qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
3,600,139 | <blockquote>
<p>Solve <span class="math-container">$y''-y'-y=\cos x$</span>.</p>
</blockquote>
<p>After first solving the homogeneous equation we know that the solution to it is <span class="math-container">$$y(x)=a\sin(x)+b\cos(x).$$</span></p>
| Ankita Pal | 739,790 | <p>I have used D-operators method for this.
<span class="math-container">$$y''-y'-y=\cos x\\
\implies(D^2-D-1)y=\cos x.$$</span>
Let <span class="math-container">$$y=e^{mx}\\
\implies Dy=me^{mx}\\
\implies D^2y=m^2e^{mx}.$$</span>
<span class="math-container">$$\therefore m^2-m-1=0\\
\implies m=\frac{1\pm\sqrt{1+4}}{2}... |
3,762,306 | <p>I have the equation <span class="math-container">$$(e^x-1)-k\arctan(x) = 0$$</span> where <span class="math-container">$0<k \leq \frac 2\pi$</span> and I was wondering how I would go about starting to determine the amount of real roots of this equation. So far I have just manipulated the equation to get different... | Z Ahmed | 671,540 | <p>If <span class="math-container">$k>0$</span></p>
<p>Let <span class="math-container">$$f(x)=e^x-1-k \tan^{-1} x \implies f'(x)=e^x-\frac{k}{1+x^2}, f''(x)=e^{x}+\frac{k}{(1+x^2)^2}>0$$</span>
<span class="math-container">$\implies$</span> <span class="math-container">$f'(x)$</span> is an increasing function <... |
7,354 | <p>some of my students refer to there being an invisible $-1$ in front of the expression $-(x + 4)$ or in the exponent of $x$. While it is not phrased mathematically, I am ok with them saying this because it reminds them to distribute fully before simplifying etc. It got me thinking though is there a reason to not te... | Dirk | 3,398 | <p>Actually, I don't think that it is quite right that there really "is an invisible $-1$" in front of $-x$. I would say that one <strong>defines the symbol $-x$ to be the one that fulfills $x + (-x) = 0$</strong> and then one shows that $(-1) x =-x$.</p>
<p>Ok, you don't teach that at high-school level, but the messa... |
7,354 | <p>some of my students refer to there being an invisible $-1$ in front of the expression $-(x + 4)$ or in the exponent of $x$. While it is not phrased mathematically, I am ok with them saying this because it reminds them to distribute fully before simplifying etc. It got me thinking though is there a reason to not te... | Jack M | 115 | <p>It depends what you mean by "teach them to do it". If you mean "insist that they do it, correct them every time they don't and take off marks on a test", then no. If you mean "let them know about this trick as a way to remember how to distribute correctly, but don't insist they do it if they can remember without it"... |
1,652,381 | <p>Suppose $f$ derivable on $\mathbb R$ and that $\lim_{x\to a}f'(x)=\ell$. Show that $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\ell.$$</p>
<p>The proof of my course goes like this:</p>
<p>By mean value theorem, there is $y_h\in ]x,x+h[$ s.t. $$f(x+h)-f(x)=f'(y_h)h$$ and thus $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\t... | Dietrich Burde | 83,966 | <p>The next term is again $27$; then we had a palindromic sequence
$$
7,16,8,27,9,27,8,16,7,\ldots,
$$
repeating like this. </p>
|
2,816,365 | <p><a href="https://i.stack.imgur.com/T1W5r.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T1W5r.jpg" alt="enter image description here"></a></p>
<p>I don't understand how it went from line four to five in the proof?
Do you need to use induction? We haven't covered it yet.</p>
| Michael Hardy | 11,667 | <p>It's similar to the concept of "almost everywhere". Suppose to every subset $T\subseteq S,$ you write
$$
\mu(T) \begin{cases} =1 & \text{if } T\in F, \\ = 0 & \text{if } S\smallsetminus T\in F, \\ \text{is undefined} & \text{otherwise.} \end{cases}
$$</p>
<p>Then, according to the definition of "filter"... |
2,816,365 | <p><a href="https://i.stack.imgur.com/T1W5r.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T1W5r.jpg" alt="enter image description here"></a></p>
<p>I don't understand how it went from line four to five in the proof?
Do you need to use induction? We haven't covered it yet.</p>
| Jesus Christ is True God | 1,129,863 | <p>Let <span class="math-container">$X={1,2,3}$</span>
Choose some element from <span class="math-container">$X$</span> say
<span class="math-container">$F={{1},{1,2},{1,3},{1,2,3}}$</span>
Then every intersection of an element of <span class="math-container">$F$</span> with another element in <span class="math-contain... |
2,065,027 | <p>$$\newcommand{\gcd}{\text{gcd}}$$</p>
<blockquote>
<p>Prove: if $d=\gcd(m,n)$ so $\gcd\left(\frac{m}{d},\frac{n}{d}\right)=1$</p>
</blockquote>
<p>Intuitively it is obvious, but I am having a hard time to formalize the proof, what I have came to this:</p>
<p>$d=\gcd(m,n)$ so $d|m$ and $d|n$ therefore $m=dx$ and... | Leox | 97,339 | <p>Suppose that $gcd(\frac{m}{d},\frac{n}{d})=d'>1.$ Then $d' \mid m$ and $ d' \mid n$. It implies that $d'd \mid m$ and $ d'd \mid n$ and we find a common divisor $d d'$ which is greater then $d.$ Contradiction.</p>
|
168,495 | <p>I am being asked to calculate the area of a triangle with a side-length of 15.5 inches. </p>
<p>The formula for calculating a regular polygon's area is 1/2Pa</p>
<p>Where P is the perimeter of the polygon, and a is the apothem. I am completely lost.</p>
| André Nicolas | 6,312 | <p><strong>Hint:</strong> The calculations are too elaborate. Why do we have $c_1$, $c_2$, and $c_3$? There is one and only one constant term $c$. It can be easily eliminated, resulting in two linear equations involving the unknowns $a$ and $b$.</p>
<p>For the elimination process, solving and substituting works, but ... |
1,246,540 | <p>$$ x ^{x } = e^{\ln x^x } $$
$$ \lim_{x\rightarrow 0} x^x = \;? $$</p>
<p>I need to find the limit of x to the power of x as x approaches to 0 using l'Hopital's rule. From previous part there is a hint that I should use the first equation somehow, however I am confused how to rearrange the equation into a fraction ... | 5xum | 112,884 | <p>Use these two rules:</p>
<ul>
<li>For every $a>0, b$: $$\ln(a^b) = b\ln(a)$$</li>
<li>If the limit $$\lim_{x\to a} f(x)$$ exists and $g$ is continuous, then</li>
</ul>
<p>$$\lim_{x\to a}(g(f(x)) = g(\lim_{x\to a} f(x))$$</p>
|
130,235 | <p>I need to find </p>
<p>$$f(n) = \int^\infty_0 t^{n-1} e^{-t} dt$$</p>
<p>So I think I find the indefinate integral first? But what do I do with $n$, since I am integrating with respect to $t$?</p>
| copper.hat | 27,978 | <p>If $n>0$ is an integer, here is another approach: Let $I(x) = \int^\infty_0 e^{-x t} dt$, with $x>0$. It is straightforward to evaluate $I(x) = \frac{1}{x}$, and notice that $f(1) = I(1) = 1$. To continue, notice that $\frac{d I(x)} {d x} = \int^\infty_0 (-t) e^{-x t} dt$, and by direct computation, $\frac{d I... |
130,235 | <p>I need to find </p>
<p>$$f(n) = \int^\infty_0 t^{n-1} e^{-t} dt$$</p>
<p>So I think I find the indefinate integral first? But what do I do with $n$, since I am integrating with respect to $t$?</p>
| Sidharth Iyer | 25,645 | <p><a href="http://en.wikipedia.org/wiki/Gamma_function" rel="nofollow">$$\int^\infty_0 t^{n-1} e^{-t} dt=\Gamma(n)$$</a></p>
|
130,235 | <p>I need to find </p>
<p>$$f(n) = \int^\infty_0 t^{n-1} e^{-t} dt$$</p>
<p>So I think I find the indefinate integral first? But what do I do with $n$, since I am integrating with respect to $t$?</p>
| Sangchul Lee | 9,340 | <p>Here is also another approach, though a posteriori.</p>
<p>Tonelli's theorem enables us to exchange the order of integration and summation of a sequence of nonnegative functions. Thus for $ 0 < r < 1 $,
$$\begin{align*} \sum_{n=0}^{\infty} \frac{r^n}{n!} \int_{0}^{\infty} x^n e^{-x} \; dx & = \int_{0}^{\i... |
24,412 | <p>Trying to plot the following two functions to show points of intersection.</p>
<pre><code>2 x + y - 1 == 0,
x - y + 2 == 0
ContourPlot[{2 x + y - 1 == 0, x - y + 2 == 0}, {x, -5, 5}, {y, -5, 5}]
</code></pre>
<p>The above shows the plots, but I find it difficult to see the point of intersection. I suspect that th... | Jonathan Shock | 5,081 | <p>If you want to know the value of the points then you can use <code>Solve</code>. If you want to be able to see the intersection point highlighted then you can add a point at that value of <code>x</code> and <code>y</code>:</p>
<pre><code>eqs = {2 x + y - 1 == 0, x - y + 2 == 0}
sol = {x, y} /. Solve[eqs, {x, y}]
Sh... |
2,505,216 | <blockquote>
<p>How can I evaluate<br>
<span class="math-container">$$
\lim_{n \rightarrow \infty} \frac{n\sin n}{2n^2- 1}?
