Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
What is the area of the triangle here?
We are given that the angle of $BAD$ is $2\alpha$ and the angle of $DAC$ is $\alpha$. $|AC| = 10$, $|BD| = 6$, $|DC| = 5$ units. Find the area of the triangle $ABC$.
The answer would be $33$. We need to show that if we drop an altitude from $A$ to $BC$ at point $E$, $|AE| = 6, |E... | Draw $AE$, the angle bisector of $\angle BAD$. Now $\angle A$ is divided into $3$ equal angles. Say, $DE = x$. As $AD$ bisects $\angle CAE$, using angle bisector theorem,
$ \displaystyle AE = 2 DE = 2x$
Using formula for the length of angle bisector,
$AD^2 = AE \cdot AC - DE \cdot CD = 15x$
Now using the fact that $AE$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4387342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The solution to $z^2+(1-i)z-1=0$? I am teaching myself complex numbers, yet I can't solve this exercise:
Find all complex number solutions to $z^2+(1-i)z-1=0,$ and provide them in standard $z=a+bi$ format.
I've tried using the classic formula $\dfrac{-b \pm \sqrt{D} }{2a},$
and have found that $D=4-2i$, which leads t... | Your answer is correct. Note the principal square root of a complex number is
$$\sqrt{x+iy}=\sqrt{\frac{x+\sqrt{x^2+y^2}}{2}}+isgn(y)\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}} .$$
Applied to your case gives
$$\sqrt{1-i/2}=\frac{1}{2}\sqrt{2+\sqrt{5}}-\frac{i}{2}\sqrt{-2+\sqrt{5}} .$$
So the solution to your quadratic is
$$z=-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4389754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $\gcd(2a+4b, 2a +8b)$, if $a\equiv b \pmod{\! 5},\ 6a+11b = 5$ Given:
*
*$a$ is even
*$6a+11b=5$
*$a-b=0\pmod 5$
Q: Calculate $\gcd(2a+4b,2a+8b)$
My try:
We know there is some $i$ such that $a=2i$, plus from 3 we know there is some $j$ such that: $a-b=5j$ which means $b=a-5j=2i-5j$. From 2, we get: $34... | This helps with the last step after @Buraian's answer.
Now $\gcd(a,b) \in \{1,5\}$ because there is an integral linear combination of $a$ and $b$ that sums to $5$; in particular, $6a+11b=5$.
Meanwhile, as $a$ is even, it follows that $\frac{a}{2}$ is an integer, and as $\gcd(2,5)=1$, if $5|a$ then $5|\frac{a}{2}$. [In ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4395440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Solving $(x+1)^2-x^2=0$ two ways gives different results In mathematics, there are several method of finding the solution of a particular problem. But in this equation,
$$(x+1)^2-x^2=0$$
Method 1
Using, $(a+b)^2=a^2+2ab+b^2$,
$$x^2+2x+1-x^2=0$$
So, $x=-1/2$ is the solution.
Method 2
$$(x+1)^2-x^2=0$$
$$(x+1)^2=x^2$$
$$... | Another way to see what went wrong is to use the factorization $a^2-b^2=(a+b)(a-b)$ like this:
$$(x+1)^2-x^2=0\\(x+1+x)(x+1-x)=0\\x+1+x=0~~~~\text{ or }~~~~x+1-x=0\\x=-\frac12~~~~\text{ or }~~~~1=0$$
Of the two resulting equations, only the first gives a valid solution for $x$. So in this case there is no $a-b=0$ solut... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How to decompose $\frac{1}{(1 + x)(1 - x)^2}$ into partial fractions Good Day.
I was trying to decompose $$\frac{1}{(1 + x)(1 - x)^2}$$ into partial fractions.
$$\frac{1}{(1 + x)(1 - x)^2} = \frac{A}{1 + x} + \frac{B}{(1 - x)^2}$$
$$1 = A(1 - x)^ 2 + B(1 + x)$$
Substitute $x = 1$, $$B = \frac{1}{2}$$
Substitute $x = -1... | Hint: $$1=\frac{(1+x)+(1-x)}2.$$ This should make the exercise trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Integrate $\int_0^{\infty} \frac{\sin^2 x}{\cosh x\>+\>\cos x}\frac{dx}x $ It is known that (see for example)
\begin{align}
&\int_0^{\infty} \frac{\sin x}{\cosh x+\cos x}\frac{dx}x =\frac\pi4\\
&\int_0^{\infty} \frac{\sin^3 x}{\cosh x+\cos x}\frac{dx}x =\frac\pi8
\end{align}
I am wondering if the similar integral below... | I'm not sure how you showed the two integrals are equivalent, but the following is an evaluation of $$\int_{0}^{\infty} e^{-x} \cos (x) \tanh(x) \, \frac{\mathrm dx}{x}.$$
For $\Re(s) >0$, we have
$$ \begin{align} \int_{0}^{\infty} \tanh(t) e^{-st} \, \mathrm dt &= \int_{0}^{\infty} \frac{1-e^{-2t}}{1+e^{-2t}} \, e^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4405099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
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Limit of some probability of the empirical mean of independent random variables Let $X_{1}, X_{2}, \ldots, X_{n}$ be a sequence of independent, standard Normal, real-valued random variables, and consider the empirical mean $\hat{S}_{n}=\frac{1}{n} \sum_{i=1}^{n} X_{i}$. Since $\hat{S}_{n}$ is again a Normal random vari... | Attempting to flesh out Henry's comment:
The left-hand side is $\frac{1}{n} \log [2(1-\Phi(\delta \sqrt{n}))]$, which has the same limit as $\frac{1}{n} \log(1-\Phi(\delta \sqrt{n}))$ (if the limits exist).
From integration by parts, we have the following Mills ratio bounds for $z>0$:
$$\frac{1}{z} - \frac{1}{z^3} < \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $m$ such that the inequality has $4$ integer solutions.
Find all values of $m$ such that the inequality: $\log(60x^2+120x+10m-10)>1+3\log(x+1)$ has exactly $4$ integer solutions?
The first thing I did was to gather from the inequality;
$\log(60x^2+120x+10m-10)>1+3\log(x+1)$
$\Leftrightarrow 1+\log(6x^2+12x+m-1)>... | Consider the inequality
$$m - 2 > x^3 - 3x^2 - 9x. \tag1 $$
As $~x \to -\infty, ~$ the RHS of (1) above goes to $-\infty.$
Therefore, it is tempting, but wrong to assume that regardless of the (fixed) value of $m$ chosen, there will be an infinite number of (negative) integer solutions.
The reason that this is wrong i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the power of the matrix. Let $A = \left( {\begin{array}{*{20}{c}}
0&1&1\\
1&0&1\\
1&1&0
\end{array}} \right)$.
I want to find $A^k,$ where $k \in N$. So far I calculated $A^2, A^3, A^4,...$ but I can not see the general formula for $A^k$. Here are $A^2, A^3, A^4, A^5$.
Not sure if this leads to anything but I fou... | Hint. From the first few examples you gave, it is easy to conjecture that $A^n = a_nB+(-1)^nI$ for some sequence $(a_n)$ that starts off $1,1,3,5,11,\dots$. If this is the case, then we would have $$a_{n+1}B+(-1)^{n+1}I=A(a_nB+(-1)^nI)=a_nAB+(-1)^nA$$ for all $n$. Rearranging, and substituting the relationship $A+I=B$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the all possible values of $x$ such that $\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$ Find possible value of $x$ such that
$$\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$$
Progress: what I did was to consider a case when $x^2 -1 < 1$ $(xy < 1)$ and $3x>-1$ $(xy > -1)$ and the... | Since $$\tan\big(\arctan x+\arctan y\big)\equiv\frac{x+y}{1-xy}$$ and $\arctan$ has principal range $\left(-\frac\pi2,\frac\pi2\right),$ thus, for each $(x,y)$ and some $k\in\{-1,0,1\},$ $$\arctan x+\arctan y=\arctan\left(\frac{x+y}{1-xy}\right)+k\pi.$$
Therefore, \begin{align}&\arctan(x+1) + \arctan(x) + \arctan(x-1) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4409707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving integral $\int \frac{\sqrt{x^2 + x}}{x}dx$ (problem 36 in section $6.25$ in Tom Apostol's calculus) Integrals which involve $\sqrt{(cx + d)^2 - a^2}$ could often be simplified if we do a substitution $cx + d = a \sec t$. If we take a concrete example, $\int \frac{\sqrt{x^2 + x}}{x}dx$, then the substitution wou... | You may avoid worrying about square-roots or absolute signs by integrating as follows
\begin{align}
\int \frac{\sqrt{x^2 + x}}{x}dx
=& \int \frac{x+\frac12}{\sqrt{x^2 + x}}\>dx+
\int \frac{\frac12}{\sqrt{x^2 + x}}dx\\
=&\>\sqrt{x^2+x}+\frac12\tanh^{-1}\frac{\sqrt{x^2+x}}{x+\frac12}+C
\end{align}
which is valid for all... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4410824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $z^2-1\mid x^2z^2-1$, prove $\frac{x^2z^2-1}{z^2-1}$ is never prime, for $x$, $z$ integers such that $x>z>1$.
Given $z^2-1\mid x^2z^2-1$, prove that $\frac{x^2z^2-1}{z^2-1}$ can never be prime, assuming $x$, $z$ are integers such that $x>z>1$.
So far I have tried taking mod a lot of different numbers, but I can... | Let $p = \frac{x^2z^2-1}{z^2-1} \implies pz^2 - p = x^2z^2 - 1$
So $z^2(x^2 - p) = 1 - p$.
We know $p > 1$. So this implies $p > x^2$.
Now $p$ divides $xz-1$ or $xz+1$ but not both.
But $p > x^2 > x > z \implies p > xz - 1$ and $p > xz + 1$.
which means $p \nmid xz-1$ and $p \nmid xz+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Let $\left,\left$ be real sequences satisfied as following conditions： $\left<a_i\right>, \left<b_i\right>$ be real sequences with
$$a_1^2+a_2^2+\cdots+a_n^2=1，\\
b_1^2+b_2^2+\cdots+b_n^2=1，\\
a_1b_1+a_2b_2+\cdots+a_nb_n=0.$$
Prove that $(a_1+a_2+\cdots+a_n)^2+(b_1+b_2+\cdots+b_n)^2\leq n$.
My attempt：
I try to prove i... | Set $\vec{e}=(1/\sqrt{n},...,1/\sqrt{n})^t$, $\vec{a}=(a_1,...,a_n)^t$ and $\vec{b}=(b_1,...,b_n)^t$.
