Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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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... | $$I=\int_0^\pi x \frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx\overset{IBP}=4\int_0^\frac{\pi}{2}\ln(\sqrt{\sin x}+\sqrt{1+\sin x})dx$$
$$\overset{\large x\to \frac{\pi}{2}-2x}=8\int_0^\frac{\pi}{4}\ln\left(\sqrt{\cos(2x)}+\sqrt 2 \cos x\right)dx\overset{\cos x\to x}=8\int_\frac{1}{\sqrt 2}^1\frac{\op... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 2
} |
How to compute $\lim\limits_{x\to 0} (-1)^{n+1} \frac{\int_0^x \frac{t^{2n+2}}{1+t^2}dt}{x^{2n+1}}$? Consider the function
$$f(x)=(-1)^{n+1} \frac{\int_0^x \frac{t^{2n+2}}{1+t^2}dt}{x^{2n+1}}$$
How do we compute $\lim\limits_{x\to 0} f(x)$?
Here is my attempt
If $t\geq 0$ then $\frac{t^{2n+2}}{1+t^2} \leq t^{2n+2}$ and... | You can expand ${1\over 1+t^2}$ as a geometric series as $x \rightarrow 0$.
$$\lim_{x \rightarrow 0} {1\over x^{2n+1}}{\int_{0}^{x}{t^{2n+2} \over {1+t^2}}dt}\space =\space\lim_{x \rightarrow 0} {1\over x^{2n+1}}\sum_{k=0}^{\infty}{(-1)^k}\int_{0}^{x}{t^{2n+2k+2}dt}$$
$$\lim_{x \rightarrow 0} {1\over x^{2n+1}}\sum_{k=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4512058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing $\int_{0}^{\pi} \ln (\sin x+2) d x$ and $\int_{0}^{\pi} \ln (2-\sin x) d x$ I first encountered this integral
$$
I=\int_{0}^{\pi} \ln (\sin x+2) d x
$$
several months ago without any idea and had tried many methods such as integration by parts, substitution and Fourier series etc. but all are in vain. Today, ... | To evaluate the second integral, we may use their product.
$$\int_{0}^{\pi} \ln (\sin x+2) d x+ \int_{0}^{\pi} \ln (2-\sin x) d x = 2\int_{0}^{\pi/2} \ln (4-\sin^2 x) d x $$
Using my post,
$$\begin{aligned}\int_{0}^{\pi} \ln (2-\sin x) d x &=2 \pi \ln \left(\frac{2+\sqrt{3}}{2}\right) -\left(-\pi \ln 2+\frac{2 \pi}{3} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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If $f(x) =ax^3+bx^2+cx+d$ is a cubic equation with roots $\alpha,\beta,\gamma.$ Is there a way to find $\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha?$ Suppose $f(x) = ax^3 + bx^2 + cx + d$ is a cubic equation with roots $\alpha, \beta, \gamma.$ Then we have:
$\alpha + \beta + \gamma= -\frac{b}{a}\quad (1)$
$\alpha\beta +... | Quantity is not-symmetric. This results in rather complex formula:
$$\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha=t_{1}\,t_{3}^2-{{b\,t_{3}^2}\over{3\,a}}+t_{2}^2\,t_{3}-{{2\,b\,
t_{2}\,t_{3}}\over{3\,a}}-{{2\,b\,t_{1}\,t_{3}}\over{3\,a}}+\\{{b^2\,
t_{3}}\over{3\,a^2}}-{{b\,t_{2}^2}\over{3\,a}}+t_{1}^2\,t_{2}-{{2\,b
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If the matrix A is given, then find the value of $A^{2022}$ If it is given that matrix $ A = \begin{bmatrix}
\dfrac{5}{2} & \dfrac{3}{2}\\
-\dfrac{3}{2} & -\dfrac{1}{2}
\end{bmatrix}$ then find the value of $A^{2022}$.
Here is my try on it : $$\\$$
If we look at the pattern while multliplying A with itself it ... | The characteristic polynomial of $A$ is
\begin{align}
\det(A-\lambda I)=\left|\begin{array}{c}
\frac{5}{2}-\lambda & \frac{3}{2} \\ -\frac{3}{2} & -\frac{1}{2}-\lambda
\end{array}\right|\\
= (5/2-\lambda)(-1/2-\lambda)+9/4 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
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... | You just use the trigonometrical formulae: $sina=\dfrac{2tan(a/2)}{1+tan^{2}(a/2)}$ and
$cosa=\dfrac{1-tan^{2}(a/2)}{1+tan^{2}(a/2)}$ and set $\,\,\,tan(a/2)=\dfrac{1}{cot(a/2)}=-\dfrac{2}{3}$.
If we put this value in the original fraction we obtain :
value=$\dfrac{102}{41}$. If my numerical calculations are correct!
| {
"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": 1
} |
What is wrong in this solution of $\sec x + \csc x = 2 \sqrt{2}$
Find number of solutions in the interval $[0,2\pi]$ of the equation - $$\csc x + \sec x = 2 \sqrt{2}.$$
$⇒ \dfrac{1}{\sin x}+ \dfrac{1}{\cos x }= 2 \sqrt2$
$⇒ \dfrac{\sin x +\cos x}{\sin x \cos x} = 2 \sqrt 2 ⇒ (\sin x+\cos x)^2=8\sin^2x\cos^2x⇒1+2\sin... | Hint:
Instead of squaring $$\dfrac {\sin x + \cos x}{\sin x \cos x} = 2\sqrt{2}$$ clear fractions to get $$\sin x + \cos x = 2\sqrt{2}\sin x \cos x$$ and note that you have a double angle identity hidden on the RHS. Then you can use another identity to get the LHS and find the relationship between the two.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"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... | Set $\cos y=p$, then you can get this equation in sine rule you wrote:
$$
10\sin4y=3\sin 3y
\\
40\sin y\cos y\cos2y=9\sin y-12\sin^3 y
\\
40p(2p^2-1)=9-12(1-p^2)=12p^2-3
\\
80p^3-12p^2-40p+3=(4p-3)(20p^2+12p-1)=0
$$
Then, you can see $\cos B=\frac34$, and the only thing left is the calculation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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 problem is not easy, but you can simplify the work a bit by observing that you can divide both sides by $x^3$. It can be shown that $x=0$ is not a solution to the original polynomial, hence this operation is allowed. We get
$$x^3-\frac 1{x^3}+3\left(x^2+\frac 1{x^2}\right)=0$$
Now, use the two identities below
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4534232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\lim _{x \to \infty }\frac{1 - 2x^2}{x^2 + 3}\ = -2$ I solved it using the definition of limit to the $\left|\frac{-7}{x^2 +3}\right| < \varepsilon$
Then $\varepsilon > \frac{7}{x + 3} > \frac{7}{x^2 +3}$
Then $x > \frac{7}{\varepsilon} - 3$
But my teacher said this was wrong and I have to fix it to
$x > \sqr... | We have that for $0<\varepsilon <\frac 7 3$
$$\left|\frac{1 - 2x^2}{x^2 + 3}\ -(-2)\right|<\varepsilon \iff \left|\frac{7}{x^2 + 3}\ \right|<\varepsilon \iff x^2>\frac 7 \varepsilon -3$$
which leads to
$$x>\sqrt{\frac 7 \varepsilon -3}$$
or also, as you teacher suggests, for any $\varepsilon>0$
$$x>\sqrt{\left|\frac 7 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4538068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How do I show $\frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$ for $n \ge 1$? So this was part of a bigger induction proof problem, but I'm stuck at what I think is the last step. I need to show that
$\frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$, for $n \ge 1$
Would it be enough to show that the limit as i... | There are lots of ways to prove this.
One way is to square both sides and
cross-multiply.
Another way is to write
$\dfrac{2n+1}{2n+2}
=1-\dfrac{1}{2n+2}
$
and
$\dfrac{\sqrt{n+1}}{\sqrt{n+2}}
=\sqrt{\dfrac{n+1}{n+2}}
=\sqrt{1-\dfrac{1}{n+2}}
$.
Squaring both sides,
you want
$1-\dfrac{1}{n+2}
\ge (1-\dfrac{1}{2n+2})^2
=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4538193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
$\displaystyle\int\limits_{\frac{\pi}{2}}^{\pi}\dfrac{1}{(\sin\theta - 2\cos\theta)^2}\,d\theta$: singularity in Weierstrass sub Could I use the integration by Weierstrass substitution or tangent-half angle substitution for this case? Because I see on Wikipedia, when the lower limit was $0$, they made the definite inte... | THASsing, as I like to call the process, is perfectly fine here as there are no poles (which solve $\sin x=2\cos x$ or $\tan x=2$) on the interval of integration. After simplifying you get
$$\int_1^\infty\frac{1+t^2}{2(t^2+t-1)^2}\,dt=\left[-\frac t{2(t^2+t-1)}\right]_1^\infty=\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4543032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Squeeze Theorem for $(2^n + 3^n + 4^n)^{\frac{1}{n}}$ How to find this limit when $n \rightarrow \infty$.
$$(2^n + 3^n + 4^n)^{\frac{1}{n}}.$$
My Idea:
We need to squeeze this sequence.
$$ (4^n)^{\frac{1}{n}}\le (2^n + 3^n + 4^n)^{\frac{1}{n}} \le (3\cdot4^n)^{\frac{1}{n}}$$
So that we obtain
$$ 4\le (2^n + 3^n + 4^n)^... | I have another idea that might be useful:
$$
(2^n+3^n+4^n)^\frac{1}{n}=\frac{4}{4}(2^n+3^n+4^n)^\frac{1}{n}=4\left(\frac{2^n+3^n+4^n}{4^n}\right)^\frac{1}{n}=4\left(\left(\frac{2}{4}\right)^n+\left(\frac{3}{4}\right)^n+1\right)^\frac{1}{n}
$$
Taking limits you get $4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4544007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
$\epsilon$-$\delta$ argument Given $x \in \left ( \frac{1}{10}, \frac{3}{10} \right)$. Find $\delta > 0$ so that
$$0< \left| x- \frac{1}{5} \right| < \delta \Rightarrow \left| \frac{1}{x} - 5\right| < \frac{1}{10}.$$
How can I answer that question?
