Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Taylor series of $\frac{x-2}{\sqrt{x^2-4x+8}}$ at $x = 2$. Where is the mistake? $$\frac{x-2}{\sqrt{x^2-4x+8}} \quad \text{at}\ x_0=2$$
Let $t = x-2$. then as $x\to 2$, $t\to 0$ and we can find the Maclaurin Series in terms of $t$.
\begin{align*}
\frac{x-2}{\sqrt{x^2-4x+8}} &=\frac{x-2}{\sqrt{(x-2)^2+4}}\\
&=\frac{t}{\... | There's no error in what you did, you only "wasted" two orders of magnitude in your remainder term.
The terms of the textbook's answer look different from yours, but are in fact the same, since
$$(2k)!! = 2^k\cdot k!$$
for one definition of the double factorial of even natural numbers (there's another definition that d... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Can Every Irrational, but Non-Transcendental be Expressed as a Repeating Continued Fraction? So, as far as I understand any rational number can be expressed as a continued fraction of finite length
for example:
$$\frac {7}{17} = \cfrac {1}{2 + \cfrac{1}{2 + \cfrac{1}{3}}}$$
which for convenience I will express as $[0,2... | If a simple continued fraction repeats but does not terminate then it converges to a second-degree algebraic number. ("Simple" means all of the numerators are $1$.)
First, look at the case where the repeating part begins immediately:
$$
x = a_1 + \cfrac 1 {a_2 + \cfrac 1 {a_3 + \cfrac 1 {\ddots \cfrac{}{a_n + \cfrac 1 ... | {
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"timestamp": "2023-03-29T00:00:00",
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How can we show $\cos^6x+\sin^6x=1-3\sin^2x \cos^2x$? How can we simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?
One reasonable approach seems to be using $\left(\cos^2x+\sin^2x\right)^3=1$, since it contains the terms $\cos^6x$ and $\sin^6x$. Another possibility would be replacing all occurrences of $\sin^2x$ by $1... | $$(\cos^2 x+\sin^2x)^3=1^3=\cos^6x+3\cos^4x\sin^2x+3\cos^2x\sin^4x+\sin^6x$$
$$\cos^6x+\sin^6x=1-3\cos^2x\sin^2x(\cos^2x+\sin^2x)$$
so
$$\cos^6x+\sin^6x=1-3\cos^2x\sin^2x$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 11,
"answer_id": 7
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Show that $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$ by induction I have some with proving by induction. I cannot find a solution for the inductive step: $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$
I already did the induction steps:
Basis: P(1) = $1^3 = (1(1+1)/2)^2$ (This is true)
Inductive step: Assume $P(k) = ((k)(k+1)/2)^2$... | Take your "to be proven" equation and divide through by $(k+1)^2$ and multiply through by $4$. This gives $k^2+4(k+1)=(k+2)^2$ as the equation to be proven, and this is trivial.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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System of four equations of four variables including second powers. I've been tasked with solving the following system of equations and it seems like I am stuck:
$$a-x^2=y$$$$a-y^2=z$$$$a-z^2=t$$$$a-t^2=x,$$where $a$ is a real number, for which $0\leq a\leq 1$. I thought the best way would be to subtract some equations... | from the first equation we get
$$a-(a-x^2)^2=z$$ then we obtain
$$a-(a-(a-x^2)^2)^2=t$$
$$a-(a^2+(a-x^2)^4-2a(a-x^2)^2)=t$$ and finally we obtain
$$a-(a^2+(a-x^2)^4-2a(a-x^2)^2)^2=x$$
and now you can try to get $x$
and this is what i'm get:
$$a^8+a^7 \left(-8 x^2-4\right)+a^6 \left(28 x^4+24 x^2+6\right)+a^5 \left(-56 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\forall n\in\mathbb N$ prove that at least one of the number $3^{3n}-2^{3n}$,$3^{3n}+2^{3n}$ is divisible by $35$. $3^{3n}-2^{3n}=27^n-8^n=(27-8)(27^{n-1}+27^{n-2}\cdot 8+...+27^1\cdot8^{n-2}+8^{n-1})$
If $n$ is even,
$3^{3n}+2^{3n}=27^n+8^n=(27+8)(27^{n-1}-27^{n-2}\cdot 8+...-27^1\cdot8^{n-2}+8^{n-1})$
If $n$ is eve... | if n is odd
$3^{3n} + 2^{3n} = (27+8)(27^{n-1} - 27^{n-2}8 +\cdots + 8^{n-1})$
and $35|(3^{3n} + 2^{3n})$
if $n$ is even, $n = 2k:$
$(3^{3n}-2^{3n}) = (3^{n}-2^{n})(3^{2n} + 3^n2^n + 2^{2n})$
$(3^{n}-2^{n}) = (3^{2k} - 2^{2k}) = (3^2 - 2^2)(3^{2k-1} +\cdots + 2^{2k-1})\\
5|(3^{2k}-2^{2k})$
What is left?
showing that ... | {
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"timestamp": "2023-03-29T00:00:00",
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Find possible number of triangles with integer sides for a given inradius inradius = $5$
possible triangle sides
$(25, 20, 15)$
$(37, 35, 12)
(39, 28, 17)$
...
Find formula to find other possible sides of triangles.
| Let $a$, $b$, $c$ be the integer lengths of the triangle sides, and $r$ the inradius, which I suppose to be integer too. It is well known that $r(a+b+c)$ is twice the area of triangle, so by squaring Heron's formula we have:
$$
r^2(a+b+c)={1\over4}(a+b-c)(a-b+c)(-a+b+c).
$$
The three factors in parentheses on the right... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Help with $\int \cos^6{(x)} \,dx$ Problem:
\begin{eqnarray*}
\int \cos^6{(x)} dx \\
\end{eqnarray*}
Answer:
\begin{eqnarray*}
\int \cos^4{(x)} \,\, dx &=& \int { \cos^2{(x)}(\cos^2{(x)}) } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=& \int { \frac{(1+\cos(2x))^2}{4} } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=&
\int { \fra... | You could also employ the binomial theorem
\begin{align}
\cos^6x&=\frac1{64}(e^{ix}+e^{-ix})^6\\
&=\frac1{64}(e^{i6x}+6e^{i4x}+15e^{i2x}+20+15e^{-i2x}+6e^{-i4x}+e^{-i6x})
\\
&=\frac1{32}(\cos(6x)+6\cos(4x)+15\cos(2x)+10).
\end{align}
Which now is easy to integrate.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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summation of a infinite series i want to know if this infinite sum $\sum_{n=1}^{\infty}$ $\frac{a(a+1)(a+2)......(a+n-1)}{b(b+1)(b+2)......(b+n-1)}$ converges or diverges ? where a>0 and b>a+1.
if the sum converges what is the sum i.e where it will converge? i need a concrete explanation
i found some inequalities $\fra... | Let $u_n=\dfrac{a(a+1)\ldots(a+n-1)}{b(b+1)\ldots(b+n-1)}$
We have $\dfrac{u_{n+1}}{u_n}=1-\dfrac{b-a}{n}+o(\dfrac{1}{n})$
Hence by Raabe Duhamel :
If $b-a>1$ the infinite sum converge
If $b-a<1$ the infinite sum diverge.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How to use quaternions express number How to use quaternions express number like 49 as sums of four squares
based on their factorizations?
I just start to learn quaternion in my math class,
I tired to find my class note :
Quaternions look like a + b i + c j + d k .
Hmmmm ... i guess a,b,c,d are integers.Real numbers ... | Quaternions $q=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$ have a "norm" defined by
$$ |a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}|^2 = a^2+b^2+c^2+d^2. $$
This is a generalization of $|a+bi|=a^2+b^2$ defined for complex numbers. Indeed, it is still multiplicative: we have $|xy|=|x||y|$ for quaternions $x$ and $y$.
How can we ... | {
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Prove $4a^2+b^2+1\ge2ab+2a+b$ Prove $4a^2+b^2+1\ge2ab+2a+b$
$4a^2+b^2+1-2ab-2a-b\ge0$
$(2)^2(a)^2+(b)^2+1-2ab-2a-b\ge0$
Any help from here? I am not seeing how this can be factored
| High school solution:
$$
4a^2+b^2+1-2ab-2a-b=\frac{(2a-1)^2+(2a-b)^2+(b-1)^2}{2}\ge 0.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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Show that $(1+\frac{x}{n})^n \rightarrow e^x$ uniformly on any bounded interval of the real line.
Show that $(1+\frac{x}{n})^n \rightarrow e^x$ uniformly on any bounded interval of the real line.
I am trying to argue from the definition of uniform convergence for a sequence of real-valued functions, but am struggling... | To find your sequence $(a_n)$ where $|(1 + x/n)^n - e^x| \leqslant a_n \to 0$ -- proving uniform convergence on any bounded interval -- use the inequality $\ln(1+y) \leqslant y$.
We have for $0 \leqslant y < 1$,
$$1+y \leqslant e^y = \sum_{k=0}^{\infty} \frac{y^k}{k!} \leqslant \sum_{k=0}^{\infty} y^k = \frac1{1-y},$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2036694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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"answer_id": 0
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how to find mod 100 in the following case what is the remainder when $3^{999}$ divided by 100?
Is there is any short method to find remainder ?
i tried
$3^{1}$ divided by 100=3
$3^{2}$ divided by 100=9
$3^{3}$ divided by 100=27
$3^{4}$ divided by 100=81
but unable to proceed further
| There are some good more-direct methods in the other answers, but let me just help you "proceed further".
Note that "$x \bmod 100$" effectively means the remainder of $x$ when divided by $100$, and $x \equiv y \bmod 100$ means that $x$ and $y$ have the same remainder when divided by $100$.