WTS: $$(\vec{e}\cdot \vec{a})^2 + (\vec{e}\cdot \vec{b})^2 \leq 1 \, . \tag{1}$$
Proof 1: Setting $\vec{e}\cdot \vec{b}=\cos\theta$, where $\theta$ is the angle between $\vec{b}$ and $\vec{e}$. Since $\vec{a}$ is ortho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Maximum value with inequality This is a problem that my friend and I are working on for olympiad training.
Let $a, b, c$ be real numbers in the interval $[0,1]$ that satisfy $ab+c \leq 1$. What is the maximum value of $a+b+c?$
I'm guessing the maximum is at $a,b=1$ and $c=0$, where we have $a+b+c=2$. As for proving t... | 1)$ab+c\le 1;$ $a, b, c \in [0,1];$
$d:=a+b+c$
The $2$ expressions are symmetric in $a$ and $b$. It follows that $a=b$ for $d_{max}$.
2)We want to find the maximum of
$d=2a+c,$ with $a^2+c \le 1$.
3)$a^2+d-2a \le 1;$
$(a-1)^2-1 +d \le 1;$
$d \le 2 - (a-1)^2.$
4)Maximal value for $d_{max}=2$ at $a=1;$
Hence $a=b=1$,and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4413345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Prove that $\int_{0}^{2\pi}f(x)\cos(kx)dx \geq 0$ for every $k \geq 1$ given that $f$ is convex. Given $f: [0, 2\pi] \to \mathbb{R}$ convex function, prove that for every $k\geq1$
\begin{align}
\int_{0}^{2\pi}f(x) \cos (kx)dx \geq 0
\end{align}
I am completely stumped. What I have tried to do is return the query for $k... | Without loss of generality we may assume that $f$ is twice continuously differentiable on $[0,2\pi].$ Then
$$ \int_0^{2\pi}f(x)\cos(kx)\,dx=\left. f(x)\frac {\sin(kx)} k\right |_0^{2\pi} -\int_0^{2\pi} f'(x)\frac {\sin(kx)} k\,dx= -\int_0^{2\pi} f'(x) \frac {\sin(kx)} k\,dx.$$
Next, again integraiting by parts, this eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4414089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 4
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Volume between paraboloid $x^2 +y^2 -4a(z+a)=0$ and sphere $x^2 + y^2 +z^2 =R^2$ I'm trying to obtein the volume via triple integral but think I'm setting the wrong radius. The solid in particular is bounded by the sphere $x^2 + y^2 +z^2 =R^2$ and above the parabolloid $x^2 +y^2 -4a(z+a)=0$ (consedering $R>a>0$). I'm s... | Note that it is not advisable to integrate in cylindrical coordinates because the $z$-limit could be either on the sphere or the parapoloid depending on $a$. Instead, integrate in spherical coordinates with the limits $\theta\in (\cos^{-1}\frac{R-2a}R,\pi)$. The volume is then
\begin{align}
V=&\>2\pi\int_{\cos^{-1}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\prod\limits_{k=1}^{n-1}\cot^2(\frac{6k+1}{6n}\pi)=2 - (-1)^n.$
Let $u, v \in \Bbb C \setminus \{0\}$ such that $u^{2n} + v^{2n} − u^n v^n = 0$ where $n \in \Bbb N^+$.
(a) Show that $$u = v \left(\cos\left(\frac{6k+1}{3n}\pi\right)\pm i\sin\left(\frac{6k+1}{3n}\pi\right)\right)$$ for $k=0,1,2,\ldots, n-1.$
... | Here is an answer that I have got. However, it is indeed more complicated than I expected, and it requires a property from one of my previous questions. See $$z^{2n} - 2 a^n z^n \cos (nθ) + a^{2n}= \prod_{k=0}^{n-1}\left[z^2-2az\cos\left(\theta+\frac{2\pi k}{n}\right)+a\right]$$
for any positive integer a.
Now, puttin... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Logic question - just to clear up meaning of 'implies' Okay so;
I have the following statement
$p \lor \neg q \Rightarrow q \lor \neg q$
I simplify the right hand side using the complement law to get
$p \lor \neg q \Rightarrow T $
I think that $\neg q \lor p$ is the same as saying $q \Rightarrow p$ ,
so does q imply p ... | *
*
so does q imply p imply true?
Note that this phrasing is potentially ambiguous, because the conditional $(\to)$ is not associative, that is, $$(A → B) → C \:\not\equiv\: A → (B → C).$$ \begin{array}{ccc|c@{}c@{}c@{}ccc@{}ccc@{}ccc@{}ccc@{}ccc@{}c@{}c@{}c}
a&b&c&(&(&(&a&\rightarrow&b&)&\rightarrow&c&)&\leftrighta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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What's $\Pr(X=0|X=Y)$? Let $X$ and $Y$ be independent random variables that take values in $\{0,1\}$.
My question is: how to solve our $\Pr(X=0 | X=Y)$? We only know that $\Pr(Y=0) = \Pr(Y=1) = 0.5$.
My attempt:
$$\begin{split}
\Pr(X=0|X=Y) &= \frac{
\Pr(X=0)
}{
\begin{split}
&\Pr(X=0,Y=0)\\
+&\Pr(X=1,Y=1)
\end{spl... |
Let $X$ and $Y$ be independent random variables that take values in
$\{0,1\}.$
We know that $\Pr(Y=0) = \Pr(Y=1) = 0.5\tag2.$
What is $\Pr(X=0 | X=Y) \;?$
Your new attempt is all good.
Note that armed only with the information in the first line but without statement $(2),$ $X$ and $Y$ attaining the same value, i.e., ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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why is $ 2 = \frac{5}{1+\frac{8}{4+\frac{11}{7 + \frac{14}{10 + \dots}}} } $ Why is $ 2 = \cfrac{5}{1+\cfrac{8}{4+\cfrac{11}{7 + \cfrac{14}{10 + \ddots}}} } $
where the sequences $5,8,11,14,\dots$ and $1,4,7,10,\dots$ are of the form $5 + 3 n$ and $1 + 3n$.
(This converges on both even and uneven iterates)
I was surpri... | Assume the RHS of the equation as $A_n$ so we have the following recurrence relation:
$$
A_n = \frac{3n + 2}{3n - 2 + A_{n + 1}}.
$$
Claim:
$$A_n = \frac{n + 1}{n}.$$
Proof:
Step of the induction:
$$A_{n + 1} = \frac{n + 2}{n + 1} \Rightarrow A_n = \frac{3n + 2}{3n - 2 + \frac{n + 2}{n + 1}} = \frac{n + 1}{n}.$$
Base o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4435750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 1
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Explanation for a solution: Howard Anton, Elementary Linear Algebra I am currently working with the 1st edition of Howard Anton's "Elementary Linear Algebra". I tried the following problem:
Excercise Set 1.2 (p. 17), Problem 12:
For which values of $a$ will the following system have no solutions? Exactly one solution? ... | You cannot just look at the final product if you did not carefully note steps in which you were assuming facts about the value of $a$. So let us take a careful look at the Gaussian elimination process.
Starting from
$$\left(\begin{array}{rrr|r}
1 & 2 & -3 & 4\\
3 & -1 & 5 & 2\\
4 & 1 & a^2-14 & a+2
\end{array}\right)$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Consider $E = \mathbb{Q}(\sqrt{2}, \sqrt{7} )$. Here are the following questions and my respective answers follow below. I hope you have suggestions or if there are any mistakes I hope you help me fix them.
*
*Find a basis for $E$ over $\mathbb{Q}(\sqrt{2}).$
*Find a basis for $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q... | 1, 2, 3 look good.
Regarding 4, you can simplify by noticing that $p(x)=x^2-2 \in \mathbb Q[x]$ is a fixed polynomial under $\phi$ having $\sqrt 2$ as a root, while $\sqrt 7$ is not a root.
Or if you want to avoid using polynomials...
You have $\left(\sqrt 2\right)^2 - 2 = 0$. So applying $\phi$ you should have
$$0= \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding pattern in matrix inverse I want to see if any pattern that could be formulated exist across the rows (or equivalently , columns) of the matrix inverse (in the middle , let's name it $A^{-1}$) so that an analytical formula that relates $o_n$ and $b_n$ (perhaps along with other b's) exists (for any positive int... | Let $S$ be the $n\times n$ matrix with $S_{ij}=\begin{cases}1,&i=j+1\\0,&\text{otherwise}\end{cases}$.
The matrix being inverted is $A=I+wS+iS^2(I-S)^{-1}$, hence $A^{-1}=f(S)$ with $$f(z)=\frac{1-z}{1-(1-w)z-(w-i)z^2}.$$
Next, we find the partial fraction decomposition of $f(z)$, say $$f(z)=\frac{\alpha_+}{1-\lambda_+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4437959",
"timestamp": "2023-03-29T00:00:00",
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If $\cos\frac \pi{n+1}$ is a root of the equation $8x^3+4x^2-4x-1=0$, then find n $(n\in\mathbb N)$
If $\cos\frac \pi{n+1}$ is a root of the equation $8x^3+4x^2-4x-1=0$, then find n $(n\in\mathbb N)$
My Attempt:
Let $\theta=\frac\pi{n+1}$, therefore,
$$8\cos^3\theta+4\cos^2\theta-4\cos\theta-1=0$$
Also, $\cos3\theta=... | If you know that a polynomial has a root of the form $\cos\left(\frac{\pi}{n+1}\right)$ then you have to think in roots of unity and cyclotomic polynomials. The idea is to find the right change of variable that transforms your polynomial into a cyclotomic polynomial. Since $8x^3+4x^2-4x-1=(2x)^3+(2x)^2-2(2x)-1$ it seem... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\frac{2(1+4a^2)}{(12x-1)^3}\leqslant \frac{(1-a)^4}{[12x-(1-a)^2]^3}+\frac{(1+a)^4}{[12x-(1+a)^2]^3}$ for $0\leq a<\frac13$ and $x>\frac{(1+a)^2}8$
How to prove the inequality below？
$$\frac{2(1+4a^2)}{(12x-1)^3}\leqslant \frac{(1-a)^4}{[12x-(1-a)^2]^3}+\frac{(1+a)^4}{[12x-(1+a)^2]^3}$$ holds for all $0\leqslant a<\f... | Note that
$$(1 - a)^4 + (1 + a)^4 - 2(1 + 4a^2) = 4a^2 + 2a^4 \ge 0.$$
It suffices to prove that
$$\frac{(1 - a)^4 + (1 + a)^4}{(12x-1)^3}\leqslant \frac{(1-a)^4}{[12x-(1-a)^2]^3}+\frac{(1+a)^4}{[12x-(1+a)^2]^3}$$
or
\begin{align*}
&\frac{1}{(12x - 1)^3}\\
\le\,& \frac{(1 - a)^4}{(1 - a)^4 + (1 + a)^4}\cdot \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4443744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Laurent series $\ \ \ \ \ $ Trying to compute first few terms of the Laurent series for:
$$ \frac{e^z}{z^2(z^2+1)} =\sum_{n=-2}^\infty c_n z^n = \quad ?$$
I know the expansion of $$e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$$ and I could have $\frac{e^z}{z^2}$ expanded. But how do I fit all together and be able to see wh... | Via the Cauchy product,
\begin{align}
\frac{e^z}{z^2+1}
&= e^z\cdot\frac{1}{1-(-z^2)} \\
&= \left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)\left(\sum_{n=0}^\infty (-z^2)^n\right) \\
&= \left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)\left(\sum_{n=0}^\infty (iz)^{2n}\right) \\
&= \left(\sum_{n=0}^\infty \frac{z^n}{n!}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4444142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Analytic solution inverse kinematics - different solutions with different calculation steps I have the following problem:
$$5\cos \theta_1+3\sqrt{3}\sin \theta_1=4\qquad\qquad\textbf{(I)}$$
$$5\sin \theta_1-3\sqrt{3}\cos \theta_1=6\qquad\qquad\textbf{(II)}$$
I have two seemingly correct paths to a solution, however onl... | For this system, I would set $\sigma=\sin \theta_1,\chi=\cos\theta_1$ and work those out rather than throwing yourself into inverse trig functions of irrationals so early.