My Idea
Let's do a little observation:
$$\left| \frac{1}{x} - 5\right| ... | Your idea is correct. As you may directly verify, you have the implication $$0< \left| x- \frac{1}{5} \right| < \frac1{255} \;\;\Rightarrow\;\; \left| \frac{1}{x} - 5\right| < \frac{1}{10}.$$ Sometimes I prefer to avoid double inequalities (i.e. $A<x<B$), since they can be difficult to look at if there is a longer sequ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4544725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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To prove: $e^{\csc^2\theta}\sin^2\theta+e^{\sec^2\theta}\cos^2\theta\gt e^2$ Question:
Statement I: If $0\lt\theta\lt\frac\pi2$, then $e^{\csc^2\theta}\sin^2\theta+e^{\sec^2\theta}\cos^2\theta\gt e^2$
Statement II: AM of positive numbers is always greater than or equal to their GM.
A) Both the statements are true and S... | The inequality is false as stated. However, we can show that $$e^{\csc^2\theta}\sin^2\theta+e^{\sec^2\theta}\cos^2\theta\geq e^2 \tag{1}$$
for $0 < \theta < \frac\pi 2$. Using the AM-GM inequality,
$$e^{\csc^2\theta}\sin^2\theta+e^{\sec^2\theta}\cos^2\theta\geq 2 e^{\frac{1}{2\sin^2\theta \cos^2\theta}} \sin\theta\cos\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4545604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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how to find $ x^3+y^3$ if $(\sin^{-1}x)^2+(\cos^{-1}y)^2 = \frac{\pi^2}{2}$ I have to find $ x^3+y^3$ if $$(\sin^{-1}x)^2+(\cos^{-1}x)^2 = \frac{\pi^2}{2}$$
My book has arbitrarily declared that $x=y$, I have an idea, but I have no way to check if it's conceptually right as we just assumed $x=y$ even when we did this ... | You're idia would only work if it's sin(x) and cos(x)
but there are arcsin(x) and arccos(x)...
Trigonometric Way
We can use the trigonometric identity that relates arcsin and arccos:
$$ \arccos(x) = \frac{\pi}{2} - \arcsin(x) $$
Now we can say:
$$
\begin{align*}
\sin^{-1}(x)^{2} + \cos^{-1}(x)^{2} &= \frac{\pi^{2}}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4546672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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determining the start velocity and angle based on the goals(End point) angle and coordinates. ========
I'm trying to figure out the start angle and velocity of an parabola, When i know only the start coordinates and the end point coordinates and the angle which it reaches the endpoint.
I'm tasked with figuring this out... | The equation of motion of a projectile starting from the origin, and shot at an angle $\theta $ above the horizon, at an initial velocity $\mathbf{v_0}$ is
$\mathbf{P}(t) = \mathbf{v_0} t + \dfrac{1}{2} \mathbf{g} t^2 $
Now, $\mathbf{P}(t) = ( x(t), y(t) ) $ and $\mathbf{v_0} = v_0 (\cos \theta, \sin \theta) $, and $ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the value of given integral The given integral is :$$\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$$
I tried solving it with some trigonometric identities but comes out it does not work here.
However solution g... | First you should simplify the term and then bild the antiderivative, what is really easy after simplifying the term:
$$
\begin{align*}
y &= \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}} ~\operatorname{d}\theta\\
F(\theta) &= \int_{0}^{\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a_n$ be a sequence of positive real numbers satisfying $a_n$ < $\frac{1}{2}a_{n-1}$ for all $n \ge 2$. Prove that it converges to zero. Let $a_n$ be a sequence of positive real numbers satisfying $a_n$ < $\frac{1}{2}a_{n-1}$ for all $n \ge 2$. Use an appropriate theorem and the fact that $\lim_{n\to\infty} (\fra... | EDIT
One possible approach consists in noticing that:
\begin{align*}
0 < a_{n} = \frac{a_{n}}{a_{n-1}}\times\frac{a_{n-1}}{a_{n-2}}\times\ldots\frac{a_{2}}{a_{1}}\times a_{1} < a_{1}\times\left(\frac{1}{2}\right)^{n-1}
\end{align*}
Consequently, due to the arithmetic properties of numerical sequences, one gets the desi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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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|... | Though you agreed to use a more appropriate general argument (see comments), let us finish your $\epsilon-\delta$-proof (assuming $K\ne0$).
You were stuck on how to relate the inequality
$$\left|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}\right|\le|K|\left|\frac{xK^2-yK^2+xy^2-yx^2}{(K^2+y^2)(K^2+x^2)}\right|$$
to $|x−y|.$ J... | {
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"timestamp": "2023-03-29T00:00:00",
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Whats the probability for person A to win the dice game? The rules: Two people rolls a single dice. If the dice rolls 1,2,3 or 4, person A gets a point. For the rolls 5 and 6, person B gets a point. One person needs a 2 point lead to win the game.
This is a question taken from my math book. The answer says the probabil... | As you showed, person $A$ can win the first round of two rolls with probability $4/9$. However, as Fishbane pointed out in the comments, person $A$ can also win if both person $A$ and person $B$ fail to win in the first round and person $A$ obtains two points in a subsequent round before person $B$ does. Hence, the p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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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 us first find $a$.
Let $O$ be the center of the regular polygon. Then, a circle centered at $O$ circumscribes the polygon. Each side is a chord of the circle and hence the perpendicular from the center bisects the side. Also, $\triangle AOM\cong\triangle BOM$ so $\angle AOM=\angle BOM=\dfrac{\pi}{n}$. Now, in $\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Find the smallest possible value for $x^2-3x+2y^2+4y+2$ Find the smallest possible value for $x^2-3x+2y^2+4y+2$.
I know this: $x^2-3x+2y^2+4y+2=(x-\frac{3}{2})^2+2(y+1)^2-\frac{9}{4}$ so the smallest value is $-\frac{9}{4}$.
Now we also know this: $x^2-3x+2y^2+4y+2=(x-1)^2-1-x+2(y+1)^2=(*)(x-1)^2+2(y+1)^2-(1+x)$
My que... | $x^2-3x+2y^2+4y+2=0$ is an ellipse, so what do you mean by "the smallest possible value of $x^2-3x+2y^2+4y+2$"?
Anyway, since you have two distinct values of $y=y(x)$ for the given $x: x=\frac{3}{2}$ such that one of them (the negative one) is minimum, I will write only the positive value of $y(x) : x=\frac{3}{2}$, lea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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... | Let $x = a+c$ and $y=b+d$. Then, we have $xy = (a+c)(b+d) = ab + ad + bc + cd$ which is the quantity we want to bound. So, we want to show that if $x+y = 2$, then $xy \leq 1$.
We just use basic algebra now: $x=2-y$, so $xy = y(2-y)$. Now you can rearrange, $y(2-y) = -(y-1)^2 +1$. Since $(y-1)^2 \geq 0$, it follows that... | {
"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": 0
} |
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... | Taking logarithm, then using $\sin x\sim x$ and $\ln(1+x)\sim x$ when $x\to0$, we have
\begin{align*}
\frac1x\cdot\ln\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)&\sim\frac1{\sin x}\cdot\left(\frac{a^{\sin x}+b^{\sin x} }{2}-1\right)\\
&=\frac12\frac{a^{\sin x}-1}{\sin x}+\frac12\frac{b^{\sin x}-1}{\sin x}\\
&\righta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4561863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$3$-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... | We need to prove that $$\sum_{cyc}\frac{a^2b^2}{3+c}\leq\frac{3}{4},$$ where $a$, $b$ and $c$ are positives such that $a^2+b^2+c^2=3$.
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, the condition does not depend on $w^3$ and we need to prove that
$$4\sum_{cyc}a^2b^2(a+3)(b+3)\leq3\prod_{cyc}(a+3)$$ and since... | {
"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": 0
} |
How to Evaluate the Integral? $\int_{0}^{1}\frac{\ln\left( \frac{x+1}{2x^2} \right)}{\sqrt{x^2+2x}}dx=\frac{\pi^2}{2}$ I am trying to find a closed form for
$$
\int_{0}^{1}\ln\left(\frac{x + 1}{2x^{2}}\right)
{{\rm d}x \over \,\sqrt{\,{x^{2} + 2x}\,}\,}.
