Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Proving $\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3}$. How would I show $\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3}$?
I tried starting from the LHS, and rationalising and what-not but I can't get the result...
Also curious to how they got the LHS expression from considering the right.
| We have $$\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3};$$ that is,
$$3+\sqrt{13+4\sqrt{3}}=4+2\sqrt{3},$$ I.e.,
$$\sqrt{13+4\sqrt{3}}=1+2\sqrt{3},$$ which is
$$13+4\sqrt{3}=13+4\sqrt{3}.$$ Now just work backwards.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\frac{1}{2^n}\binom{n}{n}+\frac{1}{2^{n+1}}\binom{n+1}{n}+...+\frac{1}{2^{2n}}\binom{2n}{n}=1$: short proof? The identity $\frac{1}{2^n}\binom{n}{n}+\frac{1}{2^{n+1}}\binom{n+1}{n}+...+\frac{1}{2^{2n}}\binom{2n}{n}=1$ arises from a question on probability in my textbook. A proof by induction on $n$, which exploits the... | Suppose you are flipping a coin until you get $n+1$ heads or $n+1$ tails. For $k=0,\dots,n$, what is the probability that you are done after exactly $n+1+k$ flips?
If the last flip was a tails, this means that in the former $n+k$ flips there were exactly $n$ tails, and this happens with probability $\frac{1}{2}\cdot\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
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Simplify $\frac1{\sqrt{x^2+1}}-\frac{x^2}{(x^2+1)^{3/2}}$ I want to know why
$$\frac1{\sqrt{x^2+1}} - \frac{x^2}{(x^2+1)^{3/2}}$$
can be simplified into
$$\frac1{(x^2+1)^{3/2}}$$
I tried to simplify by rewriting radicals and fractions. I was hoping to see a clever trick (e.g. adding a clever zero, multiplying by a cle... | $\sqrt x$ is just a shorthand for $x^{1/2}$. Hence we can multiply the two halves of the first fraction in the first term by $x^2+1$:
$$\frac1{\sqrt{x^2+1}}-\frac{x^2}{(x^2+1)^{3/2}}=\frac{x^2+1}{(x^2+1)^{3/2}}-\frac{x^2}{(x^2+1)^{3/2}}$$
and the target expression follows.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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In expansion of $(1+x+x^{2}+x^{3}+....+x^{27})(1+x+x^{2}+x^{3}+....x^{14})^{2}$ . Find the coefficient of $x^{28}$ I am not able to apply binomial theorem here
$(1+x+x^{2}+x^{3}+....+x^{27})(1+x+x^{2}+x^{3}+....x^{14})^{2}$
Please help me to find the coefficient of$ x^{28}$
Any help will be appreciated.
| Express is as a product of three factors $F_1,F_2,F_3$:
$$(1+x+x^{2}+x^{3}+....+x^{27})(1+x+x^{2}+x^{3}+....x^{14})^{2}=(1+\cdots +x^{27})(1+\cdots +x^{14})(1+\cdots +x^{14}).$$
Make up a table of powers of $x$:
$$\begin{array}{c|c|lcr}
m& n & \text{$F_1$} & \text{$F_2$} & \text{$F_3$} \\
\hline
& 1 & 14 & 14 & 0 \\
& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is the double integral equal to the area? I have to compute the double integral $\int_0^{\sqrt{\frac{\pi}{2}}}\int_x^{\sqrt{\frac{\pi}{2}}} 2\sin(y^2)dydx$.
I drawed the region $0\leq x\leq \sqrt{\frac{\pi}{2}}$ and $x\leq y\leq \sqrt{\frac{\pi}{2}}$. This is a triangle with height and base equal to $\sqrt{\frac{\pi}{... | Hint. Note that the double integral over that triangle can be written as iterated integrals in two ways.
\begin{align*}
\int_{x=0}^{\sqrt{\frac{\pi}{2}}}\left(\int_{y=x}^{\sqrt{\frac{\pi}{2}}} 2\sin(y^2)dy\right)dx
&=\int_{y=0}^{\sqrt{\frac{\pi}{2}}}\left(\int_{x=0}^{y} 2\sin(y^2)dx\right)dy\\
&=\int_{y=0}^{\sqrt{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve in positive integers, $ x^6+x^3y^3-y^6+3xy(x^2-y^2)^2=1$ Solve in positive integers, $$ x^6+x^3y^3-y^6+3xy(x^2-y^2)^2=1$$
My attempt :
$ x^3y^3+x^6-y^6+3xy(x^2-y^2)^2$
$=x^3y^3+3x^2y^2(x^2-y^2)+3xy(x^2-y^2)^2+(x^2-y^2)^3$
$=(xy+(x^2-y^2))^3$
$=(x^2+xy-y^2)^3 = 1$
so $x^2+xy-y^2 = 1$
Please suggest how to proceed.... | Hint:
You should keep the cube
$$(x^2+xy-y^2)^3 = 1$$
$$(x^2+xy-y^2)^3 -1^3 = 0$$
You should factorize with the formula $a^3-b^3= (a-b) (a² + ab + b²))$
Edit:
Then solve the equations.But as pointed by Alex, $ a^2+a=-1$ has no solution in $ Z $, so only $a=1$ need to be solved.
| {
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"url": "https://math.stackexchange.com/questions/2412268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving/disproving $∀n ∈ \text{positive integers}$, $\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5$? Is my proof getting anywhere for proving/disproving $∀n ∈ \text{postive integers}$, $$\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5\text{?}$$
$$\left\lceil{\frac{4n^2}{n^2}}+\frac{1}{n^2}\right \rceil = 5$$
$$\lef... | $$\left\lceil \frac{4n^2+1}{n^2} \right\rceil=5\tag{1}$$
means
$$4 < \frac{4n^2+1}{n^2} \le 5\tag{2}$$
so try to prove (2). This should be less error prone that transforming expressions with ceilings in them.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Sequence : $a_k a_{k+2} +1 = a^2_{k+1}$ Determine all integers $b$ for which there exists a sequence $a_0, a_1, a_2,\ldots $ of rational numbers
satisfying $a_0=0, a_2=b$ and $a_k a_{k+2} +1 = a^2_{k+1}$ for $k=0, 1, 2, \ldots$
that contains at least one nonintegral term.
My attempt :
$a_0=0, a_2=b$, so $a_0a_2+1=a^... | Notice that $a_1 = \pm 1$. For simplicity, let us write $c = b a_1$. Then consider a sequence $(f_n)$ that solves the recurrence relation
$$ f_{n+2} = c f_{n+1} - f_n, \qquad f_0 = a_0, \quad f_1 = a_1. $$
If we consider matrices
$$ A_n = \begin{pmatrix} f_{n+2} & f_{n+1} \\ f_{n+1} & f_n \end{pmatrix}, \qquad
P = \beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integration of $\int_{0}^{\infty}\frac{dw}{1+\left ( \frac{w}{B} \right )^4}$ with the help of signal properties.
Find the integral
$$\int_{0}^{\infty}\frac{dw}{1+\left ( \frac{w}{B} \right )^4}$$
where $B$ is a constant.
This integration i tried by normal method that gives the result $\frac{\pi B}{2\sqrt2}$ but ... | Just want to share a "smart" method.
(Too long for comment)
Here is a fast way to evaluate the indefinite integral.
Sub $ t=Bx$,
Then,
\begin{align*}
\int \frac 1{1+x^4}\,dx&=\frac 12\left(\int \frac {1+x^2}{1+x^4}\,dx+\int \frac {1-x^2}{1+x^4}\,dx\right)\\
&=\frac 12\left(\int \frac {\color{red}{\left(\frac 1 {x^2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2419814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Laurent Series - What am I doing wrong? I want to obtain the Laurent series around the origin of:
$$ f(z) = \frac{1}{z^2 \sinh z}$$
My plan is to obtain first the laurent series of: $f(z) = \frac{1}{\sinh z}$ and then divide the terms by $z^2$
So I know that: $\frac{1}{\sinh z} = \ln(z + \sqrt{1+z^2}) = \ln(\frac{z}{\... | It's the inverse of the hyperbolic sine that has that expression (yet another reason to write the inverse as $\arg\sinh{z}$). $1/\sinh{z}=\operatorname{csch}{z}$ is
$$ \frac{2}{e^z-e^{-z}} = 2\left( 2z+\frac{2}{3!}z^3 + \frac{2}{5!}z^5 + \dotsb \right)^{-1}, $$
and then take out a factor of $z$ and use the binomial exp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?
How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?
I started my proof with Suppose $\epsilon > 0$ and $m>?$ because I plan to do scratch work and fill in.
I started with our conergence definition, i.e. $\lvert a_n -... | $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14
$
$\begin{array}\\
\dfrac{\sqrt {n^2 +2}}{4n+1}
&\gt \dfrac{n}{4n+1}\\
&= \dfrac{n+1/4-1/4}{4n+1}\\
&=\dfrac14- \dfrac{1}{4(4n+1)}\\
\end{array}
$
so
$\dfrac{\sqrt {n^2 +2}}{4n+1}-\dfrac14
\gt - \dfrac{1}{4(4n+1)}
$.
Since
$(n+1/n)^2
=n^2+2+1/n^2
\gt n^2+2
$,
$\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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Find all triples $(a, b, c) \in \mathbb{R} : ab + c = bc + a = ca + b = 4$ Any help on this question would be great. Ive found that (1,1,3), (3,1,1), (1,3,1) but am confused on how to find more anymore (if there are any).Any help would be appreciated.
| If $a b+c=4$ then $a=\frac{4-c}{b}$ assuming that $b\not=0$ and then $b c+a =4$ which is $b c+\frac{4-c}{b}=4$ so $c(b-\frac{1}{b})+\frac{4}{b} = 4$ so $c= \frac{4-\frac{4}{b}} { b-\frac{1}{b}} = \frac{4}{b+1}$ assuming that $b\not -1,1$ and to the final equation we get that $c a+b=4$ which is $c(\frac{4-c}{b})+b=4$ wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Probability that a point is closer to a side than a diagonal So I have a rectangle in which a point is randomly chosen. One side is $a$ and the other is $b=a\sqrt3$. I am supposed to find the probability that a point is closer to a side than to the closest diagonal.
I have found the probability that the point is close... | The red areas are of points closer to a side of the rectangle. Its area is the sum of the triangles, two acutangle and two obtusangle.
The two acutangle are each $1/3$ of the equilateral triangle having side $1$ and half diagonals. A diagonal is $\sqrt{3+1}=2$ so half diagonal is $1$ (that's why they are equilateral).... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Prove $\lim_{x\to 2} \sqrt{x^2+5} = 3$ As stated in the title, I need to prove that $\lim_{x\to 2} \sqrt{x^2+5} = 3$ using only the precise definition of a limit.
