Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Determine the number of ordered triple $(x, y, z)$ of integer numbers (negatives and positives) satisfying $|x| + |y| + |z| \le 6$ Determine the number of ordered triple $(x, y, z)$ of integer numbers (negatives and positives) satisfying $|x| + |y| + |z| \le 6$
I know that final answer is 377, but how?
Edit:
Drawing f... | Let $a_{n,m}$ be the number of $n$-tuples of integers so that
$$
\sum_{k=1}^n|x_k|\le m\tag1
$$
Pretty simply, we have
$$
a_{1,m}=2\binom{m}{1}+\binom{m}{0}\tag2
$$
Furthermore, we have the recurrence
$$
\begin{align}
a_{n+1,m}
&=\overbrace{\quad a_{n,m}\quad\vphantom{\sum_1^m}}^\text{$0$ in position $n+1$}+\overbrace{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Prove: if $c^2+8 \equiv 0$ mod $p$ then $c^3-7c^2-8c$ is a quadratic residue mod $p$. I want to show:
If $c^2+8 \equiv 0$ mod $p$ for prime $p>3$, then $c^3-7c^2-8c$ is a quadratic residue mod $p$.
I have calculated that $c^3-7c^2-8c \equiv -7c^2-16c \equiv 56- 16c \equiv 8(7-2c) \equiv c^2 (2c -7)$, so it should be ... | We have that
$$(c +1)^2 = c^2 + 2c + 1 = c^2 + 2c + 8 -7 = (c^2 + 8) + (2c - 7) \equiv 2c - 7 \mod p.$$
This proves the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find limit supremum from 3 sequence theorem Find limit $$\limsup_{n\rightarrow \infty} \sqrt[n]{\left| \left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right) \right)^p\right |} $$
dependency on value of $p$
I think that $p$ doesn't matter there (by doesn't matter I mean that I can compute limit... | Hint
Compose Taylor series for large values of $n$ to get
$$2-2 \cos\left(\frac{1}{n}\right)-\frac 1n{\sin \left(\sin \left(\frac{1}{n}\right)\right)}=\frac{1}{4 n^4}-\frac{7}{72 n^6}+O\left(\frac{1}{n^8}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding place of the nine digits The nine digits 1, 2, 3, ... .., 9 are placed in the nine triangles of the attached figure in such a way that the digits around each circle add up as indicated. Calculate the value of N.
| The total of the digits from $1$ to $9$ is $45$. Four digits on the left side of the diagram add to $16$; four on the right add to $25$. That leaves $N = 45 - 16 - 25 = 4$.
EDIT: Let's find the rest.
With $4$ assigned, the additional four digits in the total $32$ must add up to $28$. The only possibility is $5,6,8,9$... | {
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"url": "https://math.stackexchange.com/questions/3163261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the angle in an isosceles triangle Let triangle $\Delta ABC$ have $AB=AC$. Then we draw the angle bisector from $B$ to $AC$ intersecting at $D$. Find the angle $\angle BAC$ if $BC=AD+BD$.
My attempts:
I know that the answer is 100° but I couldn't prove that if you extended $AD$ to a point $E$ so it is equal to $BC... | In the standard notation we obtain:
$$\frac{AD}{DC}=\frac{AB}{BC}=\frac{c}{a}$$ and $$AD+DC=AC=b,$$ which gives
$$AD=\frac{bc}{a+c},$$ $$DC=\frac{ab}{a+c}$$ and
$$BD^2=AB\cdot BC-AD\cdot DC=ac-\frac{b^2ac}{(a+c)^2}=ab-\frac{ab^3}{(a+b)^2},$$
which gives $$BD=\frac{a\sqrt{b(a+2b)}}{a+b}.$$
Id est, by the given we obtain... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3165650",
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"source": "stackexchange",
"question_score": "3",
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Solution of this Diophantine Equation
If $x$ and $y$ are prime numbers which satisfy $x^2-2y^2=1$, solve for $x$ and $y$.
My attempt:
$x^2-2y^2=1$
$\implies (x+\sqrt{2}y)(x-\sqrt{2}y)=1$
$\implies (x+\sqrt{2}y)=1$ and $(x-\sqrt{2}y)=1$
$\implies x=1$ and $y=0$
Clearly $x$ and $y$ are not prime numbers . Why is my so... | What about
\begin{align*}&x^2-2y^2=1\tag{1}\\\iff & x^2-1=(x+1)(x-1)=2y^2\end{align*}
Since $2\mid (x+1)(x-1)$, we conclude that both $(x+1)$ and $(x-1)$ have to be even, and hence $$4\mid 2y^2\implies 2\mid y^2\implies 2\mid y$$ and since $y$ is prime, $\color{red}{y=2}$. Can you end it now?
From (1), it follows im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3166201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Prove independence of two random variables. Let two random variables $\xi_1$ and $\xi_2$ be given. They are independent and have a standard normal distribution. Proove that $\frac{\xi_1^2 - \xi_2^2}{\sqrt{\xi_1^2+\xi_2^2}}$ and $\frac{2\xi_1\xi_2}{\sqrt{\xi_1^2+\xi_2^2}}$ are independent.
I was given a little hint that... | according to
Given that $X,Y$ are independent $N(0,1)$ , show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ are independent $N(0,\frac{1}{4})$
they are independent.
but if you want to continue your way:
$X=r\cos(\theta) \hspace{.5cm}
Y=r\sin(\theta) \hspace{.5cm}
X,Y \sim normal(0,1) \Leftrightarro... | {
"language": "en",
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Proving difference between two functions is small when $x$ is small... I'm really struggling to prove the following claim and I was wondering whether anyone could help me.
Claim:$$ 0<|x| \leq 10^{-4} \implies \bigg{|} \frac{x}{e^x-1} - \frac{1}{\sum_{k=1}^3 \frac{x^{k-1}}{k!}} \bigg{|} \leq 10^{-12}$$
My attempt at a... | Write $|\frac{x}{e^x-1} - \frac{x}{\sum_{k=1}^3 \frac{x^k}{k!}}|
= |x||\frac{\sum_{k=1}^3 \frac{x^k}{k!} - (e^x-1)}{(e^x-1)\sum_{k=1}^3 \frac{x^k}{k!}}| $
You can use the $|x|<10^-4$ bound given, and also instead an easier inequality for the numerator using the series sum for $e^x$. Then it is clear that the denomina... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3167693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$ What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$
They all are positive terms so arithmetic mean is greater than equal to geometric mean.
$$ \sec^6 x +\csc^6 x + \sec^6... | You want to find the least value of $f(x)=\sec^6(x)+\csc^6(x)+\sec^6(x)\csc^6(x)$. You found that $f(x) \geq g(x)=3(\sec(x)\csc(x))^4$. In addition, the minimum value of $g(x)$ is $48$. Therefore, you can conclude that $f(x) \geq 48$ for all $x$. But why would you expect there to exist some $x$ such that $f(x)=48$, whe... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all values of $c$ for the Mean Value Theorem's $f'(c)$ of the function $f(x) = 2x^3-6x^2-90x+6$ of the range $[-5, 8]$ I need to find all values of $c$ for the $f'(c)$ of the function $f(x) = 2x^3-6x^2-90x+6$ of the range $[-5, 8]$ such that $f'(c) = -10$. I have already found that $f'(c) = \frac{-130}{13}=-10$.
I... | You got the wrong answer because you set $f'(c)=\frac{f'(b)-f'(a)}{b-a}$ instead of $\frac{f(b)-f(a)}{b-a}.$
Otherwise your approach was correct.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Any hints on how to compute this integral? Could anyone please give me a hint on how to compute the following integral?
$$\int \sqrt{\frac{x-2}{x^7}} \, \mathrm d x$$
I'm not required to use hyperbolic/ inverse trigonometric functions.
| Write $y(x):=\sqrt{\frac {x-2} {x^7}}$.
Note that $$y'(x)= \frac {7-3x}{x^8} \frac 1 y $$
Hence $$\frac d {dx} x^n y=n x^{n-1} y + x^{n-8} \frac {7-3x} y.$$
Do the ansatz $$F(x)=\sum_{n=0}^k a_nx^ny ~~~~\text{ and } ~~~~F'(x)=y(x). $$
We get
$$y\sum_{n=1}^{k}a_n nx^{n-1} +\frac 1 y\sum_{n=0}^{k} a_n x^{n-8} ({7-3x})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170871",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Why does $\sin(x) - \sin(y)=2 \cos(\frac{x+y}{2}) \sin(\frac{x-y}{2})$? Why does this equality hold?
$\sin x - \sin y = 2 \cos(\frac{x+y}{2}) \sin(\frac{x-y}{2})$.
My professor was saying that since
(i) $\sin(A+B)=\sin A \cos B+ \sin B \cos A$
and
(ii) $\sin(A-B) = \sin A \cos B - \sin B \cos A$
we just let $A=\frac... | Following your notation, let $A=\dfrac{x+y}{2}$ and $B=\dfrac{x-y}{2}$.
Note that $A+B=x$ and $A-B=y$.
Now, $\sin x=\sin(A+B)=\sin A\cos B+\cos A\sin B$ and $\sin y=\sin(A-B)=\sin A\cos B - \cos A\sin B$ from your professor's advice.
