Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to prove Vector Angles?
Any vector $\mathbf{v}$ is a unit vector if $\|\mathbf{v}\| = 1$.
Let $\mathbf{x}$, $\mathbf{y}$, and $\mathbf{z}$ be unit vectors, such
that $\mathbf{x} + \mathbf{y} + \mathbf{z} = \mathbf{0}$. Show that
the angle between any two of these vectors is $120^\circ$.
I know how to prove th... | We have that $$(\mathbf{x} + \mathbf {y} + \mathbf {z})^2 =0$$ $$\Rightarrow \mathbf {x}^2 + \mathbf {y}^2 + \mathbf {z}^2 +2 (\mathbf {x}\cdot \mathbf {y} + \mathbf {y}\cdot \mathbf {z} + \mathbf {z}\cdot \mathbf {x}) =0$$ $$\Rightarrow (\mathbf {x}\cdot \mathbf {y} + \mathbf {y}\cdot \mathbf {z} + \mathbf {z}\cdot \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2132975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Showing matrices in $SU(2)$ are of form $\begin{pmatrix} a & -b^* \\ b & a^*\end{pmatrix}$ Matrices $A$ in the special unitary group $SU(2)$ have determinant $\operatorname{det}(A) = 1$ and satisfy $AA^\dagger = I$.
I want to show that $A$ is of the form $\begin{pmatrix} a & -b^* \\ b & a^*\end{pmatrix}$ with complex n... | Using @Omnomnomnom's suggestion $AA^\dagger =A^\dagger A$, we first obtain the relations
\begin{align}
AA^\dagger: r &= -\frac{su^*}{t^*}\ , \ u= -\frac{tr^*}{s^*} \\
A^\dagger A: r &= -\frac{tu^*}{s^*}\ , \ u= -\frac{sr^*}{t^*} \ .
\end{align}
Noticing the common factor $\frac{-t}{s^*}$ for $r_{A^\dagger A}$ and $u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Evaluate the integral $\int \frac{x^2(x-2)}{(x-1)^2}dx$
Find $$\int \frac{x^2(x-2)}{(x-1)^2} dx .$$
My attempt:
$$\int \frac{x^2(x-2)}{(x-1)^2}dx = \int \frac{x^3-2x^2}{x^2-2x+1}dx $$
By applying polynomial division, it follows that
$$\frac{x^3-2x^2}{x^2-2x+1} = x + \frac{-x}{x^2-2x+1}$$
Hence $$\int \frac{x^3-2x^2}{... | HINT: write your Integrand in the form $$x- \left( x-1 \right) ^{-1}- \left( x-1 \right) ^{-2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
Find the value of $\binom{2000}{2} + \binom{2000}{5} + \binom{2000}{8} + \cdots \binom{2000}{2000}$
Find the value of $\binom{2000}{2} + \binom{2000}{5} + \binom{2000}{8} + \cdots \binom{2000}{2000}$
I've seen many complex proofs. I am looking for an elementary proof. I know the fact that $\binom{2000}{0} + \binom{... | For $n\ge0$ let
$$a_n=\binom n0+\binom n3+\binom n6+\cdots=\sum_{k=0}^\infty\binom n{3k},$$
$$b_n=\binom n1+\binom n4+\binom n7+\cdots=\sum_{k=0}^\infty\binom n{3k+1},$$
$$c_n=\binom n2+\binom n5+\binom n8+\cdots=\sum_{k=0}^\infty\binom n{3k+2};$$
we seek the value of $c_{2000}.$ Observe that
$$a_n+b_n+c_n=2^n$$
and, f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
If $m$th term and $n$th term of arithmetic sequence are $1/n$ and $1/m$ then the sum of the first $mn$ terms of the sequence is $(mn+1)/2$
If $m$th term and $n$th term of arithmetic sequence are $1/n$ and $1/m$ respectively then prove that the sum of the first $mn$ terms of the sequence is $(mn+1)/2$.
My Attempt ;
$$... | I am assuming equations created by you as equation (1) and (2) respectively.
From equation (1),
$a = \frac 1n - (m - 1)d$
Put value of a in equation (2),
$\frac 1n - (m - 1)d + (n - 1)d = \frac 1m$
$\implies (-m + 1 + n - 1)d = \frac 1m - \frac 1n$
$$\implies (n - m)d = \frac {n - m}{mn}$$
$$\implies d = \frac 1{mn}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2138109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
In right triangle $ABC$ ($\angle A=90$), $E$ is a point on $AC$.Find $AE$ given that...
In right triangle $ABC$ ($\angle A=90$), $AD$ is a height and $E$ is a point on $AC$ so that $BE=EC$ and $CD=CE=EB=1$. $\color {red} {Without}$ using trigonometric relations find $AE$.
I do have a solution USING trigonometric re... | Here is a proof with only Pythagoreans:
(as I see, Michael Rozenberg used also the altitude theorem of right triangles)
We have 5 eqns in 5 variables AE, AC, BA, BD, AD:
Obviously
$AE = AC - CE = AC - 1;$
and 4 Pythagoreans:
$$
AC^2 = - AB^2 + (BD+DC)^2 = - AB^2 + (BD+1)^2;\\
AC^2 = AD^2 + DC^2 = AD^2 + 1;\\
AD^2 + BD... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2138416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove, by contradiction, that, if $cx^2 + bx + a$ has no rational root, then $ax^2 + bx + c$ has no rational root. Proposition: Suppose that $a$, $b$, and $c$ are real numbers with $c \not = 0$. Prove, by contradiction, that, if $cx^2 + bx + a$ has no rational root, then $ax^2 + bx + c$ has no rational root.
Hypothesis... | Since you have proved that $ap^2+bpq \neq 0$ and $cq^2 \neq 0$, you are very close to the answer.
Notice that $p(ap+bq)=-cq^2$. But $p(ap+bq) \neq 0 \Rightarrow p \neq 0$.
Divide both sides by $p^2$. Then you get $a+b\frac qp=-c\frac {q^2}{p^2}.$
This shows that $cx^2+bx+a=0$ has a rational root $\frac qp$. WHICH is co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve an integral $\int\frac{\cos^3 x}{\sin^3 x+\cos^3 x}dx$ Solve an integral $$\int\frac{\cos^3 x}{\sin^3 x+\cos^3 x}dx$$
I tried to divide the numerator and denominator by $\cos^4 x$ to get $\sec x$ function but the term ${\sin^3 x}/{\cos^4 x}$ gives $\tan^2 x\sec^2 x\sin x$. How to get rid of $\sin x$ term?
| Let $$I = \int \frac{\cos^3x}{\sin^3x+\cos^3x} dx$$ $$I_1 = \int\frac{\sin^3x+\cos^3x}{\sin^3x+cos^3x}dx = x + C$$ and $$I_2 = \int\frac{\cos^3x-\sin^3x}{\sin^3x+\cos^3x}dx$$Then $$I = \frac{I_1 + I_2}{2}$$
$$I_2 = \int\frac{(\cos x-\sin x)(1+\frac{\sin2x}{2})}{(\sin x+\cos x)(1-\frac{\sin 2x}{2})}dx$$
Now substitute $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
A series involve combination I want find another Idea to find sum of $\left(\begin{array}{c}n+3\\ 3\end{array}\right)$ from $n=1 ,to,n=47$
or $$\sum_{n=1}^{47}\left(\begin{array}{c}n+3\\ 3\end{array}\right)=?$$ I do it first by turn $\left(\begin{array}{c}n+3\\ 3\end{array}\right)$ to $\dfrac{(n+3)(n+2)(n+1)}{3!}=\dfr... | By the well known hockey stick identity
$$ \sum_{n=0}^{47}\binom{n+3}{3} = \binom{47+3+1}{3+1} $$
and the problem is trivial from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$
Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$.
I expanded the brackets and applied AM-GM on all of the eight terms to get :
$$\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big) \geq 3\... | More way.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that
$$(3u^2+2a^2)(3+2b^2)(3+2c^2)\geq125u^6$$ or
$$27u^6+18u^4(9u^2-6v^2)+12u^2(9v^4-6uw^3)+8w^6\geq125u^6$$ or $f(w^3)\geq0$, where
$$f(w^3)=2w^6-18u^3w^3+16u^6-27u^4v^2+27u^2v^4.$$
But $f'(w^3)=2w^3-18u^3<0$, which says that $f$ is a de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
$\tan^{-1}(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots)+ \cot^{-1}(x^4-\frac{x^8}{2}+\frac{x^{12}}{4}-\cdots)=\pi/ 2$
Find $x$ with $0<|x|<2$ such that
$$\tan^{-1}(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots)+ \cot^{-1}(x^4-\frac{x^8}{2}+\frac{x^{12}}{4}-\cdots)=\pi/ 2$$
My try:
$(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots)=\alp... | What I can come up with is as below
$\tan^{-1}(\alpha) =A \implies \tan(A)=\alpha$
$\cot^{-1}(\beta) =B \implies \cot(B)=\beta$
Also, we have
$A+B=\frac{\pi}{2}$
Add $\frac{\pi}{2}$ to both sides
$A+B+\frac{\pi}{2}=\pi$
Take $\tan$ from both sides and use the famous formula of $\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Compute $\lim\limits_{x \to 0} \frac{ \sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x }}{x}$
Compute $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x }}{x}$
Original question was to solve $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+ x } - \sqrt[3]{1-x }}{x}$ and it was solved by adding and subtracting 1 i... | $$\lim _{ x\to 0 }{ \frac { { (1+\sin x) }^{ \frac { 1 }{ 3 } }-{ (1-\sin x) }^{ \frac { 1 }{ 3 } } }{ x } } \frac { \left( { (1+\sin x) }^{ \frac { 2 }{ 3 } }+{ (1+\sin x) }^{ \frac { 1 }{ 3 } }{ (1-\sin x) }^{ \frac { 1 }{ 3 } }+{ (1-\sin x) }^{ \frac { 2 }{ 3 } } \right) }{ \left( { (1+\sin x) }^{ \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
concurrence of three lines in a quadrilateral Prove that the lines joining the midpoints of opposite sides of a quadrilateral and the line joining the midpoints of the diagonals of the quadrilateral are concurrent.
| Let $E=\frac{A+B}{2}$ and $F=\frac{C+D}{2}$ be the midpoints of two opposite sides.
Let $G=\frac{A+D}{2}$ and $H=\frac{C+B}{2}$ be the midpoints of another two opposite sides.
