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1

The Weierstrass approximation theorem says that continuous real-valued functions on the unit interval can be uniformly approximated arbitrarily closely by polynomials. That is, for every continuous function $f:[0,1]\to\mathbb{R}$ and each $\varepsilon>0$, there is a polynomial $p$ such that $|f(x)-p(x)|<\varepsilon$ for all $x\in[0,1]$. An elementary ...


0

I conjecture $$ 4n+3 \in \{4x^2-4xy+7y^2 \in P :x,y\geq 1\}$$ $$\iff$$ $$\frac{(2-\sqrt{3})^{n+1}+(2+\sqrt{3})^{n+1}}{4n+3}\in Z $$ Will complete this as time allows since another answer is already accepted. I think it can be proven using theory of binary quadratic forms or the equivalent.


0

Notice that $$\ln\left(1-|\mathcal{X}|^{-\alpha n}\right)^{2^{nC}\left(1-|\mathcal{X}|^{-\alpha n}\right)}=2^{nC}\left(1-|\mathcal{X}|^{-\alpha n}\right)\ln\left(1-|\mathcal{X}|^{-an}\right).$$ Now we have to treat the cases $2^C/\mathcal X^a\gt 1$; $2^C/\mathcal X^a= 1$; $2^C/\mathcal X^a\lt 1$.


1

One application of real analysis is showing certain differential or integral equations have unique solutions. For instance consider the equation $f(x)=e^x + \int _0 ^x 0.5* sin(t) *f(t) dt$ we will use real analysis to show that this equation has a unique solution. define $T(f)(x) = e^x + \int _0 ^x 0.5* sin(t) f(t) dt$ Under the metric induced by the ...


0

The condition given implies that $-K \leq {d \over dx} \log |f(x)| \leq K$ wherever $f(x)$ is nonzero. But the set of $x$ for which $f(x)$ is nonzero is an open set, and therefore is a countable union of intervals if nonempty. If $I$ is one such interval and $x_0$ is an endpoint of $I$, then $\lim_{x \rightarrow x_0} \log |f(x)| = -\infty$ since $f(x_0) = ...


0

Let it be that order $(X,<)$ has the greatest upper bound property. Then order $(X,>)$ has the least upper bound property and (as you said) consequently $(X,>)$ has the greatest upper bound property. The last statement means exactly that order $(X,<)$ has the least upper bound property.


3

The statement you are trying to prove is false. Consider, for example, $X = [0,1] \cup [2,3] \subset \Bbb R$, with the usual metric. In this space, the set $[0,1]$ is both open and closed. In fact, because $X$ has such a subset, we say that it is disconnected.


2

For example, consider the metric space $X = \{0,1\}$ with $d(0,1) = 1$. The set $E = \{0\}$ is a counterexample.


0

If $\lim\limits_{x\to\infty}f'(x)$ exists, then $\lim\limits_{x\to\infty}f(x)=L-\lim\limits_{x\to\infty}f'(x)$ also exists. Then $$ \begin{align} \lim\limits_{x\to\infty}f'(x) &=\lim\limits_{x\to\infty}(f(x+1)-f(x))\\ &=\lim\limits_{x\to\infty}f(x+1)-\lim\limits_{x\to\infty}f(x)\\ &=0\tag{1} \end{align} $$ If $\lim\limits_{x\to\infty}f(x)$ ...


1

You are correct about 1 and 2. Note also that 2 implies 4, since $$ \lim_{x \to \infty} f'(x) = L - \lim_{x \to \infty} f(x) $$ (assuming the latter limit exists). Note that for arbitrary functions $g,h$: if $\lim_{x \to \infty} g(x)$ and $\lim_{x \to \infty} h(x)$ both exist, then $$ \lim_{x \to \infty} [g(x) \pm h(x)] = \lim_{x \to \infty} g(x) \pm ...