For any given $\varepsilon \gt 0$, there exists a $\delta = $
Such that $0 \lt \lvert x-2 \rvert \lt \delta \Rightarrow \lvert \sqrt{x^2+5} \rvert \lt \vare... | You're doing fine! So, in order to have$$\left|\sqrt{(x-2)^2+4(x-2)+9}-3\right|<\varepsilon,$$all you need is to have\begin{multline}\left|\sqrt{(x-2)^2+4(x-2)+9}-3\right|.\left|\sqrt{(x-2)^2+4(x-2)+9}+3\right|<\\<\varepsilon.\left|\sqrt{(x-2)^2+4(x-2)+9}+3\right|,\end{multline}or, in other words,$$\bigl|(x-2)^2+4(x-2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Finding a general formula for two squares whose sum = 2 Presumably, a formula for u & v that generates all the possible u & v such that $u^2 + v^2 = 2$.
With the limitation that the problem must be solved without using roots or fractional powers assuming that u & v are only required to be real numbers.
I'm not exactly... | As $(1-t^2)^2+(2t)^2=(1+t^2)^2$, we can parametrize $a^2+b^2=1$ by letting $a=\frac{1-t^2}{1+t^2}$, $b=\frac{2t}{1+t^2}$.
Now to get to $u^2+v^2=2$ without introducing $\sqrt 2$, just take $u=a+b$, $v=a-b$ (which makes $u^2+v^2=a^2+2ab+b^2+a^2-2ab+b^2=2(a^2+b^)=2$), so take
$$\tag1 u = \frac{1-t^2+2t}{1+t^2},\qquad v ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to determine the $\lim_{n\to \infty} \frac{1+2^2+\ldots+n^n}{n^n}=1$. I stuck to do this,
$$\lim_{n\to \infty} \frac{1+2^2+\ldots+n^n}{n^n}=1.$$
The only thing I have observed is $$ 1\le \lim_{n\to \infty} \frac{1+2^2+\ldots+n^n}{n^n}$$ I am unable to get its upper estimate so that I can apply Sandwich's lemma.
| $$\lim_{n \rightarrow \infty} \frac{1^1 + 2^2 + \dots + (n-1)^{n-1} + n^n}{n^n} = \lim_{n \rightarrow \infty} \left(\frac{1^1}{n^n} + \frac{2^2}{n^n} + \dots + \frac{(n-1)^{n-1}}{n^n} + 1 \right) $$
Let's examine that second to last term:
$$
\begin{align}
\lim_{n \rightarrow \infty} \frac{(n-1)^{n-1}}{n^n} &= \lim_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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find the total number of possible way to reach to a particular sum suppose you have given a sum like : 5.
we have to find the total number of possible way to reach to 5.
for example
1 + 1 + 1 + 1 + 1 = 5
2 + 1 + 1 + 1 = 5
1 + 2 + 1 + 1 = 5
1 + 1 + 2 + 1 = 5
1 + 1 + 1 + 2 = 5
2 + 2 + 1 = 5
2 + 1 ... | By Stars and Bars the number of ordered $k-$tuples of positive integers that sum to $n$ is $\binom {n-1}{k-1}$. It follows that your answer is $$\sum_{k=1}^n\binom {n-1}{k-1}=\sum_{k=0}^{n-1}\binom {n-1}k=2^{n-1}$$
Direct Proof: List $n$ $*'s$ with blanks between them. There are $n-1$ blanks. your decompositions ar... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\mathbb{Q}$ is dense. Let $a$ and $b$ be the real numbers with $a<b$. Prove that there are integers $m$ and $n\neq 0$ so that
\begin{align*}
a<\frac{m}{n}<b.
\end{align*}
Is my attempt correct?
Proof
Let $x=(b-a)>0$ and $y=1>0$, then there is a positive integer $n$ such that
\begin{align*}
n(b-a)>1\implies ... | I have included some minor corrections in red.
If $na > 0$, there is a least integer $m$ such that $na < m$. This means the below is true
$$m-1 \color{red}{\leq} na < m$$
\begin{align*}
na<m \color{red}{\leq}na+1<nb\implies na<m<nb
\end{align*}
Similar for $na=0$.
If $na < 0$ and therefore bounded above, there is a gre... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the following nonlinear system of equations:$ xy+xz+yz=12 , xyz=2+x+y+z$
Solve the following system of equations in $\Bbb R^+$:
$$
\left\{
\begin{array}{l}
xy+yz+xz=12 \\
xyz=2+x+y+z\\
\end{array}
\right.
$$
I did as follows.
First equation yields to: $(x+y+z)^2-x^2-y^2-z^2=24$. Now, using second equat... | I will show that the only positive solution is indeed $(2,2,2).$ $$\dfrac{xyz}{xy+yz+zx} = \dfrac{1}{6}+\dfrac{x+y+z}{12} = \dfrac{\sqrt{xy+yz+zx}}{12\sqrt{3}}+\dfrac{x+y+z}{12}\geq\dfrac{xyz}{4(xy+yz+zx)}+\dfrac{3xyz}{4(xy+yz+zx)} = \dfrac{xyz}{xy+yz+zx}$$, by AM-GM inequalities provided that all of them are non-negat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Classification Of Conic Section
Classify $$x^2-4xy+y^2+8x+2y-5=0$$
So the eigenvalues are $\lambda_{1}=3,\lambda_{2}=-1$ so the eigenvectors are $v_{1}=\begin{pmatrix} \frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}}\end{pmatrix}$ and $v_{2}=\begin{pmatrix} \frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{pmatrix}$
So We have $$\be... | You don't have to find the reduced equations to classify the conic:
*
*$\;\begin{vmatrix}1&-2&4\\-2&1&1\\4&1&-5\end{vmatrix}\ne 0$, so the conic is non-degenerate.
*The quadratic form $x^2-4xy+y^2$ has signature $(1,1)$:
$$x^2-4xy+y^2=(x-2y)^2-3y^2.$$
As a conclusion, the conic is a hyperbola.
For an ellipse, th... | {
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"url": "https://math.stackexchange.com/questions/2427693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the inequality satisfying the given conditions. (Multi-choice-multi-correct question):
If the roots of the equation $ax^2+bx+c=0$, $a\ne0$ are imaginary and $a+c<b$,($a,b,c$ are real numbers) then:
*
*$a+4c< 2b$
*$a+b+c<0$
*$4a+c<2b$
*$4a+c< 2b$ if $a<0$ and $4a+c>2b$ if $a>0$
My attempt... | Roots are imaginary numbers $\implies b^2- 4ac < 0\implies b^2 < 4ac \le(a+c)^2\implies b^2 - (a+c)^2 < 0\implies (b-a-c)(b+a+c) < 0$
Since $b-a-c > 0$, it must be true that $a+b+c < 0$.
Thus $2)$ is true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving convergence of series $\sum_{k=1}^n {\sqrt{k^3+1}-\sqrt {k^3-1}}$ How to prove convergence of the following?
$$\sum_{k=1}^n {\sqrt{k^3+1}-\sqrt {k^3-1}}$$
Thanks!
| $\sqrt{k^3+1}-\sqrt {k^3-1}=\frac{(\sqrt{k^3+1}-\sqrt {k^3-1})(\sqrt{k^3+1}+\sqrt {k^3-1})}{(\sqrt{k^3+1}+\sqrt {k^3-1})}=\frac{2}{\sqrt{k^3+1}+\sqrt {k^3-1}}\leq \frac{2}{\sqrt{k^3+1}}\leq \frac{2}{\sqrt{k^3}}=\frac{2}{k^{1+\frac{1}{2}}}$
Then apply comparison test.
| {
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"answer_id": 1
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$\{a,b,c\}\subset \Bbb R$, $a\not =b$ and $a^2(b+c)=b^2(a+c)=2010$. Find $c^2(a+b)$
Consider that $\{a,b,c\}\subset \Bbb R$, with $a\not =b$, and it is known that $a^2(b+c)=b^2(a+c)=2010$.
Find $c^2(a+b)$.
A question from the third phase of OBM 2010 (Brazilian Math Olympiad). Sorry if it is a duplicate. My developmen... | We have
\begin{eqnarray*}
a^2b +a^2 c =2010 \\
a b^2+b^2 c =2010.
\end{eqnarray*}
Subtract these equations ( and use $ a \neq b$) we have $ab+bc+ca=0$ so
\begin{eqnarray*}
c= - \frac{ab}{a+b}. \\
\end{eqnarray*}
Now multiply the first equation by $b^2$ and the second by $a^2$ and subtract ( and again use $ a \neq b... | {
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Limit as $x$ approaches infinity of a ratio of exponential functions. The problem appears as follows:
Use algebra to solve the following:
$$\lim_{x\to \infty} \frac{2^{3x+2}}{3^{x+3}}$$
The result is infinity which makes intuitive sense but I can’t get it to yield algebraically.
Attempt:
$$\frac{2^{3x+2}}{3^{x+3}} = \f... | Hint: $\displaystyle\;\frac{2^{3x+2}}{3^{x+3}}=\frac{2^2}{3^3} \cdot \frac{\left(2^3\right)^x}{3^x}=\frac{4}{27} \cdot \left(\frac{8}{3}\right)^x \gt \frac{4}{27} \cdot 2^x\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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To prove $\lim_{x\to 0}\frac{e^x-e^{x\cos x}}{x-\sin x}=3$ I came across this question to prove the given limit
$$\lim_{x\to 0}\frac{e^x-e^{x\cos x}}{x-\sin x}=3$$
First I tried using LHospital's rule directly.
Then I tried using expansion of $e^x$ and then using LHospital's rule but I am getting stuck.
| I solved it using Taylor expansion:
First we calculate some ugly derivatives:
$$(e^{x\cos(x)})' = (\cos(x) - x\sin(x))e^{x\cos(x)},$$
$$(e^{x\cos(x)})'' = [(-2\sin(x) - x\cos(x))+(\cos(x)-x\sin(x))^2] e^{x\cos(x)}$$
and
$$(e^{x\cos(x)})''' = [(-3\cos(x) + x\sin(x))+(\cos(x)-x\sin(x))^3+2(\cos(x)-x\sin(x))(-2\sin(x)-x\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2436145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help clarifying the steps to find the derivative of $y=(3x+1)^3(2x+5)^{-4}$ For the problem $y=(3x+1)^3(2x+5)^{-4}$ do I use the chain, quotient and product rules? If so how do I know what parts to break up and where the rules apply? For instance would I consider $f(x)$ to be $(3x+1)^3$ and $g(x)$ to be $(2x+5)^{-4}$... | Instead of the "quotient rule" you could write the function as $y= (3x+ 1)^3(2x+ 5)^{-4}$ and use the product rule: $y'= [(3x+ 1)^3]'(2x+ 5)^{-4}+ (3x+ 1)^3[(2x+ 5)^{-3}]'$. Now, using the "chain rule", $[(3x+ 1)^3]'= 3(3x+ 1)^2(3)= 9(3x+ 1)^2$ and $[(2x+ 5)^{-4}]'= (-4)(2x+ 5)^{-5}(2)= -8(2x+ 5)^{-5}$ so that $y'= 9(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2436600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof by induction (Combinatorics) This is my first question on here so please bear with me, thank you.