$$</span></p>
</blockquote>
<hr>
<p>Unsuccessful <strong>attempt</strong>:</p>
<p>In the expression <span class="math-container">$\frac{n\sin n}{2n^2 - 1}$</span>, I divided the numerator... | Invisible | 721,644 | <p>With your work:</p>
<p><span class="math-container">$L=\lim\limits_{n\to\infty}\frac{n\sin n}{2n^2-1}=\lim\limits_{n\to\infty}\frac{\frac{\sin n}n}{\color{red}{\underbrace{2-\frac1{n^2}}_{\in [1,2)\ \forall n\geqslant 1}}}$</span></p>
<p>As mentioned:
<span class="math-container">$$-1\leqslant\sin n\leqslant 1\Bi... |
2,223,788 | <p>$R$ (1 is not assumed to be in $R$) is a prime right Goldie ring (finite dimensional and ACC on right annihilators) which contains a minimal right ideal. Show that $R$ must be a simple Artinian ring. </p>
<p>This appeared in a past paper, the first 2 parts which I managed to prove were "Every essential right ideal ... | rschwieb | 29,335 | <p>By primeness of $R$, the minimal right ideal of $R$ must be faithful, so the ring is actually right primitive, and its Jacobson radical is zero.</p>
<p>There exists, then, a collection of maximal regular right ideals $\{T_i\mid i\in I\}$ such that $J(R)=\cap T_i=\{0\}$. This allows the natural embedding of $R_R\hoo... |
1,318,884 | <blockquote>
<p>Let S be a piecewise smooth oriented surface in <span class="math-container">$\mathbb{R}^3$</span> with positive oriented piecewise smooth boundary curve <span class="math-container">$\Gamma:=\partial S$</span> and <span class="math-container">$\Gamma : X=\gamma(t), t\in [a,b]$</span> a rectifiable para... | Steven Alexis Gregory | 75,410 | <p>Let's just do this directly. Let $g = \gcd(m,n)$. We need to prove that
$\operatorname{lcm}(m,n) = \dfrac{mn}{g}$.</p>
<hr>
<p>STEP $0$. (Preliminary stuff.)</p>
<p>DEFINITION $1$. $L = \operatorname{lcm}(m,n)$ if and only if</p>
<pre><code> 1. L is a multiple of m and of n.
2. If C is a multiple of m and of n,... |
2,042,476 | <p>In this question, I'm assuming the definition of Riemann Integrability.<br>
a) Produce an example of a sequence $f_n \rightarrow 0$ pointwise on $[0,1]$ where $\lim_{n\to\infty}\int_0^1f_n$ does not exist.<br>
b) Produce an example of a sequence $g_n$ with $\int_0^1g_n\to0$ but $g_n(x)$ does not converge to zero for... | Marcus M | 215,322 | <p>For part $a$, your answer works, but none of the integrals exist. Try to see if you can alter your example so that each $f_n$ is integrable, but the limit of the integrals doesn't exist.</p>
<p>For $b$, try to think about each $g_n$ being an indicator function of a set; if the indicator functions get narrow quite ... |
110,040 | <p>Can someone answer how can I find the matrix $A$ from the matrix equation:$$A^+BA=C$$ where $B$ and $C$ are known square matrices, $A^+$ denotes the Hermitian conjugate, and we are given a constraint: $\det(A)=a$ where $a$ is a known constant.</p>
| draks ... | 19,341 | <p>Using J.D.'s result, I think it's possible to solve it via Superoperator formalism:
$$
A=EAD \Rightarrow {\rm vec}A = (D^T\otimes E){\rm vec}A,
$$
where ${\rm vec}A$ is a vector, with all coloumns stacked. So $A$ is an eigenvector of $(D^T\otimes E)$ with eigenvalue $+1$.</p>
|
110,040 | <p>Can someone answer how can I find the matrix $A$ from the matrix equation:$$A^+BA=C$$ where $B$ and $C$ are known square matrices, $A^+$ denotes the Hermitian conjugate, and we are given a constraint: $\det(A)=a$ where $a$ is a known constant.</p>
| Community | -1 | <p>First, J.D.'s result $$A = B^{-1} B^{+} A (C^{+})^{-1} C$$ is valid for both singular and non-singular matrices if you use the Moore-Penrose inverse in place of the regular inverse.</p>
<p>Then, as suggested by draks, the Kronecker-Vec formalism can be employed to find an eigenvector $\vec{a} = {\rm vec}A$ associat... |
1,379,899 | <p>I'm trying to achieve a better conception of what it means to "divide out" a variable/number, because I'm currently have a lot of trouble justifying to myself why it actually works the way it does in certain contexts. I apologize if this is too much of an elementary question but I can't seem to come to a satisfying ... | goblin GONE | 42,339 | <p>I won't attempt to explain this to a clever $5$ year old; but anyway, you are (tacitly) looking for the following definition/theorem pair.</p>
<blockquote>
<p><strong>Definition.</strong> Let $X$ denote a set non-empty set. Then given an equivalence relation $\sim$ on $X$, call $\sim$ <em>regular</em> iff there e... |
3,646,694 | <p>Let <span class="math-container">$M$</span> be an elliptic element of <span class="math-container">$SL_2(\mathbb R)$</span>. Then it is conjugate to a rotation <span class="math-container">$R(\theta)$</span>. Note that we can calculate <span class="math-container">$\theta$</span> in terms of the trace of <span class... | Rew | 707,748 | <p>Given three roots <span class="math-container">$\alpha, \beta, \gamma$</span> of a polynomial, it can generally be written as
<span class="math-container">$$(x-\alpha)(x-\beta)(x-\gamma)=0$$</span></p>
<p>Note how each of the roots contribute in making the equality work. On expanding the braces, </p>
<p><span cla... |
3,646,694 | <p>Let <span class="math-container">$M$</span> be an elliptic element of <span class="math-container">$SL_2(\mathbb R)$</span>. Then it is conjugate to a rotation <span class="math-container">$R(\theta)$</span>. Note that we can calculate <span class="math-container">$\theta$</span> in terms of the trace of <span class... | Jack D'Aurizio | 44,121 | <p>Since by Vieta's formulas the sum of the roots is <span class="math-container">$-2$</span>, there is not much to do here: given that two roots are <span class="math-container">$\alpha$</span> and <span class="math-container">$\alpha^2+2\alpha-4$</span>, the third one has to be <span class="math-container">$-\alpha^2... |
3,341,784 | <p>How do you get to the general equation of <span class="math-container">$y = ax^2 + bx + c$</span> for a parabola?</p>
<p>I have not found any resources that show how to get to that equation.</p>
<p>This equation is shown 1 minute 30 into this video on Simpsons Rule <a href="https://www.youtube.com/watch?v=vpfy3sGw... | Rory Daulton | 161,807 | <p>Since you show little work of your own, I'll just give an outline and let you fill in the details. If you need more detail, show some more work of your own then ask.</p>
<p>Your question involves two parts. First, given a parabola with a horizontal directrix, show that an equation of the form <span class="math-cont... |
3,341,784 | <p>How do you get to the general equation of <span class="math-container">$y = ax^2 + bx + c$</span> for a parabola?</p>
<p>I have not found any resources that show how to get to that equation.</p>
<p>This equation is shown 1 minute 30 into this video on Simpsons Rule <a href="https://www.youtube.com/watch?v=vpfy3sGw... | Narasimham | 95,860 | <p>The relationship is basic coming in next after the linear relation. It better to learn fundamentals as such. If anyone asks to prove that <span class="math-container">$ y=ax+b$</span> for linear relations, then...you see?</p>
<p>It is the integration step coming next after the straight line</p>
<p><span class="math-... |
4,264,856 | <p>I filled in from the definition of a Stirling number of the second kind that the following holds.
<span class="math-container">$${2n\brace 2} = \frac{1}{2} \sum_{i=0}^{2} (-1)^i \binom{2}{i} (2-i)^{2n}$$</span></p>
<p>And I've visually confirmed in Desmos that the following equality 'appears' to hold.</p>
<p><span c... | Paradox | 969,042 | <p>I guess the most straight-forward way is to expand the summation:<br />
<span class="math-container">\begin{align}
{2n\brace 2} = &\frac{1}{2} \sum_{i=0}^{2} (-1)^i \binom{2}{i} (2-i)^{2n} = \\
& \frac{1}{2} \Bigg( \Big[(-1)^0 \binom{2}{0} (2-0)^{2n}\Big] + \Big[(-1)^1 \binom{2}{1} (2-1)^{2n}\Big] + \Big[(-1... |
171,602 | <p>In all calculus textbooks, after the part about successive derivatives, the $C^k$ class of functions is defined.
