Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to solve $x^3 \equiv 1 \pmod{37}$ We are asked to solve $x^3 \equiv 1 \pmod{37}$. I know that the answer is $10$ since $27\cdot37 = 999$ and $10^3 = 1000$ but how do I show this rigorously? If it helps, we are given the primitive roots of $37$ which are $2, 5, 13, 15, 17, 18, 19, 20, 22, 24, 32$, and $35$. But I am... | Just for fun another way.
Because of $\mathbb F_{37}$ is a field the equation in $\mathbb F_{37}[x]$ $$x^3=1\iff (x-1)(x^2+x+1)=0$$
must have three roots so, since $1$ is clearly root, one can find out the other two ones solving $$x^2+x+1=0$$ We have
$$x+y=-1=36\\xy=1$$ Exploring $$35+1\\34+2\\33+3\\32+4\\31+5\\30+6\\2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2708291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
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Proving distributive law of natural numbers Is my proof correct?
If we define multiplication for natural numbers as
$a \times S(b) = (a \times b) + a$
$a \times 0 = 0$
And addition as
$a + 0 = a$
$a + S(b) = S(a+b)$
Where $S(n)$ is the successor function of $n$ (and assume we've already proven commutative property, etc... | Perfect ... though you may want to note that you need associativity of addition, since:
$a \times (b + S(c)) = a \times (S(b + c)) = (a \times (b+c)) + a = \color{red}{(}(a \times b) + (a \times c)\color{red}) + a$
while:
$(a \times b) + (a \times S(c)) = (a \times b) + \color{red}((a \times c) + a\color{red})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2708849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve an equation with integer variables I am trying to find all the integers $(n,m)$ such that $$2^{2n+1} + 2^n + 1 = m^2$$.
With a simple python program, I find that $n=0$, $n=4$ are the only solutions less than $50$. However with the precision of number I can not get accurate results when $n > 50$. Does any one have... | Suppose $n > 2$. We have $m^2 -1 = (m-1)(m+1)\equiv 0 \mod 2^n$, so one of $m-1$ and $m+1$ is divisible by $2^{n-1}$ (the other is divisible by $2$ but no higher power of $2$). By symmetry $m \to -m$, we may assume it is $m-1$. Thus
$m = 1 + k 2^{n-1}$. With this substitution the equation becomes
$$ 2^{2n} \left(2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2709676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
solving $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$ I have to solve $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$
Dividing by $dx$ we have
$x + xy^2 + yy' + yy'x^2=0$
From where,
$$\frac{yy'}{1+y^2}+\frac{x}{1+x^2}=\frac{y\,dy}{1+y^2}+\frac{x\,dx}{1+x^2}=\\ =\frac{d(y^2+1)}{1+y^2}+\frac{d(x^2+1)}{1+x^2}= \frac{1}{2}d\ln(1+y^2)+\... | it is $$-\frac{y'(x)}{\frac{1+y(x)^2}{y(x)}}=\frac{x}{1+x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
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Triangle inequality with the exradii $r_{a}$, $r_{b}$, $r_{c}$, the medians $m_{a}$, $m_{b}$, $m_{c}$ Given a triangle with the exradii $r_{a}$, $r_{b}$, $r_{c}$, the medians $m_{a}$, $m_{b}$, $m_{c}$. Show that
$$r_{a}^{2}+ r_{b}^{2}+ r_{c}^{2}\geq 3\sqrt{3}. S+ \left ( m_{a}- m_{b} \right )^{2}+ \left ( m_{b}- m_{c}... | Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, $x$, $y$ and $z$ they are positives and we need to prove that
$$\sum_{cyc}\frac{4S^2}{(b+c-a)^2}\geq3\sqrt{3xyz(x+y+z)}+2\sum_{cyc}(m_a^2-m_am_b)$$ or
$$\sum_{cyc}\frac{xyz(x+y+z)}{x^2}+\frac{1}{2}\sum_{cyc}\sqrt{\left(4x(x+y+z)+(y-z)^2\right)\left(4y(x+y+z)+(x-z)^2\right)}\geq... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Multiplicative group of integers modulo $p$ $\mathbb{Z}/7\mathbb{Z}=\{1,2,3,4,5,6\}$.
$6\times 6=1~{\rm mod}~ 7$ implies $6$ is an element of order $2$; however, we know that $\mathbb{Z}/7\mathbb{Z}\cong C_7$, not containing an element of order $2$.
I found it incredibly confusing, what have I missed? Any help will be... | The multiplicative group modulo a prime $n$ and a group opperator of multiplication and the element $0$ removed is equivalent to an additive group modulo $n-1$ with zero in.
$A=\mathbb Z/7\mathbb Z-\{0\}^{\times}\cong B=\mathbb Z/6\mathbb Z^{+}$
$1 \to 0$ because $1\times a = a \to 0 + a = a$.
$5 \to 1$ because then $A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2712491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Factoring $x^6+x^5+x^4+x^3+x^2+x+1$ into two cubics On page 587 of Dummit and Foote, the authors demonstrate producing irreducible polynomials over $\mathbb{F}_p$ recursively. This is after Proposition 18 which states "The polynomial $x^{p^n}-x$ is precisely the product of all the distinct irreducible polynomials in $\... | We want to find $(x^3+ax^2+bx+c)(x^3+dx^2+ex+f)$. We have little choice for $c$ and $f$, as they must be $1$. Also, changing $x$ into $1/x$ and multiplying by $x^6$ should leave the thing fixed: since we get
$$
(x^3+bx^2+ax+1)(x^3+ex^2+dx+1)
$$
we need either $a=b$ and $d=e$ or $a=e$ and $b=d$. This is because of uniqu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all integer solutions of: $\;\frac{1}{m}+\frac{1}{n}-\frac{1}{mn^2}=\frac{3}{4}$ I found the following problem from the 10th Iranian Mathematical Olympiad in Crux Magazine.
Find all integer solutions of $$\frac{1}{m}+\frac{1}{n}-\frac{1}{mn^2}=\frac{3}{4}$$
Initially it looked like a typical quadratic problem, h... | The other answers may be more elegant, but here is a way to continue from the point you reached. Assume $m$ and $n$ are integers. From
$$n = \frac{-2m \pm 2\sqrt{m^2-3m+4}}{4-3m}$$
you know that $-2m \pm 2\sqrt{m^2-3m+4}$ must be a multiple of
the integer $4-3m$; in particular, it must be an integer.
Therefore $2\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 4
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Is this proof of the irrationality of the square root of $3$ valid? Assume that $\sqrt{3}=\frac{a}{b}$ where $\frac{a}{b}$ is in its simplest possible form. So $3=\frac{a^2}{b^2}$, hence $3b^2=a^2$. If $b$ is even then $a^2$ and $a$ are also even. That means they have a common factor of $2$. But this is impossible sinc... | Short version:
$$c^2\bmod4=c\bmod2$$
so that with $a,b$ not both even,
$$(3a^2-b^2)\bmod4\ne0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Define an isomorphism from $\mathbf{R}^2 $to $S$, where $S=\operatorname{span} .... $(image) Not a clue what to do with this question. We only briefly touched on isomorphisms in class,and I can't find any example online that uses transformations on two basis vectors, as in this question. Help a plebe out?
| For part $a)$ You can define a Linear map $T:\mathbf{R^2}\to S$ as follows
$$T\left(c_1\begin{pmatrix}1\\0\\\end{pmatrix}+c_2\begin{pmatrix}0\\1\\\end{pmatrix}\right) = c_1\begin{pmatrix}3\\1\\4\\\end{pmatrix}+c_2\begin{pmatrix}2\\-7\\1\\\end{pmatrix}$$
The remaining task is then to show that with the above definition ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2719811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Horizontal asymptotes of function defined as an inequation I was given as a homework:
Suppose $f$ is a function of $x$ and, $\forall x\in \mathbb{R}$, it holds
$$
\left| f(x)-\frac{1}{x}\right| \le \frac{2x^2+x|x|+2}{x^2+1}
$$
Find the horizontal asymptotes of $f(x)$.
I tried to use the Squeeze Theorem, rewritting the ... | There will be no hints on how to solve it because the condition you gave is not enough to completely determine horizontal asymptotes. For $x < 0$, the bounding expression becomes $$\frac{2x^2+x|x|+2}{x^2+1} = \frac {x^2 + 2}{x^2+1} = 1 + \frac 1{x^2+1}$$
which converges to $1$ as $x \to -\infty$, while for $x > 0$, it ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2722136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving the trig identity $\frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta}$ without cross-multiplying I need to prove the following identity.
$$\frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta}$$
I want to prove it by deduction rather than cross multiplying.
| You could use the half angle formulae .
$\large\frac{1-\cos\theta}{\sin\theta}$
$=\large\frac{2\sin^2\frac\theta2}{2\sin\frac\theta2.cos\frac\theta2}$
$= \large\frac{2\sin\frac\theta2}{2\cos\frac\theta2}$
multiply both numerator and denominator by $\cos\frac\theta2$
$=\large\frac{ 2\sin\frac\theta2\cos\frac\theta2}{2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2722902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 7
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Verifying Isomorphism between finite fields On page 587 in Dummit and Foote, we are given that $f_1(x)=x^4+x^3+1$ and $f_2(x)=x^4+x+1$ are two irreducible quartics over $\mathbb{F}_2$.
The authors then claim that a "simple calculation" verifies that $\alpha (x)=x^3+x^2$ is a root of $f_2(x)$ in $\mathbb{F}_{16}=\mathbb... | The answer is easy. We have
$$
x^{12}+x^8+x^3+x^2+1=
(x^8 + x^7 + x^6 + x^5 + x^4 + x^2 + 1)(x^4 + x^3 + 1)=0
$$
because $x^4+x^3+1=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Proving Schwarz inequlity. Is the following proof correct?
The Schwarz inequality states:
$$
x_1 \cdot y_1 + x_2 \cdot y_2 \leq \sqrt{(x_1^2+x_2^2)\cdot(y_1^2+y_2^2)}
$$
From
$$
(x_1^2+x_2^2)\cdot(y_1^2+y_2^2)
$$
We can deduce:
$$
(x_1^2+x_2^2)\cdot(y_1^2+y_2^2) = (x_1y_1+x_2y_2)^2 + (x_1y_2-x_2y_1)^2
$$
We can rewrite... | Yes it is correct even if from here
$$(x_1^2+x_2^2)\cdot(y_1^2+y_2^2) = (x_1y_1+x_2y_2)^2 + (x_1y_2-x_2y_1)^2$$
we can deduce directly that
$$(x_1^2+x_2^2)\cdot(y_1^2+y_2^2) - (x_1y_1+x_2y_2)^2=(x_1y_2-x_2y_1)^2 \geq 0$$
This proof is nice but it is limited to n=2, for a general proof we can start from
$$\sum x_iy_i\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Elementary geometry problem about two squares.
