Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides.the length of each perpendicular is equal to a units.The distance between their feet being equal to b units.Find the area of the rhombus
| Set the $A,B,C,D \,\,$ vertices of the rhombus in $(m,0) $, $(0,n)\,\, $, $(-m,0)\,\, $, and $(0, -n) \,\, $, respectively. The slope of the side $AB\,\, $ is $-n/m \,\, $. So the slope of the perpendicular to $AB\,\, $, drawn from its opposite vertex $C $, is $m/n\,\, $. Since this perpendicular line pas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Simplifying fraction with factorials: $\frac{(3(n+1))!}{(3n)!}$ I was trying to solve the limit:
$$\lim_{n \to \infty} \sqrt[n]{\frac{(3n)!}{(5n)^{3n}}}$$
By using the root's criterion for limits (which is valid in this case, since $b_n$ increases monotonically):
$$L= \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \... | $$ (3(n+1))! \neq 3 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 4 \cdots 3 \cdot (n+1) $$ To make it clear what the problem is, let's write the right-hand side with brackets: $$ (3 \cdot 2) \cdot (3 \cdot 3) \cdot (3 \cdot 4) \cdots (3 \cdot (n+1)) $$ That's just multiplying all the positive multiples of 3 less than $ 3(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2109192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Ask about beautiful properties of $e$ One of students asked me about "some beautiful properties (or relation) of $e$". Then I list like below
\begin{align}
& e \equiv \lim_{x \to \infty} \left(1+\frac{1}{x} \right)^x\\[10pt]
& e = \sum_{k=0}^\infty \frac{1}{k!}\\[10pt]
& \frac{d}{dx} (e^x) = e^x\\[10pt]
& e^{ix} = \co... | Here are some more relations which might be pleasing.
From section 1.3 of Mathematical Constants by S.R. Finch:
*
*A Wallis-like infinite product is
\begin{align*}
e=\frac{2}{1}\cdot\left(\frac{4}{3}\right)^{\frac{1}{2}}
\cdot\left(\frac{6\cdot 8}{5\cdot 7}\right)^{\frac{1}{4}}
\cdot\left(\frac{10\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
Is it possible to evaluate $\lim_{x\rightarrow 0}(-1+\cos x)^{\tan x}$? Let's say we have the following limit:
$$\lim_{x\rightarrow 0}(-1+\cos x)^{\tan x}$$
Would the following solution be correct?
The solution is incorrect, please see the correction of @YvesDaoust
\begin{align}
\lim_{x \rightarrow 0}(-1+\cos x)^{\ta... | You can reduce 1 step that's looking odd.
You can write simply.
$$\lim_{x\to 0} \left[-1\left(1-\cos x \right)\right]^{\tan x}$$
$$\lim_{x\to 0} \left[-1\left(2\sin^2 \frac x2\right)\right]^{\tan x}$$
$$\lim_{x\to 0} \left(-2\right)^{\tan x} . \left(\sin^2 \frac x2 \right)^{\tan x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $\alpha $ and $\beta$ are roots of equation $a\tan\theta +b \sec\theta=c$. Prove that $\tan(\alpha + \beta)=\frac{2ac}{a^2-c^2}$ If $\alpha $ and $\beta$ are roots of equation $a\tan\theta +b \sec\theta=c$. Prove that $\tan(\alpha + \beta)=\frac{2ac}{a^2-c^2}$
i have converted tan to sin and cos and reached to $\si... | We can write the quadratic equation as: $$a\tan \theta + b\sec \theta = c $$ $$\Rightarrow a\tan \theta + b\sqrt {\tan^2 \theta + 1} = c $$ $$\Rightarrow (a^2-b^2)\tan^2 \theta -2ac \tan \theta +(c^2-b^2)=0$$
Now $$\tan (\alpha +\beta ) =\frac {\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}. $$ Can you take it fro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$ Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$
My Attempt,
$$L.H.S=8\cos^3(\frac {\pi}{9}) - 6\cos(\frac {\pi}{9})$$
$$=2\cos(\frac {\pi}{9}) [4\cos^2(\frac {\pi}{9}) - 3]$$
$$=2\cos(\frac {\pi}{9}) [2+2\cos(\frac {2\pi}{9}) - 3]$$
$$=2\co... | Hint:
$$\cos(3x) = 4\cos^3(x) - 3\cos(x).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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How to show that $F(x)=F(xq)+xq F(xq^2)$ Define$$\begin{align*}F(x) & =1+\dfrac {xq}{1-q}+\dfrac {x^2q^4}{(1-q)(1-q^2)}+\dfrac {x^3q^9}{(1-q)(1-q^2)(1-q^3)}+ \cdots\\ & =1+\sum\limits_{n=1}^{\infty}\dfrac{x^nq^{n^2}}{(q;q)_{n}}\end{align*}$$
Then $F(x)$ also satisfies $F(x)=F(xq)+xq\cdot F(xq^2)$.
Question: How would ... | Using the convention $(q;q)_0 = 1$, we can write
$$F(x) = \sum_{n = 0}^\infty \frac{x^n q^{n^2}}{(q;q)_n}$$
So
$$F(xq) = \sum_{n = 0}^\infty \frac{x^n q^{n^2 + n}}{(q;q)_n}$$
and
$$xq F(xq^2) = xq\sum_{n = 0}^\infty \frac{x^n q^{n^2+2n}}{(q;q)_n} = \sum_{n = 0}^\infty \frac{x^{n+1}q^{n^2+2n+1}}{(q;q)_n} = \sum_{n = 1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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The Formula to this Sequence Series What is the formula for following sequence?
$$\frac12 + \frac12 \cdot \frac34 + \frac12 \cdot \frac34 \cdot \frac56 + ... + \frac12 \cdot \frac34 \cdot \frac56 ... \frac{2n - 1}{2n}$$
This is a question from my Calculus class about sequences.
| $$
\begin{align}
&\, 1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1) = (2n-1)!! \,,\quad\, 2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n) = (2n)!! \\[2mm]
&\, S_{\small n} = \frac{1}{2}+\frac{1\,\cdot\,3}{2\,\cdot\,4}+\frac{1\,\cdot\,3\,\cdot\,5}{2\,\cdot\,4\,\cdot\,6}+\,\cdots\,+\frac{1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)}{2\,\cdot\,4\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find the determinant of order $100$ Find the determinant of order $100$:
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\en... | Using this matrix determinant lemma I am getting $(1+5\cdot100\cdot(-\frac{1}{6}))(-6)^{100}$.
Note about general size, one needs to think about the determinant of anti-diagonal matrix to compute it. My guess is that it equals $(1+5n(-\frac{1}{6}))(-1)^{\frac{n(n-1)}{2}}(-6)^{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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What is the maximal value of the expression $\overline{abc}-(a^3+b^3+c^3)?$ Let $\overline{abc}$ be $3$-digits number. What is the maximal value of the expression $\overline{abc}-(a^3+b^3+c^3)?$
Computer gives the answer $396$ for the number $620$ but I have no idea how to do it by hand.
| You can write $\overline{abc} = 100a + 10 b + c$. Then $$\overline{abc} - (a^3 + b^3 + c^3) = a(100-a^2) + b(10 - b^2) + c(1-c^2).$$ You just need to find the digits $a,b,c$ that maximize each term individually.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2123059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate: $\lim_{x\to 0}{\frac{\tan x-\sin x}{x^3}}$
Calculate: $\lim{\frac{\tan x-\sin x}{x^3}}$ as $x\to 0$
I have solved it using a way that gives a wrong answer but I can't figure out why:
$$\lim{\frac{\tan x-\sin x}{x^3}}\\
=\lim{\frac{\tan x}{x^3}-\lim\frac{\sin x}{x^3}}\\
=\lim{\frac{1}{x^2}}\cdot\lim{\fra... | 1) With l'Hospital:
$$\lim_{x\to0}\frac{\tan x-\sin x}{x^3}\stackrel{\text{l'H}}=\lim_{x\to0}\frac{\frac1{\cos^2x}-\cos x}{3x^2}\stackrel{\text{l'H}}=\lim_{x\to0}\frac{\frac{2\sin x}{\cos^3x}+\sin x}{6x}=$$
$$=\lim_{x\to0}\frac16\frac{\sin x}x\left(\frac2{\cos^3x}+1\right)=\frac16\cdot1\cdot(2+1)=\frac12$$
2) Without l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the matrix of a linear transformation3. Let $f$$ \begin{pmatrix}
\begin{pmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{pmatrix}
\end{pmatrix}
$=
$ \begin{pmatrix}
x_1+2x_2+x_3 \\
x_1-2x_2 \\
x_1+x_3 \\
3x_1-4x_2 \\
\end{pma... | Here is another way to do it:
$B$ is the representation of the principle vectors of B in the standard basis.
$B^{-1}$ is then the representation of the principle vectors in the standard basis as represented in the basis of $B.$
We can do a similar exercise for $C.$
$fB$ would take the principle vectors in $B$ to a matr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_{x\to 0^+}\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}$ I need to find the following limit: $$\lim_{x\to 0^+}\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}$$
I started this way: $$\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}=e^{\frac{1}{x}\cdot \ln\left[\frac {\sin x}{x}\right]}$$
So it's enough to find: $$\l... | Hints:
*
*Write
$$\left(\frac{\sin x }{x} \right)^\frac{1}{x} = \exp\left( \frac{1}{x} \ln \left(\frac{\sin x }{x} \right)\right).$$
*$\exp$ is continuous, $\lim_{x\to 0} \frac{\sin x }{x}=1$ and $\ln 1=0$.
*Evaluate $\lim_{x\to 0}\frac{1}{x} \ln \left(\frac{\sin x }{x} \right)$ by using L'Hospital.
For step 3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Polynomials $P(x,y)$ with nonnegative integer coefficients such that $P(x,y) \equiv 1 \text{ (mod } x+y-1)$ and $P(1,1) = n$. In 1971 Richard Guy sent a letter to Neil Sloane outlining some integer sequences. One of these sequences, A279196, was added to the OEIS by Neil only in December of 2016:
A279196: Number of po... | I think I figured it out (but I would still love to see a proof that this technique is exhaustive!)
