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 |
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1,267,395 | <p>Julie is required to pay a 2 percent tax on all income over 3,000. She also has to pay 2.5 percent on all income over 20,000. She earned more than 20,000 and paid 992.50 what was her total income</p>
| maths525 | 235,883 | <p>I think I follow what you're asking. Your question states that for income over \$3,000, the tax rate is 2%, and that for income over \$20,000, the tax rate is 2.5%. Is that correct?</p>
<p>(I doubt that the question is asking you to calculate taxes on her first \$17,000 and her remaining income minus the \$17,000 separately, as most questions of this kind bear resemblance to U.S. income tax, which uses tax brackets, and in which you only pay in one bracket.)</p>
<p>If so, this is what the question is asking: knowing that Julie's income was more than \$20,000, how much did she make if she paid \$992.50 in taxes?</p>
<p>We know that Julie was taxed 2.5% of her income, but we don't know her income, so we'll call her income $x$.</p>
<p>Now, the way to solve this question is to solve the equation is which we multiply Julie's income, $x$, by her tax rate, 2.5%, and get a product which is her taxes paid, \$992.50.</p>
<p>$0.025x = 992.50$, and solve for $x$.</p>
<p>Does this clarify how to approach this kind of question?</p>
|
1,631,589 | <p>Consider the sequence $\{\frac{x^n}{n!}\}_n$ for any number $x$.</p>
<p>By choosing $m>x$ and letting $n>m$ , show that:</p>
<p>$\frac{x^n}{n!} < \frac{x^n}{m^n} < \frac{m^m}{(m-1)!}$</p>
<p>Am using the squeeze theorem , but unable to start third inequality.</p>
| Archis Welankar | 275,884 | <p>Rhs using simple combinatorics formula simplifies to $\frac{(n)(n-1)^2(n-2)}{8}$ now lhs ${n\choose 2}=\frac{(n)(n-1)}{2}$ so we can write lhs as $\frac{(n^2-n)!}{(2!).(\frac{n^2-n-4}{2})}$ now treat $n^2-n=y$ so it becomes $\frac{(n^2-n)(n^2-n-2)}{8}=\frac{(n)(n-1)^2(n-2)}{8}=Rhs$ </p>
|
1,457,063 | <p>I am utterly confused on how to solve this problem. I found a lemma that says $|A\cup B|=|A|+|B|$ is true if the two sets are disjoint which makes sense, but how do I prove the entire statement. </p>
| Bernard | 202,857 | <p>If the sets are not disjoint, in the right-hand side $\lvert A\rvert+\lvert B\rvert$, formula, the elements of $A\cap B$ are counted twice.</p>
|
1,115,645 | <p>I understand that a primitive polynomial is a polynomial that generates all elements of an extension field from a base field. However I am not sure how to apply this definition to answer my question. Can someone explain to me how I need to start please?</p>
| Bernard | 202,857 | <p>A primitive polynomial root is also the minimal polynomial of a <code>primitive root of unity</code> in $\mathbf F_7$. Let $\xi$ be a root of $f$. The field $\mathbf F_7(\xi)$ is the field $\mathbf F_{49}$ and its nonzero elements form a group of order $48$.
It suffices to show $xi$ has order $48$. Anyway its order can only be a divisor of $48$, i.e. $1,2,4,8,16, 3,6,12,24,48$.</p>
<p>Let's compute the powers of $\xi$ from its minimal polynomial. I give the details only for one of them: from $\xi^2=-\xi-3$, we deduce $$\xi^4 =(\xi+3)^2=\xi^2+6\xi+2=5\xi-1$$. Similarly
\begin{alignat*}{4}
\xi^8&=3,&\qquad&\xi^{16}=2, &\qquad&\xi^{24}=-1,&\qquad&\xi^{48}=1.
\end{alignat*}
The order cannot be $3, 6$ or $12$, since otherwise we would have $\xi^{24}=1$.</p>
<p>Thus $\xi$ is a primitive root of unity in $\mathbf F_{49}$, which proves $f$ is a primitive polynomial in $\mathbf F_7[x]$.</p>
|
1,765,538 | <p>If $N$ is the set of all natural numbers, $R$ is a relation on $N \times N$, defined by $(a,b) \simeq (c,d)$ iff $ad=bc$, how can I prove that $R$ is an equivalence relation ?</p>
| marwalix | 441 | <p>Let's do it step by step:</p>
<p>$$\forall (a,b)\in \Bbb{N}\times \Bbb{N}^{*},\, ab=ab$$</p>
<p>So $(a,b)\mathcal{R}(a,b)$ and $\mathcal{R}$ is <strong>reflexive</strong>.</p>
<p>Assume now that $(a,b)\mathcal{R}(c,d)$ i.e $ad=bc$. Multiplication is commutative so we can write $cb=da$ and this gives $(c,d)\mathcal{R}(a,b)$. The relation is <strong>symmetric</strong></p>
<p>Now take $(a,b)\mathcal{R}(c,d)$ and $(c,d)\mathcal{R}(e,f)$ ; this means $ad=bc$ and $cf=de$.
Multiply the first equality by $f\neq 0$ to get $afd=bcf$ and the second by $b\neq 0$ to get $bcf=bed$. So we have $afd=bed$ and keeping in mind $d\neq 0$ we have $af=be$ i.e $(a,b)\mathcal{R}(e,f)$ and the relation is <strong>transitive</strong></p>
<p>It is therefore an equivalence relation</p>
|
2,426,892 | <blockquote>
<p>Between which two integers does <span class="math-container">$\sqrt{2017}$</span> fall? </p>
</blockquote>
<p>Since <span class="math-container">$2017$</span> is a prime, there's not much I can do with it. However, <span class="math-container">$2016$</span> (the number before it) and <span class="math-container">$2018$</span> (the one after) are not, so I tried to factorise them. But that didn't work so well either, because they too are not perfect squares, so if I multiply them by a number to make them perfect squares, they're no longer close to <span class="math-container">$2017.$</span> How can I solve this problem?</p>
<p>Update:
Okay, since <span class="math-container">$40^2 = 1600$</span> and <span class="math-container">$50^2 = 2500$</span>, I just tried <span class="math-container">$45$</span> and <span class="math-container">$44$</span> and they happened to be the answer - but I want to be more mathematical than that... </p>
| Duncan Ramage | 405,912 | <p>$44^2 = 1936 < 2017 < 2025 = 45^2$.</p>
<p>Really, I don't think there's much to this one except for "try squaring small integers until you find the right ones".</p>
|
2,426,892 | <blockquote>
<p>Between which two integers does <span class="math-container">$\sqrt{2017}$</span> fall? </p>
</blockquote>
<p>Since <span class="math-container">$2017$</span> is a prime, there's not much I can do with it. However, <span class="math-container">$2016$</span> (the number before it) and <span class="math-container">$2018$</span> (the one after) are not, so I tried to factorise them. But that didn't work so well either, because they too are not perfect squares, so if I multiply them by a number to make them perfect squares, they're no longer close to <span class="math-container">$2017.$</span> How can I solve this problem?</p>
<p>Update:
Okay, since <span class="math-container">$40^2 = 1600$</span> and <span class="math-container">$50^2 = 2500$</span>, I just tried <span class="math-container">$45$</span> and <span class="math-container">$44$</span> and they happened to be the answer - but I want to be more mathematical than that... </p>
| kingW3 | 130,953 | <p>Knowing the answer lies between $40$ and $50$ the rational thing to do would be to try the number in the middle of $40$ and $50$ i.e $45$. </p>
<p>If you don't feel comfortable with multiplying $45\cdot 45$ you can use the
formula for $(40+5)^2=40^2+2\cdot 40\cdot 5+5^2=1600+400+25=2025$</p>
<p>Now you can use that $45^2-44^2=(45-44)(45+44)=89$ so one can see that $45^2>45^2-7=2017>45^2-89$.</p>
<p>It helps speed up a little of calculation for contests and such.</p>
|
2,426,892 | <blockquote>
<p>Between which two integers does <span class="math-container">$\sqrt{2017}$</span> fall? </p>
</blockquote>
<p>Since <span class="math-container">$2017$</span> is a prime, there's not much I can do with it. However, <span class="math-container">$2016$</span> (the number before it) and <span class="math-container">$2018$</span> (the one after) are not, so I tried to factorise them. But that didn't work so well either, because they too are not perfect squares, so if I multiply them by a number to make them perfect squares, they're no longer close to <span class="math-container">$2017.$</span> How can I solve this problem?</p>
<p>Update:
Okay, since <span class="math-container">$40^2 = 1600$</span> and <span class="math-container">$50^2 = 2500$</span>, I just tried <span class="math-container">$45$</span> and <span class="math-container">$44$</span> and they happened to be the answer - but I want to be more mathematical than that... </p>
| Mr. Brooks | 162,538 | <p>You can also use the <em>least</em> significant digits to get your bearings. Since $2016 \equiv 16 \pmod{100}$, if $2016$ is a perfect square, then $n$ in $n^2 = 2016$ is an integer satisfying $n \equiv 4, 6 \pmod{10}$. Clearly $n = 4$ or $6$ is too small.</p>
<p>Then, working our way up, we get $(196, 256), (576, 676), (1156, 1296), (1936, 2116)$, the last two corresponding to $44$ and $46$. Of course $45^2 \neq 2017$, but maybe it's $2025$. Notice then that $2116 - 2025 = 91$ and $91$ is the $45$th odd number. Likewise, $2025 - 1936 = 89 = 2 \times 44 + 1$, so it checks out.</p>
<p>So the answer is $44 < \sqrt{2017} < 45$.</p>
|
39,790 | <p>I'm teaching a programming class in Python, and I'd like to start with the mathematical definition of an array before discussing how arrays/lists work in Python.</p>
<p>Can someone give me a definition?</p>
| GEL | 10,825 | <p>I'd go with comparing an array to a matrix. That way when you introduce arrays of arrays, the mental leap will be easier for your students to make.</p>
|
930,949 | <p>Given that the circle C has center $(a,b)$ where $a$ and $b$ are positive constants and that C touches the $x$-axis and that the line $y=x$ is a tangent to C show that $a = (1 + \sqrt{2})b$</p>
| James | 81,163 | <p>When you are confused in this way you need to take a step back. In mathematics many words are used in different ways in different contexts. For instance, in one topology class I took, a "map" was defined to be a continuous function. Then later in a differential geometry class, "map" was the broad term and a "function" was defined as a map with co-domain $\mathbb{R}$.</p>
<p>This things aren't defined in this way to confuse you, English just only has so many synonyms and so words have to pull multiple duties sometimes.</p>
<p>You are correct that there is something called the predicate calculus, however it isn't really relevant here and so you can forget about it for the moment.</p>
<p>As for "for every x in the set of $D$, $P(x)$", this is an abbreviation for the statement "For every x, if $x\in D$ then $P(x)$". Thus the predicate that you are quantifying the free variables out of is "If $x\in D$ then $P(x)$".</p>
<p>This certainly has $x$ as a free variable, your question is why is $D$ not free too? Basically by stipultion, your professor is declaring that $D$ is a name for a set. Which set? Well that is important for saying whether the statement is true of false, but it is not important for deciding whether or not we have a predicate.</p>
<p>To contrast, you aren't asking why we are allowed to use $P(x)$ here without specifying exactly what it is, but it doesn't matter because we have declared that it is a predicate so everything works out. Similarly we declare that $D$ is a set and the collection of symbols "for every x in the set of $D$, $P(x)$" is a statement, independent of exactly what $D$ and $P(x)$ are.</p>
|
4,038,392 | <p>This question is inspired by the problem <a href="https://projecteuler.net/problem=748" rel="nofollow noreferrer">https://projecteuler.net/problem=748</a></p>
<p>Consider the Diophantine equation
<span class="math-container">$$\frac{1}{x^2}+\frac{1}{y^2}=\frac{k}{z^2}$$</span>
<span class="math-container">$k$</span> is a squarefree number. <span class="math-container">$A_k(n)$</span> is the number of solutions of the equation such that <span class="math-container">$1 \leq x+y+z \leq n$</span>, <span class="math-container">$x \leq y$</span> and <span class="math-container">$\gcd(x,y,z)=1$</span>. This equation has infinite solutions for <span class="math-container">$k=1$</span> and <span class="math-container">$k>1$</span> that can be expressed as sum of two perfect squares.</p>
<p>Let <span class="math-container">$$A_k=\lim_{n \to \infty}\frac{A_k(n)}{\sqrt{n}}$$</span></p>
<p><span class="math-container">\begin{array}{|c|c|c|c|c|}
\hline
k& A_k\left(10^{12}\right)& A_k\left(10^{14}\right)& A_k\left(10^{16}\right)& A_k\left(10^{18}\right)& A_k \\ \hline
1& 127803& 1277995& 12779996& 127799963& 0.12779996...\\ \hline
2& 103698& 1037011& 10369954& 103699534& 0.1036995...\\ \hline
5& 129104& 1291096& 12911049& 129110713& 0.129110...\\ \hline
10& 90010& 900113& 9000661& 90006202& 0.0900062...\\ \hline
13& 103886& 1038829& 10388560& 103885465& 0.103885...\\ \hline
17& 86751& 867550& 8675250& 86752373& 0.086752...\\ \hline
\end{array}</span></p>
<p>From these data, it seems that these limits converge. I wonder if it is possible to write them in terms of known constants or some sort of infinite series. For Pythagorean triples, there are about <span class="math-container">$\frac{n}{2\pi}$</span> triples with hypotenuse <span class="math-container">$\leq n$</span>.</p>
<p>The Python code to calculate <span class="math-container">$A_1(n)$</span>:</p>
<pre><code>def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
N = 10 ** 14
cnt = 0
for a in range(1, 22000):
a2 = a * a
for b in range(1, a):
if (a + b) % 2 == 1 and gcd(a, b) == 1:
b2 = b * b
x = 2 * a * b * (a2 + b2)
y = a2 * a2 - b2 * b2
z = 2 * a * b * (a2 - b2)
if x + y + z > N:
continue
cnt += 1
print(cnt)
</code></pre>
<p>The Python code to calculate $A_2(n)$:</p>
<pre><code>def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
N = 10 ** 14
cnt = 1 # (1, 1, 1)
for a in range(1, 22000):
a2 = a * a
a4 = a2 * a2
for b in range(1, a):
if gcd(a, b) == 1:
b2 = b * b
b4 = b2 * b2
x = 2 * a * b * (a2 + b2) - (a4 - b4)
y = 2 * a * b * (a2 + b2) + (a4 - b4)
z = 6 * a2 * b2 - (a4 + b4)
if x > 0 and y > 0 and z > 0 and x + y + z <= N and gcd(x, gcd(y, z)) == 1:
cnt += 1
x = (a4 - b4) - 2 * a * b * (a2 + b2)
y = 2 * a * b * (a2 + b2) + (a4 - b4)
z = (a4 + b4) - 6 * a2 * b2
if x > 0 and y > 0 and z > 0 and x + y + z <= N and gcd(x, gcd(y, z)) == 1:
cnt += 1
print(cnt)
</code></pre>
| individ | 128,505 | <p>To solve the Diophantine equation.</p>
<p><span class="math-container">$$ \frac{ 1 }{ x^2 } +\frac{ 1 }{ y^2 } = \frac{ q }{ z^2 } $$</span></p>
<p>It is necessary to use the solution of the following equation.</p>
<p><span class="math-container">$$a^2+b^2=qc^2$$</span></p>
<p>There are solutions when the coefficient can be represented as the sum of squares. <span class="math-container">$q=t^2+k^2$</span></p>
<p><span class="math-container">$$a=-tp^2+2kps+ts^2$$</span></p>
<p><span class="math-container">$$b=kp^2+2tps-ks^2$$</span></p>
<p><span class="math-container">$$c=p^2+s^2$$</span></p>
<p>And the solution itself can be presented in this form.</p>
<p><span class="math-container">$x=ac$</span></p>
<p><span class="math-container">$y=bc$</span></p>
<p><span class="math-container">$z=ab$</span></p>
|
2,805,975 | <p>For matrices $A, B$, I would like to show and understand the intuition behind the following identity
$$
(A+B)^{-1} = A^{-1} - (A+B)^{-1} B A^{-1}
$$
assuming the inverses exist.</p>
| Jakobian | 476,484 | <p>$$ A^{-1}-(A+B)^{-1}BA^{-1} = (I-(A+B)^{-1}B)A^{-1} = (A+B)^{-1}(A+B-B)A^{-1} = (A+B)^{-1}AA^{-1} = (A+B)^{-1} $$</p>
|
633,985 | <p>How can I find the values of $a$ for which the following function $f:\{0,1,\dots,m-1\} \rightarrow \mathbb{Z}_m$ is bijective for a fixed $m$?
$$f(n) = \sum_{k=0}^n a^k$$</p>
| Nate | 91,364 | <p>First choose 3 points arbitrarily this gives $\dbinom{25}{3}=2300$ possibly degenerate triangles.</p>
<p>Now we need to subtract off triples that all lie in a line:</p>
<p>There are 12 lines (5 vertical, 5 horizontal, and the main diagonals) containing 5 points these contribute $12\cdot\dbinom{5}{3}=120$ degenerate triangles.</p>
<p>There are 4 lines (off diagonals) containing 4 points. These contribute $4\cdot\dbinom{4}{3}=16$ degenerate triangles.</p>
<p>Finally, there are 12 lines with exactly 3 points (3 of each with slopes 2,-2, 1/2 , -1/2) which contribute 12 total degenerate triangles.</p>
<p>So 2300 total triangles minus 120+16+12 degenerate triangles gives 2152 nondegenerate triangles.</p>
|
3,653,148 | <p>Let <span class="math-container">$w$</span> be a primitive 5th root of unity. Then the difference equation <span class="math-container">$$x_nx_{n+2}=x_n-(w^2+w^3)x_{n+1}+x_{n+2}$$</span> generates a cycle of period 5 for general initial values:
<span class="math-container">$$u,v,\frac{u-(w^2+w^3)v}{u-1},\frac{uv-(w^2+w^3)(u+v)}{(u-1)(v-1)},\frac{v-(w^2+w^3)u}{v-1},u,v, ...$$</span> </p>
<p>For equations of the form <span class="math-container">$$x_nx_{n+2}=w^{a+b}x_n-(w^a+w^b)x_{n+1}+x_{n+2},\text{ for }w^a+w^b\ne 0$$</span> with the same globally periodic property, I can show that the only possible periods are 5,8,12,18 and 30.</p>
<p>Curiously, the 'same' equation works for all these periods:
<span class="math-container">$$x_nx_{n+2}=w^5x_n-(w^2+w^3)x_{n+1}+x_{n+2},$$</span>
where <span class="math-container">$w$</span> is a primitive <span class="math-container">$p$</span>th root of unity for <span class="math-container">$p=5,8,12,18,30$</span>. Is this a fluke or is there a way of seeing why this 'family' of equations always generate cycles? </p>
| Pavel Kozlov | 143,912 | <p>Another simple case periodic difference equation is the next one, with zero coefficient at <span class="math-container">$x_{n+1}$</span>:
<span class="math-container">$$x_nx_{n+2}=ax_n+x_{n+2}.$$</span></p>
<p>Simplicity of this case expressed in existence of nice closed form description for sequence members. Of course given sequence could have only cycles with even period <span class="math-container">$2T$</span>.</p>
<p>If <span class="math-container">$T=2k+1$</span> then
<span class="math-container">$$\frac{1}{x_{-2k-2}}-\frac{1}{x_{2k}}=\frac{1}{u}\left (a^{k+1}-\frac{1}{a^k}\right )+\sum_{i=0}^{k} (-a)^i+\frac{\sum_{i=0}^{k-1} (-a)^i)}{a^k}$$</span></p>
<p>Coefficient at <span class="math-container">$1/u$</span> don't vanish if <span class="math-container">$a=-1$</span>, so existence of cycle with period <span class="math-container">$T$</span> is equivalent to the system of equations:
<span class="math-container">$$a^{2k+1}=1, \tag{1}$$</span>
<span class="math-container">$$1+(-a)^{k+1}+\frac{1+(-a)^k}{a^k}=0. \tag{2}$$</span></p>
<p>Multiplying second equation by <span class="math-container">$a^k$</span> and recalling equation <span class="math-container">$(1)$</span> we get
<span class="math-container">$$a^k+(-1)^{k+1}+1+(-1)^ka^k=0,$$</span>
or
<span class="math-container">$$a^k(1+(-1)^k)=(-1)^k-1,$$</span>
which is impossible while <span class="math-container">$a$</span> is some root of unity.</p>
<p>If <span class="math-container">$T=2k$</span> then
<span class="math-container">$$x_{-2k}=\frac{-u}{-a^k-u(\sum_{i=0}^{k-1} (-a)^i)}=\frac{a^ku}{1-u(\sum_{i=0}^{k-1} (-a)^i)}=x_{2k},$$</span></p>
<p>so <span class="math-container">$a$</span> should be some primitive <span class="math-container">$T$</span>-th root of unity - the only case when periodicity can be reached.</p>
|
1,949,966 | <h2>Q 1a</h2>
<p>Is it possible to define a number $x$ such that $|x|=-1$, where $|\cdot|$ means absolute value, in the same manner that we define $i^2=-1$?</p>
<p>I have no idea if it makes sense, but then again, $\sqrt{-1}$ used to not be a thing either.</p>
<p>To be more explicit, I want as many properties to hold as possible, e.g. $|a|\times|b|=|a\times b|$ and $|a|=|-a|$, as some properties that seem to hold for all different types of numbers (or in some analogous way).</p>
<hr>
<h2>Q 1b</h2>
<p>If we let the solution to $|x|=-1$ be $x=z_1$, and we allow the multiplicativeness property,</p>
<p>$$|(z_1)^2|=1$$</p>
<p>Or, further,</p>
<p>$$|(z_1)^{2n}|=1\tag{$n\in\mathbb N$}$$</p>
<p>Note that this does not mean $z_1$ is any such real, complex, or any other type of number. We used to think $|x|=1$ had two solutions, $x=1,-1$, but now we can give it the solution $x=e^{i\theta}$ for $\theta\in[0,2\pi)$. Adding in the solution $(z_1)^{2n}$ is no problem as far as I can see.</p>
<p>However, there result in some problems I simply cannot quite see so clearly, for example,</p>
<p>$$|z_1+3|=?$$</p>
<p>There exists no such way to define such values at the moment.</p>
<p>Similarly, let $z_2$ be the number that satisfies the following:</p>
<p>$$|z_2|=z_1$$</p>
<p>As far as I see it, it is not possible to create $z_2$, given $z_1$ and $z_0\in\mathbb C$.</p>
<p>The following has a solution, in case you were wondering.</p>
<p>$$|\sqrt{z_1}|=i$$</p>
<p>so no, I did not forget to consider such cases.</p>
<p>But, more generally, I wish to define the following numbers in a recursive sort of way.</p>
<p>$$|z_{n+1}|=z_n$$</p>
<p>since, as far as I can tell, $z_{n+1}$ is not representable using $z_k$ for $k\le n$. In this way, the nature of $z_n$ goes on forever, unlike $i$, which has the solution $\sqrt i=\frac1{\sqrt2}(1+i)$.</p>
<p>So, my second question is to ask if anyone can discern some properties about $z_n$, defining them as we did above? And what is $|z_1+3|=?$</p>
<hr>
<h2>Q 2a</h2>
<p>This part is important, so I truly want you guys (and girls) to consider this:</p>
<blockquote>
<p>Can you construct a problem such that $|x|=-1$ will be required in a step as you solve the problem, but such that the final solution is a real/complex/anything already well known. This is similar to <em><a href="https://en.wikipedia.org/wiki/Casus_irreducibilis" rel="nofollow noreferrer">Casus irreducibilis</a></em>, which basically forced $i$ to exist by establishing its need to exist.</p>
</blockquote>
<p>I am willing to give a large rep bounty for anyone able to create such a scenario/problem. </p>
<hr>
<h2>Q 2b</h2>
<p>And if it is truly impossible, why? Why is it not possible to define some 'thing' the solution to the problem, keep a basic set of properties of the absolute value, and carry on? What's so different between $|x|=-1$ and $x^2=-1$, for example?</p>
<hr>
<h2>Thoughts to consider:</h2>
<p>Now, <a href="https://math.stackexchange.com/a/1345391/272831">Lucian</a> has pointed out that there are plenty of things we do not yet understand, like $z_i\in\mathbb R^a$ for $a\in\mathbb Q_+^\star\setminus\mathbb N$. There may very well exist such a number, but in a field we fail to understand so far.</p>
<p>Similarly, the triangle inequality clearly cannot coexist with such numbers as it is. For the triangle inequality to exist, someone has to figure out how to make triangles with non-positive/real lengths.</p>
<p>As for the <a href="https://en.wikipedia.org/wiki/Norm_(mathematics)#Definition" rel="nofollow noreferrer">properties/axioms of the norm</a> I want:</p>
<p>$$p(v)=0\implies v=0$$</p>
<p>$$p(av)=|a|p(v)$$</p>
| Patrick Sheehan | 599,007 | <p>What about extending the real number absolute value to complex numbers in a different way?</p>
<p>For <span class="math-container">$z = re^{iθ}$</span>, let <span class="math-container">$|z| = re^{2iθ}$</span></p>
<p>You still get that the "new" absolute value for all reals matches the regular one.</p>
<p>With this, we have
<span class="math-container">$|i| = 1\cdot e^{2i(π/2)} = e^{iπ} = -1$</span>.</p>
|
1,706,939 | <p>Can anyone share an easy way to approximate $\log_2(x)$, given $x$ is between $0$ and 1?</p>
<p>I'm trying to solve this using an old fashioned calculator (i.e. no logs)</p>
<p>Thanks!</p>
<p>EDIT: I realize that I stepped a bit ahead. The x comes in the form of a fraction, e.g. 3/8, which is indeed between 0 and 1, but could also be written as log2(3) - log2(8). I am hoping there is a quick way to approximate this calculation to let's say 2 decimals</p>
| seoanes | 322,225 | <p>Using that $\log_2(x)=\mbox{ln}(x)/\mbox{ln}(2)$, now you can use the Taylor expansion:
\begin{equation}
\mbox{ln}(x)=\mbox{ln}\left((x-1)+1\right)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}(x-1)^k
\end{equation}
The same expansion can be used for determining the $\mbox{ln}(2)$. The first terms of the expansion are
\begin{equation}
\log_2(x)=\frac{1}{\mbox{ln}(2)}\left[(x-1)-(x-1)^2/2+...\right]
\end{equation}
This expansion is working better around $x=1$. These kind of expansion are commonly used in calculators or programming languages to compute analytic functions.</p>
<p>EDIT: Another possibility to compute the natural logarithm is to use the generalized continued fraction
\begin{equation}
\log((x-1)+1)=\cfrac{2(x-1)}{(x-1)+2-\cfrac{(x-1)^2}{3((x-1)+2)-\cfrac{4(x-1)^2}{5((x-1)+2)-\cfrac{9(x-1)^2}{7((x-1)+2)-\cdots}}}}
\end{equation}
I think this option will be converging faster than the Taylor expansion</p>
<p>As a final approximation, I would recommend you also to have a look at the Padé expansion of the logarithm. Having a fast look to the old questions in this website I found <a href="https://math.stackexchange.com/questions/698384/approximating-logs-and-antilogs-by-hand/919287#919287">Approximating Logs and Antilogs by hand</a>,
where user153012 gives next Padé approximation to compute the logarithm $\phi_3(x)\leq\mbox{ln}(x)\leq\psi_3(x)$ where the lower bound is
$$\phi_3(x)=\frac{x(60+60x+11x^2)}{3(20+30x+12x^2+x^3)},$$
and the upper
$$\psi_3(x)=\frac{x(30+21x+x^2)}{3(10+12x+3x^2)}.$$</p>
|
288,340 | <p>I am having difficulty understanding the recursive definition of a language. The problem asked how to write this non recursively. But I want to understand just how a recursive definition of a language works.</p>
<p>Recursive definition of a subset of L of $\{a,b\}^*$.</p>
<p>Basis : $a\in L$ </p>
<p>Recursive Definition : for any $x\in L$, $ax$ and $xb$ are in $L$.</p>
<p>Below is my attempt at explaining the recursive definition.</p>
<p>Starting with the basis $a$ saying that $ax$ is in it means that all strings formed such as $\{a,aa,aaa,...\}$ are present.</p>
<p>Defining all $xb$ represents $\{ab,abb,abbbb,...\}$</p>
<p>The answer I have so far is $\{a\}^*\{b\}^*$ but again it is understanding it that I am after.</p>
| Brian M. Scott | 12,042 | <p>The recursive definition says that once you have a word of $L$, you can prefix an $a$ or suffix a $b$ to get another word of $L$. You can repeat either of these operations any number of times, so you can prefix $a^n$ for any $n\ge 0$ or suffix $b^n$ for any $n\ge 0$, or both (not necessarily with the same $n$). If you start with the one given word $a$, and apply the ‘prefix $a$’ operation $m$ times and the ‘suffix $b$’ operation $n$ times, you get $a^{m+1}b^n$. (The order in which you perform the $m+n$ operations pretty clearly doesn’t matter.) Since the definition doesn’t allow you to do anything else, these are the only words that you can form:</p>
<p>$$L=\{a^{m+1}b^n:m,n\ge 0\}\;.$$</p>
<p>In other words, you can have any <strong>positive</strong> number of $a$’s followed by any <strong>non-negative</strong> number of $b$’s. A regular expression describing this language is $aa^*b^*$; the first $a$ ensures that you get at least one $a$. (An extended regular expression equivalent to this is $a^+b^*$, if you’ve been introduced to the notation $a^+$; it’s just an abbreviation for $aa^*$.)</p>
|
2,041,251 | <p>Suppose $f$ is a twice-differentiable function with $f(0) = 0$, $f\left(\frac12\right) = \frac12$ and $f'(0) = 0$. Prove that $|f''(x)| \ge 4$ for some $x \in \left[0,\frac12\right]$.</p>
<p>I've been stuck on this question for a while now without any idea on how to get started. Is it possible for someone to help me?</p>
<p>Thanks!</p>
| Community | -1 | <p>[Paramanand's answer is definitely the way to go. I just want to share another elementary approach in the $f''$-is-continuous case.]</p>
<p>If $f$ is twice <em>continuously</em> differentiable then by the (second) <em>Fundamental theorem of Calculus</em> we can write
$$
f(x)=0+\int_0^xf'(y)\, dy=\int_0^x\left(0+\int_0^yf''(z)\, dz\right)dy.