So you already have:
\begin{a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving difference equation by z transform I have the following difference equation:
$$\ \ y(k+2) - 2y(k+1) +2y(k) = x(k) \,$$
where x(k) is an input of the form $\ x(k) = cos(\pi k)\ $, we also have the initial value conditions
$$\ y(0) = 1\,$$
$$\ y(1) = 1\,$$
I got to the following equation by applying the properti... | For the difference equation:
$$y(k+2) - 2y(k+1) +2y(k) = x(k)$$
We have:
$$
\begin{align}
x(k)&= \cos(\pi k) & y(0)&=1 & y(1)&=1\
\end{align}
$$
Z-transforming both sides of the equation:
$$
\begin{align}
\mathcal{Z}\left[y(k+2) - 2y(k+1) +2y(k)\right]&=\mathcal{Z}\left[x(k)\right]\\
z^2Y(z) - z^2y(0) - zy(1) - 2zY(z) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Use generating functions to solve nonhomogeneous recurrence relation The recurrence relation is
$$a_n = a_{n-1} + a_{n-2} + n,\quad n\ge 2$$
with initial conditions $a_0 = 0$ and $a_1 = 1$.
I know I need to convert the recurrence into series and I have broken it down, but am struggling with getting it into a proper ... | Just manipulate the generating function;
\begin{align}
f(x)=x+\sum_{n=2}^\infty a_nx^n=&x+x\sum_{n=2}^\infty a_{n-1}x^{n-1}+x^2\sum_{n=2}^\infty a_{n-2}x^{n-2}+x\sum_{n=2}^\infty nx^{n-1}\\
=&x+xf(x)+x^2f(x)+x\,\frac{d}{dx}\sum_{n=2}^\infty x^{n}\\
=&x+xf(x)+x^2f(x)+x\,\frac{d}{dx}\underbrace{\left(\frac{1}{1-x}-x-1\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the sum of $\displaystyle\sum_{n=3}^\infty \frac{2^n-1}{3^n}$ My work:$$\sum_{n=3}^\infty \frac{2^n-1}{3^n}$$
$$a_1 = \frac{2^3-1}{3^3}$$
$$a_1 = \frac{7}{27}, r=\frac{2}{3}$$
$$S_N=\frac{\frac{7}{27}}{1-\frac{2}{3}} = \frac{7}{9}$$
The correct answer is $\frac{5}{6}$
| This is not a geometric series on its own, it's the difference of two of them
$$\sum_{n\ge 3} \left({2\over 3}\right)^n-\sum_{n\ge 3}\left({1\over 3}\right)^n = {8/27\over 1-2/3}-{1/27\over 1-1/3}={8\over 9}-{1\over 18}={5\over 6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\lim_{N\rightarrow\infty}\left< \psi{\left(N_k+1\right)} - \psi{\left(N_j+1\right)} \right >$ Question
What is
\begin{align}
\lim_{N\rightarrow\infty}\left< \psi{\left(N_k+1\right)} - \psi{\left(N_j+1\right)} \right >?
\end{align}
Here, $N_k$ and $N_j$ are multinomially distributed random variable with expected val... | My answer is correct. Problem solved.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How many solutions does $x^2 + 3x +1 \equiv 0\, \pmod{101}$ have? $x^2 + 3x +1 \equiv 0 \pmod{101}$. To solve this I found the determinant $D = 5 \pmod{101}$). Using the Legendre symbol,
$$\left(\frac{5}{101}\right) = \left(\frac{101}{5}\right) \equiv \left(\frac{1}{5}\right) \equiv 1,$$
$\therefore$ The equations have... | An idea to tackle this and similar questions using as small numbers and multiples of $\;101\;$ as possible (when possible, of course...).
Observe that $\;5=-96\pmod {101}\;$ , and $\;96=2^5\cdot3\;$ , so we can try to deal with these apparently easier numbers. Since we have $\;96=16\cdot6\;$, we have:
$$4^2=16=2^4\;,\... | {
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"timestamp": "2023-03-29T00:00:00",
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Solutions of $x^3 + (x+4)^2 = y^2$ I want to solve the above equation in integers. I'm pretty sure the only solutions are $(x,y) = (0, \pm 4)$ but I'm not sure how to prove it.
| When $x=0$, $y^2 = 4^2$ and hence $y=\pm 4$. Thus $(x,y) = (0,\pm 4)$ are solutions. We will show there are no other solutions.
Suppose that $x \neq 0$. We have \begin{align}
x^3 = y^2 - (x+4)^2 = (y+x+4)(y-x-4)
\end{align} Let $d$ be the greatest common divisor of $y+x+4$ and $y-x-4$. We claim that $d$ is a power o... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding all solutions to $3x + 4y \equiv 1 \pmod 7$ Find all solutions $\pmod 7$: $3x + 4y$ is congruent to $1 \pmod 7$.
I have tried writing out the various equations such as $3x + 4y = 1$, $3x + 4y = 8$, $3x + 4y = 15$, etc., but I do not know how to find the finite solution.
| Consider the equation
$$ 3x + 4y = 1 (\mod 7).$$
$ (3,4,1) = 1 | 7$, so there are $7\times 1 = 7$ solutions mod $7$. I’ll solve the equation using a reduction trick.
The given equation is equivalent to
$3x + 4y + 7z = 1$ for some $z$.
Set $$ w=\frac {3x+4y}{(3,4)} $$.
Then
$(3,4)w + 7z = 1, w + 7z = 1$.
$w_0 = -6, z_0 ... | {
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"timestamp": "2023-03-29T00:00:00",
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Limit of a sequence including infinite product. $\lim\limits_{n \to\infty}\prod_{k=1}^n \left(1+\frac{k}{n^2}\right)$ I need to find the limit of the following sequence:
$$\lim\limits_{n \to\infty}\prod_{k=1}^n \left(1+\frac{k}{n^2}\right)$$
| $$\left(1+\frac{k}{n^2} \right)\left(1+\frac{n-k}{n^2} \right)=\left(1+\frac{1}{n} +\frac{k(n-k)}{n^4} \right)$$
Then we have
$$\left(1+\frac{1}{n} \right) \leq \left(1+\frac{1}{n} +\frac{k(n-k)}{n^4} \right) \leq \left(1+\frac{1}{n} +\frac{1}{n^2} \right)=\left(1 +\frac{n+1}{n^2} \right)$$
Therefore
$$\left(1+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 2
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Limit of recursive sequence defined by $a_0=0$, $a_{n+1}=\frac12\left(a_n+\sqrt{a_n^2+\frac{1}{4^n}}\right)$ Given the following sequence: $a_0=0$, $$a_{n+1}=\frac12\left(a_n+\sqrt{a_n^2+\frac{1}{4^n}}\right),\ \forall n\ge 0.$$
Find $\lim\limits_{n\to\infty}a_n$.
| By the given recurrence, $a_{n+1}$ is a root of $x^2-a_n x-\frac{1}{4^{n+1}}$, hence:
$$ \sum_{n\geq 0} a_{n+1}(a_{n+1}-a_n) = \sum_{n\geq 0}\frac{1}{4^{n+1}} = \frac{1}{3}. \tag{1}$$
By assuming $\lim_{n\to +\infty}a_n = L$ and applying summation by parts, we get:
$$ \frac{1}{3} = L^2 - \sum_{n\geq 0}(a_{n+2}-a_{n+1})... | {
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"url": "https://math.stackexchange.com/questions/2054026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $a,b,c$ be the length of sides of a triangle then prove that $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge0$
Let $a,b,c$ be the length of sides of a triangle then prove that:
$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge0$
Please help me!!!
| $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)
=\dfrac{1}{2}[(a+b-c)(b+c-a)(a-b)^2+(b+c-a)(a+c-b)(b-c)^2+(a+c-b)(a+b-c)(c-a)^2]
≥0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2055559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Curious limits with tanh and sin These two limits can be easily solved by using De l'Hopital Rule multiple times (I think), but I suspect that there could be an easier way... Is there?
\begin{gather}
\lim_{x\to 0} \frac{\tanh^2 x - \sin^2 x}{x^4} \\
\lim_{x\to 0} \frac{\sinh^2 x - \tan^2 x}{x^4}
\end{gather}
Thanks... | There is: Taylor's formula at order $4$:
*
*$\tanh x=x-\dfrac{x^3}3+o(x^4)$, hence
$$\tanh^2 x=\Bigl(x-\dfrac{x^3}3+o(x^4)\Bigl)^2=x^2-\dfrac{2x^4}3+o(x^4).$$
*$\sin^2x=\frac12(1-\cos 2x)=\frac12\Bigl(1-1+\dfrac{4x^2}2-\dfrac{16x^4}{24}+o(x^4)\Bigr)=x^2-\dfrac{x^4}3+o(x^4)$.
Thus
$$\frac{\tanh^2 x-\sin^2x}{x^4}=\f... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
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First coin is tossed 3 times. $X$=heads after 3 tosses.Second coin is tossed $X$ times and$Y$represents total heads.Find $P(X>=2|Y =1) $ There are 2 fair coins. First coin is tossed 3 times. $X$ represents number of heads in the 3 tosses.After this second coin is tossed $X$ number of times, where $Y$ represents number ... | Can you tell me how does it appear to be $\frac{11}{18}$? Is it by simulation or something else? I'm giving the mathematical argument below. Your conclusion based on statistical reasoning is absolutely fine! There seems to be some holes in whatever procedure that gives you $\frac{11}{18}$ as the answer.
First you need ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}=a-\sqrt b$, find the value of $a+b$
$\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}=a-\sqrt b$ where $a,b$ are natural numbers. Find the value of $a+b$.
I am not able to proceed with solving this ... | Hint: To find
$$\sqrt{a+\sqrt{a+\cdots}} $$
solve the equation
$$x = \sqrt{a+x}$$
As for the answer, I get $11$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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All fractions which can be written simultaneously in the forms $\frac{7k-5}{5k-3}$ and $\frac{6l-1}{4l-3}$ Find all fractions which can be written simultaneously in the forms $\frac{7k-5}{5k-3}$ and $\frac{6l-1}{4l-3}$
for some integers $k,l$.
