Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Square root of 9 from a technical perspective. When we say $$ \sqrt{9}= x $$
then $\;x = 3,\;$ right?
So why when we square both sides it becomes different:
$$ (\sqrt{9})^2 = x^2$$
$$9 = x^2$$
Here $\;x =\pm 3.$
So, does $\;x = \pm 3\;$ in $\sqrt{9} = x?$
| Just because $A$ is something that does $P$, does not mean that $A$ is the ONLY thing that does $P$.
It is true that $3^2 = 9$ but $3$ is not the only thing that when squared will equal $9$.
so $3^2 =9$ and $w^2 = 9$ does not in any way mean that $w = 3$.
This is no more valid or reasonable than saying $(1)^2- 3(1) +7 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Find the sum of the maximum of the trigonometric polynomial coefficient somedays ago,I have studying and discussing this Maximum Trigonometric polynomial coefficient
problem
Now I want consider a simple problem:
(The sum of the maximum of the trigonometric polynomial coefficient:) Let $n$ be give postive integers If... | Solution due to @quasi:
Let $A_k = 2 - \frac{2k}{n+1}$. We have $A_1 + A_2 + \cdots + A_n = n$.
We have
\begin{align}
&\sum_{k=1}^n A_k \cos(kx) + 1\\
=\ & 2\sum_{k=1}^n \cos (kx) - \frac{2}{n+1}\sum_{k=1}^n k \cos (kx) + 1\\
=\ & 2\sum_{k=1}^n \cos (kx) - \frac{2}{n+1}\left(\sum_{k=1}^n \sin (kx)\right)' + 1\\
=\ & 2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3997663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why combining two quadratic equations of a circle and a parabola creates extra solutions for $x$ We have a parabola and a circle with the following equations and their graph placed at the end of my question.
Parabola: $y^2 = 4x -4$
Circle: $(x-2)^2 + y^2 = 9$
My goal was to calculate their intersection points so I subs... | A standard Cartesian graph deals only with real numbers "plotted against" real numbers, so in looking at the intersections of two curves in such a graph, one may not be "getting the whole story." Two distinct quadratic curves will have four intersections; since one of these (the circle) is bounded, all of the interse... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3998217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Find what a Infinite product number approaches Find what the following number approaches:
$$\frac{6}{5}\times\frac{26}{25}\times\frac{626}{625}\times\frac{390626}{390625}\times...$$
What I found is that each fraction can be simplified as:$$1+\frac{1}{5^n}$$
Where $n$ is $1,2,4,8$ and so on (instead of $1,2,3,4,\ldots$... | You have
$$\left(1+\frac{1}{5^1}\right)\left(1+\frac{1}{5^2}\right)\left(1+\frac{1}{5^4}\right)\left(1+\frac{1}{5^8}\right)\cdots$$
$$=\left(1+\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}\right)\left(1+\frac{1}{5^4}\right)\left(1+\frac{1}{5^8}\right)\cdots$$
$$=\left(1+\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4001675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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From the system of equations prove $a+b+c=0$
$a,b,c$ are distinct real numbers and $x,y$ are also real numbers. we have these equations:
$${ \begin{cases}{a^3+ax+y=0} \\ {b^3+bx+y=0} \\
{c^3+cx+y=0}\end{cases} }$$ Prove $a+b+c=0$
I added all the equations together and get:
$$a^3+b^3+c^3+(a+b+c)x+3y=0$$
It is similar... | Here's an alternative solution to Albus Dumbledore's. Note that the three equations imply that the nonzero vector $(1,x,y)$ is in the kernel of
$$\begin{pmatrix}a^3&a&1\\b^3&b&1\\c^3&c&1\end{pmatrix}$$
of determinant (after factoring)
$$(a-b)(a-c)(b-c)(a+b+c).$$
But also, the determinant must be 0 since it has nontrivi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4005297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding all natural $x$, $y$, $z$ satisfying $7^x+1=3^y+5^z$ The problem goes as follows:
Find all possible pairs of $x,y,z \in \mathbb{N}$ which satisfy the equation $7^x+1=3^y+5^z$
My first instinct was to continue by modding, but I don't think I can get anything out of it. The obvious solutions seem to be $x,y,z=1... | The two solutions you found are indeed all of them. In fact, we can solve this with a straight mod bash, which was surprising to me.
If any of $x,y,z$ are $0$, then we easily get $x=y=z=0$. If $y=1$, then $7^x=5^z+2$, and we readily get that $z=1$, so $x=1$ too. So now assume $y>1$ and $x,z>0$. Let the mod bash begin.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
USA MO 1980's inequality Let $x;y;z$ be real numbers and $x;y;z \in [0;1]$
Prove: $\dfrac{x}{y+z+1}+\dfrac{y}{z+x+1}+\dfrac{z}{x+y+1} \leq 1-(1-x)(1-y)(1-z)$
The official solution uses calculus, so I can't understand it.
I hope someone have a "simpler solution" (for middle-school student). Thank you.
| Assume $x+y+z>0$, else the inequality is easy. Do some rearranging:
\begin{align*}(1-x)(1-y)(1-z)\leq1-\sum\frac{x}{y+z+1}&=\sum\left(\frac{x}{x+y+z}-\frac{x}{y+z+1}\right) \\
&= \frac{1}{x+y+z}\sum\frac{x(1-x)}{y+z+1}
\end{align*}
Multiplying through by $x+y+z>0$, and moving everything to one side gives
$$\sum x(1-x)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
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How to find all third roots of $-i$? How to find all third roots of $-i$?
I have no idea.
Could you give me some hint?
| If you know de Moivre's law, or representation of complex numbers as exponentials, this problem is simple.
If you don't, here is an algebraic approach.
There exists some complex number $a+bi$ such that $(a+bi)^3 = -i$
Multiplying it out and combining terms.
$a^3 + 3a^2bi + 3a(bi)^2 + (bi)^3 = -i\\
(a^3 - 3ab^2) + (3a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Integral involving ratios of polynomial and square roots Is there a simple way to compute the following integral
$$I(y,\beta,\gamma)=\int_1^y \frac{x^2}{\sqrt{\left(x^2-1\right)\left(x^2-\gamma ^2\right) \left(x^2+\beta ^2\right) }}\mathrm{d}x$$
with $y>1$, $0<\gamma<1$ and $\beta>0$?
|
Define the function $\mathcal{I}:\left(0,1\right)\times\mathbb{R}_{>0}\times\left(1,\infty\right)\rightarrow\mathbb{R}$ via the improper integral
$$\mathcal{I}{\left(p,q,y\right)}:=\int_{1}^{y}\mathrm{d}x\,\frac{2x^{2}}{\sqrt{\left(x^{2}-1\right)\left(x^{2}-p^{2}\right)\left(x^{2}+q^{2}\right)}}.$$
Suppose $\left(p,q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation: $\lfloor a/2\rfloor+\lfloor a/3\rfloor+\lfloor a/5\rfloor=a$, where $a> 0$. Let $a\in\mathbb{Z}^+$, then $a$ we can write it as $a = 30k + i$, where $i\in\{0,1,2,\ldots, 29\}$ and $k\in\mathbb{Z}_0^+$. Thus substituting we have:
\begin{align*}
\lfloor(30k + i)/2\rfloor + \lfloor(30k + i)/3\rflo... | Yes, this is an excellent solution.
It would perhaps be a good idea, after obtaining
\begin{align*}
k &= i -(\lfloor i/2\rfloor + \lfloor i/3\rfloor + \lfloor i/5 \rfloor)
\end{align*}
to show that $k$ is non-negative. This is not totally trivial and, since$$\frac{1}{2}+\frac{1}{3}+\frac{1}{5}>1,$$ it is only jus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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$\langle A,B \rangle$ is the unique Sylow $2$-subgoup of $SL_2(F_3)$ Prove that the subgroup of $SL_2(F_3)$ generated by $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ and $\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ is the unique Sylow $2$-subgoup of $SL_2(F_3)$.
Proof: Let $G=SL_2(F_3)$. $|G|=24=2^3\cdot 3$. Let
... | Let $P = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$. Then $P^3 = I$ and $P^{-1} = P^2 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$. Since $| \langle P \rangle | = 3$ and $| \langle A, B \rangle | = 8$ we have $\langle P \rangle \cdot \langle A, B \rangle = \langle P, A, B \rangle = SL_2(\mathbb{F}_3)$.
Calcula... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Showing the sequence $\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$ tends to $2$ using the Epsilon-Neighbourhood definition I am given the sequence
$$a_{n+1} = \sqrt{2 a_n}, \quad a_1 = \sqrt{2}.$$
That is, the sequence
$$ \sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$$
I am aware of using ... | Let's make an observation: for each natural number $n$, your sequence can be given explicitly by $$a_n = 2^{\sum_{i=1}^n (\frac{1}{2})^i}$$
For example, $a_2 = \sqrt{2\sqrt{2}} = (2(2^{1/2}))^{1/2} = 2^{1/2} \cdot 2^{1/4} = 2^{1/2 + 1/4}$
Furthermore, we know that $${\sum_{i=1}^n \left(\frac{1}{2}\right)^i} \to 1 \ \ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Showing $\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$ and $\cos A+\cos B+\cos C=1+\frac rR$ in $\triangle ABC$
In a triangle with vertices $A$, $B$, $C$, semiperimeter $s$, inradius $r$ and circumradius $R$, prove that $$\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$$ and $$\cos A+\cos B+\cos C=1+\frac rR$$
(note: we... | The area of the triangle can be expressed as $ \frac12r(a+b+c)= \frac{abc}{4R}$. Then
\begin{align}
\frac rR =\frac12 \frac{\frac{abc}{R^3}}{\frac{a+b+c}R}
=\frac{2\sin A \sin B \sin C}{\sin A +\sin B +\sin C}\tag1
\end{align}
Note
$$\sin A +\sin B +\sin C
= 2\cos\frac A2\sin\frac A2 + 2\sin\frac {B+C}2\cos\frac{B-C}2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4019680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Writing a group as a product of its generators in MAGMA Let $G \subseteq S_8$ be generated by the elements $s=\begin{pmatrix} 1 & 2 \end{pmatrix}\begin{pmatrix} 3 & 5 \end{pmatrix}\begin{pmatrix} 4 & 6 \end{pmatrix}\begin{pmatrix} 7 & 8 \end{pmatrix}$ and $t=\begin{pmatrix} 1 & 3 & 7 & 4 \end{pmatrix}\begin{pmatrix} 2 ... | The code below will do what you want - in fact the answer is rather easy! This method will work provided that the group is not too big - say maximum order about $10^7$ or maybe $10^8$. Although not really necessary for solving this problem, it computes a presentation of the group $G$.
