Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
counting Number of matrices We have a $2 \times 2$ matrix.
We are given the trace of the matrix as $N$.
Also, all elements of the matrix are greater than or equal to $1$.
And, the determinant of matrix is $\geq 1$.
QUESTIONS:
If given $N$, what are number of $2 \times 2$ matrices that can be formed in consideration to ... | $$A = \left[\begin{array}{cc}x & \frac{x(N-x)-D}{y} \\ y & N - x\end{array}\right]$$
This is the prototype of your matrix where I'm assuming that $N, x, y$ and $D$ are all natural number.
It's easy to show that $\det(A) = D$.
A first constraint is: $x \in \{1, \ldots , N-1\}$.
The second constraint is: $D < x(N-x)$.
Ha... | {
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Is $(A+B)^2 = A^2 + B^2$ if $A$ and $B$ are matrices If $A$ and $B$ are matrices is $(A+B)^2 = A^2 + B^2$?
I thought because, $AB + BA = AB - AB = 0$.
| $$(A+B)^2 = (A+B)(A+B) = A^2 + AB + BA + B^2 \neq A^2 + B^2$$
$(A+B)^2 = A^2 + B^2$ if and only if $AB + BA = 0 \iff AB = -BA$, but this is not true in general.
| {
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What particular solution should I guess for this relation? Just trying to solve a non-homogeneous recurrence relation:
$$f(n) = 2f(n-1) + n2^n$$
$$f(0) = 3$$
Characteristic equation is:
$$f(n) - 2f(n-1) = 0$$
$$a-2 = 0$$
$$a = 2$$
Homogeneous solution is:
$$f_H(n) = b_0\cdot (2)^n$$
Right-hand side is:
$$n2^n$$
What ... | Just solve the bugger by using generating functions. Define $F(z) = \sum_{n \ge 0} f(n) z^n$, adjust indices so there aren't subtractions:
$$
f(n + 1) = 2 f(n) + (n + 1) 2^{n + 1}
$$
Multiply by $z^n$, sum over $n \ge 0$, recognize:
\begin{align}
\sum_{n \ge 0} f(n + 1) z^n &= \frac{F(z) - f(0)}{z} \\
\sum_{n \ge 0}... | {
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Fair die rolled twice, random variables A fair die is tossed twice. Let $d_1=\text{value of die on roll 1}$ and $d_2=\text{value of die on roll 2}$ Let $X=d_1+d_2$, the sum of the faces; $Y=\max\left\{d_1,d_2\right\}$, the maximum of the two faces; and $Z = |d_1-d_2|$.
What is the value of $X, Y, Z,$ and $ W = XZ$ for... | With the sample space $\Omega=\left\{1,2,3,4,5,6\right\}^2=\left\{\left(1,1\right),\left(1,2\right),\dotsc,\left(6,5\right),\left(6,6\right)\right\}$, we can easily find what is being asked.
The range of a Random Variable is the set of values it can assume, so:
$\mathrm{Range}\left(X\right)=\left\{2,3,\dotsc,11,12\righ... | {
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Finding circumcentre Tangents are draw from $P(2,3)$ to $x^2+y^2=4$ meeting at $Q,R$ on circle. Parallelogram $PQSR$ is completed. Find the circumcentre of triangle $QSR$.
My attempt: Clearly, the parallelogram is a rhombus. $S$ will be mirror image of $P$ in chord of contact. But, I want to know a good way of finding... | Here's an overview ...
Note that the circumcenter (call it $C$) of $\triangle PQR$ is easy to find: It's the intersection of $\overline{PS}$ (a line through the origin) and the (horizontal) perpendicular bisector of (vertical) segment $\overline{PQ}$ with known endpoints. Now, the circumcenter you want (call it $D$) is... | {
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Are there contradictions in math? Someone told me that math has a lot of contradictions.
He said that a lot of things are not well defined.
He told me two things that I do not know.
*
*$1+2+3+4+...=-1/12$
*what is infinity $\infty$?
Since I am not a math specialist and little. How to disprove the previous two?
H... | There is already great general answer by Ittay Weiss, so I will try a different approach. In fact, I will try to explain a bit the infinite sum you stated. As for the infinity, one could write a lot about it (mainly because there are multiple infinities, each with different properties), and I couldn't even hope to fit ... | {
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Definite Integral $\int_{-\infty}^{\infty}\frac{x\sin x}{(x^2+a^2)(x^2+b^2)}\,\mathrm{d}x$ $$\int_{-\infty}^{\infty}\frac{x\sin x}{(x^2+a^2)(x^2+b^2)}\,\mathrm{d}x$$
This is easy to evaluate with complex analysis but is there an elementary way (substitution, partial fractions, integration by parts)?
| Note
$$\int_0^\infty e^{-xt}\cos(at)dt=\frac{x}{x^2+a^2}$$
and hence
$$\int_0^\infty e^{-xt}(\cos(at)-\cos(bt))dt=-\frac1{a^2-b^2}\frac{x}{(x^2+a^2)(x^2+b^2)}.$$
Thus
\begin{eqnarray}
\int_{-\infty}^\infty \frac{x\sin x}{(x^2+a^2)(x^2+b^2)}dx&=&2\int_{0}^\infty \frac{x\sin x}{(x^2+a^2)(x^2+b^2)}dx\\
&=&-\frac2{a^2-b^2}... | {
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How to integrate $(x^2 - 1)/(x^2 + 1)$? I have gone until separating $(x^2 - 1)/(x^2 + 1)$ into $x^2/(x^2 + 1)$ - $1/(x^2 + 1)$. The latter fraction I can substitute by $\tan u$, but how to solve the first fraction and how does it all come together?
| $$ \int \frac{x^2 -1}{x^2 + 1} dx = \int \frac{x^2 +1 - 1 - 1}{x^2+1}dx = \int dx - 2 \int\frac{dx}{x^2+1} = x - 2 \arctan x + C $$
| {
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Solving integral $ \int \frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}\:\mathrm{d}x $ there is integral $$ \int \frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}\:\mathrm{d}x$$
i am trying to separate this :
$$=\int \mathrm{d}x -\int \frac{\mathrm{d}x}{1+x+\sqrt{1+x+x^2}} $$ but have no idea about second
| Decompose the integrand as follows
\begin{align}
&\int \frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}dx\\
=& \int \frac{2x+1}{2\sqrt{x^2+x+1}}+\frac1{2\sqrt{x^2+x+1}}+\frac1x\bigg(\frac1{\sqrt{x^2+x+1}}
-1\bigg)\ dx\\
= &\ \sqrt{x^2+x+1}+\frac12\sinh^{-1}\frac{2x+1}{\sqrt3}
-\ln \bigg(\frac x2+1+\sqrt{x^2+x+1}\bigg)+C
\e... | {
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$\int_0^\frac{\pi}{2}\cos ^2x\log(\tan x)dx.$ Evaluate $\int_0^\frac{\pi}{2}\cos ^2x\log(\tan x)dx.$
Sidenote:Via mathalpha I know that answer is $-\pi/4$ but do not know how to derive that.
| Note that
$$
\begin{align}
I(a)
&=\int_0^{\pi/2} \cos^2 x \tan^a xdx&\\
&=\int_0^{\pi/2} \sin^a x \cos^{2-a}xdx&(\mbox{definition of beta function})\\
&=\frac{1}{2}B\left(\frac{1+a}{2},\frac{3-a}{2}\right)& \left(B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\right)\\
&=\frac{\Gamma\left(\frac{1+a}{2}\right)\Gamma\left(... | {
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Proving that $(a+b+c)^n=a^n + b^n + c^n$ Suppose that $(a+b+c)^3=a^3 + b^3 + c^3$.
For what positive integer values of n is it true that $(a+b+c)^n=a^n + b^n + c^n$. Any hint will be much appreciated
| For all odd n it is true.
$(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)$. Now according to the hypothesis $a+b=0 $ or $c+b=0 $ or $a+c=0 $. Say $a+b=0 $ then substituting it is true for all odd n. Similarly we can check for all other cases.
| {
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Minimum value of $\sqrt{(1+1/y)(1+1/z)}$ If $y,z > 0$ and $y + z = c$ where $c$ is a constant, then what's the minimum value of
$$\sqrt{\left(1+\frac1y\right)\left(1+\frac1z\right)}$$
I am having a hard time solving this.
| This can also be taken on using single-variable calculus by using the constraint to insert $ \ z \ = \ c \ - \ y \ $ into
$$ \left[ \ \frac{y \ (c-y) \ + \ (c-y) \ + \ y \ + \ 1}{y \ (c-y)} \ \right]^{1/2} \ = \ \left[ \ \frac{1 \ + \ c \ + \ cy \ - \ y^2}{y \ (c-y)} \ \right]^{1/2} \ \ . $$
Differentiating with respec... | {
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Find the length of $y=x^2$ from $x=0$ to $x=1$ $$L=\int_0^1 \sqrt{1+4x^2} dx = \frac{1}{4}\left(\theta+2\sqrt{5}\right)$$
It tells you to use $2x=\sinh\theta$.
I've tried the best I could, but I can never get the right side of the equation.
Can you help me?
| Using $2x=\sinh\theta$ yield $x=\frac{1}{2}\sinh\theta$ and $dx=\frac{1}{2}\cosh\theta\,d\theta$. The limit of integral for $0<x<1$ equals $0<\theta<\sinh^{-1}2$. Hence
\begin{align}
\int_0^1 \sqrt{1+4x^2}\,dx&=\int_0^{\sinh^{-1}2} \sqrt{1+4\left(\frac{1}{2}\sinh\theta\right)^2}\frac{1}{2}\cosh\theta\,d\theta\\
&=\frac... | {
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Number of integer solutions by generating functions method I'm stuck in the middle of a problem and not sure where to go next. The original problem is:
Find the number of integer solutions to the equation
$$2x + 3y + 4z + w + s + t = n$$ with $$0 \le w \le 2$$ $$2 \le s \le 5$$ $$0 \le t \le 3$$
Now I was able t... | Hint: you can factor $1-x^2$ and $1-x^4$.
| {
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$A^3 = I$ ($A$ is real Symmetric matrix). Does it imply that $A = I$? Question of our assignment
$A$ is a $3×3$ real symmetric matrix such that $A^3 = I$ (Identity matrix). Does it imply that $A = I$?