To get the LHS, $\sin x-\sin y = 2\cos A\sin B$. And that's it. Replace $A,B$ in terms... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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What are the integer coeffcients of a cubic polynomial having two particular properties? Let $f(x) = x^3 + a x^2 + b x + c$ and $g(x) = x^3 + b x^2 + c x + a\,$ where $a, b, c$ are integers and $c\neq 0\,$. Suppose that the following conditions hold:
*
*$f(1)=0$
*The roots of $g(x)$ are squares of the roots of $f... | Let $u$, $v$ and $w$ be the roots of $f$, so that $u^2$, $v^2$ and $w^2$ are the roots of $g$. Then comparing the coefficients of
$$(x-u)(x-v)(x-w)=f(x)=x^3+ax^2+bx+c,$$
$$(x-u^2)(x-v^2)(x-w^2)=g(x)=x^3+bx^2+cx+a,$$
yields the equations
\begin{eqnarray*}
a&=&-u-v-w&=&-u^2v^2w^2,\\
b&=&uv+uw+vw&=&-u^2-v^2-w^2,\\
c&=&-uv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3173125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Solving $\tan (2x) = 6\cos^2(x) - 4\sin(x)\cos(x) - 2\sin^2(x)$ I need to solve the following trigonometric equation:
$$\tan (2x) = 6\cos ^2(x) - 4\sin (x)\cos (x) - 2\sin ^2(x)$$
My attempt:
$$\frac{\sin(2x)}{\cos(2x)} = 3(\cos(2x)+1) - 2\sin(2x) -2(1-\cos^2(x))$$
$$\frac{\sin(2x)}{\cos(2x)} = 3\cos(2x)+3 - 2sin(2x) ... | You find correctly that
$$
\tan2x=4\cos2x-2\sin2x+2
$$
Multiplying by $\cos2x$ leads to
$$
4\cos^22x-2\sin2x\cos2x+2\cos2x-\sin2x=0
$$
that can be rewritten as
$$
2\cos2x(2\cos2x-\sin2x)+(2\cos2x-\sin2x)=0
$$
so
$$
(2\cos2x+1)(2\cos2x-\sin2x)=0
$$
This splits into
$$
\cos2x=-\frac{1}{2}\qquad\text{or}\qquad \sin2x=2\co... | {
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"url": "https://math.stackexchange.com/questions/3176171",
"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left(e^{(1+\frac{k}{n})^2} - \frac{3e^{(1 + \frac{3k}{n})}}{2\sqrt{1 + \frac{3k}{n}}}\right).$ Find
$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left(e^{(1+\frac{k}{n})^2} - \frac{3e^{(1 + \frac{3k}{n})}}{2\sqrt{1 + \frac{3k}{n}}}\right)$$
Choices:
| Let $u=\sqrt{1+3x}$, then $\frac{du}{dx}=\frac{3}{2u}$
\begin{align}
\frac32 \int_0^1 \frac{\exp(1+3x)}{\sqrt{1+3x}} \, dx&= \frac32 \int_1^2\frac{\exp(u^2)}{u}\cdot \frac{2u}{3}\, du = \int_1^2 \exp(u^2)\,du
\end{align}
Also, let $v=1+x$,
$$\int_0^1 \exp((1+x)^2) \, dx = \int_1^2 \exp(v^2) \, dv$$
Hence, the two integ... | {
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"url": "https://math.stackexchange.com/questions/3176671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Calculate the limit of the following recurrent series Find the limit of the series : $\ q_{n+1,}=q_n + \frac{2}{q_n}$ with $q_0=1,n\ge 0$.
| I will show that
$q_n$ is unbounded
and that
$\frac12+\sqrt{2n-\frac74}
\lt q_n
<3+2\sqrt{ 2 n+\frac14}
$.
Moreover,
if,
as these inequalities make plausible,
$\lim_{n \to \infty} \dfrac{q_n}{\sqrt{n}}
$
exists and this limit is $c$,
then $c = 2$,
so that
$\lim_{n \to \infty} \dfrac{q_n}{\sqrt{n}}
=2
$.
Here we go.
$q_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3179916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 2
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Distributing $60$ identical balls into $4$ boxes if each box gets at least $4$ balls, but no box gets $20$ or more balls How many different ways can the balls be placed if each box gets at least $4$ balls each, but no box gets $20$ or more balls?
I was thinking about finding all the possible ways which every box gets a... | What you have done thus far is correct.
As you observed, after distributing four balls to each box, we are left with $60 - 4 \cdot 4 = 44$ balls to distribute. If we let $x_i, 1 \leq i \leq 4$, represent the number of additional balls we distribute to the $i$th box, then
$$x_1 + x_2 + x_3 + x_4 = 44 \tag{1}$$
Equation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Laurent series of $\sin z/(1 - \cos z)$ I have trouble solving this exercise: find the first three terms of the Laurent series of $\sin z/(1 - \cos z)$ centered at $z=0$.
I have found the first two. I proved that at $z=0$ we have a first order pole and the first one I calculated the residue. I also thought that the se... | You know that laurent series of $sin(z) = z - \frac{z^3}{6} + \frac{z^5}{120} + O (z^{7})$.
Then, the laurent series of $ 1-cos(z)= \frac{z^2}{2}-\frac{z^4}{24}+O(z^{6})$
Overall you have $\frac{z - \frac{z^3}{6} + \frac{z^5}{120} + O (z^{7})}{\frac{z^2}{2}-\frac{z^4}{24}+O(z^{6})}$.
Now look at the denominator: $(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving the independence of an equation, from certain variables
Assume that $x, y, z$ are three distinct nonzero real numbers satisfying the equation
$$x^3+y^3+m(x+y)=y^3+z^3+m(y+z)=z^3+x^3+m(z+x)$$
for some real $m$. Prove that
$$K=\left(\frac{x-y}{z}+\frac{y-z}{x}+\frac{z-x}{y}\right)\left(\frac{z}{x-y}+\frac{x}{y-z... | $$
x^3 + y^3 + m(x+y) = y^3 + z^3 + m(y+z) = z^3 + x^3 + m(z+x)
$$
holds for some $m \in \mathbb R$, if and only if
$$
\begin{cases}
x^3 - y^3 = -m(x-y) \\
y^3 - z^3 = -m(y-z)
\end{cases}
$$
holds for some $m \in \mathbb R$. If we presume that $x,y,z$ are pairwise distinct, this is equivalent to saying that, considerin... | {
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Find the partial sum formula of $\sum_{i=1}^n \frac{x^{2^{i-1}}}{1-x^{2^i}}$ Given next series:
$$\frac{x}{1 - x^2} + \frac{x^2}{1 - x^4} + \frac{x^4}{1 - x^8} + \frac{x^8}{1 - x^{16}} + \frac{x^{16}}{1 - x^{32}} + ... $$
and $|x| < 1$. Need to derive $S_n$ formula from series partial sums.
I could only find that $... | By induction you can proof easily $$\sum\limits_{k=1}^n\frac{x^{2^{k-1}}}{1-x^{2^k}} = \frac{1}{1-x^{2^n}}\sum\limits_{k=1}^{2^n-1}x^k$$ and with $\enspace\displaystyle \sum\limits_{k=1}^{2^n-1}x^k = \frac{x-x^{2^n}}{1-x}\enspace$ the formula is complete.
| {
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"url": "https://math.stackexchange.com/questions/3188448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\sin (\sin x -x) =\sin x - x + o(x^5)$
Prove that $\sin (\sin x -x) =\sin x - x + o(x^5)$
In the task which I do I need record $\sin (\sin x-x)$ in a way to have $ax^3$. So: $$\sin (\sin x -x)=\sin x -x +r(\sin x -x)=x-\frac{x^5}{6}+\frac{x^5}{120}+r(x)-x +r(\sin x -x)$$ I know that $r(x)=o(x^5)$ and it i... | I would be stupid about it and work formally, in the ring $\Bbb Q[[x]]/(x^6)$.
Now, in that ring, $\sin(x)=x-\frac{x^3}6+\frac{x^5}{120}$. Let me write $g(x)=\sin(x)-x$, which is divisible by $x^3$. Then $\sin(g(x))=g(x)+g(g(x))$, but since $g(g(x))$ is divisible by $x^9$, we can ignore it. Thus in $\Bbb Q[[x]]/(x^6)$,... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Hard combinatorics question The origin of the coordinates is a pixel. After 1 second, it splits into two particles, one
shifts to the left and the other to the right. This process is repeated every second, and the two particles at the same point are mutually destroyed (so that, for example,
two seconds later, two parti... | This is called Gould's sequence, or Dress' sequence, or Glaisher's sequence; it appears in the OEIS. The $n$-th term is the number of odd entries entries in row $n$ of Pascal's triangle. The specific operation in OP is Rule 90 for cellular automata, and we are counting how many $1$'s there are in every stage. Much m... | {
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"url": "https://math.stackexchange.com/questions/3191969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding rank-$1$ matrix
Let $$S = \frac{1}{12} \begin{pmatrix} 1 & 10 & 1 \\ 5 & 2 & 5\\ 1 & 2 & 9\end{pmatrix}$$ Find a rank-$1$ matrix $R$ so that $$ M = S + R $$ will have the same eigenvalues as $S$ and all the diagonal elements equal.
I have found the eigenvalues of $S$ to be: $$\{ 1, \frac{\sqrt{2}}{3},\frac{-\... | I'm assuming you meant the three diagonal elements of $S+R$ are equal.
Let $R = u v^\top$. You have $5$ equations to solve for $6$ variables $u_1,u_2,u_3,v_1,v_2,v_3$ (of course there is redundancy here): the coefficients of $\lambda^0$ to $\lambda^2$ in $\det(S+R-\lambda I) - \det(S-\lambda I)$ are $0$, $S_{11} + R_{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find Taylor Polynomial order 5 of $f(x) = \frac{1}{(1-x)}$, at $a = 0$ Find Taylor Polynomial order 5 of $f(x) = \frac{1}{(1-x)}$ , at $a = 0$
So I start of with:
$f(0) = \frac{1}{(1-0)}=1$
$f'(0) = \frac{1}{(1-0)^2 }= 1$
$f''(0) = \frac{2}{(1-0)^3 }= 2$
$f'''(0) = \frac{6}{(1-0)^4 }= 6$
$f^{(4)}(0) = \frac{24}{(1-0)^... | i see a problem. The taylor series in this case is given by :
$$f(x)=f(a)(x-a)+f'(a)(x-a)+f''(a)\frac{(x-a)^2}{2!}+f'''(a)\frac{(x-a)^3}{3!}+...$$
I think you forgot the $!$...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the correct way to solve the equation: $x^4-x^3+x^2-x+1=0$ Given the equation: $x^4-x^3+x^2-x+1=0$ we need to find both its real and complex roots. What is the easiest and correct method for solving the equation?
Here is my approach, but it gives wrong result on the end. Since the equation is symmetric we can g... | The solutions in your textbook are wrong; you can plug them in to verify this yourself.
The easiest way to solve is to note that if $x\neq-1$ then
$$\frac{x^5+1}{x+1}=x^4-x^3+x^2-x+1,$$
and so writing $x=re^{\theta i}$ quickly yields $r=1$ and $\theta=\tfrac k5\pi$ with $k$ odd.
Your approach is also fine; you get the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3200151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solving $3x^2 - 4x -2 = 0$ by completing the square I can't understand the solution from the textbook (Stroud & Booth's "Engineering Mathematics" on a problem that involves solving a quadratic equation by completing the square.