0

It seems the following. Proposition. Each sequence of real numbers contains a monotonic subsequence. Proof. Let $\{a_n:n\in\Bbb N\}$ be a sequence of real numbers. Consider the following 2-coloring of the edges of the graph $K_{\Bbb N}$ (that is the graph whose vertices are natural numbers and each two different vertices are joined by an edge). Let ...


3

1. The intuition behind the formulation of induction principle is that you can "prove" that a certain property $Q$ is not "a valid" property for the natural numbers. For instance, suppose we have the property $Q(n):=n\text{ have decimal part}$. Now, if we define $P(n):=\text{no } Q(n)=n\text{ have no decimal part}$. Thus, using induction, cleary $P(0)$ is ...


4

Hint: If $\lim f'(x) = M$, then $\lim f(x) = L-M$ Use the MVT: $f(x+1) - f(x) = f'(\xi)$ with $x < \xi < x+1$.


1

As Daniel Fischer pointed out in the comments, the claim is false and a counter example is $f(x) = x^3\sin(\frac{1}{x})$.


9

Part 1 You can't necessarily assume that these rogue numbers exist, however, under the first four axioms, you can't say definitively that they don't exist, which is really the point. Suppose that in addition to the expected natural numbers $0,1,2,$ etc., there are numbers $a,b,$ and $c$ so that $a++=b,\ b++=c,$ and $c++=a$. This does not contradict any of ...


2

For your first question, it's not that we assume that rogue elements exist, it's that we cannot assume they do not exist. If we wish that they do not exist, we must choose a definition that allows us to prove it. Regarding your second question, in the language of Peano arithmetic, everything is a natural number. Thus, the only questions we can ask in the ...


0

I would just look at $(N+1)^2-1$. The hint probably wants you to say that for $a\in\mathbb N$, $\{ab\} = \{a\{b\}\}$, and find a value of $k$ that will do the trick.


3

The support is contained in a interval of the form $[-m,m]$ for $m$ large enough. The function is uniformly continuous on $A=[-m-1,m+1]$ since $A$ is compact. Take $\epsilon >0$. From uniform continuity on $A$ for there exists $\delta>0$ such that for $|x-y|< \delta$ it implies that $ |f(x) - f(y)| < \epsilon$. We prove continuity on ...


0

Supposed the modified Poincaré inequality fails. Then there is a sequence of functions $u_n$ such that $$\int_\Omega |u_n|^2dx=1\tag{1}$$ and $$\int_\Omega|\nabla u_n|^2dx+\Big(\int_\Omega |u_n|^{\frac{2}{m}}dx\Big)^m\to 0\tag{2}$$ Since the mean of $u_n$ is controlled by $\int_\Omega |u_n|^2dx=1$, it is uniformly bounded; passing to a subsequence we ...


1

Let $\phi:B(0,1) \to X$ be given by $\phi(x) = \begin{cases} \tan (\|x\| { \pi \over 2 }){x \over \|x\|}, & x \neq 0 \\ 0, & x=0 \end{cases}$. We see that $\phi^{-1} (x) = \begin{cases} {2 \over \pi}\arctan (\|x\| ){x \over \|x\|}, & x \neq 0 \\ 0, & x=0 \end{cases}$, hence $\phi$ is a bijection. Hence $\operatorname{card} B(0,1) = ...


0

Use the $\varepsilon$/$\delta$-definitions of continuity and uniform continuity. (The $\delta$ that works in the support works outside as well.)


3

how can this be done WITHOUT complex analysis? $\quad$ All integrals of the form $~\displaystyle\int_0^\infty\frac{x^{k-1}}{(x^n+a^n)^m}dx~$ can be evaluated by substituting $x=at$ and $u=\dfrac1{t^n+1}$ , then recognizing the expression of the beta function in the new integral, and lastly employing Euler's reflection formula for the $\Gamma$ function ...


