qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
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1,677,035 | <p>I'm new to this website so I apologize in advance if what I'm going to ask isn't meant to be posted here.</p>
<p>A bit of background though: I haven't been to school in 6 years and the last level I've graduated was Grade 7 due to financial problems, as well as my mom frequently being in and out of the hospital. I am now 18 and I wish to go to college as soon as I can, but I need to be caught up on all the math I've missed (I have been studying these past few years with what's available on the internet, but I don't think it's enough).</p>
<p>So my question is, are there any good, easy to understand, high school math books suited for my situation? I learn better with a teacher who can explain the lesson, but since I don't have one I'd prefer books that aren't too difficult, but at the same time provide everything necessary for high school level math and more. I used to be a bright student so I'm sure I can do this on my own with the right material.</p>
<p>Again, if this question isn't meant to be on this site I'd be more than willing to delete it asap! That's all. Thank you for reading. :)</p>
| Soham | 242,402 | <p>Try out this book.I am sure it will help!!</p>
<p><a href="http://www.amazon.in/Challenge-Thrill-Pre-College-Mathematics-B-J-Venkatchala/dp/8122409806" rel="nofollow">http://www.amazon.in/Challenge-Thrill-Pre-College-Mathematics-B-J-Venkatchala/dp/8122409806</a></p>
|
1,677,035 | <p>I'm new to this website so I apologize in advance if what I'm going to ask isn't meant to be posted here.</p>
<p>A bit of background though: I haven't been to school in 6 years and the last level I've graduated was Grade 7 due to financial problems, as well as my mom frequently being in and out of the hospital. I am now 18 and I wish to go to college as soon as I can, but I need to be caught up on all the math I've missed (I have been studying these past few years with what's available on the internet, but I don't think it's enough).</p>
<p>So my question is, are there any good, easy to understand, high school math books suited for my situation? I learn better with a teacher who can explain the lesson, but since I don't have one I'd prefer books that aren't too difficult, but at the same time provide everything necessary for high school level math and more. I used to be a bright student so I'm sure I can do this on my own with the right material.</p>
<p>Again, if this question isn't meant to be on this site I'd be more than willing to delete it asap! That's all. Thank you for reading. :)</p>
| Crocogator | 710,663 | <p>You can start Higher Algebra by Hall and Knight.
Elementary Number Theory by David Burton. (Great for theory, might wanna do Justin Stevens)
Circles by Dimitri would be good for beginners.
Books by Titu Andreescu are really nice for some advanced prep.</p>
<p>These books are great for Olympiad level preparation.</p>
|
4,521,661 | <p>Calls arrive according to Poisson process with parameter <span class="math-container">$$</span>. Lengths of the calls are iid with cdf <span class="math-container">$F_x(x)$</span>. What is the probability distribution of the number of calls in progress at any given time?</p>
<p>I am confused, is the answer then just the pdf of Poisson, that is, <span class="math-container">$P(X=x) = e^{-\lambda}\frac{\lambda^x}{x!}$</span>?</p>
<p>I feel like I am missing something. I assumed the <span class="math-container">$$</span> is the number of the calls and I am not sure how to use this with the cdf of the lengths of the calls.</p>
| Daniel S. | 362,911 | <p>You have an m/g/infinity queue. The distribution of the number of customers in that queue is known to be Poisson</p>
<p>See for instance Kleinrock book</p>
|
2,113,777 | <p>I have the following IVP (Initial value problem, Cauchy-Problem), and I do not know how to solve this.</p>
<p>$$y'=e^{-x}-\frac{y}{x} \qquad \qquad y(1)=2$$</p>
<p>I hope you can help me, cause I really do not know how to start.</p>
<p>Thank you! :)</p>
| projectilemotion | 323,432 | <p>Start by writing:
$$\frac{dy}{dx}+\frac{y}{x}=e^{-x} \tag{1}$$
This is a first order linear ODE. Hence, we can either use an <a href="https://en.wikipedia.org/wiki/Integrating_factor" rel="nofollow noreferrer">integrating factor</a> or use <a href="https://en.wikipedia.org/wiki/Variation_of_parameters#First_order_equation" rel="nofollow noreferrer">Variation of parameters</a> (see "First order equation").</p>
<p>If you want to use an integrating factor, it will be:</p>
<p>$$\mu(x)=e^{\int \frac{1}{x}~dx}=x$$</p>
<p>Multiply equation $(1)$ by $\mu(x)$:</p>
<p>$$x\frac{dy}{dx}+y=\frac{x}{e^x}$$</p>
<p>If you still cannot solve it, feel free to let us know.</p>
|
2,317,625 | <p>How do you compare $6-2\sqrt{3}$ and $3\sqrt{2}-2$? (no calculator)</p>
<p>Look simple but I have tried many ways and fail miserably.
Both are positive, so we cannot find which one is bigger than $0$ and the other smaller than $0$.
Taking the first minus the second in order to see the result positive or negative get me no where (perhaps I am too dense to see through).</p>
| Steven Alexis Gregory | 75,410 | <p>$6-2√3 \approx2(1.732) = 6 - 3.464 = 2.536$
$3√2-2 \approx 3(1.414) - 2 = 4.242 - 2 = 2.242$</p>
<p>This implies that $\dfrac 52$ is between the two quantities:</p>
<p>\begin{align}
\dfrac{49}{4} &> 12 \\
\dfrac 72 &> 2\sqrt 3 \\
\dfrac{12}{2} &> 2\sqrt 3 + \dfrac 52 \\
6 - 2\sqrt 3 &> \dfrac 52
\end{align}</p>
<p>and</p>
<p>\begin{align}
\dfrac{81}{4} &> 18 \\
\dfrac 92 &> 3\sqrt 2 \\
\dfrac 52 &> 3\sqrt 2 - 2
\end{align}</p>
<p>It follows that $6-2√3 > 3\sqrt 2 - 2$.</p>
|
583,030 | <p>I have to show that the following series convergences:</p>
<p>$$\sum_{n=0}^{\infty}(-1)^n \frac{2+(-1)^n}{n+1}$$</p>
<p>I have tried the following:</p>
<ul>
<li>The alternating series test cannot be applied, since $\frac{2+(-1)^n}{n+1}$ is not monotonically decreasing.</li>
<li>I tried splitting up the series in to series $\sum_{n=0}^{\infty}a_n = \sum_{n=0}^{\infty}(-1)^n \frac{2}{n+1}$ and $\sum_{n=0}^{\infty}b_n=\sum_{n=0}^{\infty}(-1)^n \frac{(-1)^n}{n+1}$. I proofed the convergence of the first series using the alternating series test, but then i realized that the second series is divergent.</li>
<li>I also tried using the ratio test: for even $n$ the sequence converges to $\frac{1}{3}$, but for odd $n$ the sequence converges to $3$. Therefore the ratio is also not successful.</li>
</ul>
<p>I ran out of ideas to show the convergence of the series.</p>
<p>Thanks in advance for any help!</p>
| Jean-Claude Arbaut | 43,608 | <p>If you sum two successive terms (for indices $2n-1$ and $2n$), you get</p>
<p>$$\frac{3}{2n+1} - \frac{1}{2n} = \frac{6n-2n-1}{2n(2n+1)}= \frac{4n-1}{2n(2n+1)}$$</p>
<p>And its sum is not convergent, thus your series is not either.</p>
|
41,676 | <p>Really stuck on this one....</p>
<p>$\displaystyle f(x) = \frac{x - \sin{x}}{x^{2}}$ for $x \neq 0$ and $0$ when $x = 0$</p>
<p>Using the definition of the derivative, find $f'(0)$</p>
<p>I know the definition is $$ \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$</p>
<p>The way I did it was to say $$\lim_{h\to 0}\frac{\frac{(x+h)-\sin(x+h)}{(x+h)^2} - \frac{x - \sin x}{x^2}}{h}$$ $$=\lim_{h\to 0}\frac{\frac{1-\cos(x+h)}{(x+h)^2} - 2\left(\frac{(x+h) - \sin(x+h)}{(x+h)^3}\right)}{1}$$ (using L'Hôpitals Rule) which is $$ \frac{1-\cos(x)}{x^2} -\frac{2(x-\sin x)}{x^3}.$$ But then we cant use this to find $f'(0)$ because the denominator it 0!!! </p>
<p>Where am I going wrong?</p>
| Isaac | 72 | <p>You misapplied the definition of the derivative. If you want to find $f'(0)$, you cannot apply the formula for $f$ when $x\neq 0$ in $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$, then substitute $x=0$; you first have to substitute $x=0$ to get $$\lim_{h \to 0} \frac{f(0+h)-f(0)}{h},$$ then apply the definition for $f$, noting that $f(0)=0$ by definition.</p>
|
41,676 | <p>Really stuck on this one....</p>
<p>$\displaystyle f(x) = \frac{x - \sin{x}}{x^{2}}$ for $x \neq 0$ and $0$ when $x = 0$</p>
<p>Using the definition of the derivative, find $f'(0)$</p>
<p>I know the definition is $$ \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$</p>
<p>The way I did it was to say $$\lim_{h\to 0}\frac{\frac{(x+h)-\sin(x+h)}{(x+h)^2} - \frac{x - \sin x}{x^2}}{h}$$ $$=\lim_{h\to 0}\frac{\frac{1-\cos(x+h)}{(x+h)^2} - 2\left(\frac{(x+h) - \sin(x+h)}{(x+h)^3}\right)}{1}$$ (using L'Hôpitals Rule) which is $$ \frac{1-\cos(x)}{x^2} -\frac{2(x-\sin x)}{x^3}.$$ But then we cant use this to find $f'(0)$ because the denominator it 0!!! </p>
<p>Where am I going wrong?</p>
| Community | -1 | <p>So here you have $$f'(0)= \lim_{h \to 0} \frac{f(0+h)-f(0)}{h} = \lim_{h \to 0} \biggl[\frac{h - \sin{h}}{h^{3}}\biggr]$$</p>
<p>Keep applying the L hospitals rule or else use expansion for the $\sin$ function which is given by $$\sin{h} = h -\frac{h^{3}}{3!} + \frac{h^{5}}{5!} - \cdots $$</p>
|
351,846 | <p>The following problem was on a math competition that I participated in at my school about a month ago: </p>
<blockquote>
<p>Prove that the equation $\cos(\sin x)=\sin(\cos x)$ has no real solutions.</p>
</blockquote>
<p>I will outline my proof below. I think it has some holes. My approach to the problem was to say that the following equations must have real solution(s) if the above equation has solution(s):</p>
<p>$$
\cos^2(\sin x)=\sin^2(\cos x)\\
1-\cos^2(\sin x)=1-\sin^2(\cos x)\\
\sin^2(\sin x)=\cos^2(\cos x)\\
\sin(\sin x)=\pm\cos(\cos x)\\
$$</p>
<p>I then proceeded to split into cases and use the identity $\cos t = \sin(\frac{\pi}{2} \pm t\pm y2\pi)$ to get </p>
<p>$$
\sin x=\frac{\pi}{2} \pm \cos x\pm y2\pi\\
$$</p>
<p>and the identity $-\cos t = \sin(-\frac{\pi}{2}\pm t\pm y2\pi) $ to get </p>
<p>$$
\sin x=- \frac{\pi}{2}\pm \cos x \pm y2\pi.\\
$$</p>
<p>where $y$ is any integer. I argued that $y=0$ was the only value of $y$ that made any sense (since the values of sine and cosine remain between $-1$ and $1$). Therefore, the above equations become </p>
<p>$$
\sin x=\frac{\pi}{2} \pm \cos x\implies \sin x \pm \cos x=\frac{\pi}{2}\\
$$</p>
<p>and </p>
<p>$$
\sin x=- \frac{\pi}{2}\pm \cos x\implies \cos x\pm\sin x= \frac{\pi}{2}.\\
$$</p>
<p>Then, by a short optimization argument, I showed that these last two equations have no real solutions.</p>
<p>First, does this proof make sense? Second, if my proof makes sense, then I feel that it was not very elegant nor simple. Is my approach the best, or is there a better (i.e., more elegant, shorter, simpler) proof?</p>
| avm | 133,055 | <p>After spending many hours trying to find the real solutions of this equation, here are the ways of proving that it has no real solutions that I found:</p>
<p>For $\sin(\cos(x))=\cos(\sin(x))$ to be true, both $\cos(x)$ and $\sin(x)$ have to be equal to $\frac{\pi}{4}$ since $\cos(x)$ and $\sin(x)$ take same value in this number. (Note that I'm talking about the terms inside the sine on the left hand and the cosine on the right hand)</p>
<p>The first way to find such $x$ is by setting up the following system of equations:
$$\cos(x)=\frac{\pi}{4}$$
$$\sin(x)=\frac{\pi}{4}$$</p>
<p>We sum both equations to get:
$$\cos(x)+\sin(x)=\frac{\pi}{2}$$
Raising both sides of the equation by the exponent 2 we get:
$$\cos^2(x)+2\cos(x)\sin(x)+\sin^2(x)=\frac{\pi^2}{4}$$
Which can be reduced to
$$1+2\cos(x)\sin(x)=\frac{\pi^2}{4}$$
$$1+\sin(2x)=\frac{\pi^2}{4}$$
Solving for $x$ we get:
$$x=\frac{1}{2}\sin^{-1}(\frac{\pi^2}{4}-1)$$
Which is impossible to perform since the value inside the inverse sine should be between $-1$ and $1$
Yet another way of proving it is by using the identity $\cos(x)=\sin(x+\frac{\pi}{2})$ so that we have:
$$\sin(\cos(x))=\sin(\sin(x)+\frac{\pi}{2})$$
$$\cos^2(x)-2\cos(x)\sin(x)+\sin^2(x)=\frac{\pi^2}{4}$$
And in the end we have:
$$1-\sin(2x)=\frac{\pi^2}{4}$$
Which will lead to the same conclusion as we did before.</p>
<p>And yet another way to prove it is by writting the equations as follows:
$$\sqrt{1-\sin^2(x)}=\sin(x)+\frac{\pi}{2}$$
Which is the same as:
$$\sin^2(x)+\frac{\pi}{2}\sin(x)+\frac{\pi^2-4}{8}=0$$
Performing a change of variable $u=\sin(x)$:
$$u^2+\frac{\pi}{2}u+\frac{\pi^2-4}{8}=0$$
Both roots of the polynomial are complex so, again, there is no such real $x$ that can satisfy the above mentioned equation. </p>
|
1,203,179 | <p>The problem I have is this:</p>
<p>Use suitable linear approximation to find the approximate values for given functions
at the points indicated:</p>
<p>$f(x, y) = xe^{y+x^2}$ at $(2.05, -3.92)$</p>
<p>I know how to do linear approximation with just one variable (take the derivative and such), but with two variables (and later on in the assignment, three variables) I'm a bit lost. Do I take partial derivatives and combine then somehow? Can someone guide me through a problem of this type? </p>
<p>Thank you in advance.</p>
| Bernard | 202,857 | <p>Yes. Denoting the partial derivatives by $f'_x$ and $f'_y$, the formula is:
$$f(x_0+h,y_0+k)=f(x_0,y_0)+f'_x(x_0,y_0)h+f'_y(x_0,y_0)k+o\bigl(\lVert(h,k)\rVert\bigr).$$</p>
|
3,038,965 | <p>Here's the question I'm puzzling over:</p>
<p><span class="math-container">$\textbf{Find the perpendicular distance of the point } (p, q, r) \textbf{ from the plane } \\ax + by + cz = d.$</span></p>
<p>I tried bringing in the idea of a dot product and attempted to get going with solving the problem, but I'm heading nowhere. That is:</p>
<p><span class="math-container">$\text{The direction vector of the normal of the plane } = (a\textbf{i}+b\textbf{j}+c\textbf{k}) \text{, where } \\\textbf{i}, \textbf{ j},\textbf{ k } \text{ are unit vectors.}$</span> </p>
<p>This dotted with the direction vector of the point (position vector, precisely) should equal 0. Am I right? And will this method work even?</p>
| Kemono Chen | 521,015 | <p><span class="math-container">$$\frac{d}{dx}\frac1{1-x}=\frac1{(1-x)^2}.$$</span></p>
|
279,808 | <p>I was working on a way of calculating the square root of a number by the method of x/y → (x+4y)/(x+y) as shown by bobbym at <a href="https://math.stackexchange.com/questions/861509/">https://math.stackexchange.com/questions/861509/</a></p>
<p>I tried to do it via functions on mathematica, everything seems correct. Why am I not seeing 2.5 as the answer? How can I fix it?
<a href="https://i.stack.imgur.com/TXA8S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TXA8S.png" alt="code" /></a></p>
| user1066 | 106 | <p>As pointed out in the <a href="https://math.stackexchange.com/a/861534">linked</a> answer at <em>Mathematica Stack Exchange</em>, this method of calculating square roots is described in <strong>Square Roots From Anywhere</strong> by Terry A. Goodman and John Bernard, <em>Mathematics Teacher</em>, Vol 73, May 1979, p344-345. The JSTOR link is <a href="https://www.jstor.org/stable/27961659" rel="nofollow noreferrer">here</a>.</p>
<p>To calculate <span class="math-container">$\sqrt n $</span>, "starting with two counting numbers <em>a</em> and <em>b</em>":</p>
<p><span class="math-container">$$
\frac{a}{b}\to \frac{a + n b}{a+b}
$$</span></p>
<p>This may be implemented recursively with <a href="https://reference.wolfram.com/language/ref/Sow.html" rel="nofollow noreferrer"><code>Sow</code></a> and <a href="https://reference.wolfram.com/language/ref/Reap.html" rel="nofollow noreferrer"><code>Reap</code></a> (and <a href="https://reference.wolfram.com/language/ref/Slot.html#24637" rel="nofollow noreferrer">#0</a>) as follows, where <em>i</em> is the number of iterations:</p>
<pre><code>fun1[n_,a_,b_,i_]:=(j=i;If[j>0, --j;
#0[Sow[(Numerator[#] + n Denominator[#])/(Numerator[#] + Denominator[#])]]
]&[a/b]//Reap)[[2,1]]
func[81,1, 2,27]//Short
</code></pre>
<p><a href="https://i.stack.imgur.com/YmIAt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YmIAt.png" alt="enter image description here" /></a></p>
<p>Alternatively, with <a href="https://reference.wolfram.com/language/ref/FixedPointList.html" rel="nofollow noreferrer"><code>FixedPointList</code></a>:</p>
<pre><code>fun2[n_,a_,b_,tolerance_]:=FixedPointList[
(Numerator@# + n Denominator@#)/(Numerator@# +Denominator@#)&,
a/b, SameTest -> (Abs[#1 - #2] < tolerance &)]
fun2[81,1,2, 0.1]//Short
</code></pre>
<p><a href="https://i.stack.imgur.com/JmhLt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JmhLt.png" alt="enter image description here" /></a></p>
<p>To append a decimal representation of the final estimate:</p>
<pre><code>fun1d[n_,a_,b_,i_]:=(j=i;If[j>0, --j;
#0[Sow[(Numerator[#] + n Denominator[#])/(Numerator[#] + Denominator[#])]],Sow[N[#]]
]&[a/b]//Reap)[[2,1]]
</code></pre>
<p>Applying the above to the OP example (find <span class="math-container">$\sqrt 4$</span> with <em>a</em> = 1, <em>b</em> = 1):</p>
<pre><code>fun1d[4,1,1,5]
</code></pre>
<p><span class="math-container">$$
\left\{\frac{5}{2},\frac{13}{7},\frac{41}{20},\frac{121}{61},\frac{365}{182},2.00549\right\}
$$</span></p>
<p>As pointed out in the <a href="https://www.jstor.org/stable/27961659" rel="nofollow noreferrer">linked</a> paper:</p>
<p><span class="math-container">$$\frac{a+b n}{a+b} = \frac{\frac{a}{b}+n}{\frac{a}{b}+1}$$</span></p>
<p>and the method may also be (recursively) implemented as follows:</p>
<pre><code>fun3[n_,a_,b_,i_]:=(j=i;If[j>0, --j;#0[Sow[(# + n)/(# + 1)]]]&[a/b]//Reap)[[2,1]]
fun3[81,1,2,5]
</code></pre>
<p><a href="https://i.stack.imgur.com/zu7P1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zu7P1.jpg" alt="enter image description here" /></a></p>
<p>Or, using <a href="https://reference.wolfram.com/language/ref/FixedPointList.html" rel="nofollow noreferrer"><code>FixedPointList</code></a>:</p>
<pre><code>fun4[n_,a_,b_,tolerance_]:=FixedPointList[(# + n)/(# +1)&, a/b,
SameTest -> (Abs[#1 - #2] < tolerance &)]
fun4[81,1,2,0.001]//Short
</code></pre>
<p><a href="https://i.stack.imgur.com/WEJLs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WEJLs.jpg" alt="enter image description here" /></a></p>
|
3,561,664 | <p>I did part of this question but am stuck and don't know how to continue</p>
<p>I let <span class="math-container">$x= 2k +1$</span></p>
<p>Also noticed that <span class="math-container">$x^3+x = x(x^2+1)$</span></p>
<p>therefore
<span class="math-container">$4m+2 = 2k+1((2k+1)^2+1)$</span></p>
<p>I simplified this and ended up with</p>
<p><span class="math-container">$4m+2 = 8k^3+12k^2+8k+2$</span></p>
<p>I don't know how to continue from and prove that <span class="math-container">$x^3+x$</span> has remainder 2 when divided by 4</p>
| Robert Lewis | 67,071 | <p><span class="math-container">$x = 2k + 1; \tag 1$</span></p>
<p><span class="math-container">$x ^3 = 8k^3 + 12k^2 + 6k + 1; \tag 2$</span></p>
<p><span class="math-container">$x ^3 + x = 8k^3 + 12k^2 + 6k + 1 + (2k + 1)$</span>
<span class="math-container">$ = 8k^3 + 12k^2 + 8k + 2 = 4(2k^3 + 3k^2 + 2k) + 2, \tag 3$</span></p>
<p>that is,</p>
<p><span class="math-container">$x^3 + x \equiv 2 \mod 4, \tag 4$</span></p>
<p>which by the <a href="https://en.wikipedia.org/wiki/Euclidean_division" rel="nofollow noreferrer">Euclidean division</a> implies the remainder of <span class="math-container">$x^3 + x$</span> when divided by <span class="math-container">$4$</span> is <span class="math-container">$2$</span>.</p>
|
3,561,664 | <p>I did part of this question but am stuck and don't know how to continue</p>
<p>I let <span class="math-container">$x= 2k +1$</span></p>
<p>Also noticed that <span class="math-container">$x^3+x = x(x^2+1)$</span></p>
<p>therefore
<span class="math-container">$4m+2 = 2k+1((2k+1)^2+1)$</span></p>
<p>I simplified this and ended up with</p>
<p><span class="math-container">$4m+2 = 8k^3+12k^2+8k+2$</span></p>
<p>I don't know how to continue from and prove that <span class="math-container">$x^3+x$</span> has remainder 2 when divided by 4</p>
| J. W. Tanner | 615,567 | <p>If <span class="math-container">$x$</span> is odd, then <span class="math-container">$x\equiv1$</span> or <span class="math-container">$3\pmod4$</span>; in the former case <span class="math-container">$x^3+x\equiv2\pmod4$</span>, and in the latter case <span class="math-container">$x^3+x\equiv30\equiv2\pmod4.$</span></p>
|
246,071 | <p>How do I solve the following equation?</p>
<p>$$x^2 + 10 = 15$$</p>
<p>Here's how I think this should be solved.
\begin{align*}
x^2 + 10 - 10 & = 15 - 10 \\
x^2 & = 15 - 10 \\
x^2 & = 5 \\
x & = \sqrt{5}
\end{align*}
I was thinking that the square root of 5 is iregular repeating 2.23606797749979 number. 2.236 multipled by itself equals 5ish.</p>
<p>I've also seen another equation like this:
\begin{align*}
x^2 & = 4 \\
x^2 + 4 & = 0 \\
(x - 2)(x + 2) & = 0 \\
x & = 2 \text{ or } -2
\end{align*}
So I guess I could near the end of my equation do the following:</p>
<p>$$x^2 + 5 = 0$$</p>
<p>and then go from there?</p>
<p>Is my first attempt at solving correct?</p>
| beauby | 49,048 | <p>When you say "What we want to prove is ...", you're actually wrong, you have to "update" the $n$ to $n+1$ on every fraction, not only the one you add.</p>
|
1,386,683 | <p>I posted early but got a very tough response.</p>
<p>Point $A = 2 + 0i$ and point $B = 2 + i2\sqrt{3}$ find the point $C$ $60$ degrees ($\pm$) such that Triangle $ABC$ is equilateral. </p>
<p>Okay, so I'll begin by converting into polar form:</p>
<p>$A = 2e^{2\pi i}$ and $B = 4e^{\frac{\pi}{3}i}$</p>
<p>$\overline{AB} = \sqrt{13}$</p>
<p>How should I find a point with length $\overline{BC} = \overline{AC} = \sqrt{13}$ and the sufficient angle?</p>
| Community | -1 | <p>Let $z=re^{i\theta}$. Then this equals:</p>
<p>$$\lim_{r\to 0} \frac{(re^{-i\theta})^2}{re^{i\theta}}=\lim_{r\to 0} \frac{r^2e^{-2i\theta}}{re^{i\theta}}=\lim_{r\to 0} r e^{-3i\theta}=0$$</p>
|
4,495,044 | <p>Edit: There is an answer at the bottom by me explaining what is going on in this post.</p>
<p>Define a function <span class="math-container">$f : R \to R$</span> by <span class="math-container">$f(x) = 1$</span> if <span class="math-container">$x = 0$</span> and <span class="math-container">$f(x) = 0$</span> if <span class="math-container">$x \ne 0$</span>. I was attempting to prove that <span class="math-container">$\lim_{x \to 0; x\in R}f(x)$</span> is undefined. The following is my proof.</p>
<p>Proof: Suppose that <span class="math-container">$\lim_{x\to 0; x \in R}f(x)=L$</span>. Then for every <span class="math-container">$\varepsilon > 0$</span> there exists a <span class="math-container">$\delta > 0$</span> such that for all those <span class="math-container">$x \in R$</span> for which <span class="math-container">$|x-0|<\delta$</span> we have that <span class="math-container">$|f(x)-L|<\varepsilon$</span>. But <span class="math-container">$|0| < \delta$</span> and by the Archimedean property we know that for <span class="math-container">$\delta > 0$</span> there exists an integer <span class="math-container">$n>0$</span> such that <span class="math-container">$0<|\frac 1 n| < \delta$</span>. This is a contradiction, as <span class="math-container">$f(0) = 1$</span> and <span class="math-container">$f(\frac 1 n)= 0$</span> and both are less than <span class="math-container">$\delta$</span>.</p>
<p>Is the proof correct?</p>
<p>Edit: Here a limit is defined using adherent points and not limit points. If we were to use limit points then, <span class="math-container">$\lim_{x \to 0; x\in R\setminus \{0\}}f(x)=0$</span>. I have updated the question with correct notation.</p>
<p>Edit 2: Most textbooks define limits using limit points. In which case you we would have that <span class="math-container">$\lim_{x \to 0}f(x)=\lim_{x \to 0; x\in R\setminus \{0\}}f(x)$</span>. We are considering the definition of the limit where limits are defined using adherent points. Where it really matters whether we are considering <span class="math-container">$lim_{x\to 0; x\in R\setminus \{0\}}f(x)$</span> or <span class="math-container">$lim_{x\to 0; x\in R}f(x)$</span>.</p>
<p>Edit 3: This is the definition of convergence of a function at a point in the book Analysis 1 by Terence Tao. Let <span class="math-container">$X$</span> be a
subset of <span class="math-container">$R$</span>, let <span class="math-container">$f : X → R$</span> be a function, let <span class="math-container">$E$</span> be a subset of <span class="math-container">$X$</span>, <span class="math-container">$x_0$</span>
be an adherent point of <span class="math-container">$E$</span>, and let L be a real number. We say that f
converges to <span class="math-container">$L$</span> at <span class="math-container">$x0$</span> in <span class="math-container">$E$</span>, and write <span class="math-container">$\lim_{x \to x_0;x\in E} f(x) = L$</span>, iff <span class="math-container">$f$</span>, after
restricting to <span class="math-container">$E$</span>, is ε-close to <span class="math-container">$L$</span> near <span class="math-container">$x_0$</span> for every <span class="math-container">$\varepsilon > 0$</span>. If <span class="math-container">$f$</span> does not
converge to any number <span class="math-container">$L$</span> at <span class="math-container">$x_0$</span>, we say that <span class="math-container">$f$</span> diverges at <span class="math-container">$x0$</span>, and leave
<span class="math-container">$\lim_{x\to x_0;x\in E} f(x)$</span> undefined.
