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 |
|---|---|---|---|---|
3,134,991 | <p>If nine coins are tossed, what is the probability that the number of heads is even?</p>
<p>So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.</p>
<p>We have <span class="math-container">$n = 9$</span> trials, find the probability of each <span class="math-container">$k$</span> for <span class="math-container">$k = 0, 2, 4, 6, 8$</span></p>
<p><span class="math-container">$n = 9, k = 0$</span></p>
<p><span class="math-container">$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$</span> </p>
<p><span class="math-container">$n = 9, k = 2$</span></p>
<p><span class="math-container">$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$</span> </p>
<p><span class="math-container">$n = 9, k = 4$</span>
<span class="math-container">$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$</span></p>
<p><span class="math-container">$n = 9, k = 6$</span></p>
<p><span class="math-container">$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$</span></p>
<p><span class="math-container">$n = 9, k = 8$</span></p>
<p><span class="math-container">$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$</span></p>
<p>Add all of these up: </p>
<p><span class="math-container">$$=.64$$</span> so there's a 64% chance of probability?</p>
| Vasili | 469,083 | <p>Your approach is good also, you probably made a mistake in calculations. The number of favorable outcomes is
<span class="math-container">$$\binom{9}{0}+\binom{9}{2}+\binom{9}{4}+\binom{9}{6}+\binom{9}{8}=1+36+126+84+9=256$$</span>
The number of all possible outcomes is <span class="math-container">$512$</span> thus the probability to get even number of heads is <span class="math-container">$0.5$</span>.</p>
|
3,134,991 | <p>If nine coins are tossed, what is the probability that the number of heads is even?</p>
<p>So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.</p>
<p>We have <span class="math-container">$n = 9$</span> trials, find the probability of each <span class="math-container">$k$</span> for <span class="math-container">$k = 0, 2, 4, 6, 8$</span></p>
<p><span class="math-container">$n = 9, k = 0$</span></p>
<p><span class="math-container">$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$</span> </p>
<p><span class="math-container">$n = 9, k = 2$</span></p>
<p><span class="math-container">$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$</span> </p>
<p><span class="math-container">$n = 9, k = 4$</span>
<span class="math-container">$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$</span></p>
<p><span class="math-container">$n = 9, k = 6$</span></p>
<p><span class="math-container">$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$</span></p>
<p><span class="math-container">$n = 9, k = 8$</span></p>
<p><span class="math-container">$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$</span></p>
<p>Add all of these up: </p>
<p><span class="math-container">$$=.64$$</span> so there's a 64% chance of probability?</p>
| Sri-Amirthan Theivendran | 302,692 | <p>The probability generating function of a Binomiall random variable <span class="math-container">$X\sim \text{Bin}(n, 1/2)$</span> with probability of success <span class="math-container">$1/2$</span> is given by
<span class="math-container">$$
g_{X}(t)=Et^X=\sum_{k=0}^nP(X=k)t^k=\sum_{k=0}^n\binom{n}{k}\frac{t^k}{2^n}=\frac{1}{2^n}(1+t)^n
$$</span>
In particular the probability that <span class="math-container">$X$</span> is even is given by
<span class="math-container">$$
\sum_{0\leq k\leq n\, k\,{\text{even}}}P(X=k)=\frac{g(1)+g(-1)}{2}=\frac{1+0}{2}=\frac{1}{2}.
$$</span></p>
|
780,895 | <p>A collection of black and white balls are to be arranged on a straight line such that each ball has at least one neighbor of different color. If there are 100 black balls, then the maximum number of white balls that allows such an arrangement is? </p>
| mm-aops | 81,587 | <p>$200$. easy to see you can't have more cause every black ball can be a neighbour of at most $2$ white balls. to obtain such an arrangement just put a white ball on the left and on the right of each black one, put them in a row, it's evident such an arrangement satisfies your conditions.</p>
|
455,979 | <p>Suppose we have three 6-sided die that all share the same common bias:</p>
<p>For a single dice: let the probability of rolling a 2 or $P(2) = 2{\times}P(1$), let the probability of rolling a 3 or $P(3) = 3{\times}P(1)$, and so on...</p>
<p>Such that:
$P(2) = 2P(1), P(3) = 3P(1), P(4) = 4P(1), P(5)=5P(1), P(6)=6P(1)$</p>
<p>Given that the sum of all the probabilities is 1, we can get that $P(1) = 1/21, P(1) {\approx} .0476$</p>
<p>My question are as follow:</p>
<p>A) Given two die are rolled what is the probability that the sum of the faces of the die is 3? How about 6?</p>
<p>B) Given that three die are rolled what is the probability that the sum of the faces of the die is 9? What about 10?</p>
<p>I made this problem myself after adapting a problem I saw that involved a single dice that had the bias described above. I have a good idea of how the probability is calculated, I just wanted to see how other people approach the problem, just to make sure I am doing it the most efficient way possible. I was also wondering if there is a frequentist method of solving the problem.</p>
<p>Also does anyone have Diagrams of how a PDF of would look like for 2 or 3 die rolled?</p>
| Stephen Herschkorn | 27,997 | <p>I think the best way to handle this is via the probability generating function $\pi(z) = Ez^X = \frac1{21}\sum_{k=1}^6 k z^k$ for the value $X$ from a roll of a single die. The probability generating function for the sum of $n$ independent rolls is $[\pi(z)]^n$. Multiply out the polynomilas (a CAS - <em>e.g.</em>, Mathematica® or Maple - will help) and read off the coefficents.</p>
|
64,130 | <p>This is an arithmetic follow-up to my previous question <a href="https://mathoverflow.net/questions/64112/does-there-exist-a-non-trivial-semi-stable-curve-of-genus-1-with-only-4-singular">Does there exist a non-trivial semi-stable curve of genus >1 with only 4 singular fibres</a> </p>
<p>Let $k$ be an algebraically closed field and let $f:X\longrightarrow \mathbf{P}^1_{k}$ be a semi-stable curve. Let $s$ denote the number of singular fibres. If $X$ is non-isotrivial and of positive genus, we have that $s>2$ (Beauville and Szpiro). As Angelo stated in my previous question, for genus >1 and $k=\mathbf{C}$, Sheng-Li Tan has shown that $s>4$.</p>
<p>Now, let $S=\textrm{Spec} \mathbf{Z}$ and let $X\longrightarrow S$ be a (regular) semi-stable arithmetic surface. Let $s$ be the number of singular fibres. Fontaine has shown that $s>0$ if $X$ is of positive genus. </p>
<p><strong>Question.</strong> Let $g>0$ be an integer. Does there exist a semi-stable arithmetic surface $X\longrightarrow S$ of genus $g$ with precisely one singular fibre?</p>
<p>I expect the answer to be yes for $g=1$ but no for $g>1$.</p>
<p><strong>Example.</strong> The modular curve $X_1(\ell)$ ($\ell$ big enough) has semi-stable reduction over Spec $\mathbf{Z}[\zeta_{l}]$. This model has precisely one singular fibre. Note that the modular curve $X_1(l)$ does not have semi-stable reduction over $\mathbf{Z}$.</p>
| Qing Liu | 3,485 | <p>There are some classical examples of such surfaces. For any prime number $p\ge 11$ different from 13, the modular curve $X_0(p)$ has good reduction away from $p$, and semi-stable reduction at $p$ (equal to the union of two projective lines intersecting at supersingular $j$'s). This is proved by Deligne-Rapoport (see also Bouw-Wewers: <a href="http://arxiv.org/abs/math/0210363">Stable reduction of modular curves</a>, Prog. In Math. <b>224</b> (2004)). </p>
<p>There are however some constraints for such curves over $\mathbb Z$. If $p$ is the unique semi-stable fiber, then Brumer-Kramer (Manuscripta Math.(2001)) showed that $p\ne 2, 3, 5, 7$, and Schoof
(Compos. Math. (2005)) showed that $p\ne 13$. </p>
|
2,736 | <p><a href="https://mathoverflow.net/questions/18989/generating-classical-groups-over-finite-local-rings">Generating Classical Groups over Finite Local Rings</a> asks a question that, according to the poster's own 'answer' <a href="https://mathoverflow.net/a/19098/2383">https://mathoverflow.net/a/19098/2383</a>, is not what was actually meant. I edited the question to reflect the stated intention (changing the words "semisimple elements" to "tori").</p>
<p>One rejection said:</p>
<blockquote>
<p>This edit does not make the post even a little bit easier to read, easier to find, more accurate or more accessible. Changes are either completely superfluous or actively harm readability.</p>
</blockquote>
<p>I have the impression that this rejection was based on the idea that I was trying to substitute a synonym for "semisimple elements" merely to improve the wording, whereas actually I was (intentionally) changing the meaning (so that it agreed with the poster's stated intentions—5 years' interval suggesting that he or she was not going to make the edit him- or herself).</p>
<p>That brings me to the next rejection, which said:</p>
<blockquote>
<p>There is already an answer; rather than edit out the question that was answered, you should add the clarification so answer is not dinged as wrong.</p>
</blockquote>
<p>This seems like a very clear action plan to me, and I agree that doing this is better than making my original edit. However, given the first rejection, I am reluctant to try again to make this edit in case there is any penalty for appearing 'argumentative'. Is it reasonable for me to try again?</p>
<p>EDIT: I was too slow, and, in the meantime, the post was edited in a better way, simply inlining asm's clarification. I didn't pay enough attention and submitted my in-progress edit anyway, but I guess that it will (properly) get re-rejected.</p>
| Community | -1 | <p>Yes, you can, and likely should, try again. It could be seen as problematic to submit the exact same proposal again, yet since it is substantially different and informed by the feedback you got this seems perfectly fine.</p>
<p>Under certain circumstances, rather not in this case, I would even support submitting the exact same proposal; it is not as if reviewers are infallible. Note that one user even approved your original edit, so it was a split decision. </p>
<p>While it is in principle possible to get banned from suggesting edits, quite a bit more is needed for this than trying anew once in case of a rejected edit. </p>
|
991,878 | <p>How can it be proven that a cycle of length k is an even permutation if and only if k is odd?
I know it can be done using the fact that a permutation which exchanges two elements but leaves the rest unchanged is an odd permutation.</p>
| Macavity | 58,320 | <p>This is to show that $x=1$ always works. So we need to show that
$$f(a)=(a-1)-\log a \ge 0, \quad \forall a> 0$$</p>
<p>$$f'(a) = 1-\frac1a = \begin{cases} < 0, && a < 1 \\ > 0, && a> 1 \end{cases}$$</p>
<p>So the function is decreasing for $a < 1$ and increasing for $a> 1$, hence it achieves its minimum when $a=1$, i.e. $f(1)=0$. Hence $f(a) \ge 0$, so $x=1$ is always a solution.</p>
|
2,002,201 | <p>simplify <span class="math-container">$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$</span>.</p>
<blockquote>
<p>1.<span class="math-container">$90^{\frac{3}{2}}$</span></p>
<p>2.<span class="math-container">$106\sqrt{41}$</span></p>
<p>3.<span class="math-container">$4\sqrt{41}$</span></p>
<p>4.<span class="math-container">$504$</span></p>
<p>5.<span class="math-container">$508$</span></p>
</blockquote>
<p><strong>My attempt</strong>:I do like this but I didn't get any of those five.</p>
<p><span class="math-container">$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}={\sqrt{45+4\sqrt{41}}}^3-\sqrt{45-4\sqrt{41}}^3$</span></p>
<p><span class="math-container">$=(\sqrt{45+4\sqrt{41}}-\sqrt{45-4\sqrt{41}})(45+4\sqrt{41}+45-4\sqrt{41}+(\sqrt{45+4\sqrt{41}})(\sqrt{45-4\sqrt{41}})$</span></p>
<p>Now I do the nested radicals formula and I get <span class="math-container">$254\sqrt{41}$</span> which is none of those where did I mistaked?</p>
| Ennar | 122,131 | <p>When I see expression where both $\alpha = a+b\sqrt{n}$ and $\beta =a-b\sqrt n$ occur, I immediately calculate $\alpha + \beta = 2a$ and $\alpha\beta = a^2-nb^2$ since they are guaranteed to be integers (more precisely, the minimal polynomial of both of them is $x^2 - (\alpha+\beta)x+\alpha\beta$ which might be helpful in some cases; if you are not familiar with the term, just ignore this remark). So, we have $$x = \sqrt{\alpha^3}-\sqrt{\beta^3}\implies x^2 = \alpha^3 -2\sqrt{(\alpha\beta)^3}+\beta^3\\
\implies x^2 = (\alpha + \beta)(\alpha^2-\alpha\beta+\beta^2)-2\sqrt{(\alpha\beta)^3}\\
\implies x^2 = (\alpha + \beta)((\alpha-\beta)^2+\alpha\beta)-2\sqrt{(\alpha\beta)^3}$$</p>
<p>Now, in your case $\alpha\beta = 1369 = 37^2$, so we have $$x^2 = 90((2\cdot 4\sqrt{41})^2+1369)-2\cdot 37^3 = 258064\implies x= 508$$</p>
|
2,002,201 | <p>simplify <span class="math-container">$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$</span>.</p>
<blockquote>
<p>1.<span class="math-container">$90^{\frac{3}{2}}$</span></p>
<p>2.<span class="math-container">$106\sqrt{41}$</span></p>
<p>3.<span class="math-container">$4\sqrt{41}$</span></p>
<p>4.<span class="math-container">$504$</span></p>
<p>5.<span class="math-container">$508$</span></p>
</blockquote>
<p><strong>My attempt</strong>:I do like this but I didn't get any of those five.</p>
<p><span class="math-container">$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}={\sqrt{45+4\sqrt{41}}}^3-\sqrt{45-4\sqrt{41}}^3$</span></p>
<p><span class="math-container">$=(\sqrt{45+4\sqrt{41}}-\sqrt{45-4\sqrt{41}})(45+4\sqrt{41}+45-4\sqrt{41}+(\sqrt{45+4\sqrt{41}})(\sqrt{45-4\sqrt{41}})$</span></p>
<p>Now I do the nested radicals formula and I get <span class="math-container">$254\sqrt{41}$</span> which is none of those where did I mistaked?</p>
| Mathew Mahindaratne | 525,941 | <p>Considering all positive values, $45\pm4\sqrt{41}$ can be written as $45\pm4\sqrt{41}=(\sqrt{41}\pm2)^2$, thus, the given expression can be simplified as follows: $$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}=\left(\sqrt{45+4\sqrt{41}}\right)^3-\left(\sqrt{45-4\sqrt{41}}\right)^3=\left(\sqrt{\left(\sqrt{41}+2\right)^2}\right)^3-\left(\sqrt{\left(\sqrt{41}-2\right)^2}\right)^3=\left(\sqrt{41}+2\right)^3-\left(\sqrt{41}-2\right)^3$$
Now recall, $a^3-b^3=(a-b)(a^2+ab+b^2)$. Thus, $$\left(\sqrt{41}+2\right)^3-\left(\sqrt{41}-2\right)^3=\left((\sqrt{41}+2)-(\sqrt{41}-2)\right)\left((\sqrt{41}+2)^2+(\sqrt{41}+2)(\sqrt{41}-2)+(\sqrt{41}-2)^2\right)=\left(\sqrt{41}+2-\sqrt{41}+2\right)\left(45+4\sqrt{41}+41-4+45-4\sqrt{41}\right)=4(45+41-4+45)=508$$</p>
|
4,510,384 | <p>In exercise 2.13 of page 43 of the book <a href="https://rads.stackoverflow.com/amzn/click/com/0134746759" rel="nofollow noreferrer" rel="nofollow noreferrer">Mathematical Proofs: A Transition to Advanced Mathematics</a> the reader is asked to state the logical negation of some statements. Of these, I find the authors' answer to one of them baffling.</p>
<p>The statement to negate is:</p>
<blockquote>
<p>"Two sides of the triangle have the same length."</p>
</blockquote>
<p>The authors' negation of the statement is:</p>
<blockquote>
<p>"The sides of the triangle have different lengths".</p>
</blockquote>
<p>Am I mistaken in assuming that when negating a statement, one is supposed to state what previously presumed false as true and vice versa? If one assumes the proposition "Two sides of the triangle have the same length." to be true, is it erroneous to conclude that the negation would be "The sides of the triangle have different lengths" or (exclusively) 'Three sides of the triangle have the same length'? I thank your aid in advance.</p>
| Dan Christensen | 3,515 | <p>Let <span class="math-container">$x,y$</span> and <span class="math-container">$z$</span> be the sides of a triangle. We have: <span class="math-container">$x\neq y,
~ x\neq z,~ y\neq z$</span></p>
<p>Let <span class="math-container">$len(s)$</span> be the length of side <span class="math-container">$s$</span>.</p>
<p>We have two sides being equal in length:</p>
<p><span class="math-container">$~~~~~~len(x)=len(y)~ \lor ~len(x)=len(z) ~ \lor ~ len(y)=len(z)$</span></p>
<p>(<strong>EDIT:</strong> All three sides being equal in length is not ruled out.)</p>
<p>Applying De Morgan's Law and the elimination of double negations (<span class="math-container">$\neg\neg$</span>), we can obtain its negation:</p>
<p><span class="math-container">$~~~~~~len(x)\neq len(y) ~\land~ len(x)\neq len(z) ~\land ~ len(y)\neq len(z)$</span></p>
<p>In words, we have <strong>NO</strong> two sides being equal in length.</p>
|
3,032,950 | <p>I have the following Cauchy problem. I do not know where to start, so I would appreciate any help and tips.
<span class="math-container">$$\frac{\partial^2 Y(t, x)}{\partial t^2} = 9\frac{\partial^2 Y(t,x)}{\partial x^2} - 2Z(t,x)$$</span>
<span class="math-container">$$\frac{\partial^2 Z(t, x)}{\partial t^2} = 6\frac{\partial^2 Z(t,x)}{\partial x^2} - 2Y(t,x)$$</span>
<span class="math-container">$$Y(0, x) = \cos(x),\ Z(0, x) = 0$$</span>
<span class="math-container">$$\frac{\partial Y(t, 0)}{\partial t} = \frac{\partial Z(t, 0)}{\partial t} = 0$$</span></p>
| Yuri Negometyanov | 297,350 | <p><span class="math-container">$$\begin{cases}
2Y(t,x) = 6\dfrac{\partial^2 Z(t,x)}{\partial x^2} - \dfrac{\partial^2 Z(t, x)}{\partial t^2}\\[4px]
2\dfrac{\partial^2 Y(t, x)}{\partial t^2} = 18\dfrac{\partial^2 Y(t,x)}{\partial x^2} - 4Z(t,x),
\end{cases}$$</span>
so
<span class="math-container">$$\dfrac{\partial^2}{\partial t^2}\left(6\dfrac{\partial^2 Z(t,x)}{\partial x^2} - \dfrac{\partial^2 Z(t, x)}{\partial t^2}\right) = 9\dfrac{\partial^2}{\partial x^2}\left(6\dfrac{\partial^2 Z(t,x)}{\partial x^2} - \dfrac{\partial^2 Z(t, x)}{\partial t^2}\right) - 4Z(t,x),$$</span>
<span class="math-container">$$\begin{cases}\dfrac{\partial^4 Z(t,x)}{\partial t^4} - 15\dfrac{\partial^4 Z(t,x)}{\partial t^2\partial x^2} + 54\dfrac{\partial^4 Z(t,x)}{\partial x^4}-4Z(t,x) = 0\\[4pt]
Z(0, x) = 0\\[4pt]
6\dfrac{\partial^2 Z(0,x)}{\partial x^2} - \dfrac{\partial^2 Z(0, x)}{\partial t^2} = 2\cos(x)\\[4pt]
6\dfrac{\partial^3 Z(t,0)}{\partial t\partial x^2} - \dfrac{\partial^3 Z(t, 0)}{\partial t^3} = 0\\[4pt]
\dfrac{\partial Z(t, 0)}{\partial t} = 0
\end{cases}$$</span></p>
<p><span class="math-container">$$\begin{cases}\dfrac{\partial^4 Z(t,x)}{\partial t^4} - 15\dfrac{\partial^4 Z(t,x)}{\partial t^2\partial x^2} + 54\dfrac{\partial^4 Z(t,x)}{\partial x^4}-4Z(t,x) = 0\\[4pt]
Z(0, x) = 0,\quad \dfrac{\partial^2 Z(0,x)}{\partial x^2} = \dfrac13\cos(x)\\[4pt]
\dfrac{\partial Z(t, 0)}{\partial t} = 0,\quad\dfrac{\partial^3 Z(t, 0)}{\partial t^3} = 0.
\end{cases}$$</span>
And it reqires four conditions more.</p>
|
3,243,503 | <p>If <span class="math-container">$x + y = 2c$</span>, find minimum value of
<span class="math-container">$ \sec x +\sec y $</span> if <span class="math-container">$x,y\in(0,\pi/2)$</span>, in terms of <span class="math-container">$c$</span>.</p>
<p>I was able to solve by differentiating the equation and got the answer as 2secc.