Consider the picture above, where $ABCD$ and $EFGC$ are squares with areas respectively $A$ and $B$. Find the shaded area.
Well, I observerd the following: If the intersection between $BG$ and $EF$ is the midpoint of $EF$, then the area will be $\dfrac{A}{2} + \dfrac{3B}{... | $$ \triangle EHB \sim \triangle FHG
\\ \implies \frac{FH}{EH} = \frac{FG}{EB} = \frac{\sqrt B}{\sqrt A - \sqrt B}$$
So $$ EH = FH ( \frac {\sqrt A -\sqrt B} {\sqrt B} ) $$
But we also know that
$$ FH + EH = EF = \sqrt B $$
So
$$FH(1+ \frac {\sqrt A -\sqrt B} {\sqrt B} ) = \sqrt B $$
$$ \implies FH(\sqrt B +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Can I apply regular derivative instead of implicit describing rectangle diagonal?
The sides of this rectangle increase in such a way that dz/dt=1 and
dx/dt=3dy/dt. At the instant when x=4 and y=3, what is the value of
dx/dt?
Rectangle with diagonal z
I solve it like this:
$\dfrac{dz}{dy}=(\sqrt{x^2+y^2})'=\dfrac{... | The problem with your answer is that your first derivative only takes the derivative with respect to $y$. $x$ is essentially ignored. $\frac{dz}{dy}$ is not the same as $\frac{\partial z}{\partial y}$.
I would, indeed, start with $z^2 = x^2 + y^2$ because it is just a much simpler problem. Starting the way you did i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to factor this polynomial over $\mathbb{F}_2$ On page 587 in Dummit and Foote, the authors say the polynomial $\frac{x^{16}-x}{x(x-1)(x^2+x+1)}$ can be factored into quartics over $\mathbb{F}_2$ as $(x^4+x^3+x^2+x+1)(x^4+x^3+1)(x^4+x+1)$.
I am having trouble seeing this. When I divide the polynomial $x^{16}-x$ by $... | Over $\mathbb Z$ we have
$$
\begin{align}
x^{16}-x &= x(x^{15}-1)
\\&= x \Phi_1(x) \Phi_3(x) \Phi_5(x) \Phi_{15}(x)
\\&= x (x - 1) (x^2 + x + 1) (x^4 + x^3 + x^2 + x + 1) (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)
\end{align}
$$
and so your polynomial is $(x^4 + x^3 + x^2 + x + 1) (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)$.
You... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding unit digit of $f(10)$
If we define $$f(x)=\left\lfloor \frac {x^{2x^4}}{x^{x^2}+3}\right\rfloor$$ and we have to find unit digit of $f(10)$
I had tried approximation, factorization and substitutions like $x^2=u$ but it proved of no use. Moreover the sequential powers are feeling the hell out of me. Can someo... | First substitute in $10$ for $x$
$$f(10)=\left\lfloor \frac {10^{2\cdot 10^4}}{10^{10^2}+3}\right\rfloor\\
=\left\lfloor \frac {10^{20000}}{10^{100}+3}\right\rfloor$$
Now ask Alpha
or ask Python
and the answer is $3$
Added in response to the comment: You can do long division in base $10^{100}$. Unfortunately the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725831",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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In any triangle, if $\frac {\cos A+2\cos C}{\cos A+2\cos B}=\frac {\sin B}{\sin C}$, then the triangle is either isosceles or right-angled In any triangle, if $\dfrac {\cos A+2\cos C}{\cos A+2\cos B}=\dfrac {\sin B}{\sin C}$, prove that the triangle is either isosceles or right angled.
My Attempt:
Given:
$$\dfrac {\cos... | Alternate approach:
Cross multiply
$\cos A \sin C + \sin 2C = \sin B \cos A + \sin 2B$
$\cos A (\sin C - \sin B) - (\sin 2B - \sin 2C)=0$
$\cos A * 2* \sin \frac{C-B}{2} \cos \frac{B+C}{2} - 2* \sin (B-C) \cos(B+C)=0$
$\cos A * 2* \sin \frac{C-B}{2} \cos \frac{B+C}{2} + 2* \sin (B-C) \cos A=0$
$2\cos A [\sin \frac{C-B}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Integrate $\int\frac{1}{x^3+1}dx$ The problem is, as stated:
$$\int\frac{1}{x^3+1}dx$$
I tried using substitution: $t^3 = x^3 + 1$ but didn't get far with that. I also tried setting: $t = x^3 + 1$, with no luck again.
I tried partial decomposition but I didn't know how to integrate $$\int\frac{1}{x^2-x+1}$$ and I kept ... | Method 1
$$\int \frac {dx}{x^2-x+1}=\frac {4}{3}\int \frac {dx}{1+\frac {4(x-1/2)^2}{3}}$$
Put $u=x-1/2$
Hence $$\frac 43\int \frac {dx}{1+\frac {4(x-1/2)^2}{3}} =\frac 43\int \frac {du}{1+\frac {4u^2}{3}}$$
And note that $$\int \frac {dx}{1+x^2}=\arctan x$$
I hope you can take it from here
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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"answer_id": 2
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Show that there are no rational solutions on the ellipse $2x^2 + 3y^2 = 1$ Show that there are no rational solutions on the ellipse $2x^2 + 3y^2 = 1$.
Want to make sure that my proof is correct.
Suppose there are rational solutions.
Note $2x^2 + 3y^2 = 1 \iff 2x^2 + 3y^2 = z^2\ |\ x,y.z \in \mathbb{Z}$
Since taking $(\... | You correctly arrived at the following: If there are rational points on this ellipse then the equation$$2x^2+3y^2=z^2\tag{1}$$
has integer solutions $\ne(0,0,0)$. But starting from here the argument is more intricate than you assumed.
If there are integer solutions of $(1)$ then after dividing them by a suitable power... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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In any $\triangle ABC$, prove that: $\frac {\cos B-\cos C}{\cos A +1}=\frac {c-b}{a}$ In any $\triangle ABC$, prove that: $\dfrac {\cos B-\cos C}{\cos A +1}=\dfrac {c-b}{a}$
My Attempt:
$$\begin{align}
\text{R.H.S.}&=\dfrac {c-b}{a} \\[4pt]
&=\frac {a\cos B+b\cos A-a\cos C-c\cos A}{b\cos C+c\cos B} \\[4pt]
&=\dfrac {a(... | Obvious if $B=C$
Otherwise using this $$\dfrac{\cos B-\cos C}{\sin C-\sin B}=\cot\dfrac A2$$
$$\dfrac{1+\cos A}{\sin A}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Proving determinant with variables I have a problem that asks:
Prove that det$\begin{pmatrix}
1 && 1 && 1 \\
a && b && c \\
a^2 && b^2 && c^2 \end{pmatrix} = (c-a)(c-b)(b-a)$
I was thinking of solving it like normal and see if I can end up with $(c-a)(c-b)(b-a)$
What I did:
det$\begin{pmatrix}
b && c \\
b^2 && c^2 \e... |
which simplifies to $\;\; (b-a)c^2 + (a-c)b^2 + (c-b)a^2$
Hint: $\;(b-a)c^2 + (a-c)b^2 + (c\color{red}{-a+a}-b)a^2\,=\,\ldots$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Taking the derivative of $x^4\sin(x)\cos(x)$, which step is wrong? I'm trying to take the derivative of $x^4\sin(x)\cos(x)$ and I keep getting the wrong answer.
My steps:
$$\frac {d}{dx}[x^4\sin(x)\cos(x)]$$
Apply product rule:
$$\frac {d}{dx}[x^4](\sin(x)\cos(x)+x^4\frac {d}{dx}[\sin(x)\cos(x)]$$
Simplify first part:
... | $$\dfrac{d}{dx}(x^4\cdot \sin(x)\cdot cos(x))$$
Let $u = x^4$ and $v = \sin(x) \cdot \cos(x)$. From the well known formula we know
$$
\begin{align}
\dfrac{d}{dx}(u \cdot v) & = u \dfrac{dv}{dx} + v \dfrac{du}{dx} \\
& = x^4 \dfrac{d}{dx}(\sin(x) \cdot \cos(x)) + \sin(x) \cdot \cos(x) \cdot \dfrac{d}{dx}(x^4) \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
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Probability Problem Inquiry In an examination, the probability of getting a credit is 1/3. If four students are selected at random, what is the probability that at least one of them got a credit?
From the book:
$P = 1 - F$
$P$ = 1 - (2/3)(2/3)(2/3)(2/3) = 65/81
The book solved the problem by considering the complement ... | Since the probability for each student receiving credit is the same, the Binomial distribution applies. The probability of exactly $k$ successes in $n$ trials, each of which has probability $p$ of success is
$$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$
where $p^k$ is the probability of $k$ successes, $(1 - p)^{n -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
For which values of m is the improper integral convergent For which values of m is the improper integral convergent:
$\int_0^\infty \frac{1}{(x^2+x)^mln^2(1+2x)}tg(\frac{x^4}{x^7+x^4+1})dx$
I've been at this one for a while, here's what I've got so far:
$\int_0^\infty \frac{1}{(x^2+x)^mln^2(1+2x)}tg(\frac{x^4}{x^7+x^4+... | Note that
$$\int_0^\infty \frac{1}{(x^2+x)^m \ln^2(1+2x)}\tan\left(\frac{x^4}{x^7+x^4+1}\right)dx
\\=\int_0^1 \frac{1}{(x^2+x)^m \ln^2(1+2x)}\tan\left(\frac{x^4}{x^7+x^4+1}\right)dx\\+\\+\int_1^\infty \frac{1}{(x^2+x)^m \ln^2(1+2x)}\tan\left(\frac{x^4}{x^7+x^4+1}\right)dx$$
and for $x\to 0^+$
$$\frac{1}{(x^2+x)^m \ln^2... | {
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"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
} |
How do I find $f(x)$ when provided $g(x)$ and $(f\circ g)(x)$? Let $g(x) = 2x^2−2x−1$.
Find a function $f$ such that $(f\circ g)(x) = −8x^6 + 24x^5 −8x^4 − 24x^3 + 6x^2 + 10x + 6.$
$f(x) =$
| Let's suppose that $f(x)=Ax^3+Bx^2+Cx+D$.
Because $2^3=8$, we can see that $A=-1$.