In other words, we have <span class="math-container">$lim_{x\to x_0;x\in E} f(x) = L$</span> iff for every <span class="math-container">$\varepsilon > 0$</span>,
there exists a <span class="math-container">$\delta > 0$</span> such that <span class="math-container">$|f(x) − L| ≤ ε$</span> for all <span class="math-container">$x \in E$</span> such that
<span class="math-container">$|x − x0| < \delta$</span>.</p>
<p>There are a two other things he defines that are used in this definition. That of <span class="math-container">$\varepsilon$</span> closeness and local <span class="math-container">$\varepsilon$</span> closeness. For those wanting to read those, <a href="https://lms.umb.sk/pluginfile.php/111477/mod_page/content/5/TerenceTao_Analysis.I.Third.Edition.pdf" rel="nofollow noreferrer">Here is the link</a>. It is on page 221.</p>
<p>Edit 4: It might be useless defining limits without limit points. But that is the definition for which I am trying to prove this.</p>
| Seeker | 1,050,393 | <p>Tao defines limits using adherent points and not limit points. My previous comments regarding the use of adherent points and limit points throughout this post are completely wrong as I didn’t understand it too well myself at the time either.</p>
<p>But now that I do, here is what is going on in this post. Tao mentions that this is a general definition(a user also commented saying that there are two competing definitions of limits of functions. Unfortunately the answer that comment was under has been deleted. But I think some users with a lot of reputation can still see deleted answers?). He defines a limit as follows(paraphrasing because he used a few simpler concepts to help understand the definition).</p>
<p>Definition: Let <span class="math-container">$X$</span> be a subset of <span class="math-container">$R$</span>. Let <span class="math-container">$f:X\to R$</span> be a function. Let <span class="math-container">$E$</span> be a subset of <span class="math-container">$X$</span>. Let <span class="math-container">$x_0$</span> be an adherent point of <span class="math-container">$E$</span>. Let <span class="math-container">$L$</span> be a real number. We say that <span class="math-container">$f$</span> converges to <span class="math-container">$L$</span> near <span class="math-container">$x_0$</span> and write <span class="math-container">$$\lim_{x\to x_0;x\in E}f(x)=L$$</span> if for every <span class="math-container">$\varepsilon >0$</span> there exists a <span class="math-container">$\delta >0$</span> such that for all those values of <span class="math-container">$x\in E$</span> for which <span class="math-container">$|x-x_0|<\delta$</span>, we have that <span class="math-container">$|f(x) - L|<\varepsilon$</span>.</p>
<p>Notice the Specification under the limit sign that <span class="math-container">$x\in E$</span>. He then remarks that most authors define a limit using limit points. Which in the above notation of a limit would become <span class="math-container">$$\lim_{x\to x_0;x\in E\setminus \{x_0\}}f(x)=L $$</span></p>
<p>Notice the specification that <span class="math-container">$x\in E\setminus \{x_0\}$</span>.</p>
<p>Now to my post, I originally asked if the following limit diverges. <span class="math-container">$$\lim_{x\to 0;x\in R}f(x) $$</span> Here we are working with adherent points. In the sense that <span class="math-container">$x$</span> can take on the value <span class="math-container">$0$</span>. Notice the specification that <span class="math-container">$x\in R$</span>.</p>
<p>This was the first example after defining the limit of a function given by Tao as to what happens to limits if consider <span class="math-container">$x_0$</span> to be an adherent point(meaning that we consider the limit <span class="math-container">$$\lim_{x\to x_0;x\in E}f(x) $$</span>) or a limit point(meaning that we consider the limit <span class="math-container">$$\lim_{x\to x_0;x\in E\setminus \{x_0\}}f(x) $$</span>). If we consider <span class="math-container">$0$</span> above to be a limit point, meaning that we consider the limit <span class="math-container">$$\lim_{x\to 0;x\in R\setminus \{0\}}f(x)$$</span> Then that limit converges to <span class="math-container">$1$</span>. But if we consider <span class="math-container">$0$</span> to be an adherent point, meaning that we consider the limit <span class="math-container">$$\lim_{x\to 0;x\in R}f(x)$$</span> then the limit is undefined.</p>
<p>As you probably have figured out by now, saying adherent point and limit point when describing whether the value of <span class="math-container">$x\in E$</span> or <span class="math-container">$x\in E\setminus \{x_0\}$</span> is confusing as limit points are also adherent points. Although, once you get it, it’s clear what it means.</p>
<p>At the time I did not understand this, but then eventually I did. I though I should type up an explanation for the whole confusion I caused that day to others and myself.</p>
|
2,879,883 | <p>Suppose that $f$ and $g$ are differentiable functions on $(a,b)$ and suppose that $g'(x)=f'(x)$ for all $x \in (a,b)$. Prove that there is some $c \in \mathbb{R}$ such that $g(x) = f(x)+c$.</p>
<p>So far, I started with this:</p>
<p>Let $h'(x)=f'(x)-g'(x)=0$, then MVT implies $\exists$ c $\in \mathbb{R}$ such that $h'(c) = \frac{h(b)-h(a)}{b-a} =0$. Then $h'(c)=0 \implies h(c)=c$</p>
<p>After this i'm not sure where to go, or if this is correct at all, any hints?
This is also my first post in Latex so sorry if there's any mistakes!</p>
| Lev Bahn | 523,306 | <p>Fixing your proof a little bit, I can extend your proof.</p>
<p>Suppose that your domain of function is (a,b). And let's choose any $\beta\in (a,b)$.</p>
<p>So for any $x\in (a,b)$ $\exists \alpha\in (\beta,x)$ (or $\alpha \in(x,\beta)$ if $x<\beta$) such that $h'(\alpha)=\frac{h(x)-h(\beta)}{x-\beta}=0$.</p>
<p>Thus,</p>
<p>$$\frac{h(x)-h(\beta)}{x-\beta}=0 \implies h(x)-h(\beta)=0\\ \implies f(x)-g(x)=f(\beta)-g(\beta)\\ \implies f(x)=g(x)+f(\beta)-g(\beta)$$</p>
<p>Thus, the $c$, we were finding is $f(\beta)-g(\beta)$.</p>
|
2,879,883 | <p>Suppose that $f$ and $g$ are differentiable functions on $(a,b)$ and suppose that $g'(x)=f'(x)$ for all $x \in (a,b)$. Prove that there is some $c \in \mathbb{R}$ such that $g(x) = f(x)+c$.</p>
<p>So far, I started with this:</p>
<p>Let $h'(x)=f'(x)-g'(x)=0$, then MVT implies $\exists$ c $\in \mathbb{R}$ such that $h'(c) = \frac{h(b)-h(a)}{b-a} =0$. Then $h'(c)=0 \implies h(c)=c$</p>
<p>After this i'm not sure where to go, or if this is correct at all, any hints?
This is also my first post in Latex so sorry if there's any mistakes!</p>
| mfl | 148,513 | <p>You say that </p>
<blockquote>
<p>Let $h'(x)=f'(x)-g'(x)=0$, then MVT implies $\exists$ c $\in
> \mathbb{R}$ such that $h'(c) = \frac{h(b)-h(a)}{b-a} =0$. Then
$h'(c)=0 \implies h(c)=c.$</p>
</blockquote>
<p>This is not correct. If $h'(c)=0$ then you have $h(b)=h(a).$ </p>
<p>But you are in the correct way. Instead of $a,b$ consider $x,y\in [a,b], x\ne y.$ Then you have </p>
<p>$$h'(c) = \frac{h(y)-h(x)}{y-x} =0.$$ So $h(x)=h(y)$ from where you get that $h$ must be constant.</p>
|
2,601,088 | <p>I'm new to the group theory and want to get familar with the theorems in it, so I choose a number $52$
to try making some obseveration on all group that has this rank. Below are my thoughts. I don't know if there is any better way to think of these (i.e., an experienced group theorist would think), and I still have some questions not being solved. So I post this.</p>
<p>Let $G$ be a group and $|G|=52$. Since $52$ in not prime, then there is no theorem guarantee that $G$ must be cyclic. On the other hand, since $\Bbb Z_{52}$ is cyclic, so what we can know is only that "there exists a group $G$ with rank $52$ that is cyclic." Besides, $52$ is not of the form $pq$, where $p,~q$ are prime numbers, so again we can not use such theorem to say that $G$ must be cyclic. (Also, $52$ is not of the form $p^2$, so we're unable to conclude that $G$ is abelian anyway.)</p>
<p>So there remains a question: can $G$ be non-cyclic? I don't know how to answer it.</p>
<p>Next, move on to the obseverations on its subgroup. If $G$ is cyclic (i.e., isomorphic to $\Bbb Z_{52}$), then by a theorem that I don't know its name, $G$ has <em>exactly</em> one nontrival proper subgroup of rank $2$ (also $4,~13,~26$, respectively). By Lagrange Theorem we know that these $4$ subgroup, plus the $\{e\}$ and $G$, are the all $6$ subgroups of $G$. However, since I don't know whether $G$ can be non-cyclic, so I discuss such case below. The prime divisors of $52$ are $2,~13$. By Cauchy theorem, there exists at least one element that has order $2$ and $13$ respectively. So we know that there are two subgroups of $G$ with rank $2$ and $13$ respectively (but maybe $G$ has more subgroup with these ranks). However, how about subgroup with rank $26$ and $4$, etc.? Is their a way to tell whether such subgroup exists?</p>
| Mark Bennet | 2,906 | <p>You can take a direct product of cyclic groups of order $26$ and $2$ - this is abelian, but has no element of order $52$ so can't be cyclic.</p>
<p>Look out on your travels for a structure theorem on abelian groups which tells you how to find all the different ones. If the prime factorisation of the order of $G$ is made up of distinct primes, the only possible abelian group is cyclic. If $G$ is divisible by the square (or higher power) of some prime, there will possible abelian groups which are not cyclic.</p>
|
2,371,108 | <p>Cubic equations of the form $ax^3+bx^2+cx+d$ can be solved in various ways. Some are easy to easy to factor in a pair, for some the roots can be found out by trial-and-error, some are one-of-a-kind, some can be reduced to a quadratic equation. A compilation of all possible ways to solve cubic equations would be very helpful for students and learners. </p>
| Soha Farhin Pine | 331,967 | <h2>One-of-a-kind cubic equations</h2>
<blockquote>
<ol>
<li><blockquote>
<p>$x^3-6x^2+12x-8=0 $</p>
</blockquote></li>
<li><blockquote>
<p>$\implies x^3-2^3-6x^2+12x=0$</p>
</blockquote></li>
<li><blockquote>
<p>$a^3-b^3=(a-b)[(a-b)^2+3ab]\\\therefore x^3-2^3=(x-2)[(x-2)^2+3\times x\times2]$</p>
</blockquote></li>
<li><blockquote>
<p>$6x^2+12x=6x(x-2)$</p>
</blockquote></li>
<li><blockquote>
<p>$\implies (x-2)[(x-2)^2+6x]-6x(x-2)=0$</p>
</blockquote></li>
<li><blockquote>
<p>$\implies(x-2)[(x-2)^2+6x-6x]=0$ </p>
</blockquote></li>
<li><blockquote>
<p>$\implies (x-2)(x-2)^2=0\\ \implies(x-2)^3=0 \\ \implies x-2=0$ </p>
</blockquote></li>
<li><blockquote>
<p>$\implies \boxed{x=2}$</p>
</blockquote></li>
</ol>
</blockquote>
|
22,839 | <p>Is it possible to have the text generated by <code>PlotLabel</code> (or any other function) aligned to the left side of the plot instead of in the center?</p>
| kglr | 125 | <pre><code> Labeled[Plot[Sinc[x], {x, 0, 9}], Style["plot label", "Section"], {{Top, Left}}]
</code></pre>
<p><img src="https://i.stack.imgur.com/3sCtU.png" alt="enter image description here"></p>
<p>Also</p>
<pre><code>Panel[Plot[Sinc[x], {x, 0, 9}], Style["plot label", "Section"], {{Top, Left}},
Appearance -> "Frameless"]
</code></pre>
<p><a href="https://i.stack.imgur.com/CuvQj.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/CuvQj.jpg" alt="enter image description here"></a></p>
|
355,740 | <p>Today in class we learned that for exponential functions $f(x) = b^x$ and their derivatives $f'(x)$, the ratio is always constant for any $x$. For example for $f(x) = 2^x$ and its derivative $f'(x) = 2^x \cdot \ln 2$</p>
<p>$$\begin{array}{c | c | c | c}
x & f(x) & f'(x) & \frac{f'(x)}{f(x)}\\ \hline
-1 & \frac{1}{2} & 0.346 & 0.693 \\
0 & 1 & 0.693 & 0.693 \\
1 & 2 & 1.38 & 0.693\\
2 & 4 & 2.77 &0.693&
\end{array}$$ </p>
<p>So as you can see, the ratio is the same and this is true for all functions of the form $b^x$ and its derivative. So my question is, why is the ratio always constant? Is there some proof or logic behind it or is <em>just like that</em>? Furthermore, what's the use of knowing this?</p>
<p>EDIT:</p>
<p>It seems that I have missed the fairly simple </p>
<p>$$\require{cancel}\frac{f^\prime(x)}{f(x)}=\frac{\cancel{{b^x}}\ln\,b}{\cancel{{b^x}}}=\ln\,b$$</p>
<p>But what's the use knowing and learning this? Will this reduce a step in the future or help solve a much harder problem more easily? </p>
| Hui Yu | 19,811 | <p>Other answers have already clarified the definition of the characteristic of a ring. But I may add one definition that is completely <em>element-free</em>, so hopefully this would make things clearer that characteristics do not depend on the <em>number</em>.</p>
<p>$\mathbb{Z}$ is initial in the category $\operatorname{Ring}$ (this simply means that for any ring $R$, there is a unique ring homomorphism $\mathbb{Z}\xrightarrow{\phi_R} R$), thus a ring $R$ uniquely defines a subring of $\mathbb{Z}$, namely, $\operatorname{ker}(\phi_R)$. </p>
<p>This is a subgroup of $(\mathbb{Z},+)$, hence it is generated by some $d\in\mathbb{Z}$, and we can define $d$ to be the characteristic of $R$.</p>
<p>You can easily show that this definition is equivalent to yours, but in this definition we have not mention any <em>special element</em>, thus characteristics is a property of the rings themselves.</p>
|
20,314 | <p>Hi all.
I'm looking for english books with a good coverage of distribution theory.
I'm a fan of Folland's Real analysis, but it only gives elementary notions on distributions.
Thanks in advance.</p>
| Akhil Mathew | 344 | <p>Many books on PDE or functional analysis (e.g. Taylor's) will have a detailed coverage of distributions.</p>
|
20,314 | <p>Hi all.
I'm looking for english books with a good coverage of distribution theory.
I'm a fan of Folland's Real analysis, but it only gives elementary notions on distributions.
Thanks in advance.</p>
| Tom Leinster | 586 | <p>Friedlander and Joshi's <em>Introduction to the Theory of Distributions</em> is short, elegant and efficient.</p>
|
20,314 | <p>Hi all.
I'm looking for english books with a good coverage of distribution theory.
I'm a fan of Folland's Real analysis, but it only gives elementary notions on distributions.
Thanks in advance.</p>
| Anthony Pulido | 8,479 | <p>I agree with Johannes's comment, but despite this, one book that might fit your criteria is <em>Theory of distributions</em> by M.A. Al-Gwaiz. I haven't looked at it for some months, but it made the following standard texts more accessible:</p>
<ul>
<li>Friedlander and M. Joshi's <em>Introduction to the Theory of Distributions.</em></li>
<li>Hörmander's <em>The Analysis of Linear Partial Differential Operators.</em></li>
</ul>
<p>A book that I haven't looked at thoroughly, but you might find interesting, is <em>Guide to Distribution theory and Fourier transforms</em> by Robert S. Strichartz. I once took a class with the author, whose verbal explanatory style is complete and who is also a clear writer.</p>
|
1,048,668 | <p>Let $f\colon (a,b) \to \mathbb{R}$ a non constant differentiable function. </p>
<p>Is the following statement true:</p>
<p>If $f$ has a local maximum <em>and</em> a local minimum then $f$ also does have an inflection point.</p>
<p>If so, how to prove it, if not, what would be a counterexample?</p>
<p><em>Remark</em></p>
<p>If there are $a_0,b_0 \in [a,b]$ with $a_0 < x_0 < b_0$ such that $f|_{(a_0,x_0)}$ is convex and $f|_{(x_0,b_0)}$ concave or vice versa, then $(x_0,f(x_0))$ is called inflection point of $f$ (Amann Escher Analysis I, p349).</p>
<p>Sometimes <em>stricly</em> concave (convex) is used in the definition. Does this change the theorem?</p>
<p>If the theorem is not true, does it hold if one allows only smooth functions or even more restrictive only polynomial ones (excluding linear functions)?</p>
| hmakholm left over Monica | 14,366 | <p>Consider this:</p>
<p>$$ f(x) = \begin{cases} (x+1)(x+3)-2 & \text{when }x\le -1 \\
2x & \text{when } -1 < x < 1 \\
2-(x-1)(x-3) & \text{when } x \ge 1 \end{cases} $$</p>
<p>(Graphed by <a href="http://www.wolframalpha.com/input/?i=piecewise+[%7b%7b+%28x%2B1%29%28x%2B3%29-2%2C+x%3C-1%7d%2C+%7b2x%2C+-1+%3C%3D+x+%3C%3D+1%7d%2C+%7b+2-%28x-1%29%28x-3%29%2C+x%3E1+%7d%7d]+for+-4%3Cx%3C4" rel="nofollow noreferrer">Wolfram Alpha</a>)</p>
<p>This has local minimum and maximum at $x=-2$ and $x=2$ respectively, but would you consider it to have an inflection point? Different defintions would give different answers.</p>
<p>In the definition you quote, every point in $[-1,1]$ would be an inflection point if you have "convex" and "concave"; but no points would be inflection point if you have "strictly convex/concave".</p>
<hr>
<p>A counterexample to the claim in the non-strict case would be the indefinite integral of the <a href="https://en.wikipedia.org/wiki/Weierstrass_function" rel="nofollow noreferrer">Weierstrass function</a>. It's easy to see that it must have local maxima and minima <em>somewhere</em> (e.g. you can find closed intervals where the endpoints are certainly not maxima, but a continuous function always attains its supremum over a closed interval). On the other hand the Weierstrass function itself is <a href="https://math.stackexchange.com/questions/42326/nowhere-monotonic-continuous-function">nowhere monotone</a>, so its integral cannot be convex or concave on <em>any</em> interval, which again excludes the existence of inflection points.</p>
<p>Even requiring that $f$ is twice differentiable won't do; according to <a href="https://math.stackexchange.com/questions/622076/continuity-rightarrow-intermediate-value-property-why-is-the-opposite-not-tr/622083#622083">this anwswer</a> there exist everywhere differentiable yet nowhere monotone functions, and (with an offset to make sure they cross the $x$ axis) their indefinite integrals would be counterexamples.</p>
|
662,403 | <p>I'm working on a homework assignment concerning convex optimization and I came across a problem involving the convexity of the function and the convexity of the domain of the function.</p>
<p>Consider the function $f : [0,1]^3 \in R$ with the following form
$$
f(x,y,z) = xlnx + ylnz + zlnz + \alpha ( x + y + z - 1)
$$</p>
<p>The gradient</p>
<p>$$
\nabla f(x,y,z) = (lnx + \alpha + 1)\vec{i} + (lny + \alpha + 1)\vec{j} + (lnz + \alpha + 1)\vec{k}
$$</p>
<p>The hessian matrix:</p>
<p>$$
\left[
\begin{array}{ccc}
\frac{1}{x} & 0 & 0 \\
0 & \frac{1}{y} & 0 \\
0 & 0 & \frac{1}{z} \\
\end{array}
\right]
$$</p>
<p>Since for the interior of the function of which is $[0,1]$ does it follow that the epigraphs that are formed would be positive semidefinite since the hessian matrix is positive definite?</p>
<p>And therefore would the domain of the function also be convex since the cartesian product of the epigraphs would result in a convex set?</p>
| Michael Grant | 52,878 | <p>As requested, my comments promoted to an answer, with a little extra commentary.</p>
<p>Here's a basic definition of convexity: a function $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is convex if its domain $\mathop{\textbf{dom}} f$ is a convex set and if
$$f(\alpha x+(1-\alpha)y)\leq \alpha f(x)+(1-\alpha) f(y) \quad \forall x,y\in\mathop{\textbf{dom}} f,~0\leq\alpha\leq 1.$$
So proving the convexity of a function's domain is a fundamental part of proving convexity of the function itself. For instance, the function
$$g(x)=x^2 \qquad \mathop{\textbf{dom}} g = (-\infty,1] \cup [1,\infty)$$
is <em>not</em> convex; $x^2$ is itself the quintissential convex function, but the domain is not convex. On the other hand, the function
$$h(x)=1/x, \qquad \mathop{\textbf{dom}} h=(0,+\infty)$$
<em>is</em> convex, even though $1/x$ is not convex if considered over the entire real line.</p>
<p>In many cases, the domain of a function is trivial; that is, $\mathbb{R}^n$. Artificially restricted domains like $g$ above, or even your example, are frankly not common. More common are domains restricted to the region where the function is well defined, like $-\ln x$, or when considering a convex branch of a function, like $h$ above.</p>
<p>In your case, the domain is almost handed to you. The set $[0,1]^3$ is trivially convex, being the Cartesian product of intervals. But you must also confirm that the function is defined and finite on that interval. In this case, this is true if you use the $0\cdot\ln 0=0$ convention, which is reasonable to do since it's the limiting behavior of $x\ln x$. So the domain of $f$ truly is $[0,1]^3$.</p>
<p>A common practice in convex analysis, by the way, is to define functions on the <em>extended real number line</em> $[-\infty,+\infty]$. Then the domain of a convex function simply becomes the set of points over which the function is finite. For instance:
$$g(x) = \begin{cases} x^2 & x \leq 1 \text{ or } x \geq +1 \\ +\infty & -1<x<+1 \end{cases} \qquad
h(x) = \begin{cases} 1/x & x>0 \\ +\infty & x \leq 0 \end{cases}$$
Your function becomes
$$f(x,y,z)=\begin{cases} x\ln x+(y+z)\ln z+\alpha(x+y+z−1) & (x,y,z)\in[0,1]^3\\ +\infty & \text{otherwise} \end{cases}$$
It takes a bit of care to work with the extended real number line; you need to be careful not to do $\infty-\infty$ or $0\cdot \infty$. The reward is that all of this domain discussion disappears. In fact, a slightly modified secant test for convexity still works: an extended real function $f:\mathbb{R}^n\rightarrow[-\infty,+\infty]$ is convex iff
$$f(\alpha x+(1-\alpha)y) \leq \alpha f(x) + (1-\alpha) f(y) \qquad \forall x,y\in\mathbb{R}^n, ~0<\alpha<1$$
Try this on $h(x)$ above, for instance: it works even if $x$ or $y$ is negative.</p>
|
618,339 | <p>Could somebody please check my solution?</p>
<p>I want to check, whether $\sum\limits_{n=1}^{\infty}\frac{(1+\frac{1}{n})^n}{n^2}$ converges or diverges.</p>
<p>Using the Comparison test:</p>
<p>Let $a_n = \frac{(1+\frac{1}{n})^n}{n^2},~ b_n=\frac{1}{n^2}$</p>
<p>Since $\sum\limits_{n=1}^{\infty}\frac{1}{n^2}$ converges and $\lim\limits_{n \rightarrow \infty} \frac{a_n}{b_n}= \lim\limits_{n \rightarrow \infty} \frac{(1+\frac{1}{n})^n}{n^2} \frac{n^2}{1} = \lim\limits_{n \rightarrow \infty} (1+\frac{1}{n})^n = e$. </p>
<p>Since $0<e<\infty$, $\sum\limits_{n=1}^{\infty}\frac{(1+\frac{1}{n})^n}{n^2}$ converges.</p>
| ncmathsadist | 4,154 | <p>Using the limit comparison test,
$$ {1\over n^2}\left(1 + {1\over n}\right)^n \sim {e\over n^2},$$
your series converges.</p>
|
214,486 | <p><a href="https://i.stack.imgur.com/rZXpG.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/rZXpG.gif" alt="enter image description here"></a></p>
<p>I made it by another software, and met some problems to change it into MMA code.</p>
<pre><code>f[x_] := Graphics[
Line[AnglePath[{90 °, -90 °}[[
1 + Nest[Join[#, {0}, Reverse[1 - #]] &, {0}, x]]]]]];
f /@ Range[5]
</code></pre>
<p>The effect is weird.</p>
<p>It has two affine rules</p>
<p><span class="math-container">$(x,y)\to(0.5x-0.5y,0.5x+0.5y)$</span> and <span class="math-container">$(x,y)\to(-0.5x-0.5y+1,0.5x-0.5y)$</span></p>
<p>for example: </p>
<pre><code>g[{x_, y_}] := Block[
{}, Return[{{0.5 x - 0.5 y, 0.5 x + 0.5 y}, {-0.5 x - 0.5 y + 1,
0.5 x - 0.5 y}}]
]
h[x_] := Flatten[g /@ x] // Partition[#, 2] &
NestList[h, {{0, 0}}, 13] // ListPlot
</code></pre>
<p>gives <a href="https://i.stack.imgur.com/qX8Xq.png" rel="noreferrer"><img src="https://i.stack.imgur.com/qX8Xq.png" alt="enter image description here"></a></p>
<p>So,I know how to plot still picture, But I have no idea about let it animate.</p>
| Vitaliy Kaurov | 13 | <p><a href="https://i.stack.imgur.com/w2kuu.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/w2kuu.gif" alt="enter image description here"></a></p>
<p>A simple way to make Dragon Curve is using <code>AnglePath</code>. Define a function that generates points for the Dragon curve:</p>
<pre><code>dragonPTS[k_]:=AnglePath[{Pi/2,-Pi/2}[[1+Nest[Join[#,{0},Reverse[1-#]]&,{0},k]]]]
</code></pre>
<p><code>k</code> is an integer number of iterations. Try it out:</p>
<pre><code>Graphics[Line[dragonPTS[10]]]
</code></pre>
<p><a href="https://i.stack.imgur.com/DmS8h.png" rel="noreferrer"><img src="https://i.stack.imgur.com/DmS8h.png" alt="enter image description here"></a></p>
<p>Now generate a list of the transitions:</p>
<pre><code>Table[Graphics[Line[dragonPTS[k]]], {k, 1, 10, 1}]
</code></pre>
<p><a href="https://i.stack.imgur.com/jUXB5.png" rel="noreferrer"><img src="https://i.stack.imgur.com/jUXB5.png" alt="enter image description here"></a></p>
<p>or animate:</p>
<pre><code>Manipulate[Table[Graphics[Line[dragonPTS[k]]], {k, 1, n, 1}], {n, 1, 10}]
</code></pre>
<p><a href="https://i.stack.imgur.com/9DcsN.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/9DcsN.gif" alt="enter image description here"></a></p>
<p>To make it a bit cleaner - as in the top animation image - you can try:</p>
<pre><code>Manipulate[
Row[Table[Graphics[Line[dragonPTS[k]],ImageSize->100{1,1}],{k,1,n,1}]],
{n,1,10,1},Paneled->False,AppearanceElements->None]
</code></pre>
<p>Also see numerous interactive apps at Demonstrations Project:</p>
<p><a href="https://demonstrations.wolfram.com/search.html?query=Dragon" rel="noreferrer">https://demonstrations.wolfram.com/search.html?query=Dragon</a></p>
|
3,807,550 | <p>I am stuck in a true/false question. It is</p>
<p>In a finite commutative ring, every prime ideal is maximal.</p>
<p>The answer says it's false.</p>
<p>Well what I can say is (Supposing the answer is right)</p>
<p><span class="math-container">$(1)$</span> The ring can't be Integral domain since finite integral domain is a field.</p>
<p><span class="math-container">$(2)$</span> There can't be unity in the ring since in that case the result would be true.(By the Theorem that if <span class="math-container">$R$</span> is a commutative ring with unity then an ideal <span class="math-container">$I$</span> is prime iff <span class="math-container">$R/I$</span> is Integral Domain)</p>
<p><span class="math-container">$(3)$</span> All the elements are zero divisors since if there is at least one non- zero divisor, there will be a unity and so <span class="math-container">$(2)$</span> would follow.</p>
<p>So at the end, I am in search of a finite commutative with all elements as zero -divisors, having no unity and obviously a prime ideal in it which is not maximal.</p>
<p>What kind of strange looking ring is this (if possible) ? Any hints??</p>
| rschwieb | 29,335 | <p>There is no counterexample, because even if the ring has no identity, the quotients by primes must have identity.</p>
<p><a href="https://math.stackexchange.com/a/2882463/29335">Every nonzero finite ring without zero divisors has a multiplicative identity</a>, so the quotient would in fact be a finite domain with identity, and hence a field.</p>
|
422,799 | <p>Maschke's theorem says that every <em>finite-dimensional</em> representation of a finite group is completely reducible. Is there a simple example of an infinite-dimensional representation of a finite group which is not completely reducible?</p>
<p>EDIT: As mentioned in the answers, there is actually no finite-dimensional caveat in Maschke's theorem. It seems that I just got unlucky, in the that the first couple of references I found included a finite-dimensional assumption.</p>
| Anonymus | 258,276 | <p>Take $Z_2\times Z_2$, with generators A and B. Consider the field K(t), where K is the field of two elements. Let $\phi(A)=\left[ {\begin{array}{cc}
1 & 1 \\ 0 & 1 \ \end{array} } \right]$ and $\phi(B)=\left[ {\begin{array}{cc}
1 & t\\ 0 & 1 \ \end{array} } \right]$, where $t$ is a variable. Then you got a representation of the Klein-group in $GL(2,K(t))$ and it is not completely reducible. </p>
|
4,101,974 | <p>I'm trying to understand what does this matrix operator norm means and what it does to matrix A. <span class="math-container">$${{\left\| A \right\|}_{1,\,\infty }}:={{\max }_{{{\left\| x \right\|}_{\infty }}=1}}{{\left\| Ax \right\|}_{1}}$$</span> Can somebody help with the explanation and maybe an example?</p>
| Paul Frost | 349,785 | <p>Munkres defines a word as finite sequence <span class="math-container">$s = (x_1,\dots,x_n)$</span> of elements <span class="math-container">$x_i \in G_{\alpha_i}$</span> and says that such a sequence represents <span class="math-container">$x \in G$</span> if <span class="math-container">$x = \prod_{i=1}^n x_i = x_1\dots x_n$</span>. As an adhoc notation let us write <span class="math-container">$\prod s = \prod_{i=1}^n x_i$</span>. Then he defines two reduction operations shortening sequences <span class="math-container">$s$</span> of length <span class="math-container">$n$</span> to sequences <span class="math-container">$s'$</span> of length <span class="math-container">$n-1$</span> such that <span class="math-container">$\prod s = \prod s'$</span>. One of these operations is the omission of <span class="math-container">$x_i$</span> if <span class="math-container">$x_i = 1$</span>.</p>
<p>At the beginning he does not say anything about <span class="math-container">$n$</span> and thus one might think that we only deal with sequences of length <span class="math-container">$n \ge 1$</span>. However, after he defined the reduction operations he explicitly makes the convention that the sequence <span class="math-container">$\emptyset$</span> of length <span class="math-container">$0$</span> (i.e. the empty word) is regarded as a reduced word which represents <span class="math-container">$1$</span>. This is common practice; for example in elementary calculus one frequently defines the empty sum as <span class="math-container">$0$</span> and the empty product as <span class="math-container">$1$</span> (i.e. as the neutral elements with respect to additon and multiplication).</p>
<p>Let us come to your question. The empty word always represents <span class="math-container">$1$</span>. But there may be other reduced words <span class="math-container">$s$</span> representing <span class="math-container">$1$</span> and then the representation of <span class="math-container">$1$</span> as a reduced word is not unique. This is what Munkres means. But admittedly the formulation "the representation of <span class="math-container">$1$</span> by the empty word is unique" may be misleading. He should better have said "the representation of <span class="math-container">$1$</span> by a reduced word is unique, and this is the representation by the empty word".</p>
<p>Here is an example. Let <span class="math-container">$G = \mathbb Z \oplus \mathbb Z$</span>. Let <span class="math-container">$G_1$</span> be the subgroup of all pairs <span class="math-container">$(a,0)$</span> and <span class="math-container">$G_2$</span> the subgroup of all pairs <span class="math-container">$(0,b)$</span>. Their intersection only contains the additive identity element <span class="math-container">$(0,0)$</span>. Clearly each element <span class="math-container">$(a,b)$</span> of <span class="math-container">$G$</span> is represented by the word <span class="math-container">$((a,0),(0,b))$</span>. But all words <span class="math-container">$((a,0),(0,b),(-a,0),(0,-b))$</span> with <span class="math-container">$a,b \ne 0$</span> are reduced and represent <span class="math-container">$(0,0)$</span>.</p>
|
4,528,629 | <p>When doing an exercise about linear representations of finite groups I stumbled upon this Isomorphism in the comments of another <a href="https://math.stackexchange.com/questions/308680/basic-identity-of-characters?rq=1">post</a> which I was not aware of.</p>
<p>In this context <span class="math-container">$V$</span> and <span class="math-container">$W$</span> are finite dimensional Vector complex vector spaces.