> G<s,t> := PermutationGroup< 8 |... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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In a course there's 10 boys and 10 girls and we need to divide them into 3 different groups.. We have $3$ groups $A,B,C$.
We have to divide all the boys and the girls into these three groups such that in group $C$, there's exactly $6$ or $7$ members, there's atleast one girl, and the number of the boys must be greater... | It seems that the question is only about the possible combinations for group $C$.
This is where the answer 50610 comes from:
$$\binom{10}{5}\cdot\binom{10}{1} + \binom{10}{6}\cdot\binom{10}{1} + \binom{10}{4}\cdot\binom{10}{2} + \binom{10}{5}\cdot\binom{10}{2} + \binom{10}{4}\cdot\binom{10}{3} = 50610$$
For example: to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x+2y=8$, find the minumum of $x+y+\frac{3}{x}+\frac{9}{2y}$ (x, y ∈ R+) Question : If $x+2y=8$, find the minumum of $x+y+\dfrac{3}{x}+\dfrac{9}{2y}$ $(x, y \in \mathbb{R}^+)$
I tried to use $x=8-2y$, and I thought I can plug in. But I think I'm not doing it well. I thought about AM-GM, and how can I expand this us... | The AM-GM inequality, as in the post tag, gives us a lower bound.
Indeed,
$$x+\frac{3}{x} + y +\frac{9}{2y}=2\frac{x+\frac{3}{x}}{2} + 2\frac{y +\frac{9}{2y}}{2}.$$
Since $x,y>0$ we can apply the AM-GM inequality to obtain
$$2\frac{x+\frac{3}{x}}{2} + 2\frac{y +\frac{9}{2y}}{2}\geq 2\sqrt{3}+2\sqrt{\frac{9}{2}}=2(\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve $x^3-1=2y^2$ integer solution find the equation integer solution
$$x^3-1=2y^2$$
I have read this proof:proof, have simple methods? Because this problem is a Olympiad problem. So maybe have other methods to solve it
since $x>0.y>0$ so
$$(x-1)(x^2+x+1)=2y^2$$ since
$(x-1,x^2+x+1)=1$ or$3$.so
(1):if $x-1=2u^2,x^2+x... | As suggested in the proof you link, it is easier to pass to $\Bbb{Z}[\sqrt{-2}]$ right from the start:
If $x$ and $y$ are integers satisfying $x^3-1=2y^2$, then
$$x^3=1+2y^2=(1+y\sqrt{-2})(1-y\sqrt{-2}),$$
where the two factors on the right hand side are coprime because their sum is $2$ and their product is odd. Then b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4026096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Prove that $\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$ Prove that for every $a,b,c \in \mathbb{R}^{+}$ We have $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$$
Unfortunately i can just prove that :
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{b^2}{b^2+ac} <3$$
like this :
$$a^2+bc>a^2 \i... | let $p=\frac{a^2}{a^2+bc},q=\frac{b^2}{b^2+ac},r=\frac{c^2}{c^2+ab}$ then easy to see $p,q,r\in (0,1)$ Now notice that $$\frac{(1-p)(1-q)(1-r)}{pqr}=1$$ so $$p+q+r=1+pq+qr+rp-2pqr$$ It remains to prove $$pq+qr+rp-2pqr\le 1$$ $$\iff \underbrace{p(1-q)(r-1)}_{\le 0}+\underbrace{q(1-r)(p-1)}_{\le 0}+\underbrace{(1-p)(q-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4029795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Modular quadratic equation question- Where did I go wrong? \begin{gather}
\frac{N^2+N}{2} \equiv 0 \pmod 4 \\
N^2+N\equiv 0\\
(N+\frac{1}{2})^2-\frac{1}{4}\equiv 0\\
4(N+\frac{1}{2})^2-1\equiv 0\\
2^2(N+\frac{1}{2})^2\equiv 1\\
(2N+1)^2\equiv 1 \\
2N+1 \equiv 1\\
2N=0\\
N=2,4,6,8...\\
2N+1 \equiv 3\\
2N=2\\
N=1,3,5,7..... | $(2N+1)^{2} \equiv 1 $
$2N+1 \equiv 1$
This step is wrong, for example: $(2*3+1)^2 \equiv 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4030136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Close-form for triple integral $ \int_0^c \int_0^b \int_0^a \sqrt{x^2+y^2+z^2} dx dy dz$ I am able to work out the double integral
$$\int_0^b \int_0^a \sqrt{x^2+y^2} dx dy $$
with brute-force (i.e. integrating $x$, then $y$) to arrive at the close-form result
$$\frac13ab\sqrt{a^2+b^2} +\frac16a^3\sinh^{-1}\frac ba +\fr... | A start
Spherical coordinates. $(\rho,\theta,\phi)$. They are related to rectangular coordinates $(x,y,z)$ by:
\begin{align}
x &= \rho\sin\theta\cos\phi\\
y &= \rho\sin\theta\sin\phi\\
z &= \rho\cos\theta
\end{align}
and in reverse by
\begin{align}
\rho &= \sqrt{x^2+y^2+z^2}\\
\theta &=\arccos\frac{z}{\sqrt{x^2+y^2+z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4030756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Proof of the Moreau Envelope of $l_1$ norm Given function $| \cdot | : \mathbb{R} \rightarrow \mathbb{R}_{+}$ and $\alpha > 0$, its Moreau envelope $e_{\alpha}|\cdot|: \mathbb{R} \rightarrow \mathbb{R}_{+}$, reads:
\begin{equation}
e_{\alpha}|\cdot|\left(x\right) = \inf_{y \in \mathbb{R}}{\bigg\{|y| + \frac{1}{2\alpha}... | Suppose $y\ge 0$. Observe that
\begin{align}
|y|+\frac{1}{2\alpha}|x-y|^2 =&\ \frac{x^2+y^2}{2\alpha}+\frac{xy}{\alpha}+y = \frac{x^2-(x+\alpha)^2}{2\alpha} + \frac{1}{2\alpha}\left(y^2+2(x+\alpha)y+(x+a)^2 \right)\\
=&\ \frac{x^2-(x+\alpha)^2}{2\alpha} +\frac{(y+x+\alpha)^2}{2\alpha}.
\end{align}
If $x+\alpha>0$, then... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate an absolute monster integral $\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$ I want to figure out a way to evaluate $$\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$$
I tried to substitute $x = u^2$ and cancel the square root in the denominator, getting $$2\int\limits... | \begin{align}J&=\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x\\
&\overset{y=x^2}=2\int_0^1 \frac{\ln(1-y^2+y^4)}{1+y^2}dy\\
K&=\int_0^1 \frac{\ln\left(\frac{1-x^2+x^4}{x^2}\right)}{1+x^2}dx\\
&\overset{z=\frac{2}{x+\frac{1}{x}}}=\int_0^1 \frac{\ln\left(\frac{4-3z^2}{z^2}\right)}{2\sqrt{1-z^2}}dz\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
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How to Evaluate $\int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4}dx$? How to evaluate
$$ \int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4} \,dx \approx -0.162858 \tag{1}$$
The integral arises in the computation of
$$\left( \sum_{n=1}^{\infty} \frac{(-1)^n}{n}\right)\left(\sum_{n=1}^{\infty} \frac{(-1)^n}{4n-1}\right)$$
as
$$ \scripts... | By expanding the denominator and integrating term-by-term we obtain:
$$
-\sum _{n=0}^{\infty } \frac{(-1)^n }{4 n+3}H_{n+\frac{3}{4}},
$$
where $H_n$ is the $n$th harmonic number, in the formula above--the harmonic number of fractional order.
Not sure if it is useful, but the sum can be performed with the help of, e.g.... | {
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"url": "https://math.stackexchange.com/questions/4038542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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} |
Find $\int_0^4(g\circ f\circ g)(x)\mathrm{d}x$ where $f(x)=\sqrt[3]{x+\sqrt{x^2+1/27}}+\sqrt[3]{x-\sqrt{x^2+1/27}}$, $g(x)=x^3+x+1$
Let $$f(x)=\sqrt[3]{x+\sqrt{x^2+\frac{1}{27}}}+\sqrt[3]{x-\sqrt{x^2+\frac{1}{27}}}$$
and
$$g(x)=x^3+x+1$$
then, find $$\int_0^4(g\circ f\circ g)(x) \mathrm dx$$
My attempt:
Let $\display... | Note that $f^{-1}(x) = \dfrac{x^3+x}{2}\\$
$\therefore g(x) = 2f^{-1}(x)+1\\$
$\therefore g(f(g(x))) = 2f^{-1}(f(g(x)))+1\\$
and I'm sure you know $f^{-1}(f(x)) = x \\$
$\therefore g(f(g(x))) = 2g(x)+1\\$
$$\therefore \int_0^4g(f(g(x)))dx = \int_0^4(2g(x)+1)dx = \int_0^4(2x^3+2x+3)dx\\$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4038943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Express $S =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\ldots+\frac{1}{(n-2)\cdot(n-1)}+\frac{1}{n\cdot(n+1)}$ in terms of $n$.
Express $S =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\ldots+\frac{1}{(n-2)\cdot(n-1)}+\frac{1}{n\cdot(n+1)}$ in terms of $n$.
Here's what I have done so far:
$$\begin{align*}
S &=\left(\frac{1}{1}-\frac... | It seems the expression $S$ needs to be revised somewhat. If we take a look at the first two summands
\begin{align*}
\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\cdots
\end{align*}
we see the left-hand factor in the denominator is odd and the right-hand factor is even. Taking $n$ summands of this kind we have
\begin{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4039484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving differential equation with boundary conditions $y=2$ at $x=0$ I need to find a solution to this differential equation $(1+x^2) $$\tfrac{\mathrm{d}y}{\mathrm{d}x} = x-xy^2$, with boundary condition $y=2$ at $x=0$
I collect all terms on the RHS to get $$\tfrac{\mathrm{d}y}{\mathrm{d}x} = \frac{x}{1+x^2} (1-y^2)$... | $$\int\frac{dy}{1-y^2}=\int\frac{x\,dx}{1+x^2}$$
now you can do the left a few ways but we can use PFD:
$$\int\frac{dy}{1-y^2}=-\frac12\int\left(\frac{1}{y-1}-\frac{1}{y+1}\right)dy=\frac12\ln\left|\frac{y+1}{y-1}\right|$$
now do RHS:
$$\int\frac{x}{x^2+1}dx=\frac12\ln(x^2+1)+C_1$$
now equate:
$$\ln\left(\frac{y+1}{y-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve determinants with $|AB| = |A||B|$ How should I figure out the following determinants?