If so, why? If not, give an example.
Any help will be appreciated.
| For any vector $x$, let $y=(A-I)x$. Since $A$ is symmetric and $A^3=I$, we have
\begin{align*}
\|Ay\|^2 + \|A^2y\|^2 + \|y\|^2
&=y^T\left[A^TA+(A^2)^TA^2+I\right]y\\
&=y^T\left(A^2+A^4+I\right)y\\
&=y^T(A^2+A+I)y\\
&=y^T(A^2+A+I)(A-I)x\\
&=y^T(A^3-I)x\\
&=0.
\end{align*}
Therefore $\|y\|$ must be zero, i.e. $y=0$ or $A... | {
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How many different systems are there? A Football team has 10 players besides the Keeper, who are allocated into 3 positions, the defense, the center and the attack. A system is the allocation of the players into these positions.
For example the system 4 4 2 means that there are 4 players at the defense, 4 players at th... | How about enumerating all possibilities?
Starting with 1 defender (the minimum) and 4 attackers (the maximum), we get:
*
*1-5-4
*1-6-3
*2-4-4
*2-5-3
*2-6-2
*3-3-4
*3-4-3
*3-5-2
*3-6-1
*4-2-4
*4-3-3
*4-4-2
*4-5-1
*5-1-4
*5-2-3
*5-3-2
*5-4-1
Hmm. That is not what you have. ;)
| {
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Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$. Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$.
I made quite a few... | WLOG, let us suppose $f(x)=a(x-u)^2-v$ for $a>0$ and $0< u\le 1$, $v\ge 0$.
Then the premise says $1\ge au^2-v\ge -1$, $1\ge a(1-u)^2-v\ge -1$, $1 \ge a(1+u)^2-v\ge -1$. We need to show that $v\le \frac{3}{2}$.
The above inequalities implies that $1\ge au$ and $2\ge a(1+2u)$.
If $u\in (0,1/2]$, then
$$
v\le 1+au^2\le ... | {
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For any integer n greater than 1, $4^n+n^4$ is never a prime number. For any integer n greater than 1, show that $4^n+n^4$ is never a prime number.
My approach: I tried to use mathematical induction. But somehow couldn't manage to prove it. Is it to be done by induction or can it be done by another method?
| This is a nice identity. Try to complete the square: $4^n+n^4=(2^n)^2+(n^2)^2=(2^n+n^2)^2-2\cdot2^n\cdot n^2$.
Now, clearly we may assume that $n$ is odd, else the sum is even and larger than $2$. But if $n$ is odd, then $2\cdot 2^n$ is itself a square, and we can continue: Say $n=2k-1$, then $(2^n+n^2)^2-2\cdot2^n\cd... | {
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Calculate the inequality $a,b,c$ are the sides of a triangle.Then show that
$$(a+b+c)^3 > 27(a+b-c)(b+c-a)(c+a-b)$$ Also give the case where equality holds i.e. $$(a+b+c)^3=27(a+b-c)(b+c-a)(c+a-b)$$
I tried triangle inequality, AM-GM inequality in many way but cannot help my cause.
| By AM-GM
$$(a+b+c)^3=\left(\sum_{cyc}a\right)^3=\left(\sum_{cyc}(a+b-c)\right)^3\geq$$
$$\geq\left(3\sqrt[3]{\prod_{cyc}(a+b-c)}\right)^3=27(a+b-c)(a+c-b)(b+c-a).$$
Done!
| {
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Calculate $\lim_{n \to \infty} \frac{1^4 + 3^4 + \ldots + (2n-1)^4}{n^5}$ Is my solution correct?
By Faulhaber's formula, $1^4 + 2^4 + \ldots + n^4 = \frac{6n^5+15n^4+10n^3-n}{30} $.
$$\frac{1^4 + 3^4 + \ldots + (2n-1)^4}{n^5} = \frac{1}{n^5} \left[\sum_{k=1}^{2n} k^4 - \sum_{k=1}^{n} (2k)^4 \right] = \frac{1}{n^5} \le... | It is correct, of course, but it would've perhaps been much simpler if one were to just recognize the expression as the Riemann sum of $\displaystyle\int_0^1(2x)^4dx$.
| {
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Working out the value of $a^4+b^4$ If $ab = 2$ and $a+b = 5$ then calculate the value of $a^4+b^4$
My approach:
$$a^4+b^4 = (a+b)^4-4a^3b-6a^2b^2-4ab^3$$
$$=(5)^4-6(ab)^2-4ab.a^2-4ab.b^2$$
$$=(5)^4-6(24)-4ab(a^2-b^2)$$
$$=(5)^4-6(24)-8(a+b)(a-b)$$
$$=(5)^4-6(24)-8(5)(a-b)$$
I am a little stuck now and any help will be ... | You could start with $a+b$ and $ab$,
*
*Note first $(a+b)^2 - 2ab = a^2+ b^2 $
*Note that $(a^2+b^2)^2 - 2(ab)^2 = a^4 + b^4$
As an asside, i use this set of numbers to factorise $b^n-a^n$. Part of the process involves creating a sequence $T_n = a^n+b^n$. You find that $T_{n+1}=(a+b)T_n - ab T_{n-1}$
Applying $... | {
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Finding rational points on a non-trivial algebraic curve The curve in question is defined as the set of all points $(x,y)$ where both $x$ and $y$ are in $(0,1)$, and that satisfy the following:
$$ \left( x^4 - 6 x^2 + 1 \right) \left( y^4-6y^2+1
\right) -64 x^2 y^2 = 0$$
I do not know how I could identify even a sing... | The equation
$$
\left( x^4-6x^2+1 \right) \left( y^4-6y^2+1 \right) - 64 x^2 y^2 = 0
$$
has no rational solutions; indeed the only solutions for which both
$X := x^2$ and $Y := y^2$ are rational are those with $x^2 = y^2 = -1$.
Bruno Joyal already noted that $X$ and $Y$ satisfy
$$
\left( X^2 - 6X + 1 \right) \left( ... | {
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Calculation of sum of infinite products $\sum_{n=1}^\infty n*\frac{r^n}{n!}$ Is there any analytical solution to this sum of infinite products??
$$\sum_{n=1}^\infty n\cdot \frac{r^n}{n!}$$
and
$$\sum_{n=1}^\infty n^2\cdot \frac{r^n}{n!}$$
| Hint for $1$:
$$\sum_{n=1}^\infty n\cdot \frac{r^n}{n!} = r \sum_{n=1}^\infty \frac{r^{n-1}}{(n-1)!} = r \sum_{n=0}^\infty \frac{r^n}{n!}$$
Hint for $2$:
$$\sum_{n=1}^\infty n^2\cdot\frac{r^n}{n!} = \sum_{n=1}^\infty (n(n-1) + n)\cdot\frac{r^n}{n!} = \sum_{n=1}^\infty n(n-1)\cdot\frac{r^n}{n!}+\sum_{n=1}^\infty n\cdot... | {
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Prove that $\lim_{n\to\infty} (\sqrt{n^2+n}-n) = \frac{1}{2}$ Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$
Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it sp... | (Yet another solution - sorry!)
$$
\left(1 + \frac1{2n}\right)^2 = 1 + \frac1n + \frac1{4n^2} > 1 + \frac1n,
$$
therefore
$$
\sqrt{1 + \frac1n} < 1 + \frac1{2n},
$$
therefore
$$
\sqrt{n^2 + n} - n = n\left(\sqrt{1 + \frac1n} - 1\right) < \frac12,
$$
but also
$$
\sqrt{n^2 + n} - n = \frac1{\sqrt{1 + \frac1n} + 1} > \fra... | {
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Point on $z = \frac{1}{xy}$ closest to origin Where $x>0$ and $y>0$.
I want to work with the square of the distance formula from the origin, so I went with $f(x,y) = x^2 + y^2 + \frac{1}{(xy)^2}$.
Then I found the first partial derivatives:
$f_x = 2x - \frac{2}{x^3y^2}$ and $f_y = 2y - \frac{2}{y^3x^2}$.
Set them equal... | The surface you are describing can be described as $xyz = 1$. Since this is symmetric in $x$ versus $y$ versus $z$, any point where $|x| = |y| = |z|$ would be an extremum in distance from the origin. Visually, we can see that that such points are minima.
Therefore the solutions must be $x,y,z = (1,1,1), (1,-1,-1), (-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Value of $ \cos 52^{\circ} + \cos 68^{\circ} + \cos 172^{\circ} $? I am a little weak in trigonometry. I have two questions:
*
*Find the value of $\cos 52^{\circ} + \cos 68^{\circ} + \cos 172^{\circ} $
*Find the value of $\sin 28^{\circ}+ \cos 17^{\circ} + \cos 28^{\circ} + \sin 17^{\circ} $
I am asking these quest... | Note that,
$$
\cos(\alpha\pm \beta)=\cos \alpha\cos \beta\mp\sin \alpha\sin \beta
$$
and
$$
\cos(180^\circ-\alpha)=-\cos\alpha.
$$
Hence
$$
\begin{align}
\cos52^\circ+\cos68^\circ+\cos172^\circ&=\color{blue}{\cos(60^\circ-8^\circ)+\cos(60^\circ+8^\circ)}+\color{red}{\cos(180^\circ-8^\circ)}\\
&=\color{blue}{2\cos60^\ci... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/783695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Telescoping trigonometric series: $\sin(x) \sec(3x) + \cdots+ \sin(3^{n}x) \sec(3^{n+1} x)$ How would you evaluate the sum of this series:
$$\sin(x) \sec(3x) + \sin(3x) \sec(3^{2}x) + \sin(3^{2}x) \sec(3^{3} x) + \cdots+ \sin(3^{n}x) \sec(3^{n+1} x)$$
I can tell it is a telescoping series but I am not able to get that ... | Lab Bhattacharjee has provided the necessary components to evaluate the series.
