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 |
|---|---|---|---|---|
1,299,032 | <p>The following problem is from Taylor's PDE I. I do not get why $D \det(I)B=\frac{d}{d t}\det(I+t B)|_{t=0}$ hold. Since $\det(I)=1$ and $D$ is $n\times n$ matrix, the left side seems to be a matrix, while the right side is clearly a scalar. Could anyone tell me how I should correctly interpret $\det(I)$ in this case... | Robert Lewis | 67,071 | <p>Hoping I'm not completely off-base here! I think that mjqxxx had it right in his comment, i.e. that $D \det(I)$ refers to the linear map from $M_n(\Bbb C) \to \Bbb C$ which is the derivative of $\det: M_n(\Bbb C) \to \Bbb C$ at the point $I \in M_n(\Bbb C)$; then $D\det(I) B$ makes sense as the directional derivati... |
3,542,485 | <p>Suppose my birthday fell on a Sunday. What is the maximum number of years I may have to wait till my birthday falls on a Sunday again?</p>
<p>My intuition says 11 years:</p>
<ul>
<li><p>At the earliest, the next Sunday is in 6 years time. Reason: before I can encounter another Sunday, there must be at least one le... | Geoffrey Trang | 684,071 | <p>Assuming that one is not born on February 29, then the following possibilities can occur:</p>
<ul>
<li>The day of the week repeats after 5 years. For example, March 1, 2015, and March 1, 2020, are both Sundays.</li>
<li>The day of the week repeats after 6 years. For example, April 1, 2012, and April 1, 2018, are bo... |
1,489,362 | <p>$$\lim_{x\to\infty} \left(\frac{3x^2}{\sqrt{4x^2+x+1}+\sqrt{x^2+x+1}}-x\right)$$</p>
<p>I need help in finding this limit.</p>
| egreg | 62,967 | <p>Rewrite it in the form
$$
\lim_{x\to\infty} \biggl(
\frac{3x^2\bigl(\sqrt{4x^2+x+1}-\sqrt{x^2+x+1}\,\bigr)}{3x^2}-x
\biggr)
$$
and simplify $3x^2$. Now change $x=1/t$ so you get
$$
\lim_{t\to0^+}
\frac{\sqrt{4+t+t^2}-\sqrt{1+t+t^2}-1}{t}
$$
which is the derivative at $0$ of the function
$$
f(t)=\sqrt{4+t+t^2}-\sqr... |
1,489,362 | <p>$$\lim_{x\to\infty} \left(\frac{3x^2}{\sqrt{4x^2+x+1}+\sqrt{x^2+x+1}}-x\right)$$</p>
<p>I need help in finding this limit.</p>
| mathlove | 78,967 | <p>Multiplying the first term by $$\frac{\sqrt{4x^2+x+1}-\sqrt{x^2+x+1}}{\sqrt{4x^2+x+1}-\sqrt{x^2+x+1}}$$gives$$\begin{align}&\lim_{x\to \infty}\left(\frac{3x^2}{\sqrt{4x^2+x+1}+\sqrt{x^2+x+1}}-x\right)\\\\&=\lim_{x\to\infty}\left(\sqrt{4x^2+x+1}-\sqrt{x^2+x+1}-x\right)\\\\&=\lim_{x\to\infty}x\left(\sqrt{4... |
4,361,626 | <p><span class="math-container">$4$</span> balls are randomly distributed into <span class="math-container">$3$</span> cells (<span class="math-container">$3^4=81$</span> possibilities of equal probability).</p>
<p>What is the probability that there is a cell that contains exactly <span class="math-container">$2$</span... | Lubin | 17,760 | <p>It occurs to me that you may be looking at the offending equation
<span class="math-container">$$
(J+a)(J+b)=(J+ab)
$$</span>
and making too much of it. In particular, how is “<span class="math-container">$(J+a)(J+b)$</span>” to be understood? To multiply these two cosets, you don’t take all possible products, one f... |
1,642,427 | <p>If the stem of a mushroom is modeled as a right circular cylinder with diameter $1$, height $2$, its cap modeled as a hemisphere of radius $a$ the mushroom has axial symmetry, is of uniform density,and its center of mass lies at center of plane where the cap and stem join, then find $a$.</p>
<p>I really need help.<... | user311100 | 311,100 | <p>The arc length should also be $\pi$. The curve "converges" to a straight line, but is actually an infinite union of semicircles of 0 length. You cannot approximate the arc length like that, it is similar to the fallacious proofs that $\pi=4$. In fact, there are $2^n$ semicircles each with diamater $\frac{1}{2^n}$, t... |
510,488 | <p>Using the fact that $x^2 + 2xy + y^2 = (x + y)^2 \ge 0$, show that the assumption $x^2 + xy + y^2 < 0$ leads to a contradiction...
So do I start off with...</p>
<p>"Assume that $x^2 + xy + y^2 <0$, then blah blah blah"?</p>
<p>It seems true...because then I go $(x^2 + 2xy + y^2) - (x^2 + xy + y^2) \ge 0$.
It... | peterwhy | 89,922 | <p>By completing square,</p>
<p>$$x^2+xy+y^2 = x^2+2x\frac{y}{2}+\frac{y^2}{4} + \frac{3y^2}{4} = \left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}\ge 0$$</p>
|
510,488 | <p>Using the fact that $x^2 + 2xy + y^2 = (x + y)^2 \ge 0$, show that the assumption $x^2 + xy + y^2 < 0$ leads to a contradiction...
So do I start off with...</p>
<p>"Assume that $x^2 + xy + y^2 <0$, then blah blah blah"?</p>
<p>It seems true...because then I go $(x^2 + 2xy + y^2) - (x^2 + xy + y^2) \ge 0$.
It... | Abel | 71,157 | <p>Another way to see this is as follows:</p>
<p>Suppose $x^2+xy+y^2 <0$, then $xy<-x^2-y^2< 0$, hence $2xy<xy<-x^2-y^2$ and thus $x^2+2xy+y^2<0$, which is absurd.</p>
|
510,488 | <p>Using the fact that $x^2 + 2xy + y^2 = (x + y)^2 \ge 0$, show that the assumption $x^2 + xy + y^2 < 0$ leads to a contradiction...
So do I start off with...</p>
<p>"Assume that $x^2 + xy + y^2 <0$, then blah blah blah"?</p>
<p>It seems true...because then I go $(x^2 + 2xy + y^2) - (x^2 + xy + y^2) \ge 0$.
It... | Clayton | 43,239 | <p>Assume $x^2+xy+y^2<0$. Adding and subtracting $xy$ on the left-hand side gives $x^2+2xy+y^2-xy=(x+y)^2-xy<0$, and therefore $0\leq(x+y)^2<xy$. Conversely, $x^2+xy+y^2<0$ implies $xy<-(x^2+y^2)\leq0$. Combining these, we have $$xy<0<xy,$$ a clear contradiction.</p>
|
510,488 | <p>Using the fact that $x^2 + 2xy + y^2 = (x + y)^2 \ge 0$, show that the assumption $x^2 + xy + y^2 < 0$ leads to a contradiction...
So do I start off with...</p>
<p>"Assume that $x^2 + xy + y^2 <0$, then blah blah blah"?</p>
<p>It seems true...because then I go $(x^2 + 2xy + y^2) - (x^2 + xy + y^2) \ge 0$.
It... | Brian M. Scott | 12,042 | <p>Suppose that $x^2+xy+y^2<0$; then $x^2+2xy+y^2<xy$, so $(x+y)^2<xy$. Subtracting $3xy$ from both sides of the original inequality, we see that $x^2-2xy+y^2<-3xy$, so $(x-y)^2<-3xy$. Squares are non-negative, so on the one hand $xy>0$, and on the other hand $-3xy>0$ and therefore $xy<0$.</p>
|
46,726 | <p>In many proofs I see that some variable is "fixed" and/or "arbitrary". Sometimes I see only one of them and I miss a clear guideline for it. Could somebody point me to a reliable source (best a well-known standard book) which explains, when and how to use both in proofs?</p>
<p>EDIT: A little add-on to the question... | Qiaochu Yuan | 232 | <p>Both "arbitrary" and "fixed" are just shorthand for a universal quantifier. When I say something like "fix $\epsilon > 0$" it means I am about to prove a statement that is true for all $\epsilon > 0$ (and thereby prove that some function is continuous, for example) but I don't want to actually write out "for all ... |
1,330,376 | <p>In textbooks and online tutorials I see that the remainder is calculated by using a new unknown variable on the same interval. For example we take the Taylor polynomial $T_n(a)$ but find the remainder $R(x)$ with a new variable $z$ inside it. See <a href="http://www.millersville.edu/~bikenaga/calculus/remainder-term... | KittyL | 206,286 | <p>Consider the easy case where $f(x)=f(c)+f'(z)(x-c)$. If you are in Calc 2, you should have learned linear approximation. The linear approximation of $f(x)$ around point $c$ is $f(x)\approx f(c)+f'(c)(x-c)$. See the following picture:</p>
<p><img src="https://i.stack.imgur.com/G4avH.png" alt="enter image description... |
1,630,074 | <p>My brother needs help coming up with a formula for a problem that I already did but failed to write out the formula for. </p>
<p>The problem is:</p>
<p>Consider a circle with the point (5,4) and a radius of 3.
