Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How find $\max _{z: \ |z|=1} \ f \left( z \right)$ for $f \left( z \right) = |z^3 - z +2|$ Let $f : C \mapsto R $, $f \left( z \right) = |z^3 - z +2|$. How find $\max _{z: \ |z|=1} \ f \left( z \right)$ ?
| i am going to parametrize the unit circle by $z = \cos t + i \sin t.$ we have $$\begin{align}|z^3 - z + 2|^2 &= (\cos 3t - \cos t + 2)^2 +(\sin 3t - \sin t)^2 \\&=\cos^2 3t + \cos^2t+4-2\cos 3t \cos t+4\cos 3t-4\cos t \\ &+\sin^2 3t + \sin^2 t-2\sin 3t \sin t\\
&=6-2\cos 2t+4\cos 3t-4 \cos t\end{align}$$
the critical ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Area of an equilateral triangle
Prove that if triangle $\triangle RST$ is equilateral, then the area of $\triangle RST$ is $\sqrt{\frac34}$ times the square of the length of a side.
My thoughts:
Let $s$ be the length of $RT$. Then $\frac s2$ is half the length of $\overline{RT}$. Construct the altitude from the $S$ t... | From AoPS wiki,
Method 1: Dropping the altitude of our triangle splits it into two triangles. By HL congruence, these are congruent, so the "short side" is $\frac{s}{2}$. Using the Pythagorean theorem, we get $s^{2}=h^{2}+\frac{s^{2}}{4}$, where $h$ is the height of the triangle. Solving, $h=\frac{s \sqrt{3}}{2}$. (no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Help in evaluating limit of the given function.
Evaluate the limit $$\lim_{x\to \infty} \left(\cfrac{x^2+5x+3}{x^2+x+2}\right)^x $$
I'm not sure how to evaluate this limit. This is what I've done yet:
$$\begin{align}
\lim_{x\to \infty} \left\{1 + \left(\cfrac{x^2+5x+3}{x^2+x+2} - 1\right) \right\} ^x\\
=\ \lim_{x\... | $$\begin{align}\lim_{x\to\infty}\left(\frac{x^2+5x+3}{x^2+x+2}\right)^x&=\lim_{x\to\infty}\left(1+\frac{1}{\frac{x^2+x+2}{4x+1}}\right)^{\frac{x^2+x+2}{4x+1}\cdot\frac{x(4x+1)}{x^2+x+2}}\\&=\lim_{x\to\infty}\left(\left(1+\frac{1}{\frac{x^2+x+2}{4x+1}}\right)^{\frac{x^2+x+2}{4x+1}}\right)^{\frac{4+\frac 1x}{1+\frac 1x+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1255779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
range of $m$ such that the equation $|x^2-3x+2|=mx$ has 4 real answers. Find range of $m$ such that the equation $|x^2-3x+2|=mx$ has 4 distinct real solutions $\alpha,\beta,\gamma,\delta$
To show how I got the wrong answers.
From $|x^2-3x+2|=mx$
I got the two case $x^2-3x+2=mx$ when $x>2 $ or $ x<1$
and $x^2-3x+2=-mx$ ... | There is some positive value $m$ such that $y=mx$ is tangent to $y=-(x^2-3x+2)$.
This value must make $0$ the discriminant of the equation
$$x^2-3x+2=-mx$$
That is, $$m^2-6m+1=0$$
The least root of this equation is $$3-2\sqrt2$$
So $0<m< 3-2\sqrt 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1259271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many ordered pairs are there in order for $\frac{n^2+1}{mn-1}$ to be an integer? For how many ordered pairs of positive integers like $(m,n)$ the fraction
$\frac{n^2+1}{mn-1}$
is a positive integer?
| We have:
$$n^2+1=kmn-k$$
so we have $n$ divides $k+1$ we can write $k+1=nt$ so that $$n^2+1=(nt-1)(mn-1)$$
but if $m,t,n>1$ we have $(nt-1)(mt-1)\geq (2n-1)^2>n^2+1$impossible
if either $t=1$ or $m=1$ in the two cases $n-1$ divides $n^2+1$ but we know that $n-1=\gcd(n^2+1,n-1)=1$ or $2$, so that $n=1$, $n=2$ or $n=3$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to factor the polynomial $x^4-x^2 + 1$ How do I factor this polynomial: $$x^4-x^2+1$$
The solution is: $$(x^2-x\sqrt{3}+1)(x^2+x\sqrt{3}+1)$$
Can you please explain what method is used there and what methods can I use generally for 4th or 5th degree polynomials?
| Actually you have:
$$x^4-x^2+1=x^4+2x^2+1-3x^2=(x^2+1)^2-(\sqrt3 x)^2 $$
and use the identity $a^2-b^2=(a-b)(a+b)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Derivatives - optimization (minimum of a function) For which points of $x^2 + y^2 = 25$ the sum of the distances to $(2, 0)$ and $(-2, 0)$ is minimum?
Initially, I did $d = \sqrt{(x-2)^2 + y^2} + \sqrt{(x+2)^2 + y^2}$, and, by replacing $y^2 = 25 - x^2$,
I got $d = \sqrt{-4x + 29} + \sqrt{4x + 29}$, which derivative d... | For better readability, $$S=\sqrt{29+4x}+\sqrt{29-4x}$$
$$\dfrac{dS}{dx}=\dfrac2{\sqrt{29+4x}}\cdot4-\dfrac2{\sqrt{29-4x}}\cdot4$$
For the extreme values of $S,$ we need $\dfrac{dS}{dx}=0\implies29+4x=29-4x\iff x=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1262073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Area enclosed by an equipotential curve for an electric dipole on the plane I am currently teaching Physics in an Italian junior high school. Today, while talking about the electric dipole generated by two equal charges in the plane, I was wondering about the following problem:
Assume that two equal charges are placed... | Here is another method based on the curve-linear coordinates introduced by Achille Hui. He introduced the following change of variables
$$\begin{align}
\sqrt{(x+1)^2+y^2} &= u+v\\
\sqrt{(x-1)^2+y^2} &= u-v
\end{align} \tag{1}$$
Then solving for $x$ and $y$ we shall get
$$\begin{align}
x &= u v\\
y &= \pm \sqrt{-(u^2-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1262174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 2
} |
How to find a basis of a linear space, defined by a set of equations Problem
Find a basis of the intersection $P\cap Q$ of subspaces $P$ and $Q$ given by:
$$ P:
\begin{cases}
x_1 - 2 x_2 + 2 x_4=0,\\
x_2 - x_3 + 2 x_4 = 0
\end{cases}
\qquad Q:
\begin{cases}
-2 x_1 + 3 x_2 + x_3 -6 x_4=0,\\
x_1 - x_2 - x_3 + 4 x_4 = 0
\... | Having:
$$
\begin{cases}
x_1 = 2 x_3 - 6 x_4,\\
x_2 = x_3 - 2 x_4\\
\end{cases}
$$
We could set
1) For the first element of basis:
\begin{split}
x_3=1,\quad x_4=0:\\
x_1 = 2\cdot 1 - 6\cdot 0 = 2,\\
x2 = 1 - 2 \cdot 0 = 1
\end{split}
So, we get: (2, 1, 1, 0).
2) Second element of basis
\begin{split}
x_3=0,\quad x_4=1:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1262342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating trigonometric limit: $\lim_{x \to 0} \frac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2}$
Evaluate $\lim_{x \to 0} \cfrac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2} $
This is what I've tried yet:
$$\begin{align} & \cfrac{x(\tan 2x - 2\tan x)}{4\sin^4 x} \\
=&\cfrac{x\left\{\left(\frac{2\tan x}{1-\tan^2 x} \right) - 2\... | Well, let's try something different from using power series expansions. Here, we simplify using trigonometric identities to reveal that
$$\frac{x\tan 2x-2x \tan x}{(1-\cos 2x)^2}=\frac{2x}{\sin 4x }=\frac{1}{2\text{sinc}(4x)}$$
where the sinc function is defined as $\text{sinc}(x)=\frac{\sin x}{x}$.
The limit as $x \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1263968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Differentiate $\sin^{-1}\left(\frac {\sin x + \cos x}{\sqrt{2}}\right)$ with respect to $x$
Differentiate $$\sin^{-1}\left(\frac {\sin x + \cos x}{\sqrt{2}}\right)$$ with respect to $x$.
I started like this: Consider $$\frac {\sin x + \cos x}{\sqrt{2}}$$, substitute $\cos x$ as $\sin \left(\frac {\pi}{2} - x\right)$,... | An alternative approach is to use Implicit Differentiation:
\begin{equation}
y = \arcsin\left(\frac{\sin(x) + \cos(x)}{\sqrt{2}} \right) \rightarrow \sin(y) = \frac{\sin(x) + \cos(x)}{\sqrt{2}}
\end{equation}
Now differentiate with respect to '$x$':
\begin{align}
\frac{d}{dx}\left[\sin(y) \right] &= \frac{d}{dx}\left[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
show $\sum_{k=0}^n {k \choose i} = {n+1 \choose i+1}$ show for n $\geq i \geq 1 : \sum_{k=0}^n {k \choose i} = $ ${n+1} \choose {i+1}$
i show this with induction:
for n=i=1: ${1+1} \choose {1+1}$ = $2 \choose 2$ = 1 = $0 \choose 1$ + $1 \choose 1$ = $\sum_{k=0}^1 {k \choose 1}$
now let $\sum_{k=0}^n {k \choose i} = $ $... | We can also use the recurrence from Pascal's Triangle and telescoping series:
$$
\begin{align}
\sum_{k=0}^n\binom{k}{i}
&=\sum_{k=0}^n\left[\binom{k+1}{i+1}-\binom{k}{i+1}\right]\\
&=\sum_{k=1}^{n+1}\binom{k}{i+1}-\sum_{k=0}^n\binom{k}{i+1}\\
&=\binom{n+1}{i+1}-\binom{0}{i+1}\\
&=\binom{n+1}{i+1}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
For which values of $a,b,c$ is the matrix $A$ invertible? $A=\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}$
$$\Rightarrow\det(A)=\begin{vmatrix}b&c\\b^2&c^2\end{vmatrix}-\begin{vmatrix}a&c\\a^2&c^2\end{vmatrix}+\begin{vmatrix}a&b\\a^2&b^2\end{vmatrix}\\=ab^2-a^2b-ac^2+a^2c+bc^2-b^2c\\=a^2(c-b)+b^2(a-c)+c^2(b-a)... | suppose the matrix is singular. then so is its transpose, which must annihilate some non-zero vector $(h,g,f)$
this gives three equations:
$$
fa^2 + ga+ h =0 \\
fb^2 + gb+ h =0 \\
fc^2 + gc+ h =0 \\
$$
we know from the algebra of fields that a quadratic equation can have at most two roots, so the three values $a,b,c$ c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
} |
Solve by using substitution method $T(n) = T(n-1) + 2T(n-2) + 3$ given $T(0)=3$ and $T(1)=5$ I'm stuck solving by substitution method:
$$T(n) = T(n-1) + 2T(n-2) + 3$$ given $T(0)=3$ and $T(1)=5$
I've tried to turn it into homogeneous by subtracting $T(n+1)$:
$$A: T(n) = T(n-1) + 2T(n-2) + 3$$
$$B: T(n+1) = T(n) + 2T(n-... | suppose we look for a constant solution $t_n = a.$ then $a$ must satisfy $a = a+2a + 3.$ we pick $a = -3/2.$ make a change a variable $$a_n = t_n + 3/2, t_n = a_n - 3/2.$$ then $a_n$ satisfies the recurrence equation $$a_n= a_{n-1}+ 2a_{n-2}, a_0 = 9/2, a_1=13/2.$$ now look for solutions $$a_n = \lambda^n \text{ wher... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1273673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How does $-\frac{1}{x-2} + \frac{1}{x-3}$ become $\frac{1}{2-x} - \frac{1}{3-x}$ I'm following a solution that is using a partial fraction decomposition, and I get stuck at the point where $-\frac{1}{x-2} + \frac{1}{x-3}$ becomes $\frac{1}{2-x} - \frac{1}{3-x}$
The equations are obviously equal, but some algebraic mani... | $$ -\frac{1}{x-2} = \frac{1}{-(x-2)} = \frac{1}{-x+2} = \frac{1}{2-x}$$ and
$$ \frac{1}{x - 3} = \frac{1}{-3 + x} = \frac{1}{-(3 - x)} = -\frac{1}{3-x}$$
Thus $$ -\frac{1}{x-2} + \frac{1}{x - 3} = \frac{1}{2-x} - \frac{1}{3-x} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
} |
Finding a nullspace of a matrix. I am given the following matrix $A$ and I need to find a nullspace of this matrix.
