Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Evaluate $\lim\limits_{x\to\ 0}(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)})$
Evaluate $$\lim\limits_{x\to\ 0}\left(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)}\right)$$
It is easy with L'Hospital's rule, but takes too much time to calculate derivatives. What are some shorter methods?
| The problem as corrected by the OP is now more symmetric. We can consider cases characterized as
$$
L = \lim_{x \to 0} \frac{1}{f(x)g^{-1}(x)}-\frac{1}{f^{-1}(x)g(x)}
$$
where smooth $f(x) = x + \varepsilon_f x^m + o(x^{m+1}), g(x) = x + \varepsilon_g x^n + o(x^{n+1})$ with integers $m, n > 1$. Then $f^{-1}(x) = x - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
Prove an iff condition for the existence of LU decomposition I'm asked to prove that the matrix
$$A=\begin{pmatrix}
a & b & 0\\
c & d & e\\
0 & 1 & g
\end{pmatrix}$$
has a $LU$ decomp iff $a\not=0$ and $ad\not=bc$. However, I do not believe this to be true since we have
$$\frac{1}{2}\begin{pmatrix}
1 & 0 & 0\\
1 & 1 & ... | Given: $A=\begin{bmatrix}
a &b &0 \\
c&d &e \\
0 &1 &g
\end{bmatrix}$, we use the fact the the LU decomposition is possible if no row exchanges are required.
$\Rightarrow :$ Assuming $A$ has an $LU$ decomposition, we start by doing an elimination step. If $a\ne 0$, then the multipliers are $m_{21}=\frac{c}{a}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $f(x) = \frac{\sin^{-1} x}{\sqrt{1- x ^2}}$, then evaluate $(1-x^2)f''(x) - xf(x)$
$f(x) = \dfrac{\sin^{-1} x}{\sqrt{1- x ^2}}$
Differentiating the given function, we get
$f'(x) = \dfrac{1 + \dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}}{1-x^2}$
which can also be written as
$f'(x) = \dfrac{1 + xf(x)}{1-x^2}$
Differentiatin... | Your question has probably a typo, here's how.
Given $$f(x)=\frac{\sin^{-1}x}{\sqrt{1-x^2}}$$
Using quotient rule,
$$f'(x)=\frac{1+\frac{x\sin^{-1}x}{\sqrt{1-x^2}}}{1-x^2}$$
Re arranging this we get,
$$(1-x^2)f'(x)=1+x\times \frac{\sin^{-1}x}{\sqrt{1-x^2}}$$
$$(1-x^2)f'(x)=1+xf(x)$$
We get finally, $$(1-x^2)f\color{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int\frac{d\theta}{1+x\sin^2(\theta)}$ We have to evaluate $$\int\frac{d\theta}{1+x\sin^2(\theta)}$$
Here is all my steps:
$$\tan\frac{\theta}{2}=\omega\Rightarrow \sin(\theta)=\frac{2\omega}{1+\omega^2}\Rightarrow 1+x\sin^2{(\theta)}=1+\frac{x 4\omega^2}{(\omega^2+1)^2}$$
Therefore: $$\int\frac{d\theta}{1+... | Instead of the half-angle substitution I would try using
$$\frac{1}{1+x\sin^2\theta} = \frac{1}{(1+x)\sin^2\theta + \cos^2\theta}= \frac{\sec^2\theta}{(1+x)\tan^2\theta + 1}.$$
Now substitute $u=\sqrt{|1+x|}\tan \theta$ (I don't see a way to avoid splitting the cases $x>-1$ and $x<-1$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Calculus differentiable gra Hey guys so stuck on a calculus question. So far all I know is that $d$ should equal $4$. I then got the derivative of the a b c d function to $3ax^2 + 2bx + c$ , subbed in $(0,4)$ to get $c$, which was also $4$. Just wondering if I'm on the right track and where I should go from here.
A ... | you have $$ f(x) = \begin{cases}
x+4 &x\leq 0 \\
ax^3 +bx^2 +cx +d &0<x<4 \\
4-x &x\geq 4
\end{cases}$$
making $f$ continuous at $x=0$ and at $x = 4$ requires $$d = 4,\quad 64a+16b+4c+d= 0\tag 1$$
we also have $$f'(x) = \begin{cases}
1 &x < 0 \\
3ax^2 +2bx +c & 0<x<4 \\
-1 &x\geq 4
\end{cases}$$
making $f'$ co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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An unusual limit involving $e$ From MathWorld I have the following quote:
$e$ is given by the unusual limit $$\lim_{n \to \infty}\left(\frac{(n + 1)^{n + 1}}{n^{n}} - \frac{n^{n}}{(n - 1)^{n - 1}}\right) = e\tag{1}$$ Now if we put $$a_{n} = \frac{(n + 1)^{n + 1}}{n^{n}}$$ then the above result says that $$\lim_{n \to \... | For $n > 1$, we can write
$$a_n - a_{n-1} = \biggl(1+\frac{1}{n}\biggr)^n\cdot \biggl(n+1 - (n-1)\frac{n^{2n}}{(n^2-1)^n}\biggr).\tag{1}$$
Bernoulli's inequality says on the one hand that
$$\frac{n^{2n}}{(n^2-1)^n} = \biggl(1 + \frac{1}{n^2-1}\biggr)^n \geqslant 1 + \frac{n}{n^2-1},$$
so
$$n+1 - (n-1)\frac{n^{2n}}{(n^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim\limits_{n\to\infty}y_n$ if $y_1=\frac{x}{2},y_n=\frac{x}{2}+\frac{y^2_{n-1}}{2},0\le x \le 1,n=2,3,...$ Is it a good approach to use induction? If $0\le x \le 1$ then $0\le y_1 \le \frac{5}{8}$. Suppose that $$0\le y_n \le \frac{5}{8}$$ and prove $$0\le y_{n+1} \le \frac{5}{8}$$ If $$y_{n+1}=\frac{x}{2}+\fra... | Since the only possible limits are $1\pm\sqrt{1-x}$, let $z_n=1-y_n$, find the recursion for $z_n$, and check the consequences of $z_n^2<1-x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral using contour integration Here is the integral I want to evaluate:
$$\int_{0}^{2\pi} \frac{dx}{a+b \cos x }, \quad a>b >0$$
Apparently there are limitations as to what values $a, b$ are supposed to take but let us not concern about this. Since using the sub $u =\tan \frac{x}{2}$ (the Weiersstrass sub) results ... | You have a pole at $z=\dfrac{\sqrt{a^2-b^2}-a}b$ and one at $z=-\dfrac{\sqrt{a^2-b^2}+a}b$. The pole at $z=\dfrac{\sqrt{a^2-b^2}-a}b$ has is simple and has residue $\dfrac{1}{2\sqrt{a^2-b^2}}$. The other pole is outside the circle and therefore does not matter.
The final answer is therefore ${4\pi}\dfrac{1}{2\sqrt{a^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
inequalities with fraction problem x $\frac{1}{x} > \frac{2x} {x^2 +2}$ solving this inequalities:
My long solution (wrong) :
multiplying $(x^2 + 2)^2 (x)^2 \dots$ (multiplying square of each denominator, getting rid of the > or < 0)
$x (x^2 +2)^2 > 2x(x^2) ( x^2 +2)$
$x(x^4+4x^2 +4)>2x^2(x^2+2)$
$x^5+4x^3+4x>2x^4+4x^2... | We need $$\dfrac1x-\dfrac{2x}{x^2+2}>0$$
$$\iff\dfrac{x^2+2-2x^2}{x(x^2+2)}>0$$
As $x^2+2>0$ for real $x,$ multiplying both sides by $x^2(x^2+2)$
$$\iff x(2-x^2)>0\iff\{x-(-\sqrt2)\}\cdot x\cdot (x-\sqrt2)<0$$
Clearly, we need to check for $(-\infty,-\sqrt2);(-\sqrt2,\sqrt2);(\sqrt2,\infty)$
As the product of three ter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Show that $n^n<(n!)^2$ I want to show that $\lim\limits_{n \to \infty}\frac{n^n}{(n!)^2}=0$
But I have absolutely no idea besides that $\frac{n^n}{(n!)^2}=\frac{n}{1}\cdot \frac{n}{2}\cdot ...\cdot \frac{n}{(n-1)^2}\cdot \frac{n}{n^2}$
Help me please.
| Let's check ratio of $a_n$ and $a_{n+1}$:
$$a_n = \frac{n^n}{n!^2}$$
$$a_{n+1} = \frac{(n+1)^{n+1}}{(n+1)!^2}$$
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^{n+1}}{(n+1)!^2} : \frac{n^n}{n!^2} =
\frac{(n+1)^{n+1}}{n^n}\frac{n!^2}{(n+1)!^2} = \left(1+\frac{1}{n}\right)^n \frac{n!^2(n+1)}{n!^2(n+1)^2} = \frac{1}{n+1}\left(1+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Asymptotic expansion computation I found a paper in which the following expression
$$\log\left(1 - \frac{1}{x}\right) + \frac{1}{x} + \frac{A}{x^2}+\epsilon x = 0$$
Where $\epsilon$ is a constant of order $10^{-2}$ to $10^{-5}$, $A$ is a constant of order unity, is approximated for $\frac{1}{x}<<1$ as
$$\epsilon x^4 +... | $$\log\left(1 - \frac{1}{x}\right) + \frac{1}{x} + \frac{A}{x^2}+\epsilon x =-\frac1x-\frac1{2x^2}-\frac1{3x^3}+O\left(\frac1{x^4}\right)+ \frac{1}{x} + \frac{A}{x^2}+\epsilon x=0$$
then
$$x^3\left(\left(A-\frac12\right)\frac1{x^2}-\frac1{3x^3}+O\left(\frac1{x^4}\right)+ \frac{A}{x^2}+\epsilon x\right)=(A-\frac12)x-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1327442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find out all solution of trigonometric equation $\tan \theta = - \frac{\sqrt 3}{3}$ $\tan \theta = - \frac{\sqrt 3}3$
I thought it was $2\pi\over3$ and $5\pi\over3$ but I was wrong
please help
| For any $a\in\Bbb R$,
$$\tan\theta=a\iff \theta=\arctan a+ n\pi,\, n\in\Bbb Z$$
$\arctan x$ is the unique number $y$ in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ such that $\tan y=x$.