I was already able to show that the dimension of the two Vector spaces or rather Tensors are the same. As such it should be sufficient to construct an injective or surjective map between them.</p>
<p>However I tried to construct one such as:
<span class="math-container">$$
\phi: (v_1 \oplus w_1) \cdot (v_2\oplus w_2) \in Sym^2(V \oplus W) \mapsto (v_1 \cdot v_2 ) \oplus ((v_1+v_2) \otimes (w_1 + w_2)) \oplus (w_1 \cdot w_2) \in Sym^2(V)\oplus (V\otimes W)\oplus Sym^2(W)
$$</span>
However it seems to me as if this map is not injective nor surjective since <span class="math-container">$\phi((v_1 \oplus w_1) \cdot (v_2\oplus w_2)) = \phi((v_1 \oplus w_2) \cdot (v_2\oplus w_1))$</span>.</p>
<p>Any tips for constructing such a map or should I be trying a different approach?</p>
| Nicolas | 1,015,842 | <p>This follows immediately from the definition (performing the matrix multiplication and employing linearity of the expectation). Let <span class="math-container">$\mathbf{X} \in \mathbb{R}^{m, n} $</span> and
<span class="math-container">$\mathbf{A}\in \mathbb{R}^{d, m}$</span>
<span class="math-container">$$
\begin{align*}
\mathbb{E}[ \mathbf{A}\mathbf{X}]
&= \mathbb{E}\!\left[ \left[ \sum_{k = 1}^{m} a_{i, k}x_{k, j}\right]_{ \substack{1 \leqslant i \leqslant d\\[2pt]
1\leqslant j\leqslant n}}\right]
\\[5pt]
&=
\left[\mathbb{E}\!\left[ \sum_{k = 1}^{m} a_{i, k}x_{k, j}\right] \right]_{ \substack{1 \leqslant i \leqslant d\\[2pt]
1\leqslant j\leqslant n}}
\\[5pt]
&=
\left[ \sum_{k = 1}^{m} a_{i, k}\mathbb{E}[ x_{k, j}] \right]_{ \substack{1 \leqslant i \leqslant d\\[2pt]
1\leqslant j\leqslant n}}
= \mathbf{A}\mathbb{E}[ \mathbf{X}]
.\end{align*}
$$</span>
As stated in the comment, you mimic the computation of the case where <span class="math-container">$m=n=1$</span> and apply it to each component of the matrix resulting from the matrix product.</p>
|
1,392,209 | <blockquote>
<p>Evaluate the limit $$\lim_{x \to 0}\left( \frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$$</p>
</blockquote>
<p>My attempt </p>
<p>So we have $$\frac{1}{x^{2}}-\frac{\cos^{2}x}{\sin^{2}x}$$</p>
<p>$$=\frac{\sin^2 x-x^2\cos^2 x}{x^2\sin^2 x}$$
$$=\frac{x^2}{\sin^2 x}\cdot\frac{\sin x+x\cos x}{x}\cdot\frac{\sin x-x\cos x}{x^3}$$</p>
<p>Then I have $3$ limits to evaluate
$$\lim_{x \to 0}\frac{x^2}{\sin^2 x}=\left(\lim_{x \to o}\frac{x}{\sin x}\right)^2=1^2=1$$</p>
<p>$$\lim_{x \to 0}\frac{\sin x+x\cos x}{x}=\lim_{x \to 0}\left(\frac{\sin x}{x} + \cos x\right)=1+1=2$$</p>
<p>Now I'm having trouble with the last one which is
$$\lim_{x \to 0}\frac{\sin x-x\cos x}{x^3}=?$$</p>
<p>Thanks for any help. </p>
| Nathan Smith | 209,892 | <p>I think this limit would be considerably easier using Taylor Series instead of LHR.</p>
<p>$ \frac{\sin^2(x)-x^2\cos^2(x)}{x^2\sin^2(x)} \approx \frac{(x-\frac{x^3}{6})^2-x^2(1-\frac{x^2}{2})^2}{x^2(x)^2}=\frac{\frac{2x^4}{3}+O(x^5)}{x^4} \rightarrow \frac{2}{3} $</p>
|
2,706,872 | <p>I'm working on my latest linear algebra assignment and one question is as follows: </p>
<p>In $\mathbb R^3$ let <em>R</em> be the reflection over the null space of the matrix </p>
<p><em>A</em> = [4 4 5]</p>
<p>Find the matrix which represents <em>R</em> using standard coordinates. </p>
<p>I am familiar with the fact that the matrix for the orthogonal projection onto a subspace is given by [<em>P</em>] = <em>A</em>(<em>$A^TA$)$A^T$</em>, where <em>A</em> is a matrix with columns that form a basis for the space in question. However, I am not familiar with the concept of a "reflection over the null space". How does this compare with the matrix for the orthogonal projection onto the null space? Is it a similar process?
Thanks for any help.</p>
| trancelocation | 467,003 | <p>First of all, the formula should be $$P = B(B^TB)^{-1}B^T$$ where the columns of $B$ form of a basis of $ker(A)$. </p>
<p>Think geometrically when solving it. Points are to be reflected in a plane which is the kernel of $A$ (see third item):</p>
<ul>
<li>find a basis $v_1, v_2$ in $ker(A)$ and set up $B = (v_1 \, v_2)$</li>
<li>build the projector $P$ onto $ker(A)$ with above formula</li>
<li>geometrically the following happens to a point $x = (x_1 \, x_2 \, x_3)$ while reflecting in the plane $ker(A)$: $x$ is split into two parts - its projection onto the plane and the corresponding orthogonal part of $x$. Then flip the direction of this orthogonal part: $$x = Px + (x - Px) \mapsto Px - (x-Px) \rightarrow x \mapsto Px - (I-P)x = (2P-I)x$$
So, the matrix looked for is
$$2P-I$$</li>
</ul>
|
1,663,113 | <p>I'm having a mind wrenching question that I just cannot answer. It's been a while since I was at the school bench so I wonder if anyone can help me out? :)</p>
<p>We have 10 students with 5 cakes each to be shared amongst each other.
The students can give the cakes out, but they can’t give a piece to a person who gives them a piece (and vice versa). They also can't give more than one piece for the same person. </p>
<p>How can they distribute the cakes so that everyone gets as many cakes as possible? </p>
| Mike Earnest | 177,399 | <p>Place the kids in a circle, numbered 1 to 10. Have the odd numbered kids pass four cakes, one to each of the four closest students on their right (so 1 passes to 2,3,4,5). Have the even numbered students pass five cakes to the five students on their left (so 2 passes to 3,4,5,6,7).</p>
<p>The odd students receive 5 cakes, and the evens receive 4. This is optimal; there are 45 pairs of students, so only 45 cakes can be passed, meaning only 4.5 cakes can be received on average.</p>
|
3,143,670 | <p>I'm not entire sure how to proceed on this question. I believe I am supposed to use a triangle inequality with epsilons and <span class="math-container">$m$</span>, <span class="math-container">$n \geq N$</span> to get <span class="math-container">$N_1$</span> and <span class="math-container">$N_2$</span> before setting it to <span class="math-container">$\max\{N_1, N_2\}$</span>. The absolute value in the question is making this tricky for me. I'm wondering how to go about this. Thanks!</p>
| String | 94,971 | <p>You must show that for <span class="math-container">$m,n\geq N$</span> we have:
<span class="math-container">$$
||x_m-y_m|-|x_n-y_n||\leq\varepsilon
$$</span>
from the triangle inequality applied twice in two different versions we know that:
<span class="math-container">$$
\begin{align}
||x_m-y_m|-|x_n-y_n||
&\leq|x_m-y_m+x_n-y_n|\\
&\leq|x_m-x_n|+|y_m-y_n|
\end{align}
$$</span>
and the last two terms can be made as small as you like by the assumption of <span class="math-container">$\{x_n\},\{y_n\}$</span> both being Cauchy. The two triangle inequalities used here are:</p>
<ol>
<li><span class="math-container">$||x|-|y||\leq|x-y|$</span></li>
<li><span class="math-container">$|x+y|\leq|x|+|y|$</span></li>
</ol>
|
947,254 | <p>The problem is part (b):</p>
<p><b>1.4.7.</b> A pair of dice is cast until either the sum of seven or eigh appears.</p>
<p> <b>(a)</b> Show that the probability of a seven before an eight is 6/11.</p>
<p> <b>(b)</b> Next, this pair of dice is cast until a seven appears twice or until each of a six and eight has appeared at least once. Show that the probability of the six and eight occurring before two sevens is 0.546.</p>
<p>I would like to try to solve this problem using Markov chains, but I'm encountering a dilemma. To calculate the probability, I would need to multiply down the branches that lead to a terminating state, and then sum all of those branches. But I have loops in my diagram, so I'm not sure how to account for the fact that I could remain in a state for an indefinite number of rolls:</p>
<p>[I only drew the branches corresponding to rolling a 6, but there are of course the two other branches (and sub-branches) for rolling a 7 or 8.]</p>
<p><img src="https://i.stack.imgur.com/Cy6S6.jpg" alt="enter image description here"></p>
<p>If that's hard to read, here <a href="https://i.imgur.com/T8BE5ix.jpg" rel="nofollow noreferrer">is a higher resolution</a>. This is my chain of reasoning: We start out in a state of not having a 6, 7, or 8 yet. We could stay here indefinitely. Rolling a 6 takes us to the next state. We could also stay here indefinitely, or roll an 8 and get an accept state. Or we could roll a 7. At that state, we could roll another 7 and get an accept state or roll and 8 or indefinitely roll a 6 (or any other number). All of those probabilities are noted in the transitions.</p>
<p>How do I account for these possibilities?</p>
| Yuval Filmus | 1,277 | <p>We can think of the experiment as follows. At the start, we have a biased three-sided coin that outputs $6,7,8$ with probabilities $5/16,3/8,5/16$; we don't care about the other outcomes, so we can just ignore them. After we see $6$, we don't care about $6$, so the probabilities of $7,8$ are $6/11,5/11$.</p>
<p>Here are the possible runs of the game:</p>
<ol>
<li>$6,8$ or $8,6$: probability $2\cdot 5/16 \cdot 5/11$.</li>
<li>$6,7,8$ or $8,7,6$: probability $2\cdot 5/16 \cdot 6/11 \cdot 5/11$.</li>
<li>$6,7,7$ or $8,7,7$: probability $2\cdot 5/16 \cdot 6/11 \cdot 6/11$.</li>
<li>$7,6,8$ or $7,8,6$: probability $2\cdot 3/8 \cdot 5/16 \cdot 5/11$.</li>
<li>$7,6,7$ or $7,8,7$: probability $2\cdot 3/8 \cdot 5/16 \cdot 6/11$.</li>
<li>$7,7$: probability $3/8\cdot 3/8$.</li>
</ol>
<p>Summing up the relevant cases, the probability that both $6,8$ appear before $7$ appears twice is $4225/7744$.</p>
|
3,689,513 | <p>I've been stuck on one of my homework numbers.
The number precise that the following equation is a non-linear equation of order 1 with x>0.</p>
<p><span class="math-container">$$y' + {{y\ln(y)}\over x}= xy$$</span></p>
<p>So far, I tried 2 different methods to solve them. As suggested by internet (link below): Bernoulli, and as an inexact one. However, I get stuck every time. </p>
<p><a href="https://www.symbolab.com/solver/ordinary-differential-equation-calculator/y'%2B%5Cfrac%7By%5Ccdot%20ln%5Cleft(y%5Cright)%7D%7Bx%7D%20%3D%20xy" rel="nofollow noreferrer">Bernoulli sample</a></p>
<p>**As a small side note, can someone explain to me the following from the link: how, by passing ln(y) to the right side, we get y^1+1 ? </p>
<p>For the Bernoulli equation, when I try to solve it manually, I get in a kind of infinite integral, no matter how many time I integrate.</p>
<p>Am I missing something?</p>
| Jan Eerland | 226,665 | <p>Well, we have the following first-order nonlinear ordinary differential equation:</p>
<p><span class="math-container">$$\text{y}'\left(x\right)+\frac{\text{y}\left(x\right)\ln\left(\text{y}\left(x\right)\right)}{x}=x\text{y}\left(x\right)\tag1$$</span></p>
<p>This can be rewritten in the following form:</p>
<p><span class="math-container">$$x^2\text{y}\left(x\right)-\text{y}\left(x\right)\ln\left(\text{y}\left(x\right)\right)-x\text{y}'\left(x\right)=0\tag2$$</span></p>
<p>Now, let <span class="math-container">$\text{R}\left(x,\text{y}\right)=x^2\text{y}-\text{y}\ln\left(\text{y}\right)$</span> and <span class="math-container">$\text{S}\left(x,\text{y}\right)=-x$</span>. This is not an exact equation, because:</p>
<p><span class="math-container">$$\left(\frac{\partial\text{R}\left(x,\text{y}\right)}{\partial\text{y}}=x^2-\ln\left(\text{y}\right)-1\right)\ne\left(-1=\frac{\partial\text{S}\left(x,\text{y}\right)}{\partial\text{y}}\right)\tag3$$</span></p>
<p>Find an integrating factor <span class="math-container">$\mu(\text{y})$</span> such that <span class="math-container">$\mu(\text{y})\text{R}\left(x,\text{y}\right)+\text{y}'\left(x\right)\text{S}\left(x,\text{y}\right)=0$</span> is exact. This means:</p>
<p><span class="math-container">$$\frac{\partial}{\partial\text{y}}\left(\mu(\text{y})\text{R}\left(x,\text{y}\right)\right)=\frac{\partial}{\partial x}\left(\mu(\text{y})\text{S}\left(x,\text{y}\right)\right)\tag4$$</span></p>
<p>Which gives:</p>
<p><span class="math-container">$$\frac{\text{d}\mu(\text{y})}{\text{d}\text{y}}\left(\text{y}x^2-\text{y}\ln(\text{y})\right)+\mu(\text{y})\left(x^2-\ln(\text{y})-1\right)=-\mu(\text{y})\tag5$$</span></p>
<p>Isolte <span class="math-container">$\mu(\text{y})$</span> to the left-hand side:</p>
<p><span class="math-container">$$\frac{\partial\mu(\text{y})}{\partial\text{y}}\cdot\frac{1}{\mu(\text{y})}=-\frac{1}{\text{y}}\tag6$$</span></p>
<p>Integrate both sides with respect to <span class="math-container">$\text{y}$</span>:</p>
<p><span class="math-container">$$\ln\left|\mu(\text{y})\right|=-\ln\left|\text{y}\right|\space\Longrightarrow\space\mu(\text{y})=\frac{1}{\text{y}}\tag7$$</span></p>
<p>So:</p>
<p><span class="math-container">$$x^2-\ln\left(\text{y}\left(x\right)\right)-\frac{\text{d}\text{y}\left(x\right)}{\text{d}x}\cdot\frac{x}{\text{y}\left(x\right)}=0\tag8$$</span></p>
<blockquote>
<p>I let you finish.</p>
</blockquote>
|
4,086,485 | <blockquote>
<p>We can regard <span class="math-container">$\pi_1(X,x_0)$</span> as the set of basepoint-preserving homotopy classes of maps <span class="math-container">$(S^1,s_0)\rightarrow(X,x_0$</span>). Let <span class="math-container">$[S^1,X]$</span> be the set of homotopy classes of maps <span class="math-container">$S^1\rightarrow X$</span>, with no conditions on basepoints. Thus there is a natural map <span class="math-container">$\Phi:\pi_1 (X,x_0)\rightarrow[S^1,X]$</span> obtained by ignoring basepoints. Show that <span class="math-container">$\Phi$</span> is onto if <span class="math-container">$X$</span> is path-connected, and that <span class="math-container">$\Phi([f])=\Phi([g])$</span> iff <span class="math-container">$[f]$</span> and <span class="math-container">$[g]$</span> are conjugate in <span class="math-container">$\pi_1(X,x_0)$</span>. Hence <span class="math-container">$\Phi$</span> induces a one-to-one correspondence between <span class="math-container">$[S^1,X]$</span> and the set of conjugacy classes in <span class="math-container">$\pi_1(X)$</span>, when <span class="math-container">$X$</span> is path-connected.</p>
</blockquote>
<p>We think of <span class="math-container">$\pi(X, X_0)$</span> as homotopy classes of basepoint preserving maps <span class="math-container">$(S^1, s_0) \rightarrow (X, x_0)$</span>. Recall that a map is basepoint preserving iff <span class="math-container">$f(s_0) = x_0$</span>.</p>
<p>Define <span class="math-container">$[S^1, X]$</span> to be homotopy classes of maps <span class="math-container">$S^1 \rightarrow X$</span> with no condition on basepoints.
The exercise asks us to show that the map <span class="math-container">$\Phi: \pi_1(X, x_0) \rightarrow [S^1, X]$</span> is an onto map if <span class="math-container">$X$</span> is path-connected.</p>
<p>I don't think is true. Consider a wedge of circles. This is path-connected. Now consider all loops based at <span class="math-container">$x_0$</span>, a point on the left circle that is not on the right circle. In the figure, this is in blue. These loops are in <span class="math-container">$\pi(X, x_0)$</span>. Now, on the other hand, consider a loop on the circle on the right hand side. In the figure, this loop is in pink. It seems impossible that any blue loop can be homotped into a pink loop.</p>
<p>Indeed, by reading ahead, we know that the fundamental group of the wedge of circles is <span class="math-container">$W \equiv \mathbb Z * \mathbb Z \simeq \langle a, b\rangle$</span>. All loops on the left circle will be of the form <span class="math-container">$a^n$</span>, while a loop on the right circle will be of the form <span class="math-container">$b^m$</span>. The map <span class="math-container">$\Phi$</span> that produces all the <span class="math-container">$a^n$</span> cannot produce a <span class="math-container">$b^m$</span>, and thus the map <span class="math-container">$\Phi$</span> cannot be surjective.</p>
<p>What am I missing?</p>
<p><a href="https://i.stack.imgur.com/4WUVa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4WUVa.png" alt="enter image description here" /></a></p>
| Paul Frost | 349,785 | <p>Loops based at <span class="math-container">$x_0$</span> <em>can be (freely) homotopic</em> to the pink loop. As an example consider the loop starting at <span class="math-container">$x_0$</span> which travels along the upper half of the left circle until it reaches the intersection of both circles, then travels once along the pink circle and travels back along the upper half of the left circle to <span class="math-container">$x_0$</span>.</p>
|
3,417,001 | <blockquote>
<p>I am wondering if the ring of polynomials <span class="math-container">$F[x]$</span> with coefficients in the field <span class="math-container">$F$</span> is ever isomorphic to <span class="math-container">$\mathbb{Z}$</span> for some field <span class="math-container">$F$</span>. </p>
</blockquote>
<p>For all the fields I've examined, such as <span class="math-container">$F = \mathbb{C}$</span> or <span class="math-container">$F = \mathbb{R}$</span>, it's true that both <span class="math-container">$\mathbb{C}[x]$</span> and <span class="math-container">$\mathbb{R}[x]$</span> cannot be isomorphic to <span class="math-container">$\mathbb{Z}$</span>. Is it true that <span class="math-container">$F[x]$</span> is not isomorphic to <span class="math-container">$\mathbb{Z}$</span> for any field <span class="math-container">$F$</span> ? If so, is there a nice way to prove this in general? </p>
<p>Thanks! </p>
| lhf | 589 | <p>No such fields exist.</p>
<p>If <span class="math-container">$\phi: F[x] \to \mathbb Z$</span> is a ring homomorphism, then so is its restriction to <span class="math-container">$F$</span>. This restriction must be an injection. Therefore, <span class="math-container">$\phi(F)$</span> is a subfield of <span class="math-container">$\mathbb Z$</span>, which has no subfields. </p>
|
2,544,261 | <p>The question:</p>
<blockquote>
<p>Find values of $x$ such that $2^x+3^x-4^x+6^x-9^x=1$, $\forall x \in \mathbb R$.</p>
</blockquote>
<p>Notice the numbers $4$, $6$ and $9$ can be expressed as powers of $2$ and/or $3$. Hence let $a = 2^x$ and $b=3^x$.</p>
<p>\begin{align}
1 & = 2^x+3^x-4^x+6^x-9^x \\
& = 2^x + 3^x - (2^2)^x + (2\cdot3)^x-(3^2)^x\\
& = 2^x + 3^x - (2^x)^2 + 2^x\cdot3^x-(3^x)^2 \\
& = a+b-a^2+ab-b^2 \\
0 & = a^2-ab+b^2-a-b+1
\end{align}</p>
<p>\begin{align}
0 & = a^2-ab+b^2-a-b+1 \\
& = 2a^2-2ab+2b^2-2a-2b+2 \\
& = (a^2-2ab+b^2)+(a^2-2a+1)+(b^2-2b+1) \\
& = (a-b)^2 + (a-1)^2 + (b-1)^2
\end{align}</p>
<p>This is where I am stuck. I am convinced that this factorisation could help solve the question, but I don't know how. Also, once we find values for $x$, we must prove that there are no further values of $x$. Could someone complete the question?</p>
| Servaes | 30,382 | <p>A sum of squares equals zero if and only if each of the squares equals zero. So you get $a=b=1$.</p>
|
3,191,233 | <p>Why is <span class="math-container">$ce^λ=1$</span> equal to <span class="math-container">$c=e^{-λ}$</span>?</p>
| Michael Stachowsky | 337,044 | <p>You are looking for a solution to your differential equation between those two times. In general you are going to have to recompute things every time step. From the theory of ODEs, your state equation is solved thus:</p>
<p><span class="math-container">$$\vec{x}((k+1)T) = \vec{x}[k+1] = e^{AT}\vec{x}(kT) + \int_{kT}^{(k+1)T}e^{A((k+1)T-\tau}d\tau + Bu(kT)$$</span></p>
<p>The key to the entire thing is that your sampling rate, T, is what governs the bounds on the integrals. Here we went from a lower bound of <span class="math-container">$kT$</span> to an upper bound of <span class="math-container">$(k+1)T$</span>. So in your question you go from <span class="math-container">$k\times0.1$</span> to <span class="math-container">$(k+1)\times0.1$</span>. Hence, in your casee, <span class="math-container">$T = 0.1$</span></p>
|
2,113,413 | <p>A man on the top of the tower, standing in the seashore finds that a boat coming towards him makes 10 minutes to change the angle of depression from $30$ to $60$. How soon will the boat reach the seashore.