Let $J=\frac{A+C}{2}$ and $K=\frac{B+D}{2}$ be the midpoints of diagonals.
The midpoint of section $EF$ is a point $\frac{E+F}{2}=\frac{A+B+C+D}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $\gcd(x,y)=1$, and $x^2 + y^2$ is a perfect sixth power, then $xy$ is a multiple of $11$ This is a problem that I don't know how to solve:
Let $x, y, z$ integer numbers such that $x$ and $y$ are relatively primes and $x^2+y^2=z^6$ . Show that $x\cdot y$ is a multiple of $11$.
| It's not complete answer
By computer search, it seems to be valid for co-prime $x$ and $y$.
Now, I'm trying to give the general solution of $(x,y,z)$.
\begin{align*}
(a+bi)^3 &= a(a^2-3b^2)+b(3a^2-b^2)i \\[7pt]
(a^2+b^2)^3 &=
\underbrace{a^2(a^2-3b^2)^2}_{\Large{m^2}}+
\underbrace{b^2(3a^2-b^2)^2}_{\Large{n^2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Sum of a series using squeeze theorem How do I find using the Squeeze theorem
$$\lim_{n\to \infty}\sum_{k=1}^n \frac{1}{\sqrt{n^2+k}} \;,$$
using the fact that
$$ \lim_{n\to \infty}\frac{n}{\sqrt{n^2+n}}=1.$$
Thank you very much for your help,
C.G
| Note that
$$\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\leq \frac{n}{\sqrt{n^2+1}}$$
because $\sqrt{n^2+1}\leq\sqrt{n^2+k}$ for $k\geq1$.
On the other hand, by a similar reasoning, using that $\sqrt{n^2+n}\geq\sqrt{n^2+k}$ for $k\geq1$:
$$\frac{n}{\sqrt{n^2+n}}\leq\frac{1}{\sqrt{n^2+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Let $\omega$ be a complex number such that $\omega^7 = 1$ and $\omega \neq 1$. Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Then $\alpha$ and $\beta$ are roots of the quadratic
[x^2 + p... | method is due to Gauss. This book 1875. $x^2 + x + 2.$
gp-pari to check; note a^8 = a
? x = a + a^2 + a^4
%1 = a^4 + a^2 + a
? q = x^2 + x + 2
%2 = a^8 + 2*a^6 + 2*a^5 + 2*a^4 + 2*a^3 + 2*a^2 + a + 2
?
If we switched to one of the real numbers $$ t = \omega + \omega^6, $$ we would have a root of
$$ t^3 + t^2 - 2 t -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Factor $9(a-1)^2 +3(a-1) - 2$ I got the equation $9(a-1)^2 +3(a-1) - 2$ on my homework sheet. I tried to factor it by making $(a-1)=a$ and then factoring as a messy trinomial. But even so, I couldn't seem to get the correct answer; they all seemed incorrect.
Any help would be greatly appreciated.
Thank you so much in... | $x=a-1\\
9x^2+3x-2=0\\
\Delta=9+72=81\\
\sqrt{\Delta}=9\\
x=\frac{-3 \pm 9}{18} = \pm \frac{1}{2}-\frac{1}{6}\\
a=x+1=\pm \frac{1}{2}-\frac{1}{6}+1 = \pm \frac{1}{2}+\frac{5}{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Inequality of arithmetic and geometric mean I did a proof for inequality below, anyone has a other proof?
Let $a$ and $b$ be positive real numbers, and $t$ the parameter. Prove that:
$$a+b\geq 2\sqrt{1-t^2}\sqrt{ab}+(a-b)t$$
| Let $f (t)=Rhs $ of inequality . So differentiating and setting it equal to $0$ gives $\frac {2t}{\sqrt {1-t^2}}=\frac {a-b}{\sqrt {ab}} $ squaring both sides and solving we have $t=\frac {a-b}{a+b} $ thus putting the value of $t $ in original equation and simplifying we get it as $a^2+b^2$ now both $a,b $ are positive... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to find the maximum and minimum value of $\left|(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2\right|$ (where $|z_1|=|z_2|=|z_3|=1$)? How to find the maximum and minimum value of $\left|(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2\right|$ (where $|z_1|=|z_2|=|z_3|=1$ are complex numbers.) ?
My try:
$$\begin{align}\left|(z_1-z_... | Given
$$(z_1-z_2)+(z_2-z_3)+(z_3-z_1)=0$$
and
$$\left|(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2\right|=\\
\left|z_1^2-2z_1z_2+z_2^2+z_2^2-2z_2z_3+z_3^2+z_3^2-2z_3z_1+z_1^2\right|=\\
2\left|z_1^2+z_2^2+z_3^2-z_1z_2-z_2z_3-z_3z_1\right|=\\
2\left|z_1(z_1-z_2)+z_2(z_2-z_3)+z_3(z_3-z_1)\right|=\\
2\left|z_1(z_1-z_2)+z_2(z_2-z_3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
How to show $(a + b)^n \leq a^n + b^n$, where $a, b \geq 0$ and $n \in (0, 1]$? Does anyone happen to know a nice way to show that $(a+b)^n \le a^n+b^n$, where $a,b\geq 0$ and $n \in (0,1]$? I figured integrating might help, but I've been unable to pull my argument full circle. Any suggestions are appreciated :)
| Assume that
$a \ge b$.
If $a = b = 0$,
the result is immediate.
If $a > 0$,
divide
$(a+b)^n \le a^n+b^n
$
by
$a^n$
to get
$(1+b/a)^n \le 1+(b/a)^n
$.
Since
$b \le a$,
$0 \le b/a \le 1$,
so this becomes
$(1+x)^n \le 1+x^n
$
where
$x = b/a$.
Let $f(x)
=1+x^n-(1+x)^n
$.
$f(0) = 0$
and
$f(1)
=2-2^n
\ge 0
$
since
$0 < n \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
A formula for $\sin(\pi/2^n)$ May be this a duplicate, but I did not find any question related.
I found the following formula, but there was no proof of it:
$$2\sin\left(\frac{\pi}{2^{n+1}}\right)=\sqrt{2_1-\sqrt{2_2+\sqrt{2_3+\sqrt{2_4+\cdots\sqrt{2_n}}}}}$$
where
$$2_k=\underbrace{222\cdots222}_{k\text { times}}.$$
(... | Here's my repeated half-angle approach (I know, this is definitely not a great way to deal with it, but still am posting it here. This is my first answer, here in this website, so please bear with me..):
We know
$2\cos^2 \theta =1+\cos 2\theta\implies \cos \theta =\sqrt{\frac{1+\cos 2\theta}{2}}.$
Taking positive sign ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2151405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation : $\tan \theta + \tan 2\theta + \tan 3\theta = \tan \theta \tan 2\theta \tan 3\theta $ I've been having some trouble solving this equation. (The solution in my book is given as $ \frac {n \pi}{3}, n \in Z $)
Here is what I've done
$$\frac {\sin \theta}{\cos \theta} + \frac {\sin 2\theta} {\cos 2\thet... | We have
$$\tan (A+B+C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1-(\tan A \tan B + \tan B \tan C + \tan C \tan A)}$$
and the given condition implies that $\tan (6\theta) = 0$. Thus $\theta = \frac{n\pi}{6}$, $n \in \mathbb{Z}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to take the integral? $\int \frac{x^2-3x+2}{x^2+2x+1}dx$ $$\int \frac{x^2-3x+2}{x^2+2x+1}dx$$
So after all I had
$$ \frac{-5x+1}{(x+1)^2} = \frac{A}{(x+1)} + \frac{B}{(x+1)^2}$$
and of course $$ \int xdx $$
but it is easy to solve, I do not know how to act with devided things, probably solve the system, or is there... | \begin{align*}
\frac{x^2 - 3x + 2}{x^2 + 2x + 1} &= 1 + \frac{1-5x}{(x+1)^2} \\
&= 1 + \frac{A}{x+1} + \frac{B}{(x+1)^2}
\end{align*}
Then
\begin{align*}
\frac{1-5x}{(x+1)^2} &= \frac{A}{x+1} + \frac{B}{(x+1)^2} \\
&= \frac{A(x+1)+B}{(x+1)^2} \\
1-5x &= Ax+(A+B)
\end{align*}
So $A = -5$ and $B=6$.
Therefore
\begin{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factorise an ideal into a product of prime ideals So for the ideal $I = (20+\sqrt{-5})_{R}$ when $K=\mathbb{Q}(\sqrt{-5})$, how do I factorise this into a product of prime ideals. Do you start by taking the norm of $I$ and decomposing it into a product of primes? So
$$N(I) = 405 = 3^{4} \times 5.$$
If so, where do I go... | Recall that the primes $ 3, 5 $ split as $ 3R = \mathfrak p \mathfrak p' $ and $ 5R = (\sqrt{-5})^2 $ in $ R = \mathcal O_K = \mathbf Z[\sqrt{-5}] $. The norm suggests that the ideal $ I = (20 + \sqrt{-5}) $ factors as $ \mathfrak p^i \mathfrak p'^j (\sqrt{-5}) $, where $ i + j = 4 $. It follows upon division that
$$ \... | {
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"timestamp": "2023-03-29T00:00:00",
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Find $\frac{\sum_\limits{k=0}^{6}\csc^2\left(x+\frac{k\pi}{7}\right)}{7\csc^2(7x)}$ Find the value of
$\dfrac{\sum_\limits{k=0}^{6}\csc^2\left(x+\dfrac{k\pi}{7}\right)}{7\csc^2(7x)}$
when $x=\dfrac{\pi}{8}$.
The Hint given is: $n\cot nx=\sum_\limits{k=0}^{n-1}\cot\left(x+\dfrac{k\pi}{n}\right)$
I dont know how it come... | For future reference with this problem being tagged complex-numbers we
show how to evaluate the sum using residues. Suppose we are interested
in
$$S(n) = \sum_{k=0}^{n-1} \csc^2\left(x+\frac{k\pi}{n}\right)
= \sum_{k=0}^{n-1} \frac{2}{1-\cos\left(2x+\frac{2k\pi}{n}\right)}.$$
where we take $x$ to be a real number.
With... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving Saalschutz Theorem I saw this in a pdf, and I'm wondering
Questions:
*
*How do you prove Saalschutz Theorem:
$$_3F_2\left[\begin{array}{c,c}-x,-y,-z\\n+1,-x-y-z-n\end{array}\right]=\dfrac {\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(z+x+n+1)}{\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(x+y+z+n+1)}\tag... | The development of the Saalschütz's identity that I know proceeds along the following path.