So let's substract $-(2x^2-2x-1)^3=-8x^6+24x^5-12x^4-16x^3+6x^2+6x+1$ from $−8x^6 + 24x^5 −8x^4 − 24x^3 + 6x^2 + 10x + 6$. The result is $4x^4-8x^3+4x+5$.
$2^2=4$, so $B=1$. Now substract $(2x^2-2x-1)^2$ from $4x^4-8x^3+4x+5$. The... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$y^2+5xy+6x^2-9x-4y=0$ where $x$ and $y$ are integers Need some help solving this for integers $x$ and $y$:
$$
y^2+5xy+6x^2-9x-4y=0
$$
I managed to make something like this:
$$
(y+3x-4)(y+2x)=x\\
(y+3x)(y+2x-3)=y
$$
Find integers for $x$ and $y$ that satisfy the equations above.
But, what do I do next, or is this a bad... | Hint: written as a quadratic in $\,y\,$, the discriminant of $\;y^2+(5x-4)y+3x(2x-3)\,$ must be a perfect square: $\;\Delta_y=(5x-4)^2-12x(2x-3)=x^2 - 4 x + 16=(x-2)^2+12\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2738734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to prove that $x'Bx=0$ for all $n\times 1$ vectors $x$ if and only if $B'=-B$?
Theorem. $x^TBx=0$ for all $n \times 1$ vectors $x$ if and only if $B^T=-B$.
Then how to prove this?
| Suppose
$B^T = -B; \tag 1$
then
$(x^TBx)^T = x^TB^Tx = -x^TBx; \tag 2$
since $x^TBx$ is a scalar, that is, a $1 \times 1$ matrix, we have
$x^TBx = (x^TBx)^T, \tag 3$
whence (2) yields
$x^TBx = -x^TBx \Longrightarrow x^TBx = 0; \tag 4$
likewise, if
$x^TBx = 0 \tag 5$
for all $x$, then
$y^TBx + x^TBy = x^TBx + y^TBx + x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2739226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve this equation for c? I want to solve the equation
$$
-\frac{a}{2}\left(c+\sqrt{c^2+4}\right)=-\frac{a-1}{2}\left(c-\sqrt{c^2+4}\right)
$$
for $c$, where $a$ is just a constant.
What I get is
$$
\frac{c-\sqrt{c^2+4}}{c+\sqrt{c^2+4}}=\frac{a}{a-1}.
$$
I think there now is some "trick" to solve this for $c$.
| By distributing,
$$-\frac{ac}{2}-\frac{a\sqrt{c^2+4}}{2}=-\frac{ac}{2}+\frac{a\sqrt{c^2+4}}{2}+\frac{c}{2}-\frac{\sqrt{c^2+4}}{2}$$
Well,
$$0=(a-\frac{1}{2})\sqrt{c^2+4}+\frac{c}{2}$$
$$\frac{1}{1-2a}=\frac{\sqrt{c^2+4}}{c}$$ with assumption that $c \ne0.$
Squaring, $$\frac{1}{(1-2a)^2}=\frac{c^2+4}{c^2}=1+\frac{4}{c^... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
The biggest $\gcd(11n+4, 7n+2)$? I am having problems solving this problem with greatest common divisors:
What is the greatest common divisor of $11n+4$ and $7n +2$?
I tried Euclidean algorithm, and I tried to deduce the answer and I tried to incorporate
$$\gcd(a,b)\cdot\mathrm{lcm}(a,b)=|ab|$$
| \begin{align}
(11n+4) &= 1(7n+2) + (4n+2) \\
(7n+2) &= 2(4n+2) + (-n-2) \\
(4n+2) &= -4(-n-2) - 6
\end{align}
This suggests that the biggest gcd must be a divisor of $6$.
Such a value should occur when $n=-2$.
\begin{align}
11(-2)+4 &= -18 \\
7(-2)+2 &= -12 \\
\gcd(-18, -12) &= 6
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}$, then $\frac{\sin A}{k+1}=\frac{\sin B}{k}=\frac{\sin C}{k-1}$
If $$\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}$$
show that
$$\frac{\sin A}{k+1}=\frac{\sin B}{k}=\frac{\sin C}{k-1}$$
where $k$ is an integer such that $2 < k \neq 4$, and where $a$, $b$, $c... | Let $m$ be the common value of the initial fractions, so that
$$\begin{align}
a + b &= ( 2 k + 1 ) m \tag{1} \\
b + c &= ( 2 k - 1 ) m \tag{2}\\
c + a &= ( 2 k \phantom{+1\;\,}) m \tag{3}
\end{align}$$
We can combine these equations to get
$$\begin{align}
\phantom{-}(1)-(2)+(3):&\quad 2 a = 2(k+1)m \\
(1)+(2)-(3):&\qua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2742306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
attempt to solve a Bernoulli equation I tried solving the Bernoulli equation $y'-y\tan x = y^4 \cos x $ by equating the left hand side to 0 and finding the homogeneous solution, however the equation turned out to be too complex and without the ability to isolate x. Any hints on how to solve this?
| $y' - y\tan(x) = y^4\cos(x)$
divide throughout by $y^4$
$\frac{y'}{y^4} - \frac1{y^3}\tan(x) = \cos(x)$
let $z= \frac{1}{y^3}\implies z' = \frac{-3}{y^4}y'$
$\frac{-z'}{3} -z\tan(x) = \cos(x)$
$z' + 3\tan(x)\,z = -3\cos(x)$
it is now a Linear differential equation
Integrating factor , $I=e^{\int3\tan(x)}=e^{\ln(\sec^3(... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Beads and necklaces We have a several number of beads in a box and want to make 3 necklaces.
We firstly take some out of the box, put $3/4$ of them to the 1st necklace and equally share the rest $1/4$ to the 2nd and 3rd.
Then we take some more out of the box, put $1/4$ of them for the 2nd necklace and equally share t... | The last result implies two equations:
\begin{align}
\frac{\frac{11}{24}Α+\frac{7}{24}x-\frac{2}{24}y}{\frac{11}{24}Α-\frac{8}{24}x-\frac{5}{24}y}=&\; \frac32,\\[2ex]
\frac{\frac{11}{24}Α-\frac{8}{24}x-\frac{5}{24}y}{\frac{2}{24}Α+\frac{1}{24}x+\frac{7}{24}y}=&\; \frac21.
\end{align}
Solving this system for $x$ and $y$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2745921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Largest cone that can be inscribed in a sphere
Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is $\frac{8}{27}$ of the volume of the sphere.
My Attempt
r : radius of the base
h : height of the cone
$$
x=\sqrt{R^2-r^2}\implies h=R+x=R+\sqrt{R^2-r^2}\\
V(r)=\frac{1}{3}\pi r^2h=... | I would write
$$V=\frac{1}{3}\pi r^2h$$ then we get
$$r^2+x^2=R^2$$ and $$x+R=h$$ and so $$V(x)=\frac{1}{3}\pi(R^2-x^2)(x+R)$$
Or you can write: $$h=x+R$$ and $$x=\sqrt{R^2-r^2}$$ then we have
$$V=\frac{1}{3}\pi r^2(\sqrt{R^2-r^2}+R)$$
$$V'(r)=1/3\,{\frac {\pi\,r \left( 2\,{R}^{2}-3\,{r}^{2}+2\,\sqrt { \left( R-r
\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Arrangement of red and white balls with no adjacent red balls. I have $5$ white balls and $3$ red balls. Putting these $8$ balls in a row with no adjacent red balls, what is the number of arrangements?
My approach:
I tried the inclusive-exclusive method, by getting all the possible arrangements and then subtracting fro... | Method 1: The approach @JMoravitz wrote in the comments is the easiest way to handle this problem. Place the white balls in a row. This creates six spaces, four between successive white balls and two at the ends of the row.
$$\square w \square w \square w \square w \square w \square$$
To ensure that the red balls are... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Figure out all primes p and q such that Figure out all primes p and q such that
$p^3$ + 19$q^3$ + 2018 is the cube of a prime.
| Let r be the prime such that
$p^3 + 19q^3 + 2018 = r^3$
Since 2 is the only even number prime but r does not equal to 2 (trivial), by observing the even-odd of the LHS and RHS, either $p=2$ or $q=2$ .
If $p = 2$, $2^3 + 19q^3 + 2018 = r^3$
Let $s^3 = 19q^3$
Since 19 is not a perfect cube, s or $s^2$ cannot be a ration... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove the following determinant Prove the following:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|=4abc$$
My Attempt:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Using $R_1\to R_1+R_2+R_3$
$$\left |
\begin{matrix}
2(b+c)&2(a+c)&2(a+b) \\
b&(c... | Let me give a solution without using the rule of Sarrus for fun. $\require{cancel}$
\begin{align}
& \begin{vmatrix}
b+c&a&a \\
b&c+a&b \\
c&c&a+b \\
\end{vmatrix} \\
&= \frac{1}{abc}\, \begin{vmatrix}
a(b+c)&ab&ca \\
ab&b(c+a)&bc \\
ca&bc&c(a+b) \\
\end{vmatrix} \\
&= \frac{1}{abc}\, \begin{vmatrix}
0&-2bc&-2bc \\
ab&b... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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$f(x)=2x+\cot^{–1}x+\log (\sqrt{1+x^2}−x)$ increasing or decreasing in $\mathscr{R}$.
Show that $f(x)=2x+\cot^{–1}x+\log(\sqrt{1+x^2}−x)$ is increasing in $\mathscr{R}$.
My Attempt
$$
f'(x)=2-\frac{1}{1+x^2}+\frac{\frac{x}{\sqrt{1+x^2}}-1}{\sqrt{1+x^2}−x}\\
=2-\frac{1}{1+x^2}+\frac{\frac{x-\sqrt{1+x^2}}{\sqrt{1+x^2}}... | HINT:
For $x\neq0$, $$1+2x^2>\sqrt{1+x^2}\impliedby1+4x^2+4x^4>1+x^2\impliedby 4x^4+3x^2>0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2752064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $f_{10}$ of the sequence defined by $f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$ Find $f_{10}$ of the sequence defined by $$f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$$ given $f_0=0$
My Approach:
Letting $f_n=2^n b_n$ we get
$$b_{n+1}=\frac{4b_n}{5}+\frac{3 \sqrt{1-b_n^2}}{5}$$
Now letting $b_n=\... | Simply writing it out by hand quickly yields
$$f_1=\frac{6}{5},\qquad f_2=\frac{96}{25},\qquad f_3=\frac{936}{125},\qquad f_4=\frac{384}{25},$$
and the rest should only take a few minutes. Here's a few more, it's not that hard:
$$f_5=\frac{3744}{125},\qquad f_6=\frac{1536}{25},\qquad f_7=\frac{14976}{125},\qquad f_8=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
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What is the solution of this definite integral: $\int_{0}^{2\pi} \ln\big(C-D\cos(f)\big)\cos(nf)df$ I encountered this integral in my calculations:
$$\int_{0}^{2\pi} \ln\big(C-D\cos(f)\big)\cos(nf)df$$
Here, $n$ is a natural number, $C, D$ are constants, such that $C\gt D$.