<img src="https://i.stack.imgur.com/iHWdl.jpg" alt="enter image description here"></p>
<p>My Attempt,
We know, Speed $=\frac {dist. }{time }$
$=\frac {CD}{10}$.</p>
<p>Also, $Tan 60=\frac {AB}{BC}$
$\sqrt {3}=\frac {AB}{BC}$
$AB=\sqrt {3} BC$.</p>
<p>Now, what do I have to do further? </p>
| Vishnu V.S | 397,349 | <p>Let's consider $AB$ to be $h$. Then,
$$
BD=\frac{h}{\tan30^\circ}=h\sqrt{3} \\
BC=\frac{h}{\tan60^\circ}=\frac{h}{\sqrt{3}}\\
CD=BD-BC=\frac{2h}{\sqrt{3}}
$$
If the speed of the boat is $x$ metres per second,
$$
CD=\frac{2h}{\sqrt{3}}=600x
\implies \frac{h}{x\sqrt{3}}=300
$$
Time in which the boat travels distance $BC$ is :
$$
\frac{BC}{x}=\frac{h}{x\sqrt{3}}=300
$$
Hence, the boat takes $300$ seconds or $5$ minutes to reach the shore. In total, it takes $15$ minutes to cover distance $BD$</p>
|
2,797,717 | <p>Prove that exist function $\varphi :\left( {0,\varepsilon } \right) \to \mathbb{R}$ such that
$$\mathop {\lim }\limits_{x \to {0^ + }} \varphi \left( x \right) = 0,\mathop {\lim }\limits_{x \to {0^ + }} \varphi \left( x \right)\ln x = - \infty .$$
I think $\varphi \left( x \right) = \frac{1}{{\ln \left( {\ln \left( {\Gamma \left( x \right)} \right)} \right)}}$ is fine. But i can't check that.
If we replace $ln$ by $f\left( x \right)$ such that $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = - \infty$, can we find $\varphi$?</p>
| Kavi Rama Murthy | 142,385 | <p>Let $f$ be such that $\lim_{x\to0+} f(x)=-\infty$ Let $\phi (x)=\frac 1 {\sqrt |f(x)|}$. Then $\phi (x) \to 0$ and $\phi (x) f(x) \to -\infty$ (because $|\phi (x) f(x)|=\sqrt {|f(x)|}$ and $\phi (x) f(x)$ has the same sign as $f(x))$. This answers the second part and the first part is a special case when $f(x)=\ln (x)$.</p>
|
3,552,555 | <p>Let <span class="math-container">$S$</span> be the set of all column matrices
<span class="math-container">$
\begin{bmatrix}
b_1 \\
b_2 \\
b_3
\end{bmatrix}
$</span>
such that <span class="math-container">$b_1,b_2,b_3 \in \mathbb{R}$</span> and the system of equations (in real variables)
<span class="math-container">$$\begin{align*}
-x+2y+5z &=b_1 \nonumber\\
2x-4y+3z &=b_2 \nonumber\\
x-2y+2z &=b_3
\end{align*}$$</span>
has at least one solution.Then, which of the following system(s)(in real variables) has (have) at least one solution for each
<span class="math-container">$$
\begin{bmatrix}
b_1 \\
b_2 \\
b_3
\end{bmatrix} \in S?
$$</span></p>
<p><strong>A.</strong> <span class="math-container">$x+2y+3z=b_1$</span>, <span class="math-container">$4y+5z=b_2$</span> and <span class="math-container">$x+2y+6z=b_3$</span></p>
<p><strong>B.</strong> <span class="math-container">$x+y+3z=b_1$</span>, <span class="math-container">$5x+2y+6z=b_2$</span> and <span class="math-container">$-2x-y-3z=b_3$</span></p>
<p><strong>C.</strong> <span class="math-container">$-x+2y-5z=b_1$</span>, <span class="math-container">$2x-4y+10z=b_2$</span> and <span class="math-container">$x-2y+5z=b_3$</span></p>
<p><strong>D.</strong> <span class="math-container">$x+2y+5z=b_1$</span>, <span class="math-container">$2x+3z=b_2$</span> and <span class="math-container">$x+4y-5z=b_3$</span></p>
<p>Can anyone please help me with this problem? I am really clueless how to proceed.</p>
| nonuser | 463,553 | <p>So <span class="math-container">$$\frac a{b+\sqrt c}+\frac d{\sqrt c} = q,\;\;\;\;\; q\in \mathbb Q$$</span></p>
<p>Let <span class="math-container">$x=\sqrt{c}\notin \mathbb Q$</span>, then <span class="math-container">$$\frac a{b+x}+\frac d{x} = q\;\;\;\;/\cdot x(b+x)$$</span></p>
<p><span class="math-container">$$ax+bd+dx= cq+bxq$$</span></p>
<p>so <span class="math-container">$$ \boxed{x(a+d-bq) = cq-bd}$$</span></p>
<p>Now if <span class="math-container">$a+d\ne bq$</span> then <span class="math-container">$$ x = {cq-bd\over a+d-bq}$$</span> which means that <span class="math-container">$x$</span> is rational. A contradiction. So <span class="math-container">$a+d=bq$</span> and then also <span class="math-container">$cq=bd$</span>. Now the conclusion.</p>
|
3,800,521 | <p>Let <span class="math-container">$x=\tan y$</span>, then
<span class="math-container">$$
\begin{align*}\sin^{-1} (\sin 2y )+\tan^{-1} \tan 2y
&=4y\\
&=4\tan^{-1} (-10)\\\end{align*}$$</span></p>
<p>Given answer is <span class="math-container">$0$</span></p>
<p>What’s wrong here?</p>
| lab bhattacharjee | 33,337 | <p>From <a href="https://math.stackexchange.com/questions/523625/showing-arctan-frac23-frac12-arctan-frac125/523626#523626">showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$</a>
<span class="math-container">$$2\arctan x=\begin{cases}\arctan\dfrac{2x}{1-x^2}\text{ if } x^2\le1\\\pi+\arctan\dfrac{2x}{1-x^2}\text{ if } x^2>1\end{cases}$$</span></p>
<p>Again, from <a href="https://math.stackexchange.com/questions/672575/proof-for-the-formula-of-sum-of-arcsine-functions-arcsin-x-arcsin-y">Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $</a></p>
<p><span class="math-container">$$\sin^{-1}(\sin2y)=\begin{cases}2y\text{ if } -\dfrac\pi2\le2y\le\dfrac\pi2\\\pi-2y\text{ if } 2y>\dfrac\pi2\iff \tan y>1\\ -\pi-2y\text{ if } 2y<-\dfrac\pi2\iff \tan y<-1\end{cases}$$</span></p>
|
345,310 | <p>This is computed based on the following recursive formula <span class="math-container">$$w_n=\frac{\lambda_nw_{n+1}+\mu_nw_{n-1}+1}{\lambda_n+\mu_n}$$</span> where: <span class="math-container">$n$</span> is the inital state, State <span class="math-container">$0$</span> is absorbing, <span class="math-container">$\lambda_n$</span> and <span class="math-container">$\mu_n$</span> are the up and down rates respectively and <span class="math-container">$$\sum_{n=0}^\infty\prod_{j=1}^n\frac{\mu_j}{\lambda_j}$$</span>diverges (to make extinction certain). To get the recursion started, we need <span class="math-container">$w_0=0$</span> and <span class="math-container">$$w_1=\frac{1}{\mu_1}\sum_{n=0}^\infty\prod_{j=1}^n\frac{\lambda_j}{\mu_{j+1}}$$</span>The derivation of the last formula can be found in S. Karlin's classic book "A first course in stochastic processes". The last step of his proof requires showing that <span class="math-container">$$\lim_{n\to\infty}\prod_{j=1}^n\frac{\lambda_j}{\mu_j}(w_n-w_{n+1)}=0$$</span>To prove that is, according to Karlin, "more involved but still possible" (but he does not do it). How does one prove that the last limit must equal to <span class="math-container">$0$</span>?</p>
| Iosif Pinelis | 36,721 | <p>Clearly, the expected time <span class="math-container">$w_n$</span> till extinction from the initial state <span class="math-container">$n$</span> is nondecreasing in <span class="math-container">$n$</span>. So, if <span class="math-container">$w_1=\infty$</span>, then <span class="math-container">$w_n=\infty$</span> for all natural <span class="math-container">$n$</span>, so that the difference <span class="math-container">$w_{n+1}-w_n$</span> makes no sense, and hence
the desired conclusion
<span class="math-container">\begin{equation}
\lim_{n\to\infty}\prod_{j=1}^n\frac{\lambda_j}{\mu_j}(w_n-w_{n+1)}=0 \tag{1}
\end{equation}</span>
makes no sense either. </p>
<p>It remains to consider the case <span class="math-container">$w_1<\infty$</span>. Then the equation
<span class="math-container">\begin{equation}
w_n=\frac{\lambda_nw_{n+1}+\mu_nw_{n-1}+1}{\lambda_n+\mu_n} \tag{2}
\end{equation}</span>
(together with the condition <span class="math-container">$w_0=0$</span>)
implies that <span class="math-container">$w_n<\infty$</span> for all natural <span class="math-container">$n$</span>.
For natural <span class="math-container">$j$</span>, let then
<span class="math-container">$$h_j:=w_j-w_{j-1},\quad\pi_j:=\prod_{i=1}^{j-1}\frac{\lambda_i}{\mu_i}
$$</span>
(so that <span class="math-container">$\pi_1=1$</span>), and
<span class="math-container">$$u_j:=\pi_j h_j.$$</span>
Then (2) can be rewritten as
<span class="math-container">$$u_{n+1}=u_n-\frac{\pi_n}{\mu_n},
$$</span>
which implies
<span class="math-container">\begin{equation}
u_n=u_1-\sum_{j=1}^{n-1}\frac{\pi_j}{\mu_j}. \tag{3}
\end{equation}</span></p>
<p>Also, the equality
<span class="math-container">$$w_1=\frac{1}{\mu_1}\sum_{n=0}^\infty\prod_{j=1}^n\frac{\lambda_j}{\mu_{j+1}}$$</span>
can be rewritten as
<span class="math-container">$$u_1[=h_1=w_1]=\sum_{j=1}^\infty\frac{\pi_j}{\mu_j}.
$$</span>
This and (3) imply
<span class="math-container">$$\pi_n h_n=u_n=\sum_{j=n}^\infty\frac{\pi_j}{\mu_j}.
$$</span>
Now the case condition <span class="math-container">$w_1<\infty$</span> implies <span class="math-container">$\pi_n h_n\underset{n\to\infty}\longrightarrow0$</span>, which means that the desired conclusion (1) holds. </p>
|
2,138,009 | <p>Let $f(z)=(1+i)z+1$. Then $f(z)=\sqrt 2 e^{i\pi/4}z+1$ and thus $f=t\circ h\circ r$ where $t$ is the translation of vector $1$, $r$ the rotation of center $0$ and angle $\pi/4$ and $h$ the homothetic of parameter $\sqrt 2$. I found a fix point $z=\frac{-1}{2}+\frac{i}{2}$.</p>
<p>1) What if $f\circ f\circ f\circ f\circ f$ ? It looks to be an translation in the direction of $i$ and an homothetic of parameter 4, but is there an easy way to prove it (with out calculation) ?</p>
<p>2) Let $z_0=i$, $z_1=\sqrt 3+2i$ and $z_2=\sqrt 3-2i$, and consider the triangle $T=\Delta (z_0,z_1, z_3)$. Let $g_n=\underbrace{f\circ ...\circ f}_{n\ times}$. What is the smallest $n$ s.t. $$Area(g_n(T))\geq 100 Area(T) \ \ ?$$</p>
<p>First, $Area(T)= 2\sqrt 3$. I think that $g_n$ is the same triangle with length of side bigger of $n\sqrt 2$, and thus, I would says that $$Area(g_n(T))=\frac{n4\sqrt 2\cdot n\sqrt 2\sqrt 3}{2}=4\sqrt 3n^2.$$
Therefore, $$4\sqrt 3n^2\geq 200\sqrt 3\implies n^2\geq 50\implies n\geq 8.$$
We conclude that the smallest $n$ is $n=8$. Is it correct ? </p>
| lhf | 589 | <p><em>Hint:</em> If $f(z)=az+b$, with $a\ne1$, then $f^n(z)=a^n z + b\dfrac{a^n-1}{a-1}$.</p>
|
329,600 | <p>Let $U,V,W$ be vector spaces over $F$ and $S: U \to V$, and $T: V \to W$ linear maps.</p>
<p>(a) Show that if $S$ and $T$ are isomorphisms, then $T\circ S$ is an isomorphism, too.</p>
<p>(b) Show that if $U$ is isomorphic to $V$ and $V$ is isomorphic to $W$, then $U$ is isomorphic to $W$.</p>
| xavierm02 | 10,385 | <p>$a)$</p>
<p>$S\in GL(U,V), T\in GL(V,W)$</p>
<p>Since $GL(U,V) \subset L(U,V)$ and $GL(V,W) \subset L(V,W)$, $S\in L(U,V) $ and $t\in L(V,W)$ from which you can deduce $T\circ S \in L(U,W)$.</p>
<p>Then you must know that given two vector spaces $A$ and $B$, $\forall f \in L(A,B), [f\in GL(A,B) \Leftrightarrow f \text{ bijective}\Leftrightarrow f \text{ injective and surjective} \Leftrightarrow Ker(f)=\{0_A\} \text{ and } Im(f)=B]$</p>
<p>$\forall x \in U ,[(T\circ S)(x)=0_W \Leftrightarrow S(x)=T^{-1}(0_W) \Leftrightarrow S(x)=0_V \Leftrightarrow x=S^{-1}(0_V)\Leftrightarrow x = 0_U]$ which means that $\boxed{Ker(T\circ S) = \{0_U\}}$</p>
<p>Let $z\in W$</p>
<p>Since $T\in GL(V,W), \exists y \in V, T(y)=z $</p>
<p>Since $S\in GL(U,V), \exists x \in U, S(x)=y $</p>
<p>So $\forall z \in W, \exists x\in U, T(S(x))=z$ which means $W \subset Im(T\circ S)$</p>
<p>Since we already know that $Im(T\circ S)\subset W$,</p>
<p>$\boxed{Im(T\circ S)= W}$</p>
<p>So we get that $T\circ S\in GL(U,W)$</p>
<hr>
<p>$b)$</p>
<p>Since $U$ and $V$ are isomorphic, $GL(U,V)\not = \emptyset$</p>
<p>Since $V$ and $W$ are isomorphic, $GL(V,W)\not = \emptyset$</p>
<p>So you can take $S\in GL(U,V)$ and $T\in GL(V,W)$</p>
<p>By $a)$, you get that $T\circ S \in GL(U,W)$</p>
<p>So $GL(U,W)\not = \emptyset$ so $U$ and $W$ are isomorphic.</p>
|
1,178,361 | <p>The surface with equation $z = x^{3} + xy^{2} $ intersects the plane with equation $2x-2y = 1$ in a curve. What is the slope of that curve at $x=1$ and $ y = \frac{1}{2} $</p>
<p>So I put $ x^{3} + xy^{2} = 2x - 2y - 1 $</p>
<p>We have $ x^{3} + xy^{2} - 2x + 2y + 1 $</p>
<p>Do I then differentiate wrt x and y simultaneously?</p>
<p>I know how to differentiate at a point with directional derivatives. But how do I go about the above question considering the fact that direction isn't mentioned.</p>
<p>Maybe I'm going completely down the wrong route... any help is hugely appreciated !</p>
| kobe | 190,421 | <p>The statement is true, because under the given assumptions, $f(f^{-1}(A)) = A$ for all $A \subseteq Y$. Indeed, let $A \subseteq Y$. If $y \in A$, then $y = f(x)$ for some $x \in X$ (since $f$ is onto), which implies $x \in f^{-1}(A)$ and hence $y \in f(f^{-1}(A))$. Conversely, if $y \in f(f^{-1}(A))$, then $y = f(z)$ for some $z \in f^{-1}(A)$. So $f(z) \in A$, i.e., $y \in A$. Therefore $y \in A$. </p>
|
1,282,489 | <p>I have a simple problem that I need to solve. Given a height (in blue), and an angle (eg: 60-degrees), I need to determine the length of the line in red, based on where the green line ends. The green line comes from the top of the blue line and is always 90-degrees.</p>
<p>The height of the blue line is variable.
The angle of the blue line is variable.</p>
<p>Also, I do not know the length of the green dashed-line. Is there a way to figure out the length of the red line without knowing the length of the green?</p>
<p>-Adam</p>
<p>Any help would be much appreciated!</p>
<p>-Adam </p>
<p><img src="https://i.stack.imgur.com/3ccLl.png" alt="enter image description here"></p>
| Rob | 241,939 | <p>Since you're dealing with right triangle, you can just use cosine function:</p>
<p><span class="math-container">$$\cos(\theta)=\frac{blue}{red}$$</span></p>
<p>Or, substituting <span class="math-container">$60^o$</span> for <span class="math-container">$\theta$</span> and 10 for <em>blue</em>:</p>
<p><span class="math-container">$$\cos(60^o)=\frac{10}{red}$$</span></p>
<p>Thus, solving for the length of the red side:</p>
<p><span class="math-container">$$red=\frac{blue}{\cos(\theta)}$$</span></p>
<p>Or</p>
<p><span class="math-container">$$red=\frac{10}{\cos(60^o)} = 20$$</span></p>
|
427,835 | <p>Which website/journal/magazine would you recommend to keep up with advances in applied mathematics?
More specifically my interest are:</p>
<ul>
<li>multivariate/spatial interpolation</li>
<li>numerical methods</li>
<li>computational geometry</li>
<li>geostatistics</li>
<li>etc</li>
</ul>
<p>I am looking for a fairly high-level and broad ranging source of info. </p>
| user35959 | 35,959 | <p>For multivariate/spatial interpolation (I'm interested in RBFs and meshfree methods), I see things published in SIAM Journal of Numerical Analysis, Mathematics of Computation (Math. Comp), Foundations of Computational Mathematics (FoCM), Constructive Approximation, and Journal of Approximation Theory.</p>
|
172,131 | <p>Given <span class="math-container">$P$</span>, a polynomial of degree <span class="math-container">$n$</span>, such that <span class="math-container">$P(x) = r^x$</span> for <span class="math-container">$x = 0,1, \ldots, n$</span> and some real number <span class="math-container">$r$</span>, I need to calculate <span class="math-container">$P(n+1)$</span>.</p>
<p>Can this be done without Lagrange interpolation?</p>
| Michael Hardy | 11,667 | <p>Via <a href="http://en.wikipedia.org/wiki/Lagrange_polynomial" rel="nofollow">Lagrange polynomials</a> you can fit a finite number of points exactly, and then plug $n+1$ into the polynomial you get.</p>
<p>I'm not sure that fully answers the question, even though it gives you a way to find the right number in every particular case. That is because there's the further question of whether the values of $P(n+1)$ follow some general pattern resulting from the particular form of the exponential function $x\mapsto r^x$ and the fact that it's $n+1$, the next number in that arithmetic sequence, rather than some other number.</p>
|
2,286,540 | <p>$(x,y)=(x,\sqrt{x})$
$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
$=\sqrt{x^4-5x^2+9}$</p>
<p>$g(x)=x^4-5x^2+9$ , $g'(x)=4x^3-10x=0$ $: x=0, x=+-\sqrt{10}/2$</p>
<p>The question is: Should I put those x-values in g(x) or the orginal graph,$
y=\sqrt{x}$. To me, it's logical to put it into g(x), but in an example it was reverse</p>
| imranfat | 64,546 | <p>An alternative way is the use of derivative. At any point $(a,\sqrt{a})$, the value of the derivative is $\frac{1}{2\sqrt{a}}$ and so its perpendicular value at that point would be $-2\sqrt{a}$ The perpendicular line follows $y-\sqrt{a}=-2\sqrt{a}(x-a)$ and we want this point to pass through $(3,0)$ so substitute this into the equation: $0-\sqrt{a}=-6\sqrt{a}+2a\sqrt{a}$ which simplifies to $5\sqrt{a}=2a\sqrt{a}$. Squaring and solving leads to $a=5/2$ from where...</p>
|
4,098,682 | <p>I am trying to prove this following theorem about multiplying left cosets.</p>
<blockquote>
<p>Let <span class="math-container">$H \subset G$</span> a subgroup and <span class="math-container">$G/H$</span> the set of left cosets of <span class="math-container">$H$</span> in <span class="math-container">$G$</span>. We can define a group structure on <span class="math-container">$G/H$</span> by setting <span class="math-container">$aH \cdot bH = abH$</span>, which is well-defined if and only if <span class="math-container">$H$</span> is a normal subgroup of <span class="math-container">$G$</span>.</p>
</blockquote>
<p>Here is my attempt.</p>
<blockquote>
<p>Suppose that <span class="math-container">$H$</span> is normal in <span class="math-container">$G$</span>. Then <span class="math-container">$aHa^{-1} = H$</span> for all <span class="math-container">$a \in G$</span>, i.e., <span class="math-container">$aH = Ha$</span> for all <span class="math-container">$a \in G$</span>. We then have
<span class="math-container">\begin{align*}
aH \cdot bH = (Ha)(bH) = H(ab)H = (ab)HH = (ab)H
\end{align*}</span>
for all <span class="math-container">$a \in G$</span>. Conversely, if <span class="math-container">$H$</span> is not normal, there exists <span class="math-container">$a \in G$</span> and <span class="math-container">$h \in H$</span> such that <span class="math-container">$aha^{-1} \not \in H$</span>. This operation, if it were well-defined, would give
<span class="math-container">\begin{align*}
(aH)(a^{-1} H) = (aa^{-1})H = eH = H.
\end{align*}</span>
Taking <span class="math-container">$ah \in aH$</span> and <span class="math-container">$a^{-1} e = a^{-1} \in a^{-1} H$</span>, we get <span class="math-container">$aha^{-1} \in (aH)(a^{-1} h)$</span> but <span class="math-container">$aha^{-1} \not \in H$</span>, a contradiction.</p>
</blockquote>
<p>I worry I could be waving my hands a bit too much. How does this look? I would appreciate any criticisms.</p>
| Mark Saving | 798,694 | <p>This isn't what they're looking for. Recall that <span class="math-container">$S = \{aH | a \in G\}$</span> is the set of left cosets, where <span class="math-container">$aH := \{ab | b \in H\}$</span>. We are trying to show that the multiplication operator <span class="math-container">$\cdot : S^2 \to S$</span> defined by <span class="math-container">$aH \cdot bH = (ab) \cdot H$</span> is in fact a well-defined function.</p>
<p>What does it mean to be well-defined in this case? It means each input corresponds to exactly one output. The problem is that the same coset <span class="math-container">$s \in S$</span> may be represented both as <span class="math-container">$aH$</span> and as <span class="math-container">$cH$</span> where <span class="math-container">$a \neq c$</span>. Thus, our definition of <span class="math-container">$\cdot$</span> may <em>a priori</em> assign <span class="math-container">$s \cdot g$</span> to different values depending on whether we write <span class="math-container">$s = aH$</span> or <span class="math-container">$s = cH$</span>, even though both are true.</p>
<p>To illustrate this point further, consider the group <span class="math-container">$\mathbb{Z} = G$</span> (under addition) and the subgroup <span class="math-container">$2\mathbb{Z} = H \subseteq G$</span>. Let's say we tried to define a function <span class="math-container">$f : S \to \mathbb{Z}$</span> by <span class="math-container">$f(n \mathbb{Z}) = n$</span>. This would not be a well-defined function because we would have <span class="math-container">$0 = f(0H) = f(2H) = 2$</span>, since <span class="math-container">$0H = 2H$</span>.</p>
<p>What we must show is the following: suppose <span class="math-container">$H$</span> is normal. Then for all <span class="math-container">$a, b, c, d \in G$</span>, if <span class="math-container">$aH = cH$</span> and <span class="math-container">$bH = dH$</span> then <span class="math-container">$(a \cdot b) H = (c \cdot d) H$</span>.</p>
<p>And we must also show that if for all <span class="math-container">$a, b, c, d \in G$</span> such that <span class="math-container">$a H = cH$</span> and <span class="math-container">$bH = dH$</span>, we have <span class="math-container">$(a \cdot b) H = (c \cdot d) H$</span>, then <span class="math-container">$H$</span> is normal.</p>
|
783,502 | <p>Here in my exercise I have to study the function and draw its graph. Can you please tell me what's the best method to do this, because I don't think that's reasonable to use the input output method, it's quite imprecise.
$$f(x)={|x+1|\over x}$$</p>
<p>Thank you!!!</p>
| Caleb Stanford | 68,107 | <p>Here is another way.</p>
<ul>
<li><p>First, draw the graph of $\frac{x + 1}{x}$.