Determine if a vertical line segment with the points (3,5) being one end of the segment and (3,1) being t... | Balarka Sen | 117,002 | <p>The current answer already explains why the proposed homotopy cannot work. Let me give a geometric interpretation of the two-step homotopy on the linked answer. </p>
<p>Trying to contract $S^\infty$ to $(1, 0, 0,\cdots)$ directly using straightline homotopy cannot possibly work: The situation is the same as that of... |
2,006,993 | <p>Let $(a_{n,k})_{n, k \in \mathbb N} \subset \mathbb C$ be a series satisfying</p>
<p>$$
\sum_{n=0}^\infty \left| \sum_{k=0}^\infty a_{n,k}\right| \lt \infty
$$
and
$$
\sum_{k=0}^\infty \left|a_{n,k}\right| \lt \infty \qquad \forall n\in \mathbb N.
$$
Does this imply that $\sum_{k=0}^\infty \sum_{n=0}^\infty a_{n,k... | Victor Chen | 289,695 | <p>The condition
$$
\sum_{n=0}^\infty \left| \sum_{k=0}^\infty a_{n,k}\right| \lt \infty
$$
is not sufficient, see the counter-example below.</p>
<p>Define a serie $\{a_{n,k}\}$, where
$$
a_{n,k} = \cases{\frac{1}{n^2}-1 & if $n=k$, \\ 1 & if $n \ne k, k=1$, \\ 0 & else.}
$$
Then we can easily verify that ... |
1,004,056 | <p>Pick a piece of paper and a pen. Put the pen on a starting point and begin to draw an arbitrary curve and don't withdraw your hand until you reached the starting point. You can meet your curve during drawing in some <em>one point</em> intersections but not more (like a line intersection). For example the left curve ... | Laars Helenius | 112,790 | <p>Probably. We can create a graph $G$ from the drawing by assigning each region to be colored a vertex. Then if $u$ and $v$ are vertices of $G$, $(u,v)$ is an edge if and only if the regions represented by $u$ and $v$ share a face.</p>
<p>This should result in a bipartite graph which is always two colorable, which in... |
1,004,056 | <p>Pick a piece of paper and a pen. Put the pen on a starting point and begin to draw an arbitrary curve and don't withdraw your hand until you reached the starting point. You can meet your curve during drawing in some <em>one point</em> intersections but not more (like a line intersection). For example the left curve ... | Community | -1 | <p>Colour a point black if the <a href="https://en.wikipedia.org/wiki/Winding_number" rel="nofollow">winding number</a> of the curve around the point is odd.</p>
|
1,004,056 | <p>Pick a piece of paper and a pen. Put the pen on a starting point and begin to draw an arbitrary curve and don't withdraw your hand until you reached the starting point. You can meet your curve during drawing in some <em>one point</em> intersections but not more (like a line intersection). For example the left curve ... | Cyriac Antony | 120,721 | <p>I shall try to give an intuitive colouring method that doesn't require much mathematical background.
I though I clearly understood why this method of colouring works; but now I doubt it.
Anyway I shall try to give a colouring method for any such drawing using only 2 colours without presenting proof for validity (I ... |
33,743 | <p>I have a lot of sum questions right now ... could someone give me the convergence of, and/or formula for, $\sum_{n=2}^{\infty} \frac{1}{n^k}$ when $k$ is a fixed integer greater than or equal to 2? Thanks!!</p>
<p>P.S. If there's a good way to google or look up answers to these kinds of simple questions ... I'd lo... | N. S. | 9,176 | <p>The easiest way to prove that $1+\frac{1}{2^2}+...+\frac{1}{n^2}$ is convergent, is proving by induction that</p>
<p>$$ 1+\frac{1}{2^2}+...+\frac{1}{n^2} \leq 2 -\frac{1}{n} \,.$$</p>
<p>If $k >2$, convergence follows immediately from $\frac{1}{j^k} \leq \frac{1}{j^2}$. </p>
<p>I always find fascinating the fa... |
854,671 | <p>So I'm a bit confused with calculating a double integral when a circle isn't centered on $(0,0)$. </p>
<p>For example: Calculating $\iint(x+4y)\,dx\,dy$ of the area $D: x^2-6x+y^2-4y\le12$.
So I kind of understand how to center the circle and solve this with polar coordinates. Since the circle equation is $(x-3)^2... | mesel | 106,102 | <p>I think this question has already good answers,</p>
<p>It is enough to show that there exist a month starting on Sunday.</p>
<p>Let $x$ be the starting day of the year then next month's first day is $x+31\equiv x+3$ and next month's first day, $x+3+28\equiv x+3$...</p>
<p>At the and we have, $$x,x+3,x+3,x+6,x+1,... |
2,898,889 | <p>Let $\ A $ be a real matrix of $\ 3 \times 3 $ with the characteristic polynom: $\ p(t) = t^3 + 2t^2 -3t $</p>
<p>Prove: the matrix $\ A^2 + A - 2I $ is similar to $\ D = \begin{bmatrix} 0 & 0 & 0 \\ 0 &4 & 0 \\ 0 & 0 & - 2 \end{bmatrix} $</p>
<p>$\ p(t) = t^3 +2t^2 -3t = t(t^2 + 2t -3) = t... | Math Lover | 348,257 | <p>Hint: $$\frac{s}{(s+1)^2-2^2} = \frac{s}{(s-1)(s+3)} = \frac{1}{4(s-1)}+\frac{3}{4(s+3)},$$
and $$\mathcal{L}(e^{at}) = \frac{1}{s-a}.$$</p>
<hr>
<p>For the textbook solution, it is useful to write the given fraction as
$$\frac{s}{(s+1)^2-2^2} = \frac{s+1-1}{(s+1)^2-2^2} = \frac{s+1}{(s+1)^2-2^2} - \frac{1}{2}\cd... |
2,898,889 | <p>Let $\ A $ be a real matrix of $\ 3 \times 3 $ with the characteristic polynom: $\ p(t) = t^3 + 2t^2 -3t $</p>
<p>Prove: the matrix $\ A^2 + A - 2I $ is similar to $\ D = \begin{bmatrix} 0 & 0 & 0 \\ 0 &4 & 0 \\ 0 & 0 & - 2 \end{bmatrix} $</p>
<p>$\ p(t) = t^3 +2t^2 -3t = t(t^2 + 2t -3) = t... | Mohammad Riazi-Kermani | 514,496 | <p>$$\mathcal{L}^{-1} \left\{ \dfrac{s}{(s + 1)^2 - 4} \right\}$$</p>
<p>$$=\mathcal{L}^{-1} \left\{ \dfrac{s+1}{(s + 1)^2 - 4} \right\}-\mathcal{L}^{-1} \left\{ \dfrac{1}{(s + 1)^2 - 4} \right\}$$</p>
<p>$$=e^{-t} \cosh(2t) - \dfrac{1}{2}e^{-t}\sinh(2t)$$</p>
|
708,633 | <p>My question is as in the title: is there an example of a (unital but not necessarily commutative) ring $R$ and a left $R$-module $M$ with nonzero submodule $N$, such that $M \simeq M/N$?</p>
<p>What if $M$ and $N$ are finitely-generated? What if $M$ is free? My intuition is that if $N$ is a submodule of $R^n$, then... | Ehsaan | 78,996 | <p>One can show that Noetherian modules are <strong>Hopfian</strong>, that is, any surjective endomorphism is injective. In particular if $R$ is a left Noetherian ring, then the free left $R$-module $R^n$ is Noetherian too hence Hopfian.</p>
<p><strong>Proposition.</strong> Let $R$ be any ring. Then the following are ... |
708,633 | <p>My question is as in the title: is there an example of a (unital but not necessarily commutative) ring $R$ and a left $R$-module $M$ with nonzero submodule $N$, such that $M \simeq M/N$?</p>
<p>What if $M$ and $N$ are finitely-generated? What if $M$ is free? My intuition is that if $N$ is a submodule of $R^n$, then... | Slade | 33,433 | <p>If $R$ is a commutative ring, $M,N$ are finitely generated free $R$-modules of the same rank, and $\varphi: M\to N$ is surjective, then $\varphi$ is injective.</p>
<p>Note that the Noetherian hypothesis is unnecessary! We cannot use Nakayama, but we can use the same <em>proof technique</em> as Nakayama:</p>
<p>We... |
851,959 | <p>In Halmos's Naive Set Theory about well-ordering set, it states that if a collecton <span class="math-container">$\mathbb{C}$</span> of well-ordered set is a chain w.r.t continuation, then the union of these sets is a well-ordered set. However, I cannot see why it is well-ordered. That is, I cannot see why each of i... | Mauro ALLEGRANZA | 108,274 | <p>You have to re-read all the passage from page 67-on :</p>
<blockquote>
<p>We shall say that a well ordered set $A$ is a <em>continuation</em> of a well ordered set $B$, if, in the first place, $B$ is a subset of $A$, if, in fact, $B$ is an initial segment of $A$, and if, finally, the ordering of the elements in $... |
2,624,272 | <p>It is well known that, in order to track constant reference values with zero asymptotic error, in the presence of uncertainty, an integrator is required in the open loop (either by the plant or the controller itself).</p>
<p>Lets for simplicity assume the plant $P$ has no integral components.</p>
<p>In order to tr... | JMJ | 295,405 | <p>The exponential function increases very, <em>very</em> rapidly, so in practice you never need to track an exponential reference. If you tried, you would quickly hit the saturation limits of your actuators.</p>