$$A =
\begin{pmatrix}
2&1&4&-1 \\
1&1&1&1 \\
1&0&3&-2 \\
-3&-2&-5&0
\end{pmatrix}$$
I have found a row reduced form of this matrix, which is:
$$A' =
\begin{pmatrix}
1&0&3&-2 \\
0&1&-2&3 ... |
Since $x_1=2x_4-3x_3$ and $x_2=2x_3-3x_4\Rightarrow$
if $(x_1,x_2,x_3,x_4)\in$ nullspace($A$): $$(x_1,x_2,x_3,x_4)=(2x_4-3x_3,2x_3-3x_4,x_3,x_4)=x_3(-3,2,1,0)+x_4(2,-3,0,1)$$ So Nullspace$(A)=\langle (-3,2,1,0),(2,-3,0,1) \rangle$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Inequality used in the proof of Kolmogorov Strong Law of Large Numbers I'm trying show that convergence follows.
$$\sum_{k \geq 1} \frac{\sigma^2_k}{k^2} < \infty \Rightarrow \lim_{M \rightarrow \infty}\frac{1}{M^2}\sum_{k \leq M} \sigma^2_k=0.$$
Let's consider $D_k = \sum_{n \geq k} \displaystyle\frac{\sigma_n^2}{n^2... | \begin{align*}
\sum_{k=1}^M k^2(D_k-D_{k+1})
&= \sum_{k=1}^M k^2 D_k - \sum_{k=1}^M k^2 D_{k+1} \\
&= \sum_{k=1}^M k^2 D_k - \sum_{k=2}^{M+1} (k-1)^2 D_k \\
&= D_1 + \sum_{k=2}^M (k^2-(k-1)^2) D_k - M^2 D_{M+1} \\
&= D_1 + \sum_{k=2}^M (2k-1) D_k - M^2 D_{M+1} \\
&= \sum_{k=1}^M (2k-1) D_k - M^2 D_{M+1} \\
&\le \sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1278042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find $x$ and $y$ If $\frac{\tan 8°}{1-3\tan^{2}8°}+\frac{3\tan 24°}{1-3\tan^{2}24°}+\frac{9\tan 72°}{1-3\tan^{2}72°}+\frac{27\tan 216°}{1-3\tan^{2}216°}=x\tan 108°+y\tan 8°$, find x and y. I am unable to simplify the first and third terms. I am getting power 4 expressions. Thanks.
| HINT:
$$\tan(3\cdot8^\circ)=\dfrac{3\tan 8^\circ-\tan^38^\circ}{1-3\tan^28^\circ}$$
Now, $$\frac{\tan 8^\circ}{1-3\tan^28^\circ}-y\tan8^\circ=\dfrac{(1-y)\tan 8^\circ-(-3y)\tan^38^\circ}{1-3\tan^28^\circ}$$ which will be a multiple of $\tan(3\cdot8^\circ)$ if
$$\dfrac{1-y}{-3y}=\dfrac31\iff y=-\dfrac18$$
$$\implies\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
First order differential equation integrating factor is $e^{\int\frac{2}{x^2-1}}$ So i got the first order ode
$$(x^2-1)\frac{dy}{dx}+2xy=x$$
I divided both sides by $x^2-1$
$$\frac{dy}{dx}+\frac{2}{x^2-1}xy=\frac{x}{x^2-1}$$
in the form $y' + p(x)y = q(x)$
So that means the integrand is...
$$e^{\int\frac{2}{x^2-1}}$$
... | your equation $$(x^2 - 1) \frac{dy}{dx} + 2x y = x $$ is an exact differential equation. the reason is it can be written as $$\frac d{dx}\left((x^2 - 1) y\right) = x $$ on integration gives you $$(x^2 - 1)y = \frac12 x^2 + c \to y = \frac{x^2}{2(x^2 - 1)} + \frac c{x^2 - 1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1281383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Multivariable limit which should be simple ! How to calculate the following limit WITHOUT using spherical coordinates?
$$
\lim _{(x,y,z)\to (0,0,0) } \frac{x^3+y^3+z^3}{x^2+y^2+z^2}
$$
?
Thanks in advance
| Let $\epsilon \gt 0$. If $(x,y,z)$ is close enough to $(0,0,0)$ but not equal to it, then $|x^3|\le \epsilon x^2$, with similar inequalities for $|y^3|$ and $|z^3|$. It follows that
$$\frac{|x^3+y^3+z^3|}{x^2+y^2+z^2}\le \frac{|x^3|+|y^3|+|z^3|}{x^2+y^2+z^2}\le \frac{\epsilon (x^2+y^2+z^2)}{x^2+y^2+z^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1282418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving vectors as a basis in $E^{m}$ Show that if the vectors $a_{1}$, $a_2$, $\cdots$, $a_m$, are a basis in $E^{m}$, the vectors $a_{1}$, $a_2$, $\cdots$, $a_{p-1}$, $a_{q}, a_{p+1}, \cdots,a_{m}$, also are a basis if and only if $y_{p,q} \neq 0$, where $y_{p,q}$ is defined by the following tableau:
\begin{matrix}
1... | Remark: Your problem has nothing to be with the rightmost column
$(y_{1,0},y_{2,0},\dots,y_{m,0})^T$. I'll omit that column from the matrix.
Settings
Let $A = \begin{bmatrix}\mathbf{a}_1&\mathbf{a}_2&\cdots&\mathbf{a}_m&|\mathbf{a}_{m+1} & \cdots&\mathbf{a}_n\end{bmatrix}$ be the original coefficient matrix in $E^{m \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1283312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maclurin Series. (Approximation) Given that $y=\ln \cos x$, show that the first non-zero terms of Maclurin's series for $y=-\frac{x^2}{2}-\frac{x^4}{12}$. Use this series to find the approximation in terms of $\pi$ for $\ln 2$.
My question is how to determine value of $x$ which is suitable?
| We have
\begin{align}
\ln(\cos(x)) & = \dfrac{\ln(\cos^2(x))}2 = \dfrac12 \cdot \ln(1-\sin^2(x)) = - \dfrac12 \sum_{k=1}^{\infty} \dfrac{\sin^{2k}(x)}k\\
& = -\dfrac12 \sin^2(x) - \dfrac14 \sin^4(x) - \cdots\\
& = -\dfrac12 \left(x-\dfrac{x^3}{3!} + \mathcal{O}(x^5)\right)^2-\dfrac14 \left(x + \mathcal{O}(x^3)\right)^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to find cotangent? Need to find a $3\cot(x+y)$ if $\tan(x)$ and $\tan(y)$ are the solutions of $x^2-3\sqrt{5}\,x +2 = 0$.
I tried to solve this and got $3\sqrt{5}\cdot1/2$, but the answer is $-\sqrt{5}/5$
| Since by Vieta's formulas one has
$$\tan x+\tan y=-\frac{-3\sqrt 5}{1}=3\sqrt 5,\ \ \ \tan x\tan y=\frac{2}{1}=2,$$
one has
$$3\cot(x+y)=3\cdot\frac{1}{\tan(x+y)}=3\cdot\frac{1-\tan x\tan y}{\tan x+\tan y}=\frac{3(1-2)}{3\sqrt 5}=-\frac{\sqrt 5}{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Proving simple trigonometric identity I need help with this one
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \mathrm{tan}^2\alpha} - \cos\alpha = \sin \alpha
$$
I tried moving sin a on the other side of the eqation
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\... | $\left(\dfrac{\sin^2 \alpha}{\sin \alpha+\cos\alpha}-(\sin\alpha+\cos\alpha)\right)+\dfrac{\sin \alpha+\cos\alpha}{1-\tan^2\alpha}\\=\left(\dfrac{\sin^2 \alpha-\sin^2 \alpha+\cos^2\alpha}{\sin \alpha-\cos\alpha}\right)+\dfrac{\sin \alpha+\cos\alpha}{1-\tan^2\alpha}\\=\dfrac{\cos^2\alpha}{\sin \alpha-\cos\alpha}+\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the integer x: $x \equiv 8^{38} \pmod {210}$ Find the integer x: $x \equiv 8^{38} \pmod {210}$
I broke the top into prime mods:
$$x \equiv 8^{38} \pmod 3$$
$$x \equiv 8^{38} \pmod {70}$$
But $x \equiv 8^{38} \pmod {70}$ can be broken up more:
$$x \equiv 8^{38} \pmod 7$$
$$x \equiv 8^{38} \pmod {10}$$
But $x \equiv... | You made a small slip up when working $\bmod 2$, this is because $0^n$ is always congruent to $0$ no matter what congruence you are working with . I repeat this step once again.