The series is seen as:
\begin{align}
\sum_{k=0}^{n} \frac{ \sin(3^{k} x) }{ \cos(3^{k+1} x)} &= \frac{1}{2} \sum_{k=0}^{n} \left( \tan(3^{k+1} x) - \tan(3^{k} x) \right) \\
&= \frac{1}{2} \left( \tan( 3^{n+1} x) - \tan(x) \right).
\end{alig... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how can we get from $x \equiv 4 \pmod 9$ to $x \equiv 1 \pmod 3$ I want to solve the system:
$$ x \equiv 1 \pmod 3 , x \equiv 2 \pmod 5, x \equiv 3 \pmod 7, x \equiv 4 \pmod 9, x \equiv 5 \pmod {11}$$
The numbers $3,5,7,9,11$ are not pairwise coprime,especially the numbers $3,9$ are not coprime.
According to my notes, ... | Since 3 is a factor of 9, $x \equiv_9 4 \Rightarrow x \equiv_3 4 \Rightarrow x \equiv_3 1$
| {
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"question_score": "1",
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Calculate $\lim_{x \to \infty} x - \sqrt{x^2 + 2x}$ without derivations How can I calculate $\displaystyle \lim_{x \to \infty} x - \sqrt{x^2 + 2x}$?
Here is what I've done so far:
Multiplying by $\displaystyle \frac{x + \sqrt{x^2 + 2x}}{x + \sqrt{x^2 + 2x}}$
I got $\displaystyle \frac {-2x}{x+\sqrt{x^2 + 2x}}$
Multi... | Note that $x - \sqrt{x^2 + 2x} = x - \sqrt{(x+1)^2 - 1}$. For large $x$, the second term is approximately $x+1$. That is to say, for $x > -1$, $$x - \sqrt{(x+1)^2 - 1} = x - (x+1)\sqrt{1 - \frac{1}{(x+1)^2}}$$ and the limit of the term in the square root is clearly $1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Interesting Base summation contest math problem The problem is as follows:
Let $N_b=1_b+2_b+\cdots+100_b$ where $b$ is an integer greater than $2$. Compute the number of values of $b$ for which the sum of the squares of the digits of $N_b$ is at most $512$.
I originally tried to find the maximum case for which this hel... | Observe that $$N_b= 101_b\left(\frac{100_b}{2}\right)=\frac{b^4+b^2}{2}=b^2\left(\dfrac{b^2+1}{2}\right).$$
If $b$ is odd,
$$b^2\left(\frac{b^2+1}{2}\right)=b^2\left(\frac{b-1}{2}b+\frac{b+1}{2}\right)=\frac{b-1}{2}b^3+\frac{b+1}{2}b^2+0b+0,$$
so $N_b = \left(d_3d_2d_1d_0\right)_b$ where $d_3=\dfrac{b-1}{2}d_2=\dfrac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find a basis of a nullspace of a matrix
Find a basis of nullspace of S.
I have already found a parameterized solution of $S*x=0$. But I dont Know how to get a basis out of that.
| Note that
$$
S\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\\0\end{bmatrix}
$$
if and only if
\begin{align*}
x_1 &= 8x_4+2x_5-x_6+x_7\\
x_2 &= x_4-x_5+2x_6+x_7\\
x_3 &= -x_4+2x_5+3x_6-x_7
\end{align*}
It follows that the vectors in $\DeclareMathOperator{Null}{Null}\Null(S)$ ar... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to derive the least squares solution for linear regression? We have $n$ dots: $(x_1,y_1)\cdots (x_n,y_n)$.
We know that if we use the Least squares method we will get a line $y=mx+b$ that giving the minimal value for the function $w=\sum_{i=1}^n (mx_i+b-y_i)^2$.
I need to prove that $w$ maintains:
$$b=\frac{\sum_{i... | This post is an adjunct to the solution of @SA-255525, who solved the problem via the calculus. Another option is to use linear algebra.
Start with a crisp definition. We have a sequence of measurements of the form $\left\{ x_{k}, y_{k} \right\}_{k=1}^{m}$ and the trial function is $y(x) = c_{0} + c_{1}x$.
This implies... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the points in the graph (my solution) - high school. my math problem is on Swedish so i'll try my best to translate it so you can understand. I'd appreciate it if someone could point out if I did anything wrong and if there is anything that I should add that's essential to these kind of problems.
(I'm a high schoo... | My solution above is correct.
Answer:
$A(0.2)$
$B(1.6)$
$C(2.4)$
$D(3.5;1.75)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/793638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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matrix inverse and elementary row operations I'm trying to see if I can find a relationship between the inverse of 2 matrices.
Matrix A is a $n$ x $n$ matrix.
Matrix B is a $(n-1)$ x $(n-1)$ matrix
The transformation from A to B can be described in the following 2 steps:
1) Take the last 2 rows of A and add them togeth... | I don't think there would be an easy way to relate the two. In going from $A$ to $B$ we lose information about the structure of $A$ which is important in determining the inverse of $A$.
In the $2 \times 2$ case, the $B$ matrix is just the sum of the entries. This just tells us that
$$A=\begin{bmatrix}a_{11} & a_{12} ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Expected value of number of draws We have $5$ number in a bag: $(1,3,5,7,9)$. We draw one from the bag and then put it back. We do this until the sum of the numbers can be divided by $3$. Whats the expected value of the number of draws?
My idea was to solve it with Markov-chains:
States: $0,1,2$. So numbers$\mod 3$.
Th... | We use your idea, but partly for ease of typing we use different notation. Say that we are in State $1$ if the sum so far is congruent to $1$ modulo $3$, and that we are in State $2$ if the sum is congruent to $2$. Let $m$ be the mean waiting time. Let $a_1$ be the additional mean waiting time, given we are in State $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find the sum: $1^{\frac{1}{3}}+2^{\frac{1}{3}}+3^{\frac{1}{3}}+ . . . +(50)^{\frac{1}{3}}$ Can some one help me to find the sum of the following expression?
$$1^{\frac{1}{3}}+2^{\frac{1}{3}}+3^{\frac{1}{3}}+ . . . +(50)^{\frac{1}{3}}$$
| You can bound the sum for any upper limit b.
$\frac{3 b^{4/3}}{4}<\sum _{n=1}^b \sqrt[3]{n}< \frac{3}{4} \
\left(-1+\sqrt[3]{1+b}+b \sqrt[3]{1+b}\right)$
Plugging in b = 50 you would get
$ 138.151<\sum _{n=1}^{50} \sqrt[3]{n}<141.097 $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $ -3\left(x-\lfloor x \rfloor \right)^2+2(x-\lfloor x \rfloor )+a^2=0$ has no integral solution, then $a$ is if $a$ is real number and $\displaystyle -3\left(x-\lfloor x \rfloor \right)^2+2(x-\lfloor x \rfloor )+a^2=0$.has no real integral solution,
Then all possible values of $a$ lie in the interval..
(where $\lflo... | Since $$\frac{1- \sqrt{1+3a^2}}{3}<0,\quad \forall a\neq 0$$
then you only need to find $a$ such that
$$\frac{1+\sqrt{1+3a^2}}{3}>1\iff \sqrt{1+3a^2}>2\iff a^2\ge 1\iff a\le -1 \vee a\ge 1.$$
So, your equation has no solutions for $a\le -1 \vee a\ge 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{3}{\sqrt{7}}$
Let $a,b,c>0$ such that
$$\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}=\dfrac{1}{3}.$$
Show that
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{3}{\sqrt{7}}.$$
My try: since
$$\dfrac{1}{2+a^2}=\dfrac{1}{2}\left(1-\dfrac{a^2}{a^2+2}\right)$... | An analytical proof is proposed below :
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Partial fraction decomposition,how? I need to decompose this fraction:
$${x^2+1\over (x-1)^3\cdot(x+3)}$$
I tried to write it up like this: $${A\over (x-1)}+{B\over (x-1)^2}+{C\over (x-1)^3}+{D\over (x+3)}$$
But now i get $$A\cdot(x-1)^5\cdot (x+3) +B\cdot (x-1)^4\cdot (x+3) +C\cdot(x-1)^3\cdot (x+3) +D\cdot (x-1)^6 ... | Your first step is correct, in fact to find the partial fraction decomposition the first thing you need to do is:
$$\dfrac{x^2+1}{(x-1)^3 (x+3)} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{C}{(x-1)^3} + \dfrac{D}{x+3}$$
Now what you need to do is to sum the fractions on the left side in this way:
$$\dfrac{x^2+1}{(x-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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log and poisson-like integral Here is a fun looking one some may enjoy.
Show that:
$$\int_{0}^{1}\log\left(\frac{x^{2}+2x\cos(a)+1}{x^{2}-2x\cos(a)+1}\right)\cdot \frac{1}{x}dx=\frac{\pi^{2}}{2}-\pi a$$
| I understand that this an old question but I would like to share my 2 cents. First of all we recall the following Fourier identities:
Lemma 1: Let $x \in [-\pi, \pi]$ then
$$\sum_{n=1}^{\infty} \frac{\sin nx}{n} = \frac{\pi-x}{2} \Rightarrow \sum_{n=1}^{\infty} \frac{\cos n x}{n^2} = \frac{\pi^2}{6} -\frac{\pi x}{2} ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Finding the function of x in a differential equation $y=f(x)$ If the graph of $y=f(x)$ contains the point $(0,2)$, $\frac{dy}{dx}=\frac{e^{2x}}{y}$ and $f(x) > 0$ for all $x$, then $f(x) = ?$
$dy/dx = \frac{e^{2x}}y$
$y\cdot dy = e^{2x}dx$
(The following step confuses me, because of the $dy$).