It is required to use $|AB| = |A||B|$ to figure them out.
(1) $D_1 = \begin{vmatrix}
1+x_1y_1 & 1+x_1y_2 & \dots & 1+x_1y_n \\
1+x_2y_1 & 1+x_2y_2 & \dots & 1+x_2y_n \\
\vdots & \vdots & \ddots & \vdots \\
1+x_ny_1 & 1+x_ny_2 & ... | You can use matrix rank inequality:
$$r(AB) \le \min\{r(A),r(B)\}$$
For $D_1$ you wrote your matrix as $AB$, and both ranks of $A$ and $B$ are $\le 2$ so its rank is $\le 2$ by the above formula. Therefore if $n \ge 3$ then $D_1 = 0$. For $n=1, 2$ you can calculate manually. You get
$$|1+x_1y_1| = 1+x_1y_1, \quad \begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and $P$ is greater than 2? Let $P$ be a point chosen at random on the line segment between the points $(0,1)$ and $(3,4)$ on the coordinate plane. What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and... | The area of the triangle is given by the function $\displaystyle A(X) = \frac{3}{2}(X+1)$, where X is uniformly distributed in $[0,3]$, so
\begin{equation}
\mathbb{P}(A(X) \geq 2) = \mathbb{P}\left(X\geq\frac{1}{3} \right) = \frac{1}{3} \left(3 - \frac{1}{3} \right) = \frac{8}{9}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Inequality $a^2+2b^2+8c^2\geq2a(b+2c)$ Can someone prove this inequality for the real numbers a,b,c?
$$a^2+2b^2+8c^2\geq2a(b+2c)$$
I have tried simple manipulation of the terms to get quadratic expressions, but since one cannot factor the $4ac$ on the right side, I abandoned that approach. Then I tried turning it into ... | Your inequation is equivalent to
$$(a-b-2c)^2+(b-2c)^2 \ge 0 \tag{1}$$
The advantage of expression (1) is that it allows to clearly obtain the limit cases where there is an equality sign in (1) instead of a $">"$ symbol, i.e., iff
$$a=2b=4c$$
Edit: One can wonder how I have found expression (1). If you happen to know ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4052272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
studying the function $f(x)= \sqrt{x^2+x}-x$ I'm studying the function $$f(x)= \sqrt{x^2+x}-x$$
The domain is $(- \infty, -1] \cup [0, + \infty)$.
In particular I'm making some mistakes about the sign of the
first derivative $$ \frac{2x+1-2 \sqrt{x^2+x}}{2 \sqrt{x^2+x}}$$
The first derivative is never zero.
$ \frac{2x+... | $$ \frac{2x+1-2 \sqrt{x^2+x}}{2 \sqrt{x^2+x}}>0\iff 2x+1-2\sqrt{x^2+x}>0\iff 2x+1>2\sqrt{x^2+x}\iff \begin{cases}2x+1>0\\ x^2+x> 0\\ (2x+1)^2>4(x^2+x)\end{cases}\iff \begin{cases}x>\frac{-1}{2}\\x> 0 \vee x< -1\\ \forall x \end{cases}\iff x> 0 $$ Be careful to exclude $0$ in order to have the denominator not equal to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Express the sum of a geometric series in the form $p + q\sqrt2$ I am working through a maths text book as a hobby and am stuck on the following problem:
An arithmetic series and a geometric series have r as a common difference and the common ratio respectively. The first term of the arithmetic series is 1 and the first... | $$\frac{1}{1-\sqrt{1/8}} = \frac{\sqrt{8}}{\sqrt{8} - 1} = \frac{\sqrt{8}( \sqrt{8} + 1)}{(\sqrt{8} - 1)(\sqrt{8} + 1)} = \frac{8 + \sqrt{8}}{8 - 1} = \frac{8}{7} + \frac{2\sqrt{2}}{7}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4055229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show limit exists if and only if $m+n>2$. Let $m,n \in \mathbb{N}$.
Show that the limit $\lim_{x \to (0,0)} \frac{x^ny^m}{x^2+y^2}$ exists if an only if $n+m>2$.
Attempt:
$(\Leftarrow)$. If $n+m>2$ without loss of generality assume $n \geq m$. Then $|\frac{x^ny^m}{x^2+y^2}|\leq |\frac{x^ny^m}{x^2}|=|x^{n-2}y^m|\leq (\s... | Your proof seems to be correct! Another more intuitive way would be to use polar coordinates in this situation (which always is a good thing to consider when you have a $x^2+y^2$ term somewhere): $$\lim \limits_{(x,y) \to 0} \frac{x^ny^m}{x^2+y^2}=\lim \limits_{r\to 0}\frac{(r\cos(\theta))^n(r\sin(\theta))^m}{(r\cos(\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A Euclidean algorithm problem Suppose that, after running the Euclidean algorithm on two integers $a$ and $b$, we find that $r_n=3$, where $r_n$ is the last remainder in the Euclidean algorithm. Furthermore, we find that $r_{n-1}=6$ and $r_{n-2}=21$. What can we say about $a$ and $b$?
Obviously, by the definition of th... | In the Euclidean algorithm, performed on the starting couple of positive integers $(n,m)$
the remainders follow the sequence
$$
\eqalign{
& n = r_{\, - 2} =
\left\lfloor {{n \over m}} \right\rfloor m + r_{\,0} \cr
& m = r_{\, - 1} =
\left\lfloor {{m \over {r_{\,0} }}} \right\rfloor r_{\,0} + r_{\,1} \cr ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
3-variable 3-equations problem
For positive reals $x,y,z$, if $xy+yz+xz=\sqrt{\frac pq}$, given
$$x^2+xy+y^2=2$$
$$y^2+yz+z^2=1$$
$$z^2+zx+x^2=3$$
find $p-q$, where $p$ and $q$ are relatively prime integers.
I tried grouping the required sum by adding all the equations but it led me to nowhere. I tried applying basic... | If you like a more algebraic answer, consider the following linear combination of the three equations in the order written above $(x-y)(1)+(y-z)(2)+(z-x)(3)$. The LHS of the resultant equation vanishes and we get a linear relation:
$$ 2(x-y)+(y-z)+3(z-x)=0\iff z=\frac{x+y}{2}$$
Under this light we see that $\Pi=xy+xz+y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4058061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Diagonal element of the resolvent of bi-infinite tridiagonal Laurent operator For $\alpha, \beta, \gamma \in \mathbb{C}$ consider the bi-infinite tridiagonal Laurent operator $T$ with $\beta $ on the diagonal given by.
\begin{pmatrix}
\dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \\
\dots & \alpha ... | As we explain in the Appendix here: https://arxiv.org/abs/2206.09879
this is a standard calculation:
"First, $\sigma(T)$ is the image of the symbol curve
$
a(z) = \alpha z^{-1} + \beta + \gamma z $
for $z \in \mathbb{T}$. Since $T$ is invertible it holds that $ a(z) \neq 0$ for all $z \in \mathbb{T}$ and therefore,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4059125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
For which values of $a$ does this system of equations $\mathbf{{not}}$ have a unique solution? Here's my system of linear equations:
$\begin{cases}
x + 2y + 2z = 1\\x + ay + 3z = 3\\x + 11y +az = 0\\
\end{cases}$
Thus I have the augmented matrix $\left[\begin{array}{ccc|c}1&2&2&1\\1&a&3&3\\1&11&a&0\end{array}\right]$
B... | The easiest thing is to use determinants: the system has a unique solution if and only if the matrix is invertible, i.e., if and only if the matrix has determinant not equal to $0$. So you can compute the determinant (either in your original or row-reduced matrix) in terms of $a$, set it equal to $0$, and solve for $a$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4059453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Evaluating $\int_0^1 (1-x^2)^n dx$ In an exercise I'm asked the following:
a) Find a formula for $\int (1-x^2)^n dx$, for any $n \in \mathbb N$.
b) Prove that, for all $n \in \mathbb N$: $$\int_0^1(1-x^2)^n dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$$
I used the binomial theorem in $a$ and got:
$$\int (1-x^2)^n dx = \sum_{k... | My idea here is to use the Beta function:
$$ \int_0^1 t^{m}(1 - t)^n \;dt = \frac{\Gamma(m+1)\Gamma(n+1)}{\Gamma(m+n+2)}. $$
Where the Gamma function satisfies $\Gamma(n + 1) = n!$ if $n$ is an integer, and also $\Gamma$ satisfies the recurrence $\Gamma(n + 1) = n\Gamma(n)$.
So if we substitute $t = x^2$, $dt = 2x\; dx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 9,
"answer_id": 5
} |
Area of surface over $xy$-plane In my calculus book, it is claimed that the area $dS$ below, can be expressed as $$\frac{1}{\cos\gamma}dxdy.$$
Here $\vec{n}$ is perpendicular to the surface element, and $\vec{k}$ is given as $(0,0,1)$.
I understand how this can be concluded if either the area element's $x$- or $y$-comp... | WLOG we can translate things around so that the base of our rectangular prism has vertex $(0, 0, 0), (a, 0, 0), (0, b, 0), (a, b, 0)$ and $(0, 0, 0)$ is the vertex with the smallest $z$-coordinate where the plane cut the rectangular prism.
Also, WLOG we can assume our normal vector $n = \begin{pmatrix} p \\ q \\ r \end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Drawing chips from a bag A bag with contains $5$ chips that are either red or blue chips. The number of red chips is a random variable $X$ where $X$ can be either $2, 3$ or $4$ with equal probability. Given that when $5$ chips are drawn with replacement, the result is: $3$ of the draws are red and $2$ are blue, what is... | Your approach is correct but you made a mistake in calculation. Also I will suggest that your working should be using fractions instead of converting into decimals.
If $Y$ is the event of $3$ red and $2$ blue in $5$ draws with replacement,
$\small P(X = 4 | Y) = \displaystyle \frac{P(X = 4 \ \cap \ Y)}{P(Y)}$
$\small P... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4067784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Calculation of a hard integral $\int^{1/2}_0\frac{1}{x^2-x+1} \ln\frac{1+x}{1-x}\,dx $ Calculate
$$\int^{1/2}_0\frac{1}{x^2-x+1} \cdot \ln\frac{1+x}{1-x}\,dx $$
I tried to subsitute $x=1/2-t$ so $dx=-dt$ but I just complicated more my problem.
| Mathematica gives:
$$\frac{\pi \log \left(\frac{9}{8}\right)-3 i \left(\text{Li}_2\left(-\frac{1}{2}
(-1)^{2/3}\right)+\text{Li}_2\left(\frac{1}{1-\sqrt[3]{-1}}\right)-\text{Li}_2\left(\frac{1}{1+\sqrt[3]{-1}}\right)+\text{Li}_2\left(\frac{1}{1-(-1)^{2/3}}\right)-\text{Li}
_2\left(\frac{1}{1+(-1)^{2/3}}\right)+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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The inequality $\,2+\sqrt{\frac p2}\leq\sum\limits_\text{cyc}\sqrt{\frac{a^2+pbc}{b^2+c^2}}\,$ where $0\leq p\leq 2$ is: Probably true! Provably true? Let $p$ be a positive parameter in the range from $0$ to $2$.
