Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to factorize $a^2-b^2-a+b+(a+b-1)^2$? The answer is $(a+b-1)(2a-1)$ but I have no idea how to get this answer.
| $$a^2-b^2-a+b+(a+b-1)^2\Leftrightarrow (a-b)(a+b)-(a-b)+(a+b-1)^2\\\implies(a-b)(a+b-1)+(a+b-1)^2=\boxed{(a+b-1)(2a-1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Showing $\lvert \frac{x^2-2x+3}{x^2-4x+3}\rvert\le1\Rightarrow x\le0$
$\lvert \frac{x^2-2x+3}{x^2-4x+3}\rvert\le1\Rightarrow x\le0$
How is the proof. If I separate the denominator with triangle inequality,
$\lvert \frac{x^2-2x+3}{x^2-4x+3}\rvert\le \frac{\lvert x^2-2x+3\rvert}{\lvert x^2-2x+3 \rvert-\lvert 2x\rvert}\... | WLOG let $$\dfrac{x^2-2x+3}{x^2-4x+3}=\cos y$$ where $y$ is real
$$(1-\cos y)x^2+2x(2\cos y-1)+3(1-\cos y)=0$$
$$\implies x=\dfrac{1-2\cos y\pm\sqrt{(2\cos y-1)^2-3(1-\cos y)^2}}{(1-\cos y)}$$
$$=\dfrac{1-2\cos y\pm\sqrt{\cos^2y+2\cos y-2}}{(1-\cos y)}$$
Now the denominator is non-negative.
Hence we need $$\sqrt{\cos^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2064501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the interval in which $m$ lies so that the expression $\frac{mx^2+3x-4}{-4x^2+3x+m}$ can take all real values, $x$ being real
Find the interval in which $m$ lies so that the expression $\frac{mx^2+3x-4}{-4x^2+3x+m}$ can take all real values, $x$ being real.
I don't know how to proceed with this question. I h... | Momo's answer is correct except for the values $m=1$ and $m=7$.
Let $f$ be given by
$$f(x)=\frac{mx^2+3x-4}{-4x^2+3x+m}\;\;$$
If $m = 1$, then
\begin{align*}
f(x)&=\frac{x^2+3x-4}{-4x^2+3x+1}\\[4pt]
&=-\frac{x^2+3x-4}{4x^2-3x-1}\\[4pt]
&=-\frac{(x-1)(x+4)}{(x-1)(4x+1)}\\[4pt]
&=-\frac{x+4}{4x+1},\;\;x\ne 1\\[4pt]
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2064581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
How to integrate $\int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$ in a faster way? $\displaystyle \int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$
$\displaystyle \int_a^{(a+b)/2} (x-a)(x-\frac{a+b}{2})(x-b)\, dx=\frac{1}{64}(b-a)^4$
Instead of expanding the integrand, or doing integration by part, is there any faster way to ... | Hint. One may use the fact that
$$
4uv=(u+v)^2-(u-v)^2 \tag1
$$ giving
$$
\begin{align}
4\int_a^b (x-a)(x-b)\:dx=4\int_a^b \left(x-\frac{a+b}2\right)^2\:dx-\int_a^b (b-a)^2\:dx \tag2
\end{align}
$$ which is easier to evaluate.
Then, one has $$
\begin{align}
&4\int_a^b (x-a)(x-b)\left(x-\frac{a+b}2\right)\:dx
\\\\&=\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
Common proof for $(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} $ I'm asking for an alternative (more common?) proof of the following equality, more specifically an alternative proof for the inductive step:
$$(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} (x\neq 1)$$
This is how I proved it... | In the inductive step, you assume that
$$(1+x)(1+x^2)(1+x^4)...(1+x^{2^n}) = \dfrac{1-x^{2^{n+1}}}{1-x}$$
Which means that
$$(1-x)(1+x)(1+x^2)(1+x^4)...(1+x^{2^n}) = 1-x^{2^{n+1}}$$
Multiply by $(1+x^{2^{n+1}})$ and you get
$$(1-x)(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})(1+x^{2^{n+1}}) = (1-x^{2^{n+1}})(1+x^{2^{n+1}})$$
Usin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Solve $\left(\sqrt{\sqrt{x^2-5x+8}+\sqrt{x^2-5x+6}} \right)^x + \left(\sqrt{\sqrt{x^2-5x+8}-\sqrt{x^2-5x+6}} \right)^x = 2^{\frac{x+4}{4}} $
Solve $$\left(\sqrt{\sqrt{x^2-5x+8}+\sqrt{x^2-5x+6}} \right)^x + \left(\sqrt{\sqrt{x^2-5x+8}-\sqrt{x^2-5x+6}} \right)^x = 2^{\frac{x+4}{4}} $$
Preface; I think there should be... | hint: Use AM-GM inequality: $a + b \ge 2\sqrt{ab}$, for this type of question, with $a,b$ are the first and second terms of the left side of the equation. The product $ab = $ constant. Specifically, with $a =
\sqrt{\sqrt{x^2-5x+8} + \sqrt{x^2-5x+6}}, b = \sqrt{\sqrt{x^2-5x+8}-\sqrt{x^2-5x+6}}$, we have equation occurs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
outcome from fair and non-fair dice, dice are chosen from other dice part2
There're 3 dice, two are fair, one is not. The non-fair dice has an occurrence of $5$ instead of $4$. That is the non-fair dice is $\{1,2,3,5,5,6\}.$