$$3\sqrt 3 \sigma +5\chi =4 \tag I$$$$5\sigma-3\sqrt 3 \chi=6\tag {II}$$
Applying $(I)* 3\sqrt 3, (II)*5$ as in your second choice, we get: $$27\sig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4444810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $c$ is a root of the equation $6x^3-6sx^2+3(s^2-t)x+3st-s^3-2u=0$
(a) Let $p,q$ and $r$ be real numbers. Given that there are numbers $a$ and $b$ such that $$a+b=p, ~~~ a^2+b^2=q, ~~~ a^3+b^3=r~~~~~~~~~~~~~~~(*)$$
Show that $3pq - p^3=2r$
Workings: $3pq-p^3 = 3(a+b)(a^2+b^2)-(a+b)^3 = 2a^3+2b^3=2r$
(b) Con... | In order to make use of (a), note that
$$\begin{aligned}
a+b&=s-c=:p\\
a^2+b^2&=t-c^2=:q\\
a^3+b^3&=u-c^3=:r
\end{aligned}$$
so that
$$ 3pq - p^3=2r$$
translates to
$$3(s-c)(t-c^2)-(s-c)^3=2(u-c^3). $$
Expand and rearrange this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4445026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Method checking to get the range of the three variables under two constraints Suppose a,b,c are real numbers and $a+b+c= 6$ , and $ab+bc+ca = 9$ , also $a<b<c$ find range of a , b, c .
My method was on eliminating c we get $a^2 + b^2 -6a - 6b +ab + 9$ = 0 so making a quadratic in a in terms of b we get a = $\frac {6-b... | We have a plane $ a + b + c = 6 $ and the hyperboloid of two sheets $ab + ac + bc = 9 $
The intersection of the two is a circle of center $(2,2,2)$ and radius $\sqrt{6}$
Hence, the vector $v = (a,b,c) $ can be described parametrically as
$ v = (2,2,2) + \sqrt{6} \left( \cos(t) \dfrac{(1, -1, 0)}{\sqrt{2}} + \sin(t) \df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4445237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Does the limit of the cubic formula approach the quadratic one as the cubic coefficient goes to $0$? The formula for solving a cubic equation of the form $ax^3+bx^2+cx+d=0$ does not seem to yield the quadratic formula for the limit $\lim _{a \rightarrow 0} \text{(cubic formula)}$.
But, if one tries the same thing with ... | When $a \to 0$ one of the roots goes to infinity, which complicates the algebraic manipulations. Instead, it is easier to show that when $d=0$ the non-zero roots reduce to the quadratic formula.
Assume WLOG $\,a=1\,$, then with the wikipedia notations for the cubic formula in the case $d=0\,$:
$$
\begin{align}
\Delta_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Find the values of a and b from arithmetic and geometric series The $1^{st}$ , $2^{nd}$ and $3^{rd}$ terms of an arithmetic series are $a, b, a^2$, where $a$ is a negative number. The $1^{st}$, $2^{nd}$ and $3^{rd}$ terms of a geometric series are $a, a^2,b$.
Find the values of $a$ and $b$ and find the sum to infinity ... | Looks mostly OK – note that $a$ is negative, so $a=-\frac 1 2$, and $b=-\frac 1 8$ (you don't need the other two cases).
The sum is $\frac{-\frac 1 2}{1 + \frac 1 2} = -\frac 1 3$, though. The formula for a geometric series is $\frac{a}{1 - r}$, and here, $r=a$. Your sum should be from $n=1$; as it stands, it includes ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4448763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $c^3 + a^3 − 2abc = 6b − 11$ Let $a, b, c$ be real numbers such that $(a + b + c)^2 = 3(ab + bc + ca + 1).$ Given,
$$a^3 + b^3 − 2abc = 6c + 2,$$
$$b^3 + c^3 − 2abc = 6a + 9,$$
Prove that $$c^3 + a^3 − 2abc = 6b − 11$$
Note that, subtracting the two equations, we get $$a^3-c^3=6(c-a)+7. $$
Moreover, we have ... | Hint: When things look cyclic, first thing you do is add them. Try adding the three cyclic-looking equations, and recall the relation
$$ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4450018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Which integrating technique should I use? Just some context: In the mathematical course, I have undertaken this year, I've just learnt how to integrate using partial fractions, substitution(not trig though, just a variable) and integrating derivatives of inverse trig functions(arcsin, arccos, arctan)
Here's is where I ... |
How would one start, by always trying to factor out (might be a long process though if it does not work out)?
I would not recommend it precisely because of the reasons you mention, it doesn't always work out, and even if it does, you then have to do partial fractions which can be a very long and tedious process. The ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4450994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding minimum value of $x^2+y^2+xy+x-4y+9$
What is the minimum value of $f(x,y)=x^2+y^2+xy+x-4y+9$ ?
I tried completing squares,
$$x^2+y^2+xy+x-4y+9=\frac12(x^2+2xy+y^2+x^2+2x+1+y^2-8y+16+1)=\frac12[(x+y)^2+(x+1)^2+(y-4)^2+1]$$But not sure how to continue.
| Your work is already very useful. Since the function is continuous and bounded from below (your calculations show that the expression is greater or equal than $\frac 12$), it has a global minimum. Since this is a differentiable function in an open set, this global minimum must be a stationary point, i.e.
$$
2x +y +1=0,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4452791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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What did I do wrong in solving $\int\sec^{-1} x\,{dx}$? I used integration by parts:
let u=$\sec^{-1}\,x$, dv=dx,
then du=$\frac{1}{|x|\sqrt{x^2-1}}$, v=x.
I = $x\sec^{-1}\,x\;-\;\int\frac{x}{|x|\sqrt{x^2-1}}dx\\$
Integration of $\int\frac{x}{|x|\sqrt{x^2-1}}dx$:
Let x=$\sec\theta$, then dx = $\sec\theta\tan\theta\,d\t... | The anti-derivative valid over all domain $|x|\ge 1$ is obtained as follows
\begin{align}
\int \sec^{-1}x \ dx
=&\ x\sec^{-1}x - \int \frac x{|x|\sqrt{x^2-1}}dx\\
=&\ x\sec^{-1}x - \int \frac 1{\sqrt{|x|^2-1}}d(|x|)\\
=&\ x\sec^{-1}x - \cosh^{-1}(|x|)\
=x\sec^{-1}x - \ln\left(|x|+\sqrt{x^2-1}\right)
\end{align}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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There are $3$ urns $A,B$ and $C$. Urn $A$ contains $4$ red balls and $3$ black balls. Urn $B$ contains $5$ red balls and $4$ black balls. Urn $C$ There are $3$ urns $A,B$ and $C$. Urn $A$ contains $4$ red balls and $3$ black balls. Urn $B$ contains $5$ red balls and $4$ black balls. Urn $C$ contains $4$ red balls and $... | "Now, this selection can be made in three different ways"... No. It cannot. You can select red from A, red from B, and black from C in one way (each with a probability of success). So $$\mathsf P(A, B, \overline C)=\dfrac{4\cdot 5\cdot 4}{7\cdot 9\cdot 8}$$
The "three ways" a black ball may be drawn from an urn is ac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4457347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral of this weird function $ \int \frac{1}{(x^2+x+1)^2}dx $ I put this equation into Symbolab and it produced me a very complex result
Basically this is the result but I believe there is a simpler way to solve this question
$$
\frac{2}{3\sqrt{3}}\left(2\arctan \left(\frac{2x+1}{\sqrt{3}}\right)+\sin \left(2\arctan... | Substitute $y=x+\frac12$\begin{align}
&\int \frac{1}{(x^2+x+1)^2}dx\\
=& \int \frac{1}{(y^2+\frac34)^2}dy
= \int \frac{2}{3y}\ d\bigg( \frac{y^2}{y^2 +\frac34}\bigg)
\overset{ibp}=\frac{2y}{3(y^2+\frac34)}+\frac23\int \frac1{y^2+\frac34}dy\\
=&\ \frac{8y}{3(4y^2+3)}+ \frac4{3\sqrt3}\tan^{-1}\frac{2y}{\sqrt3}+C
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4459298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}=\cot\left(\frac{3\pi}{2}-\alpha\right)$ Prove that $$\dfrac{\sin4\alpha}{1+\cos4\alpha}\cdot\dfrac{\cos2\alpha}{1+\cos2\alpha}=\cot\left(\dfrac{3\pi}{2}-\alpha\right)$$
The RHS is equal to $\tan\alpha,$ so we are to show $$\dfrac{\sin4\... | Recall that $\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$, and $\cos(x) = \frac{e^{ix}+e^{-ix}}{2},$ where $i = \sqrt{-1}$. Then:
$$X = \frac{\sin(4\alpha)}{1+\cos(4\alpha)}\cdot\frac{\cos(2\alpha)}{1+\cos(2\alpha)}=
\frac{1}{i}\cdot\frac{(e^{i4\alpha}-e^{-i4\alpha})(e^{i2\alpha}+e^{-i2\alpha})}{(2 + e^{i4\alpha}+e^{-i4\alpha}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4461361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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The integral $\int_{0}^{1} \frac{ \log (1-x)}{1+x^2}dx$ Recently a very interesting result
$\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x=0$
has been proved in a more than elegant way. See Show that $\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathr... | Perhaps a cleaner way is to consider the integral
$$
\int_1^\infty \frac{\ln(u-1)}{1+u^2}du.
$$
First, perform the substitution $u\rightarrow (u+1)/(u-1) = 1 + 2/(u-1)$ to get
$$
\int_1^\infty \frac{\ln(u-1)}{1+u^2}du = \int_1^\infty \frac{\ln\left(\frac{2}{u-1}\right)}{1+u^2}du = \frac{\pi}{4}\ln2 -\int_1^\infty \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4462909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given a triangle ABC inscribed in the unit circle , the 3 vertices could be described via 3 complex number, namely, $a$, $b$, and $c$.
