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Question: <p>In a proof I found in a book regarding the ODE <span class="math-container">$\dot{x}(t) = A(t)x(t)$</span> where <span class="math-container">$A$</span> is piecewise continuous on a compact interval I, it is assumed that the maximum <span class="math-container">$K = \max_I ||A(t)||$</span> exists. In my opinion this doesn't have to be the case. Take, for example, <span class="math-container">$I = [-1,1]$</span> and <span class="math-container">$A(t) = \frac{1}{t}$</span> for <span class="math-container">$t > 0$</span> and <span class="math-container">$A(t) = 0$</span> for <span class="math-container">$t \leq 0$</span>. Is there a mistake in the book or am I just wrong somewhere?</p>
Answer: <p>The reason for the discrepancy is that the book is
probably using a different definition of a piecewise continuous function which does not allow your example to hold.</p>
<p>Consider following definition of a piecewise
continuous function. Let <span class="math-container">$D_1, \ldots, D_m$</span> be
a finite partition of <span class="math-container">$D \subseteq R^d$</span>.
Let <span class="math-container">$\phi_k : D_k \rightarrow \{0,1\}$</span> be defined
such that <span class="math-container">$\phi_k({\bf x}) = 1$</span> if <span class="math-container">${\bf x} \in D_k$</span> and
<span class="math-container">$\phi_k({\bf x}) = 0$</span> if <span class="math-container">${\bf x} \notin D_k$</span>.
Let <span class="math-container">$f_k : R^d \rightarrow R$</span> be a function
with the property that <span class="math-container">$f_k$</span> is continuous on the closure of <span class="math-container">$D_k$</span>.
The function
<span class="math-container">$f = \sum_{k=1}^m f_k \phi_k $</span> is called piecewise
continuous on D.</p>
<p>If D is bounded then
<span class="math-container">$D_1, \ldots, D_m$</span> are bounded. Therefore
With this definition it follows
f is bounded on D when D is compact.</p>
|
https://math.stackexchange.com/questions/2973787/why-can-we-assume-that-a-piecewise-continuous-function-has-a-maximum
|
Question: <p>I have been attacking this question but i got stuck. Please give me some hint.</p>
<p>Let $1\leq p<\infty$ and $\{f_n\}\subset L^p$. Suppose $f_n\to f$ in measure and $\|f_n\|_p\to\|f\|_p$, then $f_n\to f$ in $L^p$ norm.</p>
Answer: <p>It's not easy. The key idea is to mimic the proof of Dominated Convergence Theorem. In order to do that, first establish the following inequality:
$$\lvert x+y\rvert^p\le\gamma_p(\lvert x\rvert^p+\lvert y\rvert)^p,\qquad\forall x,y\in\mathbb C$$
for some constant $\gamma_p$. Then apply <a href="http://en.wikipedia.org/wiki/Fatou's_lemma#Convergence_in_measure" rel="nofollow">Fatou's lemma</a> to $g_n=\gamma_p(\lvert f_n\rvert^p+\lvert f\rvert^p)-\lvert f_n-f\rvert^p$, which shows that $\limsup_n\int\lvert f_n-f\rvert^p\le0$.</p>
<p>For a reference, see Walter Rudin's <em>Real and Complex Analysis</em>, Exercise 3.17.</p>
|
https://math.stackexchange.com/questions/1162919/convergence-in-measure-and-convergence-of-norm-implies-convergence-in-lp
|
Question: <blockquote>
<p>It's the follow metric: <span class="math-container">$d(x,y)= ||x|| +||y||$</span> if <span class="math-container">$x$</span> and <span class="math-container">$y$</span> don't lie on a line through the origin. And otherwise <span class="math-container">$d(x,y)= ||x-y||$</span>.</p>
<p>I think the answer is no, because I tried it with <span class="math-container">$\mathbb{Q}^{2}$</span> as countable and that didn't work. But I don't know how to prove it that it isn't true in general.</p>
</blockquote>
Answer: <p>HINT: You're right: it's not. You can prove it by finding an uncountable family of pairwise disjoint, non-empty open sets. I've added a further hint in the spoiler-protected block below.</p>
<blockquote class="spoiler">
<p> Consider open rays leaving the origin.</p>
</blockquote>
|
https://math.stackexchange.com/questions/1319739/is-mathbbr2-separable-for-the-railmetric-or-sncf-metric-or-post-office
|
Question: <p>Let <span class="math-container">$$f(x) =
\begin{cases}
\dfrac{1}{q^2}, & \text{if $x=\dfrac{p}{q} $ is rational and in lowest terms;} \\[2ex]
0, & \text{if $x$ is irrational}
\end{cases}$$</span>
Where is f continuous? Is f differentiable anywhere?</p>
<p>My attempt: I can prove that <span class="math-container">$\lim_{x \to c}f(x)=0$</span> for any point real value c.</p>
<p>for each positive number <span class="math-container">$\epsilon$</span>, the set <span class="math-container">$(c-1,c) \cup(c,c+1)$</span> contains only finitely many rational numbers p/q with <span class="math-container">$1/q^2 \geq \epsilon$</span> (namely , <span class="math-container">$q^2 \leq 1/\epsilon$</span>). Then I let <span class="math-container">$\delta$</span> be the distance between <span class="math-container">$c$</span> and the closest such rational number. Then for all x satisfying <span class="math-container">$0<|x-c|<\delta$</span>, we have <span class="math-container">$0\leq f(x)=f(p/q)=1/q^2 \leq \epsilon$</span>.</p>
<p>Thus, I prove that <span class="math-container">$\lim_{x \to c}f(x)=0$</span> for any point real value c. So this means f(x) is continuous at every irrational number, and discontinuous at every rational number.</p>
<p>For differentiability, I only know f(x) is not differentiable at every rational number, because its discontinuous there. But how to prove f(x) also not differentiable at every irrational number?</p>
<p>Hint: For any irrational number c, there are infinitely many fractions p/q in lowest terms such that <span class="math-container">$|c-p/q|<1/q^2$</span>. But I don't know this hint is is used in the continuous part ot differentiable part, since I already proved the continuous part.</p>
Answer: <p>Let's fix an irrational number <span class="math-container">$\gamma$</span> and a sequence of rationals <span class="math-container">$\frac{p_n}{q_n} \to \gamma$</span>. Due to <a href="https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem" rel="nofollow noreferrer">Dirichlet's theorem</a> we can also satisfy the inequality
<span class="math-container">$$
\left| \frac{p_n}{q_n} - \gamma \right| \le \frac{1}{q_n^2}.
$$</span>
Then we have
<span class="math-container">$$
\left| \frac{f\left( \frac{p_n}{q_n} \right) - f(\gamma)}{\frac{p_n}{q_n} - \gamma} \right| = \left| \frac{\frac{1}{q_n^2}}{\frac{p_n}{q_n} - \gamma} \right| \ge 1. \;\;\;\;\;\;\;\; (1)
$$</span></p>
<p>On the other side, if one chooses irrationals <span class="math-container">$\gamma_n \to \gamma $</span>
<span class="math-container">$$
\frac{f(\gamma_n) - f(\gamma)}{\gamma_n - \gamma} = 0.
$$</span>
The last equality shows that the only possible value for the derivative is <span class="math-container">$0$</span>, but this contradicts (1).</p>
|
https://math.stackexchange.com/questions/4178091/a-question-on-differentiability-of-a-variant-of-thomae-function
|
Question: <p>Suppose that <span class="math-container">$f$</span> and <span class="math-container">$g$</span> are differentiable functions on an open interval <span class="math-container">$I$</span> and that <span class="math-container">$p \in I$</span>. If <span class="math-container">$\lim_{x \to p} f(x) = \lim_{x \to p} g(x) = 0$</span> and if</p>
<p><span class="math-container">$$\lim_{x \to p} \frac{f'(x)}{g'(x)}$$</span></p>
<p>exists and equals a real number <span class="math-container">$l$</span> then</p>
<p><span class="math-container">$$\lim_{x \to p} \frac{f(x)}{g(x)}=l$$</span></p>
<p>Proof: Fix a real number <span class="math-container">$a > l$</span>. By the definition of limit there is a number <span class="math-container">$q > p$</span> such that, if <span class="math-container">$p < x< q$</span>, then <span class="math-container">$ \frac{f'(x)}{g'(x)}<a$</span>. I don't know how to derive this.</p>
Answer: <p>You have <span class="math-container">$\lim_{x\to p}\big( \text{something}\big) = \ell.$</span></p>
<p>And you have <span class="math-container">$a>\ell.$</span></p>
<p>That implies that there is some open interval about <span class="math-container">$p$</span> for which, if <span class="math-container">$x$</span> is in that open interval and <span class="math-container">$x\ne p,$</span> then <span class="math-container">$\big(\text{something}\big) <a. $</span></p>
<p>Similarly if <span class="math-container">$b<\ell$</span> then there is some open interval about <span class="math-container">$p$</span> for which, if <span class="math-container">$x$</span> is in that open interval and <span class="math-container">$x\ne p,$</span> then <span class="math-container">$\big(\text{something})>b.$</span></p>
<p>This is a consequence of the definition of <em>limit</em>: No matter how small <span class="math-container">$\varepsilon>0$</span> is, there is some open interval about <span class="math-container">$p$</span> such that if <span class="math-container">$x$</span> is in that open interval and <span class="math-container">$x\ne p,$</span> then <span class="math-container">$\big(\text{something}\big)$</span> differs from <span class="math-container">$\ell$</span> by less than <span class="math-container">$\varepsilon.$</span> Generally as <span class="math-container">$\varepsilon$</span> gets smaller, that open interval about <span class="math-container">$p$</span> must get smaller.</p>
|
https://math.stackexchange.com/questions/4178423/ask-a-step-in-proof-of-lhospitals-rule
|
Question: <p>Let <span class="math-container">$f \in C[0,1]$</span>. If <span class="math-container">$\{y_i\} \subset f([0,1])$</span> is increasing, there exists an monotonous <span class="math-container">$\{x_i\} \subset [0,1]$</span> such that <span class="math-container">$f(x_i)=y_i$</span>?</p>
<p>Intuitively, from the function image, it is true. and how to prove that?</p>
Answer:
|
https://math.stackexchange.com/questions/4181505/if-y-i-subset-f0-1-is-increasing-there-exists-an-monotonous-x-i
|
Question: <p>***How to find the this limit without using L'Hopital's Rule.</p>
<p><span class="math-container">$\lim _{x\to 0^+}\frac{\left(e^x-e^{-x}\right)}{\sin x}$</span></p>
Answer: <p>Hint:</p>
<ul>
<li><p>Use maclaurin series for <span class="math-container">$e^x$</span> and <span class="math-container">$e^{-x}$</span></p>
</li>
<li><p>then apply <span class="math-container">$$\lim_{x\to 0} \dfrac{\sin x}{x}=1$$</span></p>
</li>
</ul>
|
https://math.stackexchange.com/questions/3982661/calculate-lim-x-to-0-frac-leftex-e-x-right-sin-x
|
Question: <p>Hi I was wondering if it were possible to skip undergraduate and apply to a graduate school. I have been taking some math courses through CTY a somewhat accredited program offered by Johns Hopkins university. They offer up to real analysis 1 and complex analysis 1. If I take up to there, would it be possible to skip undergraduate having self studied the higher level courses i.e topology/further analysis courses.</p>
Answer: <p>If you plan to study further for Msc or Phd, you could and I think should just enroll in an integrated Bsc+Msc program. You can either</p>
<ul>
<li>request that you take an exam of the courses you feel like you have learned enough about without taking the course itself,</li>
<li>or take an otherwise extraordinary amount of courses parallel in your first few semesters without attending the lectures.</li>
</ul>
<p>Schools tend to accept both kind of requests with relative ease (compared to applying to gradschool without a degree), and you can negotiate with the related departments whether they support your plan or not in advance even before applying.
This way you can trim down a year or two from an otherwise 4-5 years program essentially completing your Master's in only 2-4 years. This is a fairly common occurence, even with students that have no former college education but are just hardworking and talented.</p>
<p>You would get admitted as easily as any Bsc student, but would finish with a masters degree in essentially what is the timeframe for a Bachelors program. And you would receive both degrees without any hussle or worrying about getting admitted to the gradschool.</p>
<p>TL;DR: you should ask your targeted schools whether they support trimming down Bsc+Msc years instead of trying to get them to apply someone to a higher degree program without a bachelors.</p>
<p>Edit: format.</p>
|
https://math.stackexchange.com/questions/4177654/would-it-be-possible-to-skip-undergraduate-college-with-self-study-online-course
|
Question: <p><span class="math-container">$$\lim_{n\to\infty}\left(1+\dfrac{1}{n+1}\right) ^{n+1} =e$$</span></p>
<p>Could someone explain me why this is trivial ? Maybe subsitution such that <span class="math-container">$n+1=m$</span> which would give us the definition of <span class="math-container">$e$</span> ? </p>
Answer: <p>Let
<span class="math-container">$$a_n = \left( 1 + \frac{1}{n}\right)^n$$</span></p>
<p>Since this sequence converges to <span class="math-container">$e$</span>, any subsequence of it also converges to <span class="math-container">$e$</span>.</p>
<p>Set <span class="math-container">$k_n=n+1$</span>. Then, the corresponding subsequence is
<span class="math-container">$$a_{k_n}=\left(1+\dfrac{1}{n+1}\right) ^{n+1}$$</span>
and hence it converges to <span class="math-container">$e$</span>.</p>
|
https://math.stackexchange.com/questions/3115588/why-is-lim-limits-n-to-infty-left1-dfrac1n1-rightn1-e-trivial
|
Question: <p>I want to formally prove that this limit is equal to <span class="math-container">$0$</span></p>
<p><span class="math-container">$\lim_{x \to 0} \frac{2}{x^{3}} e^{-\frac{1}{x^{2}}}$</span></p>
<p>can anybody help me?</p>
Answer: <p>This is equivalent to</p>
<p><span class="math-container">$$\lim_{t\to\infty}t^{3/2}e^{-t}=0$$</span> as it is well-known that the exponential increases faster than any polynomial.</p>
|
https://math.stackexchange.com/questions/3705034/resolution-of-a-limit
|
Question: <p>The definition of Thomae function is
<span class="math-container">$$g(x) =
\begin{cases}
\dfrac{1}{q}, & \text{if $x=\dfrac{p}{q}$} \\[2ex]
0, & \text{if $x$ is irrational}
\end{cases}$$</span>
where <span class="math-container">$p\in \mathbb{Z}, q\in \mathbb{Z}^+$</span> and <span class="math-container">$\gcd(p,q)=1.$</span></p>
<p>I know how to prove <span class="math-container">$g$</span> is Riemann integrable on [−1, 1], but how to prove this integral <span class="math-container">$\int_{-1}^1 g(x)dx$</span> is 0?</p>
<p>My attempt: for any partition, the smallest Riemann sum is 0, simply because we can always choose representative point to be irrational. But how to complete the proof <span class="math-container">$\int_{-1}^1 g(x)dx=0$</span>?</p>
Answer:
|
https://math.stackexchange.com/questions/4189450/a-question-on-the-riemann-integrable-of-thomae-function
|
Question: <p>My book says that if functions $f$ and $g$ are both onto then $f\circ g$ and $g\circ f$ may or may not be onto. </p>
<p>Why is this so? Would someone please help me understand this, maybe with an example or diagrammatically? My book states that$ f\circ g$ and $g\circ f$ may or may not be onto. I think this might be related to the domains and codomains of g and f which may not be equal. I am aware that there is a similar looking question and therefore I'm clarifying mine. With respect to the domains and codomains would someone please explain why the composite functions may or may not be onto?</p>
Answer: <p>The composition of two surjective functions is always surjective: Let $f: X \to Y$ and $g: Y \to Z$ be functions and $z \in Z$. Then since g is surjective, there exists $y$ such that $g(y)=z$ and similarly there exists $x$ such that $f(x)=y$. Then $g(f(x))=z$ and your composition is surjective.</p>
<p>You can do the same argument for injective functions.</p>
|
https://math.stackexchange.com/questions/2913101/explanation-of-composition-of-two-onto-functions
|
Question: <p>I'm working on the following question:</p>
<blockquote>
<p>Which of the following three properties imply <span class="math-container">$f$</span> is monotone (it maybe none or more than 1):</p>
<ol>
<li><span class="math-container">$f(x + y) = f(x)f(y)$</span></li>
<li><span class="math-container">$f(x - y) = f(x)f(y)$</span></li>
<li><span class="math-container">$f$</span> is periodic and non-decreasing.</li>
</ol>
</blockquote>
<p>I think 3 works, I'm not sure how to evaluate 1 and 2.</p>
Answer: <p>(1) does not imply monotonicity. Take <span class="math-container">$g$</span> to be a discontinuous (also non-integrable, non-monotone) solution to <a href="https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation" rel="nofollow noreferrer">Cauchy's Functional Equation</a>:
<span class="math-container">$$g(x + y) = g(x) + g(y).$$</span>
Then <span class="math-container">$f = \exp \circ g$</span> is a non-monotone function satisfying
<span class="math-container">$$f(x + y) = f(x)f(y).$$</span></p>
<p>(2) For such a function <span class="math-container">$f$</span>, first note that <span class="math-container">$f(0 - 0) = f(0)f(0)$</span>, which implies <span class="math-container">$f(0) = 0$</span> or <span class="math-container">$f(0) = 1$</span>. If <span class="math-container">$f(0) = 0$</span>, then
<span class="math-container">$$0 = f(x - x) = f(x)f(x) \implies f(x) = 0$$</span>
for all <span class="math-container">$x$</span>, which tells us that <span class="math-container">$f$</span> is constant (and hence monotone). Otherwise, <span class="math-container">$f(0) = 1$</span>, and similarly we see that <span class="math-container">$f(x) = \pm 1$</span> for all <span class="math-container">$x$</span>. The only monotone functions possible are constant. So, the question is, are there any non-constant solutions?</p>
<p>Suppose <span class="math-container">$f$</span> is a non-zero solution. I claim that <span class="math-container">$f$</span> is a homomorphism from the group <span class="math-container">$(\mathbb{R}, +)$</span> into the group <span class="math-container">$(\{-1, 1\}, \cdot)$</span>. If <span class="math-container">$x, y \in \mathbb{R}$</span>, then
<span class="math-container">$$f(y) = f(x + y - x) = f(x + y)f(x) \implies f(x + y) = f(y) f(x)^{-1} = f(x) f(y).$$</span>
If <span class="math-container">$f$</span> is not constantly <span class="math-container">$1$</span>, then <span class="math-container">$f$</span> maps onto <span class="math-container">$\lbrace -1, 1\rbrace$</span>, and <span class="math-container">$f^{-1}\lbrace 1 \rbrace$</span> is a group of index <span class="math-container">$2$</span> in <span class="math-container">$(\mathbb{R}, +)$</span>, which <a href="https://math.stackexchange.com/questions/103262/is-there-a-proper-subgroup-h-of-mathbbr-such-that-mathbbrh-is">doesn't exist</a>. Thus, the only solutions are constant solutions <span class="math-container">$0$</span> and <span class="math-container">$1$</span>, both of which are monotone. So <strong>yes</strong>, (2) implies monotonicity.</p>
<p>(3) This also implies monotonicity, as it specifies non-decreasing. Such functions are also constant, making me wonder if the question was about the functions being constant, not monotone?</p>
|
https://math.stackexchange.com/questions/2935065/possible-properties-of-a-monotone-function
|
Question: <p>I'm working on the following problem:</p>
<blockquote>
<p>A real valued function <span class="math-container">$f$</span> defined on <span class="math-container">$\mathbb{R}$</span> has the property that <span class="math-container">$(\forall \epsilon>0)(\exists\delta>0)$</span> s.t. <span class="math-container">$|x-1| \geq \delta \implies |f(x)-f(1)| \geq \epsilon$</span></p>
</blockquote>
<p>The choices are either: <span class="math-container">$f$</span> is unbounded, <span class="math-container">$\lim_{|x|\rightarrow \infty}|f(x)|=\infty$</span>, or <span class="math-container">$\int^{\infty}_0|f(x)|dx=\infty$</span>.</p>
<p>The key says the answer is <span class="math-container">$\lim_{|x|\rightarrow \infty}|f(x)|=\infty$</span>. I'm having trouble seeing this immediately though. In fact I'm also unable to distinguish it from <span class="math-container">$f$</span> is unbounded - doesn't the second choice imply the first?</p>
Answer: <p>Given <span class="math-container">$\epsilon >0$</span> there exists <span class="math-container">$\delta >0$</span> such that</p>
<p><span class="math-container">$$|x-1| \geq \delta \implies |f(x)-f(1)| \geq \epsilon.$$</span></p>
<p>In other words</p>
<p><span class="math-container">$$x\in (-\infty,1-\delta)\cup (1+\delta,\infty)\implies f(x)\in (-\infty,f(1)-\epsilon)\cup (f(1)+\epsilon,\infty).$$</span></p>
<p>As a consequence we get</p>
<p><span class="math-container">$$|x|\ge 1+\delta \implies |f(x)|\ge \min\{|f(1)-\epsilon|,|f(1)+\epsilon|\}.$$</span></p>
<p>This shows that <span class="math-container">$$\lim_{|x|\to \infty} |f(x)|=\infty.$$</span></p>
<blockquote>
<p>I'm having trouble seeing this immediately though. In fact I'm also unable to distinguish it from <span class="math-container">$f$</span> is unbounded - doesn't the second choice imply the first?</p>
</blockquote>
<p>Note that <span class="math-container">$f(x)=e^x$</span> is unbounded but <span class="math-container">$\lim_{|x|\to \infty} |f(x)|$</span> doesn't exist. So both statements are not the same. Of course <span class="math-container">$\lim_{|x|\to \infty} |f(x)|=\infty$</span> implies that <span class="math-container">$f$</span> is unbounded. </p>
|
https://math.stackexchange.com/questions/2937282/epsilon-delta-exercise
|
Question: <p>The radius of convergence <span class="math-container">$r$</span> can be calculated for every power series <span class="math-container">$\sum_{k=0}^\infty a_k z^k$</span> with <span class="math-container">$a_k\in \mathbb C$</span> and <span class="math-container">$z\in \mathbb C$</span> by using the Cauchy Hadamard formula:</p>
<p><span class="math-container">$r = \limsup_{k\to\infty} |\frac{1}{a_k}|^{1/k}$</span> .</p>
<p>In every textbook I find the power series starting from <span class="math-container">$k=0$</span>. </p>
<p>Can the Cauchy Hadamard formula also be used for a power series <span class="math-container">$\sum_{k=1}^\infty a_k z^k$</span> directly (ie. without changing the index of the power series before calculating <span class="math-container">$r$</span>)?</p>
Answer: <p>Certainly. The two series converge for the same points <span class="math-container">$z$</span> and have the same radii of convergence. You can thing of the second as the first with <span class="math-container">$a_0=0$</span>. </p>
|
https://math.stackexchange.com/questions/2941938/cauchy-hadamard-formula-and-starting-index-of-power-series
|
Question: <p>How can i prove uniform convergence on <span class="math-container">$E=[0, \frac{\pi}{2}]$</span> ?