$$
I have done trig substitution and it results in
$$
\int_{0}^{1... | Let
$$x=y^2,\quad y^2+1=t,\tag1$$
then
$$I=\int\limits_0^1 \dfrac{\ln\left(\dfrac{x+1}{2x^2}\right)}{\sqrt{x^2+2x}}dx
=\int\limits_0^1 \dfrac{\ln\left(\dfrac{y^2+1}{2y^4}\right)}{\sqrt{y^4+2y^2}}\, 2y\,\text dy
=\int\limits_1^2 \dfrac{\ln\left(\dfrac{t}{2(t-1)^2}\right)}{\sqrt{t^2-1}}\,\text dt,$$
$$I=I_1-2I_2-I_3,$$
w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4568778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 5,
"answer_id": 1
} |
Use the substitution $u=\frac{1}{\sqrt{3}}\tan x$ to find $\int\frac{dx}{3-2\sin^2x}$ Question:
Use the substitution $u=\frac{1}{\sqrt3}\tan x$ to find $$\int\frac{dx}{3-2\sin^2x}$$
My Working:
Let $u=\frac{1}{\sqrt3}\tan x$, then
\begin{align}
\frac{du}{dx}&=\frac{1}{\sqrt3}\cdot\sec^2x+0=\frac{\sec^2x}{\sqrt3}\\
du&=... | Your substitution $u=\frac{1}{\sqrt 3}\tan x$ works well as below:
$$
\begin{aligned}
\int \frac{dx}{3-2 \sin ^2 x} =& \int \frac{\frac{\sqrt{3} d u}{3 u^2+1}}{3-2\left(\frac{3 u^2}{3 u^2+1}\right)} \\
=& \sqrt{3} \int \frac{d u}{3 u^2+3} \\
=& \frac{1}{\sqrt{3}} \tan ^{-1} u+C \\
=&\frac{1}{\sqrt{3}} \tan ^{-1}\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4569523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
In $\triangle ABC$, $BD$ is a median, $\angle DAB=15$ and $\angle ABD=30$. Find $\angle ACB$. As title implies, the goal is to find the measure of the missing angle in the following figure. While I have solved this, which I'll show below as an answer, I'm not quite sure if my answer is accurate, so I'm posting this her... |
WLOG, let $AD=DC=1$
Using sine rule yields
$$
\begin{aligned}
\frac{A B}{\sin 135^{\circ}}&=\frac{1}{\sin 30^{\circ}}
\Rightarrow A B =\sqrt{2}
\end{aligned}
$$
Using cosine rule gives
$$
\begin{aligned}
BC^2 &=2^2+\sqrt{2}^2-2(2) \sqrt{2} \cos 15^{\circ} \\
&=6-4 \sqrt{2} \cos 15^{\circ} \\
&=6-2(\sqrt{3}+1) \\
B C &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4569632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Volume of the region determined by $x^2 + y^2 + z^2 \leq 10, z \geq 2$ This is question 13 on page 294 of Vector Calculus by Marsden and Tromba.
Find the volume of the region determined by $x^2 + y^2 + z^2 \leq 10, z \geq 2.$
I have attempted it as follows.
The region can be described by using spherical polar coordin... | What you have computed is the volume of a larger region, which is the region that you have been given plus the cone whose vertex is located at the origin and whose base is the the intersection of the solid sphere centered at the origin with radius $\sqrt{10}$ and the plane $z=2$.
I suggest the use of cylindrical coordi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4574446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integral Involving Harmonic Numbers: $\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx$ (Motivation) In an attempt to answer this question, I got stuck on evaluating a certain integral. I have made up the following conjecture:
$$\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx = \frac{... | Mathematica 13.1 produces
Integrate[Log[x^4 - 1]/(x^2 - 1), {x, Sqrt[3], Infinity}]//FullSimplify
$$\frac{1}{32} \left(16 \left(\text{Li}_2\left(\frac{1}{2} \left(1-\sqrt{3}\right)\right)+\text{Li}_2\left(\sqrt{3}-1\right)+\text{Li}_2\left(\left(-\frac{1}{2}+\frac{i}{2}\right) \left(\sqrt{3}-1\right)\right)+\text{Li}_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4576184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Laurent Series of $(z+1)^2\sin(z)$ centred at $0$ How am I to write the Laurent series of $(z+1)^2\sin(z)$ centred at $0$? I know
$$(z+1)^2 \sin(z) = (z^2+2z+1) \sum_{k = 0}^{\infty} \frac{(-1)^k z^{2k+1}}{(2k+1)!} \\ \Rightarrow (z+1)^2 \sin(z) = \sum_{k = 0}^{\infty} \frac{(-1)^k z^{2k+3}}{(2k+1)!} + \sum_{k = 0}^{\... | We derive a series representation of $(z+1)^2\sin(z)$ evaluated at $z=0$ by calculating the coefficients $a_n$ in
\begin{align*}
\color{blue}{(z+1)^2\sin(z)}&=(z+1)^2\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)!}\\
&\,\,\color{blue}{=\sum_{n=0}^{\infty}a_nz^n}\tag{1}\\
&=\sum_{n=0}^{\infty}a_{2n}z^{2n}+\sum_{n=0}^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4578024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that: $\frac{1}{c}\sqrt{h} \Big\|\begin{pmatrix} f(0) \\f(h) \end{pmatrix} \Big\|_2 \le \Big\|f\Big\|_{L^2} \le c\sqrt{h}\ldots$ Let $\mathcal P:=\left\{f:\Omega \rightarrow \mathbb R \space \vert \space f(x)=ax+b, a,b \in \mathbb R \right\}$ with $\Omega:=[0,h], h\in \mathbb R_+$. I want to show this inequality ... | I made a mistake writing down the $L^2$-Norm. We have:
$$\vert \vert f \vert\vert_{L^2} = \left( \int_0^h \vert f(x) \vert^2dx \right)^{\frac{1}{2}} $$
Therefore the inequality becomes:
$$\frac{1}{c}\sqrt{a^2h^3+2bh^2+2b^2h}\le \sqrt{\frac{a^2h^3}{3}+abh^2+b^2h} \le c\sqrt{a^2h^3+2bh^2+2b^2h}$$
Let $c=\sqrt{4}=2$
$$\im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Verify that $\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})\le\frac{1}{8}$ in a general triangle $\triangle ABC$ So, this problem is inspired by a contest preparation problem I saw back in Japan, and it is as follows:
In a general triangle $\triangle ABC$, show that $\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C... | By your work
$$\prod_{cyc}\sin\frac{\alpha}{2}=\prod_{cyc}\sqrt{\frac{(a-b+c)(a+c-b)}{4bc}}=\frac{\prod\limits_{cyc}(a+b-c)}{8abc}\leq\frac{1}{8}$$
because $$abc-\prod\limits_{cyc}(a+b-c)=\sum_{cyc}(a^3-a^2b-a^2c+abc)=$$
$$=\sum_{cyc}(a^3-abc-b^2c-a^2c+2abc)=\sum_{cyc}(a-b)^2\left(\frac{a+b+c}{2}-c\right)=$$
$$=\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4583104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why can't I use the form $\frac1{x^2+1}$ from the derivative of $\arctan(x)$ to convert the integral form in this situation? In the question that solves the integral $\displaystyle\int\frac1{6x^2 + 36x + 78} \,\mathrm{d}x$, I first tried to solve it by changing the denominator in a form of $\dfrac1{x^2 + 1}$ to apply $... | Basically, you messed up in your substitution:
$$\int \frac{1}{6x^2+36x+78}dx = \frac{1}{6} \int \frac{1}{(x+3)^2+2^2}dx = \frac{1}{24} \int \frac{1}{(\frac{x+3}{2})^2+1}dx$$
Let $u=\frac{x+3}{2}$, Then $du = \frac{dx}{2}$ So:
$$\frac{1}{24} \int \frac{1}{(\frac{x+3}{2})^2+1}dx = \frac{1}{24} \int \frac{1}{(u)^2+1} \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to solve $\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n$ Question:
$$\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n=?$$
My work:
$\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n=\lim_{n\to\infty}\left(\dfrac{n^2(1+5/n+3/n^2}{n^2(1+1/n+2/n^2)}\right)^n=\lim_{n\to\infty}\left(\dfrac{1... | I've also posted this answer with the thought that some readers might find it useful.
Using high school algebra trick, we have:
$$
\begin{align}&\lim_{n\to\infty}\left(\frac{n^2+5n+3}{n^2+n+2}\right)^n\\
=~&\lim_{n\to\infty}\left(1+\frac{4n+1}{n^2+n+2}\right)^n\\
=~&\lim_{n\to\infty}\left(\left(1+\frac {1}{\frac{n^2+n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Prove that $2\cdot 3^x +1= p^y$ has no solution Prove that the Diophantine equation of $3$ variables $(x,y,p)$
$$2\cdot 3^x +1= p^y$$
has no solution where $x,y\in\mathbb{N}_+$, $x\ge2, y\ge2$ and $p$ is a prime number.
I found that $y$ cannot be even, if $y =2k$ then $4| (p^k -1)(p^k+1) =2\cdot3^x$ which is a contradi... | Note that
$$2\cdot 3^x = p^y - 1 = (p - 1)(p^{y-1} + p^{y-2} + \ldots + p + 1) \tag{1}\label{eq1A}$$
With $p$ being odd and $p \gt 3$, then
$$p - 1 = 2(3^{a}) \; \; \to \; \; p = 2(3^{a}) + 1 \tag{2}\label{eq2A}$$
for some integer $a \ge 1$. Next, using the lifting-the-exponent (LTE) lemma, we have
$$x = \nu_3(p^y - 1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4586098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Indefinite Integral of ${\tan(x)}^{p/q}$ I have seen a lot of videos in which people integrate functions like $\sqrt{\tan(x)}$, $\sqrt[3]{\tan^2(x)}$, etc. I was wondering if there was a closed-form expression for
$$\int (\tan{x})^{\frac{p}{q}} dx$$
Where $p,q$ are integers, and $q \neq 0$. Eventually, I would like to ... | Here is a more general result. Let $a \in \mathbb{R}\backslash\left\{-1,-3\right\}$ and let the integral in question be $I$. For this answer, we will suppose the following:
$$\cos^2(x) =\text{ } _2F_1\left(1, \frac{1+a}{2}; \frac{3+a}{2};-\tan^2(x)\right)-\frac{2\tan^2(x)}{3+a}\text{ }_2F_1\left(2, \frac{3+a}{2}; \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4588784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Fair Value of a Dice Game 1 die. Up to 3 rolls. Your winnings are equal to the value of a single roll. You can stop after roll 1 or roll 2. If you proceed to the next roll (i.e the 2nd or 3rd), you forfeit the previous value (no memory). Therefore, if you roll 3 times, the value of the 3rd roll is what you win.
What is... | The value of a single roll is $3.5$
With two rolls, keep the first if it is more than $3.5$, so the value is $$\frac12\times5+\frac12\times3.5=4.25$$
With three rolls, keep the first if it is more than $4.25$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4589195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving the Diophantine equation $k^2 = a^2 - 4b + 8$ for $k,a,b$ We have the system of Diophantine equations in 4 variables
$$\begin{align}
a & = p+q \\
b-2 &= pq \tag 1
\end{align}$$
We have
$$\begin{align}
(p-q)^2 & = (p+q)^2 - 4pq \\
& = a^2 - 4(b-2) \tag 2
\end{align}$$
Putting $p-q = k$, we need to solve the 3-v... | If $a$ and $b$ are integer constants in your question, then $p$ and $q$ are obviously restricted and the pairs $(p,q)$ or $(q,p)$ are the roots of the quadratic equation $x^2-ax+(b-2)=0$ therefore $(p,q)$ is not always a pair of integers.
