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Question about an integral trick This is an excerpt from brilliant.org on integral tricks:
A similar method to the above is to reverse the interval of
integration: to "integrate backwards." For a function $f$ and real
numbers $a < b$,
$\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx.$
Instead of the function being c... | If we let $x=a+b-y$ then when $x=a$, $y=a+b-a=b$ and when $x=b$, $y=a+b-b=a$ and $dx=-dy$, so
$$\int_a^bf(x)dx=\int_b^af(a+b-y)(-dy)=\int_a^bf(a+b-y)dy=\int_a^bf(a+b-x)dx$$
Where running the integral backwards changes its sign and we renamed the variable of integration back to $x$. Then
$$\begin{align}\int_a^bf(x)dx&=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2657821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is function $f(x) = \frac{x^{2}-2}{x-\sqrt{2}}$ continuous for all $x$?
Is function $f(x) = \frac{x^{2}-2}{x-\sqrt{2}}$ continuous for all $x$?
I understand that the function
$$ g(x) = \frac{x^{2}-1}{x-1} $$
has one discontinuity for $x=1$.
[Added later] In general my question is for the function
$$
\frac{x^{2}-a... | $\require{cancel}$
Indeed, $$f(x) = \frac{x^{2}-2}{x-\sqrt{2}}$$ has a discontinuity at $x=\sqrt 2$ (the point at which the domominator is 0). But it is removable.
$$f(x) = \frac{x^2 - 2}{x-\sqrt 2} =\frac{x^2 - (\sqrt 2)^2}{x-\sqrt 2}= \frac{(x+\sqrt 2)(\cancel{x-\sqrt 2})}{\cancel{x-\sqrt 2}} = {x+\sqrt 2}$$ when $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2658760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How do you find the roots of the negative polynomial $(-x^3 - x + 1)$ to further use for partial fractions? I am trying to factor the polynomial $p(x)=-x^3-x+1$ to use in partial fractions, but I am getting stumped on how to break this up into roots. Any tips will be helpful! My teacher did something similar incorpora... | Cubic equations in the form $x^3+px+q=0$ can be solved by the substitution $x=rf(\theta)$ so
$$r^3\left[f(\theta)\right]^3+rf(\theta)p+q=0$$
Then multiply by $4/r^3$ to get
$$4\left[f(\theta)\right]^3+\frac{4p}{r^2}f=\frac{-4q}{r^3}$$
Choose the sign of $r=\pm\sqrt{\frac{4|p|}3}$ such that $qr\le0$. Then
$$4\left[f(\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Possible remainders of a perfect square when divided by $3,5,6$. A perfect square will have in its prime factorization all the primes having an even power, so that the square root will simply divide each by $2$.
Let the two constituents of a perfect square number ($p$) be : $mm$, i.e. same positive integer is repeated ... | If $p$ is an odd prime $>3$, then
*
*$p\equiv \pm 1\mod 3$, so $p^2\equiv 1\mod 3$;
*$p\equiv \pm 1,\pm 2\mod 5$, so $p^2\equiv \pm 1\mod 5$;
*$p\equiv \pm 1\mod 6$, so $p^2\equiv 1\mod 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $577^3$ the smallest cube that is expressible as the sum of $3$ positive cubes in $6$ different ways? A Taxi Cab Number is a number $\text{Ta}(n)$ that can be written as the sum of two cubes in $n$ different ways. More formally, they are known as Hardy-Ramanujan Numbers or, as Ramanujan had called them, Magic Number... | No.
$$\begin{align}216^3 &=24^3+ 144^3+ 192^3\\ &=26^3 +102^3+208^3\\ &=30^3 +164^3+ 178^3\\ &=48^3 +76^3+ 212^3\\ &= 102^3 +117^3 +195^3\\ &=108^3 +144^3+ 180^3\end{align}$$
Find more here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $A$, $B$ and $C$ are the angles of a triangle then find the value of $\Delta$ I'll state the question from my book below:
If $A$, $B$ and $C$ are the angles of a triangle, then find the determinant value of
$$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}... | In the standard notation we obtain:
$$\Delta=\sum_{cyc}\sin^2\alpha(\cot\beta-\cot\gamma)=\sum_{cyc}\frac{4S^2}{b^2c^2}\left(\frac{\frac{a^2+c^2-b^2}{2ac}}{\frac{2S}{ac}}-\frac{\frac{a^2+b^2-c^2}{2ab}}{\frac{2S}{ab}}\right)=$$
$$=S\sum_{cyc}\frac{a^2+c^2-b^2-a^2-b^2+c^2}{b^2c^2}=2S\sum_{cyc}\frac{c^2-b^2}{b^2c^2}=2S\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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what is the best way to solve $\int\frac{\cos^2x\,\mathrm dx}{\sin^2x+4\sin x\cos x}$? I have the integral:
$$\int\frac{\cos^2x\,\mathrm dx}{\sin^2x+4\sin x\cos x}$$
of course there is $u = \tan\frac{x}{2}$
But I want to avoid it if possible, also neither $\cos x = u$ nor $\sin x = u$ made the life easier :(
| I must say Lab bhattachargee shew a method that is good, here is my method which may also be used. First we split the numerator using $\sin^2 x + \cos ^2 x = 1$.
$$\begin{align}
I &=\int\frac{ dx}{\sin^2x+4\sin x\cos x}-\int\frac{\sin^2x dx}{\sin^2x+4\sin x\cos x}\\
&= \int \frac{\csc^2 x}{1+4 \cot x}dx - \int \frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2666075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Prove that $24|n^2-1$, if $(n,6)=1$ $n^2-1 = (n-1)(n+1)$
Then $24|(n-1)(n+1)$
$(n,6)=1$: $\exists a,b\in\mathbb{Z}$ that $n = 6\cdot a+b$
Investigate the residues, which arise when dividing the number n by two and three:
$\frac{6\cdot a+b}{3} = \frac{6\cdot a}{3}+\frac{b}{3} = 2\cdot a+\frac{b}{3}$
$\frac{6\cdot a+b}{2... | If $\gcd(n,6)=1$, then there is some integer $k$ such that $n=6k\pm 1$. We compute $$n^2-1=(6k\pm 1)^2-1 = 36k^2\pm 12k = 12k (3k\pm 1)$$
Now, if $k$ is even, then $24|12k$ already. If instead $k$ is odd, then $3k\pm 1$ is even, so again $24|12k(3k\pm 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2667849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Prove $f(x) = \frac{1}{1+x^2}$ is uniformly continuous on $\mathbb{R}$ Prove $f(x) =\frac{1}{1+x^2}$ is uniformly continuous on $\mathbb{R}$.
My attempt...
Proof
$$\left| f(x) - f(y) \right| = \left| \frac{1}{1+x^2} - \frac{1}{1+y^2}\right| = \frac{\left|x+y\right|}{\left(1+x^2\right)\left(1+y^2\right)}\left|x-y\ri... | Big Hint: Use this and the fact that $\vert f'(x)\vert <1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Show that $f(x)=x^{5/3}-kx^{4/3}+k^2x$ is increasing for $k\neq0$ So to show the function is increasing/decreasing we differentiate and show it is more than zero/less than zero:
We have
$$f(x)=x^{5/3}-kx^{4/3}+k^2x$$
Hence,
$$f'(x)=\frac{5}{3}x^{2/3}-\frac{4k}{3}x^{1/3}+k^2$$
But how do I show
$$\frac{5}{3}x^{\frac{2}{... | Let
$$y = x^{\frac{1}{3}}$$
So,
$$\frac{5}{3}y^{2} - \frac{4k}{3}y + k^{2} > 0 \Longleftrightarrow 5y^{2} - 4ky + 3k^{2} > 0$$
and
$$(-4k)^{2} - 4.5.3k^{2} = 16k^{2} - 60k^{2} < 0.$$
Therefore, $\frac{5}{3}y^{2} - \frac{4k}{3}y + k^{2} > 0$. Complete the solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove $\lim\limits_{n\to∞}{\sum\limits_{x=0}^n\binom nx(1+{\rm e}^{-(x+1)})^{n+1}\over\sum\limits_{x=0}^n\binom nx(1+{\rm e}^{-x})^{n + 1}}=\frac 13$ I am trying to find limit of the following function:
$$
\lim_{n\rightarrow \infty}\frac{\sum\limits_{x = 0}^{n}\binom{n}{x}\left[1 + \mathrm{e}^{-(x+1)}\right]^{n + 1}}{... | $\def\e{\mathrm{e}}$For $n \in \mathbb{N}_+$, denote$$
S_n = \sum_{k = 0}^n \binom{n}{k} (1 + \e^{-(k + 1)})^{n + 1},\ T_n = \sum_{k = 0}^n \binom{n}{k} (1 + \e^{-k})^{n + 1}.
$$
First,\begin{align*}
S_n &= \sum_{k = 0}^n \binom{n}{k} (1 + \e^{-(k + 1)})^{n + 1} = \sum_{k = 0}^n \binom{n}{k} \sum_{j = 0}^{n + 1} \binom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2674981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
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Proving $f(x,y) = |xy| + a(x^2 + y^2)$ is convex for $a \ge 1/2$ I am working on a problem to prove $$
f(x,y) = |xy| + a(x^2 + y^2)
$$
is convex for $a \ge 1/2$.
My approach is to show that the Hessian is PSD for the cases where $x \not= 0, y \not= 0$.
However, this approach breaks down for cases when $x$ or $y$ are ze... | $\begin{array}\\
f(x,y)
&= |xy| + a(x^2 + y^2)\\
&= |xy| + (a-\tfrac{1}{2})(x^2+y^2)+\tfrac{1}{2}|x|^2+\tfrac{1}{2}|y|^2\\
&= (a-\tfrac{1}{2})(x^2+y^2) + \tfrac{1}{2}(|x|+|y|)^2\\
\end{array}
$
The first term, $(a-\tfrac{1}{2})(x^2+y^2)$, is clearly convex.
The second term is the composition of the (outer) convex incr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2675784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factorize $\det\left[\begin{smallmatrix}yz-x^2&zx-y^2&xy-z^2\\zx-y^2&xy-z^2&yz-x^2\\xy-z^2&yz-x^2&zx-y^2\end{smallmatrix}\right]$ using factor theorem Factorize and prove that
$$
\Delta=\begin{vmatrix}
yz-x^2&zx-y^2&xy-z^2\\
zx-y^2&xy-z^2&yz-x^2\\
xy-z^2&yz-x^2&zx-y^2
\end{vmatrix}\\=\frac{1}{4}(x+y+z)^2\Big[(x-y)^2+(... | Note that
$$
M:=
\begin{bmatrix}
yz-x^2&zx-y^2&xy-z^2\\
zx-y^2&xy-z^2&yz-x^2\\
xy-z^2&yz-x^2&zx-y^2
\end{bmatrix}
$$
is the matrix of $2\times 2$ cofactors of the matrix:
$$
N:=
\begin{bmatrix}
x & y & z\\
y & z & x\\
z& x & y \\
\end{bmatrix}.
$$
Then as usual $MN=(\det N) I$, so that $\det M \det N =(\det N)^3$. As ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2675951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Find all real solutions for $x$ in $2(2^x−1)x^2+(2^{x^2}−2)x=2^{x+1}−2$.