I tried to find solution in Table of integral... | For $r<1$, we have
$$ \log{(1+r^2-2r\cos{\theta})} = \log{(1-re^{i\theta})(1-re^{-i\theta})} \\
= \log{(1-re^{i\theta})}+\log{(1-re^{-i\theta})} \\
= -\sum_{k=1}^{\infty} \frac{1}{k}r^k (e^{ik\theta}+e^{-ik\theta}) \\
= -2\sum_{k=1}^{\infty} \frac{r^k}{k} \cos{k\theta}, $$
provided that we choose all the logarithms to ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Partial fractions integral. Compute $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$
Evaluate $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$.
I directly approached this with partial fractions and rewritten $x^6+1=(x^2)^3+1=(x^2+1)(x^4+x^2+1)$.
Therefore the integral is:
$$\int_{0}^{1} \frac {x^4+1}{(x^2+1)(x^4+x^2+1)}dx=-2\int_{0}^{1} \... | Set $x^2=y$ $$\dfrac{y^2+1}{y^3+1}=\dfrac A{y+1}+\dfrac{By+C}{y^2-y+1}$$
$$y^2=y^2(A+B)+y(B+C-A)+A+C$$
$\implies A+C=1\iff C=1-A, B+C=A\implies B=A-C=2A-1,$
$1=A+B=3A-1\iff A=\dfrac23, B=\dfrac13,C=\dfrac13$
$$\implies\dfrac{x^4+1}{x^6+1}=\dfrac1{3(x^2+1)}+\dfrac{x^2+1}{3(x^4-x^2+1)}$$
Now $\dfrac{x^2+1}{x^4-x^2+1}=\df... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Integrating $\int_0^{2\pi} \frac1{a\cos(t)+b\sin(t)+c}dt$ with $\sqrt{a^2+b^2}=1Let $a,b,c\in\mathbb{R}$ such that $\sqrt{a^2+b^2}=1<c$.
Find $I:=\int_{0}^{2\pi} \frac{1}{a\cos(t)+b\sin(t)+c}dt$.
So, I set $z=e^{it}$ and then $dz=ie^{it}dt$ and so $\frac{dz}{iz}=dt$. Therefore,
$\begin{align*} I & =\frac{1}{i}\int_{|z|... | Note, we have
\begin{align}
\frac{a}{\sqrt{a^2+b^2}}\cos(t)+\frac{b}{\sqrt{a^2+b^2}}\sin(t) = \cos(t-\delta)
\end{align}
where
\begin{align}
\delta=\cos^{-1}\frac{a}{\sqrt{a^2+b^2}}.
\end{align}
Hence it follows
\begin{align}
\int^{2\pi}_0 \frac{dt}{c+a\cos(t)+b\sin(t)} = \int^{2\pi}_0 \frac{dt}{c+\cos(t-\delta)} =\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2760900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Rewriting the nth term of a series in the form $ar^{n-1}$ In the formula below, where does the $\frac{4}{3}$ come from and what happened to the $3$? How did they get the far right answer? Taken from Stewart Early Transcendentals Calculus textbook.
$$\sum^\infty_{n=1} 2^{2n}3^{1-n}=\sum^\infty_{n=1}(2^2)^{n}3^{-(n-1)}=\... | $$\frac{4^n}{3^{n-1}}=\frac{4^{1+(n-1)}}{3^{n-1}}$$
$$=\frac{4^1\cdot 4^{n-1}}{3^{n-1}}$$
$$=4\cdot \frac{4^{n-1}}{3^{n-1}}$$
$$=4\cdot\big(\frac{4}{3}\big)^{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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To find the smallest value of 'k' for the following equation
Let $\mathrm a,b$ are positive real numbers such that for $\mathrm a - b = 10$, then the smallest value of the constant $\mathrm k$ for which $\mathrm {\sqrt {x^2 + ax}} - {\sqrt{x^2 + bx}} < k$ for all $\mathrm x>0$, is?
I don't get how to approach this p... | An elementary way ($x>0$):
$$f(x) = \sqrt{x^2 + ax} - \sqrt{x^2+bx} = \frac{x^2 + ax - (x^2+bx)}{\sqrt{x^2 + ax} + \sqrt{x^2+bx}} =\frac{(a - b)x}{x\sqrt{1 + \frac{a}{x}} + x\sqrt{1 + \frac{b}{x}}}= \frac{(a - b)}{\sqrt{1 + \frac{a}{x}} + \sqrt{1 + \frac{b}{x}}} < \frac{a-b}{2}= 5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find the integral $\int\frac{dx}{\sqrt{x^2-a^2}}$
Evaluate the integral of $\frac{1}{\sqrt{x^2-a^2}}$
Put $x=a\sec\theta\implies dx=a\sec\theta\tan\theta d\theta$
$$
\begin{align}
& \ \ \ \int \frac{dx}{\sqrt{x^2-a^2}} \\
&=\int \frac{a\sec\theta\tan\theta d\theta}{\sqrt{a^2\sec^2\theta-a^2}} \\
&=\int \frac{a\sec\... | This is an elementary integral, using an inverse hyperbolic function.
By the formula for inverse functions, $$\cosh'(x)=\sinh(x)=\sqrt{\cosh^2(x)-1}$$ lets you establish
$$\text{arcosh}'(x)=\frac1{\sqrt{x^2-1}}.$$
It is known that
$$\text{arcosh}(x)=\log(x+\sqrt{x^2-1})$$ though the form $\text{arcosh}$ is fine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find all integral ordered pairs $(n,k)$ such that $\left\lfloor\frac{n^2+18n+10}{2}\right\rfloor = k^2$. I had two problems that I want to solve. The first one was easy, but the second one... not so much:
First Problem:
Find all values of $n$ such that$$\frac{(n+1)(n+9)+8n+1}{2} = n^2.$$
My Attempt:
$$(n+1)(n+9) = ... | The floor function seems annoying so lets get rid of it, letting $n=2m$ we get:
$\frac{(2m+1)(2m+9)+16m +1}{2}=k^2$.
So $4m^2+36m+10=2k^2$ or $(2m+9)^2 = 2k^2+71$
letting $n=2m+1$ we get
$\frac{(2m+2)(2m+10)+16m+9-1}{2}=k^2$
So $4m^2+40m+28 = 2k^2$ or $(2m+10)^2=2k^2+72$
Hence the problem reduces to finding all value... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I solve this Combination with indistinguishable-objects problem? This question is not about how to solve the problem, but is about why doesn't my solution work.
A bowl has $2$ red, $2$ green, and $2$ blue marbles. How many combinations are possible if we take $3$ random marbles at a time?
I know that the answ... | You can essentially break the possibilities down into 3 cases:
*
*all the colors are the same; that cannot happen so it yields 0 possibilities
*two marbles have the same color and the last one is different, as much possibilities as ordered couples of colors (6 possibilities).
*all marbles are different, 1 possibil... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Finding the value of $6b-3a$
Consider $f(x)= [x]^2 - [x+6]$ and $g(x)= 3kx^2+ 2x + 4(1-3k)$ where
$[x]$ denotes the floor function.
Let $A= \{x ~|~ f(x)= 0\}$ and $k \in [a,b]$ for which every element
of set A satisfies the inequality $g(x)\ge 0$.
1) The set $A$ is equal to ?
2) The value of $6b- 3a =$?
The fi... | You need for $k$ to satify:
$$k (3x^2-12) =3k(x-2)(x+2)\geq -2(x+2)$$ for all $x \in \Bbb A= [-2, -1) \cup [3,4)$.
You can then consider the three cases:
*
*$x=-2$ then the inequality is verified.
*For all $x \in (-2, -1)$, as $x-2 <0$ and $x+2>0$ the inequality is equivalent to:
$$k \leq \frac{2}{3(2-x)}$$
and as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2766405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find a corresponding eigenvector for each eigenvalue Let A =$\begin{pmatrix} 6 & 1 \\ 2 & 7 \end{pmatrix}$
(a) Find the eigenvalues of A.
(b) Find a corresponding eigenvector for each eigenvalue in part (a).
My attempt
a) Eigenvalues:
$$\begin{vmatrix} 6-\lambda& 1 \\ 2 & 7-\lambda \end{vmatrix}=0 \Rightarrow \lambda^2... | If $v$ is an eigenvector to your matrix, so is $\xi \cdot v$ for any $\xi \in \mathbb{R} \setminus \left\{ 0\right\}$. This can be extended to a complex setting.
Moreover if $u$ and $v$ are eigenvectors to the same eigenvalue, all their linear combinations (except $0$) are as well eigenvectors to that eigenvalue. This ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2767115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Inequality with $abc=1$, Given 3 positive numbers $a, b, c$ satisfying $abc= 1$
Prove: $$\frac{1}{a^{5}+ b^{5}+ c^{2}}+ \frac{1}{b^{5}+ c^{5}+ a^{2}}+ \frac{1}{c^{5}+ a^{5}+ b^{2}}\leq 1$$
My opinion:
Let: $x= \frac{a}{b}, y= \frac{b}{c}, z= \frac{c}{a}$
We have to prove:
$$\sum \frac{x^{2}y^{5}z^{5}}{x^{7}z^{5}+x^{2}y... | By Muirhead we obtain:
$$\sum_{cyc}\frac{1}{a^5+b^5+c^2}\leq\sum_{cyc}\frac{1}{a^4b+ab^4+c^3ab}=\sum_{cyc}\frac{1}{ab(a^3+b^3+c^3)}=\frac{a+b+c}{a^3+b^3+c^3}\leq1.$$
For the proof of the last inequality we can use also AM-GM:
$$a^3+b^3+c^3=\frac{1}{9}\sum_{cyc}(5a^3+2b^3+2c^3)\geq\frac{1}{9}\sum_{cyc}9\sqrt[9]{\left(a^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What will be the possible permutations of set S={1,2,3,4}. And how many are of those odd and even? I have solved it as per my knowledge and understanding.