Now $AD$ is an altitude, $D$ is the foot of $AD$ on $BC$.
Prove: $$D = \frac{a+b+c}2 - \frac{bc}{2a}$$
So far my progress is --
*
*the circumcentre of triangle $ABC$ is just $O = 0$.
... | Instead of treating the points as arbitrary $\mathbb{R}^2$ vectors, let's make use of them being complex numbers. If $z_1, z_2 \in \mathbb{C}$ are interpreted as vectors, then:
*
*They are perpendicular if the quotient between them is a pure imaginary number.
*They are parallel (or overlapping) if the quotient betw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Why Lagrange Multiplier Doesn't Work? Question: Find maximum value $f(x,y,z) = xy + zy + xz - 4xyz$ subject to constraint $x + y + z = 1$ and $x,y,z \geq 0$.
$$
g(x,y,z) = x + y + z - 1
$$
and
$$
\nabla g(x,y,z) \neq 0,
\qquad
\nabla g(x,y,z) = \langle 1,1,1 \rangle.
$$
When we apply Lagrange Multiplier Method, we fi... | You may use KKT conditions instead.
Or you may apply Lagrange Multiplier for the following equivalent problem:
Find the maximum of $F(a, b, c) = a^2b^2 + b^2c^2 + c^2a^2 -4a^2b^2c^2$ subject to $a^2+b^2+c^2 = 1$.
(Note: We have only one constraint. )
Let
$$L(a, b, c) := a^2b^2 + b^2c^2 + c^2a^2 -4a^2b^2c^2 + \lambda (a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Where is my mistake in evaluating $\lim_{x \to -\infty} \frac{\sqrt{x^2+4}}{x}$? Given is $\lim_{x \to -\infty} \dfrac{\sqrt{x^2+4}}{x}$
I divide numerator and denominator by x to the largest degree in the denominator and I get
$$\lim_{x \to -\infty} \frac{\sqrt{1+\frac{4}{x^2}}}{1}=\frac{\sqrt{1}}{1}=1.$$
But the answ... | The mistake is when you are shortening out $x$. $x$ is negative, but the square root is positive. Thus:
$$\begin{align}
\lim_{x \to-\infty} \frac{\sqrt{x^2+4}}{x}
&= \lim_{x \to-\infty} \frac{|x|\sqrt{1+4/x^2}}{-|x|} \\
&= \lim_{x \to-\infty} \frac{\sqrt{1+4/x^2}}{-1} \\
&\to -1
\end{align}$$
where $x=-|x|$ in the de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4466047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can we show that $\int_0^\infty\frac{\arctan(t)\cdot\ln(1+t^2)}{t(t^2+a^2)}\mathrm{d}t=\frac{\pi}{2a^2}\cdot\ln^2(1+a)$ without complex analysis? $\newcommand{\d}{\,\mathrm{d}}\newcommand{\Res}{\operatorname{Res}}\newcommand{\Arg}{\operatorname{Arg}}$Consider the function: $$\begin{align}I:(0,\infty)&\to(0,\infty)\... | Utilize the integral
\begin{align}
&\int_0^\infty\frac{\ln x}{t^2+(x+1)^2}\overset{x\to\frac{1+t^2}x} {dx}\\
=&\int_0^\infty \frac{\ln(1+t^2)-\ln x}{t^2+(x+1)^2}dx
=\frac12 \int_0^\infty\frac{\ln (1+t^2)}{t^2+(x+1)^2}dx
=\frac1{2t}\tan^{-1}t\ln(1+t^2)
\end{align}
to integrate
\begin{align}
&\int_0^\infty\frac{\tan^{-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given a permutation $\sigma = (13)(254)$, state $\sigma^2$. Given a permutation $\sigma = (13)(254)$, state $\sigma^2$.
$\sigma = (13)(254), \sigma^2=(13)(254)(13)(254) = (13)(13)(254)(254) = (425)
$
Or, in two row format, get:
$$ \sigma = \begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & ... | We have
$$\begin{align}
\sigma^2&=(13)(254)(13)(254)\\
&=(13)(13)(254)(254)\\
&=(254)^2\\
&=(245)
\end{align}$$
because
$$\begin{align}
1&\xrightarrow{(254)}1\xrightarrow{(254)}1,\\
2&\xrightarrow{(254)}5\xrightarrow{(254)}4,\\
3&\xrightarrow{(254)}3\xrightarrow{(254)}3,\\
4&\xrightarrow{(254)}2\xrightarrow{(254)}5,\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Compute $f^{(2020)}(0)$ Problem :
Let
$$f(x)=\frac{x}{(x+1)(1-x^2)}$$
Then find $f^{(2020)}(0)$.
My Attempt :
From partial fraction decomposition,
$$f(x)=\frac{1}{4(1-x)}+\frac{1}{4(x+1)}-\frac{1}{2(x+1)^2}$$
and,
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n,\quad \frac{1}{1+x}=\sum_{n=0}^\infty (-1)^nx^n, \quad\frac{1}{(1+x... | We have
$$f(x) = \frac14 \cdot \frac{1}{1-x}
+ \frac14 \cdot \frac{1}{1 + x}
+ \frac12 \cdot \left(\frac{1}{1 + x}\right)'.$$
By observing the first several derivatives to see the pattern:
$$\left(\frac{1}{1-x}\right)' = \frac{1}{(1-x)^2},
\left(\frac{1}{1-x}\right)'' = \frac{2}{(1-x)^3},
\left(\frac{1}{1-x}\right)'''... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4471123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Answer:
$a+b = 7, ab = 2$
$$\begin{align}
(a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt]
a^6 + b^6 &= (a+b)^6 - (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \... | Alternatively, let $x_n = a^n + b^n$. Then $x_{n+2}=7x_{n+1}-2x_n$ and so we get
$x_0=2$
$x_1=7$
$x_2=7x_1-2x_0=45$
$x_3=7x_2-2x_1=301$
$x_4=7x_3-2x_2=2017$
$x_5=7x_4-2x_3=13517$
$x_6=7x_5-2x_4=90585$
This is easy to do by hand.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 5
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Showing that switching the order of variables gives same sum so as to simplify the required expression for exact sum If $S(x,y)$ = $\sum_{y=0}^{\infty}$$\sum_{x=0}^{\infty}$$\frac{(x+y +xy)}{(5^x(5^x +5^y))}$ , then if we want to show that $S(x,y) = S(y,x)$ so as to get the simplification by adding both to get the exac... | $S(x,y)$ = $\sum_{y=0}^{\infty}$$\sum_{x=0}^{\infty}$$\frac{(x+y +xy)}{(5^x(5^x +5^y))}$
By interchanging $x$ and $y$ we get
$S(x,y)$ = $\sum_{y=0}^{\infty}$$\sum_{x=0}^{\infty}$$\frac{(x+y +xy)}{(5^y(5^x +5^y))}$
$\implies 2S = \sum_{y=0}^{\infty}$$\sum_{x=0}^{\infty}$$\frac{(x+y +xy)}{(5^x +5^y)}\left(\frac{1}{5^x}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is this relationship between trigonometric and hyperbolic function? In the following, I don't understand how they put $\tan{\phi} = \sinh{\frac{\psi}{\sqrt{2}}}$.
Is there a relationship?
| This isn't asserting a relationship. It's simply setting up a substitution that trades circular functions for hyperbolics in order to escape the square root. It's really no different than having $\sqrt{4-x^2}$ and setting $x=2\sin\theta$.
In the exercise, first note that the expression (ignoring $d\theta$) can be rewri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4477813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Primes representable by either $x^2+36y^2$ or $4x^2+9y^2$ Is there a simple criterion for primes that are representable by either $x^2 + 36 y^2$ or $4x^2 + 9y^2$?
This is not my area of expertise, so any pointers appreciated.
I had a look in the Cox book "Primes of the form $x^2 + n y^2$", but if those specific forms ... | Sure, for $p \equiv 1 \pmod 4,$ there is an expression $u^2 + 36 v^2 = p$ precisely when $x^4 + 3$ factors into four linear factors $\pmod p,$ that is four distinct roots. Let me check whether it is enough to have one root. This is from Kenneth S. Williams and D. Liu, Tamkang Journal of Mathematics, volume 25,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proofs for couple of details when proving that $n^7 + 7$ is not a perfect square for any integer $n$.
Prove that $n^7 + 7$ is not a perfect square for any integer $n$.
The proof given for the problem is the following
Assume $n > 0$ since $n = 0$ and $n = -1$ are easy and for $n \le -2$ the expression is negative. Su... | We first show that $\nu_{11}(a^2+11^2)$, and thus, $\nu_{11}(n+2)$, is at most $2$. Now, on the one hand, for $\nu_{11}(a^2+11^2)$ to be positive at least $2$ in the first place, $a^2$ and thus $a$ itself must also be a multiple of $11$, say $a=11c$ for some integer $c$. So $a^2+11^2$ can be written $a^2+11^2 = (11c)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Trying to solve $|2x-15| = -x^2 - 5x -8$ My first instinct was to take the positive and negative of the right hand side, resulting in
$2x-15 = -x^2 - 5x - 8$, and $2x-15 = x^2 + 5x + 8$, which results in the first giving me two real answers using the quadratic equation, and the second being two imaginary solutions. The... | I would recommend you to notice that
\begin{align*}
x^{2} + 5x + 8 = \left(x^{2} + 5x + \frac{25}{4}\right) + \frac{7}{4} = \left(x + \frac{5}{2}\right)^{2} + \frac{7}{4} \geq \frac{7}{4}
\end{align*}
Hence we conclude that $-x^{2} - 5x - 8 < 0$ for every possible value of $x$.
Based on such considerations, we conclude... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Find another method of solving the shaded area. The diagram shows a square of side length $10\;\rm cm$. A quarter circle, of radius $10\;\rm cm$, is drawn from each vertex of the square. Find the exact area of the shaded region.
And This is my answer.
The answer is right, but I am searching for other ways.