Find all real solutions for $x$ in $2(2^x−1)x^2+(2^{x^2}−2)x=2^{x+1}−2$.
I started off by dividing $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2$ by $2$, and I got $(2^x-1)x^2 + (2^{x^2-1}-1)x = 2^x -1$. I tried dividing by $2^x-1$ on both sides wh... | $\begin{align*} 2\left (2^{x}-1 \right )x^{2}+\left(2^{x^{2}}-2\right)x&=2^{x}.2-2\\ 2\left(2^{x}-1\right)x^{2}-\left(2^{x^{2}}-2\right)x&=2(2^{x}-1) \end{align*}$
Let $2^{x}-1=a$ and $2^{x^{2}}-2=b$ and equation becomes:
$2ax^{2} + bx=2a$ or $2ax^{2}+bx-2a=0$. It's a quadratic equation. The quadratic equation has real... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2677239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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What is the area of the triangle ABC?
$ABC$ is an equaliteral triangle.
Suppose $DB=4$, $DA=4\sqrt{3}$ and $DC=8$.
Find the area of the triangle $ABC$.
| Reflect $D$ in $AB$ to obtain $P$. Reflect $D$ in $BC$ to obtain $Q$. Reflect $D$ in $CA$ ro obtain $R$. Then $[APBQCR]=2[\triangle ABC]$.
Note that $\triangle APR$, $\triangle BQP$ and $\triangle CRQ$ are $30^\circ$- $120^\circ$-$30^\circ$ isosceles triangles.
$[\triangle APR]=\frac{1}{2}(4\sqrt{3})^2\sin 120^\circ $
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2677611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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Definite integrals question $$\int^{\pi/2}_{0} \frac{1}{\sin^4x + \cos^4 x} \,dx$$
First I solve the indefinite integral $$\int \frac{1}{\sin^4x + \cos^4 x} \,dx=\frac{\sqrt2}{2}\arctan\Big(\frac{\tan2x}{\sqrt2}
\Big)+C$$ by the universal substitution $\tan2x=y$.
My question is: when I do this substitution should I bou... | One way to address the original integral is to say that
$$\begin{align}\int_0^{\pi/2}\frac1{\sin^4x+\cos^4x}dx&=\int_0^{\pi/2}\frac1{\frac14(1-\cos2x)^2+\frac14(1+\cos2x)^2}dx\\
&=2\int_0^{\pi/2}\frac1{1+\cos^22x}dx=\int_0^{\pi}\frac1{1+\cos^2y}dy\\
&=\int_0^{\pi}\frac{dy}{\sin^2y+2\cos^2y}\end{align}$$
Now,
$$\int_0^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2679478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove that $ \frac{(99)!!}{(100)!!} < \frac{1}{10}$ How to prove that $ \dfrac{(99)!!}{(100)!!}=\dfrac{1\cdot3\cdot5\cdot7\cdot9 \cdots99}{2\cdot4\cdot6\cdot8\cdot10\cdots100} < \dfrac{1}{10}$
Any hint to prove it?
| In general the ratio
$$ r_n = \frac{(2n-1)!!}{(2n)!!}$$
is related to the central binomial coefficient for which many results are known.
In particular, we have that
$$ (2n-1) (2n-3) \cdots 1 = \frac{(2n) (2n-1)\cdots 1}{(2n)(2n-2) \cdots 2} =2^{-n} \frac{(2n) (2n-1)\cdots 1}{n(n-1) \cdots 1} = 2^{-n} \frac{(2n)!}{n!}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2681690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Why is $|x^2 - a^2| = |x - a||x + a|$? By the definition of absolute value, we have
\begin{equation}
|x^2 - a^2| =
\begin{cases}
(x - a)(x + a) \ if \ x^2 \geq a^2\\
-(x - a)(x + a) \ if \ x^2 \leq a^2.
\end{cases}
\end{equation}
How do we conclude that the right hand side equals $|x - a||x + a|$?
| Assume that $a\ge0$.
Note that $x^2\ge a^2$ if and only if $x\le -a$ or $x\ge a$.
If $x\le-a$, then $x+a\le0$ and $x-a\le0$. So $|x+a||x-a|=[-(x+a)][-(x-a)]=(x+a)(x-a)$.
If $x\ge a$, then $x+a\ge0$ and $x-a\ge0$. So $|x+a||x-a|=(x+a)(x-a)$.
We also have $x^2\le a^2$ if and only if $-a\le x\le a$. In such case, $x+a\ge0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2681771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral with fourth power in denominator How can I find the antiderivate of $$\int \frac{1}{(x-a)^4+(x-b)^4} dx$$ with $$a<b$$
My try was to rewrite the integral as $$\int \frac{\frac{1}{(x-a)^2(x-b)^2}}{\frac{(x-a)^2}{(x-b)^2}+\frac{(x-b)^2}{(x-a)^2}} dx$$ which is equal to $$\int \frac{\frac{1}{(x-a)^2(x-b)^2}}{(\fr... | Let $d=\frac{a+b}{2}$ and substitute $x=u+d$ to make the integrand
$$\frac{1}{(u+c)^4+(u-c)^4}.$$
Where $c=(b-a)/2.$ Factor out a $c^4$:
$$\frac{1}{c^4((\frac{u}{c}+1)^4 + (\frac{u}{c}-1)^4)}.$$
Substitute $v=\frac{u}{c}$ to get
$$\frac{c}{c^4((v+1^4)+(v-1)^4)}.$$
Ignoring the $c$'s, we need to integrate
$$\frac{1}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2682341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Difference in total number of positive integers $x, y, z$ determined from an equation.
Find all positive integers $x, y, z$,such that $$x+2y+3z=12$$
Using the formula (got in almost similar thread) I got ${k−1 \choose n−1}=55$ solutions are available from the equation where $k=12$ and $n=3$
After reading some si... | Maybe you can get to fifty-five solutions if you ditch the restriction that $x, y, z$ have to be positive:
$$\ldots \\
8 + 2 \times 2 + 3 \times 0 = 12 \\
0 + 2 \times 0 + 3 \times 4 = 12 \\
0 + 2 \times 3 + 3 \times 2 = 12 \\
0 + 2 \times (-3) + 3 \times 6 = 12 \\
-1 + 2 \times 5 + 3 \times 1 = 12 \\
\ld... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2682963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Incorrect in solving $\frac{a}{b} - \frac{a}{c} = 1$ for $c$ I have this:
$$\frac{a}{b} - \frac{a}{c} = 1$$ Solve for $c$. Then,
$$\frac{a}{b} - \frac{a}{c} = 1 \cdot bc$$
$$ = ac - ab = bc$$
$$ = a(c - b) = bc$$
$$ c = \frac{bc}{a} + b$$ This is my final result.
But the correct result is:
$$c=\frac{ab}{a-b}$$
What... | Your algebra is correct but $$c = \frac{bc}{a} + b$$ is solving $c$ in terms of $a$, $b$, and $c$ which is sort of cyclic. We want to see $c$ in terms of only $a$ and $b$.
You have to isolate $c$ by keeping all terms involving $c$ on the left side and other terms on the right side.
Note that $$ ac - ab = bc\implies c(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Finding $\frac{\partial^6 f}{\partial x^4 \partial y^2}$
Find $$\frac{\partial^6 f}{\partial x^4 \partial y^2}(0,0)$$ Of the $f(x,y)=\frac{1}{1-x^2y}$
$$\frac{1}{1-x^2y}=\sum_{n=1}^{\infty}(x^2y)^n$$ as $|x^2y|<1$
So we have $\frac{1}{1-x^2y}\approx1+x^2y+x^4y^2$
But how should we continue from here?
| Use the below statement before starts to differentiate.
$$\frac{\partial^6 f(x,y)}{\partial x^4 \partial y^2}=\frac{\partial^4 \left(\frac{\partial^2 f(x,y)}{\partial y^2}\right)}{\partial x^4}$$
Calculate $\displaystyle \frac{\partial^2 f(x,y)}{\partial y^2}$ for first.
$$\displaystyle \frac{\partial f(x,y)}{\partial ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2686731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $5x \equiv 15 \pmod{25}$, then definitely $x \equiv 3 \pmod{25}$. Is this true or false? I don't understand how to prove this statement true or false.
If $5x \equiv 15 \pmod{25}$, then definitely $x \equiv 3 \pmod{25}$.
All I know is that $\frac{5}5x \equiv \frac{15}5 \pmod{\frac{25}5}$ = $x \equiv 3 \pmod{5}$, ... | The congruence $5x \equiv 15 \pmod{25}$ means that there exists $t \in \mathbb{Z}$ such that
$$5x = 15 + 25t$$
Dividing each side of the equation by $5$ yields
$$x = 3 + 5t$$
Therefore, if
$$x \equiv 3 \pmod{5}$$
it satisfies the congruence $5x \equiv 15 \pmod{25}$. There are five such equivalence classes modulo $25... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2687784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Verfication of deduction made using the Cauchy-Schwarz inequality Is the following proof correct?
Show that $$16\leq(a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)$$ for all positive numbers $a,b,c,d$.
Proof. Let $\mathbf{R}^4$ be the inner product space with the inner product defined as in the ... | While OP's proof is fine, it may be worth noting that the result also follows directly from the AM-HM inequality by simply rewriting the given relation as:
$$
\frac{a+b+c+d}{4} \;\ge\; \frac {4}{\cfrac{1}{a}+\cfrac{1}{b}+\cfrac{1}{c}+\cfrac{1}{d}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2689137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Calculating $\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$. I have a difficulty in calculating this limit:
$$\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3},$$
I have tried $\tan x = \frac{\sin x}{\cos x}$, then I unified the denominator of the numerator of the given limit problem finally I got $$\lim_{x \rig... | $$\frac{\tan{x}-\sin{x}}{x^3}=\frac{x+\frac{1}{3}x^3+o(x^3) - (x-\frac{1}{6}x^3+o(x^3))}{x^3}=\frac{\frac{1}{2}x^3+o(x^3)}{x^3}=\frac{1}{2} +\frac{o(x^3)}{x^3}\stackrel{x\rightarrow 0}{\longrightarrow}\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2690311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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On finding $\sup\left\{3(-1)^n-\frac{1}{n^2+1}\right\}$ How does one find $\sup(A)$ where $A = \left\{3(-1)^n-\frac{1}{n^2+1}: n \in \mathbb{N}\right\}$?