Please check my answer and tell me is correct or not....
$$\frac{43}{31},\f... | We have that $$\frac{7k-5}{5k-3}=\frac{6l-1}{4l-3}\iff kl+8k+l=6.$$ That is, if $k\ne -1,$
$$l=2\frac{3-4k}{k+1}=-2\left(4-\frac{7}{k+1}\right)=-8+\frac{14}{k+1}.$$ Since $l$ has to be an integer $k+1$ must divide $14.$ So, we have that $k\in\{-15,-8,-3,-2,0,1,6,13\}.$
Note that $k\ne -1$ since if $k=-1$ the equation ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Simplifying $\frac { \sqrt2 + \sqrt 6}{\sqrt2 + \sqrt3}$? Is there anything else you can do to reduce it to something "nicer" other than multiplying it by $\dfrac {\sqrt3 - \sqrt2}{\sqrt3 - \sqrt2}$ and get $\sqrt 6 -2 + \sqrt {18} - \sqrt {12}$? The reason I think there's a nicer form is because the previous problem i... | You are correct. Multiply by the conjugate which is $\sqrt{2}-\sqrt{3}$
After multiplying by the conjugate we have $\frac{2-\sqrt6+\sqrt{12}-\sqrt{18}}{-1}$ which gives $-2+\sqrt6-\sqrt{12}+\sqrt{18}$. The only further simplification is as follows: Simplify $\sqrt{12}$ and $\sqrt{18}$ and we have $-2+\sqrt6-2\sqrt3+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Polynomials and the IMO I have been stuck on this problem, as I am unable to proceed properly. The problem is as follows:
If $f(x) = (x+2x^2+\cdots nx^n)^2 = a_2x^2 + a_3x^3 +\cdots a_{2n}x^{2n},$ prove that $$a_{n+1} + a_{n+2} +\cdots +a_{2n} = \binom{n+1}{2}\frac{5n^2+5n+2}{12}$$
I tried expanding the LHS but o... | Hint: Induction works. Let $S_n$ be the sum on the LHS of the desired inequality. Then
$$
S_1=1=\binom{1+1}{2}\frac{5\cdot 1^2+5\cdot 1+2}{12}.
$$
From
\begin{aligned}
&\quad(x+2x^2+\cdots+nx^n+(n+1)x^{n+1})^2\\
&=(x+2x^2+\cdots+nx^n)^2+2(n+1)(x+2x^2+\cdots+nx^n)x^{n+1}+(n+1)^2x^{2(n+1)}\\
\end{aligned}
we have
$$
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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show $\frac{y}{x^2+y^2} $ is harmonic except at $y=0,x=0$ Let $f(z)=u(x,y)+iv(x,y) $
where $$ f(z)=u(x,y)=\frac{y}{x^2+y^2}$$
show $u(x,y)$ is harmonic except at $z=0$
Attempt
$$ u=\frac{y}{x^2+y^2}=y(x^2+y^2)^{-1} $$
Partial derivatives with x
$$\begin{aligned}
u_x&= y *(x^2+y^2)^{-2}*-1*2x
\\ &= -y*2x(x... | $$\begin{align*}&u_x=-\frac{2xy}{(x^2+y^2)^2}\;&,\;\;&u_{xx}=-\frac{2y(x^2+y^2)-8x^2y}{(x^2+y^2)^3}=\frac{6x^2y-2y^3}{(x^2+y^2)^3}\\{}\\
&u_y=\frac{x^2+y^2-2y^2}{(x^2+y^2)^2}=\frac{x^2-y^2}{(x^2+y^2)^2}\;&,\;\;&u_{yy}=\frac{-2y(x^2+y^2)-4y(x^2-y^2)}{(x^2+y^2)^3}=\frac{-6x^2y+2y^3}{(x^2+y^2)^3}\end{align*}$$
and now jus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Does it converge? $\sum_{n=1}^{\infty} \arcsin^n\frac{1}{n}$ Does it converge? $\sum_\limits{n=1}^{\infty} \arcsin^n\frac{1}{n}$
I'm trying to solve it with the ratio test and using L'hopital, but it does not seem to go well. Can anyone help please?
| It converges drastically.
Since,
for $0 \le x \le \pi/2$
we have
$1 \ge \dfrac{\sin x}{x}
\ge \dfrac{2}{\pi}
$,
so that
$1 \ge \dfrac{ x}{\arcsin x}
\ge \dfrac{2}{\pi}
$.
Therefore
$1
\le \dfrac{\arcsin x}{ x}
\le \dfrac{\pi}{2}
$,
so
$\arcsin x
\le \dfrac{\pi x}{2}
$.
In particular,
$\arcsin \dfrac1{n}
\le \dfrac{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve quadratic equations when the coefficients are complex and real? I needed to solve this: $$x^2 + (2i-3)x + 2-4i = 0 $$
I tried the quadratic formula but it didn't work. So how do I solve this without "guessing" roots? If I guess $x=2$ it works; then I can divide the polynomial and find the other root; but... | The quadratic equation works for real and imaginary coefficients (but maybe without as much geometric intuitiveness when there are imaginary coefficients). You can "check" this by trying to use the quadratic equation and then substituting back in your answers to see if we get $0$. We have $a=1$, $b=2i-3$, $c=2-4i$: $$x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
"answer_id": 1
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Finding the position vector of the centre of a circle of intersection. If there are two spheres with vector equations:
$(r-a)^2=A^2$ and $(r-b)^2=B^2$
where, intuitively, "a" and "b" are vectors representing the centre of the spheres and "A" and "B" are scalars representing the radius of the spheres,
Assuming that they... | Consider any plane containing the sphere's centers $\mathbf a$ and $\mathbf b$. If there is an intersecting circle, it will also contain two points from this circle. Let us consider one and call it $\mathbf c$.
This triangle is illustrated here as follows:
As you can see, the sides of the triangles coincide with the r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given a positive number n, how many tuples $(a_1,...,a_k)$ are there such that $a_1+..+a_k=n$ with two extra constraints The problem was: Given a positive integer $n$, how many tuples $(a_1,...,a_k)$ of positive integers are there such that $a_1+a_2+...+a_k=n$. And $0< a_1 \le a_2 \le a_3 \le...\le a_k$. Also, $a_k-a_1... | Every solution is a tuple in order of increasing integers. Also, $a_k=a_1$ or $a_k=a_1+1$. Therefore, we have all of the $a_1$s at the beginning and $a_1+1$s at the end. We can say there are $l$ instances of $a_1$ and thus $k-l$ instances of $a_1+1$. Since the sum of the tuples is $n$, we find:
$$l*a_1+(k-l)*(a_1+1)=n$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Finding the value of $f'(0)$ If $f$ is a quadratic function such that $f(0)=1$ and
$$\int\frac{f(x)}{x^2(x+1)^3}dx$$
is a rational function, how can we find the value of $f'(0)$? I am totally clueless to this. Any tip on how to start? If you wish to give details, then many thanks to you.
| Let $f(x) = ax^2 + bx + c.$ As $f(0) = 1,$ we have that $f(x) = ax^2 + bx + 1.$ Then $$I=\int \frac{f(x) dx}{x^2(x+1)^3} = \int \frac{a}{(x+1)^3}+\frac{b}{x(x+1)^3}+\frac{1}{x^2(x+1)^3}\,dx \\\stackrel{Partial Fractions}{=} -\frac 1x + \frac{b-a-1}{2 (1 + x)^2} + \frac{-2 + b}{1 + x} + (b-3)\log(x) - (b-3)\log(1 + x) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Choose $a, b$ so that $\cos(x) - \frac{1+ax^2}{1+bx^2}$ would be as infinitely small as possible on ${x \to 0}$ using Taylor polynomial $$\cos(x) - \frac{1+ax^2}{1+bx^2} \text{ on } x \to 0$$
If $\displaystyle \cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \cdots $
Then we should choose $a, b$ in a such way that it's T... | The quick-and-dirty method:
\begin{align*}
f(x) = \frac{1 + a x^2}{1 + bx^2} &= (1 + a x^2) \left( 1 - b x^2 + b^2 x^4 - b^3 x^6 + \cdots \right) \\&= 1 - (b - a) x^2 + (b^2 - ab) x^4 - (b^3 - a b^2) x^6 + \cdots
\end{align*}
We want $b - a = \frac{1}{2}$ and $b (b-a) = \frac{1}{24}$, so that (at least) the first three... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
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How many ways can 8 teachers be distributed among $4 $ schools? There are several ways that the teachers can be divided amongst $4$ schools, namely here are the possible choices I came up with:
$1) 1 1 1 5$
$2) 1 1 2 4$
$3) 1 1 3 3$
$4) 1 2 2 3$
$5) 2 2 2 2$
now given the fact that say $2213$ is the same as $1 2 2 3$ i... | With teachers and schools distinct and every school must have at least one teacher assigned to it, I would approach via inclusion-exclusion.
Let the schools be labeled $a,b,c,d$. Let the teachers be labeled $1,2,3,\dots,8$.
Let event $A$ be the event that school $a$ has no teachers assigned to it. Similarly $B,C,D$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Find the maximum power of $24$ in $(48!)^2$? Find the maximum power of $24$ in $(48!)^2$ ?
How to approach for such questions ?
| Divide $48$ by $2$ repeatedly and add the quotient omitting any remainder,
$48/2=24, 24/2=12, 12/2=6, 6/2=3, 3/2=1$(with remainder as $1$). Therefore adding the quotients,i.e, $24+12+6+3+1=46$ $\Longrightarrow$ there are $46$ $2$s in $48!$ and $92$ $2$s in $(48!)^2$.