It seems that (at least the small) values can be built up recursively:
To create a list of polynomials for $a(n)$, for each term (on the right) and polynomial (on the left) in $a(n-1)$, take a term from the left-hand sid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the maximum of the value a such $\sum_{cyc}\sqrt{\frac{x}{ax+y+z}}\le 3\sqrt{\frac{1}{2+a}}$ Find the maximum of the value $a$ such foy any real postive numbers $x,y,z$ have
$$\sqrt{\dfrac{x}{ax+y+z}}+\sqrt{\dfrac{y}{ay+z+x}}+\sqrt{\dfrac{z}{az+x+y}}\le3\sqrt{\dfrac{1}{2+a}}$$
I conjecture $a>0?$
| It's wrong for $a\rightarrow0^+$. Try $x\rightarrow+\infty.$
By the way, just by Jensen your inequality is true for all $a\geq\frac{3}{4}.$
Also, by Vasc's LCF Theorem it's enough to prove your inequality for $z=y$
and since our inequality is homogeneous, we can assume $y=z=1$,
which gives a minimal value of $a$, fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Second-order linear differential equation of the form $x^2 y'' + (ax-b)y' - ay =0$ I need to solve the following differential equation
\begin{equation}
x^2 y'' + (ax-b)y' - ay =0
\end{equation}
with $a,b>0$, $x\geq 0$ and $y(0)=0$. The power series method will fail since there is a singularity at $x=0$, while the form ... | $$x^2 y'' +(ax-b)y'-ay=0
$$
Start by removing the leading behavior for small $x$. Near $x=0$ the leading terms in the equation are
$$
-by'-ay \approx 0
$$
(this is valid as long as in the end, as $x\to 0$, $y''$ does not grow as fast as $y'/x^2$ or $y/x^2$, and $y'$ does not grow as fast as $\frac{y}x$).
The solutio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Power series expansion of a holomorphic function One must show that equality$$\sum_{n=0}^\infty n^2z^n = \frac{z(z+1)}{(1-z)^3} $$ holds for $z\in\mathbb C$ and $|z|<1$.
I tried to expand $f(z):=\frac{z(z+1)}{(1-z)^3}$ into power series where $ z_0=0$ is the center and coefficients given by $a_n=\frac{1}{2\pi i}\int_{... | $\sum_{n \ge 0} n^2z^n$ is absolutely convergent for $|z| < 1$ and so the following manipulations are valid for $|z| < 1$ :
$(1-z)^3(z+4z^2+9z^3+16z^4+\ldots)
\\ = (1-z)^2(z-z^2+4z^2-4z^3+9z^3-9z^4+16z^4+\ldots)
\\ = (1-z)^2(z+3z^2+5z^3+7z^4+\ldots)
\\ = (1-z)(z-z^2+3z^2-3z^3+5z^3-5z^4+7z^4+\ldots)
\\ = (1-z)(z+2z^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find the locus of the point of intersection of the tangents of the ellipse $b^2x^2+a^2y^2=a^2b^2$ at the points $\phi$ and $\frac{\pi}{2}+\phi$. - "a" and "b" are constants.
- "x" and "y" are variables.
- pi=22/7
- ø= an angle
I try this question at tangent method. But perhaps I do some mistake.
I take the tangents wh... | Let $(x',y')$ be the point of intersection, then the equation of the polar line (i.e. the chord in this case) is
$$\frac{x'x}{a^2}+\frac{y'y}{b^2}=1 \tag{1}$$
End points of the chord:
$$(a\cos \phi,b\sin \phi) \: \text{ and } \: (-a\sin \phi,b\cos \phi)$$
Hence the equation of the chord is
\begin{align*}
\frac{y-b\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Solve the inequality $\sin(x)\sin(3x) > \frac{1}{4}$
Find the range of possible values of $x$ which satisfy the inequation $$\sin(x)\sin(3x) > \frac{1}{4}$$
SOURCE : Inequalities (PDF)( Page Number 6; Question Number 306)
One simple observation is that both $x$ and $3x$ have to positive or negative simultaneously. I ... | $\require{cancel}$
\begin{eqnarray}
\sin x\sin3x &>&\dfrac14\\
2\sin x\sin3x&>&\dfrac12\\
\cos2x-\cos4x&>&\dfrac12\\
\cos2x-2\cos^22x+1&>&\dfrac12~~~~~~~~~~~~~~~~~~,~~~~~\cos2x=t\\
4t^2-2t-1&<&0
\end{eqnarray}
with $\Delta=20$ so $t=\dfrac{2\pm2\sqrt{5}}{8}=\dfrac{1\pm\sqrt{5}}{4}$ are roots and then
$$\dfrac{1-\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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How can you prove $\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$ without much effort?
I will keep it short and take only an extract (most important part) of
the old task.
$$\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$$
What I have done is a lot work and time consuming, I have "simply" s... | Since you want $n+1$ to appear as a factor at the end, I would just leave that as a factor:
$$
\frac{n(n+1)(2n+1)+6(n+1)^2}6 = \frac {n(2n+1) + 6(n+1)} 6 (n+1).
$$
That becomes
$$
\frac {2n^2 + n + 6n + 6} 6 (n+1) = \frac{2n^2 + 7n+6} 6 (n+1) = \frac{(n+2)(2n+3)} 6 (n+1)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2137292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
} |
Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$ Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$
My Attempt:
$$L.H.S=\sec^2 20^\circ + \sec^2 40^\circ +\sec^2 80^\circ$$
$$=\dfrac {1}{\cos^2 20°} +\dfrac {1}{\cos^2 40°} +\dfrac {1}{\cos^2 80°}$$
$$=\dfrac {\cos^2... | Let $z = \cos 20^\circ\,$ then by the triple angle formula $\frac{1}{2}=\cos 60^\circ = 4 z^3 - 3 z$ $\iff 8z^3-6z-1=0$.
By the double angle formula $\cos 40^\circ = 2 z^2 - 1\,$ and $\cos 80^\circ = 2(2z^2-1)^2-1=8z^4-8z^2+1\,$. But $8z^4=z\cdot8z^3=z(6z+1)$ per the previous equation, so $\cos 80^\circ = -2z^2+z+1\,$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2137609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Computing: $\lim_{x\to\infty}\frac{\sqrt{1-\cos^2\frac{1}{x}}\left(3^\frac{1}{x}-5^\frac{-1}{x}\right)}{\log_2(1+x^{-2}+x^{-3})}$ Find the following limit:
$$\lim_{x\to\infty}\frac{\sqrt{1-\cos^2\frac{1}{x}}\left(3^\frac{1}{x}-5^\frac{-1}{x}\right)}{\log_2(1+x^{-2}+x^{-3})}$$
I'm not sure whether my solution is corre... | $\begin{array}\\
\dfrac{\sqrt{1-\cos^2\frac{1}{x}}
(3^\frac{1}{x}-5^\frac{-1}{x})}{\log_2(1+x^{-2}+x^{-3})}
&=\dfrac{\sin(1/x)
(e^\frac{\ln 3}{x}-e^\frac{-\ln 5}{x})}{(1/\ln 2)\ln(1+x^{-2}+x^{-3})}\\
&=\dfrac{(1/x+O(1/x^3))
((1+\ln 3/x+O(1/x^2)-(1-\ln 5/x+O(1/x^2))}{(1/\ln 2)(x^{-2}+x^{-3}+O(x^{-4})}\\
&=\dfrac{(1/x+O(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\frac{\sum_{k=0}^6 \csc^2(a+\frac{k\pi}{7})}{7\csc^2(7a)}$ The question is to evaluate $$\frac{\sum_{k=0}^{6}\csc^2(a+\frac{k\pi}{7})}{7\csc^2(7a)}$$ where $a=\pi/8$ without looking at the trigonometric table.
I tried to transform the $\csc^2$ term to $\cot^2$ term and use addition formula but it made the p... | The result of the summation is:
$\frac{\sum_{k=0}^{6}\csc^2(a+\frac{k\pi}{7})}{7\csc^2(7a)}=7$.
The summation, after some calculation, is reduced to:
$\frac{(A E F+B D F+C D E)+D E F}{7 D E F}=7$
where
$A=4-2\sqrt{2}+(2\sqrt{2}-2)cos(\frac{\pi}{7})$,
$B=-4+2\sqrt{2}+(2-2\sqrt{2})sin(\frac{\pi}{14})$,
$C=-4+2\sqrt{2}+(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
If $n$ is a positive integer such that $n^3 − n − 6 = 0$, then, for every positive integer $m$ with $m \not = n$, $m^3 − m − 6 \not = 0$. Prove, by contradiction, that, if $n$ is a positive integer such that $n^3 − n − 6 = 0$, then, for every positive integer $m$ with $m \not = n$, $m^3 − m − 6 \not = 0$.
Proposition... | Let $f(x)=x^3-x-6.$ Let $n$ be any real number such that $f(n)=0$. Then $f(x)=(x-n)(x^2+Ax+B)$ for all $x,$ for some constants $A, B.$
$$\text {So }\quad x^3-x-6=x^3+x^2(A-n)+x(B-nA)+(-nB)$$ for all $x.$ This requires $0=A-n$ and $-1=B-nA ,$ so $A=n$ and $B=nA-1=n^2-1.$
Now if $f(m)=0$ with $m\ne n$ then $0=m^2+Am+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Probabilty of a certain run in a 1000 coin flips So a run of $x$ heads has the probability $0.5^x$.
Thefore a run of $9$ would be $0.5^9 = 0.001953125$.
But what is the chance of that run of $9$ occuring in $1000$ coinflips?
Can anyone please explain how to work this out?
|
Here is an answer based upon generating functions. We start with a generating function for words of a two character alphabet $V=\{T,H\}$ which counts words with no consecutive equal characters at all.
These words are called Smirnov or Carlitz words. See example III.24 Smirnov words from Analytic Combinatorics by Phi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve system using $1+j+j^{2}=0$ Exercise :
Solve the following system:
$$
(S):\begin{cases}
A+B+C=2^{n} & \\
A+j\;B+j^{2}\;C=(−1)^{n}\;j^{2n} & \\
A+j^{2}\;B+j\;C=(−1)^{n}\;j^{n} & \\
\end{cases}
$$ Using the coefficients $1,j,j^{2}$ and $1+j+j^{2}=0$
Solution :
\begin{aligned}
A&=\dfrac{1}{3}\left( 2^{n}+(-1)^{n}j^... | Hint: In order to get $A$ add all equations note that the coefficients of B and C will be equal to zero. So you directly get $A$. Then Multiply the second equation with $j$ and subtract the second and third equation, don't forget to plug in $A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove this conjecture $x_{n}(n\ge 4)$ can't be an integer.