Start with the known identity about the sum of the product of three binomials:
$$ \bbox[lightyellow] {
\begin{gathered}
F(m,n,r,s)\quad \left| {\;0 \leqslant \text{integers}\,m,n} \right.\quad = \hfill \\
= \sum\limits_{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2155025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What is the sum of the solutions to $6x^3+7x^2-x-2=0$ What is the easy way to solve the problem?
The sum of the solutions to $6x^3+7x^2-x-2=0$ is:
$$A) \ \frac{1}{6}$$ $$B) \ \frac{1}{3}$$ $$C) \ \frac{-7}{6}$$ $$D) -2$$ $$E) \text{ none of above}$$
| For a cubic equation of the form $ax^3 + bx^2 + cx +d$, the sum of all the roots is given by $\frac{-b}{a}$. Here, $b=7,a=6$ so sum of the roots $= \frac{-7}{6}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a^2+b^2+c^2+d^2+e^2=5$ so $\sum\limits_{cyc}\frac{1}{7-2a}\leq1$.
Let $a$, $b$, $c$, $d$ and $d$ be non-negative numbers such that $a^2+b^2+c^2+d^2+e^2=5$. Prove that:
$$\frac{1}{7-2a}+\frac{1}{7-2b}+\frac{1}{7-2c}+\frac{1}{7-2d}+\frac{1}{7-2e}\leq1$$
The equality occurs also for $a=2$ and $b=c=d=e=\frac{1}{2}$... | We must to prove that $f(a_1)+f(a_2)+f(a_3)+f(a_4)+f(a_5) \le 1$ ,
for non-negative $a_1+a_2+a_3+a_4+a_5=5$
$$f(x)=\dfrac{1}{7-2\sqrt{x}}$$
Since $f''(x)=\dfrac{(7-6\sqrt{x})}{2(2x-7\sqrt{x})^3}$ , we only need to consider the inequality in case $0< a_1=a_2=a_3=a_4=t^2 \le 1 \ , \ a_5=5-4t^2$
Clearly
$g(t)=\dfrac{4}{7... | {
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"timestamp": "2023-03-29T00:00:00",
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Which of the following numbers is greater? Which of the following numbers is greater?
Without using a calculator and logarithm.
$$7^{55} ,5^{72}$$
My try
$$A=\frac{7^{55} }{5^{55}×5^{17}}=\frac{ 7^{55}}{5^{55}}×\frac{1}{5^{17}}= \left(\frac{7}{5}\right)^{55} \left(\frac{1}{5}\right)^{17}$$
What now?
| Note: $7^2<2\cdot 5^2$ and $5>2^2$
$7^{55}<7\cdot 5^{54}\cdot 2^{27}<5^{55}\cdot 2^{28}<5^{69}<5^{72}$ as required
With an extra jink into factors of $3$, we can show $7^{55}<5^{67}$
Extra notes: $3^3>5^2$ and $5^5>3\cdot2^{10}$
$7^{55}<7\cdot 5^{54}\cdot 2^{27}<5^{54}\cdot 2^{30}<5^{52}\cdot 2^{30}\cdot 3^{3}<5^{67}... | {
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What will happen to the roots of $ax^2 + bx + c = 0$ if the $a \to 0$? Exercise:
What will happen to the roots of the quadratic equation
$$ax^2 + bx + c = 0$$
if the coefficient $a$ approaches zero while the coefficients $b$ and $c$ are constant, and $b \neq 0$?
Attempt:
$\lim\limits_{a \to 0}{(ax^2 + bx + c)} =... | If $b=0$, then the roots are (if real), $\pm\sqrt{-c/a}$. If $c\ne0$, both roots tend to infinity (positive and negative).
If $b\ne0$, it's not restrictive to assume $b>0$ (otherwise multiply by $-1$).
If $c=0$, the roots are $0$ and $-b/a$, the latter tending to infinity ($\infty$ if $a$ approaches $0$ from the negat... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality for positive real numbers less than $1$: $8(abcd+1)>(a+1)(b+1)(c+1)(d+1)$ If $a,b,c,d$ are positive real numbers, each less than 1, prove that the following inequality holds:
$$8(abcd+1)>(a+1)(b+1)(c+1)(d+1).$$
I tried using $\text{AM} > \text{GM}$, but I could not prove it.
| First, note that $2^0(a+1) \ge a + 1$.
Then we prove by induction that $2^{k-1}(x_1\dots x_k+1) > \prod_{i=1}^k(x_i + 1)$. For the ease of notation, there I will focus on the case $k = 4$.
$$(d+1)(a+1)(b+1)(c+1)\le 4(d+1)(abc+1)< 4(abcd+abc+1+d)< 8(abcd+1)$$
The last inequality is provided by
$$abc+d< abcd+1 \Leftarrow... | {
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"timestamp": "2023-03-29T00:00:00",
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Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$. Find $(a+b+c)$ Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$.
Find $(a+b+c)$.
I computed the whole product ;If $(a+b+c)=x\implies (1+x)(1+\frac{bc+ca+ab}{abc})=16$. Unable to view how to p... | By $AM \ge GM$ inequality,$$(1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge \left(4\sqrt[4]{abc}\right)\left(4\sqrt[4]{\frac{1}{abc}}\right)=16$$and equality holds when $1=a=b=c$.
| {
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"timestamp": "2023-03-29T00:00:00",
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equation $x^4+ax^3-6x^2+ax+1 = 0$ has two distinct positive roots Finding parameter of $a$ for which equation $x^4+ax^3-6x^2+ax+1 = 0$
has two distinct positive roots
Attempt: writing equation $\displaystyle \bigg(x^2+\frac{1}{x^2}\bigg)+a\bigg(x+\frac{1}{x}\bigg)-6=0\;,$ where $x\neq 0$
So $\displaystyle \bigg(x+\fr... | With $\displaystyle \bigg(x+\frac{1}{x}\bigg)=t$ the roots occur when $t^2+at-8=0$.
$f(x)=x^4+ax^3-6x^2+ax+1 = 0$ shows $f(0)=1>0$ and $\displaystyle \lim_{x\to+\infty} f=+\infty$, which show $f$ has two distinct positive roots if there is $x_0>0$ for which $f(x_0)<0$. Let $\displaystyle \bigg(x_0+\frac{1}{x_0}\bigg)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2163770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the following limit I need to evaluate the following limit, however, in doing so, I let $\sqrt{1+c}=1$ which I came to undertand that it's not valid. My procedure was the following:
$$\lim_{c\to0}\left(-\ln(c)\sqrt{1+c}-\ln\left(\frac{1+\sqrt{1+c}}{1-\sqrt{1+c}}\right)+\ln\left(\frac{1+\sqrt2}{1-\sqrt2}\right... | You can write this as
$$-\ln(c)\sqrt{1+c}-\ln\left(\frac{1+\sqrt{1+c}}{1-\sqrt{1+c}}\right)+\ln\left(\frac{1+\sqrt2}{1-\sqrt2}\right) \\
= -\ln(c)\sqrt{1+c} + \ln c- \ln c -\ln\left(\frac{1+\sqrt{1+c}}{1-\sqrt{1+c}}\right)+\ln\left(\frac{1+\sqrt2}{1-\sqrt2}\right) \\ = \ln c ( 1 - \sqrt{1 + c}) + A(c) $$
where
$$A(c)... | {
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"timestamp": "2023-03-29T00:00:00",
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Let X~geometric (1/3), and let Y=|X-5|. Find the range n and PMF of Y. Let X~geometric (1/3), and let Y=|X-5|. Find the range n and PMF of Y.
Here is my trial
If $x=0$, $P(Y=|0-5|)=P(Y=5)=\left(\frac{2}{3}\right)^5 \frac{1}{3}$
If $x=1$, $P(Y=|1-5|)=P(Y=4)=\left(\frac{2}{3}\right)^4 \frac{1}{3}$
If $x=2$, $P(Y=|2-5|)=... | Range of $Y$ is $(0,\infty)$.
$$P(Y=k) \;=\; (2/3)^{k-1} * (1/3).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Calculate the image and a basis of the image (matrix)
What's the image of the matrix? What's the basis of the image?
$M=\begin{pmatrix}
-1 & 1 & 1\\
-2 & -3 & 6\\ 0 & -1 & 1 \end{pmatrix}$
First transposed the matrix:
$M^{T}=\begin{pmatrix}
-1 & -2 & 0\\
1 & -3 & -1\\
1 & 6 & 1
\end{pmatrix}$
Now we u... | The image of a matrix is the same as its column space. To find column space, you first find the row echelon form of the given matrix (do not transpose it). The definition of row-echelon form is:
*
*Rows with all zero's are below any nonzero rows
*The leading entry in each nonzero row is a one
*All entries below ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $\cos 20^{\circ} + \cos 100^{\circ} + \cos {140^{\circ}} = 0$ Assume $A = \cos 20^{\circ} + \cos 100^{\circ} + \cos 140^{\circ}$ . Prove that value of $A$ is zero.
My try : $A = 2\cos 60^{\circ} \cos 40^{\circ} + \cos 140^{\circ}$ and I'm stuck here
| $\cos 20^{\circ} + \cos 100^{\circ} = \cos (60^{\circ}-40^{\circ}) + \cos (60^{\circ}+40^{\circ}) = \cos 40^{\circ}$ by using the addition formulae for $\cos$.
Then $\cos 140^{\circ} = \cos (180^{\circ} - 40^{\circ}) = -\cos (40^{\circ})$. So $$\cos 20^{\circ} + \cos 100^{\circ} + \cos 140^{\circ} = \cos 40^{\circ} - ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$\lim_{x\to0}\,(a^x+b^x-c^x)^{\frac1x}$
Given $a>b>c>0$, calculate$\displaystyle\,\,\lim_{x\to0}\,(a^x+b^x-c^x)^{\frac1x}\,$
I tried doing some algebraic manipulations and squeeze it, but couldn't get much further.
| $$\lim_{x\to0}(a^x+b^x-c^x)^\frac{1}{x}=[1^\infty]=\exp\lim_{x\to 0}(a^x+b^x-c^x-1)\frac{1}{x}\boxed=\\(a^x+b^x-c^x-1)\frac{1}{x}=(a^x-c^x+b^x-1)\frac{1}{x}=c^x\cdot\frac{\left(\frac{a}{c}\right)^x-1}{x}+\frac{b^x-1}{x}\\ \boxed =\exp \lim_{x\to 0}\left(c^x\cdot\frac{\left(\frac{a}{c}\right)^x-1}{x}+\frac{b^x-1}{x} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2170630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Two different expansions of $\frac{z}{1-z}$ This is exercise 21 of Chapter 1 from Stein and Shakarchi's Complex Analysis.
