qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
3,029,708 | <p>Suppose there is a vector <span class="math-container">$U \in \mathbb{R}^n$</span>. How would you find the derivative of:</p>
<p><span class="math-container">$$
F(U)=trace\left(diag(U) A\ diag(U) \right)
$$</span>
where <span class="math-container">$A \in \mathbb{R}^{n \times n} \succ 0 $</span> and where <span class="math-container">$diag(\cdot)$</span> creates a diagonal matrix with <span class="math-container">$(\cdot)$</span> on the leading diagonal. Where the derivative is taken with respect to the vector <span class="math-container">$U$</span>, i.e.</p>
<p><span class="math-container">$$
{\partial F(U) \over \partial U } \\
$$</span></p>
<p>I am more interested in the method used. Thanks in advance.</p>
| greg | 357,854 | <p>Here are two useful identities involving the Hadamard <span class="math-container">$(\odot)$</span> product of a matrix and two vectors.<br>
<span class="math-container">$$\eqalign{
A\odot xy^T &= {\rm Diag}(x)\cdot A\cdot{\rm Diag}(y) \\
{\rm Tr}\Big(A\odot xy^T\Big) &= x^T\left(I\odot A\right)y \\
}$$</span>
Note that <span class="math-container">$(I\odot A)$</span> preserves of the main diagonal of <span class="math-container">$A$</span> but sets all off-diagonal terms to zero.
<br>It is a diagonal matrix and therefore <span class="math-container">$\;(I\odot A)^T=(I\odot A)$</span></p>
<p>These identities can be used to re-write the function
<span class="math-container">$$\eqalign{
F &= {\rm Tr}\Big({\rm Diag}(u)\cdot A\cdot{\rm Diag}(u)\Big) \\
&= u^T\left(I\odot A\right)u \\
}$$</span>
and then calculate its differential and gradient
<span class="math-container">$$\eqalign{
dF &= du^T\left(I\odot A\right)u \;+\; u^T\left(I\odot A\right)du \\
&= 2u^T\left(I\odot A\right)du \\
&= g^Tdu \\
\frac{\partial F}{\partial u}
&= g \;=\; 2\left(I\odot A\right)u \\
}$$</span></p>
|
3,215,381 | <p>When we write <span class="math-container">$\mathbb{Z}_3$</span>, does it mean <span class="math-container">$\mathbb{Z}/3\mathbb{Z}$</span>? Also, does <span class="math-container">$3\mathbb{Z}_3$</span> mean <span class="math-container">$0 \pmod 3$</span>?</p>
| PrincessEev | 597,568 | <p><span class="math-container">$\Bbb Z_n$</span> is a bit of an ambiguous notation without further context. I've seen it used both as a means of denoting the integers mod <span class="math-container">$n$</span>, and as a means of denoting the <span class="math-container">$n$</span>-adic integers - two very different things as you might imagine.</p>
<p><span class="math-container">$\Bbb Z / n \Bbb Z$</span> however is less ambiguous from my experience: it denotes the quotient set of the integers mod the set of the integer multiples of <span class="math-container">$n$</span>, i.e. it is the integers mod <span class="math-container">$n$</span>. I've never seen it used to denote a <span class="math-container">$p$</span>-adic entity.</p>
<p>So in a <span class="math-container">$p$</span>-adic context you'll probably see the subscript notation when speaking of the <span class="math-container">$p$</span>-adic integers and to avoid confusion in said context you should see the quotient set notation whenever modular arithmetic and such is relevant.</p>
|
1,464,747 | <p>I am trying to solve this question:</p>
<blockquote>
<p>How many ways are there to pack eight identical DVDs into five indistinguishable boxes so that each box contains at least one DVD?</p>
</blockquote>
<p>I am very lost at trying to solve this one. My attempt to start this problem involved drawing 5 boxes, and placing one DVD each, meaning 3 DVDs were left to be dropped, but I am quite stuck at this point.</p>
<p>Any help you can provide would be great. Thank you.</p>
| André Nicolas | 6,312 | <p>You are nearly finished. We have three DVD to dump into boxes. Maybe we put all $3$ in the same box. Maybe we use a $2$-$1$ split. Or maybe we use a $1$-$1$-$1$ split. Since the boxes are indistinguishable, we have a total of three possibilities.</p>
|
58,060 | <p>I have been looking at Church's Thesis, which asserts that all intuitively computable functions are recursive. The definition of recursion does not allow for randomness, and some people have suggested exceptions to Church's Thesis based on generating random strings. For example, using randomness one can generate strings of arbitrarily high Kolmogorov complexity but this is not possible with recursion alone.</p>
<p>However, these exceptions are not true <em>functions</em>. They generate multiple outputs, which collectively have some property. A recursive function takes an inputs and outputs a single unique answer. So some people do not consider these random coin-flips to be true exceptions to Church's Thesis.</p>
<p>My question is whether it is possible to use randomness to get something which is still essentially deterministic like a function, but is non-recursive. </p>
<p>For example, if we had a sequence of functions $F^r_i(n)$, which are recursive functions relative to a random tape oracle $r$, which have the property that for some function $g(n)$, we have $F^r_i(n) = g(n)$ with probability approaching $1$ as $n \rightarrow \infty$ (the probability taken over the random tapes). Furthermore, $g$ would not be recursive itself. </p>
<p>Here, I am suggesting relativizing to $r$, rather than having $r$ as an input, because one might need arbitrarily many random cells. This <em>could</em> be thought of as allowing to "intuitively compute" $g$.</p>
| Joseph O'Rourke | 6,094 | <p>I asked <a href="https://cstheory.stackexchange.com/questions/1263/truly-random-number-generator-turing-computable">a related question at CS Theory</a>, which ended with this question:</p>
<blockquote>
<p>Is it the case that a TM [Turing Machine] with access to a pure source of randomness (an oracle?), can compute a function that a classical TM cannot?</p>
</blockquote>
<p>I received a detailed and knowledgeable reply from <a href="http://www.liafa.jussieu.fr/%7Elbienven/" rel="noreferrer">Laurent Bienvenu</a>, which I found so
enlightening that it may be worth quoting the relevant portions here:</p>
<blockquote>
<p>From a computability perspective, the answer is 'Yes and No.' If you are given access to a random source as an oracle (where the output is presented as an infinite binary sequence), with probability 1 you will get a Martin-Löf random oracle, and as we saw earlier [earlier in his own answer], Martin-Löf random implies non-computable, so it suffices to output the oracle itself! Or if you want a function <span class="math-container">$f:N \rightarrow N$</span>, you can consider the function <span class="math-container">$f$</span> which for all <span class="math-container">$n$</span> tells you how many zeroes there are among the first <span class="math-container">$n$</span> bits of your oracle. If the oracle is Martin-Löf random, this function will be non-computable.</p>
<p>But of course you might argue that this is cheating: indeed, for a different oracle we might get a different function, so there is a non-reproducibility problem. Hence another way to understand your question is the following: Is there a function <span class="math-container">$f$</span> which is non-computable, but which can be "computed with positive probability," in the sense that there is an Turing machine with access to a random oracle which, with positive probability (over the oracle), computes <span class="math-container">$f$</span>. The answer is 'No,' due to a theorem of Sacks whose proof is quite simple. Actually it has mainly been answered by Robin Kothari [citing another answer]: if the probability for the TM to be correct is greater than <span class="math-container">$\frac{1}{2}$</span>, then one can look for all <span class="math-container">$n$</span> at all the possible oracle computations with input <span class="math-container">$n$</span> and find the output which gets the "majority vote", i.e. which is produced by a set of oracles of measure more than <span class="math-container">$\frac{1}{2}$</span> (this can be done effectively). The argument even extend to smaller probabilities: suppose the TM outputs f with probability <span class="math-container">$\epsilon >0$</span>. By Lebesgue's density theorem, there exists a finite string <span class="math-container">$\sigma$</span> such that if we fix the first bits of the oracle to be exactly <span class="math-container">$\sigma$</span>, and then get the other bits at random, then we compute <span class="math-container">$f$</span> with probability at least 0.99. By taking such a <span class="math-container">$\sigma$</span>, we can apply the above argument again.</p>
</blockquote>
|
2,764,221 | <p>Let $A$ be a symmetric invertible $n \times n$ matrix, and $B$ an antisymmetric $n \times n$ matrix. Under what conditions is $A+B$ an invertible matrix? In particular, if $A$ is positive definite, is $A+B$ invertible? </p>
<p>This isn't homework, I am just curious. Assume all matrices have entries in $\mathbb{R}$. </p>
<p><strong>Edit to include context:</strong> </p>
<p>This question comes from a question that popped up in my research on string theory. One is interested in (pseudo)-Riemannian manifolds equipped with a two-form gauge field, modelling a background in which a closed string is moving. The metric, $g$, is a symmetric covariant 2-tensor, while the $b$-field is an antisymmetric covariant 2-tensor. The metric is non-degenerate and therefore invertible. Choosing local coordinates for the manifold, we can express the metric and $b$ field as $n \times n$ matrices, say $A$ and $B$, where $A$ is invertible. There is an operation on string backgrounds called T-duality which, in this simplified context, acts by inverting the matrix $E = A + B$, and so I am therefore interested in which scenarios this procedure works. I am mainly interested in the context where $A$ is real, invertible and positive definite (positive eigenvalues), corresponding to a Riemannian metric $g$, although I have tried to be a bit more general in the wording of the question. </p>
<p><strong>Where to start:</strong> The main issue I have is that I don't really have any criteria for when the sum of two matrices is invertible. Certainly if the determinant is non-zero then I will be happy, but the determinant is not additive, so I don't know how to approach this. In two dimensions I can construct a counterexample whenever A has negative determinant, but the situations I really care about have $det(A)>0$. I would like to find a general criterion for when $A+B$ is invertible. </p>
| N. S. | 9,176 | <p>Pick $B$ any anti-symmetric matrix which is not nilpotent, and $\lambda \neq 0$ an eigenvalue of $B$.</p>
<p>Set $$A=-\lambda I$$</p>
|
3,399,195 | <p>So I've seen various questions with the limit 'equal' to <span class="math-container">$\infty$</span> or that the limit doesn't exist in a case where the function tends to <span class="math-container">$\infty$</span>.</p>
<p>For example, the limit of <span class="math-container">$\sqrt{x}$</span> as <span class="math-container">$x$</span> tends to <span class="math-container">$\infty$</span>. Is the answer <span class="math-container">$\infty$</span> or that the limit doesn't exist?</p>
<p>Obviously the function tends to <span class="math-container">$\infty$</span> as <span class="math-container">$x$</span> tends to <span class="math-container">$\infty$</span> but I don't know what to give as an answer.</p>
<p>I've seen similar questions where the function tends to <span class="math-container">$\infty$</span> as <span class="math-container">$x$</span> tends to a certain value where the answer has been that the limit doesn't exist. I've also seen where, in a similar situation, the limit has been 'equal' to <span class="math-container">$\infty$</span>.</p>
<p>So which is the one to use? What's the difference? Thanks!</p>
| Clark Driscoll | 1,064,598 | <p>If you could show some examples of the problems you talking about that would help greatly.</p>
<p>For cases where <span class="math-container">$x \to \infty$</span>, if the limit is increases with out bounds then you would say the limit equals infinity.</p>
<p><span class="math-container">$\lim_{x\to\infty}\sqrt{x}=\infty$</span></p>
<p><span class="math-container">$\lim_{x\to\infty} x^2=\infty$</span></p>
<p>For the instances of the limits that approach a finite value and end up being 'does not exist'.</p>
<p>In these cases the we have to be approaching the same value from the left and the right side of the finite value. Say you are approaching from the left and heading towards negative infinity and from the right you are approaching positive infinity. In this case you would say the limit 'does not exist'. Now if it was approaching positive infinity on both the left and right, you would be able to say the limit equals infinity.</p>
<p>For some examples</p>
<p><span class="math-container">$\lim_{x\to2}\frac{x}{x^2-4}\Rightarrow$</span> Does Not Exist</p>
<p><span class="math-container">$\lim_{x\to0}\frac{1}{x^2}=\infty$</span></p>
|
3,009,362 | <p>I need to find
<span class="math-container">$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}$$</span></p>
<p>Looking at the graph, I know the answer should be <span class="math-container">$\frac{20}{17}$</span>, but when I tried solving it, I reached <span class="math-container">$0$</span>.</p>
<p>Here are the two ways I approached this:</p>
<p>WAY I:</p>
<p><span class="math-container">$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35} =
\lim_{x\rightarrow -5} \frac{\require{cancel} \cancel{x^2}(2- \frac{50}{x^2})} {\require{cancel} \cancel{x^2}(2+ \frac{3}{x}-\frac{35}{x^2})}
=\frac{2-2}{\frac {42}{5}}=0
$$</span></p>
<p>WAY II:
<span class="math-container">$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35} =
\lim_{x\rightarrow -5} \frac{\require{cancel} \cancel{2}(x^2- 25)} {\require{cancel} \cancel{2}(x^2+ \frac{3}{2}x-\frac{35}{2})}
=\lim_{x\rightarrow -5} \frac{{\require{cancel} \cancel{(x-5)}}(x+5)}{{\require{cancel} \cancel{(x-5)}}(x+3.5)}= \frac{-5+5}{-5+3.5}=0
$$</span></p>
<p>What am I doing wrong here?</p>
<p>Thanks!</p>
| Peter Szilas | 408,605 | <p>Numerator</p>
<p><span class="math-container">$2x^2-50=2(x-5)(x+5)$</span>.</p>
<p>Denominator</p>
<p><span class="math-container">$2x^2+3x -35 =(2x-7)(x+ 5)$</span></p>
<p><span class="math-container">$\dfrac{2(x-5)(x+5)}{(2x-7)(x+5)}=$</span></p>
<p><span class="math-container">$\dfrac{2(x-5)}{2x+7}.$</span></p>
<p>Take the limit <span class="math-container">$x \rightarrow -5.$</span></p>
<p>Try to factorize the original expression.The term <span class="math-container">$(x+5) $</span> cancels out .</p>
|
3,169,142 | <p>The question goes like this:</p>
<p>If <span class="math-container">$f(x)$</span> is a non-constant, continuous function defined on a closed interval <span class="math-container">$[a,b]$</span> Then by the Extreme Value Theorem, there exist an absolute minimum <span class="math-container">$m$</span> and an absolute maximum <span class="math-container">$M$</span>. </p>
<p>Based on this, I need to show that the range of <span class="math-container">$f$</span>, <span class="math-container">$\{f(x) \mid a \le x \le b\}$</span>, is the interval <span class="math-container">$[m, M]$</span>.</p>
<p>Thanks in advance!</p>
| Simon Goodwin | 364,836 | <p>Consider the Intermediate value theorem, let <span class="math-container">$x\in [m,M]$</span> and then since there are <span class="math-container">$x_1,x_2$</span> such that <span class="math-container">$f(x_1)=m, f(x_2)=M$</span> we get the desired result.</p>
|
1,843,662 | <p>Let C be that part of the circle $z=e^{i\theta}$, where $0\le\theta\le\frac\pi2$. Evaluate $\int_{c}\frac{z}{i}dz$.</p>
<p>This is my first time posting my question here. I'm really poor in writing English. for that reason please understand my bad explanation. proceed to the main issue I have no idea on solving this problem and also the way how to approach the answer. There is no solution manual which means that i'm not sure what i'm solving and answer. So please i need your any advice, suggest, soulution or etc.</p>
| J.-E. Pin | 89,374 | <p>According to Eilenberg [1, Chap. IV, Prop. 1.1], the following result holds:</p>
<blockquote>
<p><strong>Proposition</strong>. For any nonempty subset $L$ of $A^*$, the following conditions are equivalent:</p>
<ol>
<li>for all $u, v \in L$, $u^{-1}L = v^{-1}L$,</li>
<li>the minimal automaton of $L$ has a single final state,</li>
<li>$L$ is recognized by a deterministic automaton with a single final state that is accessible.</li>
</ol>
</blockquote>
<p>[1] S. Eilenberg, Automata, Languages and Machines, Volume A, Academic Press (1974)</p>
<p>See also my <a href="https://cstheory.stackexchange.com/q/36074">answer</a> to the related question <a href="https://cstheory.stackexchange.com/questions/18271/ndfa-with-same-initial-accepting-states">(N)DFA with same initial/accepting state(s)</a> on cstheory.</p>
|
912,217 | <p>Let $X$ be a R.V whose pdf is given by
$$f(x)=p\frac{1}{\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(x-\mu_1)^2}{2\sigma_1^2}\right)+
(1-p)\frac{1}{\sqrt{2\pi\sigma_2^2}}\exp\left(-\frac{(x-\mu_2)^2}{2\sigma_2^2}\right)$$</p>
<p>clearly $X\sim pN(\mu_1,\sigma_1^2)+(1-p)N(\mu_2,\sigma_2^2)=N(p\mu_1+(1-p)\mu_2,p^2\sigma_1^2+(1-p)^2\sigma_2^2)$</p>
<p>therefore <strong><em>$\kappa=E(e^X)-1=e^{p\mu_1+(1-p)\mu_2+\frac{1}{2}(p^2\sigma_1^2+(1-p)^2\sigma_2^2)}-1$.</em></strong></p>
<p>However, if I let $Y=e^X$, then
Let $G(y)$ be the cdf of $Y$, then
$G(y)=P(Y<y)=P(e^X<y)=p(X<\ln(y))=F_X(\ln(y))$
there fore $g(y)=f(\ln(y))\frac{1}{y}$
so
the pdf $g(y)$ of $Y$ is given by </p>
<p>$$g(y)=p\frac{1}{y\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(\ln(y)-\mu_1)^2}{2\sigma_1^2} \right) + (1-p) \frac{1}{y\sqrt{2\pi\sigma_2^2}} \exp\left(-\frac{(\ln(y)-\mu_2)^2}{2\sigma_2^2}\right)$$</p>
<p>i.e $Y=p\times \mathrm{logNormal}(\mu_1,\sigma_1^2)+(1-p)\times \mathrm{logNormal} (\mu_2,\sigma_1^2)$</p>
<p>where $\mathrm{logNormal}(\mu_1,\sigma_1^2)$ means a log-normal distribution.</p>
<p>hence <strong><em>$\kappa=E(Y)-1=pE(\mathrm{logNormal}(\mu_1,\sigma_1^2))+(1-p) E(\mathrm{logNormal} (\mu_2,\sigma_2^2))=
pe^{\mu_1+\frac{\sigma_1^2}{2}}+(1-p)e^{\mu_2+\frac{\sigma_2^2}{2}}-1$.</em></strong></p>
<p>My question is : why I have different values for $\kappa$? Could some one explain for me where I am wrong ? thank you for your time.</p>
| Snufsan | 122,989 | <p>Note that if $f$ is differentiable in $x$ then $f'(x)$ exsists, i.e. the limit exist, which exactly means that the side-limits are equal: $f'(x)=f'_{-}(x)=f'_{+}(x)$</p>
|
912,217 | <p>Let $X$ be a R.V whose pdf is given by
$$f(x)=p\frac{1}{\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(x-\mu_1)^2}{2\sigma_1^2}\right)+
(1-p)\frac{1}{\sqrt{2\pi\sigma_2^2}}\exp\left(-\frac{(x-\mu_2)^2}{2\sigma_2^2}\right)$$</p>
<p>clearly $X\sim pN(\mu_1,\sigma_1^2)+(1-p)N(\mu_2,\sigma_2^2)=N(p\mu_1+(1-p)\mu_2,p^2\sigma_1^2+(1-p)^2\sigma_2^2)$</p>
<p>therefore <strong><em>$\kappa=E(e^X)-1=e^{p\mu_1+(1-p)\mu_2+\frac{1}{2}(p^2\sigma_1^2+(1-p)^2\sigma_2^2)}-1$.</em></strong></p>
<p>However, if I let $Y=e^X$, then
Let $G(y)$ be the cdf of $Y$, then
$G(y)=P(Y<y)=P(e^X<y)=p(X<\ln(y))=F_X(\ln(y))$
there fore $g(y)=f(\ln(y))\frac{1}{y}$
so
the pdf $g(y)$ of $Y$ is given by </p>
<p>$$g(y)=p\frac{1}{y\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(\ln(y)-\mu_1)^2}{2\sigma_1^2} \right) + (1-p) \frac{1}{y\sqrt{2\pi\sigma_2^2}} \exp\left(-\frac{(\ln(y)-\mu_2)^2}{2\sigma_2^2}\right)$$</p>
<p>i.e $Y=p\times \mathrm{logNormal}(\mu_1,\sigma_1^2)+(1-p)\times \mathrm{logNormal} (\mu_2,\sigma_1^2)$</p>
<p>where $\mathrm{logNormal}(\mu_1,\sigma_1^2)$ means a log-normal distribution.</p>
<p>hence <strong><em>$\kappa=E(Y)-1=pE(\mathrm{logNormal}(\mu_1,\sigma_1^2))+(1-p) E(\mathrm{logNormal} (\mu_2,\sigma_2^2))=
pe^{\mu_1+\frac{\sigma_1^2}{2}}+(1-p)e^{\mu_2+\frac{\sigma_2^2}{2}}-1$.</em></strong></p>
<p>My question is : why I have different values for $\kappa$? Could some one explain for me where I am wrong ? thank you for your time.</p>
| Paramanand Singh | 72,031 | <p>An alternative way of looking at this problem (different from your class notes):</p>
<p>What happens when $f'(x) < 0$ even at a single point $c$? Clearly this would mean that the there will be an interval $I$ containing $c$ such that $f(x) > f(c) $ for all $x \in I, x < c$ and $f(x) < f(c)$ for all $x \in I, x > c$. This goes contrary to the increasing nature of $f(x)$. Hence we must have $f'(x) \geq 0$. It should be noted that the equality can occur as can be seen with $f(x) = x^{3}$, $f'(x) = 0$ for $x = 0$ yet the function is strictly increasing everywhere.</p>
|
2,832,374 | <p>I'm given the following definition asked to prove the following theorem:</p>
<p>Definition: Let $X$ be a set and suppose $C$ is a collection of subsets of $X$. Then $\cup \mathbf{C}=\{x \in X : \exists C\in \mathbf{C}(x\in C)\}$</p>
<p>Theorem: Let $\mathbf{C,D}$ be collections of subsets of a set $X$. Prove that $\cup ( \mathbf{C} \cup \mathbf{D}) = (\cup \mathbf{C}) \cup (\cup\mathbf{D})$ </p>
<hr>
<p>By my reading of the definition, I run into two problems:</p>
<p>Firstly I think there is a type error since $ ( \mathbf{C} \cup \mathbf{D}) = \cup\{\mathbf{C}, \mathbf{D}\}$ is not a collection of subsets of $X$ (i.e. it is not a set whose elements are subsets of $X$); instead it is a set whose elements are collections of subsets of $X$.</p>
<p>Second, if we ignore the type error and plug into the definition, we get:
$\cup \{\mathbf{C}, \mathbf{D}\} =\{x\in X:\exists C\in \{\mathbf{C}, \mathbf{D}\}(x\in C)\}=\{x\in X:x\in \mathbf{C} \lor x \in \mathbf{D}\}$, however, since $\boldsymbol{C},\boldsymbol{D}$ are collections, all their elements are sets. Since $x$ is not a set, $x\notin\boldsymbol{C}\land x\notin\boldsymbol{D}$. Thus $\cup \{\mathbf{C}, \mathbf{D}\}=\emptyset$ </p>
<p>However this can't be right since the other side of the equality; $(\cup \mathbf{C}) \cup (\cup\mathbf{D})\neq \emptyset$ in general. What am I missing?</p>
| Community | -1 | <p>Suppose $f_n \to f$ with $\Vert f_n \Vert \leq 1$, then also $\Vert f_n \Vert \to \Vert f \Vert$ by the reverse triangle inequality, and limits preserve inequalities.</p>
|
3,156,570 | <p>I need to evaluate the following limit:
<span class="math-container">$$
\lim_{x\downarrow 0} \dfrac{(1 - e^x)^{-1}}{x^c}
$$</span>
for different values of the constant <span class="math-container">$c$</span>.</p>
<p><em>What I've tried thus far:</em></p>
<p>We have that
<span class="math-container">$$
\lim_{x\downarrow 0} \dfrac{(1 - e^x)^{-1}}{x^c} = \lim_{x\downarrow 0}\dfrac{1}{x^c(1 - e^x)}
$$</span>
Now I know that <span class="math-container">$1 - e^x \to 0$</span> as <span class="math-container">$x\to 0$</span>. If <span class="math-container">$c\geq 0$</span> then <span class="math-container">$x^c\to 0$</span> as <span class="math-container">$x\to 0$</span> so we have that
<span class="math-container">$$
\lim_{x\downarrow 0}\dfrac{1}{x^c(1 - e^x)} = \infty
$$</span>
What I'm not so sure about is what happens when <span class="math-container">$c<0$</span>. If <span class="math-container">$c< 0$</span> then we can write <span class="math-container">$x^c$</span> as <span class="math-container">$x^{-\gamma}$</span> where <span class="math-container">$\gamma = \left|c\right|$</span>. I know that <span class="math-container">$\lim_{x\downarrow 0}x^{-\gamma}\to \infty$</span>, but I'm not sure whether <span class="math-container">$x^{-\gamma}\to \infty$</span> quicker than <span class="math-container">$1 - e^x\to 0$</span> as <span class="math-container">$x\to 0$</span>. I need to know because right now I'm not sure what the term <span class="math-container">$x^c(1 - e^x)$</span> does as <span class="math-container">$x\downarrow 0$</span> when <span class="math-container">$c <0$</span>.</p>
<p><strong>Question:</strong> </p>
<p>How do I evaluate
<span class="math-container">$$
\lim_{x\downarrow 0}\dfrac{1}{x^c(1 - e^x)}
$$</span>
when <span class="math-container">$c< 0$</span>?</p>
| farruhota | 425,072 | <p>Write it as:
<span class="math-container">$$\lim_{x\to 0}\dfrac{1}{x^c(1 - e^x)}=\lim_{x\to 0}\dfrac{x^{-c}}{1 - e^x}\stackrel{L'H}=\lim_{x\to 0}\dfrac{-cx^{-c-1}}{-e^x}=\\
\lim_{x\to 0}\dfrac{c}{x^{c+1}}=\begin{cases}-1, c=-1\\ \ \ \ 0, c<-1\\
-\infty, -1<c<0\\
+\infty, c>0\\
\end{cases}$$</span>
If <span class="math-container">$c=0$</span>, it is <span class="math-container">$\pm \infty$</span> depending on the side of approach.</p>
|
3,314,561 | <p>Consider the triangle <span class="math-container">$PAT$</span>, with angle <span class="math-container">$P = 36$</span> degres, angle <span class="math-container">$A = 56$</span> degrees and <span class="math-container">$PA=10$</span>. The points <span class="math-container">$U$</span> and <span class="math-container">$G$</span> lie on sides TP and TA respectively, such that PU = AG = 1. Let M and N be the midpoints of segments PA and UG. What is the degree measure of the acute angle formed by the lines MN and PA?</p>
<p>It would be very helpful if anyone had a solution using complex numbers to this problem.</p>
| Blazej | 155,834 | <p>Reason 1. Even if you actually care only about smooth solutions, it some cases it is much easier to first establish that a weak solution exists and separately show that the structure of the PDE actually enforces it to be smooth. Existence and regularity are handled separately and using different tools.</p>
<p>Reason 2. There are physical phenomena which are described by discontinuous solutions of PDEs, e.g. hydrodynamical shock waves.</p>
<p>Reason 3. Discontinuous solutions may be used as a convenient approximation for describing macroscopic physics neglecting some details of the microscopic theory. For example in electrodynamics one derives from the Maxwell equations that the electric field of an electric dipole behaves at large distances in a universal way, depending only on the dipole moment but not on the charge distributions. On distances comparable to the dipole size these microscopic details start to become important. If you don't care about these small distances you may work in the approximation in which dipole is a point-like object, with charge distribution given by a derivative of the delta distribution. Even though the actual charge distribution is given by a smooth function, it is more convenient to approximate it by a very singular object. One can still make sense of the Maxwell equations, and the results obtained this way turn out to be correct (provided that you understand the limitations of performed approximations).</p>
<p>Reason 4. It is desirable to have "nice" spaces in which you look for solutions. In functional analysis there are many features you might want a topological vector space to have, and among these one of the most important is completeness. Suppose you start with the space of smooth functions on, say, <span class="math-container">$[0,1]$</span> and equip it with a certain topology. In this case it is completely natural to pass to the completion. For many choices of the topology you will find that the completed space contains objects which are too singular to be considered as bona fide functions, e.g. measures or distributions. Just to give you an example of this phenomenon: if you are interested in computing integrals of smooth functions, you are eventually going to consider gadgets such as <span class="math-container">$L^p$</span> norms on <span class="math-container">$C^{\infty}[0,1]$</span>. Once you complete, you get the famous <span class="math-container">$L^p$</span> spaces, whose elements are merely equivalence classes of functions modulo equality almost anywhere. Space of distributions on <span class="math-container">$[0,1]$</span> may be constructed very similarly: instead of <span class="math-container">$L^p$</span> norms you consider the seminorms <span class="math-container">$p_f$</span> given by <span class="math-container">$p_f(g)= \int_{0}^1 f(x) g(x) dx$</span> for <span class="math-container">$f,g \in C^{\infty}[0,1]$</span>. If you can justify to yourself that it is interesting to look at this family of seminorms, then distibutions (and also weak solutions of PDEs) become an inevitable consequence.</p>
|
122,468 | <p>I know how to find the number of solutions to the equation:</p>
<p><span class="math-container">$$a_1 + a_2 + \dots + a_k = n$$</span></p>
<p>where <span class="math-container">$n$</span> is a given positive integer and <span class="math-container">$a_1$</span>, <span class="math-container">$a_2$</span>, <span class="math-container">$\dots$</span>, <span class="math-container">$a_n$</span> are positive integers. The number of solutions to this equation is:</p>
<p><span class="math-container">$$\binom{n - 1}{k - 1}$$</span></p>
<p>This can be imagined as <span class="math-container">$n$</span> balls arranged on a straight line and selecting <span class="math-container">$k - 1$</span> gaps from a total of <span class="math-container">$n - 1$</span> gaps between them as partition boundaries. The <span class="math-container">$k - 1$</span> partition boundaries divide the <span class="math-container">$n$</span> balls into <span class="math-container">$k$</span> partitions. The number of balls in the <span class="math-container">$i$</span>th partition is <span class="math-container">$a_i$</span>.</p>
<p>Now, I don't know how to find the number of solutions to the same equation when we have an additional constraint: <span class="math-container">$$0 < a_1 \leq a_2 \leq \dots \leq a_k.$$</span> Could you please help me?</p>
| Marc van Leeuwen | 18,880 | <p>While there is no closed form for the number of partitions of $n$ into $k$ parts, these numbers are not hard to compute. First let's get rid of the awkward initial strict inequality: by setting $a'_i=a_i-1$ we get $0\leq a'_1\leq\cdots\leq a'_k$ and $a'_1+\cdots+a'_k=n-k$. This means we want to count weak partitions of $m=n-k$ into $k$ parts, where weak means the parts can be $0$. Now forming a Young diagram by putting $a'_i$ squares in row $k+1-i$, we are counting Young diagrams with $m$ squares that fit in the first $k$ rows. This is the same as counting Young diagrams with $m$ squares and columns of length at most $k$, or counting partitions of $m$ into parts (arbitrarily many) of size at most $k$.</p>
<p>Now such partition problems have easy generating series. In this case the number we want is the coefficient of $X^m$ in the product of formal power series:
$$
\frac1{(1-X)(1-X^2)\cdots(1-X^k)} = \prod_{i=1}^k\frac1{1-X^i}.