But i would like to know solution with trigonometry as base or without differentiating the equation.</p>
| lab bhattacharjee | 33,337 | <p><span class="math-container">$$u=\sec x+\sec y=\dfrac{4\cos\dfrac{x+y}2\cos\dfrac{x-y}2}{\cos(x-y)+\cos(x+y)}$$</span></p>
<p><span class="math-container">$$u=\dfrac{4\cos C\cdot t}{2t^2-1+\cos2c}$$</span></p>
<p>which is a quadratic equation in <span class="math-container">$t=\cos\dfrac{x-y}2$</span></p>
<p>As <span class="math-container">$t$</span> is real, the discriminant must be <span class="math-container">$\ge0$</span></p>
<p>See <a href="https://math.stackexchange.com/questions/3242973/proving-1-cos-x-over2-sin-x-frac43/3243092#3243092">Proving ${1+\cos x\over2+\sin x} < \frac43$</a></p>
|
1,598,545 | <p>Maybe I am not well versed with the actual definition of mean, but I have a doubt. On most resources, people say that arithmetic mean is the sum of $n$ observations divided by n. So my first question: </p>
<blockquote>
<p>How does this formula work? Is there any derivation to it? If not,
then while creating this definition, what was the creator thinking?</p>
</blockquote>
<p>Okay, so using my intuition, I thought that it is the value that lies in the centre. And it worked for some cases, like the mean of $1$ , $2$ and $3$ is $2$ , which is the central value. But, lets imagine a number line from numbers $0$ to $9$. Now, I choose $3$ numbers, say $1$, $8$ and $9$. By the formula, I get the mean is equal to $6$. But, if mean really is a central value, shouldn't it be $5$(I know we call $5$ the median in this case)? But it seems like the mean is getting closer to $8$ and $9$, which means it is not central? So my final question?</p>
<blockquote>
<p>Have I imagined mean incorrectly? What kind of central value really
mean is?</p>
</blockquote>
| Ahmed S. Attaalla | 229,023 | <p>I suppose how you would come up with mean:</p>
<p>Suppose we have played some soccer games, and here is the list of the goals in the soccer games we played:</p>
<p>$${1,1,1,1,1,9}$$</p>
<p>And we want to get a sense of how we did as a whole or how much we scored per game as a whole.</p>
<p>Well looking at the list we see that we played for $6$ games and in that $6$ games we scored $14$ goals, so our rate as a whole would be about:</p>
<p>$$2.33 goals/game$$</p>
<p>Which we see is a pretty good measure of how we did as a whole. But of course in this case median is a better measure of how we would do on average, as it is not influenced by outliers, meaning that that $9$ was very unusual so it doesn't influence the median. We look at many tools to find an average, like mean, median,... and use what we find best in the situation. </p>
|
761,286 | <p>let $G$ be an infinite group of the form $G_1 \oplus G_2 \oplus \dots \oplus G_n$ where each $G_i$ is a <strong>non trivial</strong> group and $n>1$. Prove that $G$ is not cyclic.</p>
<p><strong>Attempt</strong> : Let $G = G_1 \oplus G_2 \oplus \dots \oplus G_n$ be cyclic.</p>
<p>then $\exists ~g =(g_1,g_2,......,g_n)$ such that $g$ is generator of $G$ if and only if :</p>
<p>$(i)~~ g_1$ generates $G_1, ~g_2$ generates $ G_2$, and so on.</p>
<p>$(ii)~\gcd(|g_i|,|g_j| =1)$ whenever $i \neq j$ .</p>
<p>Since a group of prime order is cyclic, this means that if we take $G_i$ such that $|G_i|=p_i$ , where $p_i$ is a prime number and for any $i \neq j,~\gcd (p_i,p_j) =1$ , hence, if we take $G_i$'s in this fashion such that $g_i$ of prime order generates $G_i$, $G$ should turn out to be cylic?</p>
<p>Where could I be making a mistake? Thank you for the help.</p>
| ml0105 | 135,298 | <p>It would be $\Omega(n^{2})$ time, but not $O(n^{2})$ time. The intuition for this is that you multiply the complexities of inner loops.</p>
<p>Consider as well, straight from the definition of Big-O. Suppose your algorithm was $O(n^{2})$. Then $n * n\sqrt{n} \leq C * n^{2}$, for some positive constant $C$. We then get that $\sqrt{n} \leq C$, which we know not to be true.</p>
<p>You can derive the runtime function $T(n)$ for your algorithm yourself, but these are good tricks to eye-ball it.</p>
<p>Edit: Now that you've clarified the inner loop runs in $\sqrt{n}$ time, you would be correct that $n\sqrt{n} \in O(n^{2})$. </p>
<p>Here is a really good tutorial on basic Big-O analysis of code: <a href="http://www.dreamincode.net/forums/topic/125427-determining-big-o-notation/" rel="nofollow">http://www.dreamincode.net/forums/topic/125427-determining-big-o-notation/</a></p>
|
3,845,475 | <p>Here's what I'm tasked with showing:</p>
<p>Let <span class="math-container">$(a_n)$</span> be a convergent sequence with <span class="math-container">$a_n\rightarrow a$</span> as <span class="math-container">$n\rightarrow\infty$</span>. By the Algebraic Limit Theorem, we know that <span class="math-container">$(a_n^2)\rightarrow a^2$</span>. Now prove this using the definition of convergence.</p>
<p>In doing so, I have the following:</p>
<p>Let <span class="math-container">$\epsilon>0$</span> be given. Choose some <span class="math-container">$N\in\mathbb{N}$</span> so that for all <span class="math-container">$n\geq N$</span>, <span class="math-container">$|a_n^2-a^2|<\epsilon$</span>. By algebra, <span class="math-container">$|a_n^2-a^2|=|a_n-a||a_n+a|$</span>. Consider <span class="math-container">$|a_n+a|$</span>. By the triangle inequality, <span class="math-container">$|a_n+a|\leq|a_n|+|a|$</span>, thus <span class="math-container">$|a_n|$</span> is bounded by some <span class="math-container">$M\in\mathbb{N}$</span>.</p>
<p>I know I'm trying to choose <span class="math-container">$M$</span> so that <span class="math-container">$|a_n-a|<\epsilon+M+|a|$</span>. Where should I take it from here?</p>
| Ned | 67,710 | <p>Your argument (the second one) seems to miss the FIRST ace aspect of the situation. The <span class="math-container">$3*51!$</span> orderings in the numerator have the spade ace immediately following another ace but some have yet a different ace occurring before both of them, so they shouldn't be counted.</p>
|
2,299,466 | <p>For the first-order language with vocabulary $(E)$ (the binary relation $E$ which holds if two vertices have an edge) together with a set $G$ of vertices, I've been told that the property "a symmetric graph is connected" cannot be axiomatized by any set of first-order sentences. </p>
<p>I think the proof involved taking two constants, let's say $x,y \in G $ and building an infinite set of sentences $T$ that states "there is no path of length $n$ between $x$ and $y$" for each $n$ (where a path is, for some vertices $(v_1,...,v_n) \in G$, $[E(x, v_1) \wedge ... \wedge E(v_n, y)]$). By compactness, I see that $T$ is satisfiable. However, I don't see how $T$ shows the impossibility of constructing another set of sentences $T'$ which is satisfied by connected graphs. </p>
<p>I do understand the use of compactness to show that a theory whose models have arbitrarily large domains also has a model with an infinite domain, but I don't understand its use here. I don't have any experience with ultraproducts, so answers using that concept may be lost on me.</p>
| Eric Wofsey | 86,856 | <p>Your description of $T$ is not complete; here's what it should be. Suppose there exists a first order axiomatization $T_0$ of connected graphs. Now let $T$ be the union of $T_0$ and your sentences "there is no path of length $n$ between $x$ and $y$" for each $n$. Any finite subset of this $T$ has a model, since you can find a connected graph with points $x$ and $y$ satisfying any finite subset of your sentences. Therefore $T$ is satisfiable. A model of $T$ is then a connected graph (because it satisfies $T_0$) with elements $x$ and $y$ with no path of any length between them. This is a contradiction. Therefore the set $T_0$ cannot exist, and there is no first order axiomatization of connected graphs.</p>
|
813,517 | <p>Deduce that the next integer greater than $(3+\sqrt 5)^n$ is divisible by $2^n$</p>
<p>I tried expanding it by binomial theorem but got nothing</p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>Let $a=3+\sqrt5, b=3-\sqrt5$</p>
<p>So, $a,b$ are the roots of $x^2-6x+4=0\implies t_{n+2}=6t_{n+1}-ta_n$ where $t_m=(3+\sqrt5)^m$</p>
|
813,517 | <p>Deduce that the next integer greater than $(3+\sqrt 5)^n$ is divisible by $2^n$</p>
<p>I tried expanding it by binomial theorem but got nothing</p>
| Mark Bennet | 2,906 | <p>There is a trick to this kind of question, which comes in handy. Note that $$0\lt 3-\sqrt 5 \lt 1$$ so so that if $$a_n=(3+\sqrt5)^n+(3-\sqrt 5)^n$$ we know that $$a_n-(3+\sqrt 5)^n=(3-\sqrt 5)^n\lt 1$$</p>
<p>Also $a_0=0, a_1=6$ are integers.</p>
<p>Now $3+\sqrt 5+3-\sqrt 5=6$ and $(3+\sqrt 5)(3-\sqrt 5)=4$ so $3+\sqrt 5$ and $3-\sqrt 5$ are roots of the equation $$x^2-6x+4=0$$</p>
<p>And you will find that $a_{n+1}-6a_n+4a_{n-1}=0$</p>
<p>From this you can deduce that $a_n$ is an integer greater than $(3+\sqrt 5)^n$ and within $1$ of it, so the next integer greater, which you wanted to find. And the recurrence will enable you to complete your induction.</p>
<hr>
<p>I am using some facts about recurrences here:</p>
<p>If $a, b$ are roots of the equation $x^2+px+q=0$ [note that $p=-a-b, q=ab$ in this case] we have $$a^2+pa+q=0$$We multiply by $a^{n-1}$ to get $$E_a:a^{n+1}+pa^n+qa^{n-1}=0$$ and equivalently $$E_b: b^{n+1}+pb^n+qb^{n-1}=0$$ Now if we set $u_n=Aa^n+Bb^n$ we find, on taking $AE_a+BE_b$ that: $$u_{n+1}+pu_n+qu_{n-1}=0$$</p>
<p>Note that it is quite frequent in "nearest integer" problems with irrationals that there is some form of conjugate in the background.</p>
|
3,930,659 | <blockquote>
<p>Evaluate: <span class="math-container">$$ \int \frac{x^2}{\sqrt{1-x^2}}\,dx$$</span></p>
</blockquote>
<p>The solution I came across does a <span class="math-container">$u$</span>-substitution by letting <span class="math-container">$x = \sin(t)$</span>. But why <span class="math-container">$\sin(t)$</span>? It seems a lot like guessing to me - should I just guess the right <span class="math-container">$x$</span> for substitution? Why not <span class="math-container">$\cos(t)$</span>? Is there any way that I could identify that "this" expression is solved by "that" trig substitution?</p>
| Community | -1 | <p>I think that the OP's question refers to how to distinguish why to make this change of variable. Now, I will write the answer to that question.</p>
<p>Suppose you want to solve an integral of the form <span class="math-container">$$\color{blue}{\boxed{\int R(x,\sqrt{a^{2}-x^{2}})dx}}$$</span>
As in your problem, that you have <span class="math-container">$$\color{red}{\text{Example:} \quad \int \frac{x^{2}}{\sqrt{1^{2}-x^{2}}}dx}$$</span>So, this integral becomes a <strong>trigonometric integral</strong> with the change of variable <span class="math-container">$$\color{green}{\boxed{x=a\cos(t)}} \quad \color{green}{\boxed{x=a\sin(t)}}$$</span>
As the <strong>MathS</strong> users indicate in the comments. Also note that you can also perform the variable change <span class="math-container">$$\color{green}{\boxed{x=a\tanh(t)}}$$</span> and this variable change transforms the integral into a <strong>hyperbolic integer</strong>.</p>
|
2,547,933 | <p>Consider the integral $I=\displaystyle\int_{R}\int f(x,y)dx dy$ over the region $R$, given by the triangle with vertices $(0,0),(1,1)$ and $(2,0)$. </p>
<p>This is an isosceles triangle with one side lying along the $x-$axis. So, our domain is not "nice" to find the bounds for integral I assume, since even if we write $0\le x \le 2$, we can not give bounds for $y$ easily. To find this integral, the book I am reading makes the following transformation: $u = y-x$, $v=y+x$. After that, our new domain becomes a right angled triangle with the perpendicular edges lying on the $u$ and $v$ axis.</p>
<p>Finally, my question is how can we conclude this transformations? In general, setting up $u = x+y, v= x-y$ works quite nice for triangular/rectangular domains but is there a rule for this?</p>
<p>Thank you. </p>
| Reese Johnston | 351,805 | <p>It's the same reason that you can't argue that $\lim_{n \to \infty}(1 + \frac{1}{n})^n = 1$, even though $1 + \frac{1}{n} \to 1$ and $1^n = 1$. The issue is that when we say that the <em>limit</em> of an expression is a certain value, we just mean the expression gets <em>very close</em> to that value - we make no promises about <em>how fast</em>.</p>
<p>When $n$ is very large, $1 - \frac{1}{n}$ is very nearly $1$ - but a number very nearly $1$, when raised to a very large power, can be very small. For example, $0.99^{10000}$ is zero to more than forty decimal places. So that exponent of $n$ can "pull" the value away from one - for $(1 - \frac{1}{n})^n$, it pulls it down, while it pulls $(1 + \frac{1}{n})^n$ up.</p>
|
2,547,933 | <p>Consider the integral $I=\displaystyle\int_{R}\int f(x,y)dx dy$ over the region $R$, given by the triangle with vertices $(0,0),(1,1)$ and $(2,0)$. </p>
<p>This is an isosceles triangle with one side lying along the $x-$axis. So, our domain is not "nice" to find the bounds for integral I assume, since even if we write $0\le x \le 2$, we can not give bounds for $y$ easily. To find this integral, the book I am reading makes the following transformation: $u = y-x$, $v=y+x$. After that, our new domain becomes a right angled triangle with the perpendicular edges lying on the $u$ and $v$ axis.</p>
<p>Finally, my question is how can we conclude this transformations? In general, setting up $u = x+y, v= x-y$ works quite nice for triangular/rectangular domains but is there a rule for this?</p>
<p>Thank you. </p>
| zhw. | 228,045 | <p>Just to give you some intuition, note that since $1-x < 1/(1+x)$ for $0<x<1,$</p>
<p>$$ (1-1/n)^n < \frac{1}{(1 + 1/n)^n}$$</p>
<p>for $n>1.$ Now the right side $\to 1/e.$ Thus if the limit of the left side exists, it has to be $\le 1/e.$ </p>
|
1,527,197 | <p>So in the case where data points have the same variance $\sigma^2$, the estimator (in normal equation form) can be written as </p>
<p>$$\theta=(X^TX)^{-1}X^TY$$</p>
<p>I'm not sure how to derive a similar formula when the data points have different variances, and thus the covariance matrix would be</p>
<p>$$\Sigma = diag(\sigma_1^2, \sigma_2^2, ...,\sigma_n^2)$$</p>
| Bernard | 202,857 | <p>You can say more for 1):</p>
<p>Set $\displaystyle s_2n=\sum_{k=1}^{2n}(-1)^{k+1}\frac{k+1}k $. Grouping pairs of consecutive terms, you can write
$$s_{2n}=\sum_{i=1}^n\biggl(\frac{2i}{2i-1}-\frac{2i+1}{2i}\biggr)=\sum_{i=1}^n\frac{1}{2i(2i-1)}=\sum_{i=1}^n\biggl(\frac{1}{2i-1}-\frac{1}{2i}\biggr)=\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k},$$
so $\;s_{2n}$ corresponds to the <em>alternating harmonic series</em>, which converges to $\color{red}{\ln 2}$.</p>
<p>As $\;s_{2n+1}=s_{2n}+\dfrac{2n+2}{2n+1}$, we see the sums of an odd number of terms also converge, to $\color{red}{\ln 2+1}$.</p>
|
337,930 | <p>Given two polynomials</p>
<p>$$
p(x) = a_0 + a_1 x + a_2 x^2 + \ldots + a_{n-1}x^{n-1} \\
q(x) = b_0 + b_1 x + b_2 x^2 + \ldots + b_{n}x^{n}
$$</p>
<p>And the series expansion from their rational polynomial</p>
<p>$$
\frac{p(x)}{q(x)} = c_0 + c_1 x + c_2 x^2 + \ldots
$$</p>
<p>is it possible to recover the the original polynomials $a_n$, $b_n$ from only the series $c_n$ via the solution of a linear system? </p>
| Felix Marin | 85,343 | <p><span class="math-container">$\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\ket}[1]{\left\vert #1\right\rangle}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
\newcommand{\sech}{\,{\rm sech}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert #1 \right\vert}%
\newcommand{\yy}{\Longleftrightarrow}$</span>
<span class="math-container">\begin{align}
\pars{1 + x}^{R + M}&=\sum_{k = 0}^{R + M}{R + M \choose k}x^{k}
\\
\pars{1 + x}^{R}\pars{1 + x}^{M}
&=\sum_{\ell = 0}^{R}{R \choose \ell}x^{\ell}\sum_{\ell' = 0}^{M}{M \choose \ell'}x^{\ell'}\sum_{k = 0}^{R + M}\delta_{k,\ell + \ell'}
\\[3mm]&=
\sum_{k = 0}^{R + M}x^{k}\sum_{\ell = 0}^{R}{R \choose \ell}
\sum_{\ell' = 0}^{M}{M \choose \ell'}\delta_{\ell',k - \ell}
\\[3mm] & =
\left.\sum_{k = 0}^{R + M}x^{k}\sum_{\ell = 0}^{R}{R \choose \ell}
{M \choose k - \ell}\right\vert_{0 \leq k - \ell \leq M}
\end{align}</span></p>
<blockquote>
<span class="math-container">$$\color{#0000ff}{\large%
{R + M \choose k}
=
\sum_{\ell = 0 \atop {\vphantom{\LARGE A^{a}}k - M \leq\ \ell\ \leq k}}
^{R}{R \choose \ell}{M \choose k - \ell}}\,,
\qquad 0 \leq k \leq R + M
$$</span>
</blockquote>
|
337,930 | <p>Given two polynomials</p>
<p>$$
p(x) = a_0 + a_1 x + a_2 x^2 + \ldots + a_{n-1}x^{n-1} \\
q(x) = b_0 + b_1 x + b_2 x^2 + \ldots + b_{n}x^{n}
$$</p>
<p>And the series expansion from their rational polynomial</p>
<p>$$
\frac{p(x)}{q(x)} = c_0 + c_1 x + c_2 x^2 + \ldots
$$</p>
<p>is it possible to recover the the original polynomials $a_n$, $b_n$ from only the series $c_n$ via the solution of a linear system? </p>
| g------ | 161,506 | <p>Consider the $K\times K$ matrix
$$B= \left[ \begin{array}{ccccccc}
1 & 0 & 0 & \cdots & 0 & 0 & 0 \\
1 & 1 & 0 & \cdots & 0 & 0 & 0 \\
0 & 1 & 1 & \cdots & 0 & 0 & 0 \\
0 & 0 & 1 & \cdots & 0 & 0 & 0 \\
& & & \ddots & & & \\
0 & 0 & 0 & \cdots & 1 & 1 & 0 \\
0 & 0 & 0 & \cdots & 0 & 1 & 1 \\
\end{array}
\right]$$</p>
<p>When it is multiplied by $K\times 1$ vectors, which starts with k-1st row of Pascals's triangle, the result is a $K\times 1$ vector which contains the 1st $K$ elements of kth row of Pascal's triangle. This is because it effectively mimics addition of elements of k-1st row of Pascal's triangle to produce kth row of Pascal's triangle. Except that it only does it for the 1st $K$ elements of a row. In particular,
$$B \times \left[ \begin{array}{c}
1 \\
0 \\
\vdots \\
0
\end{array}
\right]
=\left[ \begin{array}{c}
1 \\
1 \\
\vdots \\
0
\end{array}
\right]
$$
$$
B\times B \times
\left[ \begin{array}{c}
1 \\
0 \\
\vdots \\
0
\end{array}
\right]
=B\times \left[ \begin{array}{c}
1 \\
1 \\
\vdots \\
0
\end{array}
\right]
=\left[ \begin{array}{c}
1 \\
2 \\
1 \\
\vdots \\
0
\end{array}
\right]
$$</p>
<p>It is an easy proof that all powers of $B$ are symmetric with respect to their <em>antidiagonal</em> (just consider $B$ as a sum of $I$ and $N=B-I$ and then look at the expansion of $(I+N)^m$). If we designate $\left[ \begin{array}{c}
1 \\
0 \\
\vdots \\
0
\end{array}
\right]$
as the <em>zeroth</em> row of Pascal's triangle, then the vector containing the 1st $K$ elements of $m$'s row of Pascal's triangle is $$B^m \times \left[ \begin{array}{c}
1 \\
0 \\
\vdots \\
0
\end{array}
\right],$$
which is the 1st column of $B^m$ (and by the symmetry, the last row of $B^m$).</p>
<p>Now set the $K$ (of this answer) equal to $n$ (of the posited question).</p>
<p>Then the left-hand side of the identity can be considered to be the dot product of $nth$ row of $B^R$ and 1st column of $B^M$, which is the $(n,1)$ element of $B^{R+M}$. Which also happens to be the right-hand side of the identity (in the posited question).</p>
|
337,930 | <p>Given two polynomials</p>
<p>$$
p(x) = a_0 + a_1 x + a_2 x^2 + \ldots + a_{n-1}x^{n-1} \\
q(x) = b_0 + b_1 x + b_2 x^2 + \ldots + b_{n}x^{n}
$$</p>
<p>And the series expansion from their rational polynomial</p>
<p>$$
\frac{p(x)}{q(x)} = c_0 + c_1 x + c_2 x^2 + \ldots
$$</p>
<p>is it possible to recover the the original polynomials $a_n$, $b_n$ from only the series $c_n$ via the solution of a linear system? </p>
| Marko Riedel | 44,883 | <p>Using a coefficient-extractor e.g. <span class="math-container">$[z^k] (1+z)^n = {n\choose k}$</span>, we
find</p>
<p><span class="math-container">$$\sum_{k=0}^n {R\choose k} {M\choose n-k}
= \sum_{k=0}^n {R\choose k} [z^{n-k}] (1+z)^M
\\ = [z^n] (1+z)^M \sum_{k=0}^n {R\choose k} z^k.$$</span></p>
<p>Now we may extend <span class="math-container">$k$</span> to infinity because the coefficient extractor
<span class="math-container">$[z^n]$</span> makes from a zero contribution from multiples of <span class="math-container">$z^k$</span> where
<span class="math-container">$k\gt n.$</span> We find</p>
<p><span class="math-container">$$[z^n] (1+z)^M \sum_{k\ge 0} {R\choose k} z^k
\\ = [z^n] (1+z)^M (1+z)^R
= [z^n] (1+z)^{R+M} = {R+M\choose n}.$$</span></p>
<p>This example is included here to illustrate the coefficient extractor
technique which also yields more sophisticated results as shown
e.g. at the following <a href="https://math.stackexchange.com/questions/3467933/">MSE
link</a>.</p>
|
622,076 | <p>Continuity $\Rightarrow$ Intermediate Value Property. Why is the opposite not true? </p>
<p>It seems to me like they are equal definitions in a way. </p>
<p>Can you give me a counter-example? </p>
<p>Thanks</p>
| user44197 | 117,158 | <p>$$ f(x) = \sin(1/x), ~~ x \gt 0$$
and $$f(0) =0$$</p>
<p>This is <em>not</em> continuous at $x=0$ but clearly satisfies the intermediate value property.</p>
|
442,950 | <p>I would like to show <span class="math-container">$\lim\limits_{r\to\infty}\int_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta=0$</span>.</p>
<p>Now, of course, the integrand does not converge uniformly to <span class="math-container">$0$</span> on <span class="math-container">$\theta\in [0, \pi/2]$</span>, since it has value <span class="math-container">$1$</span> at <span class="math-container">$\theta =0$</span> for all <span class="math-container">$r\in \mathbb{R}$</span>. </p>
<p>If <span class="math-container">$F(r) = \int_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta$</span>, we can find the <span class="math-container">$j$</span>th derivative <span class="math-container">$F^{(j)}(r) = (-1)^j\int_{0}^{\pi/2}\sin^{j}(\theta)e^{-r\sin\theta}\text d\theta$</span>, but I don't see how this is helping.</p>
<p>The function is strictly decreasing on <span class="math-container">$[0,\pi/2]$</span>, since <span class="math-container">$\partial_{\theta}(e^{-r\sin\theta})=-r\cos\theta e^{-r\sin \theta}$</span>, which is strictly negative on <span class="math-container">$(0,\pi/2)$</span>.</p>
<p>Any ideas?</p>
| cool | 79,292 | <p>It's only enough to show that</p>
<p><span class="math-container">$$ \int\limits_{0}^{\pi/2}{e^{-r\sin\theta}\text d\theta}\le \int\limits_{0}^{\pi/2}{e^{-r\frac{2}{\pi}\theta}\text d\theta}=\frac{\pi}{2r}\left(1-e^{-r}\right) \to 0 \quad (r \to +\infty)$$</span></p>
|
442,950 | <p>I would like to show <span class="math-container">$\lim\limits_{r\to\infty}\int_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta=0$</span>.</p>
<p>Now, of course, the integrand does not converge uniformly to <span class="math-container">$0$</span> on <span class="math-container">$\theta\in [0, \pi/2]$</span>, since it has value <span class="math-container">$1$</span> at <span class="math-container">$\theta =0$</span> for all <span class="math-container">$r\in \mathbb{R}$</span>. </p>
<p>If <span class="math-container">$F(r) = \int_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta$</span>, we can find the <span class="math-container">$j$</span>th derivative <span class="math-container">$F^{(j)}(r) = (-1)^j\int_{0}^{\pi/2}\sin^{j}(\theta)e^{-r\sin\theta}\text d\theta$</span>, but I don't see how this is helping.</p>
<p>The function is strictly decreasing on <span class="math-container">$[0,\pi/2]$</span>, since <span class="math-container">$\partial_{\theta}(e^{-r\sin\theta})=-r\cos\theta e^{-r\sin \theta}$</span>, which is strictly negative on <span class="math-container">$(0,\pi/2)$</span>.</p>
<p>Any ideas?</p>
| Guy Fsone | 385,707 | <h2>Very simple trick:</h2>
<p>By studying the function $[0,\frac{\pi}{2}]\ni\theta \mapsto\frac{\sin\theta}{\theta}$ that </p>
<p>$$ \color{blue}{\sin\theta \geq \frac{2}{\pi}\theta ~~ \forall \theta\in [0,\frac{\pi}{2}] } $$
therefore we get that
$$\lim_{R\to\infty}\int_0^{\frac{\pi}{2}} e^{-R\sin\theta}d\theta\leq\lim_{R\to\infty}\int_0^{\frac{\pi}{2}} e^{-\frac{2R}{\pi}\theta}d\theta =\lim_{R\to\infty}\frac{\pi}{2R}(1-e^{-R}) =0$$</p>
|
422,225 | <p>The proof uses this lemma which I understand: </p>
<p>$\mathbf {Lemma}$: Suppose $x$ and $y$ are positive real numbers such that $x>y$. If we decrease $x$ and increase $y$ by some positive quantity $E$ such that $x-E \ge y+E$, then $(x-E)(y+E) \gt xy$ . $\;$Hence, by subtracting $E$ from $x$ and adding it to $y$, we leave the average of the two numbers unchanged while increasing their product. </p>
<p>$\mathbf {Proof}:$ Suppose $a_{1}, a_{2}, a_{3}... a_{n}$ are positive real numbers with average $A$ and product $P$. If all $a_{i}$ are equal, then both the geometric mean and the arithmetic mean are equal to $A$. Let $a_{j}$ be one number closest to $A$ without being equal to $A$. Without loss of generality, let $a_{j} \lt A$ . Since the average of the numbers is $A$, there is at least one member of the set greater than $A$. Let $a_{k}$ be the greatest of these numbers. Clearly we must have $a_{k}-A \gt A-a_{j}$ since $a_j$ is closer to $A$ than any other $a_i$ not equal to $A$. We now use our lemma. Replace $a_j$ with $A$ and $a_k$ with $a_k-(A-a_j)$ . Note that $a_k-(A-a_j) \ge a_j +(A-a_j)$ , so we can apply our lemma with $(A-a_j)$ as our $E$ . By our lemma, the average of the numbers in the new set is the same, but the product is now higher. If we continue this process, we make one of the members of the set equal to $A$ with each application of the process. Hence, in some finite number of steps, we will make all the numbers equal to $A$. Thus, we prove that of all the sets of positive numbers with average $A$, the set with maximum product has all the elements equal to $A$.</p>
<p>There are three things I don't understand about this proof:</p>
<p>$1)$ I don't understand why they don't loose generality when they say to let $a_j$ be the number closest to $A$ and let $a_j \lt A$. It certainly is possible for this not to be the case, for example the set ${2, 10, 10, 10}$. The average is $8$, but the number closest to $A$ is greater than $A$, so I don't see how the proof can apply to this set. </p>
<p>$2)$ I don't see how this process makes makes the elements of the set equal to $A$. If you want $a_j+E$ and $a_k-E$ to be equal to $A$, then $a_j$ and $a_k$ have to be equidistant from $A$. </p>
<p>$3)$ if you do bring one pair of terms at a time equal to $A$, then that means you must have an equal number of therms below and above $A$.</p>
<p>Any help is appreciated, thanks!</p>
| chizhek | 165,778 | <p>The proof that uses the Lemma is rather hard-going.