Similarly, divide 48 by 3 repeatedly and add the quo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
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Natural numbers $a$ for which $ \int\frac{x^2-12}{(x^2-6x+a)^2}dx$ is a rational function possible values of $a$ for which $\displaystyle f(x) = \int\frac{x^2-12}{(x^2-6x+a)^2}dx$ is a rational function for $a\in N$
$\displaystyle \frac{d}{dx}f(x)=\frac{x^2-12}{(x^2-6x+a)}$ assuming $\displaystyle f(x) = \frac{px+q}{x^... | The polynomial $x^2-6x+a$ has two roots in the complex plane, given by
$$ \zeta_a^\pm = 3\pm\sqrt{9-a}$$
so that $x^2-6x+a = (x-\zeta_a^+)(x-\zeta_a^-)$. Let we perform the partial fraction decomposition of the integrand function given such information:
$$\begin{eqnarray*} \frac{x^2-12}{(x^2-6x+a)^2}&=&\frac{x^2-12}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to simplify an expression that does not have a common factor I am trying to simplify this expression :
$$9a^4 + 12a^2b^2 + 4b^4$$
So I ended up having this :
$$(3a^2)^2 + 2(3a^2)(2b^2) + (2b^2)^2$$
However, after that I don't know how to keep on simplifying the equation, it is explained that the answer is $(3a^2 + ... | Notice that
$$(x+y)^2=x^2+2xy+y^2$$
If $x=3a^2$ and $y=2b^2$, we end up with
$$(3a^2+2b^2)^2=(3a^2)^2+2(3a^2)(2b^2)+(2b^2)^2\\=9a^4+12a^2b^2+4b^4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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$\sum\limits_{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq 1$ proof
$\sum_\limits{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq 1$ for all $n\geq 2$
Basecase n=2
$\sum_\limits{k=1}^{2-1}\frac{1}{\sqrt{k(2-k)}}=1\geq 1$
Assumption
$\sum_\limits{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq 1$ holds for some $n$
Claim
$\sum_\limits{k=1}^{n}... | Using $\sqrt{ab}\leq \frac{a+b}{2}$ it follows that $\frac{1}{\sqrt{k(n-k)}}\geq \frac{2}{n}$, therefore lhs $\geq \frac{2(n-1)}{n}\geq1$ for $n\geq 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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If $ f(x) = \frac{\sin^2 x+\sin x-1}{\sin^2 x-\sin x+2},$ then value of $f(x)$ lies in the interval If $\displaystyle f(x) = \frac{\sin^2 x+\sin x-1}{\sin^2 x-\sin x+2},$ then value of $f(x)$ lies in the interval
assume $\sin x= t$ where $|\sin x|\leq 1$
let $\displaystyle y = \frac{t^2+t-1}{t^2-t+1}$
$\displaystyle yt... | Where DeepSea has left off, let $2t-3=-u\implies1\le u\le5$
As $u>0,$
$$y=1-\dfrac{4u}{u^2-4u+11}=1-\dfrac4{u+\dfrac{11}u-4}$$
Now $u+\dfrac{11}u=\left(\sqrt u-\sqrt{\dfrac{11}u}\right)^2+2\sqrt{11}\ge2\sqrt{11}$
$\implies\dfrac1{u+\dfrac{11}u-4}\le\dfrac1{2\sqrt{11}-4}=\dfrac{2\sqrt{11}+4}{44-4^2}=\dfrac{\sqrt{11}+2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer When checked in calculator it is 1. But how to prove it?
Also it is not a normal addition like $x+ \frac1x$ which needs direct rationalization. So I just need something to proceed.
| Let $x=(2+\sqrt 5)^{1/3}$ and $y=(2-\sqrt 5)^{1/3} .$
We have $5>x^3>4$ and $0>y^3>-1$ so $2>x>1$ and $0>y>-1,$ so $2>x+y>0.$ Therefore $$x+y\in \mathbb Z \iff x+y=1.$$ We have $xy= (x^3y^3)^{1/3}=(-1)^{1/3}=-1$ and $x^3+y^3=4.$ So we have $$(x+y)^3=x^3+y^3+3xy(x+y)=4-3(x+y).$$ So $x+y$ is a positive solution to the e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 6
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Calculate the antiderivative of a given function Consider the function $f : \left[ 0, \frac{\pi}{4} \right)$, $f(x) = \frac{(\cos x + \sin x)^n}{(\cos x - \sin x)^{n + 2}}$, where $n \in \mathbb{N}^*$. Find the antiderivative $F$ of $f$ such that $F(0) = \frac{1}{2(n + 1)}$.
I've noticed that $(\cos x - \sin x)' = -(\c... | We're trying to integrate
$$ \int f(x) dx = \int \frac{(\cos x + \sin x)^n}{(\cos x - \sin x)^{n + 2}} dx,$$
Divide numerator and denominator by $\cos^{n+2}(x)$:
$$ \int \frac{1}{\cos^2(x)}\frac{(1 + \tan x)^n}{(1 - \tan x)^{n + 2}} dx,$$
Now substitute $t = \tan(x)$; $dt = \frac{1}{\cos^2(x)}dx$:
\begin{align} \int \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Doing definite integration $\int_0^{\pi/4}\frac{x\,dx}{\cos x(\cos x + \sin x)}$ We have to solve the following integration
$$
\int_0^{\pi/4}\frac{x\,dx}{\cos x(\cos x + \sin x)}
$$
I divided both in Nr and Dr by $\cos^2 x$.
But after that I stuck.
| Let $$I(x)= \int^{\frac{\pi}{4}}_0 \frac{x}{\cos(x)(\cos(x)+\sin(x))} dx$$
$$ I(x)= \int^{\frac{\pi}{4}}_0 \frac{x}{\cos(x)(\sqrt2\sin(x+\frac{\pi}{4}))} dx$$
$$ I(x)= \int^{\frac{\pi}{4}}_0 \frac{x}{\sqrt2\cos(x)\sin(x+\frac{\pi}{4})} dx$$
Since $$ I(x)=I(\frac{\pi}{4}-x)$$
$$ I(\frac{\pi}{4}-x)=\int^{\frac{\pi}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
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Finding the numbers $n$ such that $2n+5$ is composite. Let $n$ be a positive integer greater than zero. I write
$$a_n =
\begin{cases}
1 , &\text{ if } n=0 \\
1 , &\text{ if } n=1 \\
n(n-1), & \text{ if $2n-1$ is prime} \\
3-n, & \text{ otherwise}
\end{cases}$$
The sequence goes like this $$1,1,2,6,12,-2,30,42,-5,72,90... | $n=13$ is a counter example for claim 2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Proving $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$
Show that $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$ and find the correct phase angle $\alpha$.
This is my proof.
Let $x$ and $\alpha$ be the the angles in a right triangle with sides $a$, $b$ and $c$, as shown in the figure. Then, $c=\sqrt{a^2+b^2}$. The le... | Let $a = r \cos \alpha$ and $b = r \sin \alpha$. Then,
$
a\cos x + b\sin x\\
= r\cos \alpha \cos x + r\sin \alpha \sin x\\
= r \cos (x - \alpha)
$
since we know that $\cos \theta \cos \phi + \sin \theta \sin \phi = \cos (\theta - \phi)$. Now, we already have $a = r \cos \alpha$ and $b = r \sin \alpha$. Squaring and add... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Maximum value of $|x-y|$ Given $x,y\in\mathbb R$ such that
$$5x^2+5y^2-6xy\ =\ 8$$
find the maximum value of $|x-y|$.
My attempt
$5x^2 - 6yx + (5y^2-8)\ =\ 0$
$x\ =\ \dfrac{6y\pm\sqrt{(6y)^2-4(5)(5y^2-8)}}{10} = \dfrac{6y\pm\sqrt{160-64y^2}}{10} = \dfrac{3y\pm2\sqrt{10-4y^2}}{10}$
$5y^2 - 6xy + (5x^2-8)\ =\ 0$
$y\ =\ \... | Hint: We can write $$P= 5x^2 +5y^2-6xy-8 =5 (x-y)^2+4xy-8 =5 (x-y)^2 +(x+y)^2-(x-y)^2-8=4 (x-y)^2 +(x+y)^2-8$$ Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the $\gcd[x+y+z; x^2+xy+z^2; y^2+yz+z^2; z^2+zx+x^2]$ What I have done:
There exists a non-zero integer $t$ such:
$$x+y+z=kt$$
$$x^2+xy+y^2=ut$$
$$y^2+yz+z^2=vt$$
$$z^2+zx+x^2=wt$$
$\implies$
$$(x-y)(x+y+z)=(u-v)t$$
$$(y-z)(x+y+z)=(v-w)t$$
$$(z-x)(x+y+z)=(w-u)t$$
$\implies$
$$\dfrac{x+y+z}{t}= \dfrac{u-v}{x-y}=\d... | $gcd(x+y+z,x^2+xy+z^2,y^2+yz+z^2,z^2+zx+x^2)=1$
If $(x+y+z,x^2+xy+z^2,y^2+yz+z^2,z^2+zx+x^2)$ shared a common factor, the existence of an algebraic factor would soon be apparent after trying a few numerical examples.
So I used $(x,y,z)=(17,23,31),(2,3,5),(3,7,11)$ and found $gcd=1$ in each case.
I realise this method m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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If $f(x+1)+f(x-1)= \sqrt{2}\cdot f(x)$, find the period of $f(x)$. If $f(x+1)+f(x-1)= \sqrt{2}\cdot f(x)$, then the period of $f(x)$ is?
I tried replacing $x$ with $x-1$ and stuff, but didn't lead me to anything...
| Make $x=x-1$
$$f(x)+f(x-2)=\sqrt{2}f(x-1) \tag {1}$$
Make $x=x+1$
$$f(x+2)+f(x)=\sqrt{2}f(x+1) \tag {2}$$
Now sum $(1)$ and $(2)$:
$$2f(x)+f(x+2)+f(x-2)=\sqrt{2}(f(x+1)+f(x-1))=2f(x) $$
$$f(x+2)+f(x-2)=0$$
Now make $x=x+2$ then $f(x+4)=-f(x)$.