The definition says : </p>
<blockquote>
<p>A function is of class $C^k$ if it is differentiable $k$ times and the $k$-th derivative is continuous.</p>
</blockquote>
<p>Wouldn't be more natural to defi... | Georges Elencwajg | 3,217 | <p>You can certainly consider $k$-times differentiable functions on, say, $[a,b]\subset \mathbb R$ and give them a notation, like $D^k[a,b]$.<br>
The point however is that many well-known and interesting theorems, true for $C^k[a,b]$, will fail for $D^k[a,b]$or won't even make sense. Here are three examples from elem... |
36,083 | <p>I'm trying to find a way to prove this:</p>
<p>EDIT: without using LHopital theorem.
$$\lim_{x\rightarrow 1}\frac{x^{1/m}-1}{x^{1/n}-1}=\frac{n}{m}.$$
Honestly, I didn't come with any good idea.</p>
<p>We know that $\lim_{x\rightarrow 1}x^{1/m}$ is $1$.</p>
<p>I'd love your help with this.</p>
<p>Thank you.</p>
| Emre | 9,901 | <p>Are you aware of <a href="http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule" rel="nofollow">L'Hôpital's rule</a>? It is useful in evaluating the limits of fractions such as yours.</p>
|
36,083 | <p>I'm trying to find a way to prove this:</p>
<p>EDIT: without using LHopital theorem.
$$\lim_{x\rightarrow 1}\frac{x^{1/m}-1}{x^{1/n}-1}=\frac{n}{m}.$$
Honestly, I didn't come with any good idea.</p>
<p>We know that $\lim_{x\rightarrow 1}x^{1/m}$ is $1$.</p>
<p>I'd love your help with this.</p>
<p>Thank you.</p>
| Alex Becker | 8,173 | <p>One thing about limits is that, if they exist, the "speed" at which you approach them doesn't matter. That is to say, $\lim_{x\rightarrow 1}\frac{x^{1/m}-1}{x^{1/n}-1} = \lim_{x^{1/n}\rightarrow 1}\frac{x^{1/m}-1}{x^{1/n}-1} = \lim_{y\rightarrow 1}\frac{y^{n/m}-1}{y-1}$. If you then apply L'Hopital's rule, you shoul... |
36,083 | <p>I'm trying to find a way to prove this:</p>
<p>EDIT: without using LHopital theorem.
$$\lim_{x\rightarrow 1}\frac{x^{1/m}-1}{x^{1/n}-1}=\frac{n}{m}.$$
Honestly, I didn't come with any good idea.</p>
<p>We know that $\lim_{x\rightarrow 1}x^{1/m}$ is $1$.</p>
<p>I'd love your help with this.</p>
<p>Thank you.</p>
| Zarrax | 3,035 | <p>You can rewrite the limit as
$$\lim_{x \rightarrow 1} {{x^{1\over m} - 1 \over x - 1} \over {x^{1 \over n} - 1 \over x- 1}}$$
By the quotient rule for limits this is exactly
$${\lim_{x \rightarrow 1} {x^{1 \over m} - 1 \over x - 1} \over \lim_{x \rightarrow 1} {x^{1 \over n} - 1 \over x - 1}}$$
But notice that for a... |
3,779,996 | <p>Let <span class="math-container">$(X_1,...,X_n)$</span> be a random sample with PDF <span class="math-container">$f(x;\theta) = \frac{x}{\theta}\exp(-x^2/(2\theta)), \theta > 0$</span></p>
<p>I want to show that the likelihood ratio test of <span class="math-container">$H_0 : \theta \le \theta_0$</span> against <... | tommik | 791,458 | <p>The given density is a Rayleigh. If a sufficient estimator exists, the test must be based on this estimator.</p>
<p>It is very easy to verify, via factorization theorem, that this sufficient statistic is <span class="math-container">$T=\sum_{i} X_i^2$</span></p>
<p>Now let's derive the density of <span class="math-c... |
1,357,405 | <p>Under what conditions will the cubic equation
$ax^3 + bx^2 + cx + d$ where $a,b,c,d \in \mathbb R$ yield roots which have negative real parts? (All roots must have negative real parts)</p>
<p><strong><em>Motivation</strong>:
I am studying a dynamical system i.e, Chua circuit, in $ \mathbb R^3 $and wish to analyze i... | Jack D'Aurizio | 44,121 | <p>Assume we have the monic polynomial
$$ p(x) = x^3 - s_1 x^2 + s_2 x -s_3 = (x-\xi_1)(x-\xi_2)(x-\xi_3).$$
Since $s_1=\xi_1+\xi_2+\xi_3$, $s_1< 0$ is a necessary condition for the roots of $p(x)$ to lie in $\text{Re}(z)<0$.</p>
<hr>
<p>If the discriminant of $p(x)$ is positive, then $p(x)$ has three real root... |
2,539,327 | <p>The equation of a parabola is given by:
$y= ax^2 + bx+c$</p>
<p>Why is it that when the coefficient of $x$ i.e. $a$ is positive we get an upward parabola and when it's negative we get a downward parabola? </p>
<p>Also, I <a href="https://www.desmos.com/calculator/ldyxmzfrws" rel="nofollow noreferrer">saw</a> that... | user | 505,767 | <p>In the quadratic equation "<span class="math-container">$b$</span>" and "<span class="math-container">$c$</span>" terms are correlated to an axis translation, thus we can consider the simpler case <span class="math-container">$$y=ax^2$$</span> for which is clear the role of "<span class="mat... |
2,456,263 | <h2>Question</h2>
<blockquote>
<p>In how many way can <span class="math-container">$7$</span> girls be seated at a round table so that <span class="math-container">$2$</span> particular girls are separated?</p>
</blockquote>
<h2>My Approach</h2>
<blockquote>
<p><span class="math-container">$6!$</span> ways for the gir... | Donald Splutterwit | 404,247 | <p>\begin{eqnarray*}
\frac{1}{i+1}= \int_0^1 x^i dx
\end{eqnarray*}
Sub this into the sum & interchange the order of the sum & the integral
\begin{eqnarray*}
\sum_{i=0}^{n} (-1)^i\binom{n}{i} \frac{1}{i+1}&=& \int_0^1 \sum_{i=0}^{n} (-1)^i\binom{n}{i}x^i dx \\
&=& \int_0^1 (1-x)^n dx \\
&=... |
2,456,263 | <h2>Question</h2>
<blockquote>
<p>In how many way can <span class="math-container">$7$</span> girls be seated at a round table so that <span class="math-container">$2$</span> particular girls are separated?</p>
</blockquote>
<h2>My Approach</h2>
<blockquote>
<p><span class="math-container">$6!$</span> ways for the gir... | xpaul | 66,420 | <p>\begin{eqnarray}
&&1 - \frac{1}{2}\binom{n}{1}+\frac{1}{3} \binom{n}{2}- \frac{1}{4}\binom{n}{3}+....+(-1)^n \frac{1}{n+1}\binom{n}{n}\\
&=&\sum_{k=0}^n(-1)^{k}\frac{1}{k+1}\binom{n}{k}\\
&=&\sum_{k=0}^n(-1)^{k}\frac{1}{n+1}\binom{n+1}{k+1}\\
&=&\frac{1}{n+1}\sum_{k=1}^{n+1}(-1)^{k}\b... |
1,373,927 | <p>We know that a tree with n edges have n+1 nodes.So if $|B_{n+1}|$ is the number of all possible ordered trees with n+1 nodes then its true that $C_{n+1} = |B_{n+1}|$ where $C$ is the Catalan number.Let it be $|L_k|$ the number of all possible ordered trees with n+1 nodes and k leaves and Its true that
$|B_{n+1}| = ... | Colm Bhandal | 252,983 | <p>Your thinking looks fairly sound there. All I can think of is the following complicated recurrence from looking at the leftmost sub-tree and enumerating over all possible values for its nodes and its leaves:</p>
<p>$$L_{n, k} = \sum_{i=1}^{n - 1}\prod_{j=1}^{max(k, i)}L_{i, j}L_{n - i, k - j} = \sum_{i=1}^{k}\prod_... |
867,207 | <p>The Efron-Stein inequality sais that if $X_1,\ldots,X_n$ are independent random variables on say $R^n$, and $f:R^n \rightarrow R$ s.t. $Z:=f(X_1,\ldots,X_n)$ has finite variance, then</p>
<p>$$\operatorname{Var}(X)\le \sum_{i=1}^n E[(Z-E^{(i)}[Z]]$$</p>
<p>where $E^{(i)}$ denotes conditional expectation taken w.r.... | Adam | 82,101 | <p>Any set $$R \subseteq A \times A = \{(x,y) \mid x \in A ,\,y \in A\} $$ is a relation on $A $. Since $$\{(b,c), (b,d)\} \subseteq A \times A$$ holds, it is indeed a relation on $ A$. </p>
|
4,367,330 | <p>Let <span class="math-container">$X_1, X_2,...$</span> be a sequence of independent uniform random variables on <span class="math-container">$(0,1)$</span>. Define: <span class="math-container">$$N := \text{min} \{n\geq 2: X_n < X_{n-1}\}.$$</span> Calculate <span class="math-container">$E(N)$</span>.</p>
<p>I th... | nejimban | 206,936 | <ol>
<li>Check that <span class="math-container">$$N=2+\mathbf1_{\{X_1<X_2\}}+\mathbf1_{\{X_1<X_2<X_3\}}+\cdots$$</span></li>
<li>Check that <span class="math-container">$\Bbb P(X_1<\cdots<X_k)=\frac1{k!}$</span> for any <span class="math-container">$k\ge2$</span>.</li>
<li>Take expectations & Fubini... |
412,051 | <p>Suppose
$X_i \sim N(0,1)$ (independent, identical normal distributions) </p>
<p>Then by Law of large number,
$$
\sqrt{1-\delta} \frac{1}{n}\sum_i^\infty e^{\frac{\delta}{2}X_i^2} \rightarrow \sqrt{1-\delta} \int e^{\frac{\delta}{2}x^2}\frac{1}{\sqrt{2\pi}}e^{\frac{1}{2}x^2}dx =1
$$</p>
<p>However, according to si... | doraemonpaul | 30,938 | <p>Note that since this problem has only one condition, this is in fact an under-determining problem.</p>