Thus, $a$ and $b$ cannot be considered as constants.
If $a,b,p,q$ are all integer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4591846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Parabola passing through TWO points with known tangents directions Given the points $ P_1 (0, 5), P_2 (0, 0), P_3 (3, 0) $, a parabola passes through $P_1$ and is tangent to the segment $P_1 P_2$ and passes through $P_3$ and is tangent to the segment $P_2 P_3$. Determine the vertex, and axis of symmetry of the parabol... | Following the method outlined in my OP, we have the parametric equation of the parabola,
$P(t) = A t^2 + B t + C = A' (t - \tau)^2 + B' (t - \tau) + C' $
where
$A = P_1 - 2 P_2 + P_3 = (0, 5) - 2 (0, 0) + (3, 0) = (3, 5) $
$B = -2 P_1 + 2 P_2 = - 2 (0, 5) + 2 (0, 0) = (0, -10) $
$C = P_1 = (0, 5) $
And the constant $\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Attempt to solve $\lim_{x\to +\infty} \exp{\left(\frac{x^2}{1+x}\right)} - \left(1+\frac1x\right)^{x^2}$ I tried to solve this limit. Is it correct? There exists a more straightforward way?
$$\lim_{x\to +\infty} \exp{\left(\frac{x^2}{1+x}\right)} - \left(1+\frac1x\right)^{x^2}$$
$$\lim_{x\to +\infty} \exp{\left(\frac{x... | I always try to find a solution without using $o(f(x))$ (yes, I know that De L'Hopital is equivalent to $o(f(x))$)
\begin{align*}
& \lim_{x\to +\infty}\left[ \exp{\left(\frac{x^2}{1+x}\right)} - \left(1+\frac1x\right)^{x^2}\right] = \\
& \lim_{x\to +\infty}\left[ \exp{\left(\frac{x^2}{1+x}\right)} - \exp\left(x^2\log\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find the (1,1) entry in this dynamical system? Here's the question:
Consider the dynamical system $V_{k+1}$ = $AV_k$ where
$$
A=\begin{pmatrix}
0 & 1 \\
-13 & 4\\
\end{pmatrix}
\quad \text{and} \quad
V_0=\begin{pmatrix}
1 \\
3
\end{pmatrix}
$$
Find a formula in terms of k for the $(1,1)$-entry $x_k$ of $V_k$... | Obviously,
$ V_k = A^k V_0 $
Now,
$ A^k = \alpha_0 I + \alpha_1 A \hspace{30pt} (*)$
To find $\alpha_0$ and $\alpha_1$, we use the eigenvalues of $A$.
The characteristic polynomial of $A$ is $ \lambda^2 - 4 \lambda + 13 = 0 $
whose roots (the eigenvalues) are $\lambda_1= 2 - 3 i $ and $ \lambda_2 = 2 + 3 i $
Substitute... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4596735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the value of $\int_0^1f(x)dx$
If $$f(x)=\binom{n}{1}(x-1)^2-\binom{n}{2}(x-2)^2+\cdots+(-1)^{n-1}\binom{n}{n}(x-n)^2$$
Find the value of $$\int_0^1f(x)dx$$
I rewrote this into a compact form.
$$\sum_{k=1}^n\binom{n}{k}(x-k)^2(-1)^{k-1}$$
Now,
$$\int_0^1\sum_{k=1}^n\binom{n}{k}(x-k)^2(-1)^{k-1}dx$$
$$=\sum_{k=1}^... | A devious little problem indeed! I am interested in where you found it. We in fact have a very nice formula:
$$\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} (x-k)^2 =x^2$$
It follows from:
$$F(x)=\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} (x-k)^2 \\ =\sum_{k=1}^n (-1)^{k-1}\binom{n}{k}\big(x^2-2kx+k^2\big) \\ =x^2\color{red}{\sum_{k=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 0
} |
How to find the constant $C$ such that $f(x)\geq Cx$ Problem :
Define for strictly positive $x$ :
$$f\left(x\right)=\left(\prod_{k=1}^{\operatorname{floor}\left(x\right)}\left(1+\sum_{n=1}^{k}\frac{1}{k\cdot2^{n}}\right)\right)$$
Does there exists a constant $C$ such that :
$$f(x)\geq Cx$$
I think definitely yes and $C... | Note that
\begin{equation}
f(x)=\prod_{k=1}^{\left \lfloor x \right \rfloor} \left(1+ \frac{1}{k}\sum_{j=1}^k \frac{1}{2^j}\right)= \prod_{k=1}^{\left \lfloor x \right \rfloor} \left( 1+ \frac{1-2^{-k}}{k} \right)
\end{equation}
Therefore,
\begin{align}
\ln f(x) =& \sum_{k=1}^{\left \lfloor x \right \rfloor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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What particular solution should I guess for $y'' - 4y' + 4y = (6x - 2)e^{2x}$? I have the following initial value problem
$$ y'' - 4y' + 4y = (6x-2)e^{2x},\ y(0)=2,\ y'(0)=1 $$
to which I have obtained the following solution for the homogenous ODE
$$ y_{h} = e^{2x}(2 - 3x) $$
Now I am trying to figure out the particula... | After advice from the comments I managed to solve the problem.
\begin{split}
y'' - 4y' + 4y &= 2(3x-1)e^{2x} \\
y(0) &= 2\\
y'(0) &= 1
\end{split}
$$ y'' - 4y' + 4y = 0 $$
$y_{h} = e^{rx}$, $y'_{h} = re^{rx}$, $y''_{h} = r^{2}e^{rx}$.
\begin{split}
r^{2}e^{rx} - 4re^{rx} + 4e^{rx} &= 0 \\
e^{rx}(r^{2} - 4r + 4) &= 0
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is wrong with the calculation $ \lim\limits_{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} $? We are given
$$ \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} $$
We can write
$$ L = \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} = \lim _{x \to 0} \frac {x (cos x - \sin x/x)} {x^2 \sin x} $$
It can ... | The thing is you can take a partial limit when it only involves products (or quotients). Here there is an additive $\cos(x)$ term so you cannot do that.
Remember that $\cos(x) = 1 - \frac{x^2}{2} + o(x^2)$ and $\frac{\sin x}{x} = 1 - \frac{x^2}{6} + o(x^2)$ therefore $\cos x - \frac{\sin x}{x} = -\frac{x^2}{3} + o(x^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4602507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluation of $~\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta$ $$
I:=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta
$$
My tries
$$\begin{align}
s&:=\sin\theta\\
c&:=\cos\theta\\
I&=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\the... | Method 1
Note that by defining $u=\theta-\dfrac{\pi}{2}$, we have
$$
\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta
=
\int_{\dfrac{\pi}{2}}^{2\pi+\dfrac{\pi}{2}}{\sin^2u-\cos^2u\over\cos^4 u+\sin^4 u}\mathrm du.
$$
Now, since the integrand is periodic with a period of $2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
Solution of an equation with $3$ variables I've been thinking about a solution for the following equation for integers $x, y, z$:
$$6 = (x-y)^2 + (y-z)^2 + (x-z)^2$$
A possible approach would probably be to set $(x-y)^2 = a$, $(y-z)^2 = b$ and $(x-z)^2 = c$.
Now, the euqation woud look like this:
$$6 = a + b + c$$
Henc... | Hint:
Let $x-y=a$ etc.
$\implies a+b+c=0\iff c=-?$
$$6=a^2+b^2+c^2=2a^2+2ab+2b^2\iff2a^2+a(2b)+2b^2-6=0$$
$$a=\dfrac{-2b\pm\sqrt{12(4-b^2)}}4=\dfrac{-b\pm\sqrt{3(4-b^2)}}2$$
For integer $a,$ $$b^2=4,1\implies b=\pm2,\pm1$$
We shall get the same set of values for $c,$ if we start with $b=-(c+a)$
So, $(a,b,c)\in(1,1,-2)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4606594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Where did I go wrong with this integration? $$\int_0^{\frac {1}{\sqrt 3}} \frac {1}{\sqrt {2-3x^2}}$$
$$\int_0^{\frac {1}{\sqrt 3}} \frac {1}{\sqrt {2(1-\frac32x^2)}}$$
$$\frac 1{\sqrt 2}\int_0^{\frac {1}{\sqrt 3}} \frac {1}{\sqrt {1-\frac32x^2}}$$
$$\bigg(\frac 1{\sqrt 2}\sin^{-1}{\sqrt {\frac 32 x}}\bigg)\bigg|_0^\fr... | Let $x=\frac{\sqrt{2} \sin \theta}{\sqrt{3}}$, then $dx=\frac{\sqrt{2}}{\sqrt{3}} \cos \theta d \theta$
$$
\begin{aligned} \\
I & =\int_0^{\frac{\pi}{4}} \frac{1}{\sqrt{2-2 \sin ^2 \theta}} \frac{\sqrt{2}}{\sqrt{3}} \cos \theta d \theta \\
& =\frac{1}{\sqrt{3}} \int_0^{\frac{\pi}{4}} d \theta \\
& =\frac{\pi}{4 \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Contest Math Question: simplifying logarithm expression further I am working on AoPS Vol. 2 exercises in Chapter 1 and attempting to solve the below problem:
Given that $\log_{4n} 40\sqrt{3} = \log_{3n} 45$, find $n^3$ (MA$\Theta$ 1991).
My approach is to isolate $n$ and then cube it. Observe:
\begin{align*}
\frac{\l... | I would go the following way.