We choose one dice from the three randomly, and throw it twice. Suppose the second roll in a... | The problem with your method is that you did not calculate $P(A \cap B)$ correctly. $P(A \cap B)$ is the probability that the first roll is $4$ and the second roll is $5$, but you just found the probability that the first roll is $4$. Let's break this up into two cases:
Case 1: Fair die
*
*The probability of getting... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Matrix determinant lemma derivation While reading this wikipedia article on the determinant lemma, I stumbled upon this expression (in a proof section):
\begin{equation}
\begin{pmatrix} \mathbf{I} & 0 \\ \mathbf{v}^\mathrm{T} & 1 \end{pmatrix}
\begin{pmatrix} \mathbf{I}+\mathbf{uv}^\mathrm{T} & \mathbf{u} \\ 0 & 1 \end... | This is just Gaussian elimination. We begin with $\pmatrix{I+uv^T&0\\ 0&1}$ and we want to eliminate the summand $uv^T$ in the first sub-block using row or column additions (so that the determinant is preserved):
\begin{align*}
\pmatrix{I+uv^T&0\\ 0&1}\pmatrix{I&0\\ -v^T&1}&=\pmatrix{I+uv^T&0\\ -v^T&1},\\
\pmatrix{I&u\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
How many ways can 8 teachers be distributed among $4 $ schools? There are several ways that the teachers can be divided amongst $4$ schools, namely here are the possible choices I came up with:
$1) 1 1 1 5$
$2) 1 1 2 4$
$3) 1 1 3 3$
$4) 1 2 2 3$
$5) 2 2 2 2$
now given the fact that say $2213$ is the same as $1 2 2 3$ i... | The total number of combinations for $\color\red{111}\color\green{5}$:
$$\frac{(1+1+1+5)!}{1!\times1!\times1!\times5!}\times\frac{(\color\red3+\color\green1)!}{\color\red3!\times\color\green1!}=1344$$
The total number of combinations for $\color\red{11}\color\green{2}\color\orange{4}$:
$$\frac{(1+1+2+4)!}{1!\times1!\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
proving $t^6-t^5+t^4-t^3+t^2-t+0.4>0$ for all real $t$ proving $t^6-t^5+t^4-t^3+t^2-t+0.4>0$ for all real $t$
for $t\leq 1,$ left side expression is $>0$
for $t\geq 1,$ left side expression $t^5(t-1)+t^3(t-1)+t(t-1)+0.4$ is $>0$
i wan,t be able to prove for $0<t<1,$ could some help me with this
| By breaking the polynomial into groups of three terms and completing the square, we get:
\begin{align}
& \hspace{0.36 in} t^6-t^5+t^4-t^3+t^2-t+\dfrac{2}{5}
\\
&= \left(t^6-t^5+\dfrac{1}{4}t^4\right)+\dfrac{3}{4}t^4-t^3+t^2-t+\dfrac{2}{5}
\\
&= \left(t^6-t^5+\dfrac{1}{4}t^4\right)+\left(\dfrac{3}{4}t^4-t^3+\dfrac{1}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Why is the solution to $x-\sqrt 4=0$ not $x=\pm 2$? If the equation is $x-\sqrt 4=0$, then $x=2$.
If the equation is $x^2-4=0$, then $x=\pm 2$.
Why is it not $x=\pm 2$ in the first equation?
| Obviously the answer to
$x - k = 0$ is $x = k$. One answer.
And easily, but not obviously, the answer to $x^2 - k^2 = 0$ is $x =\pm k$. Two answers (if we assume $k \ne 0$).
So Question number 1: Why does $x - k$ have one answer while $x^2 - m = 0$ has two (assuming $m > 0$)?
And obviously $x - \sqrt{4} = 0$ has ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 13,
"answer_id": 5
} |
$|z-3| + |z| + |z+3| = 12$ Let $z$ be a complex number such that $|z-3|+ |z|+ |z+3| = 12$.
If $a = \lfloor|z|\rfloor$ and $b = \lceil|z|\rceil$, where $\lfloor i\rfloor$ denote the greatest integer less than or equal to $i$ and
$\lceil i\rceil$ denotes the least integer greater or equal to $i$, then find $k = a + b$.
... | Write $z=r(\cos\phi+i\sin\phi)$. Then we have to investigate the equation
$$\sqrt{r^2-6r\cos\phi+9}+\sqrt{r^2+6r\cos\phi+9}+r=12\ ,$$
which can be written as
$$\sqrt{r^2+9}\ \bigl(\sqrt{1-t\cos\phi}+\sqrt{1+t\cos\phi}\bigr)+r=12\ ,\tag{1}$$
whereby
$$0\leq t:={6r\over r^2+9}\leq1\ .$$
The function
$$s\mapsto\sqrt{1-s}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
limit using Taylor's theorem i need to find the limit of $\lim _{x \to 0}f(x)$
$f(x)=\frac{\left (\sinh \left (x \right ) \right )^{n}-x^{n}}{\left (\sin \left (x \right ) \right )^{n}-x^{n}}$
i tried this $f(x)=\frac{(\frac{\sinh (x)}{x} )^{n}-1}{(\frac{(\sin (x)}{x})^{n}-1}$
and with Taylor's theorem $\lim _{x \t... | Let's use the usual route of standard limits combined with the use of advanced tools like Taylor series/L'Hospital's Rule if needed.