Define sequence $x_{1}=1$,and $x_{n+1}=1+\dfrac{n}{x_{n}},n\ge 1$.
I found
$$x_{2}=2,x_{3}=2,x_{4}=1+\dfrac{3}{x_{3}}=\dfrac{5}{2}, x_{5}=1+\dfrac{4}{2.5}=\dfrac{13}{5},x_{6}=1+\dfrac{5}{x_{5}}=\dfrac{38}{13}$$
$$x_{7}=1+\dfrac{6}{x_{6}}=\dfrac{58... | This answer uses that for $n\ge 4$,
$$\frac{1+\sqrt{4n-3}}{2}\lt x_n\lt \frac{1+\sqrt{4n+1}}{2}\tag1$$
The proof for $(1)$ is written at the end of this answer.
Multiplying $(1)$ by $2$, subtracting $1$ and squaring, we get
$$4n-3\lt 4(x_n^2-x_n)+1\lt 4n+1$$
Then, subtracting $1$ and dividing by $4$, we get
$$n-1\lt x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Solve $\lim_{x \rightarrow +\infty}\frac{\sqrt{x}+x}{x+\sqrt[3]{x}}$ I tried first without L'Hôpital's rule:
$$\lim_{x \rightarrow +\infty}\frac{\sqrt{x}+x}{x+\sqrt[3]{x}} =
\frac{\sqrt{x}}{\sqrt[3]{x}} \cdot \frac{1+\frac{x}{\sqrt x}}{1+ \sqrt[3] x} = \frac{\sqrt x}{\sqrt[3] x} \cdot \frac{\frac{\sqrt x+x}{\sqrt x}}{... | Hint:
$$\frac{\sqrt{x}+x}{x+\sqrt[3]{x}}=\frac{\frac1{\sqrt x}+1}{1+\frac1{\sqrt[3]{x^2}}}$$
It's that simple!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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If $f(n+1)=(-1)^{n+1}n-2f(n)$ and $f(1)=f(1986)$ find $f(1)+f(2)+f(3)+...+f(1985)$ I got a solution but it was really messy with f(1)=$\frac{1987}{2^{1985}+1}$ and an explicit formula for $f(x)$. Then the whole mess was a crocodile (With many $2^{1985}$ and $4^{1985}$ terms) when I solved it so I assume I made a mistak... | Here's a fully worked version. The sum we are looking for is ($N = 1986$)
$$
\begin{align}
S = \sum_{n = 1}^{N - 1}f(n) &= \sum_{n = 1}^{N - 1}\left((- 1)^n(n-1) - 2f(n-1)\right)\\
&= \sum_{n = 1}^{N - 1}(-1)^n(n-1) - 2\sum_{n = 1}^{N - 1}f(n-1)\ .
\end{align}
$$
If we look at the first sum on the right side and write ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Can anyone explain how the complex matrix representation of a quaternions is constructed? I am reading some properties of quaternionic matrices and I am unable to understand how can we got such matrix representation. please help in this regards.
| The ring of quaternions $\mathbb{H}$ is isomorphic to the
ring of matrices with complex entries of the form $A =\begin{pmatrix}
x & y
\\
- \bar{y} & \bar{x}
\end{pmatrix} $
For a quaternion $z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$ the isomorphism is given by:
$$
z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k} \quad \map... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Other Idea to show an inequality $\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$ $$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$$
I want to prove this by Induction
$$n=1 \checkmark\\
n=k \to \dfrac{1}{\sqrt 1}+\dfrac{1}{... | By the generalized mean inequality the harmonic mean is no larger than the quadratic mean:
$$
\require{cancel}
\cfrac{n}{\cfrac{1}{\sqrt{1}}+\cfrac{1}{\sqrt{2}}+\cdots+\cfrac{1}{\sqrt{n}}} \;\le\; \sqrt{\frac{(\sqrt{1})^2+(\sqrt{2})^2+\cdots+(\sqrt{n})^2}{n}} = \sqrt{\frac{\cancel{n}(n+1)}{2\,\cancel{n}}}
$$
$$
\implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
Is this limit $\lim\limits_{x \to\, -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} = 0$? Is this limit $\lim\limits_{x \to\, -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} = 0$?
It's in indeterminant form, 0/0 when $x$ approaches $-8$. So I used LHopital's rule and got $$-\frac{3x^{\frac{2}{3}}}{2\sqrt{1-x}}$$ plug in $-8$ it is $-2(-1)... | With substiuation $x+8=t$ we have
$$\lim_{x\to-8}\dfrac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}=\lim_{t\to0}\frac{\sqrt{9-t}-3}{\sqrt[3]{t-8}+2}=\frac{\lim_{t\to0}\frac{\sqrt{9-t}-3}{t}}{\lim_{t\to0}\frac{\sqrt[3]{t-8}+2}{t}}=\frac{(\sqrt{9-t})'\Big|_{t=0}}{(\sqrt[3]{t-8})'\Big|_{t=0}}=\frac{-\frac16}{\frac{1}{12}}=\color{blue}{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$ \lim_{n \to \infty} \frac{\sqrt{n}(\sqrt{1} + \sqrt{2} + ... + \sqrt{n})}{n^2} $ How do I find the following limit?
$$ \lim_{n \to \infty} \frac{\sqrt{n}(\sqrt{1} + \sqrt{2} + ... + \sqrt{n})}{n^2} $$
Can limit be find by Riemann sums?
$$\lim_{n\to \infty}\sum_{k=1}^{n}f(C_k)\Delta{x} = \int_{a}^{b}f(x)\,dx$$
I'm not... | By Stolz we have $$ \lim\limits_{n \to \infty} \frac{\sqrt{n}(\sqrt{1} + \sqrt{2} + ... + \sqrt{n})}{n^2}= \lim\limits_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + ... + \sqrt{n}}{n^\frac{3}{2}}= \lim\limits_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n^3}-\sqrt{(n-1)^3}}=$$
$$= \lim\limits_{n \to \infty} \frac{\sqrt{n}\left(\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
$\lim \limits_{x\to 0} \frac {4^{\sin x}-2^{\tan 2x}}{3^{\sin 2x}-9^{\tan x}}=?$ find the limit :
$$\lim_{x\to 0} \frac {4^{\sin x}-2^{\tan 2x}}{3^{\sin 2x}-9^{\tan x}}=?$$
my try :
$$\lim_{x\to 0} \frac {2^{2\sin x}-2^{\tan 2x}}{3^{\sin 2x}-3^{2\tan x}}=$$
$$\lim_{x\to 0} \frac {2^{2\sin x}-2^{\frac{\sin 2x}{\cos2x}}}... | Hint. One may write, by using a Taylor series expansion, as $x \to 0$,
$$
\begin{align}
\frac {4^{\sin x}-2^{\tan 2x}}{3^{\sin 2x}-9^{\tan x}}&= \frac {2^{\tan 2x}\cdot \left(2^{2\sin x-\tan 2x}-1\right)}{3^{2\tan x}\cdot \left(3^{\sin 2x-2\tan x}-1\right)}=\frac {2^{\tan 2x}\cdot \left(-x^3 \ln 8+o(x^3)\right)}{3^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$S=1+2i+3i^2+4i^3+\dots+(n+1)i^n$; $4\mid n$; closed form for $S$ $$S=1+2i+3i^2+4i^3+\dots+(n+1)i^n$$
where $4\mid n$. How can I simplify this exprerssion so as to obtain a general expression?
| Let $n=4m$.
$$\begin{align}
S&=\boxed{\begin{array}
&\;\;\;\;1&+2i&+3i^2&+4i^3\\
+5i^4&+6i^5&+7i^6&+8i^7\\
+\vdots\\
+(4m-3)i^{4m-4}&+(4m-2)i^{4m-3}&-(4m-1)i^{4m-2}&+4mi^{4m-1}\\
+(4m+1)i^{4m}\\
\end{array}}\\\\
&=\boxed{\begin{array}
&\;\;\;1&+2i&-3&-4i\\
+5&+6i&-7&-8i\\
+\vdots\\
+(4m-3)&+(4m-2)i&-(4m-1)&-4mi\\
+(4m+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How to calculate this sum. I am trying to solve an excercise and i come across the following sum
$ \sum_{k=1}^n \frac{(k+1)(k^3-2k+2)}{k(k+2)} $
I put it in Wolfram alpha and it says that it is equal to:
$\frac{n(2n^4+6n^3+2n^2+3n+11)}{6(n+1)(n+2)}$
but how can i prove this?
| You may notice that
$$ \frac{(k+1)(k^3-2k+2)}{k(k+2)} = \left(\frac{1}{k}-\frac{1}{k+2}\right)+2\binom{k}{2} \tag{1}$$
hence by creative telescoping and the hockey stick identity
$$ \sum_{k=1}^{n}\frac{(k+1)(k^3-2k+2)}{k(k+2)} = 2\binom{n+1}{3}+\frac{3}{2}-\frac{1}{n+1}-\frac{1}{n+2}.\tag{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2152099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $\lim_{x \rightarrow 0} \frac{e^x+e^{-x}-2}{x^2+2x}$ without using L'Hopital's rule I tried:
$$\lim_{x \rightarrow 0} \frac{e^x+e^{-x}-2}{x^2+2x} = \\
\frac{e^{x}(1+e^{-2x}-\frac{2}{e^x})}{x(x-2)} = \frac{e^x(1+e^{-2x})-2}{x(x-2)} = \frac{e^x(1+e^{-2x})}{x(x-2)} - \frac{2}{x(x-2)} = \\
\frac{1+e^{-2x}}{x} \cdot ... | Aside from not carelessly turning $x(x+2)$ into $x(x-2)$, partial fractions is the wrong tool for this job. If you are allowed to used Taylor series, you can see that
$$
e^x + e^{-x} - 2 = x^2 + O(x^x)$$
so you get the expression $$\frac{x^2}{x(x+2)} = \frac{x}{x+2} \to 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2155353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$ How to prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$? Factoring $504$ yields $2^3 \cdot 3^2 \cdot 7$. Divisibility by $8$ can be easily seen, because if $n$ is even then $8 | n^3$, else $(n^3 - 1)$ and $(n^3 + 1)$ are both even and one of these is divi... | Hint -
We have 504 =$(2^3-1)2^3(2^3+1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
Max and Min$ f(x,y)=x^3-12xy+8y^3$ First I solved $f_y=0$ then plugged in my variable into $f_x$ to get an output and then plugged that output back into $f_y$ to get a point and did this again for $f_x$ I think I screwed up on the $(-7,\sqrt{3})$
$f_y=-12x+24y^2$
$-12x+24y^2=0$
$-12x=24y^2$
$x=2y^2$
$f_x(2y^2,y)=3(2y^2... | $\hskip 2.1in$
Given function $$f(x,y) = x^3 - 12xy + 8y^3$$ $$ \Rightarrow f_x = 3x^2 - 12y, f_y = 24y^2 -12x = 0 $$
Which on solving gives two stationary points $(2,1)$ and $(0,0)$.