You have :
$$\begin{align}&\log_{4n} 40\sqrt{3} = \log_{3n} 45\\
\implies &\log_{4n} \left(3\cdot 40^2\right) = \log_{3n} 45^2=k
\end{align}$$
This leads to :
$$\begin{align}&\begin{cases}(4n)^k=3\cdot 40^2\\
(3n)^k=45^2 \end{cases}\\
\implies &\left(\frac 43\right)^k=\frac {3\cdot 40^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
How to prove $\sum_{k=1}^n{n\choose k}\frac{(-1)^{k+1}}{k}=\sum_{k=1}^n\frac{1}{k}$? I think by induction we can do it. Let
$I(n)=\sum_{k=1}^n{n\choose k}\frac{(-1)^{k+1}}{k}=\sum_{k=1}^n\frac{1}{k}.$ Then, we must show that $I(n+1)-I(n)=\frac{1}{n+1}$.
$\begin{align}
I(n+1)-I(n)&=\frac{(-1)^{n+1}}{n+1}+\sum_{k=1}^n\l... | Let $n\geq 1$ and define the sequence,
$$ a_n = \sum_{k=1}^n {n\choose k} \frac{1}{k} (-1)^{k+1} = \sum_{k=1}^{\infty} {n\choose k} \frac{1}{k} (-1)^{k+1} $$
(If $k>n$ then the binomial coefficient is zero, not changing the sum).
The generating function of this sequence is equal to,
$$ A(x) = \sum_{n=1}^{\infty} a_n x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4614504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Find $x$ such that $\sqrt{x+1} - \sqrt{1-x} = 1$ To solve this equation, I started by putting the condition $x\in [-1, 1]$, then squared a few times: $\sqrt{x+1} - \sqrt{1-x} = 1 \iff x + 1 +1-x-2\sqrt{1-x^2} =1 \iff 2\sqrt{1-x^2}=1 \iff 4(1-x^2)=1 \iff 4x^2=3 \iff x=\pm \frac{\sqrt{3}}{2}$
This, however, is not the ri... | Yet another way :
Let $\sqrt{1+x}=a, \sqrt{1-x}=b$
For real $x, a\ge0, b\ge0 $
By the given condition,
$$a-b=1$$ and $$a^2+b^2=2\implies2=(b+1)^2+b^2\iff2b^2+2b-1=0$$
$\implies(2b+1)^2=3\implies2b+1=+\sqrt3$ as $b\ge0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 4
} |
Maximizing/minimizing $f(\theta) = \sqrt{2}\cos(\theta)-4\sin(\theta)$
Assume that $f : [0, 2\pi]\rightarrow \mathbb{R}$ is a function such that $f(\theta) = \sqrt{2}\cos(\theta)-4\sin(\theta)$. Then, how can we maximize/minimize $f$?
We can re-parametrize our function $f$ by defining another function $g : [-1, 1]\ri... | $f(\theta) = \sqrt{2}\cos(\theta)-4\sin(\theta) = \sqrt{18}( \sin\alpha \cos \theta - \cos \alpha \sin \theta)$ where $\sin \alpha = \frac{\sqrt{2}}{\sqrt{18}}$ and $\cos \alpha = \frac{4}{\sqrt{18}}$ and hence $\alpha = arctan \frac{\sqrt{2}}{4}$. Thus
$$f(\theta) = \sqrt{18}\sin(\alpha - \theta) \le \sqrt{18}$$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Evaluate this double Integral in polar coordinates. $$ \iint y^2(a^2-x^2)^{0.5}dxdy $$ over $x^2+y^2\le a^2$
I have evaluated this in $x-y$ plane and got $32a^5/45$. Please help in evaluating the same in polar coordinates.
Ive tried putting $x=r\cos\theta$ and $y=r\sin\theta$ and After considering the Jacobian I am get... | Since the integrand and the region $x^2 + y^2 \le a^2$ is symmetric, it suffices to calculate the quarter.
$$
\begin{align}
\iint_{x^2 + y^2 \le a^2} y^2\sqrt{a^2 - x^2} dx dy &= 4 \iint_{0 \le x,y \land x^2+y^2 \le a^2} y^2\sqrt{a^2 - x^2} dx dy \\
&=4 \int_{0}^{a} \sqrt{a^2 - x^2} dx \int_0^{\sqrt{a^2 - x^2}} y^2 dy ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4621638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Simplifying $\frac{x^4(y-z)+y^4(z-x)+z^4(x-y)}{(y+z)^2+(z+x)^2+(x+y)^2}$ I need help to simplify this expression.
$$\frac{x^4(y-z)+y^4(z-x)+z^4(x-y)}{(y+z)^2+(z+x)^2+(x+y)^2}$$
I only know to start, I extracted the $(x-y)$ from top one, and then I don't know what's next. Extracting $(x-y)$ from the numerator:
$$(x^4y -... | The nominator is a degree 5 polynomial and the denominator degree 2 polynomial, so if the fraction should simplify into a polynomial, it would have to be of degree 3, symmetric, and with the same number of positive and negative terms. My first guess is to consider the symmetric polynomial
$$
(x-y)(y-z)(z-x)
= y (z^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4622054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Can $\,9\!\cdot\!10^n+4\,$ be a perfect square? I think $\,9\!\cdot\!10^n+4\,$ can be a perfect square, since it is $0 \pmod 4$ (a quadratic residue modulo $4$), and $1 \pmod 3$ (also a quadratic residue modulo $3$).
But when I tried to find if $\;9\!\cdot\!10^n+4\,$ is a perfect square, I didn’t succeed. Can someone ... | Assume that $9\cdot10^n+4\equiv4$ is a perfect square.
$9\cdot10^n+4\equiv4\pmod9$, so $9\cdot10^n+4$ can be represented as $(9m-2)^2=81m^2-18m+4$ or $(9m+2)^2=81m^2+18m+4$, where $m\in\mathbb{N}$.
If $9\cdot10^n+4=81m^2-18m+4$, $$10^n=9m^2-2m=m(9m-2)$$
This means that $m$ and $9m-2$ must be powers of $10$.
Clearly, $9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4622956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solving system of equations by echelon form. I have an example of a solution for a system of equations and there is a step I do not understand:
Could anyone offer some help please?
Kind regards
Stany
| $$
\begin{aligned}
\left(\begin{array}{cccc}
1 & 1 & 1 & 0 \\
3 & -1 & 1 & 20 \\
6 & 2 & -1 & 40
\end{array}\right) & \rightarrow\left(\begin{array}{cccc}
1 & 1 & 1 & 0 \\
0 & -4 & -2 & 20 \\
0 & -4 & -7 & 4
\end{array}\right) \\
& \rightarrow\left(\begin{array}{cccc}
1 & 1 & 1 & 0 \\
0 & -4 & -2 & 20 \\
0 & 0 & -5 & 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4628643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $(g \circ f)^{-1}(x) = {\left( {\frac{{x - 7}}{2}} \right)^{\frac{1}{3}}}$
If $g\left( x \right) = ax + c$, $f\left( x \right) = {x^b} + 3$, and $(g \circ f)^{-1}(x) = {\left( {\frac{{x - 7}}{2}} \right)^{\frac{1}{3}}}$ what is the value of $a+b+c$?
My approach is as follows
Given $g\left( x \right) = ax + c... | You can start by noting that
$$
(g \circ f)(x) = a(x^b+3)+c
$$
and, inverting this expression, get to
$$
(g\circ f)^{-1}(x) = \left(
\frac{x-c-3a}{a} \right)^{\frac 13}
$$
Identifying coefficients, you'll see that $a=2, b=3, c=1$, and consequentely, $a+b+c = 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4628817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
For $c>b>a$ with $c,b,a\in\Bbb N,c-b=b-a=1$, prove $\frac{a^c}{b^b}+\frac{b^c}{c^b}+\frac{c^c}{a^b}\ge\frac{a^b}{b^a}+\frac{b^b}{c^a}+\frac{c^b}{a^a}$
Given $c>b>a$, also $c-b=b-a=1$ where $c,b,a$ are Natural numbers, prove that
$$
\frac{a^c}{b^b}+\frac{b^c}{c^b}+\frac{c^c}{a^b} \geqslant \frac{a^b}{b^a}+\frac{b^b}{c^... | For $(a, b, c) = (x, x+1, x+2)$ and positive real $x$ is your inequality equivalent to
$$
2 \frac{(x+2)^{x+1}}{x^{x+1}} - \frac{x^{x+1}}{(x+1)^{x+1}} - \frac{(x+1)^{x+1}}{(x+2)^{x+1}} \ge 0
$$
or
$$
2 \ge \left( \frac{x^2}{(x+1)(x+2)}\right)^{x+1} + \left( \frac{x(x+1)}{(x+2)^2}\right)^{x+1}
$$
and that is true becau... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4633045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Averaging neighbours in a list In a list of twelve numbers, the first number is 1 and the last number is 12.
1, a, b, c, d, e, f, g, h, i, j, 12 (assume the letters represent unknown numbers)
Each other number in the list is calculated by taking the average of its neighbours and adding 1. The question is asking to fin... | if $x, y, z$ are three adjacent numbers, then $y = 1 + \frac{x+z}2$, which solves to $$z = 2y - x - 2$$
Also, let me call the values as $$1, a_1 = t, a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9, a_{10}, 12$$
Thus we can calculate
*
*$a_0 = 1 = 0t + 1$
*$a_1 = t - 0$
*$a_2 = 2t - 3$
*$a_3 = 2a_2 - a_1 - 2 = 3t-8$
*$a_4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4636295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Polynomial solutions to $ax^4+bx^2+c^2 = y^2$, with $a,b,c$ as polynomials While investigating a system involving "equal sums of like powers", an elliptic curve popped up,
$$9 (1 + 4 n^4)^2 + 30 (4 + n^2 - 24 n^4 + 4 n^6) x^2 + 5 (32 - 40 n^2 + 53 n^4) x^4 = y^2$$
Some easy rational points are,
$$x = (0,\; 1,\; n)$$
Us... | Looking at a bunch of examples, it appears that your curve (which is isomorphic to an elliptic curve) has rank $2$ over $\mathbb{Q}(n)$. Using this, one can find some other rational points of degree $\leq 6$, but not where the numerator or denominator is a quadratic. In particular, there are solutions when
$$
x = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4636431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx$ How do I evaluate $\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx$ ?