The equation is this:
$$
\begin{align}
3x^2 - 4x -2 = 0 \\
3x^2 - 4x = 2
\end{align}
$$
Now,... | Note that it is better to multiply by $3$, thus transforming the given equation as follows:
$3x^2 - 4x - 2 = 0$
$9x^2 - 12x - 6 = 0$
$(3x-2)^2 = 10$
To get the last line, you want $(3x+?)^2$ to match the $9x^2$ so you twiddle the $?$ until you match the $9x^2-12x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sets of Polynomial roots For a real number $k,$ let $A$ be the set of roots of $$x^2 + (k - 1) x - 2(k + 1) = 0,$$and let $B$ be the set of roots of $$(k - 1) x^2 + kx + 1 = 0.$$(These roots may be complex.) Find the number of values of $k$ so that $|A \cup B| = 3.$
I don't know how to start this. Can someone help?
| The roots of the two equations are
$$\frac{1}{2}\left(-(k-1) \pm \sqrt{(k-1)^2 + 8 (k+1)}\right) = \frac{1}{2}(-(k-1) \pm (k+3)) = \{2, -(k+1)\}$$
$$\frac{1}{2(k-1)} \left(-k \pm \sqrt{k^2 - 4(k-1)}\right) = \frac{1}{2(k-1)}(-k \pm (k-2)) = \{-\frac{1}{k-1}, -1\}$$
so you want
$$\{2, -(k+1), -\frac{1}{k-1}, -1\}$$
to h... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $\int \frac{6x^{3}+7x^2-12x+1}{\sqrt{x^2+4x+6}}dx$ I have previously asked here the following integral:$$\int \frac{6x^{3}+7x^2-12x+1}{\sqrt{x^2+4x+6}}dx.$$
In the meantime I have received an answer from Achille Hui here, but
there are still some points that I don't understand from his answer.
At some point th... | I have used a slightly different approach using a substitution. This makes the integral into a routine computational effort.
$$I = \int \frac{6x^{3}+7x^2-12x+1}{\sqrt{x^2+4x+6}}dx$$
$$ = \int \frac{6x^{3}+7x^2-12x+1}{\sqrt{(x+2)^2+2}}dx$$
Let $x + 2 = \sqrt2\tan\theta \implies dx = \sqrt2\sec^2\theta\ d\theta\text{ and... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$ if $x+y=12$ If $x,y\in\mathbb R^+$, $x+y=12$, what is the minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$?
I got the question from a mathematical olympiad competition (of China) for secondary 2 student, so I don't expect an "analysis" answer.
The answer should be $15$, when... | We are asked for the minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$ if $x+y=12;$
that is, the minimum value of $\sqrt{x^2+25}+\sqrt{(12-x)^2+16}.$
Let $\overrightarrow a=(x,5)$ and $\overrightarrow b=(12-x,4)$ in $\mathbb R^2,$ so $\overrightarrow a + \overrightarrow b=(12,9).$
By the triangle inequality, $|\overrighta... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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$a, b$ are positive whole numbers such that $a + b = a/b + b/a$. How many possible values can $a^4 + b^4$ be? $a, b$ are positive whole numbers such that $a + b = a/b + b/a$. How many possible values can $a^4 + b^4$ be?
I tried using $(a^2 + b^2)^2$ but I don't know what to do after.
| $a+b=\frac ab+\frac ba\iff a+b=\frac{a^2+b^2}{ab}\implies a^2b+ab^2=a^2+b^2\implies a^2\underbrace{(b-1)}_{\ge 0}+b^2\underbrace{(a-1)}_{\ge 0}=0$
Since the equation has $a,b$ both on denominator, we can reject zero as an admissible value for $a$ or $b$.
The only possibility for the sum above to be $0$ is that each ter... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplifying $\sec^2 \frac{2\pi}{7} + \sec^2 \frac{4\pi}{7} + \sec^2 \frac{8\pi}{7}$
Simplify the following expression:
$$y =\sec^2 \frac{2\pi}{7} + \sec^2 \frac{4\pi}{7} + \sec^2 \frac{8\pi}{7}$$
Note. The source asks the value of $y/3$, which, according to the instructions, has to be an integer from $0$ to $9$.
M... | Like If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha.$,
the roots of
$$t^3-21t^2+35t-7=0$$ are $\tan^2\dfrac{2n\pi}7, n=1,2,4$
Using Vieta's Formula,
$$\tan^2\dfrac{2\pi}7+\tan^2\dfrac{4\pi}7+\tan^2\dfrac{8\pi}7=\dfrac{21}1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Problem with $\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}}$ How to simplify $$\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}}?$$
Rationalise the denominator
$$\frac{\sqrt{6+4\sqrt{2}}}{4}(2-\sqrt{2})$$
This is still not simplify.
| $\sqrt{6+4\sqrt{2}} = \sqrt{(2 + \sqrt{2})^{2}} = {2+\sqrt{2}}$ ,
as $6+4\sqrt{2}= 4 +2 +2.2\sqrt{2} = (\sqrt{2})^{2} + 2 .2. \sqrt{2} + 2^ 2 = (2 + \sqrt{2})^{2}$
Now $\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}} = \frac{2+\sqrt{2}}{2(2+\sqrt{2})}= \frac{1}{2}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I evaluate this trigonometric limit involving infinite product? Tried a lot but couldn't solve it. How do I begin approaching this limit?
$ \displaystyle\lim _{n\to\infty} \displaystyle\prod_{k=3}^{n} 1 - \tan^4\left( \frac{\pi}{2^k}\right)$
| There isn't that many trick to create a horrible looking and yet doable limit.
In school, if you are asked to evaluate a series or product which involve trigonometry functions whose arguments contain terms like $\frac{(\cdots)}{2^k}$. One thing you should do is lookup the double-angle formula for various trigonometry... | {
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"timestamp": "2023-03-29T00:00:00",
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Calculating $\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$ Calculate $$\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$$
Here is my attempt:
$$\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}= \left(\frac{4\infty^2+5\infty-6}{4\infty^2+3\infty-10}\right)^{3-4\i... | Let us suppose that you want to compute accurately the value of the expression for large values of $n$ and not only the limit.
$$a_n=\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}=\left(1+\frac{2}{4 n-5}\right)^{3-4 n}$$ Take logarithms
$$\log(a_n)=(3-4n) \log\left(1+\frac{2}{4 n-5}\right)$$ and use the Taylor expan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3222029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find antiderivative $\int (2x^3+x)(\arctan x)^2dx $
Find antiderivative $$\int (2x^3+x)(\arctan x)^2dx $$
My try:
$$\int (2x^3+x)(\arctan x)^2dx =(\arctan x)^2(\frac{1}{2}x^4+\frac{1}{2}x^2)-\int \frac{2\arctan x}{1+x^2}(\frac{1}{2}x^4+\frac{1}{2}x^2)dx=(\arctan x)^2(\frac{1}{2}x^4+\frac{1}{2}x^2)-\int (\arctan x) (x... | Note that$$\frac{x^3}{1+x^2}=\frac{x^3+x}{1+x^2}-\frac x{1+x^2}=x-\frac x{1+x^2}.$$Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Partial fraction expansion of $\frac{1}{(s+1)^{2}(s-1)(s+5)}$ I'm seeking a partial-fraction expansion of $A=\frac{1}{(s+1)^{2}(s-1)(s+5)}.$
I was solve equation differential using Laplace transform, but I need use partial fraction of $A$.
| The partial fraction expansion is:
$$\frac{1}{(s+1)^2(s-1)(s+5)} = \frac{1}{24(s-1)} -\frac{1}{32(s+1)} - \frac{1}{8(s+1)^2} - \frac{1}{96(s+5)} $$
The fraction expansion can be expressed as:
$$\frac{1}{(s+1)^2(s-1)(s+5)} = \frac{A}{s-1} + \frac{B}{s+1} + \frac{C}{(s+1)^2} + \frac{D}{s+5}$$
$$ \implies A(s+1)^2(s+5) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3226373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Expectation of dependent Bernoulli sum I want to estimate the expected value of the following sum of random variables,
$$ Y = \sum_{i=1}^N X_i $$
where each $X_i$ is a Bernoulli random variable. In particular,
$$ X_1 = \begin{cases} 1, & \text{with prob. } 1/N \\
0, & \text{with prob. } 1 - 1/N \end{cases} $$
and fo... | MY TRY
We first look at $ P\{ Y=i\}$, i.e., $Y$ has exactly $i$ ones and $N-i$ zeros. The probability corresponding to ones is $N^{-i}$. The probability corresponding to zeros depends on number and size of blocks of zeros. There can be at most $i+1$ blocks of size between 0 and $N-i$, as we already have $i$ ones. Furth... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3228206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why is $\mathbf{x}$ yielding a product of $\mathbf{y}$ and the right inverse for $\mathbf{A}$? When you solve the equation $\mathbf{Ax}=\mathbf{y}$ where $\mathbf{x} \in \mathbb{R}^{5}$, $\mathbf{y} \in \mathbb{R}^{4}$ and $\mathbf{A}$ is a $4 \times 5$ matrix with a right inverse, I understand that the $\mathbf{x}$ is... | Claim. Let $V,W$ be vector spaces, $A\colon V\to W$ and $B\colon W\to V$ be linear maps with $AB=1_W$. Also suppose that $Ax=y$ for certain $x\in V$, $y\in W$ with $y\ne 0$.
Then there exists a linear map $B'\colon W\to V$ such that $AB'=1_W$ and $x=B'y$.
Proof: It suffices to find $C\colon W\to V$ such that $AC=0$ and... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding value of $K$ when applying the error bound The problem is asking to use the error bound to find a value of $n$ for which the given inequality is satisfied.
$$\left|\sqrt{1.3}-T_{n}(1.3) \right| \leq 10^{-6}, \quad a=1 $$
Now this is how I started my attempt to solve this, we use the error bound, where,
$$\left... | It seems like you're missing where the $K$ comes from in general. One form of Taylor's theorem is that, for all $x$ there is some $z$ between $a$ and $x$ such that
$$
f(x) - T_n(x) = \frac{f^{(n+1)}(z)}{(n+1)!}(x-a)^{n+1}
$$
Therefore, if you can find $K$ such that
$$
|f^{(n+1)}(z)| \leq K
$$
for all $z$ in th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $\sin(18^\circ)=\frac{a + \sqrt{b}}{c}$, then what is $a+b+c$? If $\sin(18)=\frac{a + \sqrt{b}}{c}$ in the simplest form, then what is $a+b+c$?
$$ $$
Attempt:
$\sin(18)$ in a right triangle with sides $x$ (in front of corner with angle $18$ degrees), $y$, and hypotenuse $z$, is actually just $\frac{x}{z}$, then $x =... | Let $ABCDE$ be a regular pentagon.