Make $x+4$ and get $f(x+8)=-f(x+4)=f(x)$, then the period is $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find the sum of first $n$ terms of the series: $\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots +\frac{1}{n\times(n+1)}$
I have the series $$\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots +\frac{1}{n\times(n+1)}$$
I know the following formulas: $$1+2+3+\cdots +n=\frac {n (n+1)}{2}\ta... | Hint: We can write $$\frac {1}{n(n+1)} =\frac {1}{n} -\frac {1}{n+1} $$ Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
$4x^2-5xy+4y^2=19$ with $Z=x^2+y^2$, find $\dfrac1{Z_{\max}}+\dfrac1{Z_{\min}}$ I have no idea how to approach this question at all. I've tried to find the maximum and minimum of the quadratic but i am too confused on what to do afterwards.
| *
*From the given equation $\,xy = \cfrac{4(x^2+y^2) - 19}{5} = \cfrac{4 Z - 19}{5}\,$. But for $\forall x,y \in \mathbb{R}$ the inequality holds $\,xy \le \cfrac{1}{2}(x^2+y^2)\,$ so $\,\cfrac{4 Z - 19}{5} \le \cfrac{Z}{2} \iff \boxed{Z \le \cfrac{38}{3}}\,$
*$0 \le (x+y)^2 = x^2+y^2+2xy = Z + 2 \cfrac{4 Z - 19}{5}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Maximum of $f(x) =(1+\cos x)\cdot \sin(\frac{x}{2})$ on $x \in (0, \pi)$ I've attempted to solve this question by using $$f(x) = 2\sin\frac{x}{2}\cos^2\frac{x}{2} \leq \frac{(2\sin\frac{x}{2}+\cos^2\frac{x}{2})^2}{2}$$ but it results in the wrong answer every time. Is there another way to solve this question and is the... | When you have $2 \sin \frac{x}{2} (1-\sin^2 \frac x2) = 2\sin \frac x2 - 2\sin^3 \frac x2$, where $x$ varies between $0$ and $\pi$, hence we can let $y= \sin \frac x2$, then $y$ varies between $0$ and $1$, and we are trying to find the maximum value of $2y-2y^3$. Now, we can just use ordinary differentiation, $2-6y^2=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What values can $2^j-3^k$ have? What values can $2^j-3^k$ have?
E.g.,
$$
2^2-3^1=1\\
2^2-3^0=3\\
2^3-3^1=5\\
2^4-3^2=7
$$
Can every number not divisible by $2$ or $3$ be written as $2^j-3^k$? If not, why?
| we try to find an $n$ coprime to $6$ such that $2^j-3^k$ doesn't cover all options $\bmod n$.
Since we want $2^j$ to cover a small number of cases we are going to try with $n=2^m-1$.
We find that the order of $3\bmod 511$ is $12$ and the order of $2\bmod 511$ is clearly $9$.
Thus only a small fraction of residues $\bm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Solving irrational inequalities: $\sqrt{x^2-7x+10}+9\log_4{x/8}\geq 2x+\sqrt{14x-20-2x^2}-13$
Solving the following inequalities: $$\sqrt{x^2-7x+10}+9\log_4{x/8}\geq 2x+\sqrt{14x-20-2x^2}-13$$
I have been solving some questions on inequalities lately and I have come across this question, but I can't figure out a ... | Hint:
Let $x \in \mathbb R$
If $$\sqrt{x^2-7x+10}=\sqrt{(x-2)(x-5)}$$ then $x\le 2$ or$x \ge5$
If $$\sqrt{-2(x^2-7x+10)}=\sqrt{-2(x-2)(x-5)}$$ then $2\le x \le 5$
Hence, $x=2$ or $x=5$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Graph the inequality $1<|2z-6|<2$. How to graph the inequality $1<|2z-6|<2$ in the complex plane?
My work so far is below:
I assume $z$ is complex so it can be represented as $z=a+bi$
\begin{align}
1<|2z-6|<2 \\
1<2|(a+bi)-3|<2 \\
\frac{1}{2}<|a-3+bi|<1\\
\frac{1}{2}<|a-3+bi|<1 \\
\frac{1}{2}<\sqrt{a^2-6a+9+b^2}<1 \\
\... | hint
$$1<|2z-6|<2 \Leftrightarrow\frac{1}{2}<|z-3|<1 \Leftrightarrow \frac{1}{4}<|z-3|^2<1$$
$$\frac{1}{4}<(a-3)^2+b^2<1$$
What is:
$$(a-3)^2+b^2<1?$$
and
$$(a-3)^2+b^2>1/4?$$
what about the intersection?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find $xyz$ given that $x + z + y = 5$, $x^2 + z^2 + y^2 = 21$, $x^3 + z^3 + y^3 = 80$ I was looking back in my junk, then I found this:
$$x + z + y = 5$$
$$x^2 + z^2 + y^2 = 21$$
$$x^3 + z^3 + y^3 = 80$$
What is the value of $xyz$?
A) $5$
B) $4$
C) $1$
D) $-4$
E) $-5$
It's pretty easy, any chances of solving this que... | x + y + z = 5
On squaring both sides,
$(x + y + z)^2 = 25$
$x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = 25$
$21 + 2xy + 2yz + 2zx = 25$
$2xy + 2yz + 2zx = 25 - 21$
$2xy + 2yz + 2zx = 4$
$xy + yz + zx = 2$
Also,
$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$
Putting values,
$80 - 3xyz = (5)\left[21 - (xy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find $y'$ given $y\,\sin\,x^3=x\,\sin\,y^3$? The problem is
$$y\,\sin\,x^3=x\,\sin\,y^3$$
Find the $y'$
The answer is
Can some explain how to do this, please help.
| $y \sin x^3=x\sin y^3→$ Differentiating both sides with respect to $x .$
$⇒y(\cos x^3×3x^2)+y′\sin x^3 =\sin y^3+x \cos y^3×3y^2y′ $
$⇒3x^2y \cos x^3+y′\sin x^3=\sin y^3+3y^2x \cos y^3y′$
$⇒y′[\sin x^3−3y^2x\cos y^3]=\sin y^3−3x^2y \cos x^3$
$⇒y′=\dfrac{\sin y^3−3x^2y \cos x^3}{\sin x^3−3y^2x \cos y^3}$
I was searchi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Compute $\lim \limits_{n \to \infty} \frac{\sqrt[4]{n^4+4n}\,-\,\sqrt[3]{n^3+3n}}{\sqrt[5]{n^5+1}\,-\, \sqrt[5]{n^5+n}}\cdot \frac{1}{n^2}$ I need to compute:
$\lim \limits_{n \to \infty} \frac{\sqrt[4]{n^4+4n}\,-\,\sqrt[3]{n^3+3n}}{\sqrt[5]{n^5+1}\,-\, \sqrt[5]{n^5+n}}\cdot\frac{1}{n^2}$
I tried this
$\lim \limits_{n ... | The answer is 5.
I expect you're supposed to use the generalized binomial theorem here: namely, $(1+x)^n = \sum_{k=0}^{\infty}\left(\frac{n^{\underline{k}}}{k!}x^k\right)$, where $n^{\underline{k}}$ means $1$ for $k=0$, or $n(n-1)(n-2)\cdots(n-k+1)$ otherwise.
$$\lim \limits_{n \to \infty} \frac{\sqrt[4]{1+\frac{4}{n^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Proving an alternate quadratic formula It is well known that the quadratic formula for $ax^2+bx+c=0$ is given by$$x=\dfrac {-b\pm\sqrt{b^2-4ac}}{2a}\tag1$$
Where $a\ne0$. However, I read somewhere else that given $ax^2+bx+c=0$, we have another solution for $x$ as$$x=\dfrac {-2c}{b\pm\sqrt{b^2-4ac}}\tag2$$
Where $c\ne0$... | There are a few errors in the question as posed.
First, the quadratic formula is
$$x = \frac{-b \pm \sqrt {b^2-4ac} }{2a}$$
Note the factor of $a$ in the denominator, missing from the OP. [EDIT -- I see this has been fixed now.]
Second, you write:
In fact, (2) gives solutions for $0x^2+bx+c=0$!
This is not true, and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Find linear transformation given its matrix representation
If $$A = \left[ \begin{matrix} 1 & -1 & 2 \\
-2 & 1 & -1 \\
1 & 2 & 3
\end{matrix} \right]$$ is the matrix representation of a linear transformation $T: P_2(x) \to P_2(x)$ with respect to the bases
$\{1-x,... | The relation between $A$ and $T$ is
$$A\left[a + bx + cx^2\right]_{\mathcal B} = \left[T\left(a + bx + cx^2\right)\right]_{\mathcal B'}$$
where $\mathcal B$ is the first basis, $\mathcal B'$ is the second basis, and $[ \cdot ]_{\mathcal B}$ denotes the coordinate matrix in basis $\mathcal B.$
To find $T,$ we compute as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability with balls and a box - complementary event of "exactly" There are $12$ balls in a box: $b$ blue, $y$ yellow and $3$ red balls. Three balls are randomly chosen. If the probability of choosing one blue, one yellow and one red is $3/11$, find the number of yellow balls in a box.
Attempt:
If $t$ is the total nu... | We have,
b + y + 3 = 12
b + y = 9
b = 9 - y
Probability = $\frac{\binom{9 - y}1 \times \binom{y}1 \times \binom{3}1}{\binom{12}3}$
$\frac 3{11} = \frac{\frac{(9 - y)!}{(8 - y)!.1!} \times \frac{(y)!}{(y - 1)!.1!} \times \frac{3!}{2!.1!}}{\frac{12!}{3!.9!}}$
$\frac 3{11} = \frac{\frac{(9 - y)(8 - y)!}{(8 - y)!} \times \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solve $x=\pm\frac{y'}{\sqrt{(y')^2+1}}$
$$x=\pm\frac{y'}{\sqrt{(y')^2+1}}$$
$$x=\pm\frac{y'}{\sqrt{(y')^2+1}}$$
$$x^2=\frac{(y')^2}{{(y')^2+1}}$$
$$x^2(y')^2+x^2=(y')^2$$
$$(y')^2[x^2-1]=-x^2$$
$$(y')^2=\frac{-x^2}{x^2-1}$$
$$y'=\pm \sqrt{\frac{-x^2}{x^2-1}}$$
Have I got it wrong? as there is no ODE
| Here is a solution
$$y'=\pm \sqrt {\frac{-x^2}{x^2-1}} \\ \frac{dy}{dx} =\pm \frac x {\sqrt {1-x^2}} \\ \int dy = \pm \int \frac x {\sqrt {1-x^2}} \, dx =\pm \frac 1 2 \int \frac {d(1-x^2)}{\sqrt{ 1-x^2}}=\pm \sqrt {1-x^2} \\ y=\pm \sqrt{1-x^2} +C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solve the equation: $\sin 3x=2\cos^3x$ Solve the equation :
$$\sin 3x=2\cos^3x$$
my try :
$\sin 3x=3\sin x-4\sin^3x$
$\cos^2x=1-\sin^2x$
so:
$$3\sin x-4\sin^3x=2((1-\sin^2x)(\cos x))$$
then ?
| $$2\cos^3x=3\sin x-4\sin^3x$$
Divide both sides by $\cos^3x$
$$2=3\tan x(1+\tan^2x)-4\tan^3x$$
$$\tan^3x-3\tan x+2=0$$ which is a cubic equation in $\tan x$
Clearly, one of the roots is $\tan x=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2111959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to solve a system of 3 trigonometric equations How to solve system of three trigonometric equations:
$(\sin x)^2 (\cos y)^2 = 4 \cos x \sin y\tag1$
$(\sin y)^2 (\cos z)^2 = 4 \cos y \sin z \tag2$
$1- \sqrt{\sin z}(1+\sqrt{\cos x})=\sqrt{\frac{1-\sin y}{1+\sin y}}\tag3$
and to verify that
$\sin x=\sqrt{2}(\sqrt{2}... | This is based on modular equations of degree 2. If $k, l$ are elliptic moduli with $l$ of degree 2 over $k$ then we have $$l=\frac{1-k'}{1+k'}\tag{1}$$ where $k'=\sqrt{1-k^2}$.