Can one prove that
$$2 +\sqrt{\frac p2} \;\leqslant\;\sqrt{\frac{a^2 + pbc}{b^2+c^2}}
\,+\,\sqrt{\frac{b^... | This is a partial answer.
This answer proves that for $1\leqslant p\leqslant 2$, the inequality $(1)$ holds.
Proof :
(You have already proved that for $p=2$, $(1)$ holds. The idea used in the case $p=2$ works for the case $1\leqslant p\leqslant 2$.)
W.l.o.g. assume $\,a\geqslant b\geqslant c\,$ and $\,a,b>0$. Let
$u=\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
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Problem with system of equations $(x,y,z)$ System of equations:
$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}$$
Solution:
$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}\Rightarrow\begin{cases}x^2+y^2+z^2=3\\xy+yz+zx=3\end{cases}$$
... | By the calculations you have made, your system is actually eqivalent to $$\begin{cases}x^2+y^2+z^2=3\\(x+y+z)^2=9\end{cases}$$Which is, at the same time, equivalent to the systems $$\begin{cases}x^2+y^2+z^2=3\\x+y+z=3\end{cases} \quad \begin{cases}x^2+y^2+z^2=3\\x+y+z=-3\end{cases}$$But you can check using the distance... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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What is the following limit? How can I calculate the following limit?
What do I miss here? What am I doing wrong?
$$\begin{align*}
\lim_{x\rightarrow0}\frac{6x\cdot\sin x-6\cdot\sin\left(x^{2}\right)+x^{4}}{x^{5}\left(e^{x}-1\right)}= & \lim_{x\rightarrow0}\frac{6x\cdot\sin x}{x^{5}\left(e^{x}-1\right)}-\lim_{x\rightar... | You replaced parts of the formula with their limits, so that some terms (not replaced with their limit) cancel. This may lead to wrong results, as you see.
I suggest finding equivalents for the numerator and the denominator.
*
*For the denominator, it is basid: it is known from Taylor-Young's formula that $\mathrm e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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find the roots $(2z+3)^3=\frac{1}{64}$ There are 3 roots 1 real and 2 imaginary i found one z by doing $\frac{\frac{1}{4}-3}{2}$ so $z=\frac{-11}{8}$ however there are two more complex roots which are $z=\frac{-25+i√3}{16}$ and $z=\frac{-25-i√3}{16}$ but i dont know how to get to it any help is much appricated
. than... | There is a simpler method. Just use
$$\begin{align}a^3=b^3 \Longleftrightarrow a^3-b^3=0\Longleftrightarrow (a-b)(a^2+ab+b^2)=0\end{align}$$
To find the complex roots, just solve the quadratic:
$$a^2+ab+b^2=0.$$
In your case you can take $$a=2z+3, b=\dfrac 14.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Integral of Euler. Here is an integration problem I found in an old book:
Integrate
$$\int\frac{1+x^2}{1-x^2}\frac{dx}{\sqrt{1+x^4}}$$
The integral is attributed to Euler.
My solution is
let $$y=\frac{1+x^2}{1-x^2}$$ then get
$$\frac{1}{\sqrt{2}} \int \frac{ydy}{\sqrt{y^4-1}}$$
After another substitution $z=y^2$ get
$$... | $$ I=-\int \dfrac{1}{\sqrt{x^4+1}}\dfrac{x^2+1}{x^2-1} dx=$$
$$=-\int \dfrac{1}{\sqrt{\left(x^2-1\right)^2+2x^2}}\dfrac{x^2+1}{x^2-1} dx$$
$$=-\int\frac{1}{(x^2-1) \sqrt{1+\frac{2x^2}{\left(x^2-1\right)^2}}}\dfrac{x^2+1}{x^2-1} dx dx$$
$$=-\int\frac{1}{\sqrt{1+\frac{2x^2}{\left(x^2-1\right)^2}}}\dfrac{x^2+1}{(x^2-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4077788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Trigonometric identity with sin, cos, and tan Very cool problem I was reading today!
$(1)$ $2\sin^2 A $ – $ 2\tan A \sin A + 2 $ – $ \sin^2 A $ – $ \cos^2 B = 0; $
$(2)$ $2\cos^2 B $ – $ 2\tan A \cos B + 2 $ – $ \sin^2 A $ - $ \cos^2 B = 0;$
If $\sin A ≠ \cos B$, find the value of $\boxed{\sin^2 A}$
$\\$
I realized the... | You have a sign error in (5), it should be $-2\tan A \cos B$. Thus, $$ \tan A(\sin A+ \cos B)=2$$ and $$\sin^2 A-\cos^2 B=\tan A (\sin A - \cos B) \implies \sin A+ \cos B=\tan A$$ Can you finish now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit $\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{a x+b} - 2}{x}$ equals to $\frac{5}{12}$ A friend of mine asked this question to me. It seems it's from Stewart.
Find the values of a and b such that
$\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{a x+b} - 2}{x} = \frac{5}{12}$
This is what I tried with better results.
F... | It appears from your working that you had the right idea (at least at the start) in mind and the issue was more related to the use of limit laws.
Let us fix your approach as follows. We have $$\lim_{x\to 0}(\sqrt[3]{ax+b}-2)=\lim_{x\to 0}x\cdot\frac {\sqrt [3]{ax+b} - 2}{x}=0\cdot\frac{5}{12}=0$$ which means that $$\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
The sum of ten distinct natural numbers is $62$. Show that their product is a multiple of $1440$.
The sum of ten distinct natural numbers is $62$. Show that their product is a multiple of $1440$.
I know how to prove that it is a multiple of $60$
Lets write: $a_1+a_2+...+a_{10}=62$. Assuming $a_1<a_2<\cdots <a_{10}$ w... | Experimental approach:
Sum of first 11 numbers is:
$S=\frac {11}2(11+1)=66$
$66-62=4$
So we delete 4 and our numbers are:
$1, 2, 3, 5, 6, 7, 8, 9, 10, 11$
which their product is a multiple of 1440.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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For $z_1, z_2, \ldots, z_{2021}$ the roots of $z^{2021}+z-1$, evaluate $\sum_i\frac{z_i^3}{z_i+1}$
Let $z_1$, $z_2$, $\ldots$, $z_{2021}$ be the roots of the polynomial $z^{2021}+z-1$. Evaluate
$$\frac{z_1^3}{z_1+1} +\frac{z_2^3}{z_2+1} +\frac{z_3^3}{z_3+1} +\cdots +\frac{z_{2021}^3}{z_{2021}+1} $$
I'm not really sur... | This can be done using residues and the function
$$f(z) = \frac{z^3}{z+1} \frac{q(1-z)/z+1}{z^q+z-1}
= \frac{z^2}{z+1} \frac{q+(1-q)z}{z^q+z-1}.$$
where $q\ge 4,$ an integer.
First we need to show that the poles from the term in $q$ are all simple.
We write for a pole $\rho$ the Taylor series of $g(z) = z^q+z-1$
$$g(z)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4093451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Double Integration in region $A=\{ (x,y)\in \mathbb{R}^{2} : x^{2}+y^{2}\le 1, x+y \ge 1, y \le x\}$.
Calculate
$$\iint _{A} (x^{2}+y^{2})^{-3/2} \,dx\,dy\,,$$
where $A=\{ (x,y)\in \mathbb{R}^{2} : x^{2}+y^{2}\le 1, x+y \ge 1, y \le x\}$.
I first found the intersection points that are $(1,0)$, $\left(\frac{1}{\sqrt{2... | You are trying to integrate over the the shaded region in the diagram.
There are couple of ways to simplify this -
i) Use polar coordinates to avoid splitting the integral. In polar coordinates, $x = r \cos\theta, y = r \sin\theta, x^2+y^2 = r^2$
So line $x+y=1$ can be rewritten as $r ({\cos \theta + \sin\theta}) = 1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4093606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Uncertain how to solve this trigonometric equation I am currently attempting to discover how to find the general solutions to $$\sqrt 3\tan^2x=2\tan x+\sqrt 3$$
The given solutions are $x= \frac{\pi}{3}+ \pi k$ , $\frac{5\pi}{6} + \pi k$
To solve this equation I removed the square root from $\sqrt 3\tan^2x$ leaving me ... | $$\text{Let }u=\tan x. $$
Divide through by $\sqrt{3}.$
$$ u^2 -\frac{2}{\sqrt{3}}u-1 = 0.$$
This is a quadratic equation, the solutions are:
$$u = \frac{1}{\sqrt{3}} \mp \frac{2}{\sqrt{3}}.$$
We thus have$$ \tan x = \sqrt{3} \quad \text{ or } \quad \tan x = - \frac{1}{\sqrt{3}}.$$
So $$x \in \left\{\frac{\pi}{3} + k\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4095983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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iterated function $a(n) = \lfloor n\phi + 0.5\rfloor$ Let $a(n) = \lfloor n\phi + 0.5\rfloor$ for all $n \geqslant 1$, where $\phi$ is the golden ratio. Now let
$$a(n)^k = a(a(\ldots(a(n))))$$
where we have iterated $a(n)$, $k$ times in the RHS. I am looking for a nice closed form of $a(n)^k$. I am working on this prob... | Further to my comments, a proof of $a(a(n))=a(n)+n$.
Proposition 1. $\left\lfloor \frac{a(n)}{\phi} + \frac{1}{2}\right\rfloor=n$
Given $x-1< \lfloor x\rfloor \leq x$, we have
$$n\phi - \frac{1}{2} <a(n)\leq n\phi + \frac{1}{2} \iff\\
n-\frac{1}{2\phi} <\frac{a(n)}{\phi}\leq n+\frac{1}{2\phi} \iff\\
n+\frac{1}{2}-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4096723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding value of a when the centroid of a region (p,0) is bounded by $y^2 = 4px$ and x = a The problem is basically the title, however I just got stuck when I did two solutions. Here is one of them:
$$y^2 = 4px\\
f(x) = \sqrt{4px}\\
g(x) = -\sqrt{4px}\\
\bar{x} = p = \frac{M_y}{M}\\
M_y = \int_0^ax(\sqrt{4px} + \sqrt{4... | You have a few mistakes. It is easy to see by symmetry that $M_x = 0$.