In this case, $$\tan\theta=-\frac{\sqrt{3}}{3}\iff \theta=-\frac{\pi}{6}+n\pi,\, n\in\Bbb Z,$$
because $\arctan\left(-\frac{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1327842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Finding the maximum of an expression in three variables I am trying to find the max of $\frac{(a)(a-1)(9-a)}{2} + \frac{(b)(b-1)(9-b)}{2} + \frac{(c)(c-1)(9-c)}{2}$ subject to $a+b+c=9$ over nonnegative integers. I originally tried to see if something could work with Jensen's or calculus, and I have a conjecture that I... | Without loss of generality $a \geq b \geq c$. So $c \leq 3$ and if you write down a list of all the values of $\frac{x(x-1)(9-x)}{2}$ for $0 \leq x \leq 9$ it's easy to check which pair $(a,b)$ maximises the total for each of the four possessible values of $c$, then see which of those is the greatest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving a three variable equation I have three given values, suppose a=1.86, b=2.6 and c=4.2. Now I have to figure out x,y,z such that
*
*$x\gt 0,y\gt 0$ and $z\gt 0$
*$x+y+z=1$
*$a*x\gt 1, b*y\gt 1$ and $cz\gt 1$
I need a generalized solution steps for this to implement in programming.
Thanks.
| $\left\{ {\begin{array}{*{20}{l}}
{x > \frac{1}{a} \Rightarrow x = \frac{1}{a} + X}\\
{y > \frac{1}{b} \Rightarrow y = \frac{1}{b} + Y}\\
{z > \frac{1}{c} \Rightarrow z = \frac{1}{c} + Z}
\end{array}} \right.$ where $0 < X,Y,Z$
$ \Rightarrow \frac{1}{a} + X + \frac{1}{b} + Y + \frac{1}{c} + Z = 1 \Rightarrow X + Y +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Differential equation $y\cdot y'' - (y')^2 - 1 = 0$ I'm trying to solve this equation but at the end I'm stuck and can't reache the answer. I use the substitutions $y'=p$ and $y'' = p'\cdot p$:
$$y \cdot p'p - p^2 - 1 = 0 \implies y\cdot p \frac {dp}{dy} - (p^2 + 1) = 0 \implies \int \frac {p}{p^2 + 1}dp = \int \frac {... | Maybe your $C_1$ and $C_2$ are different to the answer's. But let solve your equation:
$$
y+\sqrt{y^2-C_1^2}=\exp(\frac{x+C_2}{C_1})\\
\sqrt{y^2-C_1^2}=\exp(\frac{x+C_2}{C_1})-y\\
y^2-C_1^2=\exp(2\frac{x+C_2}{C_1})-2y\exp(\frac{x+C_2}{C_1})+y^2\\
y=\frac{1}{2}(\exp(\frac{x+C_2}{C_1})+C_1^2\exp(-\frac{x+C_2}{C_1}))
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve $x^8 \equiv 3 \pmod {13}$ I need to find all solutions to $x^8 \equiv 3 \pmod {13}$.
What I've tried: I know $2$ is a primitive root modulo $13$.
So it is equivalent to solve
$2^{8t} \equiv 2^4 \pmod {13}$
Then I get $t = 2 + 3k$.
I think I'm wrong... and if not, what are the final solutions??
| No, your answer is correct, $t = 2,5,8,11$ give respectively $x = 4,6,9,7$.
For the sake of Google, I will write out the method OP probably used, which is different from the one used in the other answer.
Knowing that $2$ is a primitive root modulo $13$ and that $3 \equiv 2^4 \pmod{13}$, we let $x = 2^t$ and we wish to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Why does $\int_0^{2\pi} (1+2\cos(x))/(5+4\cos(x))\,dx$ vanish? The standard substitution $y=\tan(x/2)$ shows that
$$ \int_0^{2\pi} \frac{1+2\cos(x)}{5+4\cos(x)}\,dx = 0. $$
What is the "real explanation" for this fact? My guess is that the "book proof" involves contour integration; is this correct? Is there an elega... | Consider the more general integral, where $d \neq 0$,
\begin{align}\tag{1}
I = \int_{0}^{2 \pi} \frac{a + b \cos x}{c + d \cos x} \, dx.
\end{align}
Now by splitting the integral into two regions, $[0, \pi)$ and $(\pi, 2 \pi]$, then in the second integral let the variable be shifted by $\pi$. This is seen by:
\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
How exactly do we do Gauss elimination?
This is a matrix:
$$\begin{bmatrix}
1 & 1 & 1\\
1 & 2 & 3\\
1 & 3 & k
\end{bmatrix}\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=
\begin{bmatrix}
3\\
6\\
4+k
\end{bmatrix}$$
Find $k$ so that it has no unique solution. Solve the equations for this value of $k$.
I found $k = 5$ by ... | You did not row reduce correctly. Since we are told that there are infinitely many solutions, we know that the final result should have free variables, so there should be at most two pivots. We should get:
$$
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 3 \\
1 & 2 & 3 & 6 \\
1 & 3 & 5 & 9 \\
\end{array}\right]
\sim
\left[\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Value of $\int_{|z+1|=2} \frac{z^2}{4-z^2}dz$ $\int_{|z+1|=2} \frac{z^2}{4-z^2}dz=-2\pi i$.
Am I correct, I used cauchy integral formula
| Since $z^2-4=(z-2)(z+2)$ and $z=-2$ is the only zero inside $|z+1|=2$. Assuming $|z+1|=2$ is positively oriented, Cauchy's Integral Formula gives
$$
\int_{|z+1|=2}\frac{z^2}{4-z^2}dz =- \int_{|z+1|=2}\frac{z^2}{z^2-4}dz = -\int_{|z+1|=2}\frac{z^2}{(z-2)(z+2)}dz =- 2 \pi i \left( \frac{z^2}{z-2} \right)_{z=-2} = 2\pi i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1335759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Taylor series expansion of function $f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}}$ Determine the Taylor series expansion of function $f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}}$.
$f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}} = \arcsin(x)\frac{1}{\sqrt{1-x^2}}$
It is known:
(1.) $\arcsin(x) = \sum_{n=0}^\infty\frac{(2n-1)!!x^{2n+1}}{2... | We have, by the extended binomial theorem:
$$ \frac{1}{\sqrt{1-x^2}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n} x^{2n},\tag{1} $$
hence by integrating termwise:
$$ \arcsin x=\sum_{n\geq 0}\frac{1}{(2n+1)4^n}\binom{2n}{n} x^{2n+1}=x\cdot\phantom{}_2 F_1\left(\frac{1}{2},\frac{1}{2};\frac{3}{2};z^2\right)\tag{2} $$
Now we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Finding value of an expression If $x^2-3x-1=0$ then find the value of $(x^6+1)/x^3$ I tried to solve the quadratic but it became too complicated any way of doing this without a calculator
| Since $x^2=3x+1$, we have $x^3=3x^2+x=10x+3$.
In the same way, $\frac{1}{x^2}=1-\frac{3}{x}$ gives $\frac{1}{x^3}=\frac{1}{x}-\frac{3}{x^2}=\frac{10}{x}-3$, hence:
$$\frac{x^6+1}{x^3} = x^3+\frac{1}{x^3} = 10\left(x+\frac{1}{x}\right) = \pm 10\sqrt{13},$$
because $x+\frac{1}{x}$ is $\pm\sqrt{\Delta}$, since the product... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Rank-reducibility of Latin squares Consider the following Latin square with rank $N= 9$:
$$
\begin{bmatrix}
5 & 3 & 1 & 2 & 4 & 7 & 6 & 8 & 9 \\
3 & 7 & 9 & 6 & 8 & 4 & 5 & 2 & 1 \\
8 & 5 & 4 & 9 & 1 & 2 & 7 & 3 & 6 \\
9 & 2 & 3 & 8 & 7 & 6 & 1 & 4 & 5 \\
7 & 4 & 5 & 3 & 9 & 1 & 2 & 6 & 8 \\
6 & 1 & 2 & 5 &... | You're essentially doing transversal prolongation in reverse.
A transversal is a set of entries with one representative from each row, each column, and each symbol. We can extend a Latin square of order $n$ with a transversal to a Latin square of order $n+1$ by (a) replacing the transversal with a new symbol, (b) addi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Evaluating $\int{ \frac{x^n}{1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}}}dx$ using Pascal inversion (Note: I apreciate very much who marked this as a duplicate but I would like an answer for why my proof is wrong)
This is my solution, I have no clue why it failed. Let's start:
define
$$I_n(m) = \int_{0}^{x} \frac{... | You may observe that
$$
\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)'=1 + x + \frac{x^2}{2} + \cdots + \frac{x^{n-1}}{(n-1)!}
$$ giving
$$
\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)-\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)'=\frac{x^n}{n!}
$$ and
$$
\begin{align}
\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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"answer_id": 0
} |
Trigonometry equation maximum Given the equation: $\cos x + \sqrt3 \sin x = a^2$ find the maximum value for $a$ for which the equation has solutions and for this case solve the equation, $a \in \mathbb{R}$.