One should observe that the relation between $\sin x, \sin y$ is such that $\sin y$ is of degree 2 over $\sin x$. This is easily verified by wr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find two numbers whose $AM+...$ Find two numbers whose $AM + GM =25$ and $AM:GM=5:3$.
My Attempt;
Given,
$\frac {AM}{GM}=\frac {5}{3} = k (let) $
$\AM=5k,
GM=3k$.
Also,
$AM+GM=25$
$5k+3k=25$
$8k=25$
$k=\frac {25}{8}$.
Am I going right? Or, is there any other simple alternative.?
| Let the numbers be $a^2$ and $b^2$.
Then we have :
$$
\frac{a^2+b^2}{2} +ab=25
\implies (a+b)^2=50
\implies a+b=5\sqrt{2}
$$
Let $AM$ be $5x$ and $GM$ be $3x$. Then, $x=\frac{25}{8}$ and the $AM$ is $\frac{125}{8}$ and the $GM$ is $\frac{75}{8}$.
Hence
$$
a^2+b^2=2AM=\frac{125}{4} \\
2ab=2GM=\frac{75}{4} \\
a^2+b^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Help Determinant Binary Matrix I was messing around with some matrices and found the following result.
Let $A_n$ be the $(2n) \times (2n)$ matrix consisting of elements $$a_{ij} = \begin{cases} 1 & \text{if } (i,j) \leq (n,n) \text{ and } i \neq j \\ 1 & \text{if } (i,j) > (n,n) \text{ and } i \neq j \\ 0 & \text{oth... | The matrix is diagonal by blocks. so the determinant is equal to the product of the determinants of each block.
So we just have to prove that:
$\det\begin{pmatrix}0 & 1 & \dots & 1 \\ 1 & 0 & \dots & 1 \\ & &\vdots \\ 1 & 1 & \dots & 0\\ \end{pmatrix}=\pm(n-1)$
This is done in various ways here, in fact it is equal to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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$\tan {\frac{A}{2}} + \tan {\frac{B}{2}} +\tan{\frac{C}{2}} \geq 4 - \sqrt {3} $
In a triangle $ABC$ with one angle exceeding $\frac {2}{3} \pi$, prove that
$\tan {\frac{A}{2}} + \tan {\frac{B}{2}} + \tan{\frac{C}{2}} \geq 4 - \sqrt {3} $
I tried expanding that half angle, applying AM-GM on various sets, using Si... | Let $ \tan\dfrac A2 =x, \tan\dfrac B2=y, \tan\dfrac C2=z,\tan \dfrac{A}{2} + \tan \dfrac{B}{2} + \tan\dfrac{C}{2} = w$. WLOG $x \ge \sqrt{3}$.
Note that we have the identity $\tan\left(\dfrac{A+B}2\right)=\tan\left(\dfrac{\pi- C}2\right)$
$$\implies\dfrac{x+y}{1-xy}=\dfrac1{z}$$
$$\implies xy+yz+zx=1 \le \frac{(y+z)^... | {
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"url": "https://math.stackexchange.com/questions/2114781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving a well-known inequality using S.O.S Using $AM-GM$ inequality, it is easy to show for $a,b,c>0$, $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge 3.$$
However, I can't seem to find an S.O.S form for $a,b,c$
$$f(a,b,c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a} - 3 = \sum_{cyc}S_A(b-c)^2 \ge 0.$$
Update:
Please not... | Let $c=\min\{a,b,c\}$. Hence,
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3=\frac{a}{b}+\frac{b}{a}-2+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-1=\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)}{ac}\geq0$$
From here we can get a SOS form:
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3=\frac{1}{6abc}\sum_{cyc}(a-b)^2(3c+a-b)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
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Find all prime solutions of equation $5x^2-7x+1=y^2.$ Find all prime solutions of the equation $5x^2-7x+1=y^2.$
It is easy to see that
$y^2+2x^2=1 \mod 7.$ Since $\mod 7$-residues are $1,2,4$ it follows that $y^2=4 \mod 7$, $x^2=2 \mod 7$ or $y=2,5 \mod 7$ and $x=3,4 \mod 7.$
In the same way from $y^2+2x=1 \mod 5$ w... | Since $(0,1)$ is a solution to $5x^2-7x+1=y^2$, it can be used to parametrize all rational solutions to $5x^2-7x+1=y^2$. That will give us:
$$x:=\frac{-2ab - 7b^2}{a^2 - 5b^2}$$
and
$$y:=\frac{a}{b}x+1$$
where $a,b\in \mathbb{Z}$, $b\neq 0$.
Since $x$ is prime, it follows that either $b=1$ or $x=b$.
*
*case $b=1$
... | {
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"url": "https://math.stackexchange.com/questions/2117054",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "10",
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Quadratic equation, find $1/x_1^3+1/x_2^3$ In an exam there is given the general equation for quadratic: $ax^2+bx+c=0$.
It is asking: what does $\dfrac{1}{{x_1}^3}+\dfrac{1}{{x_2}^3}$ equal?
| Let $\dfrac1{x^3}=y\iff x^3=?$
$$(-c)^3=(ax^2+bx)^3\iff -c^3=a^3(x^3)^2+b^3(x^3)+3ab(x^3)(-c)$$
$$\iff-c^3=\dfrac{a^3}{y^2}+\dfrac{b^3}y-\dfrac{3abc}y$$
$$\iff c^3y^2+(b^3-3abc)y+a^3=0$$ whose roots are $\dfrac1{x_1^3},\dfrac1{x_2^3}$
Can you apply Vieta's formula now?
| {
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How to use finite differences to determine an equation of a polynomial given consecutive integer $x$ and corresponding $y$ coordinates of the graph? This chart is given:
for $x=-3$, $y=-9$
for $x=-2$, $y=3$
for $x=-1$, $y=3$
for $x=0$, $y=-3$
for $x=1$, $y=-9$
for $x=2$, $y=-9$
for $x=3$, $y=3$
I found the finite diffe... | Once you discover that the third difference is constant you know that the polynomial is third degree. Denote it by $P(x)$.
We can see that the horizontal lines $y=3$ and $y=-9$ cross the graph of $y=P(x)$ three times each. Therefore the graph of the polynomial $Q(x)=P(x)-3$ has $x$-intercepts $(-2,0),\,(-1,0)$ and $(3,... | {
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Fourier Series/ fourier transform What is the Fourier series of the following piece-wise function?
$$
f(x) = \begin{cases}
0 & -1 \leq x < -0.5 \\
\cos (3 \pi x) & -0.5 < x < 0.5 \\
0 & 0.5 \leq x < 1
\end{cases}
$$
| Define problem
Piecewise function: Resolve $f(x)$ into a left, center, and right piece
$$
\begin{align}
%
l(x) &= 0, \qquad \qquad \, -1 \le x < -\frac{1}{2} \\
%
c(x) &= \cos \left( 3\pi x \right) \quad\ -\frac{1}{2} \le x \le \frac{1}{2} \\
%
r(x) &= 0, \qquad \qquad \ \ \ \frac{1}{2} \le x \le 1
%
\end... | {
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Find the value of $x$ that satisfy the equation: $3^{11}+3^{11}+3^{11} = 3^x$ I have this question:
$$3^{11}+3^{11}+3^{11} = 3^x$$
Find the value of $x$
| Answer:
$$3^{11}+3^{11}+3^{11} = 3^x\implies 3^{10}(3^{1} + 3^{1} + 3^{1}) =3^x$$
$$(3^1 + 3^1+ 3^1) = \frac {3^x} {3^{10}}$$
$$9 = \frac{3^x}{3^{10}}$$
$$3^2 = \frac {3^x}{3^{10}}$$
Hence:
$$2 = x -10$$
$$x = 12$$
| {
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Common roots of a quadratic equation If the quadratic equations, $ x^2
+ bx + c = 0$ and $ bx^2
+ cx + 1 = 0 $ have a common root then we have to prove that either $b + c + 1 = 0$ or $ b^2
+ c^2 + 1 = bc + b + c$.
I tried it a lot , but not able to proceed further .
Can anybody provide me a hint ?
| We have that both these equations have a common root. Solving these equations for $x^2$ and $x $, we get, $$x^2=\frac {b-c^2}{c-b^2} \text { and } x =\frac {bc-1}{c-b^2} $$ Now eliminating $x $, we get, $$b^3+c^3+1-3bc=0$$ $$\Rightarrow (b+c)^3-3bc (b+c) +1 -3bc =0$$ $$\Rightarrow (b+c+1)(b^2-(c+1)b +(c^2-c+1))=0$$ ... | {
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Find the determinant of order $100$ Find the determinant of order $100$:
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\en... | Starting from
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\end{vmatrix}=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
0 &0 ... | {
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finding limit with $\cos$ function occur $n$ times Finding $\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos(1-\cos(1-\cos(1-\cdots \cdots (1-\cos x))))}{x^{2^n}}$
where number of $\cos$ is $n$ times
when $x\rightarrow 0$ then $\displaystyle 1-\cos x = 2\sin^2 \frac{x}{2} \rightarrow 2\frac{x}{2} = x$
so $1-\cos (1-\cos... | Take the simple case
$$\frac{1-\cos(1-\cos x)}{x^4}$$
$$=\frac{1-\cos(1-\cos x)}{(1-\cos x)^2}\left(\frac{1-\cos x}{x^2}\right)^2$$
$$\rightarrow \frac{1}{2}\left(\frac{1}{2}\right)^2$$
Can you see how to do the induction ?