$\int y\cdot dy = \int ... | If we have $$\frac{1}{2}y^{2} = \frac{1}{2}e^{2x} + C$$ then $$y = \sqrt{e^{2x}+C'}$$ where $C'=2C$.
For the initial value problem:
$y(0) = 2 = \sqrt{e^{2(0)}+C'} \iff C' = 4 - e^{2(0)} = 4 - 1 = 3$
So $$y(x) = \sqrt{e^{2x} + 3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Double Integral $\iint_D\ (x+2y)\ dxdy$ $$\iint_D (x+2y)\ dxdy $$
If the area is range by $x=2,\ x=3,\ y=x,\ y=2x$,
how to include the lines? How limits for integral will looks like?
You mean something like this? ( I made mess)
$$\iint_D (x+2y)\ dxdy $$
If the area is range by $x=2,\ x=3,\ y=x,\ y=2x$,
how to include ... | Hint :
$$
\iint_D\ (x+2y)\ dx\,dy=\int_{x=2}^3\int_{y=x}^{2x}\ (x+2y)\ dx\,dy.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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find p,q to the expression A does not depend on x? Find $p,q$ to the expression $A = p (\cos^{8}x-\sin^{8}x) + 4(\cos^{6}x-2\sin^{6}x) + q\sin^{4}x$ does not depend upon $x$
| $$A=p((1-\sin^2x)^4- \sin^8x) + 4((1-\sin^2x)^3 - 2\sin^6x) + q\sin^4x=\\
=p(\sin^8x - 4\sin^6x + 6\sin^4x - 4\sin^2 x + 1 - \sin^8x) + 4(-\sin^6x + 3\sin^4x -3\sin^2x+1-2\sin^6x) + q\sin^4x = \\
=(-4p-12)\sin^6x + (6p+12+q)\sin^4x + (-4p-12)\sin^2x + p+4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A definite integral with tanh and sin In an old textbook, Thomas John I'Anson Bromwich asks the reader to use the expansion
$$ \tanh \left(\frac{\pi x}{2} \right) = 1 + 2 \sum_{n=1}^{\infty} (-1)^{n} e^{-n \pi x}$$
to show $$\lim_{t \to 0^{+}} \int_{0}^{\infty} e^{-tx} \tanh \left(\frac{\pi x}{2} \right) \sin (ax) \, ... | Consider the complex function $ \displaystyle f(z) = \tanh \left( \frac{\pi z}{2} \right) \frac{e^{iaz}}{1+z^{2}}$ where $a > 0$.
On the right, top, and left sides of a rectangular contour with vertices at $\pm 2N, \pm 2N + i2N$ (where N is a positive integer), $\tanh \left(\frac{\pi z}{2} \right)$ is uniformly bounded... | {
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Value of $\sin (2^\circ)\cdot \sin (4^\circ)\cdot \sin (6^\circ)\cdots \sin (90^\circ) $ How can I calculate the value of
*
*$\sin (1^\circ)\cdot \sin (3^\circ)\cdot \sin (5^\circ)\cdots \sin (89^\circ)$
*$\sin (2^\circ)\cdot \sin (4^\circ)\cdot \sin (6^\circ)\cdots \sin (90^\circ)$
My solution: Let $$A = \sin (1... | Using the identity
$$\prod_{k=1}^{n-1}\sin \left( \frac{k\pi}{n}\right) = \frac{n}{2^{n-1}} \tag{1}$$
Putting $n=180$, it gives
$$ \left(\sin (1^\circ)\sin (2^\circ)\dots \sin (89^\circ)\right)^2= \frac{180}{2^{179}} \tag{2}$$
The value of
$$\boxed{ \sin(2^{\circ})\sin(4^{\circ}) \dots \sin(90^{\circ}) = \sqrt{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/812083",
"timestamp": "2023-03-29T00:00:00",
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Prove that $(4/5)^{\frac{4}{5}}$ is irrational Prove that $(4/5)^{\frac{4}{5}}$ is irrational.
My proof so far:
Suppose for contradiction that $(4/5)^{\frac{4}{5}}$ is rational.
Then $(4/5)^{\frac{4}{5}}$=$\dfrac{p}{q}$, where $p$,$q$ are integers.
Then $\dfrac{4^4}{5^4}=\dfrac{p^5}{q^5}$
$\therefore$ $4^4q^5=5^4p^5$
I... | This is essentially the same as the standard proof that $\sqrt 2$ is irrational.
Let $z = (4/5)^{(4/5)}$. Now to calculate $z^5$:
$$
z^5 = \left(\left({4\over 5}\right)^{4\over 5}\right)^5 = \left({4\over 5}\right)^{4}
= {4^4\over 5^4}
$$
Clearly $z \neq 0$ as $0^{(5/4)} = 0 \neq 4/5$.
Let a rational number $r = p/q$... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Simplify expression, skanavi 2.001 How can I simplify this expression?
$$
\frac {\sqrt{x}+1}{x\sqrt{x} + x +\sqrt{x}} \colon \frac {1}{x^2 - \sqrt{x}}
$$
Solving:
$$
a = \sqrt x
$$
$$
\frac {a+1}{a^3 + a^2 + a} \colon \frac{1}{a^4-a} = \frac{ a(a + 1)(a-1)(a^2 + a + 1) }{a(a^2 + a + 1)} = (a +1)(a -1)=a^2-1
$$
Anser... | Notice that $x^2 - \sqrt{x} = \sqrt{x}\Big( (\sqrt{x})^3 - 1 \Big)$ and use the formula $a^3 - 1 = (a -1)(a^2+a+1)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality $\left(a-1+\frac{1}b\right)\left(b-1+\frac{1}c\right)\left(c-1+\frac{1}a\right)\leq1$ Let $a,b,c\in\mathbb{R}^*_+$, $abc=1$.
How can i show that $\left(a-1+\frac{1}b\right)\left(b-1+\frac{1}c\right)\left(c-1+\frac{1}a\right)\leq1$ ?
I got $\left(ab-b+1\right)\left(bc-c+1\right)\left(ca-a+1\right)\leq abc=1$,... | Make the substitution $a=x/y$, $b=y/z$, $c=z/x$ where $x$, $y$ and $z$ are positive reals. The inequality becomes
$$
\left(\frac xy - 1 + \frac zy\right) \left( \frac yz - 1 + \frac xz \right) \left( \frac zx - 1 + \frac yx \right) \stackrel{?}{\leq} 1
$$
or
$$ (x-y+z)(x+y-z)(-x+y+z) \stackrel{?}{\leq} xyz. $$
At most ... | {
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"question_score": "4",
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Trigonometry equation solution I have to solve this equation for all solutions
$$\sin(2x) = -\cos(2x)$$
Here are my steps
$$\sin(2x) + \cos(2x) = 0$$
$$\cos(2x)(\tan(2x) + 1) = 0$$
Upon solving these two equations, I find:
For cosine, $x = \frac{\pi}{4} + \pi n$
For both, $x = \frac{3 \pi}{8} + \pi n$
For both, $x = \f... | Note that $\;\tan 2x = \frac{\sin 2x}{\cos 2x}\;$ is undefined when $\cos 2x = 0$, i.e., when $x = \pi/4 + k\pi$, so it cannot be a solution to either the original or factored equation.
| {
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"question_score": "2",
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Proving $\cos x < 1 - \frac{x^2}{2} +\frac{x^4}{24}$ I wish to prove the following inequality for $x\ne 0$:
$$\cos x < 1 - \frac{x^2}{2} +\frac{x^4}{24}$$
Using the fact that I already prove:
$$\cos x > 1 - \frac{x^2}{2}$$
My try:
$\cos x = 1 - \frac{x^2}{2} +\frac{x^4}{24} + R_2(x)$ where $R_2(x)$ is the remainder wit... | By the symmetry of $\cos$ about $x=0$, we only need to prove it for positive $x$. Since $\cos t\gt 1-\frac{t^2}{2!}$, we have for positive $x$ that
$$\sin x=\int_0^x \cos t\,dt\gt \int_0^x \left(1-\frac{t^2}{2!}\right)\,dt=x-\frac{x^3}{3!}.$$
Do it again. We have
$$1-\cos x=\int_0^x \sin t\,dt\gt \int_0^x \left(t-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/819660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How to use Legendre symbol to find a prime which divides $ax^2+b$? I'm trying to prove that $\dfrac{x^2-2}{2y^2+3}$is never an integer if $x,y\in\mathbb{Z}$.
It can be proven if $\forall p\in\mathbb{P}\:$doesn't suffice both of the following congruences:
$$\: x^2\equiv 2 \pmod{p}$$
$$2y^2\equiv -3\pmod{p}$$
If I coul... | There are odd primes $p$ such that the congruences $x^2\equiv 2\pmod{p}$ and $2y^2\equiv -3\pmod{p}$ have solutions. For example, $p=7$ works ($x=y=3$).
OP has asked for which (odd) primes $p$ does the congruence $2y^2\equiv -3\pmod{p}$ have a solution. It is not difficult to show that this congruence has a solution ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/821185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Solving a separable differential equation Solve the differential equation:
$$y'=\frac{1-y^2}{1-x^2}$$
My book says the solution is: $$y=\frac{x+c}{cx+1},$$ where $c$ is a constant.