I'm guessing I need to find the maximum for the function and for this I have to differentiate it and solve it wh... | Since \begin{align}\cos x+\sqrt 3\sin x &=2\left( \frac 12\cos x+\frac {\sqrt 3}2\sin x \right) \\
& =2 \sin\left(x + \frac{\pi}{6}\right) = a^2\end{align}
Since $\sin x$ oscillates between $-1$ and $1$, we need $\displaystyle -1 \leq \frac{a^2}{2} \leq 1$. The maximum value of $a$ is easily seen to be $a = \sqrt{2}$ s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve $x^2-|5x-3|-x<2,\ \ x\in \mathbb{R} $
Solve $x^2-|5x-3|-x<2,\ \ x\in \mathbb{R} $
I tried $x^2-|5x-3|-x<2$ ,
case $1$ , $x^2-(5x-3)-x<2,\ x\geq 0 \\
x^2-6x+1<0 \\
3-2\sqrt2 < 3+2\sqrt2 \\
0.17<x<5.8\\
$
$x^2-(5x-3)-x<2$ ,
case $2$ , $x^2+(5x-3)-x<2,\ x< 0 \\
x^2+4x-5<0 \\
-5 < x< 1\\
$
The region common... | Case 1 is not for $x>0$ but for $5x-3>0\implies x>\frac{3}{5}$
So for case 1 you have $\frac{3}{5}<x<3+\sqrt{2}$ (since $\frac{3}{5}>3-\sqrt{2}$
For case 2 you have $x<\frac{3}{5}$, so $-5<x<\frac{3}{5}$
So the general solution is $-5<x<3+\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Can anyone prove the identity $\sum_{m=-\infty}^\infty (z+\pi m)^{-2} = (\sin z)^ {-2} $ I came across this identity in a paper on elliptic curves, and the proof wasn't provided. It really irked me, and I couldn't find an explanation anywhere else. Can anyone shed some light?
$$\sum_{m=-\infty}^\infty (z+\pi m)^{-2} =... | Following the suggestion by @Lucian, we use the Euler's infinite product representation of the sine function and write
$$\sin z=z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2\pi^2}\right)\tag 1$$
Next, taking the logarithmic derivative of $(1)$ yields
$$\cot z=\frac1z-2z\sum_{n=1}^{\infty}\frac{1}{n^2\pi^2-z^2} \tag 2$$
A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1348737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Find the value of $x$ such that $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$ Find the value of $x$, $$\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$$
Help guys please, I have tried and I got, $x=-2, x=1$, and I think it's wrong
| This should be a comment but then too long ...
Picking up on the solution given by Jan Erlang:
$$\left(x^4-8 x^2+10\right)^2 \left(x^4-8 x^2+14\right)^2+x-4=\left(x^2-x-3\right) \left(x^2+x-4\right) \left(x^{12}-24
x^{10}-x^9+228 x^8+16 x^7-1087 x^6-88 x^5+2720 x^4+191 x^3-3380
x^2-136 x+1633\right)$$
Now $x^2-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Integrating $\sqrt{1-x^2}$ without using trigonometry I am a beginning calculus student. Tonight I had a thought. Maybe I could calculate $\pi$ using integration, but no trig.
The problem is that I don't really know where to start. I thought perhaps I could use the Chain Rule for derivatives.
$~~~~~~~~~$
| Faster converging series
$\sqrt{1-x^2} = 1 - \frac{1}{1!} \frac{1}{2} x^2 - \frac{1}{2!} \frac{1}{2} \frac{1}{2} x^4 - \frac{1}{3!} \frac{1}{2} \frac{1}{2} \frac{3}{2} x^6 + \cdots$ [by Binomial theorem]
$\ = 1 - \sum_{k=1}^\infty \frac{(2k-3)!!}{2^k k!} x^{2k}$ [where $n!!$ is the double-factorial of $n$]
$\int \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Computing $\sum _{k=1}^{\infty } \frac{\Gamma \left(\frac{k}{2}+1\right)}{k^2 \Gamma \left(\frac{k}{2}+\frac{3}{2}\right)}$ in closed form What tools other than beta function you might like to use here?
$$\sum _{k=1}^{\infty } \frac{\displaystyle \Gamma \left(\frac{k}{2}+1\right)}{\displaystyle k^2 \Gamma \left(\frac{k... | Writing $\dfrac{\Gamma\left(\frac{k}{2}+1\right)}{\Gamma\left(\frac{k}{2}+\frac{3}{2}\right)}$ as an integral and exchanging summation and integration, we get
\begin{align}
\sum^\infty_{k=1}\frac{\Gamma\left(\frac{k}{2}+1\right)}{k^2\Gamma\left(\frac{k}{2}+\frac{3}{2}\right)}
&=\frac{2}{\sqrt{\pi}}\int^1_0\frac{x{\rm L... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Calculate Infinite Limit I'm trying to calculate the limit and when I get to the last step I plug in infinity for $\frac 8x$ and that divided by -4 I get - infinity for my answer but the book says 0. Where did I go wrong?
$$
\frac {8x^3-x^2}{7+11x-4x^4}
$$
Divide everything by $x^4$
$$
\frac {\frac{8x^3}{x^4}-\frac{x^2... | Suppose that $n,m\in \mathbb{N}$:
$$f(x)=\dfrac{a_{n}x^n+a_{n-1}x^{n-1}+\dots+a_{1}x+a_{0}}{b_{m}x^m+a_{m-1}x^{m-1}+\dots+b_{1}x+b_{0}}$$
We have three case:
Case $1$:
$$n< m\Longrightarrow \lim_{x\rightarrow\infty}f(x)=0$$
Case $2$:
$$m<n\Longrightarrow \lim_{x\rightarrow\infty}f(x)=\infty$$
Case $3$:
$$n=m\Longright... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Is there anyway to show $\left| {\frac{{\cos x - \cos y}}{{x - y}}} \right| \le 1$ other than taking derivatives? The purpose is to show $\left| {\frac{{\cos x - \cos y}}{{x - y}}} \right| \le 1$ for any $x,y\in\Bbb{R}$.
Taking partial derivatives with respect to $x,y$ respectively $\frac{{\partial \frac{{\cos x - \co... | Using the difference to product identity $\cos x - \cos y = -2\sin\dfrac{x-y}{2}\sin\dfrac{x+y}{2}$ along with the inequality $|\sin \theta| \le |\theta|$ gives:
$\left|\dfrac{\cos x - \cos y}{x-y}\right| = \left|\dfrac{-2\sin\frac{x-y}{2}\sin\frac{x+y}{2}}{x-y}\right| = \dfrac{2\left|\sin\frac{x-y}{2}\right|\left|\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
If eigenvalues are $\pm 1$, then matrix is orthogonal If eigenvalues of a matrix are $1$, or $-1$, or both of them, does that necessarily means that the matrix is orthogonal?
The only thing I'm certain about is that the matrix is non-singular, due to absence of zero among the eigenvalues.
| The eigenvectors with eigenvalue $1$ have to be orthogonal to the eigenvectors with eigenvalue $-1$ in order for the matrix to be orthogonal. A $2\times2$ counterexample using this idea is not difficult to construct. For example, let $M$ be a linear operator on $\mathbb{R}^{2}$ for which
$$
M\left[\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Show that $\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}\leq 2^{n}$ for all $n\in\mathbb{N}$.
Show that $$\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}\leq 2^{n}\qquad (n\in \mathbb{N}).$$
I want to show the last step, that is, the inductiv... | Another approach:
$$\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}< \frac{2\cdot 4\cdot 6\cdot \ldots \cdot 2n}{1\cdot 2\cdot 3\cdot \ldots \cdot n}=\frac{2^n n!}{n!}=2^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
When are we permitted to multiply or divide both sides of an equation by a variable? As it is said in the mathematics books (at least the one I have), we are not permitted to divide or multiply both sides of an equation by a variable, because it is possible to lose some answers. For example, in the following equation
$... | When dividing by any quantity,
or when canceling out two quantities in a ratio
(for example, canceling $x$ and $x$ to find that $\frac xx=1$),
you need to be aware of what assumptions you have to make so
that the division or canceling makes sense,
and remember that those assumptions apply to any results you get.
For ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
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Limits. Finding positive value Find a real number k such that the limit $$\lim_{n\to\infty}\ \left(\frac{1^4 + 2^4 + 3^4 +....+ n^4}{n^k}\right)$$ has as positive value.
If I am not mistaken every even $k$ can be the answer. But the answer is 5.
| First note that
$$\sum\limits_{i=1}^n i^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$
$$=\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30}$$
So now we have
$$\lim\limits_{n\to\infty} \left(\frac{1^4 + 2^4 + 3^4 +\cdots + n^4}{n^k}\right)$$
$$=\lim\limits_{n\to\infty} \left(\frac{\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove this inequality: $\frac n2 \le \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac1{2^n - 1} \le n$ $\dfrac{n}{2} \le \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2^n - 1} \le n $
I've Tried for hours but didn't got any striking idea. I don't have any efforts to show rather than induction. Please Try If you 've... | It's a standard trick. Let
$$
H_n = 1 + \frac12 +\ldots+ \frac1n.
$$
Look on $H_{2^n-1}$:
$$
n=1: H_1 = 1=1\\
n=2: H_3 = 1 + \frac12 + \frac13 < 1 + \frac12 + \frac12 = 1 + 1=2\\
n=3: H_7 = 1 + \frac12 + \frac13 + \left(\frac14+\frac15+\frac16+\frac17\right) < 1 + \frac{2}{2} + \left(\frac14+\frac14+\frac14+\frac14\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Solve the trigonometric equation: $\sin {3x} = 4 \sin^2 x$ Solve the equation $\sin{3x} = 4 \sin^2 x$.
I tried to change the $\sin{3x}$ to $3\sin x\cos x$ then solve it, but I could not find the correct answer.
| Notice, $$\sin 3x=4\sin^2 x$$ $$\implies 3\sin x-4\sin^3x=4\sin^2 x$$ $$\implies 4\sin^3x+4\sin^2 x-3\sin x=0$$$$\implies \sin x(4\sin^2x+4\sin x-3)=0$$
$$\implies \sin x(2\sin x-1)(2\sin x+3)=0$$ As you have not mentioned any condition of the unknown value $x$ hence writing the general solution for $x$ as follows $$\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the matrix $\mathbf{A}$ if $A\binom{7}{-1} = \binom{6}{2}.$ Find the $2\times2$ matrix $A$ where $A^2=A$ and
$$A\begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}.$$
I tried plugging in: $A= \begin{pmatrix}a&b\\c&d\end{pmatrix}$ but that became messy very quickly. I got the equations:
$7a... | Hint: Since you have $7a-b=6$ and $7c-d=2$ then $b=7a-6$ and $d=7c-2.$ Moreover, $A^2=A$ then $\det(A)^2=\det(A)$ which implies $\det(A)=0$ or $\det(A)=1$ then $ad=bc$ or $ad-bc=1.$ If $ad=bc$ then you get $a(7c-2)=c(7a-6)$ then $-2a=-6c$ so the coefficients of the matrix $A$ are of the form $a=3c,$ $b=7a-6=21c-6 $ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solving $6 \cos x - 5 \sin x = 8$ My attempt:
Using the formula for linear combinations of sine and cosine:
$$A \cos x+B \sin x=C \sin (x+\phi)$$
$$
\sqrt{51} \left(\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x\right) = 8
$$
$$
\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x = \frac{8}{\sqrt{51}}
$$
And... | $6^2 + 5^2 = 61 < 8^2$.