So you have in general if $f(n)=1-\cos(1-\cos(1-\cos(1-\cdots \cdots (1-\cos x))))$
$$\frac{f(n... | {
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Let $a,b,c$ be positive real numbers such that $abc =1$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$a^2+b^2+c^2\geq a+b+c$$.
Also, state the condition for equality.
My Attempt,
$a,b,c$ are real and positive numbers, then
$$(a-1)^2+(b-1)^2+(c-1)^2\ge 0$$
$$a^2-2a+1+ b^2-2b+1+c^2-2c+1\ge 0$$
$... | Note that from $$(a-1)^2+(b-1)^2+(c-1)^2 \ge 0$$we have $$a^2+b^2+ c^2 \ge 2(a+b+c)-3$$
By AM-GM $$a+b+c \ge 3 \sqrt[3]{abc}=3 \implies a+b+c-3 \ge 0$$
So $$a^2+b^2+ c^2 \ge 2(a+b+c)-3=a+b+c+(a+b+c-3) \ge a+b+c$$
EDIT
Here is a simple proof of AM-GM when $n=3$. We will prove $$\frac{a+b+c}{3} \ge \sqrt[3]{abc}$$
Let $x... | {
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Prove the polynomial $f(x) = 3x^3+x^2+2x+1155$ has no root in $\mathbb{Z}$ Prove the polynomial $f(x) = 3x^3+x^2+2x+1155$ has no root in $\mathbb{Z}$
The hint says If $f$ has a root in $\mathbb{Z}$, then $f$ has also a root in $\mathbb{Z}/2\mathbb{Z}$. I'm still confused
| If $f(x) = 0$ then $f(x) \equiv 0 \mod n$ for all $n$.
So $f(x) \equiv 0 \mod 2$.
So $3x^3+x^2+2x+1155 \equiv 0 \mod 2$
$x^3 + x^2 + 1 \equiv 0 \mod 2$
$x^3 + x^2 \equiv 1 \mod 2$
Notice $0^k \equiv 0 \mod n$ and $1^k \equiv 1 \mod n$ and and $x$ is either $x \equiv 0 \mod 2$ or $x \equiv 1 \mod2$ we have $x^k \equiv ... | {
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If:$(\sqrt{x + 9})^{\frac{1}{3}} - (\sqrt{x-9})^{\frac{1}{3}} = 3$, Find $x^2$
If:$$\sqrt[3]{(x + 9)} - \sqrt[3]{(x-9)} = 3$$
Find $x^2$
I can't seem to solve this question. Any hints or solutions is welcomed.
| Put $a=(x+9)^{1/3},b=(x-9)^{1/3}$
$$a-b=3\\(a-b)^3=3^3\\a^3-3ab(a-b)+b^3=27$$
Now you already know that $a-b=3$ so
$$a^3-9ab-b^3=27$$
Now plugging back $a,b$
$$((x+9)^{1/3})^3-9(x+9)^{1/3}(x-9)^{1/3}-((x-9)^{1/3})^3=27\\x+9-9(x^2-81)^{1/3}-(x-9)=27\\18-9(x^2-81)^{1/3}=27\\(x^2-81)^{1/3}=-1\\x^2-81=-1\\x^2=80$$
| {
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Solving a univariable identity that satisfies the following relationship:
Question: Consider the following equations:$$\begin{align*}1^2+2^2+2^3 & =3^2\\2^2+3^2+6^2 & =7^2\\3^2+4^2+12^2 & =13^2\\4^2+5^2+20^2 & =21^2\end{align*}$$
State a one variable identity that is suggested by these examples.
Since the question ... | Notice that the $2$nd and $3$rd term are respectively $2,2$ then $3,6$ then $4,12$ then $5,20$.Now dividing the $3$rd with the second you get $1,2,3,4$ respectively you can guess that $3$rd term is $n$ times bigger then $2$nd and you can see that the $2$nd is $(n+1)$.Also the $4$th term is $3$rd minus one.
Putting tha... | {
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help prove inequality when $3\left( {{a^4} + {b^4} + {c^4}} \right) + 10 = 7\left( {{a^2} + {b^2} + {c^2}} \right)$ Given the positive real numbers $a,b,c$ satisfy $3\left( {{a^4} + {b^4} + {c^4}} \right) + 10 = 7\left( {{a^2} + {b^2} + {c^2}} \right)$. Prove that $$\frac{{{a^2}}}{{b + 2c}} + \frac{{{b^2}}}{{c + 2a}} +... | By C-S $$\sum_{cyc}\frac{a^2}{b+2c}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(b+2c)}=\frac{a+b+c}{3}.$$
Thus, it remains to prove that $$a+b+c\geq\frac{16}{9}$$ or
$$\sum_{cyc}\left(a-\frac{16}{27}+\frac{1}{2}\left(3a^4-7a^2+\frac{10}{3}\right)\right)\geq0$$ or
$$\sum_{cyc}(81a^4-189a^2+54a+58)\geq0,$$
which is true becaus... | {
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How to integrate $\int_{0}^{1} \frac{1-x}{1+x} \frac{dx}{\sqrt{x^4 + ax^2 + 1}}$? The question is how to show the identity
$$ \int_{0}^{1} \frac{1-x}{1+x} \cdot \frac{dx}{\sqrt{x^4 + ax^2 + 1}} = \frac{1}{\sqrt{a+2}} \log\left( 1 + \frac{\sqrt{a+2}}{2} \right), \tag{$a>-2$} $$
I checked this numerically for several cas... | $$\int_0^1 \frac{1-x}{1+x}\frac{dx}{\sqrt{x^4+ax^2+1}}\overset{\large \frac{1-x}{1+x}=t}=\int_0^1 \frac{2t}{\sqrt{(a+2)t^4-2(a-6)t^2+(a+2)}}dt$$
$$\overset{t^2=x}=\frac{1}{\sqrt{a+2}}\int_0^1 \frac{dx}{\sqrt{x^2-2\left(\frac{a-6}{a+2}\right)x+1}}=\frac{1}{\sqrt{a+2}}\ln \left(1+\frac{\sqrt{a+2}}{2}\right)$$
| {
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Is there a systematic way to find irreducible polynomials? I am preparing for an exam and I'm looking for a systematic way to find all irreducible polynomials. I am given the following problem > Find all irreducible polynomials of degree 3 with coefficients over Z2.
I know that a polynomial is reducible if it can be re... | If $f(x)\in \mathbb Z_2[x]$ and has degree $3$,then $f(x)=x^3+ax^2+bx+c$ and we have the possible states:
$a=0,b=0,c=0 \Rightarrow f(x)=x^3$
$a=1,b=0,c=0 \Rightarrow f(x)=x^3+x^2$
$a=1,b=1,c=0 \Rightarrow f(x)=x^3+x^2+x$
$a=1,b=1,c=1 \Rightarrow f(x)=x^3+x^2+x+1$
$a=1,b=0,c=1 \Rightarrow f(x)=x^3+x^2+1$
$a=0,b=1,c=0 \... | {
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Evaluate the integral $\int \frac{x^2(x-2)}{(x-1)^2}dx$
Find $$\int \frac{x^2(x-2)}{(x-1)^2} dx .$$
My attempt:
$$\int \frac{x^2(x-2)}{(x-1)^2}dx = \int \frac{x^3-2x^2}{x^2-2x+1}dx $$
By applying polynomial division, it follows that
$$\frac{x^3-2x^2}{x^2-2x+1} = x + \frac{-x}{x^2-2x+1}$$
Hence $$\int \frac{x^3-2x^2}{... | With a shift, $t=x-1$,
$$\int \frac{(t+1)^2(t-1)}{t^2} dt=\int\left(t+1-\frac1t-\frac1{t^2}\right)dt\\
=\frac{(x-1)^2}2+(x-1)-\ln|x-1|+\frac1{x-1}+C.$$
| {
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A quadratic polynomial $f(x)$ satisfy $ f(x) = \left[\frac{f(x+1)-f((x-1)^2)}{2}\right]^2$ for all real $x$. then $f(x)$ is A quadratic polynomial $f(x)$ satisfy $\displaystyle f(x) = \left[\frac{f(x+1)-f((x-1)^2)}{2}\right]^2$ for all real $x$. then $f(x)$ is
Let $f(x) = ax^2+bx+c$, then substitute $x=0$ in functiona... | With $f(x)=ax^2$ and $f(x) = \left[\frac{f(x+1)-f((x-1)^2)}{2}\right]^2$ we get
$ax^2=\left[\frac{a(x+1)^2-a(x-1)^4}{2}\right]^2$.
Now take $x=1$ and look what happens....
| {
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Find the value of $\binom{2000}{2} + \binom{2000}{5} + \binom{2000}{8} + \cdots \binom{2000}{2000}$
Find the value of $\binom{2000}{2} + \binom{2000}{5} + \binom{2000}{8} + \cdots \binom{2000}{2000}$
I've seen many complex proofs. I am looking for an elementary proof. I know the fact that $\binom{2000}{0} + \binom{... | Let $w=\exp(2\pi i/3)$. Then $w^3=1$ and $1+w+w^2=0$, and hence
$$
(1+1)^{2000}+w(1+w)^{2000}+w^2(1+w^2)^{2000}\\=\sum_{k=0}^{2000}\Bigg(\binom{2000}{k}+\binom{2000}{k}w^{k+1}+\binom{2000}{k}w^{2k+2}\Bigg)
\\=3 \Bigg(\binom{2000}{2}+\binom{2000}{5}+\cdots+\binom{2000 }{2000}\Bigg),
$$
since
$$
1+w^{k+1}+w^{2k+2}=\left\... | {
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In right triangle $ABC$ ($\angle A=90$), $E$ is a point on $AC$.Find $AE$ given that...