<p>You can let one more dummy condition so that this becomes a just-determining problem.</p>
<p>For example, letting the dummy condition be $u(x,0)=g(x)$ , the general solution is more conventient to consider as ... |
222,312 | <p>How to prove that if you have $x^*$ such that $x^*=\text{psuedoinverse}(A) b$, and $Ay=b$, then
$$\Vert x^* \Vert_2 \leq \Vert y \Vert_2$$</p>
| Community | -1 | <p>Let me try a one liner solution. </p>
<p>$Ay=b$ $\Rightarrow$ $A(y-x^*)=0$ $\Rightarrow$ $\langle x^*,y-x^*\rangle=0$ $\Rightarrow$ $\|y\|^2=\|y-x^*\|^2+\|x\|^2\ge \|x\|^2$.</p>
<p><strong>Remark.</strong> $R=A^t(AA^t)^{-1}$, $x^*=Rb$. Then we have $\langle x^*,y-x^*\rangle=\langle Rb,y-x^*\rangle=$ $\langle (AA^t... |
3,772,371 | <p>Write the polynomial of degree <span class="math-container">$4$</span> with <span class="math-container">$x$</span> intercepts of <span class="math-container">$(\frac{1}{2},0), (6,0) $</span> and <span class="math-container">$ (-2,0)$</span> and <span class="math-container">$y$</span> intercept of <span class="math-... | Riccardo | 662,881 | <p><span class="math-container">$$\Big(x-\dfrac{1}{2}\Big)^2 = \Big(\color{red}{\frac{1}{2}}\Big(2x-1\Big)\Big)^2=\color{red}{\frac{1}{4}}\Big(2x-1\Big)^2$$</span></p>
|
3,772,371 | <p>Write the polynomial of degree <span class="math-container">$4$</span> with <span class="math-container">$x$</span> intercepts of <span class="math-container">$(\frac{1}{2},0), (6,0) $</span> and <span class="math-container">$ (-2,0)$</span> and <span class="math-container">$y$</span> intercept of <span class="math-... | fleablood | 280,126 | <p>You didn't go wrong anywhere.</p>
<p><span class="math-container">$-6(x-\frac{1}{2})^2(x+2)(x-6) = -\frac{3}{2}(2x-1)^2(x+2)(x-6)$</span></p>
<p>That is because</p>
<p><span class="math-container">$-6(x- \frac 1{2})^2 = -6(\frac 12[2x-1])^2= -6\cdot (\frac 12)^2(2x-1)^2 = -6\cdot \frac 14(2x-1) = -\frac 32(2x-1)$</s... |
372,420 | <p>Let <span class="math-container">$X$</span> be a finite ultrametric space and <span class="math-container">$P(X)$</span> be the space of probability measures on <span class="math-container">$X$</span> endowed with the Wasserstein-Kantorovich-Rubinstein metric (briefly WKR-metric) defined by the formula
<span class="... | Carlo Beenakker | 11,260 | <p>There exists a variety of measures of uniformity of a point set. See, for example, <A HREF="https://www.bme.psu.edu/labs/Yang-lab/publications%20PDF/with%20Lizeng.pdf" rel="nofollow noreferrer">On assessing spatial uniformity of particle distributions...</A> for an overview, and a critical comparison when applied to... |
372,420 | <p>Let <span class="math-container">$X$</span> be a finite ultrametric space and <span class="math-container">$P(X)$</span> be the space of probability measures on <span class="math-container">$X$</span> endowed with the Wasserstein-Kantorovich-Rubinstein metric (briefly WKR-metric) defined by the formula
<span class="... | gmvh | 45,250 | <p>Judging from your pictures, it should be sufficient to consider the root-mean-square distance <span class="math-container">$\rho$</span> of the points <span class="math-container">$\vec{x}_k$</span> from their center of mass <span class="math-container">$\vec{\mu}$</span>, divided by the radius <span class="math-con... |
577,490 | <p>I wonder if I have approached this in the right way. I'm not sure if I have interpreted the question correctly, or made correct use of the successor function. Thank you in advance</p>
<blockquote>
<p><strong>Question:</strong> Prove that $¬∃n∅ = n^+$</p>
</blockquote>
<p>My approach:</p>
<p>Prove that $¬∃n∅... | Asaf Karagila | 622 | <p>The proof is incorrect, because you assume that $n=1$. You need to assume that $n^+=\varnothing$, then conclude somehow contradiction.</p>
<p>For example,</p>
<ol>
<li>Assume $n^+=\varnothing$, then $n\cup\{n\}=\varnothing$.</li>
<li>In particular $n\in\varnothing$.</li>
<li>Contradiction.</li>
</ol>
|
2,433,765 | <p><a href="https://i.stack.imgur.com/AolxA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AolxA.png" alt="enter image description here"></a></p>
<p>In the following diagram of a triangle, $\overline{AB} = \overline{BC} = \overline{CD}$ and $\overline{AD} = \overline{BD}$. Find the measure of angle... | g.kov | 122,782 | <p><a href="https://i.stack.imgur.com/wEb4F.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wEb4F.png" alt="enter image description here"></a></p>
<p>\begin{align}
5\,\delta&=180^\circ
\end{align} </p>
<p><strong>Edir</strong></p>
<p>\begin{align}
|AB|&=|BC|=|CD|
,\\
|AD|&=|BD|
.
\e... |
1,865,086 | <p>$$T_n=2^{-n}$$</p>
<p>How can I tell if this converges? with previous questions I have just let $n = \infty$, however I'm unsure about this one.</p>
| 5xum | 112,884 | <p>You can <strong>tell</strong> (intuitively) that this converges by writing down a couple of elements and trying to find a pattern:</p>
<p>$$\frac12, \frac14, \frac18,\frac1{16}\dots$$</p>
<p>You should immediatelly see that the elements are becoming very small very fast. So, your intuition should tell you that the... |
1,865,086 | <p>$$T_n=2^{-n}$$</p>
<p>How can I tell if this converges? with previous questions I have just let $n = \infty$, however I'm unsure about this one.</p>
| operatorerror | 210,391 | <p>I think it is worth proving that $1/n\rightarrow 0$ rigorously with an epsilon-n proof at least once. After that, you can compare sequences like the one in your question to this sequence, since it is clear that
$$
2^{-n}\leq 1/n
$$
Hint: use the Archimedean property.</p>
|
1,778,037 | <p>$ 9^x-6^x=4^{x+1/2}$, solve for $x$</p>
<p>Please don't solve this problems entirely, I just want some hints. I have tried to substitute $3^x=b$ and $2^x=a$. </p>
| Niklas Rosencrantz | 5,602 | <p>Hint: $\ln (y ^x) =x\ln y$. Now you should see the first one. </p>
|
1,778,037 | <p>$ 9^x-6^x=4^{x+1/2}$, solve for $x$</p>
<p>Please don't solve this problems entirely, I just want some hints. I have tried to substitute $3^x=b$ and $2^x=a$. </p>
| Siddd | 274,503 | <p>Consider this:
The equation can be reduced to $$9^x - 6^x = 4^x.4^{1/2}$$ which is $$9^x = 6^x + 2.4^x$$ Now the RHS is divisible by $2$, but the LHS is not, for any values of $x\in \mathbb Z^+$. <br>Rest, try using $ln(y^x) = xln(y)$ as others have suggested. That should give the final answer</p>
|
3,987,911 | <p>In the game of bridge, a player is given 13 cards from a deck of 52 cards, what is the probability that he/she gets exactly one king and one queen? Furthermore what is the expected value of the number of aces he/she gets?</p>
| Community | -1 | <p>This question can be solved 'in a flash' when looked at in the right way.</p>
<p>If you look at the second term in your expression for time, you can simplify it to <span class="math-container">$$\frac{\sqrt{(x-37)^2+37^2}}{\sqrt2}.$$</span></p>
<p>We have now, in effect, made it a route with two legs travelled at th... |
778,140 | <p>I want to draw a box plot, which requires that I know the median, the lower and upper quartiles, and the minimum and maximum values of my data.</p>
<p>I understand that the quartiles are simply the value on certainly "percentile" of the cumulative frequency of the data.</p>
<p>So lower quartile = the value of the... | jsk | 147,497 | <p>If your data consist of the frequency of observations within each class, then the best that you can day is say that the quartile is somewhere within the class. To find out which class the quartile belongs to, you need to first figure out the index of the quartile. For example, if you have 100 observations in your ... |
778,140 | <p>I want to draw a box plot, which requires that I know the median, the lower and upper quartiles, and the minimum and maximum values of my data.</p>
<p>I understand that the quartiles are simply the value on certainly "percentile" of the cumulative frequency of the data.</p>
<p>So lower quartile = the value of the... | David K | 139,123 | <p>If taking the $25$th percentile of the data gives you the value $3.5$,
then the value of the first quartile is $3.5$.