$$
Multiplying a power series by $\frac1{1-X^i}$ is obtained by adding, in an increasing order, to each coefficient of $X^j$ with $j\geq i$, the coefficient of $X^{j-i}$ (the latter is in general already modified during this run). Thus the following little program (in C++) computes you number in the final entry of the array $c$:</p>
<pre><code>int m = n-k;
vector<int> c(m+1,0); c[0] = 1; // series with m+1 coefs, set to unity
for (int i=1; i<=k; ++i)
for (int j=i; j<=m; ++j)
c[j] += c[j-i];
int count = c[m];
</code></pre>
|
21,491 | <p>The question is prompted by change of basis problems -- the book keeps multiplying the bases by matrix $S$ from the left in order to keep subscripts nice and obviously matching, but in examples bases are multiplied by $S$ (the change of basis matrix) from whatever side. So is matrix multiplication commutative if at least one matrix is invertible?</p>
| Ofir | 2,125 | <p>In general, two matrices (invertible or not) do not commute. For example
$$\left(\begin{array}{cc}
1 & 1\\
0 & 1\end{array}\right)\left(\begin{array}{cc}
1 & 0\\
1 & 1\end{array}\right) = \left(\begin{array}{cc}
2 & 1\\
1 & 1\end{array}\right)
$$
$$
\left(\begin{array}{cc}
1 & 0\\
1 & 1\end{array}\right)\left(\begin{array}{cc}
1 & 1\\
0 & 1\end{array}\right) = \left(\begin{array}{cc}
1 & 1\\
1 & 2\end{array}\right)$$</p>
<p>Also, to change a basis you usually need to conjugate and not just multiply from the left (or just right).</p>
<p>What you do know is that a matrix A commutes with $A^n$ for all $n$ (negative too if it is invertible, and $A^0 = I$), so for every polynomial P (or Laurent polynomial if A is invertible) you have that A commutes with $P(A)$.</p>
|
3,394,050 | <p>I'm having trouble with this problem.</p>
<blockquote>
<p>Using logical equivalencies prove that <span class="math-container">$(p \land q)\implies (p \lor q)$</span> is a tautology.</p>
</blockquote>
| RyRy the Fly Guy | 412,727 | <p><span class="math-container">$T$</span></p>
<p><span class="math-container">$\Leftrightarrow T \vee T$</span></p>
<p><span class="math-container">$\Leftrightarrow (p \vee \neg p) \vee (q \vee \neg q) \Leftrightarrow T$</span></p>
<p>We drop the parentheses because the statement is well defined...</p>
<p><span class="math-container">$\Leftrightarrow p \vee \neg p \vee q \vee \neg q \Leftrightarrow T$</span></p>
<p><span class="math-container">$\Leftrightarrow \neg p \vee \neg q \vee p \vee q \Leftrightarrow T$</span> </p>
<p><span class="math-container">$\Leftrightarrow (\neg p \vee \neg q) \vee (p \vee q) \Leftrightarrow T$</span></p>
<p><span class="math-container">$\Leftrightarrow \neg (p \wedge q) \vee (p \vee q) \Leftrightarrow T$</span></p>
<p><span class="math-container">$\Leftrightarrow (p \wedge q) \rightarrow (p \vee q) \Leftrightarrow T$</span></p>
<hr>
<p>However, I prefer a proof via natural deduction because it is cleaner...</p>
<p><span class="math-container">$1.$</span> <span class="math-container">$p \wedge q$</span> ACP</p>
<p><span class="math-container">$2.$</span> <span class="math-container">$(p \wedge q) \vee p$</span> by disjunction introduction <span class="math-container">$1$</span></p>
<p><span class="math-container">$3.$</span> <span class="math-container">$p \vee (p \wedge q)$</span> by commutative rule <span class="math-container">$2$</span></p>
<p><span class="math-container">$4.$</span> <span class="math-container">$(p \vee p) \wedge (p \vee q)$</span> by distributive rule <span class="math-container">$3$</span></p>
<p><span class="math-container">$5.$</span> <span class="math-container">$p \wedge (p \vee q)$</span> by idempotence <span class="math-container">$4$</span></p>
<p><span class="math-container">$6.$</span> <span class="math-container">$p \vee q$</span> by conjunction elimination <span class="math-container">$5$</span></p>
<p><span class="math-container">$7.$</span> <span class="math-container">$(p \wedge q) \rightarrow (p \vee q)$</span> CP <span class="math-container">$1$</span>-<span class="math-container">$6$</span></p>
|
1,026,506 | <p>If $I_n=\int _0^{\pi }\:sin^{2n}\theta \:d\theta $, show that
$I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$, and hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $</p>
<p>Hence calculate $\int _0^{\pi }\:\:sin^4tcos^6t\:dt$</p>
<p>I knew how to prove that $I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$ ,, but I am not very good at English, what does it mean Hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $ do we need to prove this part as well or is it just a hint to use? and for the other calculation to find $\int _0^{\pi }\:\:sin^4tcos^6t\:dt$ is there something in the first part that I can do to help me solve this question because otherwise it becomes very long and it's part of the question.</p>
| Community | -1 | <blockquote>
<p>what does it mean Hence $I_n = \frac{(2n)!\pi}{(2^n n!)2}$ do we need to prove this part as well or is it just a hint to use? </p>
</blockquote>
<p>"Hence" means "from here"; it is an archaic word. It survives in mathematical English, where it means "because of what was just said", or "using what was just proved". They're saying that you should use $I_n = \frac{2n-1}{2n} I_{n-1}$ to prove $I_n = \frac{(2n)!\pi}{(2^n n!)2}$.</p>
|
3,007,443 | <p>I've heard the words "internal" and "external" generalization of concepts in category theory.</p>
<p>Specifically, i heard the idea that the concept of 'power set' has an internal and an external generalization in category theory.</p>
<p>What is the difference between these two?</p>
| Musa Al-hassy | 80,406 | <p>Categorial internalisation is about taking a statement
that involves “points”, which is the usual Set theoretic rendition,
and turning it “point free” so that it is soley rendered in the
language of category theory.</p>
<p>For example, an adjoint between preorders is a pair <span class="math-container">$f, g$</span> such that
<span class="math-container">$$∀ x, y • \quad f\,x ≤ y \;\;≡\;\; x ≤′ g\, y$$</span>
Notice the “points” <span class="math-container">$x$</span> and <span class="math-container">$y$</span> from each preorder being utilised.
However, if we move from the category Set to the category Rel, for example,
to consider relations. Then a preorder is reflexive and transitive relation;
let us use <span class="math-container">$E$</span> in-place of <span class="math-container">$≤$</span>. Then the above can be rephrased with <em>no points</em>
<span class="math-container">$$ f˘;E \;=\; E′;g˘$$</span>
Where <span class="math-container">$-;-$</span> is relational composition and <span class="math-container">$-˘$</span> is relational converse.</p>
<p>This is another form of internalisation; it is about rephrasing statements that use,
e.g., logical connectives <span class="math-container">$\forall, \Rightarrow$</span>, into forms that do not use them.
For example, <a href="https://alhassy.github.io/HeytingAlgebra/" rel="nofollow noreferrer">see this presentation of Cartesian Closed Categories in the case of preorders</a>
where properties are shown using, e.g., ∀, then later obtained without it; e.g., having
internal homs <span class="math-container">$[X, Y]$</span> the fact
<span class="math-container">$$\text{there is a unique map from $X$ to the initial object }$$</span>
Can be <em>internalised</em>, i.e., rendered without using the logical notion of existence as
<span class="math-container">$$ [X, ] \;≅\; $$</span></p>
|
2,524,890 | <p>I know that if matrix $a$ is similar to matrix $b$ then $\operatorname{trace} a=\operatorname{trace} b$.</p>
<p>Does it go to the other side?</p>
<p>Thanks.</p>
| John Doe | 399,334 | <p>The equation is actually $$(b-4)^2=16(1-a)$$</p>
<p>Your LHS is a square number, your RHS is $16 x$ where $x$ is an integer. When can your RHS be made into a square number? </p>
<p>It already has $2^4=(2^2)^2$ as a factor, so all you need is for $x$ to be a square number.</p>
<p>So $x=1-a=n^2$ for $n\in \Bbb N$. Then $b-4=\pm4n$. </p>
<p>So solutions are $$a=1-n^2\\b=4\pm4n$$
$\forall n\in \Bbb N$.</p>
|
2,524,890 | <p>I know that if matrix $a$ is similar to matrix $b$ then $\operatorname{trace} a=\operatorname{trace} b$.</p>
<p>Does it go to the other side?</p>
<p>Thanks.</p>
| nonuser | 463,553 | <p>First we see that $8\mid b^2$ so $b=4c$ for some integer $c$. Thus we get $$c^2=2c-a\;\;\;\;\Longrightarrow \;\;\;\;c\mid a$$
So $a=cd$ for some integer $d$ and we have now $$c=2-d$$
So all solution are $(a,b)= (2d-d^2, 8-4d)$, where $d\in \mathbb{Z}$.</p>
|
2,600,776 | <blockquote>
<p>A continuos random variable $X$ has the density
$$
f(x) = 2\phi(x)\Phi(x), ~x\in\mathbb{R}
$$
then</p>
<p>(<em>A</em>) $E(X) > 0$</p>
<p>(<em>B</em>) $E(X) < 0$</p>
<p>(<em>C</em>) $P(X\leq 0) > 0.5$</p>
<p>(<em>D</em>) $P(X\ge0) < 0.25$</p>
<p>\begin{eqnarray}
\Phi(x) &=& \text{Cumulative distribution function of } N(0,1)\\
\phi(x) &=& \text{Density function of } N(0, 1)
\end{eqnarray}</p>
</blockquote>
<p>I don't have a slightest clue where to start with. Can someone give me a little push. I saw some answers on same question like this but I didn't understand how should I integrate it when calculating expectation. </p>
| Gono | 384,471 | <p>It holds $$\left(\Phi^2(x)\right)' = 2\Phi(x)\varphi(x)$$</p>
<p>So we have $$P(X \le x) = \Phi^2(x)$$</p>
<p>And we can conclude $$P(X \le 0) = 0.25$$
$$P(X \ge 0) = 1 - P(X < 0) = 0.75$$</p>
<p>For the expectation we get:</p>
<p>$$\begin{align*} E[X] &= \int_0^\infty (1-\Phi^2(x)) dx - \int_{-\infty}^0 \Phi^2(x) dx \\ &= \int_0^\infty (1-\Phi^2(x)) dx - \int_{0}^\infty \Phi^2(-x) dx \\ &= \int_0^\infty (1-\Phi^2(x)) dx - \int_{0}^\infty (1-\Phi(x))^2 dx \\ &= 2 \int_0^\infty \Phi(x)(1-\Phi(x)) dx > 0\end{align*}$$</p>
|
2,600,776 | <blockquote>
<p>A continuos random variable $X$ has the density
$$
f(x) = 2\phi(x)\Phi(x), ~x\in\mathbb{R}
$$
then</p>
<p>(<em>A</em>) $E(X) > 0$</p>
<p>(<em>B</em>) $E(X) < 0$</p>
<p>(<em>C</em>) $P(X\leq 0) > 0.5$</p>
<p>(<em>D</em>) $P(X\ge0) < 0.25$</p>
<p>\begin{eqnarray}
\Phi(x) &=& \text{Cumulative distribution function of } N(0,1)\\
\phi(x) &=& \text{Density function of } N(0, 1)
\end{eqnarray}</p>
</blockquote>
<p>I don't have a slightest clue where to start with. Can someone give me a little push. I saw some answers on same question like this but I didn't understand how should I integrate it when calculating expectation. </p>
| Ant | 66,711 | <p>An intuitive way to see that A) holds is the following:</p>
<p>Forget about the $2$ in front, it won't change the sign of the integral. Now, if there was only $\phi(x)$, you would get zero, because we know what </p>
<p>$$\int _\mathbb R x\phi(x) dx = 0$$</p>
<p>Now you multiply the integrand by $\Phi(x)$. Being a cumulative distribution function, it starts from $0$ and goes to $1$. In $0$, it is $0.5$. This means that $\Phi(x)$ is bigger for $x$ positive than for $x$ negative; that is, $x > 0, y < 0 \implies \Phi(x) > \Phi(y)$. Therefore, the positive parts will have bigger weight than the negative parts. And since before they balanced out to give $0$, now the positive part wins and we have
$$\int _\mathbb R x\phi(x)\Phi(x) dx > 0$$</p>
<p>More formally, </p>
<p>$$\int _\mathbb R x\phi(x)\Phi(x) dx = \int _{-\infty}^0 x\phi(x)\Phi(x) dx + \int _0^{+\infty} x\phi(x)\Phi(x) dx > 0.5\left(\int _\mathbb R x\phi(x) dx\right) = 0$$</p>
|
3,535,088 | <p>A function <span class="math-container">$\phi:X\rightarrow Y$</span> between two topological space <span class="math-container">$(X,\tau)$</span> and <span class="math-container">$(Y,\sigma)$</span> is continuous in <span class="math-container">$x\in X$</span> if and only if for any open set <span class="math-container">$V$</span> in <span class="math-container">$Y$</span> such that <span class="math-container">$\phi(x)\in V$</span> there is a open set <span class="math-container">$U$</span> in <span class="math-container">$X$</span> such that <span class="math-container">$x\in U$</span> and <span class="math-container">$\phi(U)\subseteq V$</span>.</p>
<p>Well with this definition we consider the euclidean space <span class="math-container">$\mathbb{R}^2$</span> and the functions <span class="math-container">\begin{align}f&:\mathbb{R}^2\owns(x,y)\rightarrow(x+y)\in\mathbb{R}\\g&:\mathbb{R}^2\owns(x,y)\rightarrow(xy)\in\mathbb{R}\end{align}</span> and we prove to demonstrate that are continuous functions.</p>
<p>Therefore we consider the point <span class="math-container">$(\alpha,\beta)$</span> and its image under <span class="math-container">$f$</span> or <span class="math-container">$g$</span> and we chose a open range <span class="math-container">$(a,b)$</span> that contains <span class="math-container">$f(\alpha,\beta)$</span> and <span class="math-container">$g(\alpha,\beta)$</span>: so it result that <span class="math-container">$a<\alpha+\beta<b$</span> and <span class="math-container">$a<\alpha\beta<b$</span>.</p>
<p>Now let be
<span class="math-container">$$
\varepsilon_f=\inf\Big\{d\big((\alpha,\beta);(x,a-x)\big),d\big((x,y);(x,b-x)\big)\Big\}\\
\varepsilon_g=\inf \biggl\{d\Big((\alpha,\beta);\Big(x,\frac{a}{x}\Big)\Big),d\biggl((x,y);\Big(x,\frac{b}{x}\Big)\Big)\biggl\}.
$$</span>
If we consider the balls <span class="math-container">$B_f:=B\big((\alpha,\beta),\varepsilon_f\big)$</span> and <span class="math-container">$B_g:=B\big((\alpha,\beta),\varepsilon_g\big)$</span> that contain <span class="math-container">$(\alpha,\beta)$</span> is true that <span class="math-container">$f(B_f)\subseteq(a,b)$</span> and <span class="math-container">$g(B_g)\subseteq(a,b)$</span> and that if <span class="math-container">$(x,y)\in B_f$</span> or <span class="math-container">$(x,y)\in B_g$</span> it result that <span class="math-container">$f(x,y)\in(a,b)$</span> or <span class="math-container">$g(x,y)\in(a,b)$</span>? Then if this is true will we have proved the continuity of <span class="math-container">$f$</span> and <span class="math-container">$g$</span>?</p>
<p>Could someone help me, please?</p>
| daw | 136,544 | <p>Take <span class="math-container">$w\in L^2(\Omega)$</span>. Then
<span class="math-container">$$
\int_\Omega(u_nv_n - uv)w =
\int_\Omega u(v_n - v)w +
\int_\Omega v_n(u_n - u)w .
$$</span>
The first integral vanishes because <span class="math-container">$uw\in L^1(\Omega)$</span> and <span class="math-container">$v_n-v \rightharpoonup^* 0$</span>
in <span class="math-container">$L^\infty(\Omega)=L^1(\Omega)^*$</span>. The second one vanishes since <span class="math-container">$(v_n)$</span> is bounded in <span class="math-container">$L^\infty(\Omega)$</span>, hence <span class="math-container">$(v_nw)$</span> is bounded in <span class="math-container">$L^2(\Omega)$</span>, and <span class="math-container">$u_n\to u $</span> in <span class="math-container">$L^2(\Omega)$</span>.</p>
|
1,921,101 | <p>$∃x.P(x) \Rightarrow ∀x.P(x) $</p>
<p>How can I read this in simple English? I translated it as: There exists an element x for which P(x) implies that for all elements x, P(x) is true - but I feel like this doesn't make much sense. What am I doing wrong here?</p>
| James | 246,902 | <p>Your translation is right: suppose that there is an element $x$ that makes $P(x)$ true. Then all elements make $P(x)$ true.</p>
<p>This does sound a little weird, but just because it is weird, that doesn't make the translation incorrect. (Logic is pretty weird.) There are predicates $P(x)$ that make $\exists x. P(x) \Rightarrow \forall x. P(x)$ true. One possible $P(x)$ is a predicate that doesn't actually depend on $x$, such as the statement '2+2=4'. Do you see how $\exists x. P(x) \Rightarrow \forall x. P(x)$ is true for this $P(x)$?</p>
|
2,657,301 | <p>On the set of natural numbers$\mathbb { N} $, define the operations $a \oplus b := \max(a,b)$
and $a\otimes b := a+b$ Is $(\mathbb {N},\oplus,\otimes)$ is ring? commutative ring with unity? Field?</p>
<p>My solution :</p>
<p>1- $(\mathbb {N},\oplus) $ is abelian groub because : </p>
<p>a. It is comutative
$a \oplus b = \max(a,b)= \max(b,a) =b \oplus a$</p>
<p>b. It is associative
$(a \oplus b) \oplus c =a \oplus( b\oplus c) =\max(a,b,c)$ </p>
<p>C. The identity of element $a$is $a $
$$a \oplus a= \max(a,a)= a $$
d. The inverse also a
$a \oplus a = \max(a,a)=a $</p>
<p>$(\mathbb {N},\times) $ </p>
<p>1.it is comutative :
$a\otimes b = a+b= b+a= b\otimes a$</p>
<ol start="2">
<li><p>$\otimes $has a identity and it is 0
$a\otimes 0 = a+0=a$</p></li>
<li><p>$\otimes $ has inverse and inverse of a is -a
$a\otimes -a = a-a=0$</p></li>
</ol>
<p>Distributive law:
$a\otimes (b\oplus c) =a\otimes (max (b,c))= a+max( b,c) $</p>
<p>It is true ? If not, why?</p>
<p>Thanks </p>
| Adam | 67,429 | <p>The identity for $(\mathbb{N}, \oplus)$ should be $e=0$.</p>
|
303,933 | <p>How can I prove that
$$ \lim_{n\to\infty} \frac{(\ln(n))^a}{n^b} = 0 \;\forall a,b > 0 $$
? Intuitively it is clear to me because of the behavior of the functions. Thanks for all.</p>
<p><strong>Edit</strong> I'm not able to use L'Hopital rule. Sorry.</p>
| Julien | 38,053 | <p>Afer the appropriate change of variable ($x=\ln n$ and then $y=bx$), this boils down to
$$
\lim_{y\rightarrow +\infty}\frac{y^a}{e^y}=0.
$$</p>
<p>Let $n$ be an integer such that $n>a$.