$8^{38}\equiv 3^{38}\equiv3^{9\cdot4}3^{2}\equiv(3^{4})^{9}3^2\equiv1^9\cdot3^2\equiv9\equiv 4\bmod 5$
$8^{38}\equiv0 \bmod 2$ since it is cle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 1
} |
Solving an integral with trig substitution I'm looking to solve the following integral using substitution:
$$\int \frac{dx}{2-\cos x}$$
Let $z=\tan\frac{x}{2}$
Then $dz=\frac 1 2 \sec^2 \frac x 2\,dx$
$$\sin x=\frac{2z}{z^2+1}$$
$$\cos x =\frac{1-z^2}{z^2+1}$$
$$dx=\frac{2\,dz}{z^2+1}$$
$$\int \frac{dx}{2-\cos x} = \in... | HINT:
$$\int \frac{1}{a^2+x^2}dx=\frac1a \arctan(x/a)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to compute $\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{\pi}{48\sqrt{3}}$? $$\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{\pi}{48\sqrt{3}}$$
I have difficulty to evaluating above integrals.
First I try the substitution $x^4 =t$ or $x^4 +x^2+1 =t$ but it makes integral worse.
Using Mathematica I fou... | Here is an approach.
You may write
$$\begin{align}
\int_0^{\infty}\frac{x^4}{\left(x^4+x^2+1\right)^3}dx
&=\int_0^{\infty}\frac{x^4}{\left(x^2+\dfrac1{x^2}+1\right)^3\,x^6}dx\\\\
&=\int_0^{\infty}\frac{1}{\left(x^2+\dfrac1{x^2}+1\right)^3}\frac{dx}{x^2} \\\\
&=\int_0^{\infty}\frac{1}{\left(x^2+\dfrac1{x^2}+1\right)^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 2
} |
Find the range of a $4$th-degree function For the function $y=(x-1)(x-2)(x-3)(x-4)$, I see graphically that the range is $\ge-1$. But I cannot find a way to determine the range algebraically?
| Non-calculus Approach/ Completing Square Approach
Let $u=(x-2)(x-3)=x^2-5x+6$
$(x-1)(x-4)=x^2-5x+4=u-2$
So
\begin{align}
y&=(x-1)(x-2)(x-3)(x-4)\\&=u(u-2)\\&=u^2-2u\\&=(u-1)^2-1
\end{align}
As $(u-1)^2\ge0$ for all $u\in{\Bbb{R}}$
$y\ge0-1=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1295586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Show that $\frac {\sin x} {\cos 3x} + \frac {\sin 3x} {\cos 9x} + \frac {\sin 9x} {\cos 27x} = \frac 12 (\tan 27x - \tan x)$ The question asks to prove that -
$$\frac {\sin x} {\cos 3x} + \frac {\sin 3x} {\cos 9x} + \frac {\sin 9x} {\cos 27x} = \frac 12 (\tan 27x - \tan x) $$
I tried combining the first two or the la... | @MayankJain @user,
I don't know that the case for $n=1$ is trivial, especially for someone in a trig class currently. I will offer a proof without induction using substitution instead. Mayank, to see that this is the case for $n=1$, convert the RHS of the equation as follows:
$\dfrac{1}{2} \cdot \big{[}\dfrac{\sin 3x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1299957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to solve the difference equation $u_n = u_{n-1} + u_{n-2}+1$ Given that:
$$
\begin{equation}
u_n=\begin{cases}
1, & \text{if $0\leq n\leq1$}\\
u_{n-1} + u_{n-2}+1, & \text{if $n>1$}
\end{cases}
\end{equation}
$$
How do you solve this difference equation?
Thanks
EDIT:
From @marwalix's answer:
$$
u_n=v_n... | Write $u_n=v_n+a$ where $a$ is a constant. In that case the recurrence reads as follows
$$v_n+a=v_{n-1}+v_{n-2}+2a+1$$
So if we chose $a=-1$ we are left with
$$v_n=v_{n-1}+v_{n-2}$$
And we're back to a Fibonnacci type and in this case we have $v_0=v_1=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\sin x + c_1 = \cos x + c_2$ While working a physics problem I ran into a seemingly simple trig equation I couldn't solve. I'm curious if anyone knows a way to solve the equation:
$\sin(x)+c_1 = \cos(x)+c_2$
(where $c_1$ and $c_2$ are constants) for $x$ without using Newton's method or some other form of approximatio... | $$
A\sin x + B\cos x = \sqrt{A^2+B^2}\left( \frac A {\sqrt{A^2+B^2}}\sin x+ \frac B {\sqrt{A^2+B^2}}\cos x \right)
$$
Notice that the sum of the squares of the coefficients above is $1$; hence they are the coordinates of some point on the unit circle; hence there is some number $\varphi$ such that
$$
\cos\varphi = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Complex analysis, residues
Find the residue at $z=0$ of $f(z)=\dfrac{\sinh z}{z^4(1-z^2)}$.
I did
\begin{align}
\frac{\sinh z}{z^4(1-z^2)} & =\frac{1}{z^4}\left[\left(\sum_{n=0}^\infty \frac{z^{2n+1}}{(2n+1)!)}\right)\left(\sum_{n=0}^\infty z^{2n}\right)\right] \\[8pt]
& =\frac{1}{z^4}\left[\left(z+\frac{z^3}{6}+\cd... | The residue at $z=0$ is the coefficient of the $\frac1z$ term in the expansion
$$
\frac{\sinh(z)}{z^4(1-z^2)}=\frac1{z^4}\left(z+\frac{z^3}6+\frac{z^5}{120}+\dots\right)\left(1+z^2+z^4+\dots\right)
$$
That is the coefficient of $z^3$ term in the expansion
$$
\left(\color{#C00000}{z}+\color{#00A000}{\frac{z^3}6}+\frac{z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why does $\frac{49}{64}\cos^2 \theta + \cos^2 \theta$ equal $\frac{113}{64}\cos^2 \theta $? I have an example:
$$ \frac{49}{64}\cos^2 \theta + \cos^2 \theta = 1 $$
Then what happens next:
$$ \frac{113}{64}\cos^2 \theta = 1 $$
Where has the other cosine disappeared to? What operation happened here? Any hints please.
| $\dfrac{49}{64}\cos^2 \theta + \cos^2 \theta = \left(\dfrac{49}{64} +1\right) \cos^2 \theta = \dfrac{113}{64}\cos^2 \theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
limits and infinity I'm having trouble wrapping my head around some of the 'rules' of limits. For example,
$$
\lim_{x\to \infty} \sqrt{x^2 -2} - \sqrt{x^2 + 1}
$$
becomes
$$
\sqrt{\lim_{x\to \infty} (x^2) -2} - \sqrt{\lim_{x\to \infty}(x^2) + 1}
$$
which, after graphing, seems to approach zero. My question is how do yo... | Hint: $$\left[\sqrt {x^2 - 2} - \sqrt{x^2 + 1}\right]\dot\, \frac{\sqrt {x^2 - 2} + \sqrt{x^2 + 1}}{\sqrt {x^2 - 2} + \sqrt{x^2 + 1}} = \frac{x^2 - 2 - (x^2 + 1)}{\sqrt {x^2 - 2} + \sqrt{x^2 + 1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$\mathbb{Q}\left ( \sqrt{2},\sqrt{3},\sqrt{5} \right )=\mathbb{Q}\left ( \sqrt{2}+\sqrt{3}+\sqrt{5} \right )$ Prove that
$$\mathbb{Q}\left ( \sqrt{2},\sqrt{3},\sqrt{5} \right )=\mathbb{Q}\left ( \sqrt{2}+\sqrt{3}+\sqrt{5} \right )$$
I proved for two elements, ex, $\mathbb{Q}\left ( \sqrt{2},\sqrt{3}\right )=\mathbb{Q}\... | It is worth noting from alex.jordan and achille hui's responses that if $\theta = \sqrt{2} + \sqrt{3} + \sqrt{5}$, we explicitly have $$\begin{align*} \sqrt{2} &= \tfrac{5}{3} \theta - \tfrac{7}{72} \theta^3 - \tfrac{7}{144} \theta^5 + \tfrac{1}{576} \theta^7, \\ \sqrt{3} &= \tfrac{15}{4} \theta - \tfrac{61}{24} \theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$ Question:
Given $x + y = 12$, find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$?
Key:
I use $y = 12 - x$ and substitute into the equation, and derivative it.
which I got this
$$\frac{x}{\sqrt{x^2+4}} + \frac{x-12}{\sqrt{x^2-24x+153}} = f'(x).$$
However, aft... | the constraint is $x + y = 12.$ at a local extremum of $\sqrt{x^2 + 4} + \sqrt{y^2 + 9},$ the critical numbers satisfy $$dx + dy =0,\quad \frac{x\, dx}{\sqrt{x^2 + 4}} + \frac{y\, dy}{\sqrt{y^2 +9}} = 0 \to x\sqrt{y^2 + 9}=y\sqrt{x^2 + 4} $$ squaring the last equation we have $$9x^2 =4y^2 \to y = \pm\frac32x, x + y ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
How to solve inequality for : $|7x - 9| \ge x +3$ How to solve inequality for : $|7x - 9| \ge x + 3$
There is a $x$ on both side that's make me confused...
| We can solve the absolute value inequality by squaring both sides, then solving the resulting quadratic inequality.
\begin{align*}
|7x - 9| & \geq x + 3\\
|7x - 9|^2 & \geq (x + 3)^2\\
(7x - 9)^2 & \geq (x + 3)^2\\
49x^2 - 126x + 81 & \geq x^2 + 6x + 9\\
48x^2 - 132x + 72 & \geq 0\\
4x^2 - 11x + 6 & \geq 0\\
4x^2 - 8x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to compute determinant of $n$ dimensional matrix? I have this example:
$$\left|\begin{matrix}
-1 & 2 & 2 & \cdots & 2\\
2 & -1 & 2 & \cdots & 2\\
\vdots & \vdots & \ddots & \ddots & \vdots\\
2 & 2 & 2 & \cdots & -1\end{matrix}\right|$$
When first row is multiplied by $2$ and added to second, to $nth$ row, determina... | Your recursive approach is fine; just follow it through. Let $D_n(a,b)$ be the determinant of the matrix with diagonal elements $a$ and all other elements $b$; clearly $D_1(a,b)=a$. For $n>1$, multiplying the first row by $-b/a$ and adding it to every other row gives $0$'s in the first column (except for an $a$ in th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1308502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Finding $\lim\limits_{n\to\infty }\frac{1+\frac12+\frac13+\cdots+\frac1n}{1+\frac13+\frac15+\cdots+\frac1{2n+1}}$ I need to compute: $\displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}$.