For my perspective on the proof see my answer to the
<a href="https://math.stackexchange.com/questions/919140">related question</a>;
it also answers your question, I think.</p>
<p>I gather that the AM-GM inequality you are talking about is
$\,(a_1\cdots a_n)^{1/n}\leq(a_1+\cdots+a_n)/n\,$
for positive real numbers $a_1$, $\ldots$, $a_n$.
There is a slightly more general AM-GM inequality:</p>
<blockquote>
<p>Let $n\geq 1$ be an integer,
let $a_1,\ldots,a_n>0$,
and let $\lambda_1,\ldots,\lambda_n> 0$ satisfy $\lambda_1+\cdots+\lambda_n=1$.
Then $a_1^{\lambda_1}\cdots a_n^{\lambda_n}\leq\lambda_1a_1+\cdots\lambda_na_n$,
where the equality holds if and only if $a_1=\cdots=a_n$.</p>
</blockquote>
<p>I will give my favorite proof of the generalized AM-GM inequality.
The embarassing thing about this particular proof is that I cannot remember
whether I have seen it somewhere
or I hacked it up myself when I was fooling around thinking up different ways
(some of them extremely weird) of proving the inequality.
In case you have come across this proof, or its close relative, somewhere, anywhere,
please let me know by giving the reference in a comment to the present answer.<br>
(Yes, I know the proof by Pólya, I know the Jensen's inequality,
I know that the AM-GM inequality
is a manifestation of the concavity of the $\log$ function.)</p>
<p>All we need for the proof of the generalized AM-GM inequality
is the following inequality:$
\newcommand{\RR}{\mathbb{R}}$</p>
<blockquote>
<p>For every $x\in\RR$, $x>0$, we have
$\,x-1\geq\ln x\,$, where the equality holds iff $\,x=1$.</p>
</blockquote>
<p>The proof is simple: setting $f(x):=x-1-\ln x$ we have $f(1)=0$,
and $f'(x)=1-x^{-1}$ is (strictly) negative for $0<x<1$
and is (strictly) positive for $x>1$.</p>
<p><strong>Proof of the generalized AM-GM inequality.</strong>
$~$For every $x>0$ we have
$$
\begin{aligned}
(xa_1)^{\lambda_1}\cdots(xa_n)^{\lambda_n}
~&\:=\: x\cdot a_1^{\lambda_1}\cdots a_n^{\lambda_n}~, \\[1ex]
\lambda_1(xa_1)+\cdots+\lambda_n(xa_n)
~&\:=\: x\cdot(\lambda_1a_1+\cdots\lambda_na_n)~.
\end{aligned}
$$
This means that it suffices to prove the inequality with $xa_1$, $\ldots$, $xa_n$
in place of $a_1$, $\ldots$, $a_n$ for any $x>0$ we choose.
We choose $x=(a_1^{\lambda_1}\cdots a_n^{\lambda_n})^{-1}\!$,
that is, we can assume that $a_1^{\lambda_1}\cdots a_n^{\lambda_n}=1$
and have to prove that then $1\leq \lambda_1a_1+\cdots+\lambda_na_n$.
But this is easy:
$$
\begin{aligned}
\lambda_1a_1+\cdots+\lambda_na_n-1
~&\:=\: \lambda_1(a_1-1) + \cdots +\lambda_n(a_n-1) \\[.5ex]
~&\:\geq\: \lambda_1\ln a_1 + \cdots + \lambda_n\ln a_n \\[.5ex]
~&\:=\: 0~,
\end{aligned}
$$
where the equality holds iff $a_1=\cdots=a_n=1$.$~$ <strong>Done.</strong></p>
|
1,930,743 | <p>We have a map <span class="math-container">$f:P(X)\to P(X)$</span>, where <span class="math-container">$P(X)$</span> means the part of <span class="math-container">$X$</span> and the function is monotone (by considering inclusion "<span class="math-container">$\subseteq$</span>"). So <span class="math-container">$\forall \space A\subseteq B $</span> we have <span class="math-container">$f(A)\subseteq f(B)$</span>.
Show that this map has a fixed point.</p>
<p>This claim is used in some proofs of <a href="https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem" rel="nofollow noreferrer">Cantor–Schröder–Bernstein theorem</a>, for example, see <a href="https://proofwiki.org/wiki/Cantor-Bernstein-Schr%C3%B6der_Theorem#Proof_3" rel="nofollow noreferrer">proof 3</a> on ProofWiki (<a href="https://proofwiki.org/w/index.php?title=Cantor-Bernstein-Schr%C3%B6der_Theorem&oldid=306562#Proof_3" rel="nofollow noreferrer">current revision</a>).</p>
| Aloizio Macedo | 59,234 | <p>I'll generalize the nice answer by @Brian and give a curiosity (that you don't need to show that there exists at least one $A \subset X$ such that $A \subset f(A)$!).</p>
<p><em>Definition:</em> Given a partially ordered set $X$ and a subset $A$, $\sup(A)$ is defined as (if it exists) the element $s$ such that:</p>
<ol>
<li>$s\geq a ~\forall a \in A$.</li>
<li>$b \geq a ~\forall a \in A \implies b \geq s$.</li>
</ol>
<p>Uniqueness is clear. Note that $\emptyset$ has a $\sup$ if and only if $A$ has a minimum element.</p>
<p>Now, we have:</p>
<p><strong>Theorem:</strong> Let $X$ be partially ordered s.t. every subset has a supremum and $f:X \to X$ monotone. Then $f$ has a fixed point.</p>
<p><em>Proof:</em> Let $A=\{x \mid f(x) \geq x\}$. Take $s=\sup(A)$ (note that this is exactly the set given by Brian).</p>
<p>We prove that $f(s)=s$.</p>
<p>First, since $s$ is $\sup(A)$, it follows that $s \geq x $ for all $x \in A$. Since $f$ is monotone, we have $f(s) \geq f(x)$ for all $x \in A$. Since $f(x)\geq x$ for all $x \in A$, we have $f(s) \geq x$ for all $x \in A$. Since $s$ is $\sup(A)$, it follows that $f(s) \geq s$.</p>
<p>Now, note that monotonicity implies $f(f(s)) \geq f(s)$. Therefore, $f(s) \in A$. We then have $s \geq f(s)$, since $s$ is $\sup(A)$. </p>
<p>It follows that $s=f(s)$. $\blacksquare$</p>
<p>Note that in no point of the proof we needed $A$ to be non-empty, although it clearly is (since the set will have a minimum element by assumption).</p>
<p>Now your exercise is to adapt the proof to your case.</p>
|
3,785,982 | <p>Given the following ODE,</p>
<p><span class="math-container">$$\frac{{dy}}{{dx}}=\cos ({x})-\sin ({y})+{x}^{2}; \quad {y}\left({x}_{0}=-1\right)=y_0=3$$</span></p>
<p>I have to use the Taylor Series Method to compute the value of <span class="math-container">$y(x)$</span> at <span class="math-container">$x=-0.8$</span> with a Taylor's polynomial of second-order, with <span class="math-container">$h=x-x_0=0.1$</span>.</p>
<p>Considering all this, how should this method be applied for solving this problem?</p>
<hr />
<p><strong>My attempt at a solution.</strong></p>
<p>I'm not sure if this is the correct way to apply the method, but I have written Taylor's second-order polynomial centred at <span class="math-container">$x_0=-1.0$</span>:</p>
<p><span class="math-container">$$y(x) \approx y\left(x_{0}\right)+\left(x-x_{0}\right) y^{\prime}(x_0,y_0)+\frac{1}{2}\left(x-x_{0}\right)^{2} y^{\prime \prime}(x_0,y_0)=
\\=3.0+1.39918(x+1)+0.11333(x+1)^2 $$</span></p>
<p>And I have evaluated this <span class="math-container">$y(x)$</span> at <span class="math-container">$x=-0.8$</span>, so <span class="math-container">$y(-0.8) \approx 3.28437$</span>.</p>
<p>However, this doesn't match my textbook's solution, <span class="math-container">$3.2850$</span>, neither Wolfram-Alpha's one, <span class="math-container">$3.28687$</span>.</p>
<p>Would the method be applied this way or am I missing something?</p>
| Bernard | 202,857 | <p><strong>Hint</strong>:</p>
<p>It is simpler here to use <em>Hadamard's formula</em>:
<span class="math-container">$$\frac 1R=\limsup |a_n|^{1/n}=\limsup\Bigl(\frac{n+2}n\Bigr)^{\!n}. $$</span></p>
|
4,112,308 | <p>I was just exploring a little bit on Desmos, and was trying to figure out something somewhat interesting. I'm familiar that this is an elliptic curve, but ALL I know about them is that they are of the form <span class="math-container">$y^2=x^3+ax+b$</span>. Nothing else, really....</p>
<p>So, here's what I'm thinking. Give <span class="math-container">$a$</span> an easy value, say <span class="math-container">$-1$</span>, and let <span class="math-container">$b$</span> range over the reals. There is a specific value of <span class="math-container">$b$</span> for which the graph moves from one curve, to a curve and a closed circle-like figure. This specific value of <span class="math-container">$b$</span> makes the graph kind of.... "self-intersect" in a nice way. I've estimated it to be approximately <span class="math-container">$0.384900179459750$</span>. I put this into WolframAlpha and it didn't recognize anything important. Is this constant important? Does it have a name? What's its value? Thanks.</p>
| Math Lover | 801,574 | <p>For the curve to be self intersecting, we take the form</p>
<p><span class="math-container">$y^2 = (x-p)^2 (x-q) = x^3 - (2p+q)x^2 + (2pq+p^2)x-p^2q \ $</span>. Please note that the curve forms only for <span class="math-container">$x \geq q$</span></p>
<p>As <span class="math-container">$x^2$</span> term is zero, <span class="math-container">$2p+q = 0 \implies q = - 2p$</span>.</p>
<p>So, <span class="math-container">$2pq + p^2 = - 3p^2, -p^2q = 2p^3$</span></p>
<p>We also note that if <span class="math-container">$p \lt 0, q \gt 0$</span> but <span class="math-container">$x$</span> must be <span class="math-container">$\geq q$</span> so there is no solution (self-intersecting curve) for <span class="math-container">$p \lt 0$</span>. Also when <span class="math-container">$p = 0, q = 0$</span> and so there is no solution for <span class="math-container">$p = 0$</span> either.</p>
<p>That leads to,</p>
<p><span class="math-container">$y^2 = x^3 -3p^2 x +2p^3, p \gt 0$</span></p>
|
2,385,369 | <p>Explain why $K_{2,3}$ cannot have a Hamilton cycle.</p>
<p>I can visibly see and show why this is the case, but is there a mathematical proof or specific way of explaining how this Hamilton cycle cannot exist? Thanks a ton for all the help!</p>
| Bob Krueger | 228,620 | <p>A more general solution, and possibly the reason why you can visibly see that there is no Hamiltonian cycle, is as follows:</p>
<p>Let $S$ be the larger part of this bipartite graph. Then $|S| = 3$, $|N(S)| = 2$, and $S$ is an independent set. Whenever there is a set of vertices $S$ of a graph with $S$ an independent set and $|N(S)| > |S|$, then the graph cannot contain a Hamiltonian cycle. (To prove this, use contradiction to assume that there is a Hamiltonian cycle, then get that $|N(S)| \geq |S|$.)</p>
|
2,780,731 | <p>In school, I have recently been learning about simple differential equations. We know that the solution of $y'=y$ is $y=Ae^x$, where $A$ is a constant. But how can we know that it is the <strong>only</strong> solution? The only thing I can figure out is that $y$ is continuously differentiable. Help me, please.</p>
| Phil H | 554,494 | <p>I've heard this kind of question before. An anti-derivative will yield a definitive delta area under a graph of a function between any 2 limits. Seeing there is only one delta area, any different expressions defining it would essentially be the same. </p>
<p>The area of a right triangle $1/2xy$ or $1/2x^2 \tan \theta$ are the same with a trigonometric substitution. For the family of anti-derivatives whose only difference is C, where C makes no difference in calculating the definite integral, in reverse they only have one derivative.</p>
<p>To summarize, a function that defines areas of different regions under a graph is unique and so those different areas are themselves defined by a unique function under which they exist.</p>
|
1,876,287 | <p><strong>Question:</strong></p>
<p>Let P be a point where the normal (in the point where the x-coordinate is h) to the curve</p>
<p>$$y = e^{2x} - 2x$$</p>
<p>cuts the y-axis. Determine the y-coordinates of P when h goes to 0.</p>
<p><strong>Attempted solution:</strong></p>
<p>I first decided to draw the following image:</p>
<p><a href="https://i.stack.imgur.com/O1aQb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O1aQb.png" alt="enter image description here"></a></p>
<p>So basically, we have the function for the curve, but we need to get on the normal and then move along the normal until it reaches the y-axis.</p>
<p>Let us start by taking the derivative of the function:</p>
<p>$$y' = 2e^{2x} - 2$$</p>
<p>The point x = h is the relevant point here, so:</p>
<p>$$y'(h) = 2e^{2h} - 2$$</p>
<p>Finding the k value of the normal:</p>
<p>$$k_1 k_2 = -1 \rightarrow k_2 = \frac{-1}{2e^{2h} - 2}$$</p>
<p>The equation for the normal is then:</p>
<p>$$y - (2e^{2h} - 2) = \frac{-1}{2e^{2h} - 2}(x-h)$$</p>
<p>Simplification gives:</p>
<p>$$y = 2e^{2h} - 2 -\frac{x-h}{2e^{2h} - 2}$$</p>
<p>If h goes to 0, this goes towards infinity. </p>
<p>But that is hardly the case and cannot be true since it obviously has to cut the y-axis. Somewhere, something must have gone terribly, terribly wrong. The correct answer is $\frac{5}{4}$.</p>
| Claude Leibovici | 82,404 | <p>In the same spirit as MathInferno's answer, not using L'Hôpital's rule, consider Taylor series around $h=0$ $$e^{2h}= 1+2 h+2 h^2+O\left(h^3\right)$$ So, $$e^{2h}-2h=1+2 h^2+O\left(h^3\right)$$ $$2e^{2h}-2=4 h+4 h^2+O\left(h^3\right)$$ then $$e^{2h}-2h+\frac h{2e^{2h}-2}=1+2 h^2+O\left(h^3\right)+\frac h{4 h+4 h^2+O\left(h^3\right)}$$ $$e^{2h}-2h+\frac h{2e^{2h}-2}=1+2 h^2+O\left(h^3\right)+\frac 1{4 +4 h+O\left(h^2\right)}$$ Now, long division for the last term leads to $$e^{2h}-2h+\frac h{2e^{2h}-2}=\frac{5}{4}-\frac{h}{4}+O\left(h^2\right)$$ which shows the limit and how it is approached.</p>
|
2,481,046 | <p>I have a question that asks to show that $S^2 = \{(x,y,z) \in \mathbb{R}^3|x^2+y^2+z^2=1\}$ is a differentiable manifold. My professor says that one way to do this is to define the following 6 parametrizations of the sphere, which cover the entire sphere.</p>
<p>$\vec{\phi_{i}}:V \to \mathbb{R}^3$ where $V = \{(u,v) \in \mathbb{R}^2|u^2+v^2<1\}$</p>
<p>$\vec{\phi_{1}}(u,v) = (u,v,\sqrt{1-u^2-v^2}) \qquad (z>0)$</p>
<p>$\vec{\phi_{2}}(u,v) = (u,v,-\sqrt{1-u^2-v^2}) \qquad (z<0)$</p>
<p>$\vec{\phi_{3}}(u,v) = (u,\sqrt{1-u^2-v^2},v) \qquad (y>0)$</p>
<p>$\vec{\phi_{4}}(u,v) = (u,-\sqrt{1-u^2-v^2},v) \qquad (y<0)$</p>
<p>$\vec{\phi_{5}}(u,v) = (\sqrt{1-u^2-v^2},u,v) \qquad (x>0)$</p>
<p>$\vec{\phi_{6}}(u,v) = (-\sqrt{1-u^2-v^2},u,v) \qquad (x<0)$</p>
<p>I don't understand what these parameterizations mean at all and I don't understand what a parameterization is. From what I can read online, it's some function but I'm not sure why this specific function with $u$ and $v$ is what we're using to cover the entire sphere. Can someone explain this to me please?</p>
| choco_addicted | 310,026 | <p>Since $\displaystyle \int_1^{\infty} \frac{1}{y^2}dy$ converges and $|\frac{1}{y^2}\sin(y+\frac{1}{y})| \le \frac{1}{y^2}$ for $y\in [1,\infty)$, $\displaystyle \int_1^\infty \frac{1}{y^2}\sin\left(y+\frac{1}{y}\right)dy$ converges absolutely by the comparison test. According to <a href="http://www.wolframalpha.com/input/?i=integrate+sin(x%2B1%2Fx)+from+0+to+1" rel="noreferrer">Wolframalpha</a>,
$$
\int_0^1 \sin\left(x+\frac{1}{x}\right)dx \approx 0.3762
$$</p>
|
2,481,046 | <p>I have a question that asks to show that $S^2 = \{(x,y,z) \in \mathbb{R}^3|x^2+y^2+z^2=1\}$ is a differentiable manifold. My professor says that one way to do this is to define the following 6 parametrizations of the sphere, which cover the entire sphere.</p>
<p>$\vec{\phi_{i}}:V \to \mathbb{R}^3$ where $V = \{(u,v) \in \mathbb{R}^2|u^2+v^2<1\}$</p>
<p>$\vec{\phi_{1}}(u,v) = (u,v,\sqrt{1-u^2-v^2}) \qquad (z>0)$</p>
<p>$\vec{\phi_{2}}(u,v) = (u,v,-\sqrt{1-u^2-v^2}) \qquad (z<0)$</p>
<p>$\vec{\phi_{3}}(u,v) = (u,\sqrt{1-u^2-v^2},v) \qquad (y>0)$</p>
<p>$\vec{\phi_{4}}(u,v) = (u,-\sqrt{1-u^2-v^2},v) \qquad (y<0)$</p>
<p>$\vec{\phi_{5}}(u,v) = (\sqrt{1-u^2-v^2},u,v) \qquad (x>0)$</p>
<p>$\vec{\phi_{6}}(u,v) = (-\sqrt{1-u^2-v^2},u,v) \qquad (x<0)$</p>
<p>I don't understand what these parameterizations mean at all and I don't understand what a parameterization is. From what I can read online, it's some function but I'm not sure why this specific function with $u$ and $v$ is what we're using to cover the entire sphere. Can someone explain this to me please?</p>
| J.G. | 56,861 | <p>Or you could just use the fact that the integrand has modulus $\le 1$. The behaviour at one point, $x=0$, doesn't change the integral.</p>
|
4,276,974 | <p>I have to prove that sentence, but I'm not sure how to do that. Help!</p>
| Anonmath101 | 306,753 | <p><span class="math-container">$p+1 $</span> and <span class="math-container">$p-1$</span> are both even and one of <span class="math-container">$p-1, p, p+1$</span> is a multiple of <span class="math-container">$3$</span> but of course it cannot be <span class="math-container">$p$</span> itself. So <span class="math-container">$p-1$</span> or <span class="math-container">$p+1$</span> is divisible by <span class="math-container">$2$</span> and <span class="math-container">$3$</span>.</p>
|
76,853 | <p>I have a list of stock symbols and related information containing some entries <code>Missing["NotAvailable"]</code>. I would like to delete all nested lists which contain a NotAvaiable entry, as <em>Mathematica</em> obviously does not support these instruments anymore (see also <a href="http://reference.wolfram.com/mathematica/ref/FinancialData.html" rel="nofollow">http://reference.wolfram.com/mathematica/ref/FinancialData.html</a>).</p>
<p>The list entries are formated as follows:</p>
<pre><code>indexMaster= {"^RDM-SO", Missing["NotAvailable"], "AMEX"}
</code></pre>
<p>I tried to use the following function, but it does not work. </p>
<pre><code>instruments =
DeleteCases[indexMaster, {p__, q_String, r__} /;
StringMatchQ[q, "*NotAvailable*"] -> {p, q, r}]
</code></pre>
<p>Does anyone have an idea how to delete the <code>Missing["NotAvailable"]</code> entries? </p>
<p>Thanks</p>
| Karsten 7. | 18,476 | <p><code>Missing["NotAvailable"]</code> is not a string. Its <code>Head</code> is <code>Missing</code>, therefore you can use</p>
<pre><code>instruments = DeleteCases[indexMaster, _Missing]
</code></pre>
<blockquote>
<pre><code>{"^RDM-SO", "AMEX"}
</code></pre>
</blockquote>
|
76,853 | <p>I have a list of stock symbols and related information containing some entries <code>Missing["NotAvailable"]</code>. I would like to delete all nested lists which contain a NotAvaiable entry, as <em>Mathematica</em> obviously does not support these instruments anymore (see also <a href="http://reference.wolfram.com/mathematica/ref/FinancialData.html" rel="nofollow">http://reference.wolfram.com/mathematica/ref/FinancialData.html</a>).</p>
<p>The list entries are formated as follows:</p>
<pre><code>indexMaster= {"^RDM-SO", Missing["NotAvailable"], "AMEX"}
</code></pre>
<p>I tried to use the following function, but it does not work. </p>
<pre><code>instruments =
DeleteCases[indexMaster, {p__, q_String, r__} /;
StringMatchQ[q, "*NotAvailable*"] -> {p, q, r}]
</code></pre>
<p>Does anyone have an idea how to delete the <code>Missing["NotAvailable"]</code> entries? </p>
<p>Thanks</p>
| Jinxed | 24,763 | <p>Just use the same pattern as what you don't want to see:</p>
<pre><code>DeleteCases[indexMaster,Missing["NotAvailable"]]
(* {"^RDM-SO", "AMEX"} *)
</code></pre>
|
891,575 | <p>The circumference of a circle has length 90 centimeters, Three points on the circle divide the circle into three equal lengths. Three ants A, B, and C start to crawl clockwise on the circle, with starting from one of the three points. Initially A is ahead of B and B is ahead of C. Ant A crawls 3 centimeters per second, ant V 5 centimeters, and and C 10 centimeters. How long does it take for the three ants to arrive at the same spot for the first time?</p>