We would usually expect that the <em>number of observations</em> in each
of your frequency classes is a whole number, but
there is nothing wrong with having a quartile or median <em>value</em> that
i... |
1,644,130 | <p>I proved for a bounded set $\Omega$ and $1 \leq p \leq q \leq \infty$ that $L^{q}(\Omega) \subset L^{p}(\Omega)$. What is an example that would show strict inclusion, $ p<q$, and false if $\Omega$ is not bounded?</p>
| DanielWainfleet | 254,665 | <p>For $1\leq p<q<\infty ,\quad $ $f(x)=x^{-1/q} $ belongs to $L^p(0,1)$ but not to $L^q(0,1)$.</p>
|
959,410 | <p>I need to evaluate an expression similar to the following:</p>
<p>$\frac{\partial\mathrm{log}(a+b)}{\partial a}$</p>
<p>At this point I don't know how to proceed. $b$ is a constant so there should be some way to eliminate it.
How would you proceed in this case?</p>
<p>Actually, the original expression is much mor... | James S. Cook | 36,530 | <p>Suppose you know $\frac{d}{dx} e^x=e^x$ then define $f(x)=e^x$ to have inverse function $g(x) = \ln (x)$ to be the inverse function of the exponential function. In particular this means we assume $e^{\ln(x)}=x$ for all $x \in (0, \infty)$ and $\ln(e^x) = x$ for all $x \in (-\infty, \infty)$. The existence of $g$ is ... |
2,895,129 | <blockquote>
<p>Let $f(x) \leq g(x)$ for all $x$ in an open set $D$, and let $a$ be an interior point of $D$ such that $f(a) = g(a)$ and both $f, g$ are differentiable at $a$.
Does it follows that $f'(a)=g'(a)$?</p>
</blockquote>
<p>Is that right, and if so, why? If not then please give examples.</p>
| Rigel | 11,776 | <p>By assumption, the differentiable function $h(x) := g(x) - f(x)$ has a minimum point at $a$.
Since $a$ is an interior point, then $h'(a) = 0$, i.e. $f'(a) = g'(a)$.</p>
|
2,895,129 | <blockquote>
<p>Let $f(x) \leq g(x)$ for all $x$ in an open set $D$, and let $a$ be an interior point of $D$ such that $f(a) = g(a)$ and both $f, g$ are differentiable at $a$.
Does it follows that $f'(a)=g'(a)$?</p>
</blockquote>
<p>Is that right, and if so, why? If not then please give examples.</p>
| mengdie1982 | 560,634 | <h1>Proof</h1>
<p>I will give a proof for the problem as follows, which totally depends on some most basic facts of calculus.
According to the assumptions, <span class="math-container">$$\frac{f(x)-f(a)}{x-a}\leq \frac{g(x)-g(a)}{x-a},\tag1$$</span>where <span class="math-container">$x$</span> belongs to some right nei... |
3,163,525 | <p>How would I expand the following function as a power series, around <span class="math-container">$\eta=0$</span>?</p>
<p><span class="math-container">$$g_0(1,\eta)=\frac{\left(\frac{PV}{NkT}\right)_0-1}{4\eta}$$</span></p>
<p>Note that:</p>
<p><span class="math-container">$$\left(\frac{PV}{NkT}\right)_0=1+\frac{3... | Andrei | 331,661 | <p>Starting from your last equation: <span class="math-container">$$g_0(1,\eta)=\frac{\frac{3\eta}{\eta_c-\eta}+\sum_{k=1}^4kA_k\left(\frac{\eta}{\eta_c}\right)^k}{4\eta}$$</span>
you have:
<span class="math-container">$$g_0(1,\eta)=\frac{3}{4(\eta_c-\eta)}+\sum_{k=1}^4\frac k{4\eta_c^k}A_k\eta^{k-1}$$</span>
The terms... |
3,187,793 | <p>We know that <span class="math-container">$221 = 17*13$</span>. So we can check if the system has roots to both of those equations separately, which it does:</p>
<p><span class="math-container">$x^{5} \equiv 2$</span> mod <span class="math-container">$13$</span> has the solution <span class="math-container">$6 + 13... | Bill Dubuque | 242 | <p>Below we quickly <em>mentally</em> solve <span class="math-container">$\,x^{\large 5}\equiv 2\,$</span> by taking a <span class="math-container">$5$</span>'th root, i.e. raising both sides to power <span class="math-container">$\color{#c00}{1/5}$</span></p>
<p>Suppose <span class="math-container">$a$</span> is copri... |
2,755,091 | <p>I know that I have to prove this with the induction formula. If proved the first condition i.e. $n=6$ which is divisible by $6$. But I got stuck on how to proceed with the second condition i.e $k$.</p>
| Delta-u | 550,182 | <p>Let suppose that the result is true for $n=k$ i.e $7^k-1$ is divisible by $6$.</p>
<p>Then:
$$7^{k+1}-1=7 \times (7^k-1)+7-1=7 \times (7^k-1)+6$$
but $(7^k-1)$ is divisible by $6$ so $7 \times (7^k-1)$ is also divisible by $6$ and finally $(7^k-1)+6$ is divisible by $6$.</p>
|
2,755,091 | <p>I know that I have to prove this with the induction formula. If proved the first condition i.e. $n=6$ which is divisible by $6$. But I got stuck on how to proceed with the second condition i.e $k$.</p>
| Anastassis Kapetanakis | 342,024 | <p>We have:
$$7^n-1=(7-1)(7^{n-1}+7^{n-2}+\dots +7+1) = 6(7^{n-1}+7^{n-2}+\dots +7+1) $$</p>
<p>Which definitely is divisible by $6$.</p>
|
4,008,277 | <p><a href="https://en.wikipedia.org/wiki/Vandermonde%27s_identity#Generalized_Vandermonde%27s_identity" rel="noreferrer">Vadermonde's identity</a> states that</p>
<p><span class="math-container">$$ \sum_{k=0}^{r} {m \choose k}{n \choose r-k} = {m+n \choose r} $$</span></p>
<p>I wonder if we can use this formula or oth... | Trevor | 493,232 | <p>Nope, not unique.</p>
<p>I hadn't checked far enough. The first counterexample is at <span class="math-container">$$f(38183)=f(38185)=258840858.$$</span></p>
<p>It could plausibly still have only finitely many counterexamples, though, making the 'almost all' claim true. There are three more found so far through <spa... |
4,008,277 | <p><a href="https://en.wikipedia.org/wiki/Vandermonde%27s_identity#Generalized_Vandermonde%27s_identity" rel="noreferrer">Vadermonde's identity</a> states that</p>
<p><span class="math-container">$$ \sum_{k=0}^{r} {m \choose k}{n \choose r-k} = {m+n \choose r} $$</span></p>
<p>I wonder if we can use this formula or oth... | Barry Cipra | 86,747 | <p>This is just a comment on the counterexample in the OP's answer, noting a relatively quick way to verify that <span class="math-container">$f(38183)=f(38185)$</span> without computing each to equal <span class="math-container">$258840858$</span>.</p>
<p>A formula at the <a href="https://oeis.org/A004125" rel="norefe... |
4,008,277 | <p><a href="https://en.wikipedia.org/wiki/Vandermonde%27s_identity#Generalized_Vandermonde%27s_identity" rel="noreferrer">Vadermonde's identity</a> states that</p>
<p><span class="math-container">$$ \sum_{k=0}^{r} {m \choose k}{n \choose r-k} = {m+n \choose r} $$</span></p>
<p>I wonder if we can use this formula or oth... | Joffan | 206,402 | <p>I feel this is worth an answer, using Barry Cipra's technique, to report the first such equality over a gap of more than <span class="math-container">$2$</span>:</p>
<p><span class="math-container">$$f(113393278)=f(113393282)$$</span></p>
<p>with:<br />
<span class="math-container">$\sigma(113393278)=171694080$</spa... |
3,774,263 | <ol>
<li>Is it possible that distance (<span class="math-container">$r$</span>) or angle (<span class="math-container">$θ$</span>) contains Imaginary or Complex number?</li>
<li>If the answer is yes, how can I convert a number like that (Polar with complex argument) to Rectangular number? <br />
For example:
<strong><s... | DonAntonio | 31,254 | <p>The fastest way is to invoque the dimensions theorem, and in this case:</p>
<p><span class="math-container">$$\dim\ker x^*+\dim\text{Im}\,x^*=\dim X$$</span></p>
<p>and since <span class="math-container">$\;\dim\text{Im}\,x^*\le1\;,\;\;\dim X>1\;$</span> , we get that</p>
<p><span class="math-container">$$\;\dim\... |
1,726,745 | <p>I am looking for information in regards to a couple particular functions: </p>