I've tried as follows, but I'm not so sure.
$\displaystyle 3(-1)^n-\frac{1}{n^2+1} \le 3-\frac{1}{n^2+1} \le 3. $ So $3$ is an upper bound for $A$.
Let $\epsilon >0$... | Looks good to me. EDIT - though as Martiny pointed out in the comments, it's not. You wrote $3-\frac{1}{N^2+1} \ge 3-\frac{1}{2N^2}$ when this is not true.
You can also do this by taking every other term directly: noting that $$\lim_{n \to \infty} \left[3(-1)^{2n} - \frac{1}{(2n)^2+1} \right] = \lim_{n \to \infty} 3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solve the equation $\log_2(\frac{8}{2^x}-1)=x-2$ This equation comes from the book of Prilepko. It looks sufficiently simple but I want to know if my solution is correct.
$\log_2(\frac{8}{2^x}-1)=x-2$
I have only been able to transform into this:
$\log_2(\frac{8-2^x}{2^x})=x-2$
$\iff$ $\log_2(8-2^x)=2x-2$
$\iff$ $8-2^x... | If you want "to stretch" the logarithm method or solution to the very end here it is:
$$\log_{2} \left ( \frac{8}{2^{x}} - 1 \right ) = x-2 $$
$$\log_{2} \left ( \frac{8-2^{x}}{2^{x}} \right ) = x-2 $$
$$\log_{2} \left (8-2^{x} \right ) - \log_{2} \left ( 2^{x} \right ) = x-2 $$
$$\log_{2} \left (2^{3}-2^{x} \right ) -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2692763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Find the pmf of $Y=X^2$ Let $X$ be a random variable with the following pmf:
$$
\begin{array}{c|ccccc}
x& -2 & -1 & 0 & 1 & 2 & \\ \hline
p(x)
& 3/10 & 3/10 & 1/10 & 2/10 & 1/10 &
\end{array}
$$
Find the pmf of $Y = X^2$ and find $P(Y\ge3)$.
I am struggling to get the... | We have that the pmf of $X$ is$$p_X(x)=\begin{cases}
\frac{3}{10} & x=-2 \\
\frac{3}{10} & x=-1 \\
\frac{1}{10} & x=0 \\
\frac{2}{10} & x=1 \\
\frac{1}{10} & x=4
\end{cases}$$
Transforming this to get the pmf of $Y$ we get
$$p_Y(y)=\begin{cases}
\frac{3}{10} & y=4 \\
\frac{3}{10} & y=1 \\
\frac{1}{10} & y=0 \\
\frac{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2694636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding the the derivative of $y=\sqrt{1-\sin x}; 0A question I'm attempting is:
Find the derivative of $ y = \sqrt {1 - \sin x} ; 0 < x <\pi/2$.
I did this:
$y = \sqrt {1 - \sin x} = \sqrt {\cos^2\frac{x}{2} + \sin^2\frac{x}{2} - 2\sin \frac{x}{2}\cos \frac{x}{2}} = \sqrt { (\sin \frac{x}{2}-\cos \frac{x}{2})^2} = \... | $$y = \sqrt {1 - \sin x} ; 0 < x <\pi/2$$
$$ y' = \frac {-\cos x}{2 \sqrt {1 - \sin x}}$$
$$=\frac {-\cos x\sqrt {1 + \sin x }}{2 \cos x}$$
$$=-\frac {\sqrt {1 + \sin x}}{2}=-\frac12(\sin x/2+\cos x/2)$$
Note that $$(\sin x/2+\cos x/2)^2=1+2\sin x/2\cos x/2=1+\sin x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2698798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Factoring $x^2-y^2-z^2+2yz+x+y-z$ This is a factorization problem on polynomials. I can't find a way to solve it, neither can a math program called MathWay. Help me, please.
$$x^2-y^2-z^2+2yz+x+y-z$$
My book says that the answer is $(x+y-z)(x-y+z+1)$.
| You can consider $\,x^2+x-y^2-z^2+2yz+y-z\,$ as a quadratic in $\,x\,$, and find its roots using the quadratic formula, the discriminant being:
$$
\begin{align}
\Delta_x = 1 - 4(-y^2-z^2+2yz+y-z) &= 4y^2 - 4(2z+1)y +4z^2+4z+1 \\
&= \left(2y\right)^2- 2 \cdot 2y(2z+1)+\left(2z+1\right)^2 \\
&= \left(2y-2z-1\right)^2
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2703737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Change of Basis Given 2 Vectors and Transition Matrix
Given v$_1=\left( \begin{array}{ccc} 3\\
-4 \end{array} \right)$, v$_2=\left( \begin{array}{ccc} 2\\ 5 \end{array} \right)$, $S=\left( \begin{array}{ccc}
-1 & 7\\ 2 &-5 \end{array} \right)$
Find the vectors u$_1$, u$_2$ such that $S$ will be a
transition matrix... | Note that the inverse should be
$$S^{-1} = \left( \begin{array}{ccc}
5/9 &7/9\\
2/9 & 1/9 \end{array} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Playing cards: Probability pairs An ordinary deck of $52$ playing cards ($4$ suits, $13$ values) are shuffled. Then player $A$ takes the top two cards. Player $B$ takes the next to top cars. Let $Pr[A]$ denote the probability for player $A$ to have a pair (i.e. two cards with the same value) and $Pr[B]$ the probability... | Here is a perhaps inelegant way to show that $P(B)=P(A)$
Suppose player $A$ gets a pair with probability $\frac{3}{51}$
Then player $B$ gets a pair with probability $$\frac{48}{50}\cdot\frac{3}{49}+\frac{2}{50}\cdot\frac{1}{49}$$
since for his first card, he can either get a different rank than $A$ with probability $\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2709133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Laurent series for $f(z)=\frac1{z^3 +1}$ on $0<|z-1|<\sqrt3$ I have this complex function $$f(z)=\dfrac{1}{z^3 +1}$$ and I have to find the Laurent series for $$0<|z+1|<\sqrt{3}$$
First, I consider that $$f(z) = \dfrac{1}{(z+1)(z^2 - z +1)} = \dfrac{2-z}{3(z^2 - z+1)} + \dfrac{1}{3(z+1)}$$
The second term is ok but I d... | Let $a$ and $b$ be the complex cubic roots of $-1$, so $z^3+1=(z+1)(z-a)(z-b)$. Now you can find the partial fraction decomposition
$$
\frac{1}{z^3+1}=\frac{A}{z+1}+\frac{B}{z-a}+\frac{C}{z-b}
$$
Then
$$
\frac{1}{z-c}=\frac{1}{z+1-(c+1)}=-\frac{1}{c+1}\frac{1}{1-\dfrac{z+1}{c+1}}
$$
and then
$$
\frac{1}{z-c}=-\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2712433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $2 \cdot \sum\limits_{k = 0}^{n} \binom{2n}{2k} = 2^{2n}$ without induction I am aware of
$$
2^{2n} = (1 + 1)^{2n} =
\sum_{k = 0}^{2n} \binom{2n}{k}
$$
I then tried grouping the terms and using that $\binom{n}{k} = \binom{n}{n-k}$, to obtain
$$
\begin{align}
\sum_{k = 0}^{2n} \binom{2n}{k} &=
\binom{2n}{0} + \bi... | Note that
$$
\begin{align}
(1+1)^{2n}&=\sum_{k=0}^{2n}\binom{2n}{k}\tag{1}\\
(1-1)^{2n}&=\sum_{k=0}^{2n}\binom{2n}{k}(-1)^k\tag{2}
\end{align}
$$
by the binomial theorem. Subtracting (2) from (1) we get that
$$
2^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}(1-(-1)^k)=2\sum_{0\leq k\leq2n, \, \text{k even}}\binom{2n}{k}=2\sum_{k=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2712554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How can I solve this inequality? $ \frac{x+14|x|-10}{|4x-6|-21}>3$ First I looked the x that doesnt belong to this function.
$$|4x -6| - 21 \neq 0$$
$$ x \neq \frac{-15}{4}$$ and $$ x \neq \frac{27}{4}$$
Then I found the roots of the x
$$x = 0$$
$$x = \frac{3}{2}$$
After I found the roots I wrote the inequality like th... | Your way is good, first consider
*
*$|4x-6|-21> 0$
and we have
*
*$x<\frac32\implies -4x+6-21> 0 \implies x<-\frac {15} 4$
*$x\ge\frac32\implies 4x-27> 0 \implies x>\frac {27} 4$
then
*
*$|4x-6|-21> 0$ for $x<-\frac {15} 4$ and $x>\frac {27} 4$
*$|4x-6|-21< 0$ for $-\frac {15} 4<x<\frac {27} 4$
*$|4x-6|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$ I want to evaluate $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$$
First,I tried to evaluate like this:
$$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)\frac{dx}{1+\cos x}=\int_{0}^{\frac{\pi}{2}}x^... | We can adapt the formula derived in $(2)$ of this answer:
$$
\log(2\cos(x/2))=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\cos(kx)\tag{1a}
$$
Substituting $x\mapsto\pi-x$ in $\text{(1a)}$, we get
$$
\log(2\sin(x/2))=\sum_{k=1}^\infty\frac{-1}k\cos(kx)\tag{1b}
$$
Subtracting $\text{(1a)}$ from $\text{(1b)}$, the even terms cance... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 9,
"answer_id": 7
} |
Show convergence by using partial fractions Show that $\sum\limits_{n=1}^{\infty}{\frac{1}{n(n+1)(n+2)}}=\frac{1}{4}$
Answer:
$\sum\limits_{n=1}^{\infty}{\frac{1}{n(n+1)(n+2)}}=$ $\sum\limits_{n=1}^{\infty}{(\frac{1}{2n}-\frac{1}{n+1} + \frac{1}{2(n+2)})}$
= $1/2 -1/2 + 1/6 +1/4 -1/3 +1/8+1/6 -1/4 +1/10 +1/8 -1/5 +1/1... | We can rewrite the sum as
$$ \sum_{n=1}^{\infty} \frac{1}{2}\left( \left[ \frac{1}{n} - \frac{1}{n+1} \right] + \left[ \frac{1}{n+2} - \frac{1}{n + 1} \right] \right) $$
From this the way the sum telescopes should be clear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is the field extension $\mathbb{Q}(\sqrt{5 + \sqrt{7}})$ over $\mathbb{Q}$ a Galois extension? Let $K= \mathbb{Q}(\sqrt{5 + \sqrt{7}}) $. Is the field extension $\mathbb{Q}(\sqrt{5 + \sqrt{7}})$ over $\mathbb{Q}$ a Galois extension?