<span class="math-container">$$\sum^\infty _{n=1} {x e^{-nx}\cos(nx)}$$</span></p>
Answer: <p>Hint: For <span class="math-container">$N$</span> even, let</p>
<p><span class="math-container">$$T_N(x)=\sum_{n=N/2}^{N}xe^{-nx}\cos (nx).$$</span></p>
<p>If the convergence is uniform, then <span class="math-container">$T_N\to 0$</span> uniformly. Consider <span class="math-container">$T_N(1/N).$</span></p>
|
https://math.stackexchange.com/questions/2962462/uniform-convergence-of-sum-infty-n-1-x-e-nx-cosnx
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Question: <p>Consider a sequence <span class="math-container">$x_n$</span> with positive values such that
<span class="math-container">$\sum _{n=1}^{\infty} x_n $</span> converges .
Is the set of the subsequences <span class="math-container">$ x_{k_n}$</span> of <span class="math-container">$x_n$</span> such that
<span class="math-container">$\sum _{n=1}^{\infty} x_{k_n} =c \in R$</span> countable ?
<span class="math-container">$c$</span> is previously fixed so the corresponding series of every subsequence converges to the same value.</p>
Answer:
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https://math.stackexchange.com/questions/3035548/countability-of-a-set-of-subsequences
|
Question: <p>I realize there are several questions dealing with Theorem 1.1 from Apostol's "Mathematical Analysis", but those questions are mostly about how to prove the result or the geometrical intuition around it. My question is about the application of the theorem. The theorem states:</p>
<p>If <span class="math-container">$a\leq b+\varepsilon$</span></p>
<p>for all <span class="math-container">$\varepsilon>0$</span>, then </p>
<p><span class="math-container">$a\leq b$</span>.</p>
<p>Take an example.</p>
<p><span class="math-container">$a=2$</span>, <span class="math-container">$b=1$</span> and <span class="math-container">$e = 3$</span>; <span class="math-container">$e := \varepsilon$</span> Then, <span class="math-container">$2 \leq 1 + 3$</span>;</p>
<p><span class="math-container">$2 \leq 4$</span></p>
<p>These meet the conditions required in the theorem.</p>
<p>But, now I get the absurd result: <span class="math-container">$2 \leq 1$</span></p>
<p>What is wrong in my usage of the theorem?</p>
Answer: <p>The premise says <strong>for all</strong> <span class="math-container">$\epsilon>0.$</span> You have only shown the inequality holds for <strong>one</strong> <span class="math-container">$\epsilon$</span> (namely <span class="math-container">$\epsilon=3$</span>). But it fails for <span class="math-container">$\epsilon = 1/2,$</span> for example, so the premise does not hold and you cannot draw the conclusion.</p>
|
https://math.stackexchange.com/questions/3039044/apostol-mathematical-analysis-theorem-1-1
|
Question: <p>I am working on exercice 9.5.2 of Analysis by Zorich and I am stuck at the question b.</p>
<blockquote>
<p>a) A set <span class="math-container">$E\subset X$</span> of a metric space <span class="math-container">$(X,d)$</span> is nowhere dense in X if it is not dense in any ball, that is, if for every ball <span class="math-container">$B(x,r)$</span> there is a second ball <span class="math-container">$B(x_1,r_1)\subset B(x,r)$</span> containing no points of the set <span class="math-container">$E$</span>.
A set E is of first category in X if it can be represented as a countable union of nowhere dense sets. A set that is not of first category is of second category in <span class="math-container">$X$</span>. Show that a complete metric space is set of second category (in itself).</p>
<p>b) Show that if a function <span class="math-container">$f\in C^{(\infty)}[a,b]$</span> is such that <span class="math-container">$\forall x\in [a,b] \;\exists n\in \mathbb{N} \;\forall m>n \;(f^{(m)}(x)=0)$</span>, then <span class="math-container">$f$</span> is a polynomial.</p>
</blockquote>
<p>Here's my try: Define the sets <span class="math-container">$S_n:=\{x\in[a,b]\mid f^{(m)}(x)=0\; \forall m>n\}$</span>. Then <span class="math-container">$\cup_{n=1}^{\infty} S_n =[a,b]$</span>. As <span class="math-container">$[a,b]$</span> is a complete metric space, <span class="math-container">$S_n$</span> cannot be all nowhere dense. Define <span class="math-container">$Y:=\{x\in [a,b] \mid \text{ there exists a neighborhood of } x \text{ and } n \text{ such that } S_n \text{ is dense in that neighborhood}$</span>. I want to say that <span class="math-container">$Y=[a,b]$</span> so that I could conclude with the compactness of <span class="math-container">$[a,b]$</span>. But all I can say is that the complement of <span class="math-container">$Y$</span> in <span class="math-container">$[a,b]$</span> contains no interval.</p>
<p>Can you help me? Thanks!</p>
Answer: <p>Here is a sketch of a possible approach, which may have some holes, but it is too long for a comment. Maybe someone can patch it up.Picking up on your idea, set </p>
<p><span class="math-container">$T = \{t\in [a,b]: \forall (c,d)\ni t: f\restriction_{(c,d)}$</span> is not a polynomial<span class="math-container">$\}$</span></p>
<p>Then <span class="math-container">$T$</span> is non-empty and closed. By construction, it has no isolated points. Now we may apply the Baire theorem on <span class="math-container">$\{T\cap S_n\}$</span> to find an interval <span class="math-container">$(c,d)$</span> such that for some <span class="math-container">$n\in \mathbb N,\ (c,d)\cap T\subset S_n$</span>.</p>
<p>Now, <span class="math-container">$f$</span> is a polynomial on <span class="math-container">$(c,d)\setminus T,$</span> which is open and so contains an interval <span class="math-container">$(\alpha,\beta)$</span>, which we may take to be maximal: indeed, <span class="math-container">$(c,d)\setminus T$</span> is a countable disjoint union of intervals <span class="math-container">$\bigcup_n(a_n,b_n)$</span> and so <span class="math-container">$\{x:a',b'\in (a,b)\setminus T\ \text{and} \ b'-a'\ge b_n-a_n\}$</span> is a maximal interval in <span class="math-container">$(a,b)\setminus T$</span>.</p>
<p>Now, either <span class="math-container">$\alpha$</span> or <span class="math-container">$\beta\in T$</span>. Suppose <span class="math-container">$\alpha\in T$</span>. Then, on every interval <span class="math-container">$(c',c'')$</span> containing <span class="math-container">$\alpha,\ f$</span> is <span class="math-container">$not$</span> a polynomial. Choose <span class="math-container">$c'''$</span> such that <span class="math-container">$c'<\alpha<c'''<c''<d$</span>. Then, <span class="math-container">$f$</span> is not a polynomial on <span class="math-container">$(c''',c'').$</span> But since <span class="math-container">$(c''',c'')\subseteq (\alpha,\beta)$</span>, so we have a contradiction. </p>
<p>So, either there is no such interval <span class="math-container">$(c,d)$</span> or <span class="math-container">$T$</span> is empty. In either case, the result follows.</p>
|
https://math.stackexchange.com/questions/3042725/if-at-each-point-of-a-closed-interval-the-m-th-derivative-of-f-is-0-for-m
|
Question: <p>Show that <span class="math-container">$\left( x_{n}\right) $</span> is a Cauchy sequence, where
<span class="math-container">$$
x_{n}=\frac{\sin1}{2}+\frac{\sin2}{2^{2}}+\ldots+\frac{\sin n}{2^{n}}.
$$</span></p>
<p>We try to evaluate <span class="math-container">$\left\vert x_{n+p}-x_{n}\right\vert $</span> and to show that
this one is arbitrary small. So
<span class="math-container">\begin{align*}
\left\vert a_{n+p}-a_{n}\right\vert & =\left\vert \frac{\sin\left(
n+1\right) }{2^{n+1}}+\frac{\sin\left( n+2\right) }{2^{n+2}}+\ldots
+\frac{\sin\left( n+p\right) }{2^{n+p}}\right\vert \\
& \leq\frac{1}{2^{n+1}}+\frac{1}{2^{n+2}}+\ldots+\frac{1}{2^{n+p}}.
\end{align*}</span>
Now, <span class="math-container">$\left( \frac{1}{2^{n+j}}\right) $</span> converge all to zero. Hence
<span class="math-container">$\left\vert a_{n+p}-a_{n}\right\vert \rightarrow0$</span>.
Does it work this argument according to the definition of Cauchy sequence
<span class="math-container">$$
\forall\varepsilon>0,
\quad
\exists n_{\varepsilon}\in
\mathbb{N}
,
\quad
\forall n\in
\mathbb{N}
,
\quad
\forall p\in
\mathbb{N}
:n,p\geq n_{\varepsilon}\Rightarrow\left\vert a_{n+p}-a_{n}\right\vert
<\varepsilon?
$$</span>
My question was born from the fact that for the sequence
<span class="math-container">$$
a_{n}=1+\frac{1}{2}+\ldots+\frac{1}{n},
$$</span>
we have that
<span class="math-container">\begin{align*}
\left\vert a_{n+p}-a_{n}\right\vert & =\left\vert \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+p}\right\vert \\
& \leq\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+p}
\end{align*}</span>
and also all sequences <span class="math-container">$\left( \frac{1}{n+j}\right) _{n}$</span> converge to zero,
but this time <span class="math-container">$\left( a_{n}\right) $</span> is not a Cauchy sequence.</p>
Answer: <p>It is not sufficient that all <span class="math-container">$\left( \frac{1}{2^{n+j}}\right)$</span> converge to zero. A correct argument would be that
<span class="math-container">$$
\left\vert a_{n+p}-a_{n}\right\vert \le
\frac{1}{2^{n+1}}+\frac{1}{2^{n+2}}+\ldots+\frac{1}{2^{n+p}}
= \frac{1}{2^{n}} \left( \frac 12 + \frac 14 + \ldots + \frac{1}{2^{p}}\right) < \frac{1}{2^{n}}
$$</span>
becomes arbitrary small for large <span class="math-container">$n$</span> and <em>arbitrary</em> <span class="math-container">$p$</span>.</p>
<p>That won't work for the harmonic series.</p>
|
https://math.stackexchange.com/questions/3061670/proof-that-a-sequence-is-cauchy
|
Question: <p>Why is Completeness needed to demonstrate the Archimedean principle?</p>
<p>Could someone criticize the following proof.</p>
<p>Thanks in advance.</p>
<hr>
<p><strong>Proof</strong></p>
<p>Considering any <span class="math-container">$x \in \mathbb{R},$</span> <span class="math-container">$\lfloor{x}\rfloor \in \mathbb{N},$</span> and <span class="math-container">$$0 \leq x - \lfloor{x}\rfloor < 1$$</span></p>
<p>and </p>
<p><span class="math-container">$$x < \lfloor{x}\rfloor + 1.$$</span></p>
<p>Given that <span class="math-container">$\lfloor{x}\rfloor \in \mathbb{N},$</span> <span class="math-container">$\lfloor{x}\rfloor + 1 \in \mathbb{N}.$</span> Hence, <span class="math-container">$\exists n \in \mathbb{N}$</span> such that <span class="math-container">$x < n.$</span> </p>
Answer: <p>As for why completeness is needed? Here are two ordered fields that aren't complete, one with the Archimedean property and one without:</p>
<p><span class="math-container">$\mathbb{Q}$</span> is an ordered field with the property.</p>
<p>Construct an order on <span class="math-container">$\mathbb{R}(x)$</span> (the field of rational functions over <span class="math-container">$\mathbb{R}$</span>) as follows: a fraction <span class="math-container">$\frac{P(x)}{Q(x)}$</span> with <span class="math-container">$Q$</span> monic is positive when the leading coefficient of <span class="math-container">$P$</span> is positive. In terms of function values, a rational function <span class="math-container">$f$</span> is positive when there is some <span class="math-container">$M$</span> such that <span class="math-container">$f(x)>0$</span> for all <span class="math-container">$x>M$</span>.<br>
With this order, <span class="math-container">$\mathbb{R}(x)$</span> is an ordered field that doesn't have the Archimedean property. <span class="math-container">$x$</span> is an element that's greater than every integer, and <span class="math-container">$\frac1x$</span> is an element that's smaller than the reciprocal of every positive integer.</p>
<p>If we can construct examples without that completeness axiom in which the property doesn't hold, that means we need that axiom to prove the property.</p>
|
https://math.stackexchange.com/questions/3099533/why-is-completeness-needed-to-demonstrate-the-archimedean-principle
|
Question: <p>We already know that
<span class="math-container">$$\lim_{n \rightarrow +\infty} \int_{-1}^1 (1-x^2)^n dx = 0$$</span></p>
<p>If we have <span class="math-container">$f(x) \in C[-1,1]$</span> then prove
<span class="math-container">$$\lim_{n \rightarrow +\infty} \frac{\int_{-1}^1 f(x)(1-x^2)^n dx}{\int_{-1}^1 (1-x^2)^n dx } = f(0)$$</span></p>
<p>My thought is <span class="math-container">$\lim\limits_{n \rightarrow +\infty} \frac{\int_{-1}^1 f(x)(1-x^2)^n dx}{\int_{-1}^1 (1-x^2)^n dx } - f(0) = \lim\limits_{n \rightarrow +\infty} \frac{\int_{-1}^1 [f(x)-f(0)](1-x^2)^n dx}{\int_{-1}^1 (1-x^2)^n dx } $</span> so assume <span class="math-container">$f(0) = 0$</span>. And <span class="math-container">$\lim\limits_{n \rightarrow +\infty} \frac{\int_{-1}^1 f(x)(1-x^2)^n dx}{\int_{-1}^1 (1-x^2)^n dx } \le \lim\limits_{n \rightarrow +\infty} \frac{\int_{-1}^{-\delta} M(1-x^2)^n dx + \int_{-\delta}^\delta \epsilon (1-x^2)^n dx + \int_{\delta}^1 M(1-x^2)^n dx}{\int_{-\delta}^{\delta} (1-x^2)^n dx } $</span> where <span class="math-container">$M$</span> is the upper bounder of <span class="math-container">$|f(x)|$</span> and <span class="math-container">$\epsilon$</span> is small enough since <span class="math-container">$f(0)$</span>. It suffice to show that <span class="math-container">$\lim\limits_{n \rightarrow +\infty} \frac{\int_{-1}^{-\delta} (1-x^2)^n dx}{\int_{-\delta}^\delta (1-x^2)^n dx } =0$</span>.</p>
Answer: <p>1) Observe that
<span class="math-container">$$
c_n := \int_{-1}^1 (1-x^2)^n\, dx \geq 2\int_0^{1/\sqrt{n}}(1-x^2)^n\, dx
\geq 2\int_0^{1/\sqrt{n}}(1- n x^2)\, dx = \frac{4}{3\sqrt{n}}\,.
$$</span></p>
<p>2) If <span class="math-container">$\delta\in (0,1)$</span>, then
<span class="math-container">$$
\frac{1}{c_n}\int_{\delta}^1 (1-x^2)^n\, dx\leq \frac{(1-\delta^2)^n}{c_n}
\leq \frac{3(1-\delta^2)^n}{4\sqrt{n}} \to 0,
$$</span>
and similarly <span class="math-container">$\frac{1}{c_n}\int_{-1}^{-\delta} (1-x^2)^n\, dx \to 0$</span>.</p>
<p>3) Let <span class="math-container">$M := \max_{x\in [-1,1]} |f(x)|$</span>.
If <span class="math-container">$f(0) = 0$</span>, given <span class="math-container">$\epsilon > 0$</span> there exists <span class="math-container">$\delta\in(0,1)$</span> such that
<span class="math-container">$|f(x)| < \epsilon$</span> for every <span class="math-container">$|x| < \delta$</span>.
Then
<span class="math-container">$$
0\leq \limsup_n\frac{1}{c_n} \int_{-1}^1 |f(x)| (1-x^2)^n \, dx
\leq
\limsup_n\frac{M}{c_n}\int_{\delta\leq |x|\leq 1 } (1-x^2)^n \, dx + \epsilon
=\epsilon,
$$</span>
hence we can conclude that
<span class="math-container">$$
\lim_n\frac{1}{c_n} \int_{-1}^1 |f(x)| (1-x^2)^n \, dx = 0 = f(0).
$$</span></p>
|
https://math.stackexchange.com/questions/3118986/frac-int-fg-dx-int-g-dx-f0
|
Question: <p><strong>Problem</strong></p>
<blockquote>
<p>Let <span class="math-container">$u(x,t) = \frac{e^\frac{-x^2}{4t}}{\sqrt{4 \pi t}} $</span> for <span class="math-container">$ t > 0,
> x \in \mathbb{R} $</span>. If a > 0, prove that <span class="math-container">$u(x,t) \rightarrow 0$</span> as <span class="math-container">$t
> \rightarrow 0+$</span>, uniformly for <span class="math-container">$x \in [a, \infty ]$</span></p>
</blockquote>
<p>I proved like this, but I didn't understand well. <p></p>
<blockquote>
<p><span class="math-container">$|u(x,t)| \le u(a,t)$</span> for all <span class="math-container">$ x \in \mathbb{R} $</span> and <span class="math-container">$u(a,t) =
\frac{e^\frac{-a^2}{4t}}{\sqrt{4 \pi t}} $</span>.<br> As t -> 0+, <span class="math-container">$e^\frac{-a^2}{4t}$</span>
converges to zero much faster than <span class="math-container">$ \frac{1}{\sqrt{4 \pi t}} $</span> to
infinity. Thus u(a,t) uniformly converges to zero, so u(x,t) uniformly
converges to zero for all <span class="math-container">$ x \in \mathbb{R}
$</span>. <p></p>
</blockquote>
<p>Is my proof correct?</p>
Answer: <p>Showing that <span class="math-container">$u(a,t) \to 0$</span> is equivalent to showing that <span class="math-container">$u^{2}(a,t) \to 0$</span>. Make the substitution <span class="math-container">$s=\frac {a^{2}} {2t}$</span> and reduce to the proof to <span class="math-container">$se^{-s} \to 0$</span> as <span class="math-container">$ s\to \infty$</span> Apply L'Hopital's Rule to the ration <span class="math-container">$ \frac s {e^{s}}$</span>. </p>
|
https://math.stackexchange.com/questions/3155369/uniformly-convergence-of-heat-equation
|
Question: <p>When we shift <span class="math-container">$\sin(x)$</span> by <span class="math-container">$\pi$</span> (half the period) we get an odd function <span class="math-container">$-\sin(x)$</span>.
I was wondering if every periodic, even function can be made odd if we shift it by half it's period?</p>
<p>I guess this is not the case but I wasn't able to come up with a good counter example..</p>
Answer: <p>What you seem to be noticing is that <span class="math-container">$\sin(x)$</span> is odd, and
<span class="math-container">$\cos(x)$</span> is even and the second is a shifted version of the first by a quarter of a period. This generalizes. Suppose we have a function <span class="math-container">$f(x)$</span> defined on an interval <span class="math-container">$\ [0,L].\ $</span> Define an extension to <span class="math-container">$\ [L,2L]\ $</span> with the reversed function <span class="math-container">$\ f(2L-x).\ $</span> Extend even further to <span class="math-container">$\ [2L,4L]\ $</span> with <span class="math-container">$\ -f(x-2L).\ $</span> Finally, extend periodically to all real numbers with <span class="math-container">$\ f(x) = f(x+4nL)\ $</span> for all integer <span class="math-container">$n$</span>. The extended period <span class="math-container">$\ 4L\ $</span> function <span class="math-container">$\ f(x)\ $</span> is odd, the shifted function <span class="math-container">$\ f(x+L)\ $</span> is even, and the twice shifted function <span class="math-container">$\ f(x+2L) = -f(x)\ $</span> is also odd.</p>
<p>However, if a <strong>non-zero</strong> non-negative periodic function is even, then <strong>no</strong> shift of it will be odd because that would require it to take on negative values which we excluded by assumption. The simplest case is any non-zero constant function. It is an even function and has all periods, but no shift of it can be an odd function.</p>
|
https://math.stackexchange.com/questions/3207241/shift-of-even-periodic-function
|
Question: <p>I am trying to read <em>Mathematical Analysis I</em> by Zorich on my own. Here is an exercise that I could not solve: </p>
<blockquote>
<p>For all <span class="math-container">$l \in \mathbb{R}$</span> that is not of the form <span class="math-container">$\frac{1}{n}$</span> for some <span class="math-container">$n \in \mathbb{N}$</span>, there exists a continuous function <span class="math-container">$f: [0, 1] \Rightarrow \mathbb{R}$</span> such that <span class="math-container">$f(0) = f(1)$</span> and the graph of <span class="math-container">$f$</span> contains no horizontal chord with length <span class="math-container">$l$</span>. </p>
</blockquote>
<p>I was wondering how to prove this? Thanks in advance. </p>
Answer:
|
https://math.stackexchange.com/questions/3230799/a-problem-from-zorich
|
Question: <p>Let's say a function <span class="math-container">$f:[a,b] \to \mathbb{R}$</span> is bounded and Riemann integrable, then would there always exist a partition <span class="math-container">$P$</span> of <span class="math-container">$[a,b]$</span> such that the lower sum of <span class="math-container">$P$</span> equals the upper sum of <span class="math-container">$P$</span>?</p>
<p>My guess is "No" because the Riemann integrability implies that the infimum of upper sums equals the supremum of lower sums, but I am struggling to come up with a proof or counter-example.</p>
<p>Any help will be greatly appreciated. </p>
Answer: <p>The lower and upper sums are sums of areas of rectangles. It should be clear from a diagram that unless <span class="math-container">$f$</span> is constant (or at any rate piecewise constant) we will always have
<span class="math-container">$$\{\hbox{lower sum}\}<\{\hbox{exact area}\}<\{\hbox{upper sum}\}\ .$$</span>
For a specific example simply take <span class="math-container">$f(x)=x$</span> on <span class="math-container">$[0,1]$</span>. You should not find it hard to show that for any partition <span class="math-container">$P$</span>,
<span class="math-container">$$\{\hbox{lower sum}\}<\frac12<\{\hbox{upper sum}\}\ .$$</span></p>
|
https://math.stackexchange.com/questions/3275494/riemann-integrability-and-lower-and-upper-sums
|
Question: <p>I'm lost with the assignment from my teacher.