So your attempted answer is ok except for a digit and the fact that you get
$$
\cos(\text{something}) = \frac 8 {\sqrt{61}} >1.
$$
A cosine of a complex number can be bigger than $1$, but a cosine of a real number cannot.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Baby Rudin claim: $1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}...$ converges This sequence is a rearrangement of the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}...$. Note that at this point in the text we do not have any theorem about the convergence ... | The following argument shows that the series converges, and gives its sum:
$\hspace{.3 in}1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots=\ln 2$ $\;\;\;$so
$\hspace{.27 in}\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots=\frac{1}{2}\ln 2$. $\;\;\;$Insert... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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How to integrate this : $ \int \cot(5x) \tan(2x) \mathrm{d}x$ Methods to integrate this integral:
$$\int \cot(5x) \tan(2x) \mathrm{d}x$$
I have tried several methods, step by step, and they have led to invalid results. Helpful hints or processes are welcome.
I did this further :
$$ \cot(5x) = \frac{1}{\tan(2x+ 3x)}$... | Letting $\tan x=z$, the integral can be rewritten as $$I=\int \frac{2z(1-10z^2+5z^4)}{(1-z^4)(5z-10z^3+z^5)}dz\\=2\int\frac{1-10z^2+5z^4}{(1-z^4)(5-10z^2+z^4)}dz\\=2\int \frac{1}{1-z^4}dz-8\int \frac{1}{5-10z^2+z^4}dz\\=2J-8K$$ where $J$ can be easily evaluated. As for $K$, it can be written as $$\int \frac{1}{(z^2-5-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Is there a way to show the following equality without induction I wanted to show the following equality without using induction:
$$
\sum_{k=2}^n \frac{1}{k(k-1)} = \frac{n-1}{n}
$$
Any hint on how to do it?
| Notice, $$\sum_{k=2}^{n}\frac{1}{k(k-1)}$$ $$=\sum_{k=2}^{n}\left(\frac{1}{k-1}-\frac{1}{k}\right)$$ $$=\left(\frac{1}{2-1}-\frac{1}{2}\right)+\left(\frac{1}{3-1}-\frac{1}{3}\right)+\left(\frac{1}{4-1}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{(n-1)-1}-\frac{1}{n-1}\right)+\left(\frac{1}{n-1}-\frac{1}{n}\right)$$ $$=\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Confusion about the integral $\int dT/(1-T^2)$ From some reference on the internet we have the following real valued function and its derivative:
$$
M(T) = \frac{\sqrt{1-T^2}}{1+T}
\quad \Longrightarrow \quad \frac{dM}{dT} = - M/(1-T^2)
$$
The reverse of differentiation is integration:
$$
\frac{dM}{dT} = - M/(1-T^2) \q... | You need to pay attention to the constant of integration and don't ignore absolute value signs.
So you would have the same problem if you tried to do the same thing with this equaltiy $\frac{1}{1-x}=-\frac{1}{x-1}$
$\int \frac{1}{1-x}=-\int\frac{1}{x-1}$
$\log|1-x|=\log|x-1|+c_1$ Here you missed out the constant, also ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s... | Let $X \sim G(1-x)$, a geometric random variable with success probability $1-x$. We have
$$
\mathbb{E}[X] = \sum_{n=1}^\infty nx^{n-1}(1-x).
$$
On the other hand, we know that $\mathbb{E}[X] = 1/(1-x)$, and we deduce the formula.
We can argue that $\mathbb{E}[X] = 1/(1-x)$ in many ways. One way is to consider $N$ diffe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 15,
"answer_id": 2
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Is this problem wrongly built? Or is there a solution which I don't know how to arrive at? I was solving a Cauchy-Schwarz's inequality based problem. Given that $x^2+y^2+z^2=1$ I am supposed to show that $x+y+z \le 6$. After struggling for a while I realised that I could solve this inequality had the condition been $x^... | It seems that there is no need for Cauchy inequality. Obviously $x\leq1$. (why?). Similarly $y$ and $z$ are. So $x+y+z\leq 3$. This inequality can be strengthen since x, y, z can't take value 1 at the same time.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\int\dfrac{dx}{x^2(x^4+1)^{3/4}}$
Evaluate $$\large{\int\dfrac{dx}{x^2(x^4+1)^{3/4}}}$$
I thought of rewriting this as $$\large{\int\dfrac{dx}{x^5(1+\frac{1}{x^4})^{3/4}}}$$ and substituting $$u^4=\left(1+\frac{1}{x^4}\right)\Rightarrow u=\left(1+\frac{1}{x^4}\right)^{1/4}$$ and subsequently I got $$du=\dfrac{1}{4}\... | Let, $1+\frac{1}{x^4}=t \implies \frac{-4dx}{x^5}=dt$$$\large{\int\dfrac{dx}{x^2(x^4+1)^{3/4}}}$$ $$=\int \frac{dx}{x^5\left(1+\frac{1}{x^4}\right)^{3/4}}$$ $$=\frac{-1}{4}\int\frac{dt}{\left(t\right)^{3/4}}$$ $$=\frac{-1}{4}\int (t)^{-3/4}dt$$ $$=\frac{-1}{4}\frac{t^{1/4}}{1/4}+C$$ $$=-\left(1+\frac{1}{x^4}\right)^{1/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Using Lagrange's Method in Finding Extreme Values of $x^2 + y^2 + z^2$ for $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ (New to This Method) Did I do this hw question correctly (at least in theory, I do not expect anyone to check my algebra work)? In particular, did I solve for lambda and plug lambda back into m... | To comment on your solution I think you have a mistake when you got $x=-\frac\lambda{a^2}$ from $2x+\frac{2\lambda x}{a^2}=0$. (I will add link to the original version, in case the post will be subsequently changed.)
In fact, you can rewrite this equation as
$$2x+\frac{2\lambda x}{a^2}=2x\left(1+\frac\lambda{a^2}\righ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Integration validity of $\int\frac{1}{\sqrt{a^2 + x^2}}\,dx$ I'm just wondering if the following integration is valid.
\begin{array}{l}
\int {\frac{1}{{\sqrt {{a^2} + {x^2}} }}} dx\\
{\rm{Let }}{u^2} = {a^2} + {x^2}\\
2udu = 2xdx\\
\frac{{du}}{x} = \frac{{dx}}{u}\\
{\rm{Let }}\frac{{du}}{x} = \frac{{dx}}{u} = A\\
du = ... | Even another view on this technique could be provided via Euler substitutions. What there appeared, especially in final crucial step, was $x + u$ in the denominator, so why from the beggining assume the substitution :
$$\sqrt{a^2 + x^2} = u - x$$
Squaring both sides :
$$a^2 = u^2-2ux$$ and taking differential
$$u\mathr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Separating variables by substitution in a homogenous ODE I am brand new to ODE's, and have been having difficulties with this practice problem. Find a 1-parameter solution to the homogenous ODE:$$2xy \, dx+(x^2+y^2) \, dy = 0$$assuming the coefficient of $dy \ne 0$
The textbook would like me to use the subsitution $x =... | The variables $y$ and $u$ can be separated from each other:
$$
2u(y \, du + u \, dy) + (u^2+1 ) \, dy=0
$$
$$
2u\left( du + u \, \frac{dy} y \right) + (u^2+1 ) \, \frac{dy} y = 0
$$
$$
2u\left( \frac{du} u + \frac{dy} y \right) + \left( u + \frac 1 u \right) \, \frac{dy} y = 0
$$
$$
2 \left( \frac{du} u + \frac{dy} y \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding indefinite integral $\int \frac{x^2}{(x \sin x + \cos x)^2} dx $ I need hint in finding the integral of
$$\int \frac{x^2}{(x \sin x + \cos x)^2} dx $$
I tried dividing the term by $x^2\cos^2x$ and then substituting $\tan x$.
| From the formula
$$
D\frac fg=\frac{f'g-fg'}{g^2},
$$
we might guess that we have a quotient $f/g$ with $g=x\sin x+\cos x$, but what is then $f$?
First, we note that
$$
g'=x\cos x.
$$
Well, with the trig-one, we can write
$$
\begin{aligned}
\frac{x^2}{(x\sin x+\cos x)^2}
&=\frac{x^2\sin^2x+x^2\cos^2x}{(x\sin x+\cos x)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Show that any 2D vectors can be expressed in the form...
(a) Show that any 2D vector can be expressed in the form
$s \begin{pmatrix} 3 \\ -1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 7 \end{pmatrix},$
where $s$ and $t$ are real numbers.
(b) Let $u$ and $v$ be non-zero vectors. Show that any 2D vector can be expressed... | (a) Let $\begin{pmatrix} x\\ y\end{pmatrix}$ a 2D vector. Finding $s$ et $t$ such that $$s \begin{pmatrix} 3 \\ -1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 7 \end{pmatrix}=\begin{pmatrix} x\\ y\end{pmatrix}$$
is the same thing as solving:
$$\begin{cases}3s+2t=x\\
-s+7t=y\end{cases}$$
where $s$ and $t$ are the unknowns.
(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the least $N$ so there is no square
Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000 \cdot N$ contains no square of an integer.