Show that for $|z|<1$ one has $$\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\cdots +\frac{z^{2^n}}{1-z^{2^{n+1}}}+\cdots =\frac{z}{1-z}$$and
$$\frac{z}{1+z}+\frac{2z^2}{1+z^2}+\cdots \frac{2^k z^{2^k}}{1+z^{... | Since minimalrho has explained how to proceed with the given hint, I'll give an alternative method. The $k$th summand of the first series can be written
$$\frac{z^{2^k}}{1 - z^{2^{k}}} - \frac{z^{2^{k+1}}}{1-z^{2^{k+1}}}$$
and the $k$th summand of the second series can be written
$$\frac{2^kz^{2^k}}{1 - z^{2^k}} - \fr... | {
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"timestamp": "2023-03-29T00:00:00",
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Find all possible positive integers $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $. Proof explanation Question: Find all possible positive integers $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $.
Solution: Let's suppose that $3^{n-1}+5^{n-1}\mid 3^n+5^n$, so there is some positive integer $k$ such that $3^n+5^n=... | Perhaps an easier way to look at it: the equation
$$3^n+5^n=k(3^{n-1}+5^{n-1})$$
can be rewritten
$$5^{n-1}(5-k)=3^{n-1}(k-3)\ .$$
Therefore $5-k$ and $k-3$ must have the same sign. They can't both be negative as then $k$ would be simultaneously less than $3$ and greater than $5$, and they clearly can't both be zero, ... | {
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"timestamp": "2023-03-29T00:00:00",
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Solution of a trigonometric equation involving double sines,cosines What is the sum of all the solutions of $$\sin^2 (2\sin (x-\frac {\pi}{6}))+\sec^2 (x-\frac {\pi}{2}\tan^2 (x))=1$$ in $[0-4\pi]$ writing $\sec (..)=\frac {1}{\cos (..)} $ .. and rearranging we have $\cos (a).\cos (b)=1$ now as $\cos $ is $-1\leq \cos^... | We have $$\sin^2\left(2\sin\left(x-\dfrac\pi6\right)\right)+\tan^2\left(x-\dfrac\pi2\tan^2x\right)=0$$
For real $x,$
$\sin\left(2\sin\left(x-\dfrac\pi6\right)\right)=\tan\left(x-\dfrac\pi2\tan^2x\right)=0$
$\implies 2\sin\left(x-\dfrac\pi6\right)=m\pi\ \ \ \ (1)$ where $m$ is any integer
Now as $-1\le\sin y\le1$
$-1\le... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A hard integral Looking for a solution for an integral:
$$I(k)=\int_0^{\infty } \frac{e^{-\frac{(\log (u)-k)^2}{2 s^2}}}{\sqrt{2 \pi } s \left(1+u\right)} \, du .$$
So far I tried substitutions and by parts to no avail.
| Here is a start: $I(0) = \frac{1}{2}$
Proof:
$$I(0) = \int\limits_0^\infty \frac{\exp\left[-\frac{(\log u)^2}{2s^2}\right]}{\sqrt{2\pi} s (1+u)} \rm{d}u$$
Put $\log u = x$
\begin{align}
I(0) &= \int\limits_{-\infty}^\infty \frac{\exp\left[-\frac{x^2}{2s^2}\right]}{\sqrt{2\pi} s} \frac{e^x}{1+e^x} \rm{d}x \\
&= \int\lim... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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Substitution or comparison? For the question,
$X^4 + \frac{9}{X^4} = (X^2 - \frac{a}{X^2})^2 + b$ I have to find the values of $a$ and $b$. I tried two solutions both included the expansion of brackets.
By comparing terms I obtained $3$ for $a$ and $6$ for $b$.
However when I tried to substitute using $b$, I obtained ... | $X^4 + \frac{9}{X^4} = (X^2 - \frac{a}{X^2})^2 + b$
$x^4 + \frac {9}{x^4} = x^4 + \frac {a^2}{x^4} + b - 2a$
So by comparing $a^2 = 9$ and $b-2a = 0$.
So $a = 3$ or $a = -3$ and $b = 6$ or $b = -6$.
I'm not sure what you mean by substituting? Do you mean picking an arbitrary value for $x$ and getting two equations fo... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Does $\int _0^{\infty }\:\frac{1}{1+x^2\left(\sin x\right)^2}\ \operatorname dx$ converge? I have been trying to prove the following integral:
$$\int _0^{\infty }\:\frac{1}{1+x^2\left(\sin x\right)^2}\ dx$$
diverges (please correct me if I am mistaken).
I have tried to use different comparison tests (as this is an ... | The idea is to bound the integral below on intervals where $\displaystyle \frac{1}{1+x^2\left(\sin x\right)^2}$ has spikes, that is to say, it suffices to find some $\varepsilon_k$ such that $$\sum_{k\geq1}\int_{k\pi -\varepsilon_k}^{k\pi +\varepsilon_k}\frac{1}{1+x^2\left(\sin x\right)^2}dx$$ diverges.
On each of thes... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Minimize $A=\frac{1+2^{x+y}}{1+4^x}+\frac{1+2^{x+y}}{1+4^y}$ For $a,b>0$. Minimize $$A=\frac{1+2^{x+y}}{1+4^x}+\frac{1+2^{x+y}}{1+4^y}$$
i think we let $2^x=a;2^y=b$
Hence $A=\frac{1+ab}{1+a^2}+\frac{1+ab}{1+b^2}$
We need pro $A\geq 2$(Wolfram Alpha) but $x,y$ is a very odd number and i can't find how to prove it $\g... | we have to prove that $$(1+ab)\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}\right)\geq 2$$ and this is equivalent to $${\frac { \left( ab-1 \right) \left( a-b \right) ^{2}}{ \left( {a}^{2}
+1 \right) \left( {b}^{2}+1 \right) }}
\geq 0$$ this is right if $$ab\geq 1$$
| {
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"url": "https://math.stackexchange.com/questions/2187751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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maximum number of number of roots of $p(x) = 0$ is
Let $p(x)=x^6+ax^5+bx^4+x^3+bx^2+ax+1.$ Given that $x=1$ is a root of $p$ but $x=-1$ is not, find the maximum number of number of roots of $p$.
My attempt:
$x=0$ in not a root of $p(x)=0.$ So
$$\left(x^3+\frac{1}{x^3}\right)+a\left(x^2+\frac{1}{x^2}\right)+b\left(x+\... | This answer assumes that you want to find the maximum number of the real roots of $p(x)$.
You already have
$$(t-2)\left(t^2+(a+2)t+a-\frac 12\right)=0$$
where $t=x+\frac 1x$ (which is correct though you have a typo in the part $a(x^2+\frac{1}{x^2})=a(x+\frac 1x)^2-2\color{red}{a}$).
Let $t_{\pm}$ where $t_-\lt t_+$ be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2187861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Faster way to solve a equation. Solve the equation:
$\sqrt[3]{x-2} + \sqrt[3]{x} + \sqrt[3]{x+2} = 0$
$f(x) = \sqrt[3]{x-2} + \sqrt[3]{x} + \sqrt[3]{x+2}$
Firstly I check the amount of solutions.
*
*Graph of the function starts at the bottom and ends at the top.
*The derivative is always greater than 0, so the func... | Observe that for the given function, $f(a)=-f(-a)$ for any value of $a$.
Now put $a=0$.
Thus $f(0)=-f(0)$.
=> $2f(0)=0$
=> $f(0)=0$
=> $x=0$ is a solution.
Also, $f'(x)>0$ for all real numbers $x$.
So, the function is always increasing and thus $x=0$ is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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show that for $n=1,2,...,$ the number $1+1/2+1/3+...+1/n-\ln(n)$ is positive show that for $n=1,2,...,$ the number $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln(n)$ is positive, that it decreases as $n$ increases, and hence that the sequence of
these numbers converges to a limit between $0$ and $1$ (Euler's constant)... | Note that the sequence in question, let's call it $\alpha_n$, can be written as
$$\alpha_n = \Big(\sum^n_{k=1} 1/k\Big) - ln(n).$$
Note that
\begin{align}
ln(n) &:= \int^n_1 \frac {1} {t} dt \\\ &= \int_1^2 \frac {1}{t} dt + \int_2^3 \frac {1}{t} dt + \dots + \int^n_{n-1} \frac {1}{t} dt \\\
& \le (2-1) \cdot \frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Junior olympiad question: Minimum value of 3 digit number divided by sum of its digits I recently had a maths competition where we were given this problem. I solved the question, but I narrowed down the possibilities then did more of a guess and check method. I was hoping someone else could help me get the answer to th... | Let the number be $\overline{abc}=100a+10b+c\,$ with digits $1 \le a,b,c \le 9\,$. Then:
$$
\begin{align}
\frac{100a+10b+c}{a+b+c} & = 1 + 9\cdot\frac{11a+b}{a+b+c} \\[3px]
& \ge 1 + 9\cdot\frac{11a+b}{a+b+\color{red}{9}} \quad\quad\quad\quad\text{(*)}\\[3px]
& = 1 + 9 + 9 \cdot \frac{10a-9}{a+b+9} \\[3px]
& \ge 1 + 9 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the generating function or closed form for the recurrence relation $a_n = a_{n-1} + 4a_{n-2} + 2a_{n-3}$ I was trying to solve this recurrence relation using generating function
$a_n = a_{n-1} + 4a_{n-2} + 2a_{n-3} \qquad : \quad a_0 =1,a_1 =1,a_2 =5, $
I did in the following way
$
\begin{align*}
&G(x) = \sum_{n... | Here are two variants to derive $a_n$. The first one gives a closed form, the other one an explicit expression, which results in a nice binomial identity.
First variant: Partial fractions
In case it's easy to derive the zeros of the denominator of
\begin{align*}
G(x) = \frac{1}{1-x-4x^2-2x^3}
\end{align*}
the p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove that for all $n\ge9$, there exist natural numbers $x,y$ such that $n=2x+5y$. How would you use induction to prove this?
| If $n=9$ then $n = 2\cdot 2 + 5\cdot 1$; taking $x:= 2$ and $y:=1$ suffices.