In right triangle $ABC$ ($\angle A=90$), $AD$ is a height and $E$ is a point on $AC$ so that $BE=EC$ and $CD=CE=EB=1$. $\color {red} {Without}$ using trigonometric relations find $AE$.
I do have a solution USING trigonometric re... | Let $AE=x$.
Hence, $AD=\sqrt{(x+1)^2-1}=\sqrt{x^2+2x}$ and since $AD^2=BD\cdot DC$,
we obtain $BD=x^2+2x$, which says that $AB=\sqrt{(x^2+2x)^2+x^2+2x}$ and
$$BE=\sqrt{(x^2+2x)^2+x^2+2x+x^2}.$$
Thus,$$(x^2+2x)^2+x^2+2x+x^2=1$$ or
$$x^4+4x^3+6x^2+2x-1=0$$ or
$$(x+1)(x^3+3x^2+3x-1)=0$$ or
$$(x+1)^3=2$$ or
$$x=\sqrt[3]2-... | {
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Prove, by contradiction, that, if $cx^2 + bx + a$ has no rational root, then $ax^2 + bx + c$ has no rational root. Proposition: Suppose that $a$, $b$, and $c$ are real numbers with $c \not = 0$. Prove, by contradiction, that, if $cx^2 + bx + a$ has no rational root, then $ax^2 + bx + c$ has no rational root.
Hypothesis... | To continue what you started
$ap^2 + bpq + cq^2 = 0$ divide both sides by $p^2$.
$a + b\frac qp + c(\frac {q^2}{p^2}) = 0$
So $\frac pc$ is a rational solution to $cx^2 + bx + a = 0$. A contradiction.
Worth noting: if $w$ is a solution to $ax^2 + bx + c = 0$ ($a \ne 0; c\ne 0$) then $\frac 1w$ is a solution to $cx^2... | {
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Solve an integral $\int\frac{\cos^3 x}{\sin^3 x+\cos^3 x}dx$ Solve an integral $$\int\frac{\cos^3 x}{\sin^3 x+\cos^3 x}dx$$
I tried to divide the numerator and denominator by $\cos^4 x$ to get $\sec x$ function but the term ${\sin^3 x}/{\cos^4 x}$ gives $\tan^2 x\sec^2 x\sin x$. How to get rid of $\sin x$ term?
| I wasn't really able to come up with a better (elegant) method other than the following:
$$\int \frac{\cos^3 x}{\sin^3 x + \cos^3 x} \mathrm{d}x = \int \frac{1}{1 + \tan^3 x} \mathrm{d}x$$
Now, using the substitution, $t = \tan x \implies \frac{\mathrm{d}t}{1+t^2} = \mathrm{d}x$, we get
$$= \int \frac{1}{(1 + t^2)(1+t^... | {
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If a,b and c are sides of a triangle, then prove that the following polynomial has no real roots This is the polynomial: $$a^2x^2+(b^2+a^2-c^2)x+b^2=0$$
Now this is my progress:
Assuming l,m, and n are sides of a triangle, then $$|m-n|\lt l\lt m+n$$
Also, if a second degree polynomial in the form $kx^2+px+q$ has real r... | The following answers the What did I do wrong? part of the question
In this case, if there are no real roots, $$(b^2+a^2-c^2)^2-4a^2b^2\lt 0$$
$$b^2+a^2-c^2\lt 2ab$$
The two inequalities are not equivalent, since $b^2+a^2-c^2$ can be either positive or negative (remember that it is $0$ for right triangles by Pytha... | {
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Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$
Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$.
I expanded the brackets and applied AM-GM on all of the eight terms to get :
$$\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big) \geq 3\... | Set $g(a,b,c)=a+b+c-3$ and $f(a,b,c)=(3+2a^2)(3+2b^2)(3+2c^2)$. We shall use the Lagrange multipliers method.
We have to minimize $h(a,b,c)=f(a,b,c)-\lambda\cdot g(a,b,c)$.
$\frac{d\ h(a,b,c)}{d\ a}=0$
$\frac{d\ h(a,b,c)}{d\ b}=0$
$\frac{d\ h(a,b,c)}{d\ c}=0$
$\frac{d\ h(a,b,c)}{d\ \lambda}=0$
You obtain
$4a(3+2b^2)(3+... | {
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What is the Conjugate of $\frac{1}{4\sqrt{3}-7}$ I'm having problems finding the conjugate of $\frac{1}{4\sqrt{3}-7}$
The answer I get is as follows: ${-4\sqrt{3}-7}$
However the answer given is ${-4\sqrt{3}+7}$
Here are my workings out...
$\frac{1}{4\sqrt{3}-7} * \frac{4\sqrt{3}+7}{4\sqrt{3}+7} = \frac{4\sqrt{3}+7}{-1... | Well, after all the changes:
$$(4\sqrt3-7)(4\sqrt3+7)=48-49=-1$$
and then indeed:
$$\frac1{4\sqrt3-7}=-\left(4\sqrt3+7\right)=-4\sqrt3-7$$
and yes: the conjugate is $\;\frac1{4\sqrt3+7}\;$ , and what you say is the answer given is neither the conjugate nor the product of the original expression by its conjugate.
Added ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $\lim\limits_{x \to 0} \frac{ \sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x }}{x}$
Compute $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x }}{x}$
Original question was to solve $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+ x } - \sqrt[3]{1-x }}{x}$ and it was solved by adding and subtracting 1 i... | If you already know how to find the limit $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+ x } - \sqrt[3]{1-x }}{x}$ and also that $\dfrac{\sin x}x$ tends to 1, then you can simply use
$$
\frac {\sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x}}{x} =
\frac {\sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x}}{\sin x} \cdot \frac{\sin x}x.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Let $\omega$ be a complex number such that $\omega^7 = 1$ and $\omega \neq 1$. Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Then $\alpha$ and $\beta$ are roots of the quadratic
[x^2 + p... | Hint. Observe that
$$
\alpha\beta = \omega^4+\omega^5+\omega^6+3\omega^7+\omega^8+\omega^9+\omega^{10}
= 2+\omega^4(1+\omega+\omega+\cdots+\omega^6)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How does one show that $\sum_{k=1}^{n}\sin\left({\pi\over 2}\cdot{4k-1\over 2n+1}\right)={1\over 2\sin\left({\pi\over 2}\cdot{1\over 2n+1}\right)}?$ Consider
$$\sum_{k=1}^{n}\sin\left({\pi\over 2}\cdot{4k-1\over 2n+1}\right)=S\tag1$$
How does one show that $$S={1\over 2\sin\left({\pi\over 2}\cdot{1\over 2n+1}\right... | Using Lagrange Identities:
$$ \sum_{n=1}^{N}\sin\left(\theta n\right)=+\frac{\cot\left(\theta/2\right)}{2}-\frac{\cos\left(\theta N+\theta/2\right)}{2\sin\left(\theta/2\right)} \quad\&\quad \sum_{n=1}^{N}\cos\left(\theta n\right)=-\frac{1}{2}+\frac{\sin\left(\theta N+\theta/2\right)}{2\sin\left(\theta/2\right)}$$
One... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Other Idea to show an inequality $\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$ $$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$$
I want to prove this by Induction
$$n=1 \checkmark\\
n=k \to \dfrac{1}{\sqrt 1}+\dfrac{1}{... | $$\begin{cases}\dfrac{1}{\sqrt 1}\geq \dfrac{1}{\sqrt n}\\+\dfrac{1}{\sqrt 2}\geq \dfrac{1}{\sqrt n}\\+\dfrac{1}{\sqrt 3}\geq \dfrac{1}{\sqrt n}\\ \vdots\\+\dfrac{1}{\sqrt n}\geq \dfrac{1}{\sqrt n}\end{cases} \\\\$$
sum of left hands is $\underbrace{\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Decomposition of self-adjoint operator Suppose we have some self adjoint operator, given by either a matrix $$\begin{pmatrix}
1 & 2 \\
2 & 1
\end{pmatrix}$$ or a function $f \longmapsto xf$ on $L^2[-1,1]$. Is there a quick way of decomposing these self adjoint operators into the difference of positive operators?
| Compose the operator with the projections on the positive and negative part of the spectrum and take the difference. For your matrix $A$, you need to find the eigenvalues which are the roots of
$$ \lambda^2 - \operatorname{tr}(A)\lambda + \det(A) = \lambda^2 - 2\lambda -3 = (\lambda + 1)(\lambda - 3).$$
An eigenvector ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Use infinite series to prove Use infinite series to prove that
$$\arcsin{x}\lt \frac{x}{1-x^2},$$ for $0\lt x\lt1$.
| Let $f(x)=\frac{x}{1-x^2}-\arcsin{x}$.
Hence, $$f'(x)=\frac{1-x^2-x(-2x)}{(1-x^2)^2}-\frac{1}{\sqrt{1-x^2}}=\frac{1+x^2-\sqrt{(1-x^2)^3}}{(1-x^2)^2}=$$
$$=\frac{ x^2(x^4-2x^2+5)}{(1+x^2-\sqrt{(1-x^2)^3})(1-x^2)^2}>0.$$
Thus, $f(x)>f(0)=0$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2152374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Showing that $\sum_{n=1}^{\infty}\sum_{m=1}^{x-1} \frac{m}{n^2 x^2 - m^2}=\frac{1}{2}H_{x-1}$ Consider $(1)$, $H_n$ is the nth-harmonic number
$$\sum_{n=1}^{\infty}\left({1\over (nx)^2-1}+{2\over (nx)^2-2^2}+{3\over (nx)^2-3^2}+\cdots+{x-1\over (nx)^2-(x-1)^{2}}\right)=S\tag1$$
$x\ge2$
How does one show that $$\col... | Well, you have made a small calculation mistake. We have, $$\frac {1}{n^2x^2-1} = \frac {1}{(nx-1)(nx+1)} = \frac {1}{2}[\frac {1}{nx-1}-\frac {1}{nx+1}] $$ $$\frac {2}{n^2x^2-4} = \frac {2}{(nx-2)(nx+2)} = \frac {1}{2}[\frac {1}{nx-2}-\frac {1}{nx+2}] $$ $$\frac {3}{n^2x^2-9} = \frac {3}{(nx-3)(nx+3)} = \frac {1}{2}[\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2152635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.