6

The point of "excluding" rogue numbers is that there are larger number systems beyond the natural numbers, in which facts that we know about natural numbers no longer hold. For example, consider the non-negative real numbers with the usual operations of addition and multiplication, and with "successor" interpreted as $x \scriptsize{+\!+}\normalsize = x+1$, ...


1

If $K=0$ then problem is solved so assume $K > 0 $ , there exists a partion $P=\{x_0,x_1,...,x_n\}$ for $[0,b]$ such that $$K(x_i-x_{i-1})<1$$Assume $||f||_\infty$ and $||f'||_\infty$ be supperimum of $|f|$ and $|f'|$ on $[x_0,x_1]$ it is obvious that $$||f'||_\infty \le K||f||_\infty$$ there exists $u$ such that $x_0<u \le x_1 $ and ...


1

If $f(x) > 0$ $$\left (\vert f(x) \vert e^{-Kx} \right)'= e^{-Kx} \left (f'(x)-Kf(x) \right ) \leq 0$$ similarly if $f(x) \leq 0$ $$\left (\vert f(x) \vert e^{-Kx} \right)'= e^{-Kx} \left (-f'(x)+Kf(x) \right ) \leq 0$$ which implies $\left (\vert f(x) \vert e^{-Kx} \right) \leq \left (\vert f(0) \vert e^{-K(0)} \right) = 0$ and hence $f(x) = 0$ (The ...


1

Just consider $a_{M^2+M+1}$. Since $M^2+M+1<(M+1)^2$, we just have to prove: $$\sqrt{M^2+M+1}-M\geq\frac{1}{2},$$ that is equivalent to: $$M^2+M+1\geq \left(M+\frac{1}{2}\right)^2,$$ that is equivalent to: $$ 1\geq \frac{1}{4}.$$


1

$\cos z = \Re(e^{iz})$ and $\Re$ is a linear operator, so $\Re\int = \int\Re$.


2

A) A hint for the hint: Choose $\displaystyle \delta=\frac{1}{2K}$. Suppose that $f$ is not zero on $[0,\delta]$. Let $M={\rm Sup}\{|f(x)|, x\in [0,\delta]\}$. There exists $d\in [0,\delta]$, with $d>0$ such that $|f(d)|=M>0$. There exists $c\in ]0,\delta[$ such that $f(d)=df^{\prime}(c)$. Now use your hypothesis. B) A hint to continue: Put ...


1

The simplest way of solving this equation is the method based on DeMoivre's Formula that Lab Bhattacharjee outlined. That said, you can make your method work. You found the roots $z \pm i$ by setting the factor $z^2 + 1$ equal to zero. As Rasolnikov and 5xum noted, you should have obtained $$z^2 = \frac{1 \pm \sqrt{-3}}{2}$$ when you set the factor ...


7

Let $g(x)=\int_0^x\lvert \,f'(t)\rvert\,dt$. Then $g(x)\ge \lvert \,f(x)\rvert$, and therefore $$ 0\le g'(x)\le Kg(x), $$ for all $x\in [0,b]$. Thus $$ \mathrm{e}^{-Kx}\big(g'(x)-Kg(x)\big)\le 0,\quad\text{for all $x\in [0,b]$,} $$ and hence $$ \big(\mathrm{e}^{-Kx}g(x)\big)'\le 0,\quad\text{for all $x\in [0,b]$,} $$ which implies that $\mathrm{e}^{-Kx}g(x)$ ...


0

If you also include "uniformly bounded means," so that $E[X_n] \leq C$ for all $n$ (for some constant $C$) then you are okay. Just re-do an argument similar to your first equation (4) for the means (with some things flipped), then add it to the current (4) and take limits using (3) together with uniformly bounded means. If your last equation (4) is ...