$$
Plugging in $x=1/2$ and assuming that $|f''(z)|<4$ for $z\in[0,1/2]$ we arrive at the contradiction
$$
\frac{1}{2}=\int_0^{1/2}\int_0^yf''(z)\, dz\, dy<\int_0^{1/2}\int_0^y4\, dz\, dy=\int_0^{1/2}4y\, dy=\frac{1}{2}.
$$</p>
|
1,180,854 | <p>I was solving some probability and combination problems and I came across this one,
that I couldn't solve. Any tips appreciated.</p>
<blockquote>
<p>Chance to win with one lottery ticket is <span class="math-container">$\frac13$</span>. How many tickets must be purchased for the probability of winning at least with one ticket to be greater than <span class="math-container">$0.9$</span>?</p>
<p>Given:<br />
Chance to win: <span class="math-container">$\frac13$</span><br />
Number of tickets that have to win: <span class="math-container">$1$</span><br />
Probability of that happening: <span class="math-container">$> 0.9$</span><br />
Need to find number of tickets to buy</p>
</blockquote>
| Stefan4024 | 67,746 | <p>Assume that $z=a+bi$ then we have for the equation.</p>
<p>$$a + bi + 1 -ai +b=0$$
$$(a+b+1) + (b-a)i=0$$</p>
<p>Now set $a+b+1=0$ and $a-b=0$ to get $a=b=-\frac 12$, hence $z = -\frac 12 - \frac i2$</p>
|
2,961,796 | <p>I was trying to solve the inequality <span class="math-container">$$a-\sqrt[3]{a^3-c\cdot a^2}<b-\sqrt[3]{b^3-c\cdot b^2}$$</span> where <span class="math-container">$a>b>0$</span> and <span class="math-container">$c>0$</span>. I managed to pack the part inside the cube root: <span class="math-container">$$a-\sqrt[3]{a^2(a-c)}<b-\sqrt[3]{b^2(b-c)}$$</span> but I'm stuck after that. Can anybody help prove/disprove the inequality?</p>
| Michael Rozenberg | 190,319 | <p>We need to solve
<span class="math-container">$$\sqrt[3]{a^2(a-c)}-\sqrt[3]{b^2(b-c)}-(a-b)>0$$</span> or since <span class="math-container">$$a^2(a-c)=-b^2(b-c)=-(a-b)^3$$</span> gives
<span class="math-container">$$\frac{a^2b^2}{a^2+b^2}+(a-b)^2=0,$$</span> which is impossible, we need to solve
<span class="math-container">$$a^2(a-c)-b^2(b-c)-(a-b)^3-3(a-b)\sqrt[3]{a^2b^2(a-c)(b-c)}>0$$</span> or
<span class="math-container">$$3ab-ac-bc>3\sqrt[3]{a^2b^2(a-c)(b-c)}$$</span> or <span class="math-container">$$c<\frac{9ab(a^2-ab+b^2)}{(a+b)^3}.$$</span>
By the way, if you need to prove this inequality then it's wrong for <span class="math-container">$c\geq\frac{9ab(a^2-ab+b^2)}{(a+b)^3}.$</span></p>
|
3,789,494 | <p>I'm stuck in solving this strange and beautiful formula : <span class="math-container">$3= 3^{z}$</span> since it says 'Solve' and not 'Prove'</p>
<p>Also i really don't understand what does it means by saying <span class="math-container">$3^{z}$</span>? Will <span class="math-container">$3^z$</span> form a set ?</p>
<p>I will be very grateful if you can help me or even give me a little hint to make a headway with this problem ...</p>
<p>To be clear, find all <span class="math-container">$z \in \left\{ 3 = 3^z \right\}.$</span></p>
| mjw | 655,367 | <p>Well, if the question is asking to find all <span class="math-container">$z$</span> such that</p>
<p><span class="math-container">$$3^z = 3$$</span></p>
<p>Then</p>
<p><span class="math-container">$$\begin{aligned}
3^z &= 3e^{2 k \pi i } \\
z \log 3 &=\log 3 + 2 k \pi i \\
\end{aligned}$$</span>
<span class="math-container">$$\bbox[5px, border: 1pt solid blue]{z = 1 + \frac{2 k \pi i}{\log 3}, \quad k \in \mathbb{Z}.}
$$</span></p>
<p>There are infinitely many solutions, and they all lie on the line <span class="math-container">$\text{Re }z=1.$</span></p>
|
3,245,796 | <p>Let <span class="math-container">$f$</span> have a continuous second derivative. Prove that</p>
<p><span class="math-container">$$f(x) = f(a) + (x - a)f'(a) + \int_a^x(x - t)f''(t) dt.$$</span></p>
<p>This is a modification of exercise 6.6.4 from Advanced Calculus by Fitzpatrick. I have seen that this question has been asked here: <a href="https://math.stackexchange.com/questions/785897/proving-fx-f0-f0x-int-0x-x-t-ft-dt-for-all-x">Proving $f(x) = f(0) + f'(0)x + \int_0^x (x-t) f''(t) dt$ for all x</a>. However, there didn't seem to be a suitable answer.</p>
<p>Here is my attempt at the problem.</p>
<p>Since <span class="math-container">$f$</span> has a continuous second derivative, then the first derivative is also continuous. Therefore, by the first fundamental theorem of calculus, we have that</p>
<p><span class="math-container">$$f(x) = f(a) + \int_a^x f'(t)dt.$$</span></p>
<p>Expanding out the right-hand side of the above using integration by parts, we see that</p>
<p><span class="math-container">$$f(x) = f(a) + f'(t)t - \int_a^x tf''(t) dt.$$</span></p>
<p>This is where I am confused.</p>
| uniquesolution | 265,735 | <p>Integration by parts is not necessary.</p>
<p>Put
<span class="math-container">$$g(x)=f(x)-f(a)-(x-a)f'(a)$$</span>
and
<span class="math-container">$$h(x)=\int_a^x(x-t)f''(t)dt$$</span>
and
<span class="math-container">$$k(x)=g(x)-h(x)$$</span>
Then <span class="math-container">$k'(x)=0$</span> for all <span class="math-container">$x$</span>, hence <span class="math-container">$k(x)$</span> is a constant. In particular, <span class="math-container">$k(x)=k(a)=0$</span>.
Hence <span class="math-container">$g(x)=h(x)$</span>, whence the desired identity follows.</p>
|
1,685 | <p>Are there books or article that develop (or sketch the main points) of Euclidean geometry without fudging the hard parts such as angle measure, but might at times use coordinates, calculus or other means so as to maintain rigor or avoid the detail involved in Hilbert-type axiomatizations?</p>
<p>I am aware of Hilbert's foundations and the book by Moise. I was wondering if there is anything more modern that tries to stay (mostly) in the tradition of synthetic geometry. </p>
| Julien Narboux | 239,646 | <p>There are some axioms systems such as Birkoff axioms which assume the existence of a field from the beginning.</p>
<p>For the synthetic approach the main axiom systems are those of Hilbert and Tarski.</p>
<p>You can also use Tarski's axiom as described in W. Schwabhäuser,
W Szmielew, A. Tarski, Metamathematische Methoden in der Geometrie.</p>
|
322,140 | <p>$$\int \left ( r\sqrt{R^2-r^2} \right )dr$$</p>
<p>It looks simple. I know that the derivative of </p>
<p>$$\left (R^2-r^2 \right )^\frac{3}{2}$$</p>
<p>Is the stuff in the integral.</p>
<p>However, what about if I don't know?</p>
<p>How in general do we solve integral of</p>
<p>$$G(r)^n$$</p>
| marty cohen | 13,079 | <p>Note that $(r^2)' = 2r$, so that
$(R^2-r^2)' = -2r$
(with respect to $r$).</p>
<p>Therefore, if $u = R^2 - r^2$,
$du = -2r\ dr$,
so, for any function $f$,
$\int r f(R^2-r^2) dr
= -\frac1{2}\int (-2r) f(R^2-r^2) dr
= -\frac1{2}\int f(R^2-r^2) (-2r\ dr)
= -\frac1{2} \int f(u) du
$.</p>
<p>Putting $f(u) = \sqrt{u}$ gets the desired result.</p>
|
1,595,118 | <p>What is value of $a+b+c+d+e$? If given :</p>
<p>$$abcde=45$$</p>
<p>And $a,b, c, d, e$ all are distinct integer.</p>
<hr>
<p>My attempt :</p>
<p>I calculated, $45 = 3^2 \times 5$.</p>
<blockquote>
<p>Can you explain, how do I find the distinct values of $a,b, c, d, e$ ?</p>
</blockquote>
| Jimmy R. | 128,037 | <p>So $$45=5\cdot(\pm 3)^2\cdot(\pm1)^2$$ which implies that the only possibility (up to renaming) is $a=1, b=-1, c=3, d=-3, e=5$. </p>
|
2,749,539 | <p>I have tried for some time to solve this problem and I'm stuck, so any help would be greatly appreciated. I'm not a math guy, so I apologize if I am missing something basic.</p>
<p>I have a function $f(x)$ $$f(x) = \sum_{i=1}^x dr^{i-1}$$ where<br>
$x$ is a positive integer<br>
$d$ is an initial delta and<br>
$r$ is a factor which defines the rate at which each subsequent $f(x)$ value increase.</p>
<p>Specifically:
\begin{align}
f(1) =&\; d r^0 = d\\
f(2) =&\; f(1) + d r^1\\
f(3) =&\; f(2) + d r^2\\
f(4) =&\; f(3) + d r^3
\end{align}
And more concretely, if $d=5$, $r=1.1$, and $n=5$ then
\begin{align}
f(1) = &\;5 \\
f(2) = &\;10.5 &= &\; 5 &&+ 5(1.1)^1\quad \text{(increase by }5.5)\\
f(3) = &\;16.55 &= &\; 10.5 &&+ 5(1.1)^2\quad \text{(increase by }6.05)\\
f(4) = &\;23.205 &= &\; 16.55 &&+ 5(1.1)^3\quad \text{(increase by }6.655)\\
f(5) = &\;30.5255 &= &\; 23.205 &&+ 5(1.1)^4\quad \text{(increase by }7.3205)
\end{align}</p>
<p>Here is the problem I have not been able to solve:</p>
<p>If I am given:</p>
<p>$d$ (the initial increase)<br>
$n$ (the number of periods)<br>
and $y_n$ (the final value at $f(n)=y_n$)</p>
<p><strong>1) How can I solve for $r$?</strong></p>
<p>Right now I'm using a brute force interpolation in Python, but it seems like there must be a better way. I guess it is worth mentioning that I'm ultimately solving for $r$ programmatically in Python as opposed to using a pencil and paper, but I can write the code if you can give me the right math pointers.</p>
<p>So, as another concrete example, if I'm given $d=2$, $n=100$, and $f(100)=2000$, then how do I find that $r=1.0370626397$</p>
<p><strong>2) Is there a way to calculate $f(x)$ without knowing $f(x-1)$?</strong></p>
<p>Right now, I can only get to $f(x)$ by calculating all prior values.</p>
<p>Any help or even just advice telling me it there is no shortcut solution would be greatly appreciated.</p>
| mzp | 287,326 | <p>Notice that using the geometric progression formula we have that</p>
<p>$$f(n) = \sum_{i=1}^n dr^{i-1} = \begin{cases}\frac{d(1-r^n)}{1-r},& \quad\text{if}\;r\neq 1 \\ dn, &\quad\text{if}\;r= 1\end{cases}$$</p>
<p><em>1) How can I solve for $r$?</em></p>
<p>You would need to solve the polynomial</p>
<p>$$f(n)(1-r) = d(1-r^n).$$</p>
<p><em>2) Is there a way to calculate $f(x)$ without knowing $f(x−1)$?</em></p>
<p>Yes, simply apply the formula above.</p>
|
2,645,611 | <blockquote>
<p>Prove that:
<span class="math-container">$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$</span></p>
</blockquote>
<h3>My work so far:</h3>
<p><span class="math-container">$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$</span>
<span class="math-container">$$\frac{(n+0)!}{n!0!}+\frac{(n+1)!}{n!1!}+...+\frac{(n+n)!}{n!n!}=\frac{(2n+1)!}{n!(n+1)!}$$</span>
<span class="math-container">$$\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+...+\binom{2n}{n}=\binom{2n+1}{n}$$</span></p>
<p><em>How to prove the last equality?</em></p>
| Seewoo Lee | 350,772 | <p>Try to use the following identity:
$$
\binom{n}{k} = \binom{n-1}{k}+\binom{n-1}{k-1}.
$$</p>
|
2,645,611 | <blockquote>
<p>Prove that:
<span class="math-container">$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$</span></p>
</blockquote>
<h3>My work so far:</h3>
<p><span class="math-container">$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$</span>
<span class="math-container">$$\frac{(n+0)!}{n!0!}+\frac{(n+1)!}{n!1!}+...+\frac{(n+n)!}{n!n!}=\frac{(2n+1)!}{n!(n+1)!}$$</span>
<span class="math-container">$$\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+...+\binom{2n}{n}=\binom{2n+1}{n}$$</span></p>
<p><em>How to prove the last equality?</em></p>
| Gerry Myerson | 8,269 | <p>First, note ${n\choose0}={n+1\choose0}$, then repeatedly use ${k\choose r}+{k\choose r+1}={k+1\choose r+1}$</p>
|
2,645,611 | <blockquote>
<p>Prove that:
<span class="math-container">$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$</span></p>
</blockquote>
<h3>My work so far:</h3>
<p><span class="math-container">$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$</span>
<span class="math-container">$$\frac{(n+0)!}{n!0!}+\frac{(n+1)!}{n!1!}+...+\frac{(n+n)!}{n!n!}=\frac{(2n+1)!}{n!(n+1)!}$$</span>
<span class="math-container">$$\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+...+\binom{2n}{n}=\binom{2n+1}{n}$$</span></p>
<p><em>How to prove the last equality?</em></p>
| lab bhattacharjee | 33,337 | <p>$$\sum_{r=0}^n\binom{n+r}r$$ is the coefficient of $x^n$ in $$\sum_{r=0}^n(1+x)^{n+r}$$ which is a Finite Geometric Series and is $$=(1+x)^n\cdot\dfrac{(1+x)^{n+1}-1}{1+x-1}=\dfrac{(1+x)^{2n+1}-(1+x)^n}x$$</p>
<p>Now the coefficient of $x^{n+1}$ in $$(1+x)^{2n+1}-(1+x)^n$$</p>
<p>$$=\binom{2n+1}{n+1}-0$$</p>
|
1,808,441 | <p>I want to know for two circles why the ratio of arc length is equal to the ratio of the two central angles in geometry. It must have something to do with the concept of similarity in geometry. I have scoured the Internet looking answers but only found the ratio of circumferences to the ratio of their diameter.</p>
<p>If my question seems crude it is because I have never posted a question before.</p>
| Christian Blatter | 1,303 | <p>It is a basic feature of euclidean geometry, not present in spherical or hyperbolical geometry, that we have available a large family of "special" maps $T_{O,\rho}\,$, called <em>homotheties</em>. These maps are stretching the base space ${\mathbb E}^2$ linearly from a given center $O$ with a given factor $\rho>0$ in the obvious way. They have the property that all distances $|PQ|$ between points $P$, $Q\in{\mathbb E}^2$ are multiplied by $\rho$, and they map circles onto circles. Since lengths of curves are defined via lengths of inscribed polygons and a limiting process it follows that lengths of curves are multiplied by $\rho$ as well. This then implies that the lengths of corresponding arcs on concentric circles behave in the expected way. </p>
<p>Now the measure $\alpha$ of an angle is defined by the length of the corresponding arc on the unit circle. In this way the length of the corresponding arc on a concentric circle of radius $r$ becomes $r\,\alpha$. Furthermore it is easy to see that similarities preserve angles.</p>
|
1,528,235 | <p>Recall that <a href="http://en.wikipedia.org/wiki/Tetration" rel="noreferrer">tetration</a> ${^n}x$ for $n\in\mathbb N$ is defined recursively: ${^1}x=x,\,{^{n+1}}x=x^{({^n}x)}$. </p>
<p>Its inverse function with respect to $x$ is called <a href="http://en.wikipedia.org/wiki/Tetration#Super-root" rel="noreferrer">super-root</a> and denoted $\sqrt[n]y_s$ (the index $_s$ is not a variable, but is part of the notation — it stands for "super"). For $y>1, \sqrt[n]y_s=x$, where $x$ is the unique solution of ${^n}x=y$ satisfying $x>1$. It is known that $\lim\limits_{n\to\infty}\sqrt[n]2_s=\sqrt{2}$. We are interested in the convergence speed. It appears that the following limit exists and is positive:
$$\mathcal L=\lim\limits_{n\to\infty}\frac{\sqrt[n]2_s-\sqrt2}{(\ln2)^n}\tag1$$
Numerically,
$$\mathcal L\approx0.06857565981132910397655331141550655423...\tag2$$</p>
<hr>
<p>Can we prove that the limit $(1)$ exists and is positive? Can we prove that the digits given in $(2)$ are correct? Can we find a closed form for $\mathcal L$ or at least a series or integral representation for it?</p>
| mick | 39,261 | <p>$$\mathcal L=\lim\limits_{n\to\infty}\frac{\sqrt[n]2_s-\sqrt2}{(\ln2)^n}\tag1$$</p>
<p>Notice the resemblance with the Koenigs function</p>
<p><a href="https://en.m.wikipedia.org/wiki/Koenigs_function" rel="nofollow">https://en.m.wikipedia.org/wiki/Koenigs_function</a></p>
<p>In fact it is a Koenigs function with the variable fixed to the value $2$.</p>
<p>Since $1 < 2 < \exp(1/e)$ and the derivative is not $0$ or $1$ , the Koenigs function converges to the correct value ; your limit.</p>
|
4,180,750 | <p>Let's consider inner product space with vectors <span class="math-container">$x, y, z$</span> which satisfies:</p>
<p><span class="math-container">$$\|x+y+z\|^2 = 14$$</span></p>
<p><span class="math-container">$$\|x+y-z\|^2 = 2$$</span></p>
<p><span class="math-container">$$\|x-y+z\|^2 = 6$$</span></p>
<p><span class="math-container">$$\|x-y-z\|^2 = 10$$</span></p>
<p>I want to prove that <span class="math-container">$x$</span> is perpendicular to <span class="math-container">$y$</span>.</p>
<p><strong>My work so far</strong></p>
<p>In other words we want to prove that <span class="math-container">$\langle x, y \rangle = 0$</span>.</p>
<p>My first idea was to use Cauchy Schwarz inequality <span class="math-container">$ | \langle x , y \rangle |\le \|x\| \|y\|$</span>
and to show that our conditions force that <span class="math-container">$\|x\|\cdot\|y\| = 0$</span>. However I didn't manage to do anything sensible. Also I tried to somehow prove that under this conditions our norm has bo inducted by inner product - in other words we have that: <span class="math-container">$$2(\|x\|^2 + \|y\|^2) = \|x+y\|^2 + \|x-y\|^2$$</span></p>
<p>but also I didn't end up with something rational. Could you please give me a hand, what's the correct approach to this problem?</p>
| alepopoulo110 | 351,240 | <p>I assume you work with a real and not a complex vector space. Is that true?</p>
<p>Add the first and second one: by the <a href="https://en.wikipedia.org/wiki/Parallelogram_law#The_parallelogram_law_in_inner_product_spaces" rel="nofollow noreferrer">parallelogram identity</a> you get
<span class="math-container">$$2(\|x+y\|^2+\|z\|^2)=16$$</span>
so <span class="math-container">$\|x+y\|^2+\|z\|^2=8$</span>. Now add the third and the fourth one and in the same way you get <span class="math-container">$\|x-y\|^2+\|z\|^2=8$</span>. So <span class="math-container">$\|x+y\|^2+\|z\|^2=\|x-y\|^2+\|z\|^2$</span>, so
<span class="math-container">$$\|x-y\|^2=\|x+y\|^2$$</span>
so <span class="math-container">$\|x\|^2+\|y\|^2-2\langle x,y\rangle=\|x\|^2+\|y\|^2+2\langle x,y\rangle$</span>, so <span class="math-container">$\langle x,y\rangle=0$</span>.</p>
|
1,838,002 | <p>There is famous <a href="https://en.wikipedia.org/wiki/Quillen-Suslin_theorem" rel="nofollow">Quillen-Suslin theorem</a> which states that every finitely generated projective module over a ring of polynomials $k[x_1,...,x_n]$, where $k$ is a field, is free.</p>
<p>I have never carefully read a proof of this theorem, which is for example in the <strong>Lang's Algebra</strong>. <code>Probably it is based on Quillen's original ideas.</code>- this is not true as it was pointed out in the answer below.</p>
<p><strong>Questions:</strong> Is every finitely generated projective modules over $\mathbb{Z}[x_1,...,x_n]$ free? </p>
<p>If yes, then is the proof modification of the one given in <strong>Lang's Algebra</strong>?</p>
<p>And if yes, then how about polynomial rings over other Dedekind domains or number rings?</p>
| Mohan | 245,104 | <p>Ok Slup, here goes.<br/></p>
<p>Let $R$ be any commutative ring and let $A$ be a polynomial ring over $R$. Let $P$ be any projective module over $R$. Then Quillen (and Suslin a bit later in this generality) proved that if for every maximal ideal $\mathfrak{m}$ of $R$, $P_{\mathfrak{m}}$ is of the form $Q\otimes_{R_{\mathfrak{m}}} A_{\mathfrak{m}}$ for some projective $R_{\mathfrak{m}}$ module $Q$, then there exists a projective module $Q$ over $R$ such that $P=Q\otimes_R A$. Using this, they deduced that this always happens when $R$ is a Dedekind domain. The last part is done by the following observation. If for such a $P$, $P_f$ is free for a polynomial which is monic in one of the variables, then $P$ is free. As you can see, since any non-zero polynomial after a change of variables can be made into a monic polynomial in one of the variables if $R$ is a field, one immediately deduces Serre conjecture from this.<br/></p>
<p>Many years later, Lindel generalized this for $R$ any regular ring containing a field. </p>
|
2,097,584 | <blockquote>
<p>How many ways can the number $2160$ be written as a product of factors which are relatively prime to each other?</p>
</blockquote>
<p>I was confused by this question because couldn't we just add $1$ into the factorization every time? For example, $2160$ and $2160 \cdot 1$ would count as distinct factorizations. If that is not the case, then I tried this:</p>
<p>Note that if a prime $p$ divides a factor of $2160$, then it must be the largest power of the prime $p$ dividing $2160$. Then we see that $2160 = 2^4 \cdot 3^3 \cdot 5$ and that $2160$ can't be written as the product of four or more relatively prime factors which aren't $1$. For three factors, we just have $2^4 \cdot 3^3 \cdot 5$. For two factors, we have $(2^4 \cdot 3^3) \cdot 5, (3^3 \cdot 5),$ or $(2^4 \cdot 5) \cdot 3^3$. Finally for one factor we just have $2160$. Therefore there are $5$ different ways in which $2160$ can be written as a product of factors which are relatively prime to each other.</p>
| hmakholm left over Monica | 14,366 | <p>You're right that the problem only makes sense if factors of $1$ are not allowed.</p>
<p>Under this interpretation, your analysis seems to make sense: The number of ways to write $n$ as a product of coprime factors $>1$ (not counting the order of these factors) is exactly the number of partitions of the set of prime factors of $n$.</p>
<p>The number of partitions of a $k$-element set is the <a href="https://en.wikipedia.org/wiki/Bell_number" rel="nofollow noreferrer">Bell number</a> $B_k$. This sequence is <a href="https://oeis.org/A000110" rel="nofollow noreferrer">A000110 in OEIS</a>.</p>
<p>Since $2160$ has $3$ prime factors your answer is $B_3=5$.</p>
|
64,406 | <p>It's often said that there are only two nonabelian groups of order 8 up to isomorphism, one is the quaternion group, the other given by the relations $a^4=1$, $b^2=1$ and $bab^{-1}=a^3$. </p>
<p>I've never understood why these are the only two. Is there a reference or proof walkthrough on how to show any nonabelian group of order 8 is isomorphic to one of these? </p>
| Beginner | 15,847 | <p>Let $G$ be non-abelian group of order 8.</p>
<p>By Sylow theorem, $G$ will have subgroups of order 2.</p>
<ul>
<li><p>If $G$ has unique subgroup of order 2, then $G$ is quaternion (<em>Ref. Theorem 12.5.2, Theory of groups- Marshall Hall</em>). ($G$ can be cyclic also, but we have assumed it is non-abelian.)</p></li>
<li><p>If $G$ has more than one subgroups of order 2, then they will be $1+2k$ in number for some $k\geq 1$; so it is atleast $3$; let $H_1, H_2, H_3$ be three distinct subgroups of order $2$.</p></li>
</ul>
<p>If $H_1\triangleleft G$, then $G/H_1$ is abelian (since its order is 4), we conclude that $[G,G]\leq H_1$, hence $H_1=[G,G]$ since $G$ is non-abelian.</p>
<p>Then observe that $H_2$ can-not be normal in $G$, otherwise $H_2=[G,G]=H_1$, contradiction.</p>
<p>(So we have found a subgroup of order $2$ which is not normal in $G$.)</p>
<p>$H_2$ is a subgroup of $G$ index 4, there is a homomorphism $\phi \colon G\rightarrow S_4$, with $ker(\phi)\subseteq H_2$ (<em>See "Generalised Cayley's theorem" - Introduction to Theory of Groups, Rotman</em>).</p>
<p>But $ker(\phi)\neq H_2$ since $H_2$ is not normal, so $ker(\phi)=\{1\}$; so $G$ is isomorphic to a subgroup of $S_4$ of order $8$; it is a Sylow-2 subgroup of $S_4$; and it is $D_8$ since when considering a square with vertices labelled $"1,2,3,4"$, its symmetries in terms of permutations give $D_8\leq S_4$.</p>
<p>Conclusion:</p>
<blockquote>
<p>$G$ is either quaternion or isomorphic to Sylow-2 subgroup of $S_4$ which is dihedral group of order 8.</p>
</blockquote>
|
64,406 | <p>It's often said that there are only two nonabelian groups of order 8 up to isomorphism, one is the quaternion group, the other given by the relations $a^4=1$, $b^2=1$ and $bab^{-1}=a^3$. </p>
<p>I've never understood why these are the only two. Is there a reference or proof walkthrough on how to show any nonabelian group of order 8 is isomorphic to one of these? </p>
| Marc Olschok | 15,825 | <p>You probably meant $bab = a^3$ in the above.</p>
<p>The following steps should also work. In order not to spoil the fun for you,
I omit all the details.</p>
<p>Let $G$ be a nonabelian group of order $8$ and let $Z$ be its centre.</p>
<p>(1) From the class equation we know that $|Z|$ is divisible by $2$,
and hence $|Z|\gt1$</p>
<p>(2) If $G/Z$ were cyclic then $G$ would be abelian. Therefore only the
case $|Z|=2$ is possible and we know that $G/Z$ must be isomorphic to
the Klein group.</p>
<p>Now let $Z=\{1,z\}$ and take $a,b,c\in G$ with $G/Z = \{Z, aZ, bZ, cZ\}$.