This should be pretty simple since $\frac{x + 1}{x} = 1 + \frac{1}{x}$, and you know what $\frac{1}{x}$ looks like. You also know that $\frac{x+1}{x}$ has a zero at $x = -1$.</p></li>
<li><p>Second, since $|x + 1|$ switches the sign for $x < -1$, take the graph you just drew and reflect the portion with $x < -1$ across the $x$-axis. Keep the portion with $x > -1$ the same as it was before.</p></li>
</ul>
|
1,162,147 | <p>The definition of open set is different in metric space and topological space, though metric space is a special case of topological space. The definition in metric space seems to convey the idea that all the points isolated from outside from outside, while the definition in topological space is intended to separate different points, so I don't know how to link them.</p>
| Ittay Weiss | 30,953 | <p>A metric space has a metric function which can be used to define the notion of open set. Thus, in a metric space the notion of open space is derived from the metric. In a topological space the open sets are not derived from anything, they are given axiomatically. The axioms for a topology are chosen to capture some of the properties of the open sets in a metric space. It is incorrect to say that the definition in a topological space is intended to separate different points, nor that in a metric space the intention is to isolate points. </p>
<p>Metric spaces and topological spaces can actually be seen to coincide if in the definition of metric space you take a more general lattice for the values of the metric than the lattice $[0,\infty ]$. It is then the case that every topological space is metrizable, so the open sets of any topology are derived from a suitable metric as in the usual case. More concretely, the category of all such metric spaces and continuous mappings is equivalent (but not isomorphic) to the category of all topological spaces. That means that the two things are just different models for the same concept of topology. </p>
|
2,404,176 | <p>From the days I started to learn Maths, I've have been taught that </p>
<blockquote>
<p>Adding Odd times Odd numbers the Answer always would be Odd; e.g.,
<span class="math-container">$$3 + 5 + 1 = 9$$</span></p>
</blockquote>
<p>OK, but look at this question </p>
<p><a href="https://i.stack.imgur.com/TmYsJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TmYsJ.jpg" alt="UPSC Question"></a></p>
<p>This question was solved and the answer was 30, how it was possible? Need a valid explanation please.</p>
| cgiovanardi | 230,653 | <p>If this is a riddle, I would do : 13,1 + 7,9 + 9</p>
|
2,404,176 | <p>From the days I started to learn Maths, I've have been taught that </p>
<blockquote>
<p>Adding Odd times Odd numbers the Answer always would be Odd; e.g.,
<span class="math-container">$$3 + 5 + 1 = 9$$</span></p>
</blockquote>
<p>OK, but look at this question </p>
<p><a href="https://i.stack.imgur.com/TmYsJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TmYsJ.jpg" alt="UPSC Question"></a></p>
<p>This question was solved and the answer was 30, how it was possible? Need a valid explanation please.</p>
| Darshan Jain | 620,036 | <p>It's
<span class="math-container">$$3_3+3_3+3_3=30.$$</span>
Where <span class="math-container">$3_3$</span> is read as 3 base 3.</p>
|
1,555,697 | <p>So, I'm working out one of my assignments and I'm a little bit stuck on this problem:</p>
<blockquote>
<p>A fish store is having a sale on guppies, tiger barbs, neons,
swordtails, angelfish, and siamese fighting fish (6 kinds). How many
ways are there to choose 24 fish with at least 1 guppy, at least 2
tiger barbs, at least 3 neons, exactly 1 swordtail, at least two
angelfish, and no more than 3 siamese fighting fish?</p>
</blockquote>
<p>I feel like the way to approach this problem is to use r-combinations with repetition (because the fish store isn't running out of any of the fish) and subtract the combinations that I don't want. Here's what I've got so far:</p>
<p>We need <em>at least</em> 1 guppy, 2 tiger barbs, 3 neons, and 2 angelfish. If we set those aside (all 8 of those), we're down to $24-8$, or 16 remaining fish to choose. Add the 1 swordtail, and we only need to choose $24-9$ or 15 fish out of 5 ($(6-1)$ because there were 6 kinds, but we can only have 1 of the swordtail).</p>
<p>All this accounted for, utilizing r-combination with repetition, $\binom{n+r-1}{r}$:
$$\binom{15+5-1}{15}=3,876$$</p>
<p><strong>BUT</strong> I can't have more than 3 siamese fighting fish! How do I subtract all the groups with more than 3 siamese fighting fish?</p>
<p>Any help would be much appreciated! Also, if I've made any other errors, please point those out and I'll edit the question appropriately. We covered it in class, but it went so quick I didn't get it down in my notes. Thanks!</p>
| mathochist | 215,292 | <p>It seems like you could just sum up all the cases where you have $0$, $1$, $2$, and $3$ siamese fighting fish separately. If there are no siamese fighting fish, then we can choose $15$ out of $4$ types, if there is exactly one siamese fighting fish, then we can choose $14$ fish out of $4$ types, if there are exactly two, then we can choose $13$ fish out of $4$ types, and if there are exactly three then we can choose $12$ fish out of $4$ types. This would give </p>
<p>${15+4-1 \choose 15}+{14+4-1 \choose 14}+{13+4-1 \choose 13}+{12+4-1 \choose 12}$</p>
|
106,131 | <blockquote>
<p>Define $f:\mathbb{R}\rightarrow\mathbb{R}$ by<br>
$\
f(x) =
\begin{cases}
1/q
& \text{if } x =p/q \space(\mathrm{lowest}\space \mathrm{terms},\space\mathrm{nonzero})\\
0 & \text{if } x = 0\space\mathrm{or}\space x\not\in\mathbb{Q}
\end{cases}
$<br>
Show that f is continuous at 0 and every $x\in\mathbb{R}\setminus\mathbb{Q}$. Show that $f$ is not continuous at any nonzero rational pt. </p>
</blockquote>
<p>Attempt: (1) First I need to show that $f$ is continuous at zero. Then I need to show that $\forall\epsilon>0,\exists\delta>0$ s.t. $y\in B_{\delta}(0)$ implies $f(y)\in B_{\epsilon}(f(0))$. So I need to show $y\in B_{\delta}(0)$ implies $f(y)\in B_{\epsilon}(0)$ for $y\in\mathbb{R}$. Note $f(0)=0$. Pick $\delta =...$</p>
<p>(2) Then, I need to show that $f$ is continuous at every irrational number. Here I need to show that $\forall\epsilon>0,\exists\delta>0$ s.t. $y\in B_{\delta}(0)$ implies $f(y)\in B_{\epsilon}(f(0))$. Note that once again $f(0)=0$. Pick $\delta = ...$</p>
<p>(3) Then, I need to show that $f$ is not continuous at every nonzero rational number. Let $q\in\mathbb{Q}$. Intuitively, because $\mathbb{I}$ is dense in $\mathbb{R}$, we can construct a sequence $x_n\in\mathbb{I}$ such that $x_n\rightarrow q$. Since $x_n\in\mathbb{I}$, $\lim{x_n}=0\neq f(x)$ so clearly $f(x_n)\not\rightarrow f(x)$. </p>
| DonAntonio | 31,254 | <p>I think the following can help you a lot:</p>
<p><strong>Proposition:</strong> Let $\,\displaystyle{\left\{\frac{p_n}{q_n}\right\}}\subset\mathbb Q\,$ be a rational sequence, with $\,(p_n,q_n)=1\,\,\text{and}\,\,q_n>0\,\,,\,\forall n\,$ , and s.t.
$$\frac{p_n}{q_n}\xrightarrow [n\to\infty]{}x\in\mathbb R-\mathbb Q$$
Then $\,q_n\xrightarrow [n\to\infty]{} \infty\,$</p>
<p><strong>Proof (sketch):</strong> Suppose not. Then
$$\,\,\exists\,R\in\mathbb N\,\,s.t.\,\,\forall\,M\in\mathbb N\,\,\exists\,n_M\in\mathbb N\,\,s.t.\,\,n_m>M\,\,and\,\,q_{n_M}\leq R\,\,$$</p>
<p>We now take a closer look at the subsequence
$$\left\{\frac{p_{n_M}}{q_{n_M}}\right\}_{M=1}^\infty$$
where we choose the subindexes in such a way that $\,n_1<n_2<...\,$ (otherwise we have no subsequence at all. Fill in details here).</p>
<p>But $\,\displaystyle{\frac{p_{n_m}}{q_{n_m}}\xrightarrow [M\to\infty]{} x}\,$ (why?) , so for any
$$\epsilon>0\,\,\,\exists K_\epsilon\in\mathbb N\,\,s.t.\,\,n_M>K_\epsilon\Longrightarrow \left|\frac{p_{n_M}}{q_{n_M}}-x\right|<\epsilon\Longrightarrow $$
$$\Longrightarrow R\geq q_{n_M}>\left|\frac{p_{n_M}-q_{n_M}x}{\epsilon}\right|\geq\frac{|p_{n_M}|-R|x|}{\epsilon}\Longrightarrow $$
$$\Longrightarrow R(\epsilon+|x|)\geq |p_{n_M}|$$
which means <em>also</em> $\,\{p_{n_M}\}\,$ is bounded (take, for example, $\,\epsilon=1\,$ ), which is absurd.</p>
|
122,274 | <p>I have a question, I think it concerns with field theory.</p>
<blockquote>
<p>Why the polynomial $$x^{p^n}-x+1$$ is irreducible in ${\mathbb{F}_p}$ only when $n=1$ or $n=p=2$?</p>
</blockquote>
<p>Thanks in advance. It bothers me for several days. </p>
| Thomas Andrews | 7,933 | <p>We will use fairly liberally the result that if $q(x)\in\mathbb F_p[x]$ is irreducible, then, for any $k$, $q(x)\mid x^{p^k}-x$ if and only if $\deg q\mid k$.</p>
<p>If $q_n(x)=x^{p^n}-x+1$ is irreducible, then there is a automorphism, $\phi$ of the field $\mathbb F_p[x]/\left<q_n(x)\right>$ which sends $\bar x$ to $\bar x-1$, namely:</p>
<p>$$\phi(\alpha)=\alpha^{p^n}$$</p>
<p>for any element $\alpha$. (Where $\bar x$ is the image of $x$ from $\mathbb F_p[x]$ in this field.)</p>
<p>Then, $\phi(\bar x)=\bar x^{p^n}=\bar x-1$. So that automorphism must have order $p$: $\phi^p = 1$, the identity automorphism.</p>
<p>Now, $\phi^k(\alpha)=\alpha^{p^{kn}}$, so, in particular, $\bar x=\phi^p(\bar x)=\bar x^{p^{pn}}$, and therefore we know $0=\bar x^{p^{pn}}-\bar x$, and therefore that the polynomial $x^{p^{pn}}-x$ is divisible by $q_n(x)$.</p>
<p>Using the result above, we therefore see that $p^n=\deg q_n(x)\mid pn$. But $p^n\mid pn$ can only happen if $n=1$ or $n=2$ and $p=2$.</p>
<p>I think you can show that $q_1(x)\mid x^{p^p}-x$ pretty straight-forwardly, therefore showing that it must factor as elements of degree $p$ and degree $1$. But clearly it has no factors of degree $1$ since it has no roots in $\mathbb F_p$, so, since $\deg q_n=p$, $q_1(x)$ must be prime.</p>
<p>Then you have the last case, $x^4-x+1$ over $\mathbb F_2$, which you can brute force.</p>
|
3,396,882 | <blockquote>
<p>Let <span class="math-container">$X$</span> be a non-negative random variable, and suppose that <span class="math-container">$P(X \geq
n) \geq 1/n$</span> for each <span class="math-container">$n \in \mathbb{N}$</span>. Prove that <span class="math-container">$E(X) = \infty$</span>.</p>
</blockquote>
<p>I have been stuck with this problem for a few days now. I guess it can make some sense intuitively because you have some probability mass everywhere, and we're looking at probability of it being greater than some value. I tried to use inequalities like Markov's and Chebyshev's with no luck. I was hoping if someone can please explain to me how to answer this problem. It is coming from an introductory probability with measure theory book, and I am trying my best to get better at these kind of problems.</p>
| Henry | 6,460 | <p>Suppose we have a discrete random variable <span class="math-container">$Y$</span> with <span class="math-container">$\mathbb P(Y=n) = \frac{1}{n(n+1)}$</span> for all positive integers <span class="math-container">$n$</span> </p>
<p>then <span class="math-container">$\mathbb P(Y\ge n) = \frac1n \le \mathbb P(X \ge n)$</span> for all positive integers <span class="math-container">$n$</span> </p>
<p>and <span class="math-container">$\mathbb P(Y\ge x) \le \mathbb P(X \ge x)$</span> for all real <span class="math-container">$x$</span></p>
<p>so <span class="math-container">$X$</span> has weak first-order stochastic dominance over <span class="math-container">$Y$</span></p>
<p>making <span class="math-container">$\mathbb E[Y]\le \mathbb E[X]$</span> </p>
<p>but <span class="math-container">$\mathbb E[Y] = \sum\limits_{n=1}^\infty \frac{1}{n+1}= +\infty$</span> </p>
<p>showing <span class="math-container">$\mathbb E[X] = +\infty$</span> </p>
|
1,168,446 | <p>I have the following nonlinear differential equation (I am using $y$ as shorthand $f(x)$):</p>
<p>$$\sin(y - y') = y''$$</p>
<p>I have tried the following</p>
<p>$$\cos(y - y')(y'-y'') = y'''$$
$$-\sin(y - y')(y'-y'')^2 + \cos(y - y')(y''-y''') = y''''$$
$$-y''(y'-y'')^2 + \dfrac{y'''}{y'-y''}(y''-y''') = y''''$$
$$-y''(y'-y'')^3 + y'''(y''-y''') = y''''(y'-y'')$$</p>
<p>But this looks pretty unhelpful. Is there a better way to solve this equation?</p>
| Robert Israel | 8,508 | <p>There's not much hope of closed-form solutions. You could use numerical methods or series.</p>
|
251,705 | <p>I would like to find the residue of $$f(z)=\frac{e^{iz}}{z\,(z^2+1)^2}$$ at $z=i$. One way to do it is simply to take the derivative of $\frac{e^{iz}}{z\,(z^2+1)^2}$. Another is to find the Laurent expansion of the function.</p>
<p>I managed to do it using the first way, and the answer is $-3/(4e)$. However, I'm out of ideas as to how to find the expansion.</p>
<p>Any help is greatly appreciated.</p>
| robjohn | 13,854 | <p>Substitute $z=w+i$ to get
$$
\begin{align}
\frac{e^{iz}}{z\,(z^2+1)^2}
&=\frac{e^{iw-1}}{(w+i)\,(w^2+2iw)^2}\\
&=\frac{i}{4ew^2}\frac{e^{iw}}{(1-iw)(1-\frac{i}{2}w)^2}\\
&=\frac{i}{4ew^2}\frac{1+iw+\dots}{(1-iw)(1-iw+\dots)}\\
&=\frac{i}{4ew^2}(1+3iw+\dots)
\end{align}
$$
Thus, the residue is the coefficient of $\dfrac1w$ which is $\ -\dfrac3{4e}$.</p>
|
3,679,806 | <p>this is Probability Density Function(pdf)
if <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are independent, how to prove <span class="math-container">$X^2$</span> and <span class="math-container">$Y^2$</span> are also independent</p>
| Toni | 763,699 | <p>Here is another way, if you are not familiar with measure theory. I assume we are talking about real-valued random variables.
Let <span class="math-container">$a,b\geq0$</span>
<span class="math-container">$$P(X^2 \leq a,\,Y^2 \leq b)=P(-\sqrt{a}\leq X \leq \sqrt{a},\,-\sqrt{b}\leq Y \leq \sqrt{b}) \\= P(-\sqrt{a}\leq X \leq \sqrt{a})P(-\sqrt{b}\leq Y \leq \sqrt{b})=P(X^2 \leq a)P(Y^2 \leq b),$$</span>
where we have used the independence of <span class="math-container">$X,Y$</span> at the beginning of the second line. If <span class="math-container">$a<0$</span> or <span class="math-container">$b<0$</span>, then
<span class="math-container">$$P(X^2 \leq a,\,Y^2 \leq b)=0=P(X^2 \leq a)P(Y^2 \leq b).$$</span>
Hence, <span class="math-container">$X^2,Y^2$</span> are independent.</p>
|
1,118,259 | <p>Consider a sphere of radius $a$ with 2 cylindrical holes of radius $b<a$ drilled such that both pass through the center of the sphere and are orthogonal to one another. What is the volume of the remaining solid?</p>
<p>Can someone help me at least setting up the integral? I know that there is a similar problem but it was a sphere with one hole. </p>
| user84413 | 84,413 | <p>This answer combines the ideas of the two previous answers: </p>
<p>It implements Christian Blatter's suggestion, and makes extensive use of Achille Hui's techniques.</p>
<hr>
<p>Let $V$ be the volume of the sphere with 2 holes, let $V_{H}$ be the volume of each of the cylindrical holes, and let $V_{I}$ be the volume of the intersection of the sphere and the two cylindrical holes.</p>
<p>Let $c=\sqrt{a^2-b^2}$, and let $d=\sqrt{2b^2-a^2}$ when $a\le\sqrt{2}b$, so $c^2=b^2-d^2$ in this case.</p>
<p>Then $V_{H}=\displaystyle2\int_0^{b}2\pi x\sqrt{a^2-x^2}dx=2\pi\int_{c^2}^{a^2}\sqrt{u}du=\frac{4}{3}\pi(a^3-c^3)$ using the shell method,</p>
<p>so $\hspace{.9 in}\displaystyle \color{red}{V=\frac{4}{3}\pi a^3-2V_{H}+V_{I}=\frac{8}{3}\pi c^3-\frac{4}{3}\pi a^3+V_{I}}.$</p>
<hr>
<p>$\textbf{I)}$ If $a\ge\sqrt{2}b$, the cross-sections of the intersection perpendicular to the z-axis are the squares defined by </p>
<p>$\;\;\;\;|x|\le\sqrt{b^2-z^2},\;\; |y|\le\sqrt{b^2-z^2}\;\;$; so in this case</p>
<p>$\displaystyle V_{I}=\int_{-b}^{b}\left(2\sqrt{b^2-z^2}\right)^2dz=8\int_0^{b}(b^2-z^2)dz=\frac{16}{3}b^3$ and</p>
<p>$\hspace{1 in}\displaystyle \color{blue}{V=\frac{8}{3}\pi c^3-\frac{4}{3}\pi a^3+\frac{16}{3}b^3=\frac{4}{3}\left(2\pi c^3-\pi a^3+4b^3\right)}.$</p>
<hr>
<p>$\textbf{II)}$ If $b<a\le\sqrt{2}b$, the cross-sections of the intersection perpendicular to the z-axis are</p>
<p>the squares defined by $\;\;\;|x|\le\sqrt{b^2-z^2},\;\; |y|\le\sqrt{b^2-z^2}\;\;\;\;$ if $|z|\ge d$,</p>
<p>and the portions of these squares inside the circle $x^2+y^2=a^2-z^2\;\;\;\;$if $|z|<d$.</p>
<p>Dividing these squares into 8 equal pieces by drawing the diagonals, we get</p>
<p>$V_{I}=\displaystyle 16\int_0^{d}\left(\frac{1}{2}(a^2-z^2)\left[\frac{\pi}{4}-\cot^{-1}\frac{\sqrt{b^2-z^2}}{c}\right]+\frac{1}{2}c\sqrt{b^2-z^2}\right)dz+8\int_{d}^{b}(b^2-z^2)dz$</p>
<p>$\displaystyle\;\;\;\;=8\int_0^{d}\frac{1}{2}\left(a^2-z^2\right)\left[\frac{\pi}{4}-\cot^{-1}\frac{\sqrt{b^2-z^2}}{c}\right]dz+8c\int_0^{d}\sqrt{b^2-z^2}dz+8\int_d^{b}(b^2-z^2)dz$</p>
<p>(where the first integral corresponds to a circular sector and the second corresponds to a triangle).</p>
<p>$\textbf{A)}$ $\;\;\displaystyle8\int_d^{b}(b^2-z^2)dz=8\left[b^{2}z-\frac{z^3}{3}\right]_d^{b}=8\left(\frac{2}{3}b^3-b^{2}d+\frac{1}{3}d^3\right)=\frac{8}{3}(2b^3-3b^{2}d+d^3)$</p>
<p>$\textbf{B)}\;\;\;$ Using $z=b\sin\theta, \;\;dz=b\cos\theta d\theta$ and $\gamma=\sin^{-1}\frac{d}{b}$,</p>
<p>$\displaystyle8c\int_0^{d}\sqrt{b^2-z^2}dz=8c\int_0^{\gamma}b^2\cos^{2}\theta d \theta=8b^2c\int_0^{\gamma}\frac{1}{2}(1+\cos2\theta)d\theta=4b^2c\left[\theta+\sin\theta\cos\theta\right]_0^{\gamma}$</p>
<p>$\displaystyle\hspace{1.4 in}=4b^{2}c\left(\sin^{-1}\frac{d}{b}+\frac{cd}{b^2}\right)=4b^{2}c\sin^{-1}\frac{d}{b}+4c^{2}d$.</p>
<p>$\textbf{C)}\;\;\;$Using integration by parts with $\displaystyle u=\frac{\pi}{4}-\cot^{-1}\frac{\sqrt{b^2-z^2}}{c}, \;\;dv=(a^2-z^2)dz$,</p>
<p>$\displaystyle\;\;\;8\int_0^{d}\frac{1}{2}\left(a^2-z^2\right)\left[\frac{\pi}{4}-\cot^{-1}\frac{\sqrt{b^2-z^2}}{c}\right]dz=$</p>
<p>$\displaystyle8\left[\left(\frac{\pi}{4}-\cot^{-1}\frac{\sqrt{b^2-z^2}}{c}\right)\left(a^2z-\frac{z^3}{3}\right)\right]_0^d+8\int_0^d\left(a^2z-\frac{z^3}{3}\right)\left(\frac{cz}{a^2-z^2}\frac{1}{\sqrt{b^2-z^2}}\right)dz$</p>
<p>$\displaystyle=\frac{8c}{3}\int_0^d\frac{3a^{2}z^2-z^4}{a^2-z^2}\frac{1}{\sqrt{b^2-z^2}}dz=\frac{8c}{3}\int_0^d\left(z^2-2a^2+\frac{2a^4}{a^2-z^2}\right)\frac{1}{\sqrt{b^2-z^2}}dz$.</p>
<p>Using $z=b\sin\theta, \;\;dz=b\cos\theta d\theta$ and $\gamma=\sin^{-1}\frac{d}{b}$, we obtain</p>
<p>$\displaystyle\frac{8c}{3}\int_0^{\gamma}\left(b^2\sin^2\theta-2a^2+\frac{2a^4}{a^2-b^2\sin^2\theta}\right)d\theta$</p>
<p>$=\displaystyle\frac{8c}{3}\left[\frac{b^2}{2}(\theta-\sin\theta\cos\theta)-2a^2\theta\right]_0^{\gamma}+\frac{16a^4c}{3}\int_0^{\gamma}\frac{\sec^2\theta}{c^2\tan^2\theta+a^2}d\theta$</p>
<p>$=\displaystyle\frac{4}{3}b^2c\sin^{-1}\frac{d}{b}-\frac{4}{3}c^2d-\frac{16}{3}a^2c\sin^{-1}\frac{d}{b}+\frac{16}{3}a^4c\left[\frac{1}{ac}\tan^{-1}\left(\frac{c\tan\theta}{a}\right)\right]_0^{\gamma}$</p>
<p>$\displaystyle=\frac{4}{3}b^2c\sin^{-1}\frac{d}{b}-\frac{4}{3}c^2d-\frac{16}{3}a^2c\sin^{-1}\frac{d}{b}+\frac{16}{3}a^3\tan^{-1}\frac{d}{a}$.</p>
<p>Then $\displaystyle V_I=\left[\left(\frac{4}{3}b^2c-\frac{16}{3}a^2c\right)\sin^{-1}\frac{d}{b}-\frac{4}{3}c^2d+\frac{16}{3}a^3\tan^{-1}\frac{d}{a}\right]+\left[4c^2d+4b^2c\sin^{-1}\frac{d}{b}\right]+\;\;\;\;\;\;\left[\frac{8}{3}(2b^3-3b^2d+d^3)\right]=-\frac{16}{3}c^3\sin^{-1}\frac{d}{b}+\frac{16}{3}a^3\tan^{-1}\frac{d}{a}+\frac{8}{3}(c^2d+2b^3-3b^2d+d^3),$</p>
<hr>
<p>so $V=\displaystyle\frac{8}{3}\pi c^3-\frac{4}{3}\pi a^3-\frac{16}{3}c^3\left(\frac{\pi}{2}-\cos^{-1}\frac{d}{b}\right)+\frac{16}{3}a^3\tan^{-1}\frac{d}{a}+\frac{8}{3}\left(2b^2(b-d)\right)$</p>
<p>$\hspace{1 in}=\displaystyle\color{blue}{-\frac{4}{3}\pi a^3+\frac{16}{3}c^3\cos^{-1}\frac{d}{b}+\frac{16}{3}a^3\tan^{-1}\frac{d}{a}+\frac{16}{3}b^2(b-d)}.$</p>
<p>$\hspace{1 in}=\displaystyle\color{blue}{\frac{16}{3}\left(c^3\cos^{-1}\frac{d}{b}+a^3\tan^{-1}\frac{d}{a}-\frac{\pi}{4}a^3+b^2(b-d)\right)}.$</p>
|
368,461 | <p>Let <span class="math-container">$G=(V,E)$</span> be a finite simple graph. We say a map <span class="math-container">$p:V\to [n]:=\{1,\ldots,n\}$</span> is a <em>pseudo-coloring</em> if for all <span class="math-container">$a\neq b\in[n]$</span> there is <span class="math-container">$v\in\psi^{-1}(\{a\})$</span> and <span class="math-container">$w\in\psi^{-1}(\{b\})$</span> such that <span class="math-container">$\{v,w\}\in E$</span>. We denote the maximal number <span class="math-container">$m$</span> such that there is a pseudo-coloring <span class="math-container">$p:V\to [m]$</span> by <span class="math-container">$\psi(V)$</span>.</p>
<p>An easy argument shows that every coloring in the traditional sense is also a pseudo-coloring, which implies that <span class="math-container">$\psi(G) \geq \chi(G)$</span>.</p>
<p>Let <span class="math-container">$G, H$</span> be finite simple undirected graphs.