<p>I tried making a list and writing down the numbers, but they seem to never be the same. I know the distance formula is <em>d=rt</em>, but I don't know how to use it to solve this problem. Any help? Thanks!</p>
| Haukur Þorgeirsson | 167,969 | <p>At the start, C is at location 0, B is at location 30 and A is at location 60. It is easy to see that C will catch up to B in 6 seconds at location 60. And then they will meet again at location 60 every 18 seconds after that. When will A be at location 60? Well, he starts out there and he gets back there every 30 seconds.</p>
<p>So, B and C will be at the right place in 6, 24, 42, 60, 78 etc. seconds while A will be at the right place in 30, 60, 90 etc. seconds. We now see that they meet for the first time after <strong>60 seconds</strong>.</p>
<p>To help verify that this is correct you can see that 0 + 10 * 60 = 600; 30 + 5 * 60 = 330; 60 + 3 * 60 = 240. It is easy to see that 600, 330 and 240 all refer to the same spot on the circle. Formally expressed 600 mod 90 = 330 mod 90 = 240 mod 90 = 60.</p>
|
2,611,676 | <p>Or consider the general problem-
Find the value of n for which x^n is just greater than x!</p>
<p>I dont know even if it is possible to find the solution or not...</p>
| user326210 | 326,210 | <p>Well, if </p>
<p>$$x^n > x!$$</p>
<p>then we can take the logarithm of both sides without affecting the order:</p>
<p>$$\log(x^n) > \log(x!)$$</p>
<p>We get:</p>
<p>$$n \log(x) > \log(1) + \log(2) + \log(3) + \ldots + \log(x)$$</p>
<p>Let's divide by $\log(x)$ on both sides (if $x=1$ then $\log(x) = 0$ and we can't do this, but the problem also has no solution in that case):</p>
<p>$$n > \log_x(1) + \log_x(2) + \ldots + \log_x(x)$$</p>
<p>We can round up to get an integer:</p>
<p>$$n^\star =\left\lceil \frac{\log(1) + \log(2) + \ldots + \log(x)}{\log(x)}\right\rceil$$</p>
<hr/>
<p>We can apply this rule to the case where $x=100$. We find that the answer is:</p>
<p>$$n^\star = 79.$$</p>
<p>We can verify that this makes sense because $100!$ is equal to a 158-digit number (93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000)</p>
<p>and $100^{78}$ is a 157 digit number (a one followed by 2*78=156 zeroes), whereas $100^{79}$ is a 159 digit number, hence larger than 100!.</p>
<p>$$\begin{array}{r|l}\text{number} & \text{how many digits}\\\hline 10^{78}& 157\\10!&158\\10^{79}&159\leftarrow\end{array}$$</p>
|
1,656,145 | <p>Let the real function of two real variables$$u(x,y) =
\begin{cases}
x, & \quad \text{if } |y|>|x| \\
-x, & \quad \text{if } otherwise
\\ \end{cases} $$</p>
<p>Is there a sequence $\{(x_n,y_n)\}_{n \geq 0}$ which converge to $(0,0)$ such that $\lim_{n \to \infty} u(x_n,y_n) \not= u(0,0)$?</p>
<p>I tried to prove this by contrapositive of continuity, but I failed</p>
| Robert Israel | 8,508 | <p>Hint: $|u(x,y) - u(0,0)| = |x|$. </p>
|
3,128,352 | <p>I want to prove that when <span class="math-container">$F:K\rightarrow K[X]/\langle f\rangle $</span> is a map such that <span class="math-container">$F(a)=a+\langle f \rangle$</span>, then <span class="math-container">$F$</span> is an embedding from <span class="math-container">$K$</span> to <span class="math-container">$K[X]/\langle f \rangle$</span>, when <span class="math-container">$f\in K[X]\backslash K$</span>.</p>
<p>To show the function is an embedding do I need to check that, for <span class="math-container">$a,b\in K$</span>:</p>
<p><span class="math-container">$$F(a+b)=F(a)+F(b)$$</span></p>
<p>Which seems straightforward in this case:
<span class="math-container">$$F(a+b)=I+a+b=I+a+I+b=F(a)+F(b)$$</span></p>
<p>And the same for multiplication... And in addition check that the map is an injection?</p>
| Claude Leibovici | 82,404 | <p>For simplicity, I let <span class="math-container">$x=y+5$</span> to make
<span class="math-container">$$\frac{x^3 -2x+1}{x+7}=\frac{y^3+15 y^2+73 y+116}{y+12}$$</span> and using the long division
<span class="math-container">$$\frac{y^3+15 y^2+73 y+116}{y+12}=\frac{29}{3}+\frac{95 y}{18}+\frac{175 y^2}{216}+\frac{41 y^3}{216 (y+12)}$$</span></p>
<p>Focusing on the last term and using the classical expansion, we then have
<span class="math-container">$$\frac{ y^3}{ (y+12)}=-\sum_{n=3}^\infty (-1)^n 12^{2-n} y^n$$</span></p>
<p>If you want to make life easier, , let, for the time being, <span class="math-container">$y=12 t$</span> to make
<span class="math-container">$$\frac{41 y^3}{216 (y+12)}=\frac{82 }{3 } \frac{ t^3}{ (t+1)}$$</span></p>
|
1,521,779 | <p>I have a homework question that I want to make sure I'm getting it right.</p>
<p>This is a joint probability table for the proportions of survey respondents who smoke and who have had heart attacks.</p>
<p><kbd> </kbd><kbd>Smoker </kbd><kbd>Non-Smoker</kbd><br/>
<kbd>Heart Attack </kbd><kbd>0.03 </kbd><kbd>0.03 </kbd><br/>
<kbd>No Heart Attack</kbd><kbd>0.44 </kbd><kbd>0.50 </kbd><br/></p>
<p>If a person is a smoker, are they more likely to have had a heart attack than someone who is not a smoker?</p>
<p>So I know that $P(A|B) = \dfrac{P(A \cap B)}{P(B)}$</p>
<p>So I could think of the question like: <em>Find the conditional probability that a randomly selected person is a victim of a heart-attack, given that they’re a smoker/ non-smoker.</em></p>
<p>$ P(X = Smoker \cap Y = Heart Attack) = 0.03 $<br>
$ P(X = NonSmoker \cap Y = Heart Attack) = 0.03 $<br>
$P(Y) = 0.06$</p>
<p>That would mean that a smoker and a non-smoker would both have a $\dfrac{0.03}{0.06}$ chance of having a heart attack?</p>
<p>This defies my intuition because $\dfrac{3}{47}$ smokers get heart attacks, while only $\dfrac{3}{53}$ non-smokers get heart attacks.</p>
| poetasis | 546,655 | <p>It would be easy if <span class="math-container">$\space 2017\space $</span> were a perfect square but there are no better approaches that i know of except for limiting the search.
<span class="math-container">$$x^2+y^2=2017\implies y=\sqrt{2017-x^2}\\
\implies\bigg\lceil\sqrt{2017-\big(\big\lfloor\sqrt{2017}\;\big\rfloor\big)^2}\space \bigg\rceil=9\le
x\le \big\lfloor\sqrt{2017}\;\big\rfloor=44$$</span>
This shows a finite search and any <span class="math-container">$\space x \space $</span> that yields an integer <span class="math-container">$\space y\space $</span> is a solution.</p>
<p>Of these <span class="math-container">$\space 36\space $</span> values to test, the only <span class="math-container">$y$</span>-integers found are for <span class="math-container">$\space x=9\space$</span> and <span class="math-container">$\space x=44.\quad$</span> Since <span class="math-container">$\space x,y\space $</span> are interchangeable, this means
<span class="math-container">$\quad|x|,|y|\in\big\{9,44\big\}\quad$</span> with the valid solutions being all combinations for a total of <span class="math-container">$\space 8\space $</span> solutions.</p>
|
441,792 | <p>There are many objects in mathematics that have the term "chiral" in their name, for instance, chiral algebra by Beilinson and Drinfeld, chiral de Rham complex, chiral Koszul duality etc. Some people told me that chiral algebras are <span class="math-container">$2$</span>-dimensional analogue of associative algebras, which are considered to be <span class="math-container">$1$</span>-dimensional. However, I don't understand its precise meaning since the definition of a vertex operator algebra is so complicated. Does the term chiral has something to do with this <span class="math-container">$2$</span>-dimensionality?</p>
<p>For a vertex operator algebra <span class="math-container">$V$</span>, Yongchang Zhu constructed an associative algebra <span class="math-container">$A(V)$</span> out of <span class="math-container">$V$</span>, such that there is a bijection between the set of isomorphism classes of irreducible positive energy representations of <span class="math-container">$V$</span> and that of simple <span class="math-container">$A(V)$</span>-modules. For an associative algebra <span class="math-container">$A$</span>, Tomoyuki Arakawa calls <span class="math-container">$V$</span> to be the <em>chiralization</em> of <span class="math-container">$A$</span> if <span class="math-container">$A\simeq A(V)$</span> as associative algebras. What's the meaning of chiralization here?</p>
<p>There are some other explanations for the term chiral that I have ever heard. For example, in electromagnetism, chirality means the handedness of electromagnetic waves associated with their polarization. Some others also told me that in the <span class="math-container">$2$</span>-dimensional setting, chiral means holomorphic.</p>
<p>I want to know the geometry/physics behind the term chiral. A philosophical answer is welcome, but a mathematical/physical answer is better.</p>
| AXidenT | 88,421 | <p>A vertex operator algebra describes the algebra of local operators in the chiral part of a 2d CFT. Typically one sees a VOA described depending on a complex coordinate <span class="math-container">$z$</span>. To describe a full 2d CFT, you would typically need to also include an "anti-chiral" VOA depending on a conjugate coordinate <span class="math-container">$\bar{z}$</span>. So by considering only a single vertex algebra depending on one complex variable, you are only considering a "chiral half" of the CFT.</p>
<p>In physics more generally, people will often refer to "chiral algebras" of local operators in other types of theories in different dimensions as well, so in that field the terminology is quite broad, but relates to the chirality of a theory in a more traditional sense.</p>
<p>Mathematicians have since extended and generalized a number of things relating to vertex algebras. For example Beilinson-Drinfeld chiral algebras generalise vertex algebras, the Chiral de Rham complex and chiral differential operators are VOA versions of differential forms and differential operators etc... In the past, constructions such as this were the only examples of mathematically well-defined and well-studied chiral algebras (in the physics sense of algebras of local operators). Hence in the mathematical community, I suspect it became practice to name constructions relating to vertex algebras and BD chiral algebras "chiral" since this was the only real example of a chiral algebra of local operators they were looking at in those communities.</p>
<p>Hence in Arakawa's work for example, "chiralization" essentially means going from an associative algebra type object to a VOA type object. More generally, I believe chiralization/adding the word "chiral" to something in these communities will describe working with something an affine or loopy version of a previously known construction.</p>
<p>For example, taking the Weyl algebra of differential operators, one can consider the VOA counter-part of this assocaitive algebras which is the <span class="math-container">$\beta\gamma$</span> VOA, also known as "chiral differential operators". The modes of this VOA satisfy similar relations to the Weyl algebra and the Weyl algebra can be recovered from it in various constructions.</p>
<p>I guess the point to emphasise though is that the word "chiral" itself is often not-literal in the mathematics community, but is rather describing something like "VOA-ization" (which in BD language is chiralization).</p>
|
3,151,452 | <p>The context of the question is that a bakery bakes cakes and the mass of cake is demoted by <span class="math-container">$X$</span> such that <span class="math-container">$X \sim N(300, 40^2)$</span>. A sample of 12 cakes is taken and the mean of the sample is 292g. The question wants me to find the <span class="math-container">$p$</span>-value and test to see if the mean has changed at 10% significance.</p>
<p>So I know how to carry out the test as <span class="math-container">$\overline{X_{12}} \sim N(300,\frac{40^2}{12})$</span>, But what would the p-value I'm trying to calculate be? I know the p-value is 0.244.</p>
| farruhota | 425,072 | <p>The hypothesis testing:
<span class="math-container">$$H_0: \mu =300\\
H_1:\mu \ne 300 \\
z=\frac{\bar{x}-300}{40/\sqrt{12}}=-0.6928\\
p\text{-value}=P(z<-0.6928)=0.244 \ \\
\text{Reject $H_0$ if $p<\frac{\alpha}{2}$}: \ 0.244\not < 0.05 \Rightarrow \text{Fail to Reject} \ H_0.$$</span>
Note: <span class="math-container">$p$</span>-value calculation: </p>
<p>1) In MS Excel: <span class="math-container">$=NORM.S.DIST(-0.6928;1)$</span>. </p>
<p>2) <a href="https://www.wolframalpha.com/input/?i=Pr(z%3C%20-0.6928)" rel="nofollow noreferrer">WolframAlpha</a>.</p>
<p>3) <a href="https://www.google.com/search?q=z%20table&rlz=1C1GGRV_enUZ822UZ822&source=lnms&tbm=isch&sa=X&ved=0ahUKEwiPtYrkoYnhAhVmsYsKHRsOAYwQ_AUIDigB&biw=1440&bih=789" rel="nofollow noreferrer">Z table</a>.</p>
|
1,222,909 | <p>I was thinking to convert to cartesian coordinates and then find when the slope of the tangent line is $1$, but I get a messy equation $2\cos^2\theta -2\sin^2\theta=4\sin^2\theta\cos\theta$
I was wondering if there was an easy way as it is hard to get values from this.</p>
<p>Edit: The equation ends up simplifying to $\tan(2\theta) = 1$, but for future reference is this the best method?</p>
| goldenratio | 226,628 | <p>This is the equation of the circle of center $(0,1)$ and radius $1$. </p>
<p>$$r = 2\sin(\theta) \iff r = 2(\frac{y}{r}) \iff r^2 = 2y \iff x^2 + y^2 = 2y \iff x^2 + (y - 1)^2 = 1$$</p>
<p>Differentiate $x^2 + y^2 = 2y$ with respect to $x$..</p>
|
272,057 | <p>Let $\mu$ and $\nu$ be two probability measures on $\mathbb R^n$ with finite first moment. Denote by $d:=W_1(\mu,\nu)$, where $W_1(\cdot,\cdot)$ stands for the Wasserstein distance of order $1$. </p>
<p>My question is the following: Let $X$ be a random variable defined on some probability space (rich enough) with law $\mu$, could we find <strong>a measurable function $f:\mathbb R^n\times \mathbb R^n\to\mathbb R^n$ and a random variable $G$ independent of $X$</strong> s.t.</p>
<p>$$Y:=f(X,G)~\sim~\nu~~~~~~ \mbox{ and }~~~~~~ \mathbb E[|X-Y|]~\le ~2d~?$$</p>
<p><strong>Thought 1:</strong> Let $d_0:=\rho(\mu,\nu)$, where $\rho(\cdot,\cdot)$ denotes the Prokhorov distance. Then we have a measurable function $f_0:\mathbb R^n\times \mathbb R^n\to\mathbb R^n$ and a random variable $G_0$ independent of $X$ s.t.</p>
<p>$$Y_0:=f_0(X,G)~\sim~\nu~~~~~~ \mbox{ and }~~~~~~ \mathbb P[\{|X-Y_0|\ge2d_0\}]~\le ~2d_0.$$</p>
<p>The above construction is from the paper <em>On a representation of random variables</em> by Skorokhod, but I can't find this paper. </p>
<p><strong>Thought 2:</strong> Let $\pi(dx,dy)$ be the optimal transport plan, i.e. $\pi(A\times\mathbb R^n)=\mu(A)$ and $\pi(\mathbb R^n\times A)=\nu(A)$ for all measurable $A\subset\mathbb R^n$. Disintegration w.r.t. the first coordinate $x$, one has $\pi(dx,dy)=\mu(dx)\otimes \lambda_x(dy)$, where $(\lambda_x)_{x\in\mathbb R^n}$ denotes the r.c.p.d. (regular conditional probability distribution). But I've no idea how to recover the function $f$ using $\lambda_x$.</p>
<p>Any answer, help or comment is highly appreciated. Thanks a lot!</p>
| Iosif Pinelis | 36,721 | <p>$\newcommand{\R}{\mathbb R}
\newcommand{\B}{\mathcal B}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\renewcommand{\c}{\circ}
\newcommand{\tr}{\operatorname{tr}}$</p>
<p>The desired function $f$ and random variable (r.v.) $G$ can be built recursively, by induction, using the increasing rearrangement/<a href="https://en.wikipedia.org/wiki/Inverse_transform_sampling" rel="nofollow noreferrer">inverse transformation method</a>: If $F$ is any cumulative distribution function (cdf),
\begin{equation}
F^{-1}(u):=\inf\{x\in\mathbb R\colon F(x)\ge u\}
\end{equation}
for $u\in(0,1)$, and $U\sim\mathcal U(0,1)$ (a r.v. uniformly distributed on the interval $(0,1)$), then the cdf of the r.v. $F^{-1}(U)$ is $F$. </p>
<p>Indeed, for $j=0,\dots,n$, let $\pi_j$ be the push-forward image of the probability measure $\pi$ under the projection of $\R^n\times\R^n$ onto $\R^n\times\R^j$, so that $\pi_j(A\times B_j)=\pi(A\times B_j\times\R^{n-j})$ for $A$ in the Borel sigma-algebra $\B(\R^n)$ and $B_j$ in $\B(\R^j)$; naturally, $\R^0=\{0\}$, and we identify $\R^j\times\R^{n-j}$ with $\R^n$.
Similarly, let $\nu_j$ be the push-forward image of the probability measure $\nu$ under the projection of $\R^n$ onto $\R^j$, so that $\nu_j(B_j)=\nu( B_j\times\R^{n-j})$ for $B_j$ in $\B(\R^j)$.
Then $\pi_0 =\mu$, $\pi_n=\pi$, $\nu_0$ is the only probability measure on $\B(\R^0)=\B(\{0\})$, and $\nu_n=\nu$. </p>
<p>Write $Y=(Y_1,\dots,Y_n)$ and let $Y_{1;j}:=(Y_1,\dots,Y_j)$, with $Y_{1;0}:=0$; the r.v.'s $Y_1,\dots,Y_n$ are to be constructed.
Accordingly, for $y=(y_1,\dots,y_n)\in\R^n$ let $y_{1;j}:=(y_1,\dots,y_j)$, with $y_{1;0}:=0$. For $j=0,\dots,n$, let $G_j:=(U_1,\dots,U_j)$, where $U_1,\dots,U_n$ are independent $\mathcal U(0,1)$ r.v.'s. In particular, $G_0=0$. </p>
<p>For $j=1,\dots,n$, we are going to construct, by induction, a function $f_j\colon\R^n\times(0,1)^j\to\R^j$ such that for $Y_{1;j}:=f_j(X,G_j)$ the distribution of $(X,Y_{1;j})$ is $\pi_j$ and hence the distribution of $Y_{1;j}$ is $\nu_j$. </p>
<p>To complete the basis of induction, let $f_0(x,0):=0$ for all $x\in\R^n$, so that $Y_0=f_0(X,G_0)$.
Take now any $j=1,\dots,n$. Let
$$\R^n\times\R^{j-1}\times\B(\R)\ni(x,y_{1;j-1},C)\longmapsto
\la_{x,y_{1;j-1}}(C)$$
be a regular version of the conditional distribution of $Y_j$ given $(X,Y_{1;j-1})$ assuming that the joint distribution of $(X,Y_{1;j})$ is $\pi_j$, so that
$$\pi_j(dx\times dy_{1;j-1}\times dy_j)=\pi_{j-1}(dx\times dy_{1;j-1})\la_{x,y_{1;j-1}}(dy_j).$$
For each $(x,y_{1;j-1})\in\R^n\times\R^{j-1}$, let $F_{x,y_{1;j-1}}$ be the cdf of the probability measure $\la_{x,y_{1;j-1}}$ on $\B(\R)$, and define the function $f_j\colon\R^n\times(0,1)^j\to\R^j$ by the formula
\begin{equation}
f_j(x,u_{1;j}):=\big(y_{1;j-1},F^{-1}_{x,y_{1;j-1}}(u_j)\big)\quad\text{with}\quad
y_{1;j-1}=f_{j-1}(x,u_{1;j-1})
\end{equation}
for $(x,u_{1;j})\in\R^n\times(0,1)^j$, where the notation $u_{1;j}$ is of course quite similar to $y_{1;j}$.
Let now $Y_{1;j}:=f_j(X,U_{1;j})=f_j(X,G_j)$, which is in agreement with the definition $Y_{1;j-1}:=f_{j-1}(X,U_{1;j-1})=f_{j-1}(X,G_{j-1})$ at the previous step of the induction process. </p>
<p>Then the distribution of $(X,Y_{1;j})$ is $\pi_j$ and hence the distribution of $Y_{1;j}$ is $\nu_j$. </p>
<p>In particular, the distribution of $(X,Y)=(X,Y_{1;n})$ is $\pi_n=\pi$ and hence the distribution of $Y=Y_{1;n}$ is $\nu$. Moreover, $Y=Y_{1;n}=f_n(X,G_n)$, as desired. </p>
|
3,518,221 | <p>So I had this complex integral </p>
<blockquote>
<p>If <span class="math-container">$0 \leq y \leq 1$</span>, find the maximum value of the integral
<span class="math-container">$$
\int_0^y \left(x^4 + (y-y^2) \right)^{1/2}\, dx
$$</span></p>
</blockquote>
<p>I differentiated the integral using the leibniz rule and equated it to 0.