Now, consider hessian matrix $$H = \begin{bmatrix} 6x & -12 \\ -12 & 48y \end{bmatrix} $$
For $(2,1)$, $|H| > 0$ and $f_{xx} ( 2,1) > 0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $f_k(x)=\frac{1}{k}\left (\sin^kx +\cos^kx\right)$, then $f_4(x)-f_6(x)=\;?$ I arrived to this question while solving a question paper. The question is as follows:
If $f_k(x)=\frac{1}{k}\left(\sin^kx + \cos^kx\right)$, where $x$ belongs to $\mathbb{R}$ and $k>1$, then $f_4(x)-f_6(x)=?$
I started as
$$\begin{align... | Hint: let $f(x):=f_4(x)-f_6(x)$. Then show that $f'(x)=0$ for all $x$. Hence $f$ is constant. Furthermore: $f(0)=\frac{1}{12}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Triangular inequality extended Let $a,b,c,d,e,f\in\mathbb{R}$ all positive or zero, such that $a\leq c+d$ and $b\leq e+f$ show that:
$$\sqrt{a^2+b^2}\leq\sqrt{c^2+e^2}+\sqrt{d^2+f^2}$$
Some hint or auxiliar inequality that would help? I've done many attempts but write them will take a lot of time, I tried to use the ar... | Just $$\sqrt{c^2+e^2}+\sqrt{d^2+f^2}\geq\sqrt{(c+d)^2+(e+f)^2}\geq\sqrt{a^2+b^2}$$
If I don't see the triangle inequality I can make the following.
By C-S
$$\sqrt{c^2+e^2}+\sqrt{d^2+f^2}=\sqrt{c^2+e^2+d^2+f^2+2\sqrt{(c^2+e^2)(d^2+f^2)}}\geq$$
$$\geq\sqrt{c^2+e^2+d^2+f^2+2(cd+ef)}=\sqrt{(c+d)^2+(e+f)^2}\geq\sqrt{a^2+b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding $\lim_{n\to\infty}(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}})$ I'm trying to find $\lim_{n\to\infty}(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}})$.
*
*I tried to use the squeeze theorem, failed.
*I tried to use a sequence defined recurs... | Hint:
$$n\frac{1}{\sqrt{n^2+n}}\leq\sum\frac{1}{\sqrt{n^2+n}}\leq n\frac{1}{\sqrt{n^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
how to sum this series of combinations I have to calculate this
\begin{align}
& (1-p)^{N-1}+\frac{1}{2}C_{N-1}^{N-2}(1-p)^{N-2}p+\frac{1}{3}C_{N-1}^{N-3}(1-p)^{N-3}p^{2}+...+\frac{1}{N-2}C_{N-1}^{1}(1-p)p^{N-2}\\
& +\frac{1}{N}p^{N-1}
\end{align}
where $C_{n}^{m}=\frac{n!}{m!(n-m)!}$.
Can anyone help to sum this up? ... | The given sum can be written as:
\begin{align*}
\sum_{k=1}^{N} \frac 1k \binom{N-1}{N-k} p^{k-1}(1-p)^{N-k} = (1-p)^{N-1} \sum_{k=1}^{N} \frac 1k \binom{N-1}{N-k} \left(\frac{p}{1-p}\right)^{k-1}
\end{align*}
Set $r=\frac{p}{1-p}$ and observe that
\begin{align*}
\sum_{k=1}^{N} \frac 1k \binom{N-1}{N-k} r^k &= \sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2163670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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$\int \sec^3x dx$ in disguise I found this integral,
$$\int \sqrt{x^2+1}dx$$
on a problem list and I think it is a sneaky way of hiding a $\int \sec^3xdx$ problem but I am not sure if what I did was correct though, because of what happens at the end.
So what I did was use trig-substitution and let $u=\tan\theta$ and ... | It is just how integration works, you plugged in $x=\tan\theta$ to get $\sqrt{1+x^2}$ to equal $\sec\theta$, your answer is completely correct. Just because $\sec\theta=\sqrt{x^2+1}$ does not mean that $\int\sqrt{1+x^2}dx=\int\sec\theta d\theta$ as you recognized because of the chain rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Proving: $\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2\ge\frac{25}2$
For $a>0,b>0$, $a+b=1$ prove:$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge\dfrac{25}{2}$$
My try don't do much, tough
$a+b=1\implies\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{ab}$
Expanding: $\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfra... | I just differentiate.
$$(ab+\dfrac1{ab})(2-\dfrac1{ab})=(x+\dfrac1x)(2-\dfrac1x)=f(x)$$
and $x\leq\dfrac14$
$f'(x)=(1-\dfrac1{x^2})(2-\dfrac1x)+\dfrac1{x^2}(x+\dfrac1x)
=\dfrac{2x^3-2x+2}{x^3}
=2*\dfrac{x^3-x+1}{x^3}>0$
$LHD=4-minf(x)=4-f(\dfrac14)=4-(\dfrac14+4)(-2)=\dfrac{25}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Prove by induction that $n^4+2n^3+n^2$ is divisible by 4
I'm trying to prove by induction that $n^4+2n^3+n^2$ is divisible by 4.
I know that P(1) it's true. Then $ n=k, P(k):k^4+2k^3+k^2=4w$ it's true by the hypothesis of induction.
When I tried to prove $n=k+1, P(k+1):(k+1)^4+(k+1)^3+(k+1)^2 = 4t$,
$$k^4+4k^3+6k^2+... | You are not doing it wrongly but it's easier, with $f(n)=n^4+2n^3+n^2$ , to observe that $f(n)=(n+1)^2n^2,$ so $$f(n+1)-f(n)=(n+2)^2(n+1)^2-(n+1)^2n^2=$$ $$=(n+1)^2 ((n+2)^2-n^2)=$$ $$=(n+1)^2(4n+4)=4(n+1)^3$$ which is a multiple of $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find an indefinite integral of $\int \left | \sin x + \cos x\right | dx$ So, I've already showed that $\sin x + \cos x = \sqrt{2}\sin (x+\frac{\pi}{4})$ and $\sin x + \cos x>0$ on the interval $\left ( -\frac{\pi}{4}+\pi k; \frac{3\pi}{4}+ \pi k\right )$, $\sin x + \cos x<0$ on the interval $\left ( \frac{3\pi}{4}+\pi ... | By a shift and a scaling, this is the same as integrating $|\sin x|$.
Over the first period ($0$ to $\pi$) we have $\displaystyle\int_0^x\sin t\,dt=1-\cos x$, which is $2$ for the full period. For other values of $x$, you add $2$ as many times as necessary to return $x$ in the first period. In other words,
$$\int_0^x\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the limit of function using an epilson delta proof Find $\lim\limits_{x \to 4} \frac{\sqrt{x}-2}{x-4}$ using an epilson delta proof.
I multiplied and divided by the rational conjugate to get
$\frac{\sqrt{x}+2}{x-4}=\frac{1}{\sqrt{x}+2}$ So the limit is probably $\frac{1}{4}$.
I can't seem to figure out how to tu... |
(answer to the edited question)
We claim that $$\lim_{x\to 4}\frac{1}{\sqrt{x}+2}=\frac{1}{4}.$$
Note that $\sqrt{x}+2>1$ and so
$$0<\frac{1}{\sqrt{x}+2}<1. \tag 1$$
Also, $$x-4=(\sqrt{x}+2)(\sqrt{x}-2).\tag 2$$
Now,
$$\begin{align}
\bigg|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\bigg|&=\bigg|\frac{2-\sqrt{x}}{4(\sqrt{x}+2)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2176888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Determine the eigenvector and eigenspace and the basis of the eigenspace The yellow marked area is correct, so don't check for accuracy :)
$A=\begin{pmatrix} 0 & -1 & 0\\ 4 & 4 & 0\\ 2 & 1 & 2
\end{pmatrix}$ is the matrix.
Characteristic polynomial is $-\lambda^{3}+6\lambda^{2}-12\lambda+8=0$
The (tripple) eigenva... | You must solve $(A - 2I)\vec{x} = \vec{0}$, i.e.
$$
\begin{pmatrix}
-2 & -1 & 0\\ 4 & 2 & 0\\ 2 & 1 & 0
\end{pmatrix}
\begin{pmatrix}x\\ y\\ z\end{pmatrix}
=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}
$$
Note that all equations are multiple of each other, so let's leave the last one to use and eliminate all others. You... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2177817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Is this matrix diagonalisable or not? We will only be looking at one specific eigenvalue, eigenspace (there are three eigenvalues in total, I know two of them are fine and last one seems not but as I'm not sure I need to ask you).
We have matrix $A=\begin{pmatrix}
3 & -1 & 0\\
2 & 0 & 0\\
-2 & 2 & -1
\end{pmatr... | Another way of saying that is- no matter how many distinct eigenvalues an by n matrix has, it is diagonalizable if and only if there are n independent Eigenvectors. Of course, eigenvectors corresponding to distinct eigenvalues are necessarily independent so if an n by n matrix has n distinct eigenvalues then it must b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2179721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Summation of $n/3^n$ $$\sum_{n=1}^\infty \frac{n}{3^n}$$
How do you find the sum?