All squares a^2, b^2, c^2 lie in $Z$ and we may take $c=ab$.
Now the rest is a case by case analysis</p>
<p>(3) We cannot have $a^2 = b^2 = (ab)^2 = 1$, because then every element
of $G$ would have order $2$, forcing $G$ to be abelian. So at least one of
those squares must equal $z$. In particular $G$ has an element of order $4$.</p>
<p>(4) If two of the above squares equal $z$, then so does the third. For
instance, suppose $a^2 = z = b^2$. Then $(ab)^2 = 1$ would give
$ab=zbaz=ba$ and $G$ would be abelian.</p>
<p>So now we arrived at two possibilities (up to permutation of $a$, $b$ and $c$):</p>
<p>case (i): $a^2 = b^2 = c^2 = z$.
Then $G$ is the quaternion group.</p>
<p>case (ii) $a^2 = z$, $b^2 = 1$.
Then $bab$ lies in $aZ$. Because of (4), $bab=a$ is impossible and therefore
$bab = az = a^3$.</p>
|
2,328,505 | <p>Let $X$ be an exponential random variable with $\lambda =5$ and $Y$ a uniformly distributed random variable on $(-3,X)$. Find $\mathbb E(Y)$.</p>
<p>My attempt:</p>
<p>$$\mathbb E(Y)= \mathbb E(\mathbb E(Y|X))$$ </p>
<p>$$\mathbb E(Y|X) = \int^{x}_{-3} y \frac{1}{x+3} dy = \frac{x^2+9}{2(x+3)}$$</p>
<p>$$ \mathbb E(\mathbb E(Y|X))= \int^{\infty}_{0} \frac{x^2+9}{2(x+3)} 5 e^{-5x} \, dx$$</p>
| callculus42 | 144,421 | <p><strong>Hint:</strong> $$\int^{x}_{-3} y \frac{1}{x+3} dy=\frac{\frac12\cdot y^2}{x+3}\bigg|_{-3}^x=\frac12\cdot \frac{x^2-(-3)^2}{x+3}=\frac12\cdot \frac{x^2-9}{x+3}$$</p>
<p>At the numerator you can use the second binomial formula. </p>
|
1,439,429 | <p>Is it possible to calculate the Sagitta, knowing the Segment Area and Radius?
Alternatively, is there a way to calculate the Chord Length, knowing the Segment Area and Radius?</p>
| Rubi Shnol | 90,296 | <p>In convex functions, all chords lie above the function values.</p>
<p>You can apply gradient descent to non-convex problems provided that they are smooth, but the solutions you get may be only local. Use global optimization techniques in that case such simulated annealing, genetic algorithms etc.</p>
|
1,439,429 | <p>Is it possible to calculate the Sagitta, knowing the Segment Area and Radius?
Alternatively, is there a way to calculate the Chord Length, knowing the Segment Area and Radius?</p>
| Shiv Tavker | 687,825 | <p>Yes. The function you have is non-convex. Nevertheless, gradient descent will still take you to the global optimum as you have correctly pointed out "the function still has a single minimum". This function is quasi-convex. Gradient descent almost always work for quasi convex functions but we do not have convergence guarantees. I used almost always because you can be unlucky and end up on a saddle point.</p>
|
4,398,864 | <p>How many ways are there to select a three digit number <span class="math-container">$\underline{A}\ \underline{B}\ \underline{C}$</span> so that <span class="math-container">$A \neq B$</span>, <span class="math-container">$B \leq C$</span>, and <span class="math-container">$A < C$</span>?</p>
<hr />
<p>I found the answer as <span class="math-container">$\displaystyle\sum_{k=2}^{9} (k^2-k)=240$</span> by evaluating some big summations. I was wondering if anyone had a solution that doesn't require evaluating big sums?</p>
<p>Thanks in advance.</p>
| Keryann Massin | 450,912 | <p>If I understood your question, you want to find how many choices of <span class="math-container">$A,B,C\in\{0,1,2,3,\dots, 9\}$</span> such that <span class="math-container">$A\ne B$</span>, <span class="math-container">$B\leq C$</span> and <span class="math-container">$A<C$</span>.</p>
<p>First, <span class="math-container">$C$</span> can not be <span class="math-container">$0$</span>, otherwise we would not be able to choose <span class="math-container">$A<C$</span>.
Let us then choose <span class="math-container">$C\in\{1,2,3,\dots, 9\}$</span>. There are <span class="math-container">$9$</span> ways of choosing <span class="math-container">$C$</span>.
We then have <span class="math-container">$(C+1)-1=C$</span> ways of choosing <span class="math-container">$A<C$</span>. Finally, there are <span class="math-container">$C-1$</span> ways of choosing <span class="math-container">$B\ne A$</span> such that <span class="math-container">$B\leq C$</span>. As a result, there are :
<span class="math-container">$$\sum_{C=1}^{9}C\cdot (C-1) =240$$</span>
ways of choosing <span class="math-container">$A,B,C$</span>.</p>
|
3,487,105 | <p>I'm kinda a newbie in calculus, but what are the conditions for the power rule to happen?
For example, if we have the number <span class="math-container">$e$</span>, with the property <span class="math-container">$\frac{d}{dt}{ {e^t} } = {e^t}$</span> we can get, using the power rule, that <span class="math-container">${t \cdot e^{t-1}} = {e \cdot e^{t-1}}$</span> and then <span class="math-container">${t = e}$</span>.
Is there anything wrong?
Thanks for your answers!</p>
| Andrew Chin | 693,161 | <p>This is a mistake common to many calculus students, and it is evidence of a lack of fundamentals.</p>
<p>The power rule is used to differentiate <em>powers</em> of functions. These are functions that have some constant in the exponent (e.g. <span class="math-container">$x^2$</span>, <span class="math-container">$\sqrt{x-2}$</span>, <span class="math-container">$\sqrt[7]{3x+1}$</span>, <span class="math-container">$2x^{0.3}$</span>, etc.). </p>
<p>The power rule cannot be used to differentiate <em>exponential functions</em>. These are functions that have the variable in the exponent (e.g. <span class="math-container">$2^x$</span>, <span class="math-container">$\left(\frac12\right)^{x-1}$</span>, <span class="math-container">$5e^{3x}$</span>, etc.).</p>
<p>As you are just starting to study calculus, whenever you come across a function to differentiate and are not sure whether you can use the power rule, simply ask <em>Is the exponent simply a constant?</em> If the answer is yes, then yes, you can use the power rule. If no, then no.</p>
|
3,487,105 | <p>I'm kinda a newbie in calculus, but what are the conditions for the power rule to happen?
For example, if we have the number <span class="math-container">$e$</span>, with the property <span class="math-container">$\frac{d}{dt}{ {e^t} } = {e^t}$</span> we can get, using the power rule, that <span class="math-container">${t \cdot e^{t-1}} = {e \cdot e^{t-1}}$</span> and then <span class="math-container">${t = e}$</span>.
Is there anything wrong?
Thanks for your answers!</p>
| James Turbett | 721,003 | <p>The power rule only works for functions raised to a power, like x^3, x^4, (x+2)^5, or sqrt(x), etc. The power isn't a variable, it's a constant. When the power is a variable, like e^x, 2^x, we call that an exponential function, and you can't use the power rule to differentiate it. Think about the definition of the derivative, as f(x + h) - f(x) all over h as h goes to zero, and look at what happens for a function like x^2, x^3, x^4 (why does the derivative of x^n become n * x^(n-1)? What happens is interesting to observe), and then do the same for e^x, 2^x, and see how it's fundamentally different. Hope this helps!</p>
|
2,523,660 | <p>I'm trying to solve the next question:</p>
<p>For all $m\in I=(0,1)$ there is a subset $A_m \subseteq \mathbb{R}$ that $A_{m} = \{ a\in \mathbb{R} : a-\lfloor a \rfloor < m \} $. Find $\bigcap\limits_{m\in I} A_{m}$</p>
<p>So I think that the solution is $\bigcap\limits_{m\in I} A_m=\mathbb{Z}$, and I tried to prove it like this:</p>
<p>let $x\in\mathbb{Z}$ then $-x\leq-\lfloor x\rfloor<1-x$.</p>
<p>Therefore $0=-x+x\leq x-\lfloor x\rfloor<x+1-x=1$, so for every $m\in I, x\in A_{m} $</p>
<p>and then $x\in\bigcap\limits_{m\in I} A_{m}$, so $\mathbb{Z}\subseteq \bigcap\limits_{m\in I} A_{m}$.</p>
<p>Now I need to prove that $\bigcap\limits_{m\in I} A_{m} \subseteq \mathbb{Z}$, but how?</p>
| Robert Dixon | 322,951 | <p>The inverse projection takes x to the set {x} × Y which would only have a meaningful definition of open and closed in the power set of the product of X and Y. Topologies for power sets do exist, but I don't think you are looking for that avenue of thought.</p>
|
4,164,820 | <p>CGM = Continuous Geometric Mean. Heuristics and mathematics are described in:</p>
<ul><li><a href="https://math.stackexchange.com/questions/4162869/subset-as-arithmetic-mean-of-geometric-means-not-really">Subset as arithmetic mean of geometric means. Not really?</a></li></ul>
A shortcut to the question as presented here is:
<span class="math-container">$$
\operatorname{CGM}(\vec{r}) = \exp\left(-\!\!\!\!\!\!\int_0^1 \ln(\left|\vec{p}(t)-\vec{r}\right|^2)\,dt\right)
$$</span>
where <span class="math-container">$\,\vec{p}(t)\,$</span> is a <b>circle</b> in the plane and <span class="math-container">$\,\vec{r} = (x,y)\,$</span> is another point in the plane.
<br>A similar exercise for the circle as previously for a straight line segment leads to the following expression:
<span class="math-container">$$
\operatorname{CGM}(\xi,\eta)=
\exp\left( \frac{1}{2\pi}-\!\!\!\!\!\!\int_0^{2\pi}\ln\left|[\cos(t)-\xi]^2+[\sin(t)-\eta]^2\right|\,dt\right)
$$</span>
where dimensionless <span class="math-container">$\xi=x/R$</span> , <span class="math-container">$\eta=y/R$</span> ; <span class="math-container">$(x,y)=$</span> coordinates in the plane, <span class="math-container">$R=$</span> radius of circle.
<br>When this expression is fed into <a href="https://www.maplesoft.com/products/Maple/" rel="nofollow noreferrer">MAPLE</a> (version 8),
then surprisingly we get outcome <b>one</b>, independent of any <span class="math-container">$(x,y,R)$</span> values:
<pre>
> exp(int(ln((cos(t)-xi)^2+(sin(t)-eta)^2),t=0..2*Pi,'continuous')/(2*Pi));
<pre><code> 1
</code></pre>
</pre>
On the other hand trying to proceed by substituting polar coordinates:
<span class="math-container">$$
x = r\cos(\phi) \quad ; \quad y = r\sin(\phi) \\
\exp\left(\frac{1}{2\pi}-\!\!\!\!\!\!\int_0^{2\pi} \ln\left|\left[\cos(t)-\frac{r}{R}\cos(\phi)\right]^2
+\left[\sin(t)-\frac{r}{R}\sin(\phi)\right]^2\right|\,dt\right)
\quad \Longrightarrow \\ \operatorname{CGM}(\rho) =
\exp\left(\frac{1}{2\pi}-\!\!\!\!\!\!\int_0^{2\pi} \ln\left|\rho^2+1-2\rho.\cos(t-\phi)\right|\,dt\right)
\quad \mbox{with} \quad \rho=r/R
$$</span>
Because of circle symmetry we can choose <span class="math-container">$\,\phi=0\,$</span> without loss of generality.<br>
Care should be taken if a zero is present in the argument of the logarithm, resulting in a singularity at that place:
<span class="math-container">$$
\rho^2+1-2\rho.\cos(t)=0 \quad \Longrightarrow \quad \rho = \cos(t)\pm\sqrt{\cos^2(t)-1}
\\ \Longrightarrow \quad t = 0 \quad \mbox{and} \quad \rho = 1
$$</span>
There are two approaches to this special case.<br>The easiest one is to ignore the Cauchy principal value and remember that
the special case represents the value zero, according to the heuristics in the referenced
<a href="https://math.stackexchange.com/questions/4162869/subset-as-arithmetic-mean-of-geometric-means-not-really">Q&A</a>.
<br>The second approach is to accept the Cauchy principal value as being essential.
With our numerical method, being the <i>trapezium rule</i>, integration at the interval <span class="math-container">$[0,\Delta t]$</span> is then calculated by:
<span class="math-container">$$
\frac{f_1+f_2}{2}\Delta t \quad \mbox{with} \quad f_2 = \ln|2-2\cos(\Delta t)| \approx \ln|2-2(1-\Delta t^2/2)| = 2\ln|\Delta t|
$$</span>
On the other hand the integral at that place can be calculated more (or less) exactly:
<span class="math-container">$$
\frac{f_1+f_2}{2}\Delta t \approx \int_0^{\Delta t}\ln|2-2\cos(t)|\,dt \approx
\int_0^{\Delta t}2\ln|t|\,dt = 2(\Delta t\,\ln|\Delta t| - \Delta t) \\
\frac{f_1+2\ln|\Delta t|}{2}\Delta t \approx 2(\ln|\Delta t|-1)\Delta t \quad \Longrightarrow \quad f_1 \approx 2\ln|\Delta t|-4
$$</span>
Having programmed all this (in Delphi Pascal) we get the following output.
Mind the two bold printed values at <span class="math-container">$I(1.0)$</span> : the first one without and the second one with the Cauchy principal value.
<pre>
NUMERICALLY CONJECTURED
I(0.0) = 1.00000000000000E+0000 = 1.00000000000000E+0000
I(0.1) = 1.00000000000000E+0000 = 1.00000000000000E+0000
I(0.2) = 1.00000000000000E+0000 = 1.00000000000000E+0000
I(0.3) = 1.00000000000000E+0000 = 1.00000000000000E+0000
I(0.4) = 1.00000000000000E+0000 = 1.00000000000000E+0000
I(0.5) = 1.00000000000000E+0000 = 1.00000000000000E+0000
I(0.6) = 9.99999999999999E-0001 = 1.00000000000000E+0000
I(0.7) = 1.00000000000000E+0000 = 1.00000000000000E+0000
I(0.8) = 1.00000000000000E+0000 = 1.00000000000000E+0000
I(0.9) = 1.00000000000000E+0000 = 1.00000000000000E+0000
<b>I(1.0) = 0.00000000000000E+0000 = 0.00000000000000E+0000
I(1.0) = 9.99967575938953E-0001 = 1.00000000000000E+0000</b>
I(1.1) = 1.21000000000000E+0000 = 1.21000000000000E+0000
I(1.2) = 1.44000000000000E+0000 = 1.44000000000000E+0000
I(1.3) = 1.69000000000000E+0000 = 1.69000000000000E+0000
I(1.4) = 1.96000000000000E+0000 = 1.96000000000000E+0000
I(1.5) = 2.24999999999999E+0000 = 2.25000000000000E+0000
I(1.6) = 2.55999999999999E+0000 = 2.56000000000000E+0000
I(1.7) = 2.89000000000002E+0000 = 2.89000000000000E+0000
I(1.8) = 3.24000000000002E+0000 = 3.24000000000000E+0000
I(1.9) = 3.61000000000001E+0000 = 3.61000000000000E+0000
I(2.0) = 4.00000000000000E+0000 = 4.00000000000000E+0000
I(2.1) = 4.41000000000001E+0000 = 4.41000000000000E+0000
I(2.2) = 4.84000000000001E+0000 = 4.84000000000000E+0000
I(2.3) = 5.29000000000000E+0000 = 5.29000000000000E+0000
I(2.4) = 5.75999999999996E+0000 = 5.76000000000000E+0000
I(2.5) = 6.24999999999998E+0000 = 6.25000000000000E+0000
I(2.6) = 6.75999999999999E+0000 = 6.76000000000001E+0000
I(2.7) = 7.29000000000010E+0000 = 7.29000000000001E+0000
I(2.8) = 7.84000000000007E+0000 = 7.84000000000001E+0000
I(2.9) = 8.41000000000004E+0000 = 8.41000000000001E+0000
I(3.0) = 9.00000000000000E+0000 = 9.00000000000001E+0000
</pre>
The numerical experiments give rise to the following <b>conjecture</b> (contradictory on purpose):
<span class="math-container">$$
\operatorname{CGM}(\rho) = \begin{cases} 1 & \mbox{for} \quad \rho \le 1 \\ 0 & \mbox{for} \quad \rho = 1 \\
\rho^2 & \mbox{for} \quad \rho \ge 1 \end{cases}
$$</span>
Is there someone who can prove this conjecture analytically?<br>I have seen something with complex analysis in
<a href="https://groups.google.com/g/sci.math/c/j23Nz9Tzlzs" rel="nofollow noreferrer"><i>sci.math</i></a> a long time ago (2008) but
didn't quite understand the argument.
<p>
<h3>CGM for a Sphere</h3>
In view of further development of the theory, it seems to be advantageous to define the Continuous Geometric Mean slightly different, namely as:
<span class="math-container">$$
\operatorname{CGM}(\vec{r}) = \exp\left(- -\!\!\!\!\!\!\int_0^1 \ln(\left|\vec{p}(t)-\vec{r}\right|^2)\,dt\right)
$$</span>
For our unit circle (with Cauchy principal value) then we have instead:
<span class="math-container">$$
\operatorname{CGM}(\rho) = \begin{cases} 1 & \mbox{for} \quad \rho \le 1 \\ 1/\rho^2 & \mbox{for} \quad \rho \ge 1 \end{cases}
$$</span>
We seek to generalize the Continuous Geometric Mean for a circle to the same for the surface of a unit sphere:
<span class="math-container">$$
\operatorname{CGM}(\vec{r}) = \exp\left(-\iint \ln(\left|\vec{p}-\vec{r}\right|^2)\,dA/(4\pi)\right)
$$</span>
Expressed in spherical coordinates and specializing (without loss of generality) for <span class="math-container">$\,\vec{r} = (0,0,\rho)\,$</span>:
<span class="math-container">$$
\vec{p}-\vec{r} = \begin{bmatrix} \sin(\theta)\cos(\phi) \\ \sin(\theta)\sin(\phi) \\ \cos(\theta)-\rho \end{bmatrix}
\quad ; \quad dA = \sin(\theta)\,d\theta\,d\phi \\
\left|\vec{p}-\vec{r}\right|^2 = \sin^2(\theta)+\left[\,\cos(\theta)-\rho\,\right]^2 = 1-2\rho\cos(\theta)+\rho^2 \\
\iint \ln(\left|\vec{p}-\vec{r}\right|^2)\,dA / (4\pi)=
\frac{2\pi}{4\pi} -\!\!\!\!\!\!\!\int_0^\pi \ln\left|\rho^2+1-2\rho\,\cos(\theta)\right|\,\sin(\theta)\,d\theta
\\ \Longrightarrow \quad
\operatorname{CGM}(\rho) = \exp\left(-\frac{1}{2} -\!\!\!\!\!\!\!\int_0^{\pi} \ln\left|\rho^2+1-2\rho\,\cos(\theta)\right|\,\sin(\theta)\,d\theta\right)
$$</span>
Surprisingly enough, integration is much easier in 3-D, when compared with the 2-D case. Substitution of <span class="math-container">$\,t = \cos(\theta)\,$</span> gives a short route to the solution:
<span class="math-container">$$
-\!\!\!\!\!\!\int_0^{\pi} \ln\left|\rho^2+1-2\rho\,\cos(\theta)\right|\,\sin(\theta)\,d\theta =
-\!\!\!\!\!\!\int_{-1}^{+1}\ln\left|\rho^2+1-2\rho\,t\right|\,dt = \\
\frac{1}{2\rho}\left[\;u\ln|u|-u\;\right]_{u=1+\rho^2-2\rho}^{u=1+\rho^2+2\rho} \quad \Longrightarrow \\
\operatorname{CGM}(\rho) = \exp\left(-\left[(1+\rho)^2\ln\left|(1+\rho)^2\right|-(1-\rho)^2\ln\left|(1-\rho)^2\right|-4\rho\right]/(4\rho)\right)
$$</span>
If we make a graph of the two functions - red for 2-D, black for 3-D - then there is another surprise: the graphs coincide for large values of the normed radius <span class="math-container">$\rho$</span>.