Let $x = \sqrt{5 + \sqrt{7}} \implies x^2 = 5 + \sqrt{7} \implies (x^2-5)^2 = 7 \impl... | Hint: $K$ is Galois over $F=\Bbb{Q}[\sqrt{7}]$. Suppose $\sqrt{5-\sqrt{7}}$ belonged to $K$. What would the action of $\operatorname{Gal}(K/F)$ on $\sqrt{5+\sqrt{7}}\cdot\sqrt{5-\sqrt{7}}$ be?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2716836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $\frac{dy}{dx}$ if $y=\sin^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, $0
Find derivative of $f(x)=\sin^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, $0<x<1$
Let $x=\sin a$ and $\sqrt{x}=\cos b$
Then I'll get:
$$
y=\sin^{-1}[\sin a\cos b-\cos a\sin b]=\sin^{-1}[\sin(a-b)]\\
\implies\sin y=\sin(a-b)\\
\implies y=n\pi+(-1)... | First let $g(x) = x\sqrt{1-x} + \sqrt{x}\sqrt{1-x^2}$. Then $f(x) = \sin^{-1}(g(x))$. Now we use the chain rule, so $f'(x) = \frac{1}{\sqrt{1-g(x)^2}}g'(x)$. I let you finish, by finding $g'$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
What is the number of elements in $\{1,2,...,2018\}$ such that it can be represented by $ x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1 \:\:? $ What is the number of elements in $\{1,2,...,2018\}$ such that it can be represented by
$$ x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1 \:\:? $$
$x \in \mathbb{Z}$
Attempt :
... | For brevity let $x^6 +x^5+ x^4 +x^3+ x^2+x+1=f(x).$... For $x\in \Bbb Z$ we have:
$$(Ia).\quad |x|\geq 4\implies f(x)=(x^6+x^4+x^2)(1+1/x)+1\geq$$ $$\geq (x^6+x^4+x^2)(1-1/|x|)+1>$$ $$> x^6(1-1/|x|)\geq$$ $$\geq 4^6(1-1/2)=64^2/2=2048>2018.$$
$$(Ib).\quad |x|\leq 3\implies |f(x)|\leq |x|^6+|x|^5+|x|^4|+|x|^3+|x|^2+|x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Calculating the largest possible area of a rectangle inscribed in an ellipse So i got the equation $4x^2 + 9y^2 = 3600$
What i've done so far is:
$A= (2x)(2y) = 4xy$
Then I find the expression of $y$
$9y^2= 3600 -4x^2$
$y = \pm \sqrt{3600 -4x^2 / 9} = 2/3(\sqrt {900 - x^2} \quad 2/3(900 -x^2)^{1/2}$
Then i set
$A = 4... | There is no need of any complicated algebra. Following is a geometric way to get the answer. One advantage of this approach is you don't need to assume the largest rectangle is axis aligned with the ellipse.
Given any circle, it is well known the largest quadrilateral inscribed in it is a square. Furthermore, the area ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2719541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
How to solve the limit $\lim_{(x,y)\to(2,3)}\frac{\sin(x)+2}{(x^2-y^2+5)^2}$ How can I solve the following limit or prove that the limit does not exist in $\Bbb R$:
$$
\lim_{(x,y)\to(2,3)}\frac{\sin(x)+2}{(x^2-y^2+5)^2}?
$$
I tried to prove the limit does not exist with $y=3x/2$, $y=x+1$, but it looks like a dead end.
| Note that since $-1 \le \sin x \le 1$, we have:
$$\frac{1}{\left(x^2-y^2+5\right)^2} \le \frac{\sin x +2}{\left(x^2-y^2+5\right)^2} \le \frac{3}{\left(x^2-y^2+5\right)^2}$$
And the denominator clearly tends to $0$ (but stays positive) when $(x,y) \to (2,3)$, so...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2720049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving for $u(x,y)$ for $u_x + y u_y = -u$, $u(x,1) =x^2$ I wish to solve for $u(x,y)$ from the PDE $u_x + y u_y = -u$, $u(x,1) =x^2$. Since this is a first order linear PDE, I think I should try the method of characteristics. Let $u = u(x(s), y(s))$.
Then the char equations are $\frac{dx}{ds} = 1$, $\frac{dy}{ds} = ... | Using the fact that $x(0) = x_0$ and $y(0) = 1$, then we see that
\begin{align}
x= x_0 +s, \ \ y= e^s, \ \ \ u(x, y) = u(x_0, 1)e^{-s} = x_0^2e^{-s}.
\end{align}
Finally, since $s= \log y$ and $x_0 = x-s = x-\log y$, then we have that
\begin{align}
u(x, y) = (x-\log y)^2\exp(-\log y) = \frac{(x-\log y)^2}{y}.
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Using the Euclidean algorithm, deduce that $\gcd(x^3+2x^2+x +4;x^2+1)=1$ So, I've tried it but I keep getting to $$\frac{x^2+1} 2$$ and don't know how to proceed.
Question is Using the Euclidean algorithm, deduce that $$\gcd(x^3+2 x^2+x +4,\;x^2+1)=1.$$
| \begin{array}{c|cccc}
& x^3+2x^2+x+4 & 1 & 0 \\
-x-2 & x^2+1 & 0 & 1 \\
\hline
-\frac 12x^2 & 2 & 1 & -x-2 \\
& 1 & -\frac 12x^2 &\frac 12x^3+x^2+1
\end{array}
$$-\frac 12x^2\color{red}{(x^3+2x^2+x+4)}+
\left(\frac 12x^3+x^2+1\right)\color{r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Show that the product of any $m$ consecutive positive integers is divisible by $m!$
Show that the product of any $m$ consecutive positive integers is divisible by $m!$.
Note that we have that $\frac{n!}{m!(n-m)!} \in \mathbb{Z}$ for $0 \leq m \leq n$.
$\require{cancel} \frac{n!}{m!(n-m)!} = \frac{n\cdot(n-1)\cdot\cdo... | Both answers posted so far assume that $n \geq m$, which is not given in the problem statement.
However, this assumption can be justified as a WLOG assumption. Indeed, fix a nonnegative integer $m$. We want to prove that
\begin{align}
m! \mid n\left(n-1\right)\left(n-2\right)\cdots\left(n-m+1\right)
\qquad \text{for ea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the inverse of $5+4\sqrt[3]{2}+3\sqrt[3]{4}$? I'm trying to find the inverse of $5+4\sqrt[3]{2}+3\sqrt[3]{4}$ in $\mathbb{Z}[\sqrt[3]{2}]$. I know it is a unit, so there is an inverse, but I feel like I may be doing too much work in the wrong direction. Here's what I have so far:
Let $\alpha = 5+4\sqrt[3]{2}+3\sqr... | Mostly an observation:
$$ 5 + 4 \sqrt[3]{2}+3\sqrt[3]{4}=(1+\sqrt[3]{2}+\sqrt[3]{4})^2$$
so
$$(5 + 4 \sqrt[3]{2}+3\sqrt[3]{4})^{-1}=((1+\sqrt[3]{2}+\sqrt[3]{4})^{-1})^2=(\sqrt[3]{2}-1)^2= 1- 2\sqrt[3]{2}+\sqrt[3]{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Positive integers $a,b,$ and $c$ satisfy the equations. Find the value of $a+b+c$
Question: Positive integers $a,b,$ and $c$ satisfy the equations$$29^2-a^2=28^2-b^2=27^2-c^2$$If $a<20$, what is the value of $a+b+c$?
I first tried factoring the three equations so that$$(29-a)(29+a)=(28-b)(28+b)=(27-c)(27+c)$$But I do... | Take $a$ and $b$ first, we will have:
$a^2-b^2=29^2-28^2=57 \Rightarrow (a-b)(a+b)=57$
Because $57>0, a>0, b>0$, we will have $0<a-b<a+b<57$.
There are only two positive cases:
$57=1 \times 57= 3 \times 19$.
Because $a-b<a+b$, we will have two set of equations:
*
*Case one: ${\begin{cases}a-b=1\\a+b=57\end{cases}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Inverse Laplace transformation that is slightly different from known transformation. I have the following inverse laplace transformation:
$L^{-1} =\frac{s}{(s-3)(s-4)(s-12)}$
After looking at the laplace transformations the closest I've found is:
$\frac{ae^{at}-be^{bt}}{a-b} = \frac{s}{(s-a)(s-b)}$
I've been working ou... | Using partial-fraction expansion, we have
$$
\frac{s}{(s-3)(s-4)(s-12)} = \frac{A}{(s-3)} + \frac{B}{(s-4)} + \frac{C}{(s-12)}
$$
where $A, B$ and $C$ are obtained as follows:
$$
\begin{align}
A &= \frac{s}{(s-4)(s-12)} \Big|_{s=3} = \frac{1}{3}, \\
B &= \frac{s}{(s-3)(s-12)} \Big|_{s=4} = -\frac{1}{2}, \\
C &= \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the minimum of $λ$ such that $|a-b|^p\le λ(2-|a-b|)\left|\frac{a|a|^{p-1}}{1-|a|}-\frac{b|b|^{p-1}}{1-|b|}\right|$ for $a,b\in(-1,1)$
Let $p\ge 2$ be a give real number. Find the minimum of $λ$ such that for any $a,b\in(-1,1)$,
$$|a-b|^p\le λ(2-|a-b|)\left|\dfrac{a|a|^{p-1}}{1-|a|}-\dfrac{b|b|^{p-1}}{1-|b|}\... | First, take $(a, b) = \left( \dfrac{1}{2}, -\dfrac{1}{2} \right)$, then $λ \geqslant 2^{p - 2}$. Now it will be proved that for $λ = 2^{p - 2}$, the inequality holds. Note that $f(x) = \dfrac{x^p}{1 - x}$ is increasing on $[0, 1)$. If $ab \geqslant 0$, without loss of generality, suppose $a \geqslant b \geqslant 0$, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2729193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Show that $\int_0^{\infty} \frac{dx}{(x+\sqrt{1+x^2})^n}$ converges and is equal to $\frac{n}{n^2-1}$ Testing convergence of $$I=\int_0^{\infty} \frac{dx}{(x+\sqrt{1+x^2})^n},$$ show that $$\int_0^{\infty} \frac{dx}{(x+\sqrt{1+x^2})^n}=\frac{n}{n^2-1}$$
I thought to put $$x=\tan{z}, \text{ then } \quad I=\int_0^{\pi/2... | Using twice the fact that $\begin{aligned} & \frac{d}{d x} \frac{1}{(\tan \theta+\sec \theta)^n} = -\frac{n \sec \theta}{(\tan \theta+\sec \theta)^n}\end{aligned} $ via integration by parts, we have a beautiful recursive relation for $I$ as below:
$$
\begin{aligned}
I& = \int_0^{\infty} \frac{d x}{\left(x+\sqrt{1+x^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2729481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding the area of circles in triangle In the figure there are infinitely many circles approaching the
vertices of an equilateral triangle, each circle touching other circles and sides of the triangle. If the triangle has sides of
length $1$, how can i find the total area occupied by the circles?
| In a equilateral triangle the ratio $\frac{A_{\text{incircle}}}{A_{\text{triangle}}}$ equals $\frac{\pi}{3\sqrt{3}}$. Step 0: we draw the incircle of the original triangle.