Let $x^2$ appear before $1000N$ so:
$(x+1)^2 - x^2 > 1000 \implies x \ge 500$
Let $A = [1000N, 1000(N+1)]$
So I let $x=500$ then... | To have $(500+a)^2<1000N$ and $(500+a+1)^2\ge 1000(N+1)$ we need
$$500^2+1000a+a^2<1000N,\qquad 500^2+1000(a+1)+(a+1)^2\ge 1000(N+1) $$
and hence
$$ 250 +a+1+\left\lfloor\frac{(a+1)^2}{1000}\right\rfloor\ge N +1>N>250+a+\left\lfloor\frac {a^2}{1000}\right\rfloor$$
Therefore look for small $a$ with $\left\lfloor\frac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the sum of the series below Find the sum
$$(1\cdot2)+(1\cdot3)+(1\cdot4)+\cdots+(1\cdot2015)+(2\cdot3)+(2\cdot4)+\cdots+(2\cdot2015)+\cdots+(2014\cdot2015)$$
What I have tried...
We are looking for $$S(n)=\sum_{i=1}^{n-1}\sum_{j=i+1}^nij$$ when $n=2015$
$$S(n)=\sum_{1\le i<j\le n}ij$$ $$=\frac 12\left((\sum_{i=1... | Summing along diagonals gives a nice symmetric derivation of the solution, using binomial coefficients and minimizing messy manipulation of fractions.
$$\begin{align}
S(n)&=\sum_{i=1}^{n-1}\sum_{j=i+1}^nij
\color{lightgray}{=\sum_{1\le i<j\le n} ij}\\\\
&=\begin{cases}\begin{matrix}
\color{orange}{1\cdot 2} &+1\cdot 3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Finding the kernel of a linear map Our exercise is to find all solutions to the equation $Ax = 0$, among others for the following matrix
$$A =\begin{pmatrix} 6 & 3 & -9 \\ 2 & 1 & -3 \\ -4 & -2 & 6 \end{pmatrix}.$$
This amounts to finding the kernel, and obviously, the rows of the matrix are multiples of each other, ... | First, note that
$$\begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix} =
\begin{pmatrix} 1\\1\\1\end{pmatrix} - \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix}, $$
So your missing solution
$\left[\begin{smallmatrix} 1 \\ -2 \\ 0 \end{smallmatrix}\right] $
is, in fact, a special case of the general solution
$ \lambda \left[\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving the inequality $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$
Show that for any two positive real numbers $a$ and $b$, $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$
My attempt: $(\sqrt a-\sqrt b)^2\geq0\\\frac{a+b}{2}\geq \sqrt{ab}$
$(a-b)^2\geq0\\a^2+b^... | By setting:
$$ M_p(a,b)=\lim_{q\to p^+}\left(\frac{a^q+b^q}{2}\right)^{\frac{1}{q}}$$
we have to prove: $$2M_1\geq M_0+M_2, $$
but $M_0^2+M_2^2 = 2M_1^2$, hence the previous line follows from Cauchy-Schwarz inequality:
$$ M_0+M_2 \leq \sqrt{2}\sqrt{M_0^2+M_2^2} = 2M_1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find min & max of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$ Problem: Find the maximum and minimal value of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$.
Since $x^2 > x$ (edit $x^2 \geq x$) for all $x \in \mathbb{R}$, $f$ is bowl-ish with a minimal value in the bottom.
This is a critical point which means t... | $x^2 + y^2 =1$. So $f(x,y) = x + y + 1$
Now consider $x + y = a ,$ Then $y = x - a$
$x^2 + y^2 = 1$ So $x^2 + (x-a)^2 = 1$
$2 x^2 -2 ax + a^2 -1 = 0$
The determinant must be non negative : $4^2 - 8 a^2 + 8 >= 0 : a^2 <=2$
Which means maximum of $a$ is$\displaystyle \sqrt{2}$ and minimum is $\displaystyle -\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
} |
Prove $ \ \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2} \ = \ 0 \ $ has a solution in $ \ (-1,1) $ If $a$ and $b$ are positive numbers, prove that the equation
$$\frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2} = 0$$
has at least one solution in the interval $ \ (-1,1) \ $ .
The question is from the exercises sectio... | Proof:
Let $f(x)=\frac{a}{x^{3}+2 x^{2}-1}+\frac{b}{x^{3}+x-2} .$ We want to prove that $f$ has at least one solution in the interval
$(-1,1) .$ It suffices to prove that $f(x)=0 .$ By factorization, $f(x)=\frac{a}{(x+1)(x-a)(x-b)}+$$\frac{b}{(x-1)\left(x^{2}+x+2\right)} .$ We denote $\frac{-1+\sqrt{5}}{2}, \frac{-1-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Find $\lim\limits_{x\to 0}\frac{\sqrt{2(2-x)}-\sqrt{2(2-x)(1-x^2)}-4\sqrt{1-x}(2-\sqrt{4-x^2})}{x\sqrt{1-x}(2-\sqrt{4-x^2})}$ I can't find what is wrong when using L'Hospital's rule on this limit.
Derivative in denominator is $$\frac{-5x^3+4x^2-\sqrt{4-x^2}(6x-4)+12x-8}{2\sqrt{(1-x)(4-x^2)}}$$
and in numerator is $\fra... | using Bernoulli
\begin{align}
x \to 0 \hspace{10mm} &
{\color{Red}{(1+ax)^n \approx 1+anx} } \\
& \sqrt{1-x^2} = (1-x^2)^{\frac{1}{2}} \approx 1-\frac{1}{2}x^2 \\
& \sqrt{4-x^2}=\sqrt{4(1-\frac{x^2}{4}})=2(1-\frac{x^2}{4})^{\frac{1}{2}} \approx 2(1-\frac{1}{2} \frac{x^2}{4})=2-\frac{x^2}{4}
\end{align}
so by puttin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $(x^{2}+y^{2})$ If $x^{3}-3xy^{2}=10$, $y^{3}-3x^{2}y=30$
Find $$(x^{2}+y^{2})$$
I tried but I got nothing, any help please?
| $$(x+iy)^3 = (x^3-3xy^2) + i(3x^2y-y^3) $$
gives $(x+iy)^3 = 10-30 i $. If we multiply by the conjugate (i.e. take norms):
$$ (x^2+y^2)^3 = 10^2+30^2 = 1000 $$
we immediately get $x^2+y^2=\color{red}{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Complex number and exponent If $z = -1+ i\sqrt{3}$
Is it possible that to prove by using induction $z^{2n}+2^n\cdot z^n+2^{2n}=0$ if $n$ is not multiple of $3$.
I know other way of proving it.
| Notice, $$z=-1+i\sqrt{3}=2\left(-\frac{1}{2}+i\frac{\sqrt 3}{2}\right)$$
$$=2\left(\cos\frac{2\pi}{3}+i\sin \frac{2\pi}{3}\right)$$
Now, we have $$z^{2n}+2^nz^n+2^{2n}=\left(2\left(\cos\frac{2\pi}{3}+i\sin \frac{2\pi}{3}\right)\right)^{2n}+2^n\left(2\left(\cos\frac{2\pi}{3}+i\sin \frac{2\pi}{3}\right)\right)^n+2^{2n}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Compare $A=\frac{1.0\,000\,004}{(1.0\,000\,006)^2}$ and $ B=\frac{(0.9\,999\,995)^2}{0.9\,999\,998}$ My work:
*
*$1.0\,000\,004 = 1+\frac{4}{10^7}=1+\frac{1}{125\cdot 10^6}$
*$ (1.0\,000\,006)^2=(1+\frac{6}{10^7})^2=(1+\frac{3}{5\cdot 10^6})^2$
*$ (0.9\,999\,995)^2=(\frac{9\,999\,995}{10^7})^2=(\frac{1\,999\,999}{... | $$A=\frac {10^7\times (10^7+4)}{(10^7+6)^2}$$
$$B=\frac {(10^7-5)^2}{10^7\times (10^7-2)}$$
$$\frac AB=\frac {10^{14}(10^7+4)(10^7-2)}{(10^7+6)^2(10^7-5)^2}$$
The numerator is $10^{28}+2\cdot10^{21}-8\cdot10^{14}$
The denominator is $10^{28}+2\cdot 10^{21}-59\cdot10^{14}-60\cdot10^7+900$
It is easy to see from this whi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding $\frac{ \mathrm{d}y}{ \mathrm{d}x}$ when given that $\sqrt{1-x^2} + \sqrt{1-y^2} ={a}(x-y)$ Finding $\frac{ \mathrm{d}y}{ \mathrm{d}x}$ when given that $\sqrt{1-x^2} + \sqrt{1-y^2} = a(x-y)$. $a$ is a constant.
I have the final answer, which is $$\frac{ \mathrm{d}y}{ \mathrm{d}x} = \sqrt{\frac{1-y^2}{1-x^2}}.$$... | Given $$\sqrt{1-x^2}+\sqrt{1-y^2} = a(x-y)\;,$$ Where $a$ is constant.
Now Put $x=\sin \alpha$ and $y = \sin \beta$
So $$\sqrt{1-\sin^2 \alpha}+\sqrt{1-\sin^2 \beta} = a(\sin \alpha-\sin \beta)$$
So $$\displaystyle \cos \alpha+\cos \beta = a(\sin \alpha-\sin \beta)$$
So $$\displaystyle 2\cos \left(\frac{\alpha+\beta}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int \frac{a}{x(x^2-a^2)^{1/2}} dx$ I was trying to integrate $\int \frac{a}{x(x^2-a^2)^{1/2}} dx$ and by applying on what i saw on the formula of inverse trigonometric functions, there is formula like $\frac{1}{a}Arcsec\frac{u}{a}$ = $\int \frac{du}{u\sqrt(u^2-a^) }$ so my answer is $\frac{1}{a}Arcsec\frac{x... | $$I=\int\dfrac1{x\sqrt{x^2-a^2}}dx=\int\dfrac x{x^2\sqrt{x^2-a^2}}dx$$
Let $\sqrt{x^2-a^2}=u\implies\dfrac x{\sqrt{x^2-a^2}}dx=du$ and $x^2=u^2+a^2$
$$I=\int\dfrac{du}{u^2+a^2}=\dfrac1a\arctan\dfrac ua+K$$
Now we can prove $$\arctan y=\arcsin\dfrac y{\sqrt{1+y^2}}\text{ OR } \arcsin z=\arctan\dfrac z{\sqrt{1-z^2}}$$
an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Quadratic formula does not work If I put the equation: $5x^2-x-4 =0$ in the quadratic formula, than I get
$x = 1$ or $x = \frac{-4}{5}$
but the real zeros are: $x = -1$ or $x = \frac{4}{5}$
Can somebody explain me if the quadratic formula fails or me?
| Plug in the roots the quadratic formula gave you
$5(1)^2 - (1) - 4 $
$=5 - 5 $
$= 0$
And your other root
$5(- \frac{4}{5})^2 - (-\frac{4}{5}) - 4 $
$ = \frac{20}{5} -\frac{(16+4)}{5}$
$= 0$
And you can verify your roots work.