We have
\begin{align}
L &= \lim_{x \to 0}\frac{\sinh^{n}x - x^{n}}{\sin^{n}x - x^{n}}\notag\\
&= \lim_{x \to 0}\dfrac{\left(\dfrac{\sinh x}{x}\right)^{n} - 1}{\left(\dfrac{\sin x}{x}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is it possible to find the sum of the infinite series $1/p + 2/p^2 + 3/p^3 + \cdots + n/(p^n)+\cdots$, where $p>1$? Is it possible to find the sum of the series:
$$\frac{1}{p} + \frac{2}{p^2} +\frac{3}{p^3} +\dots+\frac{n}{p^n}\dots$$
Does this series converge? ($p$ is finite number greater than $1$)
| The exact value is $\frac{\frac{1}{p}}{(1-\frac{1}{p})^{2}}$ when $-1<\frac{1}{p}<1$. Since $p>1$ in this case, the series converges.
Proof:
$$
\frac{1}{p}+\frac{2}{p^{2}}+\frac{3}{p^{3}}...
$$
$$
=(\frac{1}{p}+\frac{1}{p^{2}}+\frac{1}{p^{3}}...)+(\frac{1}{p^{2}}+\frac{1}{p^{3}}+\frac{1}{p^{4}}...)+(\frac{1}{p^{3}}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Find the sum of all positive irreducible fractions less than 1 whose denominator is 2016. I know that we first have to use AP to find the sum of first 2015 natural nos. Then we need to subtract all those factors of 2016. But there are too many. I need help in this part.
| The prime factors of $2016$ are $2$, $3$ and $7$.
Hence use inclusion/exclusion principle:
*
*Include $\sum\limits_{n=1}^{2016}\frac{n}{2016}$
*Exclude $\sum\limits_{n=1}^{\frac{2016}{2}}\frac{2\cdot{n}}{2016}$
*Exclude $\sum\limits_{n=1}^{\frac{2016}{3}}\frac{3\cdot{n}}{2016}$
*Exclude $\sum\limits_{n=1}^{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Trigonometry and Quadratics If $\tan A$ and $\tan B$ are the roots of $x^2+px+q=0$, then prove that
$$\sin^2(A+B)+p \sin(A+B) \cos(A+B) + q \cos^2(A+B) = q$$
My Attempt:
Using the sum and product formulae we have,
$q=\tan A\tan B, $ $-p=\tan A+\tan B$
And,
$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \Right... | You are going absolutely correct.
Now proceed as follows:
$$\frac{\sin (A+B)}{\cos (A+B)}=-\frac{p}{1-q}$$
$$\implies (1-q)\sin(A+B)=-p\cos (A+B)$$
$$\implies \sin(A+B)+p\cos (A+B)=q\sin(A+B)$$
Multiplying both sides by $\sin(A+B)$, we get $$\sin^2(A+B)+p\sin(A+B)\cos (A+B)=q\sin^2(A+B)=q-q\cos^2(A+B)$$
Hence we rearra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding the numbers $n$ such that $2n+5$ is composite. Let $n$ be a positive integer greater than zero. I write
$$a_n =
\begin{cases}
1 , &\text{ if } n=0 \\
1 , &\text{ if } n=1 \\
n(n-1), & \text{ if $2n-1$ is prime} \\
3-n, & \text{ otherwise}
\end{cases}$$
The sequence goes like this $$1,1,2,6,12,-2,30,42,-5,72,90... | Just to clarify I have that both claims are trivially true and are explicit in the definition of $a_n$.
For claim 1 we have from @callus:
proof of claim 1: $a_n \leq 0$ for all $n$ unless $2n - 1$ is prime, by the definition. In that case $a_n = n(n - 1)$. So, $4a_n +
> 1 = 4(n^2 - n) + 1 = (2n-1)^2$. So $\sqrt{4a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Find coordinates of point $C$ lying on both the planes such that the perimeter of $\triangle ABC$ is minimum $A(3\mid −1\mid −1)\quad$ is a point in the plane $x+y+z=1.$
$B(3\mid 1\mid 0)\quad$ is a point in the plane $2x−y−z=5.$
Find coördinates of point C lying on both the planes such that the perimeter of △ABC is mi... | Let $C\equiv (2,t-1,-t),A\equiv (3,-1,-1),B\equiv(3,1,0)$. Here $C$ lies on line of intersection of the two given planes.
$CA+CB=\sqrt{2}\left[\sqrt{t^2-t+1}+\sqrt{t^2-2t+\dfrac{5}{2}}\right]=\sqrt{2}\left[\sqrt{(t-\dfrac{1}{2})^2+\dfrac{3}{4}}+\sqrt{(t-1)^2+\dfrac{3}{2}}\right]$
So,now the problem gets reduced to a $2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find all positive integer solutions to the equation $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$ Find all positive integers to the equation $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$
Multiply both sides with $(abc)^2$ to get $(bc)^2 + (ac)^2 = (ab)^2$.
I then tried some pythagorean triples and nothing worked so I assu... | As you noted $1/a^2+1/b^2=1/c^2 \iff b^2+a^2=\frac{b^2a^2}{c^2}$.
This implies that $a$ and $b$ are part of a pythagorean triple and can hence be written $a=k(m^2-n^2)$ and $b=2kmn$.
We therefore have $a^2+b^2=k^2(m^4-2m^2n^2-n^4+4m^2n^2)=k^2(m^2+n^2)^2$
So we need for $k^2(m^2+n^2)^2$ to divide $2k^4(m^2-n^2)m^2n^2$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to integrate $\tan(x) \tan(2x) \tan(3x)$? We have to integrate the following
$$\int \tan x \tan 2x \tan 3x \,\mathrm dx$$
In this I tried as
First split it into sine and cosine terms
Then used $\sin 2x =2\sin x \cos x$
But after that got stuck
| In addition to the other answers, we can also do like this:
Use $\tan 3x =\frac {3\cos^2 x \sin x-\sin ^3 x}{\cos^3 x-3\cos x\sin^2 x}$ ,$\tan 2x=\frac {2\cos x\sin x}{\cos^2 x-\sin^2 x}$ and substitute $u=\cos x $ to get $$I =\int \tan x\tan 2x\tan 3x dx =\int -\frac {2(4\cos^4 x -5\cos^2 x+1)}{\cos x (2\cos^2 x-1)(4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Maximum value of $|x-y|$ Given $x,y\in\mathbb R$ such that
$$5x^2+5y^2-6xy\ =\ 8$$
find the maximum value of $|x-y|$.