It's been ten minutes I tried to verify if it was correct but I'm pretty sure the book is wrong. Can someone confirm it?
| \begin{align*}
\text{Separate:} && \frac{dy}{1-y^2} & =\frac{dx}{1-x^2} \\
\text{Partial fractions:} && \frac{1}{2}\left(\int\frac{1}{1+y}dy+\int\frac{1}{1-y}dy\right) & =\frac{1}{2}\left(\int\frac{1}{1+x}dx+\int\frac{1}{1-x}dx\right) \\
\text{Integrate:} && \log\frac{1+y}{1-y} & = \log c+\log\left(\frac{1+x}{1-x}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/822680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
Help with finding eigenvectors and polynomial factorization I have a real and invertible matrix
$$ A = \left( \begin{matrix} 5 & 4 \\
4&5 \end{matrix} \right)$$
and I need to find the eigenvectors.
The characteristic polynomial is $(5-\lambda)^2 - 16$. According to Wolfram Alpha this factors to $(1-\lambda)(9-\lambda... | Well $(5-\lambda)^2-4^2$ is a difference of squares, but you really need to learn factoring.
$$\begin{pmatrix}
4&4\\4&4\\\end{pmatrix}\begin{pmatrix}x\\y\\\end{pmatrix}=0$$
gives $x=-y$ so $\begin{pmatrix}1\\-1\\\end{pmatrix}$ is one eigenvector,
and $$\begin{pmatrix}
-4&4\\4&-4\\\end{pmatrix}\begin{pmatrix}x\\y\\\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/823141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$ a \cos A + b \cos B + c \cos C = \dfrac{a+b+c}2 $ $\implies$ the triangle is equilateral? If in a triangle $ a \cos A + b \cos B + c \cos C = \dfrac{a+b+c}2 $ , then is the triangle equilateral ?
| $\displaystyle a=2R\sin A\implies\cos A=R(\sin2A)$
$\displaystyle\implies\sum a\cos A=R(\sum\sin2A)$
Using Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle, $\displaystyle\sum a\cos A=4R\prod\sin A=4R\prod2\sin\frac A2\cos\frac A2$
Again, $\displaystyle\frac{a+b+c}2=... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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how to solve this Complex Equation? I have a doubt about the resolution of this equation:
$z^2 +|z^2 -1|=\frac{1}{2}(z+\bar{z})$
I tried to replace $z$ by $x+iy$ and $\bar{z}$ by $x-iy$.
From theory, I know that $|z|^2=z \cdot \bar{z}$, but I do not think this is the case ..
My question is: how do I evaluate $|z^2 -... | $\left| z^2-1\right|\equiv \left| (x+yi)^2-1 \right|\equiv\left| x^2+2xyi-y^2-1\right|\equiv \left|[\color{green}{x^2-y^2-1}]+[\color{red}{2xy}]i\right|\equiv\sqrt{(\color{green}{x^2-y^2-1})^2+(\color{red}{2xy})^2}\equiv\sqrt{x^4+2x^2y^2-2x^2+y^4+2y^2+1}\equiv\sqrt{(x^2-2x+y^2+1)(x^2+2x+y^2+1)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/824013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$ Please help me to prove this trigonometric equation.
$\cos \left( 36^\circ \right)-\cos \left( 72^\circ \right) = \frac{1}{2}$
Thank you.
| From a recent post which relates to the above trig functions at the given angles, we have:
$\cos36^{\circ} = \sin54^{\circ} = \dfrac{\sqrt{5}+1}{4}$, and:
$\cos72^{\circ} = \sin18^{\circ} = \dfrac{\sqrt{5}-1}{4}$. Thus:
$\cos36^{\circ} - \cos72^{\circ} = \dfrac{\sqrt{5}+1}{4} - \dfrac{\sqrt{5}-1}{4} = \dfrac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/827540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$\frac{1}{\sin^2(x)}+\frac{1}{\cos^2(x)} = \cos^2(x)+\sin^2(x) ?$ This is a simple question. Since $\cos(\theta)^2 + \sin(\theta)^2 = 1$
Can I take the inverse of this $\frac{1}{\cos^2(\theta)}+\frac{1}{\sin^2(\theta)} = \frac{1}{1}$?
Finally getting $\frac{1}{\cos^2(\theta)}+\frac{1}{\sin^2(\theta)} = \cos(\theta)^2 ... | No, since: $$\dfrac{1}{\cos^2\theta}+\dfrac{1}{\sin^2\theta}\neq \dfrac1{\cos^2\theta+\sin^2\theta}.$$
$$\dfrac{1}{\sin^2\theta}+\dfrac{1}{\cos^2\theta}=\dfrac{\cos^2\theta+\sin^2\theta}{\sin^2\theta\cos^2\theta}=\dfrac{1}{\sin^2\theta\cos^2\theta}\neq1.$$
In general, $\tfrac1a+\tfrac1b=\tfrac1{a+b}$ is not true, that... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Apollonius’ Identity inner product space $||z-x||^2+||z-y||^2=\frac{1}{2}||x-y||^2+2||z-\frac{x+y}{2}||^2$
I proved it by expanding both sides and i found both sides are equal. Are there any easy way to prove it?
| In a Hilbert space you can use the Parallelogram law :
$$\|a+b\|^2 + \|a-b\|^2 = 2(\|a\|^2 + \|b\|^2)$$
So :
$$\begin{align}
&\|z - \frac{x+y}{2} \|^2\\
&= \|\frac{z}{2} - \frac{x}{2} + \frac{z}{2} - \frac{y}{2}\|^2\\
&= 2(\|\frac{z}{2} - \frac{x}{2}\|^2 + \|\frac{z}{2} - \frac{y}{2}\|^2) - \|\frac{y}{2} + \frac{x}{2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Evaluate the integrals $\int \sin{x} \cot^2{x} \,dx$ and $\int \cos{x} \cot^2{x} \,dx$. Can you please show how to evaluate the integrals $$\int \sin{x} \cot^2{x} \,dx$$ and $$\int \cos{x} \cot^2{x} \,dx.$$
I know that $\cot x=\dfrac{\cos x}{\sin x}$, which can simplify the integrands a bit.
$$\sin x\cot^2x = \frac{\c... | HINT:
$$\sin x\cot^2x=\frac{\cos^2x}{\sin x}=\frac{1-\sin^2x}{\sin x}=\csc x-\sin x$$
$$\cos x\cot^2x=\frac{\cos^3x}{\sin^2x}=\frac{\cos^3x-\cos x+\cos x}{1-\cos^2x}=-\cos x+\frac{\cos x}{\sin^2x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/830609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Bernoulli trials are run until 3 successful trials are completed. If probability of a success is 0.3 what is the expected number of failures? Bernoulli trials are run until 3 successful trials are completed. If probability of a success is 0.3 what is the expected number of failures?
My attempt:
Average number of succes... | Let $p$ be the probability of success (here $p = 0.3$), and $q = 1 - p$, the probability of failure (here $q = 0.7$). $X = k$ if there are $k - 1$ trials with exactly $2$ successes and $k - 3$ failures, and the $k$th trial is success.
The probability of exactly $2$ successes in a sequence of $k - 1$ trials is $^{k-1}\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/831697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Systems of equations with three variables Consider:
$a + b = 5$
$2a + b + c = 4$
$a - b - c = 5$
I like to use substituiton for solving systems of equations, so I firstly look at equation 1 and solve for $a$
$a + b = 5$
$a = 5 - b$
I substitute this into equation 2
$2(5-b) + b + c = 4$
$10 -2b + b + c = 4$
$-b = 4 - c ... | You have the right concept, and we can see from three equations and three unknowns that these variables are solvable (taking into account equations are linearly independent)
equation 1: $\ a+b=5$
equation 2:$\ 2a+b+c= 4$
equation 3:$\ a-b-c =5$
lets solve for b first:
from equation 1 and 3, we say
$a=5-b$ and $\ c = a-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/831767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Mass of ellipsoid
We are given the elipsoid $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} \leq 1$$ with density function $$p(x,y,z)=x^2+y^2+z^2$$ Find the mass of the elipsoid.
I used the transformation $x=ar\sin\theta \cos\phi$, $y=br\sin \theta \sin \phi$ and $z=cr\cos\theta$. Spherical coordinates but I multi... | You may integrate $\theta$ first, then $\phi$, and finally $r$.
$$I(r,\cos\theta,\phi):=(a^2r^2(1-\cos^2\theta) \cos^2\phi+b^2r^2(1-\cos^2 \theta) \sin ^2\phi+c^2r^2\cos^2 \theta)$$
$$J(r,\phi)=\int_{0}^{\pi}I(r,\cos\theta,\phi)\sin\theta d\theta=
\int_{-1}^{1}I(r,x,\phi)dx=\frac{2}{3}r^4(a^2 + b^2 + c^2 + (a^2 - b^2)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/832985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Find the orthogonal trajectories of the family of curves given by $x^2 + y^2 + 2Cy =1$. Find the orthogonal trajectories of the family of curves given by $$x^2 + y^2 + 2Cy =1.$$
The ordinary differential equation for the family of curves is given by $y'=\frac{2xy}{x^2-y^2-1}$.Therefore, the differential equation for th... | The answer at the end of your book does not seem correct. The correct one is $y^2 + x^2 - Cx + 1 = 0$.
The equation is solved most easily with the substitution (all derivatives in Lagrange notation are with respect to $x$):
$$ (y^2 + x^2 + 2Cy)' = (1)' \quad \Rightarrow \quad 2yy' + 2x + 2Cy' = 0 $$
Express $2C$ with ... | {
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"url": "https://math.stackexchange.com/questions/834911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int\frac{1}{\sec x\tan^2 x}dx$ $$\int\frac{1}{\sec x\tan^2 x}dx$$
I first thought of using the property $\int\sec^2 x = \tan x$ but figured it's no use.