If $n \geq 9$ is an integer such that
$n-1= 2x + 5y$ for some integers $x,y > 0$, then
$n = n-1 + 1 = 2x+5y + 1 = 2x' + 5y'$.
Note that $2x+5y+1 = 2x+5y+(5-4) = 2(x-2) + 5(y+1)$.
So the preceding equalities are equivalent to
$$
n = 2x' + 5y' =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2198173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Algebra Precalculus
$a> 0 < b$ For all the integer value of $a$ and $b$
$X= (a^2 + ab)-(ab^2-b)/(2a^2+b^2 -ab)$
Quantity I: $x $
Quantity II: $1.5$
(a) Quantity I $\lt$ Quantity II
(b) Quantity I $\gt$ Quantity II
(c) Quantity I $\ge$ Quantity II
(d) Quantity I $=$ Quantity II
(e) No relation
$(x^a)^c = x^c$
$x^... | I assume $a = 1$
Then,
$$x^{2b}/x^a = (x^{5a}) * (x^d)*(x^b) \implies x^{2b}/x = (x^{5}) * (x^d)*(x^b) \implies x^{2b} = (x^{6}) * (x^{d+1})*(x^{b+1}) \implies x^{2b - b - 1} = (x^{6}) * (x^{d+1})\implies x^{b - 1} = (x^{d+7}) \implies x^{b - 1 - d - 7} = 1 \implies x^{b- d - 8} = 1 \implies b = d+8$$
Therefore $b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2199433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How find this maximum of the value $\sum_{i=1}^{6}x_{i}x_{i+1}x_{i+2}x_{i+3}$? Let
$$x_{1},x_{2},x_{3},x_{5},x_{6}\ge 0$$ such that
$$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=1$$
Find the maximum of the value of
$$\sum_{i=1}^{6}x_{i}\;x_{i+1}\;x_{i+2}\;x_{i+3}$$
where
$$x_{7}=x_{1},\quad x_{8}=x_{2},\quad x_{9}=x_{3}\,.$... | For $x_i=\frac{1}{6}$ we get $\frac{1}{216}$.
We'll prove that it's a maximal value.
Indeed, let $x_1=\min\{x_i\}$, $x_2=x_1+a$, $x_3=x_1+b$, $x_4=x_1+c$, $x_5=x_1+d$ and $x_6=x_1+e$.
Hence, $a$, $b$, $c$, $d$ and $e$ are non-negatives and we need to prove that:
$$216\sum_{i=1}^6x_ix_{i+1}x_{i+2}x_{i+3}\leq\left(\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding $ \int \frac{5x^2-x-4}{x^5+x^4+1}dx$ Finding $\displaystyle \int \frac{5x^2-x-4}{x^5+x^4+1}dx$
Attempt : $\displaystyle I = \int\frac{5x^2-x-4}{x^5+x^4+1}dx = \int\frac{5x^2-x-4}{(x^2+x+1)(x^3-x+1)}dx$
because $\omega,\omega$ are the roots of $x^5+x^4+1 = 0$
so one factor is $(x-\omega)(x-\omega^2) = (x^2+x+1)$... | HINT:Using partial fraction decomposition
$$\frac{5x^2-x-4}{(x^2+x+1)(x^3-x+1)}=\frac{Ax+B}{x^2+x+1}+\frac{Cx^2+Dx+E}{x^3-x+1}$$
$${5x^2-x-4}=(Ax+B)(x^3-x+1)+(Cx^2+Dx+E)(x^2+x+1)$$
Solving gives..
$$\frac{5x^2-x-4}{(x^2+x+1)(x^3-x+1)}=\frac{-3x-3}{x^2+x+1}+\frac{3x^2-1}{x^3-x+1}$$
Another hint:
Maybe at some point you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to find the domain and range of $f(x) = \sqrt {\frac{x+1}{x+2}}$
Find the domain and range of $$f(x) = \sqrt {\frac{x+1}{x+2}}$$
I got the domain $[-1, \infty)$ but the answer contains $(-\infty, -2)$ along with it. And how to calculate range?
| We must have $$\frac{x+1}{x+2}\geq 0.$$ We consider the following cases:
Case 1. Suppose that $x+1\geq 0$ and $x+2>0$. Then $x\geq -1$ and $x>-2$. Thus,
$$SS_1=[-1,\infty).$$
Case 2. Suppose that $x+1\leq 0$ and $x+2<0$. Then $x\leq -1$ and $x<-2$. Thus,
$$SS_2=[-\infty,-2).$$
Hence, domain$=SS_1\cup SS_2$
Let $y=f(x)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2206728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proof by induction, dont know how to represent range The question asks for me to prove the following through induction:
$1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^n} \geq 1 + \frac{n}{2}$
This is my proof thus far:
Proving true for $n = 1$
\begin{align*}
1 + \frac{1}{2} &\geq 1 + \frac{1}{2}\\
\end{align*}
A... | $$
\underbrace{ 1 + \frac 1 2 + \frac 1 3 + \cdots + \frac 1 {2^k} + \frac 1 {2^{k+1}} }_{\Large\text{This is wrong.}}
$$
$$
\overbrace{ \underbrace{ \underbrace{1 + \frac 1 2 + \frac 1 3 + \cdots + \frac 1 {2^k}}_{\Large\text{The is the case }n=k.} + \frac 1 {2^k+1} + \frac 1 {2^k+2} + \frac 1 {2^k+3} + \cdots + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2210649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the limit or prove that the limit does not exist $\lim_{x \to c}x^2 + x + 1,$ for any $c \in R$
This is what I tried.
For $\epsilon > 0$ there exists $\delta >0$ such that $0< \mid x - c \mid < \delta (x \in R)$ => $\mid x^2 + x + 1 - c^2 - c - 1 \mid = \mid x^2 + x - c^2 - c \mid < \epsilon$
After that I tried... | Here is a trick:
Start by choosing $\delta \le |c|+1$, then $|x| \le |c|+|x-c| \le 2(|c|+1)$.
Then $|x^2-c^2| = |x+c||x-c| \le 3 (|c|+1) |x-c|$.
Then $|x^2+x+1 - (c^2+c+1)| \le (3|c|+2)|x-c|$.
Now choose $\delta < \min (|c|, {\epsilon \over 3|c|+2})$, then you will
obain the required bound.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that $\sum_{n=1}^\infty \frac{n^2}{(n+1)!}=e-1$ Show that:
$$\sum^\infty_{n=1} \frac{n^2}{(n+1)!}=e-1$$
First I will re-define the sum:
$$\sum^\infty_{n=1} \frac{n^2}{(n+1)!} = \sum^\infty_{n=1} \frac{n^2-1+1}{(n+1)!} - \sum^\infty_{n=1}\frac{n-1}{n!} + \sum^\infty_{n=1} \frac{1}{(nm)!}$$
Bow I will define e:
$$e^... | $$\frac{n^2}{(n+1)!} = \frac{(n+1)(n-1) + 1}{(n+1)!} = \frac{(n-1)}{n!} + \frac{1}{(n+1)!}$$
Remembering that we're summing to infinity, evaluating the first terms and paying careful attention to the indices,
$$
\begin{align}
\sum_{n=1}^\infty \left( \frac{(n-1)}{n!} + \frac{1}{(n+1)!} \right) &= \sum_{n=2}^\infty \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
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How prove this inequality $\sum\limits_{cyc}\sqrt{\frac{yz}{x^2+2016}}\le\frac{3}{2}$ Given $x,y,z$ are positive real number satisfy $xy+yz+xz=2016$. Prove that $$\sqrt{\dfrac{yz}{x^2+2016}}+\sqrt{\dfrac{xy}{z^2+2016}}+\sqrt{\dfrac{xz}{y^2+2016}}\le\dfrac{3}{2}$$
I tried
$\sqrt{\frac{yz}{x^2+2016}}=\sqrt{\frac{yz}{x^... | I believe there is something called the Purkiss Principle which would imply that in this case the maximum of $f$ is achieved when $x=y=z=\sqrt{2016/3}$. Thus, $$f(\sqrt{2016/3},\sqrt{2016/3},\sqrt{2016/3}) = 3/2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$\int_{|z-2i|=1}\frac{(z-2i+\frac{1}{2})^2 \sin(2i\pi z)}{\overline{z}^{2} (z-2i)^4} \ dz$ Let $C$ be the circle $|z -2i|=1$
How Can I Compute this Integral :
$$\int_{C}\frac{(z-2i+\frac{1}{2})^2 \sin(2i\pi z)}{\overline{z}^{2} (z-2i)^4} \ dz$$
Thank you
| I have tried solving this using cauchy integral formula as follows :
$|z-2i|= 1\Rightarrow (z-2i)(\overline{z}+2i)=1 \Rightarrow (z-2i+\frac{i}
{2}-\frac{i}{2})(\overline{z}+2i)=1$
Now we have : $$(z-2i+\frac{i}{2}-\frac{i}{2})(\overline{z}+2i)=1 \Rightarrow(z-2i+\frac{i}{2})(\overline{z}+2i)=\frac{i \overline{z}}{2}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Convergence of $1+\frac{1^2\cdot2^2}{1\cdot3\cdot5}+ \frac{1^2\cdot2^2\cdot3^2}{1\cdot3\cdot5\cdot7\cdot9}+...$ I am trying to use the ratio test, for that, I need the general formula for the series.
The general formula for the numerator is $(n!)^2$
The denominator is a sequence of odd numbers that grows by two terms e... | Lets try writing the general term
$$a_n=\frac{(n!)^2}{1\cdot 3\cdot 5\cdots (4n-5)(4n-3)}\\a_{n+1}=\frac{((n+1)!)^2}{1\cdot 3\cdot 5\cdots (4n-1)(4n+1)}\\\frac{a_{n+1}}{a_n}=\frac{((n+1)!)^2}{1\cdot 3\cdot 5\cdots(4n-1)(4n+1)}\cdot\frac{1\cdot 3\cdot 5\cdots(4n-5)(4n-3)}{(n!)^2}\\\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{(4n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2219077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Tangent Points for Common Tangent to Two Ellipses This is somewhat similar to my other question here.