The problem is that I obtain a expression containing a biquadratic in a square root and an integral. The biquadratic inside the root is always positive when I checked it and the integral was easily evaluated. The complexity of the problem kept on increasing as I started getting fractional exponents.
Please help me on this.</p>
| Claude Leibovici | 82,404 | <p>If there is no typo, I have the feeling that you face a difficult problem.</p>
<p>If
<span class="math-container">$$f(y)=\int_0^y g(x,y)\,dx$$</span> then
<span class="math-container">$$f'(y)=g(y,y)+\int_0^y \frac{\partial g(x,y)}{\partial y}\,dx$$</span> So, for your case
<span class="math-container">$$f'(y)=\sqrt{y \left(y^3-y+1\right)}+\frac{1-2y }2\int_0^y \frac{dx}{\sqrt{x^4+y-y^2}}$$</span> and the last integral involves elliptic integral of the first kind or gaussian hypergeometric function. </p>
<p>The most attractive form I found is
<span class="math-container">$$\int_0^y \frac{dx}{\sqrt{x^4+y-y^2}}=\frac{\,
_2F_1\left(\frac{1}{4},\frac{1}{2};\frac{5}{4};\frac{y^3}{y-1}\right)}{\sqrt{
\frac{1}{y}-1}}$$</span> Now, the only thing I was able to do is some numerical analysis work which shows that the zero of <span class="math-container">$f'(y)$</span> is very close to <span class="math-container">$0.888$</span>.</p>
<p>I faced the same kind of trouble with <span class="math-container">$f(y)$</span> and the best I found is
<span class="math-container">$$f(y)=y \sqrt{(1-y) y} \,
_2F_1\left(-\frac{1}{2},\frac{1}{4};\frac{5}{4};\frac{y^3}{y-1}\right)$$</span> which, evaluated for <span class="math-container">$y=0.888$</span> gives a maximum value close to <span class="math-container">$0.397$</span>.</p>
<p>A complete numerical work on <span class="math-container">$f(y)$</span> shows that the maximum value is <span class="math-container">$0.396745$</span> at <span class="math-container">$y=0.887651$</span>.</p>
|
736,684 | <p>I'm trying to figure the probability that <span class="math-container">$X < Y$</span> with:</p>
<p><span class="math-container">$$X, Y \in \mathbb R^+;\ X\in [0,5] ; \ Y \in [0,2]$$</span>
What is the law to use?</p>
| Did | 6,179 | <p><strong>IF</strong> the random variables are independent and <strong>IF</strong> they are uniformly distributed on the range you indicate then $E(Y)=1$ and $P(X\lt y)=\frac15y$ for every $y$ in $(0,2)$ hence $P(X\lt Y)=\frac15E(Y)=\frac15$.</p>
<p>More generally, if $X$ and $Y$ are independent and uniform on the intervals $(0,u)$ and $(0,v)$ respectively, with $u\geqslant v$, then $P(X\lt Y)=\frac{v}{2u}$. Finally, in the same situation but with $u\leqslant v$, $P(X\lt Y)=1-\frac{u}{2v}$.</p>
|
3,371,964 | <p>Let be <span class="math-container">$O_{2}$</span> the orthogonal group, that is, the group of reflections and rotations of <span class="math-container">$\mathbb{R}^{2}$</span>. His center is <span class="math-container">$\{ \pm I\} \simeq \mathbb{Z}_{2}$</span>. I'm having problems to study the center of the quotient <span class="math-container">$\frac{O_{2}}{\{ \pm I\}}$</span>. Someone could clarify?</p>
<p>Thanks in advance.</p>
| erFuricksen | 479,710 | <p><span class="math-container">$O_2$</span> is generated by rotations and symmetries, which means that all the elements of <span class="math-container">$O_2$</span> can be written as <span class="math-container">$R_\theta S^\epsilon$</span>, where <span class="math-container">$R_\theta $</span> is a rotation of an angle <span class="math-container">$\theta$</span>, <span class="math-container">$S= \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$</span> and <span class="math-container">$\epsilon \in \{0,1\}$</span>. Though it suffices to look for matrices in that form such that <span class="math-container">$S R_\theta S^\epsilon = \pm R_\theta S^{\epsilon+1}$</span> and <span class="math-container">$R_\alpha R_\theta S^\epsilon= \pm R_\theta S^\epsilon R_\alpha$</span> <span class="math-container">$\forall \alpha$</span>. This lies to all those matrices such that <span class="math-container">$S R_\theta=\pm R_\theta S$</span> and <span class="math-container">$R_\alpha S^\epsilon= \pm S^\epsilon R_\alpha$</span> <span class="math-container">$\forall \alpha$</span>, which are separate conditions for <span class="math-container">$\theta$</span> and <span class="math-container">$\epsilon$</span>.
Can you now find out who's the center?</p>
|
1,263,865 | <p>So I have that $700=7\cdot2^2\cdot5^2$ and I got that $3^2\equiv1\pmod2$ so then $3^{1442}\equiv1\pmod2$ also $3^2\equiv1\pmod{2^2}$ so $3^{1442}\equiv1\pmod{2^2}$ which covers one of the divisors of $700$. Im not sure if I'm supposed to use $2$ or $2^2$ and I was able to find that $3^2\equiv-1\pmod5$ so $3^{1442}\equiv-1\pmod5$, For mod $7$ I wasn't able to come up with an answer in a way like the other two, and I'm not really sure how to do this to find the least non negative residue </p>
| Anurag A | 68,092 | <p>Since $\phi(700)=240$, therefore from Euler's theorem
$$3^{240} \equiv 1 \pmod{700}$$
Now
$$1442 =240(6)+2$$
Therefore
$$3^{1442} \equiv 3^{240(6)} \cdot 3^{2} \equiv 9 \pmod{700}$$</p>
|
4,363,409 | <blockquote>
<p>Define <span class="math-container">$X_0=\alpha\in(0,1)$</span> the initial capital and <span class="math-container">$X_n$</span> as the remaining capital after each game.
A player bets <span class="math-container">$1-X_n$</span> if <span class="math-container">$X_n>1/2$</span> and <span class="math-container">$X_n$</span> if <span class="math-container">$X_n\leq 1/2$</span> such that each game is a Bernoulli<span class="math-container">$(1/2)$</span>.
Define <span class="math-container">$A_n=\{X_n\in(0,1)\}$</span>, the event that the player neither wins everything or reachs ruin. Show by induction that <span class="math-container">$P(A_n)\leq 2^{-n}$</span>.</p>
</blockquote>
<p>If <span class="math-container">$\alpha< 1/2$</span> then the player either goes broken with probability <span class="math-container">$1/2$</span> or owns <span class="math-container">$2X_n$</span> in the next game. If <span class="math-container">$\alpha >1/2$</span> then the player owns every available resource with probability <span class="math-container">$1/2$</span> or owns <span class="math-container">$2X_n -1$</span> in the next game. Should I use total law of probability and try to work with the conditionals upon the last fortune as in</p>
<p><span class="math-container">$$P(A_n) = \sum_{k=1}^m P(A_n|B_m)P(B_m)$$</span>
where <span class="math-container">$\bigcup B_m= \Omega$</span>.</p>
<p>Also if this is the right path, should I split three-ways with a <span class="math-container">$X_n=1/2$</span> case where the player reach either <span class="math-container">$0$</span> or <span class="math-container">$1$</span>?</p>
<p>I'm having a little trouble working this problem out.</p>
| Andrew D. Hwang | 86,418 | <p>In a simple mathematical model, the surface of the pond is a plane <span class="math-container">$P$</span>; the viewer's eye is a point <span class="math-container">$E$</span> "above" <span class="math-container">$P$</span>. A point <span class="math-container">$S$</span> in the scene is visible to <span class="math-container">$E$</span> as a reflection in the pond if the following conditions are met:</p>
<ul>
<li>There is a line segment from <span class="math-container">$S$</span> to a point <span class="math-container">$R$</span> (from which a light ray reflects) of the visible part of <span class="math-container">$P$</span> that does not hit any scene elements;</li>
<li>The points <span class="math-container">$S$</span>, <span class="math-container">$R$</span>, and <span class="math-container">$E$</span> lie in a plane perpendicular to <span class="math-container">$P$</span>, and the segments <span class="math-container">$SR$</span> and <span class="math-container">$RE$</span> make equal angles with <span class="math-container">$P$</span> at <span class="math-container">$R$</span>:</li>
</ul>
<p><a href="https://i.stack.imgur.com/j824U.png" rel="noreferrer"><img src="https://i.stack.imgur.com/j824U.png" alt="Reflection in a planar surface" /></a></p>
<p>The angle in the diagram is exaggerated for clarity, and it varies for different points in the scene. It may help to imagine yourself being at a point on the surface of the water looking up at angle <span class="math-container">$\theta$</span> toward the tall tree; do you see the top of the tree, or just the bank?</p>
|
4,413,093 | <p>Determine the radius of convergence of the series <span class="math-container">$\sum_{n=1}^{\infty}a_nz^n$</span> where <span class="math-container">$a_n=\frac{n^2}{4^n+3n}$</span></p>
<p>Now <span class="math-container">$\alpha=\limsup_{n\to \infty}(\vert a_n\vert)^\frac{1}{n}$</span> and so radius of convergence <span class="math-container">$R=\frac{1}{\alpha}$</span></p>
<p>So now <span class="math-container">$\alpha=\limsup_{n\to \infty}(\vert a_n\vert)^\frac{1}{n}=\limsup_{n\to\infty} (\vert \frac{n^2}{4^n+3n}\vert)^\frac{1}{n}$</span></p>
<p>Now I know <span class="math-container">$lim_{n\to\infty} n^\frac{1}{n}=1$</span> but what about the denominator?</p>
| Lorago | 883,088 | <p>The key thing there is that you <strong>don't</strong> have <span class="math-container">$f:\mathbb{R}\to\mathbb{R}$</span> for <span class="math-container">$f(x)=\sqrt{x}$</span>. Instead we usually either define it as a function <span class="math-container">$f:\mathbb{R}^+_0\to\mathbb{R}$</span>, or a function <span class="math-container">$f:\mathbb{R}\to\mathbb{C}$</span> depending on our purpose. As you say, if we want the domain to be <span class="math-container">$\mathbb{R}$</span>, then we need the codomain to be <span class="math-container">$\mathbb{C}$</span>.</p>
<p>For <span class="math-container">$f(x)=\frac{1}{x}$</span>, a similar thing applies. It is <strong>not</strong> a function with domain <span class="math-container">$\mathbb{R}$</span>, but it <strong>is</strong> a function with domain <span class="math-container">$\mathbb{R}\setminus\{0\}$</span>.</p>
|
2,208,755 | <p>I got stuck on this question: find all solutions $x$ for $a\in R$:</p>
<p>$$\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{(a^2-a+1)^3}{a^2(a-1)^2}$$</p>
<p>I see that if we simplify we get:</p>
<p>$$\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{[(x-{\frac 12})^2+{\frac 34}]^3}{[(x-{\frac 12})^2-{\frac 14}]^2}$$</p>
<p>From the expression $(x-{\frac 12})^2$, I see that if $x=x_1$ is a solution, then $x=1-x_1$ is also a solution. But in the solution to this exercise, it was stated that $x=\frac{1}{x_1}$ must also be a solution, and I don't see how.</p>
<p>[EDIT]</p>
<p>Ok, thx for the help guys. What do you think of this solution (doesn't involve any above precalculus math, and needs no long calculations)?</p>
<p>From the above we know that if $x_1=a$ is a solution, then $x_2=1-a$ is also a solution.</p>
<p>Also, from here:</p>
<p>$$\require{cancel}\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{\cancel{x^3}(x+{\frac 1x}-1)^3}{\cancel{x^3}(x+{\frac 1x}-2)}$$</p>
<p>in the expression $x+{\frac 1x}$ we see that if $x=x_1$ is a solution, then $x=\frac{1}{x_1}$ is also a solution, so $x_3=\frac{1}{a}$.</p>
<p>With these two rules we can now keep generating roots until we have 6 total.</p>
<p>If $x=x_2$ is a solution, then $x=\frac{1}{x_2}$ is also a solution, so $x_4=\frac{1}{1-a}$.</p>
<p>If $x=x_3$ is a solution, then $x=1-x_3$ is also a solution, so $x_5=\frac{a-1}{a}$.</p>
<p>Finally, if $x=x_5$ is a solution, then $x=\frac{1}{x_5}$ is also a solution, so $x_6=\frac{a}{a-1}$</p>
<p>The 6 obtained values are distinct, so they cover all the roots.</p>
<p>[EDIT2]</p>
<p>I guess this is answered. No sure whose particular answer to actually select as the right one since they're all correct, so I'll just leave it like this.</p>
| Jaideep Khare | 421,580 | <p>$$\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{(a^2-a+1)^3}{a^2(a-1)^2}$$</p>
<p>Now multiply both sides by $x^2(x-1)^2$ :</p>
<p>$$(x^2-x+1)^3=x^2(x-1)^2\frac{(a^2-a+1)^3}{a^2(a-1)^2}
\\ \implies (x^2-x+1)^3-x^2(x-1)^2\frac{(a^2-a+1)^3}{a^2(a-1)^2}=0$$</p>
<p>Without expanding, it can be written as :</p>
<p>$$x^6+a_1x^5+a_2x^4+a_3x^3+a_4x^2+a_5x+1=0$$</p>
<p>Since, product of roots of this equation is $1$, each solution's reciprocal will also be a solution to this equation.</p>
|
2,677,823 | <p>How can I precisely prove the existence of a continuous function $\rho(x)$ such that $0 \leq \rho(x) \leq 1 \forall x \in R^d $ such that $g(x) \rho(x)$ is bounded and continuous for $g(x)$ continuous?Both $g(x)$ and $\rho(x)$ are defined on $R^d$.</p>
<p>My idea was that we can choose $\rho(x)$ such that $\rho(x)g(x)$ goes exponentially to zero outside a compact set in $R^d$.But i cant argue rigorously?</p>
<p>Can hints on how could I proceed?</p>
| user284331 | 284,331 | <p>If $\displaystyle\int f$ exists, then by writing $f=u+iv$ for real $u,v$, then $\displaystyle\int u$ and $\displaystyle\int v$ exist and $\displaystyle\int f=\int u+i\int v$. Note that both $\displaystyle\int u,\int v$ are real, so $\overline{\displaystyle\int f}=\displaystyle\int u-i\int v=\int(u-iv)=\int\overline{f}$.</p>
|
2,947,953 | <p>Given the two functions <span class="math-container">$$f(x) = \ln\left(\frac{x+1}{x-1}\right)$$</span> and <span class="math-container">$$g(x) = \ln(x+1)-\ln(x-1)$$</span>
I can justify independently why <span class="math-container">$\text{dom}(f) = (-\infty, -1) \cup (1,\infty)$</span>, and <span class="math-container">$\text{dom}(g)= (1,\infty)$</span>, but I'm not sure why these two functions have different domains. Can anyone enlighten me on what I'm missing? </p>
<p>EDIT: I plan on trying to explain this to freshman students taking their first calculus course. How can I justify that these domains are different? Is it true that <span class="math-container">$f(x)=g(x)$</span>? </p>
| Mohammad Riazi-Kermani | 514,496 | <p>For this function <span class="math-container">$$f(x) = \ln\left(\frac{x+1}{x-1}\right)$$</span></p>
<p>you want the fraction to be positive so both top and bottom could be positive or both could be negative.</p>
<p>On the other hand for <span class="math-container">$$g(x) = \ln(x+1)-\ln(x-1)$$</span></p>
<p>you need both <span class="math-container">$x+1$</span> and <span class="math-container">$x-1$</span> to be positive, thus you have more restriction for the domain. </p>
|
2,947,953 | <p>Given the two functions <span class="math-container">$$f(x) = \ln\left(\frac{x+1}{x-1}\right)$$</span> and <span class="math-container">$$g(x) = \ln(x+1)-\ln(x-1)$$</span>
I can justify independently why <span class="math-container">$\text{dom}(f) = (-\infty, -1) \cup (1,\infty)$</span>, and <span class="math-container">$\text{dom}(g)= (1,\infty)$</span>, but I'm not sure why these two functions have different domains. Can anyone enlighten me on what I'm missing? </p>
<p>EDIT: I plan on trying to explain this to freshman students taking their first calculus course. How can I justify that these domains are different? Is it true that <span class="math-container">$f(x)=g(x)$</span>? </p>
| fleablood | 280,126 | <p>Here's a thought experiment:</p>
<p>Let <span class="math-container">$f: \mathbb N \to \mathbb N$</span> via <span class="math-container">$f(n) = n$</span>.</p>
<p>Let <span class="math-container">$g: \mathbb R \to \mathbb R$</span> via <span class="math-container">$g(x) = |x|$</span>.</p>
<p>Let <span class="math-container">$h: \mathbb Q \to \mathbb Q$</span> via <span class="math-container">$g(\frac mn) = m$</span> for all <span class="math-container">$\frac mn \in \mathbb Q$</span> and <span class="math-container">$\frac mn$</span> is a ratio of integers in "lowest terms".</p>
<p>Notice if <span class="math-container">$x \in \mathbb N$</span> then <span class="math-container">$g(x) = f(x) = h(x)$</span></p>
<p>Does that mean that <span class="math-container">$f(x) = g(x)$</span>? Does that mean <span class="math-container">$f(x) = h(x)$</span>? Does that mean that <span class="math-container">$g(x) = h(x)$</span>?</p>
<p>Obviously <span class="math-container">$g(x) \ne h(x)$</span> because... well a very simple counter example is <span class="math-container">$g(-\frac 23) = \frac 23$</span> and <span class="math-container">$h(-\frac 23) = -2$</span>.</p>
<p>But clearly everywhere <span class="math-container">$f(x)$</span> is defined we have <span class="math-container">$f(x) = g(x)$</span> and <span class="math-container">$f(x) = h(x)$</span> so surely that means <span class="math-container">$g(x) = h(x)$</span>. So what's the contradiction.</p>
<p>The contradiction is "<em>everywhere <span class="math-container">$f(x)$</span> is defined</em>" and indeed everywhere <span class="math-container">$f(x)$</span> is defined, we <em>DO</em> have <span class="math-container">$g(x) = h(x)$</span>.</p>
<p>But this is <em>NOT</em> true where <span class="math-container">$f(x)$</span> is <em>NOT</em> defined.</p>
<p>And <span class="math-container">$\ln \frac {x+1}{x-1} \ne \ln(x+1) - \ln (x-1)$</span> !!!!!!!<strong><em>UNLESS</em></strong>!!!!!!! <span class="math-container">$x > 1$</span>. If <span class="math-container">$x < -1$</span> then <span class="math-container">$\ln \frac {x+1}{x-1} \ne \ln (x+1) - \ln(x-1)$</span>. They are not equal because the RHS is not defined even though the LHS is.</p>
<p>So <span class="math-container">$\ln \frac {x+1}{x-1}$</span> is a <em>DIFFERENT</em> function than <span class="math-container">$\ln (x+1) - \ln{x-1}$</span> just as <span class="math-container">$f(n) = n$</span> and <span class="math-container">$g(x) = |x|$</span> are different functions. So they can have different domains.</p>
<p>In short establishing domains is <em>part</em> of the definition of a function. An <span class="math-container">$f(n)$</span> and <span class="math-container">$g(n)$</span> are different not by any arithmetic assignation, afterall <span class="math-container">$f(n) = |n|$</span> and <span class="math-container">$f(\frac ab) = a$</span> for all <span class="math-container">$\frac ab \in \mathbb N$</span> so the arithmetic "rules" don't matter, they are different *because *they have different domains.</p>
|
2,129,764 | <p>Hey guys I have a problem that I'm having trouble solving. Here is the question:</p>
<p><strong>Consider events $A, B, C$ such that $P(A\mid B) > P(A)$ and $P(B\mid C) > P(B)$. Does it follow that $P(A\mid C) > P(A)$? Either prove it to be so or provide a counterexample.</strong></p>
<p>And here is what I have so far:</p>
<p><a href="https://i.stack.imgur.com/6DdER.png" rel="nofollow noreferrer">Partial solution</a></p>
<p>I would greatly appreciate it if you guys can give a hit or a suggestion to complete the problem. </p>
| user404961 | 404,961 | <p>Your partial solution must work for an appropriate choice of $p_i$'s. However, it's probably best to think about the problem this way. </p>
<p>The problem asks you to assume that knowing $C$ makes $B$ more likely, and knowing $B$ makes $A$ more likely. Then it asks if it follows that $C$ makes $A$ more likely. </p>
<p>There are at least a couple of approaches that could yield counterexamples.</p>
<p>First, it would be possible to construct an example in which $A$ <em>never</em> occurs when $C$ does.</p>
<p>Second, you could imagine an example in which $B$ is a low-probability event which is the intersection of much more likely events $A$ and $C$.</p>
|
4,080,385 | <p>Would you please compute the behavior of the following composed generalized function?</p>
<p><span class="math-container">$g(t) = $</span> <span class="math-container">$\delta(e^t)$</span></p>
<p><strong>Is it even a valid generalized function?</strong></p>
<p>Thank you very much for your time.</p>