My attempt
$5x^2 - 6yx + (5y^2-8)\ =\ 0$
$x\ =\ \dfrac{6y\pm\sqrt{(6y)^2-4(5)(5y^2-8)}}{10} = \dfrac{6y\pm\sqrt{160-64y^2}}{10} = \dfrac{3y\pm2\sqrt{10-4y^2}}{10}$
$5y^2 - 6xy + (5x^2-8)\ =\ 0$
$y\ =\ \... | We'll prove that $|x-y|\leq\sqrt2$.
Indeed, we need to prove that
$$(x-y)^2\leq2\cdot\frac{5x^2+5y^2-6xy}{8}$$ or
$$(x+y)^2\geq0.$$
The equality occurs for $x+y=0$ and $x-y=\sqrt2$ for example.
Id est, the answer is $\sqrt2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove $\left(1+\frac1{n}\right)^{n}<\left(1+\frac1{n+1}\right)^{n+1}$ by mathematical induction I am trying to prove the following by mathematical induction:
$$\left(1+\frac1{n}\right)^{n}<\left(1+\frac1{n+1}\right)^{n+1}$$
Other proofs without induction are found here: I have to show $(1+\frac1n)^n$ is monotonically i... | Let us define
\begin{align}
t_n :=\left(1+\frac{1}{n} \right)^n.
\end{align}
We shall use the identity
\begin{align}
t_n =&\ 1+1+\frac{1}{2!}\left(1-\frac{1}{n}\right)+\frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+\ldots\\
&\ +\frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int_0^\infty \frac{\ln\left ( 2x+1 \right )}{x\left ( x+1 \right )}\mathrm dx$ How to evaluate
$$\int_{0}^{\infty }\frac{\ln(2x+1)}{x(x+1)}\,\mathrm dx?$$
I tried
$$\frac{\ln(2x+1)}{x(x+1)}=\ln(2x+1)\left (\frac{1}{x}-\frac{1}{1+x} \right)$$
but I don't know how to go on.
| This might be a late response, but consider the double integral
$$I=\int_{0}^{\infty}\int_{0}^{\infty} \frac{1}{1+x^2(2y+1)^2} \frac{x}{1+x^2} \ dy \ dx.$$
We will evaluate $I$ in two ways. Integrating with respect to $y$ first, we see that
$$\int_{0}^{\infty} \frac{1}{1+x^2(2y+1)^2} \ dy= \lim_{b \rightarrow \infty}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Find the sum of first $n$ terms of the series: $\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots +\frac{1}{n\times(n+1)}$
I have the series $$\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots +\frac{1}{n\times(n+1)}$$
I know the following formulas: $$1+2+3+\cdots +n=\frac {n (n+1)}{2}\ta... | Generally, for $a_n = a_0+(n-1)d$,$\frac{1}{a_n a_{n+1}}=\frac{1}{d}(\frac{1}{a_n}-\frac{1}{a_{n+1}})$.
Similarly, higher order fractions $\frac{1}{a_n a_{n+1} a_{n+2}}
=\frac{1}{2d}(\frac{1}{a_n a_{n+1}} - \frac{1}{a_{n+1}a_{n+2}})$.
When seeing these kinds of sequence series, you can have a try to split items cancel ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Show that $\sum\limits_{n=1}^{32}\frac1{n^2}=1 + \frac{1}{4} +\frac{1}{9} +\dots+ \frac{1}{1024}<2$ Show that
$$1 + \frac{1}{4} +\frac{1}{9} +\dots+ \frac{1}{1024} <2$$
I know that the denominators are perfect squares starting from that of $1$ till $32$. Also I know about this identity
$$\frac{1}{n(n+1)} > \frac{1}{(... | We know $$\frac{1}{n(n+1)} > \frac{1}{(n+1)^2} \Rightarrow \frac{1}{n} - \frac{1}{n+1} > \frac{1}{(n+1)^2}$$ Adding for $n =1,2,\cdots,31$, we get, $$\frac{1}{1} - \frac{1}{2} > \frac{1}{4}$$ $$\frac{1}{2}-\frac{1}{3} > \frac{1}{9}$$ $$\vdots$$ $$\frac{1}{31}-\frac{1}{32} > \frac{1}{1024}$$Adding gives us $$\frac{1}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Why do this algorithm for finding an equation whose roots are cubes of the roots of the given equation works? Let a polynomial, $p(x)$, of degree $n$ is given. Our aim is to find another polynomial, $q(x)$, whose roots are the cubes of the roots of $p(x)$.
Our algorithm go like this:
Step 1 Replace $x$ by $x^\frac{1}{3... | The canonical method to derive $q(x)$ is to eliminate $u$ between $p(u)=0$ and $u^3=x\,$, which can be done using polynomial resultants. By the definition of the resultant, it will be $0$ iff the two equations have a common root i.e. $q(x)=0$ for $x=u^3$ where $u$ is a root of $p$.