<p>The basic reason you can use the trick of adding a pole at the origin (pure integrator) to deal with po... |
3,048,063 | <p>I have to prove this preposition by mathematical induction:</p>
<p><span class="math-container">$$\left(x^n+1\right)<\left(x+1\right)^n \quad \forall n\geq 2 \quad \text{and}\quad x>0,\,\, n \in \mathbb{N}$$</span></p>
<p>I started the prove with <span class="math-container">$n=2$</span>:</p>
<p><span class... | Sri-Amirthan Theivendran | 302,692 | <p>Suppose that <span class="math-container">$(1+x)^n>1+x^n$</span> for some <span class="math-container">$n\ge 2$</span>. Then
<span class="math-container">$$
(1+x)^{n+1}=(1+x)^n(1+x)>(1+x^n)(1+x)=1+x+x^n+x^{n+1}>1+x^{n+1}
$$</span>
since <span class="math-container">$x>0$</span> where in the first inequal... |
1,874,159 | <blockquote>
<p>Given real numbers $a_0, a_1, ..., a_n$ such that $\dfrac {a_0}{1} + \dfrac {a_1}{2} + \cdots + \dfrac {a_n}{n+1}=0,$ prove that $a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n=0$ has at least one real solution.</p>
</blockquote>
<p>My solution:</p>
<hr>
<p>Let $$f(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a... | Aweygan | 234,668 | <p>Your proof looks fine. If you wanted to expand, you could add the following:</p>
<blockquote>
<p>Suppose $f(x)>0$ for all $x>0$. Then we must have
$$\int_0^1f(x)\ dx>0,$$
But we have already shown that $\int_0^1f(x)\ dx=0$, a contradiction.</p>
<p>If we assume $f(x)<0$ for all $x>0$, we ... |
402,970 | <p>I had an exercise in my text book:</p>
<blockquote>
<p>Find $\frac{\mathrm d}{\mathrm dx}(6)$ by first principles.</p>
</blockquote>
<p>The answer that they gave was as follows:</p>
<p>$$\lim_{h\to 0} \frac{6-6}{h} = 0$$</p>
<p>However surely that answer would be undefined if we let $h$ tend towards $0$, as fo... | Adriano | 76,987 | <p>When evaluating the limit:
$$\lim_{h\to 0} \frac{6-6}{h}$$
notice that (provided that $h \ne 0$), we have:
$$\frac{6-6}{h}=\frac{0}{h}=0$$
Thus, since it is possible to evaluate $\lim_{x\to a}f(x)$ even if $f(a)$ is undefined, we have:
$$\lim_{h\to 0} \frac{6-6}{h} = \lim_{h\to 0} 0 = 0$$</p>
|
236,546 | <p>$f(x)=x^4-16x^2+4$, the root of $f(x)$ is $a= \sqrt{3} + \sqrt{5}$</p>
<p>Factorise $f(x)$ as a product of irreducible polynomials over $\mathbb{Q}$, over $\mathbb{R}$ and over $\mathbb{C}$.</p>
<p>I am really confused as to how to start.</p>
| Community | -1 | <p>$R=K[X,Y]_{(X,Y)}$ is a local regular ring of dimension $2$ with maximal ideal $M=(X,Y)R$. Then $F\in M-M^2$, so $R/(F)$ is local regular of dimension $1$, hence DVR.</p>
<p><strong>Edit.</strong> At Makoto Kato request I'll sketch a proof of the following assertion: $(R,M,k)$ local regular and $F\in M-M^2$, then $... |
2,240,756 | <p>I tried rewriting $(1+x+x^2)^\frac{1}{x}$ as $e^{\frac{1}{x}\ln(1+x+x^2)}$ and then computing the taylor series of $\frac{1}{x}$ and $\ln(1+x+x^2)$ but I'm still not getting the correct answer..</p>
| Paramanand Singh | 72,031 | <p>Note that $$f(x) = (1 + x + x^{2})^{1/x} = \exp\left(\frac{\log(1 + x + x^{2})}{x}\right) = \exp\left(\frac{\log(1 - x^{3}) - \log(1 - x)}{x}\right)$$ so that we can get the series $$\frac{\log(1 - x^{3}) - \log(1 - x)}{x} = \sum_{n = 1}^{\infty}\frac{x^{n - 1} - x^{3n - 1}}{n} = 1 + \frac{x}{2} - \frac{2x^{2}}{3} +... |
735,563 | <p>For both sets C and D, provide a proof that C ∪ (D − /C) = C</p>
<p>"/C" is a set's complement of C</p>
| gt6989b | 16,192 | <p>Imagine that $A$ is diagonalizable, i.e, you can write $A = P^{-1}DP$ for some $P$ and diagonal $D$. Then,
$$
A^2 = \left( P^{-1}DP \right) \left( P^{-1}DP \right)
= \left( P^{-1}D \left(P P^{-1}\right) DP \right)
= \left( P^{-1}D^2 P \right)
$$</p>
<p>Can you take this from here?</p>
<p><strong>EDIT</st... |
735,563 | <p>For both sets C and D, provide a proof that C ∪ (D − /C) = C</p>
<p>"/C" is a set's complement of C</p>
| kleineg | 120,144 | <p>To find the diagonal form of the matrix you would calculated the eigenvalues and eigenvectors. $D$ is a diagonal matrix where the entries are the eigenvalues of $A$, $P$ is a matrix formed by the eigenvectors (each eigenvector is a column corresponding to the column its eigenvalue is in for $D$). </p>
|
4,930 | <p>This is not an urgent question, but something I've been curious about for quite some time.</p>
<p>Consider a Boolean function in <em>n</em> inputs: the truth table for this function has 2<sup><em>n</em></sup> rows.</p>
<p>There are uses of such functions in, for example, computer graphics: the so-called ROP3s (ter... | John D. Cook | 136 | <p>See <a href="http://en.wikipedia.org/wiki/Karnaugh_map" rel="nofollow">Karnaugh maps</a></p>
|
4,930 | <p>This is not an urgent question, but something I've been curious about for quite some time.</p>
<p>Consider a Boolean function in <em>n</em> inputs: the truth table for this function has 2<sup><em>n</em></sup> rows.</p>
<p>There are uses of such functions in, for example, computer graphics: the so-called ROP3s (ter... | Jose Brox | 1,234 | <p>For mor information, you should look at any book about Electronic Design Technology. When you manufacture a microchip, you want to keep it as simple as possible while integrating all the functions you are interested on. Thus, you want to minimize the number of transistors you use (i.e., to minimize the number of log... |
388,950 | <p>I've been trying to get my head around this for days. I understand what is going on with the calculation of a linear recurrence and I also understand how the characteristic is obtained.</p>
<p>What is confusion me is the general solution.</p>
<p>The general solution to the recurrence $at(n) + bt(n-1) ct(n-2) = 0$<... | Andreas Blass | 48,510 | <p>Your computation and the resulting formula look fine, but there seems to be a problem with the graph. I think you plotted the intercept 32 on the vertical axis, assuming that the vertical direction represents $F$, but then I think you plotted a slope of $5/9$ rather than $9/5$, assuming that the vertical axis repre... |
2,461,918 | <p>Often a function (real, say) is written without mentioning its domain, co-domain, just the rule $y=f(x)$ is given. In that case, how does one determine the domain, co-domain and range? For example, consider $f(x)=1/x$.</p>
| Siong Thye Goh | 306,553 | <p>The practice of not stating the domain and co-domain is a bad practice.</p>
<p>A function is only well defined only when the domain, co-domain and the rules are given. They should be pre-specified in order to define the function. The range are the values in the codomain that has a preimage. Codomain can be anything... |
1,485,463 | <p>At a school function, the ratio of teachers to students is $5:18$. The ratio of female students to male students is $7:2$. If the ratio of the female teachers to female students is $1:7$, find the ratio of the male teachers to male students. </p>
| hanish | 281,010 | <p>Lets consider there are 7 female student and 2 male student.(7:2)</p>
<p>total student = 9</p>
<p>teacher : student = 5:18</p>
<p>therefore for 18 student there will be 14 girls and 4 boy students</p>
<p>5:14:4</p>
<p>now only 1 female teacher is required for 7 female students (1:7)</p>
<p>here we have 14 fema... |
113,960 | <p>When trying to compute the (Serre-generalized) intersection number of two varieties at a closed point, I came to a need to compute the following $\operatorname{Tor}$:</p>
<blockquote>
<p>Let $k$ be an algebrically closed field, $A=k[x_1,x_2,x_3,x_4]$ and $\mathfrak m=(x_1,x_2,x_3,x_4)$. Let $M = k[x_1,x_2,x_3,x_4... | curious | 223 | <p>(I will omit the localization in this answer). </p>
<p>As $N$ has projective dimension $2$, you only need to worry about $i=0,1,2$. When $i=0$ the Tor is just tensor product, and tensoring with $N$ means setting $x_1=x_3$, $x_2=x_4$. Thus $M\otimes N = k[x_1,x_2]/(x_1^2,x_1x_2,x_2^2)$. The length is $3$.</p>
<p>Fo... |
186,417 | <p>Is it possible to write $0.3333(3)=\frac{1}{3}$ as sum of $\frac{1}{4} + \cdots + \cdots\frac{1}{512} + \cdots$ so that denominator is a power of $2$ and always different?
