you find it as the form $z=a+ib$ so $$z=\frac { 1 }{ 3x+3iy+2 } =\frac { 3x+2-3iy }{ \left( 3x+2+3iy \right) \left( 3x+2-3iy \right) } =\frac { 3x+2 }{ { \left( 3x+2 \right) }^{ 2 }+9{ y }^{ 2 } } +i\frac { -3y }{ { \left( 3x+2 \right) }^{ 2 }+9{ y }^{ 2 } } \\ $$


1

Note quite, remember that $\frac{1}{a} + \frac{1}{b} \neq \frac{1}{a+b}$. In general, the real and imaginary part of $z$ is $x$ and $y$ where $z = x+iy$. That is, you must seek to turn your expression for $z$ into the form $x+iy$. Once you have $\frac{1}{3x + 3iy + 2}$ (excellent work!) multiply the fraction by the conjugate of the denominator. In general $$\...


1

Your book is correct because in general $\frac1{a+b}\ne \frac 1a+\frac 1b$. To solve the problem, you might have tried $$\frac1{3z+2} =\frac{3\bar z+2}{(3z+2)(3\bar z+2)}=\frac{3\bar z+2}{|3z+2|^2}$$



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