| Calvin Khor | 80,734 | <p>If you mean the action of <span class="math-container">$\delta$</span> on <span class="math-container">$\exp$</span>, this is of course <span class="math-container">$1$</span>. If you mean something like the generalisation of change of variables <span class="math-container">$$\int\delta(e^t)f(t)dt := \int\delta(r) f(\log r)dr/r$$</span> which if defined for functions vanishing at <span class="math-container">$-\infty$</span>, is the identically zero distribution. Which makes sense as <span class="math-container">$e^t$</span> is never zero.</p>
|
250,454 | <p>Is there a <code>ReplaceOnce</code> which does only one replacement if possible by trying the rules sequentially in order. Consider the following as an example:</p>
<pre><code>ReplaceOnce[{"May","5","May","5"},{"May"->1,"5"->2}]
</code></pre>
<p>should produce:</p>
<pre><code>{1,"5","May","5"}
</code></pre>
<p>Similarly,</p>
<pre><code>ReplaceOnce[{"May","5","May","5"},{"5"->2,"May"->1}]
</code></pre>
<p>should produce:</p>
<pre><code>{"May",2,"May","5"}
</code></pre>
| Nasser | 70 | <p>There is probably a build in way. Too many commands and too little time :)</p>
<p>But you could always code one yourself.</p>
<pre><code>ClearAll[replaceOnce]
replaceOnce[lis_List, rules_List] := Module[{lisin = lis, n, pos},
Do[
pos = FirstPosition[lis, rules[[n, 1]]] ;
If[Not[Head[pos] === Missing],
lisin = ReplacePart[lisin, pos -> rules[[n, 2]]];
If[Not[SameQ[lisin, lis]],
Return[lisin, Module]
]
],
{n, 1, Length[rules]}
];
lis
]
</code></pre>
<p>And now</p>
<pre><code>lis = {"May", "5", "May", "5"};
rules1 = {"May" -> 1, "5" -> 2};
rules2 = {"5" -> 2, "May" -> 1};
rules3 = {"x" -> 2};
replaceOnce[lis, rules1]
(* {1, "5", "May", "5"} *)
replaceOnce[lis, rules2]
(* {"May",2,"May","5"} *)
replaceOnce[lis, rules3]
(* {"May","5","May","5"} *)
</code></pre>
|
78,641 | <p>I am interested in the relation between the property of countable chain condition (ccc) and the property of separable. Could someone recommend some papers or books about this to me? thanks in advance.</p>
| Santi Spadaro | 11,647 | <p>If you're interested in the relationship between the ccc and separability, you should read Stevo Todorcevic's survey "Chain condition methods in topology".</p>
<p><a href="http://www.sciencedirect.com/science/article/pii/S0166864198001126" rel="nofollow">http://www.sciencedirect.com/science/article/pii/S0166864198001126</a></p>
<p>It's a very convincing pamphlet on the power of chain conditions, clarifying that relationship with such beautiful theorems as:</p>
<p>1) (Todorcevic) Let X be compact Hausdorff. If every subspace of $X^2$ has the ccc then $X$ is separable.</p>
<p>2) (Rosenthal) A compact Hausdorff space $X$ is ccc if and only if every weakly compact subspace of $C(X)$ is separable.</p>
<p>Also, there's a whole book dedicated to chain conditions. It's called "Chain conditions in topology", by Wistar Comfort and Stelios Negropontis (Cambridge Tracts in Mathematics #79, Cambridge University Press).</p>
|
3,097,672 | <p>I have to find the definite integral of this:</p>
<p><span class="math-container">$$\int_2^3 \frac{dx}{(x^2-1)^{\frac{3}{2}}}$$</span></p>
<p>So let's start with the indefinite integral:</p>
<p>so <span class="math-container">$x = \sec \theta$</span> so <span class="math-container">$ dx = \sec \theta \tan \theta d \theta$</span></p>
<p>So </p>
<p><span class="math-container">$$ \frac{\sec{x} \tan{x}}{(\sec^2{x}-1)^{\frac{3}{2}}} $$</span></p>
<p><span class="math-container">$$ = \int \frac{\sec{x} \tan{x}}{\tan x^{\frac{3}{2}}}$$</span></p>
<p><span class="math-container">$$= \int \frac{\sec{x}\tan{x}}{\tan{x}^{\frac{1}{2}}}$$</span></p>
<p>But now I'm stuck...</p>
<p><strong>EDIT</strong></p>
<p>Unstuck:</p>
<p><span class="math-container">$$\int \frac{cos \theta}{sin^2 \theta} $$</span></p>
<p>Let's use <span class="math-container">$u = sin \theta$</span></p>
<p><span class="math-container">$$\int \frac{1}{u^2} du$$</span></p>
<p><span class="math-container">$$ \frac{u^-1}{-1} + c$$</span></p>
<p><span class="math-container">$$- \frac{1}{sin \theta} + c$$</span></p>
<p>So given that <span class="math-container">$ x = sec \theta$</span>:</p>
<p><span class="math-container">$$ - \frac{1}{\frac{\sqrt{x^2-1}}{x}}$$</span></p>
<p><span class="math-container">$$- \frac{x}{\sqrt{x^2-1}}$$</span></p>
<p>How does that look?</p>
| lab bhattacharjee | 33,337 | <p><span class="math-container">$$F=\dfrac{\sec x\tan x}{(\tan^2x)^{3/2}}=\dfrac{\sec x\tan x}{|\tan^3x|}$$</span></p>
<p>For <span class="math-container">$\tan x>0,$</span></p>
<p><span class="math-container">$$F=\dfrac{\cos x}{\sin^2x}=\csc x\cot x=-\dfrac{d(\csc x)}{dx}$$</span></p>
<p>What if <span class="math-container">$\tan x<0$</span></p>
|
781,776 | <blockquote>
<p>A red die, a blue die, and a yellow die (all six sided) are rolled. Given that no two of the dice land on the same number, what is the conditional probability that blue is less than yellow which is less than red?</p>
</blockquote>
<p>The Answer is a sixth. I have absolutely no idea how to do this though.</p>
| nature1729 | 29,257 | <p>Number of ways will be coefficient of $x^{15}$ in </p>
<p>$$f(x)=(1+x+x^2+x^3+x^4)(1+x+x^2+\cdots+x^{15})^2=(1-x^5)(1-x^{16})^2(1-x)^{-3}=(1-x^5-2x^{16}+2x^{21}
+x^{32}-x^{37})(1+\binom{3}{1}x+\binom{4}{2}x^2+\cdots)$$</p>
<p>Thus number of ways is $\binom{17}{2}-\binom{12}{2}=70$</p>
|
235,945 | <p>Hello please help me with these trig identities and double angles as I am not sure where I am going wrong but I keep getting the wrong answer </p>
<p>This is the problem
$$
\sin(\theta+30) = 2\cos(\theta)
$$
This is my one of my incorrect solutions</p>
<p>$$\sin(\theta +30) = 2\cos(\theta)$$
$$\sin(\theta)\cos(30) + \sin(30)\cos(\theta)=2(1 - \sin(\theta))$$
$$\sin(\theta)(\frac{\sqrt3}{2})+(\frac{1}{2})(1 - \sin(\theta))=2-2\sin(\theta)$$
$$\frac{\sin(\theta)\sqrt3+1-\sin(\theta)}{2}+2 \sin(\theta)=2$$</p>
<p>I get stuck and I am not sure what to do with this problem.</p>
<p>Please help as I am trying to self teach my A -level maths.</p>
<p>Thanks in advance</p>
| Fly by Night | 38,495 | <p>What you have written cannot be an identity. If it were then $\sin(\theta + 30)$ must equal $2\cos\theta$ for all values of $\theta$. However, while $\sin(\theta+30)$ oscillates between $-1$ and $1$, we see that $2\cos\theta$ oscillates between $-2$ and $2$. They have different ranges and so cannot possibly be identical functions. Let's assume you want to find particular values of $\theta$ for which $\sin(\theta + 30) = 2\cos\theta.$</p>
<p>The double angle formula tells us that $\sin(\theta + 30) = \sin\theta\cos(30) + \sin(30)\cos\theta.$ Two well-known values of sine and cosine are $\cos(30^{\circ}) = \sqrt{3}/2$ and $\sin(30^{\circ}) = 1/2.$ Thus:</p>
<p>$$\sin(\theta + 30) = \frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta. $$</p>
<p>It follows that $\sin(\theta+30) = 2\cos\theta$ if and only if</p>
<p>$$\frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta = 2\cos\theta \iff \frac{\sqrt{3}}{2}\sin\theta - \frac{3}{2}\cos\theta = 0 \, . $$</p>
<p>Consider the numerator: $\sqrt{3}\sin\theta - 3\cos\theta = 0 \iff \tan\theta = \sqrt{3} \iff \theta = 60^{\circ} + 180n^{\circ},$ where $n$ is any integer. The multiples of $180^{\circ}$ are added because $\tan\theta$ is periodic with a period of $180^{\circ}$; it repeats itself every $180^{\circ}$. The finally answer is then:</p>
<p>$$\theta = \ldots, -120^{\circ}, \ 60^{\circ}, \ 240^{\circ},\ldots $$
$$ \theta \in \{ (60 + 180n)^{\circ} : n \in \mathbb{Z} \}.$$</p>
|
2,677,584 | <p>I have the following question:</p>
<blockquote>
<p>Find the real values of $a$ for which the equation
$$(1+\tan^2\theta)^2 + 4a\tan\theta(\tan^2\theta + 1) + 16\tan^2\theta = 0$$
has four distinct real roots in $\left(0, \dfrac{\pi}{2}\right)$.</p>
</blockquote>
<p>I tried to solve the above equation by dividing the entire equation by $\tan^2\theta$ and then substituting $\tan\theta + \dfrac{1}{\tan\theta}$ as $y$ and then solving for $y$. Then I tried to apply the inequality $\tan\theta + \dfrac{1}{\tan\theta} \geqslant 2$ but couldn't find a proper range of values of $a$.</p>
<p>Please help. Please point out if there is any mistake in my work. Thanks in advance.</p>
| lab bhattacharjee | 33,337 | <p>$$\sec^4t+4a\frac{\sin t}{\cos^3t}+\dfrac{16\sin^2t}{\cos^2t}=0$$</p>
<p>$$0=1+4a\sin t\cos t+16(\sin^2t\cos^2t)=1+2a\sin2t+4\sin^22t$$</p>
<p>$$-a=\dfrac{1+4\sin^22t}{2\sin2t}$$</p>
<p>As $0<2t<\pi,\sin2t>0$ $$\dfrac{1+4\sin^22t}{2\sin2t}=\dfrac1{2\sin2t}+2\sin2t\ge2\sqrt{\dfrac1{2\sin2t}\cdot2\sin2t}=2$$</p>
<p>$\implies -a\ge2\iff a\le-2$</p>
<p>But if $a=-2,\sin2t=\dfrac12,\dfrac12\implies a\ne-2$</p>
|
3,005,329 | <blockquote>
<p>Suppose that <span class="math-container">$f:[0,1]\to\mathbb{R}$</span> is continuous on <span class="math-container">$[0,1]$</span>. Show that <span class="math-container">$\{\int_0^1f(x^n)dx\}_{n=1}^\infty$</span> converges to <span class="math-container">$f(0)$</span>.</p>
</blockquote>
<p>I'm not really sure to go about proving this. I know <span class="math-container">$\lim_{n\to\infty}x^n=0$</span> for <span class="math-container">$x<1$</span> so it makes sense that the sequence would converge to <span class="math-container">$\int_0^1f(0)dx=f(0)$</span>. I have no idea how to formalize this and make it into a proof though. I'm really bad at this so I'd really appreciate a thorough explanation.</p>
| hamam_Abdallah | 369,188 | <p><strong>Hint</strong></p>
<p>Let <span class="math-container">$\epsilon>0$</span> small enough.</p>
<p><span class="math-container">$$\int_0^{1-\epsilon}(f(x^n)-f(0))dx=(1-\epsilon)(f(c^n)-f(0))$$</span>
with <span class="math-container">$0\le c\le 1-\epsilon<1$</span>.
now use sequential charactersation of the continuity at <span class="math-container">$0$</span>.</p>
<p>As <span class="math-container">$c^n\to 0$</span>, For large <span class="math-container">$n$</span>,</p>
<p><span class="math-container">$$|\int_0^{1-\epsilon}(f(x^n)-f(0))dx|<\frac{\epsilon}{2}.$$</span>
On the other hand,
<span class="math-container">$$|\int_{1-\epsilon}^1(f(x^n)-f(0))dx|\le 2M\epsilon.$$</span>
with <span class="math-container">$M=\sup_{[0,1]}|f|$</span>.</p>
|
3,005,329 | <blockquote>
<p>Suppose that <span class="math-container">$f:[0,1]\to\mathbb{R}$</span> is continuous on <span class="math-container">$[0,1]$</span>. Show that <span class="math-container">$\{\int_0^1f(x^n)dx\}_{n=1}^\infty$</span> converges to <span class="math-container">$f(0)$</span>.</p>
</blockquote>
<p>I'm not really sure to go about proving this. I know <span class="math-container">$\lim_{n\to\infty}x^n=0$</span> for <span class="math-container">$x<1$</span> so it makes sense that the sequence would converge to <span class="math-container">$\int_0^1f(0)dx=f(0)$</span>. I have no idea how to formalize this and make it into a proof though. I'm really bad at this so I'd really appreciate a thorough explanation.</p>
| Jack D'Aurizio | 44,121 | <p>Since <span class="math-container">$f(x)$</span> is continuous, for any <span class="math-container">$x\in(0,1)$</span> the sequence <span class="math-container">$f(x^n)$</span> is convergent to <span class="math-container">$f(0)$</span> as <span class="math-container">$n\to +\infty$</span>.
On the other hand the continuity of <span class="math-container">$f$</span> on <span class="math-container">$[0,1]$</span> implies its boundedness, hence the dominated convergence theorem trivially applies.</p>
|
356,306 | <p>If $f:X_1 \rightarrow X_2$ and $g:X_2 \rightarrow X_3$ are homomorphisms.
If $g \circ f =0$ does it imply that $Im f \subseteq ker g$? and how to show that? do you have an example?
thanks :)</p>
| rschwieb | 29,335 | <p>Things in $Im(f)$ look like $f(x)$, any $y$ such that $g(y)=0$ is in the kernel of $g$, and $g(f(x))=0$. You have everything you need.</p>
|
89,621 | <p>All geometry in computer graphics are transformed by position * transform matrix; The issue is the fact that position is a vector with 3 components (x,y,z); And transform matrix is a 4 by 4 with one column that can be dumped(at least in my case). So my transform matrix is now a 3 by 4 matrix:<br>
axis x { x, y, z }<br>
axis y { x, y, z }<br>
axis z { x, y, z }<br>
position axis { x, y, z } </p>
<pre><code> multiplied with position vector { x, y ,z }
</code></pre>
<p>If I dump position axis this can be done with standard formula of matrix multiplication.
But it does not transform it.
I can make the position vector with 4 components { x, y, z, w } but don't know what to do with the w?
My only solution is a slow one, put position vector in a new transform matrix in position axis and multiply them. But it is computationally expensive.
How to approach such problem?</p>
| hardmath | 3,111 | <p>Here's the "math" way of looking at this. In three dimensions there are various "isometries", mappings that preserve distances between points. Some of these are linear transformations, and these can be represented in the usual way as multiplication by <a href="http://en.wikipedia.org/wiki/Orthogonal_matrix" rel="noreferrer">an orthogonal matrix</a>, real 3x3 matrix $P$ such that $P^{-1} = P^T$, the inverse is the transpose. If $P$ has determinant 1, then the mapping realized as multiplication by $P$ is a <em>rotation</em>. Otherwise the determinant can be -1, and the mapping is termed a <em>reflection</em>.</p>
<p>Now linear transformations always fix the origin. If you want an isometry that maps the origin to some other point, you need <a href="http://en.wikipedia.org/wiki/Translation_%28geometry%29" rel="noreferrer">a translation mapping</a>.</p>
<p>Combining linear transformations with translations gives the larger class of mappings we call affine transformations. The theorem here is that every isometry in Euclidean space (of whatever dimension) is an affine transformation, but not every affine transformation is necessarily an isometry (distance preserving).</p>
<p>The real 4x4 matrices being discussed here are a clever way of representing 3D affine transformations generally (and isometries in particular) with matrix multiplication in one dimension higher. The special structure of these is:</p>
<p>$$\left( \begin{array}{cc} P & v \\ 0 & 1 \end{array} \right)$$</p>
<p>where $P$ is a 3x3 matrix representing the linear transformation part of the affine mapping, $v$ is a 3x1 column representing offset of the origin, "0" denotes a 1x3 row of zeroes, and 1 is just scalar one.</p>
<p>Consider what happens if we take a 3D vector $u^T = (x,y,z)$, tack a synthetic component 1 onto the end, transpose it to a column and multiply the result by the special matrix above:</p>
<p>$$\left( \begin{array}{cc} P & v \\ 0 & 1 \end{array} \right)
\times
\left( \begin{array}{c} x \\ y \\ z \\ 1 \end{array} \right) =
\left( \begin{array}{c} {Pu + v} \\ 1 \end{array} \right) $$</p>
<p>Thus a matrix multiplication can be used to compute $Pu + v$, which amounts to applying the linear transformation to source point $u$, then adding the offset $v$ to get the destination point. If $P$ is orthogonal and has determinant 1, then we might speak of rotating the source point and adding the offset. Such an operation is often required in computer graphics to "pan" and "dolly" a virtual camera.</p>
<p>While not "space efficient" in terms of the extra dimension, this representation is convenient for programmming because a general matrix multiplication routine can be used instead of coding separate steps of adding the offset after a matrix multiplication.</p>
<p>Note also that "undoing" the represented mapping will amount to multiplying by a matrix inverse. It is left as an exercise for the reader to work out what the inverse of the 4x4 matrix is in the special (isometry) case of orthogonal $P$.</p>
|
1,821,927 | <p>Let $V = \big\{z: |z|<5,\text{Im}(z)>0 \big\}$. Let $f$ analytic in $V$, continuous in $\overline{V}$ and suppose $$\forall x \in \left[ -5,5\right]:\ f\left( x\right) \in \mathbb{R}$$
Show that $$\limsup_{n \rightarrow \infty} \root{n}\of{\frac{f^{(n)}(1)}{n!}} \le \frac{1}{4}$$
<br><br><br>
I tried expanding to power series near $z=1$ but it's impossible since $1\in \partial V$, then I tried device an analytic expansion of $f$ to the disc $|z| \leq 5$ by defining
$$g(z) = \cases{f(z), \qquad &\text{Im}(z) \geq0 \\ -f( \overline{z}), \qquad &\text{Im}(z) < 0}$$
But I cant show that $g$ is analytic in the whole disc.
How can I show that $f$ is holomorphicaly extendible to the whole disk of radius 5?
<br>Help<br><br><br></p>
| Uria Mor | 180,241 | <p>This is a slightly different approach to solve it using Morera's Theorem. </p>
<p>Of course the point in this exercise was to find an analytic continuation of $f$ to the disk, and once that is done, one can simply expand $f$ around $z=1$, to a power series $\sum \frac{f^{(n)}(1)}{n!}(z-1)^n$ and since $f\in Hol(D(0,5))$ we get $f\in Hol(D(1,4))$ which implies that $\frac{f^{(n)}(1)}{n!}(z-1)^n$ converges uniformly in a disc of radius $4$ around $1$, and by Cauchy-Hadamard rule: $$\limsup_{n \rightarrow \infty} \root{n}\of{\frac{f^{(n)}(1)}{n!}} \le \frac{1}{4}$$
<br>
So, as discussed with <strong>mercio</strong>, the "obvious" way to exted $f$ is by defining:
$$g(z) = \cases{f(z), \qquad &\text{Im}(z) \geq0 \\ \overline{f( \overline{z})}, \qquad &\text{Im}(z) < 0}$$ </p>
<p>and the "classic" way to show $g\in Hol(D(0,5))$ is using Morera's Theorem: </p>
<blockquote>
<p><em>Let $f(z)$ be continuous function on domain D. If $\int_{\partial R}f(z)dz = 0 \ $ for every closed rectangle $R$ contained in $D$ with sides parallel to the coordinate axes, then $f$ is analytic on $D$</em><br> <strong>Complex Analysis, T.W Gamelin, chapter IV.6</strong> </p>
</blockquote>
<p><br>
Now let $R \subset D(0,5)$ closed rectangle with sides parallel to the coordinate axes.
<br>
<strong>case 1:</strong> If $R$ contained in one of the upper/lower half of the plane, than by Cauchy's theorem for $f(z)$ and $\overline{f(\overline{z})}$ on the matching domains we get what we need.
<br>
<strong>case 2:</strong> If $R$ contains both positive and negative values of the imaginary axis, we write $\partial R$ = $\partial R_1 + \partial R_2$ where $\partial R_1$ is the counter clockwise directed curve on the boundary of $R \cap \{ z | Imz \geq 0 \}$, and $\partial R_2$ is the counter clockwise directed curve on the boundary of $R \cap \{ z | Imz \leq 0 \}$. And by case 1 we get:
$$\int_{\partial R}g(z)dz = \int_{\partial R_1}g(z)dz +\int_{\partial R_2}g(z)dz =0$$
<br>
Hence by Morera - $g\in Hol(D(0,5))$ and on the real axis $g \equiv f$. Getting the bound for the limit is trivial.</p>
|
99,961 | <p>I have a list with elements</p>
<pre><code>{a -> -1, b -> -2, c -> -3}
</code></pre>
<p>If I now wanted to apply a tranformation to <code>b</code> and <code>c</code> so that they would give the tranformation <code>b -> 1-10^val</code> and <code>c -> 1-10^val</code>, yielding</p>
<pre><code>{a -> -1, b -> 0.99, c -> 0.999}
</code></pre>
<p>How would I do this in <em>Mathematica</em>?</p>
| halirutan | 187 | <p>I'm not sure whether you mean something different by <code>1-b^-2</code> or you just miscalculated, because your result is not the correct result. In general, you can transform transformation-rules like this:</p>
<pre><code>{a -> -1, b -> -2, c -> -3} /.
{
(b -> val_) :> (b -> 1 - val^-2),
(c -> val_) :> (c -> 1 - val^-3)
}
</code></pre>
<p>And in case Kuba is right about what you really want, then you can use</p>
<pre><code>{a -> -1, b -> -2, c -> -3} /.
{
(b -> val_) :> (b -> 1 - 10.0^val),
(c -> val_) :> (c -> 1 - 10.0^val)
}
(* {a -> -1, b -> 0.99, c -> 0.999} *)
</code></pre>
<p>Or</p>
<pre><code>{a -> -1, b -> -2, c -> -3} /.