$M = \displaystyle \int_0^a (\sqrt{4px}+\sqrt{4px}) \ dx$
$ = 4 \sqrt p\displaystyle \int_0^a \sqrt{x} \ dx = \frac{8 \sqrt p}{3} a^{3/2}$
As you calculated, $M_y = \displaystyle \frac{8 \sqrt p}{5} a^{5/2}$
So, $p = \overline {x} = \displaystyle \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a closed-form solution to the following summation I am solving a summation that appears in a paper, it claims that
$$\sum_{n=1}^{\infty} \binom{2n}{n+k}z^n=\bigg(\frac{4z}{(1+\sqrt{1-4z})^2}\bigg)^k$$
I found this identity here in equation (66)
$$\sum_{n=0}^{\infty} \binom{2n+k}{n}z^n=1+\sum_{n=1}^{\infty} \binom{... | The identity
$$\sum_{n\ge 0} {2n+k\choose n} z^n
= \left(\frac{2}{1+\sqrt{1-4z}} \right)^k
\frac{1}{\sqrt{1-4z}}
= Q_k(z)$$
can also be proved with the Cauchy Coefficient Formula, radius of
convergence is the distance to the nearest singularity which is at
$1/4.$
We obtain
$$[z^n] Q_k(z) =
\frac{1}{2\pi i}
\int_{|z|=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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$I_n= \int_{0}^{ + \infty} \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$
*
*$x \geq 0$
*$n \geq 0$
*$f_n(x)= \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$
*$I_n= \int_{0}^{ + \infty}f_n(x) $
*What is the limit of $I_n$ ?
*What is an equivalent of $I_n$ ?
$
\begin{align*}
u'&= (-n+1) 4x^3 (1+x^4)^{-n-1}\\
u &=x^{4n-1}\\
v &= x^{4n-1} \\... | Proceed as follows
$$ \int_0^{\infty} \dfrac{ x^{2}}{ 1+x^4 } dx
\overset{x\to \frac1x}=\frac12\int_0^{\infty} \dfrac{ 1+x^2}{ 1+x^4 } dx
= \frac12 \int_0^{\infty} \dfrac{ d(x-\frac1{x})}{ (x-\frac1{x})^2+2 } dx =\frac\pi{2\sqrt2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4107233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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find all triplets $(a,b,c)$ such that $ab+bc+ca=1$ and $a^2b+c=b^2c+a=c^2a+b$. Given that $a,b,c \in R$ find all triplets $(a,b,c)$ such that $ab+bc+ca=1$ and $a^2b+c=b^2c+a=c^2a+b$.
Attempt:
Case $I:$ Exactly one of $a,b,c$ is zero. WLOG, let $a=0$. Then we have
$$bc=1$$ and
$$c=b^2c=b$$
Solving we get
$$b=c=\pm 1$$
B... | The case $a,b,c \neq 0 \implies c = \dfrac{a(1-ab)}{1-b^2}= \dfrac{ac(a+b)}{1-b^2}\implies a^2+ab=1-b^2\implies a^2+ab+b^2 = 1$. Thus it follows that: $b^2+bc+c^2 = 1 = c^2+ac+a^2\implies a^2+b^2+c^2 = 1$. But Cauchy-Schwarz inequality says: $1 = a^2+b^2+c^2 \ge ab + bc + ac = 1$. So $a = b = c = \pm \dfrac{1}{\sqrt{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let X and Y have a joint uniform distribution on the triangle with corners at $(0, 2)$, $(2, 0)$, and the origin. Find $E(Y | X = 1/2)$ so $f(x,y) = 1/2$ on the region of the triangle.
$f(x) = \int_0^{2-x}\cfrac{1}{2}dy = \cfrac{1}{2}(2-x)$
so $f(y|x) = \cfrac{1/2}{\cfrac{1}{2}(2-x)} = \cfrac{1}{2-x}$
So $E(Y|X=1/2) = ... | When $X = 1/2$, the support of the conditional density $$f_Y(y \mid x) = \frac{1}{2-x}$$ is not on $Y \in [0,2]$ as suggested by your integral, but rather, $Y \in [0, 2 - 1/2] = [0, 3/2]$. Therefore, the correct expression is $$\operatorname{E}[Y \mid X = 1/2] = \int_{y=0}^{3/2} y \cdot \frac{1}{3/2} \, dy = \frac{3}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I find the inverse of this problem u⊞v:=uv−2(u+v)+6 Let V=(2,∞). For u,v∈V and a∈R define vector addition by u⊞v:=uv−2(u+v)+6 and scalar multiplication by a⊡u:=(u−2)^a+2. It can be shown that (V,⊞,⊡) is a vector space over the scalar field R. Find the following:
the sum:
6⊞10=34
the scalar multiple:
−3⊡6=2.0156... | Before you can find an additive inverse you must find the additive identity. That is $e\square x = x$ for all $x$.
so $ex -2(e+x) +6 = x$. Solver for $e$
so $e(x-2)=3x -6$ so $e=\frac {3x-6}{x-2} = 3$
The additive identity is $3$.
Verify: $k\square 3= 3k -2(k+3) + 6 = k -6 + 6 = k$ so that is indeed the indentity.
No... | {
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"url": "https://math.stackexchange.com/questions/4115391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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simplify the expression ${\sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}}$
Simplify the expression ${\sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}}$
The problem is from a not so well renowned book for calculus in India - Concepts of Functions & Calculus - Vikas Rahi, ISBN 9780070080805. The answer for this quest... | Just write
\begin{eqnarray*} \sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}
& = & \sqrt{x -1 - 2\sqrt{x -1} +1} +\sqrt{x -1 + 2\sqrt{x -1} +1} \\
& = & \sqrt{(\sqrt{x-1}-1)^2} + \sqrt{(\sqrt{x-1}+1)^2} \\
& = & \begin{cases} 2 & 1 \leq x\leq 2 \\ 2\sqrt{x-1} & x>2 \end{cases}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4115583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation $\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}+\frac{1}{x+4}=0$ Solve the equation $$\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}=0.$$
For $x\ne -4;-3;-2;-1;0$ we have $$(x+1)(x+2)(x+3)(x+4)+x(x+2)(x+3)(x+4)+x(x+1)(x+3)(x+4)+\text{...}=0$$
Most likely that's not th... | Let $f$ be the conveniently translated function defined by :
$$f(t):=\dfrac{1}{t-2}+\dfrac{1}{t-1}+\dfrac{1}{t}+\dfrac{1}{t+1}+\dfrac{1}{t+2}$$
What I would like to underline is that a graphical representation provides a good insight for such questions. In particular, it can help to anticipate (or verify) a certain nu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
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Prove that $\left (1+\dfrac{1}{2n}\right )^n-\left (1-\dfrac{1}{2n}\right)^n\geq 1$. Let $n\in\Bbb N$, then $\left (1+\dfrac{1}{2n}\right )^n-\left (1-\dfrac{1}{2n}\right)^n\geq 1.$
Notice that,
$$\begin{align*}\left (1+\dfrac 1 {2n}\right )^n-\left (1-\dfrac{1}{2n}\right)^n&=\sum_{k=0}^n \binom{n}{k} \left (\dfrac{1}{... | At least for large values of $n$
$$a_n=\left (1+\dfrac{1}{2n}\right )^n-\left (1-\dfrac{1}{2n}\right)^n$$
$$b=\left (1+\dfrac{1}{2n}\right )^n\implies \log(b)=n \log\left (1+\dfrac{1}{2n}\right )=\frac{1}{2}-\frac{1}{8 n}+\frac{1}{24 n^2}+O\left(\frac{1}{n^3}\right)$$
$$b=e^{\log(b)}=\sqrt e \left(1-\frac{1}{8 n}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4121858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Simplify convolution integral $\int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt$ I am trying to show that $$\frac{1}{a^2+x^2} \ast \frac{1}{a^2+x^2}=\int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt = \frac{2\pi}{a(4a^2+x^2)}$$ I wrote out the integral for the convolution and it becomes a big exercise i... | Here using Fourier transform may simplify things considerably:
Let $\phi(x)=\frac{1}{1+x^2}$, and $\phi_a(x)=\frac{1}{a}\phi(a^{-1}x)=\frac{a}{a^2+x^2}$.
Recall that $\pi e^{-2\pi|t|}=\int e^{-2\pi itx}\frac{1}{1+x^2}\,dx=\widehat{\phi_1}(t)$. Hence
$$\widehat{\phi_a}(t)=\frac{1}{a}\int e^{-2\pi itx}\phi(xa^{-1})\,dx=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4122361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\sum_{k=4}^{143} \frac{1}{ \sqrt{k} + \sqrt{k+1}} = a - \sqrt{b}$, then $a$ and $b$ respectively are If $\sum_{k=4}^{143} \frac{1}{ \sqrt{k} + \sqrt{k+1}} = a - \sqrt{b}$, then $a$ and $b$ respectively are
*
*10 and 0
*-10 and 4
*10 and 4
*-10 and 0
This question is from the book, Mathematics, Class 9 (The I... | Your sum $$\sum_{k=4}^{143} \frac{1}{ \sqrt{k} + \sqrt{k+1}} = a - \sqrt{b}$$ converges to $10$ which means that $10 = a - \sqrt b$. After solving this equation we get $a = 10$ $and$ $b = 0$. Hence option(1) is correct. Your approach to rationalize it was correct. If we rationalize it we get $\sqrt(k+1) - \sqrt k$. If ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4123340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Are these the only "intersections" between the following series, and why all of them are multiples of $10$ As a web developer that programs in PhP, I enjoy running some scripts to see some of math wonders, however, PhP is limited to large calculations.
I wanted to see if there are any intersections for the following $2... | These are the numbers that are both "triangular" ($n(n+1)/2$) and "tetrahedral" ($n(n+1)(n+2)/6$). They are tabulated at https://oeis.org/A027568. You have found all of them. There are several links to the literature at the OEIS site. It's unlikely that there is a wholly elementary proof, as elliptic curves are involve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4123915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\{a,b,c\}\subset\mathbb C$ such that $a^2+b^2+c^2=ab +bc +ca$ prove that they lie on the vertices of an equilateral triangle in the complex plane
Exercise 1.11 (from Friendly Approach to Complex Analysis).
If $a$, $b$, $c$ are real numbers such that $a^2+b^2+c^2=ab+bc+ca$, then they must be equal. Indeed, doubling... | Hint.