<p><a href="https://i.stack.imgur.com/dDg8Y.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dDg8Y.jpg" alt="enter image description here" /></a></p>
<p>Confirmation is found with MAPLE, series expansion for <span class="math-container">$q=1/\rho\to 0$</span>:</p>
<pre>
> f(q) := exp(-((1+1/q)^2*ln((1+1/q)^2)-(1-1/q)^2*ln((1-1/q)^2)-4/q)/(4/q));
> series(f(q),q=0);
</pre>
<p>Output:
<span class="math-container">$$
({q}^{2}-{\frac {1}{3}}{q}^{4}+O \left( {q}^{6} \right) )
$$</span></p>
| Han de Bruijn | 96,057 | <p>Here comes a replay of <a href="https://www.cantorsparadise.com/richard-feynmans-integral-trick-e7afae85e25c" rel="nofollow noreferrer">Richard Feynman’s Integral Trick</a>,
adapted ( if not <i>improved</i> ) for our purpose.
Define:
<span class="math-container">$$
I(a)=\int_{0}^{2\pi}\ln\left(1-2a\cos x+a^2\right)\ \mathrm{d}x \quad , \quad a \ge 0
$$</span>
Then
<span class="math-container">\begin{align}
I'(a)&=\int_{0}^{2\pi}\frac{2a-2\cos x}{1-2a\cos x+a^2} \ \mathrm{d}x \\
& = \frac{1}{a}\int_{0}^{2\pi}\frac{2a^2-2a\cos x}{1-2a\cos x+a^2} \ \mathrm{d}x \\
& = \frac{1}{a}\int_{0}^{2\pi}\frac{1-1+a^2+a^2-2a\cos x}{1-2a\cos x+a^2} \ \mathrm{d}x \\
& = {2\pi \over a} + \frac{1}{a}\int_{-\pi}^{+\pi}\frac{a^2-1}{1-2a\cos x+a^2} \ \mathrm{d}x
\end{align}</span>
and making the <a href="https://en.wikipedia.org/wiki/Weierstrass_substitution" rel="nofollow noreferrer">Weierstrass substitution</a>:</p>
<hr>
<span class="math-container">$$t = \tan(x/2)$$</span>
<span class="math-container">$$t^2\cos^2(x/2)=\sin^2(x/2)=1-\cos^2(x/2)$$</span>
<span class="math-container">$$\cos^2(x/2)=\frac{1}{1+t^2}$$</span>
<span class="math-container">$$\cos(x)=2\cos^2(x/2)-1$$</span>
<span class="math-container">$$\cos x = \frac{1 - t^2}{1 + t^2},$$</span>
<span class="math-container">$$x=2\,\arctan(t)$$</span>
<span class="math-container">$$\mathrm{d}x = \frac{2 \,\mathrm{d}t}{1 + t^2}.$$</span>
<hr>
<span class="math-container">\begin{align}
I'(a)&={2\pi \over a} + \frac{2}{a}\int_{-\infty}^{+\infty}\frac{a^2-1}{(1+a^2)(1+t^2)-2a(1-t^2)}\ \mathrm{d}t \\
&={2\pi \over a} + \frac{2}{a}\int_{-\infty}^{+\infty}\frac{a^2-1}{(1-a)^2 + (1+a)^2t^2} \ \mathrm{d}t \\
&={2\pi \over a} + \frac{2}{a}\frac{a^2-1}{(1-a)^2}2\int_{0}^{\infty}\frac{dt}{1 + \left|(1+a)/(1-a).t\right|^2}
\end{align}</span>
Two cases are to be distinguished:
<span class="math-container">$$
\begin{cases} a \lt 1 & : & u = \left|(1+a)/(1-a).t\right| = (1+a)/(1-a).t \\
a \gt 1 & : & u = \left|(1+a)/(1-a).t\right| = (1+a)/(a-1).t \end{cases}
$$</span>
For <span class="math-container">$\,a\lt\ 1\,$</span>:
<span class="math-container">$$
I'(a)={2\pi \over a} + \frac{4}{a}\frac{(a+1)(a-1)(1-a)}{(1-a)^2(1+a)}\int_{0}^{\infty}\frac{du}{1+u^2}\\={2\pi \over a}-\frac{4}{a}\left[\,\arctan(u)\,\right]_{u=0}^{u=\infty}=
{2\pi \over a}-\frac{4}{a}\frac{1}{2}\pi=0 $$</span>
<span class="math-container">$$ \Longrightarrow \quad I(a) = C$$</span>
For <span class="math-container">$\,a\gt\ 1\,$</span>:
<span class="math-container">$$
I'(a)={2\pi \over a} + \frac{4}{a}\frac{(a+1)(a-1)(a-1)}{(1-a)^2(1+a)}\frac{1}{2}\pi=\frac{4\pi}{a}
$$</span> <span class="math-container">$$ \Longrightarrow \quad I(a) = 4\pi\ln(a) + C
$$</span>
where <span class="math-container">$\,C\,$</span> is an integration constant, to be determined for <span class="math-container">$\,a=0\,$</span>:
<span class="math-container">$$
C = I(a=0)=\int_{0}^{2\pi}\ln\left(1-2a\cos x+a^2\right)\ \mathrm{d}x \\ = \int_{0}^{2\pi}\ln\left(1\right)\ \mathrm{d}x = 0
$$</span>
End result:
<span class="math-container">$$
I(a)=\int_{0}^{2\pi}\ln\left(1-2a\cos x+a^2\right)\ \mathrm{d}x =
\begin{cases}0 & \mbox{for } \; a \lt 1 \\ 2\pi\,\ln\left(a^2\right) & \mbox{for } \; a \gt 1 \end{cases}
$$</span>
<hr>
The special case <span class="math-container">$\,a=1\,$</span>.
<span class="math-container">$$
I(a=1)=\int_{0}^{2\pi}\ln\left(1-2a\cos x+a^2\right)\ \mathrm{d}x \\ =
\int_{0}^{2\pi}\ln\left(2-2\cos x \right)\ \mathrm{d}x\\
=\int_{0}^{2\pi}\ln\left(4\sin^2\frac12 x\right)\ \mathrm{d}x =
2\int_{0}^{\pi}\ln\left(4\sin^2 t\right)\ \mathrm{d}t\\
=2(\ln 4)\pi+\color{red}{4\int_{0}^{\pi}\ln\left(\sin t\right)\ \mathrm{d}t}
=\color{red}{4\pi\ln 2} +8\int_{0}^{\pi/2}\ln\left(\sin t\right)\ \mathrm{d}t\\
=4\pi\ln 2 +8\left[\int_{0}^{\pi/2}\ln\left(\sin t\right)\ \mathrm{d}t + \int_{0}^{\pi/2}\ln\left(\cos t\right)\ \mathrm{d}t\right]/2\\
=4\pi\ln 2 +4\int_{0}^{\pi/2}\ln\left(\sin t\cos t\right)\ \mathrm{d}t\\
=4\pi\ln 2 +4\int_{0}^{\pi/2}\ln\left(\frac12 \sin 2t\right)\ \mathrm{d}t\\
=4\pi\ln 2 +4\ln\left(\frac12\right)\frac{\pi}{2}+4\frac12\int_{0}^{\pi}\ln\left(\sin x\right)\ \mathrm{d}x\\
=2\pi\ln 2 +2\int_{0}^{\pi}\ln\left(\sin x\right)\ \mathrm{d}x
=\color{red}{4\pi\ln 2 +4\int_{0}^{\pi}\ln\left(\sin t\right)\ \mathrm{d}t}\\
\Longrightarrow \quad \int_{0}^{\pi}\ln\left(\sin t\right)\ \mathrm{d}t = -\pi\ln 2 \\
\Longrightarrow \quad I(a=1)= 4\pi\ln 2 + 4\int_{0}^{\pi}\ln\left(\sin t\right)\ \mathrm{d}t = 0
$$</span>
|
3,328,737 | <p>For any rational number, <span class="math-container">$\frac{p}{q}$</span> , <span class="math-container">$p$</span> and <span class="math-container">$q$</span> should be integers, <span class="math-container">$q\neq0$</span> and <span class="math-container">$p,q$</span> should not have any common factors.
Now, if we have two even numbers, say <span class="math-container">$2m$</span> and <span class="math-container">$2n$</span> where <span class="math-container">$m$</span> and <span class="math-container">$n$</span> are integers.
<span class="math-container">$$\frac{\text{even}}{\text{even}}=\frac{2m}{2n}=\frac{m}{n}$$</span>
where(<span class="math-container">$\frac{m}{n}$</span>) nature is still unknown.
So, what nature does <span class="math-container">$\frac{\text{even}}{\text{even}}$</span> have, rational or irrational?</p>
<p>additional reference: <a href="https://www.quora.com/Is-2-4-rational" rel="nofollow noreferrer">https://www.quora.com/Is-2-4-rational</a></p>
| Fred | 380,717 | <p>If <span class="math-container">$m$</span> and <span class="math-container">$n$</span> are integers and <span class="math-container">$n \ne 0$</span>, then <span class="math-container">$\frac{m}{n}$</span> is rational.</p>
|
1,673,854 | <p>Taylor Series of $f(x) = \sqrt{x}$ about $c = 1$</p>
<p>I've tried doing this problem but stuck at finding a pattern..</p>
<p>Work:</p>
<p>$$T_n = \sum^\infty_{n=0}\frac{f^n(c)}{n!}(x-c)^n = f(a) + \frac{f'(c)}{1!}(x-a)^1 + \frac{f''(c)}{2!}(x-a)^2+... $$</p>
<p>So $f(x) = \sqrt{x}$</p>
<p>$$ f'(x)=\frac12x^{-\frac12}$$
$$ f''(x)=\frac{-1}4x^{-\frac{3}2}$$
$$ f'''(x)=\frac{3}8x^{-\frac{5}2}$$
and
$$f(1) = 1$$
$$f'(1) = \frac12$$
$$f''(1) = -\frac14$$
$$f'''(1) = \frac38$$</p>
<p>So far I know that its alternating so $(-1)^n$</p>
<p>But I'm having trouble with the fractions since I thought it would have been $\left(\frac12\right)^n$ but the $\frac38$ wont work with that. Have I done something wrong so far or am I just not thinking of this the right way?</p>
<p>So I've noticed that starting from the $n=3$ the numerator is $3$ then $3*5$ then $3*5*7$ and so on... but how do I account for the first 3 terms? (n = 0, 1, 2)</p>
<p>I learned that I can exclude it by simply taking it out and adding it from where the pattern starts to work so...</p>
<p>$$1+\frac{x-1}{2}-\frac{(x-2)^2}{8} + \sum^\infty_{n=3}something$$</p>
| Em. | 290,196 | <p>Let's take a smaller example. Consider
$$5! = 5\cdot4\cdot3\cdot2\cdot1.$$
Notice that this can be expressed as
$$5! = 5\cdot(4\cdot3\cdot2\cdot1\cdot) = 5\cdot4!.$$
Similarly
$$5! = 5\cdot4\cdot(3\cdot2\cdot1) = 5\cdot4\cdot3!$$
We can generalize to $k!$,
$$k! = k(k-1)(k-2)\dotsm 2\cdot1$$
and in particular
$$k! = k(k-1)(k-2)!\tag a$$</p>
<p>Now,
\begin{align*}
\lim_{n\to\infty} \frac{(2n-1)!}{(2n+1)!} &= \lim_{n\to\infty}\frac{(2n-1)!}{(2n+1)(2n)(2n-1)!}\tag b\\
&=\lim_{n\to\infty}\frac{1}{(2n+1)(2n)}\\
&= 0
\end{align*}
where in (b), I could use (a) as an aid by letting $k = 2n+1$.</p>
|
198,945 | <p>Why and how publishing a paper in proceedings?<br>
What are the difference with a "classical" journal?<br>
What's the list of the main proceedings in which one can publish?<br>
Do proceedings papers (never, sometimes, often or always) appear on mathscinet?</p>
| Andreas Blass | 6,794 | <p>Proceedings of conferences are often published as special issues of "classical" journals. But even those that are not are usually included in MathSciNet if they include a statement (often a footnote on the first page of each paper) to the effect that the papers are in final form and will not be published elsewhere. </p>
<p>Some but not all conference proceedings are refereed less thoroughly than reputable journal articles. As a result, mathematicians are sometimes suspicious about results published in conference proceedings, and administrators sometimes assign less value to such publications. Some conference proceedings have responded to this problem by explicitly saying (usually in the preface of the proceedings volume) that the papers have been refereed to the standards of such-and-such journal. Nevertheless, I would advise young (= not yet tenured) mathematicians to publish most if not all of their work in regular (and reputable, of course) journals. Once you have tenure, so that administrators' opinions are less critical for your life, it becomes reasonable to contribute more to conference proceedings.</p>
|
35,321 | <p>I need to do some simplification of an expression involving averages over a stochastic variable (in order to verify a long analytical calculation).
The easiest way to do that, I figured, were if I could implement an operator which would basically be short-hand for the averaging procedure, with all the appropriate properties. Then of course this operator would be present in the final expression, which is fine, and would enable me to compare easily with my own calculations.
So assuming I use <code>x</code> for the stochastic variable, I tried defining <code>av</code> using</p>
<pre><code>av[y_ + z_] := av[y] + av[z]
av[c_ y_] := c av[y] /; FreeQ[c, x]
av[c_] := c /; FreeQ[c, x]
</code></pre>
<p>Then when I write</p>
<pre><code>D[av[x y], y]
</code></pre>
<p>I get</p>
<pre><code>
av[x]
</code></pre>
<p>which is fine, but when I write</p>
<pre><code>D[av[Exp[-x y]], y]
</code></pre>
<p>I get</p>
<pre><code>-E^(-x y) y
</code></pre>
<p>instead of <code>-y av[Exp[-x y]]</code> as I want, i.e., the <code>av</code> is removed somehow.</p>
<p>I tried using <code>UpValue</code> for teaching <em>Mathematica</em> that it could interchange differentiation and <code>av</code>, but apparently that is not the problem.
I might be going about this entirely the wrong way, but I'd be grateful for any input. Note the builtin <code>Expectation</code> function does not accomplish it either - e.g., it doesn't handle the derivatives as a proper average operator would. For example</p>
<pre><code>h[y_] := Expectation[y, x \[Distributed] pp] (*pp unknown density*)
</code></pre>
<p>Then </p>
<pre><code>D[h[Exp[-x y]], x]
</code></pre>
<p>gives </p>
<pre><code>(Expectation^(0,1))[E^(-x y),x\[Distributed]pp] (Distributed^(1,0))[x,pp]-E^(-x y) y
(Expectation^(1,0))[E^(-x y),x\[Distributed]pp]
</code></pre>
<p>whereas I wanted</p>
<pre><code>-y h[Exp[-y x]]
</code></pre>
<p>(i.e moving the derivative inside the averaging h).
Sune</p>
| Sooner | 10,361 | <p>I think I got it.</p>
<pre><code>av /: D[av[f___], x_] := av[D[f, x]]
av[y_ + z_] := av[y] + av[z]
av[c_ y_] := c av[y] /; FreeQ[c, x]
av[c_] := c /; FreeQ[c, x]
D[av[x y], x]
(* y *)
D[av[Exp[-x y]], x]
(* -y av[E^(-x y)] *)
</code></pre>
<p>I wasn't using <code>UpValues</code> correctly before.</p>
<p>EDIT: Well, turns out there's still a problem:</p>
<pre><code>D[Log[av[Exp[-b x]]], b]
(* -((E^(-b x) x)/av[E^(-b x)]) *)
</code></pre>
<p>instead of</p>
<pre><code>-(av[(E^(-b x) x)]/av[E^(-b x)])
</code></pre>
<p>What's going on?
Sune</p>
|
1,646,135 | <p>I have the subset $\left[0,1\right] \backslash \mathbb{Q}$ in $\mathbb{R} \backslash \mathbb{Q}$. </p>
<p>Am I right in thinking that this set is open and not closed in the space given?</p>
<p>Also, how do I go about finding the interior, closure and boundary?</p>
| noctusraid | 185,359 | <p>I don't think such a function exists.
You can construct an example where all the inputs are non-zero but the function gives back 0:</p>
<p>$$A=\{1,2,3,4,5\}, B=\{1,6\},C=\{5,6\} \implies f(_\cdots)=0$$</p>
<p>You can easily construct an example where you have the same cardinalities but the intersection is non trivial and hence $f$ doesn't define a function.</p>
|
1,646,135 | <p>I have the subset $\left[0,1\right] \backslash \mathbb{Q}$ in $\mathbb{R} \backslash \mathbb{Q}$. </p>
<p>Am I right in thinking that this set is open and not closed in the space given?</p>
<p>Also, how do I go about finding the interior, closure and boundary?</p>
| Nick Graham | 547,268 | <p>You can estimate $|A \cup B \cup C|$ by computing its mean over a population where every element is three sets that meet the known conditions. This assumes that the presence of an element in a set is independent from and identically distributed to the presence of every other element in that set or other sets.</p>
<p>$$
P(e \in B \cap C | e \in A) = \frac{|A \cap B|}{|A|} \frac{|A \cap C|}{|A|}\\
P(e \in C \cap A | e \in B) = \frac{|B \cap C|}{|B|} \frac{|B \cap A|}{|B|} \\
P(e \in A \cap B | e \in C) = \frac{|C \cap A|}{|C|} \frac{|C \cap B|}{|C|} \\
$$</p>
<p>Dividing by three because we assume every element has an equal chance of being in the three sets.
$$
E(|A \cap B \cap C|) = \frac{|A|P(e \in B \cap C | e \in A) + |B|P(e \in C \cap A | e \in B) + |C|P(e \in A \cap B | e \in C)}{3} \\
= \frac{\frac{|A \cap B||A \cap C|}{|A|} + \frac{|B \cap C||B \cap A|}{|B|} + \frac{|C \cap A||C \cap B|}{|C|}}{3} \\
$$</p>
<p>I wrote up some Python code to check the estimates</p>
<pre><code>import pandas as pd
from random import randint
import numpy as np
def spread(n=1000, epochs=1000):
er = [np.abs(perc_error(n=n)) for _ in range(epochs)]
return pd.Series(er).describe()
def perc_error(n=1000):
test = random_three_sets(n=n)
real, estimated = in_all(test)
return (estimated - real) / real
randBinList = lambda n: [randint(0, 1) for b in range(1, n + 1)]
def random_three_sets(n=1000):
data = [randBinList(3) for _ in range(n)]
df = pd.DataFrame(data, columns=['A','B','C'])
return df
def in_all(rtcd):
inA = rtcd['A']
inB = rtcd['B']
inC = rtcd['C']
nA = np.sum(inA)
nB = np.sum(inB)
nC = np.sum(inC)
nAB = np.sum(inA.multiply(inB))
nBC = np.sum(inB.multiply(inC))
nAC = np.sum(inA.multiply(inC))
trueABC = np.sum(inA.multiply(inB).multiply(inC))
predictedABC = (nAB*nAC/nA + nBC*nAB/nB + nAC*nBC/nC)/3
return trueABC, predictedABC
def main():
print(spread())
</code></pre>
<p>Percent error spread with population size 100 over 1000 runs</p>
<pre><code>count 1000.000000
mean 9.797707
std 10.256113
min 0.000000
25% 3.318324
50% 7.621868
75% 13.052138
max 150.760582
</code></pre>
<p>Percent error spread with population size 1000 over 1000 runs</p>
<pre><code>count 1000.000000
mean 2.851916
std 2.119233
min 0.005230
25% 1.205157
50% 2.411512
75% 4.135936
max 11.853071
</code></pre>
|
3,621,223 | <p>I use a software called Substance Designer which has a Pixel Processor where I can assign to every pixel of a image a gray-scale value defined by a series of operations.</p>
<p>I am basically trying to generate a <a href="https://i.stack.imgur.com/6jhYa.jpg" rel="nofollow noreferrer">"normal gradient"</a> generated by the normals of a semi-ellipse with a given <strong>a</strong> and <strong>b</strong> semi-major and semi-minor axis. </p>
<p>This semi ellipse is **origin-centered ** and has the principle axes parallel to the x and y axes.</p>
<p>For all points P(x,y) with y≥0, I want to find the angle or direction θ of the outwards-facing ellipse normal that intersects that point. Both when a>b and if possible when b>a "</p>
<p><a href="https://i.stack.imgur.com/LA7x1.jpg" rel="nofollow noreferrer">Here is a visual representation of what I am after, although I only need the values for y>0</a></p>
<p><a href="https://commons.wikimedia.org/wiki/File:Evolute1.gif" rel="nofollow noreferrer">I am trying to visualize the blue tangents on the evolute when y> 0. All the points on the same normal will have the same value.</a></p>
<p>Thanks </p>
| Anonymous Coward | 770,101 | <p>This is an extended comment, to solve the following problem that came up in the comments:</p>
<p>We have an axis-aligned ellipse, centered at origin, width <span class="math-container">$2$</span>, height <span class="math-container">$2 h$</span> — so semi-major axis is <span class="math-container">$1$</span> and on <span class="math-container">$x$</span> axis, and semi-minor axis is <span class="math-container">$h$</span> and on <span class="math-container">$y$</span> axis, with <span class="math-container">$0 \lt h \lt 1$</span>. For all points <span class="math-container">$(x, y)$</span> on or outside the ellipse, <span class="math-container">$x \gt 0$</span>, <span class="math-container">$y \gt 0$</span>, we want to find the direction (angle or slope) of the ellipse normal that passes through that point.</p>
<p>The eccentricity <span class="math-container">$\epsilon$</span> of this ellipse is
<span class="math-container">$$\epsilon = \sqrt{1 - h^2} \quad \iff \quad h = \sqrt{1 - \epsilon^2}$$</span></p>
<p>Using angle parameter <span class="math-container">$\varphi$</span> the points on the ellipse can be parametrized as
<span class="math-container">$$\left\lbrace ~ \begin{aligned}
x_{\varphi}(\varphi) &= \cos\varphi \\
y_{\varphi}(\varphi) &= h \sin\varphi \\
\end{aligned} \right .$$</span>
The slope of the normal is
<span class="math-container">$$m = \frac{ -\frac{d x_\varphi(\varphi)}{d \varphi} }{ \frac{d y_\varphi(\varphi)}{d \varphi} } = \frac{1}{h}\tan\varphi$$</span>
Solving for <span class="math-container">$\varphi$</span> we find
<span class="math-container">$$\varphi = \arctan\left(m h\right)$$</span>
and reparametrizing the points on the ellipse (in the positive quadrant, <span class="math-container">$x \gt 0$</span>, <span class="math-container">$y \gt 0$</span>), we get
<span class="math-container">$$\left\lbrace ~ \begin{aligned}
\displaystyle x_m(m) &= \frac{1}{\sqrt{1 + h^2 m^2}} \\
\displaystyle y_m(m) &= \frac{h^2 m}{\sqrt{1 + h^2 m^2 }} \\
\end{aligned} \right .$$</span>
The points on the normal are on a line <span class="math-container">$y = m x + y_0$</span>, i.e.