{(key : b | c -> val_) :> (key -> 1 - 10.0^val)}
</code></pre>
|
99,961 | <p>I have a list with elements</p>
<pre><code>{a -> -1, b -> -2, c -> -3}
</code></pre>
<p>If I now wanted to apply a tranformation to <code>b</code> and <code>c</code> so that they would give the tranformation <code>b -> 1-10^val</code> and <code>c -> 1-10^val</code>, yielding</p>
<pre><code>{a -> -1, b -> 0.99, c -> 0.999}
</code></pre>
<p>How would I do this in <em>Mathematica</em>?</p>
| Kuba | 5,478 | <pre><code>list = {a -> -1, b -> -2, c -> -3}
</code></pre>
<p>The quesiton is unclear but let's say I know what you want :)</p>
<pre><code>MapAt[
1. - 10^# &,
Association[list],
List@*Key /@ {b, c}] // Normal
</code></pre>
<blockquote>
<p>{a -> -1, b -> 0.99, c -> 0.999}</p>
</blockquote>
<p>You could work on <code>Association</code> from the begining, then you can skip <code>Association[]</code> and <code>Normal</code> which makes it even more compact.</p>
<hr>
<p>Other way, if you know positions and don't want to use <code>Associations</code>:</p>
<pre><code>MapAt[
1. - 10^# &,
list,
{2 ;; 3, 2}]
</code></pre>
|
3,688,680 | <p>I know cantor set and rational numbers in <span class="math-container">$\mathbb{R}$</span> are meagre. But they are all zero measure.</p>
<p>So is there any meagre set that is non-zero measure?</p>
| Integrand | 207,050 | <ol>
<li>Comparison</li>
</ol>
<p><span class="math-container">$$\underbrace{\int_{0}^{\infty} \sin^2(x) x^{-5/2}\,dx}_{I} = \underbrace{\int_{0}^{1} \sin^2(x) x^{-5/2}\,dx}_{I_1} +\underbrace{\int_{1}^{\infty} \sin^2(x) x^{-5/2}\,dx}_{I_2} $$</span>
<span class="math-container">$I_2$</span> converges by directly comparing the integrand to <span class="math-container">$x^{-5/2}$</span>. <span class="math-container">$I_1$</span> converges because on <span class="math-container">$[0,1]$</span>, we have <span class="math-container">$\sin(x)\leq x$</span>, whence
<span class="math-container">$$
0\leq I_1\leq \int_0^1 x^2\cdot x^{-5/2}\,dx = \int _0^1 x^{-1/2}\,dx = 2
$$</span></p>
<ol start="2">
<li>Direct computation</li>
</ol>
<p>Use integration by parts with <span class="math-container">$u=\sin^2(x)$</span>:
<span class="math-container">$$
I = \left.- \frac{2}{3} \frac{\sin^2(x)}{x^{3/2}} \right|_{0}^{\infty} + \frac{2}{3} \int _{0}^{\infty} \frac{2\sin(x)\cos(x)}{x^{3/2}}\,dx
$$</span>
<span class="math-container">$$
= \frac{2}{3} \int _{0}^{\infty} \frac{\sin(2x)}{x^{3/2}}\,dx
$$</span> Put <span class="math-container">$y=2x$</span>:
<span class="math-container">$$
= \frac{4}{3\sqrt{2}} \int _{0}^{\infty} \frac{\sin(y)}{y^{3/2}}\,dy
$$</span>
<span class="math-container">$$
= \frac{4}{3\sqrt{2}}\cdot \Gamma\left(\frac{-1}{2}\right)\sin\left(\frac{-\pi}{4}\right) = \frac{4\sqrt{\pi}}{3}
$$</span></p>
<p>Source: <a href="https://dlmf.nist.gov/5.9" rel="nofollow noreferrer">https://dlmf.nist.gov/5.9</a></p>
|
632,891 | <p>I'm trying to solve this limit, for which I already know the solution thanks to Wolfram|Alpha to be $\sqrt[3]{abc}$:</p>
<p>$$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n:\forall a,b,c\in\mathbb{R}^+$$</p>
<p>As this limit is an indeterminate form of the type $1^\infty$, I've been trying to approach it by doing:</p>
<p>$$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{1}{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{1}{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}}\right)^{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}\cdot\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n}=e^{\lim_{n\rightarrow\infty}\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n}$$</p>
<p>But now when I approach that top limit this is what I get:</p>
<p>$$\lim_{n\rightarrow\infty}\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n=\lim_{n\rightarrow\infty}\frac{n\cdot a^{\frac{1}{n}}}{3}+\frac{n\cdot b^{\frac{1}{n}}}{3}+\frac{n\cdot c^{\frac{1}{n}}}{3}-n=\lim_{n\rightarrow\infty}\frac{n\cdot a^0}{3}+\frac{n\cdot b^0}{3}+\frac{n\cdot c^0}{3}-n=\lim_{n\rightarrow\infty}\frac{n}{3}+\frac{n}{3}+\frac{n}{3}-n=0$$</p>
<p>And hence the final limit should be $e^0=1$ which is clearly wrong but I honestly don't know what I did wrong, so what do you suggest me to solve this limit?</p>
| Community | -1 | <p>By Taylor series we have:
$$\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}=\frac 1 3\left(3+\frac1 n(\log a +\log b+\log c)++o\left(\frac 1 n\right)\right)=1+\frac 1 n \log\sqrt[3]{abc}+o\left(\frac 1 n\right)$$
so
$$\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n=\exp\left(n\log\left(1+\frac 1 n \log\sqrt[3]{abc}+o\left(\frac 1 n\right)\right)\right)\sim_\infty \sqrt[3]{abc} $$</p>
|
63,723 | <p>How to find $f'(a)$ where $f(x) = \sqrt{1-2x}$ ?</p>
<p>I am not too sure what to do, no matter what I do I can't get the correct answer. I know it is simple algebra but I can't figure it out.</p>
| Arturo Magidin | 742 | <p>As you surmise, you need to multiply by the conjugate; the problem is that you forgot to distribute the negative sign correctly, and you forgot to <em>divide</em> by the conjugate as well as multiply by it.
$$\begin{align*}
\lim_{h\to 0}\frac{f(a+h)- f(a)}{h} &= \lim_{h\to 0}\frac{\sqrt{1-2(a+h)}-\sqrt{1-2a}}{h}\\
&\strut\\
&=\lim_{h\to 0}\frac{\sqrt{1-2a-2h}-\sqrt{1-2a}}{h}\\
&\strut\\
&=\lim_{h\to 0}\left(\frac{\sqrt{1-2a-2h} - \sqrt{1-2a}}{h}\right)\left(\frac{\sqrt{1-2a-2h}\;+\sqrt{1-2a}}{\sqrt{1-2a-2h}\; + \sqrt{1-2a}}\right)\\
&\strut\\
&=\lim_{h\to 0}\frac{\left(\sqrt{1-2a-2h}-\sqrt{1-2a}\right)\left(\sqrt{1-2a-2h}\;+\sqrt{1-2a}\right)}{h\left(\sqrt{1-2a-2h}\;+\sqrt{1-2a}\right)}\\
&\strut\\
&= \lim_{h\to 0}\frac{(1-2a-2h) - (1-2a)}{h\left(\sqrt{1-2a-2h}\;+\sqrt{1-2a}\right)}\\
&\strut\\
&= \lim_{h\to 0}\frac{1-2a-2h-1+2a}{h\left(\sqrt{1-2a-2h}\;+\sqrt{1-2a}\right)}\\
&\strut\\
&= \lim_{h\to 0}\frac{-2h}{h\left(\sqrt{1-2a-2h}\;+\sqrt{1-2a}\right)}.
\end{align*}$$
Nothing but algebra so far. Trying to plug in $0$ for $h$ gives $\frac{0}{0}$, as expected. But there is a factor of $h$ in the numerator, and a factor of $h$ in the denominator. If we cancel them, can the resulting limit be evaluated simply by pluggin in $h=0$?</p>
|
3,069,244 | <p>Consider the following functional <span class="math-container">$\Phi:\mathbb R^n\to\mathbb R $</span>:
<span class="math-container">$$
\Phi(x)=\sum_{i=1}^{n-1}(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1).
$$</span>
The computer experiments show that it is non-negative for all <span class="math-container">$x_i\geq 0$</span>. I need to prove this. Note that we have both <span class="math-container">$\Phi(x)=0$</span> and <span class="math-container">$\nabla \Phi(x)=0$</span> for all <span class="math-container">$x$</span> with equal coordinates. The proof should be simple, but I can't manage to find it. Any ideas?</p>
| symchdmath | 626,816 | <p>This integral is more complicated than it looks and only requires comfort with hyperbolic trigonometric definitions. My initial instinct is to look at the expression inside the <span class="math-container">$\cosh$</span> to see if I can simplify it. In fact we have by the definition of the hyperbolic trigonometric functions,</p>
<p><span class="math-container">$$\tanh^{-1}(x) = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right) $$</span></p>
<p>Now using log laws we have that,</p>
<p><span class="math-container">$$\tanh^{-1}(3x) - \tanh^{-1}(x) = \frac{1}{2} \ln \left(\frac{(1+3x)(1-x)}{(1-3x)(1+x)} \right), \ \ (*)$$</span></p>
<p>We note at this point that,</p>
<p><span class="math-container">$$\sqrt{36x^4 - 40x^2 + 4} = 2\sqrt{(1-9x^2)(1-x^2)} = 2\sqrt{(1-3x)(1+3x)(1-x)(1+x)}$$</span></p>
<p>There is a lot in common between the above two equations which motivates us to press forward, we now use the following identity which we can prove just with the definitions of the hyperbolic trig functions,</p>
<p><span class="math-container">$$\cosh(x+y) = \cosh(x) \cosh(y) + \sinh(x) \sinh(y)$$</span></p>
<p>We use to simplify the <span class="math-container">$\cosh$</span> in the integral, we find that,</p>
<p><span class="math-container">$$\cosh(3x + (\tanh^{-1}(3x) - \tanh^{-1}(x))) = \cosh(3x) \cosh(\tanh^{-1}(3x) - \tanh^{-1}(x)) + \sinh(3x) \sinh(\tanh^{-1}(3x) - \tanh^{-1}(x)) $$</span></p>
<p>Using the following definitions of the hyperbolic functions,</p>
<p><span class="math-container">$$\cosh(x) = \frac{e^x + e^{-x}}{2} $$</span></p>
<p><span class="math-container">$$\sinh(x) = \frac{e^x - e^{-x}}{2} $$</span></p>
<p>We find that using <span class="math-container">$(*)$</span>, leaving the details to you,</p>
<p><span class="math-container">$$\cosh(\tanh^{-1}(3x) - \tanh^{-1}(x)) = \cosh\left(\frac{1}{2}\ln \left(\frac{(1+3x)(1-x)}{(1-3x)(1+x)} \right)\right) = \frac{1-3x^2}{\sqrt{(1-9x^2)(1-x^2)}}$$</span></p>
<p>Similarly, we have</p>
<p><span class="math-container">$$\sinh(\tanh^{-1}(3x) - \tanh^{-1}(x)) = \frac{2x}{\sqrt{(1-9x^2)(1-x^2)}} $$</span></p>
<p>Putting this all together,</p>
<p><span class="math-container">$$I = \int_{-1/3}^{1/3} \sqrt{36x^4 - 40x^2 + 4} \cosh(3x + \tanh^{-1}(3x) - \tanh^{-1}(x)) \ \mathrm{d}x $$</span></p>
<p><span class="math-container">$$I = \int_{-1/3}^{1/3} 2\sqrt{(1-9x^2)(1-x^2)} \left(\cosh(3x) \frac{1-3x^2}{\sqrt{(1-9x^2)(1-x^2)}} + \sinh(3x)\frac{2x}{\sqrt{(1-9x^2)(1-x^2)}} \right) \ \mathrm{d}x$$</span></p>
<p>Finally the integral simplifies to,</p>
<p><span class="math-container">$$I = 2 \int_{-1/3}^{1/3} (1-3x^2)\cosh(3x) \ \mathrm{d}x + 2 \int_{-1/3}^{1/3} 2x \sinh(3x) \ \mathrm{d}x $$</span></p>
<p>Which is a lot easier to compute with IBP and I will leave that to you to complete, this does indeed give the correct answer.</p>
|
153,409 | <p>Would you please tell me whether there is any wrong on this problem? given that $g$ is continuous on $[0,\infty)\rightarrow \mathbb{R}$ satisfying $\int_{0}^{x^2(1+x)}g(t)dt=x \forall x\in [0,\infty)$ then I need to find what is $g(2)$?</p>
| A.S | 24,829 | <p>Let $G(x) = \int _{0} ^{x} g(t) dt$. Since $g(t)$ is continuous, we can deduce that $G(x)$ exists and will be differentiable for $x \ge 0$.</p>
<p>Then, by your condition, $G(x^2(1+x))=x$.</p>
<p>Differentiating both sides with respect to $x$, we get $(2x(1+x)+x^2) \times g(x^2(1+x))=1$.</p>
<p>Simplifying, we get $g(x^2(1+x))= \frac{1} {3x^2+2x}$</p>
<p>Now, we set $(x^2(1+x))=2$. This is true when $x=1$ and has two imaginary roots at $x=i-1$ and $x=-i-1$.</p>
<p>We are looking for real solutions, so $x=1$ and $g(2)=\frac {1} {5}$</p>
|
1,087,874 | <p>I want to understand how I can count the terms of the expression $x^{m-1} + x^{m-2} +\ldots+ x^0$ when $x=1$.</p>
<p>The result is $m$, I dont know how to count them formally, any advice would be helpful. I'm desperated, not because it is required to do the above, but how can be done, I need to understand the subject. Sorry for my bad english.</p>
<p>PS: It is related to this limit:
$$\lim_{x \to 1} \frac{x^m-1}{x-1} = m$$</p>
<p>I dont want to use L'Hôpital's rule, I just use a simple factorization and a change of a variable.</p>
| GFauxPas | 173,170 | <p>There isn't a need to calculate anything or to use limits of any sort. As soon as you write "$x^{m-1} + \cdots + x^0$", you have <em>defined</em> the expression to have $m$ terms. There's nothing to prove.</p>
<p>To evaluate your limit, I recommend a proof by induction on $m$.</p>
|
550,188 | <p>Okay so I have an equation in my book which is as follows..
$$
\frac {a}{s(s+a)}
$$
it says "using partial fractions this can be expanded to
$$
\frac {1}{s} + \frac {-1}{s+a}
$$</p>
<p>My usual method would be to cross multiply and do something like this
$$
\frac {a}{s(s+a)} = \frac {A(s+a)}{s(s+a)} + \frac {B(s)}{s(s+a)}
$$</p>
<p>Then cancel off the denominators and solve..</p>
<p>$$
a = A(s+a) + B(s)
$$</p>
<p>usually though the a would be some constant but here I have no values to play around with.. how has he done it in the book?</p>
| Empy2 | 81,790 | <p>Let $s=0$, so $a=Aa$. Let $s=1$, so $a=A(a+1)+B$<br>
Solve for $A$ and $B$.</p>
|
2,352,721 | <h2>Question</h2>
<blockquote>
<p>Four fair six-sided dice are rolled. The probability that the sum of the results being <span class="math-container">$22$</span> is <span class="math-container">$$\frac{X}{1296}.$$</span> What is the value of <span class="math-container">$X$</span>?</p>
</blockquote>
<h2>My Approach</h2>
<p>I simplified it to the equation of the form:</p>
<blockquote>
<p><span class="math-container">$x_{1}+x_{2}+x_{3}+x_{4}=22, 1\,\,\leq x_{i} \,\,\leq 6,\,\,1\,\,\leq i \,\,\leq 4 $</span></p>
</blockquote>
<p>Solving this equation results in:</p>
<p><span class="math-container">$x_{1}+x_{2}+x_{3}+x_{4}=22$</span></p>
<p>I removed restriction of <span class="math-container">$x_{i} \geq 1$</span> first as follows-:</p>
<p><span class="math-container">$\Rightarrow x_{1}^{'}+1+x_{2}^{'}+1+x_{3}^{'}+1+x_{4}^{'}+1=22$</span></p>
<p><span class="math-container">$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$</span></p>
<p><span class="math-container">$\Rightarrow \binom{18+4-1}{18}=1330$</span></p>
<p>Now i removed restriction for <span class="math-container">$x_{i} \leq 6$</span> , by calculating the number of <strong>bad cases</strong> and then subtracting it from <span class="math-container">$1330$</span>:</p>
<p>calculating <strong>bad combination</strong> i.e <span class="math-container">$x_{i} \geq 7$</span></p>
<p><span class="math-container">$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$</span></p>
<p>We can distribute <span class="math-container">$7$</span> to <span class="math-container">$2$</span> of <span class="math-container">$x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$</span> i.e<span class="math-container">$\binom{4}{2}$</span></p>
<p>We can distribute <span class="math-container">$7$</span> to <span class="math-container">$1$</span> of <span class="math-container">$x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$</span> i.e<span class="math-container">$\binom{4}{1}$</span> and then among all others .</p>
<p>i.e</p>
<p><span class="math-container">$$\binom{4}{1} \binom{14}{11}$$</span></p>
<p>Therefore, the number of bad combinations equals <span class="math-container">$$\binom{4}{1} \binom{14}{11} - \binom{4}{2}$$</span></p>
<p>Therefore, the solution should be:</p>
<p><span class="math-container">$$1330-\left( \binom{4}{1} \binom{14}{11} - \binom{4}{2}\right)$$</span></p>
<p>However, I am getting a negative value. What am I doing wrong?</p>
<p><strong>EDIT</strong></p>
<p>I am asking for my approach, because if the question is for a larger number of dice and if the sum is higher, then predicting the value of dice will not work.</p>
| G Tony Jacobs | 92,129 | <p>There aren't too many to just count.</p>
<p>Permutations of $6+6+6+4$: $\binom41=4$</p>
<p>Permutations of $6+6+5+5$: $\binom42=6$</p>
<p>These are the only options, so your numerator must be $4+6=10$</p>
|
2,352,721 | <h2>Question</h2>
<blockquote>
<p>Four fair six-sided dice are rolled. The probability that the sum of the results being <span class="math-container">$22$</span> is <span class="math-container">$$\frac{X}{1296}.$$</span> What is the value of <span class="math-container">$X$</span>?</p>
</blockquote>
<h2>My Approach</h2>
<p>I simplified it to the equation of the form:</p>
<blockquote>
<p><span class="math-container">$x_{1}+x_{2}+x_{3}+x_{4}=22, 1\,\,\leq x_{i} \,\,\leq 6,\,\,1\,\,\leq i \,\,\leq 4 $</span></p>
</blockquote>
<p>Solving this equation results in:</p>
<p><span class="math-container">$x_{1}+x_{2}+x_{3}+x_{4}=22$</span></p>
<p>I removed restriction of <span class="math-container">$x_{i} \geq 1$</span> first as follows-:</p>
<p><span class="math-container">$\Rightarrow x_{1}^{'}+1+x_{2}^{'}+1+x_{3}^{'}+1+x_{4}^{'}+1=22$</span></p>
<p><span class="math-container">$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$</span></p>
<p><span class="math-container">$\Rightarrow \binom{18+4-1}{18}=1330$</span></p>
<p>Now i removed restriction for <span class="math-container">$x_{i} \leq 6$</span> , by calculating the number of <strong>bad cases</strong> and then subtracting it from <span class="math-container">$1330$</span>:</p>
<p>calculating <strong>bad combination</strong> i.e <span class="math-container">$x_{i} \geq 7$</span></p>
<p><span class="math-container">$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$</span></p>
<p>We can distribute <span class="math-container">$7$</span> to <span class="math-container">$2$</span> of <span class="math-container">$x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$</span> i.e<span class="math-container">$\binom{4}{2}$</span></p>
<p>We can distribute <span class="math-container">$7$</span> to <span class="math-container">$1$</span> of <span class="math-container">$x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$</span> i.e<span class="math-container">$\binom{4}{1}$</span> and then among all others .</p>
<p>i.e</p>
<p><span class="math-container">$$\binom{4}{1} \binom{14}{11}$$</span></p>
<p>Therefore, the number of bad combinations equals <span class="math-container">$$\binom{4}{1} \binom{14}{11} - \binom{4}{2}$$</span></p>
<p>Therefore, the solution should be:</p>
<p><span class="math-container">$$1330-\left( \binom{4}{1} \binom{14}{11} - \binom{4}{2}\right)$$</span></p>
<p>However, I am getting a negative value. What am I doing wrong?</p>
<p><strong>EDIT</strong></p>
<p>I am asking for my approach, because if the question is for a larger number of dice and if the sum is higher, then predicting the value of dice will not work.</p>
| jvdhooft | 437,988 | <p>In order for the sum to equal 22, either three dice equal $6$ and one equals $4$, or two dice equal $6$ and two dice equal $5$. The number of valid outcomes thus equals:</p>
<p>$${4 \choose 1} + {4 \choose 2} = 4 + 6 = 10$$</p>
<p>As such, the probability of the four dice having a sum of $22$ equals:</p>
<p>$$\frac{{4 \choose 1} + {4 \choose 2}}{6^4} = \frac{10}{1296} = \frac{5}{648} \approx 0.00772$$</p>
|
2,352,721 | <h2>Question</h2>
<blockquote>
<p>Four fair six-sided dice are rolled. The probability that the sum of the results being <span class="math-container">$22$</span> is <span class="math-container">$$\frac{X}{1296}.$$</span> What is the value of <span class="math-container">$X$</span>?</p>
</blockquote>
<h2>My Approach</h2>
<p>I simplified it to the equation of the form:</p>
<blockquote>
<p><span class="math-container">$x_{1}+x_{2}+x_{3}+x_{4}=22, 1\,\,\leq x_{i} \,\,\leq 6,\,\,1\,\,\leq i \,\,\leq 4 $</span></p>
</blockquote>
<p>Solving this equation results in:</p>
<p><span class="math-container">$x_{1}+x_{2}+x_{3}+x_{4}=22$</span></p>
<p>I removed restriction of <span class="math-container">$x_{i} \geq 1$</span> first as follows-:</p>
<p><span class="math-container">$\Rightarrow x_{1}^{'}+1+x_{2}^{'}+1+x_{3}^{'}+1+x_{4}^{'}+1=22$</span></p>
<p><span class="math-container">$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$</span></p>
<p><span class="math-container">$\Rightarrow \binom{18+4-1}{18}=1330$</span></p>
<p>Now i removed restriction for <span class="math-container">$x_{i} \leq 6$</span> , by calculating the number of <strong>bad cases</strong> and then subtracting it from <span class="math-container">$1330$</span>:</p>
<p>calculating <strong>bad combination</strong> i.e <span class="math-container">$x_{i} \geq 7$</span></p>
<p><span class="math-container">$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$</span></p>
<p>We can distribute <span class="math-container">$7$</span> to <span class="math-container">$2$</span> of <span class="math-container">$x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$</span> i.e<span class="math-container">$\binom{4}{2}$</span></p>
<p>We can distribute <span class="math-container">$7$</span> to <span class="math-container">$1$</span> of <span class="math-container">$x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$</span> i.e<span class="math-container">$\binom{4}{1}$</span> and then among all others .</p>
<p>i.e</p>
<p><span class="math-container">$$\binom{4}{1} \binom{14}{11}$$</span></p>
<p>Therefore, the number of bad combinations equals <span class="math-container">$$\binom{4}{1} \binom{14}{11} - \binom{4}{2}$$</span></p>
<p>Therefore, the solution should be:</p>
<p><span class="math-container">$$1330-\left( \binom{4}{1} \binom{14}{11} - \binom{4}{2}\right)$$</span></p>
<p>However, I am getting a negative value. What am I doing wrong?</p>
<p><strong>EDIT</strong></p>
<p>I am asking for my approach, because if the question is for a larger number of dice and if the sum is higher, then predicting the value of dice will not work.</p>
| richard1941 | 133,895 | <p>Divide the dice into two pairs. The way you can get 22 is by 10 and 12, 11 and 11, and 12 and 10. The ways are 3, 4, and 3, totaling 10. Or you have looked at the dice individually and listed the winning combinations (in lexicographic order to be sure you don't miss anything).</p>
|
4,041,842 | <p>I want to solve for <span class="math-container">$t \in \mathbb{R}, u'(t)=-u(t)\ln \lvert u(t) \rvert$</span>.</p>
<p>I defined two cases: <span class="math-container">$\mathbb{R^*_+}$</span> and <span class="math-container">$\mathbb{R^*_-}$</span>.</p>
<p>For <span class="math-container">$\mathbb{R^*_+}$</span>:</p>
<p><span class="math-container">$$\frac{du}{u}=-\ln(u(t))dt$$</span></p>
<p>And by integrating, it follows that:</p>
<p><span class="math-container">$$\ln\lvert u \rvert = -u\ln(u) + u$$</span></p>
<p>But I'm stuck here, I don't know what to do next.</p>
| Kieran Mullen | 510,314 | <p>The above solution is absolutely correct. A slightly quicker way is to substitute
<span class="math-container">$$
u(t) = e^{f(t)}
$$</span>
producing the differential equation:
<span class="math-container">$$
f'=-f
$$</span>
eventually leading to the same answer.</p>
|
643,918 | <blockquote>
<p>Let $G$ be a group and $a, b \in G$. Show that $(a*b)' = a' * b'$ if and only if $a*b = b*a$.</p>
</blockquote>
<p>While this is simple to see by intuition, I am having a hard time expressing this formally. It seems as if I want to show that $(a*b)' = a' * b'$ strictly implies $a*b = b*a$, but I'm not sure how much I am allowed to tweak with the equations, and every time I've tried to solve this, I've caught myself in assuming $a*b = b*a$, which naturally results in a foul proof. </p>
| Ulrik | 53,012 | <p>Hint: Remember that $(a*b)' = b' * a'$ is an identity that always holds.</p>
|
4,047,601 | <p>I did a question <span class="math-container">$\int_{0}^{1}\frac{1}{x^{\frac{1}{2}}}\,dx$</span>, and evaluating this is divergent integral yes? Then as a general form <span class="math-container">$\int_{0}^{1} \frac{1}{x^p}\,dx$</span>, <span class="math-container">$p \in \mathbb{R}$</span>, what values of <span class="math-container">$p$</span> can give me <span class="math-container">$\int_{0}^{1} \frac{1}{x^p}\,dx = \frac{4}{3}$</span>? This is a easy integral to calculate, make it <span class="math-container">$\int_{0}^{1}x^{-p}dx$</span> and calculate, etc. Then how do I get this to solve <span class="math-container">$p$</span>? I am using the fundamental theorem of calculus and confused here.</p>
| Alan | 175,602 | <p>It's divergent for <span class="math-container">$p\leq 1$</span>. Otherwise by the fundamental theorem of calculus, you get <span class="math-container">$\frac {x^{-p+1}} {-p+1}$</span>. Plug in your 1 and 0 and then set equal to 4/3 should let you solve for p.</p>
|
418,647 | <p>Sorry if the question is dumb. I am trying to learn representation theory of finite groups from J.P.Serre's book by myself. In section 2.6 on canonical decomposition, he says that let V be a representation of a finite group G, $W_1,...,W_h$ be the distinct irreducible representations of G, and let V = $U_1 \oplus ... \oplus U_m$ be some decomposition of V into irreducible subrepresentations. Then we can write V = $V_1\oplus ...\oplus V_h$, where $V_i$ is the direct sum of irreducible subrepresentations among $U_i$'s which are $isomorphic$ to $W_i$. This much is clear. But then he says that :</p>
<blockquote>
<p>Next, if needed, one chooses a decomposition of $V_i$ into a direct sum of irreducible representations each isomorphic to $W_i$: $$V_i = W_i \oplus ...\oplus W_i$$
The last decomposition can be done in infinitely many ways; it is just as arbitrary as the choice of a basis in a vector space.</p>
</blockquote>
<p>I am confused with this part. I understand $external$ direct sums of same spaces, but how is the $internal$ direct sum of same spaces $W_i$ defined in general? I think I might be facing some notational difficulty.