Note that this is a Q&A post and I've presented my solution below.
| Another way to evaluate the OP’s integral is using Euler’s substitutions.
By letting $\;t=x^2\,,\;$ we get that
$\displaystyle\int\frac{2x}{\left(1-x^2\right)\sqrt{x^4-1}}\,\mathrm dx=\int\frac1{(1-t)\sqrt{t^2-1}}\,\mathrm dt\;.$
Now, we will apply Euler’s first substitution that is $\,\sqrt{t^2-1}=-t+u\,$ and obtain t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4640836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Proving a trigonometric finite sum $\sum_{k=1}^N(-1)^k(\cos \frac{k\pi}{N})^{N-m}(\sin\frac{k\pi}{N})^m=(-1)^{m/2}\frac{N}{2^{N-1}}$ How to prove this following formula?
$\sum_{k=1}^N(-1)^k(\cos \frac{k\pi}{N})^{N-m}(\sin\frac{k\pi}{N})^m=(-1)^{m/2}\frac{N}{2^{N-1}}$
for m is even, and $0$ for m is odd.
If we know $\su... | We seek to show that
$$S_{N,m} = \sum_{k=1}^N (-1)^k (\cos \frac{k\pi}{N})^{N-m}
(\sin\frac{k\pi}{N})^m =
\frac{1+(-1)^m}{2} (-1)^{m/2} \frac{N}{2^{N-1}}.$$
Observe that
$$\sum_{k=N+1}^{2N} (-1)^k (\cos \frac{k\pi}{N})^{N-m}
(\sin\frac{k\pi}{N})^m
\\ = \sum_{k=1}^N (-1)^{N+k}
(-1)^{N-m} (\cos \frac{k\pi}{N})^{N-m}
(-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4644963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Computing the limit of $ \frac{1^3+2^3+\cdots+n^3}{\sqrt{4n^8 +1}} $ I had this exercise:
Compute the limit
$$ \lim_{n\to\infty} \frac{1^3+2^3+\cdots+n^3}{\sqrt{4n^8 +1}} $$
I tried two different approaches and got different answers.
Approach 1:
$$\begin{split}
\lim_{n\to\infty} \frac{1^3+2^3+\cdots+n^3}{n^4\sqrt{4 +... | In your first attempt, $n$ is also present as the number of terms. You cannot use "the limit of the sum is the sum of the limit" in that case. The same way you cannot say
$$
\lim_{n\to\infty}\sum_{k=1}^n\frac1n=\sum_k0=0.
$$
In this example the sum is $1$ for all $n$, and so the limit is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4647003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove $\dfrac{n^2}{\ln n^2-1} - \dfrac{\dfrac{n^2}{2}}{\ln \dfrac{n^2}{2}-1.1} > n$ for $n \ge 347$ Prove $\dfrac{n^2}{\ln n^2-1} - \dfrac{\dfrac{n^2}{2}}{\ln \dfrac{n^2}{2}-1.1} > n$ for $n \ge 347$. It seems to be true for all $n \ge 11$, but I only need it to be true for $n \ge 347$. I tried to manipulate the equat... | $$\frac{n^2}{\ln n^2-1}-\frac{\frac{n^2}{2}}{\ln\frac{n^2}{2}-1.1} \gt n \; \iff \;
\frac{1}{\ln n^2-1}-\frac{1}{2(\ln\frac{n^2}{2}-1.1)} \gt \frac{1}{n} \tag{1}\label{eq1A}$$
For $n \ge 347$, we have $\ln(n) \gt 2(\ln(2) + 1.1) \; \to \; \ln(n) - \ln(2) - 1.1 \gt \frac{1}{3}\left(2\ln(n) - \ln(2) - 1.1\right)$ so, us... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4647326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Distance between the point of touching in three touching circles $\textbf{Question : }$ Say you have three touching circles $\Gamma_1,\Gamma_2,\Gamma_3$ with radii $x,y,z$ and centers $A,B,C$ as per the diagram, then prove the following $$|DE|=\frac{2}{\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x}+\dfr... | From double angle formulae, we have $$2\sin^2\frac{\alpha}2=1-\cos\alpha.$$
Using cosine rule,
$$\cdots=1-\frac{b^2+c^2-a^2}{2bc}=\frac{a^2-(b-c)^2}{2bc}=\frac{(a-b+c)(a+b-c)}{2bc}$$
where $a=y+z,\ b=x+z,\ c=x+y$ in your diagram.
Thus,
$$\begin{align}2\sin^2\frac{\alpha}2&=\frac{4yz}{2(x+z)(x+y)}\\ \sin\dfrac{\alpha}2&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Proving that $0 \leq k \sin\left(\frac{2\pi}{n}\right) - \sin\left(\frac{2\pi k}{n}\right)$ I need help proving that
$$
0 \leq k \sin\left(\frac{2\pi}{n}\right) - \sin\left(\frac{2\pi k}{n}\right)
$$
For $n > 2$ and positive $k$.
I've tried all sorts of identities, and nothing have worked. Any help would be greatly app... | Assume $0 < k \le 1$ and $n \ge 4$. Put $x = \dfrac{2\pi k}{n} \implies k = \dfrac{nx}{2\pi}$. Thus you prove: $\dfrac{\sin\left(\frac{2\pi}{n}\right)}{\frac{2\pi}{n}} \le \dfrac{\sin x}{x}$. But the function $ f(x) = \dfrac{\sin x}{x}$ on $\left(0,\frac{2\pi}{n}\right]$ has $f’(x) = \dfrac{x\cdot \cos x - \sin x}{x^2}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?
|
Above is an image representing Pascals Triangle. What I want to draw attention to is the hockey stick formation, particularly, the blue hockey stick. Notice how the entries in the stick of the blue hockey stick are in arithmetic progression, and that the entry in the blade represents the sum of the entries in the stic... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "136",
"answer_count": 36,
"answer_id": 18
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Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$ Consider the sequence defined as
$x_1 = 1$
$x_{n+1} = \sin x_n$
I think I was able to show that the sequence $\sqrt{n} x_{n}$ converges to $\sqrt{3}$ by a tedious elementary method which I wasn't too happy about.
(I think I did this by showing that $\sqrt{\fr... | If your sequence is of the form
$$x_{n+1} = f(x_n)$$
with
$$f(x) = x( 1 - c x^{\alpha} + \textrm{h. o. t} )$$
then notice ( e.g. ) that
$$\frac{1}{f(x)^{\alpha}} = \frac{1}{x^{\alpha}} + \alpha c+ \textrm{h. o. t.}$$
and so
$$\frac{1}{x_{n+1}^{\alpha}} - \frac{1}{x_n^{\alpha}} \to \alpha c$$
and with Cesaro-Stolz we ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$? Could you provide a proof of Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?
| Assuming you mean $e^{ix}=\cos x+i\sin x$, one way is to use the MacLaurin series for sine and cosine, which are known to converge for all real $x$ in a first-year calculus context, and the MacLaurin series for $e^z$, trusting that it converges for pure-imaginary $z$ since this result requires complex analysis.
The Mac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "177",
"answer_count": 17,
"answer_id": 12
} |
Proving a binomial sum identity $\sum _{k=0}^n \binom nk \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}$ Mathematica tells me that
$$\sum _{k=0}^n { n \choose k} \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}.$$
Although I have not been able to come up with a proof.
Proofs, hints, or references are all welcome.
| $$S_n=\sum\limits_{k=0}^n \dfrac{(-1)^k \binom{n}{k}}{2k+1}$$
We have:
$$\begin{align}S_n&=\dfrac{(-1)^n}{2n+1}+\sum\limits_{k=0}^{n-1} \dfrac{(-1)^k \binom{n}{k}}{2k+1}\\
&=\dfrac{(-1)^n}{2n+1}+\sum\limits_{k=0}^{n-1}\left[\dfrac{n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\
&=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 8,
"answer_id": 1
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$\frac{a}{\sqrt{a+b}} + \frac{b}{\sqrt{b+c}} + \frac{c}{\sqrt{c+a}} > \sqrt{a+b+c}$ is true for positive a,b,c How do you prove that for all positive $a,b,c$ this formula holds true:
\begin{equation*}
\frac{a}{\sqrt{a+b}} + \frac{b}{\sqrt{b+c}} + \frac{c}{\sqrt{c+a}} > \sqrt{a+b+c}?
\end{equation*}
Any help will be inv... | Since
$$a+b+c=\sqrt{(a+b+c)^2}=\sqrt{a+b+c}\ \sqrt{a+b+c}\iff \dfrac{a+b+c}{\sqrt{a+b+c}}=\sqrt{a+b+c}$$
and
$$\dfrac{a}{\sqrt{a+b}}>\dfrac{a}{\sqrt{a+b+c}}$$
$$\dfrac{b}{\sqrt{b+c}}>\dfrac{b}{\sqrt{a+b+c}}$$
$$\dfrac{c}{\sqrt{c+a}}>\dfrac{c}{\sqrt{a+b+c}}$$
we have
$$\dfrac{a}{\sqrt{a+b}}+\dfrac{b}{\sqrt{b+c}}+\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/6950",
"timestamp": "2023-03-29T00:00:00",
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Find an isomorphism from the octic group $G$ to the group $G'$ Find an isomorphism from the octic group $G$ to the group $G'$:
$G = \{e, s, s^2, s^3, b, g, d, t\}$
$$e = (1)\quad s = (1234)\quad s^2= (13)(24)\quad s^3= (1432)\quad b = (14)(23)\quad g = (24)\quad d = (12)(34)\quad t = (13)$$
$G' = \{I_2, R, R^2,... | Jessica: Preserving the order of all elements isn't sufficient; you have to show that it preserves the group operation.