Then, for $y\geq 0$,
$$
e^y=\sum_{k=0}^{+\infty}\frac{y^k}{k!}\geq \frac{y^n}{n!}.
$$
So
$$
0\leq \frac{y^a}{e^y}\leq \frac{n!y^a}{y^n}=\frac{n!}{y^{n-a}}.
$$</p>
<p>Conclude with the squeeze theorem.</p>
|
478,517 | <blockquote>
<p>Construct a topological mapping of the open disk $|z|<1$ onto the whole plane.</p>
</blockquote>
<p>I represent $z=re^{i\theta}$. I thought about the bijection from $(0,1)$ to $(0,\infty)$, which is given by $x\rightarrow \dfrac1x-1$. Applying this to the norm, we will get the mapping $re^{i\theta}\rightarrow\left(\dfrac1r-1\right)e^{i\theta}$. The only problem is that the point $0$ has not been mapped to or from yet. If I map $0$ to itself, the map becomes non-continuous.</p>
| Brian M. Scott | 12,042 | <p>Don’t flip it: just expand it. To do this, you must multiply $z$ by a factor that increases without bound as $|z|\to 1$; one natural choice is $(1-|z|)^{-1}$.</p>
|
3,786,654 | <blockquote>
<p>Let <span class="math-container">$x_1, x_2, x_3 \in \Bbb R$</span>, satisfy <span class="math-container">$0 \leq x_1 \leq x_2 \leq x_3 \leq 4$</span>. If their squares form an arithmetic progression with common difference <span class="math-container">$2$</span>, determine the minimum possible value of <span class="math-container">$$|x_1-x_2|+|x_2-x_3|$$</span></p>
</blockquote>
<hr />
<p>So far, I've started with the fact <span class="math-container">$x_2^2 - x_1^2 = x_3^2 - x_2^2 = 2$</span> since we know their squares form an arithmetic progression with common difference <span class="math-container">$2.$</span> We can solve this to obtain</p>
<p><span class="math-container">$$x_2 = \pm\sqrt{x_1^2+2}, \qquad x_3 = \pm\sqrt{x_1^2+4}$$</span></p>
<p>I'm not sure how to continue. Thanks in advance for the help.</p>
| boojum | 882,145 | <p><a href="https://i.stack.imgur.com/xG0Bj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xG0Bj.png" alt="enter image description here" /></a></p>
<p>( vertical scale compressed )</p>
<p>A way to see this intuitively is to consider the parabolic curve <span class="math-container">$ \ y \ = \ x^2 \ \ . $</span> We are to choose three values of <span class="math-container">$ \ x \ $</span> so that the <em>vertical</em> spacings of their points on the curve are <span class="math-container">$ \ 2 \ $</span> units apart. We see that since the parabola becomes "steeper" as <span class="math-container">$ \ x \ > \ 0 \ $</span> increases, the total spacing <span class="math-container">$ \ x_3 - x_1 \ $</span> <em>decreases</em>. So we would want to have our three values of <span class="math-container">$ \ x \ $</span> be as large as possible in the interval, which means setting <span class="math-container">$ \ x_3 \ = \ 4 \ \ , $</span> for which we then must have <span class="math-container">$ \ x_2 \ = \ \sqrt{14} \ $</span> and <span class="math-container">$ \ x_1 \ = \ \sqrt{12} \ \ . $</span></p>
<p>We can estimate the required spacing by using the slope formula <span class="math-container">$ \ m \ = \ \frac{\Delta y}{\Delta x} \ \ : \ $</span> at a particular value of <span class="math-container">$ \ x \ \ , $</span> we take the nearby points <span class="math-container">$ \ x \ \pm \ \delta \ $</span> to write
<span class="math-container">$$ m \ \ \approx \ \ \frac{(x + \delta)^2 \ - \ (x - \delta)^2}{(x + \delta) \ - \ (x - \delta)} \ \ = \ \ \frac{4·\delta·x}{2 · \delta} \ \ = \ \ 2·x \ \ . $$</span>
The spacing between <span class="math-container">$ \ x_{i} \ $</span> and <span class="math-container">$ \ x_{i+1} \ $</span> will then be about
<span class="math-container">$ \ x_{i+1} - x_{i} \ = \ \frac{\Delta y}{m} \ = \ \frac{2}{2·x_{i}} \ = \ \frac{1}{x_{i}} \ \ , $</span> indicating that the spacing between <span class="math-container">$ \ x-$</span>values will be "tighter" as <span class="math-container">$ \ x \ $</span> increases. For purposes of comparison, with <span class="math-container">$ \ x_3 \ = \ 3 \ \ , $</span> we have roughly <span class="math-container">$ \ x_3 - x_1 \ = \ \frac{1}{\sqrt5} \ + \ \frac{1}{\sqrt7} \ \approx \ 0.825 \ \ , $</span> while for <span class="math-container">$ \ x_3 \ = \ 4 \ \ , \ x_3 - x_1 \ = \ \frac{1}{\sqrt{12}} \ + \ \frac{1}{\sqrt{14}} \ \approx \ 0.556 \ \ . $</span> This confirms our geometrical intuition.</p>
<p>Naturally, the precise answer is derived from the method used by <strong>dezdichado</strong>.</p>
|
4,068,314 | <p>I do know that double negation and LEM are equivalent, but can we prove
<span class="math-container">$$\vdash \neg \neg (p \vee \neg p)$$</span>
without using either of them, in a Fitch-style proof?</p>
| Mauro ALLEGRANZA | 108,274 | <p>Yes, we can.</p>
<p><em>Hint</em>: assume <span class="math-container">$\lnot (p \lor \lnot p)$</span> and derive a contradiction.</p>
<p>The conclusion follows by <span class="math-container">$(\lnot \text I)$</span>, which is intuitionistically valid.</p>
|
3,762,174 | <p>I have some confusion in integration . My confusion marked in red and green circle as given below<a href="https://i.stack.imgur.com/4fc1I.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4fc1I.png" alt="enter image description here" /></a></p>
<p>Im not getting why <span class="math-container">$$\int_{0}^ {x} = \int_{\frac{-1}{2}}^{y}$$</span> and <span class="math-container">$$\int_{0}^{x} = \int_{y}^{1} $$</span> ?</p>
<p>Im not getting how its derive ?</p>
| zkutch | 775,801 | <p>Hint: Try to plot these areas and look firstly from <span class="math-container">$Ox$</span>, then from <span class="math-container">$Oy$</span> axes.</p>
<p>More formally: we should prove</p>
<p><span class="math-container">$$\left\{ \begin{array}{}
0 \leqslant x \leqslant 1 \\
0 \leqslant y \leqslant x
\end{array} \right\} = \left\{ \begin{array}{}
0 \leqslant y \leqslant 1 \\
y \leqslant x \leqslant 1
\end{array} \right\}
$$</span>
and then obtain
<span class="math-container">$$\int\limits_{0}^{1}\int\limits_{0}^{x} dxdy= \int\limits_{0}^{1}\int\limits_{y}^{1}dydx$$</span>
Same for second.</p>
|
454,040 | <p>I need to know whether There exists any continuous onto map from $(0,1)\to (0,1]$</p>
<p>could any one give me any hint?</p>
| rurouniwallace | 35,878 | <p>Find a polynomial that:</p>
<ol>
<li>Crosses the x-axis at $x=0$ and $x=1$.</li>
<li>Has an absolute maximum of $f(x)=1$.</li>
</ol>
<p>$$f(x)=-4(x^2-x),x\in(0,1)$$</p>
|
1,654,545 | <p>My teacher explained this problem to us - "There are $3$ mailboxes. $3$ people put letters in at random. There is no preference for any of the $3$ mailboxes. Compute the probability that each mailbox contains $1$ letter."</p>
<p>I tried this problem on my own and got the wrong answer. I understand the teacher's solution but I don't get what went wrong with my answer. Can someone explain where I went wrong?</p>
<p>My answer: Treat each outcome as a sequence of $3$ numbers denoting the number of letters in each mailbox, e.g $300$ means $3$ in the 1st, $0$ in the second and $0$ in the 3rd mailbox.
Sample space $S = \{300, 030, 003, 111,012,021, 120,102,210,201\}$
$S$ contains $10$ sample points.
Let $A =$ event that each mailbox is chosen once.
$P(A) = 1/10$ </p>
<p>Teacher's solution: Treat each outcome as a sequence of $3$ numbers denoting the person and their corresponding chosen mailbox, e.g $312$ means the first person chose mailbox 3, the second chose mailbox 1, the third chose mailbox 3. So the sample space is $S = \{ 111, 112...\}$ and contains $27$ sample points.
Now let $A =$ event that each mailbox is chosen once.
$A =\{123,132,213,231,312,321\}$
So the answer is $P(A) = 6/27$</p>
| Jan van der Vegt | 298,403 | <p>The problem with your approach is that not every one of those outcomes have the same probability. In your sample space the probability of $300$ is not $\frac{1}{10}$, it is $\frac{1}{27}$. This is because there is no preference, you can multiply the probabilities of each event individually. The first letter has $\frac{1}{3}$ just like the second and the third. Another way of looking at it is that there are multiple ways to get to $111$ but only one way to get to $300$.</p>
|
484,117 | <p>What is the simplest $\Bbb{R}\to\Bbb{R}$ function with two peaks and a valley?</p>
<p>I have a set of points in $\Bbb{R^2}$ and I would like to fit a curve to the points, the points approximately lie on a curve like the one depicted in the following figure:</p>
<p><img src="https://i.stack.imgur.com/8tVLu.png" alt="A $\Bbb{R}\to\Bbb{R}$ function with two peaks and a valley."></p>
<p>My points are such that $a$ (the width of the valley) is almost constant while $b$ (the height of the peaks with respect to the valley) can change.</p>
<p>My informal definition of "simplest" is based on the following requirements:</p>
<ol>
<li>the function should be sufficient smooth</li>
<li>the fit should be easy to do with some off-the-shelf algorithm</li>
<li>I have just the points lying on the curve in figure, so I think that the function should smoothly go to zero to the left of the left peak and to the right of the right peak.</li>
</ol>
<p>My goal is to estimate $b$.</p>
| Bazin | 93,122 | <p>You will not find a close solution, except in some very particular cases such as $f$ is a constant. The uniqueness is not true in general for such a problem since $\sqrt{f(t)-y}$ is only Hölder continuous with exponent $1/2$. However, uniqueness holds say near $t=0, y=0$ provided $f(0)>0$, say with $f$ continuous:
$$
\frac{dy_j}{dt}=\sqrt{f(t)-y_j}, j=1,2, y_j(0)=z\Longrightarrow y_2(t)-y_1(t)=\int_0^t\bigl(\sqrt{f(s)-y_2(s)}-\sqrt{f(s)-y_1(s)} \bigr)ds
$$
which implies with $\rho(t)=\vert y_2(t)-y_1(t)\vert, t\ge 0$
$$
0=\rho(0)\le\rho(t)\le \int_0^t
\frac{\rho(s)}{\sqrt{f(s)-y_2(s)}+\sqrt{f(s)-y_1(s)}}ds.
$$
Since we may assume $f(0)-y_j(0)>0$ and by continuity that for some small enough positive $T$,
$$s\in(0,T),\quad
f(s)-y_j(s)\ge \frac{f(0)-y_j(0)}2,
$$
we get the differential inequality
$0=\rho(0)\le
\rho(t)\le C\int_0^t \rho(s) ds
$
which (Gronwall) implies $\rho\equiv 0$. </p>
|
4,644,904 | <p>Find the greatest and the least values of the function <span class="math-container">$f(x)=\sin x\sin2x$</span> on the interval <span class="math-container">$(-\infty,\infty)$</span>.</p>
<p>The solution presented is as follows:</p>
<blockquote>
<p>Represent the function <span class="math-container">$y=f(x)=\sin x\sin 2x$</span> in the form <span class="math-container">$y=\frac{\cos x-\cos 3x}{2},$</span> whence it is seen that the function is an even one and has a period <span class="math-container">$2\pi.$</span> Hence, it is sufficient to seek the greatest and the least values among the extrema on the interval <span class="math-container">$[0, \pi].$</span> Find the derivative <span class="math-container">$y'=\frac12(3\sin {3x}-\sin x)$</span> In <span class="math-container">$[0,\pi]$</span> the derivative vanishes at the points <span class="math-container">$x_1=0,x_2=\arccos \frac {1}{\sqrt{3}},x_3=\arccos(-\frac{1}{\sqrt{3}}),x_4=\pi.$</span> Compute the values of the function at these points: <span class="math-container">$y(0)=y(\pi)=0,y[\arccos(\pm\frac{1}{\sqrt{3}})]=\pm\frac {4}{3\sqrt{3}}.$</span> Hence, the least value of the function in the interval <span class="math-container">$(-\infty,\infty)$</span> is equal to <span class="math-container">$-\frac{4}{3\sqrt 3}$</span> and <span class="math-container">$\frac{4}{3\sqrt 3}.$</span></p>
</blockquote>
<p>However, in the above solution, I dont get, how do they assert "Hence, it is sufficient to seek the greatest and the least values among the extrema on the interval <span class="math-container">$[0, \pi].$</span>" ? The function is periodic and even but that only leads us to investigate it in the interval <span class="math-container">$[0, 2\pi].$</span> I don't get why investigating this function at <span class="math-container">$[0, \pi]$</span> is sufficient. I am not quite getting it.</p>
| Enkidu | 455,216 | <p>It is also periodic on <span class="math-container">$[-\pi,\pi]$</span>, and so by eveness the solutions are symmetric along <span class="math-container">$0$</span> and so we can focus on <span class="math-container">$[0,\pi]$</span>.</p>
|
259,308 | <p>The output of <code>ListPointPlot3D</code> is shown below:
<a href="https://i.stack.imgur.com/ypt73.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ypt73.png" alt="enter image description here" /></a>
I only want to connect the dots in such a way that it forms a ring-like mesh. However, when I use <code>ListPlot3D</code>, the output is like this:
<a href="https://i.stack.imgur.com/zmaI1.png" rel="noreferrer"><img src="https://i.stack.imgur.com/zmaI1.png" alt="enter image description here" /></a>
Which is not what I want.</p>
<p>May I know how I should modify the code to achieve the intended result? Thank you.</p>
<p>Here is the data I used (with fewer data points):</p>
<pre><code>{{15.,-4.33154,0.015},{15.4114,-3.89839,0.0154114},{15.7792,-3.46523,0.0157792},{16.1037,-3.03208,0.0161037},{16.3847,-2.59893,0.0163847},{16.6224,-2.16577,0.0166224},{16.8169,-1.73262,0.0168169},{16.968,-1.29946,0.0169681},{17.076,-0.866309,0.017076},{17.1408,-0.433154,0.0171408},{17.1624,-6.01122*10^-16,0.0171624},{17.1408,0.433154,0.0171408},{17.076,0.866309,0.017076},{16.968,1.29946,0.0169681},{16.8169,1.73262,0.0168169},{16.6224,2.16577,0.0166224},{16.3847,2.59893,0.0163847},{16.1037,3.03208,0.0161037},{15.7792,3.46523,0.0157792},{15.4114,3.89839,0.0154114},{14.698,-4.79395,2.99474},{15.1194,-4.31455,3.08061},{15.4564,-3.83516,3.14927},{15.7246,-3.35576,3.2039},{15.9361,-2.87637,3.24701},{16.1009,-2.39697,3.28058},{16.2264,-1.91758,3.30616},{16.3185,-1.43818,3.32492},{16.3815,-0.95879,3.33775},{16.4181,-0.479395,3.34521},{16.4301,-6.65294*10^-16,3.34766},{16.4181,0.479395,3.34521},{16.3815,0.95879,3.33775},{16.3185,1.43818,3.32492},{16.2264,1.91758,3.30616},{16.1009,2.39697,3.28058},{15.9361,2.87637,3.24701},{15.7246,3.35576,3.2039},{15.4564,3.83516,3.14927},{15.1194,4.31455,3.08061},{13.8101,-5.25915,5.85509},{14.2225,-4.73323,6.02996},{14.5189,-4.20732,6.15563},{14.7316,-3.6814,6.24579},{14.8837,-3.15549,6.31026},{14.9917,-2.62957,6.35607},{15.0675,-2.10366,6.3882},{15.1193,-1.57774,6.41015},{15.1527,-1.05183,6.42432},{15.1714,-0.525915,6.43225},{15.1774,-7.29853*10^-16,6.4348},{15.1714,0.525915,6.43225},{15.1527,1.05183,6.42432},{15.1193,1.57774,6.41015},{15.0675,2.10366,6.3882},{14.9917,2.62957,6.35607},{14.8837,3.15549,6.31026},{14.7316,3.6814,6.24579},{14.5189,4.20732,6.15563},{14.2225,4.73323,6.02996},{12.3716,-5.68073,8.48201},{12.7528,-5.11265,8.74341},{13.0014,-4.54458,8.91387},{13.1635,-3.97651,9.02496},{13.2689,-3.40844,9.09726},{13.3374,-2.84036,9.14419},{13.3815,-2.27229,9.17442},{13.4094,-1.70422,9.19355},{13.4263,-1.13615,9.20515},{13.4353,-0.568073,9.21134},{13.4382,-7.88359*10^-16,9.21329},{13.4353,0.568073,9.21134},{13.4263,1.13615,9.20515},{13.4094,1.70422,9.19355},{13.3815,2.27229,9.17442},{13.3374,2.84036,9.14419},{13.2689,3.40844,9.09726},{13.1635,3.97651,9.02496},{13.0014,4.54458,8.91387},{12.7528,5.11265,8.74341},{10.4398,-6.03187,10.7708},{10.7685,-5.42868,11.1099},{10.9656,-4.8255,11.3133},{11.0838,-4.22231,11.4352},{11.1547,-3.61912,11.5083},{11.1971,-3.01594,11.552},{11.2223,-2.41275,11.5781},{11.2372,-1.80956,11.5935},{11.2457,-1.20637,11.6022},{11.25,-0.603187,11.6066},{11.2513,-8.3709*10^-16,11.608},{11.25,0.603187,11.6066},{11.2457,1.20637,11.6022},{11.2372,1.80956,11.5935},{11.2223,2.41275,11.5781},{11.1971,3.01594,11.552},{11.1547,3.61912,11.5083},{11.0838,4.22231,11.4352},{10.9656,4.8255,11.3133},{10.7685,5.42868,11.1099},{8.09191,-6.3005,12.6302},{8.35003,-5.67045,13.033},{8.49481,-5.0404,13.259},{8.57602,-4.41035,13.3858},{8.62155,-3.7803,13.4569},{8.64707,-3.15025,13.4967},{8.66132,-2.5202,13.5189},{8.66922,-1.89015,13.5313},{8.67348,-1.2601,13.5379},{8.67557,-0.63005,13.5412},{8.67619,-8.74371*10^-16,13.5421},{8.67557,0.63005,13.5412},{8.67348,1.2601,13.5379},{8.66922,1.89015,13.5313},{8.66132,2.5202,13.5189},{8.64707,3.15025,13.4967},{8.62155,3.7803,13.4569},{8.57602,4.41035,13.3858},{8.49481,5.0404,13.259},{8.35003,5.67045,13.033},{5.42138,-6.48148,13.986},{5.5955,-5.83333,14.4352},{5.68862,-5.18519,14.6754},{5.73842,-4.53704,14.8039},{5.76505,-3.88889,14.8726},{5.77928,-3.24074,14.9093},{5.78687,-2.59259,14.9289},{5.79089,-1.94444,14.9393},{5.79297,-1.2963,14.9446},{5.79396,-0.648148,14.9472},{5.79425,-8.99486*10^-16,14.9479},{5.79396,0.648148,14.9472},{5.79297,1.2963,14.9446},{5.79089,1.94444,14.9393},{5.78687,2.59259,14.9289},{5.77928,3.24074,14.9093},{5.76505,3.88889,14.8726},{5.73842,4.53704,14.8039},{5.68862,5.18519,14.6754},{5.5955,5.83333,14.4352},{2.53472,-6.57254,14.7843},{2.6163,-5.91529,15.2601},{2.65878,-5.25803,15.5079},{2.6809,-4.60078,15.6369},{2.69241,-3.94352,15.704},{2.69841,-3.28627,15.739},{2.70152,-2.62902,15.7572},{2.70313,-1.97176,15.7666},{2.70395,-1.31451,15.7713},{2.70433,-0.657254,15.7735},{2.70444,-9.12123*10^-16,15.7742},{2.70433,0.657254,15.7735},{2.70395,1.31451,15.7713},{2.70313,1.97176,15.7666},{2.70152,2.62902,15.7572},{2.69841,3.28627,15.739},{2.69241,3.94352,15.704},{2.6809,4.60078,15.6369},{2.65878,5.25803,15.5079},{2.6163,5.91529,15.2601},{-0.452986,-6.5727,14.9932},{-0.467535,-5.91543,15.4747},{-0.475076,-5.25816,15.7243},{-0.478985,-4.60089,15.8537},{-0.481011,-3.94362,15.9207},{-0.48206,-3.28635,15.9555},{-0.482603,-2.62908,15.9734},{-0.482883,-1.97181,15.9827},{-0.483024,-1.31454,15.9873},{-0.483089,-0.65727,15.9895},{-0.483108,-9.12146*10^-16,15.9901},{-0.483089,0.65727,15.9895},{-0.483024,1.31454,15.9873},{-0.482883,1.97181,15.9827},{-0.482603,2.62908,15.9734},{-0.48206,3.28635,15.9555},{-0.481011,3.94362,15.9207},{-0.478985,4.60089,15.8537},{-0.475076,5.25816,15.7243},{-0.467535,5.91543,15.4747},{-3.42264,-6.48177,14.6043},{-3.5319,-5.8336,15.0705},{-3.58953,-5.18542,15.3164},{-3.61993,-4.53724,15.4462},{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</code></pre>
| N.J.Evans | 11,777 | <p>You can partition the data into units of three or four and create polygons. There's probably a better way to do it, but here's one option:</p>
<pre><code>dat = 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Graphics3D[
Table[
With[{dat = RotateRight[dat, 20*j]},
Table[
Polygon[{dat[[i]], dat[[i + 1]], dat[[i + 21]], dat[[i + 20]]}]
, {i, 1, 19}
]
]
, {j, 1, Length@dat/20}
]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/wR7Xe.png" rel="noreferrer"><img src="https://i.stack.imgur.com/wR7Xe.png" alt="enter image description here" /></a></p>
<p>If you prefer triangles:</p>
<pre><code>Graphics3D[
Table[
With[{dat = RotateRight[dat, 20*j]},
Table[{
Polygon[{dat[[i]], dat[[i + 1]], dat[[i + 20]]}],
Polygon[{dat[[i + 1]], dat[[i + 20]], dat[[i + 21]]}]
}
, {i, 1, 19}
]
]
, {j, 1, Length@dat/20}
]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/W534d.png" rel="noreferrer"><img src="https://i.stack.imgur.com/W534d.png" alt="enter image description here" /></a></p>
|
2,448,696 | <p>Show that $\frac{1}{n}<\ln n$, for all $n>1$ where n is a positive integer</p>
<p>I've tried using induction by multiplying both sides by $\ln k+1$ and $\frac{1}{k+1}$ but but all it does is makes it more complicated, I've tried using the fact that $k>1$ and $k+1>2$ during the inductive $k+1$ step, but I'm still stuck. </p>
<p>Looking for clues for this question.</p>
| Jihoon Kang | 452,346 | <p>Suppose $\frac1{k}<\ln(k)$. Then: $$\frac1{k+1}<\frac1{k}<\ln(k)<\ln(k+1)$$</p>
<p>Now you just have to show $\frac12<\ln(2)$.</p>
|
2,448,696 | <p>Show that $\frac{1}{n}<\ln n$, for all $n>1$ where n is a positive integer</p>
<p>I've tried using induction by multiplying both sides by $\ln k+1$ and $\frac{1}{k+1}$ but but all it does is makes it more complicated, I've tried using the fact that $k>1$ and $k+1>2$ during the inductive $k+1$ step, but I'm still stuck. </p>
<p>Looking for clues for this question.</p>
| Fred | 380,717 | <p>For $n \ge 2$ we have:</p>
<p>$1/n \le 1/2 < \ln 2 \le \ln n$.</p>
|
4,089,114 | <p>I'm a newbie for mathematics and now I'm learning PDE and stuck on that. Could anyone help me out to understand this elimination from PDE. The equation is similar to solve <span class="math-container">$$(D^2 -6DD'+9D'^2)u = y\cos x$$</span></p>
| Summand | 754,593 | <p>From Frobenius inequality we have <span class="math-container">$$\operatorname{rank}(ABC)+\operatorname{rank}(B) \geq \operatorname{rank}(AB)+\operatorname{rank}(BC)$$</span></p>
<p>Set <span class="math-container">$A=A, B=A^k, C=A$</span> so that <span class="math-container">$ABC=A^{k+2},AB=BC=A^{k+1}$</span>. But then:</p>
<p><span class="math-container">$$\operatorname{rank}(A^{k+2})+\operatorname{rank}(A^k) \geq 2\operatorname{rank}(A^{k+1})$$</span> for all <span class="math-container">$k$</span> and thus we are done.</p>
|
1,163,033 | <p>I want to calculate $ 8^{-1} \bmod 77 $ </p>
<p>I can deduce $ 8^{-1} \bmod 77$ to $ 8^{59} \bmod 77 $ using Euler's Theorem.</p>
<p>But how to move further now. Should i calculate $ 8^{59} $ and then divide it by $ 77 $ or is there any other theorem i can use ? </p>
| Fermat | 83,272 | <p>$$8\equiv 1 \bmod 7\implies 8^{-1}\equiv 8 \bmod 7$$
also
$$7\times 8\equiv 1 \bmod 11\implies 8^{-1}\equiv 7 \bmod 11$$
Now apply the chinese remainder theorem. Or without this theorem you can write
$$11\times 8^{-1}\equiv 88 \bmod 77$$
and
$$7\times8^{-1}\equiv 49 \bmod 77$$
multiplying the latter by 2 and subtracting from the previous gives us
$$3\times 8^{-1}\equiv 10 \bmod 77$$
now multiply both sides of this by $26$
$$78\times 8^{-1}\equiv 8^{-1}\equiv 260\equiv 29\bmod 77$$
i.e.,the inverse of $8$ $\bmod 77$ is $29$.</p>
|
1,561,370 | <p>Is there any graphical interface in <a href="http://gap-system.org" rel="noreferrer">GAP</a>? Something like <a href="https://www.rstudio.com/" rel="noreferrer">RStudio</a> for <a href="https://www.r-project.org/" rel="noreferrer">R</a> or <a href="http://andrejv.github.io/wxmaxima/" rel="noreferrer">WxMaxima</a> for <a href="http://maxima.sourceforge.net/" rel="noreferrer">Maxima</a>. I'm using GAP under a Linux system.