Prove that for all positive integers $n$
$$2(1+2+...+n)^4 = (1^5 + 2^5 +...+ n^5) + (1^7 + 2^7 +...+ n^7)$$
After establishing the base case, I proceeded to the induction:
$2(1+2+...+(n+1))^4 = (1^5 + 2^5 +...+ (... | Note that
$$1+2+\cdots+n =\frac{n(n+1)}{2},$$ and $$1+2+\cdots+n+1 =\frac{(n+1)(n+2)}{2}.$$
Therefore, the left-hand side of the last equation is
$$\frac{(n+1)^4}{8}\left((n+2)^4 -n^4\right)=\frac{(n+1)^4}{8}\left((n+1+1)^4 -(n+1-1)^4\right).$$
But $$(a+1)^4-(a-1)^4 = ((a+1)^2)^2-((a-1)^2)^2\\=((a+1)^2-(a-1)^2)((a+1)^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating limits without L'Hopital's I need to evaluate without using L'Hopital's Rule. They say you don't appreciate something until you lose it. Turns out to be true.
*
*$\lim_{x \to 0} \frac{\sqrt[3]{1+x}-1}{x}$
*$\lim_{x \to 0} \frac{\cos 3x - \cos x}{x^2}$
I have tried to "rationalise" Q1 by using the identi... | *
*We use the binomial theorem: $$\frac{\sqrt[3]{1+x}-1}{x}=\frac{(1+x)^{1/3}-1}{x}=\frac{(1+\frac1{3}x+\frac1{9}x^2+\ldots)-1}{x}=\frac{(\frac1{3}x+\frac1{9}x^2+\ldots)}{x}=(\frac1{3}+\frac1{9}x+\ldots)\implies\lim_{x\to0}\frac{\sqrt[3]{1+x}-1}{x}=\frac1{3}$$
2.Again, we use series expansion for $\cos$ function. We... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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Calculate the line integral $I(a,b)=\int_{x^2+y^2=R^2}\limits\ln\frac{1}{\sqrt{(x-a)^2+(y-b)^2}} ds\quad(a^2+b^2\ne R^2).$ Calculate the line integral $$I(a,b)=\int_{x^2+y^2=R^2}\limits\ln\frac{1}{\sqrt{(x-a)^2+(y-b)^2}} ds\quad(a^2+b^2\ne R^2).$$
The parametrized integral path can be given as
$$x=R\cos t,y=R\sin t,t\i... | Hint : $$W=C\sin(t)+D\cos(t)=(\sqrt{C^2+D^2})(\frac{C}{\sqrt{C^2+D^2}}\sin(t)+\frac{D}{\sqrt{C^2+D^2}}\cos(t))$$
$$W=\sqrt{C^2+D^2}\sin(t+\phi)=E\sin(t+\phi)$$
with $\cos(\phi)=\frac{C}{\sqrt{C^2+D^2}}$ and $ \sin(\phi)=\frac{D}{\sqrt{C^2+D^2}}$
so $$I=\int_0^{2\pi} \ln(A+C\sin(t)+B\cos(t))\text{d}t=\int_0^{2\pi} \ln(A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$ The problem is the following: For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$
This is what I have at the moment:
Let's call $d = \gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$
Then $d \ \vert \ 3^{1... | From $d|102$ you have $d\in\{1,2,3,17,34,51,102\}$ and because $d|3^{16}2a+10$ you have $d\notin \{3,51,102\}$. Clearly $2|d$ iff $2|a$. And with $17$ you have a lot of different subcases to deal with. If $17|a$ then $d\ne 17$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Three dice are rolled simultaneously. In how many different ways can the sum of the numbers appearing on the top faces of the dice be $9$?
Three dice are rolled simultaneously. In how many different ways can the sum of the numbers appearing on the top faces of the dice be $9$?
What I did:
I know that the maximum valu... | In these types of problems, it is not too hard to just consider case by case. Let's list out all the possibilities and how many ways to reorganize them:
$$1,2,6 \rightarrow 3!=6 \text{ ways}$$
$$1,3,5 \rightarrow 3!=6 \text{ ways}$$
$$1,4,4 \rightarrow 3!/2!=3 \text{ ways}$$
$$2,2,5 \rightarrow 3!/2!=3 \text{ ways}$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Approximate $e$ with $4$-digit chopping and $5$-degree Maclaurin-Taylor polynomial
Number $e$ is defined as
$$ e=\sum^\infty_{n=0} \frac{1}{n!}.$$
Use four-digit chopping arithmatic to compute the following
*
*Approximations to $e$
$$ e\approx \sum^5_{n=0} \frac{1}{n!}$$
Attempt:
\begin{align}
\sum^5_... | That is quite easy: you forgot the $1/ 0!$ in your calculations. Then you get something like $\mathrm e \approx2.716$. Also rounding in the end of your calculation might be the best.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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reduce a rational expression to lowest terms $$\frac{x^2 + 4x + 3}{x^2 - 2x - 3}$$
I'm coming up with $\frac{x + 3}{ x - 3}$ however it seems wrong.
$$x^2 + 4x + 3 = (x + 3) ( x + 1)$$
$$x^2 - 2x - 3 = (x - 3 )(x + 1) $$
cancel out $x + 1$ and left with $\frac{x + 3}{x - 3}$
| It's correct,
$$\require{cancel} \frac{x^2 + 4x + 3} {x^2 - 2x - 3} = \frac{(x+3)\cancel{(x+1)}} {(x-3) \cancel{(x+1)}}$$
Provided $x \neq 3$, $x \neq -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2444187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I solve $\int\frac{\sqrt{x}}{\sqrt{x}-3}\,dx?$ $$
\int \frac{\sqrt{x}}{\sqrt{x}-3}dx
$$
What is the most dead simple way to do this?
My professor showed us a trick for problems like this which I was able to use for the following simple example:
$$
\int \frac{1}{1+\sqrt{2x}}dx
$$
Substituting:
$u-1=\sqrt{x}$
be... | With $\sqrt x = u$
Let $\sqrt x = u$, then we have $x = u^2$ and $\mathrm{d}x=2u\,\mathrm{d}u$.
\begin{align}
\int\frac{u}{u-3}\times 2u\,\mathrm{d}u
&= 2\int\frac{u^2}{u-3}\,\mathrm{d}u\\
&= 2\int\frac{u^2-9+9}{u-3}\,\mathrm{d}u\\
&= 2\int\frac{u^2-9}{u-3}\,\mathrm{d}u + 18\int\frac{1}{u-3}\,\mathrm{d}u\\
&= 2\int(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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If α and β are the roots of the equation $3x^2+5x+4=0$, find the value of $α^3+β^3$ If α and β are the roots of the equation $3x^2+5x+4=0$, find the value of $α^3+β^3$
How can I factorize the expression to use the rule of sum and product of roots?
The answer is $\frac{55}{27}$
| expanding $$\frac{1}{216} \left(-5-i
\sqrt{23}\right)^3+\frac{1}{216} \left(-5+i
\sqrt{23}\right)^3$$ we obtain $$\frac{55}{27}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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How to compute symmetrical determinant I'm learning of determinants and am trying to find a trick to compute this one
\begin{pmatrix}
2 & 1 & 1 & 1 & 1\\
1 & 3 & 1 & 1 & 1\\
1 & 1 & 4 & 1 & 1\\
1 & 1 & 1 & 5 & 1\\
1 & 1 & 1 & 1 & 6
\end{pmatrix}
I expanded it out and got $349$ but I feel there must b... | Use the following rules:
*
*Adding a multiple of one row to another row does not change the determinant.
*If $B$ is obtained by multiplying a row of $A$ by a constant $c$ then $\det B=c \det A$
Start by subtracting the first row from each of the other rows, we get $$A=\begin{pmatrix}
2 & 1 & 1 & 1 & 1\\
-1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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The $n$th prime number is the smallest difference of products of distinct previous primes. For $n\geq 2$, the smallest positive prime number formed by multiplying $2, 3, \dots, p_n$ and up to $1$ subtraction / addition
$$
2 + 3 = 5 \\
2\cdot 5 - 3 = 7 \\
3 \cdot 7 - 2\cdot 5 = 11 \\
5\cdot 11 - 2\cdot 3 \cdot 7 = 13 \\... | You stopped your examples one step too early. The smallest number $>1$ obtainable from $2,3,\ldots,17$ is
$$2\cdot 5\cdot 7\cdot 11-3\cdot 13\cdot 17=107.$$
(I.e., $663$ and $770$ are the factors of $N:=2\cdot 3\cdot 5\cdot \ldots\cdot 17$ that are closest to $\sqrt N$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all n for which $2^n + 3^n$ is divisible by $7$? I am trying to do this with mods. I know that:
$2^{3k+1} \equiv 1 \pmod 7$, $2^{3k+2} \equiv 4 \pmod 7$, $2^{3k} \equiv 6 \pmod 7$ and $3^{3k+1} \equiv 3 \pmod 7$, $3^{3k+2} \equiv 2 \pmod 7$, $3^{3k} \equiv 6 \pmod 7$, so I thought that the answer would be when $... | Powers of $2$ modulo $7$ are $1,2,4,1,2,4,\cdots$ repeating every $3$ and powers of $3$ modulo $7$ are $1,3,2,6,4,5,1,\cdots$ repeating every $6$
Adding these sequence gives $n \equiv 3 \pmod{6}$ so $7$ divides $2^n+3^n$ if and only if $n=3+6m$ (for $ m \in \mathbb{N}_0$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Fill out a group table with 6 elements Let $G=\{0,1,2,3,4,5\}$ be a group whose table is partially shown below:
\begin{array}{ c| c | c | c | c |c|c|}
* & 0& 1 & 2 & 3& 4 & 5\\
\hline
0 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
1 & 1 & 2 & 0 & 4 & & \\
... | Suppose you had got this far and needed to fill out the rest:
\begin{array}{ c| c | c | c | c |c|c|}
* & 0& 1 & 2 & 3& 4 & 5\\
\hline
0 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
1 & 1 & 2 & 0 & 4 & 5 & 3 \\
\hline
2 & 2 & 0 & 1 & 5 & 3 & 4 \\
\hline
3 & 3 & 5 & 4 & ? \\
\hline
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve $2\log_bx + 2\log_b(1-x) = 4$ I need to solve $$2\log_bx + 2\log_b(1-x) = 4.$$
I have found two ways to solve the problem. The first (and easiest) way is to divide through by $2$:
$$\log_bx + \log_b(1-x) = 2.$$ Then, combine the left side: $$\log_b[x(1-x)] = 2,$$ and convert to the equivalent exponential form, $$... | You are correct to transform $2\log_b x + 2\log_b (1-x) = 4$ into $x^2-x+b^2=0$.