Step 1: we draw three incircles for equilateral triangles with side length $\frac{1}{3}$. Step $n$: we draw three incircles for equilateral triangl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2733098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solving $\sqrt{a +\sqrt{a-x}}+\sqrt{a-\sqrt{a+x}}=2x, \ a\in\Bbb{R}$
Given
$$\sqrt{a +\sqrt{a-x}}+\sqrt{a-\sqrt{a+x}}=2x$$
and $a\in\Bbb{R}$, express $x$ in terms of $a$.
I rationalised the above expression and then again rationalised which gave me :
$$\sqrt{a+\sqrt{a-x}}-\sqrt{a-\sqrt{a+x}}=\frac{1}{\sqrt{a+x}-\... | Here is a simplification, which leads to a parametrized solution.
Write the original equation as two equations
$$
\sqrt{a +\sqrt{a-x}} = x +q\\
\sqrt{a-\sqrt{a+x}}= x -q
$$
where $q$ is a function of $a$.
Double squaring both equations gives
$$
(x+q)^4-2a(x+q)^2+x+a^2-a = 0 \\
(x-q)^4-2a(x-q)^2-x+a^2-a = 0
$$
Subtra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2734359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Define $f: \Bbb R^2 \to \Bbb R$ by $f(x,y) =\begin{cases} \frac{xy^2}{x^2 + y^4}, & \text{if $(x,y) \ne 0$} \\ 0, & \text{if $(x,y)= 0$} \end{cases}$ Define $f: \Bbb R^2 \to \Bbb R$ by
$f(x,y) =\begin{cases}
\frac{xy^2}{x^2 + y^4}, & \text{if $(x,y) \ne 0$} \\
0, & \text{if $(x,y)= 0$}
\end{cases}$
I'm to show that $f... | To show that it's continuous along straight lines:
Approach the origin along the half line $(x,y) = (r\cos\theta,r\sin\theta)$, where $r>0$ and $r\to 0^+$.
\begin{eqnarray*}
\frac{xy^2}{x^2+y^2} &\rightsquigarrow&
\frac{r^3\cos\theta\sin^2\theta}{r^2\cos^2\theta+r^2\sin^2\theta} \\ \\
&\equiv& \frac{r^3\cos\theta\sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Adding $k^2$ to $1^2 + 2^2 + \cdots + (k - 1)^2$. In his book, Calculus Vol. 1, Tom Apostol mentions that adding $k^2$ to the predicate
$$A(k): 1^2 + 2^2 + \cdots + (k - 1)^2 < \frac{k^3}{3}$$
gives the inequality
$$ 1^2 + 2^2 + \cdots + k^2 < \frac{k^3}{3} + k^2.$$
Why does the RHS $$1^2 + 2^2 + \cdots + (k - 1)^2 $$ ... | I will give you an example.
1+2+3+........+99+100 is same as 1+2+.....+100.
And, I believe this is quite obvious.
Hence, both of them have the same value whatever way you decide to solve it by.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Mini-Challenge on a condition Hello I would like to purpose to you an enigma this is the following :
Let $a,b,c$ be positive real number find the condition on $abc$ and $ab+bc+ca$ and $a+b+c$ to have $a^9+b^9+c^9=3$
I have a solution using the following identity :
$$a^9+b^9+c^9= 3a^3b^3c^3−45abc(ab+bc+ca)(a+b+c)^4+ ... | To make it less ugly, calculate the lower amounts instead, or setting new variables, for example:
${\begin{cases}a+b+c=x\\ab+bc+ca=y\\abc=z\\S_{n}=a^n+b^n+c^n\end{cases}}$
In fact, I have encountered this problem in a test that gives $x=3;y=-10;z=11$ and it asks me to find $S_{5}$. Here's how I attemped it (I have gene... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Determining the eigenvalues for the linear operator $T$ on $V$ $V=M_{2x2}$
$T\begin{bmatrix}a&b\\c&d\end{bmatrix} = \begin{bmatrix}c&d\\a&b\end{bmatrix}; \tag 1$
Attempt:
I used the standard basis to find
$A = \begin{bmatrix}0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0\end{bmatrix}; \tag 2$
then
$\det(A-\lambda I)=\begin{bma... | First, a few remarks on our OP Essie Stern's work; then I'll provide my own take on this one.
Apparently we are using the mapping
$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \mapsto \begin{pmatrix} a \\b \\ c \\ d \end{pmatrix} \tag 1$
to express members of $M_{2 \times 2}$ as column vectors; the map is clearly a lin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2738787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Alternate solutions to similar integrals? So, when I started this problem I couldn't see any reasonable trig substitutions that would make it simpler, and I don't know of a better method, so I attempted to solve it using repeated use of integration by parts. Is there a much easier, better, alternate solution to integra... | Hint
$$I=\int sin^4(x)dx=\int \sin^2(x)-\sin^2(x)\cos^2(x)dx$$
$$I=\int \sin^2(x)-\frac 14\sin^2(2x)dx$$
Note that
$$\sin^2(x)=1-\cos^2(x)=1-\frac 12(\cos(2x)+1)=\frac 12(1 -\cos(2x))$$
$$I=\frac 12\int (1 -\cos(2x)) -\frac 18\int (1 -\cos(4x))dx$$
$$\boxed{I=\frac 38x-\frac 14 \sin(2x)+\frac 1{32} \sin(4x)+K} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find $y^{(y^2-6)}$? $$\frac{3}{1-3^{x-2}} + \frac{3}{1-3^{2-x}} = y$$ $$x≠2$$ $$y^{(y^2-6)} = ?$$What is $y^{(y^2-6)}$? Could you please explain to me how to solve this question step by step?
| You have
$$\begin{align}
y & = \frac{3}{1-3^{x-2}} + \frac{3}{1-3^{2-x}} \\
& = \frac{3(1-3^{2-x}) + 3(1-3^{x-2})}{(1-3^{x-2})(1-3^{2-x})} \\
& = \frac{3(1-3^{2-x} + 1-3^{x-2})}{(1 - 3^{2-x}-3^{x-2}+3^{x-2+2-x})} \\
& = \frac{3(2-3^{2-x} - 3^{x-2})}{(1 - 3^{2-x}-3^{x-2}+3^{0})} \\
& = \frac{3(2-3^{2-x} - 3^{x-2})}{(2 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2742815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Baffled with $\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}$
Calculate $$\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}$$
Personal work:
$$\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}=^{0 \over 0}\lim\limits_{x\to 0}{e^x-e^{\sin x}\cdot\cos x \over 1-\cos x}=^{0 \over 0}\lim\limits_{x\to 0}{{e^x-(e^... | Use the third-order Maclaurin formulae
$$
e^x = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3),
$$
$$
\sin x = x-\frac{x^3}{6}+o(x^3),
$$
$$
e^{\sin x} = e^{x-\frac{x^3}{6}+o(x^3)} =
1+\left(x-\frac{x^3}{6}+o(x^3)\right)+\frac{1}{2} \left(x-\frac{x^3}{6}+o(x^3)\right)^2+
\frac16 \left(x-\frac{x^3}{6}+o(x^3)\right)^3 + o\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2743545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
How to factor $a^{3} + b^{3} + c^{3} - 3abc$ into a product of polynomials The question is in the title.
This question is from "Algebra" by Gelfand.
My initial thought is that if $a$, $b$ and $c$ are $1$ or $-1$, then the polynomial evaluates to $0.$ So, maybe two of the factors will be $(a + b + c - 3)$ and $(a + b + ... |
Factor $a^3+b^3+c^3-3abc$ to a product of polynomials.
Think when $a=b=c$.
$$a^3+b^3+c^3-3abc=a^3+a^3+a^3-3a^3=0.$$
This only fits when $a=b=c$, but not when $a=b$ or $b=c$ or $c=a$.
So, you can think about $(a-b)^2+(b-c)^2+(c-a)^2$, which is $0$ if and only if $a=b=c$.
Therefore, you can reason $a^2+b^2+c^2-ab-bc-ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
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Evaluation of $\int_{0}^{\pi/6}\tan^{-1}\sqrt{2-\tan^2(x)}\mathrm dx $ $$\int_{0}^{\pi/6}\tan^{-1}\sqrt{2-\tan^2(x)}\mathrm dx $$
$1+\tan^2(x)=\sec^2(x)$
$u=\sqrt{2-\tan^2(x)}, dx={udu\over \sqrt{2-u}+(\sqrt{2-u^2})^3}$
$$\int_{\sqrt{2-\sqrt{3}/3}}^{\sqrt{2}}{udu\over \sqrt{2-u}+(\sqrt{2-u^2})^3}$$
Very messy, how do w... | We can approximate the value using composition of Taylor series for the integrand around $x=0$.
This would give
$$\tan ^{-1}\left(\sqrt{2-\tan ^2(x)}\right)=\tan ^{-1}\left(\sqrt{2}\right)-\frac{x^2}{6 \sqrt{2}}-\frac{23 x^4}{144
\sqrt{2}}-\frac{3727 x^6}{25920 \sqrt{2}}-\frac{124627 x^8}{967680
\sqrt{2}}+O\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Matrix Notation Form of Roots of a Quadratic Equation We know that the quadratic equation
$$f(x)=ax^2+bx+c=0$$
has roots
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=-\frac b{2a}\pm \frac 1a\sqrt{-\left(ac-\frac {b^2}4\right)}$$
Also, $f(x)$ can be written in matrix notation as follows:
$$f(x)=
\left(\begin{matrix}x&1\\\end{m... | I have obtained some formula with a little changed notation comparing the notation used in the question.
I've interchanged the components of $\mathbf x$ (the new vector is denoted as $\mathbf{ \hat{x}}$ ) and I've interchanged the entries on the main diagonal of $ \mathbf Q$ (new matrix is denoted as $ \mathbf M$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2750445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find the remainder when $(x+1)^n$ is divided by $(x-1)^3$. I know this question has been answered before, but I have a slightly different different question.