In right triangle $GHC$ we see $\sin 18^{\circ} = {y\over 2x} $. Let $k=y/x$.
From triangle similarity of $GFC$ and $ACE$ we have $${x\over y} = {2x+y\over a}$$
and from triangle similarity of $BFC$ and $EDC$ we have $${a\over x} = {2x+y\over a}$$
Eliminate $a = {x^2\over y}$ and we ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Kronecker Product Interpretation The algebraic expression for a Kronecker product is simple enough. Is there some way to understand what this product is?
The expression for matrix-vector multiplication is easy enough to understand. But realizing that the multiplication yields a linear combination of the columns of th... | A nice motivating example is provided in Kronecker Products & Matrix Calculus with Applications by A. Graham.
We consider two linear transformations
\begin{align*}
\boldsymbol{x}=\boldsymbol{A}\boldsymbol{z}\qquad\text{and}\qquad \boldsymbol{y}=\boldsymbol{B}\boldsymbol{w}
\end{align*}
and the simple case
\begin{align*... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Why is this integration method not valid? Let $$I=\int \frac{\sin x}{\cos x + \sin x}\ dx \tag{1}$$
Now let $$u=\frac{\pi}{2} - x \tag{2}$$ so $$I=\int \frac{\sin (\frac{\pi}{2} - u)}{\cos (\frac{\pi}{2} - u)+\sin (\frac{\pi}{2} - u)}\ du \tag{3}$$
$$=\int\frac{-\cos u}{\sin u + \cos u} \ du \tag{4}$$
$$= \int\frac{-\c... | Let's ignore the missing minus sign in line (3) (since $\mathrm{d}u = - \mathrm{d}x$). Instead let's inspect the method.
The short version: Once you establish a relation between $x$ and $u$, you have established it. You do not get to change that relation without implicitly reducing the domain of validity to the sim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3232250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
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real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Number of real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Plan
Put $x^2+6x+7=f(x)$. Then i have $f(f(x))=x$
For $f(x)=x$
$x^2+5x+7=0$ no real value of $x$
For $f(x)=-x$
$x^2+8x+7=0$
$x=-7,x=-1$
Solution given is all real solution
Help me pleas... | You are right that $f(f(x))=x$
This gives $f(x)=f^{-1}(x)$
Do you know anything about the relationship between a function and its inverse?
Try sketching both $y=f(x)$ and $y=f^{-1}(x)$ on the same axes...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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Evaluate $ \int \frac{x^4}{(2-x^2)^{3/2}}dx$
Evaluate the following integral:
$ \int \frac{x^4}{(2-x^2)^{3/2}}dx$
I've tried to apply Chebyshev theorem on the integration of binomial differentials.
We have $ m=4,a=2,b=-1,n=2,p=-3/2$.
$\frac{m+1}{n}+p$ is integer then we do the substitution
$t^2=2x^{-2}-1$, $x^2=\f... | Hint:
Another way:
Use Trigonometric substitution
As $2-x^2\ge0,$ WLOG $x=\sqrt2\sin t$
$\sqrt{2-x^2}=\sqrt2\cos t, dx=?$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why does $ \lim_{x \to 2} \frac{x^2-4}{x-2} =4 $ if x cannot be 2? I know that $ \lim_{x \to 2} \frac{x^2-4}{x-2} $ is evaluated as follows :-
$$ \lim_{x \to 2} \frac{x^2-4}{x-2} \\ = \lim_{x \to 2} \frac{(x+2)(x-2)}{x-2} \\ = \lim_{x \to 2} x+2 \\ = 2+2 \\ = 4 $$
By looking at the function $ \frac{x^2 - 4}{x - 2} $, I... | It simply means that $f(x)$ approaches $4$ from both sides as $x$ approaches $2$. The function does not need to be defined at that point, although it could. The function $f(x) = \frac{x^2-4}{x-2}$ is exactly the "same" as the function $f(x) = x+2$ at all points except $x = 2$, where there is a removable discontinuity, ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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Calculate $\int_{-\infty}^\infty{x^2\,dx\over (1+x^2)^2}$ The question:\,
Calculate $$\int_{-\infty}^\infty{x^2\,dx\over (1+x^2)^2}.$$
Book's final solution: $\dfrac\pi 2$.
My mistaken solution: I don't see where is my mistake because my final solution is $-\dfrac{\pi i}2$:
$$\begin{align}
\text{A}:\int_{-\infty}^\i... | That $\ln z$ makes no sense. That integral is equal to $2\pi i$ times the sum of the residues at the singularities of $\frac{z^2}{(1+z^2)^2}$ in the upper half-plane. There is actually only one such singularity: at $i$. So\begin{align}\int_{-\infty}^\infty\frac{x^2}{(1+x^2)^2}\,\mathrm dx&=2\pi i\operatorname{res}_{z=i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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positive root of the equation $x^2+x-3-\sqrt{3}=0$ $x^2+x-3-\sqrt{3}=0$
using the quadratic formula we get
$$x=\frac{-1+\sqrt{13+4\sqrt{3}}}{2}$$ for the positive root
but the actual answer is simply $x=\sqrt3$
I am unable to perform the simplification any help would we helpful
| Rather than pulling $\sqrt{13+4\sqrt3}=1+2\sqrt3$ magically out of a hat, write $\sqrt{13+4\sqrt3}=a+b\sqrt3,$
as suggested by Parcly Taxel in a comment. Squaring both sides, $13+4\sqrt3=(a^2+3b^2)+2ab\sqrt3.$
$13=a^2+3b^2$ and $4=2ab$ means $b=2/a$ and $13=a^2+\dfrac{12}{a^2},$ i.e., $a^4-13a^2+12=0$, i.e., $(a^2-1)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the minimum value of $\frac{a}{b + 1} + \frac{b}{a + 1} + \frac{1}{a + b}$ where $a, b > 0$ and $a + b \le 1$.
$a$ and $b$ are two positives such that $a + b \le 1$. Find the minimum value of $$\large \frac{a}{b + 1} + \frac{b}{a + 1} + \frac{1}{a + b}$$
We have that $\dfrac{a}{b + 1} + \dfrac{b}{a + 1} = \dfra... | Let $y=a+b$. Your expression is the same as $$A=\frac{y+1}{b+1}-1+\frac{y+1}{a+1}-1+\frac{1}{y}=\frac{(y+1)(y+2)}{ab+y+1}+\frac{1}{y}-2$$
Now, by $GM\leq AM$ we have $ab\leq (y/2)^2$. And therefore
$$A\geq \frac{(y+1)(y+2)}{(y/2)^2+y+1}+\frac{1}{y}-2=\frac{(y+1)(y+2)}{(y/2+1)^2}+\frac{1}{y}-2=\frac{4(y+1)}{y+2}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Determining the coefficients of $(1 + x + x^2 +\cdots+x^n)^{n-1}$ Suppose we have the following polynomials:
$$f_1(x)=(1 + x + x^2)$$
$$f_2(x)=(1 + x + x^2 + x^3)^2$$
$$f_3(x)=(1 + x + x^2 + x^3 + x^4)^3$$
$$f_4(x)=(1 + x + x^2 + x^3 + x^4 + x^5)^4$$
$$\vdots$$
$$f_{n-1}(x)=(1 + x + x^2 + x^3 +x^4+ x^5+\cdots+x^n)^{n-1... | Putting (probably far too many) details around Gerry Myerson's comment and avoiding the awkward upper bounds in IV_'s answer's sums...
\begin{align*}
f_n(x) &= \left( \sum_{k=0}^{n+1} x^{k} \right)^n \\
&= \left( \frac{1}{1-x} - \frac{x^{n+2}}{1-x} \right)^n \\
&= \sum_{k=0}^n (-1)^k \binom{n}{k} \left( \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
} |
Calculate the maximum value of $\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c}$ where $3a + 4b + 5c = 12$
$a$, $b$ and $c$ are positives such that $3a + 4b + 5c = 12$. Calculate the maximum value of $$\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c}$$
I want to know if ... | We have that $$3a + 4b + 5c = 12$$
$$\implies (a + b) + 2(c + a) + 3(b + c) = 12 \implies \sqrt{ab} + 2\sqrt{ca} + 3\sqrt{bc} \le 6$$
In addition to that, $$\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c} \le \frac{ab}{ab + 2\sqrt{ab}} + \frac{2ca}{ca + 2\sqrt{ca}} + \frac{3bc}{bc + 2\sqrt{bc}}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $a_4-a_2$
Let $a_1<a_2<a_3<a_4$ be positive integers such that $\sum_{i=1}^{4}\frac{1}{a_i}=\frac{11}{6}$. Find the value of $a_4-a_2.$
I do not know how to proceed. I have tried to simplify the summation but did not obtain anything useful. Please help.
| $\dfrac{11}{6}=\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\dfrac{1}{a_4}
\nleq\dfrac12+\dfrac13+\dfrac14+\dfrac15=\dfrac{77}{60} \quad \Longrightarrow \quad a_1=1$
$\dfrac{5}{6}=\dfrac{1}{a_2}+\dfrac{1}{a_3}+\dfrac{1}{a_4}
\nleq\dfrac13+\dfrac14+\dfrac15=\dfrac{47}{60} \quad \Longrightarrow \quad a_2=2$
$\dfrac{1}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3253866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Interesting inequality $\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$
Prove that $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[k]{\frac{1+x^k}{2}}$$ for all real $x \ge 1$ and for all positive integers $m$ and $k \le 2m+1$.