The derivation of $q(x)$ in the poste... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Find $xyz$ given that $x + z + y = 5$, $x^2 + z^2 + y^2 = 21$, $x^3 + z^3 + y^3 = 80$ I was looking back in my junk, then I found this:
$$x + z + y = 5$$
$$x^2 + z^2 + y^2 = 21$$
$$x^3 + z^3 + y^3 = 80$$
What is the value of $xyz$?
A) $5$
B) $4$
C) $1$
D) $-4$
E) $-5$
It's pretty easy, any chances of solving this que... | We have $$(x+y+z)^3=(x^3+y^3+z^3)+3x(y^2+z^2)+3y(x^2+z^2)+3z(x^2+y^2)+6xyz.$$ Hence
$$125=80+3x(21-x^2)+3y(21-y^2)+3z(21-z^2)+6xyz.$$
This leads to
$$45=63(x+y+z)-3(x^3+y^3+z^3)+6xyz.$$
This gives us $45=315-240+6xyz$, so $6xyz=-30$ and $xyz=-5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find all complex numbers $z$ satisfying the equation $z^{4} = -1+\sqrt{3}i$ Since any complex number can be of polar form. We set that $z = r\cos \theta +ir\sin \theta$. Now by de Moivre's Theorem, we easily see that $$z^{4} = r^{4}\cos4 \theta + ir^{4}\sin4 \theta$$
Since $$z^{4} = -1+\sqrt{3}i$$
We equip accordingly ... | The absolute value is very simple:
$$|z^4|=|-1+\sqrt3 i|=2$$
Thus,
$$|z|=\sqrt[4]2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Summation of $\arcsin $ series. What is $a $ if
$$\sum _{n=1} ^{\infty} \arcsin \left(\frac {\sqrt {n}-\sqrt {n-1}}{\sqrt {n (n+1)}}\right) =\frac {\pi }{a} \,?$$
Attempt: What I tried is to convert the series to $\arctan$ and then convert it telescoping series. So in terms of $\arctan $ it becomes
$$\arctan \left(\f... | We can utilize the following trigonometric identity
$$\arcsin x-\arcsin y = \arcsin(x\sqrt{1-y^2}-y\sqrt{1-x^2})$$
by putting $x=\frac{1}{\sqrt{n}}$, $y=\frac{1}{\sqrt{n+1}}$. Then we get
\begin{align}\arcsin \frac{1}{\sqrt{n}}-\arcsin \frac{1}{\sqrt{n+1}} &= \arcsin\left(\frac{1}{\sqrt{n}}\sqrt{1-\frac{1}{n+1}}-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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If $ \sin\theta + \cos\theta = \frac 1 2$, what does $\tan\theta + \cot\theta$ equal? A SAT II question asks:
If $ \sin\theta + \cos\theta = \dfrac 1 2$, what does $\tan\theta + \cot\theta$ equal?
Which identity would I need to solve this?
| $$\tan\theta+\cot\theta=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}\\=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}\\=\frac1{\sin\theta\cos\theta}\\=\frac1{\frac12+\sin\theta\cos\theta-\frac12}\\=\frac1{\frac12(\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta)-\frac12}\\=\frac1{\frac12(\sin\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Proving an alternate quadratic formula It is well known that the quadratic formula for $ax^2+bx+c=0$ is given by$$x=\dfrac {-b\pm\sqrt{b^2-4ac}}{2a}\tag1$$
Where $a\ne0$. However, I read somewhere else that given $ax^2+bx+c=0$, we have another solution for $x$ as$$x=\dfrac {-2c}{b\pm\sqrt{b^2-4ac}}\tag2$$
Where $c\ne0$... | Take (2), and rationalize the denominator:
$$\frac{-2c}{b \pm \sqrt{b^2-4ac}} = \frac{-2c}{b \pm \sqrt{b^2-4ac}}\frac{b \mp \sqrt{b^2-4ac}}{b \mp \sqrt{b^2-4ac}} = $$
$$\frac{-2c(b\mp\sqrt{b^2-4ac})}{b^2-(b^2-4ac)} = \frac{-b \mp \sqrt{b^2-4ac}}{2a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
What is the probability that 4 cards drawn randomly from a deck are all hearts, given that there are at least three hearts? Here is the question:
$4$ cards are drawn at random from a standard $52$-card deck. At least
$3$ of them are hearts. What is the probability that they are all
hearts?
I was able to obtain th... | You were doing great, only notice that the number of ways to obtain exactly three hearts is $\binom{13}{3}\times 38$, as there are $\binom{13}{3}$ ways to pick the three hearts and $39$ ways to pick the non-heart.
So the answer is $\frac{\binom{13}{4}}{\binom{13}{4}+39\binom{13}{3}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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If $a^2+b^2+c^2=3$ so $a^2b+b^2c+c^2a\geq3\sqrt[3]{a^2b^2c^2}$ Let $a$, $b$ and $c$ be non-negative numbers such that $a^2+b^2+c^2=3$. Prove that:
$$a^2b+b^2c+c^2a\geq3\sqrt[3]{a^2b^2c^2}$$
I tried Rearrangement, uvw, BW and more, but without any success.
| $$\Leftrightarrow (a^2b+b^2c+c^2a)^2 \ge 9(abc)^{4/3}=3(abc)^{4/3}(a^2+b^2+c^2)$$
$$\sum_{cyc}a^4b^2+2\sum_{cyc}a^3bc^2 \ge 3\sum_{cyc}\sqrt[3]{a^{10}b^4c^4}$$
By AM-GM :
$$a^4b^2+2a^3bc^2 \ge 3\sqrt[3]{a^{10}b^4c^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Probability with balls and a box - complementary event of "exactly" There are $12$ balls in a box: $b$ blue, $y$ yellow and $3$ red balls. Three balls are randomly chosen. If the probability of choosing one blue, one yellow and one red is $3/11$, find the number of yellow balls in a box.
