Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
First derivative meaning in this case If we have a function:
$$f(x)=\frac{x}{2}+\arcsin{\frac{2x}{1+x^2}}$$
And it's first derivative is calculated as:
$$f'(x)=\frac{1}{2}+\frac{1}{\sqrt{1-\big(\frac{2x}{1+x^2}\big)^2}}\frac{2+2x^2-4x^2}{(1+x^2)^2}=$$
$$\frac{1}{2}+\frac{2(1-x^2)}{\sqrt{\frac{(1-x^2)^2}{(1+x^2)^2}}\cdo... | If $\arctan x=y,x=\tan y$ and $-\dfrac\pi2\le y\le\dfrac\pi2$
(See definition of the principal values of inverse trigonometric functions)
$$\dfrac{2x}{1+x^2}=\cdots=\sin2y$$
$$\implies\arcsin\dfrac{2x}{1+x^2}=\arcsin(\sin2y)=\begin{cases}2y&\mbox{if }-\dfrac\pi2\le2y\le\dfrac\pi2\iff-1\le x\le1\\
\pi-2y & \mbox{if }2y>... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Prove that $|\log(1 + x^2) - \log(1 + y^2)| \le |x-y|$ I need to show that $ \forall x,y \in \mathbb R, |\log(1 + x^2) - \log(1 + y^2)| \le |x-y|$
I tried using concavity of log function: $\log(1 + x^2) - \log(1 + y^2)=\log(\frac{1 + x^2}{1 + y^2})=\log(\frac{x^2y^2}{(1 + y^2)y^2}+\frac{1}{1 + y^2}) \ge \frac{2(\log(x)... | We can write the term of interest as
$$\left|\log(1+x^2)=\log(1+y^2)\right|=\left|\int_y^x \frac{2t}{1+t^2}\,dt\right|$$
Now, the integrand $f(t)=\frac{2t}{1+t^2}$ attains its maximum value of $1$ when $t=1$. Therefore, the inequality follows immediately from the mean value theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the two other sides in a 15-30-135 triangle A triangle has angle measures of 15, 30, and 135 degrees. The side opposite the 15 angle is x feet, the side opposite the 30 angle is y feet, and the side opposite the 135 angle is 2 feet.
Find x and y without the law of sine.
| Apply sine rule in given triangle as follows
$$\frac{x}{\sin 15^\circ}=\frac{y}{\sin 30^\circ}=\frac{2}{\sin 135^\circ}$$
considering first & third,
$$\frac{x}{\sin 15^\circ}=\frac{2}{\sin 135^\circ}$$
$$x=\frac{2\sin 15^\circ}{\sin 135^\circ}=\frac{2\sin (45^\circ-30^\circ)}{\sin 45^\circ}=\frac{2\left(\frac{1}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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If $\sin x+\sin^{2} x=1$ , Find $\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2 $
If $\sin x+\sin^{2} x=1$, then the value of
$\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2 $
is equal to
$a.)\ 0 \\
b.)\ 1 \\
c.)\ 2 \\
\color{green}{d.)\ \sin^{2} x} $
$\boxed{... | HINT:
$$\sin x=1-\sin^2x\implies(\cos^2x)^2=\sin^2x=1-\cos^2x$$
$$\implies\cos^4x+\cos^2x-1=0$$
Divide $$\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2 $$ by $\cos^4x+\cos^2x-1$ to get a reminder of $O(\cos^2x)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Value of $\sum_{n=2}^\infty (-1)^n ((\sum_{k=1}^\infty 1/{k^n})-1)$ To find the value of:
$\displaystyle\sum_{n=2}^\infty (-1)^n ((\sum_{k=1}^\infty 1/{k^n})-1)$
The answer says: $1/2$
My attempt:
$=(\displaystyle\sum_{k=1}^\infty 1/{k^2})-1-(\sum_{k=1}^\infty 1/{k^3})+1+(\sum_{k=1}^\infty 1/{k^4})-1-\cdots$
$=(1+1/2^2... | You've calculated $S_2$ wrong. The sum $\sum_{n=2}^\infty(−1/2)^n=\frac1{1−(−1/2))}−1+1/2=1/6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1626104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$\iint (y^2-x^2) e^{xy} dxdy$ $\iint_G(y^2-x^2)e^{xy}dxdy$ on G = $\{ (x,y) | x \ge 0, 1 \le xy \le 4, 0 \le y-x \le 3 \}$
I have tried a substitution of
$u = xy$
$v = y-x$
which got me to $\bar G = \{ (u,v) | 1 \le u \le 4, 0 \le v \le 3 \}$
and
$\iint_\bar G \left(\frac{v}{\sqrt{v^2+4u}}\right)^{\frac12} e^u dudv$
(... | A slightly different change of variables which simplifies a lot the boundary conditions :
$\begin{cases}
s=y-x\\
p=xy\\
\end{cases}$
The Jacobian is : $-\begin{vmatrix}
-1 & 1 \\
y & x \\
\end{vmatrix}^{-1}=\frac{1}{y+x}$
$(y^2-x^2)\frac{1}{y+x}=y-x=s$
$\iint_{1 \le xy \le 4, 0 \le y-x \le 3}(y^2-x^2)e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simple Inequality of Complex Numbers, $\left| \frac{a-b}{1-\overline{a}b} \right| <1$ Exercise from Ahlfor's Complex:
Given $a,b \in \mathbb C$, with $|a| <1$, $|b|<1$, prove:
$$\left|\frac{a-b}{1-\overline{a}b}\right| <1.$$
My argument:
Lemma:
If $\alpha, \beta \in \mathbb R$ and $\alpha, \beta <1$, then $\alpha^... | We have
$$
\begin{align*}
|1 - \overline{a}b|^2 - |a - b|^2 &= \left(1 - \overline{a}b \right)\left(1 - a\overline{b}\right) - (a - b)\left( \overline{a} - \overline{b}\right) \\
&= 1 - a\overline{a} - b\overline{b} + a\overline{a}b\overline{b} \\
&= \left( 1 - |a|^2\right)\left( 1 - |b|^2\right)\\
&> 0.
\end{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solution of Differential equation $\frac{xdx-ydy}{xdy-ydx} = \sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$
Solution of Differential equation $$\frac{xdx-ydy}{xdy-ydx} = \sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$$
$\bf{My\; Try::}$ Let $x=r\sec \theta$ and $y=r\tan \theta\;,$ Then $x^2-y^2=r^2$ and $xdx-ydy=rdr$
and $$\frac{y}{x} = \frac{... | A better change of variables (in hyperbolic system of coordinates) is :
$\begin{cases}
x=r \cosh(t)\\
y=r \sinh(t)\\
\end{cases}$
$\begin{cases}
dx=dr \cosh(t)+r \sinh(t) dt\\
dy=dr \sinh(t)+r \cosh(t) dt\\
\end{cases}$
Bringing them into :
$$\frac{xdx-ydy}{xdy-ydx} = \sqrt{\frac{1+x^2-y^2}{x^2-y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How do you find the probability of A winning if the probability of getting a favourable outcome in the $r^{th}$ turn is a function of $r$? Problem:
Two players A and B are playing snake and ladder. A is at 99 and he needs 1 to win in rolling of a dice. However, he is always allowed to re-throw the dice if 6 appears.
Wh... |
The generating function for the central binomial coefficients is
\begin{align*}
\sum_{n=0}^{\infty}\binom{2n}{n}z^n=\frac{1}{\sqrt{1-4z}}\qquad |z|<\frac{1}{4}
\end{align*}
This is an application of the binomial series
\begin{align*}
(1+z)^{\alpha}=\sum_{n=0}^{\infty}\binom{\alpha}{n}z^n\qquad |z|<1, \alpha\in\ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Prove that $\frac{(2n)!}{(n!)^2}-1$ is divisible by $(2n+1)$
Prove that $$\frac{(2n)!}{(n!)^2}-1$$ is divisible by $(2n+1)\;,$ Where $n\in \mathbb{N}$ and $n>1$
$\bf{My\; Try::}$ Let $$S = \frac{(2n)!}{n!^2}-1 = \frac{2^n(2n-1)(2n-3)\cdot \cdot ........\cdot 3 \cdot 2 \cdot 1}{n!}-1$$
Now How can we prove that $2^n(... | $\dfrac{(2n)!}{(n!)^2} - 1 = \dbinom{2n}{n}-1$
This is the count of ways to arrange $n$ discrete objects into two equal piles, excluding a particular arrangement.
$\begin{array}{l:l:l} n & \binom{2n}n-1 & 2n+1
\\ 1 & 1 & 3
\\ 2 & 5 & 5
\\ 3 & 19 & 7
\\ 4 & 69 & 9
\\ \vdots & \vdots & \vdots
\end{array}$
Only occasional... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Laurent series expansion of $f$ Find the Laurent series expansion of $f(z)=\dfrac{1}{2z^2-13z+15}$ about the annulus $\dfrac{3}{2}<|z|<5$.
I did like this :
$f(z)=\dfrac{2}{7}(\dfrac{3}{3-2z}-\dfrac{1}{z-5})$
Then I took $\dfrac{3}{3-2z}=\dfrac{1}{1-\dfrac{2z}{3}}=(1-\dfrac{2z}{3})^{-1}=1+\dfrac{2z}{3}+(\dfrac{2z}{3})... | Here is a somewhat detailed description of the Laurent expansion of $f(z)$ around $z=0$.
The function
\begin{align*}
f(z)&=\frac{1}{2z^2-13z+15}
=-\frac{1}{7}\frac{1}{z-\frac{3}{2}}+\frac{1}{7}\frac{1}{z-5}\\
\end{align*}
has two simple poles at $\frac{3}{2}$ and $5$.
Since we want to find a Laurent expansion wit... | {
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"url": "https://math.stackexchange.com/questions/1634024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$f(x) - f'(x) = x^3 + 3x^2 + 3x +1; f(9) =?$ Given the following $f(x) - f'(x) = x^3 + 3x^2 + 3x +1$
Calculate $f(9) = ?$
I have tried to play with different number of derivatives. Also tried to solve it by equations.
Maybe there is some geometric meaning that could shade the light ?