My work. If $k \le 2m+1$ then $$\sqrt[k]{\frac{1+x^k}{2}}\le \sqrt[2m+1]{\frac{1+x^{2... | Incomplete answer to date
Let $$f(x,k)=\frac{x^{m+1}+1}{x^m+1}-\left(\frac{x^k+1}2\right)^{\frac1k}\tag1$$ so that $$f_x(x,k)=\frac{\partial f(x,k)}{\partial x}=\frac{x^{m-1}(x^{m+1}+(m+1)x-m)}{(x^m+1)^2}-\frac{x^{k-1}}{x^k+1}\left(\frac{x^k+1}2\right)^{\frac1k}\tag2.$$ Clearly, $$f_x(1,k)=\frac2{2^2}-\frac12=0\tag3$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 2
} |
Prove that $(x + y + z)^3 + 9xyz \ge 4(x + y + z)(xy + yz + zx)$ where $x, y, z \ge 0$.
Prove that $(x + y + z)^3 + 9xyz \ge 4(x + y + z)(xy + yz + zx)$ where $x, y, z \ge 0$.
This has become the norm now... This problem is adapted from a recent competition.
We have that $6(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) \g... | We need to prove that $$\sum_{cyc}(x^3+3x^2y+3x^2z+2xyz)+\sum_{cyc}3xyz\geq 4\sum_{cyc}(x^2y+x^2z+xyz)$$ or $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is Schur.
Your way is wrong because you took too strong estimation, that got a wrong inequality.
Here happens like the following.
Let we need to prove that $2>1$.
We ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How can I prove that $ (3/p) = -1$ if $ p \equiv \pm 5 \pmod {12}$ I know how to prove that $ (3/p) = 1$ if $ p \equiv \pm 1 \pmod {12}$ but I need to prove that $ (3/p) = -1$ if $ p \equiv \pm 5 \pmod {12}$, which the book write it as it is without explaining why after explaining the first case thoroughly. Could anyon... | A possible way is as follows. Both rely on the following fact: Let $p>3$ be prime. $-3$ is a quadratic residue modulo $p$ if and only if $p\equiv 1\pmod{6}$. Now, let me first give a self-contained proof of this fact. Let $x\equiv 2a+1\pmod{p}$ such that $x^2\equiv -3\pmod{p}$ (note that such an $a$ must indeed exist a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3260262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$a+OA\lt b+OB\lt c+OC$ when $a\lt b\lt c$ in a triangle In the triangle $\triangle ABC$ of sides $a,b,c$ let $O$ be the incenter. If $a\lt b\lt c$ then (it is easy to prove that) $OC\lt OB\lt OA$.
Prove that
$$\max \{a+OA, b+OB, c+OC\}=c+OC$$
| We need to prove that
$$c+\sqrt{\left(\frac{a+b-c}{2}\right)^2+r^2}>b+\sqrt{\left(\frac{a+c-b}{2}\right)^2+r^2}$$ and$$c+\sqrt{\left(\frac{a+b-c}{2}\right)^2+r^2}>a+\sqrt{\left(\frac{b+c-a}{2}\right)^2+r^2}.$$
We'll prove a first inequality. The second inequality can be proved by the same way.
We need to prove that
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3260407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of $\underset{\{x,y\}\to \{0,0\}}{\text{lim}}\frac{-\frac{x y}{2}+\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}}$
Find limit of $ \underset{\{x,y\}\to \{0,0\}}{\text{lim}}\frac{-\frac{x y}{2}+\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}}$
How can I do that? It is interesting due to mathematica says that
$$\underset{\{x,y\}\to \{0,0\}}... | Assuming $y\ne 0$,
\begin{align*}
&\lim_{(x,y)\to(0,0)}
\frac
{-\frac{x y}{2}+\sqrt{x y+1}-1}
{y \sqrt{x^2+y^2}}
\\[4pt]
=\;&\lim_{(x,y)\to(0,0)}
\frac{-\frac{x y}{2}+\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}}
\cdot
\frac
{\frac{x y}{2}+\sqrt{x y+1}+1}
{\frac{x y}{2}+\sqrt{x y+1}+1}
\\[4pt]
=\;&\lim_{(x,y)\to(0,0)}
\frac
{\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3261179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Matrix complicated equation Let $$A = \begin{bmatrix}
1 & 3 & 4\\
3 & 6 & 9\\
1 & 6 & 4
\end{bmatrix},$$
$B$ be a $3\times 3$ matrix and $$A \cdot A^{T} \cdot A +3B^{-1} =0$$
What would be the value of
$ \det( \operatorname{adj} (A^{-1}(B^{-1}){2B^{T}}))$ ?
| No need to find the value of matrix $B$ since
\begin{equation}
B^{-1}=-\frac{1}{3} A A^T A
\end{equation}
then
\begin{equation}
B=-\frac{1}{3}(A A^T A)^{-1} = 3 A^{-1} A^{-T} A^{-1}
\end{equation}
put into the desired system
\begin{align}
A^{-1}(B^{-1})2B^T &=A^{-1} (-\frac{1}{3} A A^T A) 2 (3 A^{-1} A^{-T} A^{-1})
\no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3262105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$. Please tell if the problem can be solved using telescoping technique or not.
If yes, how to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$ using that. It is given that $a,b \in \mathbb{R}{+},\, a\gt b,\, ... | \begin{align}
(a-b)\sum_{i=1}^n (a^{n-i}b^{i-1}) &=\sum_{i=1}^n (a^{n+1-i}b^{i-1}-a^{n-i}b^i )\\
&=a^n+\sum_{i=2}^n a^{n+1-i}b^{i-1}-\sum_{i=1}^{n-1}a^{n-i}b^i - b^n \\
&= a^n+\sum_{i=1}^{n-1}a^{n-i}b^i-\sum_{i=1}^{n-1}a^{n-i}b^i-b^n\\
&=a^n-b^n
\end{align}
You might like to read the working backward to be similar to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3264215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Is this series for Pi correct?
The idea was to use an infinite series of triangles. The red then green then the... to get the area of this sector then the area of the circle is 16 times this. If it is a unit circle than area should equal Pi.
Here is the series I got using Pythagorean’s theorem , is it correct?
$$\begi... | Note that, in
$x_{n+1}
=\sqrt{2-2\sqrt{1-\frac{x{_{n}}^{2}}{4}}}
$
if
$x_n = 2\sin(t)
$
then
$\begin{array}\\
x_{n+1}
&=\sqrt{2-2\sqrt{1-\frac{x{_{n}}^{2}}{4}}}\\
&=\sqrt{2-2\sqrt{1-\sin^2(t)}}\\
&=\sqrt{2-2\cos(t)}\\
&=2\sqrt{\dfrac{1-\cos(t)}{2}}\\
&=2\sin(\dfrac{t}{2})\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3264331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Finding the sum $\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\cdots$ The sum of series is
$${1\over(3\times5)}+{1\over(5\times7)}+{1\over(7\times9)}+\cdots$$
I used to solve this problem as its $n$th term
$${1\over (2n+1)(2n+3)}$$
Now how can I proceed??
| Let there are functions $$f(x) = \sum_{n=1}^\infty \frac{x^n}{\alpha n + \beta}$$
and $$\psi(x) = \frac{f(x,\alpha,\beta) - f(x,\alpha,\alpha + \beta)}{\alpha} = \sum_{n=1}^\infty \frac{x^n}{(\alpha n + \beta)(\alpha (n+1) + \beta)}$$
where $\alpha,\beta > 0$ and $x \in [-1,1]$.
Thereby $$\psi(x) = \frac{1}{\alpha}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3267577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Proving inequality for induction proof: $\frac1{(n+1)^2} + \frac1{n+1} < \frac1n$ In an induction proof for $\sum_{k = 1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$ (for $n \geq 1$), I was required to prove the inequality $\frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}$.
This is my attempt:
$$ \begin{align} \frac{1}{(n+1)^... | Your way is right, but you need to open the sumation from $k=2$:
$$\sum_{k=1}^n\frac{1}{k^2}=1+\sum_{k=2}^n\frac{1}{k^2}<1+\sum_{k=2}^n\frac{1}{k(k-1)}=1+\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)=2-\frac{1}{n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3268191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
How to correctly solve $\sqrt{1+x}+\sqrt{1-x}>1$? $$\sqrt{1+x}+\sqrt{1-x}>1, x\geq-1 \wedge x\leq1$$
$$\sqrt{1+x}>1-\sqrt{1-x}$$
$$1+x>1-2\sqrt{1-x}+1-x$$
$$0>-2\sqrt{1-x}+1-2x$$
$$2\sqrt{1-x}>1-2x$$
*
*Lets say both sides of inequality are positive, then we can easily $2\sqrt{1-x}>1-2x$ square and arrange. What we ... | The domain of this inequation is $[-1,1]$. On its domain, both sides are non-negative, so comparing them amounts to comparing their squares:
$$\bigl(\sqrt{1+x}+\sqrt{1-x}\bigr)^2=1+\not x+1-\not x+2\sqrt{1-x^2}>1\iff 1+2\sqrt{1-x^2}>0,$$
which is satisfied by any $x$ in the domain since $1+2\sqrt{1-x^2}\ge 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
$\sin(\frac{\pi}{n})\sin(\frac{2\pi}{n})...\sin(\frac{(n-1)\pi}{n})=\frac{n}{2^{n-1}}$ Prove that $$\sin\left(\frac{\pi}{n}\right)\sin\left(\frac{2\pi}{n}\right)\sin\left(\frac{3\pi}{n}\right).....\sin\left(\frac{(n-1)\pi}{n}\right)=\frac{n}{2^{n-1}}$$
Is there a proof without using complex numbers and $n-th$ roots of... | The following is the simplest proof I know. We have the identity
\begin{equation}
x^{2n} - 2x^n y^n \cos n\theta + y^{2n} = \bigg\{x^2 -2xy \cos \theta + y^2\bigg\}\bigg\{x^2-2xy \cos \bigg(\theta+\frac{2\pi}{n}\bigg)+y^2\bigg\}\cdots
\end{equation}
to $n$ factors adding $2\pi/n$ to each angle successively. This can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Expanding the Polynomial Using Taylor Series? Expand the polynomial
$$f(x) = x^3-2x^2-3x+5$$
In power of $(x-2)$
This might be a simple question but all we have to do to solve this question is to expand using the Taylor series with $x =2 $ , as opposed to the Maclaurin at $x=0$.
$f(x) = f(a) + \frac{f'(a)(x-a)^1}{1!}+... | Yes that's correct.
Moreover for a polynomial of degree $n$, the Taylor series of order $n$ is exact (because all the remaining derivatives are nulls), thus you have:
$$
f(x)= -1+(x-2)+4 (x-2)^2+(x-2)^3
$$
(no need for the $O\left((x-2)^4\right)$ term here, this is a true equality)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Does any (right) triangle exist such that $a^3+b^3=c^3$?