<p>1) $P(x)=\sum_{p\in\mathbb{P}}\frac{x^p}{p!}$</p>
<p>2) $Q(x)=\sum_{p\not\in\mathbb{P}}\frac{x^p}{p!}$ (assuming $0, 1$ are included powers in the series...)</p>
<p>3) $R(x)=\frac{1}{P(x)}$</p>
<p>4) $S(x)=\frac{1}{Q(x)}$</p>
... | reuns | 276,986 | <p><strong>trying to relate $P(x) = \sum_{p \in \mathcal{P}} \frac{x^p}{p!}$ to the logarithm of the Riemann zeta function</strong>, I get : </p>
<p>$$P(x) = \sum_{p \in \mathcal{P}} \frac{x^p}{p!} = \frac{x}{(2 i \pi)^2}\int_{|u|=r} \frac{1}{(u-2x)(u-x)}\left( \int_{c-i\infty}^{c+i\infty} (-\ln u)^{-s} \Gamma(s) \sum... |
254,926 | <p>I was working on a problem involving perturbation methods and it asked me to sketch the graph of $\ln(x) = \epsilon x$ and explain why it must have 2 solutions. Clearly there is a solution near $x=1$ which depends on the value of $\epsilon$, but I fail to see why there must be a solution near $x \rightarrow \infty$... | hmakholm left over Monica | 14,366 | <p>The <em>slope</em> of $\log x$ falls of as $1/x$. So for, say, $x>1/\varepsilon$ the logarithm always grows strictly <em>slower</em> than $\varepsilon x$, and the latter will eventually overtake it.</p>
<p>More precisely, if we put $a=2/\varepsilon$ then at any point to the right of $a$, the function $\varepsilo... |
1,963,214 | <p>$\displaystyle\sum_{i=0}^{k-1}2^i=2^k-1$ for all $k \in\Bbb N$.</p>
<p>Clearly, the first step here is easy. You can start with k=1 and solve to get $2^0=2^1-1$.</p>
<p>What is a bit more challenging is the induction step. I don't even know where to begin here. Where would I even begin?</p>
| Tsemo Aristide | 280,301 | <p>We have $|f(0)|\leq |0|^2$ implies that $f(0)=0$.</p>
<p>$|lim_{x\rightarrow 0}{{f(x)-f(0)}\over{|x-0|}}|=lim_{x\rightarrow 0}|{{f(x)}\over |x|}|\leq lim_{x\rightarrow 0}{{|x|^2}\over{|x|}}=lim_{x\rightarrow 0}|x|=0.$ This implies that the differential of $f$ at $0$ is $0$.</p>
|
1,145,902 | <p>I've searched for the inner product definition and I saw that it should only satisfy some conditions (axioms) and thus there could be several operations representing an inner product, one of which is the usual multiplication, or the dot product.
the question is: why is the inner product of 2 functions is defined by ... | Iulia | 205,001 | <p>The absolute value: $|\cdot|:\mathbb{R}\rightarrow \mathbb{R}_+$ is given by:
$|x|=\begin{cases} x,\text{ if }x\geq 0\\-x\text{ if }x<0\end{cases}$.
I will write: $abs(x):=|x|$. If you want to compute: $abs(abs(x))$, you know that $abs(x)\geq 0$, therefore, using the definition, $abs(abs(x))=abs(x)$.
Indeed, $abs... |
673,946 | <p>I have the function $f(x)=\frac {1}{2} \mathbf x^T Q \mathbf x$. </p>
<p>I want to use the steepest descent algorithm where $Q$ is the diagonal matrix $\begin{bmatrix}1 & 0\\0 & 20\end{bmatrix}$ and $\mathbf x = \begin{bmatrix}0.7\\-0.2\end{bmatrix}$.</p>
<p>I want to implement the ideal line search algori... | gt6989b | 16,192 | <p>One does typically implement numerical algorithms in some computing environment. Matlab or Python or Mathematica will do fine, whatever you have available, or even a raw programming language without the mathy surroundings :-), like C or C++.</p>
<p>You describe the algorithm pretty well. At each step,</p>
<ol>
<li... |
4,168,198 | <p>If M is a connected n-manifold and N is a codim 2 submanifold
then M-N is connected.</p>
<p>Is this true?</p>
<p>I want to show <span class="math-container">$H_0(M-N)=\mathbb Z$</span>. I think it's better to do in <span class="math-container">$\mathbb Z_2$</span> coefficient because we get orientability. But becaus... | Moishe Kohan | 84,907 | <p>Here is an argument which works for a connected <span class="math-container">$n$</span>-dimensional topological manifold <span class="math-container">$M$</span> and a closed subset <span class="math-container">$C\subset M$</span> homeomorphic to a manifold of dimension <span class="math-container">$\le n-2$</span>.... |
28,348 | <p>Thomson et al. provide a proof that <span class="math-container">$\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$</span> in <a href="http://classicalrealanalysis.info/documents/TBB-AllChapters-Landscape.pdf#page=95" rel="noreferrer">this book (page 73)</a>. It has to do with using an inequality that relies on the binomial... | Mark | 687 | <p>$\sqrt[n]{n}=\sqrt[n]{1\cdot\frac{2}{1}\cdot\frac{3}{2}\dots\cdot\frac{n-1}{n-2}\cdot\frac{n}{n-1}}$ so you have a sequence of geometric means of the sequence $a_{n}=\frac{n}{n-1}$. Therefore its limit is equal to $\lim_{n\to\infty}a_{n}=\lim_{n\to\infty}\frac{n}{n-1}=1$.</p>
|
2,268,225 | <p>I want to investigate the convergence behavior of $$\int_{0}^{\infty} \cos(x^r)\, dx \hspace{5mm} \textrm{and} \hspace{5mm} \int_{\pi}^{\infty} \left(\frac{\cos x}{\log x}\right)\arctan\lfloor x\rfloor dx.$$ My theoretical tools are Abel's test and Dirichlet's test:
Say I have an integral of the form $$\int_{a}^{b... | Olivier Oloa | 118,798 | <p>One may start with the standard evaluation,
$$
1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1
$$ Then by differentiating $(1)$ one gets
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}-\frac{(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2
$$ by multiplying by $x^2$ and by making $n \to +\infty$ in $(2)$, us... |
3,536,120 | <p>Textbook example says that:</p>
<p><span class="math-container">$$\sum \limits_{i=k+1}^{\infty}(1-p)^{i-1}p = \frac{(1-p)^k p}{1-(1-p)}=(1-p)^k~~~~~k=1,2,3,\cdots$$</span></p>
<p>I'm told that a geometric series identity used to obtaining the above result is:</p>
<p><span class="math-container">$$\sum \limits_{n=... | Francis Adams | 29,633 | <p>The proof so far is correct.</p>
<p>For the other direction, since <span class="math-container">$\mathbb{K}$</span> is a basis for the topology every open set <span class="math-container">$O\in\tau$</span> is a union of open sets from <span class="math-container">$\mathbb{K}$</span>, and since <span class="math-cont... |
4,434,225 | <p>Prove or disprove, "if <span class="math-container">$f$</span> is concave function, then <span class="math-container">$|f|$</span> is also concave".</p>
<p>I know the result is false for conves function, but for concave function, I guess it is true. But I am unable to do the proof.</p>
<p>I have tried to s... | MHMH | 427,446 | <p>Counterexample: <span class="math-container">$f(x)=x$</span> is concave, but <span class="math-container">$|f(x)|=|x|$</span> is convex (and it is not concave).</p>
|
2,240,616 | <p>Let $f:[0,1] \to \mathbb{R}$ be a function such that for every $a \in [0,1)$ and $b \in (0,1]$ the one-sided limits $$f(a^+)=\lim _{x\to a^+}f(x) \in \mathbb{R}$$ $$f(b^-)=\lim _{x \to b^-} f(x) \in \mathbb {R}$$ exist. </p>
<p>A) Show that $f$ is bounded. </p>
<p>B) Does $f$ necessarily achieve its maximum at som... | Fnacool | 318,321 | <p>A) Let $x_n$ be such that $|f(x_n)|\to \sup |f|$. By compactness, there exists a convergent subsequence, and furthermore, it can be chosen to be monotone. By assumption, the limit along the monotone subsequence exists. Thus $\sup |f|$ is finite. </p>
<p>B) Answered before. </p>
<p>C) WLOG $f(0)<0$. Then by cont... |
602,789 | <p>Find the fundamental group of the surface of a cube with interior of all edges removed (i.e. the space which consists of the vertices and interior of the faces of the cube. </p>
<p>Can I deformation retract the interior of each face into a cross meeting in the middle of the face made up of two diagonal lines which ... | Simon Rose | 87,590 | <p>The answer by rewritten is correct, although your intuition is also good.</p>