Do we necessarily have <span class="math-container">$$\psi(G\times H) \leq \min\{\psi(G),\psi(H)\}?$$</span></p>
<p>(By <span class="math-container">$G\times H$</span> we denote the categorical product, sometimes also referred to as the <a href="https://en.wikipedia.org/wiki/Tensor_product_of_graphs" rel="nofollow noreferrer">tensor product of graphs</a>.)</p>
| Markiian Khylynskyi | 144,883 | <p>It is not true. Let <span class="math-container">$G$</span> and <span class="math-container">$H$</span> be graphs, and let <span class="math-container">$p_{max}$</span> be maximal pseudo-coloring of the graph <span class="math-container">$H$</span>. Show that the map <span class="math-container">$p((x,y))=p_{max}(y)$</span> is pseudo-coloring of graph <span class="math-container">$G\times H$</span>. Fix some <span class="math-container">$\{u,v\}\in E(G)$</span>. For arbitrary distinct colors <span class="math-container">$a,b$</span> there exist <span class="math-container">$\{k,l\}\in E(H)$</span> such that <span class="math-container">$p_{max}(k)=a$</span> and <span class="math-container">$p_{max}(l)=b$</span>. Then the edge <span class="math-container">$\{(u,k),(v,l)\}$</span> connects colors <span class="math-container">$a$</span> and <span class="math-container">$b$</span> in <span class="math-container">$G\times H$</span>. Hence <span class="math-container">$\psi(G\times H)\geq \max\{\psi(G),\psi(H)\}$</span>.</p>
|
4,545,364 | <blockquote>
<p>Solve the quartic polynomial :
<span class="math-container">$$x^4+x^3-2x+1=0$$</span>
where <span class="math-container">$x\in\Bbb C$</span>.</p>
<p>Algebraic, trigonometric and all possible methods are allowed.</p>
</blockquote>
<hr />
<p>I am aware that, there exist a general quartic formula. (Ferrari's formula). But, the author says, this equation doesn't require general formula. We need some substitutions here.</p>
<p>I realized there is no any rational root, by the rational root theorem.</p>
<p>The harder part is, WolframAlpha says the factorisation over <span class="math-container">$\Bbb Q$</span> is impossible.</p>
<p>Another solution method can be considered as the quasi-symmetric equations approach. (divide by <span class="math-container">$x^2$</span>).</p>
<p><span class="math-container">$$x^2+\frac 1{x^2}+x-\frac 2x=0$$</span></p>
<p>But the substitution <span class="math-container">$z=x+\frac 1x$</span> doesn't make any sense.</p>
<p>I want to ask the question here to find possible smarter ways to solve the quartic.</p>
| lone student | 460,967 | <p>You can easily observe that, the expression <span class="math-container">$(x-1)^2$</span> is almost included in the polynomial <span class="math-container">$P(x):=x^4+x^3-2x+1$</span>.</p>
<p>Let's rewrite your polynomial as follows:</p>
<p><span class="math-container">$$
\begin{align}P(x)=x^4+x^3-\color{red}{x^2}+\color{red}{x^2}-2x+1\end{align}
$$</span></p>
<p>Based on this observation, we will represent the polynomial <span class="math-container">$P(x)$</span> as a <em>"quadratic"</em> polynomial:</p>
<p><span class="math-container">$$
\begin{align}P(x):&=x^4+x^3-x^2+(x-1)^2\\
&=x^2(x^2+x-1)+(x-1)^2\\
&=x^4+x^2(x-1)+(x-1)^2\\
&=\color{red}{(x-1)^2}+\color{blue}{x^2}\color{red}{(x-1)}+\color{blue}{x^4}\end{align}
$$</span></p>
<p>Letting <span class="math-container">$x-1=u$</span> and writing <span class="math-container">$P(x)=0$</span> leads to:</p>
<p><span class="math-container">$$
\begin{align}&\color{red}{u^2}+\color{blue}{x^2}\color{red}{u}+\color{blue}{x^4}=0\\
&\Delta_{\color{red}{u}}=x^4-4x^4=-3x^4\\
\implies &u_{1,2}=\frac{-x^2\pm i\sqrt 3x^2}{2}\\
\implies &u_{1,2}=x^2\left(\frac{-1\pm i\sqrt 3}{2}\right)\\
\implies &x-1=x^2\left(\frac{-1\pm i\sqrt 3}{2}\right)\\
\implies &x^2\left(\frac{-1\pm i\sqrt 3}{2}\right)-x+1=0.\end{align}
$$</span></p>
<p>Now, this is obvious that the given polynomial has no real roots.</p>
<p>Finally, you can complete the solution by using the quadratic formula to find all complex roots.</p>
|
1,890,040 | <p>This is related to <a href="https://math.stackexchange.com/questions/1888881/expanding-a-potential-function-via-the-generating-function-for-legendre-polynomi">this previous question of mine</a> where (with lots of help) I show that $$\sum_{l=0}^\infty \frac{R^l}{a^{l+1}}P_l(\cos\theta)=\frac{1}{\sqrt{a^2-2aR\cos\theta+R^2}}\tag{1}$$ by using the Legendre generating function.</p>
<p>The generating function for Legendre Polynomials is:</p>
<p>$$\Phi(x,h)=(1-2xh+h^2)^{-1/2} \quad\text{for}\quad |h|\lt 1\tag{2}$$
or
$$\Phi(x,h)=\sum_{l=0}^\infty h^l P_l(x)\quad\text{for}\quad |h|\lt 1\tag{3}$$</p>
<p>I need to show that $$\sum_{l=0}^\infty\frac{R^{2l+1}r^{-l-1}P_l(\cos\theta)}{a^{l+1}}=\frac{R}{a\sqrt{(R^2/a)^2-2r(R^2/a)\cos\theta+r^2}}\tag{4}$$ I note the striking similarity between the RHS's of $(1)$ and $(4)$ within the square root of the denominator.</p>
<hr>
<p><code>It is not necessary to read the following if you understand how to prove that the sum can be written as above; But here is some context anyway:</code>
<a href="https://i.stack.imgur.com/DIR6f.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DIR6f.png" alt="Diagram"></a>
<a href="https://i.stack.imgur.com/PkIFq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PkIFq.png" alt="Potential function 1"></a>
<a href="https://i.stack.imgur.com/CtYlI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CtYlI.png" alt="Potential function 2"></a> </p>
<hr>
<p><code>So here is my attempt:</code></p>
<p>From the RHS of equation $(4)$:</p>
<p>$$\displaystyle\begin{align}\frac{R}{a\sqrt{r^2-2r(R^2/a)\cos\theta+(R^2/a)^2}}&=\frac{R}{ar\sqrt{1-2(R^2/ar)\cos\theta+(R^2/ar)^2}}\end{align}$$</p>
<p>Changing variables to match equation $(2)$: Let $h=R^2/ar$ and $x=\cos\theta$</p>
<p>Then
$$\displaystyle\begin{align}\frac{R}{ar\sqrt{1-2(R^2/ar)\cos\theta+(R^2/ar)^2}}&=\frac{R}{ar\sqrt{1-2hx+h^2}}\\&=\frac{R}{ar}(1-2xh+h^2)^{-1/2}\\&=\frac{R}{ar}\Phi(x,h)\\&=\frac{R}{ar}\sum_{l=0}^\infty h^l P_l(x)\\&=\frac{R}{ar}\sum_{l=0}^\infty \frac{R^{2l} P_l(\cos\theta)}{(ar)^l}\\&=\sum_{l=0}^\infty \frac{R^{2l+1} P_l(\cos\theta)}{a^{l+1}r^{l+1}}\\&=\sum_{l=0}^\infty \frac{R^{2l+1}r^{-l-1} P_l(\cos\theta)}{a^{l+1}}\quad\fbox{}\end{align}$$</p>
<hr>
<p>You may be wondering what I'm asking at this point; so here is the question:</p>
<p>In the last line of the extract the textbook says "by summing the series"; and this seems to suggest that one must arrive at the RHS <em>from</em> the LHS (instead of going from RHS to LHS as I did). So is my interpretation correct (LHS $\to$ RHS) and if so <em>can</em> it be done in this order?</p>
<p>Many thanks.</p>
| Ivan Neretin | 269,518 | <p>OK, Henning Makholm did the explanation, and did it good, but I'll do my part anyway.</p>
<p>First we extend the faces of our dodecahedron just a bit beyond the edges, until it grows nice little spikes on all 12 faces and thus becomes a <a href="https://en.wikipedia.org/wiki/Small_stellated_dodecahedron" rel="nofollow noreferrer">small stellated dodecahedron</a>.
<a href="https://i.stack.imgur.com/Il8JM.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Il8JM.gif" alt="Dodecahedron to small stellated dodecahedron"></a></p>
<p>Then we grow them a bit further, until each valley between the two neighboring spikes is filled. There were as many valleys as edges in a dodecahedron, so this gives us 30 more regions and creates the thing known as <a href="https://en.wikipedia.org/wiki/Great_dodecahedron" rel="nofollow noreferrer">great dodecahedron</a>.
<a href="https://i.stack.imgur.com/cVDJa.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cVDJa.gif" alt="Small stellated dodecahedron to great dodecahedron"></a></p>
<p>Now we grow them further yet until we see new, sharper spikes where the depressions used to be. These correspond to the vertices of the original guy, so we add 20 of them and behold, here is your <a href="https://en.wikipedia.org/wiki/Great_stellated_dodecahedron" rel="nofollow noreferrer">great stellated dodecahedron</a>.</p>
<p><a href="https://i.stack.imgur.com/pOgiW.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pOgiW.gif" alt="Great dodecahedron to great stellated dodecahedron"></a></p>
<p>All in all, this gives us $1+12+30+20=63$ bounded regions.</p>
<p>So it goes.</p>
|
1,811,528 | <p>The definition of the order of an element in a group is:</p>
<blockquote>
<p>The order of an element $x$ of a group $G$ is the smallest positive integer $n$ such that $x^{n}=e$.</p>
</blockquote>
<p>Doesn't this definition assume that the integers are somehow relevant to every group? </p>
<p>All of the other definitions concerning Groups do not invoke the integers in any way.</p>
<p>I would expect that Groups would contain elements that would somehow represent exponentiation.</p>
<p>This question may reflect some deep misunderstandings on my part so please feel free to offer guidance not directly relevant to the question.</p>
| Stefan Perko | 166,694 | <p>Here is an alternative viewpoint:</p>
<p>Actually, the common definition of a group with an operation $\cdot : G\times G\to G$ is not the only possible definition of a group. In fact it is <em>biased</em>, because for any group there a alot of $n$-ary operations besides the group multiplication.
For example you have the following operations:</p>
<ul>
<li>$^{-1} : G\to G, x\mapsto x^{-1}$</li>
<li>$/ : G^2 \to G, (x,y)\mapsto xy^{-1}$</li>
<li>$G^2\to G, (x,y) \mapsto xyx^{-1}$</li>
<li>$G^3 \to G, (x,y,z) \mapsto xyz^{-1}$</li>
<li>$G^4 \to G, (x_1,x_2,x_3,x_4) \mapsto x_{1} x_{2} x_{3} x_{4}$</li>
<li>...</li>
</ul>
<p>but there is no end to this list! Furthermore the multiplication $\cdot$ is not special among all these operations. You can state the group axioms with a lot of these other operations as well (division $/$ is a prime candidate)</p>
<p>It begs the question: is there an "ultimate" operation that stands above all of the other? The answer is: Yes! (sort of):</p>
<p>Let $G^* = \{ (x_1,\dots, x_n) : x_1,\dots, x_n\in G\}$. That is the set of $n$-tuples or <em>words</em> over $G$. Then there is an operation:</p>
<p>$$\prod : G^* \to G, (x_1,\dots, x_n) \mapsto x_1\cdots x_n$$
that maps a word of symbols, that is a bunch of group elements, to their multiplication.
(Yes, it is kind of cheating because I still need the group multiplication, but maybe you get the idea).
We commonly write:
$$\prod_{i=1}^n x_i\text{ for }\prod (x_1,\dots, x_n)$$</p>
<p>The associativity says, that $\prod$ is well-defined (bracketing does not matter). The identity is also encoded in $\prod$! We have:
$$1_G := \prod ()$$
where $()$ is the empty word (the tuple of length $0$). This is a convenient definition making the recursive formula work:
$$\prod_{i=1}^n x_i = x_1 \cdot \prod_{i=2}^n x_i$$
even when $n = 1$, because:
$$x_1\cdot \prod_{i=2}^1 x_i = x_1\cdot \prod () = x_1\cdot 1_G = x_1 = \prod_{i=1}^1 x_i$$
Note, that $x^n = \prod_{i=1}^n x$.</p>
<p>This should make obvious how nonnegative integers come into play (as lengths of words that I can put into $\prod$)</p>
<p>(Even if you do not believe all of this is useful yet, you will sort of rediscover all of this when you take a look at <em>free groups</em>.)</p>
|
7,237 | <p>this came up in class yesterday and I feel like my explanation could have been more clear/rigorous. The students were given the task of finding the zeros of the following equation $$6x^2 = 12x$$ and one of the students did $$\frac{6x^2}{6x}=\frac{12x}{6x}$$ $$x = 2$$ which is a valid solution but this method eliminates the other solution of $$ x = 0$$ When the student brought it up, I explained to the student that if $x = 0, 2$ and we divide by $6x$ there is a possibility that we would be dividing by 0 which is undefined. The student, very reasonably, responded "Well, obviously I didn't know that zero was an answer when I was doing the problem". The student understands why we can't divide by 0 but is still struggling with how that connects to dividing by $x$. I went on to explain that by dividing by $x$ you are "dropping a solution" because the problem, which was quadratic, is now linear. Again, this didn't seem to click with the student. Does anyone have maybe an axiom/law/theorem that I can show the student to give a rigorous reason as to why you can't just divide by $x$?</p>
| Henry Towsner | 62 | <p>I think your student pointed out the key issue: "Well, obviously I didn't know that zero was an answer when I was doing the problem". That's exactly right: at the time the student was dividing, they didn't know whether or not x is 0, which means they didn't know whether or not it made sense to divide by x.</p>
<p>If you know you can't divide by 0, you know not to divide by things which <em>might</em> be 0, because you don't know whether or not it's vaild.</p>
|
7,237 | <p>this came up in class yesterday and I feel like my explanation could have been more clear/rigorous. The students were given the task of finding the zeros of the following equation $$6x^2 = 12x$$ and one of the students did $$\frac{6x^2}{6x}=\frac{12x}{6x}$$ $$x = 2$$ which is a valid solution but this method eliminates the other solution of $$ x = 0$$ When the student brought it up, I explained to the student that if $x = 0, 2$ and we divide by $6x$ there is a possibility that we would be dividing by 0 which is undefined. The student, very reasonably, responded "Well, obviously I didn't know that zero was an answer when I was doing the problem". The student understands why we can't divide by 0 but is still struggling with how that connects to dividing by $x$. I went on to explain that by dividing by $x$ you are "dropping a solution" because the problem, which was quadratic, is now linear. Again, this didn't seem to click with the student. Does anyone have maybe an axiom/law/theorem that I can show the student to give a rigorous reason as to why you can't just divide by $x$?</p>
| Jared | 2,204 | <p>Frankly, I think your reasoning is the best. You initially have a quadratic, thus you should <em>expect</em> to find two solutions (although you may not). By dividing by $x$, you get only a single solution: so what is the other? You clearly cannot find it by dividing by $x$, but you <em>can</em> find it by either factoring or using the quadratic equation (which should be discouraged in this case).</p>
<p>Really this is about pattern recognition. The student should recognize when they have:</p>
<p>$$
ax^2 + bx = 0
$$</p>
<p>That they should factor out $x$ to get: $x(ax + b) = 0$ and then set each factor to zero to solve: $x = 0$ and $ax + b = 0 \rightarrow x = -\frac{b}{a}$. The second part is recognizing that when you have something like $ax^2 + bx + c = dx^2 + ex + f$ that they should solve by setting the equation to zero, collecting like terms, and then attempting to factor (using the quadratic equation as a last resort).</p>
<p>If they can understand factoring and that <em>this</em> is the correct approach to solving such equations then they should have less trouble generalizing to higher degree polynomials (i.e. ones that <em>don't</em> have a nice formula like the quadratic formula).</p>
<p>The rationale for this is 1) the fact that polynomial equations, in general, have multiple solutions (so you should either find them or justify why there are fewer) and 2) we <em>really</em> only "know" how to solve linear equations so we must somehow convert our problem into a set of linear equations to be solved and the best way to do that is to attempt to factor (but explain that this is <em>not</em> always possible/"easy").</p>
<p>Granted, dividing by $x$ is one way to get <em>a</em> linear equation, but since we see that it's quadratic we should find <em>two</em> linear equations...</p>
<p>...which could be redundant like:</p>
<p>$$
x^2 = 2x - 1 \\
x^2 - 2x + 1 = 0 \\
(x - 1)^2 = 0 \\
(x - 1)(x - 1) = 0 \\
x - 1 = 0 \\
x - 1 = 0
$$</p>
<p>We got our two linear equations, but they were identical/gave the same solution.</p>
|
7,237 | <p>this came up in class yesterday and I feel like my explanation could have been more clear/rigorous. The students were given the task of finding the zeros of the following equation $$6x^2 = 12x$$ and one of the students did $$\frac{6x^2}{6x}=\frac{12x}{6x}$$ $$x = 2$$ which is a valid solution but this method eliminates the other solution of $$ x = 0$$ When the student brought it up, I explained to the student that if $x = 0, 2$ and we divide by $6x$ there is a possibility that we would be dividing by 0 which is undefined. The student, very reasonably, responded "Well, obviously I didn't know that zero was an answer when I was doing the problem". The student understands why we can't divide by 0 but is still struggling with how that connects to dividing by $x$. I went on to explain that by dividing by $x$ you are "dropping a solution" because the problem, which was quadratic, is now linear. Again, this didn't seem to click with the student. Does anyone have maybe an axiom/law/theorem that I can show the student to give a rigorous reason as to why you can't just divide by $x$?</p>
| Steven Gubkin | 117 | <p>This is more of a long comment than an answer.</p>
<p>Some of the other answers here are advocating for factoring rather than case analysis. I just want to point out that proving the theorem that for any two real numbers <span class="math-container">$ab = 0$</span> if and only if <span class="math-container">$a=0$</span> or <span class="math-container">$b=0$</span> requires the same sort of reasoning as @Aeryk advocates in the accepted answer:</p>
<p>In one direction, if we assume that <span class="math-container">$ab=0$</span>, then we consider two cases: if <span class="math-container">$a=0$</span>, we are done. If <span class="math-container">$a \neq 0$</span>, then we can divide both sides by <span class="math-container">$a$</span> to conclude that <span class="math-container">$b=0$</span>.</p>
<p>In the other direction, assuming either <span class="math-container">$a=0$</span> or <span class="math-container">$b=0$</span> quickly leads to <span class="math-container">$ab=0$</span> by multiplying both sides by the appropriate factor.</p>
<p>All of this is just to say that we cannot truly avoid case analysis through factoring: we can only postpone it.</p>
|
1,818,764 | <p>I think everything I have done is kosher, but unless I am missing an identity it is a different answer than the online quiz and wolfram alpha give.</p>
<p>I tried to use the trig substitution
$$ x=2\sin(\theta)\Rightarrow dx=2\cos(\theta)$$</p>
<p>Which yields
$$\int\frac{x^2}{\sqrt{4-x^2}}dx=\int\frac{4\sin^2(\theta)}{2\sqrt{1-\sin^2{\theta}}}2\cos(\theta)d\theta=4\int \sin^2(\theta)d\theta\\
=2\int (1-\cos(2\theta))d\theta=2\theta-\sin(2\theta)$$ </p>
<p>By the half angle formula. Then since $x=2\sin(\theta)\Rightarrow \theta=\arcsin(x/2)$ this gives a final answer of </p>
<p>$$\int\frac{x^2}{\sqrt{4-x^2}}dx=2\arcsin(x/2)-\sin(2\arcsin(x/2))+c $$</p>
<p>Is this right? If not, where did I go wrong?</p>
| Micheal Brain Hurts | 324,389 | <p>for $\displaystyle\int\dfrac{x^2}{\sqrt{4-x^2}}dx$ ;</p>
<p>Integral by parts;</p>
<p>Let be $\quad du=\dfrac{-x}{\sqrt{4-x^2}}dx$</p>
<p>$u=\sqrt{4-x^2}$</p>
<p>And $\quad(-x)=v\longrightarrow -dx=dv$</p>
<p>$\displaystyle\int\dfrac{x^2}{\sqrt{4-x^2}}dx=-x.\sqrt{4-x^2}+\displaystyle\int\sqrt{4-x^2}dx$</p>
<p>And, for $\quad\displaystyle\int\sqrt{4-x^2}dx$;</p>
<p>$x=2\sin a$</p>
<p>$\arcsin \frac{x}{2}=a$</p>
<p>$dx=2.\cos a.da$</p>
<p>$\quad\displaystyle\int\sqrt{4-x^2}dx=4\displaystyle\int \cos^2a \;da=2\displaystyle\int \dfrac{\cos 2a+1}{1} \;da=\cos 2a+2a+C=\cos (2.(\arcsin \frac{x}{2}))+2.\arcsin \frac{x}{2}+C$ </p>
<p>And all of Integral'll be;</p>
<p>$\boxed{\boxed{\displaystyle\int\dfrac{x^2}{\sqrt{4-x^2}}dx=-x.\sqrt{4-x^2}+\cos (2.(\arcsin \frac{x}{2}))+2.\arcsin \frac{x}{2}+C}}$</p>
|
121,450 | <p>I am trying to prove that the series <span class="math-container">$\sum \dfrac {1} {\left( m_{1}^{2}+m_{2}^{2}+\cdots +m_{r }^{2}\right)^{\mu} } $</span> in which the summation extends over all positive and negative integral values and zero values of <span class="math-container">$m_1, m_2,\dots, m_r$</span>, except the set of simultaneous zero values, is absolutely convergent if <span class="math-container">$\mu > \dfrac {r} {2}$</span>.</p>
<p>Any help with a proof strategy would be much appreciated.</p>
| Did | 6,179 | <p>Comparing the series with an integral, one sees that the series converges if and only if the $r$-dimensional integral
$$
I_r=\int_{\mathbb R^r}[\|x\|\geqslant1]\,\frac{\mathrm dx}{\|x\|^{2\mu}}
$$
converges. Consider the spherical coordinates $(s,\alpha)$ with $s\geqslant0$ and $\alpha$ in the sphere $S^{r-1}$. Then $\mathrm dx$ is proportional to $s^{r-1}\mathrm ds\mathrm d\alpha$. Hence , $I_r$ converges if and only if the $1$-dimensional integral
$$
\int_1^{+\infty}\frac{s^{r-1}\mathrm ds}{s^{2\mu}}
$$
converges, that is, if and only if $2\mu\gt r$.</p>
|
3,103,372 | <p>The following is a system of quadratic congruences:
<span class="math-container">$$\left\{\begin{array}{cl}x^{2}\equiv a&\pmod{3}\\x^{2}\equiv b&\pmod{7}\end{array}\right.$$</span>
If <span class="math-container">$\left(\frac{a}{3}\right)=1=\left(\frac{b}{7}\right)$</span>, where <span class="math-container">$(\ast)$</span> is the Legendre symbol, then above system has four solutions by the Chinese remainder theorem.</p>
<p>But, I'm wondering about that: Is it possible the first and second congruence has the same solution?</p>
<p>Since <span class="math-container">$\gcd(3,7)=1$</span>, I guess it is impossible, but I can't explain clearly.</p>
<p>How do I explain it?</p>
<p>Give some advice. Thank you!</p>
| Wolfgang Kais | 640,973 | <p>Let's assume that <span class="math-container">$\gcd(n,m)=1$</span> and that we have integers <span class="math-container">$\alpha,\beta \in \mathbb Z$</span> each solving one of the quadratic congruences such that</p>
<p><span class="math-container">$$\alpha^2 \equiv a \pmod n \quad \land \quad \beta^2 \equiv b \pmod m$$</span></p>
<p>Then, using the CRT twice, we can be sure that there exist <span class="math-container">$y, z \in \mathbb Z$</span> satisfying </p>
<p><span class="math-container">$$
\begin{align}
& y \equiv \alpha \pmod n \quad \land \quad y \equiv 1 \pmod m\\
\text{and also } & z \equiv 1 \pmod n \quad \land \quad z \equiv \beta \pmod m
\end{align}
$$</span></p>
<p>For the product <span class="math-container">$x:=yz$</span>, it follows that</p>
<p><span class="math-container">$$
\begin{align}
&x^2 \equiv y^2z^2 \equiv \alpha^21^2 \equiv a \pmod n\\
\text{and also } & x^2 \equiv y^2z^2 \equiv 1^2\beta^2 \equiv b \pmod m
\end{align}
$$</span></p>
<p>So, if both quadratic congruences are solvable individually and <span class="math-container">$\gcd(n,m)=1$</span>, then both quadratic congruences are also solvable simultaneously.</p>
|
3,103,372 | <p>The following is a system of quadratic congruences:
<span class="math-container">$$\left\{\begin{array}{cl}x^{2}\equiv a&\pmod{3}\\x^{2}\equiv b&\pmod{7}\end{array}\right.$$</span>
If <span class="math-container">$\left(\frac{a}{3}\right)=1=\left(\frac{b}{7}\right)$</span>, where <span class="math-container">$(\ast)$</span> is the Legendre symbol, then above system has four solutions by the Chinese remainder theorem.</p>
<p>But, I'm wondering about that: Is it possible the first and second congruence has the same solution?</p>
<p>Since <span class="math-container">$\gcd(3,7)=1$</span>, I guess it is impossible, but I can't explain clearly.</p>
<p>How do I explain it?</p>
<p>Give some advice. Thank you!</p>
| Servaes | 30,382 | <p>Let <span class="math-container">$x$</span> be any solution to <span class="math-container">$x^2\equiv a\pmod{3}$</span> and <span class="math-container">$y$</span> any solution to <span class="math-container">$y^2\equiv b\pmod{7}$</span>. Then <span class="math-container">$x+3m$</span> is also a solution for any integer <span class="math-container">$m$</span>, and <span class="math-container">$y+7n$</span> is also a solution for any integer <span class="math-container">$n$</span>. So it suffices to show that there exist integers <span class="math-container">$m$</span> and <span class="math-container">$n$</span> such that
<span class="math-container">$$x+3m=y+7n,$$</span>
or equivalently <span class="math-container">$3m+7(-n)=y-x$</span>. Such <span class="math-container">$m$</span> and <span class="math-container">$n$</span> exist because <span class="math-container">$3$</span> and <span class="math-container">$7$</span> are coprime. Concretely, as
<span class="math-container">$$3\times(-2)+7\times(1)=1,$$</span>
we can take <span class="math-container">$m=-2(y-x)$</span> and <span class="math-container">$n=-1(y-x)$</span> to find that
<span class="math-container">$$x-6(y-x)=y-7(y-x)=7x-6y,$$</span>
satisfies both congruences simultaneously.</p>
|
122,546 | <p>There is a famous proof of the Sum of integers, supposedly put forward by Gauss.</p>
<p>$$S=\sum\limits_{i=1}^{n}i=1+2+3+\cdots+(n-2)+(n-1)+n$$</p>
<p>$$2S=(1+n)+(2+(n-2))+\cdots+(n+1)$$</p>
<p>$$S=\frac{n(1+n)}{2}$$</p>
<p>I was looking for a similar proof for when $S=\sum\limits_{i=1}^{n}i^2$</p>
<p>I've tried the same approach of adding the summation to itself in reverse, and I've found this:</p>
<p>$$2S=(1^2+n^2)+(2^2+n^2+1^2-2n)+(3^2+n^2+2^2-4n)+\cdots+(n^2+n^2+(n-1)^2-2(n-1)n$$</p>
<p>From which I noted I could extract the original sum;</p>
<p>$$2S-S=(1^2+n^2)+(2^2+n^2-2n)+(3^2+n^2-4n)+\cdots+(n^2+n^2-2(n-1)n-n^2$$</p>
<p>Then if I collect all the $n$ terms;</p>
<p>$$2S-S=n\cdot (n-1)^2 +(1^2)+(2^2-2n)+(3^2-4n)+\cdots+(n^2-2(n-1)n$$</p>
<p>But then I realised I still had the original sum in there, and taking that out mean I no longer had a sum term to extract.</p>
<p>Have I made a mistake here? How can I arrive at the answer of $\dfrac{n (n + 1) (2 n + 1)}{6}$ using a method similar to the one I expound on above? <strong>I.e following Gauss' line of reasoning</strong>?</p>
| Fakemistake | 173,351 | <p>This answer uses the formula for the sum of odd numbers, which is
<span class="math-container">$$i^2=\sum_{k=1}^i 2k-1$$</span>
First insert this into the considered formula
<span class="math-container">$$s_n\overset{\mathrm{def}}{=}\sum_{i=1}^ni^2=\sum_{i=1}^{n}\sum_{k=1}^i (2k-1)$$</span></p>
<p>Definitely in the style of Gauss is writing out the sum for each <span class="math-container">$i$</span> and using one row for each <span class="math-container">$i$</span>:
<span class="math-container">\begin{align}
s_n&=\color{red}{1}\\
&+\,\color{red}{1}+\color{blue}{3}\\
&+\,\color{red}{1}+\color{blue}{3}+\color{green}{5}\\
&+\,\color{red}{1}+\color{blue}{3}+\color{green}{5}+7\\
&\;\vdots\\
&+\,\color{red}{1}+\color{blue}{3}+\color{green}{5}+7+\ldots+2n-1
\end{align}</span>
Now take the sum column by column beginning with the last to obtain
<span class="math-container">$$s_n=1\cdot (2n-1)+2\cdot (2n-3)+3\cdot (2n-5)+\ldots+\cdot \color{green}{(n-2)\cdot 5}+\color{blue}{(n-1)\cdot 3}+\color{red}{n\cdot 1}$$</span>
which is exactly the sum
<span class="math-container">$$s_n=\sum_{i=1}^n i\cdot(2n-(2i-1))=\sum_{i=1}^n i((2n+1)-2i)$$</span>
To get the final result, you only have to do simple calculations, you see here:
<span class="math-container">$$s_n=(2n+1)\sum_{i=1}^n i-2\sum_{i=1}^n i^2=(2n+1)\frac{n(n+1)}{2}-2s_n$$</span>
Thus
<span class="math-container">$$3s_n=\frac{n(n+1)(2n+1)}{2}\qquad \Rightarrow\qquad s_n=\frac{n(n+1)(2n+1)}{6}$$</span></p>
|
122,546 | <p>There is a famous proof of the Sum of integers, supposedly put forward by Gauss.</p>
<p>$$S=\sum\limits_{i=1}^{n}i=1+2+3+\cdots+(n-2)+(n-1)+n$$</p>
<p>$$2S=(1+n)+(2+(n-2))+\cdots+(n+1)$$</p>
<p>$$S=\frac{n(1+n)}{2}$$</p>
<p>I was looking for a similar proof for when $S=\sum\limits_{i=1}^{n}i^2$</p>
<p>I've tried the same approach of adding the summation to itself in reverse, and I've found this:</p>
<p>$$2S=(1^2+n^2)+(2^2+n^2+1^2-2n)+(3^2+n^2+2^2-4n)+\cdots+(n^2+n^2+(n-1)^2-2(n-1)n$$</p>
<p>From which I noted I could extract the original sum;</p>
<p>$$2S-S=(1^2+n^2)+(2^2+n^2-2n)+(3^2+n^2-4n)+\cdots+(n^2+n^2-2(n-1)n-n^2$$</p>
<p>Then if I collect all the $n$ terms;</p>
<p>$$2S-S=n\cdot (n-1)^2 +(1^2)+(2^2-2n)+(3^2-4n)+\cdots+(n^2-2(n-1)n$$</p>
<p>But then I realised I still had the original sum in there, and taking that out mean I no longer had a sum term to extract.</p>
<p>Have I made a mistake here? How can I arrive at the answer of $\dfrac{n (n + 1) (2 n + 1)}{6}$ using a method similar to the one I expound on above? <strong>I.e following Gauss' line of reasoning</strong>?</p>
| Patrick Sheehan | 599,007 | <p>This has a few parts, but should bridge the gap between what you were looking for (hopefully?) and the answers from Tyler and Pedro.</p>
<p>If I remember right, that method of Gauss' has a geometric interpretation that Gauss provided along with it. Namely that the sum 1+2+...+n can be visualized as a "staircase" of rectangles. Flipping the sum and adding it to the first one is putting a new rectangle on each of the original ones, but this time in descending order, so 1 will have n above it, 2 will have n-1, etc. Since the bottom portion of increasing by 1 each time the top part decreases by 1, every rectangle now has height n+1. There are n of them, so together they cover an area equal to n*(n+1). That uses two "staircases" though, so one "staircase" amounts to n*(n+1)/2, which is also (n^2+n)/2. (Note that in this last form the rectangle is a square with a single extra row of units at the top.)</p>
<p>So Gauss' method was to realize that you could put two staircases together to make a rectangle with an easy to calculate area and then divide by 2 to get back what you really wanted. Taking that thinking up to the sum of squares question then would potentially involve some 3D versions of staircases. You definitely can make some 3D staircase analogues. Put a 4x4x1 slab under a 3x3x1 slab under a 2x2x1 slab under a 1x1x1 cube and align them so they all would fit neatly in the corner of a box. There's your 3D version of a staircase made of sums of squares.</p>
<p>Now that we're up a dimension though, 2 sets of these stairs aren't going to fit together the way we want to make a nice convenient rectangular prism for us to easily find the volume of and then divide to get what we want. Turns out we need 3 of them.</p>
<p>That's where Tyler and Pedro's answers come in. They show how you can build a cube using 3 "3D-staircases", 3 "2D-staircases", n singles and an extra single. Their formula shows that in going from a cube of size n^3 to a cube of size (n+1)^3, the larger cube is made by adding 3 face-extensions of size nxnx1, 3 new outside edges of size 1x1xn, and a new corner of size 1x1x1. This means you can "peel" any cube down into a sum of these added layers. The three sets of faces make three 3D staircases, the three sets of edges make three 2D staircases, the new corners sum to n, and the last 1x1x1 cube is the "seed" all the layers were wrapped around.</p>
<p>So, that's how Gauss' method connects to this problem and also how Tyler and Pedro's answers are relevant to that discussion. In terms of why the algebra isn't working out, I think this breakdown should provide some ideas, first and foremost that you'll need 3 sums rather than 2 and that you'll probably need the 2D staircases as well.</p>
<p>I hope it goes well (or went well, if you resolved this long ago). Your question generated a lot of cool responses! Thanks!</p>
|
1,747,696 | <p>First of all: beginner here, sorry if this is trivial.</p>
<p>We know that $ 1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2 $ .</p>
<p>My question is: what if instead of moving by 1, we moved by an arbitrary number, say 3 or 11? $ 11+22+33+44+\ldots+11n = $ ?