Does any right triangle exist such that $a^3+b^3=c^3$? Does any triangle exist such that $a^3+b^3=c^3$?
I'm stuck on this problem; I tried applying the Pythagorean theorem in three dimensions, but in vain. Any tips?
| If $a^2+b^2=c^2,$ with $a,b,c$ positive, you'd have: $c>a,b>0$ and thus $ca^2>a^3$ and $cb^2>b^3.$ So you get:
$$c^3=c\cdot c^2=c(a^2+b^2)=ca^2+cb^2>a^3+b^3.$$
So $c^3>a^3+b^3.$
You can prove more generally for any triple $(a,b,c)$ of positive reals, there is at most one positive $n$ such that $a^n+b^n=c^n.$ If $a^n+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3276249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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$\lim_{n\to \infty} \ \frac{1}{n} \Bigl[(a+\frac{1}{n})^2+(a+\frac{2}{n})^2+\cdots+(a+\frac{n-1}{n})^2\Bigr]$ without L'Hopital Find limit:
$$a_n = \lim_{n\to \infty} \ \frac{1}{n} \Bigl[\Bigl(a+\frac{1}{n}\Bigr)^2+\Bigl(a+\frac{2}{n}\Bigr)^2+\cdots+\Bigl(a+\frac{n-1}{n}\Bigr)^2\Bigr]$$
I tried limiting it with $$n\c... | $\lim_{n\to \infty} \ \frac{1}{n} \Bigl[\Bigl(a+\frac{1}{n}\Bigr)^2+\Bigl(a+\frac{2}{n}\Bigr)^2+\cdots+\Bigl(a+\frac{n-1}{n}\Bigr)^2\Bigr] =\displaystyle\int_{a}^{a+1} x^2\, dx=\dfrac{x^3}{3}|_a^{a+1}=\dfrac{(a+1)^3-a^3}{3}=a^2+a+\dfrac{1}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Show that $3^{7^n}+5^{7^n}\equiv 1 \pmod{7^{n+1}}$ Show that $3^{7^n}+5^{7^n}\equiv 1 \pmod{7^{n+1}}$
I tried to induct on n:
For $n = 0$ we have $3+5 = 8$ and $8 \equiv 1 \pmod{7^{n+1}}$.
Suppose it is true for $n = k$:
$$3^{7^k}+5^{7^k}\equiv 1 \pmod{7^{k+1}}$$
so $3^{7^k}+5^{7^k}=7^{k+1}*q_1+1$
For $n = k+1$:
$$3^{... | It is known that
$$(a+b)^7=a^7+b^7+7(a^6b+ab^6)+21(a^5b^2+a^2b^5)+35(a^4b^3+a^3b^4)$$ and $(A+B)^7= {A^7+B^7}$ in fields of characteristic $7$. We apply induction.
►$3^{7^1}+5^{7^1}=80312=1639\times7^2+1\iff3^{7^1}+5^{7^1}\equiv1\pmod{7^2}$
►$3^{7^n}+5^{7^n}\equiv1\pmod{7^{n+1}}\iff3^{7^n}+5^{7^n}=7^{n+1}m+1$ assumed ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
What is the correct solution of $\sqrt[7]{(-\sqrt{3}-i)^5}$? $\sqrt[7]{(-\sqrt{3}-i)^5}=(-\sqrt{3}-i)^\frac{5}{7}=
2^\frac{5}{7}(\cos(\frac{5}{7}\alpha)+i\sin(\frac{5}{7}\alpha)=$
$\tan\alpha=\frac{-1}{-\sqrt{3}} \implies \alpha=\frac{\pi}{6}+2k\pi$
$=2^\frac{5}{7}(\cos(\frac{5\pi}{42}+\frac{2k\pi}{7})+i\sin(\frac{5\pi... | $(-\sqrt3-i)^5$ is a specific complex number, but it can be written in multiple ways, making use of the fact that $e^{i2\pi k}=1$ for any $k\in\mathbb{Z}$: Since
$$-\sqrt3-i=-2\left(\sqrt3+i\over2\right)=-2\left(\cos\left(\pi\over6\right)+i\sin\left(\pi\over6\right)\right)=-2e^{i\pi/6}$$
we have
$$(-\sqrt3-i)^5=-2^5e^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Solve for $x,y,z \in \mathbb{R^+}$ ,$x^2+y^2+z^2=xyz+4$ and $xy+yz+zx=2(x+y+z)$
Find all positive real numbers such that
$$x^2+y^2+z^2=xyz+4$$
And $$xy+yz+zx=2(x+y+z)$$.
I substitute $x,y,z$ by $(a+\frac{1}{a}),(b+\frac{1}{b}),(c+\frac{1}{c})$ respectively where $abc=1$ which comes from the first relation. Puttin... | After homogenization we obtain:
$$\frac{(x^2+y^2+z^2)(xy+xz+yz)}{2(x+y+z)}=xyz+\frac{(xy+xz+yz)^3}{2(x+y+z)^3}.$$
We'll prove that
$$\frac{(x^2+y^2+z^2)(xy+xz+yz)}{2(x+y+z)}\geq xyz+\frac{(xy+xz+yz)^3}{2(x+y+z)^3}.$$
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v>0$, and $xyz=w^3$.
We see that it's enough to prove t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Special Factorization Consider the natural numbers that are sum of a perfect square plus the product of consectutive natural numbers. For example, $97 = 5^{2} + 8\cdot 9$. What is the smallest multiple of 2019 that is not as described above?
Someone can help me? Thank you in advance.
| consider the following theorem:
theorem: An integer greater than one can be written as a sum of two squares if and only if its prime decomposition contains no prime congruent to $3(mod4)$ raised to an odd power.
Now, observe that
$$N = m^{2} + n\cdot (n + 1) \Leftrightarrow$$
$$4N = 4m^{2} + 4n^{2} + 4n \Leftrightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3289430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\sum^{\infty}_{n=1}{\frac{2(n+1)\cdot3^n}{\sqrt[3]{n!}}}$ is Convergent
Show that $\sum^\infty_{n=1}\frac{n+1}{\sqrt[3]{n!}}\cdot(x^n+x^{-n})$ is uniformly convergent, when $-\frac{1}{3}\leq x\leq3$.
So I`m trying to show with Weierstrass M-test:
$|\frac{n+1}{\sqrt[3]{n!}}\cdot(x^n+x^{-n})|\leq|\frac{n+1}{... | Ratio test:
$\lim_{n\to\infty} \frac{\frac{2(n+2)3^{n+1}}{\sqrt[3]{(n+1)!}}}{\frac{2(n+1)3^n}{\sqrt[3]{n!}}}$= $\lim_{n\to\infty}\frac{3\cdot(n+2)\cdot\sqrt[3]{n!}}{(n+1)\sqrt[3]{(n+1)!}}$=$\lim_{n\to\infty}\frac{3\cdot(n+2)}{\sqrt[3]{n+1}(n+1)}$=$3\cdot\lim_{n\to\infty}\sqrt[3]{\frac{(n+2)^3}{(n+1)^4}}$=$0 <1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many matrices satisfy this equality? How many matrices $A\in\mathcal{M}_{3\times 3} (\mathbb{N})$ satisfy this equality?
$$\begin{pmatrix}
1 \ \ 2 \ \ 4
\end{pmatrix}\cdot A=\begin{pmatrix}
3 \ \ 2 \ \ 1
\end{pmatrix}$$
I tried with examples and I found just one but I want to know how to approach this exercise.The ... | Let $A=[x_{i,j}]$ Then we get
$$x_{1,1} +2 x_{2,1} +4 x_{3,1} = 3$$
$$ x_{1,2} + 2x_{2,2} + 4 x_{3,2} = 2 $$
$$x_{1,3} + 2 x_{2,3} + 4 x_{3,3} = 1$$
We shall count the number of solutions for each equation.
First equation has 2 solutions:
Since $4>3$ we have that $x_{3,1}=0$. Similarly $x_{2,1}<2$. If $x_{2,1}=1$ then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $a,b,c$ be the sides of a triangle. The find maximum value $ \frac{A}{S^{2}}$ Let $a,b,c$ be the sides of a triangle, $A$ is the area and $S$ is the semi-perimeter $(a+b+c)/2$.
Find the maximum value $\frac{A}{S^{2}}$.
My Approach:
Method 1:
Applying AM-GM inequality on $S,S-a,S-b,S-c$
$$\frac{4S-2S}{4} \ge \sqrt... | Your first way does not give a solution because the equality
$$s=s-a=s-b=s-c$$ is impossible.
By the way, by AM-GM
$$\frac{A}{s^2}=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s^2}\leq\frac{\sqrt{s\left(\frac{s-a+s-b+s-c}{3}\right)^3}}{s^2}=\frac{1}{3\sqrt3}.$$
The equality occurs for $$s-a=s-b=s-c$$ or $$a=b=c,$$ which says that we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$ I'm trying to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$. I am getting x = 4 but the book gets x = 81. What am I doing wrong?
\begin{align*}
\log_{3}(x) & = \log_{\frac{1}{3}}(x) + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{\log_{3}(\frac{1}{3})} + 8\\
\log_{3}... | First, $\log_{1/3}x = \log_{3}\frac{1}{x}$; hence,
$$ \log_{3}x - \log_{3} \left( \frac{1}{x} \right) = \log_{3} \left(\frac{x}{\frac{1}{x}} \right) = \log_{3}x^{2} = 8$$
which is equivalent to
$$ 3^{8} = \left( 3^{4} \right)^{2} = x^{2}.$$
And so,
$$x = 3^{4} = 81.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is every prime contained in the set $\,\{n\in\mathbb{N}:n|10^k-1\}\cup\{2,5\}\,$ for $k=1,2,3,...$?
Is every prime contained in the set
$\,\{n\in\mathbb{N}:n|10^k-1\}\cup\{2,5\}\,$ for $k=1,2,3,...$?
Let $p\notin\{2,5\}$ be a prime number. Then $\frac{1}{p}$ has a decimal period of length at most $p-1$. Denote $\fr... | Yes: it's true that every prime $p \neq 2$ or $5$ divides some number of the form $10^k-1$.
There is a simple way to prove this fact.