<p>The specific deformation that you consider will give you the correct fundamental group, but as rewritten says, the loops do not commute so you will get $\mathbb{Z} * \cdots * \mathbb{Z}$ for some number of copies of $\mathbb{Z}$. What ... |
324,307 | <p><a href="http://en.wikipedia.org/wiki/Tucker%27s_lemma" rel="nofollow">Tucker's Lemma is here.</a></p>
<p>Let's stay within the 2D case for now. A standard proof is constructive:</p>
<p>(1) Pick an arced edge on the boundary of the circle. Note its labeling (for example, (1, 2)).</p>
<p>(2) Walk into the circle... | Cameron Buie | 28,900 | <p>Your conclusion doesn't follow. For example, let $b=c=1,d=0$. Then $ad+b=cb+d$ doesn't imply <em>anything at all</em> about $a$.</p>
<hr>
<p>Let $$f(x)=a_1x+b_1,\;g(x)=a_2x+b_2,\;h(x)=a_3x+b_3,$$ where $a_1\neq 1$ or $b_1\neq 0$, and suppose that $fg=gf$ and $fh=hf$. Your work shows that this means $$a_1b_2+b_1=a_... |
4,214,631 | <p>I'm trying to understand a paragraph from the article <a href="https://www.sciencedirect.com/science/article/pii/S0022123600936875" rel="nofollow noreferrer">Christ, Kiselev: Maximal functions associated to filtrations</a>. After the proof of Theorem 1.1., the authors deduce a theorem of Menshov as a corollary:</p>
... | Sangchul Lee | 9,340 | <p>For each <span class="math-container">$m \in \mathbb{N}$</span>, by applying the inequality to the residual tail <span class="math-container">$n \mapsto c_n \mathbf{1}_{\{n\geq m\}},$</span> we get</p>
<p><span class="math-container">$$ \left\| \sup_{N\in\mathbb{N}} \left| \sum_{n=m}^{N} c_n\phi_n \right| \right\|_{... |
54,088 | <p>I am studying KS (Kolmogorov-Sinai) entropy of order <em>q</em> and it can be defined as</p>
<p>$$
h_q = \sup_P \left(\lim_{m\to\infty}\left(\frac 1 m H_q(m,ε)\right)\right)
$$</p>
<p>Why is it defined as supremum over all possible partitions <em>P</em> and not maximum? </p>
<p>When do people use supremum and w... | Michael Hardy | 11,667 | <p>The set of all negative numbers has a sup, which is $0$, but not a max.</p>
<p>$0$ is the smallest number that no negative number can exceed.</p>
|
2,392,841 | <p>Assuming the series of functions $\{\phi_{n}(x)\}$ and the function $\phi(x)$ are in Schwartz space $S(R)$,
now my question is this,</p>
<p>Does $\lim_{n\to\infty} \phi_{n}(x)\to \phi(x)$ (in Hilbert space $L^2[-\infty,\infty]$)
imply $\lim_{n\to\infty}\phi_{n}(x)\to \phi(x)$ (in semi-norms topology of Schwartz spa... | Viktor Vaughn | 22,912 | <p>All right, I think I see how to use the direct approach now. Since $Z(e_i) = Z(P_i)$, then $\sqrt{(e_i)} = \sqrt{P_i} = P_i$. Since $A$ is Noetherian, then $(e_i) = P_i^{r_i}$ for some $r_i$, and since the ideals $P_i$ are pairwise comaximal, then so are $P_i^{r_i}$ (c.f., exercise 1.13 in Atiyah-MacDonald). (One... |
2,392,841 | <p>Assuming the series of functions $\{\phi_{n}(x)\}$ and the function $\phi(x)$ are in Schwartz space $S(R)$,
now my question is this,</p>
<p>Does $\lim_{n\to\infty} \phi_{n}(x)\to \phi(x)$ (in Hilbert space $L^2[-\infty,\infty]$)
imply $\lim_{n\to\infty}\phi_{n}(x)\to \phi(x)$ (in semi-norms topology of Schwartz spa... | Viktor Vaughn | 22,912 | <p>I don't yet see how to get the direct approach to work, so here's an answer that uses the theory of Artinian rings. This is basically just Theorem 8.7 in Atiyah-MacDonald, but I'll try to add a little extra exposition.</p>
<p>First note that since $A$ is a finite-dimensional $k$-algebra, then it is Artinian: each i... |
2,417,029 | <p>I have been told that a line segment is a set of points.
How can even infinitely many point, each of length zero, can make a line of positive length?</p>
<p>Edit:
As an undergraduate I assumed it was due to having uncountably many points.
But the Cantor set has uncountably many elements and it has measure $0$.</p>
... | John Wayland Bales | 246,513 | <p>At that level, you can only define length in terms of line segments, which should not present a problem.</p>
<p>To approach the idea of "length" of a point you could use the idea of probability.</p>
<p>You might use an example such as this. Suppose we randomly choose a number $x$ in the open interval $(0,1)$.</p>
... |
2,417,029 | <p>I have been told that a line segment is a set of points.
How can even infinitely many point, each of length zero, can make a line of positive length?</p>
<p>Edit:
As an undergraduate I assumed it was due to having uncountably many points.
But the Cantor set has uncountably many elements and it has measure $0$.</p>
... | Community | -1 | <p>Elsewhere in comments, you've suggested you're familiar with topology. <em>In these terms</em>, I think I can describe more precisely what the issue is.</p>
<p>Consider the usual subspace topology on the interval $[0,1]$ of the real line, and also the discrete topology on the same set of points.</p>
<p>The notion ... |
1,072,302 | <p>let $f$ and $g$ be two functions from $[0,1]$ to $[0,1]$ with $f$ strictly increasing. Which of the follwing is true?</p>
<blockquote>
<blockquote>
<p>(a). If $g$ is continuous, then $f\circ g$ is continuous.</p>
<p>(b). If $f$ is continuous, then $f\circ g$ is continuous.</p>
<p>(c). If... | Brandon Humpert | 30,417 | <p>For (a), consider $g(x) = x$ and $f$ any discontinuous increasing function (for example $f(x) = \frac{1}{2} + \frac{1}{2}x, x > 0$ and $f(0) = 0$).</p>
<p>For (b), consider $f(x) = x$ and $g$ any discontinuous function.</p>
<p>For (d), consider $g(x) = 0$ and $f$ any discontinuous increasing function.</p>
|
1,775,378 | <p>So this question is inspired by the following thread: <a href="https://forums.factorio.com/viewtopic.php?f=5&t=25008">https://forums.factorio.com/viewtopic.php?f=5&t=25008</a></p>
<p>In it, the poster is examining an $8$-belt balancer (more on that to come) which he shows fails to satisfy a desirable proper... | Kieren Pearson | 297,523 | <p>This might be off from what you are talking about because I am slightly unclear as to what you mean by a 'universally throughput unlimited' Balance.</p>
<p>In my proof I attempt to prove that:</p>
<blockquote>
<p>For all natural numbers {n, m} there exists a perfect balance that will take n inputs and m outputs ... |
83,921 | <p>Consider $J = \sum_{i=0}^{N}y_{i-1}x_{i}y_{i+1}$ where $+$ and $-$ in the indices are mod $N+1$. Let $x_{i} = 1 - y_{i} \in \{0,1\}$. What are some of the tools useful and relaxation techniques available to maximize $J$ or any other symmetric multivariate polynomial?</p>
| Terry Loring | 6,133 | <p>I have not proven the following to work, but I've done similar calculations when debugging software in my 2009-10 work with Matt Hastings. If these seem fast enough, it should not be too much work to validate them.</p>
<p>(1) I would take small, random perturbations of the identity matrix and then do four or five ... |
394,489 | <p>$$\int^L_{-L} x \sin(\frac{\pi nx}{L})$$</p>
<p>I've seen something like this in Fourier theory, but I'm still not sure how to approach this integral. Wolfram Alpha gives me the answer, but no method. Integrate by parts? Substitution?</p>
| Cameron Buie | 28,900 | <p>If $n=0,$ this is simple, so suppose not. Use the substitution $$u=\frac{n\pi}lx,$$ so that $$x=\frac{l}{n\pi}u,$$ and so $$\int_{-l}^lx\sin\left(\frac{n\pi}lx\right)\,dx=\frac{l^2}{n^2\pi^2}\int_{-n\pi}^{n\pi}u\sin u\,du,$$ then integrate by parts.</p>
|
2,426,394 | <p>I'm sorry to bother you with this easy problem. But I'm working alone and totally confused.