Thanks.... |
Area of closed figure contains square $EFGH$ and 4 segments congruent to $EF$
Area of segment can be found as difference of sector and triangle:
$$S_{seg}=\frac12 AF^2 \frac{\pi}{6}-\frac12 AF^2 \sin \frac{\pi}{6}=\frac{\pi-3}{12}AF^2$$
Square side can be found in many ways, for example with cosine rule
$$AB^2=AG^2+BG... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the measure of the relationship $\frac{1}{r_1} - \frac{1}{r} $in the figure below In In a right triangle $ABC$ ($A=90°$) with inradio $r$, cevian $AD$ is drawn in such a way that the inradium of $ABD$ and $ADC$ are equal to $r1$.If $AD=2$, calculate $\frac{1}{r1}-\frac{1}{r}$ (Answer:0,5).
My progress:
$\triangle... | Here's what seems to be an unnecessarily-complicated solution.
Define $b:=|AC|$, $c:=|AB|$, $d:=|AD|$, $p:=|BD|$, $q:=|CD|$. Let $r$ be the inradius of $\triangle ABC$, and let $s$ be the common inradius of $\triangle ABD$ and $\triangle ACD$.
We know $$\text{inradius}\cdot \text{perimeter} = 2\,\text{area}$$ so we ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Polynomials and Girard If the polynomial $P(x)=x^3 - 3x^2 -7x -1$ has roots $a,b,c$, find the value of $(\frac{1}{a-b} + \frac{1}{b-c} + \frac{1}{c-a})^2$.
My attempt:
I developed the expression and by Girard I was able to simplify the numerator by finding an integer. But the denominator couldn't
|
by Girard I was able to simplify the numerator by finding an integer
This leaves the denominator to calculate, which, as noted in a comment, is actually the discriminant of the cubic (for verification, its value is $1300$ per WA). The following is a shortcut to calculate the denominator without using the cubic discri... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\epsilon-N$ for $\lim\limits_{n \to \infty} \sqrt{n^{2} +3n-3} -n = \frac{3}{2}$ First, I tried to use the triangle inequality only once to find an N:
$$
\left | \sqrt{n^2+3n-3}-n-\frac{3}{2} \right | \leqslant \left | \sqrt{n^2+3n-3}-n \right | + \left | \frac{3}{2} \right | = \epsilon
$$
$$
N=\left \lfloor \frac{(... | You cannot “choose” $\varepsilon$, forget it. You want to show that, for every $\varepsilon>0$, the inequality
$$
\Bigl|{\textstyle\sqrt{n^{2} +3n-3}} -n - \frac{3}{2}\Bigr|<\varepsilon
$$
is satisfied for all $n$ greater that some integer $N$ (depending on $\varepsilon$).
Your idea of applying the triangle inequality ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show the function defined on $[0,1] \times [0,1]$ via $\frac{x^2-y^2}{(x^2+y^2)^2}$ if $(x,y) \neq 0$ and $0$ otherwise is not integrable. Let $f: [0,1] \times [0,1] \rightarrow \Bbb{R}$ be defined via
$$f(x,y)=
\begin{cases}
\frac{x^2-y^2}{(x^2+y^2)^2} &(x,y) \neq (0,0) \\
(0,0) & \text{otherwise}
\end{cases}$$
So I k... | I claim that
$$\int_0^1 \int_0^1 f(x,y) dx dy \neq \int_0^1 \int_0^1 f(x,y) dy dx.$$
We compute the LHS:
\begin{align*}
\int_0^1 \int_0^1 f(x,y) dx dy &= \int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dx dy\\
&= \int_0^1 -\frac{x}{x^2+y^2} \bigg\vert_0^1 dy\\
&=-\int_0^1 \frac{1}{1+y^2}dy\\
&=- \tan^{-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that for all natural numbers $n \geq 2$, $n^2 \geq n + 2$ Prove that for all natural numbers $n \geq 2$, $n^2 \geq n + 2$.
In solving this proof, I tried two methods, first working backward and then using the contradiction approach. i wish to solve this proof directly or using a contradiction however I am stuck o... | Perhaps the quickest way:
If $n \ge 2$ then $n^2 = n \times n \ge n\times 2 = n + n \ge n + 2$.
If that was too slick:
We now that $n^2 > n$ in general. But by how much and when? $n^2 - n = n(n-1)$. As $n \ge 2$ and $n-1 \ge 1$ then $n^2 - n \ge 2\times 1$ and $n^2 \ge n + 2$.
A cute thing happens when we try a proo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving the Diophantine system $pqr=a^4$, $p+q+r=b^4$ I am trying to find solutions of the following system of diophantine equations:
$$\left\{\begin{array}{rcl}pqr&=&a^4\\p+q+r&=&b^4\end{array}\right.$$
where $a$, $b$, $p$, $q$ ans $r$ are positive integers such that $\gcd(p,q,r)$ is not divisible by $\theta^4$, $\the... | We have the below Identity:
$(b^2+c^2-a^2)^2+(2ab)^2+(2ac)^2=(b^2+c^2+a^2)^2$
Let:
$p=(b^2+c^2-a^2)^2$
$q=(2ab)^2$
$r=(2ac)^2$
$p+q+r=(a^2+b^2+c^2)^2$
To make (RHS) a fourth power we take:
$b^2+c^2=8a^2$ ---$(1)$
Hence,
$p+q+r=(9a^2)^2=(3a)^4$
$(pqr)=((b^2+c^2-a^2)(2ab)(2ac))^2$
=$(2a)^4(bc(b^2+c^2-a^2))^2$
Since, $(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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Why can't I take the 3rd root of both sides to simplify the implicit differentiation of $\tan^3\left(xy^2+y^{\:}\right)=x$? The task is to find the implicit derivative of $$\tan^3\left(xy^2+y^{\:}\right)=x$$
I can calculate
$$ \frac{dy}{dx}\mathrm{\:of\:}\tan ^3\left(xy^2+y\right)=x $$
as
$$\frac{1-3y^2\tan ^2\left(xy^... | It does yield the same result
$$
\frac{1-3y^2x^{\frac{2}{3}}\sec ^2\left(xy^2+y\right)}{3x^{\frac{2}{3}}\sec ^2\left(xy^2+y\right)\left(2xy+1\right)}
$$
we also have
$$
x^{1/3} = \tan\left(xy^2+y^{\:}\right)
$$
or
$$
x^{2/3} = \tan^2\left(xy^2+y^{\:}\right)
$$
insert into the above
$$
\frac{1-3y^2\tan^2\left(xy^2+y^{\:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Expected value in coin flipping process You flip a coin, and if the result is tails, you lose. If the result is heads, you get to play again. What is the expected value of throws before you lose?
My Approach
The expected value is the sum of all the outcomes multiplied by their respective probabilities:$$\sum_{i=1}^{n}V... | This is an Arithmetic-Geometric Series. Let $S=\displaystyle \sum_{i=1}^{N}\frac{i}{2^i}$. Then,
$$S=\displaystyle\frac {1}{2^1}+\frac {2}{2^2}+\frac{3}{2^3}+…+\frac{N}{2^N}.\tag{1}$$
Now, divide the whole equation by $2$ and arrange it by shifting the terms by a step: $$\frac{S}{2}=\frac {1}{2^2}+\frac {2}{2^3}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4497470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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For an arbitrary $n$, evaluate $\left(z^n+\frac{1}{z^{n}}\right)\left(z^n +\frac{1}{z^n}+1\right)$ Let me put you in context, I have the following problem:
Let $z$ be a complex number such that $$\left(z+\frac{1}{z}\right)\left(z+\frac{1}{z}+1\right)=1.$$
For an arbitrary $n$, evaluate
$$\left(z^n +\frac{1}{z^n}\right... | Your answer has something wrong, especially the case where $n$ is divided by $5$. Let me explain why.
From $\left(z+\frac{1}{z}\right)\left(z+\frac{1}{z}+1\right)=1$ we have $z^5=1$ and $z\neq 1$. So $z$ is a root of unity $5$, so $z=\zeta^m$, where $\zeta=e^{\frac{2\pi i}5}$ and $m\in\{1,2,3,4\}$. Now it is easy to ch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can we prove $\sqrt{2+\sqrt{2+....\sqrt{2+\sqrt{2}}}} = 2\cos\left(\frac {\pi }{2^{n+1}}\right)$ without induction I wanted to know the proof without induction / substitution method for the equation
$$\underbrace {\sqrt{2+\sqrt{2+...\sqrt{2+\sqrt{2}}}} }_{\text{n-times}}= 2\cos\left(\frac {\pi }{2^{n+1}}\right)$$
... | Let's assume the problem. Then, let $x_n=2\cos \left( \dfrac {\pi}{2^{n+1}} \right).$ We can easily get $x_{n+1}=\sqrt{2+x_n}, x_1=\sqrt{2}$. Then, we can change the problem:
Prove that if $x_{n+1}=\sqrt{2+x_n}, x_1=\sqrt{2}$, then $x_n=2\cos \left( \dfrac {\pi} {2^{n+1}} \right)$.
We also know that $\displaystyle \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Conjectured closed form of $\int_0^{\infty} \frac{1}{(x^2+a^2)^{n}} \, \mathrm d x$ Consider
$$ I = \int_0^{\infty} \frac{1}{(x^2+a^2)^{n}} \, \mathrm d x $$
where $a > 0$ is a constant. We can evaluate this using the Leibnitz theorem for specific values of $n$. Mathematica can solve it for specific values of $n$ as we... | $$J(b)=\int_0^\infty \frac{1}{x^2+b}dx=\frac{\pi}{2\sqrt{b}}$$
$$I=\frac{(-1)^{n-1}}{(n-1)!}\frac{d^{n-1}}{db^{n-1}}J(b)|_{b=a^2}$$
$$\frac{d^{n-1}}{db^{n-1}}J(b)=\frac{\pi}{2}(-1)^{n-1}\frac{(2n-3)!!}{2^{n-1}}\frac{1}{b^{\frac{2n-1}{2}}}$$
$$I=\frac{\pi}{2}\frac{1}{(n-1)!}\frac{(2n-3)!!}{2^{n-1}}\frac{1}{a^{2n-1}}$$
w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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Show that $a^3+5a$ is an integer I've been given the following task. Let
$$a = \sqrt[3]{1+\sqrt{\frac{152}{27}}}-\sqrt[3]{-1+\sqrt{\frac{152}{27}}}$$
Show that $a^3+5a$ is an integer.
I tried calculating it by hand but the small page of my copybook is not large enough for the very long calculations. Is there a trick I ... | Let $x=1+\sqrt{\frac{152}{27}},~~y=-1+\sqrt{\frac{152}{27}}$
$$x^{\frac{1}3}y^{\frac{1}3}=\frac{5}{3}$$
$$a=x^{\frac{1}3}-y^{\frac{1}3}$$
$$\begin{align}
a\left(x^{\frac{2}3}+x^{\frac{1}3}y^{\frac{1}3}+y^{\frac{2}3}\right)&=\left(x^{\frac{1}3}-y^{\frac{1}3}\right)\left(x^{\frac{2}3}+x^{\frac{1}3}y^{\frac{1}3}+y^{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How does $a^y = x+\sqrt{x^2+1}$ imply $a^{-y} = -x+ \sqrt{x^2+1}$ and then $x=\sinh(y\ln a)$? I want to understand the steps in an argument given for the following question.