Thanks</p>
| Olexandr Konovalov | 70,316 | <p>There were several attempts, one started as Max Neunhöffer's GAP package <a href="http://www.gap-system.org/Packages/xgap.html" rel="nofollow noreferrer">XGAP</a> which works only on Linux, others as external software (see at the bottom of <a href="http://www.gap-system.org/Packages/undep.html" rel="nofollow noreferrer">Undeposited Implementations for GAP</a> page). However, none of them is currently in the active development.</p>
<p>As @Bernard remarked above, one could use GAP via SageMath (careful about components and packages installed, that's not identical), and there is also a work in progress on a <a href="https://jupyter.org/" rel="nofollow noreferrer">Jupyter</a> kernel for GAP - when that will be ready, that would be the answer number one on this question.</p>
<p>Often the question is not really why one may be interested in a GUI for GAP, but which functionality could such a GUI add to GAP? One of the unique features of the <a href="http://www.gap-system.org/Packages/xgap.html" rel="nofollow noreferrer">XGAP package</a> is its functionality to work with lattices of subgroups. Although GAP has <a href="http://www.gap-system.org/Manuals/doc/ref/chap39.html#X7E5DF287825EE7BA" rel="nofollow noreferrer"><code>DotFileLatticeSubgroups</code></a> to export the lattice to <a href="http://www.graphviz.org/" rel="nofollow noreferrer">GraphViz</a>, with XGAP one could not only visualise them, but also explore interactively. </p>
<p>I would like to post here some screenshots (made on Ubuntu 14) to illustrate some of XGAP functionality. This is how the starting screen looks like:</p>
<p><a href="https://i.stack.imgur.com/Ludu0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ludu0.png" alt="Starting screen"></a></p>
<p>After entering</p>
<pre><code>GraphicSubgroupLattice( DihedralGroup(16) );
</code></pre>
<p>it opens new window. Diamonds correspond to normal subgroups, and conjugacy classes of subgroups correspond to circles grouped together. Right-click on a subgroup opens small pop-up window with its properties:</p>
<p><a href="https://i.stack.imgur.com/b9nhm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/b9nhm.png" alt="Lattice of subgroups"></a></p>
<p>One could specify which subgroups to display and perform operations with one or several selected subgroups: </p>
<p><a href="https://i.stack.imgur.com/jdCCU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jdCCU.png" alt="Menu for exploring subgroups"></a></p>
<p>Also, one could control the layout of the lattice:</p>
<p><a href="https://i.stack.imgur.com/BPTJf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BPTJf.png" alt="Menu for lattice layout"></a></p>
<p>An interesting and useful project could be to reimplement this functionality using some more modern and cross-platform framework.</p>
|
2,526,716 | <p>Define, by structural induction, a function $f : A^* \to A^*$ that removes all occurrences of the letter $a$. For instance, we should have
$f(abcbab) = bcbb$ and $f(bc) = bc$.</p>
<p>I came up with this:</p>
<p>$f(\lambda) = \lambda$ (empty word)</p>
<p>$f(xw) = xf(w)$ if $x$ is not $a$</p>
<p>$f(w)$, otherwise.</p>
<p>But I have no idea if this is sufficient and if its a structural induction. </p>
<p>Thanks in advance</p>
| zoli | 203,663 | <p>The minimum distance is at the $x$ at which the tangent to the parabola is $4$. Where the tangent is parallel to the straight line:</p>
<p><a href="https://i.stack.imgur.com/kGTYq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kGTYq.png" alt="enter image description here"></a></p>
<p>So, one has to solve the following equation</p>
<p>$$\frac12x^{-\frac12}=4.$$</p>
<p>The solution is $$x_m=\frac1{64}.$$</p>
|
316,866 | <p>Suppose $(a_n)$ is a real sequence and $A:=\{a_n \mid n\in \Bbb N \}$ has an infinite linearly independent subset (with respect to field $\Bbb Q$). Is $A$ dense in $\Bbb R?$</p>
| user642796 | 8,348 | <p>Not necessarily. We can use recursion to construct a linearly independent set $A = \{ a_n : n \in \mathbb{N} \}$ so that $a_n \in ( n , n+1 )$ for all $n$. Such a set is clearly not dense, but even more it is a closed discrete subset of $\mathbb{R}$.</p>
<p>Even a Hamel basis $B$ for $\mathbb{R}$ may not be dense in $\mathbb{R}$ (just ensure that all elements of $B$ avoid some open interval), however unlike the simple infinite linearly independent case by the Baire Category Theorem $B$ cannot be nowhere dense (since we have $\mathbb{R} = \bigcup_{q \in \mathbb{Q}} ( q + B )$).</p>
|
1,622,779 | <p>I have been asked to calculate $\frac{1+i \tan \alpha}{1-i \tan \alpha}$, where $\alpha \in \mathbb{R}$.</p>
<p>So, I multiplied top and bottom by the complex conjugate of the denominator:</p>
<p>$\frac{(1+i \tan \alpha)(1+ i \tan \alpha)}{(1-i \tan \alpha)(1+ i \tan \alpha) } = \frac{1 + 2 i \tan \alpha + i^{2} \tan^{2} \alpha}{1-i^{2} \tan^{2} \alpha} = \frac{1 + 2i \tan \alpha - \tan ^{2} \alpha}{1 + \tan^{2} \alpha} = \frac{(1-\tan ^{2} \alpha)+2 i \tan \alpha}{\sec ^{2} \alpha} =\displaystyle \frac{1-\tan^{2} \alpha }{\sec ^{2} \alpha} + 2\sin\alpha\cos\alpha i$</p>
<p>Is this all that there is to do? Have I simplified it enough?</p>
| Shaswata | 68,110 | <p>Well what you really should've tried is multiplying the numerator and denominator by $\cos \alpha$ which would give you,
$$\frac{\cos\alpha +i \sin \alpha}{\cos\alpha -i \sin \alpha}=\frac{e^{i\alpha}}{e^{-i \alpha}}=e^{i 2 \alpha} = \cos 2\alpha + i \sin 2\alpha$$</p>
|
732,996 | <p><img src="https://i.stack.imgur.com/kXJEt.png" alt="enter image description here"></p>
<p>Hi! I am working on some ratio and root test online homework problems for my calc2 class and I am not sure how to completely solve this problem. I guessed on the second part that it converges, but Im not sure how to solve of the value that it converges to. If someone could possibly help me with this problem it would be greatly appreciated. </p>
| Ted Shifrin | 71,348 | <p>You aren't expected to figure out what the value of the series is (although in time you might figure out it has something to do with $e+e^{-1}$). Did you actually compute $\rho$? What is $$\lim_{n\to\infty} \frac{(2n)!}{(2n+2)!}?$$</p>
|
605,772 | <p>Solving $x^2-a=0$ with Newton's method, you can derive the sequence $x_{n+1}=(x_n + a/x_n)/2$ by taking the first order approximation of the polynomial equation, and then use that as the update. I can successfully prove that the error of this method converges quadratically. However, I can't seem to prove this for the residual, and this is likely a simple problem in arithmetic:</p>
<p>\begin{align*}
|x_{n+1}^2 - {a}| &= \left|\frac{1}{4}\Big(x_n+\frac{a}{x_n}\Big)^2 - {a}\right| \\
&= \left|\frac{1}{4}\Big(x_n^2+2a +\frac{a^2}{x_n^2}\Big) - {a}\right| \\
&= \left|\big(\frac{1}{2}x_n\big)^2-\frac{1}{2}a +\big(\frac{a}{2x_n}\big)^2\right| \\
&= \frac{1}{4}\left|x_n^2-2a +\big(\frac{a}{x_n}\big)^2\right| \\
&= \frac{1}{4}\left|\big(x_n+\frac{a}{x_n}\big)^2-2a +\big(\frac{a}{x_n}\big)^2\right|
\end{align*}</p>
<p>I get stuck here, as well as trying other expansions/factorizations. Is there a way to have this simplify?</p>
| Felix Marin | 85,343 | <p>$\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\fermi}{\,{\rm f}}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\half}{{1 \over 2}}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ket}[1]{\left\vert #1\right\rangle}%
\newcommand{\ol}[1]{\overline{#1}}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
\newcommand{\sech}{\,{\rm sech}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{x_{n + 1} = \half\pars{x_{n} + {a \over x_{n}}}\,,\quad\lim_{n \to \infty} = \root{a}}$.</p>
<blockquote>
Let's define $\epsilon_{n} \equiv x_{n} - \root{a}$ and we assume $x_{n}$'s are 'close' $\pars{~\verts{\epsilon_{n}} \ll \root{a}~}$ to $\root{a}$. Then,
\begin{align}
\epsilon_{n + 1} + \root{a}
&=
\half\pars{\epsilon_{n} + \root{a} + {a \over \epsilon_{n} + \root{a}}}
=
\half\pars{\epsilon_{n} + \root{a} + {\root{a} \over 1 + \epsilon_{n}/\root{a}}}
\\[3mm]&\approx
\half\bracks{\epsilon_{n} + \root{a}
+
\root{a}\pars{1 - {\epsilon_{n} \over \root{a}} + {\epsilon_{n}^{2} \over a}}}
=
\root{a} + {\epsilon_{n}^{2} \over 2\root{a}}
\end{align}
</blockquote>
<p>$$
\color{#0000ff}{\large\epsilon_{n + 1}}
\approx
\color{#0000ff}{\large{\epsilon_{n}^{2} \over 2\root{a}}}
$$
That is what is usually means by "$\tt\mbox{the error decreases quadratically}$". Notice that the worst scenario is $a \gtrsim 0$.</p>
<blockquote>
Also,
$$
\verts{x_{n + 1}^{2} - a} = \verts{x_{n + 1} - \root{a}}\
\verts{x_{n + 1} + \root{a}}
\approx
\verts{\epsilon_{n + 1}}\pars{2\root{a}}
$$
</blockquote>
<p>$$
\color{#0000ff}{\large\verts{x_{n + 1}^{2} - a}}
\approx \color{#0000ff}{\large\epsilon_{n}^{2}}
$$</p>
|
135,426 | <p>$$\frac{d}{dq}\int_{s_{1}-z-q}^{z-s_{1}} \varphi(w) \, dw$$</p>
<p>(if it helps, in my setting $\varphi$ is the CDF of some arbitrary uniform distribution). So I want to get a nice expression for this integral and it seems to suggest FTC, but I tried a change of variable and ended up with a $q$ inside the integrand which was not nice. Any help much appreciated.</p>
| Jyrki Lahtonen | 11,619 | <p>You may be familiar with Leibniz' result on an alternating monotonically decreasing (in absolute value) series: the partial sum is above (resp. below) the sum, if the last included term was positive (resp. negative). That gives us a clue. Compute
$$
\frac12-\frac13+\frac14-\frac15+\frac16-\frac17+\frac18=\frac{307}{840}\approx0.365<3/8.
$$
If you have not heard of Leibniz, you may simply observe that $-1/9+1/10<0$, $-1/11+1/12<0$, $-1/13+1/14<0$, $\ldots$ </p>
|
3,498,199 | <p>Suppose if a matrix is given as</p>
<p><span class="math-container">$$ \begin{bmatrix}
4 & 6\\
2 & 9
\end{bmatrix}$$</span></p>
<p>We have to find its eigenvalues and eigenvectors.</p>
<p>Can we first apply elementary row operation . Then find eigenvalues.</p>
<p>Is their any relation on the matrix if it is diagonalized or not.</p>
| CroCo | 103,120 | <p>First compute the eigenvalues as follows:
<span class="math-container">$$
\begin{align}
\det(\lambda I-A)&=\det\left(
\begin{bmatrix}
\lambda & 0 \\
0 & \lambda
\end{bmatrix}
-
\begin{bmatrix}
4 & 6 \\
2 & 9
\end{bmatrix}
\right)
= \det \left(
\begin{bmatrix}
\lambda-4 & -6 \\
-2 & \lambda-9
\end{bmatrix}
\right) \\
&=
(\lambda-4)(\lambda-9)-(-2)(-6) \\
&=
\lambda^2-13\lambda + 24
\end{align}
$$</span>
The eigenvalues are then
<span class="math-container">$$
\begin{align}
\lambda_{1,2} &= \frac{13 \pm \sqrt{13^2-4\times 24} }{2} \\
&= \frac{13\pm \sqrt{73}}{2}
\end{align}
$$</span>
Now to compute the eigenvector for <span class="math-container">$\lambda_1$</span>, we need to solve the following
<span class="math-container">$$
\begin{align}
\begin{bmatrix}
\lambda_1 - 4 & -6 \\
-2 & \lambda_1 -9
\end{bmatrix}
&=
\begin{bmatrix}
\frac{13 + \sqrt{73} }{2} - 4 & -6 \\
-2 & \frac{13 + \sqrt{73} }{2} -9
\end{bmatrix} \\
&=
\begin{bmatrix}
\frac{5 + \sqrt{73} }{2} & -6 \\
-2 & \frac{-5 +\sqrt{73} }{2}
\end{bmatrix}
\end{align}
$$</span>
Apply Gaussian elimination (i.e. <span class="math-container">$R_2=R_2+2\frac{2}{5+\sqrt{73}}R_1)$</span>, we get
<span class="math-container">$$
\begin{bmatrix}
\frac{5 + \sqrt{73} }{2} & -6 \\
0 & 0
\end{bmatrix}
$$</span>
So, <span class="math-container">$x_2$</span> is a free variable and let <span class="math-container">$x_2=s$</span>, we
<span class="math-container">$$
\begin{align}
\frac{5 + \sqrt{73} }{2} x_1 = 6x_2 \implies x_1 = \frac{12}{5+\sqrt{73}}s
\end{align}
$$</span>
Finally,
<span class="math-container">$$
x=
\begin{bmatrix}
x_1 \\ x_2
\end{bmatrix}
=
\begin{bmatrix}
\frac{12}{5+\sqrt{73}} s \\ s
\end{bmatrix}
=
s
\begin{bmatrix}
\frac{12}{5+\sqrt{73}} \\ 1
\end{bmatrix}
$$</span>
The eigenvector for <span class="math-container">$\lambda_1$</span> is
<span class="math-container">$$
v_1 =
\begin{bmatrix}
\frac{12}{5+\sqrt{73}} \\ 1
\end{bmatrix}
$$</span>
Now you can compute the second eigenvector in the same scenario. </p>
|
2,697,069 | <p>Two series of functions are given in which I cannot figure out how to find $M_n$ of the second problem. $$1.\space \sum_{n=1}^{\infty} \frac{1}{1+x^n}, x\in[k,\infty)\\ 2. \space \sum_{n=1}^{\infty} (\cos x)^n, x\in(0,\pi)$$.. </p>
<p>I have determined the $M_n$ for problem no. $1.$ [$\space|\sum_{n=1}^{\infty} \frac{1}{1+x^n}|<|\sum_{n=1}^{\infty} \frac{1}{1+k^n}|<\sum_{n=1}^{\infty} \frac{1}{k^n}$] </p>
<p>From problem no. $2.$, since $-1\leq \cos x\leq1$, therefore for higher $n$ the values of $\cos x$ will lie between $[-1,1]$ and in $(0,\pi)$ $\cos x$ is decreasing. But is it correct to choose $n$ as $M_n$, so that $$|f_n(x)|=|(\cos x)^n|<n,$$ where $n$ is decreasing. </p>
<p>I am not sure what the $M_n$ should be. Any help or suggestion please? Any help is greatly appreciated.</p>
| Rebellos | 335,894 | <p>It is : $y=\frac{2}{5}x$ so simply by substituting :</p>
<p>$$\frac{\frac{2}{5}x}{3} = \frac{z}{4} \Leftrightarrow \frac{x}{z} = \frac{15}{8}$$</p>
|
2,416,424 | <p>It is known that the collection of finite mixtures of Gaussian Distributions over $\mathbb{R}$ is dense in $\mathcal{P}(\mathbb{R})$ (the space of probability distributions) under convergence in distribution metric.</p>
<p>I'm interested to know the following:</p>
<p>Let $P_X$ be a random variable with finite $p$ th moment i.e. $\mathbb{E}_{P_X}[|X|^p]<\infty$, and $P_{X_n}\stackrel{d}{\to} P_X$ where $P_{X_n}$ are mixtures of Gaussian distributions. Then, suppose $X_n \sim P_{X_n}$ and $X\sim P_X$, does it follow that
$$\mathbb{E}[|X_n|^p] \to \mathbb{E}[|X|^p]$$? $\quad (*)$</p>
<p><strong>My attempt:</strong></p>
<p>I've been able to show that
$$\liminf_{n \to \infty} \mathbb{E}[|X_n|^p] \geq \mathbb{E}[|X|^p]$$
In fact this didn't even use the mixtures part. However I'm having difficulty showing the upper bound. Here are some steps:
\begin{eqnarray}
\mathbb{E}[|X_n|^p] &=& \mathbb{E}[|X_n|^p1_{\{|X_n|\leq A\}}] + \mathbb{E}[|X_n|^p1_{\{|X_n|>A\}}] \\
&=& \mathbb{E}[|X_n|^p1_{\{|X_n|\leq A\}}] + A^pPr[|X_n|\geq A] + \mathbb{E}[(|X_n|^p-A^p)1_{\{|X_n|>A\}}]
\end{eqnarray}
Let $f(x)=x1_{0\leq x\le A} + A1_{x\geq A}$ which is a continuous and bounded function. Then the first two terms of RHS are equal to $\mathbb{E}[f(|X_n|^p)]$, which by definition of weak convergence, will converge to $\mathbb{E}[f(|X|^p)]$ as $n \to \infty$. Now we may use MCT as $A\to \infty$ to get $\mathbb{E}[|X|^p]$. Hence in a nutshell, we need to show:
\begin{equation}
\limsup_{A\to\infty} \limsup_{n \to \infty} \mathbb{E}[(|X_n|^p-A^p)1_{\{|X_n|>A\}}] \leq 0
\end{equation}
However I do not know how to proceed from here.</p>
<p>I'd like to point out that I don't know the answer to the question I asked in $(*)$ but my guess is that this is true. The reason is because I read a similar result in Soren Asmussen's book. The result states that for distributions over non-negative reals, we have phase type distributions not only being weak dense but also the moments converging. But in the provided proof, they said "It can be easily shown that the moments converge".</p>
<p><strong>Update:</strong>
Sincerest apologies for I forgot one very essential condition. The mixtures are not any mixtures but specific ones. Namely, I'm looking at distributions $P_X$ such that $P_{X_n}$ has the form $\sum_{k=1}^n \alpha_k \mathcal{N}(\mu_k;P_k)$ ($\mathcal{N}(\mu_k;P_k)$ stands for Gaussian distribution with mean $\mu_k$ and variance $P_k$.) where given $P<\infty$, $\sum_{k=1}^n \alpha_k =1$, $\alpha_k >0$ and $\sum_{k=1}^n\alpha_k P_k = P$. </p>
<p><strong>Update 2:</strong> The reference I mentioned earlier is Soren Asmussen "Applied probability and queues", 2nd edition, page 84.</p>
<p><strong>Update 3:</strong> Looks like I misinterpreted Asmussen. What he wanted to say was that for any distribution $P_X$ with finite $p$th moment, <strong>there exists</strong> a sequence of phase type distributions $P_X^k$ such that $P_X^k\stackrel{d}{\to}P$ and $E_{P_X^K}[|X|^p] \to E_{P_X}[|X|^p]$. This doesn't mean any weak converging mixture will have moment convergence as the answer points out.</p>
| Theo Bendit | 248,286 | <p>Finding the roots is a good idea when your inequality is in the form $f(x) > 0$, where $f$ is a <strong>continuous</strong> function (the definition of "continuous" is a little technical, but it means, in essence, that the graph can be drawn without lifting pencil off paper).</p>
<p>There is a theorem about continuous functions called the Intermediate Value Theorem, that supports this. A simple, equivalent formulation of the Intermediate Value Theorem states that, if $f(x)$ is continuous over an interval $a \le x \le b$ (or indeed the whole real line), and $f(a) < 0 < f(b)$ (or $f(b) < 0 < f(a)$), then there is a root $c$ of $f$ between $a$ and $b$.</p>
<p>In other words, if the graph begins on one side of the $x$-axis and ends on the other side of the $x$-axis, then it must pass through the $x$-axis somewhere in between.</p>
<p>How is this helpful? Well, it means that the intervals where $f(x) > 0$ must be between roots. You'll never get a dip into negative values without there being a root there too.</p>
<p>Of course, all of this requires $f(x)$ to be continuous. If $f(x)$ has some kind of jump, or an assymptote, then $f(x)$ can easily stop being positive and start being negative, without there being a root in between. For example, $f(x) = 1/x$.</p>
<hr>
<p>In your case, we have $f(x) = \frac{2^x}{2^x - 8} - 2$. This function has one asmyptote, where $2^x - 8 = 0$, or equivalently, $x = 3$. Everywhere else it's continuous. According to your own (correct) working, this function has one root: when $2^x = 16$, or equivalently, at $x = 4$. Let's piece this together to form the solution.</p>
<p>For $x > 4$, the function is continuous, but the function has no roots. This means, by the Intermediate Value Theorem, it is either all positive or all negative (or else there'd be a root in between). Since $f(5) < 0$ (plug it into your calculator), we have $f(x) < 0$ for $x > 4$.</p>
<p>For $3 < x < 4$, the function is also continuous, and there are no roots. Similarly, it's all positive or all negative. Trying a value in this range (e.g. $f(3.5)$) yields a positive value, so $f(x) > 0$ for $3 < x < 4$.</p>
<p>For $x < 3$, the function is continuous and has no roots. You know the drill. Since $f(2) = -3 < 0$, the function is negative for all $x < 3$.</p>
<p>Finally, note that $f(4) = 0$ (as we found earlier) and $f(3)$ does not exist. This clears up all the possibilities. Thus, the only interval on which $f(x) > 0$ is $3 < x < 4$. In interval notation, $x \in (3, 4)$.</p>
<p>I know that this is a far wordier approach to this question than you've probably seen before, but you asked about a general rule, and the Intermediate Value Theorem is about the closest thing we have to a general rule with (mostly) continuous inequalities.</p>
|
3,547,529 | <p>I did the following: I set <span class="math-container">$3^m+3^n+1=x^2$</span> where <span class="math-container">$x\in\Bbb{N}$</span> and assumed it was true for positive integer exponents and for all whole numbers x so that I can later on prove it's invalidity with contradiction. Since <span class="math-container">$3^m+3^n+1$</span> is odd we can write <span class="math-container">$3^m+3^n+1 = (2k+1)^2$</span> for <span class="math-container">$k\in\Bbb{N}$</span>. After a while I can't seem to prove it and I don't have any ideas on how else to approach this problem. </p>
<p>It would be very helpful if someone could give me a brief explanation on a possible proof since I'm currently practicing for a math competition. </p>
<p>Thanks in advance </p>
| Yiorgos S. Smyrlis | 57,021 | <p>If <span class="math-container">$A(y)>0$</span>, such that
<span class="math-container">$$
a\big(A(y)\big)^{2n}+b\big(A(y)\big)^{2n-1}+\cdots+q=y, \tag{1}
$$</span>
then <span class="math-container">$A(y)\to\infty$</span>, as <span class="math-container">$yas\to\infty$</span>, and hence <span class="math-container">$\big(A(y)\big)^{2n}\gg \big(A(y)\big)^{2n-1}$</span>. So from <span class="math-container">$(1)$</span> we obtain <span class="math-container">$A(y)\approx\Big(\frac{y}{a}\Big)^{\frac{1}{2n}}$</span>. In particular, there exists a function <span class="math-container">$e_+(y)$</span>, such that
<span class="math-container">$$
A(y)=\Big(\frac{y}{a}\Big)^{\frac{1}{2n}}+e_+(y), \quad |e_+(y)|\ll y^\frac{1}{2n}.
$$</span>
To get a better estimate for <span class="math-container">$e_+$</span>, we have
<span class="math-container">$$
a\bigg(\Big(\frac{y}{a}\Big)^{\frac{1}{2n}}+e_+(y)\bigg)^{2n}+b\bigg(\Big(\frac{y}{a}\Big)^{\frac{1}{2n}}+e_+(y)\bigg)^{2n-1}+\cdots+q=y,
$$</span>
or
<span class="math-container">$$
\bigg(y+2na\Big(\frac{y}{a}\Big)^{\frac{2n-1}{2n}}e_+(y)+\frac{2n(2n-1)}{2}a\Big(\frac{y}{a}\Big)^{\frac{2n-2}{2n}}e_+^2(y)+\cdots\bigg)+\bigg(b\Big(\frac{y}{a}\Big)^{\frac{2n-1}{2n}}+b\Big(\frac{y}{a}\Big)^{\frac{2n-2}{2n}}e_+(y)+\cdots\bigg)\\+\cdots+q=y.