These are the restrictions implicated by the original equation:
*
*$b$ is a positive number not equal to $0$ nor $1$
*$x>0 \iff 0<x$
*$1-x>0 \iff 1>x \iff x<1$
I personally felt that it was easiest to wrap my mind around this using ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2458109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $(a+2b+\frac{2}{a+1})(b+2a+\frac{2}{b+1})\geqslant16$ if $ ab\geqslant1$ $a$, $b$ are positive real numbers.
Show that $$\left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right)\geqslant16$$
if
$$ ab\geqslant1$$
Should I use AM-GM inequality? If yes, where?
| Expand to get:
$$ab + 2a^2 + \frac{2a}{b+1} + 2b^2 + 4ab + \frac{4b}{b+1} + \frac{2b}{a+1} + \frac{4a}{a+1} + \frac{4}{(a+1)(b+1)} \ge 9 + \frac{2a^2 + 2a + 4ab + 4b + 2b^2 + 2b + 4ab + 4a + 4}{(a+1)(b+1)} \ge 9 + 7 = 16$$
The last inequality follows as:
$$2a^2 + 2a + 4ab + 4b + 2b^2 + 2b + 4ab + 4a + 4 \ge 7(a+1)(b+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve system of equations: $\sqrt2 \sin x = \sin y, \sqrt2\cos x = \sqrt5\cos y$ $\sqrt2 \sin x = \sin y, \sqrt2\cos x = \sqrt5\cos y$
I tried to use tangent half-angle substitution, tried to sum these two equations and get this
$\sin(x+ \pi/4) = \sqrt3 / \sqrt2 \sin(y+\tan^{-1}(\sqrt5))$
I stuck.
Any Help is appreciat... | Square both equations, then add them up. $2 \sin^2 x = \sin^2 y$ and $2 \cos^2 x = 5 \cos^2 y$, so $2 = \sin^2 y + 5 \cos^2 y$, and hence $1 = 4 \cos^2 y$. From here, $\cos y = \pm \frac1{2}$, and you should substitute back in to check if both work. You'll also get multiple values of $y$ from this, which you need to ch... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For what pair of integers $(a,b)$ is $3^a + 7^b$ a perfect square. Problem: For what pair of positive integers $(a,b)$ is $3^a + 7^b$ a perfect square.
First obviously $(1,0)$ works since $4$ is a perfect square, $(0,0)$ does not work, and $(0,1)$ does not work, so we can exclude cases where $a$ or $b$ are zero for the... | The other case is $1 + 2 \cdot 7^x = 3^y,$ or
$$ 3^y - 1 = 2 \cdot 7^x. $$
Assume $x \geq 1.$ Then both sides are divisible by $7,$ giving
$$ 3^y \equiv 1 \pmod 7, $$
$$ y \equiv 0 \pmod 6. $$ Then $3^y - 1$ is divisible by
$$ 3^6 - 1 = 728 = 8 \cdot 7 \cdot 13. $$
However, then $2 \cdot 7^x$ is divisible by $13... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find coordinates of area's center of mass. So we have area $D:{(x^{2}+y^{2})^{2}=2a^{2}xy, x>0, y>0}$. $x= r \cdot cos(\varphi)$ and $y= r \cdot sin(\varphi)$
I want to find the center of mass by formula $y_{c} = \frac{M_{ox}}{m}$, where $M_{ox}=\iint_{D}(\gamma \cdot y \cdot dxdy)$ gamma is density and $m=\iint_{D}(\g... | \begin{align}
M_x
&= \int_0^{\pi/2}d\theta\int_0^{a\sqrt{\sin2\theta}}\gamma r^2\sin\theta\,dr \\
&= \gamma\int_0^{\pi/2}d\theta \dfrac13r^3\sin\theta\Big|_0^{a\sqrt{\sin2\theta}} \\
&= \gamma a^3\dfrac{2\sqrt{2}}{3}\int_0^{\pi/2}\sin^{5/2}\theta\cos^{3/2}\theta d\theta \\
&= \gamma a^3\dfrac{\sqrt{2}}{3}2\int_0^{\pi/2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find area of triangle ABD Let ABC be a right-angled triangle $B=90$. Let in a triangle pick a point $D$ inside the triangele such that $AD= 20, DC=15, DB=10$ and $AB=2BC$. What is the area of triangle $ABD$?
Thanks!
| Let $BC=a$ and $AB=2a$.
With $B$ as origin let the co-ordinates of $D$ be $(x,y)$. The required area of triangle $ABD$ is $ax$.
Applying Pythagoras' Theorem $$10^2=x^2+y^2 \text{ ... (1)}$$ $$15^2=(a-x)^2+y^2 \text{ ... (2)}$$ $$20^2=x^2+(2a-y)^2 \text{ ... (3)}$$
Subtract (1) from (2) and (3) in turn : $$125=a(a-2x... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving inhomogeneous heat equation Solve the inhomogeneous heat equation $u_t=c^2u_{xx} +\sin(5\pi x)$ for all $0 < x < 1, t > 0$ subject to homogeneous Dirichlet boundary conditions $u(t,0)=u(t,1)=0$ and initial condition $u(0,x) =4\sin(3\pi x)+9\sin(7\pi x)$.
Solved it as
$$v(x,t)=\sum_{n=1}^{\infty}\bigg[\frac{\int... | In order to solve non-homogeneous heat equations, assume that $w(x,t)=u(x,t)+\phi(x)$, where $\phi$ is a function yet to be determined.
So, we've $u_{xx} = w_{xx} - \phi''$ and $u_{t} = w_{t}$. If we substitute this in to our equation, then we have to solve the new problem
$\begin{cases}
w_t - c^2 w_{xx} - c^2 \phi'' -... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Semifactorial Identity I was wondering if anyone had any insight on how to prove the following identity:
For all $m \in \mathbb{N}$ $$ \frac{1}{2m-1} + \frac{2m-2}{(2m-1)(2m-3)} + \frac{(2m-2)(2m-4)}{(2m-1)(2m-3)(2m-5)} + \cdots + \frac{(2m-2)!!}{(2m-1)!!} = 1$$
I attempted to rewrite and simplify the left hand side ... | Starting from
$$\sum_{k=1}^m \frac{(2m-2)!!/(2m-2k)!!}{(2m-1)!!/(2m-2k-1)!!}$$
and using
$$(2n)!! = 2^n n!
\quad\text{and}\quad
(2n-1)!! = \frac{(2n)!}{2^n n!}$$
we get for the sum
$$\sum_{k=1}^m \frac{2^{m-1} (m-1)!}{2^{m-k} (m-k)!}
\frac{(2m-2k)! / 2^{m-k} / (m-k)!}{(2m)! / 2^{m} / m!}
\\ = \frac{(m-1)! m!}{(2m)!}
\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2470085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How do we minimize $\left(x+a+b\right)\left(x+a-b\right)\left(x-a+b\right)\left(x-a-b\right)$? Find the minimum value of the following function, where a and b are real
numbers.
\begin{align} f(x)&=\left(x+a+b\right)\left(x+a-b\right)\left(x-a+b\right)\left(x-a-b\right) \end{align}
Note: The solution should not contain ... | Just another solution in the same spirit as lab bhattacharjee's answer.
$$ f(x)=\left(x+a+b\right)\left(x+a-b\right)\left(x-a+b\right)\left(x-a-b\right) $$
$$f(x)=x^4-2\left( a^2+1\right) x^2+(a^2-1)^2$$ Complete the square
$$f(x)=\left(x^2-(a^2+1)\right)^2-(a^2+1)^2+(a^2-1)^2=\left(x^2-(a^2+1)\right)^2-4a^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2472757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving that $\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}> \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Prove that $\dfrac{ab}{c^3}+\dfrac{bc}{a^3}+\dfrac{ca}{b^3}> \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$, where $a,b,c$ are different positive real numbers.
First, I tried to simplify the proof statement but I got an even mo... | \begin{eqnarray*}
a^4(b^2-c^2)^2+b^4(a^2-c^2)^2+c^4(b^2-a^2)^2+a^2b^2c^2((a-b)^2+(c-b)^2+(a-c)^2) \geq 0.
\end{eqnarray*}
Rearrange to
\begin{eqnarray*}
a^4b^4+b^4c^4+c^4a^4 \geq a^2b^2c^2(ab+bc+ca).
\end{eqnarray*}
Now divide by $a^3b^3c^3$ and we have
\begin{eqnarray*}
\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{ b^3} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Identity with sums of binomials Choose any natural numbers $a \ge 1$ and $b \ge 1$. Then prove that the identity holds:
\begin{equation*}
\frac{a b }{(a+b)^2} {\Large(}\tbinom{a+b}{a} {\Large)}^2 = \\
\sum_{n=1}^{a}\sum_{m=1}^{b} {\Huge[} {\Large(}\tbinom{n+m-2}{n-1} {\Large)}^2 - \frac{ m (a-n)(b-m) }{n(a-n+b-m... | The answer can be derived from two answers to this question: There, the topic is that the partial derivatives of
$f(x,y)=(1+x^2+y^2-2x-2y-2xy)^{-\frac{1}{2}}$
satisfy
\begin{equation*}
\frac{\partial ^{m+n}}{\partial x^{m} \partial y^{n}} f (0,0)= (n+m)! \tbinom{n+m}{n}
\end{equation*}
Now this relation is shown direc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2474963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Proving equivalence of norms in $\mathbb{R}^2$
Let $\lVert \cdot \rVert_*:\mathbb{R}^2 \to \mathbb{R}, (x,y) \rightarrow \sqrt{x^2+2xy+3y^2}$ be a norm.