My Attempt: $\dis... | Hint
The numerator is $H_n$ and the denominator is $H_{2n+1}-\frac12H_n$.
Also,
$$\frac{H_n}{H_{2n+1}-\frac12H_n}=\frac1{-\frac12+\frac{H_{2n+1}}{H_n}}$$
and
$$H_n\sim\ln n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
The perpendicular distance from the origin to point in the plane The plane $3x-2y-z=-4$ is passing through $A(1,2,3)$ and parallel to $u=2i+3j$ and $v=i+2j-k$.
The perpendicular distance from the origin to the plane is $r.n = d$ but how to determine the point (Call it N) on the plane and what's the coordinate of the po... | There is a general procedure & formula derived in Reflection formula by HCR to calculate the point of reflection $\color{blue}{P'(x', y', z')}$ of the any point $\color{blue}{P(x_{o}, y_{o}, z_{o})}$ about the plane: $\color{blue}{ax+by+cz+d=0}$ & hence the foot of perpendicular say point $N$ is determined as follows
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving easy first-order linear differential question. Question
Solve $y'=2x(1+x^2-y)$.
My attempt
Rearranging gives $y'+2xy=2x(1+x^2)$. Thus, the integrating factor is $e^{\int2x\,dx}=e^{x^2}$ and multiplying the equation throughout by this gives
$e^{x^2}y'+2xe^{x^2}y=2xe^{x^2}(1+x^2)\Rightarrow\dfrac{d}{dx}{e^{x^2}y}... | make a change of variable $$1+x^2 - y = u, \quad y =1+x^2 - u, y' = 2x-u' $$ then the de $y' = 2x(1+x^2 - y)$ is turned into $$2x-u' = 2xu $$ multiplying by $e^{x^2},$ we get $$e^{x^2}(u'+2xu) = \left(e^{x^2}u\right)' = 2xe^{x^2}$$ on integration yields $$e^{x^2}u = e^{x^2} + c\to u = 1+ce^{-x^2},\quad y = x^2 -ce^{-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\lim\limits_{x\to\infty}(\sin\sqrt{x+1}-\sin\sqrt{x})$ When using Maclaurin series, the limit is
$$\lim\limits_{x\to\infty}\frac{1}{\sqrt{x+1}+\sqrt{x}}=0$$
If we expand the expression with two limits
$$\lim\limits_{x\to\infty}\sin\sqrt{x+1}-\lim\limits_{x\to\infty}\sin\sqrt{x}$$
it diverges.
Which solution i... | Hint:
Mean value theorem implies there exist $c\in]x,x+1[$ such that
$$\sin(\sqrt{x+1})-\sin(\sqrt{x})=\frac{\sqrt{x+1}-\sqrt{x}}{2\sqrt{c}}\cos\sqrt{c}$$
Then $$\left|\sin(\sqrt{x+1})-\sin(\sqrt{x})\right|\le\left(\frac{\sqrt{x+1}}{2\sqrt{x}}-\frac{\sqrt{x}}{2\sqrt{x+1}}\right)|\cos\sqrt{c}|\le\frac{1}{2}\left(\sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
} |
What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer... | $$\boxed{\text{If the boxed statement is true, then the square root of two is irrational.}}$$
Lemma. The boxed statement is true.
Proof. Assume for a contradiction that the boxed statement is false. Then it has the form "if $S$ then $T$" where $S$ is false, but a conditional with a false antecedent is true.
Theorem. Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
"answer_count": 19,
"answer_id": 14
} |
Proving a formula using another formula These questions are from the book "What is Mathematics":
Prove
formula 1: $$1 + 3^2 + \cdots + (2n+1)^2 = \frac{(n+1)(2n+1)(2n+3)}{3}$$
formula 2: $$1^3 + 3^3 + \cdots + (2n+1)^3 = (n+1)^2(2n^2+4n+1)$$
Using formulas 4 and 5;
formula 4: $$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n... | I would go like this: $(2k+1)^2 = 4k^2+4k+1 \Rightarrow 1+3^2+5^2+\cdots + (2n+1)^2=\displaystyle \sum_{k=0}^n (2k+1)^2=4\displaystyle \sum_{k=0}^n k^2 + 4\displaystyle \sum_{k=0}^n k + (n+1)= 4\cdot\dfrac{n(n+1)(2n+1)}{6}+4\cdot\dfrac{n(n+1)}{2}+(n+1)=...$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Simplify $\prod_{k=1}^5\tan\frac{k\pi}{11}$ and $\sum_{k=1}^5\tan^2\frac{k\pi}{11}$ My question is:
If $\tan\frac{\pi}{11}\cdot \tan\frac{2\pi}{11}\cdot \tan\frac{3\pi}{11}\cdot \tan\frac{4\pi}{11}\cdot \tan\frac{5\pi}{11} = X$ and $\tan^2\frac{\pi}{11}+\tan^2\frac{2\pi}{11}+\tan^2\frac{3\pi}{11}+\tan^2\frac{4\pi}{11}+... | The main building block of our solution will be the formula
\begin{align*}\prod_{k=0}^{N-1}\left(x-e^{\frac{2k i\pi}{N}}\right)=x^N-1.\tag{0}
\end{align*}
It will be convenient to rewrite (0) for odd $N=2n+1$ in the form
\begin{align*}
\prod_{k=1}^{n}\left[x^2+1-2x\cos\frac{\pi k}{2n+1}\right]=\frac{x^{2n+1}-1}{x-1}. \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Sum with Generating Functions Find the sum
$$\sum_{n=2}^{\infty} \frac{\binom n2}{4^n} ~~=~~ \frac{\binom 22}{16}+\frac{\binom 32}{64}+\frac{\binom 42}{256}+\cdots$$
How can I use generating functions to solve this?
| If $f(z) = \sum\limits_{n=0}^\infty a_n z^n$ converges for $|z| < \rho$, then for any $m \ge 0$,
$$\frac{z^m}{m!} \frac{d^m}{dz^m} f(z) = \sum_{n=0}^\infty a_m\binom{n}{m} z^n
\quad\text{ for } |z| < \rho.$$
Apply this to $$f(z) = \frac{1}{1-z} = \sum_{n=0}^\infty z^n,$$ we get
$$\sum_{n=2}^\infty \binom{n}{2} z^n = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
An inequality with $\sum_{k=2}^{n}\left(\frac{2}{k}+\frac{H_{k}-\frac{2}{k}}{2^{k-1}}\right)$
show that
$$\sum_{k=2}^{n}\left(\dfrac{2}{k}+\dfrac{H_{k}-\frac{2}{k}}{2^{k-1}}\right)\le 1+2\ln{n}$$where $ n\ge 2,H_{k}=1+\dfrac{1}{2}+\cdots+\dfrac{1}{k}$
Maybe this $\ln{k}<H_{k}<1+\ln{k}$?
| We have, by partial summation:
$$\begin{eqnarray*} \sum_{k=2}^{n}\frac{H_k}{2^{k-1}}&=&H_n\left(1-\frac{1}{2^{n-1}}\right)-\sum_{k=2}^{n-1}\left(1-\frac{1}{2^{k-1}}\right)\frac{1}{k+1}\\&=&1-\frac{H_n}{2^{n-1}}+\sum_{k=2}^{n}\frac{2}{k\, 2^{k-1}}\tag{1}\end{eqnarray*}$$
hence:
$$ \sum_{k=2}^{n}\frac{H_k-\frac{2}{k}}{2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find $\lim_{x \to 0}\frac{\cos(ax)-\cos(bx) \cos(cx)}{\sin(bx) \sin(cx)}$ How to find $$\lim_\limits{x \to 0}\frac{\cos (ax)-\cos (bx) \cos(cx)}{\sin(bx) \sin(cx)}$$
I tried using L Hospital's rule but its not working!Help please!
| Before using L'Hospital, turn the products to sums
$$\frac{\cos(ax)-\cos(bx)\cos(cx)}{\sin(bx)\sin(cx)}=\frac{2\cos(ax)-\cos((b-c)x)-\cos((b+c)x)}{\cos((b-c)x)-\cos((b+c)x)}.$$
Then by repeated application
$$\frac{2a\sin(ax)-(b-c)\sin((b-c)x)-(b+c)\sin((b+c)x)}{(b-c)\sin((b-c)x)-(b+c)\sin((b+c)x)},$$
and
$$\frac{2a^2\c... | {
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"url": "https://math.stackexchange.com/questions/1317215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How prove this $\cot(\pi/15)-4\sin(\pi/15)=\sqrt{15}$ I need some help with this demonstration, please
I have tried with some identities but nothing.
I wanted to use this $$\sin(\pi/15)\cdot \sin(2\pi/15)\cdots\sin(7\pi/15)=\sqrt{15}$$
| We may prove:
$$ \cos\frac{\pi}{15}-4\sin^2\frac{\pi}{15}=\sqrt{15}\sin\frac{\pi}{15} $$
by squaring both sides. By setting $\theta=\frac{\pi}{15}$, that leads to:
$$ \frac{13}{2}-2\cos(\theta)-\frac{15}{2}\cos(2\theta)+2\cos(3\theta)+2\cos(4\theta) = \frac{15}{2}-\frac{15}{2}\cos(2\theta)$$
or to:
$$ -\cos(\theta)+\co... | {
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"url": "https://math.stackexchange.com/questions/1317960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Evaluating a sum (begginer) I am stuck on this evalutaion number.
So the sum is $$\sum_{k=1}^n{1\over (k+1)^2}-{1\over k^2}$$
I can't find a way :(. Do we use Comparasion or partial sums?
Thanks for the help!
| $$\sum_{k=1}^n\ \left(\frac{1}{(k+1)^2} - \frac{1}{k^2}\right)$$
$S_1 = \frac{1}{2^2} - \frac{1}{1^2}$
$S_2 =\left(\frac{1}{2^2} - \frac{1}{1^2}\right)+\left(\frac{1}{3^2}-\frac{1}{2^2}\right)$
$S_3 = \left(\frac{1}{2^2} - \frac{1}{1^2}\right)+\left(\frac{1}{3^2}-\frac{1}{2^2}\right)+\left(\frac{1}{4^2}-\frac{1}{3^2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve $y''+y'+7y=t$? How do I solve $y''+y'+7y=t$ where $y(0)=0$ and $y'(0)=0$ $(t\geq 0)$?