$$</span>
The assumption <span class="math-container">$|e_+(y)|\ll y^\frac{1}{2n}$</span> implies now that
<span class="math-container">$$
e_+(y)=-\frac{b}{2an}+d_+(y)
$$</span>
where <span class="math-container">$\lim_{y\to\infty} d_+(y)=0$</span>.</p>
<p>Repeating this argument for <span class="math-container">$B(y)<0$</span>, with
<span class="math-container">$$
a\big(B(y)\big)^{2n}+b\big(B(y)\big)^{2n-1}+\cdots+q=y,
$$</span>
we obtain that
<span class="math-container">$$
B(y)=-\Big(\frac{y}{a}\Big)^{\frac{1}{2n}}+e_-(y), \quad |e_+(y)|\ll y^\frac{1}{2n}.
$$</span>
and similarly we obtain that
<span class="math-container">$$
e_-(y)=-\frac{b}{2an}+d_-(y)
$$</span>
where <span class="math-container">$\lim_{y\to\infty} d_-(y)=0$</span>.</p>
<p>Hence
<span class="math-container">$$
A(y)+B(y)=-\frac{b}{na}+d_-(y)+d_+(y)\to -\frac{b}{na}.
$$</span></p>
|
3,547,529 | <p>I did the following: I set <span class="math-container">$3^m+3^n+1=x^2$</span> where <span class="math-container">$x\in\Bbb{N}$</span> and assumed it was true for positive integer exponents and for all whole numbers x so that I can later on prove it's invalidity with contradiction. Since <span class="math-container">$3^m+3^n+1$</span> is odd we can write <span class="math-container">$3^m+3^n+1 = (2k+1)^2$</span> for <span class="math-container">$k\in\Bbb{N}$</span>. After a while I can't seem to prove it and I don't have any ideas on how else to approach this problem. </p>
<p>It would be very helpful if someone could give me a brief explanation on a possible proof since I'm currently practicing for a math competition. </p>
<p>Thanks in advance </p>
| Paramanand Singh | 72,031 | <p>Let's replace <span class="math-container">$A(y), B(y) $</span> by <span class="math-container">$A, B$</span> to simplify typing and let <span class="math-container">$A>0>B$</span> and we write <span class="math-container">$C=-B$</span> so that <span class="math-container">$C>0$</span>. Then we have <span class="math-container">$$y=aA^{2n}+bA^{2n-1}+\dots$$</span> and <span class="math-container">$$y=aC^{2n}-bC^{2n-1}+\dots$$</span> From these equations we get <span class="math-container">$$y/A^{2n}\to a, y/C^{2n}\to a$$</span> so that <span class="math-container">$A/C\to 1$</span>.</p>
<p>Subtracting these equations we get <span class="math-container">$$a(A^{2n}-C^{2n})+b(A^{2n-1}+C^{2n-1})+\dots=0$$</span> or <span class="math-container">$$a(A-C) (A^{2n-1}+A^{2n-2}C+\dots+C^{2n-1})+b(A^{2n-1}+C^{2n-1})+\dots =0$$</span> Dividing the above equation by <span class="math-container">$C^{2n-1}$</span> we get <span class="math-container">$$a(A-C)\{1+(A/C)+(A/C)^2+\dots+(A/C)^{2n-1}\} +b\{1+(A/C)^{2n-1}\} + \text{ (terms tending to zero)} =0$$</span> Letting <span class="math-container">$y\to \infty $</span> in above equation we get <span class="math-container">$$2na\lim_{y\to\infty} (A-C) +2b=0$$</span> or <span class="math-container">$$A-C\to-\frac{b} {na} $$</span> as <span class="math-container">$y\to\infty $</span>.</p>
<p>One has to observe that when we divide by <span class="math-container">$C^{2n-1}$</span> the terms like <span class="math-container">$A^{r} / C^{2n-1}$</span> tend to <span class="math-container">$0$</span> for <span class="math-container">$r<2n-1$</span> because we can write it as <span class="math-container">$(A/C) ^r(C^r/C^{2n-1})$</span>.</p>
<hr>
<p>The argument above is entirely elementary and simple. We just have to understand that <span class="math-container">$A, C$</span> are functions of <span class="math-container">$y$</span> which are strictly increasing as <span class="math-container">$y\to\infty $</span> and <span class="math-container">$$A\to\infty, C\to\infty, A/C\to 1$$</span> as <span class="math-container">$y\to\infty $</span>. Further they satisfy the relation <span class="math-container">$F(A) =y=F(-C) $</span>.</p>
<p>In general most of the algebraic limits do not involve anything more algebraic manipulation. </p>
|
2,571,909 | <p>$$\left|\frac{-10}{x-3}\right|>\:5$$</p>
<ul>
<li>Find the values that $x$ can take. </li>
</ul>
<p>I know that</p>
<p>$$\left|\frac{-10}{x-3}\right|>\:5$$
and
$$\left|\frac{-10}{x-3}\right|<\:-5$$</p>
| haqnatural | 247,767 | <p>$$\left| \frac { -10 }{ x-3 } \right| >\: 5\\ \frac { 10 }{ \left| x-3 \right| } >5\\ \left| x-3 \right| <2\\ -2<x-3<2\\ 1<x<5\\ \left( 1;5 \right) -\left\{ 3 \right\} \\ \\ $$</p>
|
1,789,373 | <p>I'm trying to figure out why the following is true:</p>
<p>Let $ \kappa $ be an uncountable, regular cardinal. Suppose we turn it into a group (i.e. there are operations $ (\cdot, ^{-1}, e) $ with which $ \kappa $ is a group. My aim is to prove that the set</p>
<p>$$ \{ \alpha \in \kappa : \alpha \text{ is a subgroup of } \kappa\}$$</p>
<p>is a club. It's rather obvious that it is closed, however I'm not quite sure why it would be unbounded. Suppose we have an $ \alpha \in \kappa $ which is a subgroup. Somehow we need to enlarge $ \alpha $ so that it will still be both a group and an ordinal.</p>
<p>We can close $ \alpha + 1 $ under the group operations, but we have no warranty that it will be an ordinal (it wont be equal to $ \kappa $ since $ \kappa $ is regular).</p>
<p>I don't see any way to approach this. I would appreciate a hints</p>
| zyx | 14,120 | <p>The geometry is the same as in 3 dimensions, when taking the plane section $x+y+z = c$ of a sphere $x^2+y^2+z^2=C$. Extrema of $z$ are when $x=y$. The only remaining thing to understand is why the answer for the parameters given in the problem does not involve square roots.</p>
<p>Let $n$, which equals $4$ in the posted problem, be the number of variables other than the last one, which I will continue to call "$e$".</p>
<p>Max/min $e$ in $(n+1)$ dimensions occur at the two solutions of </p>
<p>$na+e=c$ and $na^2+e^2=C$, so $(c-e)^2 = n(C - e^2)$ or $$(n+1)e^2 - 2ce + (c^2 - nC) = 0 .$$</p>
<p>To get a problem with extremal $e$ rational in the parameters, </p>
<p>$c^2 - (n+1)(c^2 - nC^2) = n((n+1)C - c^2)$ must be a perfect square. </p>
<p>There is no special perfect square expression that appears in the algebra, and the problem was just based on a choice of small integer values that lead to a square.</p>
|
4,506,026 | <p>Consider the set of equations:
<span class="math-container">$$
\begin{cases}
x^2 &= -4y-10\\y^2 &= 6z-6\\z^2 &= 2x+2\\
\end{cases}$$</span></p>
<p>With <span class="math-container">$x,y,z$</span> being real numbers.</p>
<p>By adding the three equations, after simple manipulations, we easily obtain
<span class="math-container">$$
(x-1)^2+(y+2)^2+(z-3)^2=0
$$</span>
Which yields
<span class="math-container">$$
\begin{cases}
x&=1\\y&=-2\\z&=3\\
\end{cases}
$$</span>
However, plugging that in we find that the solution is invalid. How can one reason about this? In other words, <em>why</em> is it invalid? What causes it to be invalid?</p>
<p><a href="https://i.stack.imgur.com/EQMj6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EQMj6.jpg" alt="enter image description here" /></a></p>
| Sambo | 454,855 | <p>This is how I think about it. When you use your initial set of three equations, which I'll call (A), to obtain your final equation, which I'll call (B), what you're saying is the following: "If <span class="math-container">$x,y,z$</span> satisfy (A), then they satisfy (B)".</p>
<p>Importantly, this reasoning only gives you a one-sided implication (A implies B), and not necessarily an equivalence (A if and only if B). That is, the converse ("If <span class="math-container">$x,y,z$</span> satisfy (B), then they satisfy (A)") may not be true. Indeed, you've found an example where this converse is false: you found <span class="math-container">$x,y,z$</span> which satisfy (B) but not (A).</p>
<p>Here's a simpler example. Start with the equality <span class="math-container">$x=2$</span>, then square both side to get <span class="math-container">$x^2=4$</span>. What this tells us is: "If <span class="math-container">$x=2$</span>, then <span class="math-container">$x^2=4$</span>". But the converse is not true: the equation <span class="math-container">$x^2=4$</span> is also satisfied by <span class="math-container">$x=-2$</span>.</p>
|
3,182,802 | <p>Show that if <span class="math-container">$ \sigma $</span> is a solution to the equation <span class="math-container">$ x^2 + x + 1 = 0 $</span> then the following equality occurs:</p>
<p><span class="math-container">$$ (a +b\sigma + c\sigma^2)(a + b\sigma^2 + c\sigma) \geq 0 $$</span></p>
<p>I looked at the solution in my textbook and it says I should multiply the parentheses and take into account that <span class="math-container">$ \sigma + \sigma^2 + 1 = 0 $</span>.
I tried factoring the rest but I just can't seem to manage to solve it?</p>
<p>Maybe I messed up at multiplying the parentheses? Here's what I got:</p>
<p><span class="math-container">$$ a^2 + ab\sigma^2 + ac\sigma + ab\sigma + b^2\sigma^3 + bc\sigma^2 + ac\sigma^2 + bc\sigma^4 + c^2\sigma^3 $$</span> </p>
| Donald Splutterwit | 404,247 | <p>First note that <span class="math-container">$ \sigma ^3=1$</span>. So the expression can be rewritten as
<span class="math-container">\begin{eqnarray*}
a^2+b^2+c^2+(ab+bc+ca)(\sigma+\sigma^2)=a^2+b^2+c^2-(ab+bc+ca) \\ =\frac{1}{2}((a-b)^2+(b-c)^2+(c-a)^2)
\end{eqnarray*}</span>
which is clearly non-negative.</p>
|
35,688 | <p>I'm looking for a fun (not too many tedious calculations) calculus one problem that uses the concept that, after subsitution, you have two integrals of diffent functions with different limits, but equal area. For example:</p>
<p><a href="http://www.wolframalpha.com/input/?i=int%20%28sin%28%28pi%5E2%29/x%29%29/%28x%5E2%29%20from%20pi%20to%20pi/2" rel="nofollow">Function one</a></p>
<p><a href="http://www.wolframalpha.com/input/?i=int%20%28sin%20x%29%2a%28-1/pi%5E2%29%20from%20pi%20to%202%20pi" rel="nofollow">Function two</a></p>
<p>Those two regions are the same. I'm thinking of a problem that asks the student to prove two regions are of equal area, and once they set up the integrals they can see it is a case of subsitution. </p>
<p>Do you know of such a problem?</p>
| Shai Covo | 2,810 | <p>How about calculating the area, $S$, of one quarter of the unit circle? </p>
<p>First note that $S = \int_0^1 {\sqrt {1 - x^2 } \,{\rm d}x} $. The substitution $x=\sin(u)$ (hence, ${\rm d}x = \cos(u) \,{\rm d}u$) gives
$$
\int_0^1 {\sqrt {1 - x^2 } \,{\rm d}x} = \int_0^{\pi /2} {\sqrt {1 - \sin ^2 (u)} \cos (u)\,{\rm d}u} = \int_0^{\pi /2} {\cos ^2 (u)\,{\rm d}u}.
$$
The integral on the right can be computed as follows.
$$
\int_0^{\pi /2} {\cos ^2 (u)\,{\rm d}u} = \int_0^{\pi /2} {\frac{{1 + \cos (2u)}}{2}\,{\rm d}u} = \frac{\pi }{4} + \frac{1}{2}\int_0^{\pi /2} {\cos (2u)\,{\rm d}u} = \frac{\pi }{4},
$$
where the last equality follows from
$$
\int_0^{\pi /2} {\cos (2u)\,{\rm d}u} = \frac{1}{2}\int_0^\pi {\cos (t)\,{\rm d}t} = 0.
$$
Thus we have shown that $S=\pi/4$, which is a non-trivial result.</p>
<p>EDIT: As another example, using the the substitution $y=e^{-x}$, we have
$$
\int_0^\infty {e^{ - x} \,{\rm d}x} = \int_1^0 {y\frac{{\,{\rm d}y}}{{ - y}}} = \int_0^1 {1\,{\rm d}y} = 1.
$$</p>
|
3,613,235 | <p>I know such integral: <span class="math-container">$\int_0^{\infty}\frac{\ln x}{e^x}\,dx=-\gamma$</span>. What about the integral <span class="math-container">$\int_0^{\infty}\frac{\ln x}{e^x+1}\,dx$</span>? </p>
<p>The answer seems very nice: <span class="math-container">$-\frac{1}{2}{\ln}^22$</span> but how it could be calculated? I tried integration by parts but the limit <span class="math-container">$\displaystyle{\lim_{x\to 0}\ln x\ln(1+e^{-x})}$</span> doesn't exist. Or I can also write the following equality
<span class="math-container">$$\int_0^{\infty}\frac{\ln x}{e^x+1}\,dx=\lim\limits_{t\to 0}\frac{d}{dt}\left(\int_0^{\infty}\frac{x^t}{e^x+1}\, dx\right)$$</span>
but I don't know what to do next. </p>
| Mark Viola | 218,419 | <p><strong>HINT:</strong></p>
<p>Note that we have</p>
<p><span class="math-container">$$\frac1{\sqrt n+\sqrt{n+2}}=\frac{\sqrt{n+2}-\sqrt n}{2}$$</span></p>
<p>Now telescope.</p>
|
3,029,446 | <p>If </p>
<ul>
<li><p><span class="math-container">$A$</span> is <span class="math-container">$m \times n$</span> (<span class="math-container">$m<n$</span>), and its rows are independent</p></li>
<li><p><span class="math-container">$B$</span> is <span class="math-container">$n \times p$</span> (<span class="math-container">$p<n$</span>), and its columns are independent</p></li>
<li><p>We also know <span class="math-container">$m\ge n$</span>.</p></li>
</ul>
<p>does <span class="math-container">$AB$</span> have linearly independent columns?</p>
<p>Or what additional requirements are needed for <span class="math-container">$AB$</span> to have linearly independent columns?</p>
| AnyAD | 107,693 | <p>For this we need <span class="math-container">$AB $</span> to have rank <span class="math-container">$p $</span> (since <span class="math-container">$p$</span> columns). All we are garanteed however is that the rank of <span class="math-container">$AB$</span> is less orequal to the minimum of <span class="math-container">$\{m,n,p\}$</span>. So this will fail for example when <span class="math-container">$m <p$</span>.</p>
|
4,468,112 | <p>Let <span class="math-container">$a,b\in\mathbb R$</span> with <span class="math-container">$a<b$</span>, <span class="math-container">$$\mathcal D_{[a,\:b]}:=\{(t_0,\ldots,t_k):k\in\mathbb N\text{ and }a=t_0<\cdots<t_k\}$$</span> and <span class="math-container">$$\mathcal T_\varsigma:=\{(\tau_1,\ldots,\tau_k):\tau_i\in[t_{i-1},t_i]\text{ for all }i\in\{1,\ldots,k\}\}\;\;\;\text{for }\varsigma=(t_0,\ldots,t_k)\in\mathcal D_{[a,\:b]}.$$</span> Moreover, let <span class="math-container">$f:[a,b]\to\mathbb R$</span> be continuous and <span class="math-container">$g:[a,b]\to\mathbb R$</span> be of bounded variation. We can show that <span class="math-container">$$\int_a^bf\:{\rm d}g:=\lim_{\substack{|\varsigma|\to0+\\\varsigma\in\mathcal D_{[a,\:b]}\\\tau\in\mathcal T_\varsigma}}S_{\varsigma,\:\tau}(f,g)$$</span> is well-defined, where <span class="math-container">$$|\varsigma|:=\max_{1\le i\le k}(t_i-t_{i-1})\;\;\;\text{for }\varsigma=(t_0,\ldots,t_k)\in\mathcal D_{[a,\:b]}$$</span> and <span class="math-container">$$S_{\varsigma,\:\tau}(f,g):=\sum_{i=1}^kf(\tau_i)(g(t_i)-g(t_{i-1}))\;\;\;\text{for }\varsigma=(t_0,\ldots,t_k)\in\mathcal D_{[a,\:b]}\text{ and }\tau\in\mathcal T_\varsigma.$$</span></p>
<blockquote>
<p>Assuming that <span class="math-container">$g$</span> is differentiable (not necessarily <em>continuously</em> differentiable), are we able to show that <span class="math-container">$$\int_a^bf\:{\rm d}g=\int_a^bf(s)g'(s)\:{\rm d}s\tag1?$$</span></p>
</blockquote>
<p>Let <span class="math-container">$\varsigma=(t_0,\ldots,t_k)\in\mathcal D_{[a,\:b]}$</span>. By the mean value theorem, there is a <span class="math-container">$\tau\in\mathcal T_\varsigma$</span> with <span class="math-container">$$S_{\varsigma,\:\tau}(f,g)=\sum_{i=1}^kf(\tau_i)g'(\tau_i)(t_i-t_{i-1})=S_{\varsigma,\:\tau}(fg',\operatorname{id}_{[a,\:b]})\tag2,$$</span> but does the right-hand side tend to the right-hand side of <span class="math-container">$(1)$</span> as <span class="math-container">$|\varsigma|\to0+$</span>? This is clearly the case when <span class="math-container">$g'$</span> is continuous though ...</p>
| Sangchul Lee | 9,340 | <p>The following claims hold:</p>
<blockquote>
<p><strong>Theorem 1.</strong> Let <span class="math-container">$g$</span> be absolutely continuous on <span class="math-container">$[a, b]$</span>, and let <span class="math-container">$f$</span> be continuous on <span class="math-container">$[a, b]$</span>. Then <span class="math-container">$fg'$</span> is Lebesgue integrable and
<span class="math-container">$$ \int_{a}^{b} f(t) \, \mathrm{d}g(t) = \int_{a}^{b} f(t)g'(t) \, \mathrm{d}t. $$</span></p>
</blockquote>
<p>The proof is quite simple. Starting from the fact that <span class="math-container">$\int_{c}^{d} g'(t) \, \mathrm{d}t = g(d) - g(c)$</span> for any interval <span class="math-container">$[c, d] \subseteq [a, b]$</span> (where we utilized the <a href="https://en.wikipedia.org/wiki/Absolute_continuity" rel="nofollow noreferrer">absolute continuity</a> of <span class="math-container">$g$</span>), the statement holds when <span class="math-container">$f$</span> is a step function. Then the desired claim follows by choosing a sequence of step functions that converges uniformly to <span class="math-container">$f$</span>. This reasoning in fact shows that the statement remains true if <span class="math-container">$f$</span> is a uniform limit of step functions, i.e., <a href="https://en.wikipedia.org/wiki/Regulated_function" rel="nofollow noreferrer">regulated functions</a>.</p>
<p>Also, if the left-hand side is understood as the <a href="https://en.wikipedia.org/wiki/Lebesgue%E2%80%93Stieltjes_integration" rel="nofollow noreferrer">Lebesgue–Stieltjes integral</a> sense, then the equality holds for any <span class="math-container">$f$</span> that is integrable with respect to the measure <span class="math-container">$|g'(t)| \, \mathrm{d}t$</span>. This is because the above statement reduces to an instance of the <a href="https://en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem" rel="nofollow noreferrer">Radon–Nikodym theorem</a>. This opens up possibilities of further generalizing the above claim.</p>
<p>We also have:</p>
<blockquote>
<p><strong>Theorem 2.</strong> Let <span class="math-container">$g$</span> be differentiable everywhere on <span class="math-container">$[a, b]$</span>, and let <span class="math-container">$f$</span> be both Riemann–Stieltjes integrable with respect to <span class="math-container">$g$</span> and Lebesgue integrable on <span class="math-container">$[a, b]$</span>. Then <span class="math-container">$fg'$</span> is Henstock–Kurzweil integrable and
<span class="math-container">$$ \int_{a}^{b} f(t) \, \mathrm{d}g(t) = \int_{a}^{b} f(t)g'(t) \, \mathrm{d}t, $$</span>
where the right-hand side is in <a href="https://en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral" rel="nofollow noreferrer">Henstock–Kurzweil integral</a> sense.</p>
</blockquote>
<p>The Henstock–Kurzweil integral can be thought as a non-absolutely convergent version of the Lebesgue integral, in the sense that a function <span class="math-container">$f$</span> is Lebesgue integrable if and only if both <span class="math-container">$f$</span> and <span class="math-container">$|f|$</span> is Henstock–Kurzweil integrable.</p>
<p>The proof is also straightforward, assuming basic theory of all the types of integral mentioned above. Indeed, for each <span class="math-container">$\varepsilon > 0$</span>, we do the following:</p>
<ul>
<li><p>Let <span class="math-container">$\varepsilon' = \frac{\varepsilon}{\int_{a}^{b} |f|+2}$</span>. Since <span class="math-container">$f$</span> is Lebesgue integrable, this is still a positive number.</p>
</li>
<li><p>Choose a number <span class="math-container">$\delta_1 > 0$</span> such that <span class="math-container">$\left| S(f, g, \dot{\mathcal{P}}) - \int_{a}^{b} f \, \mathrm{d}g \right| < \varepsilon'$</span> for any <a href="https://en.wikipedia.org/wiki/Partition_of_an_interval" rel="nofollow noreferrer">tagged partition</a> <span class="math-container">$\dot{\mathcal{P}}$</span> with mesh size <span class="math-container">$ < \delta_1$</span>. This is possible since <span class="math-container">$f$</span> is Riemann–Stieltjes integrable with respect to <span class="math-container">$g$</span>.</p>
</li>
<li><p>Choose a gauge <span class="math-container">$\delta_2$</span> satisfying the straddle lemma for <span class="math-container">$g$</span> and <span class="math-container">$\varepsilon'$</span>: For any <span class="math-container">$t \in [a, b]$</span> and <span class="math-container">$u, v \in [a, b]$</span> satisfying <span class="math-container">$t - \delta_2(t) < u < t < v < t + \delta_2(t)$</span>,
<span class="math-container">$$ |g(v) - g(u) - g'(t)(v - u)| \leq \varepsilon' (v - u). $$</span></p>
</li>
<li><p>Choose a gauge <span class="math-container">$\delta_3$</span> such that <span class="math-container">$\left| S(|f|, \dot{\mathcal{P}}) - \int_{a}^{b} |f| \right| < 1$</span> for any <span class="math-container">$\delta_3$</span>-fine tagged partition <span class="math-container">$\dot{\mathcal{P}}$</span>. This is possible because <span class="math-container">$|f|$</span> is Lebesgue integrable and hence Henstock–Kurzweil integrable.</p>
</li>
</ul>
<p>Now let <span class="math-container">$\delta(t) = \min\{\delta_1, \delta_2(t), \delta_3(t)\} $</span>. Then <span class="math-container">$\delta_3$</span> is also a gauge, and for any <span class="math-container">$\delta$</span>-fine tagged partition <span class="math-container">$\dot{\mathcal{P}}$</span> of <span class="math-container">$[a, b]$</span>,</p>
<p><span class="math-container">\begin{align*}
\left| \int_{a}^{b} f \, \mathrm{d}g - S(fg', \dot{\mathcal{P}}) \right|
&\leq \left| \int_{a}^{b} f \, \mathrm{d}g - S(f, g, \dot{\mathcal{P}}) \right|
+ \left| S(f, g, \dot{\mathcal{P}}) - S(fg', \dot{\mathcal{P}}) \right| \\
&< \varepsilon' + \sum_{i} |f(\tau_i)| \cdot |g(t_i) - g(t_{i-1}) - g'(\tau_i)(t_i - t_{i-1})| \\
&\leq \varepsilon' + \varepsilon' \sum_{i} |f(\tau_i)| (t_i - t_{i-1}) \\
&\leq \varepsilon' + \varepsilon' \left( 1 + \int_{a}^{b} |f| \right) \\
&= \varepsilon.
\end{align*}</span></p>
<p>Therefore the desired claim follows.</p>
|
962,691 | <p>I'm trying to integrate $ \int_0^1\frac {u^2 + 1}{u - 2}du$</p>
<p>I've calculated that this equates to $ [\frac{u^2}{2}+2u +5ln(u-2)]_0^1 $</p>
<p>But then I have to evaluate $ln(-1)$ and $ln(-2)$ which are obviously not defined in the real plane. I have drawn the graph and I know for certain that this integral exists. Any guidance on what I'm missing would be great.</p>
| BaronVT | 39,526 | <p>Recall that the antiderivative of $\frac{1}{u}$ is $\ln|u|$, not $\ln u$</p>
|
323,665 | <p>Given the base case <span class="math-container">$a_0 = 1$</span>, does <span class="math-container">$a_n = a_{n-1} + \frac{1}{\left\lfloor{a_{n-1}}\right \rfloor}$</span> have a closed form solution? The sequence itself is divergent and simply goes {<span class="math-container">$1, 2, 2+\frac{1}{2}, 3, 3+\frac{1}{3}, 3+\frac{2}{3}, 4, 4+\frac{1}{4}, 4+\frac{2}{4}, 4+\frac{3}{4}, . . .$</span>} and so forth. It seems like it should be easy but I can't seem to find a solution. Any suggestions?</p>
| Ale De Luca | 7,320 | <p>A similar, possibly simpler closed form is the following: set <span class="math-container">$$b_n=\left\lfloor\frac{1+\sqrt{8n-7}}{2}\right\rfloor,$$</span>
then <span class="math-container">$$a_n=\frac{b_n+1}{2}+\frac{n-1}{b_n}.$$</span>
It is not hard to derive this from the observation that whenever <span class="math-container">$n-1$</span> is a triangular number <span class="math-container">$k(k-1)/2$</span>, one has <span class="math-container">$a_n=k$</span>.</p>
|
1,649,907 | <p>Please kindly forgive me if my question is too naive, i'm just a <em>prospective</em> undergraduate who is simply and deeply fascinated by the world of numbers.</p>
<p>My question is: Suppose we want to prove that $f(x) > \frac{1}{a}$, and we <em>know</em> that $g(x) > a$, where $f,g$ and $a$ are all positive and $a$ is a nonzero real number.