Attempt:
If $t$ is the total nu... | Forget all of that. You don't need to write $t$ in terms of $b$ and $y$, notice that you can always write $P(A)$ as
\begin{align*}
P(A) = 3! \cdot \frac{b}{12} \cdot \frac{y}{11} \cdot \frac{3}{10} = \frac{3}{11}
\end{align*}
(since the order in which you select the balls matters, and there are 3! ways to pick 1 of eac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all $x$ such that $x^6 = (x+1)^6$.
Find all $x$ such that $$x^6=(x+1)^6.$$
So far, I have found the real solution $x= -\frac{1}{2}$, and the complex solution $x = -\sqrt[3]{-1}$.
Are there more, and if so what/how would be the most efficient find all the solutions to this problem? I am struggling to find the res... | We can simplify the problem by substituting $x=y-\frac{1}{2}$.
$$x^6 = \left(x+1\right)^6$$
$$\left(y-\frac{1}{2}\right)^6 = \left(y+\frac{1}{2}\right)^6$$
If we expand both sides and then collect the terms, the even powers of $y$ drop out and only the odd powers remain.
$$6y^5+5y^3+\frac{3}{8}y=0$$
You can factor out ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2109609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Sum of series $\sum \limits_{k=1}^{\infty}\frac{\sin^3 3^k}{3^k}$ Calculate the following sum: $$\sum \limits_{k=1}^{\infty}\dfrac{\sin^3 3^k}{3^k}$$
Unfortunately I have no idea how to handle with this problem.
Could anyone show it solution?
| We can use the following identity (leaving for you to prove)
$$
\sin^3 (3^x) = \frac {1}{4} ( 3 \sin(3^x) - \sin (3^{x+1}) )
$$
That is,
$$
\frac {\sin^3( 3) }{3} = \frac {1}{4} ( \sin (3) - \frac {1}{3} \sin (3^2) )
$$
$$
\frac {\sin^3( 3^2) }{3^2} = \frac {1}{4} (\frac {1}{3} \sin (3^2) - \frac {1}{3^2} \sin (3^3) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2109942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 1
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Differentiability in $(0,0)$ of $f(x,y)=\frac{x^4-y^3}{x^2+y^2}$ Let the function $f$ be defined as $$f(x,y)=\frac{x^4-y^3}{x^2+y^2}$$
If $(x,y)=(0,0)$ then $f$ is equal to zero.
My problem is to prove that this function isn't differentiable at the point $(0,0)$.
My solution:
First idea is maybe to show that $f$ is no... |
We consider the function
\begin{align*}
f(x,y)=\begin{cases}
\frac{x^4-y^3}{x^2+y^2}&\qquad\text{if }(x,y)\ne (0,0)\\
0&\qquad\text{if }(x,y)=(0,0)
\end{cases}
\end{align*}
As already stated by OP we see the partial derivatives exist.
\begin{align*}
\frac{\partial f}{\partial x}(0,0)&=\lim_{h\rightarrow 0}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2111558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Explain why $(a−b)^2 = a^2 −b^2$ if and only if $b = 0$ or $b = a$. This is a question out of "Precalculus: A Prelude to Calculus" second edition by Sheldon Axler. on page 19 problem number 54.
The problem is Explain why $(a−b)^2 = a^2 −b^2 $ if and only if $b = 0$ or $b = a$.
So I started by expanding $(a−b)^2$ to $(a... | Hint: $(a−b)^2 = a^2 −b^2 \iff (a−b)^2 - a^2 +b^2=0 \iff 2b(b-a)=0\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
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Prove that $\lim_{n\to\infty}\frac{6n^4+n^3+3}{2n^4-n+1}=3$ How to prove, using the definition of limit of a sequence, that:
$$\lim_{n\to\infty}\frac{6n^4+n^3+3}{2n^4-n+1}=3$$
Subtracting 3 and taking the absolute value of the function I have:
$$<\frac{n^3+3n}{2n^4-n}$$
But it's hard to get forward...
|
Subtracting 3 and taking the absolute value of the function I have:
$$<\frac{n^3+3n}{2n^4-n}$$
Then, since $n^2 \gt 3$ and $2n^3-1 \gt n^3$ for $n \gt 1\,$:
$$
\require{cancel}
\frac{n^3+3n}{2n^4-n} = \frac{\cancel{n}(n^2+3)}{\cancel{n}(2n^3-1)} \lt \frac{n^2+n^2}{n^3} = \frac{2}{n}
$$
You should hopefully be able to... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove that $\sum\limits_{cyc}\sqrt[4]{a^2+bc+ca}\geq3\sqrt[4]{ab+ac+bc}$ Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$\sqrt[4]{a^2+bc+ca}+\sqrt[4]{b^2+ca+ab}+\sqrt[4]{c^2+ab+bc}\geq3\sqrt[4]{ab+ac+bc}$$
I tried Holder and more, but without any success.
I didn't think that my delirium would interesting t... | Now, I have a solution.