<span class="math-container">$$y_m(m) = m x_m(m) + y_0(m)$$</span>
and rearranging and simplifying, we find that
<span class="math-container">$$y_0(m) = \frac{(h^2 - 1) m}{\sqrt{1 + h^2 m^2}}$$</span>
If point <span class="math-container">$(x, y)$</span> is on the normal with slope <span class="math-container">$m$</span>, it too fulfills the equation:
<span class="math-container">$$y = m x + y_0(m) = m x + m \frac{h^2 - 1}{\sqrt{h^2 m^2 + 1}}$$</span>
To solve this for <span class="math-container">$m$</span> (remembering that <span class="math-container">$\theta = \arctan m$</span>) we need to solve a quadric equation:
<span class="math-container">$$\begin{aligned}
\left( h^2 x^2 \right) & m^4 \\
+ \left( - 2 h^2 x y \right) & m^3 \\
+ \left( x^2 - h^4 + h^2 y^2 + 2 h^2 - 1 \right) & m^2 \\
+ \left( - 2 x y \right) & m \\
+ \left( y^2 \right) & ~ = 0 \\
\end{aligned}$$</span>
This has zero to four real roots <span class="math-container">$m$</span>. At and outside the ellipse, at <span class="math-container">$x \gt 0$</span>, <span class="math-container">$y \gt 0$</span>, there is exactly one real root (because the normals for the positive quadrant only intersect at <span class="math-container">$y \le 0$</span>).</p>
<p>Unfortunately, there is no simple algebraic expression for the solutions <span class="math-container">$m$</span> that fulfill above.</p>
<p>For what it is worth, Maple suggests the following operations to find the four candidate real roots:</p>
<pre><code>t36 = x * x
t33 = y * y
t43 = h * h
t47 = t43 * t43
t48 = t43 * t47
t83 = (t48 + t43) * t36
t32 = t33 * t33
t82 = 3 * t32
t81 = 3 * t33
t26 = 3 * t36
t80 = -3 * t47
t79 = 6 * t48
t49 = t47 * t47
t78 = -3 * t49
t77 = 3 * t49
t74 = -18 * t47
t73 = 12 * t48
t24 = t43 * t33
t64 = t47 - 2 * t43 + 1
t16 = -t24 - t36 + t64
t65 = t47 * t36
t17 = t65 * t82
t35 = t36 * t36
t67 = t43 * t35
t18 = t67 * t81
t21 = t32 * t80
t22 = 3 * t24
t57 = y * t33
t31 = t57 * t57
t23 = t48 * t31
t25 = -3 * t35
t51 = t48 * t48
t28 = t43 * t51
t30 = t43 * t49
t54 = x * t36
t34 = t54 * t54
t45 = cbrt(-1 + 6 * t67 + 18 * t65 - 15 * t47 + 20 * t48 - 15 * t49 - t51 + 6 * t43 + 6 * t30 + t34 + t21 + t22 + t23 + t25 + t26 + t17 + t18 + t36 * t77 + t32 * t78 + t32 * t79 + t35 * t80 + t30 * t81 + 6 * y * sqrt(3 * t17 + 3 * t18 + 3 * t21 + 3 * t22 + 3 * t23 + 3 * t25 + 24 * t28 - 84 * t51 + 168 * t48 - 84 * t47 + 24 * t43 - 3 * t49^2 - 3 + (-2 * t31 - 70) * t77 + 3 * t64 * t34 + 3 * (t31 + 56) * t30 + 3 * (t78 + t73 + t74 + 12 * t43) * t35 + (-3 * t51 + 12 * t30 - 18 * t49 + t73) * t82 + (3 * t28 - 18 * t51 + 45 * t30 - 60 * t49 + 45 * t48 + t74 + (3 * t48 - 6 * t47) * t35 + 21 * (t30 - 4 * t49 + t79 - 4 * t47 + t43) * t36) * t81 + (3 + 3 * t51 - 18 * t30 + 45 * t49 - 60 * t48 + 45 * t47 - 18 * t43 + (t77 - 6 * t48) * t32) * t26) * h * x + (48 * t83 - 12 * t49 - 12 * t47 - 96 * t65 + 18 * t48) * t33 - 12 * t83)
t68 = t16 * t16 / t45
t46 = sqrt((t33 + ((2 * t16) + t45 + t68) / (3 * t43)) / t36)
t66 = t16 / t43
t71 = y / x
t72 = (2 * t71 / t43 + (t66 * y + t57) / t54) / t46
t19 = t71 / 2
t70 = t19 + t46 / 2
t69 = t19 - t46 / 2
t60 = (2 * t33 - ((t45 + t68)/t43 - 4 * t66) / 3) / t36
t62 = sqrt(t60 - 2 * t72) / 2
t61 = sqrt(t60 + 2 * t72) / 2
m1 = t70 + t61
m2 = t70 - t61
m3 = t69 + t62
m4 = t69 - t62
</code></pre>
<p>Above, <code>cbrt(...)</code> is the cube root operation (same as <code>pow(..., 1.0/3.0)</code>), and <code>sqrt(...)</code> is the square root operation. If the argument of the square root is negative (but note that it could be <em>effectively zero</em>, due to rounding errors and such, even if strictly speaking negative; you should round those to positive zero instead), then those candidate solutions do not exist. Remember that you do still need to verify that <span class="math-container">$m \gt 0$</span>, to get the desired result, <span class="math-container">$\theta = \arctan m$</span>.</p>
<p>(Also remember that if <span class="math-container">$y = 0$</span> and <span class="math-container">$x \gt 0$</span>, then <span class="math-container">$\theta = 0$</span>; if <span class="math-container">$y \gt 0$</span> and <span class="math-container">$x = 0$</span>, then <span class="math-container">$\theta = 90^o$</span>. The rest of the quadrants are symmetric.)</p>
<p>Because of this, it might be worth the effort to consider <em>approximations</em> to the problem; for example, instead of a real ellipse, consider approximating the ellipse using one or two cubic polynomials (perhaps with four cubic Bézier curves). Cubic approximations should yield at least two significant digits of precision (less than 1% of error), and would definitely suffice for human perception and "visual arts". I am not sure if that approach yields any better answers, but it would be the avenue I would explore next.</p>
|
1,661,439 | <p>I'm trying to take the Lagrange polynomial $P_3(x)$ that passes through four points- $(x_1,y_1), (x_2,y_2), (x_3,y_3)$, and $(x_4,y_4)$, and integrate it (maybe deriving Simpson's rule in the process!).
All the x-values are a distance h apart from each other.
Once I did some simplifications using h, here's what I ended up with:
$$P_3(x) = {(x-x_2)(x-x_3)(x-x_4)y_1\over -6h^3} + {(x-x_1)(x-x_3)(x-x_4)y_2\over 2h^3} + {(x-x_1)(x-x_2)(x-x_4)y_3\over -2h^3} + {(x-x_1)(x-x_2)(x-x_3)y_4\over 6h^3}$$
I'm trying to find a substitution to make these integrals easier. When I choose a u-substitution, should I try to make the integrands go from -1 to 1? And how would I scale the u-substitution to make that happen?
Throw out the next step that you think I should take!
Thanks!</p>
| Carl Christian | 307,944 | <p>While Acheille Hui has outlined a good strategy I would like to continue with your approach.</p>
<p>First of all, I can confirm that your polynomial passes through the four given values, hence you have the correct expression. </p>
<p>You can always change the interval of integration from, say, $[a,b]$ to $[-1,1]$ by a linear transformation, specifically
\begin{equation}
x = \phi(t) = a \frac{1-x}{2} + b \frac{1+x}{2},
\end{equation}
where I have chosen the peculiar representation to stress the fact that
\begin{equation}
\phi(-1) = a, \quad \phi(1) = b
\end{equation}
We also see that
\begin{equation}
\phi'(t) = \frac{b-a}{2}.
\end{equation}
It follows that for any continuous function $f : [a,b] \rightarrow \mathbb{R}$ we can write
\begin{equation}
\int_a^b f(x)dx = \int_{-1}^1 f(\phi(t)) \phi'(t) dt
\end{equation}</p>
<p>In your case we could write
\begin{equation}
x_1 = a, \quad x_2 = a + h, \quad x_3 = a + 2h, \quad x_4 = a + 3h = b
\end{equation}
so that
\begin{equation}
\phi(t) = a \frac{1-x}{2} + (a+3h) \frac{1+x}{2} = a + 3h \frac{1+x}{2}
\end{equation}
and
\begin{equation}
\phi(-1) = x_1, \quad \phi\left(-\frac{1}{3}\right) = x_2, \quad \phi\left(\frac{1}{3}\right) = x_2, \quad \phi\left(1\right) = x_3
\end{equation}
The substitution will allow you to move your integral to the interval $[-1,1]$. At this point, I would recommend reducing the polynomials to standard form by carrying all the multiplications.</p>
<p>You mention Simpson's rule. It is difficult to apply it in this context, because the number of natural sub-intervals is an odd number (specifically 3), rather than an even number. The simplest way that I know to rediscover Simpson's rule is to seek numbers $a$, $b$, and $c$ such the quadrature rule
\begin{equation}
\int_{-h}^h f(x) dx = a f(-h) + b f(0) + c f(h)
\end{equation}
is exact for all polynomials of degree less than or equal to $2$. By using \begin{equation}
f(x) = 1, \quad f(x) = x, \quad \text{and}\quad f(x) = 2
\end{equation}
you will assemble a linear system of dimension $3$ for the unknowns $a$, $b$ and $c$. Once you have solved it you will be able to verify that your new quadrature rule also integrates $f(x) = x^3$ correctly. </p>
|
1,926,839 | <p>I have to find a function $f\in C^\infty(\mathbb{R})$ which, for fixed $a,b\in \mathbb{R}$, $a<b$, is identically 1 for $x\le a$, identically $0$ for $x\ge b$ and decreases in $a\le x\le b$. I've tried many times to write $f$ in $a\le x\le b$ using an exponential, but it didn't work. Can you help me?</p>
| Daniel Robert-Nicoud | 60,713 | <p><strong>Hint:</strong> Use the function
$$\varphi(x) = \cases{0 & if $x\le0$,\\e^{-\frac{1}{x^2}} & if $x\ge0$}$$
as a building block. You can easily prove that $\varphi\in C^\infty(\mathbb{R})$.</p>
|
3,640,307 | <p>The extended reals are taken to be the set <span class="math-container">$\mathbb{R}\cup\{+\infty,-\infty\}$</span>. But is there a <em>natural</em> way to define <span class="math-container">$+\infty$</span> and <span class="math-container">$-\infty$</span> as sets, in a pure set-theoretic theme, following the construction of <span class="math-container">$\mathbb{R}$</span> as equivalence classes of Cauchy sequences, <em>not</em> Dedekind’s cuts as outlined in <a href="https://math.stackexchange.com/q/459067/673223">this post</a>.</p>
<p>I emphasise a <em>natural</em> way so that properties <span class="math-container">$(-1)\times (+\infty)=-\infty$</span> and <span class="math-container">$(-1)\times (-\infty)=+\infty$</span> follow as consequences.</p>
| J.G. | 56,861 | <p>Roughly speaking, the Cauchy definition identifies each real with the Cauchy sequences of rationals that converge to it, whereas the Dedekine definition identifies each real with the set of rational numbers less than it (or an ordered pair of that set, together with the set of greater rationals). Since you asked about the former, <span class="math-container">$\pm\infty$</span> is at first problematic because Cauchy sequences in <span class="math-container">$\Bbb Q$</span> don't <span class="math-container">$\to\infty$</span>.</p>
<p>We'll have to consider something other than Cauchy sequences, but whatever sequences we do consider need to be split into equivalence classes by a suitable equivalence relation. The usual relation considers sequences equivalent if they differ by a null sequence. We can use this to partition sequences diverging to <span class="math-container">$\pm\infty$</span> if we really want. Do we use all of them, or just a subset that reminds us of Cauchy sequences? Well, since any sequence of rationals is Cauchy iff it has a real limit, we could have omitted "Cauchy" from the original definition anyway. So it seems natural to identify <span class="math-container">$\infty$</span> with the set of all sequences in <span class="math-container">$\Bbb Q$</span> that satisfy <span class="math-container">$\forall q\in\Bbb Q\exists N\in\Bbb N\forall n>N(a_n>q)$</span>. We can handle <span class="math-container">$-\infty$</span> similarly, or just multiply the previous sequences by <span class="math-container">$-1$</span> elementwise.</p>
|
4,329,888 | <p>I have a problem of understanding how to find shaded regions in Complex Plane.</p>
<p><span class="math-container">\begin{array}{l}
|z-2i|\ \geqslant \ |z+6+4i|\\
\\
\sqrt{x^{2} +( y-2)^{2}} =\sqrt{( x+6)^{2} +( y+4)^{2}}\\
x^{2} +( y-2)^{2} =( x+6)^{2} +( y+4)^{2}\\
x^{2} +y^{2} -4y+4=x^{2} +12x+36+y^{2} +8y+16\\
12x+12y+48=0\\
y=-x-4\\
\end{array}</span></p>
<p><img src="https://i.stack.imgur.com/CrMh6.png" alt="Sketch without shaded regions" /></p>
<p>I can sketch this in complex plane, But I'm confused in how to get shaded regions like this questions.</p>
| Gary | 83,800 | <p>You can continue by changing the order of summation and then splitting the sum into two parts:
<span class="math-container">\begin{align*}
& \sum\limits_{n = 0}^\infty {\frac{{( - 1)^{n + 1} \pi ^{2n + 1} }}{{(2n + 1)!}}\left( {\sum\limits_{k = 0}^{2n + 1} {( - 1)^k \binom{2n + 1}{k}(z + 1)^k } } \right)}
\\ & = \sum\limits_{k = 0}^\infty {\left( {\sum\limits_{n = \left\lceil {\frac{{k - 1}}{2}} \right\rceil }^\infty {\frac{{( - 1)^{n + k + 1} \pi ^{2n - k + 1} }}{{(2n - k + 1)!}}} } \right)\frac{{\pi ^k }}{{k!}}(z + 1)^k }
\\ & = \sum\limits_{k = 0}^\infty {\left( {\sum\limits_{n = k}^\infty {\frac{{( - 1)^{n + 2k + 1} \pi ^{2n - 2k + 1} }}{{(2n - 2k + 1)!}}} } \right)\frac{{\pi ^{2k} }}{{(2k)!}}(z + 1)^{2k} } \\ & \quad + \sum\limits_{k = 0}^\infty {\left( {\sum\limits_{n = k}^\infty {\frac{{( - 1)^{n + 2k + 2} \pi ^{2n - 2k} }}{{(2n - 2k)!}}} } \right)\frac{{\pi ^{2k + 1} }}{{(2k + 1)!}}(z + 1)^{2k + 1} }
\\ & = \sum\limits_{k = 0}^\infty {\left( {( - 1)^{k + 1} \sum\limits_{j = 0}^\infty {\frac{{( - 1)^j \pi ^{2j + 1} }}{{(2j + 1)!}}} } \right)\frac{{\pi ^{2k} }}{{(2k)!}}(z + 1)^{2k} } \\ & \quad + \sum\limits_{k = 0}^\infty {\left( {( - 1)^k \sum\limits_{j = 0}^\infty {\frac{{( - 1)^j \pi ^{2j} }}{{(2j)!}}} } \right)\frac{{\pi ^{2k + 1} }}{{(2k + 1)!}}(z + 1)^{2k + 1} }
\\ & = \sum\limits_{k = 0}^\infty {\left( {( - 1)^{k + 1} \sin \pi } \right)\frac{{\pi ^{2k} }}{{(2k)!}}(z + 1)^{2k} } + \sum\limits_{k = 0}^\infty {\left( {( - 1)^k \cos \pi } \right)\frac{{\pi ^{2k + 1} }}{{(2k + 1)!}}(z + 1)^{2k + 1} } \\ & = \sum\limits_{k = 0}^\infty {( - 1)^{k + 1} \frac{{\pi ^{2k + 1} }}{{(2k + 1)!}}(z + 1)^{2k + 1} } .
\end{align*}</span></p>
|
23,502 | <p><em>Edit: I wrote the following question and then immediately realized an answer to it, and moonface gave the same answer in the comments. Namely, $\mathbb C(t)$, the field of rational functions of $\mathbb C$, gives a nice counterexample. Note that it is of dimension $2^{\mathbb N}$.</em></p>
<p>The following is one statement of Schur's lemma:</p>
<blockquote>
<p>Let $R$ be an associative unital algebra over $\mathbb C$, and let $M$ be a simple $R$-module. Then ${\rm End}_RM = \mathbb C$.</p>
</blockquote>
<p>My question is: are there extra conditions required on $R$? In particular, how large can $R$ be?</p>
<p>In particular, the statement is true when $\dim_{\mathbb C}R <\infty$ and also when $R$ is countable-dimensional. But I have been told that the statement fails when $\dim_{\mathbb C}R$ is sufficiently large.</p>
<p>How large must $\dim_{\mathbb C}R$ be to break Schur's lemma? I am also looking for an explicit example of Schur's lemma breaking for $\dim_{\mathbb C}R$ sufficiently large?</p>
| Manny Reyes | 778 | <p>The idea behind Schur's Lemma is the following. The endomorphism ring of any simple $R$-module is a division ring. On the other hand, a finite dimensional division algebra over an algebraically closed field $k$ must be equal to $k$ (this is because any element generates a finite dimensional subfield over $k$, which must be equal to $k$). </p>
<p>Thus, when $R$ is an algebra over an algebraically closed field $k$, the endomorphism ring of a finite dimensional simple module is a finite dimensional division algebra over $k$ and hence is equal to $k$. </p>
<p>On the other hand, let $D$ be any division algebra over a field $k$, which we no longer assume to be algebraically closed. Then $D_D$ is a simple module, and $\text{End}_D(D)\cong D$. This allows us to break Schur's Lemma two different ways. If $D$ is infinite-dimensional and $k$ algebraically closed, the endomorphism ring $\text{End}_D(D)$ will also be infinite dimensional over $k$, hence not isomorphic to $k$. We can easily come up with such $D$, even commutative examples. For instance, $k(x)$ will be an infinite dimensional division algebra over $k$. On the other hand, if $k$ is not algebraically closed, we can take $D$ to be a finite field extension, and we get a $\text{End}_D(D)\cong D$ not isomorphic to $k$.</p>
|
23,502 | <p><em>Edit: I wrote the following question and then immediately realized an answer to it, and moonface gave the same answer in the comments. Namely, $\mathbb C(t)$, the field of rational functions of $\mathbb C$, gives a nice counterexample. Note that it is of dimension $2^{\mathbb N}$.</em></p>
<p>The following is one statement of Schur's lemma:</p>
<blockquote>
<p>Let $R$ be an associative unital algebra over $\mathbb C$, and let $M$ be a simple $R$-module. Then ${\rm End}_RM = \mathbb C$.</p>
</blockquote>
<p>My question is: are there extra conditions required on $R$? In particular, how large can $R$ be?</p>
<p>In particular, the statement is true when $\dim_{\mathbb C}R <\infty$ and also when $R$ is countable-dimensional. But I have been told that the statement fails when $\dim_{\mathbb C}R$ is sufficiently large.</p>
<p>How large must $\dim_{\mathbb C}R$ be to break Schur's lemma? I am also looking for an explicit example of Schur's lemma breaking for $\dim_{\mathbb C}R$ sufficiently large?</p>
| Victor Protsak | 5,740 | <p>It all depends on what you call "Schur's Lemma". If M is a simple module over a ring R then D=End<sub>R</sub>M is always a division ring (think of it as a weak Schur's lemma). The question is, can the endomorphism ring be pinned down more concretely, for example, if R is an algebra over a field k? In the usual Schur's lemma for finite groups, k=C and D=C. More general versions of Schur's lemma assert that D is algebraic over k (so D=k if k is algebraically closed).</p>
<p>If R is an affine (i.e. finitely-generated) commutative algebra over a field k and I is a maximal ideal of R then Hilbert's Nullstellensatz is asserting that R/I is algebraic over k. Since M=R/I is a simple module and R/I=End<sub>R</sub>M, you may interpret the statement as a version of Schur's lemma: R/I=End<sub>R</sub>M is algebraic over k. It is also known that every prime ideal of R is an intersection of maximal ideals.</p>
<p>Now if R is a noncommutative algebra over k one can ask whether the analogous properties hold: the endomorphism ring of a simple module is algebraic over k ("endomorphism property", implying the usual Schur's lemma when k is algebraically closed) and every prime ideal of R is an intersection of primitive ideals ("R has radical property"). This is true in a range of situations and constitutes <em>noncommutative Nullstellensatz</em>. Duflo proved that the endomorphism property for R[x] implies Nullstellensatz (endomorphism + radical) for R. </p>
<ol>
<li><p>As Kevin pointed out, if the dimension of M over k is smaller than the cardinality of k then the endomorphism property holds for R, and by Duflo, full Nullstellensatz holds for R.</p></li>
<li><p>In general, some extra assumptions are necessary, but Nullstellensatz holds for the Weyl algebra A<sub>n</sub>(k), the universal enveloping algebra U(g) of a finite-dimensional Lie algebra g, and the group algebra k[G] of a polycyclic by finite group. (The first two cases use Quillen's lemma also mentioned by Kevin). </p></li>
</ol>
<p>This is described in chapter 9 of McConnell and Robson, Noncommutative Noetherian rings. </p>
|
2,662,033 | <p>My question is how to prove that $(X,d)$ is complete if and only if $(X,d')$ is complete.</p>
<p>I have that $d$ and $d'$ are strongly equivalent metrics and I have used this to show that a sequence $x_{n}$ is Cauchy in $(X,d)$ if and only if it is Cauchy in $(X,d')$.</p>
<p>I have the definition of complete as:
"A metric space $X$ is complete if every Cauchy sequence in $X$ is convergent in $X$."</p>
<p>Since this is an if and only if statement I know I need to prove it both ways. </p>
<p>I am wondering if you also use the definition of strongly equivalent metrics to prove the completeness of the metrics or if you need to prove the convergence of the Cauchy sequence. A bit confused on how to approach this.</p>
| José Carlos Santos | 446,262 | <p><strong>Hint:</strong> You are assuming that $d$ and $d'$ are strongly equivalent, right?! Now, given a sequence $(x_n)_{n\in\mathbb N}$ of elements of $X$, prove that</p>
<ol>
<li>if $x\in X$, then $(x_n)_{n\in\mathbb N}$ converges to $x$ in $(X,d)$ if and only if it converges to $x$ in $(X,d')$;</li>
<li>$(x_n)_{n\in\mathbb N}$ is a Cauchy sequence in $(X,d)$ if and only if it is a Cauchy sequence in $(X,d')$.</li>
</ol>
|
3,318,130 | <p>Let <span class="math-container">$(a_j)_{j \in \mathbb{N}}$</span> and <span class="math-container">$(b_j)_{j \in \mathbb{N}}$</span> two real valued sequences such that <span class="math-container">$a_j \nearrow +\infty$</span> and <span class="math-container">$b_j \nearrow +\infty$</span>.