Look at the remainders of $10^k-1$ when divided by $p$. Clearly there are only finitely many remainders, namely $0,1,2, \dots, p-1$. Since the numbers of the form $10^k-1$ are infinitely... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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If $a,b,c$ are positives such that $a+b+c=\pi/2$ and $\cot(a),\cot(b),\cot(c)$ is in arithmetic progression, find $\cot(a)\cot(c)$
Suppose $a,b,c$ are positive reals such that $a+b+c=\pi/2$ and $2\cot(b)=\cot(a)+\cot(c)$.
Find $\cot(a) \cdot \cot(c)$.
I have tried writing $b=\pi/2-a-c$ into the given equation and pla... | As
$$\cot(x+y) = \frac{\cot x \cot y - 1}{\cot x + \cot y}$$
$$\cot(a+c) = \cot\left(\frac{\pi}{2}-b\right) = \tan(b) = \frac{1}{\cot b}= \frac{\cot a\cot c -1}{\cot a+ \cot c}$$
$$\frac{1}{\cot b} = \frac{\cot a\cot c - 1}{2\cot b}$$
So,
$$\cot a \cot c = 1+2 = 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Minimizing distance between an ellipse and a point Problem
An ellipse has the formula:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
What is the shortest distance from the ellipse to the point $P = (a,0)?$
Attempted solution:
I have previously solved the same problem with P = (1,0) and P = (2,0). For P = (1,0), the minimum whe... | The problem may be solved with the method of Lagrange multiplier.
$$L = (x-a)^2 + y^2 - \lambda \left(\frac{x^2}{9} + \frac{y^2}{4} -1\right)$$
$$ 0 = \frac{\partial L}{\partial x} = 2(x-a) - \frac{2}{9} \lambda x$$
$$ 0 = \frac{\partial L}{\partial y} = 2y - \frac{1}{2} \lambda y \phantom{00000}$$
From the second equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Find the number of roots of the polynomial $P(z)=z^5+2z^3+3$ in the closed unit disk $\{z\colon |z|\le 1\}$.
Find the number of roots of the polynomial $P(z)=z^5+2z^3+3$ in the closed unit disk $\{z\colon |z|\le 1\}$.
My solution:
Set $f(z)=3$.
For every $0<\delta<1$ and $|z|=\delta$, we have $|P(z)-f(z)|\le \delta^... | Let $z^5+2z^3+3=0$ with $|z|\le 1. $ Then $3=|-3|=|z^5+2z^3|=|z|^3\cdot |z^2+2|\le |z^2+2|\le |z|^2+|2|=|z|^2+2\le 3.$
So $|z|=1$. So let $z=e^{it}$ with $t\in \Bbb R.$ Then $-3=Re (e^{5it}+2e^{3it})=\cos 5t+2\cos 3t.$
Since $\cos 5t\ge -1$ and $2\cos 3t\ge -2,$ therefore $\cos 5t=\cos 3t=-1$. This implies $z^5=z^3=-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Writing a proper inductive argument The algebra is simple for some reason I am having a hard time setting up the argument. Here is my attempt
Suppose we have:
$\sum_{k=1}^n (2k-1)^2 = \frac{n(4n^2-1)}{3}$ for all $n \in \mathbb{N}$
Let see if this holds true for $n = 1$
LHS: $= 1$
RHS $= 1$
Thus this formula holds for... | Given the statement in this question, I will assume that by the natural numbers, you mean the set of positive integers. If you instead meant the set of nonnegative integers, you need to prove that the statement holds when $n = 0$ as well, as shown by Antoine Mathys in the comments.
In writing a proof by mathematical i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3300135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove That $3^n + 8^n$ is Not Divisible by $5$ (Using Induction) Prove that $3^n+8^n$ is not divisible by 5.
I know that this can be proved by using congruence and I am providing the proof by congruence below. But is there any way to Prove It By Induction.
The proof by congruence goes like this:
$3\equiv 3\pmod 5 \\ 3^... | Yes, you can do it by induction as well. Note that:
$3^{n+1}+8^{n+1}=3^n\times 3+8^n\times (3+5)=(3^n+8^n)\times 3+5\times 8^n$
By induction hypothesis the first term is not divisible by $5$ while the second term is obviously divisible by $5$. The required result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3300548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 0
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Jacobi symbol:$(\frac{7m^2-1}{18m^2+1})=(\frac{25m^2}{18m^2+1})=1$? By definition of Jacobi symbol, $(\frac{q}{n})=1$ if there exists $x$ such that $x^2 \equiv q \pmod n$ where $n$ is odd. So, $(\frac{7m^2-1}{18m^2+1})$ is equivalent to-
$x^2 \equiv 7m^2-1 \pmod {18m^2+1} \implies x^2+18m^2+1 \equiv 7m^2-1+18m^2+1 \pmo... | If $a\equiv b\pmod n$ then $\left(\dfrac a n\right)=\left(\dfrac b n\right)$.
Now $7m^2-1\equiv 7m^2-1 +(18m^2+1)\equiv 25m^2\pmod{18m^2+1}.$
Therefore $\left(\dfrac{7m^2-1}{18m^2+1}\right)=\left(\dfrac{25m^2}{18m^2+1}\right).$
Furthermore, $25m^2=(5m)^2$, so $\left(\dfrac{25m^2}{18m^2+1}\right)=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3301041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove that $11 | 10^{2n+1}+1$ for all $n\in \mathbb{N}\cup \{0\}$. $$11 | 10^{2n+1}+1\;\;\; \forall n\in \mathbb{N}\cup\{0\} \tag{$\star$}$$
My proof of $(\star)$ is as follows:
\begin{align}
10^{2n+1}+1
&= 10\cdot10^{2n}+1 \\
&= (11-1)\cdot10^{2n}+1 \\
&= 11\cdot10^{2n}-10^{2n}+1 \\
&= 11\cdot10^{2n}-\left(10^{2n}-1\... | Use modular arithmetic:
$$10\equiv-1\mod11$$
$$10^2\equiv1\mod11$$
$$10^{2n}\equiv1\mod11$$
$$10^{2n+1}\equiv-1\mod11$$
$$11|10^{2n+1}+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
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shortest distance of $x^2 + y^2 = 4$ and $3x + 4y = 12$ $x^2 + y^2 = 4$ and $3x + 4y = 12$
line from origin with gradient $\frac{4}{3}$ intersect circle at $(\frac{6}{5} , \frac{8}{5}) $.
intersect $3x + 4y = 12$ at $(\frac{36}{25}, \frac{36}{25})$.
so distance is $\sqrt{{(\frac{36}{25}- \frac{6}{5})}^2 + {(\frac{36}{2... | Let $$x_c^2+y_c^2=4$$ and $$3x_s+4y_s=12$$ then $$d=\sqrt{(x_c-x_s)^2+(y_c-y_s)^2}$$ you can eliminate two variables.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Integral $ \int_0^\infty \frac{\ln x}{(x+c)(x-1)} dx$ I've been trying to solve the following integral for days now.
$$P = \int_0^\infty \frac{\ln(x)}{(x+c)(x-1)} dx$$
with $c > 0$. I figured out (numerically, by accident) that if $c = 1$, then $P = \pi^2/4$. But why? And more importantly: what's the general solution o... | A different approach using polylogarithms
For this solution the following identities are used
$\displaystyle\text{Li}_2\left(z\right)=-\int _0^z\frac{\ln \left(1-t\right)}{t}\:dt$, $\displaystyle \int _0^1\frac{c\ln ^n\left(x\right)}{1-cx}\:dx=\left(-1\right)^nn!\text{Li}_{n+1}\left(c\right)$, $\displaystyle \text{Li}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
$\int_0^\infty \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx$ Trying to compute Integral $\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx$
I was facing:
\begin{align}J=\int_0^\infty \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx\end{align}
I want to prove that $\displaystyle J=0$, or equivalently, that... | Let $
K=\int_0^\infty \frac{\ln(1+x+\sqrt{2x})}{x^2+1}dx$
and note that $J+K=\int_0^\infty \frac{\ln(x^2+1)}{x^2+1}\ {dx}$
\begin{align}
J-K=&\int_{0}^\infty \frac{\ln({1+x-\sqrt{2x}})-\ln({1+x+\sqrt{2x}})}{x^2+1}\ \overset{x=t^2}{dx}\\
=&\int_{-\infty}^\infty \frac{t\ln({1+t^2-\sqrt{2}t})-t\ln({1+t^2+\sqrt{2}t})}{t^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
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Finding maximum of a given function
Show that $f(x)=\sin x(1+\cos x)$ attains its maximum at $x= \pi/3$.
I differentiated the function $f$ and got $f'(x)=\cos x(1+2\cos x)$. After equating with $0$, I got $x=\pi/2$ and $x =\pi/3 + n\pi$ with $n\neq 0$. So I did not get $x= \pi/3$ even as an extreme value.
| $$\cos x(1+2\cos x)=0$$
$$\cos x=0\ \ \ \text{or} \ \ 1+2\cos x=0$$
$$x=\frac{(2n+1)\pi}{2}\ \ \ \text{or} \ \ \cos x=-\frac12$$
$$x=\frac{(2n+1)\pi}{2}\ \ \ \text{or} \ \ \cos x=\cos\frac{2\pi}{3}$$
$$x=\frac{(2n+1)\pi}{2}\ \ \ \text{or} \ \ \ x=2n\pi\pm\frac{2\pi}{3}$$
$$x=\ldots, -\frac{\pi}{2}, \frac{\pi}{2}, \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Express $x=0.\overline{31}_5$ as a fraction in lowest terms
Given $x=0.\overline{31}_5$, find the value of $x$, expressed as a fraction in lowest terms.
I tried to change it into base $10$, but I don't think it's possible with fractions. So please help I'll appreciate it. Also I'm in 7th grade (easy solutions please)... | Updated solution:
$0.31_5 = \frac{3}{5} + \frac{1}{25} = \frac{16}{25}$.