The way I've understood the usual formula is that the first number plus the last equals the second number plus second to last, and so on.
In this case, this is also true but I can't seem to find a way to generalize it.</p>
| Community | -1 | <p>The general formula derives from the simple one.</p>
<p>The general sequence is $$a=a-b+b,a+b=a-b+2b,a+2b,\cdots a-b+nb,$$ i.e. $n$ terms from $a$ to $a+(n-1)b=a+c$.</p>
<p>Then</p>
<p>$$\sum_{k=1}^n(a-b+kb)=n(a-b)+\frac{n(n+1)}2b=n\frac{a+a+(n-1)b}2=n\frac{a+c}2.$$</p>
<hr>
<p>With $11$ to $11n$ in steps $11$,</p>
<p>$$n\frac{11+11n}2=11\frac{n(1+n)}2.$$</p>
|
290,132 | <p>Let $x,a,b$ be real numbers and $f(x)$ a (nongiven) real-analytic function.</p>
<p>How to find $f(x)$ such that for all $x$ we have $f(x)+af(x+1)=b^x$ ? </p>
<p>In particular I wonder most about the case $a=1$ and $b=e$. (I already know the trivial cases $a=-1$ and $a=0$)</p>
<p>I know how to express $f(x+1)$ into a taylor series once I have the taylor series for $f(x)$ and I assume this is related ? But does it help to find a closed form solution here ? If there is a closed form solution ? Am I on right track here or do we need to use something completely different or more general ? Does this relate to the fibonacci sequence ?</p>
| Haskell Curry | 39,362 | <p>Pick $ f(x) = \lambda e^{kx} $. Then
$$
f(x) + f(x + 1) = \lambda \left[ e^{kx} + e^{k(x + 1)} \right] = \lambda (1 + e^{k}) e^{kx}.
$$
By choosing $ k = 1 $ and $ \lambda = \dfrac{1}{1 + e} $, you get a solution.</p>
|
104,375 | <p>How I am supposed to transform the following function in order to apply the laplace transform.</p>
<p>$f(t) = t[u(t)-u(t-1)]+2t[u(t-1) - u(t-2)]$</p>
<p>I know that it has to be like this</p>
<p>$L\{f(t-t_0)u(t-t_0)\} = e^{-st_0}F(s), F(s) = L\{f(t)\}$</p>
| 000 | 22,144 | <p>This question is old, but I'd like to share my view of it. I hope this offends no one.</p>
<p>What you are dealing with are equivalence classes modulo an integer. That is, you are dealing with equivalence classes defined by $[a]=\{b \in \mathbb{Z}:b \cong a \pmod 3\}$ for an element $a$.</p>
<p>In the case of the relation of congruence modulo $n$, there are $n$ equivalence classes: $[0], [1], [2], \dots, [n-1]$. You can see that supposing $[n]$ or $[n+1]$ is silly, since $b \cong n \pmod n$ reduces to $b \cong 0\pmod 0$ and $b \cong n+1 \pmod n$ reduces to $b \cong 1 \pmod n$ (and, hence $[n]=[0]$ and $[n+1]=1$). In other words, $[0], [1], [2],\dots, [n-1]$ are all of the equivalence classes.</p>
<p>For congruence modulo $3$, we thusly have $[0], [1], $ and $[2]$. That is,
\begin{align}
[0]&=\{b \in \mathbb{Z}: b\cong 0 \pmod 3\}\\
&=\{3,6,9,\dots\}.\\
[1]&=\{b \in \mathbb{Z}: b\cong 1 \pmod 3\}\\
&=\{4,7,10,\dots\}.\\
[2]&=\{b \in \mathbb{Z}: b\cong 2 \pmod 3\}\\
&=\{5,8,11,\dots\}.
\end{align}</p>
<p>In other words, the three equivalence classes are numbers that divide $3$ with remainder $0$, all numbers that divide $3$ with remainder $1$, and all numbers that divide $3$ with remainder 2.</p>
<p>This is the beauty of equivalence classes: They partition the set they are defined on into disjoint sets which as a whole form the entire set. i.e. For any given equivalence relation $E$ on set $S$ and the equivalence classes $[a]=\{b \in S: aEb\}$, $$\bigcup_{a \in S}[a]=S.$$</p>
|
2,965,717 | <p>How would you prove that <span class="math-container">$$\displaystyle \prod_{k=1}^\infty \left(1+\dfrac{1}{2^k}\right) \lt e ?$$</span></p>
<p>Wolfram|Alpha shows that the product evaluates to <span class="math-container">$2.384231 \dots$</span> but is there a nice way to write this number? </p>
<p>A hint about solving the problem was given but I don't know how to prove the lemma.</p>
<p>Lemma : Let, <span class="math-container">$a_1,a_2,a_3, \ldots,a_n$</span> be positive numbers and let <span class="math-container">$s=a_1+a_2+a_3+\cdots+a_n$</span> then <span class="math-container">$$(1+a_1)(1+a_2)(1+a_3)\cdots(1+a_n)$$</span> <span class="math-container">$$\le 1+s+\dfrac{s^2}{2!}+\dfrac{s^3}{3!}+\cdots+\dfrac{s^n}{n!}$$</span></p>
| Jack D'Aurizio | 44,121 | <p>It can be easily shown that for any <span class="math-container">$x\in(0,1)$</span> we have</p>
<p><span class="math-container">$$ \frac{1+2x+\frac{4x^2}{3}+\frac{8x^3}{21}+\frac{16 x^4}{315}+\frac{32 x^5}{9765}}{1+x+\frac{x^2}{3}+\frac{x^3}{21}+\frac{x^4}{315}+\frac{x^5}{9765}} <1+x < \frac{1+2x+\frac{4x^2}{3}+\frac{8x^3}{21}+\frac{16 x^4}{315}+\frac{32 x^5}{9450}}{1+x+\frac{x^2}{3}+\frac{x^3}{21}+\frac{x^4}{315}+\frac{x^5}{9450}} $$</span>
and in general, by setting <span class="math-container">$D_m=\prod_{k=1}^{m}(2^k-1)$</span> and <span class="math-container">$p_n(x)=1+\sum_{k=1}^{n}\frac{x^k}{D_k}$</span>,
<span class="math-container">$$ \frac{p_n(2x)+\frac{2^{n+1}x^{n+1}}{D_{n+1}}}{p_n(x)+\frac{x^{n+1}}{D_{n+1}}}<1+x< \frac{p_n(2x)+\frac{2^{n+1}x^{n+1}}{D_{n}(2^{n+1}-2)}}{p_n(x)+\frac{x^{n+1}}{D_{n}(2^{n+1}-2)}}.$$</span>
By telescoping it follows that
<span class="math-container">$$ p_n(1)+\frac{1}{D_{n+1}}<\prod_{k\geq 1}\left(1+\frac{1}{2^k}\right)<p_n(1)+\frac{1}{D_n(2^{n+1}-2)} $$</span>
and by picking <span class="math-container">$n=4$</span> we have
<span class="math-container">$$ \frac{3326}{1395}<\prod_{k\geq 1}\left(1+\frac{1}{2^k}\right)<\frac{22531}{9450} $$</span>
such that the first figures of the middle term are <span class="math-container">$\color{green}{2.3842}$</span>.<br> By picking <span class="math-container">$n=10$</span> we get that the middle term is <span class="math-container">$\color{green}{2.384231029\ldots}$</span>.<br>
As a continued fraction
<span class="math-container">$$ \prod_{k\geq 1}\left(1+\frac{1}{2^k}\right)=\left[2; 2, 1, 1, 1, 1, 14, 1, 3, 1, 1, 6, 9, 18, 7, 1, 27,\ldots\right]$$</span>
while <span class="math-container">$e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,\ldots]$</span>.</p>
|
1,522,929 | <p>For every fixed $t\ge 0$ I need to prove that the sequence $\big\{n\big(t^{\frac{1}{n}}-1\big) \big\}_{n\in \Bbb N}$ is non-increasing, i.e.
$$n\big(t^{\frac{1}{n}}-1\big)\ge (n+1)\big(t^{\frac{1}{n+1}}-1\big)\;\ \forall n\in \Bbb N$$
I'm trying by induction over $n$, but got stuck in the proof for $n+1$:
<br/>
For n=2 its clear that follows since
$$t-1\ge 2(t^{1/2}-1)\Leftrightarrow t-1\ge 2t^{1/2}-2\Leftrightarrow t+1\ge 2t^{1/2}\Leftrightarrow t^2+2t+1\ge 4t\Leftrightarrow t^2-2t+1\ge 0\Leftrightarrow (t-1)^2\ge 0$$</p>
<p>So, we suppose that $\;\ n\big(t^{\frac{1}{n}}-1\big)\ge (n+1)\big(t^{\frac{1}{n+1}}-1\big)\;\ $ is valid. (I.H.)
<br/>
So I need to prove that:
$$(n+1)\big(t^{\frac{1}{n+1}}-1\big)\ge (n+2)\big(t^{\frac{1}{n+2}}-1\big)\ $$</p>
<p>But I have not reached anywhere helpful expanding all. Any ideas or different approaches to porve this will be appreciated.</p>
| user279043 | 224,391 | <p>All right, let's do some complex analysis! Let's integrate <span class="math-container">$e^{iz^2}$</span> over the closed contour defined in three pieces (the arrows indicating the direction of contour integration)
<span class="math-container">$$
\begin{cases}
\Gamma_1: & |z|:0\rightarrow R, & \theta=0 \\
\Gamma_2: & |z| = R, & \theta: 0\rightarrow \pi/4\\
\Gamma_3: & |z|:R\rightarrow0, & \theta=\pi/4
\end{cases}
$$</span>
which we will eventually want to take the limit <span class="math-container">$R\rightarrow\infty$</span>.</p>
<p>It can be seen that
<span class="math-container">$$
\int_0^\infty e^{ix^2}dx=\lim\limits_{R\rightarrow\infty}\int_{\Gamma_1}e^{iz^2}dz
$$</span>
and since <span class="math-container">$e^{ix^2}$</span> is an even function
<span class="math-container">$$
\int_{-\infty}^{\infty}e^{ix^2}dx=2\int_0^\infty e^{ix^2}dx
$$</span>
so we're heading in the right direction.</p>
<p>Now Cauchy's Theorem states that
<span class="math-container">$$
\oint_D f(z)dz =0
$$</span>
for <span class="math-container">$f(z)$</span> analytic in <span class="math-container">$D$</span>. Our function, <span class="math-container">$e^{iz^2}$</span>, has no singularities and is defined on the entire complex plane, so it is considered an entire function, and Cauchy's Theorem holds for our closed contour:
<span class="math-container">$$
\int_0^R e^{ix^2}dx+\int_{\Gamma_2}e^{iz^2}dz+\int_{\Gamma_3}e^{iz^2}dz=0
$$</span></p>
<p>For our second integral above, we show that it vanishes as <span class="math-container">$R\rightarrow\infty$</span> using the ML test given by
<span class="math-container">$$
\left|\int_\Gamma f(z)dz\right|\leq ML
$$</span>
where <span class="math-container">$M$</span> is a finite upper bound of <span class="math-container">$f(z)$</span> and <span class="math-container">$L$</span> is the length of the contour <span class="math-container">$\Gamma$</span>. Of course, we need to assume that <span class="math-container">$f(z)$</span> is bounded and analytic on <span class="math-container">$\Gamma$</span> for this.
In order to apply the ML test, we substitute into our integrand <span class="math-container">$z=re^{i\theta}$</span> so that
<span class="math-container">$$
z^2 = r^2e^{2i\theta} = r^2\cos(2\theta)+ir^2\sin(2\theta)
$$</span>
<span class="math-container">$$
|e^{iz^2}|=|e^{ir^2\cos(2\theta)-r^2\sin(2\theta)}|\leq e^{-R^2}=M
$$</span>
because <span class="math-container">$r=R$</span> on this contour and <span class="math-container">$\sin(2\theta)\leq1$</span>. While,
<span class="math-container">$$
L=\frac{\pi R}{4}
$$</span>
since we are looking at <span class="math-container">$1/8$</span>th of the perimeter of the circle with radius <span class="math-container">$R$</span>. By the ML test
<span class="math-container">$$
\left|\int_{\Gamma_2} e^{iz^2}dz\right| \leq e^{-R^2}\frac{\pi R}{4}
$$</span>
which goes to <span class="math-container">$0$</span> as <span class="math-container">$R\rightarrow\infty$</span>.</p>
<p>Now we want to deal with the 3rd contour integral <span class="math-container">$\Gamma_3$</span>. Fortunately, the contour we picked allows us to easily parameterize this integral, as <span class="math-container">$y=x$</span>. We will also need <span class="math-container">$z^2=(x+iy)^2=x^2-y^2+2ixy$</span>. Recalling that <span class="math-container">$dz=dx+idy$</span> the integral becomes
<span class="math-container">$$
\int_{\Gamma_3} e^{i(x^2-y^2)-2xy}(dx+idy)
=\int_{R}^{0} e^{-2x^2}dx+i\int_{R}^{0} e^{-2y^2}dy
\rightarrow-\sqrt{\frac{\pi}{8}}(1+i)\ \text{as}\ R\rightarrow0
$$</span>
from our real Gaussian integral identities.</p>
<p>Taking <span class="math-container">$R\rightarrow\infty$</span>, our results for the contour integrals in our Cauchy's Theorem equation imply that
<span class="math-container">$$
\int_0^\infty e^{ix^2}dx = \sqrt{\frac{\pi}{8}}(1+i)
$$</span></p>
<p>The integral from <span class="math-container">$-\infty$</span> to <span class="math-container">$\infty$</span> is just twice this. So boom.</p>
<p>If you want, you can rewrite <span class="math-container">$e^{ix^2}=\cos(x^2)+i\sin(x^2)$</span> and equate the real and imaginary parts in the last equation and you will get the limiting values of the Fresnel Integrals.</p>
<p>Boom.</p>
<p>Also, since
<span class="math-container">$$
(1+i)=\sqrt{2}e^{i\pi/4}=\sqrt{2e^{i\pi/2}}=\sqrt{2i}
$$</span>
we have
<span class="math-container">$$
\int_0^\infty e^{ix^2}dx = \sqrt{\frac{i\pi}{4}} = \frac{1}{2}\sqrt{-\frac{\pi}{i}}
$$</span></p>
<p>which exactly matches the well-known Gaussian integral identity
<span class="math-container">$$
\int_0^\infty e^{-\alpha x^2}dx = \frac{1}{2}\sqrt{\frac{\pi}{\alpha}}
$$</span>
with <span class="math-container">$\alpha=-i$</span>. Boom. Thus, this suggests that this identity can work for imaginary <span class="math-container">$\alpha$</span>, and possibly certain complex <span class="math-container">$\alpha$</span> with the right combination of real and imaginary parts as well as choice of contours that do not make our integrals blow up.</p>
|
825,703 | <p>I have been working with vector spaces for a while and I now take for granted what the vector space does. I feel like I dont really understand why multiplication and addition must be defined on a vector space. For example, it feels like adding two vectors and having their sum contained within the space is just a name for a vector space and I dont get what necessarily happens IF the two vector's sum arent in the space. In other words, I dont know why must addition and multiplication must be defined on a vector space, is it to take advantage of nice properties? Thanks!</p>
| 4pie0 | 29,621 | <p>The sum and multiplication of two vectors $u,v$ from a linear space $V$ has to belong to $V$ because we want to consider elements that share some common properties, in particular they all can be expressed as a linear combination of vectors that form any base of $V$.</p>
<p>It is the same as in case of groups. In example we can express $V$ as a quotient set: a set of equivalence classes. $u+v$ has to belong to some equivalence class.</p>
|
4,001,031 | <p>(For all those that it may concern, this is not a duplicate of my previous post, But starts in a similar way.)</p>
<p>A triangle with side lengths a, b, c with a height(h) that intercepts the hypotenuse(c) at (x , y) such that it is split into two side lengths, c = m + n, we can find Pythagoras theorem using the area of a right triangle and the slope equations of the height and the hypotenuse.</p>
<p><img src="https://upload.wikimedia.org/wikipedia/commons/f/fb/Pythagoras_similar_triangles_simplified.svg" alt="Text" /></p>
<p>We begin by finding the coordinate (x,y) by using the concept of the area fourmula, ab = hc:</p>
<p>x - value:</p>
<p><span class="math-container">$bx = hm$</span></p>
<p><span class="math-container">$x= \frac{hm}{b}$</span></p>
<p><span class="math-container">$x = \frac{am}{c}$</span></p>
<p>y- value:</p>
<p><span class="math-container">$ay = hn$</span></p>
<p><span class="math-container">$y = \frac{hn}{a}$</span></p>
<p><span class="math-container">$y = \frac{bn}{c}$</span></p>
<p>which gives us:</p>
<p><span class="math-container">$(\frac{am}{c}, \frac{bn}{c})$</span></p>
<p>We can also find the (x,y) coordinates using the slope equations of the height and hypotenuse.</p>
<p>Height's equation:</p>
<p><span class="math-container">$y = \frac{b}{a}x$</span></p>
<p>Hypotenuse's equation:</p>
<p><span class="math-container">$y = \frac{-a}{b}x + a$</span></p>
<p>Now we can determine the (x ,y )intercept.</p>
<p>x - value:</p>
<p><span class="math-container">$\frac{b}{a}x = \frac{-a}{b}x + a$</span></p>
<p><span class="math-container">$\frac{b}{a}x + \frac{a}{b}x = a$</span></p>
<p><span class="math-container">$x(a^2 + b^2) = a^2b$</span></p>
<p><span class="math-container">$x = \frac{a^2b}{a^2 + b^2}$</span></p>
<p>y - value:</p>
<p><span class="math-container">$y = \frac{b}{a}(\frac{a^2b}{a^2 + b^2})$</span></p>
<p><span class="math-container">$y = \frac{ab^2}{a^2 + b^2}$</span></p>
<p>Which gives us:</p>
<p><span class="math-container">$(\frac{a^2b}{a^2 + b^2}
,\frac{ab^2}{a^2 + b^2})$</span></p>
<p>Using <span class="math-container">$(\frac{bm}{c},\frac{an}{c}) ,(\frac{ab^2}{a^2 + b^2} , \frac{a^2b}{a^2 + b^2})$</span> there are 2 equalities:</p>
<p><span class="math-container">$\frac{bm}{c} = \frac{a^2b}{a^2 + b^2}$</span></p>
<p><span class="math-container">$\frac{an}{c} = \frac{ab^2}{a^2 + b^2}$</span></p>
<p>Which after isolating and eliminating c becomes:</p>
<p><span class="math-container">$a^2m = b^2n$</span></p>
<p>This gives us two expressions:</p>
<p><span class="math-container">$\sqrt{\frac{n}{m}} = \frac{a}{b}$</span></p>
<p><span class="math-container">$\sqrt{\frac{m}{n}} = \frac{b}{a}$</span></p>
<p>We substitute these into <span class="math-container">$c = m + n$</span>:</p>
<p><span class="math-container">$c = m + n$</span></p>
<p><span class="math-container">$\frac{c}{\sqrt{m}{n}} = \sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}}$</span></p>
<p><span class="math-container">$\frac{ab}{\sqrt{m}{n}}c = a^2 + b^2 $</span></p>
<p>Using <span class="math-container">$h$</span>, this can be written as:</p>
<p><span class="math-container">$1 = (\frac{\sqrt{m}{n}}{h})(\frac{a^2 + b^2}{c^2})$</span></p>
<p>or:</p>
<p><span class="math-container">$1 = (\frac{h}{\sqrt{m}{n}})(\frac{c^2 }{a^2 + b^2})$</span></p>
<p>Note, this leaves us with only two possibilities, the fractions are either inverses, or the numerator and denominator are equal. We know <span class="math-container">$h = ab/c$</span> and <span class="math-container">$h < c$</span>, so <span class="math-container">$h ≠ a^2 + b^2$</span>, the denominator can also be split into <span class="math-container">$hc, c$</span> but we know <span class="math-container">$hc = ab$</span> so <span class="math-container">$ab ≠ a^2 + b^2$</span>, and we know the side lengths <span class="math-container">$m,n$</span> are smaller than <span class="math-container">$a,b$</span> which means the numerator and denominator are equal in the case of:</p>
<p><span class="math-container">$a^2 + b^2 = c^2$</span></p>
| Community | -1 | <p>It would have been much simpler to use your initial equations as follows.</p>
<p>Since <span class="math-container">$hc=ab$</span>, you have <span class="math-container">$m=\frac{bh}{a}=\frac{b^2}{c}$</span></p>
<p>Similarly, <span class="math-container">$n=\frac{ba^2}{c}$</span>.</p>
<p>Now express <span class="math-container">$m+n=c$</span> as <span class="math-container">$\frac{a^2}{c}+\frac{b^2}{c}=c$</span>
and you have <span class="math-container">$a^2+b^2=c^2$</span>.</p>
|
1,737,674 | <p>I am trying to understand how to find all congruence classes in $\mathbb{F}_2[x]$ modulo $x^2$. How can I compute them ? Can someone get me started with this? I am having trouble understanding $\mathbb{F}_2[x] $ is it the set $\{ f(x) = a_nx^n + ...+ a_1 x + a_0 : a_i = 0,1 \} $?</p>
| ashi | 299,991 | <p>I think the correct is answer equal to $75°$</p>
<p>I did it by construction</p>
<p>First of all I made triangle $BDC$ with ur provided information and doing some calculations ie ($\measuredangle DBC=15°$, $\measuredangle BDC=120^{\circ}$ ,$\measuredangle BCD=45°$) than I continued line $CD$ till point $A$ with ur information ($2CD=AD$) as I got point $A$ , I joined it with $B$ and calculated $\measuredangle BAD$ which comes up to be $75°$</p>
|
1,737,674 | <p>I am trying to understand how to find all congruence classes in $\mathbb{F}_2[x]$ modulo $x^2$. How can I compute them ? Can someone get me started with this? I am having trouble understanding $\mathbb{F}_2[x] $ is it the set $\{ f(x) = a_nx^n + ...+ a_1 x + a_0 : a_i = 0,1 \} $?</p>
| Piquito | 219,998 | <p>In the figure below we have $$\frac ac=\frac{\sin x}{\sin 45^\circ}\qquad(1)$$ Furthermore
$$\frac{z}{\sin 15^\circ}=\frac{a}{\sin 120^\circ}\qquad(2)$$
$$\frac{2z}{\sin (120^\circ-x)}=\frac{c}{\sin 60^\circ}\qquad(3)$$</p>
<p>From $(1),(2),(3)$ we get $$1.46407\sin x=1.7320\cos x+\sin x\iff \tan x=3.73219$$</p>
<p>Thus $$\color{red}{x\approx 75.00053^\circ}$$
<a href="https://i.stack.imgur.com/oJJJk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oJJJk.png" alt="enter image description here"></a></p>
|
2,245,631 | <blockquote>
<p>$x+x\sqrt{(2x+2)}=3$</p>
</blockquote>
<p>I must solve this, but I always get to a point where I don't know what to do. The answer is 1.</p>
<p>Here is what I did: </p>
<p>$$\begin{align}
3&=x(1+\sqrt{2(x+1)}) \\
\frac{3}{x}&=1+\sqrt{2(x+1)} \\
\frac{3}{x}-1&=\sqrt{2(x+1)} \\
\frac{(3-x)^{2}}{x^{2}}&=2(x+1) \\
\frac{9-6x+x^{2}}{2x^{2}}&=x+1 \\
\frac{9-6x+x^{2}-2x^{2}}{2x^{2}}&=x \\
\frac{9-6x+x^{2}-2x^{2}-2x^{3}}{2x^{2}}&=0
\end{align}$$</p>
<p>Then I got:
$-2x^{3}-x^{2}-6x+9=0$ </p>
| Javi | 434,862 | <p>I usually check if the integer divisors of the independent term of the polynomial is a root in order to decompose it in factors. </p>
|
3,840,692 | <p>The equation is <span class="math-container">$2z^2w''+3zw'-w=0$</span></p>
<p><span class="math-container">$z_0=0$</span> is a regular singular point, so <span class="math-container">$w(z)=\sum_{n=0}^{\infty} a_nz^{n+r}$</span></p>