I tried to solve this by Laplace transformation, but I couldn't find the inverse of $1/(s^2(s^2+s+7))$.
How would I solve this?
| You have the Laplace transform, so lets do a partial fraction decomposition. We want to find $a,b,c,d$ such that $$\frac{1}{s^2(s^2 + s +7)} = \frac{a}{s} + \frac{b}{s^2} + \frac{cs + d}{s^2 + s + 7}.$$
So we have
$$as(s^2 + s +7) + b(s^2 + s +7) + cs^3 + ds^2 = 1.$$
Collecting coefficients we have
$$ (a+c)s^3 + (a+b+d... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding a polynomial by divisibility
Let $f(x)$ be a polynomial with integer coefficients. If $f(x)$ is divisible by $x^2+1$ and $f(x)+1$ by $x^3+x^2+1$, what is $f(x)$?
My guess is that the only answer is $f(x)=-x^4-x^3-x-1$, but how can I prove it?
| One way you could express this idea is through functions $g,h$ where $$f(x) = g(x)(x^2+1) \\ f(x)+1 = h(x)(x^3+x^2+1)$$ This means that $$g(x)(x^2+1)+1 =h(x)(x^3+x^2+1)$$ or that $$h(x) = \frac{g(x)(x^2+1)+1}{x^3+x^2+1}$$ So you could begin by choosing any polynomial $g$ you want so long as it forces $h$ to be a polyno... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Need a hint to evaluate $\lim_{x \to 0} {\sin(x)+\sin(3x)+\sin(5x) \over \tan(2x)+\tan(4x)+\tan(6x)}$ I know that $\sin A + \sin B + \sin C = 4\cos({A \over 2})\cos({B \over 2})\cos({C \over 2})$ when $A+B+C=\pi$. If ${x \to 0}$ then I have a half circle, right?
If it is right then I have
$\tan(2x) + \tan(4x) + \tan(6x... | In what follows I will provide full details of the idea of Abhishek Parab.
First divide top and bottom by $x,$ then
\begin{eqnarray*}
\frac{\sin x+\sin 3x+\sin 5x}{\tan 2x+\tan 4x+\tan 6x} &=&\frac{\left(
\dfrac{\sin x+\sin 3x+\sin 5x}{x}\right) }{\left( \dfrac{\tan 2x+\tan
4x+\tan 6x}{x}\right) } \\
&& \\
&=&\frac{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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Partition Generating Function a)
Let
$$P(x)=\sum_{n=0}^{\infty} p_nx^n=1+x+2x^2+3x^3+5x^4+7x^5+11x^6+\cdots$$
be the partition generating function, and let $Q(x)=\sum_{n=0}^{\infty} q_nx^n$, where $q_n$ is the number of partitions of $n$ containing no $1$s.
Then $\displaystyle\frac{Q(x)}{P(x)}$ is a polynomial. What p... | Hint: Try to understand why
$$P(x)=\frac{1}{\prod_{n=1}^{\infty}\left(1-x^n\right)},$$
and what are the corresponding expressions for $Q(x)$ and $R(x)$.
Illustration:
\begin{align*} \frac{1}{1-x}\cdot \frac{1}{1-x^2} \cdot \frac{1}{1-x^3}\cdot\ldots =\,\Bigl(&1+x^1+x^{1+1}+x^{1+1+1}+\ldots\Bigr)\\ \times\,&\Bigl(1+x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Simplifying sum equation. (Solving max integer encoded by n bits) Probably a lack of understanding of basic algebra. I can't get my head around why this sum to N equation simplifies to this much simpler form.
$$
\sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i = 2^n - 2
$$
Background
To give you some background I am trying to derive... | The thing you've asked to show isn't too hard:
\begin{align}
\sum_{i=0}^{n-2} \left(2^{-i+n-2} + 2^i\right)
&= \sum_{i=0}^{n-2} \left(2^{-i+n-2}\right) + \sum_{i=0}^{n-2}\left(2^i\right) \\
&= \sum_{i=0}^{n-2} \left(2^{-i+n-2}\right) + \left(1 + 2 + \ldots + 2^{n-2}\right) \\
&= \sum_{i=0}^{n-2} \left(2^{-i+n-2}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323931",
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"source": "stackexchange",
"question_score": "3",
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When does equality hold in this inequality? The following inequality can be proven as follows:
Let $n\geq3$ and $0=a_0<a_1<\dots<a_{n+1}$ such that $a_1a_2+a_2a_3+\dots+a_{n-1}a_n=a_na_{n+1}$. Show that
\begin{equation*}
\frac{1}{{a_3}^2-{a_0}^2}+\frac{1}{{a_4}^2-{a_1}^2}+\dots+\frac{1}{{a_{n+1}}^2-{a_{n-2}}^2}\g... | Deduced from the equality condition of Cauchy-Schwarz inequality, we have:
$$ \forall 0 \leqslant k \leqslant n-1, \, \, a_{k+3}^2 = a_k^2 + c$$
where $c > 0$ is a constant. So the equation has 3 possiblities that depends on $n$.
For example, if $n = 3K + 1$:
$$\sum_{k=0}^{K-1} (a_0 + kc)(a_1 + kc) + (a_1 + kc)(a_2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How prove $\frac{(a-b)^4+(b-c)^4+(c-a)^4}{(a+b+c)^4}=2$ Let $a,b,c\in R$,and such $ab+bc+ac=0,a+b+c\neq 0$
show that
$$\dfrac{(a-b)^4+(b-c)^4+(c-a)^4}{(a+b+c)^4}=2$$
| Hint: Express the numerator in terms of elementary symmetric polynomials, as the denominator and constraint already are. You get
$$(a-b)^4+(b-c)^4+(c-a)^4 \\= 2(a+b+c)^4 - 12(ab + bc + ca) (a+b+c)^2 + 18(ab+bc+ca)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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High School Trigonometry ( Law of cosine and sine) I am preparing for faculty entrance exam and this was the question for which I couldn't find the way to solve (answer is 0). I guess they ask me to solve this by using the rule of sine and cosine:
Let $\alpha$, $\beta$ and $\gamma$ be the angles of arbitrary triangle... | First lets consider
$$\frac{b-2a\cos\gamma}{a\sin\gamma}$$
The numerator looks similar to the RHS of the cosine law, but not quite. It would be nice to see $-2ab\cos\gamma$ instead of $-2a\cos\gamma$. So lets just multiply by $b$. Then the numerator would look like this
$$b^2-2ab\cos\gamma$$
Now all that is missing is ... | {
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"url": "https://math.stackexchange.com/questions/1326863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 5
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Sums $a_k=[\frac{2+(-1)^k}{3^k}, (\frac{1}{n}-\frac{1}{n+2}), \frac{1}{4k^2-1},\sum_{l=0}^k{k \choose l}\frac{1}{2^{k+l}}]$
Determien the sums of the following series'.
1:$\sum_{k=0}^\infty \frac{2+(-1)^k}{3^k}$
2:$\sum_{k=0}^\infty (\frac{1}{n}-\frac{1}{n+2})$
3:$\sum_{k=0}^\infty \frac{1}{4k^2-1}$
4:$\sum_{k=0}^\inf... | $(1)$ Looks good to me.
$(2)$ You are correct that this is a telescoping series and your answer is correct (but the index should start at $k=1$, not $k=0$) You could also rewrite the series as $$\begin{align}\sum_{k=1}^\infty \frac{1}{k}-\frac{1}{k+2} = \sum_{k=1}^\infty \frac{1}{k}-\sum_{k=1}^\infty \frac{1}{k+2} \\ =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1327732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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evaluate $\int_0^{2\pi} \frac{1}{\cos x + \sin x +2}\, dx $ This is supposed to be a very easy integral, however I cannot get around.
Evaluate:
$$\int_0^{2\pi} \frac{1}{\cos x + \sin x +2}\, dx$$
What I did is:
$$\int_{0}^{2\pi}\frac{dx}{\cos x + \sin x +2} = \int_{0}^{2\pi} \frac{dx}{\left ( \frac{e^{ix}-e^{-ix}}{2i... | Via corindo's rearrangement: we have
$$
I = \int_0^{2 \pi} \frac{1}{\sqrt{2}\cos(x - \pi/4) + 2}\,dx =\\
\frac 1{\sqrt{2}} \int_0^{2 \pi} \frac{1}{\cos(x - \pi/4) + \sqrt{2}}\,dx = \\
\frac 1{\sqrt{2}} \int_0^{2 \pi} \frac{1}{\cos(x) + \sqrt{2}}\,dx =\\
\frac 1{\sqrt{2}} \oint_{|z| = 1} \frac{1}{iz((z + z^{-1})/2 + \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Rotation of a line by a matrix
Give the equation of the line $\ell'$ that is obtained by rotating $\ell$: $x+2y=5$ by an angle of $\theta=\frac{1}{2}\pi$ with center point $O(0,0)$.
The rotation matrix is $\left.\begin{pmatrix}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{pmatrix}\right|_{\alpha=\frac{1... | I notice that the point $(1,2)$ is on the original line. After a rotation by $\pi/2$, this point becomes $(-2,1)$. I also notice that the original slope was $-1/2$. A rotation by $\pi/2$ has the effect of finding a line perpendicular to our starting line, so it will have slope $2$.
The equation of a line with slope $2$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Is the following a conic section All vectors are in $\mathbb{R}^3$ and only $\mathbf{r} = \left[ x; y; z \right]$ is unknown. My question is does the following system define a conic section in the $x-y$ plane and, if so, how can I find it:
$$
\begin{align}
\mathbf{r}^\mathrm{T} \left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{... | No, these are not conic sections in general.
Your first equation forces $z = 0$. For \begin{align*}
\mathbf{r}_1 = \mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \\
\mathbf{r}_2 = \mathbf{v}_1 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \text{,}
\end{align*}
with $c = 0$, your second equation reduces to ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Minimum value of $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\frac{24}{5\sqrt{5a+5b}}$ Let $a\ge b\ge c\ge 0$ such that $a+b+c=1$
Find the minimum value of $P=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\dfrac{24}{5\sqrt{5a+5b}}$
I found that the minimum value of $P$ is $\dfrac{78}{5\sqrt{15}}$ when $a=b=\dfrac{3}{8};c... | You can use calculus to find this.
$f(a,b) = \sqrt{\frac{a}{1-a}} + \sqrt{\frac{b}{1-b}} + \frac{24}{5\sqrt{5(a + b)}}$
Now, let us first find the critical point, $(x,y)$, of this function, where $f_a(x,y) = 0$ and $f_b(x,y) = 0$.