I feel it is no complex problem at... | \begin{equation}
f(x) = ax^4 + bx^3 + cx^2 + dx + e\\f'(x) = 4ax^3 + 3bx^2 + 2cx + d\\f(x) - f'(x) = ax^4 + (b-4a)x^3 + (c-3b)x^2 + (d-2c)x + e - d = x^3 + 3x^2 + 3x +1\\a=0, b=1, c=6, d=15, e=16
\end{equation}
So, the function is $f(x)=x^3 + 6x^2 + 15x + 16$ and $f(9) = 1366$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
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Legendre symbol, what is it? I am reading wiki article about Legendre symbol and I don't understand the power meaning. Can you please explain the next expression.
$$\left(\frac ap\right)\equiv a^{\frac{p-1}{2}}\pmod p$$
| Since you say you have read about Legendre symbol on Wikipedia, you should already know that $\left(\frac{a}{p}\right)$ is defined as $0$ if $p\mid a$ and as $\pm1$ for $p\nmid a$, depending on whether $a$ is a quadratic residue modulo $p$ or not.
According to Euler's criterion, the congruence
$$\left(\frac{a}{p}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$ \int\frac{\sin(nx) \sin x}{1-\cos x} \,dx$ by elementary methods What is an elementary way to show that for positive integer $n$
$$
\int\frac{\sin(nx) \sin x}{1-\cos x} \,dx= x + \frac{\sin (nx)}{n} + 2 \sum_{k=1}^{n-1}\frac{\sin(kx)}{k}
$$
This cropped up when trying to answer Proving $\int_0^{\pi } f(x) \, \mathrm{... | Work backwards:
Let $$F(x)=x + \frac{\sin (nx)}{n} + 2 \sum_{k-1}^{n-1}\frac{\sin(kx)}{k}$$
Then
$$F'(x)=1 + \cos (nx) + 2 \sum_{k-1}^{n-1}\cos(kx)$$
Therefore, you need to prove that
$$\frac{\sin(nx) \sin x}{1-\cos x} = 1 + \cos (nx) + 2 \sum_{k-1}^{n-1}\cos(kx)$$
or equivalently that
$$\sin(nx) \sin(x)= (1-\cos(x)... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the polynomial $P$ of smallest degree with rational coefficients and leading coefficient $1$ such that $ P(49^{1/3}+7^{1/3})=4 $
Find the polynomial $P$ of smallest degree with rational coefficients
and leading coefficient $1$ such that $$ P(49^{1/3}+7^{1/3})=4 $$
(Source:NYSML)
My attempt
Let $$ 49^{1/3}+7^... | Your first attempt (cubing $49^{1/3}+7^{1/3}$) came close to finishing. Let $a$ be our number. Cubing, we find that
$$a^3=56+3(7^{5/3}+7^{4/3})=56+21a.$$
So if $Q(x)$ is the polynomial $x^3-21x-56$, then $Q(a)=0$. Dealing with the fact that we want $P(a)=4$ is now easy.
We sketch a proof that this polynomial $P(x)$ has... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solving Trigonometric Equation.
Solve for $\theta$ $[0°<\theta<180°]$
$$\sin2\theta + \sin4\theta=\cos\theta + \cos3\theta.$$
My solution is here:
$$\sin2\theta + \sin4\theta=\cos\theta + \cos3\theta.$$
After using the transformation formula, I got
$$\sin3\theta\cos\theta=\cos3\theta\cos\theta.$$
I could not procee... | Using Prosthaphaeresis Formulas as suggested,
$$ \sin 2\theta+\sin 4\theta = \cos \theta+\cos 3\theta \\
2\sin 3\theta \cos \theta = 2\cos 2\theta \cos \theta \\
\cos \theta \, (\sin 3\theta-\cos 2\theta) = 0 $$
If $\cos \theta=0,\theta=(2n+1)90^\circ$ where $n$ is any integer
We need $0<(2n+1)90^\circ<180^\circ\im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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finding real roots by way of complex I was given
$$x^4 + 1$$
and was told to find its real factors. I found the $((x^2 + i)((x^2 - i))$ complex factors but am lost as to how the problem should be approached.
My teacher first found 4 complex roots ( different than mine)
$$( x - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i... | change equation to
$$x^4 = -1 = e^{\pi i}$$
$$x = e^{(\frac{\pi}{4} + k\frac{\pi}{2})i}$$
where $k = 0, 1, 2, 3$. All four roots are evenly disitributed on the unit circle in complex plane.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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The sum of the following infinite series $\frac{4}{20}+\frac{4\cdot 7}{20\cdot 30}+\frac{4\cdot 7\cdot 10}{20\cdot 30 \cdot 40}+\cdots$
The sum of the following infinite series $\displaystyle \frac{4}{20}+\frac{4\cdot 7}{20\cdot 30}+\frac{4\cdot 7\cdot 10}{20\cdot 30 \cdot 40}+\cdots$
$\bf{My\; Try::}$ We can write t... | Since the question asks about $X=\frac4{20}(1+\frac7{30}(1+...)))$, consider
$$1+\frac1{10}(1+\frac4{20}(1+\frac7{30}(...)))$$
The $10,20,30$ have a factor $1,2,3$ which will become $n!$ in the denominator.
Then $\frac1{10},\frac4{10},\frac7{10}$ increase by $3/10$ each time. Take the factor $3/10$ out, and we have $... | {
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"url": "https://math.stackexchange.com/questions/1652349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Finding coefficient of polynomial? The coefficient of $x^{12}$ in $(x^3 + x^4 + x^5 + x^6 + …)^3$ is_______?
My Try:
Somewhere it explain as:
The expression can be re-written as: $(x^3 (1+ x + x^2 + x^3 + …))^3=x^9(1+(x+x^2+x^3))^3$
Expanding $(1+(x+x^2+x^3))^3$ using binomial expansion:
$(1+(x+x^2+x^3))^3 $
$= 1+3(x+... | From the OP, the coefficient of $x^{12}$ in $(x^3 + x^4 + x^5 + x^6 + \cdots)^3$ is equal to that of $x^3$ in $(1+x+x^2+x^3)^3$. This is equivalent to asking the number of ways to pick one $x^i$ below in each row so that the product of the $x^i$ picked in each row is of the form $kx^3$ for some number $k$.
$$ \require... | {
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"url": "https://math.stackexchange.com/questions/1654126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Evaluation of $\int_{0}^{1}\frac{\arctan x}{1+x}dx$
Evaluation of $$\int_{0}^{1}\frac{\tan^{-1}(x)}{1+x}dx = \int_{0}^{1}\frac{\arctan x}{1+x}dx$$
$\bf{My\; Try::}$ Let $$I = \int_{0}^{1}\frac{\tan^{-1}(ax)}{1+x}dx$$
Then $$\frac{dI}{da} = \frac{d}{da}\left[\int_{0}^{1}\frac{\tan^{-1}(ax)}{1+x}\right]dx = \int_{0}^{1... | I have another way, although the method in the other answer is pretty sweet and mine a bit longer, I decided to post it anyway.
Substitute $\tan^{-1}x=t$, The integral changes as;-
$$\int_{0}^{\pi/4}\frac{t\sec^2 tdt}{1+\tan t}$$
Apply by parts in this to get,
$$=t\log_e|1+\tan t|-\int_{0}^{\pi/4}\log_e|1+\tan t|dt$$
L... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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the maximum value of $(a+\frac1b)(b+\frac1c)(c+\frac1d)(d+\frac1a)$
If $a, b, c, d \in [\frac12, 2]$ and $abcd =1$, find the maximum value of
$$(a+\frac1b)(b+\frac1c)(c+\frac1d)(d+\frac1a)$$
Thought: $$\begin{split}(a+\frac1b)(b+\frac1c)(c+\frac1d)(d+\frac1a) &= \frac{(ab+1)(bc+1)(cd+1)(da+1)}{abcd} \\ & \stackrel... | This problem is not new: I have found it together with a solution (written in latex + chinese) in :
http://kuing.orzweb.net/viewthread.php?action=printable&tid=3286
Here is one of the interesting solutions presented there, with a slight personal adaptation.
We begin by a lemma that will be used in the sequel:
The conve... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Among the following, which is closest to $\sqrt{0.016}$? Among the following, which is closest in value to $\sqrt{0.016}$?
A. $0.4$
B. $0.04$
C. $0.2$
D. $0.02$
E. $0.13$
My Approach:
$(\frac{16}{1000})^\frac{1}{2} = (\frac{4}{250})^\frac{1}{2} = \frac{2}{5\cdot\sqrt{10}} = \frac{\sqrt{2}}{5\cdot\sqrt{5}}$
But I don't... | Though squaring each and checking works equally as well, here's another way to figure it out:
$$
\sqrt{0.016}=\frac{\sqrt{1000}}{\sqrt{1000}}\sqrt{0.016}=\frac{\sqrt{16}}{\sqrt{1000}}=\frac{4}{\sqrt{1000}}\approx\frac{4}{\sqrt{1024}}=\frac{4}{\sqrt{2^{10}}}=\frac{4}{2^{5}}=\frac{4}{32}=\frac{1}{8}=0.125
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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The area between two curves - is my answer correct?
Find the area between a parabola $y=x^2-2$ and an asymptote to $y=\sqrt{x^2+4x}+2x $ for $x \to -\infty$. As an answer, write the area multiplied by 6.
This is a very easy problem, but I made some careless mistake at first and my answer was incorrect. However, the a... | We have $\sqrt{x^2+4x}+2x\sim x-2+\frac{2}{x}+ O(\frac{1}{x^2})$ as $x\rightarrow -\infty$. Thus the asymptote is $y=x-2$. The points of intersection are the solutions of $x^2-2=x-2,$ i.e. $x=0\,$ and $x=1$. And therefore the answer is
$$ 6 \Big|\int_0^1 (x^2-2 - (x-2)) dx\Big|= 6 \Big|\int_0^1 (x^2- x) dx\Big| =
6\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Distributing identical objects to different people. What is the number of ways in which we can distribute 12 identical oranges among 4 children such that every child gets at least one and no child gets more than 4 ?
Till now ,My attempts have focused on first finding out all such ways in each person gets at least one o... | The number of ways $12$ oranges can be distributed to four children if each child gets at least one orange is the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 = 12 \tag{1}$$
in the positive integers. A particular solution of equation 1 in the positive integers corresponds to the placement of three addit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Constant length of segment of tangent
Prove that the segment of the tangent to the curve $y=\frac{a}{2}\ln\frac{a+\sqrt{a^2-x^2}}{a-\sqrt{a^2-x^2}}-\sqrt{a^2-x^2}$ contained between the $y$-axis and the point of tangency has a constant length.