By using of the Ji Chen's lemma ( https://artofproblemsolving.com/community/c6h194103 )
it's enough to prove that
$$\sum_{cyc}(a^2+bc+ca)\geq3(ab+ac+bc),$$
$$\sum_{cyc}(a^2+bc+ca)(b^2+ca+ab)\geq3(ab+ac+bc)^2$$ and
$$\prod_{cyc}(a^2+bc+ca)\geq(ab+ac+bc)^3,$$ which are obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Proving a well-known inequality using S.O.S Using $AM-GM$ inequality, it is easy to show for $a,b,c>0$, $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge 3.$$
However, I can't seem to find an S.O.S form for $a,b,c$
$$f(a,b,c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a} - 3 = \sum_{cyc}S_A(b-c)^2 \ge 0.$$
Update:
Please not... | $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge 3 \Leftrightarrow a^2c + b^2a + c^2b \ge 3abc$$
So we use these S.O.S forms:
*
*$a^3 + b^3 + c^3 - 3abc = \frac{1}{2}(a+b+c)\sum_{cyc}(a-b)^2$.
*$a^3 + b^3 + c^3 - a^2c - b^2a - c^2b = \frac{1}{3}\sum_{cyc}a^3 + a^3 + c^3-3a^2c = \frac{1}{3}\sum_{cyc} (2a+c)(c-a)^2$.
He... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
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Find $\sqrt{1.1}$ using Taylor series of the function $\sqrt{x+1}$ in $x^{}_0 = 1$ with error smaller than $10^{-4}$ I should find $\sqrt{1.1}$ using Taylor series of the function $\sqrt{x+1}$ in $x^{}_0=1$ with error smaller than $10^{-4}$.
The first derivatives are
$$f'(x)=\frac{1}{2\sqrt{x+1}}$$
$$f''(x)=\frac{-1}... | Hint:$$f(x_0+h)=f(x_0)+f'(x_0)h+f''(x_0).\dfrac{h^2}{2!}+f^3(x_0).\dfrac{h^3}{3!}+f^4(x_0).\dfrac{h^4}{4!}+o(h^5)\\h=0.1 \\
\to f(x_0+h)=f(x_0)+f'(x_0)(0.1)+f''(x_0).\dfrac{(0.1)^2}{2!}+f^3(x_0).\dfrac{(0.1)^3}{3!}+f^4(x_0).\dfrac{(0.1)^4}{4!}+o((0.1)^5)$$ so in your case $f^4(x_0).h^4$ is in order of $10^{-4}$
and the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to evaluate $\int_0^\pi \cos(x) \cos(2x) \cos(3x) \cos(4x)\, dx$ Is there an easy way to evaluate the integral $\int_0^\pi \cos(x) \cos(2x) \cos(3x) \cos(4x)\, dx$?
I know that I can plugin the $e$-function and use the linearity of the integral. However this would lead to 16 summands which I really dont want to cal... | There is a well-known identity which says
$$\cos A + \cos B = 2\cos\left(\frac{A-B}{2}\right)\cos\left(\frac{A+B}{2}\right)$$
If we put $\frac{A-B}{2} = x$ and $\frac{A+B}{2}=2x$ then we get $A=3x$ and $B=x$, so $$ \cos x \cos 2x \equiv \frac{1}{2}(\cos x+\cos 3x) $$
We can repeat this for $\cos 3x$ and $\cos 4x$. Solv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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Find all prime solutions of equation $5x^2-7x+1=y^2.$ Find all prime solutions of the equation $5x^2-7x+1=y^2.$
It is easy to see that
$y^2+2x^2=1 \mod 7.$ Since $\mod 7$-residues are $1,2,4$ it follows that $y^2=4 \mod 7$, $x^2=2 \mod 7$ or $y=2,5 \mod 7$ and $x=3,4 \mod 7.$
In the same way from $y^2+2x=1 \mod 5$ w... | Completing the square and dividing by $5$, we have
$$ (10 x - 7)^2 - 20 y^2 = 29$$
Thus $z = 10 x - 7$ and $w = 2 y$ are solutions of the Pell-type equation
$$ z^2 - 5 w^2 = 29$$
The positive integer solutions of this can be written as
$$\pmatrix{z\cr w\cr} = \pmatrix{9 & 20\cr 4 & 9\cr}^n \pmatrix{7\cr 2\cr} \ \text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
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Coefficient of $x^2$ in $(x+\frac 2x)^6$ I did $6C4 x^2\times (\dfrac 2x)^4$ and got that the coefficient of $x^2$ is $15$, but the answer is $60$, why? Did I miss a step?
| It's much easier [for me] to see where the $x^2$ terms are after rewriting: $$(x + \dfrac{2}{x})^6 = (\dfrac{x^2 + 2}{x})^6 = \dfrac{1}{x^6}(x^2 +2)^6$$
Then we only have $x^2$ terms where the expansion of the binomial gives $x^8$ (because we're dividing everything by $x^6$). That term is $15(x^2)^4 \cdot 2^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Prove that $\frac {\sec (16A) - 1}{\sec (8A) - 1}=\frac {\tan (16A)}{\tan (4A)}$ Prove that:$$\frac {\sec (16A) - 1}{\sec (8A) - 1}=\frac {\tan (16A)}{\tan (4A)}$$.