Taking partial derivatives and setting them to 0, you will see that the critical point is... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Quick way to solve the system $\displaystyle \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} = \frac{65}{36}$, $xy-x+y=118$. Consider the system
$$\begin{aligned} \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} & = \frac{65}{36}, \\ xy -x +y & = 118. \end{aligned}$$
I have solved... | Consider the equation $$\left( \dfrac{3}{2} \right)^{x-y} - \left( \dfrac{2}{3} \right)^{x-y} = \dfrac{65}{36}.$$ Let $u=2^{x-y}$ and $v=3^{x-y}$ then we have $36u^2+65uv-36v^2=0.$ Hence $$u=\dfrac49v,\,\,\,\text{or}\,\,\,\,u=-\dfrac94v$$ For real solutions, take the first one and then $\dfrac{u}{v}=\left(\dfrac23\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$ Find the cubic equation which has a root $$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$$
My attempt is
$$x^3=2-\sqrt{3}+3\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(... | $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}\\
\implies x^3=2-\color{red}{\sqrt3}+2+\color{red}{\sqrt3}+3(2-\sqrt3)(2+\sqrt3)(\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}})\\
\implies x^3=4+3.1.\color{red}{x}\\
\implies x^3-4-3x=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 6
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Substituting the value $x=2+\sqrt{3}$ into $x^2 + 1/x^2$ My teacher gave me a question which I am not able to solve:
If $x=2+\sqrt{3}$ then find the value of $x^2 + 1/x^2$
I tried to substitute the value of x in the expression, but that comes out to be very big.
| Since everyone else has answered with various shortcut methods, let me just show you how you could simply substitute in directly and get an answer cleanly without things ever getting 'too big'.
We know that $x=2+\sqrt{3}$, so we can square this using the usual $a^2+2ab+b^2$ binomial formula: $x^2=2^2+2(2)(\sqrt3)+(\sqr... | {
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"source": "stackexchange",
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Finding the sum of the series $\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \ldots$ Deteremine the sum of the series
$$\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \ldots$$
So I first write down the $n^{th}$ term $a_n=\frac{\frac{n(n+1)}{2}}{n!}=\frac{n+1}{2(n-1)!}$.
So from there I can write the series as $$1+\frac{... | HINT: Split
$$\frac{n+1}{2(n-1)!}= \frac{1}{2(n-2)!}+\frac{1}{(n-1)!}$$
and use the fact that $\sum_{n=0}^{\infty} \frac{1}{n!}=e$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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Product of Matrices I Given the matrix
\begin{align}
A = \left( \begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix} \right)
\end{align}
consider the first few powers of $A^{n}$ for which
\begin{align}
A = \left( \begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix} \right)
\hspace{15mm} A^{2} = \left( \begin{matrix} 7 & 6 \\ 9 & 10 \end{m... | In answer to part 2, the general form is $$M^n=\frac{1}{5}\begin{pmatrix} 2\times 4^n+3 \times(-1)^n & 2\times 4^n-2(-1)^n\\3\times 4^n+3(-1)^{n+1} & 3\times 4^n+2(-1)^{n}\\\end{pmatrix}$$
But you know this already because it has already been posted!
(it just took me longer to write out in MathJax)
| {
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How can I prove irreducibility of polynomial over a finite field?
I want to prove what $x^{10} +x^3+1$ is irreducible over a field $\mathbb F_{2}$ and $x^5$ + $x^4 +x^3 + x^2 +x -1$ is reducible over $\mathbb F_{3}$.
As far as I know Eisenstein criteria won't help here.
I have heard about a criteria of irreducibility... | I think that the criterion that you allude to is the following:
Assume that $X^p - a$ has no roots in $\Bbb F _{p^n}$. Then $X^{p^m} - a$ is irreducible in $\Bbb F _{p^n}[X]$ $\forall m \ge 1$.
In any case, you can't use it here.
1) Let us see how to prove that the second polynomial (call it $P$) is reducible. First,... | {
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"url": "https://math.stackexchange.com/questions/1343450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
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Find the value of this series what is the value of this series $$\sum_{n=1}^\infty \frac{n^2}{2^n} = \frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots$$
I really tried, but I couldn't, help guys?
| Hint:
$$\sum_{n=1}^{\infty} \frac{n^2}{2^n} = \sum_{i=1}^{\infty} (2i - 1) \sum_{j=i}^{\infty} \frac{1}{2^j}.$$
Start by using the geometric series formula on $\displaystyle \sum_{j=1}^{\infty} \frac{1}{2^j}$ to simplify the double series into a singular series. Then you will have a series that looks like $\displaystyl... | {
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"url": "https://math.stackexchange.com/questions/1343721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Approximating $\tan61^\circ$ using a Taylor polynomial centered at $\frac \pi 3$ : how to proceed? Here's what I have so far...
I wrote a general approximation of $f(x)=\tan(x)$ , which then simplified a bit to this:
$$\tan \left(\frac{61π}{180}\right) + \sec^2\left(\frac{61π}{180}\right)\left(\frac{π}{180}\right) + \t... | There are many ways to approximate (even very accurately) functions close to a point.
The simplest is Taylor expansion; in the case of the tangent, assuming $b<<a$, the expansion is $$\tan(a+b)=\tan (a)+ \left(\tan ^2(a)+1\right)b+ \left(\tan ^3(a)+\tan (a)\right)b^2+$$ $$
\left(\tan ^4(a)+\frac{4 \tan ^2(a)}{3}+\fr... | {
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Show that in any triangle, we have $\frac{a\sin A+b\sin B+c\sin C}{a\cos A+b\cos B+c\cos C}=R\left(\frac{a^2+b^2+c^2}{abc}\right),$ Show that in any triangle, we have $$\frac{a\sin A+b\sin B+c\sin C}{a\cos A+b\cos B+c\cos C}=R\left(\frac{a^2+b^2+c^2}{abc}\right),$$
where $R$ is the circumradius of the triangle.
Here is... | Since the sine theorem implies:
$$\sum_\text{cyc}a\sin A = \frac{1}{2R}\sum_\text{cyc}a^2 \tag{1}$$
we just need to prove:
$$ \sum_\text{cyc} a \cos A = \frac{abc}{2R^2}=\frac{2\Delta}{R}\tag 2$$
that is trivial since twice the (signed) area of the triangle made by $B,C$ and the circumcenter $O$ is exactly $aR\cos A$:
... | {
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Square roots equations I had to solve this problem:
$$\sqrt{x} + \sqrt{x-36} = 2$$
So I rearranged the equation this way:
$$\sqrt{x-36} = 2 - \sqrt{x}$$
Then I squared both sides to get:
$$x-36 = 4 - 4\sqrt{x} + x$$
Then I did my simple algebra:
$$4\sqrt{x} = 40$$
$$\sqrt{x} = 10$$
$$x = 100$$
The problem is that when... | Method $\#1:$
As for real $a,\sqrt a\ge0\ \ \ \ (1)$
$(\sqrt x+\sqrt{36-x})^2=36+2\sqrt{x(36-x)}\ge36$
$\implies\sqrt x+\sqrt{36-x}\ge6\ \ \ \ (2)$ or $\sqrt x+\sqrt{36-x}\le-6\ \ \ \ (3)$
Finally $(1)$ nullifies $(3)$
Method $\#2:$
WLOG let $\sqrt x=6\csc2y$ where $0<2y\le\dfrac\pi2\implies\sqrt{x-36}=+6\cot2y$
$\sq... | {
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"source": "stackexchange",
"question_score": "4",
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this inequality $\prod_{cyc} (x^2+x+1)\ge 9\sum_{cyc} xy$ Let $x,y,z\in R$,and $x+y+z=3$
show that:
$$(x^2+x+1)(y^2+y+1)(z^2+z+1)\ge 9(xy+yz+xz)$$
Things I have tried so far:$$9(xy+yz+xz)\le 3(x+y+z)^2=27$$
so it suffices to prove that
$$(x^2+x+1)(y^2+y+1)(z^2+z+1)\ge 27$$
then the problem is solved. I stuck in here
| with loss of generality, assume $x\ge y $
consider $f(x, y, z) = (x^2 + x + 1)(y^2 + y + 1) (z^2 + z + 1) - 9 (z(x+y) + xy)$
*
*we first show that
$$f(x, y, z) \ge f((x+y)/2, (x+y)/2, z)$$
since
$$f(x,y,z) - f((x+y)/2, (x+y)/2, z) = (x-y)((z^2 + z + 1)(z -2 - 2xy) + 9)$$
we simply show that
$$(z^2 + z + 1)(z -2 ... | {
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How to rewrite $\pi - \arccos(x)$ as $2\arctan(y)$? I get the following results after solving the equation $\sqrt[4]{1 - \frac{4}{3}\cos(2x) - \sin^4(x)} = -\,\cos(x)$, :
$$
x_{1} = \pi - \arccos(\frac{\sqrt{6}}{3}) + 2\pi n: n \in \mathbb{Z}\\
x_{2} = \pi + \arccos(\frac{\sqrt{6}}{3}) + 2\pi n: n \in \mathbb{Z}
$$
Wo... | We define $y = \arccos \left( \dfrac{\sqrt{6}}{3} \right)$. Since $0 < \dfrac{\sqrt{6}}{3} < 1$, we have $0 < y < \dfrac{\pi}{2}$, and $\dfrac{\pi}{2} < \pi - y < \pi$. Therefore,
\begin{equation*}
\tan \left( \frac{\pi - y}{2} \right) = \sqrt{\frac{1 - \cos (\pi - y)}{1 + \cos (\pi + y)}} = \sqrt{\frac{1 + \cos y}{1 -... | {
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"source": "stackexchange",
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Sum of Two Continuous Random Variables Consider two independent random variables $X$ and $Y$. Let $$f_X(x) =
\begin{cases}
1 − x/2, & \text{if $0\le x\le 2$} \\
0, & \text{otherwise}
\end{cases}$$.Let $$f_Y(y) =
\begin{cases}
2-2y, & \text{if $0\le y\le 1$} \\
0, & \text{otherwise}
\end{cases}$$. Find the probability... | Let $Z=X+Y$. We know $0\leq Z\leq 3$ but for the density of $Z$ we have three cases to consider due to the ranges of $X$ and $Y$.