What I have done: First of all I found out slope $m$ of the tangent to the... | Here are some steps:
\begin{align*}
y & =\frac{a}{2}\ln\frac{a+\sqrt{a^2-x^2}}{a-\sqrt{a^2-x^2}}-\sqrt{a^2-x^2}\\
\frac{dy}{dx} & = \frac{a}{2}\left(\frac{a-\sqrt{a^2-x^2}}{a+\sqrt{a^2-x^2}}\right) \, \frac{d}{dx}\left(\frac{a+\sqrt{a^2-x^2}}{a-\sqrt{a^2-x^2}}\right)+\frac{x}{\sqrt{a^2-x^2}}\\
& = \frac{a}{2}\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find the unknown vectors of a parallelogram and equilateral triangle I am not looking for the answers, could someone help break the questions down into simpler terms. I can find out what the answers are so that is not the goal.
Vectors $p$, $q$ and $r$ are represented on the diagram as shown:
*
*$BCDE$ is a paralle... | (a) First distribute the dot product over addition as in
$$p\cdot(q+r) = p\cdot q + p\cdot r$$
and notice that $p$ and $r$ are perpendicular and so $p\cdot r = 0$. And so we are left with $p\cdot q$. We know that the angle between $p$ and $q$ is $\frac{\pi}{3}$ and that $\|p\|=\|q\|=3$ because we are told $ABE$ is an e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Probability that $7^m+7^n$ is divisible by $5$
If $m,n$ are chosen from the first hundred natural numbers with replacement, the probability that $7^m+7^n$ is divisible by $5$ is?
$$7^m+7^n=7^m(1+7^{n-m}), n\ge m$$
The above expression is divisible by $5$ only if $n-m=4k+2$. The max value of $k$ is $24$.
So is the num... | The possibilities of ending digits of $7^m$ or $7^n$ : $9,3,1,7$
Favourable cases : $(9,1),(3,7),(1,9),(7,3)$
Probability : $\frac{4}{4\times 4} =\frac{1}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1672583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How many integer pairs (x, y) satisfy $x^2 + 4y^2 − 2xy − 2x − 4y − 8 = 0$? How many integer pairs (x, y) satisfy $x^2 + 4y^2 − 2xy − 2x − 4y − 8 = 0$?
My Attempt
Let $f(x,y)=x^2 + 4y^2 − 2xy − 2x − 4y − 8$ . So $f(x,0)=x^2 − 2x − 8$ . $f(x,0)$ has two roots $x=4 , -2$ . So (4,0), (-2,0) are solution of the given ... | HINT:
$$x^2-2x(1+y)+4y^2-4y-8=0$$
$$x=\dfrac{2(1+y)\pm\sqrt{\{2(1+y)\}^2-4(4y^2-4y-8)}}2=1+y\pm\sqrt{9+5y-3y^2}$$
We need
$$0\le9+5y-3y^2\iff3y^2-5y-9\le0$$
Now roots of $3y^2-5y-9=0$ are $\dfrac{5\pm\sqrt{25+108}}2$
Now $(x-a)(x-b)\le0, a\le b\implies a\le x\le b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1672767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Find $m$ so that $(m+4\cdot41)(m^2+4^2\cdot41^2)$ is a square number
Question: Find the minimum positive odd interger $m$ so that $(m+4\cdot41)(m^2+4^2\cdot41^2)$ is a square number.
Any suggestion will be appreciated. Thanks.
| HINT:
Since $m$ is an odd positive integer, so we must have
$$(m+4\cdot 41)(m^2+4^2\cdot 41^2)\equiv 1 \pmod 8$$
$$m^3+4^3\cdot 41^3 + m\cdot 4\cdot 41(m+4\cdot 41) \equiv 1 \pmod 8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove by induction that $n^2 < 2^n$ for all $n \geq 5$? So far I have this:
First consider $n = 5$. In this case $(5)^2 < 2^5$, or $25 < 32$. So the inequality holds for $n = 5$.
Next, suppose that $n^2 < 2^n$ and $n \geq 5$. Now I have to prove that $(n+1)^2 < 2^{(n+1)}$.
So I started with $(n+1)^2 = n^2 + 2n + 1$. Be... | Remark that
$$2^{n+1} = 2 \cdot 2^n > 2 n^2 = n^2 +n^2> n^2+2n+1 = (n+1)^2.$$
Indeed $n^2-2n-1=(n-1)^2-2>0$ for $n \geq 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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What is the maximum of $\ 2(a+b)-ab$ if we have: $\ a^2+b^2=8\ (a,b\ real)$ What is the maximum of $\quad 2(a+b)-ab\ $ if we have: $a^2+b^2=8 \ (a,b\ real) $
My work is as follows:
According to AM-GM inequality:
$$\frac{a^2+b^2}2\ge\sqrt{a^2b^2} \Rightarrow $$
$$ab\le4 $$
On the other hand , from $\ a^2+b^2... | Another, easier method(without trigonometry and differentiation):
$$a^2+b^2 = (a + b)^2 - 2ab = 8 \rightarrow ab=\frac{(a + b)^2}{2} - 4$$
Substitute in your expression. Now you have to find the maximum of the following:
$$2(a + b) - \frac{(a + b)^2}{2} + 4$$Let $t = a + b$. Since $a, b$ are arbitrary reals, then $t$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using the usual notation for a triangle write $\sin^2A$ in terms of the sides a, b and c. This is an A-level trigonometric problem.
Using the usual notation for a triangle write $\sin^2A$ in terms of the sides $a$, $b$ and $c$.
Answer: $$\frac{(a+b-c)(a-b+c)(a+b+c)(-a+b+c)}{4b^2c^2}$$
The last step of the solution requ... | $4a^2b^2-(a^2+b^2)^2+2(a^2+b^2)c^2-c^4=(2ab)^2-((a^2+b^2)^2-2(a^2+b^2)c^2+c^4))=(2ab)^2-(a^2+b^2-c^2)^2$
now you go to last step.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What's wrong with this transformation? I have the equation $\tan(x) = 2\sin(x)$ and I'd like to transform it in this way:
$$\tan(x) = 2\sin(x)
\Longleftrightarrow
\frac{\sin(x)}{\cos(x)} = 2\sin(x)
\Longleftrightarrow
\sin(x) = 2\sin(x)\cos(x)
\Longleftrightarrow
\sin(x) = \sin(2x)$$
But I'm getting a wrong result so I... | Equations $\frac{\sin x}{\cos x} = 2\sin x$ and $\sin x =2\sin x\cos x$ are indeed equivalent since for $\cos x = 0$ you have $\sin x =\pm 1$, which doesn't give solution for the second equation.
So, to solve it, we have that either $\sin x = 0$ or $\cos x = \frac 1 2$, thus solutions are given by $x = 2k\pi$, $x = \pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How to solve the non-homogeneous linear recurrence $ a_{n+1} - a_n = 2n+3 $, given that $a_0=1$? The problem: $ a_{n+1} - a_n = 2n+3 $, given that $a_0=1$
First I solved the associated homogeneous recurrence and got $a_n = A(1)^n = A$, where A is a constant, but I got stuck solving the rest. My final answer was $a_n=2n... | If $ a_{i+1} - a_i = 2i+3 $ then $\sum_{i=0}^{n-1}(a_{i+1} - a_i)=\sum_{i=0}^{n-1}(2i+3)$.
Observe that $\sum_{i=0}^{n-1}(a_{i+1} - a_i)=a_n-a_0$ and $\sum_{i=0}^{n-1}(2i+3)=n(n-1)+3n$.
Hence $a_n=1+n(n-1)+3n=(n+1)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Suppose $n$ divides $3^n + 4^n$. Show that $7$ divides $n$. Suppose $n \geq 2$ and $n$ is a divisor of $3^n + 4^n$. Prove that $7$ is a divisor of $n$.
My work so far:
I had a hypothesis that if $n| 3^n + 4^n$, then $n = 7^k$ for some $k\in\mathbb{N}$. But this is not necessarily so. Take $n = 7⋅379$, where $3^7 + 4^7 ... | First, we note that $3 \nmid 3^n + 4^n$ and $4 \nmid 3^n + 4^n$. So $3 \nmid n$ and $2 \nmid n$.
Now, reworking $3^n + 4^n \equiv 0 \pmod n$. Since n is odd, we find
\begin{equation}
3^n \equiv (-4)^n \pmod n.
\end{equation}
Since $\gcd(3, n) = 1$, we can take the inverse of $3 \mod n$, i. e. there exists $3^{-1}$ so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Methods for Integrating $\int \frac{\cos(x)}{\sin^2(x) +\sin(x)}dx$ So I've found that there's the Weierstrass Substitution that can be used on this problem but I just want to check I can use a normal substitution method to solve the equation:
$$\int \frac{\cos(x)}{\sin^2(x) +\sin (x)}dx$$
Let $u = \sin(x)$
$du = \cos(... | By using Weierstrass Substitution,
let $t = \tan{\frac{x}{2}}, \,dt = \frac{1}{2}\sec^{2}{\frac{x}{2}}\,dx$
$=> 2\cos^{2}{\frac{x}{2}}\,dt = \frac{2}{1+t^{2}}\,dt = dx$
$\because \sin{x} = \frac{2t}{1+t^{2}}, \cos{x} = \frac{1-t^{2}}{1+t^{2}}$
$\therefore \int \frac{\cos{x}}{\sin^{2}{x}+\sin{x}}\,dx = \int \frac{(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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check workings for polynomial function from graph
The diagram shows a curve with equation of the form $y = kx(x + a)^2$,
which passes through $(-2, 0)$, $(0, 0)$ and $(1, 3)$.
What are the values of $a$ and $k$.
To find $k$, I would set $f(x) = 1$.