I don't know how to start this problem and no other website I found talks about a problem like this.
| Approach 1:
$$\sum_{n=1}^\infty \frac{n}{3^n} = \frac{1}{3}\left.\sum_{n=1}^\infty \frac{d}{dx}x^{n}\right|_{x=1/3} = \frac{1}{3} \left.\frac{d}{dx} \frac{x}{1-x} \right|_{x=1/3} = \frac{1}{3} \cdot \frac{1}{(1-1/3)^2}= \frac{3}{4}.$$
Approach 2 (no derivatives):
\begin{align}
S &:= \sum_{n=1}^\infty \frac{n}{3^n}\\
S/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Noob question about $\int \frac{1}{\sin(x)}dx$ I manually integrate $\int \frac{1}{\sin(x)}dx$ as $$\int \frac{\sin(x)}{\sin^{2}(x)}dx = -\int \frac{1}{\sin^{2}(x)}d\cos(x) = \int \frac{1}{\cos^{2}(x) - 1}d\cos(x).$$
After replacing $u = \cos(x)$,
$$\int \frac{1}{u^{2} - 1}du = \int \frac{1}{u^{2} - 1}du = \frac {1} ... | The integral of $1/x$ is not $\ln(x)$, but $\ln|x|$. See What is the integral of 1/x?
Then, in your case $\frac{1}{2}\int du \left( \frac{1}{u-1} - \frac{1}{u+1} \right) = \frac{1}{2}\ln\left( \frac{|u-1|}{|u+1|} \right) + C$. When you put back $u=\cos(x)$, the expression $|u-1|$ becomes $|\cos(x)-1|=1-\cos(x)$ and $|u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Incorrectly solving the determinant of a matrix Compute $det(B^4)$, where $B =
\begin{bmatrix}
1 & 0 & 1 \\
1 & 1 & 2 \\
1 & 2 & 1
\end{bmatrix}
$
I created
$C=\begin{bmatrix}
1 & 2 \\
1 & 1 \\
\end{bmatrix}
$ and
$
D= \begin{bmatrix}
1 ... | Use row reduction to calculate the determinant:
\begin{align}
\begin{vmatrix}1&0&1\\1&1&2\\1&2&1\end{vmatrix}=\begin{vmatrix}1&0&1\\0&1&1\\0&2&0\end{vmatrix}=\begin{vmatrix}1&0&1\\0&1&1\\0&0&-2\end{vmatrix}
\end{align}
hence $\;\det B=1\cdot1\cdot(-2)=-2$, so $\;\det B^4=(\det B)^4=16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Menelaus theorem & collinear points From vertex C of the right triangle ABC height CK is dropped and in triangle ACK bisector CE is drawn. Line that passes through point B parallel to CE meets CK at point F. Prove that line EF divides segment AC in halves.
So far I have:
Construct point M on AC such that AM=MC. WE wa... | Let the line through point $E$ and orthogonal to $CA$ intersect $CA$ at point $L$. Let the line through point $F$ and orthogonal to $BC$ intersect $BC$ at point $N$. Since $AB$ is orthogonal to $CK$ and $CE$ is the angle bisector of angle $\angle \, ACK$,
$$LE = KE$$
Let $\angle \, BAC = \alpha = \angle \, CAK$. Then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
prove that $\lim_{x \to 1}\frac{x^{2}-x+1}{x+1}=\frac{1}{2}$ Please check my proof
Let $\epsilon >0$ and $\delta >0$
$$0<x<\delta \rightarrow \frac{x^{2}-x+1}{x+1}-\frac{1}{2}<\epsilon $$
$$\frac{2x^{2}-2x+2-x+1}{2x+2}<\epsilon $$
$$\frac{2x^{2}-3x+1}{2x+2}<\epsilon $$
since $\frac{2x^2-3x+1}{2x+2} <\frac{2x^{2}-3x+... | There was an error in the second line of the development.
Note that
$$\begin{align}
\left|\frac{x^2-x+1}{x+1}-\frac12\right|&=\left|\frac{2x^2-3x+1}{2(x+1)}\right|\\\\
&=\left|\frac{(2x-1)(x-1)}{2(x+1)}\right|
\end{align}$$
Now, bound $|x-1|$ by something somewhat arbitrary such as $0<x<2$. Then $1<x+1<3$ and $-1<2x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2187405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove that the spectrum of $K_n$ is $((n-1)^1, (-1)^{(n-1)})$ I am trying to prove that the spectrum of the complete graph $K_n$ is $((n-1)^1, (-1)^{(n-1)})$ (where superscripts denote multiplicities of eigenvalues, not exponents). I have part of the proof but having trouble completing it.
The adjacency matrix $A(K_n)$... | Hint: Try to use induction with the Laplace expansion of the determinant to get the characteristic polynomial of $A(K_n)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Limit without l'Hopital or Taylor series: $\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}$ find the limit without l'Hôpital and Taylor rule :
$$\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}=?$$
My Try :
$$\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}\\=\lim\limits_{x \to 0}\frac{x\cos x \sin x- \sin x\sin x}... | By a scaling of the variable, $$L:=\lim_{x\to 0}\frac{x\cos x-\sin x}{x^3}=\lim_{x\to 0}\frac{3x\cos3x-\sin3x}{27x^3}.$$
Then by the triple angle formulas,
$$3x\cos3x-\sin3x=3x\cos x(1-4\sin^2x)-3\sin x+4\sin^3x\\
=(3-4\sin^2x)(x\cos x-\sin x)-8x\cos x\sin^2x,$$
so that
$$L=\lim_{x\to0}(3-4\sin^2x)\cdot\frac L{27}-\lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2189466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Power series representation of $ f(x) = \frac{(x+2)^2}{x^2+1} $ I'm trying to find the power series representation of $ f(x) = \frac{(x+2)^2}{x^2+1} $
Here is what I tried:
$$
f(x) = \frac{((x+2)^2}{x^2+1}=(x+2)^2\frac{1}{1-(-x^2)}
$$
$$
\sum_{n=1}^\infty(-x^2)^{n-1}=1-x^2+x^4-x^6+x^8-+...
$$
$$
f(x)=(x+2)^2[1-x^2+x^4-... | I think it's better to write
$$f(x) = \frac{(x+2)^2}{x^2+1}=\frac{x^2+4x+4}{x^2+1}=1+\frac{4x+3}{x^2+1}=1+(4x+3) (1-x^2+x^4-x^6+x^8+\cdots)=4+4x-3x^2-4x^3+\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2190410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does the sequence $x_{n+1}=x_n+\frac{1}{x_n}$ converge or diverge? Set $x_1=a$, where $a > 0$ and let $x_{n+1}= x_n + \frac{1}{x_n}$. Determine if the sequence $ \lbrace X_n \rbrace $ converges or diverges.
I think this sequence diverges since Let $x_1= a > 0$ and $x_{n+1}= x_n + (1/x_n)$ be given. Let $x_1=2$ since $a... | As others have shown, $x_{n+1} \geq x_n > 0$ for all $n \in \mathbb{N}$. The important point that we will be using is the fact that each term is strictly positive (i.e., non-zero).
Suppose that $(x_n)$ is convergent. Let $L$ be its limit. $L$ is either zero or non-zero. We shall first show that $L \neq 0$. Consider the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2190592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Convergence of an infinite sequence I want to know if
$$
B=\sqrt{2+\sqrt{4+\sqrt{8+\sqrt{16+...}}}}
$$
converges to a finite value or not. $B$ can be written as
$$
B=\sqrt{2^1+\sqrt{2^2+\sqrt{2^3+\sqrt{2^4+...}}}}.
$$
However, I have no idea the next step. Any suggestion, idea, or comment is welcome, thanks!
| We have
$$B=\sqrt{2^1+\sqrt{2^2+\sqrt{2^3+\sqrt{2^4+\cdots}}}}<\sqrt{2^{2^0}+\sqrt{2^{2^1}+\sqrt{2^{2^2}+\sqrt{2^{2^3}+\cdots}}}}$$
We then note that
$$A=\sqrt{2^{2^0}+\sqrt{2^{2^1}+\sqrt{2^{2^2}+\sqrt{2^{2^3}+\cdots}}}}\\A=\sqrt{2+A\sqrt2}\implies A=\frac{\sqrt2+\sqrt{10}}2$$
Thus, $B$ is bounded above, so it converge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Surds question grade 9 I am a student and I need help answering this question.
Simplify:
$\frac{6}{\sqrt{28}}$-$\frac{9}{\sqrt{63}}$
What I did:
$\frac{6}{\sqrt{28}}$- $\frac{9}{\sqrt{63}}$
=$\frac{6}{\sqrt{7×4}}$ - $\frac{9}{\sqrt{7×4}}$
= $\frac{6}{2\sqrt{7}}$ - $\frac{9}{3\sqrt{7}} $
$\frac{6}{2\sqrt{7}}$ × $\frac... | Hint: $\frac{6}{\sqrt{28}} = \frac{3}{\sqrt{7}} = \frac{9}{\sqrt{63}}$
You made some typos in your work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Simplification of $\binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50}$ There was a post on this web site an hour ago asking for the sum of
\begin{equation*}
\binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50}
\end{equation*}
expressed as a single bi... | Solution
For any positive integer $n$ and any nonnegative integer $r \leq n$,
\begin{equation*}
\binom{n}{r} = \binom{n}{n - r} .
\end{equation*}
According to Vandermonde's Identity,
\begin{align*}
&\binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50} \\
&\qquad = \binom{50}{0}\binom{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2195269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ I am supposed to proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ for $n$ being a natural number (integer, $n > 0$).