Find the inverse of the following function:
$$f(x) = \log_a(x+\sqrt{x^2+1})$$
We find:
$$\begin{align}
a^y &= \phantom{-}x+\sqrt{x^2+1} \tag1 \\[... | Right, so what you should do is to write these things out explicitly. So:
$$a^y = \sqrt{x^2+1}+x$$
$$a^{-y} = \frac{1}{a^y} = \frac{1}{\sqrt{x^2+1}+x} = \frac{1}{\sqrt{x^2+1}+x} \cdot \frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}-x} = \frac{\sqrt{x^2+1}-x}{(x^2+1)-x^2} = \sqrt{x^2+1}-x$$
Then, notice that:
$$a^y-a^{-y} = (\sqrt{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Interesting integral $\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{2}}$ Latest Edit
Inspired by @J.G., I find a formula in general,
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{n}} &=2 \int_{0}^{\pi} \frac{d x}{(3-\cos x)^{n}} \\
&=\left.\frac{2(-1)^{n}}{(n-1) !}... | Start as J.G., then using differentiation yields
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\frac{1-\cos 2 x}{2}\right)^{2}} &=4 \int_{0}^{\frac{\pi}{2}} \frac{d x}{(3-\cos 2 x)^{2}} \\
&=2 \int_{0}^{\pi} \frac{d x}{(3-\cos x)^{2}} \\
&=-\left.2 \frac{\partial}{\partial a} \int_{0}^{\pi} \frac{d x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Do you get another method for $\int \frac{\sin x}{\sin 5 x}dx$? First of all, we use De Moivres’ Theorem to express $\sin 5x $ in terms of $\cos x$ and $\sin x$.
$$
\begin{aligned}
&\cos 5 x+i \sin 5 x\\=&(\cos x+i \sin x)^{5} \\
=& \cos^{5} x+5 i \cos ^{4} x \sin x-10 \cos^{3} x \sin ^{2} x-10 i \cos ^{2} x \sin ^{3} ... | In terms of the Gaussian hypergeometric function : if
$$I_n=\int \frac{\sin(x)}{\sin(nx)}\,dx$$then
$$I_n=\frac{i e^{i (n+1) x} \, _2F_1\left(1,\frac{n+1}{2 n};\frac{1}{2}
\left(3+\frac{1}{n}\right);e^{2 i n x}\right)}{n+1}-\frac{i e^{i (n-1) x} \, _2F_1\left(1,\frac{n-1}{2 n};\frac{1}{2}
\left(3-\frac{1}{n}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$a,b,c \in \mathbb{R}^+, \dfrac{a}{a+7b}+\dfrac{b}{b+7c}+\dfrac{c}{c+7a} \geq \dfrac{3}{8}$ How to prove
$a,b,c \in \mathbb{R}^+$
$$\dfrac{a}{a+7b}+\dfrac{b}{b+7c}+\dfrac{c}{c+7a} \geq \dfrac{3}{8}$$
my solution
this equivalent to
$$\dfrac{7(13a^2b+13b^2c+13c^2a+35a^2b^2+35b^2c^2+35b^2c^2-144abc)}{8(a+7b)(b+7c)(c+7a)}... | Your solution is correct.
Let $A$ be your expression and $B=a(a+7b) + b(b+7c) +c(c+7a)$. Then it is enought, by Cauchy Shwarz inequality, to prove $$8(a+b+c)^2\geq 3(a^2+b^2+c^2+7ab+7bc+7ca)$$
i.e.
$$5(a^2+b^2+c^2)\geq 5(ab+bc+ca)$$ which is true by again CS or AG.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find postive integer $x$ with $x>1$ and $\dfrac{x^6-1}{x-1}$ is perfect square
Find positive integer $x$ with $x>1$ and $\dfrac{x^6-1}{x-1}$ is perfect square.
My try: Let $\dfrac{x^6-1}{x-1}=y^2$, so $(x^4+x^2+1)(x+1)=y^2$.
Let $d= \gcd(x+1,x^4+x^2+1)$.
And
$d \vert x^4+x^2+1=(x^2+x+1)(x^2-x+1)$, because $d\vert x+1... | There's a simpler solution if you consider $\frac{x^6 - 1}{x - 1} = (x^2 + x + 1)(x^3 + 1)$. Let $d = \gcd(x^2 + x + 1, x^3 + 1)$, then $d \mid x^2 + x + 1 \mid x^3 - 1$ and $d \mid x^3 + 1$. So $d \mid 2$ but the number $x^2 + x + 1 = x(x + 1) + 1$ is odd. Therefore $d = 1$ and the numbers $x^2 + x + 1$ and $x^3 + 1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Integration by substitution and by partial fractions lead to different results When integrating $\int \frac{3x-2}{x+1}dx$ we can take two paths.
$\alpha)$ Let $u=x+1$, so
$$\int \frac{3x-2}{x+1}dx = \int \frac{3u-5}{u}du = 3u- 5\ln|u|=3(x+1)-5\ln|x+1|$$
$\beta)$ See that $3x-2=3(x+1)-5 \implies \frac{3x-2}{x+1}=3-\frac... | Both of them are correct, because you forget the $+C$ part.
$$\begin{align}\int \frac{3x-2}{x+1}dx&=3(x+1)-5\ln|x+1|+C_1\\
\\
\int \frac{3x-2}{x+1}dx&=3x-5\ln|x+1|+C_2\end{align}$$
Since $C_1$ and $C_2$ are arbitrary constants, if you define $C_2=3+C_1$, they are the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
find $\int_0^\infty \frac{|\cos (\pi x)|}{4x^2 - 1} dx$
Find (with proof) $\displaystyle\int_0^\infty \frac{|\cos (\pi x)|}{4x^2 - 1}dx$
It's actually not even clear that the integral converges. If there were only sines/cosines in the integral, a standard technique would be to use the trigonometric identity $\sin \th... | Since others addressed the first part of the question, I will look at the second part
$$\int_0^\infty \dfrac{\cos^2(\pi (2m+1) x)}{4x^2 - 1}dx = 0 = \int_0^\infty \dfrac{\sin^2(\pi (2m+1)x)}{4x^2-1} dx$$
To confirm or deny this statement let is consider the following integrals
$$\int_0^\infty \dfrac{\cos^2(b x)}{1+a^2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 5
} |
Evaluating $\int_0^\pi x\frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx$
How am I supposed to solve the following definite integral?
$$
\mathcal{I} = \int_0^\pi x \cdot \frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx
$$
This definite integral is solved if the minus sign is replaced by... | Evaluate \begin{align}
I=& \int_0^{\pi/2} \frac{x(\sin x- \cos x)}{\sqrt{\sin{2x}}} dx
=\int_0^{\pi/2} x\ d\bigg(- \tanh^{-1}\frac{\sqrt{2\tan x}}{1+\tan x}\bigg)\\
\overset{ibp}=&\int_0^{\pi/2} \tanh^{-1}\frac{\sqrt{2\tan x}}{1+\tan x}
\overset{t= \tan x}{dx}
=\int_0^\infty \frac1{1+t^2}\tanh^{-1}\frac{\sqrt{2t}}{1+t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
} |
How to compute $\lim\limits_{x\to 0} (-1)^{n+1} \frac{\int_0^x \frac{t^{2n+2}}{1+t^2}dt}{x^{2n+1}}$? Consider the function
$$f(x)=(-1)^{n+1} \frac{\int_0^x \frac{t^{2n+2}}{1+t^2}dt}{x^{2n+1}}$$
How do we compute $\lim\limits_{x\to 0} f(x)$?
Here is my attempt
If $t\geq 0$ then $\frac{t^{2n+2}}{1+t^2} \leq t^{2n+2}$ and... | Too much advanced but this is just for your curiosity.
Sonner or later, you will learns that
$$\int \frac{t^{2n+2}}{1+t^2}\,dt=\frac{t^{2 n+1} }{2 n+1}\left(1-\,
_2F_1\left(1,n+\frac{1}{2};n+\frac{3}{2};-t^2\right)\right)$$ where appears the Gaussian hypergeometric function.
$$\int_0^x \frac{t^{2n+2}}{1+t^2}\,dt=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4512058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving $\cos(2\theta)=\sin\left(\frac{\theta}{2}\right)$ Consider an acute angle $\theta$, which has the property that
$$\cos(2\theta)=\sin\left(\frac{\theta}{2}\right).$$
I am trying to find the value of $\theta$. Using the identity
$$\cos(2\theta)=\sin\left(\frac{\pi}{2}-2\theta\right),$$
I can write the first equat... | That guarantees that that is one of the answers.
It's justified because if $x = y$ then $f(x) = f(y)$ for any possible function.
So IF $\frac \theta 2 = \frac \pi 2 - 2\theta$ then $\sin \frac \theta 2 = \sin (\frac \pi 2 -2\theta) = \cos 2\theta$, then any solution to $\frac \theta 2 = \frac \pi 2 - 2\theta$ (there i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
If $\frac{x^2+y^2+x+y-1}{xy-1}$ is an integer for positive integers $x$ and $y$, then its value is $7$. I saw this on quora
and haven't been able to solve it.
If $\dfrac{x^2+y^2+x+y-1}{xy-1}$
is an integer for positive integers
$x$ and $y$,
then its value is $7$.
If
$y=1$ this is
$\dfrac{x^2+x+1}{x-1}
= x+2+\dfrac{3}{x... | This can be shown using Vieta Jumping. Consider the polynomial
$$P(x,y)=x^2+y^2+x+y-1-kxy+k$$
We will show that for $x,y,k\in\mathbb{N}$, $P(x,y)=0$ implies $k=7$.
$1)$ If $(x,y)$ is a solution, then $(y,x)$ is a solution
$2)$ If $y=1$ then clearly $k=7$.
$3)$ $k\geq 3$. At $k=1$ solving for $x$ in $P(x,y)=0$ gives a d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 0
} |
Prove $abc+abd+acd+bcd\le\frac{1}{27}+\frac{176abcd}{27}$ for $a+b+c+d=1$
Let $a,b,c$, and $d$ be four positive reals satisfying $a+b+c+d=1$. Show that
$$abc+abd+acd+bcd\le\frac{1}{27}+\frac{176abcd}{27}.$$
I tried the inequality between $27abc$ and $(a+b+c)^3$ but it didn't help me
| Inspired by https://math.stackexchange.com/a/1748125/823641.
Define $f(a,b,c,d) = abc+abd+acd+bcd - \frac{1}{27} - \frac{176abcd}{27}$.