How would I go by solving this question?
| $\int\dfrac{(1 - \sin^2 x)\cos x}{\sin^2 x}dx = \int\dfrac{(1 - \sin^2 x}{\sin^2 x}d(\sin x) = \int\biggl(\dfrac{1}{u^2} - 1\biggr)du = -\dfrac{1}{u} - u \quad \Rightarrow$
$\int \dfrac{\cos^3x}{\sin^2x}dx = -\dfrac{1}{\sin x} - \sin x + C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/835881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Limit of $\lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {n^2 +i} }$ Direct way leads to nowhere.
$$\lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {n^2 +i} } = \lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{n \sqrt {1 +\frac{i}{n^2}} } = $$
$$ = \lim_{n\rightarrow \infty} \frac{1}{n} \cdot \lim_... | Notice that $$\frac{n}{\sqrt{n^2+n}} =\sum_{i=1}^n \frac{1}{\sqrt{n^2+n}}\leq \sum_{i=1}^n\frac{1}{\sqrt{n^2+i}}\leq \sum_{i=1}^n \frac{1}{n} =1$$ Since $\frac{n}{\sqrt{n^2+n}}=\frac{1}{\sqrt{1+\frac{1}{n}}}\to 1$, you get that $1$ as your limit by squeeze theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/836006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Calculation of $ \lim_{x\rightarrow 1}\frac{(1-x)\cdot(1-x^2)\cdot(1-x^3)\cdots(1-x^{2n})}{\{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)\}^2} = $ Calculation of $\displaystyle \lim_{x\rightarrow 1}\frac{(1-x)\cdot(1-x^2)\cdot(1-x^3)\cdots (1-x^{2n})}{\{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)\}^2} = $
My Trial After... | Try the factorization $(1-x^n) = (1-x)(1+x+\dots+x^{n-1})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/836841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to solve this equation in general $ax^3+bx^2+cx+d=0 ?$ For example if you have $ax^2+bx+c=0$ then solution given by formula $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
Is there formula like this for $ax^3+bx^2+cx+d=0 ?$
| Yes, such formula exists, and it is called the cubic formula, however it isn't nice looking: $$\begin{align}x &= \sqrt[\displaystyle3\,]{\left(\dfrac{-b^3}{27a^3} + \dfrac{bc}{6a^2} - \dfrac{d}{2a}\right) + \sqrt{\left(\dfrac{-b^3}{27a^3} + \dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)^2 + \left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\lim_{n \rightarrow \infty} \sqrt{n+1}(\sqrt{n^3 + n} - \sqrt{n^3 + 1})$ I wish to compute the following limit
$$\lim_{n \rightarrow \infty} \sqrt{n+1}(\sqrt{n^3 + n} - \sqrt{n^3 + 1})$$
Wolfram Alpha gives the answer as $1/2$. I've tried L'Hopital's rule, but I can't seem to get it to work.
| $$ \sqrt{n+1}(\sqrt{n^3 + n} - \sqrt{n^3 + 1}) \cdot \frac{\sqrt{n^3 + n}+ \sqrt{n^3 + 1}}{\sqrt{n^3 + n}+ \sqrt{n^3 + 1}} = \frac{\sqrt{n+1}(n-1)}{\sqrt{n^3 + n}+ \sqrt{n^3 + 1}}$$ Now divide numerator and denominator by $\sqrt{n^3}.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Limit of a sequence $a_1=1;a_{n+1}=(n+1)(1+a_n)$ Let a sequence be $a_1=1;a_{n+1}=(n+1)(1+a_n)$
If $P_n=\prod_1^n(1+a_i^{-1})$
then $$\lim_{n\to\infty}P_n $$is ?
I did:
$$P_n=\prod_1^n\frac{(1+a_i)}{a_i}
=\frac{a_{n+1}}{a_1}\prod1^{n-1}\frac{(1+a_i)}{a_{i+1}}
=\frac{a_{n+1}}{a_1}\prod_1^{n-1}\frac1{i+1}=\frac{a_{n+1}}... | Let a sequence be $a_1=1;a_{n+1}=(n+1)(1+a_n)$
We have $$P_n=\prod_{i=1}^n(1+a_i^{-1})$$
then
$$
P_n
=\prod_{i=1}^n (1 + \frac{1}{a_i})
=\left (1 + \frac{1}{n(1+a_{n-1})}\right )\prod_{i=1}^{n-1} (1+a_i^{-1} )
=\left (\frac{n(1+a_{n-1}) + 1}{n(1+a_{n-1})}\right )\prod_{i=1}^{n-1} (1+a_i^{-1} )
=\left (\frac{n(1+a_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/840939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Local minimum of $f(x) = 4x + \frac{9\pi^2}{x} + \sin x$ What's the minimum value of the function
$$f(x) = 4x + \frac{9\pi^2}{x} + \sin x$$
for $0 < x < +\infty$? The answer should be $12\pi - 1$, but I get stuck with the expression involving both $\cos x$ and $x^2$ in the derivative. Taking the derivative, we have:
$$... | Important fact: the minimum of the function is NOT the same as where $f'(x) = 0$. The latter is a local minimum - a point that it is minimal compared to it's local points, but taking as a simple example the function $f(x) = x^3$ on $[-1,1]$ the function is minimised at $-1$ despite this not being a turning point.
Despi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Providing solutions for the intersection of two trigonometric functions I'm trying to find a general solution for the intersection of two trigonometric functions:
$$a(x)=500\sin \left( \frac{\pi x}{2} \right)+150$$
$$a(x)=200\cos \left( \frac{\pi }{2}\left( x+\frac{\pi }{2} \right) \right)+250$$
When graphing the equat... | Hint:
let $a=\pi/4$.
Step (1) expand $\sin(2a(x+2a))$:
$$\sin(2a(x+2a))=\sin(2ax)\cos(4a^2)+\cos(2ax)\sin(4a^2)......(1)$$
Step (2) Change variable:
$$\sin(2ax)=\frac{2t}{1+t^2}, \cos(2ax)=\frac{1-t^2}{1+t^2},t=\tan(ax)$$
Step (3) solve the resulting equation for $t$.
Step (4) solve $t=\tan(ax)$ for $x$.
| {
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"url": "https://math.stackexchange.com/questions/844256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\lim_{x \rightarrow 1} \frac{x^4-2x+1}{x-1} + \sqrt{x} =3$
Show that $\lim_{x \rightarrow 1} \frac{x^4-2x+1}{x-1} + \sqrt{x} =3$ from the definition (using $\epsilon-\delta$)
Why can't I do something like this?
We want:
$|\frac{x^4-2x+1}{x-1} + \sqrt{x}-3| = |\frac{x^4-2x+1}{x-1} + \frac{x-9}{\sqrt{x}+3}| ... | Note: This answer was posted before the condition of using the definition with $(\epsilon-\delta)$ was specified.
Factorize $x^4-2x+1$, to remove the singularity in the denominator:
\begin{aligned}
\lim_{x\to1}\frac{x^4-2x+1}{x-1} + \sqrt{x} & = \lim_{x\to1}\frac{(x-1)(x^3+x^2+x-1)}{x-1} + \sqrt{x} = \\
& = \lim_{x\to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/844603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$A$ is not similar to a diagonal matrix over the reals Let $A = \begin{bmatrix} 6 & -3 & -2 \\ 4 & -1 & -2 \\ 10 & -5 & -3 \end{bmatrix} $ then $A$ is not similar to a diagonal matrix over the reals and it is not similar to a diagonal matrix over the complexes.
We know that $A$ is similar to a diagonal matrix over the ... | *
*$A$ is diagonalisable to a matrix over $\mathbb{C}$.
*The diagonal components contain complex numbers.
*Matrix diagonals are unique (up to a permutation).
Therefore it is not diagonalisable over $\mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/844724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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system of two quadratic equations with two variables Is there a general way to solve exactly a system of this shape (the $a_i$ are constants):
$$\begin{array}{cc}a_1x^2+a_2x+a_3y^2+a_4y+a_5=0\\
a_6xy+a_7x+a_8y+a_9=0
\end{array}
$$
It comes from a geometrical problem: the first equation states two vectors have same leng... | Assume your $a_1\neq0,\,a_6\neq0$. Dividing by these, you can have the slightly simpler system with two fewer constants,
$$x^2 + b_1 x + b_2 y^2 + b_3 y + b_4 = 0$$
$$x y + b_5 x + b_6 y + b_7 = 0$$
Eliminating $y$ results in a monic quartic of form,
$$x^4+ax^3+bx^2+cx+d=0\tag1$$
The solution of the general quartic is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/845034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to establish this inequality: $(1-a)(1-b)(1-c) \geq 8abc$ for $a+b+c=1$? Let $a$, $b$, and $c$ be positive real numbers such that $a+b+c = 1$. Then how to establish the following inequality? $$ (1-a)(1-b)(1-c) \geq 8abc.$$
My effort:
Since $a+b+c =1$, we can write $$ (1-a)(1-b)(1-c) = 1 - (a+b+c) +(ab+bc+ca) - ab... | Because $x+y \ge 2\sqrt{xy}$ for any $x>0,y>0$, and $1-a = (a+b+c) - a = b+c$ (similarly for $1-b$ and $1-c$) we have
$$(1-a)(1-b)(1-c) = (b+c)(c+a)(a+b) \ge 2\sqrt{bc}\cdot 2\sqrt{ca}\cdot 2\sqrt{ab} = 8abc.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/845607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
For the matrix A as given below, which of them satisfy $A^6=I$? For the matrix A as given below, which of them satisfy $A^6=I$?
*
*$A=\left(\begin{array}{ccc}
\cos\frac{\pi}{4}&\sin\frac{\pi}{4}&0\\
-\sin\frac{\pi}{4}&\cos\frac{\pi}{4}&0\\
0&0&1
\end{array}\right)$
*$A=\left(\begin{array}{ccc}
1&0&0\\
0&cos\frac{\p... | Let us take the option (3). the matrix is a rotation matrix. The rotation is about the Y-axis. The X and Z axes are rotated by $\frac{\pi}{6}$. So $A^6$ is same as rotating X and Z axes by $\frac{\pi}{6}$ about the Y Axis, sizx times. This is same as rotating X and Z axes by $\pi$ about the Y Axis, the rotation matrix ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/845707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding Integral of $y = e^{ax}\sin (bx)$ I was solving questions like $ \int e^{2x} \sin x dx. $ I decided to find the general term for $$\int e^{ax} \sin (bx) $$ which can be directly used in questions like (above).