It is the 1378th Problem from "Problems in Mathematical Analysis" written by Demidovich. The standard answer is
$$\frac{60(1+x)^{99}(1+6x)}{(1-2x)^{41}(1+2x)^{61}}$$</p>
<p>But when I followed the rutine
$$\frac{d(P(x)/Q(x))... | Nosrati | 108,128 | <p><strong>Hint:)</strong> If you are familiar with logarithm so write
$$y=\frac{(1+x)^{100}}{(1-2x)^{40}(1+2x)^{60}}$$
and
$$\ln y=100\ln(1+x)-40\ln(1-2x)-60\ln(1+2x)$$
take derivative and simplify!</p>
|
2,426,394 | <p>I'm sorry to bother you with this easy problem. But I'm working alone and totally confused.
It is the 1378th Problem from "Problems in Mathematical Analysis" written by Demidovich. The standard answer is
$$\frac{60(1+x)^{99}(1+6x)}{(1-2x)^{41}(1+2x)^{61}}$$</p>
<p>But when I followed the rutine
$$\frac{d(P(x)/Q(x))... | mechanodroid | 144,766 | <p>You can just take the derivative of the logarithm:</p>
<p>$$y = \frac{(1+x)^{100}}{(1-2x)^{40}(1+2x)^{60}} \implies \ln y = 100\ln(1+x)-40\ln(1-2x)-60\ln(1+2x)$$</p>
<p>$$\frac{y'}{y} = (\ln y)' = \frac{100}{1+x} + \frac{80}{1-2x} - \frac{120}{1+2x}$$</p>
<p>\begin{align}y' &= y\left(\frac{100}{1+x} + \frac{8... |
4,292,822 | <blockquote>
<p>Prove linear independence of <span class="math-container">$1+x^3-x^5,1-x^3,1+x^5$</span> in the Vector Space of Polynomials</p>
</blockquote>
<p>The attempts I found online all are quite easy. You just substitute something in for <span class="math-container">$x$</span> into the equation <span class="mat... | zwim | 399,263 | <p>In fact your independence condition can be rewritten <span class="math-container">$(c-a)x^5+(a-b)x^3+(a+b+c)=0$</span></p>
<p>It has to be true for all <span class="math-container">$x$</span> so it is equivalent to solve the system <span class="math-container">$\begin{cases}c-a=0\\a-b=0\\a+b+c=0\end{cases}$</span></... |
228,283 | <p>Could someone give a reference or construct an example of closed subspace of $Y\subset L_1[0,1]$ such that $\operatorname{dist}(x,Y)$ is not attained of for any $x\notin Y$.</p>
<p>I read somewhere that $Y$ is necessarily of infinite dimension and codimension.</p>
| Bill Johnson | 2,554 | <p>The $Y$ in Mikhail's answer has codimension one. Obviously $Y$ cannot be reflexive, but $Y$ can be of any non zero finite codimension or of infinite codimension. (Let $Z$ be any separable Banach space and let $Q$ be an operator from $L_1$ to $Z$ that maps the closed unit ball of $L_1$ onto the open unit ball of $Z$.... |
3,421,969 | <p>The point <span class="math-container">$C = (1, 2)$</span> lies inside the circle <span class="math-container">$x^2 + y^2 = 9$</span>. What is the length of the shortest chord of the circle through <span class="math-container">$C$</span>?</p>
| Bernard | 202,857 | <p>It simply results from the high-school identity:
<span class="math-container">$$a^n-1=(a-1)(a^{n-1}+a^{n-2}+\dots +a+1),$$</span>
which is often used as an example of an easy, but not trivial, proof by induction.</p>
|
4,543,203 | <p>I am going through Achim Klenke's Probability Theory textbook. In Section 7.4, he discusses what it means for two measures to be singular to one another and gives the following example.</p>
<p>Let <span class="math-container">$\Omega = \{0, 1\}^\mathbb{N}$</span> and let <span class="math-container">$(\mathrm{Ber}_p... | Matija | 1,096,797 | <p>I'm also not sure what the problem is, so I'll argue why there is no problem.
Consider <span class="math-container">$\{0,1\}^{\mathbb N}$</span> equipped with the Borel algebra induced by the discrete topology (which is the power set of <span class="math-container">$\{0,1\}^{\mathbb N}$</span>, cf. this <a href="htt... |
2,787,114 | <p>Before going to my question, let me give two prelimilary definitions</p>
<blockquote>
<p><strong>Definition 1.</strong> Let $S\subseteq\mathbb{R}^n$ be a non=empty open set in $\mathbb{R}^n$ under the usual topology on $\mathbb{R}^n$ and $f:S\to \mathbb{R}^m$. Let $\mathbf{c}\in S$ and $g:U(\subseteq \mathbb{R})\... | Frank Lu | 41,622 | <p>It turns out that the answer is positive. We prove by contradiction. Assume $f$ is discontinuous at $\mathbf{c}$, then there exists $\epsilon>0$ and a sequence $(\mathbf{c}_n)_{n=1}^\infty$ in $S$ such that $\mathbf{c}_n\to \mathbf{c}$ and
$$|f(\mathbf{c}_n)-f(\mathbf{c})|>\epsilon,\quad\forall n\in\mathbb{N}.... |
481,017 | <blockquote>
<p>Find all $f(x)$ satisfying $f(f(x)) = x^2 - 2$.</p>
</blockquote>
<p>Presumably $f(x)$ is supposed to be a function from $\mathbb R$ to $\mathbb R$ with no further restrictions (we don't assume continuity, etc), but the text of the problem does not specify further. </p>
<p><strong>Possibly Helpful L... | Hugh | 77,570 | <p><strong>There is no such $f$</strong>.</p>
<p>From <a href="http://yaroslavvb.com/papers/rice-when.pdf">http://yaroslavvb.com/papers/rice-when.pdf</a> , the question of existence is determined by the theorem: </p>
<p><strong>Theorem 6.</strong> <em>Let $\mathbb{R}$ be the real line. Let $g$ be a real quadratic pol... |
1,814,474 | <p>Lets have a series
$$\sum_{n=2}^\infty\frac{\ln\left(\frac{n+1}{n-1}\right)}{\sqrt{n}}$$</p>
<p>However, I have absolutely no clue how to try to continue. I could probably use the integral criterion and integrate the problem using the residue theorem, but that is too much of a hassle. Is there an easy way to prove ... | Jack D'Aurizio | 44,121 | <p>Some preliminary lemma.<br>
<strong>Lemma 1.</strong>
$$ \sum_{n\geq 1} H_n x^n = \frac{\log(1-x)}{1-x}. $$
<strong>Lemma 2.</strong> By Lemma $1$,
$$ \sum_{n\geq 1}\frac{H_n}{n+1} x^{n+1} = \frac{1}{2}\log^2(1-x),\qquad \sum_{n\geq 1}\frac{H_n+H_{n+1}}{n+1}x^{n}=\frac{-x+\log^2(1-x)+\text{Li}_2(x)}{x}.$$
<strong>Le... |
3,671,941 | <p>In <a href="https://sites.math.northwestern.edu/~mpopa/571/chapter3.pdf" rel="nofollow noreferrer">this chapter</a>, Example 2.1 notes that we can write <span class="math-container">$\mathbb{Z}_p$</span> as <span class="math-container">$$p\mathbb{Z}_p\cup(1+p\mathbb{Z}_p)\cup\cdots\cup (p-1+p\mathbb{Z}_p).$$</span> ... | Rob Arthan | 23,171 | <p>You can think of <span class="math-container">$\Bbb{Q}_p$</span> as numbers written in base <span class="math-container">$p$</span> notation that extend infinitely to the left of the <span class="math-container">$p$</span>-ary point, but finitely to the right. So you'd think of the example:
<span class="math-contain... |
28,166 | <p>With the initial conditions: $a>b>0$;</p>
<p>I need to find $$\lim_{n\to\infty}\sqrt[n]{a^n-b^n}.$$</p>
<p>I tried to block the equation left and right in order to use the Squeeze (sandwich, two policemen and a drunk, choose your favourite) theorem.</p>
| TonyK | 1,508 | <p>Rearrange $(a^n-b^n)^{1/n}$ as $a(1-(b/a)^n)^{1/n}$. Can you do this?</p>
|
2,744,008 | <p>Since the isolated singularities are $z=k\pi, k \in \mathbb Z$, so we divided the complex plane into the disjoint annulus, i.e. $\{z: n\pi <|z|<(n+1)\pi\}, n \in \mathbb N \cup \{0\}$. On these annuli, $f$ is analytic, so has a unique Laurent series.</p>
<p>First, let's consider $\{z: 0 < |z|<\pi\}$, th... | heropup | 118,193 | <p>Yes, for a continuous uniform random variable $X$ with support on an interval $[a,b]$, where $a < b$, the probability density function of $X$ is given by $$f_X(x) = \begin{cases} \frac{1}{b-a}, & a \le X \le b, \\ 0, & \text{otherwise}. \end{cases}$$
However, this only tells us the density. To get the ex... |
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