<em>If we can show</em> that $f(x)g(x) > 1$, would that imply our required proof ?</p>
<p>EDIT: As demonstrated by various users in the solutions below, the answer is definitely <em>no</em>.
What about if we now want to prove the <em>reverse</em> inequality $f(x) \leq \frac{1}{a}$ given that $g(x) < a$, if we can show that $f(x)g(x)<1$, i guess our required result would follow ?</p>
| fleablood | 280,126 | <p>The whole thing boils done to:</p>
<p>if $a > b$ (a and b positive) then $1/a < 1/b$. The reciprical "<em>flips</em>" the inequality.</p>
<p>$f(x)g(x) < 1$ (or greater than) implies $f(x) < 1/g(x)$ (or greater than). To get a result $g(x) ?? a$ must be the <em>OPPOSITE</em> inequality because $1/g(x)$ and $1/a$ will be "flipped".</p>
<p>So if $f(x)g(x) < 1$ and $g(x) > a$ (!!!!!!<strong>GREATER</strong>!!!!!!!!) then you can conclude $f(x) < 1/a$.</p>
<p>And if $f(x)g(x) > 1$ and $g(x) < a$ (!!!!!!!<strong>LESS THAN</strong>!!!!!!!) then you can conclude $f(x) > 1/a$.</p>
<p>But if $f(x)g(x) > 1$ and $g(x) > a$ or if $f(x)g(x) < 1$ and $g(x) < a$ you can not conclude a g**d****ed thing.</p>
<p>==============</p>
<p>No. If $g(x)$ is <em>LESS</em> than $a$ but more than zero, you may conclude $f(x)g(x) > 1$ so $f(x) > 1/g(x) > 1/a.$</p>
<p>But if $g(x) > a$. We have $f(x)g(x) > 1 \implies f(x) > 1/g(x)$ and $1/g(x) < f(x)$ but we cant conclude anything. (Or we have $f(x)g(x) > f(x)a$ and $f(x)g(x) > 1$ but we don't have $f(x)a > 1$.)</p>
<p>A counter is $a = 1; g(x) = 4$ (a constant function) and $f(x) = 1/2$ $f(x) < 1/a$ while $f(x)g(x) = 2 > 1$ and $g(x) > a$.</p>
<p>===</p>
<p>Again no. If $f(x), g(x), a > 0$ then $f(x)g(x) >|=|< 1$ then $f(x) >|=|< 1/g(x)$ If we know $g(x) >|=|< a$ then $1/g(x) <|=|> 1/a$.</p>
<p>Given this we have nine possible situations:</p>
<p>1) $f(x)g(x) < 1; g(x) < a$. Then $f(x) < 1/g(x) > 1/a$. No conclusions.</p>
<p>2) $f(x)g(x) = 1; g(x) < a$. Then $f(x) = 1/g(x) > 1/a$. Conclusion: $f(x) > 1/a$.</p>
<p>3) $f(x)g(x) > 1; g(x) < a$ Then $f(x) > 1/g(x) > 1/a$. Conclusion: $f(x) > 1/a$</p>
<p>4) $f(x)g(x) < 1; g(x) = a$. Then $f(x) < 1/g(x) = 1/a$. Conclusion: $f(x) < 1/a$</p>
<p>5) $f(x)g(x) = 1; g(x) = a$. Then $f(x) = 1/g(x) = 1/a$. Conclusion: $f(x) = 1/a$.</p>
<p>6) $f(x)g(x) > 1; g(x) = a$ Then $f(x) > 1/g(x) = 1/a$. Conclusion: $f(x) > 1/a$</p>
<p>7) $f(x)g(x) < 1; g(x) > a$. Then $f(x) < 1/g(x) < 1/a$. Conclusion: $f(x) < 1/a$</p>
<p>8) $f(x)g(x) = 1; g(x) > a$. Then $f(x) = 1/g(x) < 1/a$. Conclusion: $f(x) < 1/a$.</p>
<p>9) $f(x)g(x) > 1; g(x) > a$ Then $f(x) > 1/g(x) < 1/a$. No conclusion.</p>
|
362,881 | <p>I am going to try to explain this as easily as possible. I am working on a computer program that takes input from a joystick and controls a servo direction and speed. I have the direction working just fine now I am working on speed. To control the speed of rotation on the servo I need to send it so many pulses per second using PWM. The servo that I am using takes arguments for speed between 120-150. 120 is %100 speed and 150 is %0 or stopped. 135 is %50 speed. How would I convert percentage from 0-100 into a number between 120-150 including 1/10ths? I hope this makes sense if you need me to explain further please let me know. I really don't know what tag this falls under either.</p>
| Shuhao Cao | 7,200 | <p>Depends on how you define the Dirac delta, if it is defined as a linear functional acting on an $f\in C^{\infty}_0(\mathbb{R})$:
$$
\delta[f](x) = \int^{\infty}_{-\infty} \delta(x)f(x) \,dx = f(0)
$$
Then we can see for any test function $\phi$:
$$
(e^{x}\delta(\cdot) - \delta(\cdot))[\phi](x) = \int^{\infty}_{-\infty} (e^x\delta(x)\phi(x) - \delta(x)\phi(x)) \,dx = e^0\phi(0) - \phi(0) = 0
$$
Hence $e^{x} \delta(x) = \delta(x)$.</p>
|
2,487,234 | <p>I want to prove that the following only have one solution, for $\zeta\in[0,1]$, at $\zeta =1$.</p>
<p>$$f(\zeta)=\frac{1}{1+\zeta}$$</p>
<p>$$g(\zeta) =
\frac{
(1-\zeta)(2-\zeta)\zeta
-
(2-\zeta)^2\log\left(2 - \zeta \right)
}
{\zeta(\zeta-4)(\zeta-1)^2}$$</p>
<p>These are plotted below. Note that $f(0)=1$, $\lim_{\zeta\rightarrow 0}g(\zeta)=\infty$ and $f(1)=1/2=g(1)$.</p>
<p>Both functions are convex and monotonically decreasing over this region, therefore if they intersect at $\zeta=1$ this is the only place they can? Can I shown they are both convex over $\zeta\in[0,1]$ by showing their second derivatives wrt $\zeta$ are +ve over this region? Or is this thinking too simplistic?</p>
<p><a href="https://i.stack.imgur.com/Jm6yn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jm6yn.png" alt="enter image description here"></a></p>
<p>Note, the limit as $g(\zeta)$ as $\zeta\rightarrow 1$ is proven below:
\begin{align}
\lim_{\zeta\rightarrow 1} g(\zeta)
%%$
&=
%%%
\lim_{\zeta\rightarrow 1}
\left(
\frac
{4\log(2 - \zeta) - 7\zeta - 2\zeta \log(2 - \zeta) + 3\zeta^2 + 4}
{4\zeta^3 - 18\zeta^2 + 18\zeta - 4}
\right)\\
%%%
&=
%%%
\lim_{\zeta\rightarrow 1}
\left(
\frac
{6\zeta - 2\log(2 - \zeta) - 9}
{12\zeta^2 - 36\zeta + 18}
\right)\\
%%%
&=
%%%
\frac
{6 - 9}
{12 - 36 + 18}\\
%%%
&=
%%%
\frac
{1}
{2}.
\end{align}</p>
<p>EDIT: So the second derivative of $g(\zeta)$ is pretty horrific. It does have a simplified form, but it is also pretty terrible.
<a href="http://www.wolframalpha.com/input/?i=(d%5E2%2Fdx%5E2+(+(1-x)(2-x)x+-+(2-x)%5E2%5Clog(2+-x)+)%2F(x(x-4)(x-1)%5E2))" rel="nofollow noreferrer">http://www.wolframalpha.com/input/?i=(d%5E2%2Fdx%5E2+(+(1-x)(2-x)x+-+(2-x)%5E2%5Clog(2+-x)+)%2F(x(x-4)(x-1)%5E2))</a></p>
<p>$$g'' =
\frac
{(x - 1) x (x (x (x (x (2 x - 7) - 13) + 124) - 224) + 64)
- 2 (x (x (x (x (3 (x - 12) x + 184) - 472) + 588) - 304) + 64) \log(2 - x))}
{(x - 4)^3 (x - 1)^4 x^3}$$</p>
<p>It is clear from the numerics that it is greater than $0$ in the region $\zeta\in[0,1]$.
<a href="http://www.wolframalpha.com/input/?i=(d%5E2%2Fdx%5E2+(+(1-x)(2-x)x+-+(2-x)%5E2%5Clog(2+-x)+)%2F(x(x-4)(x-1)%5E2))+%3D+0" rel="nofollow noreferrer">http://www.wolframalpha.com/input/?i=(d%5E2%2Fdx%5E2+(+(1-x)(2-x)x+-+(2-x)%5E2%5Clog(2+-x)+)%2F(x(x-4)(x-1)%5E2))+%3D+0</a></p>
| Lerigorilla | 450,010 | <p>we want to solve g(x)=f(x) for x in [0,1]
by doing the calculations we end up in this equation
$$ \ log(2-x)= \frac{4x^3-10x^2+6x}{(x+1)(2-x)^2}$$
$$h(x)=log(2-x)$$ and
$$ k(x)=\frac{4x^3-10x^2+6x}{(x+1)(2-x)^2}$$
h(x) is convex and k(x) is concave in [0,1] (1)
$$ k'(x)=\frac{-2(x^3+8x^2-17x+6)}{(x-2)^3(x+1)^2} $$
so k(1)=h(1)=0 and k'(1)=h'(1)=-1 (2)</p>
<p>from (1) and (2) we have that
$$ h(x)>=k(x)$$ for every x in [0,1] and k(x)=h(x) only for x=1</p>
|
2,399,710 | <p>I am asking for what, if any, the preferred study skills of approaching classes are like in higher math education are like <strong>in general,</strong> and I am not asking necessarily for personal anecdotes for this question (though they are welcome if it's all you have to share).</p>
<p>My question is what are the better studying methods for learning higher mathematics(i.e., subjects involving more proofs instead of calculations like Analysis, Topology, Axiomatic Set Theory, Abstract Algebra, etc.)?</p>
<p>Let me explain what I mean by giving you an example. As an undergrad at first, courses like the Calculus sequence, Differential Equations could be approached by learning how to solve problems even if you weren't able to understand the proofs(which were usually skipped over in the textbook by the professor).</p>
<p>But now, that I have taken Linear Algebra, and I am about to take Analysis I usually take a much different approach:</p>
<p><span class="math-container">$ $$\cdot$$ $</span> I start by rewriting all the definitions and theorems and proofs of a chapter and memorizing them. Then I will work through the examples, and ideally get to the exercises and finish off the chapter. And also spend time reflecting on the topics to get a more intuitive understanding of the concepts involved. I skip very little, anything if at all from the books I'm working with even if the topics are skipped in class.</p>
<p>This approach is very effective, especially when I am studying a topic for my own interests, and I am really able to understand things at a level that my peers usually have trouble following. The drawback is that I move at a much slower pace than my peers and I end up struggling in a class towards the end of a semester because I am behind on the syllabus. I end up risking a low grade though all but once I have managed an A- or above. Also am able to work out Jech's Set Theory which others have told me is out of my league but I actually find it the right amount of challenging using this approach.</p>
<p>The method followed by the math department at my school feels very superficial and not as the effective long-term. Usually, professors do not focus on the reasoning or intuition behind concepts. Tests are oriented so that we memorize by rote the proofs of the major theorems and regurgitate them on exams.</p>
<p>Let me give more concrete questions:</p>
<ol>
<li><p><em>I asked a math professor once, and he said that real mathematics is usually done where you understand 2 or 3 pages a day of a text on your first reading of it.</em> Is this true for graduate level work for the average student?</p>
</li>
<li><p>Do students who follow my immersive way of studying tend to have an advantage over those who don't when we get to grad school?</p>
</li>
<li><p>In terms of learning theory and doing exercises, how much importance is recommended to place on each?</p>
</li>
</ol>
<p><em><strong>4. Are there study techniques used in more advanced math courses (like engaging in discourse with peers, focusing more on memorization before attempting to do problem sets, taking notes in a particular way) that are more fruitful than others?</strong></em></p>
| Merkh | 141,708 | <p>You question is waay too many questions in one for this website. Just FYI. Anyway...</p>
<p>"... real mathematics is usually done where you understand 2 or 3 pages a day of a text on your first reading of it. Is this true for graduate-level work for the average student??"</p>
<ul>
<li>This completely depends on what the reading is and the person's background in it. Learning something completely new? Then yes, its probably true.</li>
</ul>
<p>"Do students who follow my immersive way of studying tend to have an advantage over those who don't when we get to grad school?"</p>
<ul>
<li>Sure, students who know more math upon entering graduate school have an advantage over the students who were content with just the details given in class. It's probably the self-motivation of the student more so than the actual knowledge that puts the student at an advantage to do well.</li>
</ul>
<p>"In terms of learning theory and doing exercises, how much importance is recommended I place on each?"</p>
<ul>
<li>Exercises confirm that theory was actually learned and understood. If you are doing exercises and you find them "easy", well then you probably have a very solid grasp of the theory. Do what feels right. Learn some theory, go back and see if you understand the theory.</li>
</ul>
<p>"Are there study techniques used in more advanced math courses (like engaging in discourse with peers, focusing more on memorization before attempting to do problem sets, taking notes in a particular way) that are more fruitful than others?"</p>
<ul>
<li>I've found that reading the book and taking notes (well) before class and then paying close attention in class is almost always sufficient to understand the material. I've also found that graduate students many times do not keep up with this regime of reading before the class, whether due to workload or general dislike for the material.</li>
</ul>
<p>In summary, do what feels right, you'll learn a lot if you stay motivated, and enjoy.</p>
|
2,399,710 | <p>I am asking for what, if any, the preferred study skills of approaching classes are like in higher math education are like <strong>in general,</strong> and I am not asking necessarily for personal anecdotes for this question (though they are welcome if it's all you have to share).</p>
<p>My question is what are the better studying methods for learning higher mathematics(i.e., subjects involving more proofs instead of calculations like Analysis, Topology, Axiomatic Set Theory, Abstract Algebra, etc.)?</p>
<p>Let me explain what I mean by giving you an example. As an undergrad at first, courses like the Calculus sequence, Differential Equations could be approached by learning how to solve problems even if you weren't able to understand the proofs(which were usually skipped over in the textbook by the professor).</p>
<p>But now, that I have taken Linear Algebra, and I am about to take Analysis I usually take a much different approach:</p>
<p><span class="math-container">$ $$\cdot$$ $</span> I start by rewriting all the definitions and theorems and proofs of a chapter and memorizing them. Then I will work through the examples, and ideally get to the exercises and finish off the chapter. And also spend time reflecting on the topics to get a more intuitive understanding of the concepts involved. I skip very little, anything if at all from the books I'm working with even if the topics are skipped in class.</p>
<p>This approach is very effective, especially when I am studying a topic for my own interests, and I am really able to understand things at a level that my peers usually have trouble following. The drawback is that I move at a much slower pace than my peers and I end up struggling in a class towards the end of a semester because I am behind on the syllabus. I end up risking a low grade though all but once I have managed an A- or above. Also am able to work out Jech's Set Theory which others have told me is out of my league but I actually find it the right amount of challenging using this approach.</p>
<p>The method followed by the math department at my school feels very superficial and not as the effective long-term. Usually, professors do not focus on the reasoning or intuition behind concepts. Tests are oriented so that we memorize by rote the proofs of the major theorems and regurgitate them on exams.</p>
<p>Let me give more concrete questions:</p>
<ol>
<li><p><em>I asked a math professor once, and he said that real mathematics is usually done where you understand 2 or 3 pages a day of a text on your first reading of it.</em> Is this true for graduate level work for the average student?</p>
</li>
<li><p>Do students who follow my immersive way of studying tend to have an advantage over those who don't when we get to grad school?</p>
</li>
<li><p>In terms of learning theory and doing exercises, how much importance is recommended to place on each?</p>
</li>
</ol>
<p><em><strong>4. Are there study techniques used in more advanced math courses (like engaging in discourse with peers, focusing more on memorization before attempting to do problem sets, taking notes in a particular way) that are more fruitful than others?</strong></em></p>
| Joe | 61,666 | <p>I found benefit in reading two books by Lara Alcock:</p>
<p>How to Study as a Mathematics Major</p>
<p>and</p>
<p>How to Think About Analysis</p>
|
1,662,876 | <p>Now we have some examples of what I mean $$\int_0^{2\pi} \sin x~dx=0$$
$$\int_0^{8\pi} \cos 4x~dx=0$$</p>
<p>$$\int_{\pi}^{2\pi} \sin^3 10x~dx=0$$</p>
<p>Looking at the graph of $f(x)=\sin (x)$ for example it makes some sense to me that $$\int_0^{2\pi} \sin x~dx=0$$ because the region below the $x$ axis will "cancel" with the part above but I don't understand how I can make claims about more general integrals like $$\int_{\pi}^{2\pi} \sin^3 10x~dx=0$$ which I just got from testing with an online integral tool.</p>
<p>Obviously there is something more deeper going on and I would like to understand when and why I can just claim that an integral will be zero as it will save me a lot of time.</p>
<p>I have had a thought but can't seem to conclude how it would be obvious that any of the less obvious integrals would be zero (like the third one I stated), I imagine that there is some kind of symmetry but I'm just not seeing it. </p>
<p>So if anyone could explain when (in a more general sense not just for the integrals above) and why this works I would appreciate it. </p>
| John Hughes | 114,036 | <p>Roughly speaking, a trig integral (sine or cosine) that includes an integer number of cycles will be zero. The same goes for $\sin nx$ or $\cos nx$. Your "area above the line is the same as area below" is a perfectly good explanation. </p>
<p>As for powers of sines, any odd power of sine can be simplified by the rule
$$
\cos 2x = 1 - 2 \sin^2 x
$$
by rewriting
$$
\sin^2 x = \frac{1 - \cos 2x}{2}
$$
For instance,
$$
\sin^3 x = \sin x \sin^2 x = \sin x \frac{1 - \cos 2x}{2}
$$
Using this, you can express any odd power of sines as a product of lower power sines and cosines (albeit with $2x$ or $3x$ or $4x$, etc., instead of $x$) and such things also integrate to zero be the periodicity-and-symmetry argument.</p>
<p>On the other hand, if you integrate $\sin (x/2)$ from $0$ to $4 \pi$, for instance, you'll get zero as well, but not if you integrate only to $2\pi$.
That's because $2\pi$ is only a half-period for this function. </p>
<p>There are other ways to get zero: integrate $\sin$ from $-a$ to $a$, for instance. These depend on the symmetry of the function rather than its periodicity, and seem somehow different form the examples you gave. </p>
<p>There are other examples that are helpful to know even though you don't get zero. For instance, look at</p>
<p>$$
Q = \int_0^{2\pi} \sin^2(x) dx.
$$</p>
<p>That's clearly the same as </p>
<p>$$\int_0^{2\pi} \cos^2(x) dx,$$</p>
<p>because $\sin$ and $\cos$ are just shifted version of each other, and we're integrating over a whole period of either one. </p>
<p>But that means that
$$
2Q =
\int_0^{2\pi} \sin^2x + \cos^2 x ~dx= \int_0^{2\pi} 1 ~dx= 2\pi,
$$
so
$Q = \pi$. </p>
<p>And in general, integrals of <strong>even</strong> powers of sines and cosines will work out like this: they're always positive, so their integrals (over any interval of length > 0) will be positive. </p>
|
1,662,876 | <p>Now we have some examples of what I mean $$\int_0^{2\pi} \sin x~dx=0$$
$$\int_0^{8\pi} \cos 4x~dx=0$$</p>
<p>$$\int_{\pi}^{2\pi} \sin^3 10x~dx=0$$</p>
<p>Looking at the graph of $f(x)=\sin (x)$ for example it makes some sense to me that $$\int_0^{2\pi} \sin x~dx=0$$ because the region below the $x$ axis will "cancel" with the part above but I don't understand how I can make claims about more general integrals like $$\int_{\pi}^{2\pi} \sin^3 10x~dx=0$$ which I just got from testing with an online integral tool.</p>
<p>Obviously there is something more deeper going on and I would like to understand when and why I can just claim that an integral will be zero as it will save me a lot of time.</p>
<p>I have had a thought but can't seem to conclude how it would be obvious that any of the less obvious integrals would be zero (like the third one I stated), I imagine that there is some kind of symmetry but I'm just not seeing it. </p>
<p>So if anyone could explain when (in a more general sense not just for the integrals above) and why this works I would appreciate it. </p>
| Surb | 154,545 | <p><strong>Hint</strong></p>
<p>If $f$ is periodic and odd, you will have $$\int_a^{a+T}f(x)dx=0$$
for all $a\in\mathbb R$.</p>
<p>For $$\int_\pi^{2\pi}\sin^3(10 x)dx$$</p>
<p>you have that $x\longmapsto \sin^3(10x)$ is odd and $\frac{\pi}{5}-$periodic. Moreover, $$\int_\pi^{2\pi}\sin^3(10x)dx=\sum_{k=0}^4\underbrace{\int_{\pi+\frac{k\pi}{5}}^{\pi+\frac{(k+1)\pi}{5}}\sin^3(10 x)dx}_{=0\ for\ all\ k}=0.$$</p>
|
1,662,876 | <p>Now we have some examples of what I mean $$\int_0^{2\pi} \sin x~dx=0$$
$$\int_0^{8\pi} \cos 4x~dx=0$$</p>
<p>$$\int_{\pi}^{2\pi} \sin^3 10x~dx=0$$</p>
<p>Looking at the graph of $f(x)=\sin (x)$ for example it makes some sense to me that $$\int_0^{2\pi} \sin x~dx=0$$ because the region below the $x$ axis will "cancel" with the part above but I don't understand how I can make claims about more general integrals like $$\int_{\pi}^{2\pi} \sin^3 10x~dx=0$$ which I just got from testing with an online integral tool.</p>
<p>Obviously there is something more deeper going on and I would like to understand when and why I can just claim that an integral will be zero as it will save me a lot of time.</p>
<p>I have had a thought but can't seem to conclude how it would be obvious that any of the less obvious integrals would be zero (like the third one I stated), I imagine that there is some kind of symmetry but I'm just not seeing it. </p>
<p>So if anyone could explain when (in a more general sense not just for the integrals above) and why this works I would appreciate it. </p>
| MPW | 113,214 | <p>If $p$ is a period for $f$, then $p$ is also a period for $f^n$.</p>
<p>If the integral is over an interval whose length is a whole number of periods for the integrand, and if the integral over a single period is zero, then that integral is also zero.</p>
<p>A period for $\sin 10x$ is $p=2\pi/10 = \pi/5$. So $\pi/5$ is also a period for $\sin^3 10x$. The interval of integration is of length $\pi$, which is $5\times\pi/5$, so is a whole number of periods of the integrand.</p>
<p>It remains to see that the integral of $f(x)\equiv\sin^3 10x$ over a single period is zero. But that follows quickly from the observation that $$f( x+\frac12 p) = (\sin 10(x+\pi/10))^3$$
$$=(\sin (10x+\pi))^3 = (-\sin 10x)^3$$
$$=-\sin^310x=-f(x)$$
since the integral over the first half of the period would then add out with the integral over the second half of the period.</p>
|
4,206,205 | <p>I came across a theorem in algebraic number theory:</p>
<blockquote>
<p><strong>Theorem</strong> Let <span class="math-container">$A$</span> be a Dedekind ring and <span class="math-container">$M, N$</span> two modules over <span class="math-container">$A$</span>. If <span class="math-container">$M_\mathfrak{p} \subset N_\mathfrak{p}$</span> for all prime ideals <span class="math-container">$\mathfrak{p} \subset A$</span>, then <span class="math-container">$M \subset N$</span>.</p>
</blockquote>
<p><em>Proof.</em> Let <span class="math-container">$a \in M$</span>. For each <span class="math-container">$\mathfrak{p}$</span> we can find <span class="math-container">$x_\mathfrak{p} \in N$</span> and <span class="math-container">$s_\mathfrak{p} \in A \setminus \mathfrak{p}$</span> such that <span class="math-container">$a/1=x_\mathfrak{p}/s_\mathfrak{p}$</span>. Let <span class="math-container">$\mathfrak{b}$</span> be the prime ideal generated by <em>the</em> <span class="math-container">$s_\mathfrak{p}$</span>. Then <span class="math-container">$\mathfrak{b}$</span> is the unit ideal, and we can write
\[
1=\sum y_\mathfrak{p}s_\mathfrak{p}
\]
with elements <span class="math-container">$y_\mathfrak{p} \in A$</span> all but a finite number of which are <span class="math-container">$0$</span>. This yields
\[
a = \sum y_\mathfrak{p}s_\mathfrak{p}a = \sum y_\mathfrak{p}x_\mathfrak{p} \in N
\]
as desired. <span class="math-container">$\square$</span></p>
<hr />
<p>Here <span class="math-container">$M_\mathfrak{p}$</span> means the localisation of <span class="math-container">$M$</span> at <span class="math-container">$\mathfrak{p}$</span>.</p>
<p>My question is not about the theorem per se, but I guess there should be some counterexamples. Chances are there are modules <span class="math-container">$M,N$</span> such that</p>
<p><span class="math-container">$$
M_\mathfrak{p} \subset N_\mathfrak{p} \text{ does not imply } M \subset N.