I hope it helps others :) (Just wanted to share my work. I will appreciate feedback/suggestions or an... | $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \large{\textbf{Method - 2}}$
Let : $I = \int e^{ax} \sin (bx) $
And say :
$$y_1 = e^{ax} \sin (bx) \\
y_2 = e^{ax} \cos (bx) $$
And : $$y'_1 = ae^{ax}\sin (bx) + e^{ax}b\cos (bx) \\
y'_2 = -e^{ax} b\sin (bx) + ae^{ax} \cos (bx) $$
Or we can also write $y'_1$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/848220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Producing a CDF from a given PDF So I have this PDF:
$$
f(x)=
\begin{cases}
x + 3 & \text{ for } -3 \leq x < -2\\
3 - x & \text{ for } 2 \leq x < 3\\
0 & \text{ otherwise}
\end{cases}
$$
To make this a CDF, I have integrated the PDF from $-\infty$ to some value, $x$.
$$
F(x)=
\begin{cases}
\frac{x^2}{2} + 3x + \fra... | Note first of all that the cdf is defined for all $x$, so a complete answer should take care of all $x$.
If $x\lt -3$, everything is easy. We have $\int_{-\infty}^x f_X(t)\,dt=0$. So $F_X(x)=0$ if $x\lt -3$.
Now for $-3\le x\lt -2$, in effect we are integrating $t+3$ from $-3$ to $x$, since there is no "mass" before $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/851854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Discrete math induction problem. I am stuck at this step in the inductive process and I was wondering if someone can help me out from where I am stuck.
Question:
if $n$ is a positive integer, prove that,
$$1\cdot2+2\cdot3+3\cdot4+\cdots+n\cdot(n+1) = \frac{n(n+1)(n+2)}3$$
Basic step is $1$, then assume $k$ is true and ... | Finishing up the induction proof,\begin{align}
1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + (k+1)(k+2) &= [1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + k(k+1)]+(k+1)(k+2) \\
&= \frac{k(k+1)(k+2)}{3}+(k+1)(k+2) \\
&= \frac{k(k+1)(k+2)}{3}+\frac{3(k+1)(k+2)}{3} \\
&= \frac{k(k+1)(k+2)+3(k+1)(k+2)}{3} \\
&= \frac{(k+1)(k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/855680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
confirmation of method: Write the vector $(2,-3)$ as a linear combination of vectors $(-1,4)$ and $(1,2)$ Write the vector $(2,-3)$ as a linear combination of vectors $(-1,4)$ and $(1,2)$
this was on an assignment rather hw so I want to be sure I'm using the right method and have the right answer.
I confirm $w = k_1u ... | It shouldn't be this matrix
$$
\left(
\begin{array}{cc|c}
-1 & 4 & 2 \\
1 & 2 & -3
\end{array}
\right)
$$
since this corresponds to the system of equations
\begin{align*}
-k_1+4k_2 &= 2 \\
k_1+2k_2 &= -3 \\
\end{align*}
whereas you want the system of equations
\begin{align*}
-k_1+k_2 &= 2 \\
4k_1+2k_2 &= -3 \\
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/856973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How does $n+2\mid(n^2+n+1)(n^2+n+2)$ imply $n+2\mid12$? I'm trying to show that if $n+2\mid(n^2+n+1)(n^2+n+2)$, then $n+2 \mid 12$ for all $n\in \mathbb{N}$.
| Observe that $\displaystyle n^2+n=n(n+2)-(n+2)+2=(n+2)(n-1)+2$
$\displaystyle\implies (n^2+n+1)(n^2+n+2)=\{\underbrace{(n+2)(n-1)+2}+1\}\{\underbrace{(n+2)(n-1)+2}+2\}\equiv(2+1)(2+2)\pmod{(n+2)(n-1)}\equiv12\pmod{(n+2)} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/857798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving a quadratic trigonometric equation? The equation is $6 \cos^2x+\cos x=1$,
My work:
$6x^2+x-1=0$
$(3x-1)(2x+1)$
$3x-1=0 ∨ 2x+1=0$
$x=\frac{1}{3} ∨ x= \frac{-1}{2}$
But I do not know how to progress further.
| It is preferable to change the variable so let $t=\cos x$ and we solve the quadratic equation
$$6t^2+t=1$$
and we find $t=\frac13$ and $t=-\frac12$. Now
$$\cos x=\frac 13\iff \left(x=\arccos\left(\frac13\right)+2k\pi\right)\lor\left( x=2k\pi-\arccos\left(\frac13\right)\right)$$
and
$$\cos x=-\frac12\iff \left(x=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/858415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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If $0\leq a \leq 3; 0\leq b \leq 3$ and the equation $x^2 +4+3 cos(ax+b)=2x$ has at least one solution , then find the value of a+b Problem : If $0\leq a \leq 3; 0\leq b \leq 3$ and the equation $x^2 +4+3 cos(ax+b)=2x$ has at least one solution , then find the value of a+b.
Solution :
We can write the given equation... | $$x^2-2x+4 =-3\cos(ax+b)$$
$$(x-1)^2 +3 =-3\cos(ax+b)$$
$$\frac{(x-1)^2 +3}{-3} =\cos(ax+b)$$
how $-1\leq\cos{(ax+b)}\leq 1$, then $x=1$
$$\cos(a+b)=-1$$
$$\cos(a+b)=-1$$
$$a+b=\pi$$
because $a,b\leq3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/858547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Trigonometry calculation problem Problem: Calculate $\cos(\alpha+60)$, if $\sin(\alpha)=\frac{2}{3}$ and $\alpha\in[90^\circ,180^\circ]$.
I have tried following:
$$\cos(60^\circ)\cos\alpha-\sin(60^\circ)\sin\alpha=\cos(\alpha+60^\circ)\\\frac{\cos\alpha}{2}-\frac{\sqrt{3}}{3}=\cos(\alpha+60^\circ)$$
I would appreciate... | Using the identity $$\sin^2 \alpha + \cos^2 \alpha = 1$$
$$\frac{4}{9} + \cos^2 \alpha = 1$$
$$\cos^2 \alpha = \frac{5}{9}$$
$$\cos \alpha = \frac{\sqrt{5}}{3}$$
Using another identity:
$$\cos(60^o)\cos\alpha-\sin(60^o)\sin\alpha=\cos(\alpha+60^o)\\\frac{\cos\alpha}{2}-\frac{\sqrt{3}}{3}=\cos(\alpha+60^o)$$
$$\cos(\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/859544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Easier way to solve $\lim \limits_{x \to 0} \space \cot{x}-{1 \over x}$ using L'Hospital's Rule? This is what I did for:
$$\lim \limits_{x \to 0} \space \cot{x}-{1 \over x}$$
*
*Check form: $\infty - \infty$.
*Rearrange it to be a quotient:
$$\begin{align} \\
& =\lim \limits_{x \to 0} \space {\cos{x} \over \sin{x}... | Speaking of simplicity, often by L'Hospital rule can be applied in combination with so-called remarkable limits. If this exercise can do so:
$$\lim \limits_{x \to 0}( \cot{x}-{1 \over x})=\lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x\sin x} = \lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x^2}\cdot\lim \limits_{x \to 0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/860283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Find the number of integers solutions How many solutions are there to the equation $$x_1+x_2+x_3+x_4=39,$$
I) where $x_1,x_2,x_3,x_4$ are nonnegative integers,
II) where $x_1,x_2,x_3,x_4$ are nonnegative integers such that $3 \leq x_1,x_2,x_3 \leq 12$ and $0 \leq x_4$.
I) There is a 1-1 correspondence between the solut... | Assume $a$ as a real number, such that $0<a <1$
I.
The number of integral solutions should be the same as:
The coefficient of $a ^ {39}$ in
$ \left(1+a + a^{2} + a^{3} + ... \right)^4$
Which is coeffient of $a^{39}$ in the series for
$ \left(1-a\right)^{-4}$
which is $\binom{39+4}{3}$
II.
Answer is same as the coeff... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/861545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Global Max and Min Hi can someone please help me figure out how to start this question?
Temperature within a circular region $R = \{(x, y)| x^2 + y^2 \le 49\}$ is given by $T(x,y) = 4x^2 -4xy + y^2$
Find the global minimum and maximum temperatures in the region and the points where these extreme temperatures occur.
I'm... | Let $G(x,y) = x^2 + y^2 - 49$. At the interior points using critical points approach we have:
$T_x = 8x - 4y = 0$, and $T_y = 2y - 4x = 0$ both gives $y = 2x$. Thus. $T(x,2x) = 4x^2 - 4x(2x) + (2x)^2 = 0$.
At the boundary, using Lagrange Multiplier we have:$T_x = \lambda G_x$, and $T_y = \lambda G_y$ gives:
$8x - 4y =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/866434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Solve this with CBS: minimum value of $ 1/x + 4/y + 9/z $ with $x+y+z=1$ How can you see the minimum value of $ 1/x + 4/y + 9/z $ with $x+y+z=1$ using the CBS inequality?
I have seen a proof of that that use trigonometric substitutions, but I don't see as one-step the solution using the CBS inequality.
| Consider positive values $x,y,z$.
Denote
$$(a,b,c)=\left( \frac{1}{\sqrt{x}}, \frac{2}{\sqrt{y}}, \frac{3}{\sqrt{z}}\right),$$
$$(A,B,C)=\left( \sqrt{x}, \sqrt{y}, \sqrt{z}\right),$$
then (by CBS inequality)
$$(a^2+b^2+c^2)(A^2+B^2+C^2)\ge (aA+bB+cC)^2,$$
so
$$
\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right) \left(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/866858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the radius Consider the parabola $y=x^2$ and a circle which is tangent to the parabola at the points $(1,1)$ and $(-1,1)$.Find the radius of circle.