Is it possible to extract subsequences <span class="math-container">$(a_{j_i})_i$</span> and <span class="math-container">$(b_{j_i})_i$</span> such that <span class="math-container">$(\frac{a_{j_i}}{b_{j_i}})_i$</span> converges to a positive finite number? </p>
<p>Any help will be very apreciated! Thanks!!</p>
| José Carlos Santos | 446,262 | <p>Not necessarily. Suppose that <span class="math-container">$a_j=j^2$</span> and that <span class="math-container">$b_j=j$</span>. Then <span class="math-container">$\frac{a_j}{b_j}=j$</span> for each <span class="math-container">$j\in\mathbb N$</span>.</p>
|
365,808 | <p>Sorry if something like this has already been asked, I searched but I couldn't find anything similar to my question.</p>
<p>I'm a senior undergraduate and currently doing my senior thesis. My senior thesis is not original work, however it's quite demanding and I'm learning a lot of high level topics. I have been lurking around arxiv and started reading "Solved and unsolved problems in Number theory" by Daniel Shanks. My plan is to work on some open problems and play around with them so that I can try to get a publication before I graduate. My main reason for trying to get a publication is to increase my chances to get into a good graduate program (my GPA is not that great and I don't have the money to apply to many programs, so unless I publish something I'll probably only apply to safety schools).</p>
<p>With that being said if I were to do original work, how would I go about publishing? I might end up modifying a problem too much and proving something that might not be interesting, so I feel it'll get rejected from a journal for not being profound. I will also attack problems with all I know, so I might also end up using some heavy tools that aren't part of an undergraduate curriculum so I don't if i would send them to an undergraduate research journal. Maybe I could just upload on dropbox or arxiv, but then it's not a publication.</p>
<p>I have thought about asking my advisors about this, but I'll rather not since I'm aware I'm probably being overly ambitious and should probably focus on my thesis instead. Which I can agree with, hence I'll probably play around with problems on the weekends only or once a week. I'm also aware I might end up not publishing anything all, however in my mind unless I give it a shot I won't know. Either way I'll have fun and end up learning a lot about research so I don't see a downside.</p>
<p>(In case my background is relevant, my senior thesis is about perfectoid spaces. I've already taken a graduate course on commutative algebra, have taken a basic course on p-adic analysis,started learning about point free topology, already know the basics of category theory, still learning more about algebraic geometry, will learn about adic spaces soon/already know a bit about krull valuations, learning about homological algebra through weibel's book, started reading szamuely's galois theory book, will have to learn about etale cohomology soon, will also learn some things from almost mathematics, etc.)</p>
| David White | 11,540 | <p>I'm adding this answer in response to Annie Lee's question, because it's too long to fit in a comment.</p>
<p>Publishing as a grad student should definitely be done in consultation with an advisor. Unlike publishing as an undergraduate, a grad student's first papers serve as their introduction to the experts in their field, and largely determine whether they get a good postdoc. For this reason, it's very wise to have papers on arxiv before hitting the job market, but in many fields it's ok if these papers haven't been published yet (as long as the advisor's letter certifies that the work is solid). It's important not to put out anything shoddy, because your reputation really matters at that stage of your career. There's also the danger of getting scooped, e.g., if your PhD thesis has two parts, and you post the first part to arxiv a full year before the second, someone else might come along and do the second part before you can. An advisor will help you determine how likely this is to happen, and the pros and cons of advertising your work before it's fully finished. That's an important conversation anyway, so you can determine how much to say at conferences. An advisor can also talk to other experts in the field to encourage them not to try to prove what you are trying to prove in your thesis.</p>
<p>All that said, I think it's very valuable for grad students to have an early introduction to the world of publishing, as I previously wrote <a href="https://mathoverflow.net/a/218269/11540">here</a>. If you've never submitted a paper, responded to a referee report, etc. this can be anxiety inducing to do during your first job, while trying to juggle a million other things. Also, having an actual publication shows postdocs that you have what it takes to see a project through to completion. I was fortunate to write a paper as a second year grad student, in a totally different area than my main research (homotopy theory), and this gave me both confidence to get me through the hard times during my PhD research, and experience in choosing a journal, choosing an editor, when to keep polishing and when to submit, etc. Pro tip: the time to submit is almost always sooner than you think, because some degree of polishing can be done after the referee report, and the referee will always have suggestions for things to change (so, sending something too perfect can lead the referee to suggest big changes instead of obvious small changes).</p>
<p>In order to help give grad students this important experience with early publications, I joined the editorial board of the <a href="https://www.gradmath.org/" rel="noreferrer">Graduate Journal of Mathematics</a>, as I wrote about <a href="https://mathoverflow.net/a/314202/11540">here</a>. If you work out some interesting result or new example, and publishing it separately won't harm your PhD thesis, please feel free to submit to our journal! I think it's the only one of its kind, unlike the plethora of undergrad journals I mentioned in my other answer. To submit, you just pick an editor on the editorial board and email the pdf.</p>
|
4,049,293 | <p>I am learning about the cross entropy, defined by Wikipedia as
<span class="math-container">$$H(P,Q)=-\text{E}_P[\log Q]$$</span>
for distributions <span class="math-container">$P,Q$</span>.</p>
<p>I'm not happy with that notation, because it implies symmetry, <span class="math-container">$H(X,Y)$</span> is often used for the joint entropy and lastly, I want to use a notation which is consistent with the notation for entropy:
<span class="math-container">$$H(X)=-\text{E}_P[\log P(X)]$$</span></p>
<p>When dealing with multiple distributions, I like to write <span class="math-container">$H_P(X)$</span> so it's clear with respect to which distribution I'm taking the entropy. When dealing with multiple random variables, it thinks it's sensible to make precise the random variable with respect to which the expectation is taken by using the subscript <span class="math-container">$_{X\sim P}$</span>. My notation for entropy thus becomes
<span class="math-container">$$H_{X\sim P}(X)=-\text{E}_{X\sim P}[\log P(X)]$$</span></p>
<p>Now comes the point I don't understand about the definition of cross entropy: Why doesn't it reference a random variable <span class="math-container">$X$</span>? Applying analogous reasoning as above, I would assume that cross entropy has the form <span class="math-container">\begin{equation}H_{X\sim P}(Q(X))=-\text{E}_{X\sim P}[\log Q(X)]\tag{1}\end{equation}</span>
however, Wikipedia makes no mention of any such random variable <span class="math-container">$X$</span> in the article on cross entropy. It speaks of</p>
<blockquote>
<p>the cross-entropy between two probability distributions <span class="math-container">$p$</span> and <span class="math-container">$q$</span></p>
</blockquote>
<p>which, like the notation <span class="math-container">$H(P,Q)$</span>, implies a function whose argument is a pair of distributions, whereas entropy <span class="math-container">$H(X)$</span> is said to be a function of a random variable. In any case, to take an expected value I need a (function of) a random variable, which <span class="math-container">$P$</span> and <span class="math-container">$Q$</span> are not.</p>
<p>Comparing the definitions for the discrete case:
<span class="math-container">$$H(p,q)=-\sum_{x\in\mathcal{X}}p(x)\log q(x)$$</span>
and
<span class="math-container">$$H(X)=-\sum_{i=1}^n P(x_i)\log P(x_i)$$</span></p>
<p>where <span class="math-container">$\mathcal{X}$</span> is the support of <span class="math-container">$P$</span> and <span class="math-container">$Q$</span>, there would only be a qualitative difference if the events <span class="math-container">$x_i$</span> didn't cover the whole support (though I could just choose an <span class="math-container">$X$</span> which does).</p>
<p>My questions boil down to the following:</p>
<ol>
<li><p>Where is the random variable necessary to take the expected value which is used to define the cross entropy <span class="math-container">$H(P,Q)=-\text{E}_{P}[\log Q]$</span></p>
</li>
<li><p>If I am correct in my assumption that one needs to choose a random variable <span class="math-container">$X$</span> to compute the cross entropy, is the notation I used for (1) free of ambiguities.</p>
</li>
</ol>
| robjohn | 13,854 | <p><strong>Fermat's Little Theorem</strong></p>
<p>In particular, this polynomial admits a lot of simplification.
<span class="math-container">$$
\begin{align}
3n^5+5n^3+7n
&\equiv2n^3+n&\pmod3\tag{1a}\\
&\equiv2n+n&\pmod3\tag{1b}\\
&\equiv0&\pmod3\tag{1c}
\end{align}
$$</span>
Explanation:<br />
<span class="math-container">$\text{(1a)}$</span>: reduce coefficients <span class="math-container">$\bmod3$</span><br />
<span class="math-container">$\text{(1b)}$</span>: apply <a href="https://en.wikipedia.org/wiki/Fermat%27s_little_theorem" rel="nofollow noreferrer">Fermat's Little Theorem</a><br />
<span class="math-container">$\text{(1c)}$</span>: <span class="math-container">$3n\equiv0\pmod3$</span></p>
<p><span class="math-container">$$
\begin{align}
3n^5+5n^3+7n
&\equiv3n^5+2n&\pmod5\tag{2a}\\
&\equiv3n+2n&\pmod5\tag{2b}\\
&\equiv0&\pmod5\tag{2c}
\end{align}
$$</span>
Explanation:<br />
<span class="math-container">$\text{(2a)}$</span>: reduce coefficients <span class="math-container">$\bmod5$</span><br />
<span class="math-container">$\text{(2b)}$</span>: apply <a href="https://en.wikipedia.org/wiki/Fermat%27s_little_theorem" rel="nofollow noreferrer">Fermat's Little Theorem</a><br />
<span class="math-container">$\text{(2c)}$</span>: <span class="math-container">$5n\equiv0\pmod5$</span></p>
<p>Thus, <span class="math-container">$3n^5+5n^3+7n\equiv0\pmod{15}$</span>.</p>
<hr />
<p><strong>Binomial Polynomials</strong></p>
<p>In general, any polynomial that takes only integer values can be written as an integer combination of <a href="https://math.stackexchange.com/a/62813">binomial polynomials</a>:
<span class="math-container">$$
\frac{3n^5+5n^3+7n}{15}=24\binom{n}{5}+48\binom{n}{4}+32\binom{n}{3}+8\binom{n}{2}+\binom{n}{1}\tag3
$$</span></p>
<hr />
<p><strong>The Computatation of the Coefficients</strong></p>
<p>For a degree <span class="math-container">$d$</span> polynomial <span class="math-container">$P$</span>,
<span class="math-container">$$
P(n)=\sum_{k=0}^dc_k\binom{n}{k}\tag4
$$</span>
We can compute <span class="math-container">$c_n$</span> inductively, using
<span class="math-container">$$
P(n)-\overbrace{\sum_{k=0}^{n-1}c_k\binom{n}{k}}^{\substack{\text{using the $c_k$}\\\text{computed}\\\text{previously}}}=c_n\overset{\substack{1\\[3pt]\downarrow\\[3pt]\,}}{\binom{n}{n}}+\overbrace{\sum_{k=n+1}^dc_k\binom{n}{k}}^{\substack{\text{the binomial}\\\text{coefficients}\\\text{are $0$}}}\tag5
$$</span></p>
<p><span class="math-container">$c_0=P(0):$</span>
<span class="math-container">$$
c_0=\frac{3\cdot0^5+5\cdot0^3+7\cdot0}{15}=0
$$</span>
<span class="math-container">$c_1=P(1)-c_0\binom{1}{0}:$</span>
<span class="math-container">$$
c_1=\frac{3\cdot1^5+5\cdot1^3+7\cdot1}{15}-0\binom{1}{0}=1
$$</span>
<span class="math-container">$c_2=P(2)-c_0\binom{2}{0}-c_1\binom{2}{1}:$</span>
<span class="math-container">$$
c_2=\frac{3\cdot2^5+5\cdot2^3+7\cdot2}{15}-0\binom{2}{0}-1\binom{2}{1}=8
$$</span>
<span class="math-container">$c_3=P(3)-c_0\binom{3}{0}-c_1\binom{3}{1}-c_2\binom{3}{2}:$</span>
<span class="math-container">$$
c_3=\frac{3\cdot3^5+5\cdot3^3+7\cdot3}{15}-0\binom{3}{0}-1\binom{3}{1}-8\binom{3}{2}=32
$$</span>
<span class="math-container">$c_4=P(4)-c_0\binom{4}{0}-c_1\binom{4}{1}-c_2\binom{4}{2}-c_3\binom{4}{3}:$</span>
<span class="math-container">$$
c_4=\frac{3\cdot4^5+5\cdot4^3+7\cdot4}{15}-0\binom{4}{0}-1\binom{4}{1}-8\binom{4}{2}-32\binom{4}{3}=48
$$</span>
<span class="math-container">$c_5=P(5)-c_0\binom{5}{0}-c_1\binom{5}{1}-c_2\binom{5}{2}-c_3\binom{5}{3}-c_4\binom{5}{4}:$</span>
<span class="math-container">$$
c_5=\frac{3\cdot5^5+5\cdot5^3+7\cdot5}{15}-0\binom{5}{0}-1\binom{5}{1}-8\binom{5}{2}-32\binom{5}{3}-48\binom{5}{4}=24
$$</span>
Thus, the converted polynomial is
<span class="math-container">$$
\frac{3n^5+5n^3+7n}{15}=0\binom{n}{0}+1\binom{n}{1}+8\binom{n}{2}+32\binom{n}{3}+48\binom{n}{4}+24\binom{n}{5}
$$</span></p>
|
3,841,806 | <p>Using spherical coordinates I have to find the volume of a cone <span class="math-container">$z=\sqrt{x^2+y^2}$</span> inscribed in a sphere <span class="math-container">$(x-1)^2+y^2+z^2=4.$</span></p>
<p>I can`t find <span class="math-container">$\rho$</span> because the center of sphere is displaced from the origin.</p>
<hr />
<p>I tried solving it using Mathematica, but i did something wrong somewhere
<a href="https://i.stack.imgur.com/CtRvq.png" rel="nofollow noreferrer">enter image description here</a></p>
| David G. Stork | 210,401 | <p>Not an answer (so please don't downvote), but a figure that should help you visualize the problem:</p>
<p><a href="https://i.stack.imgur.com/IWExy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IWExy.png" alt="enter image description here" /></a></p>
<p>Clearly you want to integrate over <span class="math-container">$\theta$</span> from <span class="math-container">$0 \to 2 \pi$</span>, and <span class="math-container">$\phi$</span> from <span class="math-container">$0 \to \phi_0$</span>, where the limit is based on the half angle of the cone.</p>
<p>The radius <span class="math-container">$\rho$</span> is a function of both these variables, and can be derived from the Pythagorean theorem.</p>
|
3,841,806 | <p>Using spherical coordinates I have to find the volume of a cone <span class="math-container">$z=\sqrt{x^2+y^2}$</span> inscribed in a sphere <span class="math-container">$(x-1)^2+y^2+z^2=4.$</span></p>
<p>I can`t find <span class="math-container">$\rho$</span> because the center of sphere is displaced from the origin.</p>
<hr />
<p>I tried solving it using Mathematica, but i did something wrong somewhere
<a href="https://i.stack.imgur.com/CtRvq.png" rel="nofollow noreferrer">enter image description here</a></p>
| Doug M | 317,162 | <p><span class="math-container">$x=\rho\cos\theta\sin\phi\\
y=\rho\sin\theta\sin\phi\\
z=\rho\cos\phi$</span></p>
<p>Plug these substitutions into the given equations.</p>
<p><span class="math-container">$x^2 - 2x + 1 + y^2 + z^2 = 4$</span></p>
<p><span class="math-container">$\rho^2 - 2\rho\cos\theta\sin\phi - 3 = 0$</span></p>
<p>We have <span class="math-container">$\rho$</span> as a quadratic, so use the quadratic formula.</p>
<p><span class="math-container">$\rho = \cos\theta\sin\phi +\sqrt {\cos^2\theta\sin^2\phi +3}$</span></p>
|
1,534,694 | <p>I tried to solve for the following limit: </p>
<p>$$\lim_{x\rightarrow \infty} (e^{2x}+x)^{1/x}$$
and I reached to the indeterminate form:
$${4e^{2x}}\over {4e^{2x}}$$
if I plug in, I will get another indeterminate form! </p>
| Claude Leibovici | 82,404 | <p>Another way, considering $$A=(e^{2x}+x)^{1/x}$$ Taking logarithms $$\log(A)=\frac 1x \log(e^{2x}+x)=\frac 1x \left(\log(e^{2x})+\log(1+\frac x {e^{2x}})\right)=\frac 1x \left(2x+\log(1+\frac x {e^{2x}})\right)$$ $$\log(A)=2+\frac 1x \log(1+\frac x {e^{2x}})\approx 2+\frac 1x \frac x {e^{2x}}=2+\frac 1 {e^{2x}}$$</p>
|
1,138,212 | <p>I am given $f(x) = 1 + x - \frac{sin(x)}{(x e^x)} $ and am asked to solve this for when x ≃ 0.</p>
<p>I'm doing the following steps but am getting stuck halfway through:</p>
<p>$$f(x) = 1 + x - \frac {x - \frac{x^3}{6} + \frac{x^5}{120}}{xe^x} $$</p>
<p>$$= 1 + x - \frac{e^{-x} (x - \frac{x^3}{6} + \frac{x^5}{120})}{x} $$</p>
<p>$$= 1 + x - \frac {(1 - x + \frac{x^2}{2} - \frac{x^3}{3!} + \frac{x^4}{4!}) (x - \frac{x^3}{6} + \frac{x^5}{120})}{x} $$</p>
<p>At this point however I'm not really sure what I should be doing next. I figure that multiplying the numerator part is not necessary, so I don't see what else to do. Could someone provide me with a hint on how to finish this?</p>
| abel | 9,252 | <p>i will take it from your last step.<br>
$\begin{align} 1 + x - \frac{e^{-x} \sin x}{x} &= 1 + x -
\left(1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots \right)
\left( 1 - \frac{x^2}{3!} + \frac{x^4}{5!} + \cdots\right)\\
&=1 + x -
\left(1 - x + x^2(\frac{1}{2!} - \frac{1}{3!}) + \cdots \right)\\
&= 2x - \frac{1}{3}x^2 + \cdots
\end{align}$</p>
|
1,138,212 | <p>I am given $f(x) = 1 + x - \frac{sin(x)}{(x e^x)} $ and am asked to solve this for when x ≃ 0.</p>
<p>I'm doing the following steps but am getting stuck halfway through:</p>
<p>$$f(x) = 1 + x - \frac {x - \frac{x^3}{6} + \frac{x^5}{120}}{xe^x} $$</p>
<p>$$= 1 + x - \frac{e^{-x} (x - \frac{x^3}{6} + \frac{x^5}{120})}{x} $$</p>
<p>$$= 1 + x - \frac {(1 - x + \frac{x^2}{2} - \frac{x^3}{3!} + \frac{x^4}{4!}) (x - \frac{x^3}{6} + \frac{x^5}{120})}{x} $$</p>
<p>At this point however I'm not really sure what I should be doing next. I figure that multiplying the numerator part is not necessary, so I don't see what else to do. Could someone provide me with a hint on how to finish this?</p>
| Jack D'Aurizio | 44,121 | <p>Since in a neighbourhood of the origin we have:
$$\frac{\sin x}{x}=1-\frac{x^2}{6}+o(x^2),\qquad e^{-x}=1-x+\frac{x^2}{2}+o(x^2)$$
it happens that:
$$\frac{\sin x}{x e^x}=1-x+\frac{x^2}{3}+o(x^2) $$
so:
$$ f(x) = 1+x-\frac{\sin x}{x e^x}\approx 2x-\frac{x^2}{3}$$
and provided that $g(x)$ is the inverse function of $f(x)$,
$$ g(x)\approx \frac{x}{2}+\frac{x^2}{24}. $$</p>
|
1,290,176 | <p>Can anybody help me with this limit?
I think the answer should be $0$ as $0$ to the power $1$ should be $0$ but it doesn't match with the book's answer.</p>
<p>$$ \lim_{x\to 0} |x|^{\lfloor\cos{x}\rfloor}$$</p>
| marwalix | 441 | <p>So the edit is right and $\lfloor\cos{x}\rfloor=0$ for $x\neq 0$ in a small neighborhood of $0$ and $\lfloor\cos{0}\rfloor=1$.</p>
<p>One has $|x|^{\lfloor\cos{x}\rfloor}=1$ is a constant function for $x\neq 0$ and the limit you're looking for is $1$</p>
|
2,878,777 | <p>Usually mathematicians consider isomorphic fields as equal fields. That is, if the $(A,+,\cdot)$ is isomorphic to $(B,\oplus,\odot)$, then I can consider those fields as equals. Thinking about it, I thought about the following interpretation:</p>
<p>Let $A$ and $B$ be two sets. I think we can interpret that $A$ and $B$, considered only as sets, are the same set if there is a bijection $\varphi:A\to B$ between them, because, just as the apparently distinct sets $\{1,\, 2,\, 3,\, \cdots\}$ and $\{I,\, II,\, III,\, \cdots\}$ can represent the set of natural numbers, $A$ and $B$, if there is a bijection between them, can be understood as distinct representations of the same set.</p>
<p>One can argue that this interpretation is wrong since there is a bijection between $\mathbb{N}$ and $\mathbb{Q}$ and these sets are clearly distinct since the first one has the well-ordering principle the second one doesn't. However $\mathbb{N}$ and $\mathbb{Q}$ are not simply sets, but an algebraic structures. For example, $\mathbb{N}$ is actually a triple $(\mathbb{N},+,\cdot)$ in which $\mathbb{N}=\{1,\, 2,\,3,\, \cdots\}$ and $+$ and $\cdot$ are binary operations defined in $\mathbb{N}$ with which we can build a well-ordering $\leq $ order in $\mathbb{N}$. So, as algebraic structures, $\mathbb{N}$ and $\mathbb{Q}$ are indeed distinct, but as simply sets we can consider them the same, because since there is an bijection $f:\mathbb{N}\to\mathbb{Q}$ I can represent all elements of $\mathbb{Q}$ with the symbols $\{1,\, 2,\, 3,\, \cdots\}$. To do this, I can simply represent an element $q\in\mathbb{Q}$ by an $n\in \{1,\, 2,\, 3,\, \cdots\}$ that satisfies $f(n)=q$. </p>
<p>If we think well it is not possible to infer that $\mathbb{N}=\{1,\, 2,\,3,\, \cdots\}$ from the axioms of Peano. What happens is that we use the symbols $\{1,\, 2,\, 3,\, \cdots\}$ to represent the natural numbers which is the same attitude as representing $q\in\mathbb{Q}$ by an $n\in \{1,\, 2,\, 3,\, \cdots\}$ satisfying $f(n)=q$. Therefore, if we consider $\mathbb{N}$ and $\mathbb{Q}$ only as sets (and not as algebraic structures), we can say that $\mathbb{N}=\mathbb{Q}$ since it is possible to represent $\mathbb{Q}$ with the same symbols used to represent the elements of $\mathbb{N}$.</p>
<p><strong>My question is:</strong> can we interpret sets of the same cardinality as <strong>distinct representations of the same set</strong>? Because if the answer is affirmative then it becomes easier to understand why we can consider isomorphic fields or isomorphic groups as equals. Thinking about groups I think it is also possible to think that if the groups $(G,\cdot)$ and $(L,\odot)$ are isomorphic, then $(G,\cdot)$ and $(L,\odot)$ are distinct representations of the same group.</p>
<p><strong>EDIT:</strong> I think I've found a way to express myself better. I can say that $\mathbb{Q}=\{1,2,3,\cdots\} $ in which the element $5$, for instance, is understood as the rational $q$ satisfying $q=f(5)$ ($f:\mathbb{N}\to\mathbb{Q}$ is a bijection). Therefore, treating $\mathbb{Q}$ and $\mathbb{N}$ only as sets (and not as algebraic structures) I believe it makes sense to say $\mathbb{Q}=\mathbb{N}=\{1,2,3,\cdots\}$.</p>
| Mohammad Riazi-Kermani | 514,496 | <p>Two sets are considered equal iff every element of the first set is an element of the second set and every element of the second set is an element of the first set.</p>
<p>Sets of the same Cardinality are not the same sets but they enjoy having a one to one correspondence between them.</p>
<p>You have answered your question when comparing the two sets , natural numbers and rational numbers.</p>
<p>They are not the same set. </p>
|
3,174,982 | <p>I tried for few primes but they do not satisfy Eisenstein criterion.Also is there any approach other than brute force with the help of which we can find that prime.</p>
| Community | -1 | <p>You can use the reducite criterium.</p>
<p>I.e., consider the polynomial over the field <span class="math-container">$\mathbb{F}_2$</span>.I.e., we get <span class="math-container">$X^3 + X^2 +1$</span>. If this reduced polynomial is irreducible over <span class="math-container">$\mathbb{F}_2$</span>, it is also irreducible over <span class="math-container">$\mathbb{Q}$</span> (this theorem is known as the reducite criterium).</p>
<p>This polynomial is irreducible over <span class="math-container">$\mathbb{F}_2$</span> (it has no roots, this is easily seen). </p>
<p>Hence, by the reducite criterium it follows that <span class="math-container">$X^3 + X^2 - 2X -1$</span> is irreducible in <span class="math-container">$\mathbb{Q}[X]$</span>.</p>
|
178,302 | <p>Assume that $H$ is a separable Hilbert space.
Is there a polynomial $p(z)\in \mathbb{C}[x]$ with $deg(p)>1$ with the following property?:</p>
<p>Every densely defined operator $A:D(A)\to D(A),\;D(A)\subset H$ with $p(A)=0$ is necessarily a bounded operator on $H$.</p>
<p>That is the polynomial-operator equation $p(A)=0$ has only bounded solution.</p>
| Robert Israel | 13,650 | <p>You mean an infinite-dimensional separable Hilbert space. The answer is no.</p>
<p>Suppose $p(z)$ has distinct roots $\alpha_1, \alpha_2$. Define a sequence $x_1, x_2, \ldots$ in the unit sphere of $H$ such that </p>
<ol>
<li>$x_1,\ldots, x_n$ are linearly independent for all $n$.</li>
<li>$\|x_i - x_{i+1}\| \to 0$ as $i \to \infty$.</li>
<li>the sequence is dense in the unit sphere of $H$.</li>
</ol>
<p>Define $A$ on the linear span of the sequence so that $A x_i = \alpha_1 x_i$ if $i$ is odd, $\alpha_2 x_i$ otherwise.</p>
<p>On the other hand, if $p$ has only one root, say $p(z) = (z - \alpha)^d$, then
with the same sequence as above take $A x_i = \alpha x_i + x_{i+1}$ for $i$ not divisible by $d$, $\alpha x_i$ otherwise.</p>
|
124,280 | <p>Show that the sequence ($x_n$) defined by $$x_1=1\quad \text{and}\quad x_{n+1}=\frac{1}{x_n+3} \quad (n=1,2,\ldots)$$ converges and determine its limit ? </p>
<p>I try to show ($x_n$) is a Cauchy sequence or ($x_n$) is decreasing (or increasing) and bounded sequence but I fail every step of all.</p>
| David Mitra | 18,986 | <p>You can easily show that $(x_n)$ is bounded below by 0 and bounded above by 1.</p>
<p>You can then show (by induction e.g.) that $(x_{2n})$ is decreasing and that $(x_{2n+1})$ is increasing. Then you can argue that the sequence $(x_{2n})$ converges to some number $L$ and that the sequence $(x_{2n+1})$ converges to some number $M$. </p>
<p>Now, since $x_{n+1} ={1-x_{n+1} x_n\over3}$, it follows that $(x_n)$ converges, to $b$, say. Then from the recursion formula, we must have $b={1\over b+3}$; solving this equation we see that $b$ is its positive solution $b={-3\over2}+{\sqrt{13}\over2}$.</p>
<p><br> </p>
<p>For the induction argument to show that $(x_{2n})$ is decreasing and $(x_{2n+1})$ is increasing:</p>
<p>Verify that $x_1<x_3$ and that $x_2>x_4$.</p>
<p>Assume that both $x_{2n-1}<x_{2n+1}$ and $x_{2n-2}>x_{2n}$ hold.</p>
<p>Then show that $x_{2n+2}<x_{2n}$. Using this result, show that $x_{2n+3}>x_{2n+1}$.</p>
|
912,426 | <p>A bag contains six chips, numbered 1 through 6. If two chips are chosen at random without replacement and the values on those two chips are multiplied, what is the probability that this product will be greater than 20?</p>
<p>I tried to solve by counting the total possibilities (36) and solving for 6 choices that worked, e.g. $4x6, 5x5, 5x6, 6x4,6x5,6x6$... so I thought the probability would be $1/6$. </p>
<p>How is my method incorrect?</p>
| Harald Hanche-Olsen | 23,290 | <p>Take logarithms:
$$\ln\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}=\sum_{k=1}^n\ln\frac{2k-1}{2k}=\sum_{k=1}^n\ln\Bigl(1-\frac1{2k}\Bigr)$$
and use $$\ln(1-x)=-x+O(x^2)$$ together with the divergence of the harmonic series.</p>
|
3,684,799 | <p>When the theory of groups is built up from its axioms, it is often necessary to establish very simple results such as</p>
<p><span class="math-container">$ax = xa \Longrightarrow a^{-1}x = xa^{-1}. \tag 1$</span></p>
<p>Thus we ask how the title question might be proved.</p>
| Disintegrating By Parts | 112,478 | <p>If <span class="math-container">$T$</span> is self-adjoint and <span class="math-container">$T^2x=0$</span>, then <span class="math-container">$Tx=0$</span> because
<span class="math-container">$$
\|Tx\|^2=\langle T^2 x,x\rangle = 0.
$$</span>
Therefore, if <span class="math-container">$T^k=0$</span> for some <span class="math-container">$k > 1$</span>, then <span class="math-container">$T=0$</span>. This doesn't rely on <span class="math-container">$T$</span> being compact, but it does rely on <span class="math-container">$T$</span> being self-adjoint.</p>
|
38,193 | <p>For simplicity, let me pick a particular instance of Gödel's Second Incompleteness
Theorem:</p>
<p>ZFC (Zermelo-Fraenkel Set Theory plus the Axiom of Choice, the usual foundation of mathematics) does not prove Con(ZFC), where Con(ZFC) is a formula that expresses that
ZFC is consistent.</p>
<p>(Here ZFC can be replaced by any other sufficiently good, sufficiently strong set of axioms,
but this is not the issue here.)</p>
<p>This theorem has been interpreted by many as saying "we can never know whether mathematics is consistent" and has encouraged many people to try and prove that ZFC (or even PA) is in fact inconsistent. I think a mainstream opinion in mathematics (at least among mathematician who think about foundations) is that we believe that there is no problem with
ZFC, we just can't prove the consistency of it.</p>
<p>A comment that comes up every now and then (also on mathoverflow), which I tend to agree with, is this:</p>
<p>(*) "What do we gain if we could prove the consistency of (say ZFC) inside ZFC? If ZFC were inconsistent, it would prove its consistency just as well."</p>
<p>In other words, there is no point in proving the consistency of mathematics by a mathematical proof, since if mathematics were flawed, it would prove anything, for instance its own non-flawedness.