$f(1) = k(x + 2)(x + 0)(x - 1)$
$3 = k(1 + 2)(x + 1)(1 - 1)$
$3 = k... |
$$y=kx(x+a)^2\quad \quad \quad (-2, 0),(0, 0) , (1, 3)$$
$$\text{for} (0,0):\quad \quad \quad f(0)=0=0\\
\text{for}(1,3):\quad \quad \quad f(1)=k(1+a)^2=3\\
\text{for}(-2,0):\quad \quad \quad f(-2)=-2k(-2+a)^2=0$$
so you have two equations with two variables:
$$\begin{cases}k(1+2a+a^2)=3\\
-2k(4-4a+a^2)=0\end{cases}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find $a+b$ in square $ABCD,$? In square $ABCD,$ $AB=1.$ $BEFA$ and $MNOP$ are congruent. $BE= a - \sqrt b.$ Where $a$ and $b$ are both primes. How to find $a+b$? I have no idea how to do this, can this be proved with simple geometry?
|
Let the side of the square be $p$ (so we can better track dimensions) and let $|\overline{BE}| = q$; and let $\theta = \angle EMP$. We have two equations:
$$\begin{align}
q + p \sin\theta + q \cos\theta &= p \\
p \cos\theta + q \sin\theta &= p
\end{align}$$
Solving for $\cos\theta$ and $\sin\theta$ gives
$$\cos\theta ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Fundamental Theorem of Calculus Confusion regarding atan According to this site,
$$ \int \frac{1}{a^2 \cos^2(x) + b^2 \sin^2(x)} \,dx =\frac{1}{ab} \arctan\left(\frac{b}{a} \tan(x)\right)$$
Thus,
$$ \int_0^{\pi} \frac{1}{a^2 \cos^2(x) + b^2 \sin^2(x)} \, dx =\frac{1}{ab} \arctan\left(\frac{b}{a} \tan(\pi)\right)-\frac{... | This very integral, with a little manipulation, is referenced in the Jeffrey paper so the nice form in that paper can be attained by observing that $\tan^{-1}(\tan x)$ has the same discontinuities as the original integral, so we may hope to cancel them out by setting
$$\frac1{ab}\tan^{-1}\left(\frac{b}a\tan x\right)=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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show this inequality with $a+b+c+d=1$ Let $a,b,c,d\ge 0$,and such $a+b+c+d=1$, show that
$$3(a^2+b^2+c^2+d^2)+64abcd\ge 1$$
use AM-GM
$$a^2+b^2+c^2+d^2\ge 4\sqrt{abcd}$$
it suffices to
$$4\sqrt{abcd}+64abcd\ge 1$$
| the inequality equals:
$3(a+b+c+d)^2(a^2+b^2+c^2+d^2)+64abcd-(a+b+c+d)^4\ge 0 $
WLOG, let $d=min(a,b,c,d)$
$3(a+b+c+d)^2(a^2+b^2+c^2+d^2)+64abcd-(a+b+c+d)^4=2d\left(\dfrac{1}{3}\sum_{cyc (a,b,c)}(a-d)^2(2a+d)+\sum_{cyc (a,b,c)} a(a-d)^2+(a+b+c-3d)((a-b)^2+(b-c)^2+(a-c)^2)+3(a^3+b^3+c^3+3abc-\sum ab(a+b))+\dfrac{a^3+b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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calculate $\sum_{1\le i\le r}\frac{i+1} { r+1}{2r-i\choose r-i}{s+i-2\choose i}+\frac{1}{r+1}{2r\choose r}$? prove the following equation :
$$\sum_{1\le i\le r}\frac{i+1} { r+1}{2r-i\choose r-i}{s+i-2\choose i}+\frac{1}{r+1}{2r\choose r}={s+2r-1\choose r} - {s+2r-1\choose r-1}$$
| I use two known properties of binomial coefficients: $${n+k \choose k} = (-1)^k {-n-1 \choose k}\tag{$i$}$$ and $$\sum_{i}{a \choose i} {b\choose c-i} = {a+b \choose c}\tag{$ii$}.$$
First,
$$
\sum_{0\le i\le r}\frac{r-i} {r+1}{2r-i\choose r-i}{s+i-2\choose i} \overset{(i)}{=}
(-1)^r \sum_{0\le i\le r-1}\frac{r-i} {r+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1699118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find a formula for $f^{(n)}(x)$ where $f(x) = \dfrac{1}{x^2-a^2}$
Let $a$ be a constant. Find a formula for $f^{(n)}(x)$ where $f(x) = \dfrac{1}{x^2-a^2}$.
After computing a few derivatives, the derivatives seem to have factorials in them sometimes and other times not. For example, $\dfrac{d^5}{dx^5} \left (\dfrac{1... | Just observe that
$$f(x)=\frac{1}{2a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right)$$
Then
$$f^{(n)}(x)=\frac{(-1)^nn!}{2a}\left[\frac{1}{(x-a)^{n+1}}-\frac{1}{(x+a)^{n+1}}\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1700939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve $ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $
Question: Solve $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 0 \le x \le 360^{\circ} $$
My attempt:
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6(\frac{1}{2} - \frac{\cos(2x)}{2}) + \sin(x)\cos(x) -(\frac{1}{2} + \frac{\cos(2x)}{2}) = 5 $$
$$ 3 - 3\cos(2x)+ \s... | $$ 6 s^2 + s c -c^2 =5 $$
Divide by c^2 . $ t=s/c,$
$$ 6 t^2 + t - 1 = 5/c^2 = 5 ( 1 +t^2) \rightarrow ;\; t^2 + t -6 =0;\; (t-2)(t+3) =0$$
$$ t= \tan x = 2,-3 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 7
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Cubic Equation Related How do I solve the following cubic equation?
$(x+1)^2(x-2)-2nx-6n=0,n \in \mathbb{N}$
$ \therefore (x+1)^2(x-2)-2n(x+3)=0$
I don't know how to solve further.
| We have
$$(x+1)^2(x-2)-2nx-6n=0,$$
i.e.
$$x^3+(-3-2n)x-2-6n=0\tag1$$
Let $f(x)=x^3+(-3-2n)x-2-6n$. Then,
$$f'(x)=3x^2-3-2n=0\iff x=\pm\sqrt{\frac{3+2n}{3}}.$$
We have
$$\begin{align}f\left(-\sqrt{\frac{3+2n}{3}}\right)\lt 0&\iff (2n+3)\sqrt{\frac{2n+3}{3}}\lt 9n+3\\&\iff (2n+3)^2\cdot\frac{2n+3}{3}\lt (9n+3)^2\\&\iff 8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove $ \frac{\sin^3(x)-\cos^3(x)}{\sin(x)+\cos(x)} = \frac{\csc^2(x) -\cot(x) -2\cos^2(x)}{1-\cot^2(x)} $
Question: Prove $$ \frac{\sin^3(x)-\cos^3(x)}{\sin(x)+\cos(x)} = \frac{\csc^2(x) -\cot(x) -2\cos^2(x)}{1-\cot^2(x)} $$
RHS: $$ \frac{\csc^2(x) -\cot(x) -2\cos^2(x)}{1-\cot^2(x)} $$
$$ ⇔ \frac{\frac{1}{\sin^2(x... | You are correct except one sign :
$$\frac{1\color{red}{-}\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{(\sin(x)-\cos(x))(\sin(x)+\cos(x))}$$
(The error starts at the very beginning.)
Now using $1=\cos^2(x)+\sin^2(x)$,
$$\begin{align}&1-\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)\\&=(1-2\cos(x)\sin(x))(1+\cos(x)\sin(x))\\&=(\cos^2(x)+\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can't solve exponential equation using logs? I can't figure out why my method isn't working.
I know it is possible to solve this using a substitution but I don't know when to use the substitution. In general when are you supposed to substitute for, say, u?
Here is how I did it;
We have the following function:
$3^{x-1}+... | Here after your first line of taking the derivatives, set $u = x-1$
$3^{x-1} - 3^{-x+1} = \frac{8}{3}$
$3^{u} - 3^{-u} = \frac{8}{3}$
Furthermore set $v = 3^u$
$v - \frac{1}{v} = \frac{8}{3}$
$\frac{v^2 - 1}{v} = \frac{8}{3}$
$v^2 - 1= \frac{8v}{3}$
$v^2 - \frac{8v}{3} - 1=0$
$\Delta = \sqrt{ \frac{64}{9} + 4 }$
$v_{1,... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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Geometric inequality involving the inradius For the $\triangle ABC$, let $T$ be the area of the triangle, $a,b,c$ its sides, $p$ the semiperimeter and $r$ the inradius. Prove the following inequality:
$$p^2\ge 2\sqrt3 T+\frac {abc}{p}+r^2.$$
| Using the known formula $pr=T$, we rewrite the inequality as follows
$$ p^2 - \frac{abc}{p} - \frac{T^2}{p^2} \ge 2 \sqrt{3} T$$
Now, using Heron's formula,
$$ \begin{align*}
\mathrm{LHS} &= \frac{1}{4} (a+b+c)^2 - \frac{2abc}{a+b+c} - \frac{(a+b-c)(a-b+c)(-a+b+c)}{4(a+b+c)} \\
&=\frac{1}{4} (a+b+c)^2 - \frac{8abc+(a+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1703460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $a$ and $b$ be the coefficient of $x^3$ in $(1+x+2x^2+3x^3)^4$ and $(1+x+2x^2+3x^3+4x^4)^4$ respectively. Let $a$ and $b$ be the coefficient of $x^3$ in $(1+x+2x^2+3x^3)^4$ and $(1+x+2x^2+3x^3+4x^4)^4$ respectively.Find $(a-b).$
I tried to factorize $(1+x+2x^2+3x^3)$ and $(1+x+2x^2+3x^3+4x^4)$ into product of two ... | The problem is a way of making the point that changing the terms of degree $k$ and higher in the argument can only change the degree $k$-or-higher terms in the result, when applying polynomial or power-series operations (such as taking the 4th power of the argument). If you are performing a calculation to get the ans... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How many solutions in integers to the following equation
What is the number of positive integer solutions $(a, b)$ to $2016 + a^2 = b^2$?
We have,
$2016 = (b-a)(b+a) = 2^5 \cdot 3^2 \cdot 7$
$b - a = 2^{t_1} 3^{t_2} 7^{t_4}$ and
$a - b = 2^{t_5} 3^{t_6} 7^{t_7}$
Thus, we must have $t_1 + t_5 = 5$ and $t_2 + t_6 = 2$ ... | First of all, $a$ and $b$ must be integers. Since
$$
a = \frac{(b+a) - (b-a)}{2},\quad b = \frac{(b+a) + (b-a)}{2}
$$
we see that for $a$ and $b$ to be integers, $(b-a)$ and $(b+a)$ must have the same parity. Since at least one of them must be even (there are five powers of $2$ to distribute between them), then both mu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$ Here is the expression:
$$2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$$
The exercise is to evaluate it.
In my text book the answer is $0$
I tried to factor the expression, but it got me nowhere.
| $$
\begin{align}
1
&=\left(\sin^2(x)+\cos^2(x)\right)^3\\
&=\sin^6(x)+3\sin^4(x)\cos^2(x)+3\sin^2(x)\cos^4(x)+\cos^6(x)\\
&=\sin^6(x)+3\sin^2(x)\cos^2(x)+\cos^6(x)\tag{1}\\\\
1
&=\left(\sin^2(x)+\cos^2(x)\right)^2\\
&=\sin^4(x)+2\sin^2(x)\cos^2(x)+\cos^4(x)\tag{2}
\end{align}
$$
Subtracting $3$ times $(2)$ from $2$ tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Prove that there is $x \in [0,1]$ such that $|f''(x)| > 4$
Assume $f$ has a continuous second derivative with $f(0) = f'(0) = f'(1) = 0$ and $f(1) = 1$. Prove that there is $x \in [0,1]$ such that $|f''(x)| > 4$.