Since there are 4 elements so no. of permutations will be
$$4! = 24$$
$$(1,2,3,4), (1,2,4,3), (1,3,2,4), (1,3,4,2), (1,4,2,3), (1,4,3,2), (2,1,3,4), (2,1,4,3), (2,3,... | There are twenty four elements in $\mathcal{S}_4$, namely $$\{\operatorname{id}, (1\ 2), (1\ 3), (1\ 4), (2\ 3),\ (2\ 4), (3\ 4), (1\ 2\ 3),\ (1\ 2\ 4), (1\ 3\ 4), (2\ 3\ 4),\ (1\ 3\ 2),\ (1\ 4\ 2),\ (1\ 4\ 3), (2\ 4\ 3),\ (1\ 2\ 3\ 4), (1\ 3\ 4\ 2), (1\ 4\ 2\ 3), (4\ 3\ 2\ 1), (2\ 4\ 3\ 1), (3\ 2\ 4\ 1), (1\ 2)(3\ 4),... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2767985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the simplest way to obtain asymptotics of $\sum_{k=0}^n \binom{n}{k} \frac{1}{(1+k)^2}$? In the middle of a proof, I have had to analyze the asymptototic behavior of
$$
\mathbb{E}\left[\frac{1}{(1+X)^2}\right] = \frac{1}{2^n}\sum_{k=0}^n \binom{n}{k}\frac{1}{(1+k)^2}\tag{1}
$$
where $X$ is Binomially distribute... | New Answer. If $p(x)$ is a non-constant polynomial of degree $d$ having no zero on $\mathbb{N}_0 = \{0,1,\cdots\}$ and $m \geq d$, then we may expand
$$ \frac{1}{p(x)} = \sum_{k=d}^{m} \frac{c_k}{(x+1)\cdots(x+k)} + r(x),$$
where $c_k$'s are constants and $r(x)$ has no pole on $\mathbb{N}_0$ and satisfies $r(x) = \math... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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A limit involving $\cos x$ and $x^2$ The question is finding the value of $L = \lim_{x \to 0}\frac{1 - \cos x \cos 2x \cos 3x}{x^2}$ if it exists . I've found the answer using taylor's formula but I'm looking for other solutions maybe using trigonometric identities .
| Alternatively:
$$\cos 3x=\cos (2x+x)=\cos 2x\cos x-\sin 2x\sin x.$$
$$L = \lim_{x \to 0}\frac{1 - \cos x \cos 2x \cos 3x}{x^2}=\lim_{x \to 0}\frac{1 - \cos x \cos 2x (\cos 2x\cos x-\sin 2x\sin x)}{x^2}=\\
\lim_{x \to 0}\frac{1 - \cos^2 x(1-\sin ^22x)+\cos x\cos 2x\sin 2x\sin x}{x^2}=\\
\lim_{x \to 0}\frac{\sin^2x + \co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
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Solve differential equation $(1+x^2) \frac{dy}{dx} - 2xy = x$ Solve differential equation $$(1+x^2) \frac{dy}{dx} - 2xy = x$$
I simplified it to $$\frac{1}{1+2y} dy = \frac{x}{1+x^2} dx$$
$$ \int \frac{1}{1+2y} dy =\int \frac{x}{1+x^2} dx $$
$$\frac{1}{2} \int \frac{2}{1+2y} dy = \frac{1}{2} \int \frac{2x}{1+x^2} dx$$
... | You got : $\quad\frac{1}{2} \ln | 1 + 2y | = \frac{1}{2} \ln | 1+x^2 | + C $
$$\frac{1}{2} \ln | 1 + 2y | - \frac{1}{2} \ln | 1+x^2 | = C $$
$$\ln | 1 + 2y | - \ln | 1+x^2 | = 2C $$
$$\ln \frac{| 1 + 2y |}{ | 1+x^2 |} = 2C $$
Thus $\quad\frac{ 1 + 2y }{ 1+x^2 }=$constant.
$$\frac{ 1 + 2y }{ 1+x^2 }=c$$
$$1+2y=c(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the differential equation: $\frac{dy}{dx} = \sqrt{ \frac{1-y^2}{1-x^2} } $ $$\frac{dy}{dx} = \sqrt{ \frac{1-y^2}{1-x^2} }$$
Simplified it to: $$ \int \frac{1}{\sqrt{1-y^2}} dy = \int \frac{1}{\sqrt{1-x^2}} dx$$
$$\implies\sin^{-1} y = ( \sin^{-1} x + C)$$
$$y = (\sin (\sin^{-1} x) \cos C_1) + \cos(\sin^{-1} x) (\... | $\cos^2(\arcsin(x))=1-\sin^2(\arcsin(x))=1-x^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2769521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Sequence $x_n$ such that $x_{n}\equiv 0\pmod {27}\Longleftrightarrow n\equiv 3\pmod 9$ Let $x_{n}$ be a sequence such that $x_{0}=9,x_{1}=89$,and such
$$x_{n+2}=10x_{n+1}-x_{n}\forall n\ge 0$$
Show that
$$x_{n}\equiv 0\pmod {27}\Longleftrightarrow n\equiv 3\pmod 9$$
Is there a very simple way to prove that a sequence... | Note the corresponding characteristic equation is
$$ \lambda^2-10\lambda+1=0 $$
which has two roots $r_1=5+2\sqrt6,r_2=5-2\sqrt6$. Therefore $x_n$ can be expressed as
$$ x_n=C_1r_1^n+C_2r_2^n. $$
Using $x_0=9,x_1=89$, it is easy to see $C_1=\frac92+\frac{11}{\sqrt6},C_2=\frac92-\frac{11}{\sqrt6}.$ So
\begin{eqnarray}
x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2772217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The ellipse of the maximum area contained between the curves $\frac{\pm 1}{x^2+c} $
The ellipse of the maximum area contained between the curves $\frac{\pm 1}{x^2+c}, c > 0 $
It seems to me that the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ of the maximum area when $ a = c; b = \frac{1}{c} $
That is, it lo... | Consider the case of the curves $y= \pm\frac{1}{x^2+1}$ (the general case can reduced to this case). The ellipse of largest area is not $2 x^2 + y^2=1$ but $ \frac{x^2}{2} + 2 y^2 =1$, tangent to the graphs at points $(\pm 1, \pm \frac{1}{2})$.
There is a family of ellipses tangent to the graph
$$\frac{x^2}{u}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2773266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Why am I getting inconsistent limit results? My objective is to solve the following limit $$\lim_{x \to \infty} \frac{a^x - b^x}{a^{x+1} - b^{x+1}},$$ where $a$ and $b$ are real constants, such that $a \gt b \gt 0$.
I (apparently) managed to solve it in two different ways, leading to two distinct answers, which of cour... | Note that after this step
$$\lim_{x \to \infty} \frac{a^x - b^x}{a^{x+1} - b^{x+1}}\frac{a^{-x}}{a^{-x}}=\lim_{x \to \infty} \frac{1 - (\frac{b}{a})^x}{a - a(\frac{b}{a})^{x+1}}$$
we need some condition to apply l'Hopital (i.e. indeterminate form $\frac{\infty}{\infty}$ or $\frac{0}{0}$) and similarly for Method 2. Tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2774152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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When does $\frac{a+b}{2}$ and $\sqrt{ab}$ have inversed tens digits and ones digits? Let $a$ and $b$ be natural numbers, and
$$A = \frac{a+b}{2}$$
$$B = \sqrt{ab}$$
It's given that $A$ and $B$ are two-digit numbers such that the tens digit of $A$ is the same as the ones digit of $B$, and the tens digit of $B$ is the ... | Since $a+b=2A$, and $ab=B^2$, it follows that $a,b$ are roots of the quadratic equation
$$t^2-(2A)t+B^2=0$$
so the discriminant
$$D=4(A^2-B^2)$$
must be a perfect square, hence $A^2-B^2$ must be a perfect square.
Also, since $A\ne B$, we get that $A^2-B^2$ is a nonzero perfect square.
\begin{align*}
\text{Then}\;\;A^2... | {
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"url": "https://math.stackexchange.com/questions/2775061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Limitations of Cramer's rule? I'm trying to find the line about which this matrix reflects a vector
$\frac{1}{2}\left(\begin{array}{cc}
-1 & -\sqrt{3} \\
-\sqrt{3} & 1 \\
\end{array}\right)$ $\left(\begin{array}{c}
x \\
y \\
\end{array}\right)$ = $\left(\begin{array}{c}
x \\
y \\
\end{array}\right)$... | Note, that the matrix has a factor $\frac{1}{2}$ in front.
So, you made a small mistake when substituting $\begin{pmatrix} x \\ y \end{pmatrix}$ into the determinant:
$$x = \frac{(\frac{1}{2})^2 \left|\begin{array}{cc}
\color{blue}{2}x & -\sqrt{3} \\
\color{blue}{2}y & 1 \\
\end{array}\right|}{(\frac{1}{2})^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2775159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An integral involving a Gaussian and an Owen's T function. Let $a,b,c \ge 0$.
In the course of solving gaussian integral of power of cdf : $\int_{-\infty}^{+\infty} \Phi(x)^n \cdot \phi(a+bx) \cdot dx$ we came across a following integral:
\begin{eqnarray}
T(a,b,c):= \int\limits_a^\infty \phi(\xi) T(\xi b,c) d\xi
\end{e... | It turns out that the answer is actually quite simple and can be obtained using elementary methods. We have:
\begin{eqnarray}
T(a,b,c)= \int\limits_{{\mathbb R}^3} 1_{\xi_0 > a} 1_{\xi_1 > \xi_0 b} 1_{c \xi_1 > \xi_2 >0} \prod\limits_{p=0}^2 \phi(\xi_p) d \xi_p
\end{eqnarray}
Now we go to spherical coordinates as follo... | {
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"source": "stackexchange",
"question_score": "4",
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Prove $\sin(\alpha -\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\frac{\alpha-\beta}{2}\sin\frac{\alpha-\gamma}{2}\cos\frac{\beta-\gamma}{2}$ Here is a problem from Gelfand's Trigonometry:
Let $\alpha, \beta, \gamma$ be any angle, show that $$\sin(\alpha -\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\left(... | Use factoring and algebra. Let $\;X:=e^{ix},\;Y:=e^{iy},\;Z:=e^{iz}.\;$
Then the following equations hold
$$\;\sin(x) = \frac{X-X^{-1}}{2i}, \; \cos(x)=\frac{X+X^{-1}}2,\;
\sin(x-y) = \frac{X^2-Y^2}{2iXY},\; \cos(x-y)=\frac{X^2+Y^2}{2XY}.$$
Summing and factoring these equations using a Computer Algebra System gives... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Exact value of a convergent series: $\sum_{n=1}^\infty \frac{1}{n^3+6n^2+8n}$ I have a series
$$\sum_{n=1}^\infty \frac{1}{n^3+6n^2+8n}.$$
I know the series converges because
$$\frac{1}{n^3+6n^2+8n}\le \frac{1}{n^3}, $$
since $p=3>1$, I know that $\sum 1/n^3$ converges. But I am not sure how to figure out what it conv... | Just another way using partial sums.