Hence such a proof would not actually improve our trust in mathematics (or ZFC, following the particular instance).</p>
<p>Now here is my question: Does the observation (*) imply that the only advantage of the Second Incompleteness Theorem over the first one is that we now have a specific sentence
(in this case Con(ZFC)) that is undecidable, which can be used to prove theorems like
"the existence of an inaccessible cardinal is not provable in ZFC"?
In other words, does this reduce the Second Incompleteness Theorem to a mere technicality
without any philosophical implication that goes beyond the First Incompleteness Theorem
(which states that there is some sentence <span class="math-container">$\phi$</span> such that neither <span class="math-container">$\phi$</span> nor <span class="math-container">$\neg\phi$</span> follow from ZFC)?</p>
| Henry Towsner | 8,991 | <p>While it's not directly a philosophical benefit, the Second Incompleteness Theorem is quite useful for giving concrete unprovability results: if we want to prove that theory T does not prove theorem X, it suffices to show that X implies the consistency of T. For instance, Harvey Friedman has a number of results showing that some theorem implies the well-foundedness of some ordinal notation, where the ordinal notation, in turn, is known to imply the consistency (indeed, 1-consistency) of the theory.</p>
|
38,193 | <p>For simplicity, let me pick a particular instance of Gödel's Second Incompleteness
Theorem:</p>
<p>ZFC (Zermelo-Fraenkel Set Theory plus the Axiom of Choice, the usual foundation of mathematics) does not prove Con(ZFC), where Con(ZFC) is a formula that expresses that
ZFC is consistent.</p>
<p>(Here ZFC can be replaced by any other sufficiently good, sufficiently strong set of axioms,
but this is not the issue here.)</p>
<p>This theorem has been interpreted by many as saying "we can never know whether mathematics is consistent" and has encouraged many people to try and prove that ZFC (or even PA) is in fact inconsistent. I think a mainstream opinion in mathematics (at least among mathematician who think about foundations) is that we believe that there is no problem with
ZFC, we just can't prove the consistency of it.</p>
<p>A comment that comes up every now and then (also on mathoverflow), which I tend to agree with, is this:</p>
<p>(*) "What do we gain if we could prove the consistency of (say ZFC) inside ZFC? If ZFC were inconsistent, it would prove its consistency just as well."</p>
<p>In other words, there is no point in proving the consistency of mathematics by a mathematical proof, since if mathematics were flawed, it would prove anything, for instance its own non-flawedness.
Hence such a proof would not actually improve our trust in mathematics (or ZFC, following the particular instance).</p>
<p>Now here is my question: Does the observation (*) imply that the only advantage of the Second Incompleteness Theorem over the first one is that we now have a specific sentence
(in this case Con(ZFC)) that is undecidable, which can be used to prove theorems like
"the existence of an inaccessible cardinal is not provable in ZFC"?
In other words, does this reduce the Second Incompleteness Theorem to a mere technicality
without any philosophical implication that goes beyond the First Incompleteness Theorem
(which states that there is some sentence <span class="math-container">$\phi$</span> such that neither <span class="math-container">$\phi$</span> nor <span class="math-container">$\neg\phi$</span> follow from ZFC)?</p>
| abo | 20,716 | <p>It is an open question whether what I have called Hilbert's ultrafinitist program is possible, that is whether a natural base theory can prove the consistency of natural stronger theories. Please see </p>
<p><a href="https://mathoverflow.net/questions/120258/is-an-ultrafinitist-hilberts-program-doomed">Is an ultrafinitist Hilbert's program doomed?</a></p>
<p>So in this sense the Second Incompleteness Theorem is not redundant: there could be natural theories which prove natural stronger theories consistent.</p>
<p>In any case I'm of the opinion that proving self-consistency is a good test for a theory; that is, if a theory can't even prove its own consistency, that is a good reason not to accept the theory.</p>
|
3,768,333 | <blockquote>
<p>Let <span class="math-container">$f: [a,b] \to R$</span> be a differentiable function of one variable such that <span class="math-container">$|f'(x)| \le 1$</span> for all <span class="math-container">$x\in [a,b]$</span>. Prove that <span class="math-container">$f$</span> is a contraction. (Hint: use MVT.) If in addition <span class="math-container">$|f'(x)| < 1$</span> for all <span class="math-container">$x \in [a,b]$</span> and <span class="math-container">$f'$</span> is continuous, show that <span class="math-container">$f$</span> is a strict contraction.</p>
</blockquote>
<p>Using MVT, <span class="math-container">$|f(x) - f(y)| = |f'(c)(x-y)| \le |x-y|$</span> for <span class="math-container">$c$</span> between <span class="math-container">$x$</span> and <span class="math-container">$y$</span>.</p>
<p>I don't know the proof for a strict contraction. I guess that I need to use the continuity of <span class="math-container">$f'$</span>, but I am not sure how to use it. Any help would be appreciated.</p>
| M Kupperman | 593,668 | <p>You're almost there! A (weak) contraction is defined as a function <span class="math-container">$f: A \to \mathbb{R}$</span> such that <span class="math-container">$$|f(x) - f(y)| \leq k | x - y| ~\forall x,y \in A,\:\:\: 0 \leq k \leq 1$$</span> (in your case <span class="math-container">$A = [a,b]$</span>.)</p>
<p>This might look a little familiar, as you have already concluded that <span class="math-container">$|f(x) - f(y)| < |x - y|$</span> (by removing the middle inequality). We just need to make sure that <span class="math-container">$k = |f(x)|$</span> meets the required bounds. Combine <span class="math-container">$0 \leq |f(x)|$</span> (a property of absolute value) with the given <span class="math-container">$|f(x)| \leq 1$</span> for all <span class="math-container">$x \in [a,b]$</span>, you can conclude <span class="math-container">$0 \leq |f(x)| \leq 1$</span>, so <span class="math-container">$1\geq k \geq |f(c)| \geq 0$</span> (by continuity and <span class="math-container">$c \in[a,b]$</span>).</p>
<p>So you have effectively shown that <span class="math-container">$f$</span> is a weak contraction.</p>
<p>Showing that if <span class="math-container">$|f(x)| < 1$</span>, then <span class="math-container">$f$</span> is a strict contraction follows a similar path. @bitesizebo is correct in noting that <span class="math-container">$f'$</span> continuous is essential here. A strict contraction replaces the weak inequality with a strong inequality. A function is a strict contradiction if
<span class="math-container">$$|f(x) - f(y)| < k | x - y| ~\forall x,y \in A,\:\:\: 0 \leq k \leq 1$$</span> (in your case <span class="math-container">$A = [a,b]$</span>.)</p>
<p>You know that <span class="math-container">$|f(x) - f(y) \leq |f(c)(x-y)|$</span> by MVT. You can separate the RHS of the equation and get <span class="math-container">$|f(c)(x-y)| = |f(c)||x-y|$</span>. Using the fact that <span class="math-container">$f'$</span> is continuous and <span class="math-container">$|f'(x)| < 1$</span>, you need to prove that <span class="math-container">$|f'(c)| < 1$</span> (level of detail here depends on your class), as @bitesizebo says.</p>
<p>There are a few ways to show that <span class="math-container">$f$</span> satisfies <span class="math-container">$\sup_{[a,b]} |f'(c)| < 1$</span>. My favorite method is a bit more broad than @Michael Hardy's approach and proves the Extreme Value Theorem along the way. I use prefer the 'fact' that the image of a compact set under a continuous function is also compact. This works in any topological space, not just <span class="math-container">$\mathbb{R}$</span>. So <span class="math-container">$f'([a,b])$</span> is a closed set, so it must attain all of it's limit points (the limit of any sequence of numbers in <span class="math-container">$f'([a,b])$</span> must also be a point in <span class="math-container">$f'([a,b])$</span>). So the infimum and supremum of <span class="math-container">$f'([a,b])$</span> must be obtained and be in the set. So as <span class="math-container">$|f'(c)| < 1$</span> for all <span class="math-container">$c \in[a,b]$</span>, you must have also hit the supremum and infimum of <span class="math-container">$f'([a,b])$</span>. So <span class="math-container">$\sup_{[a,b]}|f'(c)| < 1$</span>. You can fill in the algebraic details and inequalities.</p>
<p>Once you have <span class="math-container">$\sup_{[a,b]}|f'(c)| < 1$</span>, you can plug this in to the statement we got from the MVT, and conclude <span class="math-container">$|f(x) - f(y) < k |y-x|$</span> where <span class="math-container">$|f'(c)| < \sup_{[a,b]}|f'(c)| \leq k < 1$</span> (I've never seen the assumption <span class="math-container">$k>0$</span> used to define a strict contraction since it could give a 'strongest' possible contraction step, although I don't have my copy of Rudin handy), and conclude that <span class="math-container">$f$</span> is a contraction map.</p>
|
4,117,409 | <blockquote>
<p>Prove or disprove: if for every <span class="math-container">$n\in\Bbb{N}, |a_{n+1}-a_n|<\frac{1}{n^2}$</span> then <span class="math-container">$a_n$</span> converges.</p>
</blockquote>
<p>I think this is true, and tried using Cauchy's theorem - I take some <span class="math-container">$\varepsilon > 0$</span>. There exists such <span class="math-container">$N$</span> s.t <span class="math-container">$\varepsilon > \frac{1}{N^2}$</span>. So, for every <span class="math-container">$m,n>N$</span>, we get that:
<span class="math-container">$|a_m-a_n|\leq|a_m-a_{m-1}|+...+|a_{n+1}-a_n|<\frac{1}{m^2}+...+\frac{1}{n^2}<\frac{m-n}{N^2}$</span>. Now I think the righthand side tends to <span class="math-container">$0$</span>, but this feels like cheating. Am I correct or is there something I'm missing?</p>
| José Carlos Santos | 446,262 | <p>Take <span class="math-container">$\varepsilon>0$</span>. Since the series <span class="math-container">$\sum_{n=1}^\infty\frac1{n^2}$</span> converges, there is some <span class="math-container">$N\in\Bbb N$</span> such that <span class="math-container">$\sum_{n=N}^\infty\frac1{n^2}<\varepsilon$</span> . So, if <span class="math-container">$m\geqslant n\geqslant N$</span>,<span class="math-container">\begin{align}|a_m-a_n|&\leqslant\frac1{(m-1)^2}+\frac1{(m-2)^2}+\cdots+\frac1{n^2}\\&<\sum_{k=N}^\infty\frac1{k^2}\\&<\varepsilon.\end{align}</span>So, your sequence converges, since it is a Cauchy sequence.</p>
|
210,735 | <p>The Cantor set is closed, so its complement is open. So the complement can be written as a countable union of disjoint open intervals. Why can we not just enumerate all endpoints of the countably many intervals, and conclude the Cantor set is countable?</p>
| Carl Mummert | 630 | <p>The reason that does not work is that there are some points in the Cantor set which are not the endpoint of any interval that is removed during the construction of the Cantor set. In the proof that is suggested in the question, it would be necessary to show that every point in the set is one of these endpoints, but that just isn't true. The proof that the Cantor set is uncountable already shows there has to be at least one such "non-endpoint" point, because the set of endpoints is countable, as the question above points out. In fact we can give a specific example: the number $1/4$ is in the usual middle-thirds Cantor set, but it is not an endpoint of any interval that is removed, because all those endpoints are rationals whose denominator can be written as a power of $3$. </p>
|
210,735 | <p>The Cantor set is closed, so its complement is open. So the complement can be written as a countable union of disjoint open intervals. Why can we not just enumerate all endpoints of the countably many intervals, and conclude the Cantor set is countable?</p>
| Adam Rubinson | 29,156 | <p>Just because the open intervals approach a number, doesn't mean the number is the endpoint of one of the intervals. Consider:</p>
<p><span class="math-container">$$D_1= \bigcup\limits_{\substack{n~\in~\mathbb{N},~n~\text{odd} \\}} \left(\frac{1}{n+1}, \frac{1}{n}\right). $$</span></p>
<p>This looks like:</p>
<p><a href="https://i.stack.imgur.com/Bp6Yk.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Bp6Yk.jpg" alt="enter image description here" /></a>
<span class="math-container">$$$$</span>
Next, let
<span class="math-container">$$D_2= \bigcup\limits_{\substack{n~\in~\mathbb{N},~n~\text{odd} \\}} \left(\frac{-1}{n}, \frac{-1}{n+1}\right). $$</span></p>
<p>This looks like:</p>
<p><a href="https://i.stack.imgur.com/LBhxA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LBhxA.jpg" alt="enter image description here" /></a></p>
<p>Now let <span class="math-container">$D = D_1 \cup D_2.\ $</span> Then <span class="math-container">$0 \in D^c\ $</span> and <span class="math-container">$0$</span> is a limit point of <span class="math-container">$D^c$</span>. But <span class="math-container">$0$</span> is not the end-point of any of the intervals in <span class="math-container">$D$</span>. The same goes for members of the Cantor set- <em>most</em> members of the Cantor set, it turns out.</p>
<p>Just because the intervals "close in" on <span class="math-container">$0\ $</span> doesn't make <span class="math-container">$0$</span> an endpoint of one of the intervals.</p>
|
4,249,794 | <p>i have to determine whether this graph is bipartite or not:</p>
<p><img src="https://i.stack.imgur.com/AjxQzl.png" alt="" /></p>
<p>I have found an answer but i am not sure about it. If we divide the vertices set into <span class="math-container">$\{a,d,c,h\}$</span> and <span class="math-container">$\{b,f,e,g\}$</span>, then it fulfills bipartite properties.
Is it correct ?</p>
| AKSHAY V | 963,523 | <p>"The given graph is bipartite by the fact that there are no odd cycles in it". If you just want to prove it theoretically, this statement should be enough.</p>
<p>If you want to find out how it is bipartite then your sets of {a,d,c,h} and {b,f,e,g} are correct as an example.</p>
|
1,612,808 | <p>Suppose that $X$ is a finite $G$-set. A group $G$ is of prime power if $|G|=p^n$ for $p$ prime.</p>
<p>The fixed point set $X_G=\{x\in X : gx=x$ $\forall g\in G\}$.</p>
<p>I'm asked to prove that $|X|=|X_G|$ (mod $p$), but I'm unsure of how I should start.</p>
| Dor Marciano | 78,839 | <p>Remember for each $x \in X$, denoting $G_x = \{g\in G : gx=x\}$ and $O_x = \{y\in X | \exists g\in G : gx = y\}$, a well known theorem in Group Theory claims that $|O_x| = \frac{|G|}{|G_x|}$.</p>
<p>Notice that $|O_x|=1$ iff $x\in X_G$. If $|O_x|>1$ then, since $|O_x| \mid |G|$, we have that $p \mid |O_x|$. Now, since the orbits are a partition of $X$, we have that $|X|=\sum_i |O_i|$. </p>
<p>Taking this equation mod p, all $O_i$'s such that $|O_i|>1$ disappear, since $p \mid |O_i|$. We are only left with those such $|O_i|=1$. How many do we have of those? Exactly $|X_G|$ (as $|O_x|=1$ iff $x\in X_G$). So we are done.</p>
|
30,220 | <p>Jeremy Avigad and Erich Reck claim that one factor leading to abstract mathematics in the late 19th century (as opposed to concrete mathematics or hard analysis) was <em>the use of more abstract notions to obtain the same results with fewer calculations.</em></p>
<p>Let me quote them from their remarkable historical <a href="https://www.andrew.cmu.edu/user/avigad/Papers/infinite.pdf" rel="nofollow noreferrer">paper</a> "Clarifying the nature of the infinite: the development of metamathematics and proof theory".</p>
<blockquote>
<p>The gradual rise of the opposing viewpoint, with its emphasis on conceptual
reasoning and abstract characterization, is elegantly chronicled by <a href="http://mcps.umn.edu/philosophy/11_10Stein.pdf" rel="nofollow noreferrer">Stein</a> (<a href="https://web.archive.org/web/20140415224643/http://mcps.umn.edu/philosophy/11_10Stein.pdf" rel="nofollow noreferrer">Wayback Machine</a>),
as part and parcel of what he refers to as the “second birth” of mathematics.
The following quote, from Dedekind, makes the difference of opinion very clear:</p>
</blockquote>
<blockquote>
<blockquote>
<p>A theory based upon calculation would, as it seems to me, not offer
the highest degree of perfection; it is preferable, as in the modern
theory of functions, to seek to draw the demonstrations no longer
from calculations, but directly from the characteristic fundamental
concepts, and to construct the theory in such a way that it will, on
the contrary, be in a position to predict the results of the calculation
(for example, the decomposable forms of a degree).</p>
</blockquote>
</blockquote>
<blockquote>
<p>In other words, from the Cantor-Dedekind point of view, abstract conceptual
investigation is to be preferred over calculation.</p>
</blockquote>
<p><strong>What are concrete examples from concrete fields avoiding calculations by the use of abstract notions?</strong> (Here "calculation" means any type of routine technicality.) Category theory and topoi may provide some examples.</p>
<p>Thanks in advance.</p>
| O.R. | 5,506 | <p>All I see here are calculations. It only changed the nature of the object which you calculate with and its relation to the the final goal. For this reason I still can not make clear sense of the question.</p>
|
1,818,976 | <p>Let there be many numbers $a_1,a_2,a_3,\dots,a_n$.</p>
<p>I want to find the first digit of their product, i.e. of $A=a_1\times a_2\times a_3\times a_4\times \dots\times a_n$.</p>
<p>These numbers are huge and multiplying all of them exceeds the time limit.</p>
<p>Is there any shortcut to find the most significant digit of $A$ (first digit from the left)?</p>
| quid | 85,306 | <p>Your approach works fine, just apply it twice.</p>
<p>The real part of the sequence is the sequence $(\cos (2\pi r n))_n$ this is clearly bounded and thus has a convergent subsequence $(\cos (2\pi r n_k))_k$. </p>
<p>The imaginary part of the respective subsequence of the original sequence is $\sin (2\pi r n_k)_k$. This is again bounded and you find a convergent subsequence $\sin (2\pi r n_{k_l})_l$. Of course $(\cos (2\pi r n_{k_l}))_l$ is also convergent, and you are done—the complex sequence converges as real and imaginary part converge. </p>
|
4,275,780 | <blockquote>
<p>If <span class="math-container">$0<a<b$</span> and <span class="math-container">$0<c<d$</span> then <span class="math-container">$\frac{c+a}{d+a} <\frac{c+b}{d+b}.$</span></p>
</blockquote>
<p>I get to <span class="math-container">$$d+a<d+b \Longrightarrow \frac{1}{d+b} < \frac{1}{d+a}$$</span> but that inequality seems opposite of what I am trying to prove. Any advice is appreciated.</p>
| Ankit Saha | 876,128 | <p><span class="math-container">$$ \dfrac{c+a}{d+a} =\dfrac{d+a+c-d}{d+a} = 1 + \dfrac{c-d}{d+a}$$</span>
<span class="math-container">$$ \dfrac{c+b}{d+b} =\dfrac{d+b+c-d}{d+b} = 1 + \dfrac{c-d}{d+b}$$</span>
You have already proved that
<span class="math-container">$$ \frac{1}{d+b} < \frac{1}{d+a}$$</span>
As <span class="math-container">$c-d<0$</span>,
<span class="math-container">$$ \frac{c-d}{d+b} > \frac{c-d}{d+a}$$</span>
<span class="math-container">$$ \implies1+ \frac{c-d}{d+b} > 1+\frac{c-d}{d+a}$$</span>
<span class="math-container">$$ \implies \dfrac{c+b}{d+b} > \dfrac{c+a}{d+a}$$</span></p>
|
4,549,340 | <p>I have heard people say that the flight time from Fort Lauderdale to Seattle is the longest possible flight time within the continental United States. However, upon further consideration, I realized that the curvature of the Earth may cause the visible distance on a map to decrease when traveling north (the circumference of a cross section of the Earth is smaller further from the equator). A change in cross sectional circumference as one travels north can affect the true distance between Fort Lauderdale and a destination. This is not accounted for in a 2D map which is what most think of when considering flight times. With this in mind, does the Earth’s curvature affect the apparent distance (on a 2D map) between Fort Lauderdale and Seattle, and if so, is there another location in the continental United States with a longer flight time from Fort Lauderdale? In other words, suppose you were to fly around the globe across the equator. This would take longer than flying around the globe at a point north of the equator (say Seattle). This is due to the curvature of the Earth, so would this curvature also take effect when traveling from Fort Lauderdale to Seattle. The Earth is “wider” across Florida’s latitude than it is at Seattle’s. This means that it must take longer to travel from Florida to San Diego than from Maine to Seattle. My questions is if that difference could account for a change in flight time. Although Seattle is north of Fort Lauderdale and thus farther, there is added distance closer to the equator due to the Earth’s curvature. So may there exist a location south of Seattle that is further simply due to this change in cross sectional circumference?</p>
| Bruno B | 1,104,384 | <p>(Important edit: Damian is right, the induction itself is off in your proof. However, I still deem it informative to let my answer stay, for clarity about the symmetry argument that would've let you finish, were your statement true!)</p>
<p>Let's do as if you'd proven that, for <span class="math-container">$m > n \geq n_\varepsilon$</span>: <span class="math-container">$|x_n - x_m| < \varepsilon$</span>.</p>
<p>Now, if you let <span class="math-container">$n > m \geq \varepsilon$</span>, then: <span class="math-container">$|x_n - x_m| = |x_m - x_n| < \varepsilon$</span>. Basically, thanks to <span class="math-container">$|.\! |$</span>, your statement is "symmetric" in <span class="math-container">$n$</span> and <span class="math-container">$m$</span>, so it ends up being fine! (Remember, they're "mute" variables: I just named <span class="math-container">$n$</span> "<span class="math-container">$m$</span>" and <span class="math-container">$m$</span> "<span class="math-container">$n$</span>" in the first statement to arrive at the conclusion)</p>
<p>Finally, the case <span class="math-container">$n = m$</span> is trivial as <span class="math-container">$0 < \varepsilon$</span>. And thus you can conclude the sequence is indeed Cauchy.</p>
|
1,653,106 | <p>I was following a calculus tutorial that factored the equation $x^4-16$ into $(x^2 +4) (x+2)(x-2)$.</p>
<p>Why is the factorization of $x^4-16 = (x^2 + 4)(x+2)(x-2)$ rather than $(x^2 - 4)(x^2 +4)$? </p>
| 3SAT | 203,577 | <p>Notice that $x^4-16=(x^2)^2-4^2\\
=(x^2-4)(x^2+4)\\
=(x+2)(x-2)(x^2+4)\\
=(x-2)(x+2)(x+2\color{red}i)(x-2\color{red}i)$</p>
|
4,205,906 | <p>Since X and Y are independent and uniform I have the joint density function f(x,y)=1 but Im not sure where to go from there. I keep getting that the answer is f(z)=1 but this doesnt make sense since the range of z is from 0 to infinity.</p>
<p>So if I make the substitution Z=Y/X and W=X, I get Y=ZW and X=W. The 4 derivatives for the Jacobian are <span class="math-container">$\frac{dx}{dz}=0$</span>, <span class="math-container">$\frac{dx}{dw}=1$</span>, <span class="math-container">$\frac{dy}{dz}=W$</span> and <span class="math-container">$\frac{dy}{dw}=Z$</span> which give |J|=W and so f(z,w)=<span class="math-container">$1*w$</span></p>
<p>The range of W is 0,1 which and integrating W over that range just gives f(z)=1. This answer doesnt make sense to me since the range of Z is 0 to <span class="math-container">$\infty$</span> and f(z) should integrate to 1 over that range. I think I am making a mistake somewhere right after I get f(z,w)=w but Im not sure where.</p>
| MSIS | 678,294 | <p>The <span class="math-container">$p$</span>-value is the probability of obtaining the value of the Statistic, or, as pointed out below, a value as extreme as the one you obtained that you obtained while assuming your <span class="math-container">$H_0$</span>; Null Hypothesis is correct.</p>
<p>In Inferential Statistics you are interested in estimating a population parameter,
say the average age of your (human) population. You believe the average is less than, say <span class="math-container">$35$</span>. This becomes your Null Hypothesis <span class="math-container">$H_0$</span>. In order to test <span class="math-container">$H_0$</span> , you use a statistic ( A function of randomly-selected data. here the sample mean) to obtain a value <span class="math-container">$\mu$</span> from your data. Your statistic; here sample mean, has a distribution function <span class="math-container">$f$</span> corresponding to it. Using <span class="math-container">$f$</span>, you can compute the probability of finding a value like <span class="math-container">$\mu$</span> or lower. This last is your <span class="math-container">$p$</span>-value.</p>
|
330,775 | <p>Let $G$ be a Lie Group, how to prove that the tangent Space of $\operatorname{Aut}(T_e G)$ at identity is $\operatorname{End}(T_e G)$? Thanks.</p>
| Martin Brandenburg | 1,650 | <p>If $V$ is a finite-dimensional vector space, considered as a manifold, then the tangent space is $V$ at any point. And $\mathrm{Aut}$ is an open subspace of the vector space $\mathrm{End}$. This doesn't have anything to do with Lie groups.</p>
|
330,775 | <p>Let $G$ be a Lie Group, how to prove that the tangent Space of $\operatorname{Aut}(T_e G)$ at identity is $\operatorname{End}(T_e G)$? Thanks.</p>
| Julien | 38,053 | <p>Assume $E$ is a Banach space. Let $B(E)$ be the Banach algebra of bounded linear operators on $E$, equipped with the induced operator norm. You can call it the algebra of endomorphisms if you prefer. Then denote $GL(E)$ the group of invertible elements in $B(E)$, that is the group of automorphisms.</p>
<p>Fix $T_0$ in $GL(E)$. Then for very $S\in B(E)$ such that $\|S\|<\frac{1}{\|T_0^{-1}\|}$, we have $T_0+S$ in $GL(E)$ with inverse given by
$$
(T_0+S)^{-1}=T_0^{-1}(I+ST_0^{-1})^{-1}=T_0^{-1}\sum_{n\geq 0}(-ST_0^{-1})^n.