I saw the solution of this question in my book and the author has solved it by substituting $x-1=y$ and then equating the coefficients of $y^2$, $y^1$ and $y^0$ ... | $$(x+1)^n=(x-1)^3Q (x)+ax^2+bx+c $$
for $x=1$, we get
$$\boxed {2^n=a+b+c} $$
differentiating,
$$n (x+1)^{n-1}=3 (x-1)^2R (x )+2ax+b $$
with $x=1$,
$$\boxed {2^{n-1}=2a+b} $$
differentiating again
$$n (n-1)(x+1)^{n-2}=6 (x-1)S (x)+2a $$
with $x=1$, we find
$$\boxed {a=n (n-1)2^{n-3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Why A.M.$\geq $ G.M. not works here. Find the minimum value of $f(x)= x^2+4x+(4/x)+(1/x^2)$ for $x>0$
The minimum value of given function $f(x)= x^2+4x+(4/x)+(1/x^2)$ where $x>0$
(A) 9.5
(B) 10
(C) 15
(D) 20
My try
By A.M G.M inequality
$\frac {x^2+4x+(4/x)+(1/x^2)}{4} \geq \left(x^2.4x. \frac {4}{x}. \frac{1}{x^2}\r... | AM-GM works but a bit differently:
$$f(x)= x^2+4x+(4/x)+(1/x^2) = x^2+\frac{1}{x^2} + 4(x+\frac{1}{x}) \geq 2 + 4\cdot 2 = 10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2754372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$\lim_{n\to\infty}\frac{n -\big\lfloor\frac{n}{2}\big\rfloor+\big\lfloor\frac{n}{3}\big\rfloor-\dots}{n}$, a Brilliant problem I encounter a question when visiting Brilliant:
Find
$\space\space\space\space\lim_{n\to\infty}s_n$
$=\lim_{n\to\infty}\frac{n - \big \lfloor \frac{n}{2} \big \rfloor+ \big \lfloor \frac{n}{3} ... | Elementary high school approach
One can show immediately with the squeeze theorem that
$$L_1=\lim_{n\to\infty}\left(\sum_{k=1}^{2\left\lfloor \sqrt{n} \right\rfloor}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=1}^{\left\lfloor \sqrt{n} \right\rfloor}\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\right)=\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2756119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 3,
"answer_id": 1
} |
Finding line of intersection between two planes by solving a system of equations Ok, so the question is to find the line of intersection between two planes, given their equations.
$x+3y+2z=4$
$x-y-z=4$
I know there's the way of using the vector perpendicular to both normals of the planes as the direction vector of the... | Having
$$
a_1x+b_1y+c_1z= d_1\\
a_2x+b_2y+c_2z=d_2
$$
or
$$
\left[
\begin{array}{ccc}
a_1 & b_1 & c_1\\
a_2&b_2 & c_2
\end{array}
\right]
\left[
\begin{array}{c}
x\\
y\\
z
\end{array}
\right]
=
\left[
\begin{array}{c}
d_1\\
d_2
\end{array}
\right]
$$
Now choosing an invertible $2\times 2$ submatrix in $\left[
\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2757158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Given linear mapping and bases, determine the transformation matrix and the change of basis
Given is linear mapping $f:\mathbb{R}^3 \rightarrow \mathbb{R}^3$. Matrix of $f$ in terms of the ordered standard basis $B_0^3 = \left\{\vec{e_1}, \vec{e_2}, \vec{e_3}\right\}$ of $\mathbb{R}^3$ is $\,\,\,\, ... | The change of basis matrix $T_B^{B_0^3}$ is simply$$\begin{pmatrix}2&1&2\\2&1&1\\3&1&1\end{pmatrix}$$and $T_{B_0^3}^B$ is$$\begin{pmatrix}2&1&2\\2&1&1\\3&1&1\end{pmatrix}^{-1}=\begin{pmatrix}0 & -1 & 1 \\ -1 & 4 & -2 \\ 1 & -1 & 0\end{pmatrix}.$$So$$A_{BB}^f=T^B_{B_0^3}A_{B_0^3B_0^3}^fT^{B_0^3}_B=\begin{pmatrix}1 & 0 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2760817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Suppose $x, y, z$ are positive real number such that $x + 2y + 3z = 1$. Find the maximum value of $xyz^2$ I'm trying to solve this problem using Lagrange's multiplier method but I'm unable to get the value of lambda and also $z.$ kindly help me out with this problem.
I'm getting $x = 2y$
I'm stuck somewhere between th... | Alt. hint (no calculus): the following inequality holds by AM-GM, with equality iff $\displaystyle x = 2y = \frac{3}{2}z\,$:
$$
\frac{\sqrt{3}}{\sqrt[4]{2}} \cdot \sqrt[4]{xyz^2} = \sqrt[4]{x \cdot 2y \cdot \frac{3}{2}z \cdot \frac{3}{2}z} \;\le\; \frac{x+2y+3z}{4} = \frac{1}{4} \;\;\iff\;\;xyz^2 \le \frac{2}{3^2 \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How do I solve the inequality $x^{12}−x^9+x^4−x+1>0$ using intervals? My book has the question $x^{12}−x^{9}+x^{4}−x+1>0$. The solution given gives three cases, when $x \le 0$, when $0 < x \le 1$ and when $x > 1$. How did they get these intervals? What is the method used?
| Clearly for $x=0$, We have $$x^{12}-x^{9}+x^4-x+1>0$$
Now for $x\neq 0$, Using Arithmetic Geometric Inequality
$$\frac{x^{12}+x^{12}+x^{12}+1}{4}\geq \sqrt[4]{x^{12}\cdot x^{12}\cdot x^{12}\cdot 1}=|x^9|>x^9$$
$$\frac{x^4+1+1+1}{4}\geq \sqrt[4]{x^4\cdot 1 \cdot 1\cdot 1}=|x|>x$$
So $\displaystyle \frac{3x^{12}}{4}-x^9+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Simplifying the derivative of $x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}$ $$x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}$$
So I get:
$$-x^{\frac{2}{3}} \cdot \frac{1}{3} (6-x) ^{\frac{-2}{3}} + (6-x) ^{\frac{1}{3}} \cdot \frac{2}{3} x ^ {\frac{-1}{3}}$$
How does one go about simplifying this?
I guess I can pull out common... |
Pulling out this common factor is fine. We can simplify the last expression slightly more and obtain
\begin{align*}
\color{blue}{-x^{\frac{2}{3}}}&\color{blue}{ \cdot \frac{1}{3} (6-x) ^{-\frac{2}{3}} + (6-x) ^{\frac{1}{3}} \cdot \frac{2}{3} x ^ {-\frac{1}{3}}}\\
&=\frac{1}{3}x^{-\frac{1}{3}}(6-x)^{-\frac{2}{3}}(-x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Calculate $\sum\limits_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately Is it possible to calculate $\sum_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately?
The original question was find the number of solutions to $2x+y+z=20$ which I calculated to be the coefficient of $x^{20... | Alternatively:
$$\begin{align}\sum_{k=0}^{20}(-1)^k\binom{k+2}2=&\sum_{k=0}^{20}\binom{k+2}2-2\cdot \sum_{k=0}^{9}\binom{2k+3}2=\\
&{23\choose 3}-\sum_{k=0}^9 (2k+3)(k+1)=\\
&1771-2\cdot 825=121.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2772397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Determinant of a matrix of ones, whose anti diagonal elements are zero I'm trying to prove a formula I have constructed for the determinant of a general $n\times n $ real matrix $A$, given here in the case $n=5$:
$$ A = \begin{bmatrix}
1 & 1 & 1 & 1 & 0 \\
1 & 1 & 1 & 0 & 1 \\
1 & 1 & 0 & 1 & 1 \\
1 & 0 & 1 & 1 & 1 \\
... | This is easy to calculate by row reduction:
Add all rows to row 1.
$$\det(A) = \begin{vmatrix}
1 & 1 & ... & 1 & 0 \\
1 & 1 & ... & 0 & 1 \\
... & ... & ... & ... & ... \\
1 & 0 & ... & 1 & 1 \\
0 & 1 & ... & 1 & 1
\end{vmatrix}=\begin{vmatrix}
n-1 & n-1 & ... & n-1 & n-1 \\
1 & 1 & ... & 0 & 1 \\
... & ... & ... & ...... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2775676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Condition for rank to be 2. Suppose that
$$\mbox{rank}\begin{equation}
\begin{pmatrix}
1 & 1 & 2 & 2\\
1 & 1 & 1 & 3\\
a & b & b & 1
\end{pmatrix}
\end{equation} = 2$$
for some real numbers $a$ and $b$. What is the value of $b$?
*
*$0.3$
*$3$
*$1$
*$0.5$
I'm unable to get the condition on $b$ only.
| $$\begin{equation}
\begin{pmatrix}
1 & 1 & 2 & 2\\
1 & 1 & 1 & 3\\
1/3 & 1/3 & 1/3 & 1
\end{pmatrix}
\end{equation}$$
has rank $2$ thus the closest answer is $ b=0.3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2777775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluate $\int \frac {dx}{\sin \frac x2\sqrt {\cos^3 \frac x2}}$
Evaluate $$\int \frac {dx}{\sin \frac x2\sqrt {\cos^3 \frac x2}}$$
My try
Write $t=\frac x2$ and hence $dx=2dt$
To change the integral to $$\int \frac {\csc t dt}{\cos^{\frac 32} t}$$
Multiplying both bottom and top by $\csc t$ and then using $\csc^2 t... | Elaborating on answer of @JoseCarlosSantos:
By performing the substitution $t=u^2$,
$$\int\frac{dt}{(1-t^2)t\sqrt t}=\int\frac{2udu}{(1-u^4)(u^3)}=2\int\frac{du}{(1-u^4)u^2}$$
Performing partial fraction decomposition,
$$\frac1{(1-u^4)u^2}$$
$$=\frac1{u^2}+\frac{u^2}{1-u^4}$$
$$=\frac1{u^2}+\frac{u^2}{2(1+u^2)}+\frac{u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2779429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\binom{2n}{n}\geq2^n$ for all $n\in\mathbb{N}_0$. Show that $\binom{2n}{n}\geq2^n$ for all $n\in\mathbb{N}_0$.
First, observe that $\binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!}=\frac{(2n)!}{(n!)^2}=\frac{1\cdot 2\cdot 3\cdot...\cdot 2n-1\cdot 2n}{(n!)^2}=\frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n-1)\cdot(2\cdot 4\... | Another approach: Consider a line of $2n$ people, and we want to put them in two teams of $n$ people say $\color{red}{red}$ and $\color{blue}{blue}$ then what if you restrict to the first $n$ people that you put in the line, they can be whatever because regardless of the choice at most you picked $n$ of them in one tea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Determinant of $\left(\begin{smallmatrix} -2a &a+b &a+c \\ b+a& -2b &b+c \\ c+a&c+b & -2c \end{smallmatrix}\right)$ Evaluate $$D=\begin{vmatrix}
-2a &a+b &a+c \\
b+a& -2b &b+c \\
c+a&c+b & -2c
\end{vmatrix}$$
My try:
Applying $R_1 \to R_1+R_2$ we get
$$D=\begin{vmatrix}
b-a&a-b &a+b+2c \\
b+a& -2b &b+c \\
c+... | Let $p(a,b,c) $ be the determinant. Note that each term of $p$ has degree 3 (sum of degrees of $a,b,c$).