How can I find to constants $k,K \in \mathbb{R}^{>0}$ so that the following equivalence is given:$$k\lVert (x,y) \rVert_2 \leq \Vert (x,y) \rVert_* \leq K\Vert (x,y) ... | Note that
$$x^2 + 2xy + 3y^2 = \begin{pmatrix} x \\ y \end{pmatrix}^T \begin{pmatrix} 1 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}.$$
The eigenvalues of the matrix have sum 4 and product 2, so they are both stricly positive; let us call them $\lambda_1$ and $\lambda_2$. Since the matrix is symmet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Integrate $\arctan{\sqrt{\frac{1+x}{1-x}}}$ I use partial integration by letting $f(x)=1$ and $g(x)=\arctan{\sqrt{\frac{1+x}{1-x}}}.$ Using the formula:
$$\int f(x)g(x)dx=F(x)g(x)-\int F(x)g'(x)dx,$$
I get
$$\int1\cdot\arctan{\sqrt{\frac{1+x}{1-x}}}dx=x\arctan{\sqrt{\frac{1+x}{1-x}}}-\int\underbrace{x\left(\arctan{\sq... | Hint: make the substitution $x = \cos \theta$ with $0 \le \theta \le \pi$. Then $1 + x = 2 \cos^2(\theta/2)$, $1 - x = 2 \sin^2(\theta/2)$, so $\sqrt{\frac{1+x}{1-x}} = |\cot(\theta/2)|$. But by the choice of range of $\theta$, $\cot(\theta/2)$ is positive. So $\arctan \sqrt{\frac{1+x}{1-x}} = \frac{\pi}{2} - \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2477162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Proving limits without limit theorems I have to prove the following limit without using any limit theorems. I can only do so by using the Archimedean Proprety and the definition of a limit.
I have to prove:
$$\lim_{n\to\infty} \frac{n^2 - 2}{3n^2 + 1} = \frac13$$
| The Archimedean Property can be used to show that
$\tag 1 \displaystyle \lim_{n \to \infty}\frac{1}{n^2} = 0$
It is easy to show that
$\tag 2\frac{n^2 - 2}{3n^2 + 1} \lt \frac{1}{3} \text{ for all } n \ge 0$
The following statements are all equivalent for any $0 \lt \varepsilon \lt \frac{7}{3}$ when $n \gt 0$:
$\tag 3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2486254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Prove that $\binom{ap}{p} \equiv a \pmod{ap}$
Let $N = ap$ be a composite positive integer, where $a > 1$ is a positive integer and $p$ is a prime. Prove that $$\binom{ap}{p} \equiv a \pmod{ap}.$$
We have $$\binom{ap}{p} = \dfrac{ap(ap-1) \cdots (ap-(p-1))}{p!} = \dfrac{a(ap-1)(ap-2) \cdots (ap-(p-1))}{(p-1)!}.$$ If ... | using Lucas'Theorem, suppose $a\equiv k \pmod p $, we will have:
$\binom{ap}{p} \equiv\binom{k}{1} =k \equiv a \pmod p $
We also have
$\binom{ap}{p} = \dfrac{a(ap-1)(ap-2) \cdots (ap-(p-1))}{(p-1)!} = a\binom{ap-1}{p-1} \equiv 0 \pmod a$
Let $\binom{ap}{p}=Zp+a$, so $a|Z$, so
$Zp+a=Kap+a \equiv a \pmod {ap} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2486493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Rationalize denominator: $\frac{9\sqrt{2}}{1+\sqrt{2}-2\sqrt{5}-\sqrt{10}}$ $$\frac{9\sqrt{2}}{1+\sqrt{2}-2\sqrt{5}-\sqrt{10}}$$
I simplified it by making it:
$$\frac{9\sqrt{2}}{(1-\sqrt{10})(1+\sqrt{2})}$$
What do I do next? I am confused what to do because it is multiplication. All of my previous examples were either... | $$\frac{9\sqrt{2}(1+\sqrt{10})(1-\sqrt{2})}{(1-\sqrt{10})(1+\sqrt{2})(1+\sqrt{10})(1-\sqrt{2})}=$$
$$\frac{9\sqrt{2}(1+\sqrt{10})(1-\sqrt{2})}{(1-10)(1-2)}=$$
$$\sqrt{2}(1+\sqrt{10})(1-\sqrt{2})=...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$f(x)=\sqrt{\dfrac{x-\sqrt {20}}{x-3}}$, $g(x)=\sqrt{x^2-4x-12}$, find the domain of $f(g(x))$
$f(x)=\sqrt{\dfrac{x-\sqrt {20}}{x-3}}$, $g(x)=\sqrt{x^2-4x-12}$, find the domain of $f(g(x))$.
For some reason, I cannot get the correct answer. Here is what I tried.
$D_g:$
$$x^2-4x-12\ge0$$
$$(x-6)(x+2)\ge0$$
$$\boxed{(-... | Here is the complete solution:
The domain of $g(x)$:
$$x\in (-\infty, -2]\cup[6,+\infty).$$
The domain of $f(x)$:
$$\ 1)\begin{cases}\sqrt{x^2-4x-12}\le \sqrt{20} \\ \sqrt{x^2-4x-12}<3\end{cases} \text{or} \ \ 2)\begin{cases}\sqrt{x^2-4x-12}\ge \sqrt{20} \\ \sqrt{x^2-4x-12}>3\end{cases} \Rightarrow$$
$$1)\ \begin{cases... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Solve a complex number equation I am start learning complex number in my university. I don't know a lot about complex number but I know most of the facts and rules about them.
I have the equation:
$$z^{10} + (z - 1)^{10} = 0$$
It has $10$ roots and they are $z_1$, $\overline{z_1}$, $z_2$, $\overline{z_2}$,..., $z_5$, $... |
We will show
$$\sum_{k=1}^5 {\frac{1}{z_k \overline{z_k}}}=10.$$
Note that $z^{10} + (z - 1)^{10} = 0$ is equivalent to
$$\left(1-\frac{1}{z}\right)^{10}=-1=e^{-i\pi}.$$
Hence for $k=1,2,3,4,5$,
$$1-\frac{1}{z_k}=e^{i\pi(2k-1)/10}$$
which implies
$$\frac{1}{z_k \overline{z_k}}=(1-e^{i\pi(2k-1)/10})(1-e^{-i\pi(2k-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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three distinct prime factors $x$, $y$ and $10x+y$, where $x$ and $y$ are each less than 10. What is the largest possible value of $m$? The number $m$ is a three-digit positive integer and is the product of the three distinct prime factors $x$, $y$ and $10x+y$, where $x$ and $y$ are each less than 10. What is the larges... | $\{x,y\} \subset \{2,3,5,7\}$
$10x + 2$ and $10x + 5$ are not prime. $5 + 7 = 12 = 3*4;2+7 = 9=3*3 $ so $57,75,27,72$ are not prime. So possible contenders are.
$(x,y, 10x+y): = (7,3,73),(3,7,37),(5,3,53), (2,3,23)$.
$7*3*73 > 21*70 > 1400$ and not three digits.
Clearly $2< 3, 3 < 5 < 7$ and $23< 37< 53$ so $(2,3,23... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2494175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find: $\lim\limits_{n\to \infty}\arcsin \frac {1}{\sqrt{n^2+1}}+\arcsin \frac {1}{\sqrt{n^2+2}}+...+\arcsin \frac {1}{\sqrt{n^2+n}}$ Let
$$x_n=\arcsin \frac {1}{\sqrt{n^2+1}}+\arcsin \frac {1}{\sqrt{n^2+2}}+...+\arcsin \frac {1}{\sqrt{n^2+n}}$$
I would like to find $\lim\limits_{n\to \infty}x_n$
I attempted to use R... | for any $k\in\mathbb{N}$ we have
$$\arcsin \frac {1}{\sqrt{n^2+1}}\sim \frac{1}{n},\;n\to\infty$$
Indeed
$$\lim_{n\to \infty } \, \frac{ \arcsin \frac{1}{\sqrt{k+n^2}}}{\frac{1}{n}}=^{(*)}\lim_{n\to \infty } \, \frac{-\frac{n}{\sqrt{(k+n^2)^3} \sqrt{1-\frac{1}{k+n^2}}}}{-\frac{1}{n^2}}=\\=\lim_{n\to \infty } \,\frac{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2496586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solve summation of $\sum_{j=0}^{n-2}2^j (n-j)$ Question
While Solving a recursive equation , i am stuck at this summation and unable to move forward.Summation is
$$\sum_{j=0}^{n-2}2^j (n-j)$$
My Approach
$$\sum_{j=0}^{n-2}2^j (n-j) = \sum_{j=0}^{n-2}2^j \times n-\sum_{j=0}^{n-2}
2^{j} \times j$$
$$=n \times (2^{n... | $$\sum_{j=1}^{n-2}j\;2^j=2+2\times 2^2+3\times 2^3+\cdots+(n-2)2^{n-2}$$$$=[2+2^2+2^3+\cdots+2^{n-2}]+[2^2+2^3+\cdots+2^{n-2}]+[2^3+\cdots+2^{n-2}]+\cdots+2^{n-2}$$$$=2(2^{n-2}-1)+2^2(2^{n-3}-1)+2^3(2^{n-4}-1)+\cdots+2^{n-2}(2^{n-(n-1)}-1)$$ $$=(2^{n-1}-2)+(2^{n-1}-2^2)+(2^{n-1}-2^3)+\cdots+(2^{n-1}-2^{n-2})$$ $$=(n-2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability that exactly three cards in at least one of the suits in an eight card hand from a standard deck. so my solution was to...
a) find the number of hands where exactly one suit has exactly 3 cards
b) find the number of hands where exactly two suits have exactly 3 cards
c) add a) and b) then divide by c(52,8). ... | The numerator:
${13\choose 3}{13\choose 5}{13\choose 0}{13\choose 0}{4\choose 1,1,2} + {13\choose 3}{13\choose 4}{13\choose 1}{13\choose 0}{4\choose 1,1,1,1} + {13\choose 3}{13\choose 3}{13\choose 2}{13\choose 0}{4\choose 2,1,1,1} + {13\choose 3}{13\choose 3}{13\choose 1}{13\choose 1}{4\choose 2,2}+{13\choose 3}{13\ch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2498679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\cos(A+C)$ given that $A,B,C$ are angles in an acute triangle. Given that $A,B,C$ be angles in an acute triangle.
If $(5+4\cos A)(5-4\cos B)=9$ and $(13-12\cos B)(13-12\cos C)=25$
find $\cos(A+C)$.
I know $A+B+C=180^\circ$ and $\cos B=-\cos(A+C)$ and what next?
| Thanks for the challenging problem! I enjoyed working through it.