For real numbers w, x, y, and z. Prove for x,z≠0, (w/x)(y/z)=(wy)/(xz) and (w/x)+(y/z)=(wz+xy)/(xz)</p>
Answer: <p>You can write <span class="math-container">$$\frac{w}{x}+\frac{y}{z}=\frac{wz}{xz}+\frac{xy}{xz}=…$$</span>
<span class="math-container">$$\frac{w}{x}\cdot \frac{y}{z}=\frac{wy}{xz}$$</span></p>
|
https://math.stackexchange.com/questions/3355313/for-real-numbers-w-x-y-and-z-prove-for-x-z-%e2%89%a0-0-w-xy-z-wy-xz-and-w
|
Question: <blockquote>
<p>Suppose <span class="math-container">$f(x)$</span> defines on the bounded interval <span class="math-container">$I$</span>. Prove that <span class="math-container">$f(x)$</span> is uniformly continuous on <span class="math-container">$I$</span> if and only if the image of each Cauchy sequence under <span class="math-container">$f$</span> is also a Cauchy sequence.</p>
</blockquote>
<p>The "only if" part is easy. For the "if" part I can prove that <span class="math-container">$f$</span> is continuous on <span class="math-container">$I$</span>. But how to prove it's uniformly continuous? I try to prove by contradiction, but could not get a Cauchy sequence.</p>
Answer: <p>Hint: The following can be used together with some basic theory of real analysis to complete the OP's analysis.</p>
<p>Given <span class="math-container">$a. b \in \Bbb R$</span> with <span class="math-container">$a \lt b$</span>.</p>
<p>Let <span class="math-container">$f: (a,b) \to \Bbb R$</span> be a continuous function satisfying</p>
<p><span class="math-container">$$\tag 1 {\displaystyle \bigcap_{n \ge 1}} \, \overline {f\big ( \,(a,a+\frac{b-a}{2^n}] \, \big )} = \{\alpha\}$$</span></p>
<p>and</p>
<p><span class="math-container">$$\tag 2 {\displaystyle \bigcap_{n \ge 1}} \, \overline {f\big( [b-\frac{b-a}{2^n},b) \big )} = \{\beta\}$$</span></p>
<p>Then <span class="math-container">$f$</span> can be extended to a continuous function <span class="math-container">$f_♯: [a,b] \to \Bbb R$</span> by defining <span class="math-container">$f_♯(a) = \alpha$</span> and <span class="math-container">$f_♯(b) = \beta$</span>.</p>
|
https://math.stackexchange.com/questions/3381457/prove-that-fx-is-uniformly-continuous-on-i-if-and-only-if-the-image-of-eac
|
Question: <blockquote>
<p>Suppose <span class="math-container">$f(x)$</span> is differentiable on <span class="math-container">$[0,\,1]$</span>, <span class="math-container">$f(0)=0$</span>, <span class="math-container">$f(1)=1$</span> and <span class="math-container">$p_1,\,p_2,\cdots,\,p_n$</span> are <span class="math-container">$n$</span> positive real numbers. Prove there are distinct <span class="math-container">$x_1,\,x_2,\cdots,\,x_n$</span> such that
<span class="math-container">$$
\sum_{i=1}^n\frac{p_i}{f'(x_i)}=\sum_{i=1}^n p_i.
$$</span></p>
</blockquote>
<p>I can only prove some special cases.
Let <span class="math-container">$p=\sum_{i=1}^n p_i$</span>. It suffice to prove that <span class="math-container">$\sum_{i=1}^n\frac{p_i}{pf'(x_i)}=1$</span>. A proper choose is <span class="math-container">$f'(x_i)=\frac{np_i}{p}$</span>. From Darboux theorem, if <span class="math-container">$f'$</span> is large enough, these values can attain. </p>
Answer: <p><strong>Proof</strong>. <span class="math-container">$\blacktriangleleft$</span> Assume <span class="math-container">$\sum p_j = 1$</span>, otherwise replace <span class="math-container">$p_j$</span> by <span class="math-container">$p_j/p$</span> for each <span class="math-container">$j$</span>. By continuity and the Intermediate Value Theorem, there is some <span class="math-container">$y_1 \in (0,1)$</span> that <span class="math-container">$f(y_1) = p_1$</span>, then there is some <span class="math-container">$y_2 \in (y_1, 1)$</span> that <span class="math-container">$f(y_2) = p_1 + p_2$</span>. Do this <span class="math-container">$n-1$</span> times, we obtain that
<span class="math-container">$$
0 = y_0 < y_1 < \dots < y_{n-1} < y_n = 1, f(y_j) = \sum_{k = 1}^j p_k.
$$</span>
Now by the Mean Value Theorem, there is <span class="math-container">$x_j \in (y_{j-1} , y_j)$</span> where <span class="math-container">$f(y_j) - f(y_{j-1}) = (y_j - y_{j-1}) f'(x_j)$</span> for <span class="math-container">$j \leqslant n$</span>, and
<span class="math-container">$$
\frac {p_j}{f'(x_j)} = \frac {f(y_j) - f(y_{j-1})} {f'(x_j)} = y_j - y_{j-1}, j \leqslant n,
$$</span>
so sum these up and we are done. <span class="math-container">$\blacktriangleright$</span></p>
|
https://math.stackexchange.com/questions/3383736/prove-there-are-distinct-x-1-x-2-cdots-x-n-such-that-sum-i-1n-frac
|
Question: <p>Find the explicit form of
<span class="math-container">$$
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1}.
$$</span></p>
<p>Let <span class="math-container">$S(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1}$</span>. It has radius of convergence <span class="math-container">$1$</span>.</p>
<p>Let <span class="math-container">$S_1(x)=xS(x)$</span>. Then <span class="math-container">$S_1'(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(n+2)}x^{n-1}$</span> for <span class="math-container">$|x|<1$</span>.</p>
<p>Let <span class="math-container">$S_2(x)=x^3S_1'(x)$</span>. Then <span class="math-container">$S_2'(x)=\sum_{n=1}^{\infty}(-x)^{n-1}=\frac{x^2}{1+x}$</span>.</p>
<p>By integration, I obtained <span class="math-container">$S_1'(x)=\frac{1}{2x}-\frac{1}{x^2}+\frac{\ln (x+1)}{x^3}$</span>. Then how to obtain <span class="math-container">$S(x)$</span>? Or there is other method to do this problem?</p>
Answer: <p><span class="math-container">$$\log(1+x)=\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n}~~~(1)$$</span>
Multipluy by <span class="math-container">$x$</span> on both sides and integrate w.r.t. <span class="math-container">$x$</span>
<span class="math-container">$$\int x \log(1+x) dx= \sum_{n=1}^{\infty} (-1)^{n-1}\int \frac{x^{n+1}}{n} dx= \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{n+2}}{n(n+2)} .$$</span>
<span class="math-container">$$\implies S(x)=\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{n-1}}{n(n+2)}= \frac{1}{x^3} \int x \log(1+x) ~dx.$$</span>
Finally by carrying out integration by parts, we get the required sum
<span class="math-container">$$S(x)=\frac{1}{4x^3}[2x-x^2-2\log(1+x)+2x^2 \log(1+x)]$$</span></p>
|
https://math.stackexchange.com/questions/3386371/find-the-explicit-form-of-sum-n-1-infty-frac-1n-1nn2xn-1
|
Question: <blockquote>
<p>Suppose <span class="math-container">$\displaystyle f(x)=\sum_{n=0}^{\infty}2^{-n}\sin(2^nx)$</span>, evaluate
<span class="math-container">$$
\lim_{x\to0+}\frac{f(x)-f(0)}{x}.
$$</span></p>
</blockquote>
<p><span class="math-container">$f(x)$</span> converges uniformly on <span class="math-container">$\mathbb R$</span>. But I don't know how to evaluate
<span class="math-container">$$
\lim_{x\to0+}\sum_{n=0}^{\infty}2^{-n}\frac{\sin (2^nx)}{x}.
$$</span></p>
Answer: <p>After defining</p>
<p><span class="math-container">$$g_n(x) = \frac{\sin\left(2^nx\right)}{2^nx},$$</span></p>
<p>and
<span class="math-container">$$g(x) = \sum_{n=0}^{+\infty}g_n(x),$$</span></p>
<p>observe that in the interval
<span class="math-container">$$0<x<\frac{\pi}{2^N},$$</span></p>
<ol>
<li>For <span class="math-container">$0\leq j\leq m\leq N-1$</span>, <span class="math-container">$g_j(x)\geq g_m(x)$</span>.</li>
<li><span class="math-container">$g_n(x)$</span> is a monotonically decreasing function.</li>
</ol>
<p>Consider a sequence <span class="math-container">$\{a_k\}$</span>, with</p>
<p><span class="math-container">$$\{a_k\}\to 0^+.$$</span></p>
<p>For large enough <span class="math-container">$k$</span>,
<span class="math-container">$$a_k< \frac{\pi}{2^N}$$</span>
so that
<span class="math-container">\begin{eqnarray}
g\left(a_k\right) &=& \sum_{n=0}^{N-1}g_n\left(a_k\right) + \sum_{N}^{+\infty} g_n\left(a_k\right)\geq\\
&\geq& N\cdot g_{N-1}\left(a_k\right) - \sum_{n=N}^{+\infty}\frac1{2^n\pi}>\\
&>& N\cdot g_{N-1}\left(\frac{\pi}{2^N}\right) -\frac1{2^{N-1}\pi}=\\
&=&\frac{2N}{\pi} -\frac1{2^{N-1}\pi},
\end{eqnarray}</span>
where we took advantage of fact 1 in the first inequality, and of fact 2. in the second inequality. From the above chain we conclude that <span class="math-container">$g\left(a_k\right)$</span> can be made arbitrarily large by a suitable choice of <span class="math-container">$k$</span>, thus</p>
<p><span class="math-container">$$\left\{g\left(a_k\right)\right\} \to +\infty,$$</span>
and therefore
<span class="math-container">$$\lim_{x\to 0^+} \sum_{n=0}^{+\infty} \frac{\sin\left(2^nx\right)}{2^nx} = +\infty.$$</span></p>
<hr>
<p><strong>Edit</strong></p>
<p>Prompted by <a href="https://math.stackexchange.com/users/278017/jake-mirra">Jake</a>'s comment I want to make facts 1. and 2. above more explicit.</p>
<ol>
<li><p>This fact is just a consequence of <span class="math-container">$$\left|\frac{\sin\left(2^m\alpha\right)}{2^m\alpha}\right| = \left|\frac{\sin\left(2^{m-1}\alpha\right)\cos\left(2^{m-1}\alpha\right)}{2^{m-1}\alpha}\right| \leq \left|\frac{\sin\left(2^{m-1}\alpha\right)}{2^{m-1}\alpha}\right|$$</span></p></li>
<li><p>This fact can be derived by differentiation, since for <span class="math-container">$0 <\alpha<\frac{\pi}2$</span>, <span class="math-container">$$\alpha < \tan \alpha$$</span> and therefore, in such interval we get
<span class="math-container">$$\frac{d\left(\frac{\sin\alpha}{\alpha}\right)}{d\alpha}=\frac{\alpha\cos\alpha-\sin\alpha}{\alpha^2}<0$$</span></p></li>
</ol>
<p>I do hope someone can give a simpler proof of the limit in OP's question, since I could not find one, so far.</p>
|
https://math.stackexchange.com/questions/3401237/evaluate-lim-x-to0-sum-n-0-infty2-n-frac-sin-2nxx
|
Question: <p>So I have to prove if is <span class="math-container">$\mathbb{N} \times \mathbb{R}$</span> closed, open or neither?</p>
<p>Any help?</p>
Answer: <p>In what topological space? If <span class="math-container">$\mathbf{R}^2$</span>, closed. If <span class="math-container">$\mathbf{R}\times\mathbf{N}$</span> itself, both. In an arbitrary space <span class="math-container">$X$</span>, who’s to say?</p>
|
https://math.stackexchange.com/questions/3448363/is-mathbbn-times-mathbbr-closed-open-or-neither
|
Question: <p><span class="math-container">$f(x)$</span> is differentiable, and for any <span class="math-container">$x\in \mathbb R$</span>, <span class="math-container">$|f'(x)|\le \lambda |x|$</span>, then how to show <span class="math-container">$f\equiv 0$</span> ? </p>
<p>This is a question my student ask me, but I don't know how to deal it. So ask help here. Thanks for any hint or answer.</p>
Answer: <p>This is not correct: <span class="math-container">$f(x)=x^{2}$</span>, <span class="math-container">$|f'(x)|=2|x|$</span>.</p>
<p>Even that <span class="math-container">$\lambda$</span> is required to be <span class="math-container">$\lambda\in(0,1)$</span> is still not correct: <span class="math-container">$f(x)=\lambda^{2}\log(3\lambda+x^{2})$</span>, then <span class="math-container">$|f'(x)|=\dfrac{2\lambda^{2}|x|}{3\lambda+x^{2}}\leq\dfrac{2}{3}\cdot\lambda|x|$</span>.</p>
|
https://math.stackexchange.com/questions/3457881/how-to-show-f-equiv-0
|
Question: <p>Assume <span class="math-container">$f\in C^2(D)$</span>, where <span class="math-container">$D=\{(x,y)\in \mathbb R^2: x^2+ y^2 \le 1\}$</span>, if
<span class="math-container">$$
\frac{\partial^2 f }{\partial x^2} + \frac{\partial^2 f }{\partial y^2}
=e^{-x^2 -y^2}
$$</span>
how do I show
<span class="math-container">$$
\iint\limits_D(x\frac{\partial f }{\partial x}+ y \frac{\partial f}{\partial y})\,dx\,dy= \frac{\pi}{2e}
$$</span></p>
<p>I think I should use Green's Theorem, but I wasn't able to make progress. </p>
Answer: <p>Notice that the integral can be rewritten as</p>
<p><span class="math-container">$$\iint_D (x,y)\cdot \nabla f dA = \iint_D \nabla \left( \frac{x^2 + y^2}{2}\right) \cdot \nabla f dA$$</span></p>
<p>then use integration by parts (really the divergence theorem in higher dimensions) with <span class="math-container">$u=\nabla f$</span> and <span class="math-container">$dv = (x,y)$</span>:</p>
<p><span class="math-container">$$\iint_D \nabla \left( \frac{x^2 + y^2}{2}\right) \cdot \nabla f dA = \int_{\partial D} \left(\frac{x^2+y^2}{2}\right) \nabla f \cdot \mathbf{n} ds - \iint_D \left(\frac{x^2+y^2}{2}\right) \Delta f dA$$</span></p>
<p><span class="math-container">$$ = \frac{1}{2} \int_{\partial D} \nabla f \cdot \mathbf{n} ds - \iint_D \left(\frac{x^2+y^2}{2}\right) \Delta f dA$$</span></p>
<p>We can use the divergence theorem on the first integral</p>
<p><span class="math-container">$$\frac{1}{2}\iint_{D} \Delta f dA - \iint_D \left(\frac{x^2+y^2}{2}\right) \Delta f dA$$</span></p>
<p>then plug in <span class="math-container">$\Delta f = e^{-x^2-y^2}$</span> and convert to polar coordinates</p>
<p><span class="math-container">$$\frac{1}{2}\int_0^1 \int_0^{2\pi}re^{-r^2}d\theta dr-\frac{1}{2} \int_0^1 \int_0^{2\pi} r^3e^{-r^2}d\theta dr=\pi\int_0^1 re^{-r^2}dr-\pi \int_0^1 r^3e^{-r^2}dr = \frac{\pi}{2e}$$</span></p>
|
https://math.stackexchange.com/questions/3465278/how-to-show-iint-limits-dx-frac-partial-f-partial-x-y-frac-partial-f
|
Question: <p>Let <span class="math-container">$f(x)$</span> be a continuous function on <span class="math-container">$[a, b]$</span> (not necessarily derivable) and satisfy <span class="math-container">$f(a)<f(b)$</span>. At the same time, the limit <span class="math-container">$g(x)=\lim_{t \to 0} \frac{f(x+t)-f(x-t)}{t}$</span> exists for all <span class="math-container">$x\in(a,b)$</span>.</p>
<p>Prove that there is a <span class="math-container">$c\in(a,b)$</span> such that <span class="math-container">$g(c)\geqslant0$</span>.</p>
Answer:
|
https://math.stackexchange.com/questions/3544900/gx-lim-t-to-0-fracfxt-fx-tt-prove-that-there-is-a-c-ina-b
|
Question: <p>Consider the non-negative orthant
<span class="math-container">$$
R\equiv \{(x_1,...,x_n): x_i\geq 0 \text{ }\forall i\}
$$</span>
The boundary of <span class="math-container">$R$</span> is not smooth.
I'm looking for a function which can smooth the boundary of <span class="math-container">$R$</span> and which depends on only one smoothing parameter. Could you advise? I'm a beginner and any related question you may have to clarify my request are welcome. </p>
Answer: <p>I guess the question is, why would you want to do that? Since your question is somewhat vague, I'm going to imagine that you are trying to solve an optimization problem like
<span class="math-container">$$
\max_{x \in \mathbb{R}^N} f(x)
$$</span>
subject to <span class="math-container">$x \ge 0$</span> and <span class="math-container">$h(x) \le 0$</span>, where <span class="math-container">$h(x)$</span> is quasi-convex and smooth, <span class="math-container">$f(x)$</span> is strictly quasi-concave and smooth, and <span class="math-container">$\{x:x \ge 0 \text{ and } h(x) \le 0 \} \neq \varnothing$</span>. For some reason, you don't want to deal with <span class="math-container">$x \ge 0$</span>, perhaps because <span class="math-container">$N$</span> is large and <span class="math-container">$2^N$</span> of these corner cases is too computationally taxing for a numerical solver or something.</p>
<p>Take <span class="math-container">$g(x,\alpha) = -\sum_{i=1}^N ( \min\{ 0, \alpha_i-x_i\})^2$</span>. If <span class="math-container">$x_i$</span> drops below <span class="math-container">$\alpha_i$</span>, it activates a penalty that pushes you away from <span class="math-container">$x_i=0$</span>. The function <span class="math-container">$(\min\{0,\alpha_i-x_i\})^2$</span> is differentiable since <span class="math-container">$\lim_{x_i \uparrow \alpha_i } -(\alpha_i-x_i) = 0$</span> and <span class="math-container">$\lim_{x_i \downarrow \alpha_i} 0 = 0$</span>. So if you maximize the Lagrangian
<span class="math-container">$$
\mathcal{L}(x,\alpha) = f(x) - \mu h(x) + g(x,\alpha),
$$</span>
the <span class="math-container">$g(x,\alpha)$</span> function will push you away from picking corner solutions if <span class="math-container">$\mathcal{L}(x,0)$</span> has them. </p>
<p>Then you can take a sequence <span class="math-container">$\alpha_n \rightarrow 0$</span>. Since <span class="math-container">$f$</span> is strictly quasi-concave and <span class="math-container">$h$</span> is quasi-convex, there is a unique maximizer, and by Berge's theorem of the maximum it is an upper-hemi continuous correspondence in <span class="math-container">$\alpha$</span> (and so a continuous function), so that <span class="math-container">$x^*(\alpha_n) \rightarrow x^*(0)$</span>.</p>
<p>You could implement this idea of penalties a bunch of different ways. Perhaps by looking at which penalties get activated for small <span class="math-container">$||\alpha||$</span>, you can get a guess of which corners are likely to be the problem for the optimization in advance of taking <span class="math-container">$\alpha \rightarrow 0$</span>.</p>
|
https://math.stackexchange.com/questions/3606304/non-negative-orthant-smoothing
|
Question: <p>I picked up a copy of Principles of Mathematical Analysis by Rudin. It just so happened to be the second edition which is drastically different from the third and I can't find solutions to a lot of exercises.</p>
<p>I'd like some hints on how to complete the following few problems.</p>
<ol>
<li>If <span class="math-container">$x>0$</span>, <span class="math-container">$y>0$</span>, and <span class="math-container">$n$</span> is a positive integer, prove that <span class="math-container">$\sqrt[n]{x}\sqrt[n]{y} = \sqrt[n]{xy}$</span>.</li>
</ol>
<p>I can't seem to figure out where to start with this one. I assume I prove the inequality each way.</p>
<ol start="2">
<li><p>If <span class="math-container">$x>0$</span>, and r is rational (<span class="math-container">$r = n/m$</span>), define <span class="math-container">$x^r = \sqrt[m]{x^n}$</span>. Prove that <span class="math-container">$x^r = \sqrt[m]{x}^n$</span>.</p></li>
<li><p>If <span class="math-container">$x>1$</span>, prove that <span class="math-container">$x^p < x^q$</span> whenever <span class="math-container">$p < q$</span>, <span class="math-container">$p, q$</span> are rational.</p></li>
</ol>
<p>For this one I wrote p, q as fractions and used 2.</p>
Answer: <ol>
<li><p>I presume <span class="math-container">$\sqrt[n]x$</span> is defined as the unique solution of <span class="math-container">$y^n=x$</span> with <span class="math-container">$y>0$</span>
(and I presume that Rudin has proved there is a unique solution). Then to prove <span class="math-container">$\sqrt[n]{xy}=\sqrt[n]x\sqrt[n]y$</span> all you need is to verify that <span class="math-container">$\sqrt[n]x\sqrt[n]y>0$</span>
and <span class="math-container">$(\sqrt[n]x\sqrt[n]y)^n=xy$</span>.</p></li>
<li><p>Again you need that <span class="math-container">$(\sqrt[m]x)^n>0$</span> and <span class="math-container">$((\sqrt[m]x)^n)^m=x^n$</span>.</p></li>
<li><p>I would put <span class="math-container">$p$</span> and <span class="math-container">$q$</span> as fractions over a common denominator and try to
reduce to the case where they are integers.</p></li>
</ol>
|
https://math.stackexchange.com/questions/3647631/few-questions-1-4-1-5-1-6-from-rudin-2e-on-roots-of-reals
|
Question: <p>The limit location theorem states for a sequence $\{a_n\}$, if for large enough $n$ we have $a_n \leq M$, then $\lim_{n \to \infty} a_n \leq M$ (the same holds if we replace $\leq$ with $\geq$). Why does the theorem not hold with strict inequalities?</p>
Answer: <p>In general we have that even if $a_n < M$ then $\lim_{n \to \infty} a_n \leq M$, consider for example</p>
<p>$$\forall n>0 \quad a_n=\frac{n}{n+1}<1$$</p>
<p>but</p>
<p>$$\lim_{n \to \infty}a_n=\lim_{n \to \infty}\frac{n}{n+1}=1$$</p>
<p>thus the theorem you mentioned considers the more general case.</p>
|
https://math.stackexchange.com/questions/2784352/why-cant-we-replace-leq-with-in-the-limit-location-theorem
|
Question: <p>For $N\geq 1$ the discrete Fourier Coefficient of $f$ are defined by
\begin{align}
a_N(n)=\frac{1}{N}\sum_{k=1}^{N}f(e^{2\pi i k/N})e^{-2\pi i k n/N}, \quad \forall n\in \mathbb{Z}
\end{align}
Suppose $f\in C^1$ function on circle, how to prove there exist $c>0$ such that $|a_N(n)|\leq \frac{c}{|n|}$ whenever $0<|n|\leq \frac{N}{2}$.