As far as I can prove, it should be an infinite series, but I can be wrong.
In case if it can't be written using pluses only, minuses are allowed... | Richard Evans | 39,080 | <p>More generally we can do this for any fraction $\frac{m}{n}$. We know that the sum of the infinite geometric series
$$ S=a+ar+ar^2+ar^3+\dots=\frac{a}{1-r} $$
if $ |r| < 1 $. So we could take $r=-\frac{1}{n-1}$ and $a=\frac{m}{n-1}$ giving</p>
<p>$$ S=\frac{\frac{m}{n-1}}{1+\frac{1}{n-1}}=\frac{\frac{m}{n-1}}{\... |
3,839,244 | <p>Is <span class="math-container">$\{3\}$</span> a subset of <span class="math-container">$\{\{1\},\{1,2\},\{1,2,3\}\}$</span>?</p>
<p>If the set contained <span class="math-container">$\{3\}$</span> plain and simply I would know but does the element <span class="math-container">$\{1,2,3\}$</span> include <span class=... | fleablood | 280,126 | <p>No.</p>
<p>the elements inside elements of the set do not count.</p>
<p>The elments of your big set are:</p>
<ul>
<li><span class="math-container">$\{1\}$</span></li>
<li><span class="math-container">$\{1,2\}$</span></li>
<li><span class="math-container">$\{1,2,3\}$</span>.</li>
</ul>
<p>The elements of your small s... |
159,438 | <p>Can be easily proved that the following series onverges/diverges?</p>
<p>$$\sum_{k=1}^{\infty} \frac{\tan(k)}{k}$$</p>
<p>I'd really appreciate your support on this problem. I'm looking for some easy proof here. Thanks.</p>
| Pedro | 23,350 | <p>A simle example would be the term $\tan 121 / 121$. Noting that $\dfrac{\pi}{2}+2 \cdot 19 \cdot \pi \approx 120,95131$ shows why the term is much larger relative to the others. Here you can see a plot and spot the rogues. Those are the first 20k terms.</p>
<p><img src="https://i.stack.imgur.com/pMtLh.jpg" alt="ent... |
1,935,320 | <p>If $a^2-b^2=2$ then what is the least possible value of: \begin{vmatrix} 1+a^2-b^2 & 2ab &-2b\\ 2ab & 1-a^2+b^2&2a\\2b&-2a&1-a^2-b^2 \end{vmatrix}</p>
<p>I tried to express the determinant as a product of two determinants but could not do so. Seeing no way out, I tried expanding it but that ... | Jacky Chong | 369,395 | <p>Consider the matrix
\begin{align}
A=
\begin{pmatrix}
\cos \frac{2\pi}{3} & -\sin\frac{2\pi}{3}\\
\sin\frac{2\pi}{3} & \cos \frac{2\pi}{3}
\end{pmatrix}
\end{align}</p>
|
135,235 | <blockquote>
<p>if $f(3)=-2$ and $f'(3)=5$, find $g'(3)$ if,<br>
$g(x)=3x^2-5f(x)$</p>
</blockquote>
<p>the answer is -7,
I find that very hard to understand the question.
thanks </p>
| chemeng | 25,845 | <p>Since it's a homework question here are some tips. </p>
<ol>
<li>Are $f,g$ differentiable functions everywhere??If not, how can you find the derivative of $g$ at a specific point (in our case 3).</li>
<li>Find everything needed for calculating $g'(3)$ from the equation given.</li>
</ol>
|
135,235 | <blockquote>
<p>if $f(3)=-2$ and $f'(3)=5$, find $g'(3)$ if,<br>
$g(x)=3x^2-5f(x)$</p>
</blockquote>
<p>the answer is -7,
I find that very hard to understand the question.
thanks </p>
| dato datuashvili | 3,196 | <p>$g'(x)=6*x-5*f'(x)$</p>
<p>$g'(3)=6*3-5*f'(3)$=$18-25=-7$
i hope it would help you</p>
|
2,992,477 | <p>Can u help me prove that every simple module is cyclic? Ive already proved that a cyclic module is isomorphic to A quotiented by I ( I being a left ideal of A) using homomorphisms, but i cant seem to make any advance in proving this.Thx in advance.</p>
| egreg | 62,967 | <p>Suppose <span class="math-container">$M$</span> is a simple left <span class="math-container">$R$</span>-module. This means that <span class="math-container">$M\ne\{0\}$</span> and that the only submodules of <span class="math-container">$M$</span> are <span class="math-container">$\{0\}$</span> and <span class="mat... |
1,867,352 | <p>Find all the angles $v$ between $-\pi$ and $\pi$ such that</p>
<p>$$-\sin(v)+ \sqrt3 \cos(v) = \sqrt2$$</p>
<p>The answer has to be in the form of: $\pi/2$ (it must include $\pi$)</p>
<p>I have tried squaring but I get nowhere.</p>
| Zain Patel | 161,779 | <p>It's sometimes a good idea to turn a sum of sines and cosines into a single trigonometric term when solving equations of the form $a \sin x + b\cos x = c$, in this case: $$\sqrt{3}\cos v - \sin v \equiv 2\cos \left(v + \frac{\pi}{6}\right)$$</p>
<p>So, if you set $x = v + \frac{\pi}{6}$ you need only solve $\cos x ... |
1,867,352 | <p>Find all the angles $v$ between $-\pi$ and $\pi$ such that</p>
<p>$$-\sin(v)+ \sqrt3 \cos(v) = \sqrt2$$</p>
<p>The answer has to be in the form of: $\pi/2$ (it must include $\pi$)</p>
<p>I have tried squaring but I get nowhere.</p>
| DonAntonio | 31,254 | <p>Divide the equation by $\;2\;$ :</p>
<p>$$\frac1{\sqrt2}=-\frac12\sin x+\frac{\sqrt3}2\cos x=\sin\frac\pi3\cos x-\cos\frac\pi3\sin x=\sin\left(\frac\pi3-x\right)\implies$$</p>
<p>$$\frac\pi3-x=\begin{cases}\cfrac\pi4\\{}\\\cfrac{3\pi}4\end{cases}\implies \ldots$$</p>
<p>Observe this is similar to the other answer... |
3,450,692 | <p>How to solve this equation?</p>
<p><span class="math-container">$$
\frac{dy}{dx}=\frac{x^{2}+y^{2}}{2xy}
$$</span></p>
| Dr. Sonnhard Graubner | 175,066 | <p>It is <span class="math-container">$$\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}\right)$$</span>,lets do <span class="math-container">$$u=\frac{y}{x}$$</span> then <span class="math-container">$$ux=y$$</span> and <span class="math-container">$$y'=u'x+u$$</span></p>
|
3,450,692 | <p>How to solve this equation?</p>
<p><span class="math-container">$$
\frac{dy}{dx}=\frac{x^{2}+y^{2}}{2xy}
$$</span></p>
| user577215664 | 475,762 | <p>Another approach:
<span class="math-container">$$\frac{dy}{dx}=\frac{x^{2}+y^{2}}{2xy}$$</span>
<span class="math-container">$$2xydy=x^2dx+y^2dx$$</span>
<span class="math-container">$$xdy^2-y^2dx=x^2dx$$</span>
<span class="math-container">$$\frac {xdy^2-y^2dx}{x^2}=dx$$</span>
<span class="math-container">$$d\left... |
426,264 | <ol>
<li><p>Given one sine wave in time domain, I want to find its frequency. Because I observe only a very small part of the sine wave ~1 cycle, FFT methods have a poor spectral resolution. </p></li>
<li><p>Has there been work that bounds the error on the frequency estimate? </p></li>
</ol>
<p>Thanks a ton</p>
| Christian Blatter | 1,303 | <p>I don't think that FFT is of any use here. You are given a data set $(t_k,x_k)$ $\>(1\leq k\leq N)$ and want to know the best fit for a function of the form
$$x(t)=A\cos(\omega t)+B\sin(\omega t)+C\ ,$$
where $A$, $B$, $C$, $\omega$ are parameters to be determined. This is a standard problem of numerical analysis... |
102,402 | <p>I wanted to make a test bank of graphs of linear equations for my algebra classes. I want the $y$-intercept of each graph to be an integer no less than $-10$ and no greater than $10$. Generally, you want these graphs to be small, so i've decided on a $20 \times 20$ grid (10 units from the origin). Additionally, i ... | Gerry Myerson | 8,269 | <p>Here's a simple upper bound. There are 21 points on the $y$-axis, and there are 420 points in your grid that are not on the $y$-axis, so there are at most $21\times420$ lines. Of course, there are actually fewer, because lines with 3 or more points are getting counted more than once. Two random integers less than $n... |
245,623 | <p><em>For the following vectors $v_1 = (3,2,0)$ and $v_2 = (3,2,1)$, find a third vector $v_3 = (x,y,z)$ which together build a base for $\mathbb{R}^3$.</em></p>
<p>My thoughts:</p>
<p>So the following must hold:</p>
<p>$$\left(\begin{matrix}
3 & 3 & x \\
2 & 2 & y \\
0 & 1 & z
\end{matrix}\... | egens | 50,866 | <p>There is more general solution, that assumes finding normalized basis of given linear subspace and then complement it to full basis by solving several homogeneous systems.</p>
<p><strong>Basis normalisation.</strong>