1

$C$ contour:the upper half of the circle \begin{array}{l} i \in C; - i \notin C \\ {\mathop{\rm Re}\nolimits} s\left( {f;i} \right) = \mathop {\lim }\limits_{z \to i} \left\{ {\left( {z - i} \right)\frac{{e^{iz} }}{{1 + z^2 }}} \right\} = \frac{{e^{ - 1} }}{{2i}} \\ \end{array}


4

$C$ contour:the upper half of the circle $$f\left( z \right) = \frac{{e^{iz} }}{{1 + z^2 }}$$ \begin{array}{l} \oint\limits_C {\frac{{e^{iz} }}{{1 + z^2 }}dz} = \int\limits_{ - R}^{ + R} {\frac{{e^{ix} dx}}{{1 + x^2 }}} + \int\limits_\Gamma {\frac{{e^{iz} dz}}{{1 + z^2 }}} = \pi e^{ - 1} \\ \Rightarrow \int\limits_{ - R}^{ + R} {\frac{{\cos \left( ...


0

Note that $\lim_{n\to\infty}f_n(x)=0$ for all $x\in\mathbb{R}$ and that one could argue that $$1=\lim_{n\to\infty}\int f_n(x)~dx=\int\lim_{n\to\infty} f_n(x)~dx=0$$ if $\int F(x)~dx<\infty$, using an appropriate convergence result.


3

If we don't require continuity at $a$, the answer is an easy "no". The answer is still "no" if we require continuity, but not quite as easily. If we look at the function $g\colon (0,1) \to \mathbb{R}$, $$g(x) = (-1)^n\quad\text{ if } \frac{1}{(n+1)!} \leqslant x < \frac{1}{n!},$$ and consider $$f(x) = \int_0^x g(t)\,dt,$$ we have an almost-example: ...


5

Step One. Define the recursive sequence $$ a_0=\sqrt{2}, \quad a_{n+1}=\sqrt{2}^{a_n},\,\,n\in\mathbb N. $$ Step Two. Show that $\{a_n\}$ is increasing (inductively), and upper bounded by $2$ (also inductively). Step Three. Due to Step Two the sequence $\{a_n\}$ is convergent. Let $a_n\to x$. Clearly, $\sqrt{2}<x\le 2$. But $a_{n+1}=\sqrt{2}^{a_n}\to ...


5

The cosine function does not vanish on the semicircle as $R \to \infty$; in fact, it does the opposite. You need to either 1) take the real part of $e^{i x}$ in the upper half plane, or 2) use $\cos{x} = (e^{i x}+e^{-i x})/2$ and use both the upper and lower half planes, respectively.


2

Define $x_1=\sqrt{2}$ and $x_{n+1}=\sqrt{2}^{x_n}$. Prove by induction that $x_n \leq x_{n+1} \leq 2$. As the sequence is bounded and increasing, it is convergent, and the limit is between $x_1=\sqrt{2}$ and $2$. Finish the proof by observing that $$\sqrt{2}^x=x$$ has an unique solution on the interval $[\sqrt{2}, 2]$. For the last part, as well as for ...


18

We can define $x=\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}...}}}}$ as follows: Let $x_1 = \sqrt 2$ and $x_{n+1} = (\sqrt 2)^{x_{n}}$ We can show $x_n \lt 2\ \forall n$ by induction, since if $y \lt 2$, then $(\sqrt 2)^y \lt 2$. And $x_n$ is clearly monotonically increasing, so $x_n \to x$. But $$x_{n+1} = (\sqrt 2)^{x_{n}}$$ so taking limits, we ...


0

Well I don't think the claim is right. This is my counter-example. Would appreciate it if someone could verify it. Define $g : [0, \infty) \to \Bbb R$ by $g(t)= \sin 3 \pi t$. Now let $t = \dfrac{1 }{2}$. Now, $$ f(\frac 1 2) = \inf \{s \ge 0 \ | \ \sin 3 \pi s \gt \frac 1 2 \} = \frac 1 {18}$$ Now I believe $f$ is continuous at $ \dfrac 1 2 $ (which ...