<strong>Hint:</strong>
Write
\begin{align}
a_N(n)[1-e^{2\pi i\ell n/N}]=\frac{1}{N}\sum_{k=1}^{N}[f(e^{2\pi i k/N})-f(e^{2\pi i(k+\ell)/N})]e^{-2\pi i k n/N}
\end{align}
and choose $\ell$ so that $\ell n/N$ nearly 1/2.</p>
<p><strong>Source:</strong> Fourier Analysis An Introduction, Elias M.Stein & Rami Sacharchi. Chapter 7 problem 2. </p>
<p>I am not understand <strong>Hint</strong> of the book. How we prove LHS=RHS in the hint.</p>
<p><strong>My approach for proof the hint.</strong></p>
<p>If $\ell=0$, it's so clear. </p>
<p>Suppose $1\leq \ell< N$ and $\ell\in \mathbb{Z}$,
\begin{align*}
f(e^{2\pi i(k+\ell)/N})e^{-2\pi i k n/N}=
\begin{cases}
f(e^{2\pi i(k+\ell)/N})e^{-2\pi i k n/N}, \quad 1\leq k\leq N-\ell\\
f(e^{2\pi i(k-(N-\ell))/N})e^{-2\pi i k n/N}\quad, N-\ell< k\leq N
\end{cases}
\end{align*}
and we get
\begin{align*}
\frac{1}{N}\sum_{k=1}^{N} f(e^{2\pi i(k+\ell)/N})e^{-2\pi i k n/N}&=\frac{1}{N}\sum_{k=1}^{N-\ell} f(e^{2\pi i(k+\ell)/N})e^{-2\pi i k n/N}\\&+\frac{1}{N}\sum_{k=N-\ell+1}^{N}f(e^{2\pi i(k-(N-\ell))/N})e^{-2\pi i k n/N}\\
&=\frac{1}{N}\sum_{k=\ell+1}^{N}f(e^{2\pi i k/N})e^{-2\pi i(k-\ell)n/N}+\frac{1}{N}\sum_{k=1}^{\ell}f(e^{2\pi ik/N})e^{-2\pi i (k-l)n/N}\\
&=\frac{1}{N}\sum_{k=1}^{N}f(e^{2\pi i k/N})e^{-2\pi i(k-\ell)n/N}\\
&=a_n(N)e^{2\pi i \ell n/N}
\end{align*}</p>
<p>If $-1\leq \ell <-N$
\begin{align*}
f(e^{2\pi i(k+\ell)/N})e^{-2\pi i k n/N}=
\begin{cases}
f(e^{2\pi i(k-(-\ell-N))/N})e^{-2\pi i k n/N}, \quad 1\leq k\leq-\ell\\
f(e^{2\pi i(k+\ell)/N})e^{-2\pi i k n/N}\quad, -\ell< k\leq N
\end{cases}
\end{align*}
\begin{align*}
\frac{1}{N}\sum_{k=1}^{N} f(e^{2\pi i(k+\ell)/N})e^{-2\pi i k n/N}&=\frac{1}{N}\sum_{k=1}^{-\ell} f(e^{2\pi i(k-(-\ell-N))/N})e^{-2\pi i k n/N}\\&+\frac{1}{N}\sum_{k=-\ell+1}^{N}f(e^{2\pi i(k+\ell)/N})e^{-2\pi i k n/N}\\
&=\frac{1}{N}\sum_{k=N+\ell+1}^{N}f(e^{2\pi i k/N})e^{-2\pi i(k+\ell)n/N}+\frac{1}{N}\sum_{k=1}^{N+\ell}f(e^{2\pi ik/N})e^{-2\pi i (k+\ell)n/N}\\
&=\frac{1}{N}\sum_{k=1}^{N}f(e^{2\pi i k/N})e^{-2\pi i(k+\ell)n/N}\\
&=a_n(N)e^{2\pi i \ell n/N}
\end{align*}</p>
Answer:
|
https://math.stackexchange.com/questions/2750867/a-nn-leq-fraccn
|
Question: <p>Let $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ defined by \begin{equation*}f: \begin{pmatrix}x \\ y\end{pmatrix}\rightarrow \begin{pmatrix}u \\ v\end{pmatrix}=\begin{pmatrix}x(1-y) \\ xy\end{pmatrix}\end{equation*} </p>
<p>I want to determine th eimage $f(\mathbb{R}^2)$. </p>
<p>$$$$ </p>
<p>We have that
\begin{equation*}f\begin{pmatrix}1 \\ 0\end{pmatrix}=\begin{pmatrix}1 \\ 0\end{pmatrix} \ \text{ und } \ f\begin{pmatrix}0 \\ 1\end{pmatrix}=\begin{pmatrix}0 \\ 0\end{pmatrix}\end{equation*} </p>
<p>Does this mean that $\text{im}(f)=\left \{\begin{pmatrix}1 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 0\end{pmatrix}\right \}$ ? Or do we have to do something else in this case where we don't have a linear map? </p>
Answer: <p>You can notice that the map is invertible in certain conditions.</p>
<p>If $(u,v)\in\mathbb R^2$ then $f(x=u+v,y=\dfrac v{u+v})=(u,v)$ except for $v=-u$.</p>
<p>Indeed these points cannot be reached since $x(1-y)=-xy\implies x=0$ </p>
<p>And when $x=0$ then the whole axis is transformed to origin point : $f(0,y)=(0,0)$.</p>
<p>So the image $f(\mathbb R^2)$ is whole $\mathbb R^2$ except for the line $v=-u$ but still including origin.</p>
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https://math.stackexchange.com/questions/2822614/determine-the-image-of-the-map
|
Question: <p>I am relevantly new to this concept of Bounded variation.
I want to know is the function $\sqrt {1-x^2} $, $x\in (-1,1) $ of bounded variation? How should I proceed. Is there any graphical interpretation of functions of bounded variation?</p>
Answer: <p>As for the specific question, $f(x)=\sqrt{1-x^2}$ it is a function of bounded variation over the closed interval $[-1,1]$ (I don't know why the restriction here to the open interval, but if taken as a limit, the variation over $(-1,1)$ would also be finite, hence of bounded variation.</p>
<p>A simple proof relies on the following characterization:</p>
<blockquote>
<p>$f$ is of bounded variation over $[a,b]$ $\iff$ there exist non decreasing functions $g,h$ over $[a,b]$ such that $f=g-h$. (*)</p>
</blockquote>
<p>So, consider
$$g(x)=\left\{\begin{array}{ccc}\sqrt{1-x^2}&\text{if}&x<0\\1&\text{if}&x\ge0\\\end{array}\right.$$
and
$$h(x)=\left\{\begin{array}{ccc}0&\text{if}&x<0\\1-\sqrt{1-x^2}&\text{if}&x\ge0,\\\end{array}\right.$$
which happen to be non decreasing and verify the condition $f=g-h$.</p>
<p>As for a graphical interpretation, I find easier to say what a function of unbounded variation is: it's graph is such that if you count the vertical 'distance traveled' going from $a$ to $b$ (or backwards), you can say that you moved over any finite number. For instance, with the function
$$f(x)=\sin \left(\frac1x\right)$$
(for $x\neq0$ and $0$ for $x=0$; [graph])1 you can move leftwards from $x_1=\frac2{\pi}$, where $f$ value is $1$, to $x_2=\frac2{3\pi}$ while the function value will go down monotonously from $1$ to $-1$: you will have gone $2$ units down.</p>
<p>Now if we go on leftwards to $x_3=\frac2{5\pi}$ while $f$ goes up until reaching again the value $1$. You have moved $2$ units up now, and this means you traveled a total of $4$ units vertically: that is, the total variation of $f$ from $x_1$ to $x_3$ is 4.</p>
<p>Now, since taking $x_n=\frac2{(2\cdot n-1)\pi}$ gives an infinite sequence of points $0<\dots<x_3<x_2<x_1$ such that the variation between to consecutive points is always $2$, you can make the variation as large as you want considering enough many points (or, in the language used for the definition, <em>taking a partition that is fine ('thin') enough</em>.</p>
<hr>
<p>ONE MORE COMMENT: The thing with this example you are asking about, is that you cannot use the following usual result:</p>
<blockquote>
<p>If $f$ has bounded continuous derivative over an interval, then it is of bounded variation over that interval.</p>
</blockquote>
<p>But your function has unbounded derivative over $(-1,1)$, so that's why we needed to use other ways to prove bounded variation.</p>
<hr>
<p>(*) The proof of this property actually gives a procedure for finding such $f$ and $g$.</p>
<p>First of all, let's remember that the <em>total variation</em> of $f$ over $[a,b]$ is defined as
$$V_a^b(f)=\sup \left\{\sum_{i=1}^n |f(x_i)-f(x_{i-1})|\colon \{x_0,\ldots,x_n\}\in\mathscr P\right\},$$
where $\mathscr P$ is the set of all possible partitions of $[a,b]$. So $f$ is said to be of <em>bounded variation</em> over $[a,b]$ iff $V_a^b(f)<\infty$.</p>
<p>The central part of the proof (which I omit although is not difficult) is to see that the function $V\colon [a,b]\rightarrow \mathbb R$
$$V(x)=V_a^x(f)$$
is non-decreasing.</p>
<p>Then, you can take $g(x)=V(x)$ and $h(x)=V(x)-f(x)$, which can also be proven to be non-decreasing. That of course implies $g(x)-h(x)=f(x)$.</p>
<p>In this example I saw that $f(x)$—being increasing from $-1$ to $0$—gave itself the total variation over this interval: think that in that case $|f(x_i)-f(x_{i-1})|=f(x_i)-f(x_{i-1})$, and so
$$\sum_{i=1}^n |f(x_i)-f(x_{i-1})|=f(x_1)-f(x_0)+f(x_2)-f(x_1)+$$
$$+f(x_3)-f(x_2)+\cdots+f(x_{n-1})-f(x_{n-2})+f(x_n)-f(x_{n-1})=f(x_n)-f(x_0).$$</p>
<p>That is, for $x<0$, $V(x)=f(x)-f(-1)=f(x).$</p>
<p>Then, from $0$ to $1$ $f$ was decreasing, so $V_0^x(f)=f(0)-f(x)$. I could have defined here then</p>
<p>$$g(x)=V(x)=\left\{\begin{array}{ccc}\sqrt{1-x^2}&if&x<0\\1-\sqrt{1-x^2}&if&x\ge0,\\
\end{array}\right.$$
and so $h$ would have been
$$h(x)=g(x)-f(x)=\left\{\begin{array}{ccc}0&if&x<0\\1-2\sqrt{1-x^2}&if&x\ge0,\\\end{array}\right.$$</p>
<p>Of course, this is not the only way, since you can see that what I did above is different. I haven't think that much at the moment, but let's say that I found more natural to take account of the 'increasing' variations in $g$ and of the 'decreasing' variations in $h$, and that's why I kept constant $g$ after $x=0$. This had the effect of making $h=g-f$ constant when $f$ was increasing and take account of the variation when $f$ was decreasing. Just two of many ways to pick $g$ and $h$ when using this property.</p>
<p>As for the reciprocal ($f=g-h$ with $g,h$ non-decreasing implies $f$ is of bounded variation) is immediate from the observation that monotone functions are of bounded variation and from the fact that $V_a^b(cf)=|c|V_a^b(f)$ and $V_a^b(f+g)\le V_a^b(f)+V_a^b(g)$.</p>
|
https://math.stackexchange.com/questions/2823531/is-this-function-a-function-of-bounded-variation
|
Question: <p>$f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous and $K\subset\mathbb{R}$ is compact.<br>
$\{Q_i : i\in I\}$ is an open cover for $f(K)$.<br>
Does it follow that $U_i:= f^{-1}(O_i)$ is open for each $i \in I$? Is $\{U_i:i\in I \}$ an open cover for $K$?<br>
Intuitively, the answer to both of these questions seems yes. However, it seems conceivable that this is wrong and I would like to know for sure.</p>
Answer: <p>Given a continuous function, the preimage of an open set is open.</p>
|
https://math.stackexchange.com/questions/2791505/q-i-i-in-i-text-is-open-rightarrow-u-i-f-1o-i-text-is-open
|
Question: <p>So I have a problem that's asking me to compute the intersection of all sets $S(k)$ such that</p>
<p>$$ S(k) = \left(\frac{k-1}{k}, \frac{2k+1}{k}\right], k \ge 1 $$</p>
<p>My answer is $(1,2)$, but the book's answer is $(0,2)$. Maybe I'm just dumb, but I do not understand why the book's answer would be correct.</p>
<p>Any help would be appreciated.</p>
Answer:
|
https://math.stackexchange.com/questions/2824115/trouble-with-intersection-of-infinitely-many-sets
|
Question: <p>I've got a task, where a set A and a map g were given.
The task was to calculate g(A).
And I don't know what to do exactly, so I'd appreciate if s.o. could explain it with an example or give me a link, where it's explained.</p>
<p>For example let A = { x in IR | |x-3|= 2 } and g (x) = (x+2)/x.</p>
<p>So if I want to calculate g (A) shall I look at ( |x-3|+2)/|x-3| and not look at the whole domain but at {1,5} ? </p>
<p>Thank you </p>
Answer: <p>You want to answer the question "what values are mapped to by $g$, from elements in $A$?". So, once you know what $A$ is, in the simplest case you just apply $g$ to the elements of $A$.</p>
<p>As you have already implicitly stated, after simplification we have $A = \{1,5\}$. Therefore, the set we seek is $g(A) = \{g(1),g(5)\}=\left\{3,\frac{7}{5}\right\}$.</p>
<p>Things could have been more complicated, for example if $A$ were not a finite set or if $g$ were a more complicated function: very often in such situations it can be helpful to draw the graph of $g$ (if possible) and mark on the $x$-axis "where $A$ is". In this situation that's not really necessary (but not unhelpful!), since $A$ is finite.</p>
|
https://math.stackexchange.com/questions/1530276/how-do-i-calculate-a-set-in-a-map
|
Question: <p>I am reading a paper on Markov chains and am trying to prove a lemma left to the reader that I need for the Ergodic Theorem for Markov Chains (though the lemma requires no knowledge of any of this). The statement of the lemma is as follows:</p>
<p>Let $\{a_n\}$ be a bounded sequence and suppose for $\{n_k\}$ a sequence of integers with $\lim\limits_{k\to\infty}\frac{n_k}{n_{k+1}}=1$ we have $\lim\limits_{k\to\infty} \frac{a_1+...+a_{n_k}}{n_k}=a$. Then $\lim\limits_{n\to\infty} \frac{a_1+....+a_n}{n}=a$.</p>
<p>We can choose $K$ large enough to guarantee that $\left|\frac{a_1+...+a_{n_K}}{n_K}-a\right|<\epsilon$ and choose $n$ even larger than $n_K$ so that</p>
<p>$\left| \frac{(a_1+...+a_n)}{n}-a\right|\leq\left|\frac{a_1+...+a_{n_K}}{n_K}-a\right|+\left|\frac{a_{n_{K+1}}+...+a_n}{n}-a\right|$.</p>
<p>The first absolute value is then less than $\epsilon$ and the second is less than or equal to</p>
<p>$\frac{(M+a)(n-n_K)}{n}$.</p>
<p>This is the piece that I am having trouble showing is less than $\epsilon$. I would greatly appreciate some help making the last leap.</p>
Answer: <p>Let $C \le a_n \le D$ for all $n$. By using $a_n -C$ if necessary, we assume $a_n \ge 0$ (that is, $C=0$). For each $n\in \mathbb N$, there is $k$ so that $ n_k\le n <n_{k+1}$. Then </p>
<p>$$\begin{split}
\frac{a_1 +\cdots + a_n}{n} &\le \frac{a_1 + \cdots + a_n}{n_k} \\
&\le \frac{a_1+\cdots +a_{n_k}}{n_k} + \frac{a_{n_k +1} + \cdots + a_n}{n_k} \\
&\le \frac{a_1+\cdots +a_{n_k}}{n_k} + D\frac{n_{k+1} - n_k}{n_k}\\
&= \frac{a_1+\cdots +a_{n_k}}{n_k} + D\left(\frac{n_{k+1}}{n_k}-1\right)
\end{split}$$
Thus
$$\limsup_{n\to \infty}\frac{a_1 +\cdots a_n}{n} \le \limsup_{k\to \infty}\frac{a_1+\cdots a_{n_k}}{n_k} + D\left(\frac{n_{k+1}}{n_k}-1\right) = a.$$
On the other hand,
$$\begin{split}
\frac{a_1 + \cdots +a_n}{n} &\ge \frac{a_1 + \cdots + a_n}{n_{k+1}}\\
&\ge \frac{a_1 + \cdots + a_{n_k}}{n_{k+1}}\\
&= \frac{a_1 + \cdots + a_{n_k}}{n_k}\frac{n_k}{n_{k+1}}.
\end{split}$$
Thus
$$\liminf_{n\to \infty} \frac{a_1 + \cdots +a_n}{n} \ge \liminf_{k\to \infty}\frac{a_1 + \cdots + a_{n_k}}{n_k}\frac{n_k}{n_{k+1}} = a.$$</p>
<p>Thus you are done. </p>
|
https://math.stackexchange.com/questions/1521861/for-a-bounded-sequence-a-n-lim-limits-k-to-inftya-1-a-n-k-n-k-a
|
Question: <p><strong>EDIT:</strong> Sorry. I basically was confused by that just valid mathematical transforming could lead into a undefined behavior.
I have to admit, my question has not much to do with the zero of a function, i've just used one method of the "zero calculating tool" set to reshape my function. (i explain that a bit more below)</p>
<p>I have a function $\,f(x)=x^3 + x^2 + 1\,$. I can now reshape $\,1\,$ to $\,x^0\,$:</p>
<p>\begin{align}
f(x)&= x^3 + x^2 + x^0
\end{align}</p>
<p>and factor one x out to
\begin{align}
x\left(x^2 + x + x^{-1}\right)
\end{align}</p>
<p>A product of zero remains zero. Therefore, either
\begin{align}
x=0
\end{align}
or
\begin{align}
x^2 + x + \dfrac{1}{x} = 0
\end{align}</p>
<p>But $\,\dfrac{1}{x}\,$ is not defined, as it could be $0$ and divide through $0$ is not valid. <strong>My confusion</strong> arises primarily because reformatting/reshaping a function normally doesn't change its value. It's simply another way to write it, hence it should be the exact same. But as i showed above, it is not. The beginning function is a normal grade 3 function. But simply transforming can obviously change the whole validity of a function. That's what i don't understand.</p>
<p><strong>My Question:</strong> </p>
<ul>
<li>Where is my mistake?</li>
<li>Is it expected that applying valid mathematical methods to a function correctly can result in not defined things?</li>
</ul>
Answer: <p>The function</p>
<p>$$f(x)= x^3 + x^2 + 1$$ </p>
<p>has domain $x \in \mathbb{R}$ when otherwise unspecified. It is not the same function as</p>
<p>$$g(x)= x \left(x^2 + x + \frac 1x\right) $$</p>
<p>whose domain is considered to be $x \in \mathbb{R}\setminus\{0\}$.</p>
<p>So $f(x) = g(x)$ if $x\ne0$.</p>
<p>If you wish to utilize the form of $f(x)$ where you factor out an $x$ but retain the original mapping, you could employ some extra notation... Perhaps...</p>
<p>$$f(x) = \begin{cases}
x \left(x^2 + x + \frac 1x\right) & x \ne 0 \\
1 & x = 0 \\
\end{cases}$$</p>
<p>It appears that in your post, you've manipulated the function to a case that is valid on only part of the original domain of definition. <em>That is, when you created the $1/x$ term, you made the assumption that $x\ne 0$.</em></p>
<p>To avoid future confusion, be 1) aware of the domain and 2) aware of any assumptions you make during algebraic manipulations. </p>
|
https://math.stackexchange.com/questions/1536458/transforming-x3-x2-1-implies-x-leftx2-x-dfrac1x-right-imp
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Question: <p>Would this even assist math in the way that $i$ did? Or is this just outright pointless and/or too exclusive to call for a definition?</p>
Answer: <p>It would require us to give up many of the usual rules of algebra -- for example the cancellation rule which says that
$$ \frac{a\times b}{a} = b $$
whenever both sides exist.</p>
<p>However, taking $(a,b)$ first to be $(0,1)$ and $(0,2)$ we would get
$$ \frac00 = \frac{0\times 1}{0} = 1 \qquad\text{and}\qquad
\frac00 = \frac{0\times 2}{0} = 2 $$</p>
<p>Since $1$ is manifestly not the same as $2$ (and if we <em>make</em> them the same, all of arithmetic ceases to be useful), your extended system would need to be one that doesn't satisfy the cancellation rule.</p>
<p>A lot of other such rules go the same way, so in the end you may have something you <em>call</em> $\frac00$, but the fraction bar in that expression would no longer denote anything resembling division anyway.</p>
|
https://math.stackexchange.com/questions/1558737/if-i2-1-at-one-point-undefined-then-why-not-define-frac00
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Question: <p>Let $f \colon [a, b] →\mathbb R$ be a continuous function. Then prove that the graph of $f$,
$$\operatorname{Graph}(f) := \{\,(y, x) \in \mathbb R^2\mid y = f(x), x \in [a, b]\,\}$$ is a Jordan region, and it has Jordan content $0$.</p>
<p>So I need to show that there exist an $\epsilon >0$ such that the volume is $<\epsilon$.</p>
Answer: <p>Note that $f$ is <em>uniformly</em> continuous. Thus given $\epsilon>0$, there exists $\delta>0$ such that yadda yadda.
With $n=\lfloor\frac{b-a}{2\delta}\rfloor+1$ we see that
the graph is contained in
$$\bigcup_{k=1}^n[x_k-\delta,x_k+\delta]\times [f(x_k)-\epsilon,f(x_k)+\epsilon] $$
where $x_k=a+\frac {2k-1}{2n}(b-a)$. This shows that the Jordan content is $$\le n\cdot 2\delta\cdot 2\epsilon\le ((b-a)+2\delta)2\epsilon .$$
As we may assume wlog. that $\delta<1$, this expression becaomes arbitrarily small, showing that the Jordan content is indeed $0$.</p>
|
https://math.stackexchange.com/questions/1374570/show-the-graph-is-jordan-region-with-volume-0
|
Question: <p>Is there an infinite closed subset $X$ of the unit circle in $\mathbb C$ such that the squaring map induces a bijection from $X$ to itself?</p>
Answer: <p>Let us think of $S^1$ as $\mathbb{R}/\mathbb{Z}$, so we want an infinite closed subset $X$ on which multiplication by $2$ is a bijection. Suppose you have such an $X$; write $T:X\to X$ for the multiplication by $2$ map. Each element $x\in X$ determines a biinfinite binary expansion $f_x:\mathbb{Z}\to\{0,1\}$, such that $T^n(x)=\sum_{k=1}^\infty f_x(k+n)2^{-k}$ for each $n\in \mathbb{Z}$. Say that a finite string of $0$s and $1$s is <em>admissible</em> if it appears as a sequence of consecutive values of some $f_x$ (i.e., if it appears as a sequence of consecutive digits in the binary expansion of some element of $X$). For each $n$, let $A_n\subseteq\{0,1\}^n$ consist of those sequences $s$ such that both $0^\frown s$ and $1^\frown s$ are admissible (where $^\frown$ is string concatenation). If $A_n$ is empty for some $n$, that means that given a sequence of $n$ consecutive digits in any $f_x$, all of the preceding digits are uniquely determined. By pigeonhole, for each $x$, some sequence of $n$ digits must appear infinitely often in the restriction of $f_x$ to $\mathbb{N}$, and it follows that every $f_x$ is periodic. Furthermore, there is a uniform bound on the periods of all the $f_x$ (because if some particular $s\in\{0,1\}^n$ appears infinitely often in $f_x|_\mathbb{N}$, that determines the period of $f_x$, and there are only finitely many different such $s$). So there are only finitely many different $f_x$, so $X$ is finite. This is a contradiction.</p>
<p>Thus each $A_n$ is nonempty. By König's lemma, it follows that there exists an infinite string $s:\mathbb{N}\to\{0,1\}$ such that every initial segment of $s$ is in the appropriate $A_n$. But then since $X$ is closed, the numbers $y_0$ and $y_1$ whose binary expansions are $0^\frown s$ and $1^\frown s$, respectively, are in $X$. Since $T(y_0)=T(y_1)$, this is a contradiction.</p>
<p>Thus no such $X$ exists.</p>
|
https://math.stackexchange.com/questions/1539215/infinite-closed-subset-of-s1-such-that-the-squaring-map-is-a-bijection
|
Question: <p>Here's an exercise in normed spaces that I can't get my head around.