Suppose having $m$ linear independent vectors $\tilde{v}_1..\tilde{v}_m$ in $R^n$. Linear independe... |
3,845,144 | <p>Let's say a number <span class="math-container">$n$</span> is insertable if for every digit <span class="math-container">$d$</span>, if we insert <span class="math-container">$d$</span> between any two digits of <span class="math-container">$n$</span>, then the obtained number is a multiple of <span class="math-cont... | N. S. | 9,176 | <p>The following solution is using the eigenvalues in an implicit way.</p>
<p><span class="math-container">$$A^3=2A^2+A-I_3$$</span></p>
<p>Which can be found by inspection or by calculating the characteristic polynomial.</p>
<p>Then, by long division
<span class="math-container">$$X^n=Q(X)(X^3-2X^2-X+1)+aX^2+bX+c$$</s... |
120,910 | <p>$f(x) = 2\sin x \hspace{10pt}(0 \leq x \leq \pi)$<br>
$g(x) = -\sin x \hspace{10pt}(0 \leq x \leq \pi)$</p>
<p>Rectangle ABCD is enclosed between the above functions' graphs (its edges are parallel to the axes).</p>
<p>How would I go about finding the maximum perimeter of ABCD?</p>
<p>I'm really clueless about th... | David Mitra | 18,986 | <p>If the left vertical side of the rectangle passes through $x=a$ then the height of the rectangle is $2\sin a +\sin a=3\sin a$. The right vertical side of the rectangle then passes through $\pi-a$ (draw a picture). So the width of the rectangle whose left vertical side passes through $x=a$ is $(\pi-2a)$. The perimet... |
3,264,935 | <p>For a given natural number <span class="math-container">$k$</span>, I'm going to call a subset <span class="math-container">$T$</span> of the plane <span class="math-container">$\Bbb{R}^2$</span> a <span class="math-container">$k$</span>-traversal if, for any <span class="math-container">$x \in \Bbb{R}$</span>,
<spa... | GEdgar | 442 | <p>Vincent's example... <span class="math-container">$T$</span> is the union of two graphs <span class="math-container">$y=\frac{|x-1|}{2}$</span> and <span class="math-container">$y=-\frac{|x+1|}{2}$</span>.</p>
<p><a href="https://i.stack.imgur.com/1gMR6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.... |
686,848 | <p>This is probably an elementary question about fields, but I think it is a little tricky. </p>
<blockquote>
<p>Can we make the integers $\mathbb{Z}$ into a field?</p>
</blockquote>
<p>Let me be more precise. Is it possible to make $\mathbb{Z}$ into a field so that the underlying additive group structure is the us... | Elchanan Solomon | 647 | <p>The multiplicative structure in $\mathbb{Z}$ is determined by the additive structure. We start with the axiom of the multiplicative identity: $1 \times n = n$ for all $n \in \mathbb{Z}$. The axiom of distributivity implies
$$2 \times n = (1 + 1) \times n = (1 \times n) + (1 \times n) = n + n = 2n$$</p>
<p>This det... |
849,025 | <p>Say I have 100 people each with a height, weight, and age. I make a regression that predicts age based on height and weight. Now, I would like to update that model when I meet someone new. I don't want to just re-process 101 people though--I want to take the model that I already have and incorporate the new person i... | Michael Hardy | 11,667 | <p>Here I am thinking I should remember something about this, or maybe I shouldn't. We have a matrix "equation":
$$
\begin{bmatrix} 1 & h_1 & w_1 \\ \vdots & \vdots & \vdots \\ 1 & h_n & w_n \end{bmatrix} \begin{bmatrix} 7 \\ 0.08 \\ 0.06 \end{bmatrix} \overset{\text{?}} = \begin{bmatrix} a_1 \... |
109,973 | <p>I have stuck solving this problem of financial mathematics, in this equation:
$$\frac{(1+x)^{8}-1}{x}=11$$</p>
<p>I'm stuck in this eight grade equation:
$$x^{8}+8x^{7}+28x^{6}+56x^{5}+70x^{4}+56x^{3}+28x^{2}+9x-11=0$$</p>
<p>But I cannot continue past this. This quickly lead me to find a general way of solving eq... | GeoffDS | 8,671 | <p>First off, let me say that the polynomial you gave does not match up with your original problem. So, I will solve it based on your original problem and not the polynomial you gave. Perhaps you typed it wrong.</p>
<p>Second off, since this is financial mathematics, there are most likely assumptions that go beyond ... |
2,442,705 | <p>Like the title says, I'm wondering if there's a non-brute-force way to determine $x$ such that
$$45\le x<200,$$
$$x\bmod5\equiv0,$$
$$x\bmod8\equiv1,$$
$$x\bmod12\equiv1$$
I know I can simply enumerate all integers in $[45, 200)$ to get the answer (145), but I'm wondering if there's a more elegant solution.</p>
| Caleb Stanford | 68,107 | <p>Well:</p>
<ul>
<li><p>Since the number is $0$ mod $5$, we don't have to check <em>all</em> integers in $[45,200)$, but only the multiples of $5$: $45, 50, 55, \ldots$</p></li>
<li><p>But we also know $x \mod{8} = 1$, which happens one in every 8 multiples of $5$. So $x \mod 40 = 25$. Now we only have to check $65, ... |
2,294,585 | <p>I can't figure out why $((p → q) ∧ (r → ¬q)) → (p ∧ r)$ isn’t a tautology. </p>
<p>I tried solving it like this: </p>
<p>$$((p ∧ ¬p) ∨ (r ∧ q)) ∨ (p ∧ r)$$
resulting in $(T) ∨ (p ∧ r)$ in the end that should result in $T$. What am I doing wrong?</p>
| egreg | 62,967 | <p>We want to see this is not a tautology. This is the case if we find a choice of $p$, $q$ and $r$ so that</p>
<ol>
<li>$(p → q) ∧ (r → ¬q)$ is false</li>
<li>$p ∧ r$ is true</li>
</ol>
<p>In order for $(p → q) ∧ (r → ¬q)$ to be true, both $p → q$ and $r → ¬q$ must be true. Let's choose $q$ true. Then, as $¬q$ is fa... |
825,848 | <p>I'm trying to show that the minimal polynomial of a linear transformation $T:V \to V$ over some field $k$ has the same irreducible factors as the characteristic polynomial of $T$. So if $m = {f_1}^{m_1} ... {f_n}^{m_n}$ then $\chi = {f_1}^{d_1} ... {f_n}^{d_n}$ with $f_i$ irreducible and $m_i \le d_i$.</p>
<p>Now I... | nombre | 246,859 | <p>There is another proof using the properties of the determinant.</p>
<p><strong>Claim: $\chi$ divides $m^n$ in $k[X]$.</strong></p>
<p>To see this, fix a basis of $V$ and consider the matrix $M$ of $T$ in said basis. We see $\mathcal{M}_n(k),\mathcal{M}_n(k[X])$ as a subrings of $\mathcal{M}_n(k(X))$ given the natu... |
428,291 | <p>What is the solution of the following system?</p>
<p>$$
\begin{align}
a \cdot e-b \cdot d & =\alpha \\
a \cdot f-c \cdot d & =\beta \\
b \cdot f-c \cdot e & =\gamma
\end{align}
$$</p>
<p>Where the unknowns $a,b,c,d,e,f$ are real numbers and $\alpha, \beta, \gamma$ are fixed real numbers.</p>
<p>I trie... | Robert Israel | 8,508 | <p>The resultant of the first two equations with respect to variable $a$ is
$bdf - cde + \alpha f - \beta e$. You can solve this and the third equation as
linear equations in $e,f$ (assuming $\alpha c - \beta b \ne 0$).</p>
|
2,340,606 | <p>$$x^2-2(3m-1)x+2m+3=0$$
Find the sum of solutions. It says that the sum equals to $-1$. I just can't wrap my head around this? Any help? Thx</p>
| Robert Z | 299,698 | <p>In a quadratic equation $ax^2+bx+c=0$ the sum of the solutions (if any) is equal to $-b/a$. In fact, if $x_1$ and $x_2$ are the solutions then
$$ax^2+bx+c=a(x-x_1)(x-x_2)=ax^2-a(x_1+x_2)x+a(x_1\cdot x_2)$$
and by comparing the coefficients we get $x_1+x_2=-b/a$.</p>
<p>In your case $x_1+x_2=2(3m-1)$. It seems that... |
317,981 | <p>Prove the following:</p>
<p>$$\lim_{n \to \infty} \displaystyle \int_0 ^{2\pi} \frac{\sin nx}{x^2 + n^2} dx = 0$$</p>
<p>How would I prove this? I know you have to show your steps, but I'm literally stuck on the first one, so I can't. </p>
| Fly by Night | 38,495 | <p>We can come up with some bounds for this integral $I$. First notice that
$$|I|=\left|\int_0^{2\pi} \frac{\sin (nx)}{x^2+n^2} \operatorname{d}\!x \right| \le \int_0^{2\pi} \left|\frac{\sin (nx)}{x^2+n^2}\right| \operatorname{d}\!x \le \int_0^{2\pi} \frac{1}{\left|x^2+n^2\right|} \operatorname{d}\!x \le \int_0^{2\pi} ... |