The quadratic formula is correct.
I recommend looking at the proof of the quadratic formula if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Prove that the trigonometric equation has no solution Prove that the trigonometric equation $$\frac{\sin^3 x}{1-\sin x}+\frac{\cos^3 x}{1-\cos x}=-1$$ has no solution. I tried applying $T2's$ lemma to contradict but could only do so for the first and third quadrant values if $x$. There must be some good proof without t... | $$\frac{\sin^3x}{1-\sin x}+\frac{\cos^3x}{1-\cos x}=-1\tag1$$
Let $f(x)$ be the LHS of $(1)$. Then, by AM-GM inequality, we have
$$\begin{align}f(x)&=\frac{(1-\sin x)(-\sin^2x-\sin x-1)+1}{1-\sin x}+\frac{(1-\cos x)(-\cos^2x-\cos x-1)+1}{1\cos x}\\&=-\sin^2x-\sin x-1+\frac{1}{1-\sin x}-\cos^2x-\cos x-1+\frac{1}{1-\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 0
} |
using half identities to find exact value of each trigonometric expression a) $\cos{({105}^{°})}$
b) $\sin{(\frac{3\pi}{8})}$
c) $\cot{({67.5}^{°})}$
please do explain how you are able to get the answer as I'm still confused about this topic... Thank you
| Notice, $$\cos 105^\circ=\cos(90^\circ+15^\circ)=-\sin 15^\circ=-\sqrt{\frac{1-\cos 30^\circ}{2}}=-\sqrt{\frac{1-\frac{\sqrt 3}{2}}{2}}=-\sqrt{\frac{2-\sqrt 3}{4}}$$ $$=-\sqrt{\frac{4-2\sqrt 3}{8}}= -\frac{\sqrt 3-1}{2\sqrt 2}=\frac{1-\sqrt 3}{2\sqrt 2}$$ $$\implies \sin\frac{3\pi}{8}=\sin 67.5^\circ=\sin(90^\circ-22.5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Differentiate $y= x(4-9x^4)^4$ So the textbook tells me to do the following question but I can't seem to get the right answer. (answer: $(4-9x^4)^3(4-45x^4)$
This is what I've done:
apply the product rule
$u = x$
$u' = 1$
$v = (4-9x^4)^4$
$v' = -144x^3 (4-9x^4)^3$
I used the product rule and got: $(4-9x^4)^3 (-153x^4+... | $y = x\left(4-9x^4\right)^4$
Use the product rule, $\left(uv\right)' = u'v + uv'$, setting:
$u = x$, $v = \left(4-9x^4\right)^4$.
Then,
$u' = 1$
Using the chain rule on $v$, we get:
$v' = 4\left(4-9x^4\right)^3 \cdot \left(-36x^3\right) = -144x^3\left(4-9x^4\right)^3$
Now, $\left(uv\right)' = u'v + uv' = \left(4-9x^4\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solve this trigonometric equation $ \sin2x-\sqrt3\cos2x=2$ Solve equation: $$ \sin2x-\sqrt3\cos2x=2$$
I tried dividing both sides with $\cos2x$ but then I win $\frac{2}{\cos2x}$.
| Notice, we have $$\sin 2x-\sqrt 3\cos 2x=2$$ $$\frac{1}{2}\sin 2x-\frac{\sqrt 3}{2}\cos 2x=1$$ $$\sin 2x\cos\frac{\pi}{3}-\cos 2x\sin \frac{\pi}{3}=1$$
$$\sin\left(2x-\frac{\pi}{3}\right)=1=\sin \frac{\pi}{2}$$
Now, writing the general solution for $x$, we get $$2x-\frac{\pi}{3}=2n\pi+\frac{\pi}{2}$$
$$2x=2n\pi+\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
} |
Explanation of a method to compute $\sum_{k \le n} k^2$ I was searching for methods to compute $\sum_{k\le n} k^2$. I stumbled across this (which is an answer provided by Gareth Rees to this question).
"..Represent $k^2$ in terms of falling powers (easy by inspection in this case, but you can use Stirling subset numbe... | Instead of falling powers I might use Binomial Coefficients:
$$ k^2 = 2\, \frac{k(k-1)}{2} + k = 2\binom{k}{2} + \binom{k}{1}$$
Now if we add over all $k$ we can get other binomial coefficients:
$$ \sum_{k \leq n} \binom{k}{2} = \binom{n+1}{3} \text{ and } \sum_{k \leq n} \binom{k}{1} = \binom{n+1}{2}$$
This feels jus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
sequence is cauchy how to prove that the recursive sequence $a_0\ge 0$, $a_{n+1}=\frac{3(1+a_n)}{3+a_n}$ is a cauchy sequence? The sequence seems to be bounded and if the sequence is monotonic increasing (I still dont know if it is..), it is convergent, then the sequence must be cauchy. But how to prove with the defini... | Define $f(x) = \dfrac{3 + 3x}{3 + x}, x \geq 0$.
It is easily seen that
$$f'(x) = \frac{3(3 + x) - (3 + 3x)}{(3 + x)^2} = \frac{6}{(3 + x)^2} \in \left(0, \frac{2}{3}\right]$$
for all $x \geq 0$.
It then follows by the mean value theorem that, for every positive integer $n$:
\begin{align*}
& \left|a_{n + 1} - a_n\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to solve $\displaystyle\lim_{x\to 0}\tfrac{\sqrt{x+25}-5} {\sqrt{x+16}-4}$ \begin{eqnarray}
\\&\lim_{x\to 0}\frac{\sqrt{x+25}-5} {\sqrt{x+16}-4}
\end{eqnarray}
Undefined limit
\begin{eqnarray}
\frac{0} {0}
\end{eqnarray}
| METHOD 1:
One quick way is to write
$$\sqrt{x+a^2}=|a|\sqrt{1+x/a^2}=|a|\,\left(1+\frac12 \left(\frac{x}{a^2}\right)+O\left(x\right)^2\right)$$
Then, we have
$$\frac{\sqrt{x+25}-5}{\sqrt{x+16}-4}=\frac45 +O(x)\to \frac45$$
METHOD 2:
Using L'Hospital's Rule gives
$$\lim_{x\to 0}\frac{\sqrt{x+25}-5}{\sqrt{x+16}-4}=\lim_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Solve trigonometric inequality $ \sin x \sin 2x - \cos x \cos 2x > \sin 6x $ Solve this trigonometric inequality:
$$ \sin x \sin 2x - \cos x \cos 2x > \sin 6x $$
My steps:
$$ \cos x \cos 2x - \sin x \sin 2x < - \sin 6x $$
$$ \cos 3x < \sin (-6x)$$
$$ \cos 3x < \cos (\frac{\pi}{2}+6x) $$
From this we get:
$$ 3x > \d... | $$\cos { \left( 3x \right) < } -\sin { \left( 6x \right) } \\ \cos { \left( 3x \right) < } -2\sin { \left( 3x \right) \cos { \left( 3x \right) } } \\ \cos { \left( 3x \right) } \left( 1+2\sin { \left( 3x \right) } \right) <0$$
$$1.\cos { \left( 3x \right) } >0\quad and\quad 1+2\sin { \left( 3x \right) } <0\\ 2.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find all integer values for $m$ such that $x_1,x_2\in\mathbb{Z}$ We have $(m+1)x^2-(2m+1)x-2m=0$ where $m\in\mathbb{R}\backslash\left\{-1\right\}$. We need to find all integer values for $m$ such that both roots of the equation are integers.
Here is all my steps:
*
*$x_{1,2}=\frac{(2m+1)\pm\sqrt{12m^2+12m+1}}{2(m+1)... | I'm sorry, but I find as the only answers $m=0$ or $m=-2$ or $m=-1$ (though in the latter case there are not two solutions).
To see this, note that if $m=-1$, it' not a quadratic equation, and the solution is $x=-2$.
If $m\neq -1$, let $x_1, x_2$ be the roots; use Vieta's relations:
$$x_1x_2=-\frac{2m}{m+1},\quad x_1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find $\int \frac{5x^4+4x^5}{(x^5+x+1)^2}$ $\displaystyle \int \frac{5x^4+4x^5}{(x^5+x+1)^2}$
Since in the denominator of the integrand,$(x^5+x+1)^2$ is there.So the answer must be in the form $\displaystyle \frac{f(x)}{(x^5+x+1)}$, but i could not figure out what $f(x)$ i should suppose.I could not imagine any other wa... | Let $$\displaystyle I = \int\frac{5x^4+4x^5}{(x^5+x+1)^2}dx = \int\frac{\left[\left(5x^4+1\right)+(4x^5-1)\right]}{(x^5+x+1)^2}dx$$
So $$\displaystyle I = \underbrace{\int\frac{5x^4+1}{(x^5+x+1)^2}dx}_{J}+\underbrace{\int \frac{4x^5-1}{(x^5+x+1)^2}dx}_{K}$$
So for Calculation of
$$\displaystyle J= \int \frac{5x^4+1}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Use calculus to find the area bounded by the circle $x^2+y^2-2x-2y-23=0$ and the pair of lines $x^2+2xy+y^2-7x-7y+12=0$ Use calculus to find the area bounded by the circle $x^2+y^2-2x-2y-23=0$ and the pair of lines $x^2+2xy+y^2-7x-7y+12=0$.
I tried to solve the two equations by subtracting them.
$2xy+35=5x+5y$,on furth... | The lines are $x+y=3$ and $x+y=4$. These are parallel lines. The equation of the circle is $(x-1)^2+(y-1)^2=25$. Hence you may rotate the lines so that they are parallel to the $x$-axis. You may also place the origin at the center of the circle. The distance between these lines is $1/\sqrt2$ and the distance between $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\arctan (x) + \arctan(1/x) = \frac{\pi}{2}$
How can I show that $\arctan (x) + \arctan(1/x) =\frac{\pi}{2}$?