We have $(b-a)^2+(c-a)^2 = (b-a)(c-a)$ so
$$
c-a = (b-a)\left(\frac{1\pm\sqrt{3}i}{2}\right)
$$
then
$|b-a| = |c-a|$ and $\angle(b-a,c-a) = \pm\frac{\pi}{3}$
NOTE
The property $a^2+b^2+c^2 - a b- a c- b c = 0$ is translation invariant.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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How does this vandermonde identity proof works? Recently I have taken a look at this vandermonde identity proof Inductive Proof for Vandermonde's Identity?.
\begin{align*}
\binom{m + (n+1)}r &= \binom{m+n}r + \binom{m+n}{r-1}\\
&= \sum_{k=0}^r \binom mk\binom n{r-k} + \sum_{k=0}^{r-1} \binom mk\binom{n}{r-1-k}... | Do you mean this step?
$\displaystyle\sum_{k=0}^r {m \choose k}{n \choose r-k} = {m \choose r} + \sum_{k=0}^{r-1}{m\choose k}{n\choose r-k}$
Also for your second question, as $0! \equiv 1$ we'll have ${n+1\choose 0} = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4131889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove that $\lim_{n\to\infty} \frac{a_n}{a_{n+1}} = p$
Given a sequence $\{a_n\}_{n\in \mathbb N}$ of real numbers such that
$$\begin{align}
\lim_{n\to\infty}\frac{a_na_{n+1} - a_{n-1}a_{n+2}}{a_{n+1}^2 - a_na_{n+2}} &= p + q && (1)\\[1mm]
\lim_{n\to\infty} \frac{a_n^2 - a_{n-1}a_{n+1}}{a_{n+1}^2 - a_na_{n+2}} &= pq &... | HINT:
Call
$$
X_n=\frac{a_na_{n+1}-a_{n-1}a_{n+2}}{a_{n+1}^2-a_na_{n+2}}\\
Y_n=\frac{a_{n}^2-a_{n-1}a_{n+1}}{a_{n+1}^2-a_na_{n+2}}
$$
then
$$
\lim_nX_n=p+q\\
\lim_nY_n=pq
$$
thus
$$
\lim_n\frac{X_n}{Y_n}=\frac{p+q}{pq}
$$
but
$$
\frac{X_n}{Y_n}
=\frac{a_na_{n+1}-a_{n-1}a_{n+2}}{a_{n}^2-a_{n-1}a_{n+1}}\;.
$$
Then you ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Evaluation of $\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=2-2\ln(2)$ I came across the following statements
$$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=2-2\ln 2 \qquad \tag{1}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(3 n+1)}=3-\frac{3 \ln 3}{2}-\frac{\pi}{2 \sqrt{3}} \qquad \tag{2}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(4 n+1)}=4-\frac{\... | Here is another method to add to the list. Consider the summation
$$\sum\limits_{n=1}^{\infty }{\frac{1}{n\left( 2n+1 \right)}}$$
Note that
$$\psi \left( -z \right)+\gamma \underset{z\to n}{\mathop{=}}\,\frac{1}{z-n}+{{H}_{n}}+\sum\limits_{k=1}^{\infty }{\left( {{\left( -1 \right)}^{k}}H_{n}^{k+1}-\zeta \left( k+1 \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4145392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
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Definite integral involving the exponential of an algebraic function We define the following integral for $\omega>0$ and $\lambda\neq0$ such that $\omega^2>\lambda$:
$$\begin{align}
f(\omega,\lambda)
&:= \int_\omega^\infty e^{-x\sqrt{x^2-\lambda}} \; \varphi(x) \, \text{d}x
\\[6pt]
\tag{1}
&=\frac{1}{\sqrt{2\pi}}\int_\... | This is not a closed form solution.
As you already found it, I did not find anything for such integrands.
A possible solution would be a Taylor expansion around $\lambda=0$. This would give for the integrand
$$e^{-\sqrt{x^4-\lambda x^2}-\frac{x^2} 2}=e^{-\frac{3 }{2}x^2}\Bigg[1+\frac 12 \lambda+\sum_{n=2}^\infty\frac {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to prove that a Local minimum is Absolute minimum in $R^3$ Just trying to solve this question:
$f(x,y,z) = x^2 + y^2 +3z^2 -xy +2xz+ yz$.
Found the only critical point of the function and explain why she is an absolute minimum.
We learn at class how to found that a point is a critical point so i found her and its $... | I don't know if there's some immediate way to know that the local minimum is a global minimum, and I would like to see such answers if there are.
I would do some algebra to rewrite $f$ as:
$f(x,y,z) = 0.5(x-y)^2 + 0.5(y+z)^2 +(x+z)^2-0.5x^2+1.5z^2$
Then, the only potentially problematic term is $-0.5x^2$.
But minimizin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4150553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $X \sim E(\lambda)$, find the density function of $Y = X^2+1$.
If $X \sim E(\lambda)$, find the density function of $Y = X^2+1$.
My try: $$F_Y(x) = P(X^2+1 < x) = \mathrm P(-\sqrt{x-1} < X < \sqrt{x-1}) = \mathrm P(X < \sqrt{x-1}) -\mathrm P(X < -\sqrt{x-1})$$
Now since $$F_X(x) = \begin{cases}\ 1-e^{-\lambda x},... | \begin{align}
F_Y(x) & = \Pr(X^2+1\le x) \\[8pt]
& = \begin{cases} \Pr(X\le \sqrt{x-1} ) & \text{if } x\ge1, \\
0 & \text{if } x<1
\end{cases} \\[8pt]
& = \begin{cases}
1 - e^{-\lambda\sqrt{x-1}} & \text{if } x\ge 1 \\
0 & \text{if } x<1 \quad \text{(not “if $\sqrt{x-1} < 0$”)}
\end{cases}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4154133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Finding which complex numbers satisfy $\left|z + 1\right| + \left|z - 1\right| = 4$ This problem is from the book Complex Analysis with Applications by Asmar. Currently I am stuck at, well, everything, as I don't know know how to manipulate the expression $\left|z + 1\right| + \left|z - 1\right| = 4$ into anything mean... | Let $z = x + iy$, then $|z+1| = \sqrt{(x+1)^2 + y^2}$ and $|z-1| = \sqrt{(x-1)^2 + y^2}$.
$|z+1| = 4 - |z-1|$
$(x+1)^2 + y^2 = 16 + (x-1)^2 + y^2 - 8\sqrt{(x-1)^2 + y^2}$
$x - 4 = -2\sqrt{(x-1)^2 + y^2}$
$x^2 + 16 - 8x = 4(x-1)^2 + 4y^2$
$12 = 4y^2 + 3x^2$
$\frac{x^2}{4} + \frac{y^2}{3} = 1$ which is formula of ellipse... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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If $(\sin^{-1}x)^3 + (\sin^{-1}y)^3 + 3\sin^{-1}x\sin^{-1}y = 1$ which of the following would be true If $(\sin^{-1}x)^3 + (\sin^{-1}y)^3 + 3\sin^{-1}x\sin^{-1}y = 1$ which of the following is true:
A) $\frac{\sin ^{-1} x}{1+\cos ^{-1} y}=\frac{\pi}{2}$
B) $\frac{x+y}{\sin 1}=-2$
C) $\sin ^{-1} x+\sin ^{-1} y=1$
D) $\... | Consider the Gauss identity
\begin{align}
a^3+b^3+c^3-3abc &= (a+b+c)(a^2+b^2+c^2-ab-ac-bc) \\
& = (a+b+c) \frac{1}{2} ((a-b)^2+(a-c)^2+(b-c)^2).
\end{align}
If $a^3+b^3+c^3-3abc=0$, plugin $a=\sin^{-1}(x)$, $b=\sin^{-1}, c=-1$ we have the hypothesis and consequently two cases.
Case 1 $a+b+c=0$ then
$\sin^{-1}(x)+ \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding Taylor's series of the function: $\frac{e^{a \sin^{-1}x}}{\sqrt{1-x^2}}$ Show that $$\frac{e^{a \sin^{-1}x}}{\sqrt{1-x^2}}=1+\frac{ax}{1!}+\frac{(a^2+1^2)x^2}{2!}+\frac{a(a^2+2^2)x^3}{3!}+\frac{(a^2+1^2)(a^2+3^2)x^4}{4!}+\cdots$$
My attempt:
I integrated the function and got $\frac{e^{a \sin^{-1}x}}{a}$ then I ... | Hint
You face a problem of composition of series.
Start with
$$\sin ^{-1}(x)=x+\frac{x^3}{6}+\frac{3 x^5}{40}+O\left(x^7\right)$$
$$a\sin ^{-1}(x)=ax+a\frac {x^3}{6}+a\frac{3 x^5}{40}+O\left(x^7\right)$$ Now, use
$$e^{a \sin ^{-1}(x)}=\exp\Big[ax+a\frac {x^3}{6}+a\frac{3 x^5}{40}+O\left(x^7\right) \Big]$$ when done, wo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 2
} |
synthetic divide polynomial $(2x^3+7x^2-13x-3) \div (2x-3)$ I need to figure out how synthetic division works. The problem is $$ (2x^3+7x^2-13X-3) \div (2x-3)$$ I can do the long division.
$$\require{enclose}
\begin{array}{r}
x^2+5x+1 \\
(2x-3) \enclose{longdiv}{2x^3+7x^2-13x-3}\\ \underline{-2x^3+3x^2} \phantom... | When you do synthetic division, you're not dividing by $2x-3$, you're dividing by $x - 3/2$.
Note that $2(x-3/2) = 2x-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Use Implicit Differentiation to find $\frac{d^2y}{dx^2}$? Given a system of equation,
\begin{align*}
x &= t^2 + 2t \\
y &= 3t^4 + 4t^3
\end{align*}
I want to find $\frac{d^2 y}{dx^2}$ at $(x,y) = (8, 80)$. Then, $\partial_x(y) = \frac{d y}{dt} \frac{dt}{dx}$. By chain rule,
\begin{align*}
\partial_x^2(y) &= \partial_x ... | You do not have to use implicit differentiation. $x$ and $y$ are explicitly given in terms of parameter $t$.
$\displaystyle \frac{dy}{dt} = 12t^3 + 12t^2, \frac{dx}{dt} = 2t + 2$.
From the above, and chain rule, you get: $\displaystyle \frac{dy}{dx} = \frac{dy}{dy} \cdot \frac{dt}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4166548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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$f(f(x))=a^3\left(x^2-(2+b)x+2b-\frac2a\right)\left(x^2-(2+b)x+2b-\frac ba\right)$, $a\ne0$ has exactly one real zeroes $5$.