314,416 | <p>Part of what it means to be a functor between two categories is to have a map of morphisms e.g. $F$ sends $f: A \to B$ to $Ff: FA \to FB$.</p>
<p>Suppose $F$ is a functor from a category to itself, and that category has (some or all) exponential objects. $B^A$ intuitively corresponds to an object of morphisms $A \t... | user1551 | 1,551 | <p>Read the Wiki page again. It says $M$ is normal if and only if $M=U\Lambda U^\ast$ for some diagonal matrix $\Lambda$ and some unitary matrix $U$. Note that $\Lambda$ may be <strong><em>non-real</em></strong>. However, in a SVD $M=U\Sigma V^\ast$, $\Sigma$ is a <strong><em>real</em></strong> diagonal matrix. So, the... |
13,922 | <p>How to sum up this series :</p>
<p>$$2C_o + \frac{2^2}{2}C_1 + \frac{2^3}{3}C_2 + \cdots + \frac{2^{n+1}}{n+1}C_n$$
</p>
<p>Any hint that will lead me to the correct solution will be highly appreciated.</p>
<p>EDIT: Here $C_i = ^nC_i $</p>
| Aryabhata | 1,102 | <p>For an elementary method which does not use calculus:</p>
<p>Notice that $\displaystyle \dfrac{{n \choose k}}{k+1} = \dfrac{1}{n+1} {n+1 \choose k+1}$
</p>
<p>Thus your sum is </p>
<p>$$\sum_{k=0}^{n} \dfrac{1}{n+1} {n+1 \choose k+1} 2^{k+1} = \dfrac{\sum_{k=0}^{n+1} {n+1 \choose k} 2^k -1}{n+1} = \dfrac{3^{n+1} ... |
13,922 | <p>How to sum up this series :</p>
<p>$$2C_o + \frac{2^2}{2}C_1 + \frac{2^3}{3}C_2 + \cdots + \frac{2^{n+1}}{n+1}C_n$$
</p>
<p>Any hint that will lead me to the correct solution will be highly appreciated.</p>
<p>EDIT: Here $C_i = ^nC_i $</p>
| Bill Dubuque | 242 | <p><strong>REMARK</strong> $\ $ The various approaches are all equivalent. Namely, suppose that we desire to prove without calculus the identity arising from integrating the binomial formula, viz. $$\rm (1 + x)^{n+1}\ =\ 1 + \sum_{k=1}^{n+1}\: \frac{n+1}{k+1} {n\choose k}\ x^{k+1}$$
</p>
<p>Comparing coefficients red... |
1,711,673 | <p>Why are the inverses of 2-cycles of the symmetric permutation group $S_3$ the elements themselves rather than their reverses?</p>
<p>E.g. why is $(1,2)^{-1}=(1,2)$ and not $(2,1)$?</p>
| Jack D'Aurizio | 44,121 | <p>$$\frac{1-\cos^3 x}{x^2} = \frac{2\sin^2\frac{x}{2}}{x^2}\cdot\left(1+\cos x+\cos^2 x\right)=\frac{1}{2}\cdot\left(\frac{\sin(x/2)}{x/2}\right)^2\cdot \left(1+\cos x+\cos^2 x\right)$$</p>
<p>and now no factor is troublesome in a neighborhood of zero, since:
$$ \frac{\sin z}{z}=\sum_{n\geq 0}\frac{(-1)^n z^{2n}}{(2n... |
1,711,673 | <p>Why are the inverses of 2-cycles of the symmetric permutation group $S_3$ the elements themselves rather than their reverses?</p>
<p>E.g. why is $(1,2)^{-1}=(1,2)$ and not $(2,1)$?</p>
| Clement C. | 75,808 | <p>To "remove" the cancellation <em>at $0$</em>, one can use any of the "usual suspects" (basic limits, L'Hôpital, you name it; I would go for Taylor expansions, to low order). For instance, from $\cos u = 1- \frac{u^2}{2} + o(u^2)$ and $(1+u)^3 = 1+3u+o(u)$ at $0$, you get
$$
\frac{1-\cos^3 x}{x^2} = \frac{1-(1-\frac... |
4,093,406 | <p>Which of the equations have at least two real roots?
<span class="math-container">\begin{aligned}
x^4-5x^2-36 & = 0 & (1) \\
x^4-13x^2+36 & = 0 & (2) \\
4x^4-10x^2+25 & = 0 & (3)
\end{aligned}</span>
I wasn't able to notice something clever, so I solved each of the equations. The first one h... | acernine | 331,233 | <p>This is a case of a "hidden quadratic" in that your equations are all really quadratic equations in a different variable. To see this, write <span class="math-container">$u=x^2$</span>. Then for example your first equation becomes
<span class="math-container">$$u^2-5u-36=0.$$</span>
The temptation is then ... |
2,549,834 | <p>Ive been looking at this problem and trying to use examples online to try to solve it but I get stuck. </p>
<p>It says to use mathematical induction to prove
(1/1*4)+(1/4*7)+(1/7*10)+ ... + (1/(3n-2)(3n+1)) = n/(3n+1)</p>
<p>I solve for n=1 and substitute k for n, but I don’t really know what to do after that step... | Andres Mejia | 297,998 | <p><strong>Hint:</strong> $\big||f(x)|-|f(x_0)|\big| \leq |f(x)-f(x_0)|$ by the <a href="https://math.stackexchange.com/questions/127372/reverse-triangle-inequality-proof">reverse triangle inequality</a>.</p>
|
1,761,189 | <p>Suppose that voters arrive to a voting booth according to a Poisson process with rate $\lambda = 100$ voters per hour. The voters will vote for two candidates, candidate $1$ and candidate $2$ with equal probability $\left(\frac{1}{2}\right)$. I am assuming that we start the voting at time $t=0$ and it continues inde... | MJ73550 | 331,483 | <p>You have :</p>
<p>$$\mathbb{P}(A_4=k|N_{10}= 1000)=\frac{\mathbb{P}(A_4=k,N_{10}=1000)}{\mathbb{P}(N_{10}=1000)}$$</p>
<p>$$\mathbb{P}(N_{10}=1000)=\frac{(10\times 100)^{1000}}{1000!}e^{-10\times 100}$$</p>
<p>You have :
$$A_t = \sum_{i=1}^{N_t} \xi_i$$
where $\xi_i$ are independent and identically distributed $$... |
1,166,524 | <p>I am looking for all irreducible polynomials of degree 5 in $\mathbb{F}_{17}$
with have the form h(y) = $y^5+C$.</p>
<p>I think there aren't any irreducible polynomials of this form because for every C
I can find an element of $\mathbb{F}_{17}$ as a root.</p>
<p>What would be the fastest way to find irreducible po... | Brent Kerby | 218,224 | <p>You are correct that there no irreducible polynomials of the form $h(y)=y^5+C$ over $\Bbb F_{17}$ because every such polynomial has a root in $\Bbb F_{17}$. An easy way to see this is to notice that $\Bbb F_{17}^*$ is a cyclic group of order 16, which is relatively prime to 5, so that $y \mapsto y^5$ is an automorph... |
1,133,581 | <p>$\lim\limits_{(x,y) \to (0,0)} \frac{\log(1-x^2y^2)}{x^2y^2}$</p>
<p>in polar coordinates:</p>
<p>$\lim\limits_{r \to 0} \frac{\log(1-r^2\cos^2t \sin^2t)}{r^2\cos^2t \sin^2t}$</p>
<p>The first limit exists if the second is independent from $t$. But if $t$ is $k \pi/2$ that fraction does not exist. So I could argu... | user2820579 | 141,841 | <p>In case you are still interested. <a href="https://rads.stackoverflow.com/amzn/click/com/1493927116" rel="nofollow noreferrer" rel="nofollow noreferrer">Abbot's</a> has this problem as Example 3.3.7. For each point <span class="math-container">$x\in(0,1)$</span> consider the open interval <span class="math-container... |
4,246,461 | <p>I am an economics undergrad who wants to develop a firm grasp over mathematics, so will I be able to understand Bourbaki's books on my own without needing additional books? What about the books titled analysis released by Roger Godement?</p>
| hm2020 | 858,083 | <p><strong>Question:</strong> "What mathematical education should I have in order to study the books released under the pseudonym N. Bourbaki?"</p>
<p><strong>Answer:</strong> For the algebra and commutative algebra books I recommend the algebra book of Lang and the commutative algebra books of Atiyah-Macdona... |
42,016 | <p>I am looking for algorithms on how to find rational points on an <a href="http://en.wikipedia.org/wiki/Elliptic_curve">elliptic curve</a> $$y^2 = x^3 + a x + b$$ where $a$ and $b$ are integers. Any sort of ideas on how to proceed are welcome. For example, how to find solutions in which the numerators and denominator... | Robin Chapman | 4,213 | <p>There is a whole industry devoted to this. The basic method is by
<em>descent</em>, which is a formalized version of the infinite descent proofs
of Fermat and Euler. It helps if there are rational 2-torsion points
but it's not essential. Chapter X in Silverman's <em>The Arithmetic of Elliptic
Curves</em> is called "... |