We have $f(1/4,1/4,1/4,1/4) = 0$.
We shall show that $f(1/4,1/4,1/4,1/4)$ is the maximum among all $a,b,c,d$ satisfying the conditions.
Suppose not, and assume that another value of $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\frac{4-5\sin\alpha}{2+3\cos\alpha}$
Calculate $$\dfrac{4-5\sin\alpha}{2+3\cos\alpha}$$ if $\cot\dfrac{\alpha}{2}=-\dfrac32$.
My first approach was to somehow write the given expression only in terms of the given $\cot\frac{\alpha}{2}$ and just put in the value $\left(-\dfrac{3}{2}\right)$. Now I don't thi... | We have $$\cot{\frac{\alpha}{2}}=\frac{\sin \alpha}{1-\cos \alpha}=-\sqrt{\frac{1+\cos \alpha}{1-\cos \alpha}}=-\frac{3}{2} \implies \sin \alpha=1.5\cos \alpha -1.5;$$ $$3\sqrt{1-\cos \alpha}=2\sqrt{1+\cos \alpha}$$
Thus, $9(1-\cos \alpha)=4(1+\cos \alpha) \implies \cos \alpha= \frac{5}{13}$
$$\frac{4-5\sin \alpha}{2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Calculate $\sin^5\alpha-\cos^5\alpha$ if $\sin\alpha-\cos\alpha=\frac12$ Calculate $$\sin^5\alpha-\cos^5\alpha$$ if $\sin\alpha-\cos\alpha=\dfrac12$.
The main idea in problems like this is to write the expression that we need to calculate in terms of the given one (in this case we know $\sin\alpha-\cos\alpha=\frac12$).... | using the hint: $x^5-y^5 = (x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$
Let's create the parts:
since $x^2+y^2=1$ we have:
$(x^2+y^2)^2 = x^4+2x^2y^2+y^4 = 1$ (2)
We have a big part of it but we are missing $xy$ so
we know $x-y = 1/2$ squaring both side: $x^2 + y^2 -2xy = 1/4$ since $x^2+y^2=1$ then $xy=3/8$ (1).
using (2)
$(x-y)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
} |
Triangle ABC with $\angle{ACB} = 3\angle{ABC }$ and $AB = \frac{10}{3}BC$. Find $\cos{A}\cos{B}\cos{C}$ assume that $\angle{ABC}=y, \angle{ACB}=3y, BC=3x,$ and $AB=10x$ then by sine rule, i obtain following
$ \frac{10x}{\sin{3y}}=\frac{AC}{\sin{y}}=\frac{3x}{\sin{4y}}$
by cosine rule in try to figure out AC
$AC=\sqr... | From the first equation:
$$10\sin(4y)-3\sin(3y)=0$$
Let's expand everything in terms of $\sin y$ and $\cos y$:
$$\begin{align}\sin(4y)&=\sin(2(2y))\\&=2\sin2y\cos2y\\&=4\sin y\cos y(1-2\sin^2y)\\&=4\sin y\cos y-8\sin^3 y\cos y\\\sin(3y)&=\sin(y+2y)\\&=\sin y\cos2y+\cos y\sin2y\\&=\sin y(1-2\sin^2y)+2\sin y\cos^2y\\&=\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Simplifying the determinant of a matrix. Suppose $$A = \begin{pmatrix} 1+a_{1}+a_{1}b_{1}+b_{2} & 1+a_{1} & 1 & 0\\ a_{2}+a_{2}b_{1}+b_{3} & 1+a_{2} & 1 & 1\\ a_{3}+a_{3}b_{1} + b_{4} & a_{3} & 1 & 1\\ a_{4} + a_{4}b_{1} & a_{4} & 0 & 1\end{pmatrix}$$ Show that $$\det(A) = \det\begin{pmatrix} b_{1}-b_{2}+b_{4} & 1+a_{1... | First, for any integer $n\geq 1$ and any numbers $A_1,...,A_n,B_1,...,B_n$ we have
$$
\det\left(I_n+\begin{pmatrix}A_1 & B_1 & 0 & \cdots &0 \\
A_2 & B_2 & 0 & \cdots &0 \\
\vdots & \vdots & \vdots & \ddots &\vdots \\
A_n & B_n & 0 & \cdots &0 \\\end{pmatrix}\right)=\det\begin{pmatrix}1+A_1 & B_1 \\
A_2 & 1+B_2\end{pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Maximum length of points of tangency of an ellipse and a variable circle Let ellipse $4x^2+16y^2=64$ and circle $x^2+y^2=r^2$ have a common tangent touching at A and B respectively. The maximum length of AB can be?
(A) 4
(B) 3
(C) 2
(D) 5
I tried solving the question using parametric coordinates, using slope form of ta... |
The dual conics are $16X^2+4Y^2=1$ and $r^2X^2+r^2Y^2=1,$ and they intersect when r is between $2$ and $4$ making the common tangents $$\pm x \sqrt{r^2-4}/(2 \sqrt3 r) \pm y \sqrt{16-r^2}/(2\sqrt3 r) + 1=0$$ and $$A: (\pm(r \sqrt{r^2-4})/(2 \sqrt3),\pm(r \sqrt{16-r^2})/(2 \sqrt3))$$ $$B: ((\pm 8\sqrt{r^2-4})/(\sqrt3 r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4532053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Proving using Mathematical Induction from my discrete math class This is a practice exercise from our class about proving inequalities using mathematical induction. I've been stuck on the last step for quite a while now.
This is the Question. "Prove that $\sum_{k=1}^n\frac{1}{k^2}=\frac{1}{1^2}+\frac{1}{2^2}+\cdots+\fr... | In step3, you added a new term $\frac{1}{c+1}$ (while $n=c+1$) to the both sides of the induction hypothesis ($\sum_{k=1}^{c}\frac{1}{k^2} \le 2-\frac{1}{c}$) and ended with:
$\sum_{k=1}^{c+1}\frac{1}{k^2} \le 2-\frac{1}{c}+\frac{1}{(c+1)^2} \le ...
\le 2-\frac{1}{c+1}-\frac{1}{c(c+1)^2}$
Because $n=c+1$ and $n>1$, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4532680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the all real roots of the polynomial $x^6+3 x^5+3 x-1=0$ in closed form
Find the all real roots of the polynomial
$$x^6+3 x^5+3 x-1=0$$
in exact form.
WolframAlpha gives only numerical results. I've asked a few similar questions before. The source of the problem comes from the algebra precalculus workbook (but n... | This is such a cursed question.
The only elementary method I can think of is to assume the existence of a factorization of the form $$P(x) = x^6 + 3x^5 + 3x - 1 = (x^2 + kx \pm 1)(x^4 + ax^3 + bx^2 + cx \mp 1) \tag{1}$$ and then after tediously expanding and equating coefficients, we require
$$(a,b,c) = (3-k, 1-3k+k^2,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4534232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\sin^n (2x) + (\sin^n x - \cos^n x)^2 \leq 1$
Prove that $\sin^n (2x) + (\sin^n x - \cos^n x)^2 \leq 1$.
Let $a = \sin x, b = \cos x.$ Then we want to show $2^n a^n b^n +a^{2n}-2a^n b^n + b^{2n}\leq 1,$ which follows once we show that $(2^n-2)a^n b^n \leq \sum_{j=1}^{n-1}{n\choose j} a^{2(n-j)}b^{2j}$. I ... | Assume that $n$ is a positive integer.
Let $a = \sin x$ and $b = \cos x$.
Then $a^2 + b^2 = 1$ and $|2ab| \le a^2 + b^2 = 1$.
Let $$f(n) := (2^n-2) a^n b^n + a^{2n} + b^{2n}.$$
We have $f(1) = a^2 + b^2 = 1$ and $f(2) = 2a^2b^2 + a^4 + b^4 = (a^2 + b^2)^2 = 1$.
First, we have
\begin{align*}
&f(2n) - f(2n+1)\\
=\,& (2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4535557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Reducing $ax^6-x^5+x^4+x^3-2x^2+1=0$ to a cubic equation using algebraic substitutions
Use algebraic substitutions and reduce the sextic equation to the cubic equation, where $a$ is a real number:
$$ax^6-x^5+x^4+x^3-2x^2+1=0$$
My attempts.
First, I tried to use the Rational root theorem, when $a$ is an integer $x=\pm... | You are on the right track.
Let,
$$P(x)=ax^6-x^5+x^4+x^3-2x^2+1$$
We observe that, $0$ is not a possible root of $P(x)$. Therefore, we can divide all terms of the polynomial by $x^2\, (x\neq 0)\,:$
$$
\begin{align}\frac {P(x)}{x^2}&=ax^4-x^3+x+\left(x-\frac {1}{x}\right)^2\\
&=x^2\left(ax^2-\left(x-\frac 1x\right)\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4538322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Finding the limit in terms of a without using L'Hospitals Find $\;\lim_\limits{x \to 0} \dfrac{1-\cos(ax)}{1-\sqrt{1+x^2}}\;$ in terms of a without using L'Hospitals Rule.
I first graphed the function and noticed that that limit tends to go towards $-a^2$
I also tried this approach:
$\lim_\limits{x \to 0} \dfrac{1-\cos... | Thank to @Wang YeFei for the hint.
$\lim_\limits{x \to 0} \dfrac{1-\cos(ax)}{1-\sqrt{1+x^2}}=\lim_\limits{x \to 0} \dfrac{2 \cdot \frac{1-\cos(ax)}{2}}{1-\sqrt{1+x^2}} = \lim_\limits{x \to 0} \dfrac{2\sin^2(\frac{ax}{2})}{1 - \sqrt{1+x^2}} $
Then we proceed to multiply the numerator and denominator by $1 - \sqrt{1+x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4541362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding the value of given integral The given integral is :$$\displaystyle\int_0^1 \dfrac{\sin{\theta}(\cos^2{\theta} -\cos^2{\frac{\pi}{5}})(\cos^2{\theta} -\cos^2{\frac{2\pi}{5}})}{\sin{5\theta}} \, d\theta$$
I tried solving it with some trigonometric identities but comes out it does not work here.
However solution g... | Note
$$ \sin(5\theta)=\sin \theta(1+2\cos2\theta+2\cos4\theta), \cos^4\theta=\frac18(3+4\cos2\theta+\cos4\theta) $$
from here.