We must also have that $f'$ is continuous by differentiability. Therefore, by Rolles theorem there exist... | $$f(\frac{1}{2}) = f(0) + f'(0)(\frac{1}{2}) + \frac{f''(\theta_1)}{2}(\frac{1}{2})^2 = f''(\theta_1)\frac{1}{8}$$
$$f(\frac{1}{2}) = f(1) - f'(1)(\frac{1}{2}) + \frac{f''(\theta_2)}{2}(\frac{1}{2})^2 = 1 + f''(\theta_2)\frac{1}{8}$$
Here, $0\le \theta_1 \le 1/2$, and $1/2 \le \theta_2 \le 1$
Combining both, we have
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
differentiation problem If $x^{13}y^{7}=(x+y)^{20}$ , then $\frac{dy}{dx}$
directly doing it makes it very complicated so, I did this $\left(\frac{x}{y}\right)^{13}=\left(1+\frac{x}{y} \right)^{20}$.
following are the options for solution
(a) $\frac{y^2}{x^2}$ (b)$\frac{x^2}{y^2}$ (c)$\frac{x}{y}$ (d)$\frac{y}{x}$
th... | I differentiated directly,
$$13x^{12}y^7+7x^{13}y^6y'=20(x+y)^{19}(1+y')$$
$${13x^{13}y^7\over {x}}+{7x^{13}y^7y'\over y}={20(x+y)^{20}(1+y')\over {(x+y)}}$$
$$(x+y)(13y+7xy')=20xy(1+y')$$
Cancelling $x^{13}y^7$ and rearranging we get,
$$y'={13y^2-7xy\over {13xy-7x^2}}$$
Divide the numerator and the dinominator by$x^2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1711839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Compute $e^{-x^2} * e^{-x^2}$ How to compute the convolution of $e^{-x^2}$ with itself?
$$e^{-x^2} * e^{-x^2} = \int_{\mathbb R} e^{-(x-y)^2} e^{-y^2}dy = e^{-x^2}\int_{-\infty}^{\infty} e^{2xy - 2y^2} dy$$
I can't solve it. I tried integration by parts so neglecting $e^{-x^2}$:
$$2\int_{-\infty}^{\infty}(2y^2 - xy) e^... | $$ \left( y - \frac{x}{2} \right)^2 = y^2 - xy + \frac{x^2}{4} $$
$$ \left( y - \frac{x}{2} \right)^2 - \frac{x^2}{4} = y^2 - xy $$
$$ 2 \left( y - \frac{x}{2} \right)^2 - \frac{x^2}{2} = 2 y^2 - 2 xy $$
$$ \frac{x^2}{2}-2 \left( y - \frac{x}{2} \right)^2 = 2 xy- 2 y^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Inequality $(x-1)(y-1)(z-1)\geq 8$ provided that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$ How can I prove that if $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1,$$ then $(x-1)(y-1)(z-1) \geq 8$?
Edit:
$x,y,z \in \mathbb R_{>0} $
Thanks
| We have $x\left(\frac1x+\frac1y+\frac1z\right)=x\implies x-1=\frac xy+\frac xz\overset{AM-GM}{≥}\frac{2x}{\sqrt{yz}}$ and similarly for $y$ and $z$. So we have:
$$
(x-1)(y-1)(z-1)\geq \frac{2x}{\sqrt{yz}}\frac{2y}{\sqrt{zx}}\frac{2z}{\sqrt{xy}}=8
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim_{x\to0}\frac{(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}-e^2}{x^2}$ Find $\lim_{x\to0}\frac{(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}-e^2}{x^2}$
My attempt:
$\lim_{x\to0}(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}=e^2$
So the required limit is in $\frac{0}{0}$ form.
Then i used L hospital form.
$\lim_{x\to0}\frac{e^2(\frac... | Note that $$\tan\left(\frac{\pi}{4} + x\right) = \frac{1 + \tan x}{1 - \tan x}$$ and therefore
\begin{align}
L &= \lim_{x \to 0}\dfrac{\left(\tan\left(\dfrac{\pi}{4} + x\right)\right)^{1/x} - e^{2}}{x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\left(\dfrac{1 + \tan x}{1 - \tan x}\right)^{1/x} - e^{2}}{x^{2}}\notag\\
&= \lim_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Does the method of substitution always work for solving linear congruence systems? By substitution, I mean if I had this example:
$$ x \equiv 1 \pmod{5}$$ $$x \equiv 2 \pmod{6}$$
Then I would solve it by solving this equation...
$$ x = 1 + 5k \equiv 2 \pmod{6}$$
$$k \equiv 4 \pmod{6}$$
From this, I can substitute k b... | DEFINITION $1.$ Let's say that the system of linear congruences
\begin{align}
x &\equiv a \pmod A \\
x &\equiv b \pmod B
\end{align}
is consistent if $a \equiv b \mod{\gcd(A,B)}$.
THEOREM $2.$ The system of linear congruences
\begin{align}
x &\equiv a \pmod A \\
x &\equiv b \pmod B
\end{align}
has ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1717302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that a positive integer is composite Let $ n $ be a positive integer that can be written as a sum of two relatively prime squares in two distinct ways, that is $ n = a^2 + b^2 = c^2 + d^2 $ so that $ gcd(a, b) = gcd(c, d) = 1, $ then $ n $ is composite.
So I have successfully factored $ n $ into $ \displaystyle \... | What you ask is equivalent to showing that if $n$ is not composite then $n$ is the sum of two co-prime squares in just one way, or in no way.
The cases $n=1$ is trivial.
If $n$ is prime, suppose $n=a^2+b^2=c^2+d^2.$ with each of $a,b,c,d$ positive and less than $n.$ We will take all congruences modulo $n.$
We have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1717885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Divergent sum of factorials Is it possible to get an exact value of the sum (using divergent series summation methods)
$$ \sum_{n=0}^\infty~ \frac{(n+k)!}{n!} \quad?$$
where $k$ is a positive integer.
The only other divergent sum of factorials I have seen is $\sum_{n=0}^\infty(-1)^nn!$.
Does anyone know any useful tech... | We have
$$
\sum_{n=0}^\infty \frac{(n+k)!}{n!}x^n=\frac{d^k}{dx^k}\sum_{n=0}^\infty x^{n+k}=\frac{d^k}{dx^k}\frac{x^k}{1-x}=\frac{k!}{(1-x)^{k+1}}
$$
for $|x|<1$ and may use the elementary Ramanujan summation (definition) (or simply linearity) and obtain your result, which corresponds to $k!(\sum_{n=0}^\infty 1)^{k+1}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integration Using Inverse Trig Functions
$$\int \frac{1}{(36-4x^2)^{0.5}} dx$$
You can immediately see that,
$$6^2 = 36$$
And,
$$(2x)^2 = 4x^2$$
Using the integration technique as follows:
$$\int \frac{1}{(a^2-x^2)^{0.5}} dx = \arcsin(\frac{x}{a})+C$$
We get,
$$\int \frac{1}{(36-4x^2)^{0.5}} dx = \int \frac{1}{(6^2-(... | Here's the issue. The general form is $$\int\frac{1}{\sqrt{a^2-x^2}}\,dx=\arcsin\left(\frac{x}{a}\right)+C,$$
but your integral is
$$\int\frac{1}{\sqrt{6^2-(2x)^2}}\,dx.$$
The $2x$ is different than $x$. The difference can be eliminated by making the substitution $2x=u$, so $dx=du/2$, and we get
$$\frac{1}{2}\int\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1721143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Does the sequence $\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}$ converge? I'm trying to determine if this sequence converges as part of answering whether it's monotonic:
$$
\left\{\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}\right\}
$$
First, I tried expanding it a bit to see if I could remove common factors i... | As far as monotonicity is concerned, write the sequence recursively:
\begin{align}
a_1 =&\ 1 \\
a_{n} =&\ a_1\frac{n}{2n + 1}
\end{align}
Then look at the difference between $a_{n + 1}$ and $a_{n}$:
\begin{align}
a_{n + 1} - a_{n} =&\ a_n\frac{n+1}{2(n + 1) + 1} - a_n\\
=&\ a_n\left(\frac{n+1}{2(n + 1) + 1} - 1\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
} |
$P$ is a point on ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(a>b)$ and $S$ and $S'$ are its focii If $P$ is a point on ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(a>b)$ and $S$ and $S'$ are its focii.
$\angle PSS'=\alpha$ and $\angle PS'S=\beta$, then prove that:
$$
\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac... | Another answer is, let the angle $\theta=\angle POS'$, the following identity is straightforward:
$$
\tan\left(\frac{180-\beta}{2}\right) = \sqrt{\frac{1+e}{1-e}} \tan\left(\frac{\theta}{2}\right)
$$
and
$$
\tan\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1-e}{1+e}} \tan\left(\frac{\theta}{2}\right)
$$
Hence,
$$
\tan\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Parametrize a curve, Multivariable Calculus I am stuck on what seems to be an easy exercise.
We have $f(x,y) = x^2 + 4xy + y^2 \mbox{ for all } (x,y)$ in $\mathbb{R}^2.$
Now we are supposed to find a parametrization of the intersection curve between $f(x,y)$ and $z = x + 3y.$
I've been stuck for hours now, anyone got... | Setting 'z= z' gives $x^2+ 4xy+ y^2= x+ 3y$. The first thing I would do is "complete the square" on the left: $x^2+ 4xy+ 4y^2- 3y^2= (x+ 2y)- 3y^2= x+ 3y$. Now, let u= x+ y and v= y then x+ 3y= u- v+ 3v= u+ 2v so the equation becomes $u^2- 3v^2= u+ 2v$ or $u^2- u- 3(v^2+ (2/3)v)= 0$ and complete the square again: $u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Which rules are used to make function like one in Laplace Transformations table? I have function like this:
$$\frac{s^2+3s+3}{(2s^2+7s+7)} $$
It needs to be brought to the level of Laplace Transformations from table, like these two:
$$\frac{a}{(s-b)^2 + a^2} $$
$$\frac{s-b}{(s-b)^2 + a^2} $$
I have the solution for th... | $$\begin{align}
\frac{s^2+3s+3}{(2s^2+7s+7)}&=\frac12 \frac{2s^2+6s+6}{(2s^2+7s+7)}\\\\
&=\frac12 \frac{(2s^2+7s+7)-(s+1)}{(2s^2+7s+7)}\\\\
&=\frac12 -\frac12\frac{s+1}{2s^2+7s+7}\\\\
&=\frac12-\frac14\frac{s+1}{s^2+(7/2)s+(7/2)}\\\\
&=\frac12-\frac14\frac{s+1}{(s+7/4)^2+(7/2)-(49/16)}\\\\
& =\frac12-\frac14\frac{s+(7/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluation of the integral $\int{\frac{x+\sin{x}}{1+\cos{x}}\mathrm{d}x}$ by parts I have to evaluate the following integral by parts: $$\int {\dfrac{x+\sin{x}}{1+\cos{x}}}\mathrm{d}x $$
So I tried to put:
$ u = x + \sin{x}$ $~\qquad\rightarrow \quad$ $\mathrm{d}u=\left(1+\cos{x}\right) \mathrm{d}x$
$\mathrm{d}v = \dfr... | $\int \dfrac{x + \sin x}{1+\cos x}dx\\
\int \dfrac{x + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2\cos^2 \frac{x}{2}} dx\\
\int \frac{1}{2}x\sec^2\frac{x}{2} + \tan \frac{x}{2} dx\\
\int \frac{1}{2}x\sec^2\frac{x}{2}dx + \int\tan \frac{x}{2} dx$
Now we do integration by parts on the $1^{st}$ integral. we won't evaluate the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1734036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Probability of selecting specific balls from an urn? An urn contains seven red balls, seven white balls, and seven blue balls. A sample of 5 balls is drawn at a random without replacement. What is the probability that the sample contains three balls of one color and two of another?