Consider the two ellipses given by the equations
\begin{equation}
\frac{x^2}{2^2} + \frac{(y-1)^2}{1^2} = 1
\end{equation}
and
\begin{equation}
\frac{x^2}{1^2} + \frac{(y-4)^2}{(1/2)^2} = 1.
\end{equation}
How do I... | Let's introduce the following new variables:
$$2u=x\ \text{ and }\ y-1=v.$$
With these new variables, we have
$$u^2+v^2=1\ \text{ and }\ u^2+(v-3)^2=\frac14$$
that is, we have two circles as shown in the figure below.
We have similar triangles and we can see that $OD=6$. Also, by the Pythagorean theorem $DC=\frac{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proof: if $a$ and $b$ are integers, then $a^2-4b-3\neq 0$. I was wondering if someone could take the time to look over this proof and make sure it is correct. I greatly appreciate the help.
Proposition: If $a$ and $b$ are integers, then $a^2-4b-3\neq 0$.
Proof: Assume $a,b\in\mathbb{Z}$ and, for contradiction's sake, $... | Yes, your proof is correct! Also, when you discorver that a has a remainder of 3 when divided by 4, you can go straight to the fact that a is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove $\prod_1^\infty (1+p_n)$ converges Let $p_{2n-1} = \frac{-1}{\sqrt{n}}$, and $p_{2n} = \frac{1}{n}+\frac{1}{\sqrt{n}}$.
Prove $\prod_1^\infty (1+p_n)$ converges.
By numerical simulations, it appears to converge (to something around $0.759$). However, I'm not sure how to prove this. I know we can skip the first t... | Note that
$$\prod\limits_{k=2}^{2n}(1+p_k) = \prod\limits_{k=2}^{n}(1+p_{2k-1})(1+p_{2k}) = \prod\limits_{k=2}^{n}\left(1-\dfrac{1}{\sqrt{k}}\right)\left(1+\dfrac{1}{k} +\dfrac{1}{\sqrt{k}}\right) = \prod\limits_{k=2}^{n} \left(1- \dfrac{1}{k\sqrt{k}}\right)$$
It is also known that for any sequence $\{a_k\}$ such that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find the solution of the $x^2+ 2x +3=0$ mod 198 Find the solution of the $x^2+2x+3 \equiv0\mod{198}$
i have no idea for this problem i have small hint to we going consider $x^2+2x+3 \equiv0\mod{12}$
| Here's a more-or-less generalizable, manual way of finding all of the solutions:
First, as P. Vanchinathan does, change variable to $a := x + 1$, which transforms the equation into one with zero linear term:
$$a^2 + 2 \equiv 0 \pmod {198} .$$
(This step is option, but reduces the amount of later work.)
Now, we exploit ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_0^{2\pi}\frac{\cos(\theta)}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \ \text{ for }\ (A^2+B^2) <<1$ I need to evaluate the definite integral $\int_0^{2\pi}\frac{\cos(\theta)}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \ \text{ for }\ A<<1, B<<1, (A^2+B^2) <<1,$
For unresricted (... | HINT: set $$t=\tan(x/2)$$, $$\sin(x)=\frac{2t}{1+t^2}$$, $$\cos(t)=\frac{1-t^2}{1+t^2}$$ and $$dx=\frac{2dt}{1+t^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove $\sin(x) = x + O(x^3)$? I was doing an exercise in which I have to find the rate of convergence of
$$\lim\limits_{h\to 0}\dfrac{\sin h}{h} = 1$$
and the answer is $O(h^2)$. I don't understand why. The only thing I have found is that $$\sin(x) = x + O(x^3)$$ when $x$ tends to zero, and with that the exerci... | Start with
$\sin' = \cos
$,
$\cos' = -\sin
$,
$\sin(0) = 0$,
$\cos(0) = 1$,
and
$\sin^2+\cos^2 = 1$.
For small $t$,
$1 \ge \cos(t)
\ge 0
$
so
$\sin(x)
=\int_0^x \cos(t)dt
\le x
$.
Therefore
$1-\cos(x)
=\int_0^x \sin(t) dt
\le \int_0^x t dt
= \frac{x^2}{2}
$
so
$\cos(x)
\ge 1-\frac{x^2}{2}
$.
Therefore
$\sin(x)
=\int_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
$\int ^{\infty}_0 \frac{\ln x}{x^2 + 2x+ 4}dx$ We have to find the integration of
$$\int ^{\infty}_0 \frac{\ln x}{x^2 + 2x+ 4}dx$$
In this I tried to do substitution of $x=e^t$
After that got stuck .
| Let $I$ be defined by the integral
$$I=\int_0^\infty \frac{\log(x)}{x^2+2x+4}\,dx \tag1$$
and let $J$ be the contour integral
$$J=\oint_{C}\frac{\log^2(z)}{z^2+2z+4}\,dz \tag2$$
where the contour $C$ is the classical "key-hole" contour for which the keyhole coincides with the branch cut along the positive real axis. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
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Proving the inequality $0\leq \frac{\sqrt{xy}}{1-p}\frac{x^{\frac{1}{p}-1}-y^{\frac{1}{p}-1}}{x^{\frac{1}{p}}-y^{\frac{1}{p}}} \leq 1$ Suppose $p\in(0,1)$. How might one show that
\begin{equation}\tag{1}
0\leq \frac{\sqrt{xy}}{1-p}\frac{x^{\frac{1}{p}-1}-y^{\frac{1}{p}-1}}{x^{\frac{1}{p}}-y^{\frac{1}{p}}} \leq 1
\end{... | I've figured out the correct integral representation to use here. For $a\in(-1,1)$, consider the following integral representations:
\begin{align*}
\frac{x^a-y^a}{x-y} &= \frac{\sin(a\pi)}{\pi}\int_{0}^{\infty}\frac{t^a}{(x+t)(y+t)}dt\\
\text{and}\qquad ax^{a-1} &= \frac{\sin(a\pi)}{\pi}\int_{0}^{\infty}\frac{t^a}{(x+t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2234190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve$\int\frac{x^4}{1-x^4}dx$ Question: Solve $\int\frac{x^4}{1-x^4}dx.$
My attempt:
$$\int\frac{x^4}{1-x^4}dx = \int\frac{-(1-x^4)+1}{1-x^4}dx = \int 1 + \frac{1}{1-x^4}dx$$
To integrate $\int\frac{1}{1-x^4}dx,$ I apply substitution $x^2=\sin\theta.$ Then we have $2x \frac{dx}{d\theta} = \cos \theta.$ which implies... | By doing long division, you'll get $$-\int \left(-\frac{1}{2(x^2+1)}-\frac{1}{4(x+1)}+\frac{1}{4(x-1)}+1\right) dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find the constant term in the expansion of $(x^2+1)(x+\frac{1}{x})^{10}$ I can't solve this problem. How to solve it?
The Problem is
"Find the constant term in the expansion of $
\left({{x}^{2}\mathrm{{+}}{1}}\right){\left({{x}\mathrm{{+}}\frac{1}{x}}\right)}^{\mathrm{10}}
$"
| $f(x)=(x^2+1)(x+\dfrac{1}{x})^{10}$
We can rewrite $f(x)$ like below:
$f(x) = x^2(x+\dfrac{1}{x})^{10} + (x+\dfrac{1}{x})^{10}$
In the first term, the power of $x$, must be $-2$ in the parenthesis, so the when it's multiplyed by $x^2$ the power of $x$ becomes $0$. And we know that:
$(x+\dfrac{1}{x})^{10}=\sum_{k=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Triangle - length of the sides - proof a, b and c are the lengths of the sides of a triangle. Prove that
$$a^2+b^2 \ge \frac{1}{2}c^2$$
Let $\gamma$ be the angle between sides a and b. then:
$$a^2 + b^2 - 2ab\cos(\gamma) = c^2$$
Hence we need to prove that
$$a^2+b^2 \ge \frac{1}{2}c^2$$
$$2a^2+2b^2 \ge a^2+b^2 -... | You're already there. Your last line is
$$a^2+b^2+2ab\cos(\gamma) \geq (a-b)^2$$
And so you only have to note $(a-b)^2\geq 0$ to see that
$$a^2+b^2+2ab\cos(\gamma) \geq 0$$
Note that we can make this inequality strict ($>$) since $\cos(\gamma)=-1$ can't happen, for one angle is then stretched, so we could've taken $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Question on infinite geometric series. Problem- The sum of the first two terms of an infinite geometric series is 18. Also, each term of the series is seven times the sum of all the terms that follow. Find the first term and the common ratio of the series respectively.
My approach- Let $a+ar+ar^2+\dots$ be the serie... |
The sum of the first two terms of an infinite geometric series is 18.
$$
S(a, r) = \sum_{k=0}^\infty a r^k = \frac{a}{1-r} \\
a + a r = 18 \quad (*)
$$
Also, each term of the series is seven times the sum of all the terms
that follow.
$$
a r^n = 7 \sum_{k=n+1}^\infty a r^k \quad (**)
$$
Find the first term and t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding out the numbers under given conditions Let $M$ be a $2$ digit number $ab$, $N$ is a $3$ digit number $cde$ , and $X=M\times N$ is such that $9(X)=abcde$.The question is to find out the ratio $\frac NM$
I tried to solve it using trial and error and examined a number of cases but couldn't reach the answer so far.... | Rewrite $9(X) = abcde$ as $9MN = abcde = 1000ab + cde = 1000M + N$, then divide by $M$ to get $9N = 1000 + \frac{N}{M}$ or $\frac{N}{M} = 9N - 1000$. Notice then that $\frac{N}{M}$ must be a whole number, call it $k$.
Replace $N$ with $kM$ to get $9kM = 1000 + k$. Since the left hand side is divisible by $k$, the right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A question about Taylor expansion Is this statement true? Statement: Let $n$ be a positive integer. Consider the Taylor expansion of $\sqrt[n]{1+x}$ to the $k$th order, that is,
\begin{gather*}
\sqrt[n]{1+x}=\sum_{j=0}^{k}\binom{\frac{1}{n}}{j}x^j+o(x^k), \qquad \text{as $x\to 0$.}
\end{gather*}
Then the Taylor polyno... | This is actually a very good exercise in combinatorics and in multiplying polynomials so I will post an answer.