I tried to let $x = \tan(u)$. Then
$$ \arctan(\tan(u)) + \arctan(\tan(\frac{\pi}{2} - x)) = \frac{\pi}{2}$$
but it does not seem useful.
I'd appreciate most a proof that gives intuition and / or uses geome... | $\arctan(a)+\arctan(b)
=\arctan(\frac{a+b}{1-ab})
$.
Therefore
$\arctan(x)+\arctan(1/x)
=\arctan(\frac{x+1/x}{1-1})
=\arctan(\frac{x+1/x}{0})
=\pi/2
$.
If this bothers you,
$\begin{array}\\
\arctan(x-c)+\arctan(1/x)
&=\arctan(\frac{x-c+1/x}{1-(x-c)/x)})\\
&=\arctan(\frac{x-c+1/x}{c/x})\\
&=\arctan(\frac{x^2-cx+1}{c})\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Why does the diophantine equation $x^2+x+1=7^y$ have no integer solutions? This following Problem is from Pell equation chapters exercise
Let $y>3$ positive integer numbers, show that following diophantine equation
$$x^2+x+1=7^y\tag{1}$$ has no integer solutions.
I tried write the equation
$$(2x+1)^2+3=4\cdot 7^y$$... | A sketch of my thoughts:
Let $\nu_7(n)=\max\{m\in\mathbb{N}:7^m\mid n\}$. A good idea may be to prove that if $\nu_7(x^2+x+1)=\nu_7(x^3-1)-\nu_7(x-1)=y>3$ then $x$ has to be large, say $x\geq 7^{y-1}$. In such a case, however, $x^2+x+1$ is too big to be just $7^y$ and it must have some other prime factor.
By this way, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 1
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$\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ equals $\frac{\pi}{5050}$ For $a\geq2$,if the value of the definite integral $\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ equals $\frac{\pi}{5050}$.Find the value of $a$.
Substituting $x-\frac{1}{x}=t$ does not seem to work here,what is th... | Let $$\displaystyle I = \int_{0}^{\infty}\frac{1}{a^2+\left(x-\frac{1}{x}\right)^2}dx = \int_{0}^{\infty}\frac{1}{x^2+\frac{1}{x^2}+(a^2-2)}dx\;,$$ Where $a^2-2 = k \geq 0$
So $$\displaystyle I = \int_{0}^{\infty}\frac{x^2}{x^4+kx^2+1}dx = \frac{1}{2}\int_{0}^{\infty}\frac{(x^2+1)+(x^2-1)}{x^4+kx^2+1}dx$$
So we get $$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
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Prove that $(1)\frac{1}{2}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{5}{6}$ $(2)2e^{-1/4}<\int_{0}^{2}e^{x^2-x}dx<2e^2$ Prove that $(1)\frac{1}{2}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{5}{6}$
$(2)2e^{-1/4}<\int_{0}^{2}e^{x^2-x}dx<2e^2$
I tried to prove it but my answer is not correct.
For first part,As $0\leq x\leq2\... | For the second problem, we note that
$$-\frac14\le x^2-x\le2$$for $x\in[0,2]$. Note that we can easily find the minimum by taking the derivative and setting it to $0$, while we merely check the endpoints to find the maximum.
Inasmuch as the exponential function is monotonically increasing, then
$$e^{-\frac14}\le e^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Rationalization of a fraction Can someone please explain how i can rationalize the fraction $\frac{2 +\sqrt{3}}{(2-\sqrt{3})^3}$ so that i obtain an answer that has an exponent of 6. I basically need to compare this value with $\dfrac{(10)^6}{(3)^6}$ to see which is greater. Thanks in advance for helping out!
| \begin{align*}
\frac{2+\sqrt{3}}{(2-\sqrt{3})^3}&=\frac{2+\sqrt{3}}{(2-\sqrt{3})^3}\cdot\frac{(2+\sqrt{3})^3}{(2+\sqrt{3})^3}\\
&=\frac{(2+\sqrt{3})^4}{(2^2-3)^3}\\
&=\frac{(2+\sqrt{3})^4}{1}\\
&=(2+\sqrt{3})^4
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What are the Eigenvectors in the following matrix? I have the matrix A:
\begin{bmatrix}
4 & 2 & 2\\
2 & 4 & 2\\
2 & 2 & 4\\
\end{bmatrix}
I found $\lambda I_n - A$ to be:
\begin{bmatrix}
(\lambda -4) & -2 & -2\\
-2 & (\lambda -4) & -2\\
-2 & -2 & (\lambda -4)\\
\end{bmatrix}
Applying Sarrus' Rule, and with help in a pr... | An easier way to check for eigenvalues is to factorize
$$
x^3 - 12x^2 + 36x -32 = \left(x-8\right)\left(x-2\right)^2
$$
Then to get the eigenvectors, input the eigenvalues one by one to
$$
(\lambda I - A) \cdot \begin{bmatrix}S_1\\S_2\\S_3\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}
$$
and solve for $S_1$, $S_2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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The set of real values of $x$ satisfying the equation $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$ The set of real values of $x$ satisfying the equation $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$,(where $[]$ denotes the greatest integer function) belongs to the interval $(a,\frac{b}{c}]$,where $a,b,... | Here are some hints.
Note that as $x$ increases, the sum can get no larger - it is decreasing but not strictly so, because it is sometimes constant. Also with $x=1$ the sum is equal to $7$ so we must have $x \gt 1$ for a sum as low as $5$.
Now if $x\gt 1$ we have immediately that $\left[\frac{3}{x}\right]+\left[\frac{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Trying to solve the trig equation $\sqrt{3+4\cos^2(x)}=\frac{\sin(x)}{\sqrt 3}+3\cos(x)$ The equation is
$$\sqrt{3+4\cos^2(x)}=\frac{\sin(x)}{\sqrt 3}+3\cos(x)$$
My solution goes like this
$$
\begin{cases}
3+4\cos^2(x)=\frac{\sin^2(x)}{3}+\frac{6}{\sqrt 3}\sin(x)\cos(x)+9\cos^2(x) \\
\frac{\sin(x)}{\sqrt 3}+3\cos(x) \g... | You made a mistake in factorising the following equation:
$$ 4t^2 -3\sqrt3 t -3 = 0$$
$$ t = \frac{3\sqrt 3 \pm \sqrt{75}}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding the maximum of a function on $ \Bbb{S}^{7} $. I'm trying to find the maximum of the function
$$2 a^2 h+\sqrt{3} a d f+\sqrt{3} a e g+2 b^2 h-\sqrt{3} b d g+\sqrt{3} b e f\\+2 c^2
h+\sqrt{3} c d^2+\sqrt{3} c e^2-\sqrt{3} c f^2-\sqrt{3} c g^2\\-d^2 h-e^2 h-f^2
h-g^2 h-2 h^3$$
on the sphere $$a^2+b^2+c^2+d^2... | (Update: I just realize that this is very similar to @Michael 's earlier solution.)
Your expression ($=:\Phi_0$) can be simplified using the following measures:
$$a=r\cos\alpha,\quad b=r\sin\alpha,\quad d=s\cos\beta,\quad e=s\sin\beta,\quad f=t\cos\gamma,\quad g=t\sin\gamma\ .$$
After some computation this results in
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Summation of the reciprocals of the product of consecutive integers It is well known that there is a closed formula for:
$$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{(n)(n + 1)}$$
And likewise for:
$$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \cdots + \frac{1}{(n)(n + 1)(n+2)}$$
I am... | Notice
$$\begin{align}
\frac{1}{\prod\limits_{j=0}^{k}(i+j)} = \frac{1}{k}\left(\frac{(i+k)-i}{\prod\limits_{j=0}^{k}(i+j)}\right)
&= \frac{1}{k}\left[\frac{1}{\prod\limits_{j=0}^{k-1}(i+j)}-\frac{1}{\prod\limits_{j=1}^{k}(i+j)}\right]\\
&= \frac{1}{k}\left[\frac{1}{\prod\limits_{j=0}^{k-1}(i+j)}-\frac{1}{\prod\limits_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1417844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How do you factor $\frac{2x^2-x-1}{x^2-9} \cdot \frac{x+3}{2x+1}=$? \begin{align}
& \frac{2x^2-x-1}{x^2-9} \cdot \frac{x+3}{2x+1}= \frac{2x^2-x-1}{(x-3)(x+3)} \cdot \frac{x+3}{2x+1} \\[10pt]
= {} & \frac{2x^2-x-1}{(x-3)} \cdot \frac{1}{2x+1}= \frac{2x^2-x-1}{(x-3)} \cdot \frac{1}{2x+1}= \frac{2x^2-x-1}{(x-3)(2x+1)}
\en... | HINT:
$2x^2-x-1=(x-1)(2x+1)$
$x^2-9=(x+3)(x-3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Evaluation of $ \int\frac{(1+x)\left[(1-x+x^2)\cdot (1+x+x^2)+x^2\right]}{1+2x+3x^2+4x^3+3x^4+2x^5+x^6}dx$
Evaluation of $\displaystyle \int\frac{(1+x)\left[(1-x+x^2)\cdot (1+x+x^2)+x^2\right]}{1+2x+3x^2+4x^3+3x^4+2x^5+x^6}dx$
$\bf{My\; Try::}$ We can write $$1+2x+3x^2+4x^3+3x^4+2x^5+x^6 $$
$$= (1+x)+(x+x^2)+2(x^2+x^... | Well, you're really almost done. All you have to do is realize that the denominator factors more: $$1+x+2x^2+2x^3+x^4+x^5=(1+x)(1+x^2)^2$$
This of course gives you
$$I=\int\frac{1}{1+x}\,dx=\ln|1+x|+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that $\sin x+2x\ge\frac{3x(x+1)}{\pi}$ for all $x\in [0,\frac{\pi}{2}]$ Question: Prove that $\sin x+2x\ge\frac{3x(x+1)}{\pi}$ for all $x\in [0,\frac{\pi}{2}]$
How did $\pi$ come in the expression?