$0.\overline{31}_5 = 0.31_5 + 0.0031_5 + 0.000031_5 + \cdots$. Now using the formula for an infinite geometric series:
$$0.\overline{31}_5 = \frac{0.31_5}{1 - 0.01_5} = \frac{16/25}{1-1/25} = \frac{16}{24} = \frac{2}{3}.$$
Old solution:
$x = 0.3_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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When going from $(x+2)^2=5$ to $x+2=\pm \sqrt{5}$, why isn't there also a $\pm(x+2)$? Say I am solving the following equation:
$$(x+2)^2 = 5$$
$$x + 2 = \pm \sqrt{5}$$
$$x = -2 \pm \sqrt{5}$$
However, when I took the positive and negative square root of $5$ in the second line, I did not take the positive and negative s... | Hint: Better is to write
$$(x+2)^2-\sqrt{5}^2=0$$ and this is, using that $$a^2-b^2=(a+b)(a-b)$$
$$(x+2-\sqrt{5})(x+2+\sqrt{5})=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Finding the best solution to an inconsistent system $A\mathbf{u} = \mathbf{b}$. Let $A = \begin{bmatrix}
-1 & 1 \\
2 & -1 \\
1 & 1
\end{bmatrix}$ and $\mathbf{b} = \begin{bmatrix}
1 \\
2 \\
0
\end{bmatrix}$.
1. Find a "best solution" to the inconsistent system $A\mathbf{u} = \mathbf{b}$.
2. Find the orthogonal projecti... | As you said $A\mathbf{u} = \mathbf{b}$ is an inconsistent system that has no solution. So the best you can hope for is to find a $\mathbf{u}$ that makes $A\mathbf{u}$ "close to" $\mathbf{b}$. Or to say the same thing you can try to make $A\mathbf{u} - \mathbf{b}$ "as close to $\mathbf{0}$ as possible".
So the problem b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Decompose $33+11\sqrt{-7}$ into irreducible integral elements of $\mathbb{Q}(\sqrt{-7})$ This is exercise I.3.1 in Neukrich's Algebraic Number Theory. Writing
$$ 33+11\sqrt{-7} = 11(3+\sqrt{-7}), $$
I have managed (by trying small values) to find
$$ 11 = (2+\sqrt{-7})(2-\sqrt{-7}) $$
and
$$ 3+\sqrt{-7} = 2 \left(\frac{... | I find that simplifying the notation helps with the calculations. This is what I did:
Since $-7 \equiv 1 \hspace{2pt}\mathrm{mod}\hspace{2pt} 4$ the ring of integers is $\mathbf{Z}[\alpha]$, where $\alpha :=\tfrac{1}{2}(1 + \sqrt{-7})$, and we know that $\{1, \alpha\}$ is an integral basis and that $\alpha^2 -\alpha +2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Euclidean distance of two uniform random variables Two random variables $X$ and $Y$ are uniformly distributed, the pdfs of which are given by $f_{X}\left(x\right) = f_{Y}\left(y\right) = 1/r$. I am trying to obtain $Z = \sqrt{X^2 + Y^2}$.
I tried the approach shown below, but I want to avoid having the $\tan^{-1}$ term... | Looking to the problem geometrically, it is clear that the pdf /cdf cannot but be defined piecewise
and it is easy to derive the pdf from the area of the segment of the circular annulus between $z$ and $z+dz$ intercepted by the square,
divided by the area of the whole square ($r^2$), i.e.
$$
p(z\,;\,r) = {1 \over {r^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3311052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Help on residue: $3^x + 22^y \equiv 15^z \equiv 15 \pmod{40} $ I am reading this note (click here and go to page 1994 for detail, in the proof of lemma 8), and found-
$$3^x + 22^y \equiv 15^z \equiv 15 \pmod{40} $$
now, I can derive $3^x + 22^y \equiv 15^z \ \pmod{40} $ but I cann't figure out how $3^x + 22^y \equiv 15... | If the assumptions of your prior question hold here, then we know that $z$ is odd.
But for odd $z$, $15^z\equiv z \pmod {40}$.
To see this, We note that $15^2=225\equiv 25\pmod {40}$ and $$15^3\equiv 15\times 25\equiv 15\pmod {40}$$ Then $$15^4=15^3\times 15\equiv 15^2\equiv 25 \pmod {40}$$ and $$15^5\equiv 15^4\tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating a Trigonometric Integral without Substitutions I have been tasked with evaluating the integral $$I=\int\frac{\sin(2x)+\sin(4x)-\sin(6x)}{\cos(2x)+\cos(4x)+\cos(6x)+1}dx$$ After substituting first $u=2x$ and then $v=\cos(u)$ and a messy partial fraction decomposition, I get the answer $$\frac{4 \log(\cos(x)) ... | First express the integrand in the form
$$\displaystyle\frac{4\sin 3x \sin 2x \sin x}{4\cos 3x \cos 2x \cos x}.$$
For example, the denominator requires $\cos 2x+\cos 4x\equiv 2\cos 3x \cos x$, $\cos 6x+1\equiv 2\cos^2 3x$, and $\cos 3x+\cos x\equiv 2\cos 2x\cos x$.
Thus the integrand is $\tan 3x\tan 2x\tan x$.
What hel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Inverse of $y = Ax^3 + Bx + C$ I know there already is a question and a bunch of excellent answers in How to find the inverse of $y=x^3-5x^2+3x+c$.
But here I am having a more general (and thus complicated) form to invert. I am stuck in obtaining the inverse, $g^{-1}$, of a polynomial
\begin{align}
y = g\left(x\right) ... | Consider that you need to solve for $x$ the cubic equation
$$ -\frac{x^3}{8r} - r\left(1 + \frac{1}{\sqrt{2}}\right)x + \left(2\sqrt{2}r^2-y\right)=0$$
Applying the method for cubic equations, you would find
$$\Delta=-\frac{27 \left(2 \sqrt{2} r^2-y\right)^2}{64 r^2}-\frac{1}{2}
\left(1+\frac{1}{\sqrt{2}}\right)^3 r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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joint and conditional probability This is table of conditional probability with $X$ is for red coin, and $Y$ for blue coin.
$$\begin{array}{rc}
& X_{\text{red}} \\
Y_{\text{blue}} & \begin{aligned}
\ P (X \mid Y)&& X=\text{Tail} && X=\text{Head}
\\Y=\text{Head} &&\frac{6}{10} && \frac{4}{10}
\\ Y=... | \begin{align}
P(X=H, Y=H) &= P(Y=H) P(X=H|Y=H)\\
&= 0.45 \cdot \color{red}{0.4}
\end{align}
For the table of conditional probability.
Summing up the rows give you $1$, because given that $Y$ has occured, and the probability of possible outcome of $X$ must sum to $1$.
$$\sum_xP(X=x|Y=y)=1$$
However, for the columns,
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding the length of a Curve The following problem is from the book, Calculus and Analytical Geometer by Thomas and Finney. It is relatively
early in the book, so I would expect the integration to be easy.
Find the length of the curve:
$$ 9x^2 = 4y^3$$
from $(0,0)$ to $\left(2\sqrt{3},3\right)$.
Answer:
The... | It's possible to solve the integral but it's quite unwieldy if you don't use the right substitution.
$$\begin{aligned}
I&=\int_0^{2\sqrt{3}}\sqrt{1+\left(\frac{4}{9x^2}\right)^{\frac{1}{3}}}\ \mathrm{d}x\\
&=\int_1^4\sqrt{u}\ \mathrm{d}u\\
&=14/3
\end{aligned}$$
With the substitution $u=\left(\frac32x\right)^{2/3}+1$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3313433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Determine an explicit form for the generating function and a closed form given a recurrence Let $f_n = 4f_{n-1} - 4f_{n-2}$ be a recurrence relation for $f$ with initial conditions $f_0 = 1$ and $f_1 = 4$. I want to determine an explicit form for the generating function of $f$, $F(x)$ as well as a closed form for $f_n$... | Define the generating function $F(z) = \sum_{n \ge 0} f_n z^n$, multiply the recurrence shifted by $2$ by $z^n$ and sum over $n \ge 0$, recognize some sums:
$\begin{align*}
\sum_{n \ge 0} f_{n + 2} z^n
&= 4 \sum_{n \ge 0} f_{n + 1} z^n - 4 \sum_{n \ge 0} f_n z^n \\
\frac{F(z) - f_0 - f_1 z}{z^2}
&= 4 \frac{F(z) - f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the min value of $\frac{1}{x+\frac{1}{y+\frac{1}{z}}}$, if $x\ne y \ne z$ and $x,y,z\in {1,2,3,4,5}$ My answer is $\frac{5}{29}$, I just use logic to substitute numbers in the expression, but I can't prove my answer If this expression be minimum then the denominator should be the greatest, so I just let x=5 and I... | $$
f(x,y,z) = \frac{1}{x+\frac{1}{y+\frac{1}{z}}}
$$
is decreasing in $x$ and $z$, and increasing in $y$. Since $x,z$ are distinct integers we have either
$$
x < z \implies x \le 4, y \ge 1, z \le 5 \implies f(x, y, z) \ge f(4, 1, 5) = \frac{6}{29}
$$
or
$$
z < x \implies x \le 5, y \ge 1, z \le 4 \implies f(x, y, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Generating function of ordered partitions What is exactly generating function of ordered partitions and how can I get number of ordered partitions from that?
Example:
$$ 4 = 1+1+1+1 \\ = 2+2 \\ = 1+1+2 \\ = 1+2+1 \\ = 2+1+1 \\ = 1+3 \\ = 3+1 \\ = 4 $$
so we have $8$ partitions. I was thinking about exponential generat... | We can do this by generating functions which you want to.
For partition $n$,
If we separate $n$ into $n$ numbers, we can generate the function $x+x^2+\cdots+x^n$ and the coefficient of $x^n$ is the number of combination.
If we separate $n$ into $\left(n-1\right)$ numbers, we can generate the function $\left(x+x^2+\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Find $y$ in $\sqrt{4+(y-6)^2}+\sqrt{16+(y-3)^2}=3\cdot\sqrt{5}$ I would like to solve this equation for $y$. Any tips? It seems like you cant really do it analytically?
$$\sqrt{4+(y-6)^2}+\sqrt{16+(y-3)^2}=3\cdot\sqrt{5}$$
| You can factor out $2$ at the second summand.
$$\sqrt{\underbrace{4+(y-6)^2}_{=5}}+2\cdot \sqrt{\underbrace{4+\left(\frac{y-3}2\right)^2}_{=5}}=\sqrt{5}+2\cdot\sqrt{5}$$
Now we see that the following equations has to be true at the same time.
*
*$(y-6)^2=1 \Rightarrow y_1=7,y_2=5$
*$\left(\frac{y-3}2\right)^2=1\R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3320253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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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.