What I have so far:
Basis: $n = 1$
\begin{align}
3^{2 \cdot 1-1} + 2^{2 \cdot 1-1} & = 3^1 + 2^1\\
... | $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2195460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Compute : $\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$ Question: Compute this integral
$$\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$$
My Approach:
$$\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$$
$$=\int\frac{x+2}{\sqrt{x^2+5x}+6}\times \frac{{\sqrt{x^2+5x}-6}}{{\sqrt{x^2+5x}-6}}~dx$$
$$\int\frac{(x+2)(\sqrt{x^2+5x})}{x^2+5x-36}~dx~~- \underb... | all of these answers are so stupendously complicated yo! here's a much better method:
as stated, we need to deal with integral of 1/(sqrt(x^2+5x)+6)
write x^2+5x= (x+5/2)^2 - (5/2)^2
now substitute x+5/2 = 5y/2
now, our main focus after taking the constants out becomes the integral of
1/(sqrt(y^2 - 1) + 12/5)
put arcs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
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Proof that $y^2=x^3+21$ has no integral solution with elementary methods? I tried to prove that $$y^2=x^3+21$$ has no integral solution in the way as shown here : http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf
My work so far : $x$ cannot be even because we would have $y^2\equiv 5\mod 8$ , which is... | This will not be a complete solution. Note that from your equations 1) and 3), two of the three possible cases of $x \pmod{3}$ can be eliminated:
If $x \equiv 0 \pmod{3}$, then from 1), $y^2 + 6 \equiv 0 \pmod{27} \longrightarrow y^2 \equiv 21 \pmod{27}$, which is impossible.
If $x \equiv 2 \pmod{3}$, then from 3), $y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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Show that for $p$ to be odd prime and $p \equiv 3$ mod $4$, then $x^2+y^2 = p$ has no integer solution Show that for $p$ to be odd prime and $p \equiv 3$ mod $4$, then $x^2+y^2 = p$ has no integer solution. I have no idea how can i apply quadratic reciprocity to the equation $x^2+y^2 = p$ or should use other method. ... | Simplest way is to note that if $x = 2k + m; m = 0,1$ and $y= 2j + n; n=0,1$ then $x^2 + y^2 = 4(k^2 + j^2 + km + jn) + m^2 + n^2 \equiv m^2 + n^2 \equiv 0,1,2 \mod 4 \not \equiv 3\mod 4$
$p$ being prime has nothing to do with it.
... or $x \equiv 0, \pm 1, 2 \mod 4 \implies x^2 \equiv 0,1 \mod 4$ so $x^2 + y^2 \equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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$\sqrt{x^{2}+\frac{1}{n}}$ converges uniformly to $\left|x\right|$ Im trying to prove that $f_{n}\left(x\right)=\sqrt{x^{2}+\frac{1}{n}}$ converges uniformly to $f(x) = \left|x\right|$ in $[-1,1]$.
So for evary $\varepsilon$ exists $N \in \mathbb{N}$ s.t for all $n>N$ and for all $x \in [-1,1]$ $\left|f_{n}\left(x\... | If $a,b\ge 0,$ then $\sqrt {a+b} \le \sqrt a + \sqrt b.$ Proof: Square both sides. We conclude
$$\sqrt {x^2+1/n} \le \sqrt {x^2}+\sqrt {1/n},$$
which will lead to your result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using 3 equations, find $2a-b+c$. From the problem, I extract the following 3 equations.
*
*$a+b+c=10$
*$ab+bc+ca=31$
*$abc=30$
The question is to find $(2a-b+c)$.
Using the equation $(a+b+c)^2 = a^2 + b^2 + c^2 +2 (ab+bc+ca)$, I found $a^2+b^2+c^2=38$.
Using the equation $a^3+b^3+c^3 - 3abc = (a+b+c)(a^2 + b^2... | By Viete $a$, $b$ and $c$ are roots of the following equation:
$$x^3-10x^2+31x-30=0$$ or
$$(x-2)(x-5)(x-3)=0$$
and the rest for you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Trouble integrating $e^{ax}\cos bx$ and $e^{ax}\sin bx$ I have to integrate the following indefinite integrals $$\int e^{ax}\sin bxdx~~;~\int e^{ax}\cos bx$$ The procedure I used is the same for both integrals:
*
*Make the change of variables $bx=t$
*Use partial integration twice with $dv=e^{at/b}dt$ until the ini... | Let, $I=\displaystyle\int{e^{ax}\sin{bx}}\ dx$
now by applying by parts rule
\begin{align*}
I&= e^{ax}\int \sin{bx}\ dx-\int\left[\frac{d}{dx}(e^{ax})\cdot\int\sin{bx}\ dx\right]\ dx\\
&= e^{ax}\left(-\frac{\cos{bx}}{b}\right)-\int{ae^{ax}}\cdot\left(-\frac{\cos{bx}}{b}\right)\ dx\\
&=-\frac{e^{ax}\cos{bx}}{b}+\frac{a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \le \frac{n}{n+1}$? I have a series $$a_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \quad n \ge 1$$
For example, $a_3 = \frac{1}{4}+\frac{1}{5}+\frac{1}{6}$.
I need to prove that for $n \ge 1$:
$$a_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \l... | Your inequality can be greatly improved. The sequence given by $a_n = H_{2n}-H_{n}$ is increasing (it is enough to compute $a_{n+1}-a_n$) hence $H_{2n}-H_n\leq \lim_{n\to +\infty}\left(H_{2n}-H_n\right) = \color{red}{\log 2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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find a power series for $f(x)=\ln(x^2+1)$. (a) Find a power series representation for
$f(x) = \ln(1 + x)$.
$$f(x) = \sum _{n=1}^{\infty } \frac{\left(-1\right)^{n-1}x^n}{n}$$
What is the radius of convergence, $R$?
$R = 1$
(b) Use part (a) to find a power series for
$f(x) = x \ln(1 + x)$.
$$f(x)=\sum _{n=2}^{\infty... | From (a),
\begin{equation}
f(x) = \ln (1 + x) = \sum_{n=1}^\infty \frac{\left(-1\right)^{n-1}\left(x^n\right)}{n}.
\end{equation}
Then, (c) becomes
\begin{equation}
\ln (x^2 + 1) = \ln(1 + x^2) = f(x^2) = \sum_{n=1}^\infty \frac{\left(-1\right)^{n-1}\left(x^{2n}\right)}{n}.
\end{equation}
| {
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"source": "stackexchange",
"question_score": "1",
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How have I computed the integral $\int \sin^{3}(3x)\cos(3x)\,dx$ incorrectly? $\displaystyle \int \sin^{3}(3x)\cos(3x)\,dx $
$u = 3x $
$du = 3\,dx $
$3\displaystyle\int \sin^{3}(u)\cos(u)\,du$
$3\displaystyle\int \sin^{2}(u)\sin(u)\cos(u)\,du$
$3\displaystyle\int (1 - \cos^{2}(u))\sin(u)\cos(u)\,du$
$v = \cos(u)$
$dv... | You determined $du=3\,dx$, but substituted $dx = 3\,du$. So your answer is off by a factor of nine.
$$
\int \sin^3(3x) \cos(3x)\,dx = \int \sin^3(u) \cos(u)\cdot \frac{1}{3}\,du
= \frac{1}{3}\int \sin^3(u) \cos(u)\,du
$$
Then proceed as you did. You'll get
$$
\int \sin^3(3x) \cos(3x)\,dx =- \frac{1}{6} \cos^2(3x) -\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $\sqrt[3]{{x\sqrt {x\sqrt[3]{{x \ldots \sqrt x }}} }} = 2$ with 100 nested radicals I have seen a book that offers to solve the following equation:
$$\underbrace {\sqrt[3]{{x\sqrt {x\sqrt[3]{{x \ldots \sqrt x }}} }}}_{{\text{100 radicals}}} = 2$$
The book also contains the answer:
$$x = {2^{\left( {\frac{{5 \ti... | Essentially what is written there is just a big product of $x^{a_n}$ where $a_n$ changes from term to term. The terms $a_n$ follow the following sequence:
$$\frac{1}{3}, \frac{1}{3\cdot 2}, \frac{1}{3^2\cdot 2}, \frac{1}{3^2\cdot 2^2},\ldots, \frac{1}{3^{50}\cdot 2^{50}}$$
The equation you have now is
$$\prod_{n=1}^{10... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Identify $\lim\limits_{x \to +\infty } x^2 \left(\sqrt{x^4+x+1}-\sqrt{x^4+x+5}\right)$
Identify $$\lim\limits_{x \to +\infty } x^2 \left(\sqrt{x^4+x+1}-\sqrt{x^4+x+5}\right)$$
My Try :
$$\sqrt{x^4+x+1}=\sqrt{x^4(1+\frac{1}{x^3}+\frac{1}{x^4})}=x^2\sqrt{(1+\frac{1}{x^3}+\frac{1}{x^4})}$$
Now : $$\frac{1}{x^3}+\frac{1}... | $$
\begin{aligned}
\lim _{x\to \infty }\left(x^2\:\left(\sqrt{x^4+x+1}-\sqrt{x^4+x+5}\right)\right)
& = \lim _{x\to \infty }\left(-\frac{4x^2}{\sqrt{x^4+x+1}+\sqrt{x^4+x+5}}\right)
\\& = \lim _{x\to \infty }\left(-\frac{4x^2}{x^2\sqrt{1+\frac{x}{x^4}+\frac{1}{x^4}}+x^2\sqrt{1+\frac{x}{x^4}+\frac{5}{x^4}}}\right)
\\& \a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Solve $\frac{2}{\sin x \cos x}=1+3\tan x$ Solve this trigonometric equation given that $0\leq x\leq180$
$\frac{2}{\sin x \cos x}=1+3\tan x$
My attempt,
I've tried by changing to $\frac{4}{\sin 2x}=1+3\tan x$, but it gets complicated and I'm stuck. Hope someone can help me out.
| Multiply by LHS and RHS by $\sin x\cos x$ :
$$3\sin ^2x +\sin x\cos x -2 =0 \implies 3\cos^2 x-\sin x\cos x -1 =0$$
$$\implies \frac32 (2\cos ^2 x -1) -\frac{2\sin x \cos x }{2}=\frac{-1}{2} \implies 3\cos {2x} - \sin{2x}=-1$$
Now, squaring both the sides, you get :
$$5 \sin^2{2x}-\sin{2x}-4=0 \implies \sin {2x} =1~ \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Find UMVUE of $6\theta^2$ given $f(x\mid\theta) = \frac{1}{2\theta^2} e^{\frac{-\sqrt{x}}{\theta}} I_{(0,\infty)}(x)$ Given $X_1, X_2,\ldots, X_n$ are i.i.d rvs with pdf $f(x\mid\theta) = \frac{1}{2\theta^2} e^{\frac{-\sqrt{x}}{\theta}} I_{(0,\infty)}(x)$ for $\ \theta > 0$.