Using
$$ \cos^2{\frac{\pi}{5}}+\cos^2{\frac{2\pi}{5}}=\frac{3}{4}, \cos^2{\frac{\pi}{5}}\cos^2{\frac{2\pi}{5}}=\frac{1}{16} $$
one has
\begin{eqnarray}
&&(\cos^2{\theta} -\cos^2{\frac{\pi}{5}})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the first derivative of $y=(x^4-1)\sqrt[3]{x^2-1}$
Find the first derivative of $$y=(x^4-1)\sqrt[3]{x^2-1}$$
We can write the function as $$y=(x^4-1)\left(x^2-1\right)^\frac13$$ For the derivative we have $$y'=4x^3\left(x^2-1\right)^\frac13+\dfrac13\left(x^2-1\right)^{-\frac23}2x(x^4-1)\\=4x^3\left(x^2-1\right)^... | The two formulas are equivalent.
$$\begin{align}
&\frac{2}{3}x\left(x^{2}-1\right)^{\frac{1}{3}}\left(7x^{2}+1\right)-\frac{2x\left(7x^{4}-6x^{2}-1\right)}{3\sqrt[3]{\left(x^{2}-1\right)^{2}}}
\\&=\frac{2x}{3\left(x^{2}-1\right)^{\frac{2}{3}}}\left(\left(7x^{2}+1\right)\left(x^{2}-1\right)^{\frac{1}{3}}\cdot\left(x^{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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A smarter (not bashy) way to solve this roots of unity problem? (Mandelbrot) Let $\xi = \cos \frac{2\pi}{5} + i \sin \frac{2pi}{5}$ be a complex fifth root of unity. Set $a = 20\xi^2 + 13 \xi, b = 20\xi^4 + 13\xi^2, c = 20\xi^3 + 13\xi^4, \text{and } d = 20\xi + 13\xi^3$. Find $a^3 + b^3 + c^3 + d^3$
Immediately what c... | You're not using the most important parts of the question: namely that $\xi$ is a fifth root of unity, and also that conveniently, all the coefficients are either $20$ and $13$.
Calculating $a^3 + b^3 + c^3 + d^3$ directly and collecting like coefficients, we have:
$$a^3 + b^3 + c^3 + d^3 = 20^3(\xi^6 + \xi^{12} + \xi^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4549205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Show a function is continuous on $\mathbb{R}$ by any method Let $f(x)=\frac{Kx}{K^2+x^2}$ where K is some constant, show this is continuous on $\mathbb{R}$.
Here are my scratch work in looking for a delta. let $x,y\in \mathbb{R} $ WTS $|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}|<\epsilon,\forall \epsilon>0$ whenever $|x-y|... | Since you are asking a solution by any method, you can try to use the limit definition of continuity. Recall that a function $f$ is continuous at a point $c$ of its domain if $$\lim_{x\rightarrow c} f(x) = f(c)$$
So for any $c\in \mathbb{R}$ we have $\lim_{x\rightarrow c}Kx = Kc$ and $\lim_{x\rightarrow c} K^2+x^2 = K^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4550150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Sum of the areas of all n-sided polygons inscribed within each other Imagine you take an $n$-sided regular polygon with side length $x$, and connect the midpoints of each of its sides to construct a smaller but identical $n$-sided regular polygon within it.
Now perform the same process with the smaller polygon, and the... | Let the side length of the outermost polygon and side length of the first constructed polygon be $x_0$ and $x_1$ respectively. The external angles of the polygons will each be $\frac{2\pi}{n}$, so the internal angles will each be $\pi - \frac{2\pi}{n}$. Consider the triangle shown in the diagram below:
Applying the co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln\big(\frac mn \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}\dots\big)$ page 371 in ‘Synopsis Of Elementary Results In Pure Mathematics’ contains the following result,
$$\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln \left(\frac{m}{n} ... | \begin{align}
&\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln x}\,dx \\
=&\int_0^1 \frac{x^{m -1}-x^{n -1}}{\ln x}\sum_{k\ge 0} (-x^p)^k \,dx\\
=&\ \sum_{k\ge 0}(-1)^k
\int_0^1 \frac{x^{m +pk-1}-x^{n +pk -1}}{\ln x}
\> \ \overset{u=-\ln x}{dx}\\
=&\ \sum_{k\ge 0} (-1)^k
\int_0^\infty \frac{e^{-(n +pk)u}-e^{-(m +pk)u}}{u}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Spivak, Calculus, Ch. 22: How do we compute the limit $\lim\limits_{n\to\infty} \frac{(n+1)^{\frac{n+1}{n}}}{n}$? My question is simply how do we compute the limit
$$\lim\limits_{n\to\infty} \frac{(n+1)^{\frac{n+1}{n}}}{n}$$
I know the limit is $1$, both from the context below and because I checked in Maple.
Here is th... | Another way:
Using $$ \lim_{n \to \infty} \frac{1}{n} \, \ln\left(1 + \frac{1}{n}\right) = 0$$ and $\frac{\ln n}{n} < 1$ then
$$ \frac{1}{n} \, (n+1)^{\frac{n+1}{n}} = \frac{n+1}{n} \, e^{\left( \ln n + \ln\left(1 + \frac{1}{n}\right) \right)/n} = \left(1 + \frac{1}{n}\right) \, e^{\frac{1}{n} \, \ln\left(1 + \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove $\sum^{n+1}_{j=1}\left|\cos\left(j\cdot x\right)\right|\geqslant \frac n4$ How do you prove that $\displaystyle\sum^{n+1}_{j=1}\left|\cos\left(j\cdot x\right)\right|\geqslant \dfrac{n}{4}$, where $x\in\mathbb{R}$?
I tried mathematical induction, but it doesn't work. I also have some ideas on complex number soluti... | By periodicity $\left|\cos(jx)\right|=|\cos (j(\pi+x))|=|\cos (j(\pi-x))|$ and the fact that $\cos$ even, it's enough to assume $0 <x \le \pi/2$ as the inequality is obvious for $x=0$.
Note that $|\cos x|+|\cos 2x| \geqslant \dfrac{\sqrt{2}}{2} $ with equality at $x=\pi/4$ (splitting into $0 \leqslant x \leqslant \pi/4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
} |
Compute the line integral of $x^2 + y^2$ on curve $|x|+|y|=1$
Compute the integral
$$\int_{\gamma} f(x,y)ds$$
where $\gamma=\{|x|+|y|=1\}$ and $f(x,y)=x^2+y^2$
Note, we have not covered any major theorems on evaluating line integrals, which has been presented is to parametrize the curve then integrate with respect to... | $\gamma=\gamma_1+\gamma_2+\gamma_3+\gamma_4$ is the obvious sum of line segments: 1) $\gamma_{1}$ joins $(1,0)$ and $(0,1)$. 2) $\gamma_{2}$ joins $(0,1)$ and $(-1,0)$. 3) $\gamma_{3}$ joins $(-1,0)$ and $(0,-1)$. 4) 2) $\gamma_{4}$ joins $(0,-1)$ and $(1,0)$.
Then, $I=\int_{\gamma}(x^2+y^2)ds=\sum_{i=1}^4I_{i}$ where ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove or disprove $ab+bc+cd+da\leq1$ if $a+b+c+d=2$ Non-negative real numbers $a,b,c,d$ are such that $a+b+c+d=2$. Prove or disprove that
$$ab+bc+cd+da\leq1$$
I see there are multiple equality cases, where $(a,b,c,d)$ is for example $(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2})$, $(\frac{3}{4},\frac{1}{2},\frac{1... | We have: $ab+bc+cd+da = (a+c)(b+d) \le \dfrac{((a+c)+(b+d))^2}{4}= \dfrac{(a+b+c+d)^2}{4}= 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4557878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Can we find the integral solutions of $x(3y-5)=y^2+1$?
$x(3y-5)=y^2+1$ has no integral solution. True or False?
Answer False, as $(-1,1)$ satisfies it.
I checked on WolframAlpha that the given hyperbola has four integral solutions. $(-1,1),(-1,-4),(5,13),(5,2)$.
I wonder if we can find that without hit and try.
My At... | Hint. This is a linear equation in $x$, so
$$ x = \frac{y^2+1}{3y-5} = \frac{1}{3}\frac{3y^2+3}{3y-5} = \frac13\frac{y(3y-5)+(5/3)(3y-5)+3+25/3}{3y-5} $$
Therefore
$$ 9x = 3y+5+\frac{34}{3y-5} \implies \frac{34}{3y-5}=9x-3y-5.$$
The right hand side of the equation is an integer, if $x$ and $y$ are integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4559135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show that for any $n$, there are infinitely many cubes of the form $2^na - 9$.
Show that for any $n$, there are infinitely many cubes of the form $2^na - 9$.
Progress: We use induction on $n.$ For $n=1$ it works. Say it works for $n-1$. We will show for $n$. Note that $2^na-9$ is $$2^{n-1}2a-9.$$ If we have infinite ... | Suppose we have some positive $b$ for which $2^nb-9=u^3$ for some positive integer $u$. Notice that $u$ must be odd. Note then that
\begin{equation}
2^n(b+3u^2+3u2^n+(2^{n})^2)-9=u^3+3u^22^n+3u(2^n)^2+(2^n)^3=(u+2^n)^3,
\end{equation}
and so $a=b+3u^2+3u2^n+(2^n)^2$ also satisfies the equation $2^na-9=v^3$ for some int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4561289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Find $\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}$ without using derivatives This limit is proposed to be solved without using the L'Hopital's rule or Taylor series:
$$
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}},
$$
where $a>0$, $b>0$ are some c... | Let's prove this expression:
Given positive $a$ and $b$:
$$\lim_{n\rightarrow\infty}\bigg(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\bigg)^n=\sqrt{ab}$$.
In order to show this we are going to use the following:
$$\lim_{n\rightarrow\infty}n(\sqrt[n]{a}-1)=\ln a$$ thus by AM-GM
$$\frac{1}{n}\ln\sqrt{ab}\leq\ln\frac{1}{2}(\sqrt[n]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4561863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$3$-var inequality: $\frac{bc}{\sqrt{a}+3}+\frac{ca}{\sqrt{b}+3}+\frac{ab}{\sqrt{c}+3} \leq \frac{3}{4}$ for $a+b+c=3$. Problem: Let $a,b,c$ be positive numbers satisfied $a+b+c=3$. Prove that $$\dfrac{bc}{\sqrt{a}+3}+\dfrac{ca}{\sqrt{b}+3}+\dfrac{ab}{\sqrt{c}+3} \leq \dfrac{3}{4}$$
I've tried U.C.T method but it doesn... | Fact 1: It holds that, for all $x \ge 0$,
$$\frac{1}{3 + \sqrt x} \le \frac13 - \frac{13}{96}x + \frac{5}{96}x^2.$$
(Note: Letting $x=y^2$,
we have $\mathrm{RHS} - \mathrm{LHS} = \frac{y(5y^2 + 25y + 32)(y-1)^2}{96(y+3)}\ge 0$.
The RHS comes from the Taylor approximation of the LHS around $x=1$,
that is $\frac{1}{3 + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4562635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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Fractions in Questions and Answers
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