It reads as follows:
"Let X be a compact space equipped with norm d. If X is countable, then the set of isolated points in X is both open and dense."
Just point me in the right direction.
Thanks a bunch!</p>
Answer: <p>You've already figured out why the set of isolated points $I$ is open.</p>
<p>To show $I$ is dense: Because your space is countable, you can enumerate the <em>non</em>-isolated points $x_1, x_2,$ and so on. Define $U_0 = X$ and for all natural numbers $n$, let $U_{n+1} = U_n \setminus \{x_n\}$. Use induction to show $U_n$ is open and dense for all $n$. Then use the Baire Category theorem. </p>
|
https://math.stackexchange.com/questions/1596301/isolated-points-within-a-compact-space
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Question: <p>Let $S= \{(x,y): y<2x+1\}$.Then prove that $S$ is open in $\mathbb{R}\times\mathbb{R}$.</p>
<p>To prove that we must show that</p>
<blockquote>
<p>for every $(u,v)\in S $ there exists $r>0$ such that the ball $B[(u,v),r]$ is contained in $S$</p>
</blockquote>
<p>Or, equivalently,</p>
<blockquote>
<p>For all $(u,v)$ such that $v<2u+1$ there exists $r>0$ such that:<br>
for all $(x,y)$, $\sqrt{(x-u)^2+(y-v)^2}<r\Longrightarrow y<2x+1$ </p>
</blockquote>
<p>How do we prove that inequality?</p>
Answer: <p>$f(x,y)=2x+1-y$ the set is $f^{-1}(]0,+\infty[)$ and is open since $f$ is continue and the inverse image of an open set by a continuous map is open.</p>
|
https://math.stackexchange.com/questions/1539580/prove-that-a-particular-set-is-open
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Question: <p>How to write this in math notation
$$A=\{2,4,...,2n,...\}
\\B=\{3,6,9,...,3n,...\}$$
what is $A\cup B$ and $A\setminus B$</p>
<p>by inspection, I can see that $A\cup B=\{2,3,4,6,8,9...\}=\{2,4,6...3,6,9...\}$ that is even set or multiple of 3 but I do not know how to write this in math notation
$ A\setminus B=\{2,4,8,10...\}$ ( element in A but not in B)</p>
<p>Can someone please help me to write this in proper way to use math notation of set.</p>
Answer: <p>It depends on what set-builder notation you use, but something like this:
$\begin{align}A & = \{k:\Bbb N^+\mid \exists n\in\Bbb N^+\;k=2n\} & = \{2,4,6,\ldots\}\\ B & = \{k:\Bbb N^+\mid \exists n\in\Bbb N^+\;k=3n\} & = \{3,6,9,\ldots\}\\ A\cap B & = \{k:\Bbb N^+\mid \exists n\in\Bbb N^+\;(k=6n)\} & = \{6,12,18\ldots\}\\ A\cup B & = \{k:\Bbb N^+\mid \exists n\in\Bbb N^+\;(k=2n\vee k=3n)\} & = \{2,3,4,6,8,9,10,\ldots\}\\ A\setminus B & = \{k:\Bbb N^+\mid \exists n\in\Bbb N^+\;(k=2n)\wedge \forall n\in\Bbb N^+\; (k\neq 3n)\} & = \{2,4,8,10,14,\ldots\}\end{align}$</p>
|
https://math.stackexchange.com/questions/1457643/set-operation-for-union-and-intersection
|
Question: <p>$f(x)=x^2\sin^2(1/x)$ for $x\neq0$ and $f(0)=0$ for $x=0$</p>
<p>How to show $f(x)$ is Lipschitz continuous on $[-a,a]$.</p>
<p>I tried to use Mean Value Theorem, but couldn't finalized it.</p>
Answer: <p>If you want to find an $K > 0$ such that $|f'(x)| \leq K$, then $K$ is not that difficult to calculate. In fact, $|f'(x)| = \left|2x\sin^2\left(\frac{1}{x}\right) - \sin\left(\frac{2}{x}\right)\right|\leq 2|x|+1 \leq 2a+1$. So $K = 2a+1$.</p>
|
https://math.stackexchange.com/questions/1598200/how-to-show-fx-x2-sin21-x-is-lipschitz-continuous
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Question: <p>I am trying to prove that: Consider the sequence $a_n = \sup_{x\in S}|f_n(x) - f(x)|$. Then $f_n$ converges to $f$ uniformly if and only if $a_n$ tends to $0$. But I can't prove that if $f_n$ converges to $f$ uniformly, then $a_n$ tends to $0$. I have not been able to go from $|f_n(x) - f(x)|<\epsilon$ to $\sup_{x\in S}|f_n(x) - f(x)|<\epsilon$ using the inequalities in the supposition. How do I do this in detail using the inequalities in the supposition?</p>
Answer: <p>Suppose $f_n$ converges to $f$ uniformly. </p>
<p>Then for all $\epsilon>0$, there exists an $N$ such that for all $x \in S$ and all $n >N$, $|f_n(x) -f(x)|< \frac{\epsilon}{2}$.</p>
<p>Then $|f_n(x) -f(x)|$ is bounded above by $\frac{\epsilon}{2}$ for all $x$.</p>
<p>Suppose $c_n = \sup_\limits{x\in S}\{|f_n(x) -f(x)|\}$. Then $c_n$ is the least upper bound of the set above, so $c_n \leq \frac{\epsilon}{2} < \epsilon$.</p>
<p>So there exists an $N$ such that for all $n>N$, $c_n < \epsilon$.</p>
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https://math.stackexchange.com/questions/1276263/uniform-convergence-and-maximum-of-an-absolute-difference
|
Question: <p>Let $T$ be a nonzero linear functional on a normed space $X$.
Is it then true that on each ball $B(x,r)$, $f$ is not constant?</p>
Answer: <p>Sure, you can consider for the sake of clarity $B(0,r)$. Take $x_1$ and $x_2$ such that $||x_1||,||x_2|| < \frac{r}{4}$ and $T(x_1) \neq 0$ $T(x_2) \neq 0$ (these elements exist because $T$ is nonzero), then $||x_1+x_2||<r$, so $x_1+x_2 \in B(0,r)$.</p>
<p>Now apply $T$ to the new element:</p>
<p>$$T(x_1+x_2) = T(x_1) + T(x_2) \neq T(x_1)$$</p>
<p>So the functional is not constant. The argument holds for any ball of radius r composing with a translation.</p>
|
https://math.stackexchange.com/questions/1564480/is-a-nonzero-linear-functional-is-not-constant-on-each-ball
|
Question: <p>How to find $f(\mathbb{R}^2)$ if $f:\mathbb{R}^2 \to \mathbb{R}^2$</p>
<p>$$f(x,y)=(e^{x+y}+e^{x-y},e^{x+y}-e^{x-y} )$$</p>
Answer:
|
https://math.stackexchange.com/questions/1609597/range-of-a-function-of-two-variable
|
Question: <blockquote>
<p>Prove that there exists a smallest positive number $p$ such that $\cos(p) = 0
$.</p>
</blockquote>
<p>I think I'm supposed to use either Rolle's theorem or the Mean Value Theorem but I'm not sure how, any help would be appreciated. Thanks!</p>
Answer: <p>$\cos$ is a continous function, the set $\{0\}$ is closed, and therefore $\cos^{-1}(\{0\})$ is a closed set.</p>
<p>We conclude that $\cos^{-1}(\{0\})\cap[0,\infty)$ is closed. This set is equal to $\cos^{-1}(\{0\})\cap (0,\infty)$ (because $\cos(0)\neq 0$).</p>
<p>This set is bounded from below and must therefore have a greatest lower bound $\alpha$. Recall that the greatest lower bound of a set is always an adherent point for the set. Since $\cos^{-1}(\{0\})\cap(0,\infty)$ is closed we conclude $\alpha\in \cos^{-1}(\{0\})\cap(0,\infty)$. Therefore the set $\cos^{-1}(\{0\})\cap (0,\infty)$ has a minimum as desired.</p>
|
https://math.stackexchange.com/questions/2155270/analysis-prove-that-there-exists-a-smallest-positive-number-p-such-that-co
|
Question: <p>Let <span class="math-container">$z$</span> be an upper bound for <span class="math-container">$A \subset \Re$</span> such that for all <span class="math-container">$\epsilon > 0$</span> there is an <span class="math-container">$x \in A$</span> such that <span class="math-container">$x > z − \epsilon$</span>. Then <span class="math-container">$z = \sup A$</span>. How to prove this statement?</p>
<p>My attempt is to prove the statement by contradiction. Namely by assuming that <span class="math-container">$z \neq \sup A$</span> and finding that the required properties for <span class="math-container">$z$</span> are no longer met.</p>
<p>I assume <span class="math-container">$z \neq \sup A$</span>, then either <span class="math-container">$z < \sup A$</span> or <span class="math-container">$z > \sup A$</span>. If <span class="math-container">$z < \sup A$</span> then <span class="math-container">$z$</span> is no longer an upper bound, a contradiction. And so <span class="math-container">$z \geq \sup A$</span>. Now let <span class="math-container">$z > \sup A$</span>. Then we can find an <span class="math-container">$\epsilon < z - \sup A$</span> so that <span class="math-container">$z - \epsilon > \sup A$</span> which means there is no <span class="math-container">$x \in A$</span> such that <span class="math-container">$x > z - \epsilon$</span> since <span class="math-container">$z - \epsilon$</span> is greater than <span class="math-container">$\sup A$</span>, a contradiction. And so <span class="math-container">$z \geq \sup A$</span> and <span class="math-container">$z \ngtr \sup A$</span> which means <span class="math-container">$z = \sup A$</span>. Which concludes the proof.</p>
<p>I feel I should add that this is not part of any course I'm taking. Also I found an equivalent question <a href="https://math.stackexchange.com/questions/1003402/an-upper-bound-u-is-the-supremum-of-a-if-and-only-if-for-all-epsilon-0">here</a>, however I only understand the first part of the proof given by OP.</p>
Answer: <p>Let <span class="math-container">$z$</span> be an upper bound for <span class="math-container">$A \subset \Re$</span> such that for all <span class="math-container">$\epsilon > 0$</span> there exists an <span class="math-container">$x \in A$</span> such that <span class="math-container">$x > z - \epsilon$</span>. We prove <span class="math-container">$z = \sup A$</span> by contradiction. Let <span class="math-container">$z \neq \sup A$</span>, then <span class="math-container">$z < \sup A$</span> or <span class="math-container">$z > \sup A$</span>. If <span class="math-container">$z < \sup A$</span> then <span class="math-container">$z$</span> is not an upper bound, a contradiction. Therefore <span class="math-container">$z \geq \sup A$</span>. If <span class="math-container">$z > \sup A$</span> then there exists an <span class="math-container">$\epsilon < z - \sup A$</span> and therefore <span class="math-container">$z - \epsilon > \sup A$</span> meaning there is no <span class="math-container">$x \in A$</span> such that <span class="math-container">$x > z - \epsilon$</span>, a contradiction. And so <span class="math-container">$z \ngtr \sup A$</span>, therefore <span class="math-container">$z = \sup A$</span>.</p>
|
https://math.stackexchange.com/questions/4993999/let-z-be-an-upper-bound-for-a-subset-re-such-that-for-all-epsilon-0-t
|
Question: <p>I am not sure if this is true, but intuitively it seems that if a function is strictly increasing and it is also continuous...it is differentiable. </p>
<p>It may be because there are no bumps like in the absolute value. </p>
Answer: <p>Not necessarily. Counterexample:
$$
f(x)=\begin{cases}
x & \text{if }x<0,\\
2x & \text{if }x\ge 0.\end{cases}
$$
Is continuous, strictly increasing but not differentiable at $x=0$.</p>
|
https://math.stackexchange.com/questions/130457/continuous-and-strictly-increasing-implies-differentiable
|
Question: <p>All right, so my Professor claims that it is impossible to find a vector field that has both div and curl zero and yet vanishes at infinity (that is, becomes the zero vector).</p>
<p>I don't know if it is correct. I decide to trust my professor and check for myself.<br>
So I conclude that,<br> if the vector field is <span class="math-container">$\vec{F} =\vec{F_1} \hat{i}+\vec{F_2} \hat{j}+\vec{F_3} \hat{k} $</span>, then,
<span class="math-container">$$\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}=0$$</span> <span class="math-container">$$\text{and}$$</span>
<span class="math-container">$$\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x},\frac{\partial F_1}{\partial z}=\frac{\partial F_3}{\partial x},\frac{\partial F_2}{\partial z}=\frac{\partial F_3}{\partial y}$$</span></p>
<p>I have no idea how to combine all of these to get the required result. We haven't been taught differential equations yet.</p>
<p>Any ideas and hints are much appreciated.</p>
Answer: <p>First of all, if <span class="math-container">$\mathrm{curl}\;F = 0$</span> then there exists a function <span class="math-container">$U$</span> such that <span class="math-container">$F = \nabla U$</span>. For some reason I can't find a proper reference, but see e.g. <a href="http://persweb.wabash.edu/facstaff/footer/courses/M225/Handouts/VectField-vs-Gradient.pdf" rel="nofollow noreferrer">here</a>. Then <span class="math-container">$\mathrm{div}F = 0$</span> means <span class="math-container">$\Delta U = 0$</span> where
<span class="math-container">$$
\Delta = \mathrm{div}(\mathrm{curl) = }\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}
$$</span>
is the <a href="https://en.wikipedia.org/wiki/Laplace_operator" rel="nofollow noreferrer">Laplace operator</a>. Solutions of <span class="math-container">$\Delta U = 0$</span> are called <a href="https://en.wikipedia.org/wiki/Harmonic_function" rel="nofollow noreferrer">harmonic functions</a>. In particular, the Liouville theorem states that a harmonic function that is bounded below (or above) is necessarily constant, the latter link provides the statement and intuition as well.</p>
<p>If that's not enough by itself, you also get the harmonic functions are real analytic and hence infinitely continuously differentiable. As a result, the order of taking derivatives does not matter, hence any partial derivative of a harmonic function is harmonic itself. In particular, any component of its gradient is a harmonic function. Applying Liouville's theorem to the latter you get non-vanishing gradient, unless it's zero of course.</p>
<p>Note that as far as I recall nothing of the facts above requires general theory of ODEs or PDEs, I think one mostly uses mean-value property of harmonic functions. And integrating zero curls should be also pretty straight forward, essentially you can define <span class="math-container">$U$</span> as a path integral of <span class="math-container">$F$</span> and show that due to curl being zero the integral only depends on the end points, not on the path itself. I guess, it was some time.</p>
|
https://math.stackexchange.com/questions/4428280/divergence-and-curl-both-zero-what-can-we-say-about-behavior-at-infinity
|
Question: <p>Let $f: R \rightarrow R$, $y_0,y_1,y_2 \in R$. We define divided differences:
$$[y_0;f]=f(y_0),$$
$$[y_0,y_1;f]=\frac{f(y_1)-f(y_0)}{y_1-y_0},$$
$$[y_0,y_1,y_2;f]=\frac{[y_1,y_2;f]-[y_0,y_1;f]}{y_2-y_0}.$$</p>
<p>Assume that for each $x \in R$ and $\varepsilon>0$ there exists a $\delta>0$ such that $$|[y_0,y_1,y_2;f]-[y_0',y_1',y_2';f]|< \varepsilon$$ for arbitrary sets $\{y_0,y_1,y_2\}$, $\{y_0',y_1',y_2'\}$ of distinct points such that $|y_i-x|<\delta$, $|y_i'-x|<\delta$ ($i=0,1,2$).</p>
<p>Is it $f$ of class $C^2$ ? </p>
Answer: <p>I'll try to provide a proof of the general case $[y_0,\ldots,y_n;f]$. This proof has many similarities to that of timur, but I'll formulate things more in terms of limits rather than using $\epsilon$s and $\delta$s. My other answer was wrong, but I leave it since it had some useful bits, one of which I'll redo here.</p>
<p>Hope I didn't overlook anything grave this time. However, I'm sure I've taken a few technical short-cuts which could need filling in: the lemma provided at the end is an attempt at that.</p>
<p><strong>Notation:</strong> Let $R^{n+1}_k\subset\mathbb{R}^{n+1}$ consist of the points $y=(y_0,\ldots,y_n)$ for which at most $k$ coefficients have the same value: i.e. $R^{n+1}_1$ are those where all $y_i$ are distinct, while $R^{n+1}_{n+1}=\mathbb{R}^{n+1}$. We then have $[\cdot;f]:R^{n+1}_1\rightarrow\mathbb{R}$ defined as
$$
[y_0,\ldots,y_n;f]=\frac{[y_1,\ldots,y_n;f]-[y_0,\ldots,y_{n-1};f]}{y_n-y_0}
\tag{1}
$$
for $y\in R^{n+1}_{1}$: generalised from the cases $n=0,1,2$ stated in the original problem.</p>
<p><strong>Definition:</strong> We say that a function $f$ is $\mathcal{F}^n$ if for any $y,y'\in R^{n+1}_1$,
$$\lim_{y,y'\rightarrow x^{[n+1]}}\Big|[y;f]-[y';f]\Big|=0\tag{2}$$
where, for $x\in\mathbb{R}$, the point $x^{[n+1]}=(x,\ldots,x)\in\mathbb{R}^{n+1}$. This is just the equivalent of the $\epsilon$-$\delta$ formulation in terms of limits.</p>
<p><strong>(i)</strong> Assume $f$ is $\mathcal{F}^n$. If we take any sequence $y^{(k)}\in R^{n+1}_1$ which converges to $x^{[n+1]}$, then $[y^{(k)};f]$ becomes a Cauchy sequence: i.e. for $p,q\ge k$, $\big|[y^{(p)};f]-[y^{(q)};f]\big|\rightarrow0$ as $n\rightarrow\infty$. Hence, $[f^{(k)};f]$ must converge. Let's denote the limit $[x^{[n+1]};f]=g_n(x)$.</p>
<p><strong>(ii)</strong> Let's prove by induction that $[y_0,\ldots,y_n;f]$ is symmetric in the $y_i$. Obviously, it holds for $n=0,1$. Assume it holds for $n<N$. Setting $n=N\ge2$, let $y=(u,v,A,w)\in R^{n+1}_1$ where $A=(a_1,\ldots,a_{n-2})$ (possibly empty). We then have
$$
\begin{split}
[u,v,A,w;f]=&\frac{[u,v,A;f]-[v,A,w;f]}{u-w}\\
=&\frac{1}{u-w}\Big([u,A,v;f]-[v,A,w;f]\Big)\\
=&\frac{1}{u-w}\left(\frac{[u,A;f]-[A,v;f]}{u-v}-\frac{[v,A;f]-[A,w;f]}{v-w}\right)\\
=&\frac{[u,A;f]}{(u-v)(u-w)}+\frac{[v,A;f]}{(v-u)(v-w)}+\frac{[w,A;f]}{(w-u)(w-v)}
\end{split}
$$
which is symmetric in $u,v,w$. From the first step, we also see that it must be symmetric in permutations of $v,A$. These symmetries generate the whole permutation group, so symmetry also holds for $n=N$.</p>
<p><strong>(iii)</strong> Again, we use induction, assuming the following holds for $n<N$: A function $f$ is $\mathcal{F}^n$ if and only if it is $C^n$, i.e. $\mathcal{F}^n=C^n$: in particular, it is also $\mathcal{F}^k$ for all $k<n$. The definition of $[y;f]$ for $y\in R^{n+1}_1$ may be extended continuously to all $y\in\mathbb{R}^{n+1}$. Moreover, $g_n(x)=f^{(n)}(x)/n!$ where $f^{(n)}$ is the $n$th derivative.</p>
<p>For $n=0,1$, the asumption holds true. So we let $n=N\ge2$.</p>
<p>In order to utilise the induction hypothesis, we first need to show that when $f$ is $\mathcal{F}^n$, it is also $\mathcal{F}^{n-1}$. To do this, we rewrite equation (1)
$$(u-v)[u,A,v;f]=[u,A;f]-[A,v;f]$$
for $A=(a_1,\ldots,a_{n-1})$. For $u=(u_1,\ldots,u_n)$ and $v=(v_1,\ldots,v_n)$, both in $R^n_1$, we then get
$$
\begin{split}
(u_1-v_1)[u_1,\ldots,u_n,v_1;f]&+(u_2-v_2)[u_2,\ldots,u_n,v_1,v_2;f]\\
&+\cdots+(u_n-v_n)[u_n,v_1,\ldots,v_n;f]\\
=&[u_1,\ldots,u_n;f]-[v_1,\ldots,v_n;f]\\
\end{split}\tag{3}
$$
where $u,v\rightarrow x^{[n]}$ leads to
$\big|[u;f]-[v;f]\big|\rightarrow0$,
i.e. the definition of $f$ being $\mathcal{F}^{n-1}$.</p>
<p>For any $y\in R^{n+1}_n$, i.e. not all $y_i$ are equal, we can rearrange the $y_i$ to make $[y;f]=[u,A,v;f]$ where $u\not=v$ and use (1) to get $[y;f]$ expressed in terms of $[u,A;f]$ and $[A,v;f]$, both of which are defined on the whole $\mathbb{R}^n$. This provides us with an extension of $[y;f]$ to all $y\in R^{n+1}_n$.</p>
<p>Finally, in equation (3), we let $u=x^{[n]}$, $v=x'^{[n]}$: by continuity, we can do that, or alternatively we can take limits $u\rightarrow x^{[n]}$ and $v\rightarrow x'^{[n]}$ (and we take these limits <em>before</em> the later limit). Dividing by $x-x'$ on both sides then gives
$$
\frac{g_{n-1}(x)-g_{n-1}(x')}{x-x'}
=[x,\ldots,x,x';f]+\cdots+[x,x',\ldots,x';f]
$$
where $x'\rightarrow x$ gives $g_{n-1}'(x)=n\cdot[g_n(x)]$. Hence, $g_n(x)=f^{(n)}/n!$.</p>
<p>The only remaining thing to point out is that the final extension of $[y;f]$ from $y\in R^{n+1}_n$ to all of $\mathbb{R}^{n+1}$ is continuous. This basically follows from the definition, but we have to use the following lemma (which I have implicitly used further up as well), to ensure the limit in (2) holds true even if the sequence even if $y\in R^{n+1}_n$ rather than $R^{n+1}_1$.</p>
<p><strong>Lemma:</strong> Given $x,v_i,u^{(i)}_j\in\mathbb{R}$ so that $v_i\rightarrow x$ and $u^{(i)}_{j}\rightarrow v_i$, then we can pick $j_i$ so that $u^{(i)}_{j_i}\rightarrow x$.</p>
<p>One use of the lemma is that if $y^{(i)}\in R^{n+1}_n$, $z^{(i,j)}\in R^{n+1}_1$, we can set $v_i=[y^{(i)};f]$, $u^{(i)}_j=[z^{(i,j)};f]$, and replace the limit of $[y^{(i)};f]$, which is taken in $R^{n+1}_n$, with a limit taken in $R^{n+1}_1$.</p>
|
https://math.stackexchange.com/questions/185848/divided-differences-and-differentiability
|
Question: <p>By the set of linear functions I mean the functions of the form <span class="math-container">$$f(x)=\alpha x,$$</span> for <span class="math-container">$\alpha\in\mathbb{R}$</span>. We clearly have for any <span class="math-container">$x,y\in[0,1]$</span> <span class="math-container">$$|f(x)-f(y)|=|\alpha||x-y|.$$</span> So for any <span class="math-container">$\epsilon>0$</span>, <span class="math-container">$\delta=\frac{\epsilon}{|\alpha|}$</span> gives uniform continuity. However this <span class="math-container">$\delta$</span> is different depending on <span class="math-container">$f$</span>, so it cannot give equicontinuity. I am having trouble figuring out how to show that there would be no <span class="math-container">$\delta$</span> that would work.</p>
Answer: <p>Here is an argument.</p>
<p>Suppose that your set of functions <em>is</em> equicontinuous on <span class="math-container">$[0,1]$</span>.