If $0\leq z\leq 1$:
\begin{eqnarray*}
f_Z(z) &=& \int_{x=0}^{z} f_X(x)f_Y(z-x)\;dx \\
&=& \int_{x=0}^{z} (1-x/2)(2-2y)\;dx \\
&=& \left[ 2x-2zx+zx^2/2+x^2/2-x^3/3 \right]_{x=0}^{z} \\
&=& 2z... | {
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Find the roots of the summed polynomial
Find the roots of: $$x^7 + x^5 + x^4 + x^3 + x^2 + 1 = 0$$
I got that:
$$\frac{1 - x^8}{1-x} - x^6 - x = 0$$
But that doesnt make it any easier.
| $$x^7 + x^5 + x^4 + x^3 + x^2 + 1 = 0$$
$$(x^7 + x^4) + (x^5 + x^3) + x^2 + 1 = 0$$
$$x^4(x^3 + 1) + x^3(x^2 + 1) + x^2 + 1 = 0$$
$$x^4(x^3 + 1) + (x^2 + 1)(x^3 + 1) = 0$$
$$(x^3 + 1)(x^4+x^2 + 1) = 0$$
from there we can continue with factoring
$$x^3+1=x^3+1^3=(x+1)(x^2-x+1)$$
and
$$x^4+x^2 + 1 =(x^2)^2+2x^2+1-x^2=(x^2... | {
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Let $Y=1/X$. Find the pdf $f_Y(y)$ for $Y$. The Statement of the Problem:
Let $X$ have pdf
$$f_X(x) =
\begin{cases}
\frac{1}{4} & 0<x<1 \\
\frac{3}{8} & 3<x<5 \\
0 & \text{otherwise}
\end{cases}$$
(a) Find the cumulative distribution function of $X.$
(b) Let $Y=1/X$. Find the pdf $f_Y(y)$ for $Y$. Hint: Co... | Crude sketch of CDF of Y based on a simulation. Perhaps helpful as a check on your work.
| {
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"question_score": "3",
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Prove that $\sum_{i=1}^{i=n} \frac{1}{i(n+1-i)} \le1$ $$f(n)=\sum_{i=1}^{i=n} \dfrac{1}{i(n+1-i)} \le 1$$
For example, we have $f(3)=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot2}+\dfrac{1}{3\cdot1}=\dfrac{11}{12}\lt 1$
If true, it can be used to prove:
Proving $x\ln^2x−(x−1)^2<0$ for all $x∈(0,1)$
Also, can you prove $f(n)\g... | For second part
$f(n)-f(n+1)= \sum\limits_{i=1}^n [\frac{1}{i(n+1-i)}-\frac{1}{i(n+2-i)}] -\frac{1}{n+1}$
$= \sum\limits_{i=1}^n [\frac{1}{i(n+1-i)}-\frac{1}{i(n+2-i)} -\frac{1}{n(n+1)}]$
$= \sum\limits_{i=1}^n [\frac{1}{i(n+1-i)}-\frac{1}{i(n+2-i)} -(\frac{1}{n}-\frac{1}{n+1})]$
$= \sum\limits_{i=1}^n [(\frac{1}{i(n+1... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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find the integral using Integration by partial Fractions Here is my work for this problem...just wanted a check over and see if i missed anything
Original Problem: $\int$ $\frac6{x^3-3x^2}$
F 6/x^3-3x^2= F 6/x^2(x-3)
6/x^2(x-3)= Ax+B/x^2+C/(x-3)
C/x^2(x-3)= (A+C)x^2+(B-3A)x-3B
B=-2
A=-2/3
C=1/3
F C/x^2-3x^2 dx= F(-2/3)... | We have the expansion
$$\frac{1}{x^3-3x^2}=\frac{A+Bx}{x^2}+\frac{C}{x-3}\tag 1$$
Multiplying both sides of $(1)$ by $x-3$ and letting $x\to 3$ reveals that $C=1/9$.
Multiplying both sides of $(1)$ by $x^2$ and letting $x\to 0$ reveals that $A=-1/3$.
Multiplying both sides of $(1)$ by $x^2$, taking a derivative with ... | {
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Find lim$_{x \to 0}\left(\frac{1}{x} - \frac{\cos x}{\sin x}\right).$ $$\lim_{x \to 0} \left(\frac{1}{x} - \frac{\cos x}{\sin x}\right) = \frac{\sin x - x\cos x}{x\sin x}= \frac{0}{0}.$$
L'Hopital's: $$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = -\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{0}{0}.$$
Once again, using L'Hopital... | $$\lim_{x\to 0} \frac{1}{x}-\frac{\cos(x)}{\sin(x)}=\lim_{x\to 0} \frac{1}{x}-\frac{1}{\tan(x)}=\lim_{x\to 0} \frac{1}{x}-\frac{1}{x+\frac{x^3}{3}+O(x^5)}=\lim_{x\to 0}\frac{1}{x}-\frac{1}{x}=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
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Bounds for $\frac{x-y}{x+y}$ How can I find upper and lower bounds for $\displaystyle\frac{x-y}{x+y}$? So I do see that
$$\frac{x-y}{x+y} = \frac{1}{x+y}\cdot(x-y) = \frac{x}{x+y} - \frac{y}{x+y} > \frac{1}{x+y} - \frac{1}{x+y} = \frac{0}{x+y} = 0$$
(is it correct?) but I don't get how to find the upper bound.
| Suppose $\dfrac{x-y}{x+y} = c$, where $c \neq -1$.
Then
$$\begin{eqnarray}
x - y &=& c(x + y) \\
x - cx &=& y + cy \\
(1 - c)x &=& (1 + c)y
\end{eqnarray}$$
Therefore $y = \dfrac{1-c}{1+c} x$.
But if $\dfrac{x-y}{x+y} = -1$, then $x - y = -1(x + y)$,
and from this we conclude that $x = 0$ and $y$ can be anything you... | {
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Generalizing the Fibonacci sum $\sum_{n=0}^{\infty}\frac{F_n}{10^n} = \frac{10}{89}$ Given the Fibonacci, tribonacci, and tetranacci numbers,
$$F_n = 0,1,1,2,3,5,8\dots$$
$$T_n = 0, 1, 1, 2, 4, 7, 13, 24,\dots$$
$$U_n = 0, 1, 1, 2, 4, 8, 15, 29, \dots$$
and so on, how do we show that,
$$\sum_{n=0}^{\infty}\frac{F_n}{10... | The difference equations given by the suggested series are:
\begin{align}
F_{n+2} &= F_{n+1} + F_{n} \\
T_{n+3} &= T_{n+2} + T_{n+1} + T_{n} \\ \tag{1}
U_{n+4} &= U_{n+3} + U_{n+2} + U_{n+1} + U_{n}
\end{align}
and so on. In general they take on the form
\begin{align}\tag{2}
\phi_{n+m} = \sum_{k=0}^{m-1} \phi_{n+m-k-1}... | {
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The maximum and minimum values of the expression Here is the question:find the difference between maximum and minimum values of $u^2$ where $$u=\sqrt{a^2\cos^2x+b^2\sin^2x} + \sqrt{a^2\sin^2x+b^2\cos^2x}$$
My try:I have just normally squared the expression and got
$u^2=a^2\cos^2x+b^2\sin^2x + a^2\sin^2x+b^2\cos^2x +2... | Write
$$\cos^2x=\frac{1+\cos 2x}{2}$$
and
$$\sin^2x=\frac{1-\cos 2x}{2}$$
Then, we have
$$u=\sqrt{A+B\cos 2x}+\sqrt{A-B\cos 2x}\tag 1$$
where
$$A=\frac{a^2+b^2}{2}$$
$$B=\frac{a^2-b^2}{2}$$
Taking the derivative of u in $(1)$ and setting the derivative equal to zero reveals
$$\frac{-B\sin 2x}{\sqrt{A+B\cos 2x}}+\fra... | {
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Prove that $\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=3/2$
Prove that $$\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=\frac{3}{2}$$
I thought of rewriting $$\cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})$$ as $$\cos^2(90^{\circ}+ (\theta +30^{\circ... | Since $\cos(2x)=2\cos^2(x)-1$, we have $$\cos^2(x)=\frac {\cos(2x)+1}2$$
Therefore, $$\cos^2(\theta)=\frac {\cos(2\theta)+1}{2}$$ $$\cos^2(\theta+120)=\frac{\cos(2\theta+240^\circ)+1}{2}$$ $$\cos^2(\theta-120)=\frac{\cos(2\theta-240^\circ)+1}{2}$$
So the original equation become: $$\frac{\cos(2\theta)+1+cos(2\theta+240... | {
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"source": "stackexchange",
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Nature of the roots of quadratic equation Here is the problem that I need to prove:
If $x$ is real and $\displaystyle{\ p = \frac{3(x^2+1)}{(2x-1)}}$, prove that $\ p^2-3(p+3) \geq 0$
Here is what I did:
\begin{align*}
p(2x-1)=3(x^2+1) \\
3x^2 - 2px + (p+3)=0 \\
b^2 - 4ac = 4(p^2-3(p+3))
\end{align*}
By inspection I c... | You have $$3x^2 - 2px + (p+3) = 0.$$ Given that $x$ is real the quadratic needs to have a discriminant $\Delta \ge 0$. So $$\Delta = 4p^2 - 12(p+3) \geq 0.$$
Dividing by $4$ yields $$\bbox[10px, border: blue 1px solid]{p^2 - 3(p+3) \ge 0.}$$ as required.
| {
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"source": "stackexchange",
"question_score": "2",
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How to solve $\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ$ Question:
$ \sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6° $
I have partially solved this:-
$$ \sin78^\circ-\sin42^\circ +\sin6^\circ-\sin66^\circ $$
$$ 2\cos\left(\frac{78^\circ+42^\circ}{2}\right) \sin\left(\frac{78^\circ-42^\circ}{2}\right) + 2\cos\le... | $\sin(5\cdot78^\circ)=\sin(360^\circ+30^\circ)=\sin30^\circ$
$\sin\{5(-66^\circ)\}=\sin(-360^\circ+30^\circ)=\sin30^\circ$
If $\sin5x=\sin30^\circ\implies5x=n180^\circ+(-1)^n30^\circ$ where $n$ is any integer
$\implies x=n72^\circ+6^\circ$ where $n\equiv-2,-1,0,1,2\pmod5$
Again, $\sin5x=16\sin^5x-20\sin^3x+5\sin x$
So,... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Minor mistake computing $\int \frac{1}{x^3+2x^2-3x} \; dx$? I'm trying to compute:
$$\int \frac{1}{x^3+2x^2-3x} \; dx$$
Until now, I did the following: Factoring:
$$x^3+2x^2-3x=x(x-1)(x+3)$$
To obtain the parcial fractions:
$$\frac{a}{x}+\frac{b}{x+3}+\frac{c}{x-1}=\frac{1}{x(x-1)(x+3)}$$
$$a(x-1)(x+3)+bx(x-1)+cx(x+3)=... | It should be $2a-b+3c=0$, not $2a-b-3c=0$.