Here's a hint though. $G^\prime$ is the set of symmetries of a square, right? Try labelling the square's corners. Then each symmetry is just a permutation of its corners...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it w... | The following proof is by Khalaf Ruhemi ( he is not a MSE member)
By partial fraction decomposition, we have
$$\frac{y}{(1+y^2)(y^2+x^2)}=\frac{1}{x^2-1}\left(\frac{y}{1+y^2}-\frac{y}{y^2+x^2}\right).$$
Integrate both sides from $y=0$ to $y=\infty$,
\begin{gather*}
\int_0^\infty\frac{y}{(1+y^2)(y^2+x^2)}\mathrm{d}y=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "814",
"answer_count": 48,
"answer_id": 29
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Explain this step How the following conversion is justified ?
$$\cos \biggl( \theta + \frac{2\pi}{3} \biggr) + \cos \biggl( \theta + \frac{4\pi}{3} \biggr) = \cos \biggl( \frac{\pi}{2} - \theta \biggr) - \cos \biggl( \frac{\pi}{3} +\theta \biggr) $$
| First, notice that
$$\cos(x+\pi) = \cos(x)\cos(\pi) - \sin(x)\sin(\pi) = \cos(x)(-1)-\sin(x)(0) = -\cos(x).$$
So, rewriting, using the above, and using that $\cos(-x)=\cos x$ you have:
\begin{align*}
\cos\left(\theta+\frac{2\pi}{3}\right) + \cos\left(\theta+\frac{4\pi}{3}\right) &= \cos\left(\left(\theta-\frac{\pi}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/10251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof of the formula $1+x+x^2+x^3+ \cdots +x^n =\frac{x^{n+1}-1}{x-1}$
Possible Duplicate:
Value of $\sum x^n$
Proof to the formula
$$1+x+x^2+x^3+\cdots+x^n = \frac{x^{n+1}-1}{x-1}.$$
| For a more mechanical proof, you could use induction. The proof then boils down to finding a common denominator:
$\frac{x^{n+1}-1}{x-1} + x^{n+1} = \frac{x^{n+1}-1+(x-1)x^{n+1}}{x-1} = \frac{x^{n+2}-1}{x-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/11703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
} |
Algebra Problem The expression $x^2-4x+5$ is a factor of $ax^3+bx^2+25$. Express the sum a+b as an integer.
Please give an explanation of how the answer
| $ax^3+bx^2+25=(x^2-4x+5)(ax+k)$ , $k$ = constant
$ax(x^2-4x+5) + k(x^2-4x+5) =0$
$ax^3 + x^2(k-4a)+x(5a-4k)+5k=0$
compare to original then $5k=25 \implies k=5$
and $(5a-4k)=0$ so $a =4$
and $b= k-4a = 5 -16 =-11$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/16284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Bounding a series from above using the integral test If $a_{n}$ is a non-negative, decreasing sequence, we know from the integral test that if $f(n)=a_{n}$ is an integrable function, then $\sum_{n=1}^{\infty} a_{n}$ and $\int_{1}^{\infty} f(x)dx$ converge/diverge together.
When the theorem is proven, it is shown that:
... | Given $\forall k \in \mathbb{N}, a_{k+1} \leq \int_{k}^{k+1} f(x)dx \leq a_{k}$,
$$\sum_{k=1}^{\infty}a_{k+1} \leq \sum_{k=1}^{\infty}\int_{k}^{k+1} f(x)dx \leq \sum_{k=1}^{\infty}a_{k}$$
so
$$\sum_{k=2}^{\infty}a_{k} \leq \int_{1}^{\infty} f(x)dx \leq \sum_{k=1}^{\infty}a_{k}$$
and, using the left part,
$$\sum_{k=1}^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve $\cos(\theta) + \sin(\theta) = x$ for known $x$, unknown $\theta$? After looking at the list of trigonometric identities, I can't seem to find a way to solve this. Is it solvable?
$$\cos(\theta) + \sin(\theta) = x.$$
What if I added another equation to the problem:
$$-\sin(\theta) + \cos(\theta) = y,$$
where $\th... | Linear equations in $\sin \theta $ and $\cos \theta $ can be solved by a
resolvent quadratic equation, using the two identities (also here):
$$\cos \theta =\frac{1-\tan ^{2}\frac{\theta }{2}}{1+\tan ^{2}\frac{\theta }{2%
}}$$
and
$$\sin \theta =\frac{2\tan \frac{\theta }{2}}{1+\tan ^{2}\frac{\theta }{2}}.$$
In this cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/20548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
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Limits and Measure Theory How would I evaluate these limits?
$$ \lim_{n \to \infty} \int_0^\infty \frac{n}{1+(nx)^2} \ dx$$
and
$$ \lim_{n \to \infty} \int_0^\infty \frac{(1+(nx)^2)}{(1+nx^2)^n} \ dx$$
| From $(1+a)^n\geq1+na\;\;\;\forall a\geq0$ with $a=nx^2$, you have:
$\frac{1+n^2x^2}{(1+nx^2)^n}\leq1$.
Furthermore, you have:$\frac{1+n^2x^2}{(1+nx^2)^n}<\frac{1+n^2x^2}{(nx^2)^n}=\frac{1}{n^nx^{2n}}+\frac{1}{n^{n-2}x^{2(n-1)}}<\frac{2}{x^2}$ if $n\geq2$ and $x>0$.
Hence, by the Lebesgue dominated convergence theorem... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\int \cos^3 x\;\sin^2 xdx$ Is this correct? I thought it would be but when I entered it into wolfram alpha, I got a different answer.
$$\int (\cos^3x)(\sin^2x)dx = \int(\cos x)(\cos^2x)(\sin^2x)dx
= \int (\cos x)(1-\sin^2x)(\sin^2x)dx.$$
let $u = \sin x$, $du = \cos xdx$
$$\int(1-u^2)u^2du = \int(u^2-u^4... | Just for the heck of it, another substitution could have been $u=\sin^3 x,$ in which case, $du = 3\sin^2x\cos x \ dx.$ Now, $1-u^{2/3} = 1-\sin^2x = \cos^2x.$ Thus,
$$
\begin{align*}
\int \cos^3x\sin^2x \ dx &= \frac{1}{3}\int \cos^2x(3\sin^2x\cos x) dx \\
&= \frac{1}{3} \int(1-u^{2/3})du \\
&= \frac{u}{3} - \frac{u^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/21589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve this recurrence relation What are some ways to solve this recurrence relation:
$a(n+1)=2 a(n) - a(n-1) -1, \text{ with }a(0)=0, a(10)=0?$
I tried to first convert this inhomogeneous equation into a homogeneous one following Wikipedia:
$b_{n}=Ab_{n-1}+Bb_{n-2}+K \,$ can be
converted into homogeneous form... | Generating functions to the rescue. Define $A(z) = \sum_{n \ge 0} a(n) z^n$,
multiply the shifted by 2 recurrence by $z^n$ and sum over $n \ge 0$. Recognizing:
\begin{align}
\sum_{n \ge 0} a(n + 1) z^n &= \frac{A(z) - a(0)}{z} \\
\sum_{n \ge 0} a(n + 2) z^n &= \frac{A(z) - a(0) - a(1) z}{z^2} \\
\sum_{n \ge 0} z^n ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Find the length of the curve from x=-2 to x=1 of $y=\sqrt{a^2-x^2}$ Homework problem I recently got:
Find the length of the curve of $y=\sqrt{a^2-x^2}$
from $x=-2$ to $x=1$. Also find the area of the surface of revolution about the $x$-axis.
Having trouble understanding this. I'm not even sure what it would look l... | The graph, as you surmised, of $y=\sqrt{a^2-x^2}$ is the upper half of the semicircle with center at $(0,0)$ and radius $a$. (Square both sides to get $y^2 = a^2-x^2$, or $x^2+y^2=a^2$, the equation of the circle described; since $y\geq 0$, it's the upper half). (Of course, you need $|a|\geq 2$ for your limits to make ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$? I know some nice ways to prove that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$. For example, see Robin Chapman's list or the answers to the question "Different methods to compute $\sum_{n=1}^{\infty} \frac{1}{n^2}$?"
Are there any nice ways to prove that $$\z... | by using Fourier series of $f(x)=1, x\in[0,1]$
$$1=\sum_{n=1,3,5,..}^\infty\frac{4}{n\pi}\sin (n\pi x)$$
take triple integrals as follows
$$\int_{0}^{1}\int_{0}^{x}\int_{0}^{x}1.dxdxdx=\int_{0}^{1}\int_{0}^{x}\int_{0}^{x} \sum_{n=1,3,5,..}^\infty\frac{4}{n\pi}\sin (n\pi x) dxdxdx$$
$$\frac{1}{6}=\sum_{n=1,3,5,..}^\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/28329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "119",
"answer_count": 15,
"answer_id": 11
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Proof by induction: $\sum\limits_{i=1}^{n} \frac{1}{n+i} = \sum\limits_{i=1}^{n} \left(\frac{1}{2i-1} - \frac{1}{2i}\right)$ How can the following be proved by induction?
$$\sum\limits_{i=1}^{n} \frac{1}{n+i} = \sum\limits_{i=1}^{n} \left(\frac{1}{2i-1} - \frac{1}{2i}\right)$$
I am out of ideas after practicing for a ... | $n=1$, check. note that
$$
\frac{1}{2(n+1)-1}-\frac{1}{2(n+1)}=\frac{(2n+2)-(2n+1)}{(2n+1)(2n+2)}=\frac{1}{(2n+1)(2n+2)}
$$
(this is what changes on the rhs for $n+1$).
on the left hand side the change is
$$
\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}=\frac{(2n+2)+(2n+1)-2(2n+1)}{(2n+1)(2n+2)}=\frac{1}{(2n+1)(2n+2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/30975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Where are these additional solutions coming from? Solve for $x$: $2\sin(2x)-\sqrt{2} = 0$ in interval $[0,2\pi)$
Step $1$: Add $\sqrt{2}$ and divide by $2$ to get $\sin(2x) = \dfrac{\sqrt{2}}{2}$
Step $2$: Set $2x$ equal to the angles where $\sin(x) = \dfrac{\sqrt{2}}{2}$:
$2x = \dfrac{\pi}{4}$ and $2x = \dfrac{3\pi... | $\sin(\theta) = \sqrt(2)/2$ iff $\theta = \pi/4 + 2k \pi$ or $\theta =3\pi/4 + 2k \pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/38583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to find roots of $X^5 - 1$? How to find roots of $X^5 - 1$? (Or any polynomial of that form where $X$ has an odd power.)
| It is easy to write down the solution in trigonometric functions, namely, $x=e^{2\pi i n/5}$ for $n=0,\ldots, 4$.