The tangent half-angle substitution works really well here. Let $a = \tan ({A\over 2})$, $b = \tan({B\over2})$, $c = \tan({C\over2})$ then $a,b,c > 0$ and
$$ \cos A = \frac{1-a^2}{1+a^2}, \quad \cos B = \frac{1-b^2}{1+b^2}, \quad \cos C = \frac{1-c^2}{1+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find Probability that even numbered face occurs odd number of times A Die is tossed $2n+1$ times. Find Probability that even numbered face occurs odd number of times
First i assumed let $a,b,c,d,e,f$ be number of times $1$ occurred, $2$ occurred and so on $6$ occurred respectively. Then we have
$$a+b+c+d+e+f=2n+1$$ i... | First of all, it should be noted that $0+3+0+1+0+1=5$ and $1+1+1+1+0+1=5$ are not equally likely. There are $\frac{5!}{3!}$ ways to get the first and $5!$ ways to get the second.
So counting these is not going to be a good approach.
Letting $P(x_2,x_4,x_6)=(3+x_2+x_4+x_6)^{2n+1}$, you then let:
$$P_2(x_2,x_4,x_6)=\fr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How prove this $\sum_{k=2}^{n}\frac{1}{3^k-1}<\frac{1}{5}$ show that $$\sum_{k=2}^{n}\frac{1}{3^k-1}<\dfrac{1}{5}\tag1$$
I try to use this well known: if $a>b>0,c>0$,then we have
$$\dfrac{b}{a}<\dfrac{b+c}{a+c}$$
$$\dfrac{1}{3^k-1}<\dfrac{1+1}{3^k-1+1}=\dfrac{2}{3^k}$$
so we have
$$\sum_{k=2}^{n}\dfrac{1}{3^... | $$\sum_{k\geq 2}\frac{1}{3^k-1}=-\frac{1}{2}+\sum_{k\geq 1}\sum_{m\geq 1}\frac{1}{3^{km}}=-\frac{1}{2}+\sum_{n\geq 1}\frac{d(n)}{3^n}\leq -\frac{1}{2}+\sum_{n=1}^{5}\frac{d(n)}{3^n}+\sum_{n\geq 6}\frac{n}{3^n}$$
leads to the sharper inequality
$$ \mathcal{T}=\sum_{k\geq 2}\frac{1}{3^k-1}\leq \frac{61}{324}.$$
We may a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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coefficient of $x^{17}$ in $(x^2+x^3+x^4+x^5+x^6+x^7)^3$ coefficient of $x^{17}$ in $(x^2+x^3+x^4+x^5+x^6+x^7)^3$ - I'm reviewing for an exam and don't understand the answer, C(11+3−1,11)−C(3,1)×C(5+3−1,5).
|
Using the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ we obtain
\begin{align*}
\color{blue}{[x^{17}](x^2+x^3+\cdots+x^7)^3}&=[x^{17}]x^6(1+x+\cdots+x^5)^3\tag{1}\\
&=[x^{11}]\left(\frac{1-x^6}{1-x}\right)^3\tag{2}\\
&=[x^{11}]\left(1-\binom{3}{1}x^6\right)\sum_{n=0}^\infty \binom{n+3-1}{n}x^n\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2504632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Help solving variable separable ODE: $y' = \frac{1}{2} a y^2 + b y - 1$ with $y(0)=0$ I am studying for an exam about ODEs and I am struggling with one of the past exam questions. The past exam shows one exercise which asks us to solve:
$$y' = \frac{1}{2} a y^2 + b y - 1$$with $y(0)=0$
The solution is given as
$$y(x) =... | write $$\int \frac{\frac{dy}{dx}}{\frac{1}{2}ay(x)^2+by(x)-1}dx=\int1dx$$
$$\frac{2\tan^{-1}\left(\frac{b+ay(x)}{\sqrt{-2a-b^2}}\right)}{\sqrt{-2a-b^2}}=x+C$$
$$y(x)=\frac{-b+\sqrt{-2a-b^2}\tan\left(\frac{1}{2}\sqrt{-2a-b^2}(x+C)\right)}{a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2504760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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If $a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4} = 0$ then relevant cubic has root between $0$ and $1$ If $a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4} = 0$ then prove the function $$P(x)=a+bx+cx^2+dx^3$$ has a root somewhere between $0$ and $1$.
If it has a root between and $0$ and $1$ then I could show that it changes ... | .Consider the polynomial defined by : $Q(x) = ax + \frac{bx^2}{2} + \frac{cx^3}{3} + \frac{dx^4}{4}$. Note that $Q(0) = 0$, since $Q(x)$ has constant term $0$. Furthermore, $Q(1) = a + \frac b2 + \frac c3 + \frac d4 = 0$. Therefore, we have located two roots of $Q$, $0$ and $1$.
Therefore, by Rolle's theorem, the deri... | {
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"url": "https://math.stackexchange.com/questions/2508352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What does $A = L + D + U$ look like? Exercise:
Consider the problem
\begin{split}
10u_1 + u_2 &= 1\\
u_1 + 10u_2 &= 10
\end{split}
with the solution $(u_1,u_2)^T = (0,1)^T$. For a general system of equations
$$Au = b$$
with an $n\times n$ matrix $A$, which consists of a lower triangular submatrix $L$, the diagonal $D$... | Your $L$, $D$, and $U$ are correct. Note that $a_{ij} = l_{ij} + d_{ij} + u_{ij}$ is true, however, two of the three terms are always zero; the upper right entry would be $a_{ij} = 0 + 0 + 1$.
In general, $\color{red}A = \color{blue}L + \color{orange}D + \color{green}U$ looks like
$$
\color{red}{\begin{pmatrix}
a_{1,1... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve for $a,b,c,d \in \Bbb R$, given that $a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac 25 =0$ Today, I came across an equation in practice mock-test of my coaching institute, aiming for engineering entrance examination (The course for the test wasn't topic-specific, it was a test of complete high school mathematics). It was havi... | It's
$$a^2-ab+\frac{b^2}{4}+\frac{3}{4}b^2-bc+\frac{1}{3}c^2+\frac{2}{3}c^2-cd+\frac{3}{8}d^2+\frac{5}{8}d^2-d+\frac{2}{5}=0$$ or
$$\left(a-\frac{b}{2}\right)^2+\frac{3}{4}\left(b-\frac{2c}{3}\right)^2+\frac{2}{3}\left(c-\frac{3d}{4}\right)^2+\frac{5}{8}d^2-d+\frac{2}{5}=0$$ or
$$\left(a-\frac{b}{2}\right)^2+\frac{3}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2512339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
A problem :$(edf)^3(a+b+c+d+e+f)^3\geq(abcdef)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{f})^3$ Hello during a problem I have this to solve :
Let $a,b,c,d,e,f$ be real positiv number such that $a\geq b\geq c\geq d\geq e\geq f\geq 1$ then :
$$(edf)^3(a+b+c+d+e+f)^3\geq(abcdef)(\frac{1}{a}+\fra... | The note of Alex says that it remains to prove that
$$(a_1+a_2+a_3+3)^3\geq a_1a_2a_3\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+3\right)^3$$ for all $a_i\geq1$.
Indeed, let $a_1=a^3$, $a_2=b^3$ and $a_3=c^3$.
Thus, it's enough to prove that
$$a^3+b^3+c^3+3\geq abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+3\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2519854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Fourier cosine transform and K-Bessel function I am looking for a reference on the link between the asymptotic expansion of a functions and its Fourier cosine transform asymptotics.
More precisely I would like to find a reference to have the expansion in zero and infinity of (for $Re(a)<\frac{1}{2}$):
$$F(y)= \int_0^{\... | To obtain the asymptotic behaviour of this integral near $y=0$, the Mellin transform method can be used. One has
\begin{align}
\mathcal{M} \left[\cos \left(z\right),s\right]&=\Gamma (s)\cos\left( \frac{\pi}{2}s \right)\\
\mathcal{M} \left[x^{-1/2}K_a\left( x^{-1} \right),1-s\right]&=2^{-5/2+s}\Gamma (\frac{s}{2}-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2521856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{n\rightarrow\infty} \frac{(n+1)^{2n^2+2n+1}}{(n+2)^{n^2+2n+1} n^{n^2}}$ Woflram gives $\frac{1}{e}$ as the limit, but I failed to obtain it. Please help.
$$\lim_{n\rightarrow\infty} \frac{(n+1)^{2n^2+2n+1}}{(n+2)^{n^2+2n+1}n^{n^2}}$$
| Consider $$A=\frac{(n+1)^{2n^2+2n+1}}{(n+2)^{n^2+2n+1}n^{n^2}}$$ and take logarithms
$$\log(A)=(2n^2+2n+1)\log(n+1)-(n^2+2n+1)\log(n+2)-n^2\log(n)$$ Now, rewrite
$$\log(n+1)=\log(n)+\log\left(1+\frac 1n \right)=\log(n)+\frac{1}{n}-\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\log(n+2)=\log(n)+\log\left(1+\frac 2n \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2522605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show recursion in closed form I've got following sequence formula:
$ a_{n}=2a_{n-1}-a_{n-2}+2^{n}+4$
where $ a_{0}=a_{1}=0$
I know what to do when I deal with sequence in form like this:
$ a_{n}=2a_{n-1}-a_{n-2}$
- when there's no other terms but previous terms of the sequence.
Can You tell me how to deal with this ... | $$\begin{align}
a_{n+2} - 2a_{n+1} + a_{n} &= 4 \cdot 2^{n} + 4 \\
a_{n+1} - 2a_{n} + a_{n-1} &= 2 \cdot 2^{n} + 4 \\
a_{n} - 2a_{n-1} + a_{n-2} &= 1 \cdot 2^{n} + 4 \\
\end{align}$$
$$
\begin{bmatrix}
1 & -2 & 1 & 0 & 0 \\
0 & 1 & -2 & 1 & 0 \\
0 & 0 & 1 & -2 & 1 \\
\end{bmatrix}
\begin{bmatrix}
a_{n + 2} \\ a_{n + 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2526695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Solve $X^2=A$, where X is a 2 by 2 matrix and A is a known matrix A = $
\begin{pmatrix}
1 & 1 \\
-1 & 1 \\
\end{pmatrix}
$
. I wrote X = $
\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix}
$. So $X^2$=$
\begin{pmatrix}
a^2+bc & ab+bd \\
... | Hints:
From top-right and bottom-left elements we get:
$$(a+b)d = (a+b)c = -1 \Rightarrow c=-d$$
The top-left and bottom-right elements give us:
$$a^2 + bc = bc + d^2 = 1 \Rightarrow a^2 = d^2 \Rightarrow a = \pm d$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2527541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Taylor inside an integral I know the following integral should be:
$$ \int_{0}^{1} \frac{dx}{\sqrt{1-x^2-\epsilon(1-x^3)}} \approx \pi/2 + \epsilon$$
for $\epsilon$ small. What I do is set $-\epsilon(1-x^3)=y$ and then since $y$ it's always greatly smaller than $(1-x^2)$ I do Taylor expansion around $y=0$:
$$ \frac{dx... | You are on safer ground factoring the $1-x$ out of the square root as follows:
$$\begin{align} I(\epsilon) &= \int_0^1 \frac{dx}{\sqrt{1-x^2-\epsilon (1-x^3)}} \\ &= \int_0^1 dx \frac{(1-x)^{-1/2}}{\sqrt{1+x-\epsilon(1+x+x^2)}} \\ &= \int_0^1 dx \, \frac{x^{-1/2}}{\sqrt{2-x-\epsilon (3-3 x+x^2)}} \\ &=2 \int_0^1 dx \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2529544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Linear Algebra exercise problem: intersection of two subspaces I’m trying to solve a linear algebra exercise, but my solution does not match the solution given on the exercise book. This is the exercise:
Given $U = \langle(1,0,1,2), (1,1,2,3)\rangle$, a subspace of $\Bbb R^4$, and $S$, solution of the system
$$\begin{... | The first row of the $4\times5$ matrix, the last entry should be $0$ rather than $-1$.