Using Mohammad Riazi-Kermani's answer, use partial fraction decomposition to get
$$a_n=\frac 1{n^3+6n^2+8n}=\frac {1}{n(n+4)(n+2)}=-\frac{1}{4 (n+2)}+\frac{1}{8 (n+4)}+\frac{1}{8 n}$$ and consider
$$S_p=\sum_{n=1}^p a_n$$ Using harmonic numbers
$$\sum_{n=1}^p \frac 1n=H_p\qquad \sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Power series of $f(x)=\ln (x^2+4)$ I am supposed to find a power series representation of
$$f(x)=\ln\left(x^{2}+4\right).$$
Then, I am to graph it and observe what happens as $n$ increases. My attempt at a solution:
$$\ln\left(x^2+4\right) = \int \frac{1}{x^2+4}\,dx = \frac{1}{4}\int \frac{1}{\frac{x^2}{4}+1}\,dx = \fr... | Your integration is not correct $$\ln\left(x^2+4\right) \ne \int \frac{1}{x^2+4}\,dx $$ You need to start with $$\ln\left(x+4\right) = \int \frac{1}{4+x}dx$$ and go from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Integrate $\sin^{-1}\frac{2x}{1+x^2}$
Integrate $\sin^{-1}\frac{2x}{1+x^2}$
The solution is given in my reference as: $2x\tan^{-1}x-\log(1+x^2)+C$.
But, is it a complete solution ?
My Attempt
$$
\int 2\tan^{-1}x \, dx=\int \tan^{-1}x \cdot 2\,dx=\tan^{-1}x\int2\,dx-\int\frac{1}{1+x^2}\int2\,dx\cdot dx\\
=\tan^{-1}x \... | We know that $\tan(\theta/2) = \frac{\sin(\theta)}{1+\cos(\theta)}$ and $\cos(\theta) = \sqrt{1-\sin^2(\theta)}$.
So, letting $x = \sin(\theta)$ $(\implies \theta = \arcsin(x))$ we have:
$\tan(\theta/2) = \frac{x}{1+\sqrt{1-x^2}} \implies \frac{\theta}{2} = \arctan\left(\frac{x}{1+\sqrt{1-x^2}}\right) \implies \arcsin(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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degree of $\sqrt{2} + \sqrt[3]{3}$ over $\mathbb Q$ How can i find the degree of the minimal polynomial $P \in \mathbb Q[x]$ such that $P(\sqrt{2} + \sqrt[3]{3}) = 0$ ?
Recently i proved that $\mathbb Q[\sqrt{2} + \sqrt{3}] = \mathbb Q[\sqrt{2}, \sqrt{3}]$ using $(\sqrt{2} + \sqrt{3})^{-1} = \sqrt{3} - \sqrt{2}$, so $2... | Hint:
Check which automorphisms of ${\mathbf Q}[\sqrt 2,\sqrt[3]{3},e^{2\pi i/3}]$ fix ${\mathbf Q}[\sqrt2+\sqrt[3]{3}]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2783025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve in prime numbers the equation $p^q+q^r=r^p$
Find all triples of prime numbers $(p,q,r)$ such that $$p^q+q^r=r^p.$$
I proved that when $r=2$, the equation becomes $$p^q+q^2=2^p.$$ Then I tried to use reciprocity laws and Fermat's little theorem. I could prove that $p\equiv 7\pmod 8$ and that $p>q$.
The equation ... | It is clear that precisely one of $p$, $q$ and $r$ must equal $2$. We'll first show that in fact $r=2$:
Observation 1: $r=2$.
If $q=2$ then reducing mod $3$ shows that
$$p^2+2\equiv r^p\pmod{3},$$
so either $p=3$ or $r=3$. Both are easily verified to be impossible.
If $p=2$ then
$$2^q+q^r=r^2,$$
but for all primes $q,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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ODE power series solution $y'+xy=1+x$ So i have been told to find the power series solution to the following ode $$y'+yx=1-x$$
Using the substitution $y=\sum_{n=0}^{\infty}a_nx^n$, i can rewrite the equation as the following;
$$\sum_{n=1}^{\infty}na_nx^{n-1}+\sum_{n=0}^{\infty}a_nx^{n+1}=1+x$$
That is
$$\sum_{n=0}^{\in... | $$ 2\cdot4\cdot6\cdots2k=2^kk! $$
$$ 1\cdot3\cdot5\cdots(2k-1)=\frac{(2k)!}{2^kk!}$$
As for your solution, you will have to separate into two series of the form $$ y=a_0y_0+y_1 $$
ADDENDUM:
Note that your textbook did not bother to find the general terms for $3\cdot5\cdot7\cdot(2n+1)$. I worked it out using the genera... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate $\int \frac {dx}{1+\sin x}$ Integrate $\int \dfrac {dx}{1+\sin x}$
My attempt:
$$\begin{align}&=\int \dfrac {dx}{1+\sin x}\\
&=\int{\dfrac{dx}{1+\dfrac{2\tan \left( \dfrac{x}{2} \right)}{1+\tan ^2\left( \dfrac{x}{2} \right)}}}\\
&=\int{\dfrac{1+\tan ^2\left( \dfrac{x}{2} \right)}{\left( 1+\tan \left( \dfrac{x... | See Tangent_half-angle_substitution
If $z = \tan \big(\frac x2 \big)$, then $dx = \dfrac{2 dz}{1+z^2}$
and $\sin(x) = \dfrac{2z}{1+z^2}$
\begin{align}
\int\frac{dx}{1+\sin x}
&= \int\frac{1}{1+\sin x} \cdot dx \\
&= \int \dfrac{1}{\left(1+\dfrac{2z}{(1+z^2)}\right)}
\cdot \dfrac{2 dz}{1+z^2} \\
&= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
How do I evaluate this integral ? How do you evaluate this integral $\int \frac{x+1}{(x^2-x+1)^3}dx$ ?
I found out:
$\int \frac{x+1}{(x^2-x+1)^3}dx = \frac{1}{2}\int \frac{2x-1}{(x^2-x+1)^3}dx + \int \frac{3}{(x^2-x+1)^3}dx
$
First integral is nicely solvable now.
I have rearrange the second one:
$
\int \frac{3}{(x^2-... | At last I have solved it myself.
$
arctg(y)=\int \frac{1}{y^2+1}dy =
$
Using per parters
$
=\frac{1}{y^2+1}dy-\int\frac{-2y^2}{(y^2+1)^2}dy=\frac{1}{y^2+1}+2\int\frac{y^2+1-1}{(y^2+1)^2}dy=\frac{1}{y^2+1}+2\int\frac{1}{y^2+1}dy-2\int\frac{1}{(y^2+1)2}dy
$
Now I know what is the $\int\frac{1}{(y^2+1)^2}$ equal to.
By p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Computing a divergent integral $\iiint r^5sin\theta cos\theta$ Problem
Evaluate
$$I = \iiint_V z \sqrt{x^2+y^2+z^2}dxdydz$$
Where V is :
$$\sqrt{x^2+y^2} \leq z\leq \sqrt{1-x^2-y^2}$$
Solution:
I must use spherical coordinates so:
$$x=r\cos\psi \sin{\theta}$$
$$y=r \sin{\psi} \sin{\theta}$$
$$z = r \cos \theta$$
$$... | There is not the slightest reason why the integral of a nice function over a compact domain should be divergent.
I'm using geographical longitude $\phi$ and geographical latitude $-{\pi\over 2}\leq\theta\leq{\pi\over2}$. The condition $z\geq\sqrt{x^2+y^2}$ then translates into $r\sin\theta\geq r\cos\theta$, or $\theta\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Verifying a Stubborn Inequality I have an inequality which arose in an unrelated problem, and it's been proving to get the better of me. I essentially know it has to be true, but cannot verify it. For integers $n_{1}, n_{2}, \ell_{1}, \ell_{2}$, I want to show that
$$8 n_{1} + 8n_{2} -\ell_{1}^{2} - \ell_{2}^{2} - 2\... | With less complicated symbols, you want to prove that $8a+8b-x^2-y^2-2xy>0$ provided $4a-x^2\ge-1$ and $4b-y^2>0$.
First notice that $4b-y^2>1$, because $4b-y^2=1$ implies $y^2\equiv 3\pmod{4}$, which is impossible. Now rewrite the expression as
\begin{align}
2(4a-x^2)+2(4b-y^2)+x^2+y^2-2xy
&=2(4a-x^2)+2(4b-y^2)+(x-y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2793615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int_0^1 \frac{x-x^2}{\sin \pi x} dx = \frac{7 \zeta(3)}{\pi^3}.$ I tried to use the series for $\sin \pi x$ and maybe find something related to $\zeta(3)$, but didn't work. I'm guessing this integral needs more than the little calculus that I know.
\begin{equation}
\int_0^1 \frac{x-x^2}{\sin \pi x} dx = \... | First, denote the integral below as $I$$$I=\int\limits_0^1dx\space\frac {x(1-x)}{\sin\pi x}$$and through integration by parts on $u=x-x^2$, then we have
$$\begin{align*}I & =-\frac 1{\pi}(x-x^2)\log\cot\left(\frac {\pi x}2\right)\,\Biggr\rvert_0^1+\frac 1{\pi}\int\limits_0^1dx\, (1-2x)\log\cot\left(\frac {\pi x}2\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2793835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 3
} |
Sum of $\sum_{n=1}^{\infty} \frac {x^n}{(n-1)!} $ the task is to find sum of $\sum_{n=1}^{\infty} \frac {x^n}{(n-1)!} $. I've tried to factor out $x$ to get $x\sum_{n=1}^{\infty} \frac {{x}^{n-1}}{(n-1)!}$ and integrate the sum, which yields $x\sum_{n=1}^{\infty} \frac {x^n}{n!} $ which I can sum as $xe^x$. Then I nee... | Note that you have to integrate the complete expression $x\sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}$ if you want to use integration. We then obtain by using integration by parts
\begin{align*}
\int \underbrace{x}_{u}\underbrace{\sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}}_{v^\prime}\,dx
&= x\sum_{n=1}^\infty \frac{x^{n}}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Inequality in number theory Prove that:
(i)$5<5^{\frac{1}{2}}+5^{\frac{1}{3}}+5^{\frac{1}{4}}$
(ii)$8>8^{\frac{1}{2}}+8^{\frac{1}{3}}+8^{\frac{1}{4}}$
(iii)$n>n^{\frac{1}{2}}+n^{\frac{1}{3}}+n^{\frac{1}{4}}$ for all integer $n\geq9$
Can raising both sides to exponent $12$ help
| HINT: For (ii) we have to prove that
$$6>2\sqrt{2}+\sqrt[4]{8}$$
This can be written as
$$6-2\sqrt{2}>\sqrt[4]{8}$$
Raise to the power $4$
we have
$$3080>2112\sqrt{2}$$
Squaring we get
$$565312>0$$
For (i) we get by $$AM-GM$$
$$\frac{5^{1/2}+5^{1/3}+5^{1/4}}{3}\geq \sqrt[3]{5^{13/12}}$$ so we have to Show that
$$3\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding factors of second order complex polynomial. It concerns with finding roots of complex polynomial:
$x^2+(i-1)x+(2+i)$. One way is to find roots by factoring, as:$(x-i)(x -(1-2i))=(x-i)(x-1+2i)=(x^2 -x +ix +i +2)=x^2 +x(i-1)+(i+2)$
But, this is a guess game for me, with no formal process to get in this case.