Note that $p(a,-a,c) = 0$, hence $a+b$ divides $p$.
Similarly we see that $a+c, b+c$ divide $p$.
Hence $p$ has the form $p(a,b,c) = k (a+b)(b+c)(a+c)$ for some constant $k$.
Compute the determinant for $a=b=c={1\ove... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Line-integral: 2-form integrated over cube
Let $$\omega=x\cos(xy)\cos(2\pi x)\ \ \ \text{d}x\wedge\text{d}y$$ Calculate $$\int_{[0,\frac{1}{4}]\times[0,2\pi]}\omega$$
Now we have:
\begin{align} \int_{[0,\frac{1}{4}]\times[0,2\pi]}x\cos(xy)\cos(2\pi x)\ \text{d}x\wedge\text{d}y & = \int_0^{\frac{1}{4}}\Big(\int_0^{2\... | It should be:
$$
\begin{align}
\int\limits_0^\frac{1}{4} \cos(2\pi x)\sin(2\pi x) \mathrm{d}x &= \int\limits_0^\frac{1}{4}\frac{1}{2}\sin(4\pi x)\mathrm{d}x \\
&= \frac{1}{2}\left[\frac{-1}{4\pi}\cos(4\pi x)\right]_0^\frac{1}{4} \\
&= \frac{1}{2}\cdot\frac{1}{4\pi}(1 - (-1)) \\
&= \frac{1}{4\pi}
\end{align}
$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2788453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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This function has two different integrals? $f(x)=∫\frac{1}{x^2}dx$
Integrating by u-substitution:
$u=x^2$
$du=2dx$
$\frac{1}{2}du = dx$
$∫\frac{1}{x^2}dx=$ $∫\frac{1}{u}\times\frac{1}{2}du$
$\frac{1}{2}$∫ $\frac{1}{u}du$
$=\frac{1}{2}ln u+c$
$=\frac{1}{2}ln x^2+c$
$=lnx+c$
Another way:
$∫\frac{1}{x^2}dx=∫x^{-2}dx $
... | If $u=x^2$, then $\mathrm du=2x\,\mathrm dx$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2788929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How I find $\gcd(\frac{a^{2m+1}+1}{a+1}, a+1)$? $$\gcd\left(\frac{a^{2m+1}+1}{a+1}, a+1\right)$$
My answer until this moment is:
$$
For: a^{2m+1}+1^{2m+1} = (a + 1)^{m+1} - 2a = (a+1)^m(a+1)-2a
$$ where
\begin{align}
(a + 1)^{m} = a^{2m}+a^{2m-1}+\cdots+1
\end{align}
So,
\begin{align}
a^{2m+1}+1 = [a^{2m}+a^{2m-1}+\... | Here is one approach. Let $b = a + 1$ so $a = b - 1$. Then, we can apply the binomial theorem to obtain the modular equivalence
$$\begin{align*} \frac{a^{2m + 1} + 1}{a + 1} & = \frac{(b - 1)^{2m + 1} + 1}{b}\\
& = \frac{1}{b} \left(1 + \sum_{k = 0}^{2m + 1} \binom{2m + 1}{k} b^{k}(-1)^{2m + 1 - k} \right)\\
& = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Need Help Finding Solution to These 5 Simple Exponent Exercises In this document, I have 5 exercises involving exponents and their five corresponding solutions. I am trying to find out why these are the solutions for these equations, as I cannot seem to get the same results as the provided solutions. The exercises ask ... | $1)$ Use the law of exponents ie $a^{-n}=\frac1{a^n}$
so $(-3)^4\div3^5=\frac{(-1)^43^4}{3^5}=3^{4-5}=3^{-1}$
$2)$ $\dfrac{2^{-1}+2^{-2}}{2^{-1}\cdot2^{-2}}=\frac{2^{-1}}{2^{-3}}+\frac{2^{-2}}{2^{-3}}= 2^2+2=6$
$3)$ note that $(a\cdot b)^c = a^c\cdot b^c$
$\dfrac{81^4\cdot4^9}{6^{16}}= \dfrac{3^{4\cdot4}\cdot2^{2\cdot9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Can all natural numbers be expressed as $\lceil \frac{3^a}{2^b} \rceil$ or $\lfloor \frac{3^a}{2^b} \rfloor$? Can any natural number $n$ be expressed as
$n=\lceil \frac{3^a}{2^b} \rceil$ or $n=\lfloor \frac{3^a}{2^b} \rfloor$ where $ a,b \in \mathbb{N}$ ?
I have found no approach to this problem. All suggestions are ap... | Yes, all natural numbers $n$ can be represented as $n=\left\lceil \frac{3^a}{2^b} \right\rceil$ or as $n=\left\lfloor \frac{3^a}{2^b} \right\rfloor$ (whichever one you want). Let's do floors, for concreteness.
For $n=\left\lfloor \frac{3^a}{2^b} \right\rfloor$ you want positive integers $a$ and $b$ such that $n \le \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Indefinite integral of $\sqrt{x^2-x}$ i was trying to compute the indefinite integral:
$$
\int\sqrt{x^2-x}dx
$$
but i got stuck: after a few (unsuccessful) attempts for some $u$-substitution, i tried integration by parts:
$$
\int\sqrt{x^2-x} \ dx=\int(x)'\sqrt{x^2-x} \ dx= \\ x\sqrt{x^2-x}-\int x(\sqrt{x^2-x})'dx= \\
=... | complete the square
$$\begin{align}
\int\sqrt{x^2-x} \ dx &=I \\
x^2-x&=\left(x-\frac{1}{2}\right)^2-\frac{1}{4} \\
\text{set} \quad u&=x-\frac{1}{2} \quad \text{then} \quad dx=du\\
I&=\int\sqrt{u^2-a} \ du \quad \text{such that} \quad a=\frac{1}{4}\\
\end{align}$$
This type of integral is well known. You should now u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
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Limit of the ratio of two modified Bessel functions Could you help me evaluate the following limit:
$$
\lim_{x\to \infty} \frac{K_1(x)}{K_0(x)}
$$
where $K_\nu$ is the modified Bessel function of the second kind of order $\nu$. Both $K_1$ and $K_0$ approaches $0$ as $x\to\infty$, so we have an indeterminate form. L'Hos... | For large values of $x$
$$K_1(x)=\sqrt{\frac{\pi }{2}}\,e^{-x}\left(\frac 1 {x^{ 1/2}}+\frac 3 8\frac 1 {x^{ 3/2}}-\frac{15}{128}\frac 1 {x^{5/2}}+O\left(\frac 1 {x^{7/2}}\right)\right)$$
$$K_0(x)=\sqrt{\frac{\pi }{2}}\,e^{-x}\left(\frac 1 {x^{ 1/2}}-\frac 1 8\frac 1 {x^{ 3/2}}+\frac{9}{128}\frac 1 {x^{5/2}}+O\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2793666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Non negative integer triplets $(x,y,z)$ in $x^2+2y^2+4z^2+5=2(x+y+xy)+4z(y-1)$
Non negative integer triplets $(x,y,z)$ in $x^2+2y^2+4z^2+5=2(x+y+xy)+4z(y-1)$
Try: Writting equation as $$4z^2-4(y-1)z+x^2+2y^2-2(x+y+xy)+5=0$$
Now if equation has real roots. Then $$z=\frac{(y-1)\pm \sqrt{2xy+2x-x^2-y^2-4}}{2}$$
So for ... | There is no way
$(x+y)^2+k^2=2x-4$ is giving you non negative values for $x,y.$
Now looking on RHS we get that $2x-4$ will give least value at $x=2$.
Also we know that $x^2>mx$ iff $x>m$. ---(1)
Now as $y$ cant be negative and $k^2$ too can't be negative.
From $x^2>2x$ for all $x>2$ which is a contradiction as least v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can we say that $\frac{a+b}{c+d} \leq \frac{a}{c} + \frac{b}{d} $ If $a,b,c,d$ are positive real numbers, can we say that $\frac{a+b}{c+d} \leq \frac{a}{c} + \frac{b}{d} $ is always true? If no, can you please give insignts on under which conditions this might be true.
Any references to a similar type of inequalities ... | After the latest edit:
Wlog. $\frac ac\le\frac bd$. Then $$\frac ac\le\frac{a+b}{c+d}\le\frac bd<\frac ac+\frac bd$$
Indeed, $\frac ac\le \frac bd$ implies $\frac{bc-ad}{cd}\ge 0$, i.e., $bc-ad\ge 0$.
Then
$$ \frac bd-\frac{a+b}{c+d}=\frac{b(c+d)-d(a+b)}{d(c+d)}=\frac{bc-ad}{d(c+d)}\ge0$$
and
$$ \frac{a+b}{c+d}-\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factorisation of $4x^4-4x^3+4x^2+2x+1$ with real coefficients I have used many mathematical trick to factorise this polynomial : $4x^4-4x^3+4x^2+2x+1$ with real coeffecients but i didn't succeed because as I see all it's root are complex , and I want if there is any suitable method to factorise it ? and thanks in adva... | Let $$ x = \frac{i t }{ \sqrt 2} $$
Your polynomial is now
$$ t^4 + i \sqrt 2 t^3 - 2 t^2 + i \sqrt 2 t + 1 $$
Divide by $t^2$ and then introduce $w = t + \frac{1}{t}$
dividing by $t^2$
$$ t^2 + i \sqrt 2 t - 2 + \frac{i \sqrt 2}{t} + \frac{1}{t^2} $$
which is
$$ w^2 + i \sqrt 2 w - 4 $$
so the roots are
$$ w =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Rearrangement inequality and minimal value of $\frac{\sin^3x}{\cos x} +\frac{\cos^3x}{\sin x}$ For $x \in \left(0, \dfrac{\pi}{2}\right)$, is the minimum value of $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x} = 1$? So considering ($\dfrac{1}{\cos x}$, $\dfrac{1}{\sin x}$) and ($\sin^3x$, $\cos^3x$), is it right to ... | Yes it is fine, indeed for $x\in(0,\pi/2)$ we have in both cases
*
*$\sin x\ge \cos x \implies \sin^3 x\ge \cos^3 x \implies \frac1{\sin x}\le \frac1{\cos x}$
then
$$\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}\ge \frac{\sin^3 x}{\sin x}+\frac{\cos^3 x}{\cos x}=1$$
*$\cos x\ge \sin x \implies \cos^3 x\ge \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Evaluate $\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}$. Problem
Evaluate $$\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}.$$
Solution
Notice that $$\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^... | Let $a=\sqrt2$. (Other values also work.)