Then, given <span class="math-container">$\epsilon >0,$</span> there exists a <span class="math-container">$\delta >0$</span> such that
<span class="math-container">$|x-y| < \delta$</span> implies <span class="math-container">$|f(x)-f(y)| < \epsilon$</span>.
But if we pick <span class="math-container">$x,y$</span> such that <span class="math-container">$|x-y| = \delta/2$</span>, then</p>
<p><span class="math-container">$$
|f(x)-f(y)| = |a||x-y| = |a|\delta/2.
$$</span></p>
<p>If choose <span class="math-container">$a$</span> large enough, say <span class="math-container">$a=4\epsilon/ \delta$</span>, then <span class="math-container">$|f(x)-f(y)| = 2 \epsilon > \epsilon$</span>, a contradiction.</p>
<p>(Note that our assumption that there exists <span class="math-container">$x,y$</span> such that <span class="math-container">$|x-y| = \delta/2$</span> is harmless.)</p>
|
https://math.stackexchange.com/questions/4438313/is-the-set-of-linear-functions-from-0-1-to-mathbbr-equicontinuous
|
Question: <p>I am investigating a strictly decreasing sequence $(a_i)_{i=0}^\infty$ in $(0, 1)$, with $\lim_{i\to\infty}a_i=0$, such that there exist constants $K>1$ and $m\in\mathbb{N}$ such that $$\frac{a_{i-1}^m}{K} \leq a_i \leq K a_{i-1}^m$$ for all $i$. Even though $K>1$, is it of the right lines to conclude that $a_i \sim \alpha^{m^i}$ for some constant $0<\alpha<1$?</p>
<p>Thanks,
DW</p>
Answer: <p>[Edit: Now that the question has been changed a day later, I removed analysis of the old version of the first inequality. Perhaps sometime I will update to fully answer the new question. The following still applies to the second inequality.]</p>
<p><s>On the other hand, f</s> For the second inequality, $$a_i\leq Ka_{i-1}^m\implies a_i\leq K^{1+m+m^2+\cdots+m^{i-1}}a_0^{m^i}\leq \left(K^{2/m}a_0\right)^{m^i}.$$ You could apply a similar inequality with each $N\in\mathbb N$ in place of $0$ to get $$a_i\leq \left(K^{2/(m^{N+1})}a_N^{1/m^N}\right)^{m^i}.$$ By choosing $N$ such that $K^{2/m}a_N<1$, you at least get $\displaystyle{a_i=O\left(\alpha^{m^{i}}\right)}$ with $\alpha=\left(K^{2/m}a_N\right)^{1/m^N}\in(0,1)$.</p>
|
https://math.stackexchange.com/questions/151811/doubly-exponential-sequence-behaviour-from-inequality
|
Question: <p>Decide whether the following statement is true or false, justify the conclusion.</p>
<p>If $f$ is defined on $\mathbb R$ and $f(K)$ is compact whenever $K$ is compact, then $f$ is continuous on $\mathbb R.$</p>
<p>Does this hold based on the Preservation of Compact Sets: Let $f : A → \mathbb R$ be continuous on $A.$ If $K ⊆ A$ is compact, then $f(K)$ is compact as well. Which means if $f$ is a continuous function on a compact set $K,$ then the range set $f(K)$ is also compact.</p>
Answer: <p>$$
f(x) = \begin{cases} 0 & \text{if } x<0, \\ 1 & \text{if } x\ge0. \end{cases}
$$</p>
<p>This is not continuous, but for all compact sets $K$ the set $\{f(x): x\in K\}$ is compact.</p>
|
https://math.stackexchange.com/questions/1518100/preservation-of-compact-sets
|
Question: <p>For a compactly supported, continuously differentiable function <span class="math-container">$f$</span>, do we have <span class="math-container">$\frac{f(x+h)-f(x)}{h}\to f'(x)$</span> uniformly? (Euclidean space)</p>
<p>I think this is true, since compact supported-ness is pretty strong condition and this is geometrically clear. But I'm not sure how to prove it. Could someone help?</p>
Answer:
|
https://math.stackexchange.com/questions/4455256/for-a-compactly-supported-continuously-differentiable-function-f-do-we-have
|
Question: <p>By a unit circle, I just mean <span class="math-container">$D = S^1$</span> i.e. a unit disk in <span class="math-container">$R^2$</span>.</p>
<p>My game is to use some arbitrary sequence <span class="math-container">$x_n = (a_n,b_n)$</span> such that <span class="math-container">$b_n = \sqrt{1-(a_n)^2}$</span> with <span class="math-container">$0 \leq a_n \leq 1$</span> and show that it converges to a boundary point.</p>
<p>I basically feel like this is the wrong approach as I have little to no experience with analysis, and I would greatly appreciate some help.</p>
<p>I think the way to do this is to show that <span class="math-container">$a_n$</span> converges to some <span class="math-container">$0 \leq a \leq 1$</span> by a properties of limits, since all terms of <span class="math-container">$a_n$</span> are in <span class="math-container">$[0,1]$</span>, so then <span class="math-container">$b_n$</span> should converge to <span class="math-container">$\sqrt{1-a^2}$</span> but I'm unsure whether or not the limit can be passed inside of the square root function since we haven't talked about that at all in class, and I heard somewhere mentioned that could be an issue.</p>
<p>Definitely though, <span class="math-container">$ 0 \leq b_n \leq 1$</span> is true from its definition from an arbitrary term of <span class="math-container">$a_n$</span>, so it too should converge to a number inside <span class="math-container">$[0,1]$</span> so I'm not quite sure how to go about showing that it does.</p>
Answer: <p>Suppose <span class="math-container">$(a,b)$</span> is a limit point. Then there are <span class="math-container">$(a_n,b_n) \in D$</span> such that <span class="math-container">$(a_n,b_n) \to (a,b)$</span>. Since <span class="math-container">$a_n^2+b_n^2 = 1$</span> for all <span class="math-container">$n$</span>, and the function <span class="math-container">$(x,y) \mapsto x^2+y^2$</span> is continuous, we see that <span class="math-container">$a^2+b^2 = 1$</span> and so <span class="math-container">$(a,b) \in D$</span>.</p>
|
https://math.stackexchange.com/questions/4439601/prove-that-a-unit-circle-contains-its-limit-points-that-it-is-closed-under-its
|
Question: <blockquote>
<p>Does the function $f(x)=\sqrt{x}\sin(1/x),x\in(0,1],f(0)=0,$ satisfy the uniform Lipschitz condition $|f(x)-f(y)|<M|x-y|^{1/2},M>0$?</p>
</blockquote>
<p>Any help is appreciated.
Thanks</p>
Answer: <p>no, take $x =\frac{1}{2n\pi +\frac{\pi}{2}}$ and $y =\frac{1}{2n\pi -\frac{\pi}{2}}$. Then $\vert f(x)-f(y)\vert$ behave like $\frac{1}{\sqrt{n}}$ and $\vert x-y \vert^\frac{1}{2}$ behaves like $\frac{1}{n}$.</p>
|
https://math.stackexchange.com/questions/86111/uniform-lipschitz-condition
|
Question: <blockquote>
<p>Suppose <span class="math-container">$z = a + bi$</span>, <span class="math-container">$w = u + iv$</span>, and
<span class="math-container">$$a = \left(\frac{|w| + u}{2}\right)^{1/2}, b = \left(\frac{|w| - u}{2}\right)^{1/2}$$</span>
Prove that <span class="math-container">$z^2 =w$</span> if <span class="math-container">$v \geq 0$</span> and that <span class="math-container">$(\bar{z})^2 = w$</span> if <span class="math-container">$v \leq 0$</span>. <strong>Conclude that every complex number (with one exception!) has two complex square roots.</strong></p>
</blockquote>
<p>I have two problems:</p>
<p>1) I completed a proof but it assumed <span class="math-container">$|w|+u > 0$</span> and <span class="math-container">$|w| - u > 0$</span> so that <span class="math-container">$a,b \in \mathbb{R}$</span>. Should I have been assuming that <span class="math-container">$a,b,u,v \notin \mathbb{R}$</span> was possible? Or is their some kind of proof that <span class="math-container">$a,b \in \mathbb{R}$</span> always? Does it matter?</p>
<p>2) What exactly does proving the first part have to do with concluding that every complex number has two roots? It seems like the real conclusion is "there are two expressions for the square root depending on whether the imaginary part is non-negative or non-positive". I've seen a couple of solution guides (yes I peeked sue me) do the same proof I did (not sure if it's a proof because of objection (1)) and then either 1) immediately conclude the bold statement without justification or 2) conclude that <span class="math-container">$z$</span> and <span class="math-container">$\bar{z}$</span> are the roots of <span class="math-container">$w$</span>. But answer (2) looks wrong, since if <span class="math-container">$v>0$</span>, <span class="math-container">$z$</span> is a square root of <span class="math-container">$w$</span> but <span class="math-container">$\bar{z}$</span> isn't and vice versa for <span class="math-container">$v < 0$</span>. Although the statement is still true since if <span class="math-container">$z$</span> is a square root then so is <span class="math-container">$-z$</span>, but then this just makes the preceding theorem seem irrelevant. Am I missing something?</p>
<p>Thanks.</p>
Answer: <p>1) This is a good point, but you can be sure that <span class="math-container">$|w| + u \ge 0$</span> and <span class="math-container">$|w|-u \ge 0$</span> because</p>
<p><span class="math-container">$(|w|+u)(|w|-u)=|w|^2-u^2=v^2 \ge 0$</span></p>
<p>If <span class="math-container">$v^2=0$</span> then <span class="math-container">$v=0$</span> and <span class="math-container">$|w|=u$</span>. If <span class="math-container">$v^2>0$</span> then <span class="math-container">$|w|+u$</span> and <span class="math-container">$|w|-u$</span> are both either strictly positive or strictly negative. But they cannot both be strictly negative because <span class="math-container">$(|w|+u)+(|w|-u)=2|w|\ge0$</span>.</p>
<p>2) I think there is an implicit assumption that positive square roots are used in the definitions of <span class="math-container">$a$</span> and <span class="math-container">$b$</span>, so <span class="math-container">$a,b \ge 0$</span>. Then we have the following cases:</p>
<ul>
<li>If <span class="math-container">$v > 0$</span> the square roots of <span class="math-container">$w$</span> are <span class="math-container">$z$</span> and <span class="math-container">$-z$</span> in the first and third quadrants</li>
<li>If <span class="math-container">$v < 0$</span> the square roots of <span class="math-container">$w$</span> are <span class="math-container">$\bar{z}$</span> and <span class="math-container">$-\bar{z}$</span> in the second and fourth quadrants</li>
<li>If <span class="math-container">$v=0$</span> then <span class="math-container">$z=\bar{z}=a=\sqrt{u}$</span> and the square roots of <span class="math-container">$w$</span> are <span class="math-container">$\pm \sqrt{u}$</span></li>
<li>The exception occurs when <span class="math-container">$w=0$</span>, when the two square roots coincide</li>
</ul>
|
https://math.stackexchange.com/questions/3499058/rudin-chapter-1-exercise-10-hang-ups
|
Question: <p>Suppose I have a deterministic sequence $\{t_n\}$ that is uniformly distributed on $[0,1]$ (for example $t_n = \{ \pi n \}$, i.e the fractional part of $\pi n$) and a decreasing function $f : \mathbb{R} \rightarrow [0,1]$.</p>
<p>I think it is reasonable to expect that
$\#\{t_k < f(k) : k \leq n \} \approx \sum_{k = 1}^n f(k)$
But I'm not sure how I could demonstrate something like that, nor what the correct error would be.</p>
<p>I know that $\#\{t_i < f(k) : i \leq n \} \approx f(k)n$ but I am unsure how I can translate that into a (rigorous) statement about $\#\{t_k < f(k) : k \leq n \} = \sum [t_k \leq f(k) \text{ and } k \leq n]$ where $[x]$ is Iverson's Bracket which is equal to $1$ if $x$ is true and $0$ if $x$ is false.</p>
<p>Edit: As chandok has pointed out, it is possible to pick a function and sequence where the above does not hold. If we require that $\#\{t_k < f(n) : k \leq n\}$ be unbounded, can we avoid "bad" functions?</p>
<p>Edit 2: The motivation for this question comes from <a href="https://math.stackexchange.com/questions/20518/estimating-a-sum-containing-a-uniformly-distributed-but-deterministic-term">Estimating a sum containing a uniformly distributed but deterministic term</a> .</p>
Answer: <p>The problem with deterministic "random" sequences is that you can easily use them to provide an object (here a decreasing function) against which they won't be "random" at all :</p>
<p>If you pick $f(k) = \inf \{t_m, 1 \leq m \leq k\} /2$, you have that $\#\{t_k < f(k) : k \leq n \} = \#\{t_k = 0 : k \leq n \}$.</p>
<p>You can easily pick "uniformly distributed" sequences where this is $0$, while $\Sigma f(k)$ gets as large as you want (in fact I think it will diverge almost surely if you pick a random uniformly distributed sequence).</p>
|
https://math.stackexchange.com/questions/21156/the-expected-value-of-a-deterministic-sequence
|
Question: <p>E.T. Jaynes writes in "Probability Theory: The Logic of Science" (Chapter 17.7, page 531):</p>
<blockquote>
<p>Let a function y = f(x) have its domain of existence in the unit square <span class="math-container">$0 \leq x, y \leq 1$</span>; we wish to compute the integral
<span class="math-container">$$\theta = \int_0^1 \mathrm{d} x f(x).$$</span>
...
Now let's take our sampling points in a nonrandomized way on a uniform grid:
divide the unit square into <span class="math-container">$\sqrt{n}$</span> steps each way, take one sampling point at each grid, and again count how many (<span class="math-container">$r$</span>) are below [the graph of <span class="math-container">$f(x)$</span>]. The maximum error we can make in each step is</p>
<p><span class="math-container">$$[\text{error in determining }f(x)]\times[\text{width of step}] = \frac{1}{2\sqrt{n}} \times \frac{1}{\sqrt{n}}=\frac{1}{2n}.$$</span>
Therefore the maximum possible error in the integral is
<span class="math-container">$$[\text{number of steps}] \times [\text{maximum error in each step}] = \frac{1}{2\sqrt{n}}.$$</span></p>
</blockquote>
<p>I don't follow this derivation, specifically:</p>
<ul>
<li>error in determening <span class="math-container">$f(x)$</span></li>
</ul>
<p>I checked <a href="http://ksvanhorn.com/bayes/jaynes/" rel="nofollow noreferrer">Unofficial Errata and Commentary for E. T. Jaynes's Probability Theory: The Logic of Science</a> and no mistakes are mentioned there, so I perhaps the derivation is correct. Why the aforementioned error terms?</p>
Answer:
|
https://math.stackexchange.com/questions/1513568/uniform-grid-estimation-for-an-integral-by-e-t-jaynes
|
Question: <p>In Rudin's POMA, the 7.29 theorem says:</p>
<blockquote>
<p>Let <span class="math-container">$\mathcal{B}$</span> be the uniform closure of an algebra <span class="math-container">$\mathcal{A}$</span> of bounded functions. Then <span class="math-container">$\mathcal{B}$</span> is a uniformly closed algebra.</p>
</blockquote>
<p>I understand the part proving that <span class="math-container">$\mathcal{B}$</span> is an algebra but not that which is uniformly closed. Shouldn't we prove that there exist <span class="math-container">$(f_n) \in \mathcal{B}$</span> s.t. <span class="math-container">$f_n \rightarrow f$</span> uniformly ?</p>
<blockquote>
<p>Proof:</p>
<p>If <span class="math-container">$f, g \in \mathcal{B}$</span> there exist uniformly convergent sequences <span class="math-container">$(f_n), (g_n)$</span> s.t. <span class="math-container">$f_n \rightarrow f$</span>, <span class="math-container">$g_n \rightarrow g$</span>, <span class="math-container">$f_n, g_n \in \mathcal{A}$</span>. Since we are dealing with bounded functions, it is easy to show that:</p>
<ul>
<li><span class="math-container">$f_n + g_n \rightarrow f+g$</span></li>
<li><span class="math-container">$f_n g_n \rightarrow fg$</span></li>
<li><span class="math-container">$c f_n \rightarrow cf$</span></li>
</ul>
<p>where <span class="math-container">$c$</span> is any constant, the convergence being uniform in each case. Hence <span class="math-container">$f+g \in \mathcal{B}, fg \in \mathcal{B}, cf \in \mathcal{B}$</span>, so that <span class="math-container">$\mathcal{B}$</span> is an algebra.</p>
<p>By theorem 2.27, <span class="math-container">$\mathcal{B}$</span> is uniformly closed.</p>
<p>Theorem 2.27:
If <span class="math-container">$X$</span> is a metric space, and <span class="math-container">$E \subset X$</span> then:</p>
<ul>
<li><span class="math-container">$\bar{E}$</span> is closed</li>
<li><span class="math-container">$E = \bar{E}$</span> iif E is closed</li>
<li><span class="math-container">$\bar{E} \subset F$</span> for every closed set <span class="math-container">$F \subset X$</span> s.t. <span class="math-container">$E \subset F$</span></li>
</ul>
</blockquote>
Answer:
|
https://math.stackexchange.com/questions/1466885/rudins-theorem-7-29
|
Question: <p>How do I show with the definition of continuity that <span class="math-container">$f : \mathbb{R} \to\mathbb{ R}$</span> defined by <span class="math-container">$f(x)= x$</span> for all <span class="math-container">$x\in\mathbb{R}$</span> is continuous?</p>
<p>I understand that this function is continuous, however I am having a hard time proving it. Especially because we have been using converging sequences to prove the continuity of functions and I am not sure if the book wants met to do that here as well.</p>
<p>If so I think it would be something like this:</p>
<blockquote>
<p>Pick an <span class="math-container">$(x_k)_{k∈\mathbb{N}}$</span> in <span class="math-container">$A$</span> that converges to <span class="math-container">$a$</span>.</p>
<p>Pick an <span class="math-container">$ε > 0$</span> , then we can find an <span class="math-container">$δ > 0$</span> such that <span class="math-container">$||f(x)-f(a)|| < ε$</span> for all <span class="math-container">$x ∈ A$</span> with <span class="math-container">$||x-a|| < δ$</span>.</p>
<p>Because <span class="math-container">$(x_k))_{k∈\mathbb{N}}$</span> converges to <span class="math-container">$a$</span>. We can find an <span class="math-container">$k_0 ∈ \mathbb{N}$</span> such that <span class="math-container">$||x_k-a|| < δ$</span> for all <span class="math-container">$k\ge k_0$</span>. Pick a <span class="math-container">$k\ge k_0$</span> then <span class="math-container">$||f(x_k)-f(a)|| < ε$</span> and thus <span class="math-container">$(f(x_k))_{k ∈ \mathbb{N}}$</span> converges to <span class="math-container">$f(a)$</span>.</p>
</blockquote>
<p>Would this be correct. Maybe someone could show me how to prove this without using converging sequences.</p>
Answer: <p>Let $f$ be the identity function $\mathbb{R} \to \mathbb{R}$. Here are six weird reasons why $f$ is continuous (number 2 will make you cry, then number 3 will restore your faith in humanity!)</p>
<ul>
<li><p>The identity trivially satisfies that the preimage of any open set is open.</p></li>
<li><p>For every $\epsilon$, pick $\delta = \epsilon$. Then $|x-y| < \delta$ implies $|f(x) - f(y)| < \epsilon$. So the identity is in fact <em>uniformly</em> continuous.</p></li>
<li><p>$f$ is differentiable: its derivative is $\lim_{h \to 0} \frac{x+h-x}{h} = 1$. Therefore it is continuous.</p></li>
<li><p>If $(x_n) \to x$, then $(f(x_n)) = (x_n) \to x = f(x)$, so $f$ preserves the convergence of sequences.</p></li>
<li><p>Consider the natural hyperreal extension which I will write $\bar{f}$ because my MathJax-foo is not good enough and I don't have access to \prescript. Then if $dx$ is infinitesimal, we have $\bar{f}(x+dx) - \bar{f}(x) = x + dx - x = dx$, which is infinitesimal.</p></li>
<li><p>$f$ is <a href="https://math.stackexchange.com/questions/821754/monotonic-surjective-function">monotone and surjective</a>. (OK, I don't know of a way of proving this that doesn't basically boil down to the epsilon-delta of method 2. But it's nice to know.)</p></li>
</ul>
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https://math.stackexchange.com/questions/2062471/show-that-the-real-valued-identity-function-is-continuous
|
Question: <blockquote>
<p><strong>Problem:</strong></p>
<p><span class="math-container">$$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$</span></p>
</blockquote>
<p><strong>My proof:</strong></p>
<p>Let <span class="math-container">$x \in A \cap (B \cup C)$</span>. Then <span class="math-container">$x \in A$</span> and <span class="math-container">$x \in (B \cup C)$</span>.</p>
<p><span class="math-container">$x \in (B \cup C)$</span> implies <span class="math-container">$x\in B$</span> or <span class="math-container">$x \in C$</span>.</p>
<p>If <span class="math-container">$x \in B$</span>, then <span class="math-container">$x \in (A \cap B)$</span>.</p>
<p>If <span class="math-container">$x \in C$</span>, then <span class="math-container">$x \in (A \cap C)$</span>.</p>
<p>Therefore <span class="math-container">$x\in (A \cap B) \cup (A \cap B)$</span>.</p>
<p><strong>EDIT:</strong> Continuing Problem</p>
<p>Let <span class="math-container">$x\in (A \cap B) \cup (A \cap C)$</span></p>
<p>Then <span class="math-container">$x\in (A \cap B)$</span> or <span class="math-container">$x\in(A \cap C)$</span></p>
<p>If <span class="math-container">$x\in (A \cap B)$</span>, then <span class="math-container">$x\in A$</span> and <span class="math-container">$x\in B$</span></p>
<p>If <span class="math-container">$x\in (A \cap C)$</span>, then <span class="math-container">$x\in A$</span> and <span class="math-container">$x\in C$</span></p>
<p>Either way, <span class="math-container">$x\in A$</span> and <span class="math-container">$x\in B$</span> or <span class="math-container">$x\in C$</span>.</p>
<p>Therefore <span class="math-container">$x\in A \cap (B \cup C)$</span></p>
<p>Planning on taking an analysis class next semester. Thought it would be best to start learning it now because I've heard that it'll be a difficult class any tips and tricks would be appreciated.</p>
Answer: <p>Your proof is good.</p>
<p>As for analysis tips: Here's a quote from my first analysis lecturer.</p>
<blockquote>
<p>"All I've ever done in my entire career is add zero, multiply by one, and use the triangle inequality."</p>
</blockquote>
<p>What he means is - when you're faced with something that looks a bit tricky, these are the first things you should try. Often there will be some fraction $\frac{a}{a}$ that you can multiply your expression by, and it makes everything a lot simpler to work with. Similarly, adding $(a-a)$ to an expression (e.g. the usual proof of the product rule) can simplify things greatly.</p>
<p>Personally what most people seem to struggle with is the entirely different approach in terms of intuition (compared to algebra). In algebra, to show that two objects are equal, you subtract them and show that the result is $0$. In analysis, to show that two objects are equal, you subtract them and show that the difference is smaller than any arbitrary real number. Many people find this sort of reasoning "fake", and say that they don't really believe it. The real reason they say that is because they don't have an intuition for it. For me the nice part is that I find it all very intuitive, because everything in analysis can be drawn. Whenever you're about to start a proof, try drawing a diagram of what you're proving and it will make a lot more sense.</p>
<p>tl;dr For analysis, add $0$, multiply by $1$, and draw pictures. Your set equality proof is fine.</p>
<p><strong>EDIT</strong></p>
<p>I liked your proof because you used words. People say "a picture is worth a thousand words", but my personal belief is "a word is worth a thousand maths symbols". This becomes especially true later on when you start to use higher level objects (in the sense that they have "big" definitions). A lot of people in my first analysis class struggled with using words because they felt that it wasn't as formal. This is entirely not true. Anywhere in maths words have precise definitions. (If they don't, you just define them before using them.) Then a proof with words ends up a lot more readable and cleaner than any proof trying to just use symbols. You seem to be fine with this already so keep it up!</p>
|
https://math.stackexchange.com/questions/2256835/basic-set-theory-proof-verification
|
Question: <blockquote>
<p>Let the Minkowski-metric on <span class="math-container">$\mathbb{R}^3$</span> be (for a p<span class="math-container">$\in \mathbb{R}^3, v,w\in T_p\mathbb{R}^3$</span>) <span class="math-container">$<v,w>_p^{Mink}=-v_1w_1+v_2w_2+v_3w_3$</span>.</p>
<p>Show that:</p>
<p>i) <span class="math-container">$H^2:=$</span>{<span class="math-container">$x\in\mathbb{R}^3|-x_1^2+x_2^2+x_3^2=-1,x_1>0$</span>} is a 2-dimensional under manifold of <span class="math-container">$\mathbb{R}^3$</span> and the Minkowski-metric induces a riemannian metric on <span class="math-container">$H^2$</span>.</p>
<p>ii) The stereographic projection with pole (-1,0,0), given as <span class="math-container">$(x_1,x_2,x_3)^T \rightarrow \frac{1}{1+x_1}(x_2,x_3)^T$</span>, is a isometry between <span class="math-container">$H^2$</span> and <span class="math-container">$B:=$</span>{<span class="math-container">$x\in\mathbb{R}^2|x_1^2+x_2^2<1$</span>} with the metric <span class="math-container">$g_{ij}^x=\frac{4}{(1-x_1^2-x_2^2)^2}\delta_{ij}$</span></p>
</blockquote>
<p>For i) I used the the regular value theorem to show that <span class="math-container">$H^2$</span> is an under manifold. But I don't know how the Minkowski-metric induces a riemannian metric. I already showed that it is symmetrical and a biliear form.