This is how I tried to solve.
$\sin x\le x$ so $\sin 2x\le 2x$
$\sin x+2x\le 3x$.
Applying AM GM inequality: $\sin ... | Just expanding mastrok's answer, over $I=\left(0,\frac{\pi}{2}\right)$ we have:
$$ \frac{d^2}{dx^2}\left(\sin x+2x\right) = -\sin x<0 \tag{1}$$
Hence $f(x)=\sin(x)+2x$ is a concave function, and since:
$$ \frac{d^2}{dx^2}\left(\frac{3}{\pi}x(x+1)\right)=\frac{6}{\pi}>0\tag{2}$$
$g(x)=\frac{3}{\pi}x(x+1)$ is a convex fu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1424145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Limit of a trigonometric rational expression How to evaluate the limit of this expression?
$$\lim_{x\to\infty} \frac{\sin^2\left( \sqrt{x+1}-\sqrt{x}\right)}{1-\cos^2\frac{1}{x}}$$
I managed to simplify the denominator into a sinus form by the Pythagorean formula, and also modified the argument of the sinus in the nume... | Observe
\begin{align}
\lim_{x\to\infty}\frac{\sin^2\left(\sqrt{x+1}-\sqrt{x}\right)}{1-\cos^2\frac{1}{x}}&=\lim_{x\to\infty}\frac{\sin^2\left(\sqrt{x+1}-\sqrt{x}\right)}{\sin^2\frac{1}{x}}\\
&=\lim_{x\to\infty}\frac{\left(\sqrt{x+1}-\sqrt{x}\right)^2\frac{\sin^2\left(\sqrt{x+1}-\sqrt{x}\right)}{(\sqrt{x+1}-\sqrt{x})^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluation of $\int\frac{\sin 2x}{(3+4\cos x)^3}\,dx$
Evaluation of $$\int\frac{\sin 2x}{(3+4\cos x)^3}\,dx$$
$\bf{My\; Try::}$ Let $$\displaystyle \int\frac{\sin 2x}{(3+4\cos x)^3}dx = 2\int\frac{\sin x\cos x}{(3+4\cos x)^3}dx$$
Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\cos^3 x.$
So we get $$\displaystyle... | Hint:
By inspection,
$$\int\frac{\sin x\cos x}{(3+4\cos x)^3}dx=-\int \frac{t}{(3+4t)^3}dt$$
which can be decomposed with
$$4t=(3+4t)-3,$$ to yield elementary functions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find a good strategy to compute$f(x) = e^x −\cos x − \sin x$ for $x$ near $0$ Find a good strategy to compute $f(x) = e^x − \cos x − \sin x$ for $x$ near 0.
In five-decimal-digit arithmetic, compute $f(0.1)$ using the straightforward method
and your better strategy, and compare the difference. (Hint: The computer can o... | With Taylor series you have
$$\begin{align} f(x) &= \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \cdots \right)\\
&-\left(1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots\right)\\
&- \left(x - \frac{x^3}{6} + \frac{x^5}{120} + \cdots\right)\end{align}$$
Now, notice that the terms after $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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To find maximum value when sum is fixed Given that $a,b,c,d,e$ and $f$ are six non-negative real numbers such that
$a+b+c+d+e+f=1$.
Find the maximum value of
$ab+bc+cd+de+ef$.
My Approach
After lot of thinking I reduced the problem to following:
Given that $a,b,c,d,e$ and $f$ are six non-negative real numbers such that... | Let $0\leq a \leq b \leq c \leq d \leq f \leq e\leq 1$
$$a+b+c+d+f+e=1 \rightarrow (a+b+c+d+f+e-1)^2=0$$
$$\rightarrow a^2+b^2+c^2+d^2+f^2+(e-1)^2+....=0$$
so we have $a^2=b^2=c^2=d^2=f^2=(e-1)^2=0$ i.e., $a=b=c=d=f=0$ and $e=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1430263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Forming a committee from $4$ gentlemen and $4$ ladies with certain conditions
From $4$ gentlemen and $4$ ladies a committee of $5$ is to be formed .
If the committee consists of $1$ president, $1$ vice president and $3$ secretaries.
What will be the number of ways of selecting the committee with at least $3$
wom... | Count the number of committees with $\color\red{3\text{ women}}$ and $\color\green{2\text{ men}}$:
$$\color\red{\binom43}\cdot\color\green{\binom42}\cdot\binom51\cdot\binom41\cdot\binom33=480$$
Subtract the number of committees where all $3$ women are secretaries:
$$\color\red{\binom43}\cdot\color\green{\binom42}\cdot\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1430501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $\sin\theta+\cos\theta=1$ prove that $\cos\theta-\sin\theta=\pm1$ So my work,
Squaring both sides $$(\sin\theta+\cos\theta)^2=1$$
$$1+2\sin\theta\cos\theta=1\ \ \ \ \ \text{-------(i)}$$
$$\sin\theta\cos\theta=0 \ \ \ \ \ \text{------(ii)}$$
So reverting back to $(i)$,
$$\sin^2\theta+\cos^2\theta+2\sin\theta\cos\t... | Let $x = \cos\theta$ and $y = \sin\theta$. Then $(x,y)$ is a point on
the unit circle. But the equation
$$\sin \theta + \cos \theta = 1$$
says that $x + y = 1$. So $(x,y)$ must be on the line given by $x + y = 1$,
that is, the line that intersects the unit circle at $(1,0)$ and $(0,1)$.
In fact, since $(x,y)$ is on tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1432165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Minimum of $\frac{1}{x+y}+\frac{1}{x+z}-\frac{1}{x+y+z}$ for $0\leq x+y,y+z,z+x\leq 1$ Let $0\leq x,y,z,x+y,y+z,z+x\leq 1$. What is the minimum of $$F(x,y,z)=\frac{1}{x+y}+\frac{1}{x+z}-\frac{1}{x+y+z}?$$
We have $F(1/2,1/2,1/2)=4/3$. Since the constraints are on $x+y,y+z,z+x$ instead of $x,y,z$, taking partial derivat... | Hint: Do the substitution $ a= x+y, b = y + z, c = z + x $, and we have $ 0 \leq a, b, c, \leq 1 $ and we want to minimize
$$ \frac{1}{a} + \frac{1}{c} - \frac{2}{a+b+c}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1432465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How can I solve this recursive function $f(n) = f(f(n+1))$? I am trying to solve this:
$$
f(n) =
\begin{cases} n - 1,& n > 5\\
f(f(n+1)),& n\leqslant 5
\end{cases}
$$
What is the technical name of this kind of function ? --> $f(f(n+1))$
A function within another function?
\begin{align}
f(1) &= f(f(1+1)... | An example will make it understand better
$ {let}$ $ n = 5$ now $ f(n)=f(f(n=1))$
so, $f(5)=f(f(5+1))$
$=f(f(6)) = f(5)$
$f(5)=f(f(5+1))$
$=f(f(6)) = f(5)$
$f(5)=f(f(5+1))$
$=f(f(6)) = f(5)$
..............
and it goes on so the function is not defined at 5
Now let $n=4$
so, $f(3)=f(f(3+1))$
$=f(f(4))$ now $f(4)$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Express $y= x^{3} - x^{2} - 5x - 3$ in its fully factorised form Don't know how to do this, please help. I have never done factorising cubic polynomials and don't know how to go about this
| $$x^3-x^2-5x-3=x^3+2x^2-3x^2-6x+x-3$$
$$x^3+2x^2+x-3x^2-6x-3=0$$
$$x(x^2+2x+1)-3(x^2+2x+1)=0$$
$$(x^2+2x+1)(x-3)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
First order differential equation involving inverses My question is to find the solutions to the following
$\frac{df(x)}{dx} = f^{-1} (x)$
where $f^{-1} (x)$ refers to the inverse of the function f. The domain really isn't important, though I am interested in either (-inf, inf) or (0, inf), so if any solutions are kn... | Here is a partial solution using power series composition.
To solve
$$
f(f'(x))=x\tag{1}
$$
take the derivative of $(1)$
$$
f'(f'(x))\,f''(x)=1\tag{2}
$$
If we can find a point $a\gt0$ so that $f'(a)=a$, we can expand $f'(a+x)$ around $x=0$. Let
$$
g(x)=f'(a+x)-a\tag{3}
$$
Applying $f'(a+x)=a+g(x)$ twice gives
$$
\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1438651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Find $n$, where its factorial is a product of factorials I need to solve $3! \cdot 5! \cdot 7! = n!$ for $n$.
I have tried simplifying as follows:
$$\begin{array}{}
3! \cdot 5 \cdot 4 \cdot 3! \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3! &= n! \\
(3!)^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\
6^3 \cdot 5^2 \cdot 4^2 \... | You already have some clever answers; here's less clever approach. You want
$$
3!\cdot5!\cdot7!=n!=7!\cdot8\cdot9\cdots\cdot n,
$$
so, cancelling 7! and computing that $3!\cdot 5!=6\cdot120=720$, you want
$$
720=8\cdot9\cdots\cdot n.
$$
Now start dividing both sides by 8, then 9, etc. until you get the answer. Dividi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1440482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 7,
"answer_id": 5
} |
Prove that $\sum^{n}_{r=0}(-1)^r\cdot \large\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{3!}{2(n+3)}$
Prove that $\displaystyle \sum^{n}_{r=0}(-1)^r\cdot \large\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{3!}{2(n+3)}$
$\bf{My\; Try::}$ We can write $$\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{n!}{r!\times (n-r)!}\time... | Your computational argument can be written a bit more simply:
$$\begin{align*}
\sum_{r=0}^n(-1)^r\frac{\binom{n}r}{\binom{r+3}r}&=3!\sum_{r=0}^n(-1)^r\frac{n!}{(n-r)!(r+3)!}\\
&=\frac6{(n+3)(n+2)(n+1)}\sum_{r=0}^n(-1)^r\binom{n+3}{r+3}\tag{1}\\
&=\frac6{(n+3)(n+2)(n+1)}\sum_{r=3}^{n+3}(-1)^{r-3}\binom{n+3}r\\
&=\frac6{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1441827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"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.