$$</span></p>
<p>But I find it pretty hard to construct an example. I try to breakdown this question into pieces:</p>
<ol>
<li>Does this coincide with local property? It's pretty close:</li>
</ol>
<blockquote>
<p>Let <span class="math-container">$\phi:M \to N$</span> be an <span class="math-container">$A$</span>-modue homomorphism (no futher property assumed on <span class="math-container">$A$</span> at this point), then the followings are equivalent:</p>
<ol>
<li><p><span class="math-container">$\phi$</span> is injective.</p>
</li>
<li><p><span class="math-container">$\phi_\mathfrak{p}:M_\mathfrak{p} \to N_\mathfrak{p}$</span> is injective for each prime ideal <span class="math-container">$\mathfrak{p}$</span>.</p>
</li>
<li><p><span class="math-container">$\phi_\mathfrak{m}:M_\mathfrak{m} \to N_\mathfrak{m}$</span> is injective for each maximal ideal <span class="math-container">$\mathfrak{m}$</span>.</p>
</li>
</ol>
</blockquote>
<p>It looks like we are talking about 2 implies 1. But the inclusion map <span class="math-container">$M \to N$</span> is not given in this theorem. Also in the proof of the theorem, Serge Lang used properties of Dedekind ring (fractional ideals form a group) explicitly. If local property is the case, why would he ignore it?</p>
<ol start="2">
<li><p>Is it possible to get a desired counterexample given by <span class="math-container">$M_\mathfrak{p} \not\subset N_\mathfrak{p}$</span> for at least one prime ideal?</p>
</li>
<li><p>Is it possible to get a desired counterexample given by <span class="math-container">$A$</span> not being Dedekind, or even wilder, <span class="math-container">$A$</span> is not Noetherian? Noetherian counterexamples will certainly be more interesting.</p>
</li>
</ol>
<p>Thanks in advance!</p>
<hr />
<p>Edit 1: I added proof of the theorem to avoid unnecessary confusion.</p>
| Stefan4024 | 67,746 | <p>In fact, the claim is true for any unital ring <span class="math-container">$R$</span>. Indeed, suppose that <span class="math-container">$M_\mathfrak p \subseteq N_\mathfrak p$</span> for all prime ideals <span class="math-container">$\mathfrak p$</span> of <span class="math-container">$R$</span>.</p>
<p>Let <span class="math-container">$a \in M$</span>. Then for any prime ideal <span class="math-container">$\mathfrak p$</span> we consider <span class="math-container">$\frac a1 \in M_\mathfrak p \subseteq N_\mathfrak p$</span>. Hence there exist <span class="math-container">$n_\mathfrak p \in N$</span> and <span class="math-container">$s_\mathfrak p,t_\mathfrak p \in R-\mathfrak p$</span>, s.t.</p>
<p><span class="math-container">$$t_\mathfrak p(s_\mathfrak pa-n_\mathfrak p) = 0$$</span></p>
<p>Now consider the ideal <span class="math-container">$I$</span> of <span class="math-container">$R$</span> generated by all the elements <span class="math-container">$t_\mathfrak ps_\mathfrak p$</span>. We claim that it is the whole ring <span class="math-container">$R$</span>. Suppose the opposite is true, then you can find a maximal ideal <span class="math-container">$\mathfrak m$</span> containing it. However, then we get <span class="math-container">$t_\mathfrak ms_\mathfrak m \in I \subseteq \mathfrak m$</span>, which is a contradiction, as <span class="math-container">$t_\mathfrak ms_\mathfrak m \in R- \mathfrak m$</span>. Hence <span class="math-container">$I=R$</span>. In particular, we can find <span class="math-container">$y_\mathfrak p \in R$</span>, s.t.</p>
<p><span class="math-container">$$1 = \sum_\mathfrak p y_\mathfrak pt_\mathfrak ps_\mathfrak p$$</span></p>
<p>where all but finitely many of the <span class="math-container">$y_\mathfrak p$</span> are <span class="math-container">$0$</span>. Therefore:</p>
<p><span class="math-container">$$a = \sum_\mathfrak p y_\mathfrak pt_\mathfrak ps_\mathfrak pa = \sum_\mathfrak p (y_\mathfrak pt_\mathfrak p)n_\mathfrak p \in N$$</span></p>
<p>Hence <span class="math-container">$M \subseteq N$</span>.</p>
<hr />
<p>From the proof we can notice that we can relax the condition and ask for <span class="math-container">$M_\mathfrak m \subseteq N_\mathfrak p$</span> for all maximal ideals <span class="math-container">$\mathfrak m$</span> of <span class="math-container">$R$</span>, which is a common feature when it comes to localizing at prime ideals.</p>
|
3,129,372 | <p>I want to compute if exists <span class="math-container">$\lim_{r\rightarrow +\infty} \int_{B(0,r)} \frac{y}{1+\sqrt{(x^2+y^2)^5}}$</span> dx dy</p>
<p>I use polar coordinates and I found <span class="math-container">$4\lim_{r\rightarrow +\infty} \int_{0}^{r} \frac{\rho^2}{1+\rho^5} d\rho$</span>.</p>
<p>This integral exists because <span class="math-container">$g(\rho)=\frac{\rho^2}{1+\rho^5} \sim \frac{1}{\rho^3}$</span> near <span class="math-container">$+\infty$</span>?</p>
| TheSilverDoe | 594,484 | <p>How many possibilities do you have for the basket <span class="math-container">$A$</span> ? You have to choose <span class="math-container">$3$</span> fruits among a total of <span class="math-container">$5$</span> fruits. By definition, the number of possibilities is <span class="math-container">${5 \choose 3}$</span>.</p>
<p>Now, once you have choosen your <span class="math-container">$A-$</span>fruits, then you have no more choice for the basket <span class="math-container">$B$</span>, you have to put the two others fruits in it. So you are done.</p>
<p>Finally the total number of possibilities is <span class="math-container">$${5 \choose 3}$$</span></p>
|
3,129,372 | <p>I want to compute if exists <span class="math-container">$\lim_{r\rightarrow +\infty} \int_{B(0,r)} \frac{y}{1+\sqrt{(x^2+y^2)^5}}$</span> dx dy</p>
<p>I use polar coordinates and I found <span class="math-container">$4\lim_{r\rightarrow +\infty} \int_{0}^{r} \frac{\rho^2}{1+\rho^5} d\rho$</span>.</p>
<p>This integral exists because <span class="math-container">$g(\rho)=\frac{\rho^2}{1+\rho^5} \sim \frac{1}{\rho^3}$</span> near <span class="math-container">$+\infty$</span>?</p>
| user298173 | 504,775 | <p>Consider filling Basket A first. For this, you have five fruit and must choose three of them. There are 5 choices for the first fruit, and once this is chosen there are 4 for the next fruit and finally 3 choices for the last fruit and so <span class="math-container">$5\cdot 4\cdot 3$</span> choices. Since the order of the fruit going in to the basket is irrelevant one must divide through the number of ways of ordering the 3 fruit to avoid overcounting. There are <span class="math-container">$3!$</span> ways to order these three fruits (namely the number of permutations of three elements) and so there are
<span class="math-container">\begin{align*}
\frac{5\cdot 4\cdot 3}{3!} = \frac{5!}{3!2!}=\begin{pmatrix} 5 \\ 3\end{pmatrix}
\end{align*}</span>
choices. For Basket B, there are only two remaining fruit and both must go into the basket and so there are no extra choices and so this is the final answer.</p>
|
1,022,523 | <blockquote>
<p>Make a series expansion of $f(z)=\dfrac{1}{z^2+z-6}$ valid in the region $2<|z|<3$.</p>
</blockquote>
<p>By partial fractions,</p>
<p>$$f(z) = \frac{1}{(z-2)(z+3)} = \frac{1}{5(z-2)}-\frac{1}{5(z+3)}.$$</p>
<p>From here, how are these fractions expanded into a geometric series?</p>
| Mister Benjamin Dover | 196,215 | <p>A very informative, brief paper is P. Deligne, <em>Quelques idées maîtresses de l'œuvre de A. Grothendieck</em> (Matériaux pour l'histoire des mathématiques au XXe siècle. Actes du colloque à la mémoire de Jean Dieudonné, Séminaires et Congrès 3), which focuses on Grothendieck's contribution to the theory of schemes, topos theory, homological algebra, and motives. (Thus it does of course not cover everything, in particular it says nothing about logic/set theory, but I guess it is agreed that this was not Grothendieck's speciality.) It can be found for instance <a href="http://www.ihes.fr/~landais/smf_sem-cong_3_11-19.pdf" rel="nofollow">here</a>.</p>
|
3,811,498 | <p>Solve, by bringing the equation to Bernoulli form:</p>
<p><span class="math-container">$$
y’ = \frac{2-xy^3}{3x^2y^2}
$$</span></p>
<hr />
<p>Therefore we want to bring it to a form like:</p>
<p><span class="math-container">$$
y’ + p(x)y = q(x)y^n
$$</span></p>
<p>So working with the equation i get:</p>
<p><span class="math-container">$$
y’ - \frac{2}{3x^2y^2} = -\frac{xy^3}{3x^2y^2} = -\frac{y}{3x}
$$</span></p>
<p>I don’t see how to get to Bernoulli equation from here...</p>
| Physor | 772,645 | <p>you have <span class="math-container">$n = -2$</span></p>
<p><span class="math-container">$$
y^\prime +\frac{1}{3x}y = \frac{2}{3x^2}y^{-2}
$$</span></p>
|
1,033,383 | <p>$ABCD$ is a rectangle and the lines ending at $E$, $F$ and $G$ are all parallel to $AB$ as shown. </p>
<p>If $AD = 12$, then calculate the length of $AG$.<img src="https://i.stack.imgur.com/OUQZ8.png" alt="enter image description here"></p>
<p>Ok, I started by setting up a system of axes where $A$ is the origin and the $x$-axis along $AB$ and the $y$-axis along $AD$. So $D(0;-12)$ AND $C(x;-12)$ and $B(x;0)$. I am really stuck now on how to proceed so can someone please help me?</p>
| Sam Dittmer | 194,283 | <p>A solution without coordinates:</p>
<p>Let $E',F'$, and $G'$ be the intersection points on $AC$ of the lines parallel to $AB$ through $E,F$, and $G$ respectively. By symmetry, $AE=ED=6$. And $\triangle ABD \sim \triangle EE'D$, so $EE' = \tfrac 12 AB$.</p>
<p>Then $\triangle ABF' \sim \triangle E'EF'$, and $EE' = \tfrac 12 AB$, so the altitude of $\triangle ABF'$ and the altitude of $E'EF'$ are in a $2:1$ ratio, i.e. $AF=2EF$, so $AF=4$ and $EF=2$.</p>
<p>Then $FF' = \tfrac 13 CD = \tfrac 13 AB$, since $\frac{AF}{AD}=\tfrac 13$, and $\triangle AFF' \sim \triangle ADC$. And $\triangle BAG' \sim \triangle FF'G'$, so comparing altitudes gives us $AG = 3GF$, so $AG=3$.</p>
|
876,310 | <p>So I <em>think</em> I understand what differentials are, but let me know if I'm wrong.</p>
<p>So let's take $y=f(x)$ such that $f: [a,b] \subset \Bbb R \to \Bbb R$. Instead of defining the derivative of $f$ in terms of the differentials $\text{dy}$ and $\text{dx}$, we take the derivative $f'(x)$ as our "primitive". Then to define the differentials we do as follows:</p>
<p>We find some $x_0 \in [a,b]$ where there is some neighborhood of $x_0$, $N(x_0)$, such that all $f(x)$ in $\{f(x) \in \Bbb R \mid x \in N(x_0)\}$ are differentiable. Then we choose another point in $N(x_0)$, let's call it $x_1$, such that $x_1 \ne x_0$. Then let $dx = \Delta x = x_1 - x_0$. Now this $\Delta x$ doesn't actually have to be very small like we're taught in Calculus 1 (in particular it's not infinitesimal, it's finite). In fact, as long as $f(x)$ is differentiable for all $x \in [-10^{10}, 10^{10}]$ we could choose $x_0 = -10^{10}$ and $x_1 = 10^{10}$.</p>
<p>Then we know that $\Delta y = f'(x_0) \Delta x + \epsilon(\Delta x)$, where $\epsilon(\Delta x)$ is some nonlinear function of $\Delta x$. If $f(x)$ is smooth, we know that $\epsilon(\Delta x)$ is equal to the sum of powers of $\Delta x$ with some coefficients, by Taylor's theorem. But of course, $\epsilon(\Delta x)$ won't be so easy to describe if $f(x)$ is only once differentiable. So we define $dy$ as $dy = f'(x_0) dx$: that is, $dy$ is the <em>linear part</em> of $\Delta y$. This has the very useful property that $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \frac{dy}{dx} = f'(x_0)$. This is then <em>not a definition</em> of the derivative, but a consequence of our definitions.</p>
<p>It can be seen from this $dy$ really depends on what we choose as $dx$, but $f'$ is independent of both. </p>
<p>This definition can be extended to functions of multiple variables, like $z = f(x, y)$ as well, by letting $\Delta x = dx,\ \Delta y=dy$ and defining $dz$ as $dz = \frac{\partial f(x_0, y_0)}{\partial x}dx + \frac{\partial f(x_0, y_0)}{\partial y} dy$. So $dz$ is the linear part of $\Delta z$. Does all of the above look correct?</p>
<p>If so, then where I'm having a problem is: <br>1) how then <em>do</em> we define the derivative of $f(x)$ if not by $f'(x_0) = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$? <br>2) how do we apply this definition of $dx$ to $\int_a^b f(x)dx$? It seems like the inherit arbitrariness of $dx$ is really going to get in the way of a good definition of the integral.</p>
| Hola_Mundo | 166,011 | <p>OP here. I think I've figured this out:</p>
<p>This definition does seem to hold for differentiation and integration.</p>
<p><strong>Differentiation</strong><br>
My worry here was that because $\Delta x = dx$ and $\Delta y$ is a function of $\Delta x = dx$, that $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$ would also be dependent on $dx$, which would make the definition $f'(x) := \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$ a circular argument (as $dx$ was defined in terms of $f'(x)$) -- but in fact, it's really only $dy$ that's defined in terms of $f'(x)$. $dx$ is just some arbitrary change in $x$, i.e., $dx = x_1 - x_0$. We need $f'(x_0)$ to be defined in terms of $x_0$, but the inherit arbitrariness of $x_1$ would make anything defined explicitly in terms of $dx$ not-well-defined. However, what <em>the limit operation</em> really does <em>is remove the arbitrariness of $x_1$</em>. That is, the smaller we make $\| x_1 - x_0 \|$, the less "arbritrary" $x_1$ is. In the limit, it has lost all of it's "arbitrariness". So $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$ does not actually depend on the value we initially select for $x_1$ <em>at all</em>. So good, this definition works for differentiation.</p>
<p><strong>Integration</strong><br>
I couldn't figure out here what some arbitrarily large interval $dx$ (weird to hear $dx$ described as "large" isn't it?) had to do with integration (and thus <em>why it would be in the integrand</em>). But again, an arbitary interval under a de-arbitrarifying process (a limit) is exactly what we need. In this case $\int_a^b f(x)dx$ really means something like $$\lim_{\| dx_i \| \to 0} \sum_{i=0}^n f(x_i)dx_i= \lim_{\max(x_{i+1}-x_i) \to 0} \sum_{i=0}^n f(x_i)(x_{i+1} - x_i)$$ where each $dx_i$ is a subinterval of $[a,b]$. Notice that each $x_{i+1} \to x_i$ and thus $x_{i+1}$ loses it's "arbitrariness" in this limit.<br>
In case you're thinking that these $dx_i$ are not defined the same as in my question, notice that in the definition I gave, $x_0$ was fixed by the problem -- our only variable that we were free to fix was $x_1$. That's exactly how these $dx_i$ are defined: for instance, the first one, $dx_0 = x_1 - x_0 = x_1 - a$ for some arbitrary $x_1 \in (a,b]$. Assuming $x_1 \ne b$, then to fill out our partition we must define another subinterval $dx_1 = x_2 - x_1$, where $x_1$ in <em>this case</em> is <strong>not</strong> arbitrary -- it's defined to be the end point of our $1^{st}$ (or $0^{th}$ maybe because of how I chose to define my partition) subinterval -- but $x_2$ <em>is</em> arbitrarily chosen from the interval $(x_1, b]$. And so on.</p>
<p>NOTE: I don't particularly like the notation I used to "define" integration above (the $\lim_{\| dx_i \| \to 0} \dots$ part), one of several reasons is that it really doesn't tell you that in the limit, $n \to \infty$. The wikipedia page on the Riemann integral doesn't have such an equation (they just describe it with words... ugh). Do you guys know a better notation?</p>
<p><strong>Conclusion</strong><br>
It seems that defining $dx:=\Delta x$ and $dy:=f'(x)dx$ does exactly what it needs to do. The relationship between the two can be used to approximate small changes in $\Delta y$, neither differentiation nor integration is defined in terms of $dy$ and though they <em>are</em> both defined in terms of $dx = x_1 - x_0$, they are <em>not dependent</em> on any values we initially select for $x_1$, and probably one of the best things is that because $dx$ and $dy$ are finite, the relationship $f'(x) = \frac{dy}{dx}$ holds for any differentiable $f$.</p>
<p>NOTE: This definition of $dx$ and $dy$ does not seem to be the same as the one used in differential geometry -- as described by Hurkyl. But then, I'm not entirely sure because I don't completely understand Hurkyl's answer. If anyone knows of a good primer on the notations and concepts he's using -- suitable for someone who's gone through the calculus sequence and linear algebra only -- I would be grateful for a link. However, even if the definitions <em>are</em> different, it doesn't mean mine is not useable -- in fact, unless you guys can come up with a situation where $dx = x_1 - x_0$ and $dy = f'(x)dx$ (where $dx$ and $dy$ can be arbitrarily large) don't do what they're supposed to, I'm just going to take them as my definition of them <em>from now on</em>.</p>
|
1,794,072 | <p>My attempt :</p>
<p>If $n$ is odd, then the square must be 2 (mod 3), which is not possible.</p>
<p>Hence $n =2m$</p>
<p>$2^{2m}+3^{2m}=(2^m+a)^2$</p>
<p>$a^2+2^{m+1}a=3^{2m}$</p>
<p>$a (a+2^{m+1})=3^{2m} $</p>
<p>By fundamental theorem of arithmetic, </p>
<p>$a=3^x $</p>
<p>$3^x +2^{m+1}=3^y $</p>
<p>$2^{m+1}=3^x (3^{y-x}-1) $</p>
<p>Which is not possible by Fundamental theorem of Arithmetic </p>
<p>Is there a better method? I even have a nagging feeling this is wrong as only once is the equality of powers of 2 and 3 $(n) $are used.</p>
| lulu | 252,071 | <p>Work $\pmod 3$. We see that $$2^n+3^n\equiv 2^n\pmod 3$$ from which we quickly deduce that your expression is only a square $\pmod 3$ if $n$ is even.</p>
<p>Now work $\pmod 5$. We see that $$2^n+3^n\equiv 2^n+(-2)^n\equiv 2^n(1+(-1)^n)\pmod 5$$ from which we deduce that your expression is only a square $\pmod 5$ if $n$ is odd.</p>
|
105,750 | <p>Given a <code>ContourPlot</code> with a set of contours, say, this:</p>
<p><a href="https://i.stack.imgur.com/cKoyo.jpg"><img src="https://i.stack.imgur.com/cKoyo.jpg" alt="enter image description here"></a></p>
<p>is it possible to get the contours separating domains with the different colors in the form of lists? </p>
<p>For example, how to extract the boundaries of the blue domain in the image above?
Or just for the sake of trial, from such a simple example:</p>
<pre><code> ContourPlot[x*Exp[-x^2 - y^2], {x, 0, 3}, {y, -3, 3},
PlotRange -> {0, 0.5}, ColorFunction -> "Rainbow"]
</code></pre>
<p><a href="https://i.stack.imgur.com/Beuzu.jpg"><img src="https://i.stack.imgur.com/Beuzu.jpg" alt="enter image description here"></a></p>
<p>The same task, let us find the lists corresponding to the blue domain boundaries.</p>
<p>To make it clear, I am not asking of how to get the lines from the function behind. This I understand. I ask of how to extract the contour lines that are generated by Mma.</p>
<p>Let us put this question another way around. Is it possible to define the areas with the same color as separate geometric regions in the sense of the computation geometry, and then work with these domains separately?</p>
| Sumit | 8,070 | <p>A simple way could be using the <code>PlotRange</code>.</p>
<pre><code>Table[
ContourPlot[x*Exp[-x^2 - y^2], {x, 0, 2}, {y, -1, 1},
PlotRange -> {i, j}, ColorFunction -> "Rainbow",
PlotLegends -> True, PlotLabel -> {i, j}]
, {i, 0.1, 0.3, 0.1}, {j, i + 0.1, 0.4, 0.1}] // TableForm
</code></pre>
<p><a href="https://i.stack.imgur.com/akAqJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/akAqJ.png" alt="enter image description here"></a></p>
<p>For a line, simply put the value</p>
<pre><code>ContourPlot[x*Exp[-x^2 - y^2] == 0.3, {x, 0, 2}, {y, -1, 1}]
</code></pre>
|
881,282 | <p>Same as above, how to simplify it. I am to calculate its $n$th derivative w.r.t x where t is const, but I can't simplify it. Any help would be appreciated. Thank you.</p>
| Deathkamp Drone | 56,720 | <p>It seems like you're right, Kasper. A trigonometric substituition would solve the equation completely. I'll write the solution I arrived at:</p>
<p>Finding inequalities is helpful to motivate the right subtituition. We know that both $\frac{1}{x^2}$ and $\frac{1}{(4-\sqrt{3}x)^2}$ are positive, so we can say that:
$$\frac{1}{x^2}<1\Leftrightarrow \left|\frac{1}{x}\right|<1\Leftrightarrow -1<\frac{1}{x}<1$$
And analogously $\displaystyle-1<\frac{1}{4-\sqrt{3}x}<1$. The substituition is starting to become obvious. One last hint is that we're trying to find two squares whose sum equals $1$. This immediately reminds us of the identity $\sin^2 a+\cos^2 a=1$. That's enough to motivate the substituition (WLOG):
$$\frac{1}{x}=\sin a \,\,\text{ and }\,\,\frac{1}{4-\sqrt{3}x}=\cos a$$</p>
<p>Which will give us:
$$\begin{align}
&\frac{1}{\displaystyle4-\frac{\sqrt{3}}{\sin a}}=\cos a \Leftrightarrow \\\\
&\frac{1}{\cos a}+\frac{\sqrt{3}}{\sin a} = 4 \Leftrightarrow \\\\
&\frac{1}{2}\sin a + \frac{\sqrt{3}}{2}\cos a = 2\sin a\cos a \Leftrightarrow \\\\
&\sin(60°+a)=\sin(2a)
\end{align}$$</p>
<p>Solving and substituting back, the solutions are: $$x_1=\csc60°=\frac{2}{\sqrt{3}},\;x_2=\csc 40°,\;x_3=\csc20°,\;x_4=-\sec10°.\;\;\blacksquare$$</p>
|
2,783,423 | <p>If I have a line formed by points A and B, how can I find the distance of another point from that line. Also, whether that line is clockwise or CCW from point A.