Here was what I tried doing.
I said... | Your denominator is correct. For the numerator, we must select one of the three colors from which three balls will be selected, select three of the seven balls of that color, select one of the two remaining colors from which two balls will be selected, and select two balls of that color. Hence, the probability of sel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1734131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
$a,b,c$ are the sides and $A,B,C$ are the angles of a triangle. If the roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are equal then,
$a,b,c$ are the sides of a $\triangle ABC$ and $A,B,C$ are the respective angles. If the roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are equal then $\sin^2 \bigl(\frac{A}{2}\... | This is not an answer but a hint and comment about your assertion you have already got that a,b,c are in H.P. Here a possible way to prove what you want with the squares of the sinus of middle angles.
►Considering the incenter you have
$$\sin^2 (\frac{A}{2})= \frac {r^2}{d_1^2}\left(= \frac{1}{\frac{d_1^2}{r^2}}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1734726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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$x_{k+1}=(x_k+c/x_k)/2$. Prove that {$x_k$} converges and find its limit. Fix any $c>0$.
Let $x_1$ be any positive number and define $x_{k+1}=(x_k+c/x_k)/2$.
a)Prove that {$x_k$} converges and find its limit.
b)Use this sequence to calculate $\sqrt5$, accurate to six decimal places.
I tried to solve $x_{k+1}=(x_k+c/x_k... | Notice for all $k \ge 1$,
$$\frac{x_{k+1}-\sqrt{c}}{x_{k+1}+\sqrt{c}}
= \frac{\frac12(x_k + \frac{c}{x_k}) - \sqrt{c}}{\frac12(x_k+\frac{c}{x_k}) + \sqrt{c}}
= \left(\frac{x_k-\sqrt{c}}{x_k+\sqrt{c}}\right)^2
$$
Using induction, it is easy to see for all $n \ge 1$, we have
$$\frac{x_n - \sqrt{c}}{x_n + \sqrt{c}} = \alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1735893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Laurent expansion Question is to write Laurent expansion of $f(z)=\dfrac{1}{(z-2)(z-1)}$ in the annulus $1<|z|<2$ based at $z=0$
I am aware of the method of partial fractions and writing expansions for both $\dfrac{1}{z-1}$ and $\dfrac{1}{z-2}$..
I am rying to do this using Laurent expansion formula..
We have $$f(z)=\s... | I think i got it...
For $1<r<2$, we have to compute $$a_n=\frac{1}{2\pi i}\int_{|z|=r}\frac{1}{(z-2)(z-1)z^{n+1}}dz$$
For $n>0$ we have $$a_{-n}=\frac{1}{2\pi i}\int_{|z|=r}\frac{z^{n-1}}{(z-2)(z-1)}dz
=\frac{1}{2\pi i}\int_{|z|=r}\frac{z^{n-1}}{(z-2)}-\frac{z^{n-1}}{(z-1)}dz$$
As $r<2$, the function $\frac{z^{n-1}}{(z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1736445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find all integer solutions to $\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$
Find all integer solutions $(x, y)$ of the equation
$$\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$$
What have done is that:
$$\frac{1}{x}= \frac{2y-3}{3y}$$
so,
$$x=\frac{3y}{2y-3}$$
If $2y-3 = +1 \text{ or } {-1}$, $x$ will be an integer, so we... | Your approach can be continued to a full solution. Let's see it in detail:
Given integers $x,y$ such that $\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$:
At least one of $x,y$ is positive, and by symmetry we can assume that $y$ is positive.
$\frac{1}{x} = \frac{2y-3}{3y}$.
Thus $x = \frac{3y}{2y-3} = 1 + \frac{y+3}{2y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the sum of $-1^2-2^2+3^2+4^2-5^2-6^2+\cdots$
Find the sum of $$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2$$
By expanding the given summation,
$$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2=-1^2-2^2+3^2+4^2-5^2-6^2+\cdots+(4n-1)^2+(4n)^2$$
$$=(3^2-1^2)+(4^2-2^2)+(7^2-5^2)+(8^2-6^2)+\cdots+[(4n-1)^2-(4n-3)^2]+((4n)^2-... | In the last line what you have enclosed into rectangular braces, namely $S$, can be rearranged as follows
\begin{align*}
S&=(1+3)+(2+4)+(5+7)+(6+8)+(9+11)+(10+12)+\cdots+[(4n-3)+(4n-1)]+[(4n-2)+4n]\\
&=\sum_{j=1}^{4n}j\\[3pt]
&=\frac{4n(4n+1)}{2}\\[3pt]
&=\boxed{\color{blue}{2n(4n+1)}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
Checking the boundedness of $A_n = \frac{x}{1-x^2} + \frac{x^2}{1-x^4} + \ldots + \frac{x^{2^n}}{1-x^{2^{n+1}}}$ Sequence is given by $$A_n = \frac{x}{1-x^2} + \frac{x^2}{1-x^4} + \ldots + \frac{x^{2^n}}{1-x^{2^{n+1}}}.$$ Please advise me how to show that this sequence is bounded above by 1. 0
| \begin{align*}
\frac{x^{2^{n}}}{1-x^{2^{n+1}}}+\color{red}{\frac{1}{1-x^{2^{n+1}}}} &=
\frac{1}{1-x^{2^{n}}} \\
\frac{x^{2^{n-1}}}{1-x^{2^{n}}}+\frac{1}{1-x^{2^{n}}} &=
\frac{1}{1-x^{2^{n-1}}} \\
& \: \; \vdots \\
\frac{x}{1-x^{2}}+\frac{1}{1-x^{2}} &=
\color{blue}{\frac{1}{1-x}} \\
A_{n} &= \color{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Normal closure of $\mathbb{Q}(\sqrt{11+3\sqrt{13}})$ over $\mathbb{Q}$ The following is a question from an undergrad course in Galois theory:
Find a normal closure $L$ of $K=\mathbb{Q}(\sqrt{11+3\sqrt{13}})$ over $\mathbb{Q}$
I know that normal extensions are splitting fields
Let: $X=\sqrt{11+3\sqrt{13}} \implies X^2... | You need to compute the zeros of $X^4-22 X^2+4=9\cdot ((\frac{X^2-11}3)^2-13)$. There is a general formula for equations of degree four which gives you
\begin{align*}
\sqrt{11+3\sqrt{13}} && -\sqrt{11+3\sqrt{13}} && \sqrt{11-3\sqrt{13}} && -\sqrt{11+3\sqrt{13}}
\end{align*}
So the normal closure of $K$ is $\Bbb Q(\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Convergence of a compound sequence let $a_n=\frac{1}{2}\sqrt{n}+\sum_{k=1}^n(\sqrt{k}-\sqrt{k+\frac{1}{2}})$ be a sequence.
Is this sequence convergent?
| Write $\sqrt{n} = \sum_{k=1}^{n} \left(\sqrt{k}-\sqrt{k-1}\right)$. Then you get:
$$a_n = \sum_{k=1}^{n}\left(\sqrt{k}-\sqrt{k+1/2}+\frac{1}{2}\sqrt{k}-\frac{1}{2}\sqrt{k-1}\right)=\sum_{k=1}^n \left(\frac 3 2\sqrt{k}-\sqrt{k+1/2}-\frac{1}{2}\sqrt{k-1}\right)$$
This represents $a_n$ as a pure series, with $a_n=\sum_{k=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve by Mathematical Induction: $2^n+1 \leq 3^n$ I'm very confused on how to prove this.
By using Mathematical Induction prove that, for all positive integers $n$ the
following inequality holds:
$$2^n + 1 ≤ 3^n$$
| For $n = 1$,
$$3^1 = 3 \ge 2^1 + 1$$
Assume that $2^k + 1 \le 3^k$ for some $k\in \mathbb N$.
For $n = k+1$,
$$\begin{align*}
3^{k+1} &= 3\cdot 3^k\\
&\ge 3\cdot(2^k+1)\\
&= 3\cdot 2^k + 3\\
&= 2\cdot2^k + 1 + 2^k + 2\\
&\ge 2^{k+1} + 1
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculating exponential limit I've been breaking my mind over this one.
Find the limit.
$\lim\limits_{n \to \infty} (\frac{n^2+3}{n^2+5 n-4})^{2n} $
I know it equals $\frac{1}{e^{10}} $ but can't figure out how to find it.