Thanks in advance.</p>
| Douglas S. Stones | 139 | <p>The following graph is a simple 7-vertex graph with an isolated vertex. It contains every possible edge subject to the constraint that it has an isolated vertex.</p>
<p><img src="https://i.stack.imgur.com/V2tk0.png" alt="A 7 vertex graph with an isolated vertex"></p>
<p>Any other 7-vertex graph with an isolated vertex would consequently be a subgraph of this graph, thus is cannot have more edges.</p>
<p>The graph above has $\binom{6}{2}=15 < 17$ edges, so a $17$-edge $7$-vertex simple graph with an isolated vertex cannot exist.</p>
|
3,005,100 | <p>Given the following formula
<span class="math-container">$$
\sum^n_{k=0}\frac{(-1)^k}{k+x}\binom{n}{k}\,.
$$</span>
How can I show that this is equal to
<span class="math-container">$$
\frac{n!}{x(x+1)\cdots(x+n)}\,?
$$</span></p>
| Batominovski | 72,152 | <p>Consider the (unique) polynomial <span class="math-container">$p(x)\in\mathbb{Q}[x]$</span> of degree at most <span class="math-container">$n$</span> such that <span class="math-container">$p(-k)=1$</span> for all <span class="math-container">$k=0,1,2,\ldots,n$</span>. Clearly, <span class="math-container">$p(x)$</span> is the constant polynomial <span class="math-container">$1$</span>. </p>
<p>However, using Lagrange interpolation, we have
<span class="math-container">$$p(x)=\sum_{k=0}^n\,p(-k)\,\frac{\prod\limits_{j\in[n]\setminus\{k\}}\,(x+j)}{\prod\limits_{j\in[n]\setminus\{k\}}\,(-k+j)}\,,$$</span>
where <span class="math-container">$[n]:=\{0,1,2,\ldots,n\}$</span>. This means
<span class="math-container">$$1=\sum_{k=0}^n\,\frac{\prod\limits_{j\in[n]\setminus\{k\}}\,(x+j)}{\prod\limits_{j\in[n]\setminus\{k\}}\,(-k+j)}=\sum_{k=0}^n\,(-1)^k\,\frac{\prod\limits_{j\in[n]\setminus\{k\}}\,(x+j)}{k!\,(n-k)!}\,.$$</span>
Multiplying both sides by <span class="math-container">$\dfrac{n!}{\prod\limits_{j\in[n]}\,(x+j)}$</span> yields
<span class="math-container">$$\frac{n!}{\prod\limits_{j=0}^n\,(x+j)}=\sum_{k=0}^n\,(-1)^k\,\left(\frac{n!}{k!\,(n-k)!}\right)\,\frac{1}{x+k}=\sum_{k=0}^n\,\binom{n}{k}\,\frac{(-1)^k}{x+k}\,.$$</span></p>
|
99,378 | <p>The following equation in $\mathbb{C}$:</p>
<p>$4z^2+8|z|^2-3=0$</p>
<p>is not algebraic and has 4 solutions : $\pm\frac{1}{2}$ and $\pm i\frac{\sqrt{3}}{2}$.
The Solve function in Mathematica only returns the 2 real values :</p>
<pre><code>Solve[4 z^2 + 8 Abs[z]^2 - 3 == 0, Complexes]
(* {{z -> -(1/2)}, {z -> 1/2}} *)
</code></pre>
<p>What am I missing ?</p>
| ubpdqn | 1,997 | <p>Specifying <code>Complexes</code> for <code>Solve</code>or <code>Reduce</code> suffices as does just doing it yourself (as alluded to by Daniel:Lichtblau):</p>
<pre><code>x + I y /.Solve[{4 (x^2 - y^2) + 8 (x^2 + y^2) - 3 == 0, 8 x y == 0}, {x, y}]
</code></pre>
<p>yield:</p>
<pre><code> {-((I Sqrt[3])/2), (I Sqrt[3])/2, -(1/2), 1/2}
</code></pre>
|
99,378 | <p>The following equation in $\mathbb{C}$:</p>
<p>$4z^2+8|z|^2-3=0$</p>
<p>is not algebraic and has 4 solutions : $\pm\frac{1}{2}$ and $\pm i\frac{\sqrt{3}}{2}$.
The Solve function in Mathematica only returns the 2 real values :</p>
<pre><code>Solve[4 z^2 + 8 Abs[z]^2 - 3 == 0, Complexes]
(* {{z -> -(1/2)}, {z -> 1/2}} *)
</code></pre>
<p>What am I missing ?</p>
| murray | 148 | <pre><code> Solve[4 z^2 + 8 z Conjugate[z] - 3 == 0, z]
(* {z -> -1/2}, {z -> 1/2}, {z -> (-I/2)*Sqrt[3]}, {z -> (I/2)*Sqrt[3]}} *)
</code></pre>
|
2,498,359 | <p>This is a basic probability question. </p>
<p>Persons A and B decide to arrive and meet sometime between 7 and 8 pm. Whoever arrives first will wait for ten minutes for the other person. If the other person doesn't turn up inside ten minutes then the person waiting will leave. What is the probability that they will meet? I am assuming uniform distribution for arrival time between 7 pm and 8 pm for both of them. </p>
| Bime | 424,324 | <p>I hope your are doing well, I tried a new method to resolve the meeting problem but I didn't find the same result.</p>
<p>We have to calculate <span class="math-container">$P(|X-Y| < 1/6)$</span> where X and Y are independents uniformly distributed between 0 and 1.</p>
<p>So : <span class="math-container">$P(|X-Y| < 1/6) = \int_{0}^{1} \int_{0}^{1} |x - y| * dx*dy$</span> </p>
<p>Then : <span class="math-container">$P(|X-Y| < 1/6) = \int_{0}^{1} \int_{y - \frac{1}{6}}^{y + \frac{1}{6}} dx*dy$</span> </p>
<p>Finally : <span class="math-container">$P(|X-Y| < 1/6) = \int_{0}^{1} \frac{1}{3} *dy = \frac{1}{3} $</span></p>
<p>Please help me to find my error.</p>
<p>Thank you </p>
|
3,516,921 | <p>Let <span class="math-container">$f : [−1, 0] → \mathbb{R}, x → x − x^2, n ∈ \mathbb{N}$</span> and let <span class="math-container">$P_n : x_0, . . . , x_n$</span> be an equal partition of <span class="math-container">$[−1, 0]$</span>.</p>
<ul>
<li>Compute the Riemann sum <span class="math-container">$S_{P_n} (f, z_1, . . . , z_n)$</span>, when <span class="math-container">$z_k$</span> is the midpoint of <span class="math-container">$[x_{k−1}, x_k]$</span> for every <span class="math-container">$k ∈ {1, . . . , n}$</span>.</li>
</ul>
<p>Im not familiar with the usage of the midpoint and cannot get this to work. All help would be appreciated.</p>
| Paramanand Singh | 72,031 | <p>You should understand the definition of a Riemann sum. It is based on notion of partition.</p>
<p>A <em>partition</em> of a closed interval <span class="math-container">$[a, b] $</span> is a finite set <span class="math-container">$P$</span> of points from the interval <span class="math-container">$[a, b] $</span> such that both end points <span class="math-container">$a, b$</span> are in <span class="math-container">$P$</span>. The elements of a partition are usually written in ascending order. Thus we say that <span class="math-container">$$P=\{x_0,x_1,x_2,\dots,x_n\}$$</span> is a partition of <span class="math-container">$[a, b] $</span> if <span class="math-container">$$a=x_0<x_1<x_2<\dots<x_n=b$$</span> The partition <span class="math-container">$P$</span> is said to be an equipartition or uniform partition if <span class="math-container">$$x_1-x_0=x_2-x_1=\dots=x_n-x_{n-1}$$</span> and clearly each of the above difference equals <span class="math-container">$h=(b-a) /n$</span> and <span class="math-container">$$x_k=x_0+kh=a+k\cdot\frac{b-a}{n}$$</span> Given any partition <span class="math-container">$$P=\{x_0,x_1,x_2,\dots, x_n\} $$</span> of <span class="math-container">$[a, b] $</span> it is usual in theory of Riemann integration to choose another set <span class="math-container">$$T=\{t_1,t_2,\dots,t_n\}$$</span>of points called <em>tags</em> such that <span class="math-container">$$t_k\in[x_{k-1},x_k],k=1,2,\dots,n$$</span> Understand that a choice of tags is always based on a specific partition.</p>
<p>Given a function <span class="math-container">$f:[a, b] \to\mathbb {R} $</span>, partition <span class="math-container">$P=\{x_0,\dots,x_n\}$</span> of <span class="math-container">$[a, b] $</span> and a corresponding set of tags <span class="math-container">$T=\{t_1,\dots,t_n\}$</span> we define a Riemann sum <span class="math-container">$$S(f, P, T) =\sum_{k=1}^{n}f(t_k)(x_k-x_{k-1})$$</span> The notation in your question for above sum is <span class="math-container">$S_{P} (f, t_1,t_2,\dots,t_n)$</span>. </p>
<hr>
<p>For the current question <span class="math-container">$[a, b] =[-1,0],f(x)=x-x^2$</span> and <span class="math-container">$P_n$</span> is an equipartition with <span class="math-container">$$x_k=a+k\cdot\frac {b-a} {n} =\frac{k-n} {n} $$</span> The tags <span class="math-container">$z_k$</span> are mid points of interval <span class="math-container">$[x_{k-1},x_k]$</span> so that <span class="math-container">$$z_k=\frac{x_{k-1}+x_k}{2}=\frac{k-1-n+k-n}{2n}=\frac{2(k-n)-1}{2n}$$</span> and the desired Riemann sum is <span class="math-container">$$S_{P_n} (f, z_1,\dots,z_n)=\sum_{k=1}^{n}f(z_k)(x_k-x_{k-1})$$</span> which equals <span class="math-container">$$\frac{1}{n}\sum_{k=1}^{n}(z_k-z_k^2)$$</span> Now you can substitute the value of <span class="math-container">$z_k$</span> and do a little bit of algebra. </p>
|
3,366,781 | <blockquote>
<p>Let <span class="math-container">$(S, +, \cdot, 0)$</span> and <span class="math-container">$(S', \oplus, \otimes, 0')$</span> be two semirings. Then <span class="math-container">$f: S\rightarrow S'$</span> is said to be a homomorphism if for all <span class="math-container">$a, b\in S,$</span> <span class="math-container">$f(a+b)=f(a)\oplus f(b)$</span>, <span class="math-container">$f(a.b)=f(a)\otimes f(b)$</span> and <span class="math-container">$f(0)=0'.$</span></p>
</blockquote>
<p>Let <span class="math-container">$\Bbb Z$</span> be a set of non negative integers and <span class="math-container">$P(\Bbb Z)$</span> be its power set. Then <span class="math-container">$(\Bbb Z, +, \cdot, 0 )$</span> and <span class="math-container">$ (P(\Bbb Z), \cup, \cap, \emptyset))$</span> are semirings, where the operations on <span class="math-container">$\Bbb Z$</span> are usual addition and multiplication, while the operations on <span class="math-container">$P(Z)$</span> are usual set union and intersection.</p>
<blockquote>
<p>Now, i wish to define a map <span class="math-container">$\phi: \Bbb Z\rightarrow P(\Bbb Z)$</span> such that <span class="math-container">$\phi $</span> is a homomorphism. Is no such homomorphism possible? If possible, how should <span class="math-container">$\phi$</span> be defined?</p>
</blockquote>
<p><strong>Edited:</strong>
Also, see a related question <a href="https://www.google.com/url?sa=t&source=web&rct=j&url=https://mathoverflow.net/questions/342038/define-a-homomorphism-of-a-set-of-graphs-to-its-power-set&ved=2ahUKEwi2tYaYn-fkAhWO63MBHa0DDNQQFjAAegQIBhAB&usg=AOvVaw1ojiTWIdWV636xCohN1NRU" rel="nofollow noreferrer">https://www.google.com/url?sa=t&source=web&rct=j&url=https://mathoverflow.net/questions/342038/define-a-homomorphism-of-a-set-of-graphs-to-its-power-set&ved=2ahUKEwi2tYaYn-fkAhWO63MBHa0DDNQQFjAAegQIBhAB&usg=AOvVaw1ojiTWIdWV636xCohN1NRU</a></p>
| Arthur | 15,500 | <p>I don't know whether this addresses your concern in general, but here is what happens in this specific case.</p>
<p>In this argument we are not applying (the negation of) uniform continuity to a sequence. We're applying it to a whole lot of different numbers separately. These numbers just also happen to form a sequence, and we use that sequence-ness in a <em>different</em> part of the argument to reach a contradiction.</p>
<p>The negation of uniform continuity says that for any <span class="math-container">$n\in \Bbb N$</span>, there are two numbers <span class="math-container">$x_n, x_n'$</span> such that <span class="math-container">$|x_n-x_n'|<\frac1n$</span> but at the same time <span class="math-container">$|f(x_n)-f(x_n')|\geq \varepsilon$</span>. There is no sequence here. Only numbers. There happens to be an <span class="math-container">$n$</span> here which we will use to make a sequence in the <em>next sentence</em> of the proof, but by then we're already done with the negation of uniform continuity. It was only ever used on numbers.</p>
|
3,908,955 | <p>Is the given series convergent or divergent? Give a reason. Show details.</p>
<p><span class="math-container">$$\sum_{n=2}^{\infty} \frac{(-i)^n}{ln \ n}$$</span></p>
<p>So maybe I'll try using the ratio test?</p>
<p>So the series converges if <span class="math-container">$$\left| \frac{z_{n+1}}{z_n} \right| < 1$$</span></p>
<p>So I have that <span class="math-container">$$z_n = \frac{(-i)^{n+1}}{ln \ (n + 1)} \cdot \frac{ln \ n}{(-i)^n}$$</span></p>
<p>So I think the lns cancel right so all we're left with is an -i in the numerator? Is that right? Is that less than 1 so does it converge?</p>
| Derek Luna | 567,882 | <p>It really is as simple as you were making it, although there are mistakes and it is badly written. You correctly realized that it is important to bound one of <span class="math-container">$|x-2|, |x+3|$</span>, namely <span class="math-container">$|x-2|$</span>.</p>
<p>It is standard to start with <span class="math-container">$\delta := 1$</span> as this gives us more information to work with for small <span class="math-container">$\epsilon$</span>, in particular when we have <span class="math-container">$0 < \epsilon < 1$</span>.</p>
<p>Remember that we are considering the limit as <span class="math-container">$x$</span> approaches <span class="math-container">$2$</span>, so we need to first look at how to restrict <span class="math-container">$x$</span>, sometimes you must do this and in this case in particular, because we need to add <span class="math-container">$5$</span> for <span class="math-container">$|x+3|$</span> while still leaving it bounded in some way (otherwise <span class="math-container">$x$</span> could just be negative number where <span class="math-container">$|x|$</span> is very large).</p>
<p>That being said, we can restrict <span class="math-container">$x$</span> around <span class="math-container">$2$</span> such that <span class="math-container">$1 < x < 3$</span> so that <span class="math-container">$|x-2| < 1:\delta_{1} $</span> and <span class="math-container">$4 < x+3 <6 \implies |x+3| < 6$</span>.</p>
<p>Hence, if we choose <span class="math-container">$\delta_{1} := 1$</span>, we see what our other term is bounded by.</p>
<p>Now we can choose <span class="math-container">$\delta_{2}$</span> so that <span class="math-container">$|x-2| < \delta_{2} = \frac{\epsilon}{6}.$</span></p>
<p>Therefore, for some arbitrary <span class="math-container">$\epsilon > 0$</span> if we take <span class="math-container">$\delta:=\min\{\delta_{1},\delta_{2}\}$</span>, then <span class="math-container">$|x^{2}+x-6| =|x-2||x+3| < \frac{\epsilon}{6}\cdot 6 = \epsilon$</span>.</p>
|
1,314,219 | <p>Is there any formula for finding the last digit of the factorials?
How to approach these type of questions?
Thanks in advance.</p>
| alkabary | 96,332 | <p>well as the comments stated, you should observe that for $n \geq 5$ we have $n!$ is a multiple of $5$. For example $$5! = 5 \times 4 \times 3 \times 2 \times 1 = 24 \times 5$$</p>
<p>and $$6 ! = 6 \times 5! = 6 \times 24 \times 5$$</p>
<p>and so on. Now you should know that any multiple of $5$ is either $0$ or $5$ for instance $5,10,15,20,25,.........$ as you can see.</p>
<p>now $$1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33$$</p>
<p>and $5! + 6! + ..........+49!$ is a mutliple of $5$ as we said.</p>
<p>However, $$5! + 6! + ..........+49!$$ is an even number because they are also a multiple of $2$ </p>
<p>And so we have $1! + 2! + 3! + 4!$ is odd </p>
<p>and $5! + 6! + ..........+49!$ is even</p>
<p>and you should know that odd + even = odd. $$(2k + 1) +2m = 2k +2m + 1 = 2(k +m) + 1$$ which is odd.</p>
<p>And so the sum is odd.</p>
<p>and so $5! + 6! + ..........+49!$ ends with $0$ because it is a multiple of $5$ and an even number.</p>
<p>and $33$ + a number that ends with $0$ is will result in a number with last digit = $3$</p>
|
947,770 | <p>Here I have a diophantine equation featuring a homogeneous polynomial:</p>
<p>$$x^2+5y^2+34z^2+2xy-10xz-22yz=0; x, y, z\in\mathbb{Z}$$</p>
<p>I have no idea how to approach this, I've tried various substitutions like $x=py, x=qz$ but then I get a non-homogeneous polynomial of 2 variables which is no better than this. Wolfram|Alpha finds a solution easily, but I still don't understand how.</p>
<p>For reference, W|A claims that the solution is $x=7n, y=3n, z=2n; n\in\mathbb{Z}$</p>
| Sandeep Silwal | 138,892 | <p>Hint: Factor as $$(2y-3z)^2+(x+y-5z)^2 = 0$$</p>
|
3,679,386 | <p>defining matrix exponentiation for natural numbers by repeated multiplication and defining it for <span class="math-container">$\frac{1}{n}$</span> by: <span class="math-container">$A^{\frac{1}{n}}$</span> is the matrix s.t. <span class="math-container">$(A^{\frac{1}{n}})^n=A$</span>.
for a rational number <span class="math-container">$a=\frac{p}{q}$</span> <span class="math-container">$A^{a}=(A^{p})^{\frac{1}{q}}$</span>
how do i prove that <span class="math-container">$A^a=(A^{\frac{1}{q}})^p$</span>?
I'm really unsure how to approach this, can i get some hints to the right direction?</p>
| fleablood | 280,126 | <p>Suppose <span class="math-container">$\sum\limits_{k=1}^{\infty} z_k$</span> converges. That means <span class="math-container">$\{\sum\limits_{k=1}^{n} z_k\}$</span> is a cauchy sequence. So for any <span class="math-container">$\epsilon >0$</span> there is an <span class="math-container">$N$</span> so that if <span class="math-container">$n,m > N$</span> then <span class="math-container">$|\sum\limits_{k=1}^{n} z_k - \sum\limits_{k=1}^{m} z_k|<\epsilon$</span>. In particular for any <span class="math-container">$n > N+1$</span>, <span class="math-container">$|\sum\limits_{k=1}^{n} z_k - \sum\limits_{k=1}^{n-1} z_k|=|z_n| < \epsilon$</span>.</p>
<p>But if it were not true that <span class="math-container">$\lim\limits_{n\to \infty} z_n = 0$</span> then there would be a value <span class="math-container">$\epsilon_{death} > 0$</span> so that for any <span class="math-container">$N$</span> there will be an <span class="math-container">$n > N$</span> where <span class="math-container">$|z_n| > \epsilon_{death}$</span>. That is contradictory to the result of paragraph 1. So it is not possilbe for <span class="math-container">$\sum\limits_{n=1}^{\infty} z_n$</span> and not have <span class="math-container">$\lim\limits_{n\to \infty} z_n = 0$</span>.</p>
|
7,130 | <p>I'm looking for an explanation on how reducing the Hamiltonian cycle problem to the Hamiltonian path's one (to proof that also the latter is NP-complete). I couldn't find any on the web, can someone help me here? (linking a source is also good).</p>
<p>Thank you.</p>
| Aryabhata | 1,102 | <p>Note: The below is a Cook reduction and not a Karp reduction. The modern definitions of NP-Completeness use the Karp reduction.</p>
<p>For a reduction from Hamiltonian Cycle to Path.</p>
<p>Given a graph $G$ of which we need to find Hamiltonian Cycle, for a single edge $e = \{u,v\}$ add new vertices $u'$ and $v'$ such that $u'$ is connected only to $u$ and $v'$ is connected only to $v$ to give a new graph $G_e$.</p>
<p>$G_e$ has a Hamiltonian path if and only if $G$ has a Hamiltonian cycle with the edge $e=\{u,v\}$.</p>
<p>Run the Hamiltonian path algorithm on each $G_e$ for each edge $e \in G$. If all graphs have no Hamiltonian path, then $G$ has no Hamiltonian cycle. If at least one $G_e$ has a Hamiltonian path, then $G$ has a Hamiltonian cycle which contains the edge $e$.</p>
|
258,332 | <blockquote>
<p>Prove that if $a^x=b^y=(ab)^{xy}$, then $x+y=1$.</p>
</blockquote>
<p>How do I use logarithms to approach this problem?</p>
| Hugo | 53,041 | <p>How about $x=y=0$ ? Am I missing something?</p>
|
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