Firstly note that the following identity holds true:
\begin{equation}
1= \sum\limits_{J=0}^{n k} \delta_{j_1+\cdots+j_n,J}
\end{equation}
Now we insert the unity into the left hand side of (1) and we expand t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find $\sin A$ and $\cos A$ if $\tan A+\sec A=4 $ How to find $\sin A$ and $\cos A$ if
$$\tan A+\sec A=4 ?$$
I tried to find it by $\tan A=\dfrac{\sin A}{\cos A}$ and $\sec A=\dfrac{1}{\cos A}$, therefore
$$\tan A+\sec A=\frac{\sin A+1}{\cos A}=4,$$
which implies
$$\sin A+1=4\cos A.$$
Then what to do?
| Together with
$$\sin A+1=4\cos A $$
you can use $\sin^2A+\cos^2A=1$ as
$$(\sin A+1)(\sin A-1)=\cos^2 A\ .$$
Putting the two together you easily get
$$4\cos A(\sin A-1)=\cos^2 A\ ,$$
and hence
$$4\sin A-4=\cos A\ .$$
You now just have to solve the linear system in $\sin A$ and $\cos A$:
$$\begin{cases}
4\sin A-4=\cos A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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The number of solutions of the equation $\tan x +\sec x =2\cos x$ lying in the interval $[0, 2\pi]$ is: The number of solutions of the equation $\tan x +\sec x =2\cos x$ lying in the interval $[0, 2\pi]$ is:
$a$. $0$
$b$. $1$
$c$. $2$
$d$. $3$
My Attempt:
$$\tan x +\sec x=2\cos x$$
$$\dfrac {\sin x}{\cos x}+\dfrac {1}{... | The answer must be 2
Since tanx and secx domain would consider every x, rejecting (2n+1)π/2 values in the interval [0,2π] such values would be rejected from the solution.
In the last step sinx=1 the solution would be 3π/2 and this would be rejected due to domain condition. The other part, that is, sinx=1/2 has solutio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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why$2 (-1 + 2^{1 + n})$ is the answer to the recurrence relation $a_{n}=2a_{n-1}+2$? $a_{0}=2$
$a_{1}=2(2)+2$
$a_{2}=2(2(2)+2)+2$
$a_{3}=2(2(2(2)+2)+2)+2$
$a_{4}=2(2(2(2(2)+2)+2)+2)+2$
$a_{5}=2(2(2(2(2(2)+2)+2)+2)+2)+2$
To simplifiy
$a_{6}=2^{6}+2^{5}...2^{1}$
so my answer is
$a_{n}=2^{n+1}+2^{n}+...2^{1}$
The correct... | The inhomogeneous recurrence relation
$$
a_n = 2 a_{n-1} + 2
$$
can be turned into a homogeneous recurrence
$$
a_n - a_{n-1} = 2 a_{n-1} + 2 - (2 a_{n-2} + 2) = 2 a_{n-1} - 2 a_{n-2} \iff \\
a_n = 3 a_{n-1} - 2 a_{n-2}
$$
and solved by the usual algorithm.
The characteristic polynomial is
$$
p(t) = t^2 - 3 t + 2
$$
wit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
$2x^2 + 3x +4$is not divisible by $5$ I tried by $x^2 \equiv 0, 1, 4 \pmod 5$ but how can I deal with $3x$?
I feel this method does not work here.
| $$5|\;(2x^2+3x+4)\iff$$ $$ \iff5|\;3(2x^2+3x+4)=(6x^2+9x+12)\iff$$ $$\iff 5|\;((6x^2+9x+12)-(5x^2+5x+10))=$$ $$=(x^2+4x+2)=(x+2)^2-2.$$ But no square is $2$ more than a multiple of $5.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2250718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
$\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2}=\frac{c^2}{2}$ without L'Hospital's rule How to show that without using L'Hospital's rule
\begin{align}
\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2}=\frac{c^2}{2}
\end{align}
I was able to show the upper bound by using th... | Denote $a=x^2-cx$ for simplicity. Then
\begin{align}
\frac{\ln(\cosh a )}{x^2}&=\frac{\ln(\cosh^2a)}{2x^2}=\color{blue}{\frac{\ln(1+\sinh^2a)}{2x^2}}=\frac12\cdot\frac{\ln(1+\sinh^2a)}{\sinh^2a}\cdot\left(\frac{\sinh a}{x}\right)^2=\\
&=\frac12\cdot\frac{\ln(1+\sinh^2a)}{\sinh^2a}\cdot\left(\frac12\cdot\left[\frac{e^a-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
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If $ab$ is a square number and $\gcd(a,b) = 1$, then $a$ and $b$ are square numbers.
Let $n, a$ and $b$ be positive integers such that $ab = n^2$. If $\gcd(a, b) = 1$, prove that there exist positive integers $c$ and $d$ such that $a = c^2$ and $b = d^2$
So far I have tried this:
Since $n^2 = ab$ we have that $n = \s... | Let $ab=c^2$ for some $c\in N$ then the result will hold if any one of the integers is 1 as 1^2=1.
So let us take a>1,b>1 and c>1. We can use prime factorization and represent the integers as follows:
a=$p_1^{d_1} * p_2^{d_2}$
b= $q_1^{e_1} * q_2^{e_2}$
and c=$k_1^{l_1} *k_2^{l_2}$
thus $ab=c^2$ becomes
$p_1^{d_1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2253940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find a non-zero integer matrix $X$ such that $XA=0$ where $X,A,0$ are all $4 \times 4$ Let $A$ be the following $4 \times 4$ matrix.
\begin{bmatrix}1&2&1&3\\1&3&2&4\\2&5&3&7\\1&4&1&5\end{bmatrix}
How can we find a non-zero integer matrix $C$ such that $CA = 0_{4 \times 4}$
Note that $0$ is a $4 \times 4$ matrix.
| Hint : The rank of $A$ is $3$ so : $$\exists P,Q \in GL_n(\mathbb R),A=P\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&0\end{bmatrix}Q$$
$$CA=0 \Rightarrow CP\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&0\end{bmatrix}Q=0\Rightarrow C\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&0\end{bmatrix}=Q^{-1}P^{-1}$$
Calcu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Solving $8yy'^2 - 2xy' + y = 0$ I'm solving the differential equation $8yy'^2 - 2xy' + y = 0$
My attempt:
We divide both sides by $x$, obtaining:
$$8\frac{y}{x}y'^2 - 2y' + \frac{y}{x} = 0$$
Then, we introduce $t = y'$, hence the differential equation becomes:
$$8\frac{y}{x}t^2 - 2t + \frac{y}{x} = 0$$
from which follo... | You lost the factor $x$ in the last term while differentiating the parametric equation.
It is even easier to multiply with $y$ and then substitute $u=y^2$ to get
$$
2u'^2-xu'+u=0\iff u=xu'-2u'^2
$$
which is a Clairaut differential equation. This has the lines
$$
u=cx-2c^2
$$
as solutions and their envelope which is th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the maximum and minimum value of $P =x+y+z+xy+yz+zx$ Let $x^2+y^2+z^2\leq27$ and $P = x+y+z+xy+yz+zx$. Find the value of $x, y, z$ such that $P$ is the maximum value and minimum value.
My attempt :
$$(x-y)^2 + (y-z)^2 + (z-x)^2 \geq 0$$
$$27 \geq x^2+y^2+z^2 \geq xy+yz+zx\tag{1}$$
$$(x+y+z)^2 \leq 3(x^2+y^2+z^2) \... | You may use the same method to find the minimum. First, we obtain a lower bound for $P$:
$$
\begin{align}
P&=x+y+z+xy+yz+zx\\
&=\frac12 [ (x+y+z+1)^2 - (x^2+y^2+z^2) - 1 ]\\
&\ge\frac12 (0 - 27 - 1)\tag{1}\\
&= -14.
\end{align}
$$
Next, note that at $\left(\frac{\sqrt{53}-1}2,-\frac{\sqrt{53}+1}2,0\right)$, we have $x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Find the exact value of $A(\beta)=8\pi-16\sin(2\beta)$ with $\tan(\beta)= \frac{1}{2}$
The picture below represents a semi-circumference of diameter [AB] and
center C. Point D belongs to the semi-circumference and it's one of
the vertices of the triangle $ABC$. Consider that BÂD = $\beta (\beta
\in ]0,\frac{\pi}{... | Given: $\tan \beta = \frac 12$, so $2\sin\beta = \cos \beta$
$\sin 2\beta = 2\sin\beta\cos\beta = \cos^2\beta = \frac 1{\sec^2 \beta} = \frac 1{1 + \tan^2 \beta} = \frac 1{ 1 + \frac 14} = \frac 45$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How can we integrate integral(s) of this type? So I was able to free $dx$ from the power. Now only wolfram can solve this Integral. How can I do this on my own?
$$r=\int_0^1\left(\frac{x^{12}}{(1-x^4)^3}+1\right)^{1/4}~dx$$
| Hint:
$\int_0^1\left(\dfrac{x^{12}}{(1-x^4)^3}+1\right)^\frac{1}{4}~dx$
$=\int_0^1\left(\dfrac{x^3}{(1-x)^3}+1\right)^\frac{1}{4}~d\left(x^\frac{1}{4}\right)$
$=\dfrac{1}{4}\int_0^1x^{-\frac{3}{4}}\left(\dfrac{x^3}{(1-x)^3}+1\right)^\frac{1}{4}~dx$
$=\dfrac{1}{4}\int_1^0(1-x)^{-\frac{3}{4}}\left(\dfrac{(1-x)^3}{x^3}+1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Parametrization of two surfaces $\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$ and $\frac{x^2}{p}+\frac{y^2}{q}=2z$. Can someone please help me to parametrize the following surfaces in terms of hyperbolic(for second it might not be possible but i need some more convenient set of parametric equation than mine ) an... | For the first: $$\frac{x^2}{a^2}\color{red}{-}\left(\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)=1$$
\begin{eqnarray}
&x&=a\cosh\theta,\\
&y&=b\cos\phi\sinh\theta,\\
&z&=c\sin\phi\sinh\theta.\\
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Computing maximum dimension of a vector subspace given that it's every element is a symmetric matrix and is closed under matrix multiplication
Q. Let $S$ be a subspace of the vector space of all $11 \times 11$ real matrices such that (i) every matrix in $S$ is symmetric and (ii) $S$ is closed under matrix multiplicati... | If $A$ and $B$ are symmetric, then $AB$ symmetric means $AB=(AB)^T=B^TA^T=BA$. All the matrices therefore commute. A symmetric
real matrix is diagonalisable, and pairwise commuting real matrices
are simultaneously diagonalisable. So the dimension of such a space is
at most $11$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Maximum of $x^3+y^3+z^3$ with $x+y+z=3$ It is given that, $x+y+z=3\quad 0\le x, y, z \le 2$ and we are to maximise $x^3+y^3+z^3$.