Then we have
$$
\begin{aligned}
&\lim_{x\to\infty}
\frac{(x+a)^a-(x-a)^a}{x^{a-1}}
\\
&\qquad=
\lim_{x\to\infty}
a\cdot \frac xa\left(\ \left(1+\frac ax\right)^a-\left(1-\frac ax\right)^a\ \right)
\\
&\qquad\qquad\text{ after forced division with $x^a$ in both numerator and den... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Arc length of ellipse in polar coordinates I have the equation of an ellipse given in Cartesian coordinates as
$\left(\frac{x}{0.6}\right)^2+\left(\frac{y}{3}\right)^2=1$ .
I need the equation for its arc length in terms of $\theta$, where $\theta=0$ corresponds to the point on the ellipse intersecting the positive x-a... | \begin{align}
r &= \frac{a b}{\sqrt{a^2\sin^2 \theta+b^2\cos^2 \theta}} \\
r' &=
\frac{ab(b^2-a^2)\sin \theta \cos \theta}
{(a^2\sin^2 \theta+b^2\cos^2 \theta)^{3/2}} \\
ds &=
\frac{ab\sqrt{a^4\sin^2 \theta+b^4\cos^2 \theta}}
{(a^2\sin^2 \theta+b^2 \cos^2 \theta)^{3/2}} d\theta \\
s &= b E
\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Compute $\int_0^1\frac {\sqrt{x}}{(x+3)\sqrt{x+3}}dx.$
Evaluate $\int_0^1\frac {\sqrt{x}}{(x+3)\sqrt{x+3}}dx.$
I tried so many substitutions but none of them led me to the right answer:
$u=\frac 1{\sqrt{x+3}}$, $u=\frac 1{x+3}$, $u=\sqrt{x}$... I even got to something like $\int_0^1 \frac {u^2}{(u^2+3)^{\frac 32}}du$... | Hint:
Integrate by parts
$$\int\sqrt x\dfrac1{(x+3)^{3/2}}dx$$
$$=\sqrt x\int\dfrac1{(x+3)^{3/2}}dx-\int\left(\dfrac{d(\sqrt x)}{dx}\int\dfrac1{(x+3)^{3/2}}dx\right)dx=?$$
Now for $\displaystyle\int\dfrac{dx}{\sqrt{x(x+3)}}$
$x(x+3)=\dfrac{(2x+3)^2-3^2}4$
set $2x+3=3\sec y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Orthonormal diagonalizable Let:
$$ A = \begin{pmatrix} 5 &2 & -1 \\ 2 & 2 & 2 \\ -1 & 2 & 5 \end{pmatrix} \in Mat_3(\mathbb{R})$$
1) Show that $0$ and $6$ are eigenvalues for $A$ and find the basis for the corresponding eigenspace.
2) Explain why $A$ is orthonormal diagonalizable and find an orthonormal basis for $\m... | It should be obvious from inspection that the two eigenspaces are orthogonal, so you just need an orthogonal basis for each one. $E_A(0)$ trivially has an orthogonal basis, and one iteration of the Gram-Schmidt process will get you one for $E_A(6)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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IVP Differential Equation I have stumbled across a very old exam question from my linear algebra course and the solutions are not available. I was wondering if my working/logic is correct and if any improvements can be made.
Let $\ B=\begin{pmatrix}
1 & -1 & 4 \\
6 & -7 & 2 \\
-3 & 1 & -6
\end{pmatr... | You can rewrite the matrix in its Jordan normal form
$$ \textbf{B} =
\textbf{V}
\textbf{J}
\textbf{V}^{-1}
$$
where
$$ \textbf{J} = \begin{bmatrix} -5 & 1 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & -2 \end{bmatrix}, \quad \textbf{V} = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -2 \\ -1 & -1 & 1 \end{bmatrix} $$
Applying the matrix expone... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there a formula that generalizes $\sin A+\sin B+\sin C = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$ (where $A+B+C=\pi$) to four angles? If $A+B+C=\pi$ then we have $$\sin A+\sin B+\sin C = 4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)$$
If $A+B+C+D=\pi$ is there a similar... | There is this similar result: if $\,A + B + C + D = 2\pi\,$ (not $\pi$), then
$$\sin(A) + \sin(B) + \sin(C) + \sin(D) = 4 \sin\Big(\frac{A+B}{2}\Big) \sin\Big(\frac{B+C}{2}\Big) \sin\Big(\frac{C+A}{2}\Big),$$ or, alternatively, both equal
$$-4 \sin\Big(\frac{A+B}{2}\Big) \sin\Big(\frac{A+C}{2}\Big) \sin\Big(\frac{A+D}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2805661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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If $2 (\cos \alpha-\cos\beta)+\cos\alpha\cos\beta=1$, then $p\tan\frac{\alpha}{2}+q\tan\frac{\beta}{2}=0$ for which choices of $p$ and $q$? Let $\alpha$ and $\beta$ be nonzero real numbers such that $2 (\cos \alpha-\cos\beta)+\cos\alpha\cos\beta=1$. Then which of the following is/are true ?
a) $\sqrt3\tan(\frac\alpha2... | Given $2(\cos \alpha-\cos \beta)+\cos \alpha\cos \beta = 1$
So $\displaystyle \cos \alpha(\cos \beta+2)=1+2\cos \beta \Rightarrow \cos \alpha =\frac{2\cos \beta+1}{2+\cos \beta}$
So $\displaystyle \frac{1}{\cos \alpha} = \frac{2+\cos \beta}{2\cos \beta +1}$
Using Componendo Dividedno, We have
$\displaystyle \frac{1+\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2808686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to prove $1+\cos2\theta+\cos4\theta+\cos6\theta+\cos8\theta=\frac{(\cos4\theta)(\sin5\theta)}{\sin\theta} $? I need help to prove that the following is true:
$$1+\cos2\theta+\cos4\theta+\cos6\theta+\cos8\theta=\frac{(\cos4\theta)(\sin5 \theta)}{\sin\theta}$$
I realize that I must evaluate the real part of this, bu... | According to the link posted as a possible duplicate and containing my answer in it, we have:
$$1+\cos{\theta}+\cos{2\theta}+...+\cos{n\theta}=\frac{\cos{\frac{n\theta}{2}}\cdot\sin{\frac{(n+1)\theta}{2}}}{\sin{\frac{\theta}{2}}} \tag{1}$$
Now
$$1 + \cos{2\theta} + \cos{4 \theta} + \cos{6 \theta} + \cos8{\theta}=\\
1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Difference equation solution My question : $y_{n+1}= ay_n+b$ ; $y_0 = \alpha$
We solved this difference equation in a class and got that $y_n = a^nc + \beta$
Can someone please explain the way how to solve it? And what is the was to solve differential equations in general? Is there some good literature with difference... | Note that it must be: $y_n=a^n\color{red}{\alpha}+\beta$.
If $a=1$, it is just an $n$-th term of an Arithmetic Progression.
Assume $a\ne 1$.
Method $1$:
$$\begin{align}y_{n+1}&= ay_n+b=\\
&=a(ay_{n-1}+b)+b= \qquad \qquad \qquad \ \ \ \ \ \ \ \ \ \ \ \ \ \ a^2y_{n-1}+(a+1)b=\\
&=a^2(ay_{n-2}+b)+(a+1)b=\qquad \qquad \qq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find a and p in the given question
I would be grateful if someone helps me with this question.
I get $p$ but to get $a$, the term with $a$ vanishes.
I checked continuity at $x=1$ for finding $p$.
$p$ comes out to be $-2$.
| As $f(x)$ is differentiable for $x\in(0,2)$ we know that the limit $$\lim_{n\to\infty}\dfrac{ax(x-1)\left(\cot^n\left(\dfrac{\pi x}4\right)\right)+px^2+2}{\left(\cot^n\left(\dfrac{\pi x}4\right)\right)+1}$$ is finite, and $f(x)$ is continuous for $x\in(0,2)$.
$$$$
Let $$L=\lim_{n\to\infty}\dfrac{ax(x-1)\left(\cot^n\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Showing that $f(x,y)=(\lvert x\rvert -\lvert y\rvert)\log\left(2x^2+\lvert y\rvert\right)$ has no local extrema I have an exercise asking to find the local extrema of $$ f(x,y)=(\lvert x\rvert -\lvert y\rvert)\log\left(2x^2+\lvert y\rvert\right)$$if they exist... Wolfram Alpha tells me they don't, but I think I've spen... | The function
$$
f(x,y) = \left(|x|-|y|\right)\log(2x^2-|y|)
$$
has symmetries along the $x$ and $y$ axis because
$$
f(x,y) = f(\pm x, \pm y)
$$
Now considering the first quadrant or
$$
f(x,y) =(x-y)\log(2x^2-y)
$$
making the change of variable $u = 2x^2+y$ we have
$$
g(x,u) = (x+2x^2-u)\log u
$$
in this coordinate sys... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to integrate the product of two or more polynomials raised to some powers, not necessarily integral This question is inspired by my own answer to a question which I tried to answer and got stuck at one point.
The question was:
HI DARLING.
USE MY ATM CARD, TAKE ANY AMOUNT OUT, GO SHOPPING AND TAKE YOUR FRIENDS FO... | Noticing that
$3 x^3-x^2+2 x-4=-(1-x)\left(3 x^2+2 x+4\right)$ and $x^2-3 x+2=(1-x)(2-x)$,
we transform the integral into
$$
I=-\int_0^1 \sqrt{\frac{1-x}{2-x}}\left(3 x^2+2 x+4\right) d x
$$
Let $t^2=\frac{1-x}{2-x}$, then $x=\frac{2 t^2-1}{t^2-1}$ and $d x=-\frac{2 t}{\left(t^2-1\right)^2} d t$, which changes the inte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2814179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Prove $\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c)$ when $a=b+c$ I want to prove this identity:
$$\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c) \qquad\text{when}\;a=b+c$$
Can somebody give a hint in the easiest way possible? I am debugging this for hours and can't get the left side to be t... | Ignoring the condition $a=b+c$, we can see that the equation can be manipulated into an interesting form:
$$\begin{align}
\sin^2 a + \sin^2 b + \sin^2 c &= 2 ( 1 - \cos a\cos b\cos c ) \\
( 1 - \cos^2 a ) + ( 1 - \cos^2 b ) + ( 1 - \cos^2 c ) &=
2 - 2 \cos a\cos b\cos c \\
1 - \cos^2 b - \cos^2 c &=
\cos^2 a - 2 \cos a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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