Here is an algebraic solution. First factor the polynomial as $$x^5-1=(x-1)(x^4+x^3+x^2+x+1).$$
The first factor gives you the obvious solution $x=1$. To find other roots, we need to solve the equation
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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For what $a$ and $b$ is $9x^4-12x^3+28x^2+ax+b$ a perfect square? If $9x^4-12x^3+28x^2+ax+b$ is a perfect square, find the value of $a$ and $b$.
This is one of my past year examination's questions, some help on it?
(The answer for this problem is $a=-16$, $b=16$.)
| $9x^4-12x^3+28x^2+ax+b=(3x^2+cx+d)^2=9x^4+6cx^3+(c^2+6d)x^2+2cdx+d^2$
So $c=-2, d=4$ and $a=2cd=-16, b=d^2=16$
| {
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"url": "https://math.stackexchange.com/questions/40021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $\sqrt{x+5} = x - 1$ I'm currently learning about radicals and simplifying them, and I came across this problem on the internet and tried to solve it:
$$\sqrt{x+5} = x - 1$$
So I used this logic:
$$
\begin{align}
\sqrt{x+5} &= x - 1 \\
x + 5 &= (x-1)^2 \\
x + 5 &= (x-1)(x-1) \\
x + 5 &= x^2 - 2x + 1 \\
0 ... | $\sqrt{x+5} = x - 1$
Because, you need to put condition $x\geq 1$
So, If you tried to plug $x = -1 < 1$, which not satified.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/41152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 5
} |
Confused by textbook solution to trig problem The following is part of a question and the solution from my textbook:
Question
Given that $2 \sin{2\theta} = \cos{2\theta}$, show that $\tan{2\theta} = 0.5$
Solution
$2 \sin{2\theta} = \cos{2\theta}$
$\Rightarrow \frac{2 \sin{2\theta}}{\cos{2\theta}} = 1$
$\Rightarrow 2 ... | If $\cos(2\theta) = 0$ then $\sin(2\theta)$ should be $1$ or $-1$.
So $2\sin(2\theta)$ could not be equal to $\cos(2\theta)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/41222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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Exercising divergent summations: $\lim 1-2+4-6+9-12+16-20+\ldots-\ldots$ I'm trying to make sense of some (assumed to be) simple exercises in divergent summation. One example I cannot resolve.
First I assume the sequence of binomialcoefficients $ \{ b_k = \binom k2 \}_{k=2\to\infty}=\{1,3,6,10,15,...\}$
Then to assign ... | The elementary Ramanujan sum of $1-2+4-6+9-\cdots$ is 1/4 and the series belongs to the class $R=2$.
In general, for $f(x)=b(x)/(1-x)$ and $b(x)$ Abel summable, the sum of $f(1)$
is $-b'(1) + b(1)/2$, with $b'(x)=xdb(x)/dx$.
Edit 1. Verification.
$$
\frac{1}{(1+x)^3(1-x)}=\frac{x}{8(1-x)}+\frac{8+7x+4x^2+x^3}{8(1+x)^3}... | {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate $\int \sqrt{(\cos{x})^2-a^2} \, dx$ How to calculate:
$$\int \sqrt{(\cos{x})^2-a^2} \, dx$$
| In SWP (Scientific WorkPlace), with Local MAPLE kernel, I got the following evaluation
$$\begin{eqnarray*}
I &:&=\int \sqrt{\cos ^{2}x-a^{2}}dx \\
&=&-\frac{\sqrt{\sin ^{2}x}}{\sin x}a^{2}\text{EllipticF}\left( \left( \cos
x\right) \frac{\text{csgn}\left( a^{\ast }\right) }{a},\text{csgn}\left(
a\right) a\right) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/48024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the rotation matrix for +15° out of the rotation matrix +60° without using trigonometric functions Find the rotation matrix for +15° out of the rotation matrix +60° without using trigonometric functions. And as much as I would love to tell you what I did so far. I don't even know where to start with this exercise.... | The only approach I can see is to take the fourth root of the matrix, that is, find a $2 \times 2$ matrix $A$, such that $A^4=\left(\begin{matrix}
1/2 & -\sqrt 3/2\\
\sqrt 3/2 & 1/2 \\
\end{matrix}\right)$. If you expect that it is of the form $\left(\begin{matrix}
a & -b\\
b & a \\
\end{matrix}\right)$ with $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/48462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
Inequality: $(x + y + z)^3 \geq 27 xyz$ Edit: $a,b,c$ and $x,y,z$ are positive, real numbers.
Since $(a-b)^2 \geq 0~$, $a^2 + b^2 - 2ab\geq0~$ and $a^2 + b^2 \geq 2ab~$. Similarly, $a^2 + c^2 \geq 2ac~$ and $b^2 + c^2 \geq 2bc~$.
Adding these inequalities together, $2(a^2 +b^2 + c^2) \geq 2(ab + ac +bc)~$ and accord... | An elementary approach, without $\text{AM} \ge \text{GM}$ is to use the identity
$$x^3 + y^3 + z^3 - 3xyz = (x+y+z)\left(\frac{(x-y)^2 + (y-z)^2 + (z-x)^2}{2}\right)$$
Thus $$\text{if } \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} \ge 0\ \text{then } (a+b+c)^3 \ge 27abc$$
by setting $x = \sqrt[3]{a}$ etc
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/48621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 8,
"answer_id": 4
} |
Simplifying trigonometric expressions, is there a unified theory? $\frac{1}{3}\cos^3 x \cos(2x)+\frac{1}{12}\sin(2x)(\sin(3x)+3\sin x)=\frac{1}{3} \cos x$
I got this as the result of a differential equation that I solved. The answer in the book is (1/3) cos(x), but after applying variation of parameters I got the expre... | Try to express everything on the $\text{l.h.s}$ in terms of $\cos$ and then see what happens. Use the following identities:
*
*$\sin{2x} = 2\cdot \sin{x} \cdot \cos{x}$
*$\sin{3x} = 3\sin{x} - 4 \sin^{3}{x}$
*$1-\sin^{2}{x}= \cos^{2}{x}$
So you have $$\frac{1}{3} \cos^{3}{x} \cdot \bigl[ 2 \cos^{2}{x}-1 \bigr]=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/49033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$ Let $a,b,c$ be positive, not equal. Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$.
I know the proof by subtracting LHS by RHS and then doing some arrangement.
But isn't there any inequality which can be used in $(1+1+1)(a^3+b^3+c^3) >(a+b+c)(a^2+b^2+c^2)$, or an e... | You can express the inequality as
$$ \frac{a^3+b^3+c^3}{3} > \frac{a+b+c}{3} \frac{a^2+b^2+c^2}{3}$$
or
$$ [m_3(a,b,c)]^3 > [m_1(a,b,c)] \; [m_2(a,b,c)]^2 $$
where $m_p$ is the generalized power (Hölder) $p$-mean. Applying the (important) property that $m_p > m_q$ whenever $p>q$ (and the values are not all equal), yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/49211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 0
} |
Find all pairs of positive integers $(x,y)$ such that $x^2-y^2=a^3$ and $x^3-y^3=b^2$ for integral $a, b$? How to find ALL pairs of positive integers $(x,y)$ such that the difference in their squares is a perfect cube and the difference in their cubes is a perfect square.
i.e.,
Positive integers $(x,y)$ such that
$x^2-... | $ x^2 - y^2 = a^3 $ is fairly trivial, because clearly for every integer solution there are integers $ p, q, r, s $ with $p$ and $q$ squarefree and coprime such that $ x + y = p q^2 r^3 $ and $ x - y = p^2 q s^3 $, so that:
$ 2 x = p q (q r^3 + p s^3) $
$ 2 y = p q (q r^3 - p s^3) $
$ a = p q r s $
and these satisfy th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/53292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Get number of elements of a square matrix given a vector that has upper right elements of that matrix Suppose I have some vectors:
*
*$[1,2,1]$ its length is $3$ wich represents a matrix like $\begin{pmatrix}
1 & 2\\
-1 & 1
\end{pmatrix}$ the complete matrix would have $4$ elements
*$[1,2,1,3,4,1]$ its len... | The function
$n\mapsto \frac12 ( 1 + 4 n - \sqrt{1+8n})$
does the job.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/55859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
A "fast" way to find the sum of the sequence $5,5.5,5.55,5.555,5.5555,\ldots $ (20 terms) My initial approach is diving the whole sum by $9$ and taking the common $5$ out which gives $$\frac{5}{9}[(10-1)+(10-0.1)+(10-0.01)+\cdots + (10-10^{-19})]$$ after some algebra this could be reduced to $$\frac{5}{9} \times [200-\... | A slower way. It solves the general case with $n$ terms. I denote your sum as
$$S_{20}=5+5.5+5.55+5.555+\ldots +5.\underset{19}{\underbrace{555\ldots 5}}$$
and the general case for $n$ as
$$S_{n}=5+5.5+5.55+5.555+\ldots +5.\underset{n-1}{\underbrace{555\ldots 5}}.$$
Since $\sum_{j=1}^{k-1}10^{j-1}=\frac{1}{90}10^{k}-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/57794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.