Help?
| Remember that $$ \left(\frac{n^2+3}{n^2+5 n-4}\right)^{2n}=\exp\left(2n \ln\left(\frac{n^2+3}{n^2+5 n-4}\right)\right).$$ To compute the limit of the inside, use L'Hospital's rule :
\begin{align*}
\lim_{n \to +\infty} 2n \ln\left(\frac{n^2+3}{n^2+5 n-4}\right)& =\lim_{n\to +\infty}\frac{\ln\left(\frac{n^2+3}{n^2+5 n-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How many ordered pairs $(x,y)$ are there such that ${1\over\sqrt x}+{1\over\sqrt y}={1\over\sqrt {20}}$, where both $x$ and $y$ are positive integers $${1\over \sqrt x}+{1\over \sqrt y}={1\over \sqrt {20}}$$
I could find one ordered pair that satisfies above equation, that is $(80,80)$. But the answer says that there a... | We have the following equation:
$$\frac{1}{\sqrt x}+\frac{1}{\sqrt y}=\frac{1}{\sqrt {20}}$$
$\sqrt{20}=2\sqrt{5}$:
$$\frac{1}{\sqrt x}+\frac{1}{\sqrt y}=\frac{1}{2\sqrt 5}$$
Multiply both sides by $\sqrt 5$:
$$\frac 1 {\sqrt{\frac x 5}}+\frac 1 {\sqrt{\frac y 5}}=\frac 1 2$$
Basically, we want answers to $\frac 1 a+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Study the convergence of the following series $\sum\limits_{n=1}^\infty \frac{n}{n^2+3} \sin(\frac{1}{\sqrt{n+2}}) $ I have to study the convergence of the following series:
$\sum\limits_{n=1}^\infty \frac{n}{n^2+3} \sin(\frac{1}{\sqrt{n+2}}) $
Is a positive series, so I should divide for $\frac{1}{\sqrt{n+2}}$
and th... | With equivalents, as it's a series with positive terms:
\begin{align*}\frac n{n^2+3}&\sim_\infty\frac1n,\quad \sin\frac1{\sqrt{n+2}}\sim_\infty\frac1{\sqrt{n+2}}\sim_\infty\frac1{\sqrt n}, \\\text{hence}\quad\frac n{n^2+3}\sin\frac1{\sqrt{n+2}}&\sim_\infty\frac 1{n^{3/2}},\quad\text{which converges.}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Whats the probability I roll a 2 on one of the dice if the sum rolled is 8? I have two dice and I want to find the probability of this situation:
Whats the probability I roll a 2 on one of the dice if the sum rolled is 8 ?
Is the answer $\frac{2}{36}$?
| Consider it intuitively
We can draw a table to show this:
$$\begin{array}{|cc|cccccc|}\hline
&&&&&D_1\\
&&1&2&3&4&5&6\\
\hline
&1&2&3&4&5&6&7\\
&2&3&4&5&6&7&\color{red}8\\
D_2&3&4&5&6&7&\color{red}8&9\\
&4&5&6&7&\color{red}8&9&10\\
&5&6&7&\color{red}8&9&10&11\\
&6&7&\color{blue}8&9&10&11&12\\
\hline\end{array}$$
The nu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find a functional equation for the generating function whose coefficients satisfy the relation Find a functional equation for the generating function whose coefficients satisfy the relation:
$\qquad{}$ $a_n = 3a_{n-1} -2a_{n-2}+2, a_0=a_1=1$
When I solve this, I get the function $g(x)-1-x=3xg(x)-2x^2g(x)$
However, the ... | Mostly you forgot to take into account the $+2$ term in the recurrence, though there is one other problem. For $n\ge2$ you have
$$a_nx^n=3a_{n-1}x^n-2a_{n-2}x^n+2x^n\;,$$
so when you sum over $n\ge 2$ you get
$$\sum_{n\ge 2}a_nx^n=3\sum_{n\ge 2}a_{n-1}x^n-2\sum_{n\ge 2}a_{n-2}x^n+2\sum_{n\ge 2}x^n\;.$$
The lefthand sid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Write $4\cos x - 3 \sin x$ in the form $k\sin(x - \alpha)$ where $0 \le \alpha \le 360$
Write $4\cos x - 3 \sin x$ in the form $k\sin(x - \alpha)$ where $0 \le \alpha \le 360$.
$4\cos x - 3 \sin x = k\sin(x - \alpha)$
=> $k(\sin x \cos \alpha - \cos x \sin \alpha)$
=> $k\cos \alpha \sin x - \sin \alpha \cos x$
Equati... | HINT:
$k\sin\alpha=-4,k\cos\alpha=-3\implies\alpha$ is in the third quadrant
See this
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Connection between quadratic residue of a number to its factors' Is it true that, If $N$ is product of two coprime numbers greater than 1. Quadratic residues of these numbers are quadratic residue of $N$ and vice versa? Can someone point me to a proof or show me if this is not the case.
In other words, if we start wit... | We state a precise version of the result you are after.
Let $m$ and $n$ be relatively prime. Then $a$ is a quadratic residue of $mn$ if and only if $a$ is a quadratic residue of $m$ and $a$ is a quadratic residue of $n$.
One direction is easy. Suppose that $a$ is a quadratic residue of $mn$. Then there exists a integer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1758711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find out all solutions of the congruence $x^2 \equiv 9 \mod 256$. I need to find all the solutions of the congruence $x^2 \equiv 9 \mod 256$.
I tried (apparently naively) to do this:
$x^2 \equiv 9 \mod 256$
$\Leftrightarrow$ $x^2 -9 \equiv 0 \mod 256$ $\Leftrightarrow$ $256 | (x-3)(x+3)$ $\Leftrightarrow$
$25... | If $(x-3)(x+3)\equiv0\pmod{2^{m+2}}$ for integer $m\ge1$
both $(x-3),(x+3)$ must be even
$\implies\dfrac{x-3}2\cdot\dfrac{x+3}2\equiv0\pmod{2^m}$
Now $\dfrac{x-3}2-\dfrac{x+3}2$ is odd $\implies$ they have opposite parity
If $\dfrac{x-3}2$ is even, $\dfrac{x-3}2\equiv0\pmod{2^m},x\equiv3\pmod{2^{m+1}}$
Can you take it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
How do I go about solving this? I have tried substitution, but it is not working for me.
$$
\int_0^\pi \frac{dx}{\sqrt{(n^2+1)}+\sin(x)+n\cos(x)}=\int_0^\pi \frac{n dx}{\sqrt{(n^2+1)}+n\sin(x)+\cos(x)}=2
$$
General form of this integral is
$$
\int_0^\pi \frac{dx}{\sqrt{(n^2+m^2)}+m\sin(x)+n\cos(x)}=\frac{2}{m}
$$
| First of all:
$$
(\sqrt{n^2+1}+\sin x+n\cos x)(\sqrt{n^2+1}-\sin x-n\cos x)=$$
$$
=n^2+1-n^2\cos^2x-2n\cos x\sin x-\sin^2 x=
$$
$$
=\cos^2x-2n\cos x\sin x+n^2\sin^2 x=(n\sin x-\cos x)^2
$$
So:
$$
\int\frac{\mathrm{d}x}{\sqrt{n^2+1}+\sin x+n\cos x}=\int\frac{(\sqrt{n^2+1}-\sin x-n\cos x)\mathrm{d}x}{(n\sin x-\cos x)^2}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Sum of the series $\frac{1}{(1-x)(1-x^3)}+\frac{x^2}{(1-x^3)(1-x^5)}+\frac{x^4}{(1-x^5)(1-x^7)}+....$ If $|x|<1$, find the sum of infinite terms of following series:
$$\frac{1}{(1-x)(1-x^3)}+\frac{x^2}{(1-x^3)(1-x^5)}+\frac{x^4}{(1-x^5)(1-x^7)}+....$$
Could someone give me hint to solve this problem. I wrote $r_{th}$ t... | $\frac{1}{(1-x)(1-x^3)}=\frac{1}{x-x^3}(\frac{1}{1-x}-\frac{1}{1-x^3})$
$\frac{x^2}{(1-x)(1-x^5)}=\frac{1}{x-x^3}(\frac{1}{1-x^3}-\frac{1}{1-x^5})$
...
So, assuming n starts from 1, $\frac{1}{(1-x)(1-x^3)}+\frac{x^2}{(1-x^3)(1-x^5)}+\frac{x^4}{(1-x^5)(1-x^7)}...=\frac{1}{x-x^3}(\frac{1}{1-x}-\frac{1}{1-x^{2n+1}})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Definite integral of a positive continuous function equals zero? Let's calculate $$\int_0^{\frac\pi 2} \frac {dx}{\sin^6x + \cos^6x}$$
We have
$$\int \frac {dx}{\sin^6x + \cos^6x} = \int \frac {dx}{1 - \frac 34 \sin^2{2x}}$$
now we substitute $u = \tan 2x$, and get
$$\int \frac {dx}{1 - \frac 34 \sin^2{2x}} = \frac{1}{... | One thing I can see: in your substitution $\;u=\tan 2x\;$ on $\;[0,\pi/2]\;$ , you get
$$0\le x\le\frac\pi2\implies 0\le 2x\le\pi\implies u=\tan2x$$
is not defined on $\;[0,\pi]\;$ , which renders the substitution incorrect.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is there a solution to this differential equation? I am trying to find a function $y(x)$ that is a solution to
$$
\left(a_3 x^3+a_1 x\right) y''(x)-\left(3 a_3 x^2+2 a_1\right) y'(x)+3 a_3\, x \,y(x)=a_0 x^4+a_2
$$
I tried using mathematica but it ran for hours without giving a solution.
Thank you.
| Notice the LHS of the ODE can be rewritten in operator form:
$$\begin{align}
LHS
&= \left[x\frac{d}{dx} - 2\right]\left[(a_1 + a_3 x^2)\frac{d}{dx} - 3a_3 x\right]y\\
&= \left[ x^3 \frac{d}{dx} x^{-2}\right]\left[ (a_1 + a_3 x^2)^{5/2} \frac{d}{dx}
\frac{1}{(a_1 + a_3 x^2)^{3/2}} \right] y
\end{align}
$$
Unwinding t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How many values does the expression $1 \pm 2 \pm 3 \pm \cdots \pm n$ take?
How many different values does the expression $1 \pm 2 \pm 3 \pm \cdots \pm n$ take?
I was wondering about this problem and didn't think it was immediately obvious. The answer can't be $2^{n-1}$ since the combinations all might not be unique. ... | To start with, you can forget about the leading term, $1$.
All that term does is to shift all the values one place to the right
on the number line.
The expression $\pm 2 \pm 3 \pm \cdots \pm n$
has the same number of distinct values as $1 \pm 2 \pm 3 \pm \cdots \pm n$.
From this point until further notice, I'll conside... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 0
} |
Prove an identity in a Combinatorics method It is a combinatorics proof. Anyone has any idea on how to prove
$$\sum \limits_{i=0}^{l} \sum\limits_{j=0}^i (-1)^j {m-i\choose m-l} {n \choose j}{m-n \choose i-j} = 2^l {m-n \choose l}\;$$
We need to prove this equation holds for all $l$.