The last two rows of the matrix correspond to equations $x-t=-2$ and $2x-z=3$. How do these equations come from the given equations? There seems to be some confusion.
Here is how the problem should be worked:
$$\begin{pmatrix}
x\\y\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2529953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to evaluate the integral $\int_0^{\pi}\frac{a^n\sin^2x+b^n\cos^2x}{a^{2n}\sin^2x+b^{2n}\cos^2x}dx$?
Evaluate the integral $$\int_0^{\pi}\frac{a^n\sin^2x+b^n\cos^2x}{a^{2n}\sin^2x+b^{2n}\cos^2x}dx.$$
I have no idea.
$$\int_0^{\pi}\dfrac{a^n\sin^2x+b^n\cos^2x}{a^{2n}\sin^2x+b^{2n}\cos^2x}dx=\int_0^{\pi/2}\dfrac{a^... | Your approach does work if you would do it correctly. Assume $a\neq 0$ and $b\neq 0$, why? Otherwise the integral is boring. Also assume that $|a|\neq |b|$ for the same reason. Let:
\begin{align}
I=\int^{\pi}_0 \frac{a^n\sin^2(x)+b^n\cos^2(x)}{a^{2n}\sin^2(x)+b^{2b}\cos^2(x)}dx = 2\int^{\pi/2}_0 \frac{a^n\sin^2(x)+b^n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2530338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Area under random walk The problem - we have random walk set of functions such that:
$$
f(0) = 0, f(x) = f(i) + \alpha_i(x-i), x \in (i,i+1], i=0,\dots,n-1
$$
where $\alpha_i \in \{-1, +1\}$ choosed uniformly with plobability $\dfrac{1}{2}$.
Need to find probability that $P(\int\limits_0^n f(x) dx = 0)$.
Thank you
| As results above justify, consider sum $S = \sum\limits_{i=0}^{n-1}(n-i+\frac{1}{2})\alpha_i = (n+\frac{1}{2})\sum\limits_{i=0}^{n-1}\alpha_i - \sum\limits_{i=0}^{n-1}i \alpha_i$
Denote as $n^{+}$ and $n^{-}$ number of steps up and down respectively. Clear, $n=n^{+}+n^{-}$. Then:
$$
(n+\frac{1}{2})\sum\limits_{i=0}^{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2532331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integration using reduction formulas issue. Show $(2n+1)I_n=2na^2I_{n-1}$ where $I_n=\int_{0}^{a}(a^2-t^2)^ndt$. I have a question where I am given $I_n=\int_{0}^{a}(a^2-t^2)^ndt$ and I am asked to show
$$(2n+1)I_n=2na^2I_{n-1}$$
I then decided to let $u=(a^2-t^2)^n$ and $\dfrac{dv}{dt}=1$. I proceeded to use integra... | If you will allow for the use of the beta function $\text{B}(m,n)$, where the beta function is defined by
$$\text{B}(m,n) = \int^1_0 x^{m - 1} (1 - x)^{n - 1} \, dx,$$
we can also solve your problem.
Starting with your integral
$$I_n = \int^a_0 (a^2 - t^2)^n \, dt,$$
where I will assume $a > 0$ and $n > - 1$, setting $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2532996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding the trace of $T^4$ and $T^2$?
Q. Let $V$ denote the (complex) vector space of complex polynomials of degree at most $9$ and consider the linear operator $T:V \to V$ defined by $$T(a_0+a_1 x+a_2 x^2+\cdots+a_9 x^9)=a_0 +(a_2 x +a_1 x^2)+(a_4 x^3+ a_5 x^4 + a_3 x^5)+(a_7 x^6 + a_8 x^7 + a_9 x^8 + a_6 x^9).$$
(a... | Hint: The matrix is diagonal by block
$$ T =
\begin{pmatrix}
A & 0 & 0 & 0 \\
0 & B & 0 & 0 \\
0 & 0 & C & 0 \\
0 & 0 & 0 & D \\
\end{pmatrix}
$$
with $$ A = I_{1,1}, B =
\begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix}, C = \begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1\\
1 & 0 & 0 \\
\end{pmatrix}, D = \begin{pmatrix} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2533800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Integrate $\sin^4(4x)/\sin^2(x)$ I would like to compute
$$\int_{0}^{2\pi} \frac{\sin^4(4x)}{\sin^2(x)} \mathrm{d}x.$$
Wolfram|Alpha is able to compute an antiderivative explicitly so I do not think use of the residue theorem is needed, but I'm interested in any approach.
| We can compute the integral by Fourier series. Consider
\begin{align}
f(x) = \frac{\sin^2(4x)}{\sin x}
\end{align}
then the sine series expansion of $f$ is given by
\begin{align}
f(x) = \sin x + \sin 3x + \sin 5x + \sin 7 x.
\end{align}
In particular, we see
\begin{align}
\int^{2\pi}_0 f^2(x)\ dx = \int^{2\pi}_0 \sin^2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the result of the root expression, is my answer correct or not? Suppose $a < 0 < b$. Then what is the result of:
$\sqrt{(a-b)^2} + \sqrt[6]{ b^6 } = ?$
I have a solution but I can't be sure if I did a mistake, because I usually do! My solution:
Call $a = -c$ for some $0 < c $. Then,
$=\sqrt{(-c-b)^2} + \sqrt[6]{ ... | That seems good to me. You should be careful that $\sqrt{(c+b)^2} = |c + b|$. But as $c$ and $b$ are both positive, $|c + b| = c+ b$.
You can actually be more direct:
$\sqrt{(a-b)^2} + \sqrt[6]{ b^6 } = |a-b| + |b|$.
As $b > 0$ we know $|b| = b$. As $a < 0$ and $b > 0$ then then $a- b < a < 0$. So $|a-b| = b -a$
So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2539919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find real solutions in $x$,$y$ for the system $\sqrt{x-y}+\sqrt{x+y}=a$ and $\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2.$
Find all real solutions in $x$ and $y$, given $a$, to the system:
$$\left\{
\begin{array}{l}
\sqrt{x-y}+\sqrt{x+y}=a \\
\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2 \\
\end{array}
\right.
$$
From a math olympiad... | Short solution:
WLOG, $a=2$ (as $x,y\propto a^2$).
By squaring the first equation and rearranging, you draw
$$\sqrt{x^2-y^2}=2-x.$$
And from the second,
$$\sqrt{x^2+y^2}=4+\sqrt{x^2-y^2}=6-x.$$
Now by summing the squares of the LHS,
$$2x^2=(2-x)^2+(6-x)^2$$
which gives$$x=\frac52.$$
Then from the first identity
$$y=\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2542647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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If $p$ is prime, $p\ne3$ then $p^2+2$ is composite I'm trying to prove that if $p$ is prime,$p\ne3$ then $p^2+2$ is composite. Here's my attempt:
Every number $p$ can be put in the form $3k+r, 0\le r \lt 3$, with $k$ an integer. When $r=0$, the number is a multiple of 3, so that leaves us with the forms $3k+1$ and $3k+... | if we have $$p\equiv 1 \mod 3$$ then we get $$p^2+2\equiv 0\mod 3$$ and with
$$p\equiv 2\mod 3$$ we get $$p^2+2\equiv 4+2\equiv 0\mod 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
$\tan(x) = 3$. Find the value of $\sin(x)$ I’m trying to figure out the value for $\sin(x)$ when $\tan(x) = 3$. The textbook's answer and my answer differ and I keep getting the same answer. Could someone please tell me what I'm doing wrong?
1.) $\tan(x) = 3$, then $\frac{\sin(x)}{\cos(x)} = 3$.
2.) Then $\cos(x) = ... | Another way to think of it is visualizing SOH-CAH-TOA. (I pronounce it "so-cuh-tow-uh".)
Use the relationships on a particular triangle to get the answer.
Tangent is opposite over adjacent (TOA), so choose your vertical side to be $3$ and your horizontal side to be $1$. This right triangle has the correct value for the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2546705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How do you factorize a polynom in $\mathbb{Z}_2$? How can you efficiently factorize a polynom (in $\mathbb{Z}_2$) ?