As a side note, if you want to solve $a(x-1)(x+3)+bx(x-1)+cx(x+3)=1$, then it's a lot easier to plug in $x=0, 1, -3$ instead of expanding.
| {
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"question_score": "1",
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Find the $\int \frac{(1-y^2)}{(1+y^2)}dy$ $\int \frac{(1-y^2)}{(1+y^2)}dy$ first I tried to divide then I got 1-$\frac{2y^2}{1+y^2}$ and i still can't integrate it.
| Hint:
$$ \int \frac{1-y^2}{1+y^2}dy = \int \frac{1}{1+y^2}dy-\int \frac{y^2}{1+y^2}dy$$
Note (using long division or otherwise): $$\int \frac{y^2}{1+y^2}dy = \int dy -\int \frac{1}{1+y^2}dy$$
Therefore:$$ \int \frac{1-y^2}{1+y^2}dy = 2\int\frac{1}{1+y^2}dy -\int dy$$
The solution should be straight forward from here.... | {
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Range of an inverse trigonometric function Find the range of $f(x)=\arccos\sqrt {x^2+3x+1}+\arccos\sqrt {x^2+3x}$
My attempt is:I first found domain,
$x^2+3x\geq0$
$x\leq-3$ or $x\geq0$...........(1)
$x^2+3x+1\geq0$
$x\leq\frac{-3-\sqrt5}{2}$ or $x\geq \frac{-3+\sqrt5}{2}$...........(2)
From (1) and (2),
domain is $x\l... | You do not have the correct domain. We must also have $-1\leq\sqrt{x^2+3x+1}\leq1$ and $-1\leq\sqrt{x^2+3x}\leq1$. In other words, $\sqrt{x^2+3x+1}\leq1$ and $\sqrt{x^2+3x}\leq1$, since they are positive.
Thus $x^2+3x+1\leq1$ (squaring is allowed since both are positive), or $x^2+3x\leq0$, this gives $-3 \leq x \leq 0... | {
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"source": "stackexchange",
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Finding $P(X < Y)$ where $X$ and $Y$ are independent uniform random variables Suppose $X$ and $Y$ are two independent uniform variables in the intervals $(0,2)$ and $(1,3)$ respectively. I need to find $P(X < Y)$.
I've tried in this way:
$$
\begin{eqnarray}
P(X < Y) &=& \int_1^3 \left\{\int_0^y f_X(x) dx\right\}g_Y(y) ... | Answer:
Divide the regions of X with respect to Y for the condition $X<Y$.
For $0<X<1$, $P(X<Y) = \frac{1}{2}$
For $1<X<2$ and $1<Y<2$ $P(X<Y) = \int_{1}^{2}\int_{x}^{2} \frac{1}{2}\frac{1}{2}dydx = \frac{1}{8}$
For $1<X<2$, and $2<Y<3$ $P(X<Y) = \frac{1}{2}.\frac{1}{2}=\frac{1}{4}$
Thus $P(X<Y) = \frac{1}{2}+ \frac{1}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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remainder of $a^2+3a+4$ divided by 7
If the remainder of $a$ is divided by $7$ is $6$, find the remainder when $a^2+3a+4$ is divided by 7
(A)$2$ (B)$3$ (C)$4$ (D)$5$ (E)$6$
if $a = 6$, then $6^2 + 3(6) + 4 = 58$, and $a^2+3a+4 \equiv 2 \pmod 7$
if $a = 13$, then $13^2 + 3(13) + 4 = 212$, and $a^2+3a+4 \equiv 2 \pmod ... | If the remainder when $a$ divided by $7$ is $b$, then $a = 7n+b$ for some integer $n$.
Hence, $a^2+3a+4 = (7n+b)^2+3(7n+b)+4$ $= 49n^2 + 14nb + b^2 + 21n + 3b + 4$ $= 7(7n^2+2nb+3n) + (b^2+3b+4)$.
So, the remainder when $a^2+3a+4$ is divided by $7$ will be the same as the remainder when $b^2+3b+4$ is divided by $7$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is the inequality $| \sqrt[3]{x^2} - \sqrt[3]{y^2} | \le \sqrt[3]{|x -y|^2}$ true? I'm having some trouble deciding whether this inequality is true or not...
$| \sqrt[3]{x^2} - \sqrt[3]{y^2} | \le \sqrt[3]{|x -y|^2}$ for $x, y \in \mathbb{R}.$
| $|x - y|^2 = (x - y)^2 $
Let $\sqrt[3]{x} = a , \sqrt[3]{y} = b$
I will study this
$$| a^2 - b^2 | \leq \sqrt[3]{|a^3-b^3|^2} $$
L.H.S
$$|a^2 - b^2|^3 = |a-b|^3 \cdot |a+b|^3= \color{red}{|a-b|^2}\cdot |a^2-b^2|\cdot |a^2 +2ab+b^2|$$
R.H.S
$$|a^3 - b^3|^2 = \color{red}{|a-b|^2} \cdot |a^2 +ab + b^2|^2 $$
So our pro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Evaluate $\iint_{R}(x^2+y^2)dxdy$ $$\iint_{R}(x^2+y^2)dxdy$$
$$0\leq r\leq 2 \,\, ,\frac{\pi}{4}\leq \theta\leq\frac{3\pi}{4}$$
My attempt :
Jacobian=r
$$=\iint_{R}(x^2+y^2)dxdy$$
$$x:=r\cos \theta \,\,\,,y:=r\cos \theta$$
$$\sqrt{x^2+y^2}=r$$
$$\int_{\theta=\pi/4}^{\theta=3\pi/4}\bigg[\int_{r=0}^{r=2}\bigg(r^2\bigg... | Switching to polar coordinates, the Jacobian is given by $ |J|$ where $$ J = \dfrac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \dfrac{\partial x}{\partial r} & \dfrac{\partial y}{\partial r} \\ \dfrac{\partial x}{\partial \theta} & \dfrac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solve the equation $4\sqrt{2-x^2}=-x^3-x^2+3x+3$ Solve the equation in $\Bbb R$:
$$4\sqrt{2-x^2}=-x^3-x^2+3x+3$$
Is there a unique solution $x=1$? I have trouble when I try to prove it.
I really appreciate if some one can help me. Thanks!
| It can be seen that $4 \sqrt{2 - x^2} = (1+x)(3 - x^2)$ leads to the expanded form, after squaring both sides and combining terms,
$$ x^6 + 2 x^5 - 5 x^4 - 12 x^3 + 19 x^2 + 18 x - 23 = 0 .$$
It is readily identified that $x=1$ is a solution and can then be factored out leading to
$$ (x-1)( x^5 + 3 x^4 - 2 x^3 - 14 x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the sum of series $\sum_{n=2}^\infty\frac{1}{n(n+1)^2(n+2)}$ How to find the sum of series $\sum_{n=2}^\infty\frac{1}{n(n+1)^2(n+2)}$ in the formal way? Numerically its value is $\approx 0.0217326$ and the partial sum formula contains the first derivative of the gamma function (by WolframAlpha).
| Partial fractions: $$\frac{1}{n(n+1)^2(n+2)} = \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n+2} \right) - \frac{1}{(n+1)^2},$$ the first part being telescoping, and the last part being related to $\zeta(2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Limit behavior of a definite integral that depends on a parameter. Let $A>0$ and $0\le \mu \le 2$. Consider a following integral.
\begin{equation}
{\mathcal I}(A,\mu) := \int\limits_0^\infty e^{-(k A)^\mu} \cdot \frac{\cos(k)-1}{k} dk
\end{equation}
By substituting for $k A$ and then by expanding the cosine in a Taylo... | Here we provide an answer for $\mu=2q/p$ where $p$ is a positive integer and $p+1\le 2q \le 2 p$. By using the multiplication theorem for the Gamma function we have shown that:
\begin{eqnarray}
&&{\mathcal I}(A,\mu) =\\
&&\frac{1}{\mu} \sqrt{\frac{(2\pi)^{2q-p}}{p 2q}} \sum\limits_{r=1}^q \left(\frac{p^{p/q}}{(2q)^2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
A basic root numbers question If $\sqrt{x^2+5} - \sqrt{x^2-3} = 2$, then what is $\sqrt{x^2+5} + \sqrt{x^2-3}$?
| Notice, we have $$\sqrt{x^2+5} - \sqrt{x^2-3} = 2\tag 1$$ let $$\sqrt{x^2+5} + \sqrt{x^2-3} = y\tag 2$$ Now, multiplying both (1) & (2), we get $$(\sqrt{x^2+5} - \sqrt{x^2-3} )(\sqrt{x^2+5} + \sqrt{x^2-3})=2y$$ $$(\sqrt{x^2+5})^2-(\sqrt{x^2-3})^2=2y$$ $$x^2+5-(x^2-3)=2y$$ $$8=2y\implies y=4$$ Hence, $$\bbox[5px, borde... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
find the complex number $z^4$
Let $z = a + bi$ be the complex number with $|z| = 5$ and $b > 0$ such that the distance between $(1 + 2i)z^3$ and $z^5$ is maximized, and let $z^4 = c + di$.
Find $c+d$.
I got that the distance is:
$$|z^3|\cdot|z^2 - (1 + 2i)| = 125|z^2 - (1 + 2i)|$$
So I need to maximize the distance... | Let $\theta$ denote the argument of $z$, then the arguments of $(1+2i)z^3$ and $z^5$ are respectively $3\theta +\arctan(2)$ and $5\theta$. Now imagine that these points lie on concentric circles, so in order to maximize the distance, the difference between the arguments must be $\pi$ (or $-\pi$ it won't matter):
$$5\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Area of a triangle with sides $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$,$\sqrt{z^2+x^2}$ Sides of a triangle ABC are $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$ and $\sqrt{z^2+x^2}$ where x,y,z are non-zero real numbers,then area of triangle ABC is
(A)$\frac{1}{2}\sqrt{x^2y^2+y^2z^2+z^2x^2}$
(B)$\frac{1}{2}(x^2+y^2+z^2)$
(C)$\frac{1}{2}... | hint: Let $a = \sqrt{x^2+y^2}, b = \sqrt{y^2+z^2}, c = \sqrt{z^2+x^2} \to a^2 = x^2+y^2, b^2 = y^2+z^2, c^2= z^2+x^2 $. Use this and Cosine Law to find $\cos^2 A$, then $\sin^2 A$, and use $S^2 = \dfrac{b^2c^2\sin^2 A}{4}$, to find $S^2$ and then take square-root to get back $S$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
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.