I know that $\sum {n \choose j}{m-... | Suppose we seek to verify that
$$\sum_{p=0}^l\sum_{q=0}^p (-1)^q
{m-p\choose m-l} {n\choose q} {m-n\choose p-q}
= 2^l {m-n\choose l}$$
where $m\ge n$ and $m-n\ge l.$
This is
$$\sum_{p=0}^l {m-p\choose m-l} \sum_{q=0}^p (-1)^q
{n\choose q} {m-n\choose p-q}.$$
Now introduce the integral
$${m-n\choose p-q} =
\frac{1}{2\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Carrying out a substituting to evaluate $\int (x + 1) (x^2 + 2 x)^5dx$ The problem is:
$$ \int { (x+1)({ x }^{ 2 } } +2x{ ) }^{ 5 }dx $$
The next step given by WolframAlpha is
$$\int { (x+1)({ x }^{ 2 } } +2x{ ) }^{ 5 }dx\\ =\quad \frac { 1 }{ 2 } \int { { u }^{ 5 }du } $$
(While I realize I am doing somthing worng)
... | Since $(x^2 + 2 x)^5 = u^5$, we can use the equation $(x + 1) \,dx = \frac{1}{2} du$ to write the integral as
$$\int \underbrace{(x^2 + 2 x)^5}_{u^5} \,\underbrace{(x + 1) \,dx}_{\frac{1}{2} du} = \int u^5 \cdot \frac{1}{2} du = \frac{1}{2} \int u^5 \,du .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the interval of convergence $\sum\limits_{n=1}^\infty \frac{(3n)!(x^n)}{(n!)^3}$ $\displaystyle \sum_{n=1}^\infty \dfrac{(3n)!(x^n)}{(n!)^3}$
I already found the interval of convergence to be $\displaystyle -\frac{1}{27} < x < \frac{1}{27}$. I am having trouble checking the endpoint of $\displaystyle x= \frac{1}{2... | Observe that
\begin{align}
\sum_{n=1}^\infty\frac{(3n)!\left(\frac{1}{27}\right)^n}{(n!)^3}
&=\sum_{n=1}^\infty\frac{(3n)!}{\left[3^n(n!)\right]^3}\\
&=\sum_{n=1}^\infty\frac{(3n)!}{\left[3\cdot6\cdot9\cdots3n\right]^3}\\
&=\sum_{n=1}^\infty\left[\frac{1\cdot2}{3^2}\cdot\frac{4\cdot5}{6^2}\cdots\frac{(3n-2)(3n-1)}{(3n)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1771148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the following limit: Find
$$\lim_{n \to \infty}\frac{1}{\sqrt{n}}\left[\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+\cdots+\frac{1}{\sqrt{2n}+\sqrt{2n+2}}\right]$$
MY TRY:
$$
\begin{align}
\lim_{n \to \infty} &\frac{1}{\sqrt{n}} \biggl[
\frac{1}{\sqrt{2}+\sqrt{4}}
+ \frac{1}{\sqrt{4}+\sqrt{6}}
+ \... | This is a classic partial sums of integrals problem. Your expression is:
$$\frac{1}{\sqrt{n}}\sum\limits_{k=1}^n \frac{1}{\sqrt{2k}+\sqrt{2k+2}}$$
write as
$$\frac{1}{n}\sum\limits_{k=1}^n \frac{1}{\sqrt{2\frac{k}{n}}+\sqrt{2\frac{k}{n}+2}}$$
then the limit is the same as the integral
$$\int_0^1\frac{1}{\sqrt{2x}+\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Solve the recurrence $a_n=3a_{n/3}+2$ given $a_0=1$ and $n$ is a power of $3$ Solve the recurrence $$a_n=3a_{n/3}+2$$ given $a_0=1$ and $n$ is a power of $3$
I am trying to study for my final using my previous quizzes, of which I got this question wrong. My instructor wants me to use the Divide and Conquer technique. I... | Consider:
$$\begin{aligned}
(3-1)(3^{k-1}+3^{k-2}+\cdots +3+1) & =(3^{k}+3^{k-1}+\cdots +3)-(3^{k-1}+3^{k-2}+\cdots +3+1) \\
&=3^k+(3^{k-1}-3^{k-1})+ ...+(3-3)+1\\
&=3^k-1
\end{aligned}$$
rearranging:
$$
3^{k-1}+3^{k-2}+\cdots +3+1=\frac{3^k-1}{3-1}
$$
Which is just the sum of a geometric series formula
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $\forall n\in\mathbb{N}$, $14^n$ can be represented as a sum of three perfect squares. Show that $\forall n\in\mathbb{N}$, $14^n$ can be represented as a sum of three perfect squares.
I checked $(\mod 7)$ and deduced that the three squares can be $1,4,2(\mod 7) $ or all divisible by $7$ and for $n \ge 2$ they... | You can use induction.
Step 1:
When $n = 1$, we have $14 = 3^2 + 2^2 + 1^2$
When $n = 2$, we have $14^2 = 12^2 + 6^2 + 4^2$
Step 2: Suppose the statment is right for all $1 \le k \le n$.
In particular, $\exists a, b, c > 0$ s.t. $14^{n-1} = a^2 + b^2 + c^2$
Then for $n + 1$,
$14^{n+1} = 14^{n-1+2} = 14^{n-1} \times 14... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to solve this double integral involving trig substitution (using tangent function)? This is a question I came across and I cannot find the answer.
By using a substitution involving the tangent function, show that $$\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx=\frac{\pi}{4}$$
My attempt
I use trig substit... | Let us start considering $$I=\int\frac{x^2-y^2}{(x^2+y^2)^2}dy$$ Defining $$y=x\tan(\theta)\implies dy=x \sec ^2(\theta )\implies I=\int \frac{\cos (2 \theta )}{x}\,d\theta=\frac{\sin (2 \theta )}{2 x}$$ Back to $x$ $$I=\frac{y}{x^2+y^2}$$ So, $$\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}dy=\frac 1 {1+x^2}$$ So, you are left w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to prove that $\sum_{i=0}^n 2^i\binom{2n-i}{n} = 4^n$. So I've been struggling with this sum for some time and I just can't figure it out. I tried proving by induction that if the sum above is a $S_n$ then $S_{n+1} = 4S_n$, but I didn't really succeed so here I am. Thanks in advance.
| I can give a mildly ugly proof by induction. Let
$$S_n=\sum_{k=0}^n2^k\binom{2n-k}n=\sum_{k=0}^n2^k\binom{2n-k}{n-k}=\sum_{k=0}^n2^{n-k}\binom{n+k}n\;,$$
and assume that $S_n=4^n=2^{2n}$. Then
$$\begin{align*}
S_{n+1}&=\sum_{k=0}^{n+1}2^{n+1-k}\binom{n+1+k}{n+1}\\
&=\sum_{k=0}^{n+1}2^{n+1-k}\left(\binom{n+k}n+\binom{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1782432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 9,
"answer_id": 0
} |
Different ways to find limit in infinity I would like to find a limit of the function $f(x)=\frac{x^3+\sqrt{x^6-1}}{4(x-2)^2}$ as $x\rightarrow \infty$ and as $x\rightarrow -\infty$.
In the first case, both numerator and denominator goes to infinity but nominator has higher exponent so $$\lim_{x\rightarrow \infty}\... | Something I learned when dealing with limits $x \to -\infty$ is that you can do a change in variable.
$$
\lim_{x \to - \infty} \frac{x^3 + \sqrt{x^6 -1}}{4(x-2)^2} \left/ x = -t, \ x \to -\infty \Rightarrow t \to \infty \right/ = \lim_{t \to \infty} \frac{-t^3 + \sqrt{t^6 -1}}{-4(t+2)^2}
$$
Now for big values of $t$ th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
About the solution to "Finding the range of $y= \sqrt x + \sqrt{3-x}"$ I was reading the solution of "Find the range of $y = \sqrt{x} + \sqrt{3 -x}$" and I had some points of confusion about the solution posted in the OP.
I wrote here my interpretation of the solution. Could you please let me know if my understanding c... | (too long for a comment)
I will have to look at the continuity later; but on the issue of checking that $\sqrt 3$ and $\sqrt 6$; is it necessary?
I think checking that both $\sqrt 3$ and $\sqrt 6$ are in the range is necessary in your "solution" because you use continuity.
My reasoning is that these values are in a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Equation with limit $\lim\limits_{n\to \infty}\sqrt{1+\sqrt{x+\sqrt{x^2+...+\sqrt{x^n}}}}=2$
I had never faced with problems like this. Give me, please, a little hint.
| We are trying to solve for $x$ where $\lim_{n\to\infty}a_n=2$ and $a_n=\sqrt{1+\sqrt{x+\sqrt{x^2+\sqrt{\dots x^n}}}}$
The trivial possible solutions are $x=0,1$, which do not work.
For $x$
$$\sqrt{1+\sqrt{x+\sqrt{x^2+\sqrt{\dots x^n}}}}=\sqrt{1+\sqrt{x}\sqrt{1+\sqrt{1+\sqrt{\frac1x+\sqrt{\frac1{x^4}+\sqrt{\dots x^n}}}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1785454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Rationalizing the denominator of $\frac{\sqrt {2}}{\sqrt {x-3}}$ ok, so im reviewing for a math test and the following question is from the practice final exam.
Rationalize the denominator in the example:
$$\frac{\sqrt {2}}{\sqrt {x-3}}$$
After multiplying both the numeration and denominator by the conjugate of the d... | $$\frac{\sqrt{2}}{\sqrt{x-3}}=\frac{\sqrt{2}\cdot\sqrt{x-3}}{\sqrt{x-3}\cdot\sqrt{x-3}}=\frac{\sqrt{2\cdot(x-3)}}{\left(\sqrt{x-3}\right)^2}=\frac{\sqrt{2x-6}}{x-3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving integration formula I want to prove the integration formula
$$
\int \frac {\sqrt {a+bu}}{u} \ du = 2 \sqrt {a+bu}+a \int { \frac {du}{u \sqrt {a+bu}} }.
$$
I tried trigonometric substitution (as $u= \frac {a \tan^2 \theta}{b}$), but it didn't seem to work. I am stuck on how to prove the integration formula.
| Let's try another substitution.
Let $x=\sqrt{a+bu}\Rightarrow u=\dfrac{x^2-a}{b} \qquad dx=b/2(a+bu)^{-1/2}du=\dfrac{bdu}{2x}\Rightarrow \dfrac{2xdx}{b}=du$
Now we have
${\Large{\int}} \dfrac{2x^2}{b}\cdot\dfrac{b}{x^2-a}dx=2{\Large{\int}} \dfrac{x^2}{x^2-a}dx=2{\Large{\int}} \dfrac{x^2-(x^2-a)}{x^2-a}+\dfrac{x^2-a}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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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.