My try:I write the general equation of circle
$(x-h)^2+(y-k)^2=r^2$
and substitute the points $(1,1)$ and $(-1,1)$ in the equation of circle,i find $h=0$.
Now furthe... | Substitute $(1,1),(-1,1)$ in $(x-h)^2+(y-k)^2=r^2$
you will get
$h^2+k^2-2h-2k+2=r^2$ and $h^2+k^2+2h-2k+2=r^2$
From here you get $h=0$, substitute this in one of the above to obtain
$k^2-2k+2=r^2$
$m_1$=slope of tangent line at $(1,1)=2$ (here use the $y=x^2$)
Where $m_2$=the slope of radial line at $(1,1)= 1-k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/869542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int_1^{\sqrt{2}}\frac{1}{x}\ln\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)dx$
Calculate the following integral:
\begin{equation}
\int_1^{\sqrt{2}}\frac{1}{x}\ln\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)dx
\end{equation}
I am having trouble to calculate the integral. I tried to use by parts method but it... | Update: Finally, a complete solution. Sorry it took so long.
Split the integral up into 3.
\begin{align}
I
&=-\int^{\sqrt{2}}_1\frac{\log{x}}{x}dx+\int^{\sqrt{2}}_1\frac{\log{((x^2-1)^2+1)}}{x}dx-\int^{\sqrt{2}}_1\frac{\log{((x-1)^2+1)}}{x}dx\\
&=-\frac{1}{8}(\log{2})^2+\frac{1}{2}\int^1_0\frac{\log(1+x^2)}{1+x}dx-\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/870751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 0
} |
Linear Algebra - elimination and linear systems By given this matrix:
\begin{pmatrix}1&1&1&0\\2&3&k&1\\3&k&5&1\end{pmatrix}
I need to find, what are the values of k the system has infinity/single/no solution.
So I did elimination and I got:
\begin{pmatrix}1&1&1&0\\0&1&k-2&1\\0&k-4&4-k&0\end{pmatrix}
As one can see, for... | You have obtained
$$\left(\begin{array}{ccc|c}
1 & 1 & 1 & 0 \\
0 & 1 & k-2 & 1 \\
0 & k-4 & 4-k & 0
\end{array}\right)$$
and correctly explained what happens for $k=4$.
Now let us have a look at the case $k\ne 4$. In this case $k-4\ne0$ and you can divide the last row by $(k-4)$:
$$\left(\begin{array}{ccc|c}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/871693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Suggestion for Computing an Integral Let $$A=\left\{(x,y,z)\in \mathbb R^3:\dfrac{x^2}{2}+\dfrac{y^4}{4}+\dfrac{z^6}{6}\leq1\right\}.$$
Then I want to compute the following integral:
$$\frac{1}{\operatorname{vol}(A)}\displaystyle\int_{\partial A}^{}\!\frac{1}{\sqrt{x^2+y^6+z^{10}}}\, d(x,y,z)$$
Should I use spherical c... | Notice that the normal vector of $A's$ boundary is $\hat{n}=\frac{(x, y^3, z^5)}{\sqrt{x^2+y^6+z^10}}$. Then we choose $F(x,y,z)=(\frac{x}{2},\frac{y}{4},\frac{z}{6})$, on the boundary, we have $$F\cdot \hat{n}=\frac{\frac{x^2}{2}+\frac{y^4}{4}+\frac{z^6}{6}}{\sqrt{x^2+y^6+z^{10}}}=\frac{1}{\sqrt{x^2+y^6+z^{10}}}$$ So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/871950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Integration problem $\displaystyle \int \frac{dx}{x(x^3+8)}$ $$\int \frac{dx}{x(x^3+8)}$$
I think I'm supposed to use partial fractions, but I am unsure of how to start the problem. Any help would be appreciated.
| Here's the set up:
$$\int \frac{dx}{x(x^3 + 8)} = \int \frac{dx}{x(x+2)(x^2 -2x + 4)} = \int \left(\frac{A}{x} + \frac{B}{x+2} + \frac{Cx + D}{x^2-2x + 4}\right) \,dx$$
Now you can solve for $A, B, C, D$.
Note: Differences and sums of cubes factor predictably: $$a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$$
In this quest... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/872036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
$\ \sqrt{x+39}-\sqrt{x+7}=4 $ So I tried to solve this problem for x
$\ \sqrt{x+39}-\sqrt{x+7}=4 $
I multiplied both sides ($\ \sqrt{m}\cdot\sqrt{n}=\sqrt{mn} $)
$\ (\sqrt{x+39}-\sqrt{x+7})^2=16 $
$\ (x+39)-2(x^2+46x+273)-(x+7) $
$\ 0x+32+(-2x^2-92x-546) $
$\ -2x^2-92x-514 $
divide the 2 out
$\ x^2+46x+257=-8 $
$\ x^2+... | One can square twice. I prefer to invert and obtain
$$\frac{1}{\sqrt{x+39}-\sqrt{x+7}}=\frac{1}{4},$$
and then by rationalizing the denominator get
$$\frac{\sqrt{x+39}+\sqrt{x+7}}{32}=\frac{1}{4},$$
or equivalently
$$\sqrt{x+39}+\sqrt{x+7}=8.$$
Now "add" to the original equation, divide by $2$. We get $\sqrt{x+39}=6$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/872183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Sum of all real solutions for $x$ to the equation $\displaystyle (x^2+4x+6)^{{(x^2+4x+6)}^{\left(x^2+4x+6\right)}}=2014.$ Find the sum of all real solutions for $x$ to the equation $\displaystyle (x^2+4x+6)^{{(x^2+4x+6)}^{\left(x^2+4x+6\right)}}=2014.$
$\bf{My\; Try::}$ Let $y=x^2+4x+6 = (x+2)^2+2\geq 2$.
So our exp. e... | As you have written if one substitutes $y=x^2 + 4x +6 = (x+2)^2+2\ge 2$, then the resulting equation is $f(y)=2014$ where $f(y)=y^{y^y}$. Due the the fact that $2014>2^{2^2}$, it accepts solutions in ${\mathbb R}$. Now note that $f(y)=y^{y^y}$ is an strictly increasing function on ${\mathbb R}_{\ge 2}$. So for $f(y)=20... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/873928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Which one is correct for $\sqrt{-16} \times \sqrt{-1}$? $4$ or $-4$? As we can find in order to evaluate $\sqrt{-16} \times \sqrt{-1}$, we can do it in two ways.
FIRST
\begin{align*}
\sqrt{-16} \times \sqrt{-1} &= \sqrt{(-16) \times (-1)}\\
&= \sqrt{16}\\
&=4
\end{align*}
SECOND
\begin{align*}
\sqr... | $$\sqrt[n]{a b} = \sqrt[n]{a} \, \sqrt[n]{b} \quad (*)$$
if $a$ and $b$ are negative, then $(*)$ works only for $n$-th roots with odd $n$,
alas $n = 2$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/876084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Easy algebra question: show equality? Let $c:=(u+d)/2$ and $-1<d<0<u$.
Show:
Can you show the steps to arrive from lhs to rhs?
| $$
\frac{1 + u}{2}(1 + 2c) + \frac{1+d}{2}(1 + c) = \frac{1 + u}{2}(1 + c) + \frac{1 + u}{2}c + \frac{1+d}{2}(1 + c)
$$
$$
=(1 + c)\biggl(\frac{1 + u}{2} + \frac{1 + d}{2} \biggr) + \frac{1 + u}{2}c = (1 + c)\biggl(1 + \frac{u + d}{2}\biggr) + \frac{1 + u}{2}c
$$
$$
= (1 + c)^2 + \frac{1 + u}{2}c
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/878144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Why $ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$? Why $ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$?
I am learning trigonometric identities one identity I have to proof is the next:
$ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$
so I t... | They are just the distributive property
$$a(b+c)=ab+ac$$
in the first one we have $a=-\sin\alpha, b= 1, c=\cos\alpha$ which comes from the terms
$$2(1+\underbrace{(-\sin\alpha)}_{a}\cdot\underbrace{1}_{b}+\cos\alpha+\underbrace{(-\sin\alpha)}_{a}\cdot\underbrace{\cos\alpha}_{c})$$
The next step is the same property w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/881273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Determining the infinite limit of a Riemann' sum I need to evaluate the following $f(x)=x^2 + 2x - 5$, on $[1,4]$, by using the riemans sum and limiting it to infinity. I have set up everything $\Delta x=\frac{3}{n}$.
$x_i=1+\frac{3i}{n}$, I would like to just check my workings against someone who actually finds this f... | My suggestion is that when you calculate an integral using Riemann sums it might be easier to find $\int_0^4 f-\int_0^1 f$ than $\int_1^4 f$ in one step -avoiding terms of the form $(h+\frac{k}{n})^c$-.
Let $R_n^{4}$ be the Riemann sum of n-terms from $0$ to $1 $ and $R_n^1$ the Riemann sum from $0$ to $1$. It should ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/881726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Factoring $x^4+x^2+2x+6$ We have to factor $x^4+x^2+2x+6$.Factoring through factor theorem is not helpful in this question as the question does not follow the integral root theorem i.e. the root of this expression is not any of the factors of 6. I am totally confused.Please help in factoring it!
| The polynomial $x^4+x^2+2x+6$ is $x^4+(x+1)^2+5\geqslant5$ for every $x$ hence it has no real root. Thus, there is no factor of degree $1$ hence one can only have two factors of degree $2$, that is, the factorization to be found is $$x^4+x^2+2x+6=(x^2+ax+b)(x^2+cx+d),$$ for some $(a,b,c,d)$. The coefficients of $x^3$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/882670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.