Let $f(x)=a(x-2)(x-b)$, where $a,b\in R$ and $a\ne0$. Also, $f(f(x))=a^3\left(x^2-(2+b)x+2b-\frac2a\right)\left(x^2-(2+b)x+2b-\frac ba\right)$, $a\ne0$ has exactly one real zeroes $5$. Find the ... | If $a > 0$, then there are two solutions to $a(x-2)(x-b) = 2$, and thus $f(f(x))$ has more than one zero. So we know $a < 0$.
Since $f(f(x))$ is a downward-opening quartic with only one zero, it takes on no positive values. Thus, the maximum of $f$ must be no larger than its smallest zero. Additionally, since $f(f(x))$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4167419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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On $\int_0^{\pi} \operatorname{Si}^n(x) \ \mathrm{d}x $ The Sine Integral, $\operatorname{Si}(x)$, is defined as $$\operatorname{Si}(x)=\int_0^x \frac{\sin t}{t} \ \mathrm{d}t $$
It follows that $$\int_0^{\pi} \operatorname{Si}(x) \ \mathrm{d}x = \bigg[x \operatorname{Si}(x) \bigg]_0^{\pi}-\int_0^{\pi} \sin x \ \mathrm... | Here is an observation that seems to hint a negative answer to OP's question.
Considering that $\pi$ does not play a particular role in the special values for $\operatorname{Si}(\cdot)$, the question would be almost equivalent to investigating the antiderivatives of $\operatorname{Si}(x)^k$'s. Now let us write
$$F(x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4169011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 0
} |
$\lim_{n \to \infty} \sum_{k=0}^n \frac{\sqrt{2n^2+kn-k^2}}{n^2}$ the problem of $k = 0$. Am I doing it right? I would like to calculate:
$$\lim_{n \to \infty} \sum_{k=0}^n \frac{\sqrt{2n^2+kn-k^2}}{n^2}$$
We have that:
$$\lim_{n \to \infty} \sum_{k=0}^n \frac{\sqrt{2n^2+kn-k^2}}{n^2} = \lim_{n \to \infty} \frac{1}{n}\... | No.
$$\sum_{k=0}^n\frac{\sqrt{2n^2+kn-k^2}}{n^2}=$$
$$\frac{\sqrt{2}}{n}+\sum_{k=1}^n\frac{\sqrt{2n^2+kn-k^2}}{n^2}$$
the first term goes to zero and the second to
$$\int_0^1\sqrt{2+x-x^2}dx=$$
$$\int_0^1\sqrt{\frac 94-(x-\frac 12)^2}dx$$
now put
$$x-\frac 12=\frac 32\sin(t)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4169545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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The quadratic equation whose roots are $\sec^2\theta$ and $\csc^2\theta$ can be
The quadratic equation whose roots are $\sec^2\theta$ and $\csc^2\theta$ can be
*
*A) $x^2-2x+2=0$
*B) $x^2-5x+5=0$
*C) $x^2-7x+7=0$
*D) $x^2-9x+9=0$
Method $1$:$$\sec^2\theta+\csc^2\theta=\frac1{\cos^2\theta}+\frac1{\sin^2\theta}\\=... | It seems that what we are asked to notice is that
$$ \sec^2 x \ + \ \csc^2 x \ \ = \ \ \frac{1}{\cos^2 x} \ + \ \frac{1}{\sin^2 x} \ \ = \ \ \frac{\sin^2 x \ + \ \cos^2 x}{\cos^2 x·\sin^2 x} \ \ = \ \ \sec^2 x · \csc^2 x \ \ , $$
a perhaps not overly-familiar trigonometry identity. This is why all of the polynomial... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the solution of equation $z^2 - 2 \bar{z} = 3$? my question is how to solve the equation $z^2 - 2 \bar{z} = 3$, where $z \in \mathbb{C}$?
I know that polar form of any complex number is $z = r \cos(\varphi) + i \cdot r \sin (\varphi)$. And similarly $\bar{z} = r \cos(\varphi) - i \cdot r \sin(\varphi)$. But now... | In a problem like this, I prefer Cartesian coordinates, where $z = x + iy.$
$z^2 - 2 \bar{z} = 3$
Therefore $(x + iy)^2 - 2(x - iy) = 3 \implies $
$(x^2 - y^2 - 2x - 3) + i(2xy + 2y) = 0 + i(0)$.
Therefore, $(2y)(x + 1) = 0$.
Therefore, either $y = 0$ or $x = -1$.
$y = 0 \implies (x^2 - 2x - 3) = 0 \implies (x - 3)(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4176452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $a$ and $b$ such that $x^3-4ax^2-8bx-3+a+b$ has a triple root $$P(x)=x^3-4ax^2-8bx-3+a+b$$
NOTE: a & b are both real and the solution is not necessarily unique
So far i've done this:
If $P(x)$ is a $3$rd degree polynomial and has a triple root, then it has the form:
$$
P(x)=(x-n)^3=x^3-3x^2 n+3xn^2-n^3.
$$
Then... | Let $(x−k)^3=x3−4ax^2+(−8b)x−(3−a−b)$
$⟹x^3−(3k)x^2+(3k^2)x−k^3=x^3−4ax^2+(−8b)x−(3−a−b)$
$⟹x^3−(3k)x^2+(3k^2)x−k^3=x^3−4ax^2+(−8b)x−(3−a−b)$
Comparing like terms
$3k=4a$
$⟹a=3k/4 $
$3k^2=−8b$
$⟹b=−3k^2/8$
$k^3=3−a−b$
$⟹k^3=3−3k/4+3k^2/8$
$⟹8k^3−3k^2+6k−24=0 $
Solving this gives $k≈1.38935$
$a≈1.04201$
$b≈−0.723859$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I prove this inequality $3(a+b+c+1)\ge 4 \left( \sqrt{\frac{a^2+1}{a+1}}+\sqrt{\frac{b^2+1}{b+1}}+\sqrt{\frac{c^2+1}{c+1}} \right)$? If $a,b,c\gt 0$ and $abc=1$, how do I prove the following inequality $3(a+b+c+1)\ge 4 \left( \sqrt{\frac{a^2+1}{a+1}}+\sqrt{\frac{b^2+1}{b+1}}+\sqrt{\frac{c^2+1}{c+1}} \right)$?
My... | Remarks: Here is a trick for the inequality
of the form $f(a) + f(b) + f(c) \ge 0$ under the constraints $a, b, c > 0$ and $abc = 1$.
Let $F(x) = f(x) + m \ln x$. If we can find
an appropriate $m$ such that $F(x)\ge 0$ for all $x > 0$, then we have $F(a) + F(b) + F(c)
= f(a) + f(b) + f(c) + m\ln (abc) \ge 0$
and we a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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If $a_n=\sqrt{1+\sqrt{2+\cdots\sqrt{n}}}$ and $\lim\limits_{n\to\infty}a_n=\ell$, prove $\lim_{n\to\infty}[(\ell-a_n)^{1/n}n^{1/2}]=\frac{\sqrt e}2$ $$a_n=\sqrt{1+\sqrt{2+\cdots\sqrt{n}}}$$
We can prove that $\{a_n\}$ is convergent
(using mathematical induction, $\sqrt {k+\sqrt{k+1+\cdots\sqrt{n}}}\leq k-1, for \ k\geq... | Let's denote for $x_i\ge 0$:
$$ [x_1, x_2,\cdots ,x_n] = \sqrt{x_1 + \sqrt{x_2 + \cdots \sqrt{x_n}}}$$
Let
$$c_{k,n} = [k, k+1, \cdots, n] \text{, for } n \ge k$$
from your inequality proved by induction we deduce:
$$c_{k,n} = [k + c_{k+1,n}] \le \sqrt{2k} \text{, for } n \ge k \ge 3$$
And injecting this enquality agai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4185099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$\frac{1-a^3}{a}=\frac{1-b^3}{b}=\frac{1-c^3}{c}, a \neq b \neq c$; What is $a^3+b^3+c^3$?
$\frac{1-a^3}{a}=\frac{1-b^3}{b}=\frac{1-c^3}{c}\\a \neq b \neq c\\a^3+b^3+c^3=$
Since $\frac{1-a^3}{a}=\frac{1-b^3}{b}$, $a^3=b^2a-\frac{a}{b}-1$.
$\therefore a^3+b^3+c^3=ab^2+bc^2+ca^2-\frac{ab+bc+ca}{abc}-1$
Since $a^3+b^3+c... | $a, b, c$ are roots of $$\frac{1-x^3}{x}=m$$
$$x^3+mx-1=0$$
Hence, we know that $a+b+c=0$ and $abc=1$ from Vieta's formula. Now you can use the identity that you stated to solve for $a^3+b^3+c^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4188295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Finding $a$ and $b$ such that the graph of $y=-2\sin3x+1$ passes through $(-\frac\pi{12},a)$ and $(b,1)$ Please look at the attached image through the link. I already figured out $b$, which I'm pretty sure is $2\pi/3$. However, I'm confused on figuring out $a$. This is what I tried:
\begin{align*}
y & = -2\sin(3x)+1\\
... | You have the correct value for $a$. Perhaps it was marked incorrect if you entered your approximation rather than the exact value of $\sqrt{2}+1$.
For $b$, you can simply let $y=1$ and solve:
$$y=-2\sin(3x)+1$$
$$1=-2\sin(3b)+1$$
$$-2\sin(3b)=0$$
$$\sin(3b)=0$$
In your diagram, it looks like $\frac{\pi}{4}<b<\frac{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Showing $\int_{0}^{2\pi}\cos(x)\cos(2x)\cos(3x)\,dx = \frac\pi2$ An integral from MIT Integration Bee:
Show that
$$I = \int_{0}^{2\pi}\cos(x)\cos(2x)
\cos(3x)\,dx = \frac\pi2$$
This integral appeared in the 2019 paper. Below is my own solution:
$$\begin{align}
I &= \int_{0}^{2\pi}\cos(x)\cos(2x)(\cos x\cos2x-\sin x\s... | Answer inspired by Aaron's answer.