Could you give me a hint on how to show the positive definiteness, please?</p>
<p>For ii) I showed that the projection is surjektive and continuous. I don't know how to preceed here, because I don't know the inverse of the projection nor how to find it.</p>
<p>Thanks for any kind of help.</p>
Answer:
|
https://math.stackexchange.com/questions/2286962/minkowski-metric-induct-riamannian-metric
|
Question: <blockquote>
<p>Let <span class="math-container">$f: [0,1] \rightarrow \mathbb{R} $</span> a continuous function and <span class="math-container">$f > 0$</span>. Let <span class="math-container">$F(x) = \int_0^x f(t)dt $</span> and let <span class="math-container">$F(1)=A$</span>.</p>
<p><strong>1</strong>. Show that <span class="math-container">$F$</span> is bijective function from <span class="math-container">$ [0,1]$</span> to <span class="math-container">$[0,A]$</span>.</p>
<p><strong>2</strong>. Show that for all <span class="math-container">$n \in \mathbb{N}^* $</span>, there is a unique subdivision: <span class="math-container">$x_0 = 0 < x_1 < ... < x_{x-1} < x_n = 1 $</span>, such that for <span class="math-container">$ k=0,1,..., n-1 $</span>, we have: <span class="math-container">$\int_{x_{k}}^{x_{k+1}} f(t)dt = \frac{A}{n}$</span>.</p>
<p><strong>3</strong>. Show that <span class="math-container">$F^{-1} (\frac{kA}{n}) = x_{k+1}$</span> and <span class="math-container">$\lim_{n \rightarrow + \infty} \frac{1}{n} \sum_{k=1}^{n} f(x_{k}) = \frac{1}{A} \int_{0}^{1}(f(x))^2 dx. $</span></p>
</blockquote>
<p><strong>My work.</strong></p>
<p><strong>1</strong>. To show <span class="math-container">$F$</span> is bijective, I need to show it is injective and surjective, which I don't know how to do for this function.</p>
<p><strong>2</strong>. <span class="math-container">$ A = \int_{0}^1f(x)dx = \sum_{k=0}^{n-1} \int_{x_{k}}^{x_{k+1}} f(x)dx $</span>.</p>
<p>I don't see how to proceed to get <span class="math-container">$\int_{x_{k}}^{x_{k+1}} f(t)dt = \frac{A}{n}$</span>.</p>
<p><strong>3</strong>. <span class="math-container">$F$</span> is bijective, so it has an inverse function.</p>
<p>From question <strong>2</strong>, <span class="math-container">$F^{-1}(\frac{kA}{n}) = F^{-1}\big(k \int_{x_{k}}^{x_{k+1}} f(t)dt \big)$</span> I don't see how to get <span class="math-container">$x_{k+1}$</span> from that.</p>
<p>For the limit, I tried to use Riemann sums but I got stuck too. Thank you for your help.</p>
Answer: <p>For (1), note that $F$ is continuous and moreover $F'(x)=f(x)>0$ so $F$ is also strictly monotone. Hence $F$ is injective and by the <a href="https://en.wikipedia.org/wiki/Intermediate_value_theorem" rel="nofollow noreferrer">Intermediate Value Theorem</a> $F$ attains all the values between $F(0)=0$ and $F(1)=A$.</p>
<p>As regards (2), by the bijectivity for any $0\leq k\leq n$ there is a unique $x_k\in [0,1]$ such that $F(x_k)=\frac{kA}{n}$.
Hence
$$\int_{x_{k}}^{x_{k+1}} f(t)dt =F(x_{k+1})-F(x_k)=\frac{(k+1)A}{n}-\frac{kA}{n}= \frac{A}{n}.$$</p>
<p>Finally for (3), we have use Riemann sums,
$$\lim_{n \rightarrow + \infty} \frac{1}{n} \sum_{k=1}^{n} f(x_{k}) =\lim_{n \rightarrow + \infty} \frac{1}{n} \sum_{k=1}^{n} f(F^{-1}(kA/n))\\=\frac{1}{A} \int_{0}^{A}(f(F^{-1}(t))) dt
=\frac{1}{A} \int_{0}^{1}(f(x))^2 dx.$$
where $t=F(x)$, $dt=f(x) dx$.</p>
|
https://math.stackexchange.com/questions/2356269/show-that-f-int-0x-ftdt-is-bijective-function-from-0-1-to-0-a
|
Question: <p>I've encountered one tiny problem - I have to find tangent plane equation of the function <span class="math-container">$\space f(x,y)=y^3 - \sqrt{1-x^2y^2} \space$</span> in the point where it intersects with <span class="math-container">$\space y$</span>-axis. So <span class="math-container">$\space f(0,y) = y^3... \space$</span> and that's not exactly one point, how am I supposed to find an equation with this? Is it OK if I fix an <span class="math-container">$\space y_0 \space$</span> and then write the equation?</p>
Answer: <p><strong>Hints:</strong> Function <span class="math-container">$f(x,y)$</span> will have the intersection point with <span class="math-container">$y$</span>-axis in a point <span class="math-container">$P=(0, y_0, 0)$</span>. In your case it will be the point <span class="math-container">$P=(0, 1, 0)$</span>. Find now the tangent plane...</p>
<p><strong>Figure:</strong></p>
<p><a href="https://i.sstatic.net/fQ43k.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/fQ43k.png" alt="enter image description here" /></a></p>
|
https://math.stackexchange.com/questions/3725351/tangent-plane-equation-in-the-point-of-intersection-with-space-y-axis
|
Question: <p>Prove that there exist infinitely many positive real numbers <span class="math-container">$r$</span> such that the equation
<span class="math-container">$2^x +3^y + 5^z = r$</span> has no solution <span class="math-container">$(x, y, z) \subseteq \mathbb{Q} \times \mathbb{Q} \times \mathbb{Q}$</span>.</p>
<p>First, I prove that set <span class="math-container">$S = \{\, 2^x +3^y + 5^z: (x, y, z) \subseteq \mathbb{Q} \times \mathbb{Q} \times \mathbb{Q} \,\}$</span> is countable; is it correct? Then what is the next step to the proof?</p>
Answer: <p>Any number not in your set $S$ is a real number $r$ that you're looking for. Can you establish that there are infinitely many real numbers not in $S$, once you know that $S$ is countable?</p>
<p>In fact, every number in your set $S$ is algebraic, and the algebraic numbers are a countable subset of $\mathbb{R}$. Any transcendental number would work as $r$, so if you produce infinitely many of those, you're done.</p>
|
https://math.stackexchange.com/questions/704990/prove-that-the-equation-has-no-solution
|
Question: <p>I read a question stating that if $z$ is complex, then $|z|\leq 1$ is a closed set. I think this is just saying that the unit disk is a closed set. Why is that so?</p>
Answer: <p>Look at the complement of the set $S = \{ z \in \mathbb{C}: |z| \leq 1 \}$. The complement is given by $S^c = \{ z \in \mathbb{C}: |z| > 1 \}$. Consider a point $z_0 \in S^c$. $|z_0| > 1$ and hence let $|z_0| = 1 + r$ where $r > 0$. Consider the ball of radius $r$ centered at $z_0$ i.e. $B_r(z_0) = \{v \in \mathbb{C}:|v-z_0| < r \}$. Clearly, $B_r(z_0) \subset S^c$. This follows from triangle inequality since $$|z_0| = |z_0 - v + v| \leq |z_0 - v| + |v| \implies |v| \geq |z_0| - |z_0 - v| > (1+r) - r = 1$$
Hence, $|v|>1 \implies v \in S^c \implies B_r(z_0) \subset S^c$. Hence, $S^c$ is open since given any point $z_0 \in S^c$ we can find a open neighborhood lying completely inside $S^c$ and hence $S$ is closed.</p>
<p>Equivalently, you can try to prove that the set $S = \{ z \in \mathbb{C}: |z| \leq 1 \}$ contains all its accumulation points. The proof of this is again not hard. Look at a subsequence converging to an accumulation point and prove that if you have $|z_n| \leq 1$, then $\displaystyle \left| \lim_{n \rightarrow \infty} z_n \right| \leq 1$. (Hint: If not, what will happen?)</p>
|
https://math.stackexchange.com/questions/57169/why-is-the-unit-disk-closed
|
Question: <p>Consider <span class="math-container">$G\in C^{1}(\mathbb{R})$</span> such that <span class="math-container">$G(0)=0$</span>. Then there exists a constant <span class="math-container">$C$</span> such that <span class="math-container">$|G(s)|\leq C|s|$</span> for all <span class="math-container">$s\in [-M,M]$</span>, where <span class="math-container">$M<\infty$</span>.</p>
<p>This statement is part of a proof in Brezis. It's clear by the extreme value theorem that we can definitely pick a uniform upper bound for <span class="math-container">$G$</span> on a closed interval. However, how would you go about picking the <span class="math-container">$C$</span> above pointwise?</p>
Answer: <p>Let <span class="math-container">$C$</span> be an upper bound for <span class="math-container">$G'$</span> on <span class="math-container">$[-M,M]$</span>, which is possible because <span class="math-container">$G'$</span> is continuous by hypothesis. By the mean value theorem, we have for every <span class="math-container">$s\in [-M,M]$</span> <span class="math-container">$$|G(s) - G(0)|=|G(s)| \leq C|s-0| = C|s| $$</span></p>
|
https://math.stackexchange.com/questions/3250395/pointwise-upper-bound
|
Question: <p>Let $f : X \to Y$ and $g: Y \to Z$ be two functions, such that $(g \circ f) : X \to Z$ is bijective</p>
<p>Prove or refute that $f$ is surjective.</p>
<p>$\\$ </p>
<p>For $f$ to be surjective, $\forall b \in Y, \exists a \in X : f(a) = b$</p>
<p>Because $(g \circ f)$ is bijective, the sets X and Z have to have the same number of elements, otherwise the function composition wouldn’t be bijective. </p>
<p>The set Y can have the same number of elements than X and Z or more. It can’t have less, because given that $g \circ f $ is bijective, in other words that X and Z have the same number of elements, the elements in X can map to a smaller number of elements in Y, but the elements in Y can’t map to more elements in Z than there are in Y.</p>
<p>So the function $f$ isn’t necessarily surjective. It could be, but it isn’s necessarily the case.</p>
<p>Are my thoughts correct or not ? If not, why ?? Also, if I am right, how could I write a mathematical proof ?</p>
<p>Thanks for your help !</p>
<p>(P.S. : English isn't my mother tongue, sorry if I made a grammar mistake)</p>
Answer: <p>Your thoughts are not bad at all, but it seems that you think of $X$, $Y$ and $Z$ as finite sets. This is enough to give a counterexample. You realized that $Y$ has to have at least the same "number" of elements as $X$. But what if $Y$ has more elements? Can $f$ be surjective?<p></p>
<p>In general, you have to be careful with </p>
<blockquote>
<p>Because $(g\circ f)$ is bijective, the sets $X$ and $Z$ have to have the same number of elements, otherwise the function composition wouldn’t be bijective. </p>
</blockquote>
<p>Define $A=[0,\infty)$ and $B=[1,\infty)$ and
$$
h:A\to B,~h(a)=a+1.
$$
Then $h$ is bijective although $B\subsetneq A$.</p>
|
https://math.stackexchange.com/questions/2348789/prove-or-refute-that-f-is-surjective-are-my-thoughts-correct
|
Question: <p>I'm trying to solve an exercise which its conclusion seems to be the title of this post. The exercise is:</p>
<blockquote>
<ol>
<li>Show that the function $h:\Bbb R\to [0,1[$ given by
$$h(t)=\begin{cases}
e^{-1/t^2} &\text{if } t\neq 0\\
0 &\text{otherwise}
\end{cases}$$
is $C^\infty$.</li>
<li>Show that the functions
$$h_+(t)=\begin{cases}
e^{-1/t^2} &\text{if } t\gt 0\\
0 &\text{otherwise}
\end{cases}\quad\text{and}\quad
h_{-}(t)=\begin{cases}
e^{-1/t^2} &\text{if } t\lt 0\\
0 &\text{otherwise}
\end{cases}$$
are $C^\infty$.</li>
<li>Show that the function $k:\Bbb R\to [0,1[$ given by $k(t)=h_-(t-b)h_+(t-a)$ is $C^\infty$ and positive for $t\in ]a,b[$.</li>
<li>Let $R$ the rentangle $]a_1,b_1[\times\cdots\times]a_n,b_n[$. Show that there is a $C^\infty$ function $g:\Bbb R^n\to [0,1[$ strictly positive on $R$.</li>
<li>Conclude that if $K$ is a compact subset of $\Bbb R^n$ and $U$ is an open neighborhood of $K$, there is a $C^\infty$ function $f:\Bbb R^n\to [0,1]$ such that $f_{|K}\equiv 1$ and its support is contained in $U$.</li>
</ol>
</blockquote>
<p>From 1.-4. I can prove that for any open and bounded set $O\subset \Bbb R^n$, there is a $C^\infty$ function with its support contained in $O$. So my first attempt was apply this to the open $U\setminus K$. Then I get a $C^\infty$ function $f$ that is $0$ (in particular) over $K$. If I just consider $\chi_K+f$ that function can fail to be $C^\infty$.</p>
<p>In a discussion on the chat, robjohn suggest <a href="http://chat.stackexchange.com/transcript/message/6200007#6200007">this</a>. It works fine, but then my question is:</p>
<blockquote>
<p>Can 5. be proved by using 1.-4.? If yes, how?</p>
</blockquote>
Answer: <p>Since $K$ is compact and $U^C$ is closed, there is a positive distance, $\Delta$, from $K$ to $U^C$.</p>
<p>We can cover each point $k\in K$ with an open cube $Q_k(\Delta/\sqrt{n})$ centered at $k$ and side $\Delta/\sqrt{n}$. Note that the entire cube is within $\frac\Delta2$ of $k$. Since $K$ is compact, choose a finite subcover of these cubes $\{Q_{k_j}(\Delta/\sqrt{n}):1\le j\le N\}$. For each cube in this subcover, define the function $f_j$ mentioned in step 4 above on $Q_{k_j}(2\Delta/\sqrt{n})$ but divide it by its minimum on $\overline{Q}_{k_j}(\Delta/\sqrt{n})$. Thus, $f_j$ is supported in $Q_{k_j}(2\Delta/\sqrt{n})$ and is $\ge1$ on $Q_{k_j}(\Delta/\sqrt{n})$.</p>
<p>$\sum\limits_{j=1}^Nf_j$ is supported in $U$ and is $\ge1$ on $K$. Then $\phi\circ\sum\limits_{j=1}^Nf_j$ is supported in $U$ and is $1$ on $K$, where</p>
<p>$$\phi(x)=\frac{h_+(x)}{h_+(x)+h_+(1-x)}$$</p>
<p>Note that $\phi\in C^\infty$, and $\phi(x)=0$ for $x\le0$ and $\phi(x)=1$ for $x\ge1$.</p>
|
https://math.stackexchange.com/questions/198748/c-infty-version-of-urysohn-lemma-in-bbb-rn
|
Question: <p>We have: </p>
<p>$|a|<1<|b| \iff |a|+|b| < 1 + |b| < 2|b|$</p>
<p>$\iff (|a|+|b|)² < (1+|b|)² < 4|b|² $</p>
<p>$ \iff|a|² + 2|a||b| + |b|² < 1 + 2|b| + |b|² < 4|b|²$</p>
<p>$ \iff 2|a||b| < 1 + 2|b| - |a|² < 3|b|² - |a|² $</p>
<p>I don't see how to proceed to get $ab \in ]-3,1[$</p>
<p>Thank you. </p>
Answer: <p>We consider the function $f(a) = a(2 - a) = ab $</p>
<p>$\forall a \in (-1,1): f'(a) = 2(1 - a) > 0 $</p>
<p>$ f(-1) < f(a) < f(1) \iff -3 < ab < 1$ </p>
|
https://math.stackexchange.com/questions/2243342/a-b-in-mathbbr2-a-b-2-and-ab-show-that-1-in-a-b
|
Question: <p>If $a < b$ then
$$((f \wedge b) -a)^-=(f-a)^-$$
where $f$ is a real valued function. I can't figure out why this holds. I know that $f^-=-(f\wedge 0)$ so I tried showing this from definitions. I would greatly appreciate any help.</p>
Answer: <p>It's easy checking cases…</p>
<p>For $f \ge b > a$ it holds: $$((f \wedge b) -a)^- = (b - a)^- = 0 = (f - a)^-$$</p>
<p>For $f < b$ it's $$((f \wedge b) -a)^- = (f - a)^-$$</p>
<p>Nothing wild…</p>
|
https://math.stackexchange.com/questions/2124370/if-a-b-then-f-wedge-b-a-f-a
|
Question: <p>Let $p\in\left[ 1,\infty\right) $ and $\left(
X_{0},\left\Vert \cdot\right\Vert _{p}\right) $ a normed linear space, where
$$
X_{0}=\left\{ \left( x_{n}\right) \in
\mathbb{R}
^{
\mathbb{N}
}\mid\exists n_{0}\in%
\mathbb{N}
,\forall n\in
\mathbb{N}
:n_{0}\geq n\Rightarrow x_{n}=0\right\} .
$$</p>
<p>Let the subspace
$$
A=\left\{ \left( x_{n}\right) \in X_{0}\mid\forall n\in%
\mathbb{N}
:\left\vert x_{n}\right\vert \leq\frac{1}{n}\right\} .