<p>then <span class="math-container">$w'(z)=\sum_{n=0}^{\infty} (n+r)a_nz^{n+r-1}$</span> and <span class="math-container">$w''(z)=\sum_{n=0}^{\infty} (n+r)(n+r-1)a_nz^{n+r-2}$</span></p>
<p>Replacing in the equation:</p>
<p><span class="math-container">$2z^2\sum_{n=0}^{\infty} (n+r)(n+r-1)a_nz^{n+r-2}+3z\sum_{n=0}^{\infty} (n+r)a_nz^{n+r-1}-\sum_{n=0}^{\infty} a_nz^{n+r}=0$</span></p>
<p><span class="math-container">$\sum_{n=0}^{\infty} 2(n+r)(n+r-1)a_nz^{n+r}+\sum_{n=0}^{\infty} 3(n+r)a_nz^{n+r}-\sum_{n=0}^{\infty} a_nz^{n+r}=0$</span></p>
<p><span class="math-container">$\sum_{n=0}^{\infty} [2(n+r)(n+r-1)a_nz^{n+r}+ 3(n+r)a_nz^{n+r}- a_nz^{n+r}]=0$</span></p>
<p><span class="math-container">$2(n+r)(n+r-1)a_n+ 3(n+r)a_n- a_n=0$</span></p>
<p><span class="math-container">$(2(n+r)(n+r-1)+3(n+r)-1) a_n=0$</span></p>
<p>At this point, how can I construct a recurrence relation since I only got <span class="math-container">$a_n$</span>?, I'd need <span class="math-container">$a_{n+1}$</span>, right?</p>
<p>I already found the indicial equation and its roots, which are <span class="math-container">$r_1=1/2$</span> and <span class="math-container">$r_2=-1$</span></p>
| Claude Leibovici | 82,404 | <p>Notice that, for this problem, Frobelius method is not required. Let <span class="math-container">$w(z)=z u(z)$</span> to end with
<span class="math-container">$$6u''+13u'=0$$</span> Just reduction of order and a quite simple problem.</p>
|
155,024 | <p>I am trying to plot a polynomial inside <code>Manipulate</code> with indexed coefficients.</p>
<p>Surprisingly, only an empty plot is generated.</p>
<p>Any help welcome.</p>
<pre><code>Block[{n = 3},
Manipulate @@ {
Column[{
Sum[a[i] x^i, {i, 0, n}],
Plot[Evaluate@Sum[a[i] x^i, {i, 0, n}], {x, -5, 5}]
}],
Sequence @@ Table[{{a[i], 0}, -2, 2}, {i, 0, n}]
}]
</code></pre>
<p>Regards
Robert</p>
| jkuczm | 14,303 | <p>On my computer simple <code>LibraryFunction</code>, that compares subsequent elements in a loop, is fastest.</p>
<pre><code>pairwiseBooleJkuczm = Last@Compile[{{data, _Real, 1}},
Table[
Boole[Compile`GetElement[data, i] === Compile`GetElement[data, i + 1]],
{i, Length@data - 1}
],
CompilationTarget -> "C",
RuntimeOptions -> {"CatchMachineIntegerOverflow" -> False, "CompareWithTolerance" -> True}
];
SeedRandom@0;
data = Developer`ToPackedArray@
Union[RandomChoice[100 Sin[Subdivide[0, 7, 40]], 10^6] +
RandomReal[{0, 10^-7}, 10^6]];
(res1 = pairwiseBoole@data) // MaxMemoryUsed // RepeatedTiming
(res2 = pairwiseBooleC@data) // MaxMemoryUsed // RepeatedTiming
(res3 = pairwiseBooleJkuczm@data) // MaxMemoryUsed // RepeatedTiming
res1 === res2 === res3
(* {0.563, 175791672} *)
(* {0.044, 15981232} *)
(* {0.013, 7990696} *)
(* True *)
</code></pre>
|
155,024 | <p>I am trying to plot a polynomial inside <code>Manipulate</code> with indexed coefficients.</p>
<p>Surprisingly, only an empty plot is generated.</p>
<p>Any help welcome.</p>
<pre><code>Block[{n = 3},
Manipulate @@ {
Column[{
Sum[a[i] x^i, {i, 0, n}],
Plot[Evaluate@Sum[a[i] x^i, {i, 0, n}], {x, -5, 5}]
}],
Sequence @@ Table[{{a[i], 0}, -2, 2}, {i, 0, n}]
}]
</code></pre>
<p>Regards
Robert</p>
| Carl Woll | 45,431 | <p>Here is an approach assuming that the <a href="http://reference.wolfram.com/language/ref/SameQ" rel="noreferrer"><code>SameQ</code></a> tolerance is not changed from its default.</p>
<p><strong>\$MachineEpsilon</strong></p>
<p><code>$MachineEpsilon</code> is the smallest number that when added to <code>1.</code> produces a different number. This means that <code>1.</code> and <code>1.+ $MachineEpsilon</code> will differ by 1 in the last bit of the base 2 representation of the mantissa. Hence the two numbers will satisfy the <a href="http://reference.wolfram.com/language/ref/SameQ" rel="noreferrer"><code>SameQ</code></a> predicate. Any larger numbers will not. This suggests the following algorithm for deciding the <a href="http://reference.wolfram.com/language/ref/SameQ" rel="noreferrer"><code>SameQ</code></a> predicate:</p>
<p><strong>Algorithm</strong></p>
<p>Let $l$ and $u$ be two numbers to be compared, with $l<u$. Then, the <a href="http://reference.wolfram.com/language/ref/SameQ" rel="noreferrer"><code>SameQ</code></a> predicate should be equivalent to:</p>
<p>$$\frac{u - l}{\left|u\right|} \le \epsilon$$</p>
<p>We can rewrite this as:</p>
<p>$$\epsilon \left|u\right| + (l-u) \ge 0$$</p>
<p>Finally, if we use a Heaviside theta function where <code>0.</code> is mapped to <code>1</code>, then we can use the following arithmetical expression:</p>
<p>$$\theta(\epsilon \left|u\right| + (l-u))$$</p>
<p>which will return 1 if $l$ and $u$ are <a href="http://reference.wolfram.com/language/ref/SameQ" rel="noreferrer"><code>SameQ</code></a>, and 0 otherwise. I'm not sure about the correctness of this algorithm, but it passes all of the tests that I could think of.</p>
<p><strong>Mathematica implementation</strong></p>
<p>Here is the Mathematica implementation:</p>
<pre><code>pairwiseSameQ[data_] := With[{u=Rest[data], l=Most[data]},
UnitStep[$MachineEpsilon Abs[u] + (l-u)]
]
</code></pre>
<p><strong>Tests</strong></p>
<p>Let's compare this function to my slow function:</p>
<pre><code>data = Developer`ToPackedArray @ Union[
RandomChoice[100 Sin[Subdivide[0,7,40]], 10^6] +
RandomReal[{0, 10^-7}, 10^6]
];
r1 = pairwiseBoole[data]; //RepeatedTiming
r2 = pairwiseSameQ[data]; //RepeatedTiming
r1 === r2
</code></pre>
<blockquote>
<p>{0.651, Null}</p>
<p>{0.0109, Null}</p>
<p>True</p>
</blockquote>
<p>We get the same results, and the function is faster than the <a href="http://reference.wolfram.com/language/ref/Compile" rel="noreferrer"><code>Compile</code></a> alternatives in the other answers. What happens if we <a href="http://reference.wolfram.com/language/ref/Compile" rel="noreferrer"><code>Compile</code></a> this approach?</p>
<p><strong>Compile</strong></p>
<p>Following @jkuczm's approach, a compiled version might look like:</p>
<pre><code>pairwiseSameQC = With[{e=$MachineEpsilon},
Last @ Compile[{{d, _Real, 1}},
Table[
Boole[e Abs[Compile`GetElement[d,i+1]]>=(Compile`GetElement[d,i+1]-Compile`GetElement[d,i])],
{i, Length@d-1}
],
CompilationTarget->"C",
RuntimeOptions->"Speed"
]
];
</code></pre>
<p>One final comparison:</p>
<pre><code>r3 = pairwiseSameQC[data]; //RepeatedTiming
r1 === r2 === r3
</code></pre>
<blockquote>
<p>{0.0016, Null}</p>
<p>True</p>
</blockquote>
<p>Now, that's pretty fast!</p>
|
254,695 | <p>The concept of dimension seems to be:</p>
<blockquote>
<p>In physics and mathematics, the dimension of a space or object is
informally defined as the minimum number of coordinates needed to
specify any point within it.</p>
</blockquote>
<p>According to <a href="http://en.wikipedia.org/wiki/Dimension_%28mathematics_and_physics%29" rel="nofollow">wikipedia.</a> But the fourth dimension seems to have also some kind of connection with <a href="http://en.wikipedia.org/wiki/Spacetime" rel="nofollow">spacetime</a> which seems to be related to <a href="http://en.wikipedia.org/wiki/Minkowski_space" rel="nofollow">Minkowski spaces</a>. I want to understand dimensionality and also learn about these issues about spacetime.</p>
<p>I'm searching for references on what I should read in order to understand this, I'm searching for a serious way (no pop-science) to understand it, I'm searching for a list of topics and also some recomendations on textbooks for it.</p>
<p>I hope I'm not asking too much, but I'm very curious to grasp this subject. I'm also realistic on this, I do not think it's something easy of fast to learn. </p>
<p><strong>EDIT</strong>: I've found some books on Minkowski spaces, I guess I'll be able to understand from those in the near future.</p>
| Nigel Galloway | 63,512 | <p>A big difference between 4d space and spacetime is its size. Consider the size (Cardinality) of all the sets which have a one to one mapping between the digits on my hand and their members to be 5. Define the Cardinality of the set of all integers as $\aleph_0$, said as aleph-null. Note that it is possible to produce a 1 to 1 mapping between the set of integers and the set of even integers. This is a property of infinite sets, they have 1 to 1 relationships with proper subsets of themselves, and these subsets have the same cardinality as the original set.</p>
<p>Note that $\aleph_0+\aleph_0=\aleph_0$. The set of Real numbers is composed of Integers + Fractions + Irrational + Transcendental numbers. The cardinality of Integers Fractions and Irrational numbers is $\aleph_0$. The cardinality of transendental numbers is $\aleph$ which is larger than $\aleph_0$ (probably $\aleph_1$ but not proven to be). $\aleph_0+\aleph_0+\aleph_0+C=C$ and this is the cardinality of the points on a line. A line is a proper subset of 2d space so the cardinality of 2d (and by extension any_d) space is C.</p>
<p>Spacetime may be imagined as your monitor consisting of C pixels each of which may represent any colour. Replacing the monitor with 3d space and the colour with time infers that spacetime has C points. This is F (probably $\aleph_2$ again not proven) because it is also the cardinality of the set of all single valued functions.</p>
<p>We have N < C < F. This makes it possible to resolve Zeno's paradoxes by having the arrow stationary in spacetime but moving in 3d.</p>
<p>This is called Relativity and would be the mathematical framework for a 3d universe moving through some other thing. Special Relativity and General Relativity continue the Relativity tag because it was thought that the universe would be explained based on this model. It turned out that realizing that the arrow and the observer have a different understanding of what is NOW is more fruitful.</p>
|
4,339,772 | <p>The problem is stated as:</p>
<blockquote>
<p>Show that <span class="math-container">$\int_{0}^{n} \left (1-\frac{x}{n} \right ) ^n \ln(x) dx = \frac{n}{n+1} \left (\ln(n) - 1 - 1/2 -...- 1/{(n+1)} \right )$</span></p>
</blockquote>
<p><strong>My attempt</strong></p>
<p>First of all, we make the substitution <span class="math-container">$1-\frac{x}{n} = t$</span>, we then have that the integral can be rewritten as:</p>
<p><span class="math-container">$\int_{1}^{0} -n t^n \ln(n(1-t)) dt = \int_{0}^{1} n t^n \ln(n(1-t)) dt$</span></p>
<p>Using logarithmic laws, we can split the integral into two seperate ones as follows:</p>
<p><span class="math-container">$\int_{0}^{1} n t^n \ln(n(1-t)) dt = \int_{0}^{1} n t^n \ln(n) dt + \int_{0}^{1} n t^n \ln(1-t) dt$</span></p>
<p>We calculate each integral from the sum above:</p>
<p><span class="math-container">$ I_1 := \int_{0}^{1} n t^n \ln(n) dt = \frac{n}{n+1}\ln(n)$</span></p>
<p><span class="math-container">$ I_2 := \int_{0}^{1} n t^n \ln(1-t) dt = -n\int_{0}^{1} t^n \sum_{k=1}^{\infty}\frac{t^k}{k} dt$</span></p>
<p>Since the radius of convergence of <span class="math-container">$\sum_{k=1}^{\infty}\frac{t^k}{k}$</span> is 1, and we are integrating from <span class="math-container">$0$</span> to <span class="math-container">$1$</span>, we can interchange the order of limit operations. Meaning, we can calculate the integral first.</p>
<p><span class="math-container">$ I_2 = -n\sum_{k=1}^{\infty}\int_{0}^{1}\frac{t^{(n+k)}}{k} dt = -\sum_{k=1}^{\infty} \frac{n}{k(n+k+1)} = \frac{-n}{n+1} \sum_{k=1}^{\infty} \frac{n+1}{k(n+k+1)}$</span></p>
<p>Using partial fraction decomposition, we have that <span class="math-container">$I_2$</span> can be written as:</p>
<p><span class="math-container">$\frac{-n}{n+1}\sum_{k=1}^{\infty} \frac{n+1}{k(n+k+1)} = \frac{-n}{n+1} \sum_{k=1}^{\infty} \frac{1}{k} + \frac{n}{n+1}\sum_{k=1}^{\infty} \frac{1}{n+k+1}$</span></p>
<p>Putting it all together we get:</p>
<p><span class="math-container">$I_1 + I_2 = \frac{n}{n+1} \left ( \ln(n) - \sum_{k=1}^{\infty} \frac{1}{k} + \sum_{k=1}^{\infty} \frac{1}{n+k+1} \right )$</span></p>
<p>Which is indeed close the the result sought, however, I don't really know what to do with the last sums, and why I did wrong in choosing <span class="math-container">$\infty$</span> as an upper limit in the summation. I see that the sum of <span class="math-container">$1/k's$</span> diverge, but how can I avoid this?</p>
<p>Thank you for any help that could help me complete the last step of this problem.</p>
| Claude Leibovici | 82,404 | <p><em>Just for your curiosity</em></p>
<p>Assuming that you enjoy the gaussian hypergeometric function, there is an antiderivative
<span class="math-container">$$I_n(x)=\int \left (1-\frac{x}{n} \right ) ^n \log(x)\, dx=\frac 1{n^n}\int(n-x)^n \log(x)\,dx $$</span>
<span class="math-container">$$I_n(x)=-\frac{(n-x)^{n+1}}{n^{n+1} (n+1) (n+2)}\Big[(n-x) \, _2F_1\left(1,n+2;n+3;\frac{n-x}{n}\right)+n (n+2) \log (x) \Big]$$</span>
<span class="math-container">$I_n(n)=0$</span> and then
<span class="math-container">$$\int_0^n \left (1-\frac{x}{n} \right ) ^n \log(x)\, dx=\frac{n (\log (n)-\psi (n+2)-\gamma )}{n+1}=\frac{n }{n+1}\left(\log (n)-H_{n+1}\right)$$</span></p>
|
3,096,115 | <p>By using a sieve created by Prime Number Tables set up by the formula PN+(PNx6) for numbers generated by 6n+or-1, takes 182 calculations to identify 170 composite numbers. Using the Sieve of Eratosthenes would take around 1600 calculations. The Prime Number Tables identify all the composite numbers on the the list of 332 possible prime numbers: </p>
<p>5,11,17,23,29,35,41,47,53,59,65,71,77,83,89,95,101,107,113,119,125,131,137,143,149,155,161...</p>
<p>7,13,19,25,31,37,49,55,61,67,73,79,85,91,97,103,109,115,121,127,133,139,145,151,157,163,169...</p>
<p>Prime Number Table for:</p>
<p>5 identifies the multiples of 5 with 67 calculations (the Sieve of E: 249)</p>
<p>7: 23 calculations (the Sieve of E: 133)</p>
<p>11: 29 calculations (the Sieve of E: 98)</p>
<p>13: 11 calculations (the Sieve of E: 75)</p>
<p>17: 17 calculations (the Sieve of E: 57)</p>
<p>19: 4 calculations (the Sieve of E: 51)</p>
<p>23: 12 calculations (the Sieve of E: 42)</p>
<p>29: 5 calculations (the Sieve of E: 34)</p>
<p>31: 1 calculations (the Sieve of E: 31)</p>
<p>41: 3 47: 3 53: 2 59: 2 calculations (the Sieve of E:0)</p>
<p>71: 2 89: 1 101: 1 calculations (the Sieve of E: 0)</p>
<p>107: 1 113: 1 131: 1 137: 1 calculations (the Sieve of E: 0)</p>
<p>2: 0 calculations (the Sieve of E:499)</p>
<p>3: 0 calculations (the Sieve of E: 332)</p>
<p>Total Calculations:</p>
<p>A Sieve using PN Tables: 187 calculations to find 166 Prime Numbers by identifying 166 composite numbers (10 of 11 duplicate of multiples of 5)</p>
<p>Sieve of Eratosthenes: 1601 calculations to find 168 Prime Numbers by identifying 832 composite numbers (769 duplication of calculations)</p>
<p><em>Note: What I am really hoping for is some help. I have tested this up to 1411. There is no reason to believe it wouldn't go to whatever number. It seems since it deals with less numbers and less calculations, it would use less memory. If you look at the tables and what I have been able to research it makes Primes numbers even more interesting for children who might then take up more interest in math. Hey, I am a guy who works in a grocery store who just likes to think about things. I need help. People keep telling me about the Sieve of Eratosthenes. I have given a comparison between the 2 sieves. Would you rather make 1600 calculations or 187?</em></p>
<p>You can check on my website: <a href="https://mrspudgetty.wixsite.com/mr-spudgetty/prime-numbers" rel="nofollow noreferrer">https://mrspudgetty.wixsite.com/mr-spudgetty/prime-numbers</a></p>
| Keith Backman | 29,783 | <p>Here is the problem with your method and its objective: You use the Sieve of Eratosthenes as a standard, and apply it to a list of <span class="math-container">$1000$</span> numbers to determine a minimum number of calculations to identify all prime numbers in that range. But you start with a sublist (Prime Number Tables) of the first <span class="math-container">$1000$</span> numbers to reduce the number of calculations by your method. You have 'hidden' the number of calculations that you used to generate your Prime Number Tables, so the comparison of getting from that starting point to the finish is inappropriate vs starting from scratch. If you add the number of calculations it took you to create your Prime Number Tables, I doubt that your method is significantly more efficient than the Sieve of Eratosthenes.</p>
|
2,688,608 | <p>Assume matrix </p>
<p>$$A=
\begin{bmatrix}
-1&0&0&0&0\\
-1&1&-2&0&1\\
-1&0&-1&0&1\\
0&1&-1&1&0\\
0&0&0&0&-1
\end{bmatrix}
$$</p>
<p>Its Jordan Canonical Form is
$$J=
\begin{bmatrix}
-1&1&0&0&0\\
0&-1&0&0&0\\
0&0&-1&0&0\\
0&0&0&1&1\\
0&0&0&0&1
\end{bmatrix}
$$</p>
<p>I am trying to find a nonsingular $P$, let $P=\begin{bmatrix}\mathbf{p}_1&\mathbf{p}_2&\mathbf{p}_3&\mathbf{p}_4&\mathbf{p}_5\end{bmatrix}$ s.t. $J=P^{-1}AP\Leftrightarrow AP=PJ$.
I came up with the Wikipedia article on JCF and I think I need to find the generalized eigenvectors so that
$AP=PJ=\begin{bmatrix}-\mathbf{p_1}&\mathbf{p_1}-\mathbf{p_2}&-\mathbf{p_3}&\mathbf{p_4}&\mathbf{p_4}+\mathbf{p_5}\end{bmatrix}$ yielding the systems
$$(A+I)\mathbf{p_1}=\mathbf{0}$$
$$(A+I)^2\mathbf{p_2}=\mathbf{0}$$
$$(A+I)\mathbf{p_3}=\mathbf{0}$$
$$(A-I)\mathbf{p_4}=\mathbf{0}$$
$$(A-I)^2\mathbf{p_5}=\mathbf{0}$$</p>
<p>I solved each of these systems making sure that the vectors $\mathbf{p_i}$ I chose are linearly independent. So I chose
$$P=\begin{bmatrix}\mathbf{p}_1&\mathbf{p}_2&\mathbf{p}_3&\mathbf{p}_4&\mathbf{p}_5\end{bmatrix}=\begin{bmatrix}1&2&-2&0&0\\1&1&2&0&1\\1&1&2&0&0\\0&0&0&1&1\\1&1&-2&0&0\end{bmatrix}$$
which even though is nonsingular I am not getting $AP=PJ$.</p>
<p>What am I doing wrong?</p>
| Will Jagy | 10,400 | <p>I got the Jordan blocks in slightly different order.</p>
<p>What you seem to be missing is the consistency part: in my</p>
<p>$$ P =
\left(
\begin{array}{rrrrr}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
1 & 0 & 1 & 0 & 0
\end{array}
\right)
$$</p>
<p>we have a place where we do have $(A+I)^2 p_3 = 0,$ <strong>but we have consistency in that $p_2 =(A+I)p_3 .$</strong> It follows automatically that $(A+I)p_2 = (A+I)^2 p_3 = 0.$</p>
<p>We also have $(A-I)^2 p_5 = 0,$ <strong>then $p_4 =(A-I)p_5.$</strong> As a result $(A-I) p_4 = 0.$</p>
<p>$$
\left(
\begin{array}{rrrrr}
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
-1 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 \\
0 & 1 & -1 & 0 & 0
\end{array}
\right)
\left(
\begin{array}{rrrrr}
-1 & 0 & 0 & 0 & 0 \\
-1 & 1 & -2 & 0 & 1 \\
-1 & 0 & -1 & 0 & 1 \\
0 & 1 & -1 & 1 & 0 \\
0 & 0 & 0 & 0 & -1
\end{array}
\right)
\left(
\begin{array}{rrrrr}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
1 & 0 & 1 & 0 & 0
\end{array}
\right) =
\left(
\begin{array}{rrrrr}
-1 & 0 & 0 & 0 & 0 \\
0 & -1 & 1 & 0 & 0 \\
0 & 0 & -1 & 0 & 0 \\
0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 1
\end{array}
\right)
$$</p>
<p>To get the Jordan form in the order they report, use my columns but permuted, $p_2p_3p_1p_4p_5$ and then correct $P^{-1}.$ We can correct $P^{-1}$ by permuting the <strong>rows</strong> to 23145.</p>
|
838,690 | <p>True or false question</p>
<p>If B is a subset of A then {B} is an element of power set A. </p>
<p>I think this is true.</p>
<p>Because B is {1,2} say A {1,2,3} then power set of includes </p>
<p>$\{\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{3,2\},\{1,2,3\},\emptyset\}$</p>
<p>Unless {B} means $\{\{1,2\}\}$</p>
| amWhy | 9,003 | <p>It is true that $B$ is in the power set of a set $A$ (we'll call the powerset $P(A)$) is the set of all <strong><em>subsets</em></strong> of $A$, so the elements of $P(A)$ include all subsets of $A$.</p>
<p>Since $B$ is given to be a subset of $A$, then it is an element in the powerset of $A$.</p>
<p>However, $\{B\}$ is a set containing the subset $B$ as its only element, and so $\{B\}$ is not in the powerset of $A$.</p>
|
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