The... | Let $z=a+bi, a,b, \in \mathbb{R}$, where $$z^2 = -8-6i$$
$$(a+bi)^2=(-8-6i)$$
$$a^2-b^2+2abi=-8-6i$$
that is we have
$$a^2-b^2=-8\tag{1}$$
$$ab=-3\tag{2}$$
Substituting $(2)$ into $(1)$,
$$a^2-\frac{9}{a^2}=-8$$
$$a^4+8a^2-9=0$$
$$(a^2-9)(a^2+1)=0$$
$$a=\pm3$$
when $a=3$, from $ab=-3$,$b=-1$.
when $a=-3$, from $ab=-3$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof that continued fractions are finite for rationals? How does one prove that the continued fraction representations of rational numbers are finite?
For every $x\in\mathbb{R}$, the (simple) continued fraction representation of $x$ is:
$$
x = [a_0; a_1, a_2, ...] = a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{...}... | it is the Euclidean Algorithm, that is all. Lots of people use "back-substitution" to finish the Extended Algorithm and find the Bezout combination, I prefer to write this as a continued fraction.
$$ \gcd( 12345, 1601 ) = ??? $$
$$ \frac{ 12345 }{ 1601 } = 7 + \frac{ 1138 }{ 1601 } $$
$$ \frac{ 1601 }{ 1138... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
Solving for the determinant only given one column of values. Given
$$ \det
\begin{bmatrix}
a & 1 & d \\
b & 1 & e \\
c & 1 & f \\
\end{bmatrix}
= 4
$$
and
$$ \det
\begin{bmatrix}
a & 1 & d \\
b & 2 & e \\
c & 3 & f \\
\end{bmatrix}
= -2
$$
I am asked to find
$$ \det
\begin{b... | You're right about the first question. For the second one, note that\begin{align}\det\begin{bmatrix} a & 4 & d \\ b & 5 & e \\ c & 6 & f \\ \end{bmatrix}&=\det\begin{bmatrix} a & 3+1 & d \\ b & 3+2 & e \\ c & 3+3 & f \\ \end{bmatrix}\\&=\det\begin{bmatrix} a & 3 & d \\ b & 3 & e \\ c & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2807634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Which values for $x$ make $x(x + 180)$ a square?
Which values for $x$ make $x(x + 180)$ a square?
$x = 12, 16, 60$ are some values.
Then maybe solving $x(x + 180) = y^2 $ would give other values?
Tried using the general formula, but that would only give 2 values each time I find a suitable $y$, not all values. And... | Note that $$y^2=x(x+180)=(x+90-90)(x+90+90)=(x+90)^2-90^2$$
and then $$90^2=(x+90)^2-y^2=(x+90+y)(x+90-y) $$
As the factors on th right have the same parity (differ by $2y$) and the left is even, we are looking for factorizations of $90^2$ into even factors. Any such factorization $90^2=(2u)(2v)$ (or simply $45^2=uv$)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2808098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Find $x+y$, if integers $x$ and $y$ satisfy the equation $y+1/x=25/3$ Find $x+y$, if $x$ and $y$ are integers and satisfy the equation $y+1/x=25/3$
so I got to the answer by placing $3$ in the $x$ cause it looked like a fraction and $8$ was left for $y$,
My question is: Is there any other/better way to solve this alge... | We have: $y = 8+\dfrac{x-3}{3x}\implies \dfrac{x-3}{3x}\in \mathbb{Z}\implies x-3=3nx\implies x-3nx=3\implies(3n-1)x=-3\implies x\mid 3 \implies x = \pm 1, \pm 3$. Since $y$ is an integer, $x = 3$. Thus $y = 8$, and $x+y = 3+8 = 11$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Cool way of finding $\cos\left(\frac{\pi}{5}\right)$ while trying to solve a problem, I stumbled upon a way of finding $\cos\left(\frac{\pi}{5}\right)$ using identities and the cubic formula. Is it possible to find other values of sine or cosine in a similar way ?
Consider
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\... | Everything begins with finding $\sin\left(\frac{\pi}{10}\right)$
Okay... Lets say $x=\frac{\pi}{10}$
$$5x=\frac{\pi}{2}$$
$$2x=\frac{\pi}{2}-3x$$
$$\sin(2x)=\sin\left(\frac{\pi}{2}-3x\right)$$
$$\sin(2x)=\cos(3x)$$
$$2\sin(x)\cos(x)=4\cos^3(x)-3\cos(x)$$
$$2\sin(x)\cos(x)-4\cos^3(x)+3\cos(x)=0$$
$$\cos(x)\left(2\sin(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
Splitting up an infinite sum I am playing with the following example trying to determine if $\frac{1}{n}-\frac{1}{\sqrt{n}}$ is in $\ell^2$.
Here is what I have,
\begin{align}
\sum\limits_{n=1}^\infty \left( \frac{1}{n}-\frac{1}{\sqrt{n}}\right)^2 &= \sum\limits_{n=1}^\infty \left(\frac{\sqrt{n}-n}{n\sqrt{n}}\right)^2\... | Put $a_n=\frac{1}{n}-\frac{1}{\sqrt{n}}, b_n =\frac{1}{n}$ and $c_n=\frac{1}{\sqrt{n}}$.
Suppose that $(a_n) \in \ell^2$. Since $(b_n) \in \ell^2$ and since $\ell^2$ is a vector space, we get that $(c_n)=(b_n)-(a_n) \in \ell^2$, a contradiction. Hence $(a_n) \notin \ell^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\frac{dy}{dx}+\sec^2(\frac{\pi}{4}-x)=0$
If $y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$, prove that $\frac{dy}{dx}+\sec^2(\frac{\pi}{4}-x)=0$
My attempts:
Attempt 1:
$y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$
$\implies y=\sqrt{\frac{\sin^2 x+\cos^2 x-2\sin x \cos x}{\sin^2 x+\cos^2 x+2\sin x \cos x}}$
$\implies y=... | In fact, the result you want is not true: it depends on the values of $x$.
Having said this, with a small modification, any of your methods should work. I'll use your third method. Note that you have
$$y=\sqrt{\frac{(1-\sin2x)^2}{\cos^22x}}\ ;$$
this gives
$$y=\left|\frac{1-\sin2x}{\cos2x}\right|\ ,$$
and the omissio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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What is the dimension of the vector space consisting of all 3 by 3 symmetric matrices? What is a basis for it? Could you please explain the following and give me a meticulous solution:"What is the dimension of the vector space consisting of all 3 by 3 symmetric matrices? What is a basis for it?"
Solution till now:
Let ... | Hint:
Any symmetric matrix $\mathbf{A}\in M_{3,3}(\mathbb{R})$ can be written as
$$\mathbf{A}=\begin{pmatrix}a&b&c\\b&d&e\\c&e&f\end{pmatrix}$$
Can you write $\mathbf{A}$ as a sum of some matrices multiplied by some scalars?
You have found that any symmetric matrix may be written $$A=a\begin{pmatrix}1&0&0\\0&0&0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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integral $\int \frac{1-a\cos x}{1+a^2-2a\cos x}dx$ I'm trying to solve this integral: $$\int \frac{1-a\cos x}{1+a^2-2a\cos x}dx.$$
My trouble: Using software, i found that if $|a|>1$, the integral is equal to $$\arctan \left(\frac{x}{2}\right)+\frac{1-a^2}{|a-1||a+1|}\arctan \left(\frac{|a+1|}{|a-1|}\tan \left( \frac{x... | Your antiderivative is correct for $\frac{|x|}2\le \pi/2$. Thus,
$$\begin{align}
\int_0^{2\pi}\frac{1-a\cos(x)}{a^2+1-2a\cos(x)}\,dx&=\int_{-\pi}^\pi \frac{1-a\cos(x)}{a^2+1-2a\cos(x)}\,dx\\\\
&=\left.\left(\frac x2+\text{sgn}(1+a)\text{sgn}(1-a)\arctan\left(\left|\frac{a+1}{a-1}\right|\tan(x/2)\right)\right)\right|_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simplify: $\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}$ Simplify:
$$\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}$$
So what I've tried was:
$$\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}=\frac{1+\cos^2x-\sin^2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}=$$
$$=\... | $$u=\frac{1+cos2x}{1-cosx}-\frac{4cos^2(\frac{x}{2})}{tan^2x}$$
Using $\tan(x/2)=t$, $\sin x=\dfrac{2t}{1+t^2}$ and $\cos x=\dfrac{1-t^2}{1+t^2}$
$$U=\dfrac{2\cos^2 x}{1-cosx}-2\dfrac{(1+\cos^2x)}{\tan^2x}$$
Then
$$U=2\dfrac{(1-t^2)^2(1+t^2)}{(1+t^2)^22t^2}-2\dfrac{\frac{2}{1+t^2}}{4t^2}(1-t^2)^2$$
Eventually:
$$U=\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2818753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Hw to find the two second order linear recursive common items? Given two sequences
$$a(n)=4a(n-1)-a(n-2), \quad a(1)=1，\; a(2)=5;$$
$$b(n)=10b(n-1)-b(n-2), \quad b(1)=1，\; b(2)=11.$$
If some $a(n)=b(m)$, $m,n\in\mathbb{N}$, exist?(Except a(1)=b(1) ) If there are no common items, how to prove it?