$$
Show that $A$ is closed, absorbing and find the set $\operatorname{int}A$.</p>
Answer:
|
https://math.stackexchange.com/questions/2227604/properties-of-a-subspace-of-sequence
|
Question: <p>An exercise from my book is as follows:</p>
<p>Assume that $x > 0$ for $x$ in $\mathbb{R}$ (Real numbers) then there is an $y$ in $\mathbb{N}$ (Natural Numbers) such that $1/y^3 < x$.</p>
<p>By the archimedean property, there exists an y in N such that $1/y < x$. How exactly would I continue on from here?</p>
Answer: <p>By Archimedean property there exists a natural number $y$ such that $xy >1$. Then $xy^3 >xy >1$. So $1/(y^3) <x$. </p>
|
https://math.stackexchange.com/questions/2125419/proof-using-archimedean-property
|
Question: <p>I wonder if a matrix of second derivatives is positive definite and symmetric and the necessary condition of existance extremum is satisfy then exist exactly one extremum? </p>
Answer: <p>Maybe if this log likelihood function is a concave function
($\frac{\partial^2 ln L(\theta)}{\partial \theta_i \partial \theta_j}= -n \frac{\partial^2 C(\theta)}{\partial \theta_i \partial \theta_j}$ with assumption is negative) and has a maximum then it must be absolute maximum?</p>
|
https://math.stackexchange.com/questions/2245651/existance-of-global-extremum
|
Question: <p>how can I get $x$ from equation</p>
<p>$ e^{(\frac{-a}{x})}\frac{x}{a} - \frac{x}{a} = b - 1 $</p>
<p>I tried to play with logaritms but I had no luck. Can I even get $a$ from this equation?</p>
Answer: <p>The equation in unsolvable for $x$, the only way is a numerical approach</p>
|
https://math.stackexchange.com/questions/2168267/get-x-from-exponential-aquation
|
Question: <p>Let a$_n$ := $\frac {(n-1)(-1)^n}{n}$. Find lim sup a$_n$ and lim inf a$_n$.</p>
<p>I see that a$_n$ can either be $\frac {n-1}{n}$ when n is even or $\frac {-(n-1)}{n}$ when n is odd.</p>
<p>$\frac {n-1}{n}$ a$_2$ = $\frac {1}{2}$, a$_3$ = $\frac {2}{3}$, a$_4$ = $\frac {3}{4}$.</p>
<p>$\frac {-(n-1)}{n}$ a$_2$ = $\frac {-1}{2}$, a$_3$ = $\frac {-2}{3}$, a$_4$ = $\frac {-3}{4}$.</p>
<p>Clearly, the trend for is $\frac {(n-1)}{n}$ is increasing, $\frac {-(n-1)}{n}$ is decreasing. I feel like I have to use the monotone convergence theorem here, but I'm unsure how to continue. Also, would there be much of a difference if it was decreasing instead of increasing?</p>
<p>Edit: Added $\frac {-(n-1)}{n}$,</p>
<p>So to find lim sup, I would take a look at the $\frac {(n-1)}{n}$ and lim inf I would take a look at $\frac {-(n-1)}{n}$ right? Since sup is the least upper bound, so one must look at the increasing portion and inf is the greatest lower bound, so one must look at the decreasing portion? Could I get an example of how to do this?</p>
Answer: <p>$a_{2n}=\frac{2n-1}{2n} \to 1$ for $n \to \infty$</p>
<p>and</p>
<p>$a_{2n-1}=-\frac{2n-2}{2n-1} \to -1$ for $n \to \infty$.</p>
<p>Now suppose that $(a_{n_k})$ is a convergent subsequence of $(a_n)$ with limit $a$. Then $|a_{n_k}| \to |a|$.</p>
<p>But since $|a_n| \to 1$, we get $a=\pm 1$.</p>
<p>Consequence: </p>
<p>$\lim \sup a_n= \max\{-1,1\}=1$ and $\lim \inf a_n= \min\{-1,1\}=-1$</p>
|
https://math.stackexchange.com/questions/2189040/how-to-find-lim-inf-lim-sup
|
Question: <p>Let ${a_n}$ be a sequence and $b_n := \frac {2a_n + (-1)^n}{|a_n| +1}$ for $n$ in $\mathbb N$. Prove that ${b_n}$ has a convergent subsequence. </p>
<p>Attempt: </p>
<p>To show that the subsequence converges, I just have to show boundedness right? Is what I have here correct? Am I missing anything?</p>
<p>$|b_n| = |\frac {2a_n + (-1)^n}{|a_n| +1}| = \frac {|2a_n + (-1)^n|}{|a_n| +1} \le \frac {2|a_n| + 1}{|a_n| +1} \le \frac {2|a_n| + 2}{|a_n| +1} = 2$</p>
Answer:
|
https://math.stackexchange.com/questions/2189158/proving-convergent-subsequence
|
Question: <p>Let $f(x) = x+\frac1x$. Show that $f(x)$ is uniformly continuous on $(0,\infty)$ for any fixed $c>0$?</p>
Answer: <p>Hint:</p>
<p>I suppose you mean $x\in[c,\infty)$
$$|\frac{1}{x}+x-\frac{1}{y}-y|=|1-\frac{1}{xy}||x-y|\leq (1+\frac{1}{xy})|x-y|\leq(1+\frac{1}{c^2})|x-y|$$</p>
<p>Hence $f(x)$ is Lipschitz, hence uniformly continuous.</p>
|
https://math.stackexchange.com/questions/1029490/uniform-continuity-of-fx-x-frac1x-on-0-infty-for-any-fixed-c0
|
Question: <p>Let $f:I \times J\rightarrow \mathbb R$ be a function of class $C^2$ in the open rectangle $I \times J \subset \mathbb R^2$. If $\frac{\partial^2f}{\partial x\partial y}\equiv 0$, prove that there exist $\phi:I \rightarrow \mathbb R$ and $\chi:J \rightarrow \mathbb R$ of class $C^2$ such that $f(x,y)=\phi(x)+\chi(y)$.</p>
<p>Solution:</p>
<p>Since $\frac{\partial}{\partial x} \left (\frac{\partial f}{\partial y} \right )=0$, then $\frac{\partial f}{\partial y}$ doesn't depend on $x$, i.e., $\frac{\partial f}{\partial y}=f(x_0,y)$, $x_0 = $ constant.</p>
<p>For a similar reason, $\frac{\partial f}{\partial x}=f(x,y_0)$.</p>
<p>Now we define $\overline{\phi}:I \rightarrow \mathbb R;\overline{\phi}=\frac{\partial f}{\partial x}f(x,y_0)$ and $\overline \chi:J \rightarrow \mathbb R; \overline{\chi}=\frac{\partial f}{\partial y}f(x_0,y)$.</p>
<p>Then $f(x,y)=f(x,y)-f(x_0,y)+f(x_0,y)-f(x_0,y_0)+f(x_0,y_o)$. By the Fundamental Theorem of Calculus, $f(x,y)=\int_{x_0}^{x}\frac{\partial f(s,y)}{\partial x}ds+\int_{y_0}^{y} \frac{\partial f(x_0,t)dt}{\partial y}+f(x_0,y_0) \Rightarrow f(x,y)=\int_{x_0}^{x} \overline{\phi}(s)ds+\int_{y_o}^{y}\overline{\chi}(t)dt +f(x_0,y_o) $.</p>
<p>If $\phi$ and $\chi$ are the primitives of $\overline{\phi}$ and $\overline{\chi}$, then</p>
<p>$f(x,y)= \phi(x)-\phi(x_0) +\chi(y) -\chi(y_0)+f(x_0,y_0)$. As you see, I didn't manage to get the result, so any help is welcome.</p>
Answer: <p>Well, the first step is fine, but $f_y = a(y)$ for some $a$ rather than $f_y = f(x_0,y)$.
Otherwise, everything looks fine: you obtained the desired representation, the only thing you need to notice is that it concerns functions $\hat\phi(x) := \phi(x) - \phi(x_0) + f(x_0,y_0)$ and $\hat\psi(y) := \psi(y) - \psi(y_0)$. </p>
|
https://math.stackexchange.com/questions/1029606/decomposition-of-a-function-as-a-sum
|
Question: <p>I am sucked by a very simple question. I want to find an element which minimizes the$\|A\|$(in $L^2$-norm ),where A is a random variable.
Then the textbook says the uniqueness will follow from strict convexity of the $L^2$ norm.</p>
<p>What does this mean? What should I do if I want to prove the uniqueness? I think it may be very easy but I just don't know how to start.</p>
Answer:
|
https://math.stackexchange.com/questions/1029825/the-strict-convexity-of-the-l2-norm
|
Question: <p>I have to prove that $\frac{e^x}{x^k} \to \infty$ for $x \to \infty$ with $k \in \mathbb N$</p>
<p>My idea is to calculate for $R \gt 0$ an $x_r$ so that for every $ x\gt x_r$ the inequation $\frac{e^x}{x^k} \gt R$ applies.
So, I struggle with solving the equation $\frac{e^{x_r}}{x_{r}^k} = R$ for $x_r$</p>
Answer: <p>As $\displaystyle e^x=\sum_{r=0}^\infty\frac{x^r}{r!}$</p>
<p>For any finite $k\ge0,$
$$\lim_{x\to\infty}\frac{e^x}{x^k}\to\infty$$</p>
|
https://math.stackexchange.com/questions/1026310/prove-divergence-of-a-series
|
Question: <p>Here is a question I asked to myself. Does there exist a function $f(z)$ satisfying the following conditions:</p>
<ul>
<li>$f(z)$ is continuous on the strip $0 \leq \text{Re}(z) \leq 1$</li>
<li>$f(z)$ is <strong>holomorphic</strong> on the strip $0 < \text{Re}(z) < 1$</li>
<li>$f(a)=0$ where $a$ is a fixed real number $0 < a < 1$</li>
<li>The real part of $f(z)$ is greater of equal to 2 for all $z = it$, ($t \in \mathbb{R}$)</li>
<li>The real part of $f(z)$ is greater of equal to 2 for all $z = 1+it$, ($t \in \mathbb{R}$)</li>
</ul>
<p>My best guess so far was $-\frac{2}{a^2}(z^2- a^2)$ but it doesn't satisfy the last condition...</p>
Answer: <p>Assuming that you mean a real differentiable function, consider, for instance,
\begin{align*}
f(x+iy)=\frac{2(x-a)^2}{\min\{a^2,(1-a^2)\}}.
\end{align*}
This function is a nonnegative real polynomial (hence infinitely times real differentiable everywhere) with $f(a)=0$ and
\begin{align*}
f(it)&=f(0)=\frac{2a^2}{\min\{a^2,(1-a)^2\}}\geq\frac{2a^2}{a^2}=2, \\
f(1+it)&=f(1)=\frac{2(1-a)^2}{\min\{a^2,(1-a)^2\}}\geq\frac{2(1-a)^2}{(1-a)^2}=2,\qquad t\in\mathbb R. \\
\end{align*}</p>
|
https://math.stackexchange.com/questions/2234149/existence-of-a-function-satisfying-the-above-condition
|
Question: <p>My professor used this term today, and I didn't get a chance to ask him what the heck it means. I can't find anything explicitly on the internet regarding the manner in which he used it.</p>
<p>for example, say we have a function $f(x)$ and a set $A$ where $f(x)$ is defined. If we say $f(x)$ is an extension by zero, we mean the function is evaluated if $x \in A$ and is $0$ if $x \notin A$?</p>
<p>Is extension by zero, in this case, saying the same thing as extension by $A?$ Furthermore, if I was to say an extension by another set $B$ with $B \cap A = \emptyset$, would I be saying $f(x)$ evaluated if $x \in B$ and $0$ otherwise? </p>
Answer:
|
https://math.stackexchange.com/questions/2234524/the-phrase-extension-by-x
|
Question: <p>Although this seems to be super basic, I cannot get my head around which steps I have to perform to transform $x^{2^{n+1}}\cdot x^{2^{n+1}}$ into $x^{2^{n+2}}$.</p>
<p>When I rewrite it as $x^{2^{{n+1}^2}}$ and the there are not braces, the power binds the strongest, that would lead to $x^{2^{n^2+2n+1}}$.</p>
<p>Thanks in advance for any help and suggestions.</p>
Answer: <p>There's a basic fact that I'm assuming that you know:
$$
x^a \cdot x^b = x^ {a+b}.
$$
In your case, $a$ and $b$ are the same, and both are $2^{n+1}$. So all you need to know is what is
$$
a + b = 2^{n+1} + 2^{n+1}?
$$
Well, it's two copies of $2^{n+1}$, so it's
$$
2^1\cdot 2^{n+1} = 2^{n+2}
$$
by the same $a$-and-$b$ fact above. And that makes
the final answer be
$$
2^{2^{n+2}}.
$$</p>
|
https://math.stackexchange.com/questions/2192160/help-transforming-a-basic-equation
|
Question: <p>I need to work out the Taylor series of $\frac{1}{1-x}$ in ascending powers of $(x-2)$ up to and including the term in $(x-2)^3$. </p>
<p>My method was to find all the derivatives (up to the third) and evaluate them at $x=0$ to give the standard geometric series formula</p>
<p>$\frac{1}{1-x} = 1 +x +x^2 + x^3 + ...$ </p>
<p>and replace all the $x$s with $(x-2)$. However, I don't know if this right or if I am meant to evaluate the derivatives at $x=2$ to get my coefficients...</p>
Answer: <p>Observe
\begin{align}
\frac{1}{1-x} = \frac{-1}{1+(x-2)} = -\sum^\infty_{n=0}(-1)^n(x-2)^n.
\end{align}</p>
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https://math.stackexchange.com/questions/2194096/taylor-series-for-frac11-x-in-ascending-powers-of-x-2
|
Question: <p>$\left( \mathsf{\mathbf{x}},\mathsf{\mathbf{y}}\right) $,
$_{[\mathsf{\mathbf{x}}=\left( x_{1},\cdots,x_{n}\right) \in R^{n}\text{
}\wedge\text{ }\mathsf{\mathbf{y}}=\left( y_{1},\cdots,y_{m}\right) \in
R^{m}]}$ $:=\left( x_{1},\cdots,x_{n},y_{1},\cdots,y_{m}\right) \in R^{n+m}$</p>
<p>$W:=\left\{ \mathsf{\mathbf{y}}\in R^{m}:\left( \mathsf{\mathbf{0}%
},\mathsf{\mathbf{y}}\right) \in V\subset R^{n+m}\right\} $</p>
<p>$V$ is open in $R^{n+m}$, then $W$ is open in $R^{m}$</p>
<hr>
<h2>proof :</h2>
<hr>
<h2>1.</h2>
<p>$V$ is open in $R^{n+m}$</p>
<p>$\rightarrow\forall\left( \mathsf{\mathbf{x}},\mathsf{\mathbf{y}}\right)
\left( \left( \mathsf{\mathbf{x}},\mathsf{\mathbf{y}}\right) \in
V\rightarrow\exists r\left( \left( 0<r\right) \wedge\left( N_{r}^{R^{n+m}%
}\left( \left( \mathsf{\mathbf{x}},\mathsf{\mathbf{y}}\right) \right)
=\left\{ \left( \mathsf{\mathbf{\hat{x}}},\mathsf{\mathbf{\hat{y}}}\right)
\in R^{n+m}:\left\vert \left( \mathsf{\mathbf{\hat{x}}},\mathsf{\mathbf{\hat
{y}}}\right) -\left( \mathsf{\mathbf{x}},\mathsf{\mathbf{y}}\right)
\right\vert <r\right\} \subset V\right) \right) \right) $</p>
<hr>
<h2>2.</h2>
<p>$\mathsf{\mathbf{y}}\in W$</p>
<p>$\rightarrow\left( \mathsf{\mathbf{0}},\mathsf{\mathbf{y}}\right) \in V$</p>
<p>since 1. </p>
<p>$\rightarrow\exists r\left( 0<r\wedge N_{r}^{R^{n+m}}\left(
\mathsf{\mathbf{0}},\mathsf{\mathbf{y}}\right) =\left\{ \left(
\mathsf{\mathbf{\hat{x}}},\mathsf{\mathbf{\hat{y}}}\right) \in R^{n+m}%
:\left\vert \left( \mathsf{\mathbf{\hat{x}}},\mathsf{\mathbf{\hat{y}}%
}\right) -\left( \mathsf{\mathbf{0}},\mathsf{\mathbf{y}}\right) \right\vert
<r\right\} \subset V\right) $</p>
<hr>
<h2>3.</h2>
<p>$\mathsf{\mathbf{\hat{y}}}\in N_{r}^{R^{m}}\left( \mathsf{\mathbf{y}%
}\right) =\left\{ \mathsf{\mathbf{\hat{y}}}\in R^{m}:\left\vert
\mathsf{\mathbf{\hat{y}-y}}\right\vert <r\right\} $</p>
<p>$\rightarrow\left( \mathsf{\mathbf{\hat{y}}}\in R^{m}\right) \wedge\left(
\left\vert \mathsf{\mathbf{\hat{y}-y}}\right\vert <r\right) $</p>
<p>since : $\left\vert \mathsf{\mathbf{x}}\right\vert ,_{\left[
\mathsf{\mathbf{x}}\in R^{k}\wedge\mathsf{\mathbf{x}}=\left( x_{1}%
,\cdots,x_{k}\right) \right] }:=\left( {\sum\limits_{i=1}^{k}x{_{{{i}}}%
^{2}}}\right) ^{1/2}$</p>
<p>$\rightarrow\left( \left( \mathsf{\mathbf{0}},\mathsf{\mathbf{\hat{y}}%
}\right) \in R^{n+m}\right) \wedge\left( r>\left\vert \mathsf{\mathbf{\hat
{y}-y}}\right\vert =\left\vert \left( \mathsf{\mathbf{0-0}}%
,\mathsf{\mathbf{\hat{y}}}-\mathsf{\mathbf{y}}\right) \right\vert =\left\vert
\left( \mathsf{\mathbf{0}},\mathsf{\mathbf{\hat{y}}}\right) -\left(
\mathsf{\mathbf{0}},\mathsf{\mathbf{y}}\right) \right\vert \right) $</p>
<p>since 2. $N_{r}^{R^{n+m}}\left( \mathsf{\mathbf{0}},\mathsf{\mathbf{y}%
}\right) =\left\{ \left( \mathsf{\mathbf{\hat{x}}},\mathsf{\mathbf{\hat{y}%
}}\right) \in R^{n+m}:\left\vert \left( \mathsf{\mathbf{\hat{x}}%
},\mathsf{\mathbf{\hat{y}}}\right) -\left( \mathsf{\mathbf{0}}%
,\mathsf{\mathbf{y}}\right) \right\vert <r\right\} $</p>
<p>$\rightarrow\left( \mathsf{\mathbf{0}},\mathsf{\mathbf{\hat{y}}}\right) \in
V$</p>
<p>$\rightarrow\mathsf{\mathbf{\hat{y}}}\in W$</p>
<hr>
<h2>4.</h2>
<p>by 3.</p>
<p>$\rightarrow\left( \mathsf{\mathbf{\hat{y}}}\in N_{r}^{R^{m}}\left(
\mathsf{\mathbf{y}}\right) \right) $....$\left( \mathsf{\mathbf{\hat{y}}%
}\in N_{r}^{R^{m}}\left( \mathsf{\mathbf{y}}\right) \right) $</p>
<p>$\rightarrow N_{r}^{R^{m}}\left( \mathsf{\mathbf{y}}\right) \subset W$</p>
<hr>
<h2>5.</h2>
<p>by 2. 3. 4.</p>
<p>$\rightarrow\exists r\left( 0<r\wedge N_{r}^{R^{m}}\left( \mathsf{\mathbf{y}%
}\right) \subset W\right) $....$\left( \mathsf{\mathbf{y}}\in W\right) $</p>
<p>$\rightarrow W$ is open in $R^{m}$</p>
Answer: <p>Yes, and no.</p>
<p>The proof looks formally correct, <strong>but</strong> still it looks a lot like smoke-and-mirror. Basically anyone that is able to actually follow the proof would probably find the statement quite obvious - which would make the proof so much easier.</p>
<p>Instead you should have written it in a way that makes it easier to follow. I don't think it even gets more compact by using a lot of "symbol-language", so that won't do as an excuse:</p>
<p>Basically you start with $y\in W$ and from definition you have $(0,y)\in V$ which means that since $V$ is open you have an $r>0$ such that whenever $|(u,v)-(0,y)|<r$ you have $(u,v)\in V$. Especially we have that whenever |$(0,v)-(0,y)|<r$ we have $(0,v)\in V$ and therefore $v\in W$.</p>
<p>That is for every $y$ in $W$ we have an $r>0$ such that whenever $|u-y|<r$ we have $u\in W$, which means that $W$ is open.</p>
<p>Perhaps it's by omitting the constructs where open sets are defined in terms of neighborhoods being a subset of the set is one thing that simplifies the proof. Anyway even by including the proof that the two definitions of openness is equivalent I'd say it would make the proof more compact <strong>and</strong> more easy to follow that yours.</p>
|
https://math.stackexchange.com/questions/2194909/is-this-proof-of-open-set-right
|
Question: <p>On p. 21, Dieudonné states that: "... if $f$ is a continuous real function in a
bounded closed interval $I$, there is a smallest root and a largest root of the
equation $f(x)=0$ in $I$." </p>
<p>I haven't been able to convince myself of this basic
statement. I keep thinking of counterexamples that have only one root, or no
roots, that qualify. Please straighten me out. Thanks in advance.</p>
Answer: <p>Let $R \subset I$ be the set of roots for $f$. Then because $f$ is continuous, $R$ is closed. Since $I$ bounded, $R$ is compact. In particular, it contains its infimum (which will be the smallest root of f in $I$) and supremum (largest root).</p>
<p>Is it clear?</p>
<p>If $f$ has no roots then $R = \emptyset$, but in that case containing any root of $f$ is a tautology. If $f$ has one root, then the root is both the smallest and largest.</p>
|
https://math.stackexchange.com/questions/2261139/question-about-statement-on-p-21-of-dieudonn%c3%a9s-infinitesimal-calculus
|
Question: <blockquote>
<p>Let $m,n \ge 0$ be positive integers. Using induction on $n$ or otherwise, show that the improper integral $$\int_0^1 x^m \left(\log x\right)^n dx$$ exists, and give a closed-form expression for it.</p>
</blockquote>
<p>I'm a bit confused by the concept of induction on an integral, any help with this would be appreciated. Thanks!</p>
Answer: <p><strong>Hint</strong></p>
<p>For $m,n>0$ and $0 <x <1$</p>
<p>$$|x^m\ln^n (x)|\leq -\ln (x) $$
and
$\int_0^1\ln (x)dx $ converges thus by comparison, the integral $I_{m,n} $ is absolutely convergent.</p>
<p>to find a recursive formula, use by parts integration.
it will be of the form</p>
<p>$$I_{m,n}=\frac {n}{m+1}I_{m,n-1} $$</p>
|
https://math.stackexchange.com/questions/2285577/use-induction-to-show-that-this-improper-integral-exists
|
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