<a href="https://i.stack.imgur.com/dpazD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dpazD.png" alt="enter image description here"></a></p>
<p>I'm not really even sure how to get started with this. I know the midpoint of A and B, but after that I am out of my depth. </p>
<p>Would appreciate advice on how to proceed.</p>
| Community | -1 | <p>$$\frac{AB}{\|AB\|}\times AZ$$ gives you the distance with a positive or negative sign depending on which side $Z$ lies.</p>
|
3,294,082 | <p>The exercise is to prove that the minimum value between <span class="math-container">$a^{1/b}$</span> and <span class="math-container">$b^{1/a}$</span> is no greater than <span class="math-container">$3^{1/3}$</span>, where <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are positive integers. As it was presented in an introductory calculus class, I tried using brute Mathematics, ploting graphs, but was unable to develop and was looking for some elegant ideas.</p>
| Jihoon Kang | 452,346 | <p>Without loss of generality, let <span class="math-container">$a \leq b$</span> and keep <span class="math-container">$b$</span> be a random fixed value, then since <span class="math-container">$a, b \geq 1$</span> and <span class="math-container">$1/b \leq 1/a$</span>,</p>
<p><span class="math-container">$$\min{(a^{1/b}, b^{1/a})}=a^{1/b}$$</span></p>
<p>Then we clearly have maximum achieved (keeping <span class="math-container">$b$</span> constant) when <span class="math-container">$a=b$</span>, or in other words, <span class="math-container">$\min{(a^{1/b}, b^{1/a})}\leq b^{1/b}$</span>. Now the question is what value of <span class="math-container">$b$</span> gives us the overall maximum?</p>
<p>Lets consider the graph <span class="math-container">$f(x)=x^{1/x}$</span> on <span class="math-container">$x>0$</span>, and see if it has any maximum points. We get (log both sides and differenciate implicitly):</p>
<p><span class="math-container">$$\frac{d}{dx}\left(x^{1/x}\right)=-x^{-2+1/x}\times (\log(x)-1)$$</span> </p>
<p>If we set this to <span class="math-container">$0$</span>, then our only solution is <span class="math-container">$\log(x)=1$</span> giving <span class="math-container">$x=e$</span> (recall <span class="math-container">$x>0$</span>).</p>
<p>Notice that if <span class="math-container">$x>e$</span>, then <span class="math-container">$\frac{d}{dx}\left(x^{1/x}\right)<0$</span> and if <span class="math-container">$0<x<e$</span>, <span class="math-container">$\frac{d}{dx}\left(x^{1/x}\right)>0$</span>, hence <span class="math-container">$x=e$</span> is a maximum point.</p>
<p>Now, we can't take <span class="math-container">$b=e$</span> as <span class="math-container">$e$</span> is not an integer, but we can try <span class="math-container">$b=2$</span> or <span class="math-container">$b=3$</span> as we know <span class="math-container">$f(x)=x^{1/x}$</span> is strictly decreasing for <span class="math-container">$x>e$</span> and strictly increasing for <span class="math-container">$0<x<e$</span>. We see that <span class="math-container">$$ 2^{1/2} < 3^{1/3} \iff 2^{6/2} < 3^{6/3} \iff 8 < 9$$</span></p>
<p>So maximum is achieved when <span class="math-container">$a=b=3$</span>, i.e <span class="math-container">$$\min{(a^{1/b}, b^{1/a})} \leq 3^{1/3}$$</span></p>
|
2,339,101 | <blockquote>
<p>There are six socks in a drawer. The socks are of two colors: black and white. If you draw two socks randomly, the probability that you get white socks is $\frac{2}{3}$. What is the probability of getting black socks, when two socks are drawn at a time?</p>
</blockquote>
<p>There is no detail about the number of white and black socks. Can anyone help me with this?</p>
| Leo163 | 185,102 | <p>The answer is $0$: out of $15$ possible unordered couples of socks, you know that $10$ are made by white socks. It means that only $5$ possible pairs of socks contain a black sock. It can only be true if there is one black sock and $5$ white socks.</p>
|
619,477 | <blockquote>
<p>Alice opened her grade report and exclaimed, "I can't believe Professor Jones flunked me in Probability." "You were in that course?" said Bob.
"That's funny, i was in it too, and i don't remember ever seeing you there."
"Well," admitted Alice sheepishly, "I guess i did skip class a lot." "Yeah, me too" said Bob. Prove that either Alice or Bob missed at least half of the classes. </p>
</blockquote>
<p>Proof:</p>
<p>Let $A$ be the set of lectures Alice attended and missed, let's assume she attended them in no particular order, similarly for Bob $B$ is the set of all lectures Bob attended and missed in random order.
Let $f$ be one-to-one and onto, we define $f:A\to B$ to be the mapping that matches the lectures Alice attended to the lectures that Bob missed and the lectures that Alice missed to the ones that Bob attended. If we consolidate the contiguous entries in the sets $A$ and $B$ into two groups, the group of lectures that Alice attended and the group of lectures she didn't attend and similarly for $B$ then the function $f$ can only be one-to-one and onto if both Alice and Bob attended the same number of lectures they missed.</p>
<p>I understand i've shown that Bob and Alice missed half the classes, how do i show that they could've missed more with this method?? </p>
| Hagen von Eitzen | 39,174 | <p>Let $n$ be the total number of lectures, $n_{Ab}$ the number of lectures Alice attended and Bob didn't, $n_{aB}$ the number of lectures Bob attended and Alice didn't, $n_{AB}$ the number they both attended and $n_{ab}$ neither of thenm attended.
Then $n=n_{ab}+n_{aB}+n_{Ab}+n_{AB}$ and Alice missed $n_{ab}+n_{aB}$, Bob missed $n_{ab}+n_{Ab}$. If we assume $n_{ab}+n_{aB}<\frac n2$ and $n_{ab}+n_{Ab}<\frac n2$, then
$$ n_{ab}-n_{AB}=(n_{ab}+n_{Ab})+(n_{ab}+n_{aB})-(n_{ab}+n_{aB}+n_{Ab}+n_{AB})<0.$$
This is only possible if $n_{AB}>0$, but they say they never met.</p>
|
3,578,357 | <p>The problem is like this : How do you solve <span class="math-container">$$ \lim _{x\to 0}\:\:\frac{x^m-sin^n(x)}{x^{n+2}} $$</span> for different values of <span class="math-container">$ n \in \Bbb N $</span>
Now, what i've started doing is to add <span class="math-container">$$ \lim _{x\to 0}\:\:\frac{x^m-x^n+x^n-sin^n(x)}{x^{n+2}} $$</span> then i split the limit into two limits like this <span class="math-container">$$ \lim _{x\to 0}\:\:\frac{x^m-x^n}{x^{n+2}} + \lim _{x\to 0}\:\:\frac{x^n-sin^n(x)}{x^{n+2}} $$</span> and i was thinking for the second limit to apply the formula : <span class="math-container">$(a-b)^n$</span> . The problem is that i don't know what to do with the first limit which has <span class="math-container">$x^m$</span>, at first i thought that i was a mistake in my textbook, but i am not sure . </p>
| marty cohen | 13,079 | <p>If <span class="math-container">$f \ne 0$</span> then,
since <span class="math-container">$(1/f)' = -f'/f^2$</span>,
<span class="math-container">$(1/f)' = -1$</span> so
<span class="math-container">$1/f = -x+c$</span> so
<span class="math-container">$f(x) = 1/(-x+c)$</span>.</p>
<p>Then
<span class="math-container">$f$</span> is injective
since <span class="math-container">$f(x) = f(y)
\implies x=y$</span>.</p>
|
3,578,357 | <p>The problem is like this : How do you solve <span class="math-container">$$ \lim _{x\to 0}\:\:\frac{x^m-sin^n(x)}{x^{n+2}} $$</span> for different values of <span class="math-container">$ n \in \Bbb N $</span>
Now, what i've started doing is to add <span class="math-container">$$ \lim _{x\to 0}\:\:\frac{x^m-x^n+x^n-sin^n(x)}{x^{n+2}} $$</span> then i split the limit into two limits like this <span class="math-container">$$ \lim _{x\to 0}\:\:\frac{x^m-x^n}{x^{n+2}} + \lim _{x\to 0}\:\:\frac{x^n-sin^n(x)}{x^{n+2}} $$</span> and i was thinking for the second limit to apply the formula : <span class="math-container">$(a-b)^n$</span> . The problem is that i don't know what to do with the first limit which has <span class="math-container">$x^m$</span>, at first i thought that i was a mistake in my textbook, but i am not sure . </p>
| Behnam Esmayli | 283,487 | <p>Note that your <span class="math-container">$f$</span> is a solution to the ODE
<span class="math-container">$$
dy/dx = y^2 \ ,
$$</span>
which can be solved using separation of variable. ALL solutions to this ODE other than the obvious solution <span class="math-container">$y = 0 $</span> are of the form
<span class="math-container">$$
y = \frac{1}{c-x} \ ,
$$</span>
for a constant <span class="math-container">$c$</span>. The claim that this is an exhaustive list follows from uniqueness of solutions to initial value problems of the type <span class="math-container">$y'=F(x,y), y(x_0) = y_0 $</span> where <span class="math-container">$F$</span> is, say, locally Lipschitz. Here <span class="math-container">$F(x,y) = y^2$</span> which is super nice!</p>
<p>Now, every solution other than the constant solution is injective (although not defined on all of <span class="math-container">$\mathbb{R}$</span>.</p>
<p><strong>Note: I actually proved more: Injective or otherwise, you cannot find a differentiable function, other than zero function, defined on all of <span class="math-container">$\mathbb{R}$</span> that satisfies <span class="math-container">$f' = f^2$</span>!</strong></p>
|
3,753,060 | <blockquote>
<p>If <span class="math-container">$\int f(x)dx =g(x)$</span> then <span class="math-container">$\int f^{-1}(x)dx $</span> is equal to</p>
<p>(1) <span class="math-container">$g^{-1}(x)$</span></p>
<p>(2) <span class="math-container">$xf^{-1}(x)-g(f^{-1}(x))$</span></p>
<p>(3) <span class="math-container">$xf^{-1}(x)-g^{-1}(x)$</span></p>
<p>(4) <span class="math-container">$f^{-1}(x)$</span></p>
</blockquote>
<p>My approach is as follows:
Let <span class="math-container">$f(x)=y$</span>, therefore <span class="math-container">$f^{-1}(y)=x$</span>, <span class="math-container">$\int f^{-1}(f(x))dx =g(f(x))$</span></p>
<p>On differentiating we get <span class="math-container">$x=g'(f(x))f'(x)$</span></p>
<p>After this step, I am not able to proceed.</p>
| Z Ahmed | 671,540 | <p>Given <span class="math-container">$y=f(x) \implies x=f^{-1}(y)$</span>
Then <span class="math-container">$$\int f^{-1}(x) dx= \int f^{-1}(y) dy= \int x dy= \int x \frac{dy}{dx} dx=\int xf'(x) dx $$</span> <span class="math-container">$$=xf(x)-\int f(x)dx=xf(x)-g(x)+C.$$</span> Lastly, we have done integration by parts.</p>
|
1,615,883 | <p>A cubic polynomial with real coefficients, $a x^3 + b x^2 + c x + d$, has either three real roots, or one real root and a pair of complex conjugate ones. If the latter happens, what is the explicit formula for this real solution, and what conditions can be placed on $a,b,c$ and $d$ to guarantee that the real root is positive?</p>
| Mike Earnest | 177,399 | <p>Your answer is correct, assuming that each of the books are placed independently, and each shelf is equally likely.</p>
<p>However, if the phrase "randomly tidied up" instead meant "choose a random arrangement of books", then you get a different answer.</p>
<ul>
<li>There are $5$ ways where they are all on the same shelf.</li>
<li>There are $5\cdot 4=20$ ways where there are two on shelf, and the other on a different shelf.</li>
<li>Three are $\binom{5}{3}=10$ ways where they are all on different shelves.</li>
</ul>
<p>Assuming each of these arrangements is equally likely, the desired probability is $\frac{5}{35}=\frac17$.</p>
|
2,555,399 | <p>The question is to find out the coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3)^{1/2}$</p>
<p>I tried using multinomial theorem but here the exponent is a fraction and I couldn't get how to proceed.Any ideas?</p>
| Jack D'Aurizio | 44,121 | <p>It is quite practical to exploit the identity
$$ (1+x)^2 (1-2x+3x^2-4x^3) = 1-5x^4-4x^5 \tag{A}$$
from which
$$\begin{eqnarray*} \sqrt{1-2x+3x^2-4x^3} &=& \frac{\sqrt{1-5x^4-4x^3}}{1+x} \\&=&\left(1+O(x^4)\right)(1-x+x^2-x^3+O(x^4))\tag{B}\end{eqnarray*}$$
and the coefficient of $x^3$ in the RHS of $(B)$ is trivially $\color{red}{-1}$.</p>
|
45,911 | <p>I've been wondering for some time now about the difference between a point and a vector. In high school, it was very important to distinguish them from each other, and we used the notation $(x,y,z)$ for points and $[x,y,z]$ for vectors. We always had to translate the point $P=(a,b,c)$ to the vector $\overrightarrow{OP} =[a,b,c]$ before we started calculating with them. </p>
<p>Now, after I started at the university, people don't seem to care anymore. My professors either say that they're the same, or that they're almost the same, and the books I have seem to share that view. The book I use for my calculus course (Colley's Vector Calculus) says, among other things, the following:</p>
<blockquote>
<p>[...] we adopt the point of view that a vector field assigns to each <em>point</em> $\textbf{x}$ in X a <em>vector</em> $\textbf{F}(\textbf{x})$ in $\mathbb{R}^n$, represented by an arrow whose tail is at the point $\textbf{x}$.</p>
</blockquote>
<p>So it seems like a point is also a vector.</p>
<p>My question is this: Do mathematicians distinguish between points and vectors, and if they do, in what circumstances?</p>
| Qiaochu Yuan | 232 | <p>A point in Euclidean space is properly regarded as an element of an <a href="http://en.wikipedia.org/wiki/Affine_space">affine space</a> rather than a vector space. That's because vector spaces have a distinguished origin, and "space" in the general sense doesn't: you can move the origin anywhere you want. Affine spaces are also constructed to have the property that the difference between two points is a vector. Because affine spaces don't have a distinguished origin, you can't add two points in an affine space, but you can take <a href="http://en.wikipedia.org/wiki/Affine_combination">affine combinations</a>. </p>
<p>There is also a more general notion of "point" as just an element of any set equipped with some kind of geometric structure, such as a point in a <a href="http://en.wikipedia.org/wiki/Topological_space">topological space</a>. </p>
|
45,911 | <p>I've been wondering for some time now about the difference between a point and a vector. In high school, it was very important to distinguish them from each other, and we used the notation $(x,y,z)$ for points and $[x,y,z]$ for vectors. We always had to translate the point $P=(a,b,c)$ to the vector $\overrightarrow{OP} =[a,b,c]$ before we started calculating with them. </p>
<p>Now, after I started at the university, people don't seem to care anymore. My professors either say that they're the same, or that they're almost the same, and the books I have seem to share that view. The book I use for my calculus course (Colley's Vector Calculus) says, among other things, the following:</p>
<blockquote>
<p>[...] we adopt the point of view that a vector field assigns to each <em>point</em> $\textbf{x}$ in X a <em>vector</em> $\textbf{F}(\textbf{x})$ in $\mathbb{R}^n$, represented by an arrow whose tail is at the point $\textbf{x}$.</p>
</blockquote>
<p>So it seems like a point is also a vector.</p>
<p>My question is this: Do mathematicians distinguish between points and vectors, and if they do, in what circumstances?</p>
| Robert Israel | 8,508 | <p>In general, mathematicians would distinguish between points and vectors in a context where that distinction is important, and might not bother to distinguish between them in a context where it isn't important. </p>
|
588,802 | <p>The problem is: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$</p>
<p>The first thing I did was use the divergence test which didn't help since the result of the limit was 0.</p>
<p>If I multiply it through, the result is $\sum_{n=1}^{\infty} \frac{1}{n^2+3n}$</p>
<p>I'm wondering if I can consider this as a p-series and simply use the largest power. In this case the power would be 2 which would mean it converges. If this is the correct way to go about this, how do I find where it converges to.</p>
| ILoveMath | 42,344 | <p>First, use estimations</p>
<p>$$ n^2 + 3n \geq n^2 \implies \frac{1}{n^2 + 3n } \leq \frac{1}{n^2} $$</p>
<p>Secondly, show that $\sum \frac{1}{n^2}$ converges. In fact, it does. More generally,</p>
<p>$$ \sum \frac{1}{n^p} \; \; \text{converges when} \; \; p > 1 $$</p>
<p>Third, use the comparison theorem: if $a_n \geq b_n $ for all $n$ and $\sum a_n$ converges, then $\sum b_n$ must converge as well (Proof?)</p>
<p>Now, as an application of this theorem, with $a_n = \frac{1}{n^2} $ and $b_n = \frac{1}{n^2 + 3n}$, we notice that your series </p>
<p>$$ \sum \frac{1}{n^2 + 3n} $$</p>
<p>must converge.</p>
|
588,802 | <p>The problem is: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$</p>
<p>The first thing I did was use the divergence test which didn't help since the result of the limit was 0.</p>
<p>If I multiply it through, the result is $\sum_{n=1}^{\infty} \frac{1}{n^2+3n}$</p>
<p>I'm wondering if I can consider this as a p-series and simply use the largest power. In this case the power would be 2 which would mean it converges. If this is the correct way to go about this, how do I find where it converges to.</p>
| Zhoe | 99,231 | <p>$$n^2 + 3n > n^2 \implies \frac{1}{n^2 +3n} < \frac{1}{n^2}$$</p>
<p>Use the Comparison Test which states that if $\sum a_n$ and $\sum b_n$ are such that $0 \le a_n \le b_n$, if $\sum b_n$ converges, then $\sum a_n$ converges. </p>
<p>Since $0 < \sum \frac{1}{n^2 +3n} < \sum \frac{1}{n^2}$ and $\sum \frac{1}{n^2}$ converges, then $\sum \frac{1}{n^2 +3n}$ converges.</p>
<p>Edit. <strong>Note:</strong> $\sum \frac{1}{n^2}$ converges since it is a p-series $$f(x) = \frac{1}{X^p}$$ with $p > 1$ and hence it converges.</p>
|
2,369,431 | <p>I took a course some years ago and in it was a treatment of how to associate a field to an abstract geometry. </p>
<p>I would very much appreciate some reading on this, as I have been unsuccessful on where to find any resources on such ideas, and I have since lost my notes!</p>
<blockquote>
<p>Is there a book which treats, thoroughly, the connection between an abstract geometry and a field?</p>
</blockquote>
<p>In particular:</p>
<blockquote>
<p>Is there a way to consider a discrete geometry, like projective Steiner triple systems, and associate to these objects a field?</p>
</blockquote>
<p>Any information would be extremely useful, introductory or extremely advanced. I would be particularly interested in anyones knowledge of resources which have connections to class field theory.</p>
<p>Thank you for your time and consideration.</p>
| Somos | 438,089 | <p>You may want to look at articles on Veblen-Wedderburn systems, Hall planes and Moufang planes to see if that helps your research. The book by Dembowski on <em>Finite Geometries</em> may interest you.</p>
|
1,379,283 | <p>In my notes it shows how to calculate by using the unit circle. But I do not know why the value of sin is the y coordinate and the value of cos is the x coordinate.</p>
| AlexR | 86,940 | <p>I'm not sure what your difficulty is, so I'll use the geometrical definition to get the requested values. This definition means that when going a distance of $\theta$ counter-clockwise along the unit circle starting at $(1,0)$, the coordinates of the point on the circle are $(\cos \theta, \sin \theta)$.</p>
<p>Note that $0$ corresponds to the point $(1,0)$ on the unit circle, i.e.
$$\sin(0) = 0, \cos(0) = 1$$
$\frac\pi2$ represents a quarter of a circle ($90^\circ$) wich corresponds to $(0,1)$ on the unit circle, i.e.
$$\sin(\frac\pi2) = 1, \cos(\frac\pi2) = 0$$
Finally, $\pi$ corresponds to half a circle ($180^\circ$) wich corresponds to $(-1,0)$. This means
$$\sin(\pi) = 0, \cos(\pi) = -1$$</p>
|
3,358,592 | <p>In the following quote, what does the notation <span class="math-container">$\{a_n\}$</span> mean?</p>
<blockquote>
<p>Дана последовательность Фибоначчи <span class="math-container">$\{a_n\}$</span>.</p>
</blockquote>
<p><strong>Translation:</strong> "You are given the Fibonacci sequence <span class="math-container">$\{a_n\}$</span>."</p>
| G Cab | 317,234 | <p>Your function can be approximated through two shifted ramp functions, as
<span class="math-container">$$
\eqalign{
& f(x) = x\,H(x) - \left( {x - 1} \right)\,H(x - 1) \approx \cr
& \approx {x \over 2}\left( {1 + {x \over {\sqrt {x^{\,2} + \varepsilon ^{\,2} } }}} \right) - {{\left( {x - 1} \right)} \over 2}\left( {1 + {{\left( {x - 1} \right)} \over {\sqrt {{\left( {x - 1} \right)}^{\,2} + \varepsilon ^{\,2} } }}} \right) \cr}
$$</span>
where<br>
<span class="math-container">$H(x)$</span> denotes the Heaviside step function;<br>
<span class="math-container">$\varepsilon <<1$</span> is a small value.</p>
<p>Concerning the square root, you might calculate itrecursively through the famous <a href="https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method" rel="nofollow noreferrer">Babylonian Method</a></p>
<p>This is an example of what you get with <span class="math-container">$\varepsilon = 0.1$</span>.</p>
<p><a href="https://i.stack.imgur.com/rsDWO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rsDWO.png" alt="Intp_Rampa_1"></a></p>
<p>It is clear that operating around the symmetric function <span class="math-container">$ x-1/2 \to y-1/2$</span> you can reduce the computations almost in half.</p>
|
285,114 | <blockquote>
<p>Find the solution of the differential equation
$$\frac{dy}{dx}=-\frac{x(x^2+y^2-10)}{y(x^2+y^2+5)}, y(0)=1$$</p>
</blockquote>
<p>Trial: $$\begin{align} \frac{dy}{dx}=-\frac{x(x^2+y^2-10)}{y(x^2+y^2+5)} \\ \implies \frac{dy}{dx}=-\frac{1+(y/x)^2-10/x^2}{(y/x)(1+(y/x)^2+5/x^2)} \\ \implies v+x\frac{dv}{dx}=-\frac{1+v^2-10/x^2}{v(1+v^2+5/x^2)} \end{align}$$ I can't seperate $x$ and $v$.</p>
| Mikasa | 8,581 | <p>Hint: It is an exact equation. Assume there is an differentiable function $f(x,y)$ such that $$f_x=x^3+xy^2-10x,~~~f_y=x^2y+y^3+5y$$ and then find the function. The solution is as $$f(x,y)=C$$</p>
|
3,897,361 | <p>Find the GS of the following system of DE's where the independent variable is <span class="math-container">$t$</span> and <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are the dependent variables</p>
<p><span class="math-container">\begin{cases}
x' = x-3y\\
y'=3x+7y
\end{cases}</span></p>
<p>I know using eigenvalues and eigenvectors or operators is one way to do this. But I wish to double check my answer using a substitution method.</p>
<p>So my work:</p>
<p>The second DE <span class="math-container">$y'=3x+7y$</span> can be rewritten as <span class="math-container">$x = \cfrac{y'}{3}-\cfrac 73y$</span></p>
<p>then <span class="math-container">$x' = \cfrac{y''}{3}-\cfrac73y'$</span></p>
<p>When we plug these values of <span class="math-container">$x$</span> and <span class="math-container">$x'$</span> into the first DE (<span class="math-container">$x' = x -3y)$</span>, we get with some rearranging</p>
<p><span class="math-container">$\cfrac{y''}{3}-\cfrac83y'+\cfrac{16}{3}y = 0$</span></p>
<p>Which has a characteristic equation of</p>
<p><span class="math-container">$\cfrac{r^2}{3}-\cfrac83r+\cfrac{16}{3} = 0$</span></p>
<p>with roots <span class="math-container">$r_1=4$</span> and <span class="math-container">$r_2 = 4$</span></p>
<p>Then the solution for <span class="math-container">$y$</span> is <span class="math-container">$y$</span> = <span class="math-container">$C_1e^{4t}+C_2te^{4t}$</span></p>
<p>Then we back sub to solve for <span class="math-container">$x$</span> using <span class="math-container">$x = \cfrac{y'}{3}$$-\cfrac73y$</span> with the solution of y we just found.</p>
<p>We get <span class="math-container">$x =-C_1e^{4t}-C_2te^{4t} + \cfrac{C_2}{3}e^{4t} =-C_1e^{4t}-C_2te^{4t} + C_3e^{4t}$</span></p>
<p>so the GS to the homo system is</p>
<p><span class="math-container">\begin{cases}
x = -C_1e^{4t}-C_2te^{4t}+C_3e^{4t}\\
y = C_1e^{4t}+C_2te^{4t}
\end{cases}</span></p>
<p>If this solution is right, then I'm confident that I understand how substitution method works for solving DE systems. (Also it would boost my confidence in using the operator method to solve this as I got the same answer as this using the operator method). I'm a little thrown off on the roots being the same but I still think my methodology is still sound. I would appreciate it if someone could tell me if I've got this right cause then I know I completely understand how to solve a system of DE's.</p>
<p>If more work is necessary to show please let me know.</p>
| Z Ahmed | 671,540 | <p>Your method is correct.</p>
<p>Another way is to write the system as <span class="math-container">$X'=MY$</span></p>
<p>Where <span class="math-container">$M$</span> is the matrix <span class="math-container">$$\begin{pmatrix} 1 & -3 \\ 3 & 7 \end{pmatrix}$$</span>
its eigen values are <span class="math-container">$m_1,m_2=4$</span> (repeated roots), so the solution for <span class="math-container">$x$</span> can be written as
<span class="math-container">$x=(A t+B)e^{4t}$</span>, putting it in first equation one can get <span class="math-container">$y$</span>. So there will be two
undetermined constants: <span class="math-container">$A,B$</span>.</p>
|
28,811 | <p>There are lots of statements that have been conditionally proved on the assumption that the Riemann Hypothesis is true.</p>
<p>What other conjectures have a large number of proven consequences?</p>
| Dan Ramras | 4,042 | <p>Resolution of singularities for algebraic varieties in positive characteristic is another example. Many statements in algebraic K-theory have been proven to follow from this conjecture. </p>
|
118,406 | <p>I have a single flat directory with over a million files. I just wanted to take a sample of the first few files but <code>FileNames</code> doesn't include a "only the first n" option, and so it took over a minute:</p>
<p><a href="https://i.stack.imgur.com/s5cBS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s5cBS.png" alt="enter image description here"></a></p>
<p>Is there a faster way?</p>
| george2079 | 2,079 | <p>a linux version using <code>find</code> to list files and <code>head</code> to take the first <code>n</code>.</p>
<pre><code>dir = "/path"
StringSplit[
RunProcess[
{"/bin/sh",
"-c",
"find "<>dir<>" -maxdepth 1 -type f | head -10"
},"StandardOutput"]]
</code></pre>
<p>note <code>find</code> does not return files in any canonical order, you can sort for example,
add <code>| sort</code> in front of <code>| head</code></p>
<p>Note <code>FileNames[][;;10]]</code> is a good bit faster by my testing, but I don't have a folder with a million files to test it on.</p>
|
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