47,181 | <p>I'll ask you to consider a situation wherein one has a series of edges for a graph, $(e_1, e_2, ..., e_N) \in E$, each with a specifiable length $(l_1, l_2, ..., l_N) \in L$, and the goal is to insure that the connected graph has a unique topology in 3-space. More specifically, I'm interested in insuring that some ... | sleepless in beantown | 8,676 | <p>You are asking about the <strong>embedding of a graph structure into 3-space</strong> $\mathbb{R}^3$. A graph structure by itself does not specify its embedding into $n$-space. In chemistry, these two different chiral instances of (tetrahedral) molecules below would be called stereo-isomers or enantiomers of each ot... |
2,470,062 | <blockquote>
<p><span class="math-container">$$\sqrt{k-\sqrt{k+x}}-x = 0$$</span></p>
<p>Solve for <span class="math-container">$k$</span> in terms of <span class="math-container">$x$</span></p>
</blockquote>
<p>I got all the way to
<span class="math-container">$$x^{4}-2kx^{2}-x+k^{2}-x^{2}$$</span>
but could not facto... | Blue | 409 | <p>(Too long for a comment.)</p>
<p>As @Leucippus has shown, the solutions are</p>
<p>$$k \quad=\quad x^2 + x + 1 \qquad\text{or}\qquad x^2 - x$$</p>
<p>Thus,
$$k + x \quad=\quad (x+1)^2 \qquad\text{or}\qquad x^2$$
so that
$$\sqrt{k+x} \quad=\quad |x+1| \qquad\text{or}\qquad |x|$$
Further,
$$k - \sqrt{k+x} \quad=\qu... |
68,899 | <p>I was rereading basic results on de Rham cohomology, and this led me inevitably to the fact that $H^q(X,\Omega^p)$ converges to $H^*(X)$ for any smooth proper variety (over any field). How does one view this spectral sequence "maturely" as a Grothendieck spectral sequence?</p>
| Mariano Suárez-Álvarez | 1,409 | <p>You probably want one (the second one...) of the two hypercohomology spectral sequences which compute $\mathbb H^\bullet(X,\Omega^\bullet)$, the hypercohomology of the de Rham complex. A reference for this is Weibel's book.</p>
<p>I doubt you can view this as a Grothendieck spectral sequence, but it has sufficient... |
1,358,574 | <p>When I searched for the derivative of the Gamma function I got something of the form:</p>
<p>$$\Gamma'(x)=\Gamma(x) \psi(x)$$</p>
<p>But from the definition of the Digamma function to me it's like writing:</p>
<p>$$\Gamma'(x)=\Gamma'(x)$$</p>
<p>And this doesn't seem very useful to me (if I'm wrong feel free to ... | Pedro | 23,350 | <p>A bit useless in light of Steven's answer, but nonetheless useful for future ideas. Recall that the Catalan numbers are equal to $\frac{1}{n+1}\binom {2n}n$ and count the number of sequences of length $2n$ of zeros and ones for which the number of ones is always at least that of zeros at any truncation of the sequen... |
2,590,537 | <p>Let $A$ be a proper subset of of $X$, $B$ be a proper subset of $Y$. If $X$ and $Y$ are connected, show that $(X\times Y)-(A\times B)$ is connected.</p>
<p>I have already seen the solution of this problem here on math.stackexchange, but my confusion is this.
Suppose $X$=(a,b) & $Y$=(c,d) and let $A$={m} &... | Dan | 500,478 | <p>For each $n-1$-vector you can append either a $0$ or a $1$ as its $n$th-coordinate to create an $n$-vector. So $A(k)=2A(k-1)$. If $A(k-1)=2^{k-1}$, then the result follows.</p>
|
400,296 | <p>There are n persons.</p>
<p>Each person draws k interior-disjoint squares.</p>
<p>I want to give each person a single square out of his chosen k, so that the n squares I give are interior-disjoint.</p>
<p>What is the minimum k (as a function of n) for which I can do this?</p>
<p>NOTES:</p>
<ul>
<li>For n=1, obv... | Kris Williams | 38,143 | <p>Here is my counter-example to the case of $n=2$ and $k=3$. Provided of course that the squares need not be the same size. You will have to trust me that all shapes are squares. However, we can see that each square overlaps all 3 of the squares of the other color.</p>
<p><img src="https://i.stack.imgur.com/p2kjF.png... |
2,070,574 | <p>I had a question </p>
<p>$$\sum_{a=1}^n \frac{1}{a^2}=?$$</p>
<p>I had learned newton's method of undetermined method
,but that doesn't work here because of negative power</p>
<p>than I saw another question </p>
<p>$$\sum_{a=1}^n \frac{1}{a(a+1)}$$</p>
<p>here we transformed it as</p>
<p>$$\sum_{a=1}^n (\frac... | Marsan | 392,650 | <p>I am not aware of any formulas for the finite partial sum... however we can compute the limit ! I think the easiest way to compute it is through trigonometric sums/estimations. Here is a sketch of the proof : </p>
<p>By using De Moivre's Formula, you can easily show that, for all real $t$ :</p>
<p>\begin{equation*... |
550,757 | <p>I don't see how $(5, 1 - 8i)$ would generate the element $5$. I think that $5$ and $1 - 8i$ would have to have a common factor but I can't find any that divide both.</p>
| TBrendle | 97,380 | <p>The prime factors of $5$ are $1+2i$ and $1-2i$ while the prime factors of $1-8i$ are $1+2i$ and $3+2i$. Therefore the GCD of $5$ and $1-8i$ is $1+2i$, or, in the language of ideals, we have $(5, 1-8i)=(1+2i)$.</p>
<p>More abstractly, you can prove that <em>any</em> ideal in the Gaussian integers must be principal, ... |
92,492 | <p>If $M$ is a combinatorial model category, it's known by the experts that there is a ''natural'' model structure on diagram categories $Hom(C,M)$, which is the <strong>projective model structure</strong>. The fibrations and weak equivalences are defined point wise. </p>
<p>There is also the one called <strong>inject... | user22479 | 22,479 | <p>This is really a comment on Igor Rivin's answer (I don't have enough rep. to comment, it seems), but one has to be attentive to whether or not the polynomial $p$ (not to be confused with the prime $p$...) is absolutely irreducible over the ground field, or at least whether its irreducible factors are absolutely irre... |
1,926,593 | <p>I am currently taking a course in Discrete Math. The first part of our lesson this week is regarding sequences. I am stuck on formulas like the ones shown in the images I attached... I was hoping someone might be able to help me learn how to solve them. :)</p>
<p>Ps: What does it mean when <span class="math-containe... | Bernard | 202,857 | <p>An inductive proof rests entirely on <em>Pascal's formula</em>. Note the above formula can as well be written:
$$\sum_{i=m}^n \binom{i}{m} = \binom{n+1}{m+1}\quad\text{or}\quad \binom{n+1}{m+1}=\sum_{k=0}^{n-m} \binom{n-k}{m}$$
We'll prove by induction on $l$ that, for all $l\;$ ($0\le l<n-m$):
$$ \binom{n+1}{m+1... |
41,000 | <p>The motivation for this question is producing a distribution that produces the gender and age of an individual when the distribution of ages depends on gender.</p>
<p>Suppose I want a distribution which, when RandomVariate is applied to it, produces two values. The first value it produces is either 0 or 1 and the ... | ciao | 11,467 | <p>I don't think there's any <em>direct</em> way to do this currently with the MM statistics/probability features - it would be nice if WRI extended things like <code>ProductDistribution</code> et. al. to allow conditioning/other logic. In any case, here's one way you could accomplish the end result:</p>
<pre><code>di... |
4,220,751 | <p>In the example in the following slide, we follow the highlighted formula. With regard to the highlight, I'm confused why the number is greater <strong>or equal</strong> to <span class="math-container">$2^{n-1}$</span>, while only need to be less than <span class="math-container">$2^n$</span> (not less than <strong>o... | Mr.Infinity | 930,710 | <p>Here is a hint: Notice that the matrix <span class="math-container">$A$</span> can be decomposed into the sum of an identity matrix and a nilpotent, that is <span class="math-container">$A=N+I$</span>, where <span class="math-container">$N^{\dim V}=N^3=0$</span>.</p>
<p>Therefore we yield <span class="math-container... |
1,648,354 | <p>We have been taught that linear functions, usually expressed in the form $y=mx+b$, when given a input of 0,1,2,3, etc..., you can get from one output to the next by adding some constant (in this case, 1).