(a) Find the UMVUE of $\ 6\theta^2$, and t... | It is no need to compute conditional expectations. Let $$T=\overline{\sqrt{X}}=\frac{\sum_{i=1}^n \sqrt{X_i}}{n}$$
You have already found its expectation:
$$\mathbb ET=\mathbb E\sqrt{X_1}=2\theta.$$
Let us find UMVUE of $6\theta^2$ as $\theta^*=c(n)T^2$. Firstly find $\mathbb E T^2$:
$$
\mathbb ET^2 = \text{Var}\,T+\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the general formula of the sequence $ a_n = 7a_{n-2} + 6a_{n-3} $ Given the equation $ a_n = 7a_{n-2} + 6a_{n-3} $, and $ a_0 = a_1 = a_2 = 1 $, how do I find the general equation?
I have tried to express this sequence in matrix form, $ Q = Xb $ as follows
$$ Q = \begin{bmatrix} a_{n+1} \\ a_{n-2} \\ a_{n-3} \... | Try instead $$Q = A_n=\begin{bmatrix} a_{n} \\ a_{n-1} \\ a_{n-2} \\ \end{bmatrix}
\hspace{1cm}
X =\begin{bmatrix} 0&7 & 6 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{bmatrix}
\hspace{1cm}
b = A_{n-1}=\begin{bmatrix} a_{n-1} \\ a_{n-2} \\ a_{n-3} \end{bmatrix}$$
Hence $A_n=X^n A_0=X^{n-3}A_3$
The matrix $X$ can be diagonalize... | {
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"question_score": "2",
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Find the minimum value of $f$ on $\{(a,b,c)\in \mathbb{R^3}\mid a+b+c=1\}$ where $f(a,b,c)=\int_{0}^1 (a+bx+cx^2)^2dx$. Find the minimum value of $f$ on $\{(a,b,c)\in \mathbb{R^3}\mid a+b+c=1\}$ where $f(a,b,c)=\int_0^1 (a+bx+cx^2)^2 \, dx$.
What I did is
$$
\int_0^1 (a+bx+cx^2)^2 \, dx=a^2+a \left( b+\frac{2c}{3} \ri... | Considering a quadric
$$x^2+\frac{y^2}{3}+\frac{z^2}{5}+\frac{yz}{2}+\frac{2zx}{3}+xy=\lambda$$
which touches the plane $x+y+z=1$ at $(a,b,c)$.
The tangent plane will be
$$ax+\frac{by}{3}+\frac{cz}{5}+
\frac{cy+bz}{4}+\frac{az+cx}{3}+\frac{bx+ay}{2}=\lambda$$
Comparing coefficients:
$$a+\frac{b}{2}+\frac{c}{3}=
\frac{a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Maximum value of $a^2c^3+b^2d^3$. Let $a,b,c,d$ be non negative real numbers such that $a^5 + b^5 \le 1$ and $c^5 + d^5 \le 1$. Find the maximum possible value of $a^2c^3 + b^2d^3$.
I tried using AM, GM and some other basic inequalities but they were of no use. Need some hints.
| It can be done easily with Holder's inequality, with $p=5/2, q=5/3$ and $u=(a^2,b^2), v=(c^3,d^3)$ then:
$$\begin{align}a^2c^3+b^2d^3 &= |(a^2c^3,b^2d^3)|_1\\&\leq |(a^2,b^2)|_p |(c^3,d^3)|_q \\&= (a^5+b^5)^{2/5}(c^5+d^5)^{3/5}\leq 1
\end{align}$$
It now suffices to find a case where it is equal to $1$, which is pretty... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Can anything interesting be said about this fake proof? The Facebook account called BestTheorems has posted the following. Can anything of interest be said about it that a casual reader might miss?
Note that \begin{align}
\small 2 & = \frac 2{3-2} = \cfrac 2 {3-\cfrac2 {3-2}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-... | Simple, the problem is the assumption the three dots aka the ellipsis, in the first equation equal the ellipsis in the second. They are not.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 4,
"answer_id": 0
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Prove by Taylor's series method that $|\sin{x}-(x-\frac{x^3}{3!}+\frac{x^5}{5!})|<\frac{1}{7!}$ for $x\in [-1,1]$ Prove by Taylor's series method that $$|\sin{x}-(x-\frac{x^3}{3!}+\frac{x^5}{5!})|<\frac{1}{7!}$$ for $x\in [-1,1]$
Attempt:
Let $f(x)=\sin{x}$, then $f^n(x)=\sin{(\frac{n\pi}{2}+x)}, ~x\in \mathbb{R}$
Then... | By the Taylor expansion with Lagrange remainder,
$$
\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}\cos\xi
$$
for some $\xi\in(-1,1)$ (actually $\xi$ can be chosen between $0$ and $x$, but it's not fully relevant; the only fact we need is that $\xi\ne0$ and $\xi\in[-1,1]$). Therefore
$$
\left|\sin x-x+\frac{x^3}{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Expand binomially to prove trigonometric identity Prompt: By expanding $\left(z+\frac{1}{z}\right)^4$ show that $\cos^4\theta = \frac{1}{8}(\cos4\theta + 4\cos2\theta + 3).$
I did the expansion using binomial equation as follows
$$\begin{align*}
\left(z+\frac{1}{z}\right)^4 &= z^4 + \binom{4}{1}z^3.\frac{1}{z} + \bino... | hint
Take now $z=e^{i\theta} $
and use Euler identity
$$\cos (x)=\frac {e^{ix}+\frac {1}{e^{ix}}}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2217925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Mass of solid inside paraboloid $z=x^2+y^2$ and beneath plane $z=1$ I need to find the mass of the solid inside paraboloid $z=x^2+y^2$ and beneath plane $z=1$ if the density is 1.
Here's what I tried:
$$
M=\int\int_R\int^1_{r^2}r^2dzdA=\int\int_Rr^2z|^1_{r^2}dA=\int\int_R(r^2-r^4)dA=\int^{2\pi}_0\int^1_0(r^2-r^4)rdrd\t... | To find the mass, the quantity we should be integrating is $1$. You appear to be integrating $r^2$!
\begin{align*} \int_{\theta = 0}^{\theta = 2\pi} \int_{r = 0}^{r = 1} \int_{z = r^2}^{z = 1} 1\times r dz dr d\theta = \int_{\theta = 0}^{\theta = 2\pi} \int_{r = 0}^{r = 1} (r - r^3)dr d\theta = \int_{\theta = 0}^{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Can someone please check if my reasoning for this proof is valid or not? I have already seen the other questions about this proof. I'm just trying a different sort of method, though I'm not sure if it's valid or not.
Context for the main question: Prove by induction $2^n\gt n^3$ for $n\ge10$
Obviously, the base case wo... | It is ok, but I think you can go faster.
If you want to prove that:
$$2n^3>(n+1)^3\Leftrightarrow 2>\left(\frac{n+1}{n}\right)^3=\left(1+\frac{1}{n}\right)^3$$
You can do:
$$n\ge10\to\frac{1}{n}\le0.1\to \left(1+\frac{1}{n}\right)^3\le (1.1)^3=1.331<2$$
| {
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Suppose $x^2+y^2=4$ and $z^2+t^2=9$ and $xt+yz=6$. What are $x,y,z,t$? Let $x^2+y^2=4$ and $z^2+t^2=9$ and $xt+yz=6$.
What are $x,y,z,t$?
| Consider
$$
(x^2+y^2)(z^2+t^2)-(xt+yz)^2=0
$$
Expanding the left-hand side you get
$$
x^2z^2+\color{red}{y^2z^2}+\color{red}{x^2t^2}+y^2t^2
-\color{red}{x^2t^2}-2xyzt-\color{red}{y^2z^2}=0
$$
that simplifies to
$$
(xz-yt)^2=0
$$
so $xz=yt$.
Now consider the linear system in the unknowns $z$ and $t$:
\begin{cases}
yz+xt... | {
"language": "en",
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"source": "stackexchange",
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Solve $a_n=7a_{n-1}-12a_{n-2}+3^n$ $a_0=1\:a_1=2$
Using generating functions I get:
$f\left(x\right)-1-2x=7xf\left(x\right)-7x-12x^2f\left(x\right)+\frac{1}{1+3x}-1-3x$
$f\left(x\right)=\frac{3}{14\left(1+3x\right)}-\frac{13}{2\left(3x-1\right)}\:+\:\frac{40}{7\left(4x-1\right)}$
Now how can I express $a_n$ from this?
| Hint (without generating functions): eliminate $3^n$ between the recurrence relations for $n$ and $n+1$, then you get a $3^{rd}$ order linear homogeneous recurrence:
$$\require{cancel}
a_{n+1}-3 a_n=7 a_n-12a_{n-1}+\cancel{3^{n+1}} - 3(7 a_{n-1}-12 a_{n-2}+ \cancel{3^n}) \\[3px]
\iff a_{n+1} = 10 a_n - 33 a_{n-1}+36 a_... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Binomial series sum of the form $\sum^{k}_{r=0}(-1)^r(2)^{k-r}\binom{20}{r}\binom{20-r}{20-k}$ the value of
$$2^k\binom{20}{0}\binom{20}{20-k}-2^{k-1}\binom{20}{1}\binom{19}{20-k}+2^{k-2}\binom{20}{2}\binom{18}{20-k} \cdots+ +(-1)^k\binom{20}{k}\binom{20-k}{20-k}$$
options:
$(a)\;\; 7$
$(b)\;\;8$
$(c)\;\; 10$
$(d)\;\... | The only possibility is D with k=1 or 19.
| {
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Find a parameterisation for $(x-8)^2 + y^2 = 16$ starting at the point $(4,0)$ and moving clockwise once around the circle. Find a parameterisation for $(x-8)^2 + y^2 = 16$ starting at the point $(4,0)$ and moving clockwise once around the circle.
My work
We can describe the circle using polar coordinates:
$x(\theta) ... | Yes, the work is correct.