The Two solution equat... | Linear difference equations are much like differential equations
To solve
$$
a_n = 4 a_{n-1}-a_{n-2},\: a_1=1,\; a_2 = 5
$$
we assume that $a_n = \alpha^n$ and substituting results in
$$
(1-4\alpha+\alpha^2)\alpha^n = 0
$$
now solving
$$
1-4\alpha+\alpha^2 = 0
$$
gives
$$
a_n = c_1(-2-\sqrt5)^n+c_1(-2+\sqrt5)^n
$$
Anal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2819582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Number of ways to distribute 10 things among 6 people given that the number of things given to two people doesn't exceed 4? Here is a more specific question:
Find the number of ways of giving $10$ identical gift boxes to 6
people : $A$, $B$, $C$, $D$, $E$, $F$ in such a way that total number
of boxes given to $A$ ... | Let $x_A$, $x_B$, $x_C$, $x_D$, $x_E$, and $x_F$ represent the number of gift boxes given to persons $A$, $B$, $C$, $D$, $E$, and $F$, respectively. Since a total of ten boxes are distributed to these six people,
$$x_A + x_B + x_C + x_D + x_E + x_F = 10 \tag{1}$$
Since $A$ and $B$ together receive at most four of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Probability that all three shirts were worn Mr. Jones has three shirts: Red, Green and White. Each day he picks randomly a red shirt with probability of $\frac{1}{2}$, green with probability of $\frac{1}{3}$ and white with probability of $\frac{1}{6}$.
What is the probability that he wears all three shirts after 6 day... | Let $R$ denote the event that he wears a red shirt in the $6$ days.
Let $G$ denote the event that he wears a green shirt in the $6$ days.
Let $W$ denote the event that he wears a white shirt in the $6$ days.
To be found is: $$P(R\cap G\cap W)=1-P(R^{\complement}\cup G^{\complement}\cup W^{\complement})=$$$$1-P(R^{\comp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Is the function $f(x,y) = \frac{xy^3}{x^4+y^2}$ a $C^1$-function? Consider the function
$$
f(x,y) = \frac{xy^3}{x^4+y^2}.
$$
with $f(0,0) = 0$.
Is this a $C^1$-function?
Firstly we compute:
\begin{align*}
& D_1f(x,y) = \frac{y^3(x^4+y^2)-xy^3(4x^3)}{(x^4+y^2)^2} = \frac{y^3(y^2-3x^4)}{(x^4+y^2)^2}, \\
& D_2f(x,y) = ... | You have
$$\left\vert \frac{y^3(y^2-3x^4)}{(x^4+y^2)^2} \right\vert \le \vert y \vert^3\frac{(y^2+3x^4)}{(x^4+y^2)^2} \le 3 \vert y \vert^3 \frac{(y^2+x^4)}{(x^4+y^2)^2}= 3 \frac{\vert y \vert^3}{x^4+y^2} \le 3\vert y \vert$$ hence
$$\lim_{(x,y) \to (0,0)} \frac{y^3(y^2-3x^4)}{(x^4+y^2)^2} = 0.$$
Using a similar argum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What is $x$ if $\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$? I need to find $x$, given that
$$\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$$
I simplified this to $x^4+14x^3+105x^2+68x-188=0$. According to Symbolab, that is not correct. That's why I'm goint to write out my attempt so you can point out where my mistake is.
My attempt... | It's $$\sqrt{1+2\sqrt{x+3}+x+3}=\frac{x+8}{3}$$ or
$$\sqrt{\left(1+\sqrt{x+3}\right)^2}=\frac{x+8}{3}$$
$$1+\sqrt{x+3}=\frac{x+8}{3}$$ or
$$\sqrt{x+3}=\frac{x+5}{3}$$ and since $x\geq-3$, it's
$$9(x+3)=(x+5)^2,$$
which gives $x=1$ or $x=-2.$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limits applications in geometric problem We have the following situation
The goal is to find $$\lim_{a \to b } \frac{a-b}{c-d} $$
Thought
As $a $ tends to $b$ then we see we are gonna have a rectangle which means that $2 \alpha $ is gonna tend to $ \frac{\pi}{2}$. In other words, $\alpha $ is gonna tend to $\frac{ \pi... |
It helps to first spell out what the construction is:
*
*start with the orthogonal "axes" through $O$ and pick point $A$ on the vertical axis;
*pick an angle $\alpha \lt \pi/4$ and draw $AB$ such that $B$ is on the horizontal axis and $\angle BAO = \alpha$;
*the (other) line through $A$ at angle $\alpha$ with $AB... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Explain this convergence among Pythagorean triplets Why do the ratios of successive values of integers $a$ and $c$, where $a^{2}+(a+1)^{2}=c^{2}$, appear to converge to
$$\frac{a_{n+1}}{a_{n}},\frac{c_{n+1}}{c_{n}}\rightarrow3+2\sqrt{2}$$
I rigorously generated all the {a,c} pairs where $a^{2}+(a+1)^{2}=c^{2}$ for $a<1... | $$\begin{align}
a^{2}+(a+1)^{2}&=c^{2} \\
2a^2+2a+1&=c^2 \\
2\left(a+\frac{1}{2}\right)^2 +\frac{1}{2}&=c^2 \\
(2a+1)^2+1&=2c^2 \\
2c^2-(2a+1)^2&=1 \\
2c^2-d^2&=1 \quad | \quad d=2a+1
\end{align}$$
The above pell equation: $2c^2-d^2=1$ factors into $(c\sqrt 2-d)(c\sqrt 2 +d)=1$. With the initial solution being $(c_0,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
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Find all possible relationships between $a$ and $b$
Find all possible relationships between $a\in \mathbb{R}$ and $b \in \mathbb{R}$ such that
$$\frac{a^2b^2(a+b)^2}{4}=a^2+ab+b^2+2$$
I can see that if $ab=2$, then we have equality. However, this was through observation only. Are there any other ways of determining... | To simplify things, lets first substitute $x=ab$ and $y=a+b$, to get
$$\frac{x^2 y^2}{4} = y^2 - x + 2\\
x^2 y^2 = 4y^2 - 4x + 8 \\
(x^2-4) y^2 + 4(x - 2) = 0\\
(x-2)((x+2) y^2 + 4) = 0$$
So either $x=2$, or $(x+2)y^2+4=0$.
So $ab=2$ or $(ab+2)(a+b)^2+4=0$.
The latter case can be rearranged and factored further, using ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is it possible to construct three cevians so that three of the parts are equivalent? Recently I have started wondering if it is possible to construct two cevians in a triangle so that at least three of four parts which the cevians divide the triangle in are equivalent. I think the answer is yes, but do not know how to ... |
Yes you can! In the triangle above, our cevians are $BE$ and $CF$. The three areas that are equal are $[BFX]$, $[AFXE]$, and $[CEX]$.
The point $X$ is defined with barycentric coordinates $(1/2, 1/4, 1/4)$. In other words, we have the ratios:
$$\frac{DX}{DA} = \frac12$$
$$\frac{EX}{EB} = \frac14$$
$$\frac{FX}{FC} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Showing that $\sin(\sqrt{4 \pi^{2}n^{2} + x})$ converges uniformly on $[0,1]$
Suppose we are considering the sequence of functions $f_{n}(x)=\sin(\sqrt{4 \pi^{2}n^{2} + x})$ and I am having trouble showing that that $f_{n}$ converges uniformly on the interval $[0,1]$.
An idea, I've tried is to consider the Taylor s... | Your sequence converges uniformly to $0$.
\begin{align}
\left|\sin\left(\sqrt{4\pi^2n^2+x}\right)\right| &= \left|\sin\left(\sqrt{4\pi^2n^2+x}\right)-\sin(2\pi n)\right| \\
&= \left|2\sin\left(\frac{\sqrt{4\pi^2n^2+x} - 2\pi n}2\right)\cos\left(\frac{\sqrt{4\pi^2n^2+x} + 2\pi n}2\right)\right|\\
&\le 2\left|\sin\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2833814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Sum of infinite series $ {1+ \frac{2}{6} + \frac{2\cdot5}{6\cdot12} + \frac{2\cdot5\cdot8}{6\cdot12\cdot18} + \cdots}$. Prove that $1+ \frac{2}{6} + \frac{2\cdot5}{6\cdot12} + \frac{2\cdot5\cdot8}{6\cdot12\cdot18} +\cdots=4^{\frac13}$
I tried it in the backward method... I rewrote $4^{\frac13}$ in this way...
$(1+3)^{... | using binomial expansion $(1-x)^{-\frac{2}{3}} =1-\frac{2}{3}(-x)+\frac{\frac{-2}{3}\times\frac{-5}{3}}{2!}(-x)^2+\frac{\frac{-2}{3} \cdot \frac{-5}{3} \cdot \frac{-8}{3}}{3!} \cdot (-x)^3 +\ldots$
when $x=\frac{1}{2}$ we get:
$(1-\frac{1}{2})^{-\frac{2}{3}}=(\frac{1}{2})^{-\frac{2}{3}}=(2)^\frac{2}{3}=(2^2)^\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2833942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Evaluate: $\lim_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}$ without L'Hopitals I have to find the limit of $\lim\limits_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}$
here is my try $\lim\limits_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}=\lim\limits_{x \to \pi/4}\frac{4(1-\c... | $\lim\limits_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}$
$= \lim_{x \to \frac{\pi}4}\dfrac{\sqrt2(\sqrt2^5 - (\sin x + \cos x)^5)}{1- \sin 2x} $
Using,
$\color{blue}{\lim _{x \to a}\frac{x^n -a^n}{x-a}=na^{n-1}}$
$= \lim _{x \to \pi/4}\dfrac{\sqrt 2 (5.\sqrt2^{4})(\sqrt 2 -(\sin x +\cos x))}{1- \sin 2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solve the equation $X^2+X=\text{a given matrix}$ I want to solve the quadratic matrix equation
$$X^2+X=\begin{pmatrix}1&1\\1&1\end{pmatrix}$$
If I put $X$ in the form
$$X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
then I find complicated equations. Is there a simple way to tackle the problem without using diagonalization... | Your method also works, though cumbersome (not asap):
$$X^2+X=1 \Rightarrow \begin{pmatrix}a&b\\c&d\end{pmatrix}^2+\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}1&1\\1&1\end{pmatrix} \Rightarrow\begin{cases}a^2+a+bc=1\\ ab+bd+b=1\\ ac+c+cd=1\\ bc+d^2+d=1\end{cases}.$$
Subtract the last from the first:
$$a^2-d^2+a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 2
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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.