$$
\begin{array}{c|l}
\text{Input} & \text{Output} \\
\hline
0 & 1\\
1 & 2\\
2 & 3
\end{array}
... | Simply Beautiful Art | 272,831 | <p>Your last table is incorrect, and as others have mentioned, tetration would be the next operation.</p>
<p>Let's try to figure out what these kinds of operations are and how they work, to begin with.</p>
<p>We will be considering $x+2$, $x\cdot2$, and $x^2$.</p>
<p>What is $x+2$? In simple terms, it means $x+1+1$... |
3,497,919 | <p><span class="math-container">$$\sum_{k=0}^{\Big\lfloor \frac{(n-1)}{2} \Big\rfloor} (-1)^k {n+1 \choose k} {2n-2k-1 \choose n} = \frac{n(n+1)}{2} $$</span></p>
<p>So I feel like <span class="math-container">$(-1)^k$</span> is almost designed for the inclusion-exclusion principle. And the left-hand side looks like s... | Mike Earnest | 177,399 | <p>Both sides of the equation are the answer to the following question:</p>
<blockquote>
<p>How many sequences of <span class="math-container">$n$</span> zeroes and <span class="math-container">$n-1$</span> ones are there where no two ones are adjacent?</p>
</blockquote>
<p>To see why <span class="math-container">$... |
1,703,332 | <p>Let $f \in L^2 \backslash L^\infty$ on some bounded domain. Let $h$ be a function such that $h(p) \to \frac 32$ as $p \to \infty$.</p>
<p>Consider $$I_p = \left(\int |f|^{h(p)}\right)^{\frac{p}{h(p)}}.$$</p>
<p>Is it true that $\lim_{p \to \infty} I_p \leq C$ exist for some finite constant $C$?</p>
<p>I don't th... | Community | -1 | <p>Turns out the problem is (a lot) easier than I thought .</p>
<p>Attach numbers to the positions : from $1$ to $n^2$ from left to right .</p>
<p>On the first column delete all the files except the first , which is useful (namely the files $n+1$ , $2n+1 , \ldots ,n^2-n+1$ )</p>
<p>Now take the useful file on the ro... |
2,013,382 | <blockquote>
<p>If $AX=\lambda X$ for every eigenvector $X\in \mathbb{R}^n$ then show that $A=\lambda I_n$.</p>
</blockquote>
<p>Given, $AX=\lambda X$ for every $X\in \mathbb{R}^n$.</p>
<p>Then $(A-\lambda I_n)X=0$ is true for every $X\in \mathbb{R}^n$.</p>
<p>But how to show $A=\lambda I_n$? Please help.</p>
| Nick | 359,825 | <p><strong>HINT:</strong>$(A-\lambda I)x=0$ for any $x\in \Bbb R^n$ implies $A-\lambda
I$ maps every $x\in \Bbb R^n$ to $0$, this is exactly the <a href="http://mathworld.wolfram.com/ZeroMap.html" rel="nofollow noreferrer">definition</a> of a zero transformation.</p>
|
3,447,872 | <p>Subtracting both equations
<span class="math-container">$$x(a-b)+b-a=0$$</span>
<span class="math-container">$$(x-1)(a-b)=0$$</span>
Since <span class="math-container">$a\not = b$</span>
<span class="math-container">$$x=1$$</span>
Substitution of x gives
<span class="math-container">$$1+a+b=0$$</span> which is con... | farruhota | 425,072 | <p>Let <span class="math-container">$x_1,x_2$</span> be the roots of the first equation and <span class="math-container">$y_1,y_2$</span> of the second.</p>
<p>By the Vieta's:
<span class="math-container">$$\begin{cases}x_1+x_2=-a \\ x_1x_2=b\end{cases} \Rightarrow x_1^2+x_2^2=a^2-2b\\
\begin{cases}y_1+y_2=-b \\ y_1y_... |
3,301,387 | <p>I don't understand what it is about complex numbers that allows this to be true.</p>
<p>I mean, if I root both sides I end up with Z=Z+1. Why is this a bad first step when dealing with complex numbers?</p>
| Community | -1 | <p>From</p>
<p><span class="math-container">$$(z+1)^6=z^6$$</span> you draw (taking the sixth root)</p>
<p><span class="math-container">$$z+1=\omega^k z$$</span> where <span class="math-container">$\omega$</span> is a primitive sixth root of unity, and <span class="math-container">$k=0,1,\cdots 5$</span>.</p>
<p>The... |
3,301,387 | <p>I don't understand what it is about complex numbers that allows this to be true.</p>
<p>I mean, if I root both sides I end up with Z=Z+1. Why is this a bad first step when dealing with complex numbers?</p>
| Nurator | 685,412 | <p>Try to understand it with real numbers first.
<span class="math-container">$$
y^2=x^2
$$</span>
does NOT mean
<span class="math-container">$$
y=x
$$</span></p>
<p>It means
<span class="math-container">$$
y=\pm x
$$</span>
instead.
With that in mind, try to understand that a complex circle has even more possible v... |
3,301,387 | <p>I don't understand what it is about complex numbers that allows this to be true.</p>
<p>I mean, if I root both sides I end up with Z=Z+1. Why is this a bad first step when dealing with complex numbers?</p>
| Michael Rozenberg | 190,319 | <p><span class="math-container">$$(z+1)^6-z^6=((z+1)^2-z^2)((z+1)^4+(z+1)^2z^2+z^4)=$$</span>
<span class="math-container">$$=(2z+1)((z+1)^4+2(z+1)^2z^2+z^4-(z+1)^2z^2)=$$</span>
<span class="math-container">$$=(2z+1)((z+1)^2+z^2-(z+1)z)((z+1)^2+z^2+(z+1)z)=$$</span>
<span class="math-container">$$=(2z+1)(z^2+z+1)(3z^2... |
1,377,927 | <p>Prove using mathematical induction that
$(x^{2n} - y^{2n})$ is divisible by $(x+y)$.</p>
<p><strong>Step 1:</strong> Proving that the equation is true for $n=1 $</p>
<p>$(x^{2\cdot 1} - y^{2\cdot 1})$ is divisible by $(x+y)$ </p>
<p><strong>Step 2:</strong> Taking $n=k$</p>
<p>$(x^{2k} - y^{2k})$ is divisible ... | bryan anthony | 966,357 | <p>When <span class="math-container">$n=1$</span>,</p>
<p><span class="math-container">$$x^{2n} - y^{2n} =x^2-y^2=(x+y)(x-y)$$</span></p>
<p>which is divisible by <span class="math-container">$(x+y)$</span>. Assume true for <span class="math-container">$n=k$</span>. Then <span class="math-container">$x^{2k}-y^{2k}$</sp... |
2,411,081 | <p>How can I show (with one of the tests for convergence , <strong>not by solving</strong>) that the integral $$\int _{1}^\infty\frac{\ln^5(x)}{x^2}dx$$ converges?</p>
| Nosrati | 108,128 | <p>We know
$$\ln u=\int_1^u\dfrac{1}{t}dt\leq\int_1^u\dfrac{t^\alpha}{t}dt=\dfrac{u^\alpha-1}{\alpha}<\dfrac{u^\alpha}{\alpha}$$
for all $\alpha>0$ so with $\alpha=\dfrac{1}{10}$:
$$\int _{1}^\infty\frac{\ln^5(x)}{x^2}dx \leqslant 10^5\int _{1}^\infty\frac{1}{x^\frac32}dx<\infty$$</p>
|
422,318 | <p>If $A$ is abelian group and $B$ is a subgroup of $A$, $B$ is normal subgroup of $A$.</p>
<p>Is it true that $B \times A/B \cong A$?</p>
<p>I ask because I was watching an online lecture from a course in abstract algebra at Harvard extension school. </p>
<p>And the lecturer (whose name is Peter) was taking about v... | DonAntonio | 31,254 | <p>Take $\,A=C_4=$ the cyclic group of order $\,4\,$ , and take $\,B=C_2\,$ , so is</p>
<p>$$C_4\cong C_2\times C_4/C_2=C_2\times C_2\;\;?$$</p>
|
422,318 | <p>If $A$ is abelian group and $B$ is a subgroup of $A$, $B$ is normal subgroup of $A$.</p>
<p>Is it true that $B \times A/B \cong A$?</p>
<p>I ask because I was watching an online lecture from a course in abstract algebra at Harvard extension school. </p>
<p>And the lecturer (whose name is Peter) was taking about v... | amWhy | 9,003 | <p>As suggested by DonAntonio:</p>
<p>We need only take $A =\mathbb Z_4$, $B = \mathbb Z_2$. Then $A/B = \mathbb Z_4/\mathbb Z_2 \cong \mathbb Z_2 = B.\;$ Clearly, $$\mathbb Z_4 \not\cong \mathbb Z_2 \times \mathbb Z_2 = B \times A/B $$</p>
<p>The same "conjecture" would fail for $A = \mathbb Z_9$ and $B = \mathbb Z... |
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