Below is a parametric plot of $\left\{8 + 4 \cos \left( \pi - t \right), 4 \sin \left(t\right)\right\}$ for $0\le t < 2\pi$.
To show that your parameterization starts at $(4,0)$ and proceeds in a clockwise manner, a sequence of values was created:
$$
\begin{array}{cccc}
k & t & x(t) & y(t) \... | {
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"question_score": "2",
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Find parametar λ if plane and line don't intersect How can I find λ, if I have line $3\lambda x-2=3y+1=\lambda z$ and plane $\lambda x-3y+2z-3=0$ and they don't intersect. The given solution says that λ=3, but I don't have any idea how can I come to that solution. Small hint would be helpful...
| Hint:
The line is the intersection of the two planes
$$
3 \lambda x - 2 = 3 y + 1 \iff 3 \lambda x - 3 y = 3 \\
3y + 1 = \lambda z \iff 3 y - \lambda z = -1
$$
Then we take the plane
$$
\lambda x - 3 y + 2z - 3 = 0
$$
into the consideration of common intersections.
This is equivalent to analyzing the solutions of the... | {
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Compute $\prod_{k=5}^{\infty}\frac{k^4-17k^2+16}{k^4-8k^2+16}$ Compute the product: $$\prod_{k=5}^{\infty}\frac{k^4-17k^2+16}{k^4-8k^2+16}$$
I was able to factor in the following manner:
$$ \frac{k^4-17k^2+16}{k^4-8k^2+16}=\frac{(k-1)(k+1)(k-4)(k+4)}{(k-2)^2(k+2)^2}$$
but what should I do now?
| As you noted we have $$\prod_{k\geq5}\frac{k^{4}-17k^{2}+16}{k^{4}-8k+16}=\prod_{k\geq0}\frac{\left(k+4\right)\left(k+6\right)\left(k+1\right)\left(k+9\right)}{\left(k+3\right)^{2}\left(k+7\right)^{2}}$$ and now using the identity $$\prod_{k\geq0}\frac{\left(k+a\right)\left(k+b\right)\left(k+c\right)\left(k+d\right)}{\... | {
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"source": "stackexchange",
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Using side length of triangle, find radius of touching circle Question:
The side lengths of a triangle are equal to lengths $8, 9$ and $10$. Find the exact value of the radius of the circle passing through the endpoints of the longest side and the midpoint of the shortest side.
I am very confused as to where to beg... | Let the triangle be $\triangle ABC$ with $AB=8$, $AC=9$, $BC = 10$, and let the midpoint of $AB$ be $M$. Then, our goal is to find the radius of the excircle of $\triangle BCM$. By the cosine law, we know that $$\cos\angle B=\frac{AB^2+BC^2-AC^2}{2\cdot AB\cdot BC}=\frac{83}{160}$$
and similarly by the cosine law, $$\c... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The maximum possible area of a yard in terms of $x$ A total of $x$ feet of fencing is to form three sides of a level rectangular yard. What is the maximum possible area of the yard, in terms of $x$?
(A) $\frac{x^2}{9}$
(B) $\frac{x^2}{8}$
(C) $\frac{x^2}{4}$
(D) $x^2$
(E) $2x^2$
I do not know the meaning of level recta... | Suppose that the side of your rectangle whose opposite side is not bordered by the fence has a length of $s$. Then the two adjacent sides have length $\frac{x-s}{2}$. The area of the rectangle can then calculated to be $s\cdot\frac{x-s}{2}=\frac{sx-s^2}{2}$, so we must maximize $sx-s^2$. With some manipulation we obtai... | {
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"url": "https://math.stackexchange.com/questions/2229349",
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Find the minimum of the value $k$ such
Let $x\geq0$, $y\geq0$, $z\ge 0$ such that
$$\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}\le\sqrt{k(x^2+y^2+z^2)+(6-k)(xy+yz+xz)}$$
Find the minimum of the real value $k$
I square the side
$$2(x^2+y^2+z^2)+2\sum_{cyc}\sqrt{(x^2+y^2)(y^2+z^2)}\le k(x^2+y^2+z^2)+(6-k)(xy+yz+xz... | Yes! For $k=4\sqrt2$ your inequality is true.
Indeed, by C-S
$$\left(\sum_{cyc}\sqrt{x^2+y^2}\right)^2\leq\sum_{cyc}(x+y+(\sqrt2-1)z)\sum_{cyc}\frac{x^2+y^2}{x+y+(\sqrt2-1)z}.$$
Thus, it remains to prove that:
$$\sum_{cyc}(x+y+(\sqrt2-1)z)\sum_{cyc}\frac{x^2+y^2}{x+y+(\sqrt2-1)z}\leq\sum_{cyc}\left(4\sqrt2x^2+(6-4\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Sine-related infinite products How do you find the infinite product for functions which share the same roots?
$$\frac{\sin(\pi x)}{\pi x}=\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)\dotsb .$$
However, this function has the exact same roots, but clearly doesn't have... | They don't have the same roots: $\cos{z}+3=0$ has complex solutions: suppose $x,y$ are real. Then
$$ \cos{(x+iy)} = \cos{x}\cosh{y}+i\sin{x}\sinh{y}, $$
so if this is $3$, we need $\sin{x}=0$ and $\cosh{y} = 3/\cos{x}$. $\sin{x}=0 \implies x = n\pi$ for some integer $n$, so $\cos{x}=(-1)^n$. $\cosh{y}$ is positive, so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2234058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the value of $(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a+1)$ if $\log_{b}a+\log_{c}b+\log_{a}c=13$ and $\log_{a}b+\log_{b}c+\log_{c}a=8$ Let $a,b$, and $c$ be positive real numbers such that
$$\log_{a}b + \log_{b}c + \log_{c}a = 8$$
and
$$\log_{b}a + \log_{c}b + \log_{a}c = 13.$$
What is the value of
$$(\log_{a}b... | Since I prefer working with exponents rather than logarithms, here is how you could tackle it that way, using your quite possibly better intuition at handling powers:
$$x + y + z = 8$$
$$u + v + w = 13$$
With:
$$ a^x = b\,,\, b^y = c\,,\, c^z = a $$
and
$$ b^u = a\,,\, c^v = b\,,\, a^w = c $$
From this you can see that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 3
} |
How to differentiate $\left(\frac{d}{dt}\left(x\right)\right)^2$? How does
$\frac{d}{dt}\left(\frac{1}{2}m\left(\frac{d}{dt}\left(x\right)\right)^2+\frac{1}{2}kx^2\right)$ = 0
equals to
m$\frac{d^2}{dt^2}\left(x\right)$ + kx = 0
| \begin{align}\frac{d}{dt}\left(\frac{1}{2}m\left(\frac{d}{dt}\left(x\right)\right)^2+\frac{1}{2}kx^2\right) &= 0 \\ \Longrightarrow \frac{1}{2} m \frac{d}{dt}\big(\frac{d}{dt}x\big)^2+\frac{1}{2}k \frac{d}{dt}x^2 &= 0 \\ \Longrightarrow m \frac{d}{dt} x\frac{d}{dt}\frac{d}{dt}x+kx \frac{d}{dt}x &= 0 ~\text{ (chain rul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Do the following series converge or diverge? Justify. $\sum_{k=0}^{\infty} \frac{3k^2 + 1}{k^3 + k^2 + 5}$ $$\sum_{k=0}^{\infty} \frac{3k^2 + 1}{k^3 + k^2 + 5}$$
Can I do this using direct comparison test?
for $k \in [1, \infty), a_k = \frac{3k^2 + 1}{k^3 + k^2 + 5} \geq 0$
for $k \in [1,\infty), a_k = \frac{3k^2 + 1}{... | It's much quicker with equivalents:
$$\frac{3k^2+1}{k^3+k^2+5}\sim_\infty\frac{3k^2}{k^3}=\frac3k,\quad\text{which diverges}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the last three digits of $383^{101}$ We have to find out $383^{101} \equiv ? \pmod {1000}$.
I know that $383^2 ≡ 689 \pmod {1000}$
$383^5≡143 \pmod {1000}$
I know that $ϕ(1000)=400 >101 $ from Euler.
It definitely can't help me.
I don't know how to continue.
I can't use the Chinese Remainder Theorem.
| The units digit of $383^1$ is $3$.
The units digit of $383^2$ is $9$ ($3 \times 3 = 9$).
The units digit of $383^3$ is $7$ ($3 \times 9 = 27$).
The units digit of $383^4$ is $1$ ($3 \times 7 = 21$).
Calculating,
$\; 383^4 \equiv 721 \pmod{1000}$
Calculating,
$\; 383^{101} \equiv 383 \cdot 721^{25} \equiv 383 \cdot (1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Applying Stokes' theorem - what surface?
$\def\d{\mathrm{d}}$Determine the integral $$\oint_L \mathbf{A} \cdot \,\d\mathbf{r},$$
where $$\mathbf{A} = \mathbf{e}_x(x^2-a(y+z))+\mathbf{e}_y(y^2-az)+\mathbf{e}_z(z^2-a(x+y)),$$
and $L$ is the curve given by the intersection between the cylinder $$\begin{cases}(x-a)^2+... | The surface can be ANY surface whose border is the curve $L$.
In this case, a possibility would be the part of the sphere inside the cylinder, which you can parametrize as follows:
\begin{cases}
x= x\\
y=y \quad \quad \quad \quad \quad \quad \text{with} \quad (x,y)\mid (x-a)^2+y^2\le a^2\\
z= \sqrt{R^2-x^2-y^2}\\
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Am I missing something quite obvious from understanding this proof?
Theorem. There exists a nonempty set of rational numbers which is bounded above in $\mathbb{Q}$ but has no least upper bound in $\mathbb{Q}$.
Here is the relevant part of the proof that I am going to ask about.
Let $k=\frac{a}{b}\in\mathbb{Q}$ be an... |
The author presents a proof purely based upon calculations within $\mathbb{Q}$. It does not rely on an embedding of $\mathbb{Q}$ in $\mathbb{R}$ and the knowledge that $\mathbb{R}$ is complete.
Insofar is the approach of the author somewhat different than yours.
Let's revisit the author's proof. We want to show there... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"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.