Required formulas from handbook:
$\cos(2x) = 2\cos^2(x) - 1.$
$\cos(3x) = 4\cos^3(x) - 3\cos(x).$
$\cos^2(x) = \frac{1}{2}[\cos(2x) + 1].$
$\cos^4(x) = \frac{1}{8}[\cos(4x) + 4\cos(2x) + 3).$
$\cos^6(x) = \frac{1}{32}[\cos(6x) + 6\cos(4x) + 15\cos(2x) + 10].$
The first thing to not... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 5
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Given $\triangle ABC$ with $m\angle A = 90^{\circ}$. $D$ is midpoint of $BC$ and $F$ is midpoint of $AB$. Given $\triangle ABC$ with $\angle A = 90^{\circ}$. $D$ is midpoint of $BC$ and $F$ is midpoint of $AB$. Points $E$ and $G$ lie on the line $AB$ such that $AE:EF = 1:1$ and $FG:GB = 1:1$ as shown on the picture. Wh... | Construct $GX\parallel AD$ ($X\in BC$). We can now use the intercept theorem and see that $AG:GB=DX:XB=4:1$. Let $XB=x$ and so $DX=4x$. We therefore have $DX+XB=5x=DB=CD$ and from here we observe $CD:DX=5:4$. However, $CD:DX=CR:RG$ by the intercept theorem and thus $CG:CR=9:5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove or disprove $\ln{(1+x^2)}\cdot\ln{(1+y^2)}\ge \ln^2{(1+xy)}$ let $x,y>0$.prove or disprove
$$\ln{(1+x^2)}\cdot\ln{(1+y^2)}\ge \ln^2{(1+xy)}$$
By this inequality it seem Cauchy-Schwarz inequality
$$(1+x^2)(1+y^2)\ge (1+xy)^2$$
But this is Log function.so How to prove it? Thanks
| It seems to me that the inequality is not correct.
Putting $y=\frac 1x$ we get,
$$\ln{\left(1+x^2\right)}\times\ln{\left(1+\frac {1}{x^2}\right)}\ge \ln^2{2}$$
But,
$$\begin{align}\lim_{x\to\infty}\left(\ln {\left(1+x^2\right)}\times \ln {\left(1+\frac {1}{x^2}\right)}\right)=0<\ln ^2 2.\end{align}$$
Evaluation of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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A question about solving quotients all. I ran into this problem and was wondering where my logic failed. I was solving an absolute value involving a quotient and went about it the following way:
$$
\begin{eqnarray}
\lvert \frac{x+1}{x-2} \rvert &<& 3\\
\Rightarrow \frac{x+1}{x-2} &>& -3 \hspace{1mm}and\hspace{1mm}\frac... | In the first method, when you go from $\dfrac{x+1}{x-2}\gt -3$ to $x+1\gt -3(x-2)$, what you're doing is multiplying both sides of the inequality by $x-2$ and you assume that $x-2\gt 0$ when doing that because otherwise the inequality sign flips to $\le$
So, assuming $x\gt 2$, you get from that inequality that $x\gt 5/... | {
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"url": "https://math.stackexchange.com/questions/4192982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Limit of sequence with greatest integer function Calculate $\lim \limits_{x \to 0} \ x^2\Big(1+2+3+....+\Big[\frac{1}{\vert x \vert}\Big]\Big)$, where [.] represents greatest integer function.
Since we know that $x-1\lt [x]\leq x,$ I tried squeezing the sequence between $\ x^2\Big(1+2+3+....+\ \frac{1}{\vert x \vert} -... | The sum of the first $n$ integers is $\frac {n(n+1)}2$. Thus, with $n=\Big[\frac{1}{\vert x \vert}\Big]$,
$$x^2\Big(1+2+3+....+\Big[\frac{1}{\vert x \vert}\Big]\Big) = x^2\frac {n(n+1)}2$$
Since, by definition of $n$, we have $n\leq \frac 1 x < n+1$
$$ x^2\frac{\left(\frac 1 x - 1\right) \frac 1 x}2< x^2\Big(1+2+3+....... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4194386",
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"source": "stackexchange",
"question_score": "1",
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How to find the maximum value of $x\cos^{-1}\left(x\right)$? I wanted to find the maximum value of $x\cos^{-1}\left(x\right)$ through differentiation, but upon on differentiation I get $$f'(x) = \arccos\left(x\right) - \frac{x}{\sqrt{1-x^2}}$$ to find the maximum I equated the function to zero: $$\arccos\left(x\rig... | \begin{align*}
f(x) &=xcos^{-1}(x) \\
f'(x) &=cos^{-1}(x) - \frac{x}{\sqrt{1-x^{2}}} \\
0 &=cos^{-1}(x) - \frac{x}{\sqrt{1-x^{2}}} \\
\frac{x}{\sqrt{1-x^{2}}} &=cos^{-1}(x) \\
cos(\frac{x}{\sqrt{1-x^{2}}}) &= x \\
\end{align*}
Now substitute $x = cos\alpha$:
\begin{align*}
cos(\frac{cos\alpha}{\sqrt{1-(cos\alpha)^{2}}}... | {
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How to evaluate $\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}$?
Evaluate $$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}.$$
I solved the problem with the Taylor series expansion of $\cos x$. Here is my solution:
$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}\\ =\lim\limits_{x\to 0}\frac{1-\{1-\frac{(7x)^2}{2!}+\frac{(7x)^4}{... | Multiplying top and bottom by $1 + \cos 7x$:
$$ \lim_{x\to 0}\frac{1-\cos 7x}{3x^2} = \lim_{x\to 0}\frac{(\sin 7x)^2}{3x^2} \frac{1}{1 + \cos 7x} = \lim_{x\to 0}\frac{(\sin 7x)^2}{(7x)^2} \frac{49/3}{1 + \cos 7x} = 1 \cdot \frac{49/3}{2} = \boxed{\frac{49}{6}}.$$
| {
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"question_score": "6",
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"answer_id": 5
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In a right triangle, the perimeter is equal to 30. How many integer values can the hypotenuse take? (Answer:2)
I did:
$a+b+h = 30\rightarrow a+b = h-30\\
a^2+b^2 = h^2 \rightarrow h = \sqrt{a^2+b^2}\\
a-b<h<a+b \rightarrow a-b < h < \sqrt{a^2+b^2}-30...\\
\text{I didn't find other relationships...}
$
| The sides of a right triangle with integer sides are of the form
$a(m^2-n^2), 2amn, a(m^2+n^2)$
for integer $a, m, n$
with $m > n$.
We therefore want
$p
=2a(m^2+mn)
=2am(m+n)$
where $p=30$ here,
or
$p/2
=am(m+n)$.
For p=30,
possible values for $a$ are
1, 3, 5, 15.
For these,
$m(m+n)
=15, 5, 3, 1
$.
If
$m(m+n)=15$
with ... | {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Calc $P=\int_C \frac{e^{-x^2y}}{\sqrt{( 1+x^2y^4 )^3}}( xy[ 2( 1+x^2y^4 )-y^3 ]dx+x^2[ e^{-x^2y}\sqrt{(1+x^2y^4)^3}+1+x^2y^4-2y^3]dy)$ Calc the integral:
$$P=\int \limits_C \dfrac{e^{-x^2y}}{\sqrt{\left( 1+x^2y^4 \right)^3}}\Bigg( xy\left[
2\left( 1+x^2y^4 \right)-y^3 \right]dx+x^2\left[ e^{-x^2y}\sqrt{\left(1+x^2y^4\... | If you parametrize the half circle as,
$r(t) = (t, \sqrt{1-t^2}), t \in (1,-1)$
$r'(t) = \left(1, - \frac{t}{\sqrt{1-t^2}}\right)$.
So the line integral is, $ \ P = \displaystyle - \int_{-1}^1 \vec F(r(t)) \cdot r'(t) \ dt$
$\vec F(r(t)) \cdot r'(t) = t \ g(t^2)$
The integrand is an odd function and its integral over $... | {
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Prove that $\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}$.
Let $a,b,c$ be positive real numbers. Prove that $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}.$$
I want to prove the inequality... | A more general inequality:
$$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 3(1+r)}{a^2+b^2+c^2+r(a+b+c)}$$
for $0\le r\le\frac{5}{2}$ is proved in Vasile Cirtoaje's "Algebraic Inequalities - Old and New Methods" using Equal-Value Theorem (Section 5.2, Problem 8, page 217).
Althoug... | {
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"question_score": "7",
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Prove that if $a,b,c$ are sides of a triangle and $s$ is the semi-perimeter, then $a^2+b^2+c^2 \geq \dfrac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg)$
Prove that, if $a,b,c$ are the sides of a triangle and $s$ is the semi-perimeter, then $a^2+b^2+c^2 \geq \dfrac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg)$.
What I Tried:- N... | After expanding, the expression is equivalent to $$13(a^3+b^3+c^3)+4(a^2b+b^2c+c^2a)+4(ab^2+bc^2+ca^2)-63abc\geq 0$$
$$\Leftrightarrow 13[a^3+b^3+c^3-3abc]+4[(a^2b+b^2c+c^2a)-3abc]+4[(ab^2+bc^2+ca^2)-3abc]\geq 0,$$ which is clear, as follows from AM-GM.
[EDIT] With more thoughts and less work, it suffices by AM-GM to p... | {
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Simplify $\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}} + \frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$ Question: Simplify $\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}} + \frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$
My Attempt:
On rationalizing both fractions separately and then adding (since denominator becomes $\... | You are right, the correct result for the sum of the two given ratios is $\sqrt{2}$. It is $\frac{\sqrt{6}}{3}$ when you take the difference. Presumably there is a typo in your book.
Revise your work and notice that
$$\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2
=2+\sqrt{3}+2\underbrace{\sqrt{(2+\sqrt{3})(2-\sqrt... | {
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"question_score": "2",
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Is my method correct to answering this question? Is there a quicker method to solve this question?
What is the smallest number that can be written as the sum of three, four and five consecutive numbers?
I encountered this question while doing my Math summer homework. I have tried to make progress on this question.
Su... | If we are talking about a non-negative integer, I have another method to propose.
Sum of three consecutive numbers = $++1++2=3+3$
Sum of four consecutive integers = $++1++2++3=4+6$
Sum of five consecutive integers = $++1++2++3++4=5+10$
Another way of formulating a solution from this step can be as follows:
$y = ... | {
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Calculate the measure of segment AB in triangle rectangle ABC For reference:
Given the triangle $ABC$, straight at $B$. The perpendicular bisector of $AC$ intersects at $P$
with the angle bisector of the outer angle $B$, then $AF \parallel BP$ ($F\in BC$) is drawn.
If $FC$ = $a$, calculate BP(x). (Answer: $\frac{a\sqr... | $BP \parallel AF \implies ∠PBF= ∠BFA=45° \text{ (alternate interior angles)}$
$ \implies ∠BAF=∠BFA \text{ (sum of angles in triangle)} $
$\implies AB=FB \text{ (sides opposite to equal angles in a triangle)}$
$\therefore BC= CF+FB=a+AB \implies AB=BC-a$
Using this, in the equation you got
$(BP - \frac1{ \sqrt{2}}BC... | {
"language": "en",
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"question_score": "9",
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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.