$$
The convergence of the Neumann series is due to the fact that $\|ST_0^{-1}\|\leq\|S\|\|T_0^{-1}\|<1$.</p>
<p>This proves that $GL(E)$ is open in $B(E)$, therefore it is a manifold modeled on $B(E)$.</p>
<p>In your case $E=T_eG$, $GL(E)=Aut(T_eG)$ and $B(E)=End(T_eG)$.</p>
<p>Actually, the same argument proves more generally that the set of invertible elements in a Banach algebra is open, hence a manifold modeled on this Banach algebra.</p>
|
322,858 | <p>Let <span class="math-container">$G$</span> be a split semisimple real Lie group in characteristic zero, and let <span class="math-container">$B=TU$</span> be a Borel subgroup with unipotent radical <span class="math-container">$U$</span> and Levi <span class="math-container">$T$</span>. Fix an ordering on the roots <span class="math-container">$\Phi^+$</span> of <span class="math-container">$T$</span> in <span class="math-container">$U$</span>, and for each root subgroup <span class="math-container">$U_{\alpha}$</span> of <span class="math-container">$U$</span>, let <span class="math-container">$u_{\alpha}: \mathbb R \rightarrow U_{\alpha}$</span> be an isomorphism.</p>
<p>For all <span class="math-container">$\alpha, \beta \in \Phi^+$</span>, there exist unique real numbers <span class="math-container">$C_{\alpha,\beta,i,j}$</span> (depending on the <span class="math-container">$u_{\alpha}$</span> and the ordering) such that for all <span class="math-container">$x, y \in \mathbb R$</span>,</p>
<p><span class="math-container">$$u_{\alpha}(x) u_{\beta}(y) u_{\alpha}(x)^{-1} = u_{\beta}(y) \prod\limits_{\substack{i,j>0\\ i\alpha + j \beta \in \Phi^+}} u_{i\alpha+j\beta}(C_{\alpha,\beta,i,j}x^iy^j)$$</span></p>
<p>I want to work out some examples with unipotent groups of exceptional semisimple groups, and am looking for table of structure constants for the root system G2. Does anyone know a reference where an ordering on the roots is chosen and these constants are written down?</p>
| Andrei Smolensky | 5,018 | <p>"Simple groups of Lie type" by R. W. Carter, a table after Section 12.4. But there only the values of <span class="math-container">$C_{\alpha\beta11}$</span> are listed. An explicit form of commutator formulas inside <span class="math-container">$U^+$</span> is given in Table IV of "Chevalley groups over commutative rings: I. Elementary calculations" by N. Vavilov and E. Plotkin, see the picture below.</p>
<p><a href="https://i.stack.imgur.com/CVNnA.png" rel="noreferrer"><img src="https://i.stack.imgur.com/CVNnA.png" alt="enter image description here"></a></p>
|
2,579,156 | <p>I found the solution of series on Wolfram Alpha
<a href="http://www.wolframalpha.com/input/?i=sum+1%2F(2k%2B1)%2F(2k%2B2)+from+1+to+n" rel="nofollow noreferrer">http://www.wolframalpha.com/input/?i=sum+1%2F(2k%2B1)%2F(2k%2B2)+from+1+to+n</a></p>
<p>$ \sum\limits_{k=1}^{n} \left(\frac{1}{2k+1} - \frac{1}{2k+2}\right) = \sum\limits_{k=1}^{n} \frac{1}{(2k+1)(2k+2)} = \frac{1}{2} \left(H_{n+\frac{1}{2}} - H_{n+1} -1 + \text{ln}(4)\right)$</p>
<p>Can someone tell how to prove this in the form of Harmonic numbers?</p>
| Claude Leibovici | 82,404 | <p>If you use polygamma functions
$$S_1=\sum\limits_{k=1}^{n} \frac{1}{2k+1}=\frac{1}{2} \left(\psi ^{(0)}\left(n+\frac{3}{2}\right)-\psi
^{(0)}\left(\frac{3}{2}\right)\right)$$
$$S_2=\sum\limits_{k=1}^{n} \frac{1}{2k+2}=\frac{1}2\sum\limits_{k=1}^{n} \frac{1}{k+1}=\frac{1}{2} (\psi ^{(0)}(n+2)+\gamma -1)$$ making
$$\sum\limits_{k=1}^{n} \left(\frac{1}{2k+1} - \frac{1}{2k+2}\right) = S_1-S_2$$ $$S_1-S_2=\frac{1}{2} \left(\psi ^{(0)}\left(n+\frac{3}{2}\right)-\psi
^{(0)}\left(\frac{3}{2}\right)\right)-\frac{1}{2} (\psi ^{(0)}(n+2)+\gamma -1)$$</p>
<p>If you look <a href="http://mathworld.wolfram.com/HarmonicNumber.html" rel="nofollow noreferrer">here</a> $$H_n=\gamma +\psi ^{(0)}(n+1)$$ making
$$S_1-S_2=\frac{1}{2} \left(H_{n+\frac{1}{2}}-H_{n+1}-H_{\frac{1}{2}}+1\right)$$
and $1-H_{\frac{1}{2}}=\log (4)-1$ and then the result.</p>
|
424,404 | <p>I want to create an algorithm/calculation that helps me figure out if the price on a used vehicle is high or low.</p>
<p>My thoughts are that to calculate this i need a critical mass of previous vehicle sales with the same characteristics; manufacturer, model, year and trim. Having this data I must figure out each sale’s raw price. (The sale price divided into the mileage the vehicle had driven). Working on with the raw price of each previously sold car, i can then calculate the average price, this being the “market price”.</p>
<p>Finally using the market price i must be able to answer my initial question if a price is too high or low compared to the market price.</p>
<p>Take notice that this doesn’t include a vehicle's color, extra equipment etc. Some colors are more sought for, and some equipment adds value to a vehicle. I would calculate this by fixed values at the end of my calculation.</p>
<p>Any help on how to construct this formula is appreciated.. Its been a long time since thinking in terms of calculations!</p>
<p>Thanks,
Kristian</p>
| Nameless | 68,482 | <p>I would also suggest running a regression. With ordinary least squares regression, you explain the observed price (of past sales) as linear combination of car characteristics. For example, an equation you estimate could be (for a particular model)
$$price_i=\beta_0+\beta_1 LowMiles_i+\beta_2 MidMiles_i+\beta_3 HighMiles_i+\beta_4 Engine_i+\varepsilon_i.$$</p>
<p>But before you jump in, you should think about which data you have and which you need to predict the price. Miles driven seems to be a major factor; instead of using it as continuous variable, I would divide the miles driven into ranges (no miles (new) is the omitted category contained in $\beta_0$, LowMiles with very few driven, etc.), and create dummy variables for that. We would expect the estimates to be $0>\beta_1\ge\beta_2\ge \beta_3$---if it doesn't come out like this, maybe you forgot some other relevant predictor. Depending on the model, there may be relevant extras that drive selling price. For example, some models offer versions with a more powerful engine, this can also be included as dummy. Or maybe some models have convertible versions, in which case you should also add a dummy for that. In order to predict the price, you simply use your estimated coefficients. With the above equation, a car of the model with few miles and a bigger engine would have a price of $\beta_0+\beta_1+\beta_4$. If the price you observe deviates, then the car may be over- or underpriced (or your statistical model is wrong).</p>
<p>You can also estimate one big equation for several models, which would also allow you to estimate the effect of "color". For example, you have model A and model B, then
$$price_i=\beta_0+\beta_1 LowMiles_i+\beta_2 MidMiles_i+\beta_3 HighMiles_i+\beta_4 Engine_i+\beta_5 B_i+\beta_6 LowMiles_i*B_i+\beta_7 MidMiles_i*B_i+\beta_8 HighMiles_i*B_i+\beta_9 Engine_i*B_i+\beta_{10} RedColor_i+\beta_{11} BlackColor+\varepsilon_i.$$
In this specification, the $\beta_{10}$ coefficient is the average price premium for a car of red color, compared to all other colors except black, after the effects of miles driven and engine are taken out. Note that this estimate will be driven by the model you have more observations for. You can prevent that by estimating the color premium for each model separately, using interactions like with miles.</p>
<p>If you want the price of model B with many miles driven an red color, you calculate $\beta_0+\beta_5+\beta_3+\beta_8+\beta_{10}$. You can see that, the more models you add, the more parameters you have to estimate. This is not a problem if you have enough data. But you can also generalize by assuming, for example, that the effect of miles driven is the same for all models (then you do not have to estimate $\beta_6-\beta_8$ in the above example). While this is implausible in general, it might be useful for models that are very similar in price and characteristics.</p>
|
113,854 | <p>I'm importing a *.pdb file containing a single protein. <em>Mathematica</em> automatically produces a plot of the protein.</p>
<p>I want to specify the color of each residue independently, in this plot. Is this possible?</p>
<p>Additionally, I would like to change the type of plot to "cartoon". How can I do this?</p>
| Jason B. | 9,490 | <p>Okay, so this is what it looks like with standard coloring of the residues:</p>
<pre><code>Import["ExampleData/1PPT.pdb", ColorFunction -> "Residue"]
</code></pre>
<p><a href="https://i.stack.imgur.com/upk6k.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/upk6k.png" alt="enter image description here"></a></p>
<p>But now, if we want to change the residue colors, we need to change the value of a certain internal color list called <code>Graphics`MoleculePlotDump`residueColorRules</code></p>
<pre><code>residuelist = {"Gly", "Pro", "Ser", "Gln", "Pro", "Thr", "Tyr", "Pro",
"Gly", "Asp", "Asp", "Ala", "Pro", "Val", "Glu", "Asp", "Leu",
"Ile", "Arg", "Phe", "Tyr", "Asp", "Asn", "Leu", "Gln", "Gln",
"Tyr", "Leu", "Asn", "Val", "Val", "Thr", "Arg", "His", "Arg",
"Tyr"};
Graphics`MoleculePlotDump`residueColorRules =
Thread[residuelist -> (ColorData[97] /@ Range[36])]
</code></pre>
<p><a href="https://i.stack.imgur.com/L6511.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/L6511.png" alt="enter image description here"></a></p>
<p>Now when we import the PDB, it uses the new color list,</p>
<p><a href="https://i.stack.imgur.com/eVRam.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eVRam.png" alt="enter image description here"></a></p>
<p>The definition of <code>Graphics`MoleculePlotDump`residueColorRules</code> will be reset when you restart the kernel.</p>
|
3,527,350 | <p>I need to calculate the spectrum of the operator <span class="math-container">$T$</span> for <span class="math-container">$f \in L^2([0,1])$</span> defined by: </p>
<p><span class="math-container">\begin{equation}
(Tf)(x) = \int_0^1 (x+y)f(y)dy.
\end{equation}</span></p>
<p>I know that <span class="math-container">$T$</span> is compact and self-adjoint so the residual spectrum is empty and the eigenvalues are real and a closed subset of <span class="math-container">$[-||T||, ||T||]$</span>.</p>
<p>So I let <span class="math-container">$\lambda$</span> be an eigenvalue so I know that <span class="math-container">$\lambda f(x) = (Tf)(x)$</span>. By differentiating twice, I found that <span class="math-container">$\lambda f''(x) = 0$</span> but I don't really know how I can continue.</p>
| operatorerror | 210,391 | <p>To continue on your path, note that <em>once you have justified</em> that <span class="math-container">$f$</span> must in fact be smooth to be an eigenvector for nonzero eigenvalue <span class="math-container">$\lambda$</span>, you can differentiate twice as you did and note that
<span class="math-container">$$
\lambda f''(x)=0
$$</span>
In general, you are a long way from being allowed to differentiate an arbitrary element of <span class="math-container">$L^2$</span>, so beware in the future.</p>
<p>So that, for <span class="math-container">$\lambda\ne 0$</span>, by solving the ode <span class="math-container">$f(x)=ax+b$</span>. Now proceed as in the other (hint, when does that system have a nontrivial solution?).</p>
<p>The point <span class="math-container">$\lambda=0$</span> is also in your spectrum however (this is of course obvious when noting that we took an infinite dimensional space to a finite dimensional one). For an explicit example, note that <span class="math-container">$Tf(x)=0$</span> where <span class="math-container">$f(x)=6x^2-6x+1$</span>. This answer was arrived at by looking for a quadratic polynomial orthogonal to both <span class="math-container">$1$</span> and <span class="math-container">$y$</span>. </p>
|
2,807,356 | <blockquote>
<p><strong>If $z_1,z_2$ are two complex numbers such that $\vert
z_1+z_2\vert=\vert z_1\vert+\vert z_2\vert$,then it is necessary that</strong> </p>
<p>$1)$$z_1=z_2$</p>
<p>$2)$$z_2=0$</p>
<p>$3)$$z_1=\lambda z_2$for some real number $\lambda.$</p>
<p>$4)$$z_1z_2=0$ or $z_1=\lambda z_2$ for some real number $\lambda.$</p>
</blockquote>
<p>From Booloean logic we know that if $p\implies q$ then $q$ is necessary for $p$.</p>
<p>For $1)$taking $z_1=1$ and $z_2=2$ then $\vert 1+2 \vert=\vert 1\vert+\vert 2\vert$ but $1\neq 2$.So,$(1)$ is false.</p>
<p>For $2)$taking $z_1=1$ and $z_2=2$ then $\vert 1+2 \vert=\vert 1\vert+\vert 2\vert$ but $2\neq 0$.So,$(2)$ is false.</p>
<p>I'm not getting how to prove or disprove options $(3)$ and $(4)?$</p>
<p>Need help</p>
| trancelocation | 467,003 | <p>$$\vert z_1+z_2\vert=\vert z_1\vert+\vert z_2\vert \Rightarrow \vert z_1+z_2\vert^2= (\vert z_1\vert+\vert z_2\vert)^2 \Rightarrow Re(z_1\bar z_2) = |z_1||z_2|$$
Now, note that $Re(z_1\bar z_2)$ is a real 2-dimensional scalar product
$$Re(z_1\bar z_2) = x_1x_2+y_1y_2 \mbox{ where } z_1 = x_1+iy_1, \, z_2 = x_2+iy_2$$
Cauchy-Schwarz tells us that
$$Re(z_1\bar z_2) = |z_1||z_2| \Rightarrow z_1z_2= 0 \mbox{ or } z_1 = \lambda z_2 \mbox{ where } \lambda > 0$$</p>
|
2,476,717 | <p>Let $f: \mathbb R^n \rightarrow \mathbb R^n$ with arbitrary norm $\|\cdot\|$. It exists a $x_0 \in \mathbb R^n$ and a number $r \gt 0 $ with </p>
<p>$(1)$ on $B_r(x_0)=$ {$x\in \mathbb R^n: \|x-x_0\| \leq r$} $f$ is a contraction with Lipschitz constant L</p>
<p>$(2)$ it applies $\|f(x_0)-x_0\| \le (1-L)r$</p>
<p>The sequence $(x_k)_{k\in\mathbb N}$ is defined by $x_{k+1}=f(x_k).$</p>
<p>How do I show that $x_k \in B_r(x_0) \forall k \in \mathbb N$?</p>
<p>How do I show that $f$ has a unique Fixed point $x_f$ with $x_f = \lim_{k \rightarrow\infty} x_k$?</p>
<p>I know this has something to do with Banach but I am totally clueless on how to prove this. Any help is welcome. Thanks. </p>
| Guy Fsone | 385,707 | <p>Hint: Prove by induction
Assume that $x_k\in B_r(x_0)$ then you have </p>
<p>$$ \|x_{k+1} -x_0\| = \|f(x_k) -f(x_{0}) +f(x_{0}) -x_0\| \\\le \|f(x_k) -f(x_{0})\|+\|f(x_{0}) -x_0\| \\ \le L
\|x_k -x_{0}\|+\|f(x_{0}) -x_0\| $$</p>
<p>That is </p>
<p>$$ \|x_{k+1} -x_0\| \le L \|x_k -x_{0}\|+\|f(x_{0}) -x_0\| .$$</p>
<p>Using the assumption that, </p>
<p>$$\|f(x_0)-x_0\| \le (1-L)r$$
we get,
$$ \|x_{k+1} -x_0\| \le L\|x_{k} -x_0\|+(1-L)r \tag{E}$$</p>
<p>Now since $x_0\in B_r(x_0)$ we assume if we Assume that $x_k\in B_r(x_0)$ .
Then from the estimate above we have
$$ \|x_{k+1} -x_0\| \le L\|x_{k} -x_0\|+(1-L)r\le (1-L +L)r= r $$</p>
<p>Hence $$x\in B_r(x_0)~~~\forall~~ k$$</p>
<blockquote>
<p>Replacing $x_k$ by $y$ in (E) in the above prove we see that for all $ y\in B_r(x_0)$ we obtain
$$ \|f(y) -x_0\| \le L \|y -x_{0}\|+(1-L)r \le Lr +(1-L)r= r .$$</p>
</blockquote>
<p>Therefore $$f :B_r(x_0) \to B_r(x_0)$$ is a contraction on closed set $B_r(x_0)$(which is therefore complete). <a href="https://en.wikipedia.org/wiki/Contraction_mapping" rel="nofollow noreferrer">From the fix point theorem</a>
$f$ has a fix point $x_f$ and its satisfies
$$x_f = \lim_{k\to\infty}x_k$$</p>
|
2,662,554 | <p>I have to use Proof by contradiction to show what if $n^2 - 2n + 7$ is even then $n + 1$ is even. </p>
<p>Assume $n^2 - 2n + 7$ is even then $n + 1$ is odd. By definition of odd integers, we have $n = 2k+1$. </p>
<p>What I have done so far:</p>
<p>\begin{align}
& n + 1 = (2k+1)^2 - 2(2k+1) + 7 \\
\implies & (2k+1) = (4k^2+4k+1) - 4k-2+7-1 \\
\implies & 2k+1 = 4k^2+1-2+7-1 \\
\implies & 2k = 4k^2 + 4 \\
\implies & 2(2k^2-k+2)
\end{align}</p>
<p>Now, this is even but I wanted to prove that this is odd(the contradiction). Can you some help me figure out my mistake? </p>
<p>Thank you.</p>
| Bram28 | 256,001 | <p>Since you are doing a proof by contradiction, your start is: Assume $n+1$ is odd. But if $n+1$ is odd, then $n$ is even, and so yo should plug in $n=2k$, rather than what you did, which was plugging in $2k+1$</p>
|
2,662,554 | <p>I have to use Proof by contradiction to show what if $n^2 - 2n + 7$ is even then $n + 1$ is even. </p>
<p>Assume $n^2 - 2n + 7$ is even then $n + 1$ is odd. By definition of odd integers, we have $n = 2k+1$. </p>
<p>What I have done so far:</p>
<p>\begin{align}
& n + 1 = (2k+1)^2 - 2(2k+1) + 7 \\
\implies & (2k+1) = (4k^2+4k+1) - 4k-2+7-1 \\
\implies & 2k+1 = 4k^2+1-2+7-1 \\
\implies & 2k = 4k^2 + 4 \\
\implies & 2(2k^2-k+2)
\end{align}</p>
<p>Now, this is even but I wanted to prove that this is odd(the contradiction). Can you some help me figure out my mistake? </p>
<p>Thank you.</p>
| marty cohen | 13,079 | <p><span class="math-container">$n^2-2n+7
=n^2+2n+1-(4n+6)
=(n+1)^2-2(2n+3)
$</span>.</p>
<p>If this is even then,
since <span class="math-container">$2(2n+3)$</span> is even,
their sum is even,
so <span class="math-container">$(n+1)^2$</span> is even
so <span class="math-container">$n+1$</span> is even.</p>
|
3,806,122 | <p>I tried using Chinese remainder theorem but I kept getting 19 instead of 9.</p>
<p>Here are my steps</p>
<p><span class="math-container">$$
\begin{split}
M &= 88 = 8 \times 11 \\
x_1 &= 123^{456}\equiv 2^{456} \equiv 2^{6} \equiv 64 \equiv 9 \pmod{11} \\
y_1 &= 9^{-1} \equiv 9^9 \equiv (-2)^9 \equiv -512 \equiv -6 \equiv 5 \pmod{11}\\
x_2 &= 123^{456} \equiv 123^0 \equiv 1 \pmod{8}\\
y_2 &= 1^{-1} \equiv 1 \pmod{8} \\
123^{456}
&\equiv \sum_{i=1}^2 x_i\times\frac{M}{m_i} \times y_i
\equiv 9\times\frac{88}{11}\times5 + 1\times\frac{88}{8} \times1 \equiv 371
\equiv 19 \pmod{88}
\end{split}
$$</span></p>
| Bill Dubuque | 242 | <p>You used an incorrect CRT formula. It should be: for coprime <span class="math-container">$\,m,n,\,$</span> and <span class="math-container">$\,c^{-1}_{\ n}:= c^{-1}\bmod n$</span></p>
<p><span class="math-container">$\qquad\begin{align} &x\equiv a\!\!\pmod{\!m}\\ &x\equiv b\!\!\pmod{\!n}\end{align}\iff x\,\equiv\, a\,n(n^{-1}_{\ m}) + b\,m(m^{-1}_{\ n})\ \ \pmod{\!mn}$</span></p>
<p>To help remember the formula note that it is easy to see it is correct since</p>
<p><span class="math-container">$\qquad\qquad\qquad\quad\ \ \bmod \color{#c00} m\!:\ \ x\,\equiv\, a\, \underbrace{n (n^{-1})}_{\large \equiv\ 1\ }\, +\, \underbrace{b\, \color{#c00}m\, (\cdots)}_{\large \color{#c00}{\equiv\ 0}}\, \equiv\, a$</span></p>
<p>i.e. <span class="math-container">$\bmod (m,n)\!:\,\ n\, n^{-1}_{\ m}\equiv (1,0),\,\ m\,m^{-1}_{\ n}\equiv (0,1)\,$</span> so the formula is simply</p>
<p><span class="math-container">$$ x\equiv (a,b) \equiv a (1,0) + b (0,1)\qquad\qquad$$</span></p>
<p>See <a href="https://math.stackexchange.com/a/3095229/242">this answer</a> for further discussion of this viewpoint (including an example with <span class="math-container">$3$</span> moduli).</p>
|
713,104 | <p>Are there any combinatorial games whose order (in the usual addition of combinatorial games) is finite but neither $1$ nor $2$?</p>
<p>Finding examples of games of order $2$ is easy (for example any impartial game), but I have not been able to think up an example with finite order where the order did not come from some sort of symmetry (for example even though Domineering is not impartial, it is easy to see that any square board will give a game of order $1$ or $2$), and such a symmetry only gives $1$ or $2$ as the possible orders.</p>
| Adola | 406,858 | <p>The hockey stick identity states that
<span class="math-container">$$\sum_{k=r}^{n} \binom{k}{r}=\binom{n+1}{r+1}$$</span></p>
<p>Note that <span class="math-container">$$1+2+...+n = \frac{n(n+1)}{2}=\binom{n+1}{2}.$$</span></p>
<p>Hence</p>
<p><span class="math-container">$$f(n)=\binom{2}{2}+\binom{3}{2}+...+\binom{n+1}{2}=\binom{n+2}{3}=\frac{(n+2)(n+1)n}{3!}$$</span></p>
|
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