Example in this answer:
$$x^{16}-x=x (x + 1) (x^2 + x + 1) (x^4 + x + 1) (x^4 + x^3 + 1) (x^4 + x^3 + x^2 + x + 1)$$
How do you do this by hand?
| Partial answer:
We know that $x-1 =x+1$, so we have:
\begin{eqnarray}
x^{16}-x &=& x(x^{15}-1)\\
&=& x(x^5-1)\underbrace{(x^{10}+x^5+1)}_{p(x)}\\
&=& x(x-1)(x^4+x^3+x^2+x+1)p(x)\\
\end{eqnarray}
Now we have to factor somehow $p(x)$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2547821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Prove that $x^5 + 4x^4 + 2x^3 + x^2 + 4x + 54$ is irreducible over $\mathbb{Z} / 7\mathbb{Z}$ I wish to show that $$f(x) = x^5 + 4x^4 + 2x^3 + x^2 + 4x + 54$$ is irreducible over $\mathbb{Z} / 7\mathbb{Z}$. Simply testing every element, it is easy to see that there are no roots and thus no linear factors, but I'm not s... | You can try case by base: 5 = 1+4 = 2+3 and $F = \mathbb{Z}/7\mathbb{Z}$ is a field so, there are three options: $f = x^5 + 4x^4 + 2x^3 +x^2 + 4x + 54$ is irreducible, it decompose int terms of degree 1 and 4 or terms of degree 2 and 3. As you checked, the second case is impossible because it has no root in $F$. Let's ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Fair and unfair coins There are 4 fair coins and 1 unfair coin that has only heads. We choose a coin and flip it three times. The result is HHH. What is the probability that the fourth flip is H?
| We would expect the probability to be greater than $\frac{1}{2}$. We must solve for $P(\text{Heads})$. We have
$$P(\text{Heads})=P(\text{Heads}|\text{fair})\cdot P(\text{fair})+P(\text{Heads}|\text{unfair})\cdot P(\text{unfair})$$
However, we cannot just say $P(\text{fair})=\frac{4}{5}$ because we are given that the fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2551233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Prove that $\sum\limits_{r=0}^\infty\,\dfrac{(-1)^r\,n}{r!\,(r+1)!}\,\prod\limits_{k=1}^r\,\left(n^2-k^2\right)=(-1)^{n+1}$
If n is a positive integer, prove that:
$$\begin{align}n - {\frac{n(n^2-1^2)}{2!}}+{\frac{n(n^2-1^2)(n^2-2^2)}{2!3!}} + \ldots\phantom{aaaaaaaaaa} &&\\
+(-1)^r{\frac{n(n^2-1^2)(n^2-2^2)\cdots(n... | $$(-1)^r{\frac{n(n^2-1^2)(n^2-2^2)...(n^2-r^2)}{r!(r+1)!}}=\dfrac{n(n-1)(n-2)\cdots(n-r)}{(r+1)!}\cdot\dfrac{-(n+1)\cdot-(n+2)\cdots-(n+r)}{r!}$$
Think of the coefficient of $x$ in $$(1+x)^n\left(1+\dfrac1x\right)^{-(n+1)}=\dfrac{x^{n+1}}{1+x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2553579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to calculate this hard integral $\int_0^{\infty} \frac{\arctan(x)\sqrt{\sqrt{1+x^2}-1}}{x\sqrt{1+x^2}}\,dx$? How to prove that $\displaystyle \int_0^{\infty} \dfrac{\arctan(x)\sqrt{\sqrt{1+x^2}-1}}{x\sqrt{1+x^2}}\,dx= \frac{1}{4}\pi^2\sqrt{2}-\sqrt{2}\ln^2\left(1+\sqrt{2}\right)$ ?
It's very difficult and I have no... | This can be reduced to a class of integral related to polylogarithm, and is related to Legendre-chi function $\chi_2$. The formula
$$\int_{0}^{\frac{\pi}{2}} \arctan(r \sin\theta) \, d\theta
= 2 \chi_{2} \left( \frac{\sqrt{1+r^{2}} - 1}{r} \right) $$
gives $$\tag{1}\int_{0}^{\frac{\pi}{2}} \arctan( \sin\theta) \ d\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2557405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Determine tranformation matrix of linear transformation
A linear transformation between polynomial spaces $f: P_2(\mathbb{R}) \to P_2(\mathbb{R})$ is given by $$f(p(x))=3 \cdot p(1)-x^2 \cdot p(0)+(x-1) \cdot p'(1)$$ Determine the transformation matrix with respect to the monomial basis $(1,x,x^2)$
I tried to approac... | Well, if you have a basis $\beta =\{1, x, x^2\}$, and we need to determine the matrix of $f$ wrt $\beta$, then let's first write out $f$ in the way that makes it the most obvious how it's transforming it's coordinates:
$$
\begin{aligned}
f(ax^2+bx+c) = f((a, b, c)) & = 3(a + b + c) - cx^2 + (2a + b)x - (2a + b) \\
& = ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sum_{n=0}^N\binom{2N-n}N2^n(n+1)=(1+2N)\binom{2N}N$ I used WolframAlpha to calculate a sum but it didn't show me the way :( Anybody has a hint or a solution for proving this sum?
$$\sum_{n=0}^N\binom{2N-n}N2^n(n+1)=(1+2N)\binom{2N}N$$
| $$\begin{align}
\sum_{n=0}^N\binom {2N-n}N2^n(n+1)
&=\sum_{n=0}^N\binom {2N-n}N\sum_{j=0}^n \binom nj(n+1)
&&\scriptsize\text{using }\sum_{j=0}^n \binom nj=2^n\\
&=\sum_{n=0}^N\binom {2N-n}N\sum_{j=0}^n \binom {n+1}{j+1}(j+1)\\
&=\sum_{n=0}^N\binom {2N-n}N\sum_{j=1}^{n+1} \binom {n+1}{j}j\\
&=\sum_{j=1}^{N+1}j\sum_{n=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2558601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Maximum and minimum value of a function.
Let us consider a function $f(x,y)=4x^2-xy+4y^2+x^3y+xy^3-4$. Then find the maximum and minimum value of $f$.
My attempt. $f_x=0$ implies $8x-y+3x^2y+y^3=0$ and $f_y=0$ implies $-x+8y+x^3+3xy²=0$ and $f_{xy}=3x^2+3y^2-1$. Now $f_x+f_y=0$ implies $(x+y)((x+y)^2+7)=0$ implies $x... | $f_x+f_y=\left(x^3+3 x y^2-x+8 y\right)+\left(3 x^2 y+8 x+y^3-y\right)=\\=x^3+3 x^2 y+3 x y^2+7 x+y^3+7 y=(x+y) \left(x^2+2 x y+y^2+7\right)=0$
$y=-x$. Substitute in $f_x=0$ and get
$9 x-4 x^3=0\to x_1=0;\;x_2=\dfrac{3}{2};\;x_3=-\dfrac{3}{2}$
and $y_1=0;\;y_2=-\dfrac{3}{2};\;y_3=\dfrac{3}{2}$
Now the Hessian, for the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2558988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Convergence of $\sum_{n=0}^{+ \infty} \frac{3}{2n² -1}$
Does $$\sum_{n=0}^{+ \infty} \frac{3}{2n² -1}$$
converge?
I did the following:
$$\frac{3}{2n²-1} = \frac{3}{2n²}\left(\frac{1}{1-\frac{1}{2n²}}\right)$$
and because $$\frac{1}{1-\frac{1}{2n²}} \to 1$$
the sequence is bounded, and hence, there exists $M \in \math... | Your series is absolutely convergent by comparison with $\sum_{n\geq 1}\frac{3}{n^2}=\frac{\pi^2}{2}$, for instance.
You may also compute it in a explicit way. If we start with
$$ \frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right) $$
and apply $\frac{d}{dz}\log(\cdot)$ to both sides, we get
$$-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Mathematical induction proof that $f(n)=\frac{1}{2}+\frac{3}{2}(-1)^n$
The function $f(n)$ for $n=0,1...$ has the recursive definition $$f(n)= \begin{cases} 2 & \text {for n=0} \\ -f(n-1)+1 & \text{for n=1,2...} \end{cases}$$
Prove by induction that the following equation holds: $$f(n)=\frac{1}{2}+\frac{3}{2}(-1)^n$... | Your last step is close, but not quite
$$ f(n+1) = \frac{1}{2} - \frac{3}{2}(-1)^n = 1 - \left(\frac{1}{2} + \frac{3}{2}(-1)^n\right) = -f(n) + 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2564657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Line integral over a non-central ellipse. I have to find the integral $$ \int_C ydx + x^2dy$$ over the curve $C$ given as a intersection of plane $z=0$ and surface $\frac{x^2}{a^2} + \frac{y^2}{b^2}=\frac{x}{a}+\frac{y}{b} $, curve $C$ is positively oriented $(a\geq b>0)$.
This is what i have this far:
given plane is $... | By Green's theorem,
$$I:=\int_C ydx + x^2dy=\iint_E (2x-1)dxdy=|E|(2\bar{x}-1)$$
where $E$ is the domain bounded by $C$ in the $xy$-plane, $|E|$ is its area and $\bar{x}$ is the $x$-coordinate of the centroid of $E$.
Since $C$ is the ellipse given by (check your algebraic manipulations!!)
$$\frac{(x-\frac{a}{2})^2}{\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2566607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the spectral decomposition of $A$ $$
A= \begin{pmatrix} -3 & 4\\ 4 & 3
\end{pmatrix}
$$
So i am assuming that i must find the evalues and evectors of this matrix first, and that is exactly what i did.
The evalues are $5$ and $-5$, and the evectors are $(2,1)^T$ and $(1,-2)^T$
Now the spectral decomposition of $A... | I think of the spectral decomposition as writing $A$ as the sum of two matrices, each having rank 1. Let $A$ be given. Then compute the eigenvalues and eigenvectors of $A$. Then
$$
A = \lambda_1P_1 + \lambda_2P_2
$$
where $P_i$ is an orthogonal projection onto the space spanned by the $i-th$ eigenvector $v_i$.
In... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
In how many ways can $3$ red, $3$ blue, and $3$ green balls be arranged so that no two balls of the same colour are consecutive (up to symmetry)? Make sequence of $9$ balls from 3 red, 3 blue, 3 green, in such a way that no two balls of the same colour are next to each other. In how many different ways can you do this?... | We have nine positions to fill with three blue, three green, and three red balls. We can fill three of the nine positions with blue balls in $\binom{9}{3}$ ways, three of the remaining positions with green balls in $\binom{6}{3}$ ways, and the remaining three positions with red balls in $\binom{3}{3}$ ways. Hence, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2569399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
derivative of $\frac 2x \sin(x^3)$ by definition The function: $$\frac 2x \sin(x^3)$$ when $x\neq0$, and $0$ while $x=0$.
I need to find if the function is derivation at $x=1$.
First step was to check if the function is continuous. there is just 1 side of limit to check (since its the same function around $x=1$, so I c... | When you have
$$\frac{\frac{2 \sin \left((h+1)^3\right)}{h+1}-2 \sin (1)}{h}$$
you must rewrite for all $h\neq0$ to
$$\frac{\sin \left(\frac{1}{2} \left((h+1)^3-1\right)\right)}{\frac{1}{2} \left((h+1)^3-1\right)}\frac{2 \left(h^2+3 h+3\right) \cos \left(\frac{1}{2} \left((h+1)^3-1\right)+1\right)}{h+1}-\frac{2 \sin (1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2570328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Why product of primes $+1$ is not divisible by any of the product primes? Given $n$ prime numbers, $p_1, p_2, p_3,\ldots,p_n$, then $p_1p_2p_3\cdots p_n+1$ is not divisible by any of the primes $p_i, i=1,2,3,\ldots,n.$ I dont understand why. Can somebody give me a hint or an Explanation ? Thanks.
| First approach:
Let $P= 2\times3\times5\times7\times11\times13.$ Then:
The next number after $P$ that is divisible by $2$ is $P+2.$
The next number after $P$ that is divisible by $3$ is $P+3.$
The next number after $P$ that is divisible by $5$ is $P+5.$
The next number after $P$ that is divisible by $7$ is $P+7.$
The ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"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.