2

Since the function $$ x \longmapsto\frac{1}{e^x-1} $$ is continuous on $(0,1]$, a potential convergence problem, concerning the integral, is for $x$ near $0$. We have $$ \frac{1}{e^x-1} \sim_{0^+} \frac1x $$ thus your integral $\displaystyle \int_0^1 \frac{1}{e^x-1} dx$ is divergent as $\displaystyle \int_0^1 \frac{1}{x} dx$ is divergent. In the same ...


0

To understand the difference between continuity and uniform continuity, it is useful to think of a particular example of a function that's continuous on $\mathbb R$ but not uniformly continuous on $\mathbb R$. An example of such a function is $f(x)=x^2$. Here to understand the failure of uniform continuity of the function, it is particularly useful to ...


0

In order to solve $\sqrt[2n]{-1}$: Draw the unit circle Draw the first solution, which is obviously $0+1i=\cos(\frac{\pi}{2})+\sin(\frac{\pi}{2})i$ Repeat $2n-1$ times: find the next solution by rotating the previous solution $\frac{\pi}{n}$ radians For example, $\sqrt[6]{-1}$: $\cos(\frac{ 3\pi}{6})+\sin(\frac{ 3\pi}{6})i$ $\cos(\frac{ ...


0

Setting $x=\dfrac1y,$ $$I=\int_0^\infty\frac1{1+x^6}dx=\cdots=\int_0^\infty\frac{x^4}{1+x^6}dx$$ $$2I=\int_0^\infty\frac{1+x^4}{1+x^6}dx=\int_0^\infty\frac{(1+x^2)^2-2x^2}{1+x^6}dx$$ $$=\int_0^\infty\frac{1+x^2}{1-x^2+x^4}dx-\frac23\int_0^\infty\frac{3x^2}{1+x^6}dx$$ ...


0

I think set theory is best way to think about infinity , cantor try to explain different types of infinity. It is also talk more in bbc dangerous knowledge series. You can also search on youtube they have some documentaries about it. I believe that you can start any topic in mathematics , it always close at infinity. Many try to think beyond it but they ...


2

One such condition is that $f$ be continuous and differentiable everywhere and that $f'$ be locally integrable and have regular variation at infinity. Let $\alpha \in \mathbb{R}$ A function $g$ is said to have regular variation of index $\alpha$ at infinity if for all $y > 0$, $$\lim_{x \to + \infty} \frac{g(xy)}{g(x)} = y^\alpha.$$ It is said to have ...


1

[OP changed the question since this answer was posted] Have you tried $f(x)=e^x$? It's a smooth nonnegative nondecreasing function with $f'(x)=e^x$ which is monotonic and nonnegative, but $$e^x\ne O(e^x/x)$$


7

Let $\displaystyle \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathcal I=\int_0^\infty\frac{1}{1+x^6} \,\mathrm dx $ $$\begin{align} I&=\frac{1}{2}\left[ \int_0^\infty \frac{(1-x^2+x^4)+x^2+(1-x^4)}{(1+x^2)(1-x^2+x^4)} \,\mathrm dx \right]\tag{1}\\ &= \frac{1}{2}\left[\int_0^\infty \frac{1}{1+x^2} \,\mathrm dx + \int_0^\infty ...


5

$$z^6=-1=e^{(2n+1)\pi i}$$ where $n$ is any integer $$\implies z=e^{\dfrac{(2n+1)\pi i}6}=\cos\dfrac{(2n+1)\pi}6+i\sin\dfrac{(2n+1)\pi}6$$ where $0\le n\le 5$ Top region of the plane, $\implies$ the ordinate has to be $>0$ $\implies\sin\dfrac{(2n+1)\pi}6>0\implies0<\dfrac{(2n+1)\pi}6<\pi\iff0<2n+1<6\implies-.5< n<2.5$ $\implies ...


5

Only your last line is incorrect. What you should write is $$z^2 = \frac{1\pm\sqrt{-3}}{2}$$



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