My attempt : if we define $f(x, y, z) =x^3+y^3 +z^3$ with $x+y+z=3$ it can be shown that,
$f(x+z, y, 0)-f(x,y,z)=3xz(x+z)\ge 0$ and thus $f(x, y, z) \le f(x+z, y, 0)$. This implies that $f... | $$
\begin{eqnarray}
&(x+y+z)^3 &=& x^3 + y^3 + z^3 + 3(x+y)(y+z)(z+x)\\
\implies & x^3 + y^3 + z^3 &=& (x+y+z)^3 - 3(x+y)(y+z)(z+x)\\
&& = & 27 - 3(x+y)(y+z)(z+x)
\end{eqnarray}
$$
Now, $x^3+y^3+z^3$ is maximum when $t = (x+y)(y+z)(z+x)$ is minimum. Now since $x$, $y$ and $z$ are each non-negative, therefore $t$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$ If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$
My Attempt:
$$\sin x + \sin^2 x=1$$
$$\sin x = 1-\sin^2 x$$
$$\sin x = \cos^2 x$$
Now,
$$\cos^8 x + 2\cos^6 x + \cos^4 x$$
$$=\sin^4 x + 2\sin^3 x +\sin^... | Hint:
$$\cos^8x+2\cos^6x+\cos^4x=(\cos^4x+\cos^2x)^2$$
Now as $\cos^2x=\sin x,\cos^4x=(\cos^2x)^2=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
$\epsilon - \delta$ proof for $\frac{x^2 - 16}{x + sin x}$ limit I'm having difficulty writing an $\epsilon - \delta$ proof for the following limit:
$\lim_{x\to 4} \frac{x^2-16}{x+\sin x} = 0$
I've factored it to $\frac{(x+4)(x-4)}{x+\sin x} = 0$
and guessed that I need $\delta = \frac{2}{5}\epsilon$ for $|x-4| < \delt... | $$|x-4| < \delta$$
$$4-\delta < x < 4+ \delta$$
$$3 - \delta< x+ \sin x < 5 + \delta$$
$$\frac{1}{5+\delta} < \frac{1}{x+\sin x} < \frac{1}{3-\delta}$$
If $\delta < 1$, $-\delta > -1$, $3-\delta > $2, $\frac{1}{3-\delta} < \frac12$
$$\left| \frac{x^2-16}{x+\sin x}\right| \leq \frac12 |x^2-16|$$
Also, if $\delta < 1$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2270291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Show that $a+b+c+\sqrt {\frac {a^2+b^2+c^2}{3}}\le4$, Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2+abc=4.$ Show that $$a+b+c+\sqrt {\frac {a^2+b^2+c^2}{3}}\le4.$$
| Let $a=\frac{2x}{\sqrt{(x+y)(x+z)}}$ and $b=\frac{2y}{\sqrt{(x+y)(y+z)}},$ where $x$, $y$ and $z$ are positives.
Hence, $c=\frac{2z}{\sqrt{(x+z)(y+z)}}$ and we need to prove that
$$\sum_{cyc}\frac{2x}{\sqrt{(x+y)(x+z)}}+\sqrt{\frac{4}{3}\sum_{cyc}\frac{x^2}{(x+y)(x+z)}}\leq4$$ or
$$\sum_{cyc}x\sqrt{y+z}+\sqrt{\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2271863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the sum $\sum_{k=1}^\infty {\frac{6^k}{\left(3^k-2^k\right) \left(3^{k+1}-2^{k+1}\right)}}$ I need to find the sum,
$$\sum_{k=1}^\infty {\frac{6^k}{\left(3^k-2^k\right) \left(3^{k+1}-2^{k+1}\right)}}$$
I have tried to break the terms into partial fractions (method of differences) but am not able to do so. How to p... | First we can try to split things into two pieces:
$$\frac{6^k}{\left(3^k-2^k\right) \left(3^{k+1}-2^{k+1}\right)} = \dfrac{A}{3^k-2^k} + \dfrac{B}{3^{k+1}-2^{k+1}}$$
So we have $A \cdot (3^{k+1}-2^{k+1}) + B \cdot (3^k-2^k) = 6^k$ which can be arranged to $3^k (3 A + B) - 2^k (2 A + B) = 6^k$.
If we make $2A+B=0$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Calculate $\sqrt{2i}$ I did:
$\sqrt{2i} = x+yi \Leftrightarrow i = \frac{(x+yi)^2}{2} \Leftrightarrow i = \frac{x^2+2xyi+(yi)^2}{2} \Leftrightarrow i = \frac{x^2-y^2+2xyi}{2} \Leftrightarrow \frac{x^2-y^2}{2} = 0 \land \frac{2xy}{2} = 1$
$$\begin{cases}
\frac{x^2-y^2}{2} = 0 \\
xy = 1\\ \end{cases} \\
=\... | Write $2i=2e^{i \pi/2+2k\pi}$. Then square root to get: $\sqrt{2} e^{i\pi/4+k\pi}$. So your roots are $\sqrt{2}e^{i\pi/4}$ and $\sqrt{2}e^{3i\pi/4}$. Which are $\pm(1+i)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Domain of $f(x)=\sqrt{\lfloor x\rfloor-1+x^2}$ I drew the number line and tested with different values, getting the correct domain $(-\infty,-\sqrt3)\cup[1,\infty)$. However, how do I solve this faster by manipulating the function?
| You must find the values of $x$ such that $\lfloor x\rfloor-1+x^2\geq0$.
The easy part:
*
*It's true for $x\geq1$, since $x^2\geq1$ and $\lfloor x\rfloor>0$.
*For $0\leq x<1$, it's false, because $\lfloor x\rfloor=0$ and $x^2<1$.
Now, the case $x<0$. First notice that for $x\in[n,n+1[$, for integer $n$, you have $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2274330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Completing the square of $x^2 - mx = 1$ is not giving me the right answer. This is my attempt
$$
\begin{align}
x^2 - mx &= 1 \\
x^2 - mx - 1 &= 0 \\
\left(x^2 - mx + \frac{m^2}{4} - \frac{m^2}{4}\right) - 1 &= 0 \\
\left(x^2 - mx + \frac{m^2}{4}\right) - \frac{m^2}{4} - 1 &= 0 \\
\left(x^2 - mx + \f... | (Not an answer, just a long comment.)
Your actual question has already been answered, but I want to point out another mistake, namely when you go from
$$\left(x - \frac{m}{2}\right)^2 = \frac{m^2 - 4}{4}$$
to
$$\sqrt{\left(x - \frac{m}{2}\right)^2} = \sqrt{\frac{m^2 - 4}{4}}.$$
At this point, there should be $\pm$ sign... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Area of triangle and determinant The area of a $\vartriangle ABC$ with given vertices $(a,a^2),(b,b^2),(c,c^2)$ is $\frac{1}{4}$ $sq. units$ and area of another $\vartriangle PQR$ with given vertices $(p,p^2),(q,q^2),(r,r^2)$ is $3$ $sq. units$.
Then what is the value of
$$
\begin{vmatrix}
(1+ap)^2 & (1... | Let $A, B, C$, $P, Q, R$ be the $6$ column vectors
$$
\begin{cases}
A^T = (1, \sqrt{2}a, a^2),\\
B^T = (1, \sqrt{2}b, b^2),\\
C^T = (1, \sqrt{2}c, c^2)
\end{cases}
\quad\text{ and }\quad
\begin{cases}
P^T = (1, \sqrt{2}p, p^2),\\
Q^T = (1, \sqrt{2}q, q^2),\\
R^T = (1, \sqrt{2}r, r^2)
\end{cases}
$$
Using identites of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $ x^4+x^3+x^2+x+1=0$ then what is the value of $x^5$ If $$x^4+x^3+x^2+x+1=0$$ then what's the value of $x^5$ ??
I thought it would be $-1$ but it does not satisfy the equation
| Well, we have $x^5-1=(x-1)(x^4+x^3+x^2+x+1)=(x-1)0=0$ so that $x^5=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to find the maximum of the value $\frac{\sin{x}+1}{\sqrt{3+2\cos{x}+\sin{x}}}$ Find the maximun of the value
$$f(x)=\dfrac{\sin{x}+1}{\sqrt{3+2\cos{x}+\sin{x}}}$$
I use wolframpha this found this maximum is $\dfrac{4\sqrt{2}}{5}$,But How to prove and how to find this value?(without derivative)
idea 1
let $\tan{\df... | We need to minimize $$\dfrac{3+2\cos x+\sin x}{(1+\sin x)^2}$$
Now WLOG let $x=\dfrac\pi2-2y$ to get $$\dfrac{3+2\sin2y+\cos2y}{(1+\cos2y)^2}$$
Using Weierstrass substitution, writing $\tan y=t$
we get $$2f(t)=(t^2+1)(t^2+2t+2)$$
Now use Second derivative test, to find the minimum value of $f(t)$ occurs at $-\dfrac12$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{x^2+xy+y^2}$ fine $x,y$ :
$$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{x^2+xy+y^2} \ \ \ : x ,y \in \mathbb{Z}$$
My Try :
$$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{x^2+xy+y^2} \\ x^2-x+1+y^2-y+1+2\sqrt{(x^2-x+1)(y^2-y+1)}=x^2+xy+y^2 \\ 2\sqrt{(x^2-x+1)(y^2-y+1)}=xy +x+y-2$$
Now ?
| You can simplify as follows
$$\begin{align}
2\sqrt{(x^2-x+1)(y^2-y+1)}&=xy +x+y-2\\
4(x^2-x+1)(y^2-y+1)&=x^2y^2+x^2+y^2+4+2x^2y+2xy^2-4xy\\&\,\,\,\,\,+2xy-4x-4y\\
4x^2y^2-4x^2y+4x^2-4xy^2+4xy+4y^2&=x^2y^2+2x^2y+x^2+2xy^2-2xy+y^2\\
3x^2y^2-6x^2y+3x^2-6xy^2+6xy+3y^2&=0\\
x^2y^2-2x^2y+x^2-2xy^2+2xy+y^2&=0\\
(xy-x-y)^2&=0
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
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.