My Attempt,
$$L.H.S= \frac {\sec (16A)-1}{\sec (8A)-1}$$
$$=\frac {\frac {1}{\cos (16A)} -1}{\frac {1}{\cos (8A)} -1}$$
$$=\frac {(1-\cos (16A)).(\cos (8A)... | $\displaystyle \frac{1-\cos^2(16 A)}{1+\cos (16A)}\times \frac{\cos 8 A}{\cos (16 A)} \times \frac{1+\cos 8A}{1-\cos^2 (8A)} = \frac{\sin^2 16A}{2\cos^2(8A)} \times \frac{\cos 8A}{\cos 16A} \times \frac{2\cos^2 4A}{\sin^2 8A}$
So $$ = \frac{\sin^2 (16 A)}{\sin (16A)} \times \frac{2\cos^2 (4A)}{\cos 16 A}\times \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Two Inequality Challenges involving $\sum_{cyc}\left(\frac{P(a,b,c)}{Q(a,b,c)}\right)^r$ for homogeneous polynomials $P$ and $Q$ of equal degree Let $F(a,b,c,r)=(P(a,b,c)/Q(a,b,c))^r$ where $P(a,b,c)$ and $Q(a,b,c)$ are homogenous polynomials with equal degrees $deg(p)=deg(q)=n$, defined for non-negative $a,b$, and $c... | Update 08/18/2019
For challenge 1: Let
$$F(a, b, c) = \frac{a^2b + 2a^2c + 2ab^2 + b^3 + 31abc}{(a+b+50c)(a+b+c)^2}.$$
Let $G(a,b,c) = F(a, b, c) + F(b, c, a) + F(c, a, b)$.
Then the minimum of $G(a, b, c)$ is not achieved at $(1, 1, 1)$ or $abc=0$.
Indeed, we have $G(1,1,1) = 37/156 \approx 0.2372$ and
$$G(a, 3-a, 0) ... | {
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Prove: If $x+y+z=xyz$ then $\frac {x}{1-x^2} +\frac {y}{1-y^2} + \frac {z}{1-z^2}=\frac {4xyz}{(1-x^2)(1-y^2)(1-z^2)}$ If $x+y+z=xyz$, prove that:
$$\frac {x}{1-x^2} +\frac {y}{1-y^2} + \frac {z}{1-z^2}=\frac {4xyz}{(1-x^2)(1-y^2)(1-z^2)}$$.
My Attempt:
$$L.H.S=\frac {x}{1-x^2}+\frac {y}{1-y^2}+\frac {z}{1-z^2}$$
$$=\f... | $\textbf{HINT:}$ Try putting $x=tan(\alpha);y=tan(\beta);z=tan(\gamma)$
Use: $$tan(2\theta)=\frac{2tan(\theta)}{1-tan^2(\theta)}$$
$\textbf{Note that $\alpha +\beta+\gamma=\pi$}$ by the condition since: $$tan(\alpha+\beta+\gamma)=\frac{\Sigma tan(\alpha)-tan(\alpha)tan(\beta)tan(\gamma)}{1-\Sigma tan(\alpha)tan(\beta)}... | {
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"source": "stackexchange",
"question_score": "8",
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Find the least number $x$ such that $ 11$ divides $x$ and sum of its digits $S(x)$ is $27$. Find the least number $x$ such that $ 11$ divides $x$ and sum of its digits $S(x)$ is $27$.
Since $S(999)=27 $ it is clear that the number of digits $n>3.$ Let $x_i$ be digits then we have two equations
\begin{cases}
x_1+x_2+\c... | When $n = 5$, $10989$ is already the least solution to the problem.
First, when $n \le 4$, no solution exists. For $n = 3$, we deduce from the sum of digits $S(x) = 27$ that $x = 999$, which is not a multiple of $11$. For $n=4$, we have $x_2 + x_4 \equiv 8 \pmod{11}$.
Since each digit is an integer from $0$ to $9$ in... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Prove the $\frac {2r+5}{r+2}$ is always a better approximation of $\sqrt {5}$ than $r$.
Prove the $\frac {2r+5}{r+2}$ is always a better approximation of $\sqrt {5}$ than $r$.
SOURCE : Inequalities (Page Number 4 ; Question Number 207)
I tried a lot of approaches, but without success.
I rewrote $\frac {2r+5}{r+2}$ as... | $$\frac{2r+5}{r+2} - \sqrt{5} = \frac{2r+5-\sqrt{5}r-2\sqrt{5}}{r+2} = \frac{(2-\sqrt{5})r - (2-\sqrt{5})\sqrt{5}}{r+2} = \frac{2-\sqrt{5}}{r+2}(r-\sqrt{5})$$
That means as long as $$\left|\frac{2-\sqrt{5}}{r+2}\right| < 1$$ holds, the proposition is true. If we assume $r > 0$ as reasonable for an approximation of a sq... | {
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"source": "stackexchange",
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How to prove that $\sum_{n \, \text{odd}} \frac{n^2}{(4-n^2)^2} = \pi^2/16$? The series:
$$\sum_{n \, \text{odd}}^{\infty} \frac{n^2}{(4-n^2)^2} = \pi^2/16$$
showed up in my quantum mechanics homework. The problem was solved using a method that avoids evaluating the series and then by equivalence the value of the serie... | Wolfram Alpha gives me the partial fraction expansion:
$$\frac{n^2}{(4 - n^2)^2} = \frac{1}{8}\left(\frac 1{n-2}-\frac{1}{n + 2}\right) + \frac{1}{4}\left(\frac{1}{(n-2)^2}+\frac{1}{(n+2)^2}\right)$$
So the first part telescopes, and the second part will be some modified version of $\zeta(2)$.
In more detail: $$\begin{... | {
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"source": "stackexchange",
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If:$(\sqrt{x + 9})^{\frac{1}{3}} - (\sqrt{x-9})^{\frac{1}{3}} = 3$, Find $x^2$
If:$$\sqrt[3]{(x + 9)} - \sqrt[3]{(x-9)} = 3$$
Find $x^2$
I can't seem to solve this question. Any hints or solutions is welcomed.
| For $\sqrt[3]{x+9}-\sqrt[3]{x-9}=3$ we have an interesting problem.
It's
$$\sqrt[3]{x+9}+\sqrt[3]{9-x}-3=0$$
and since $\sqrt[3]{x+9}=\sqrt[3]{9-x}=-3$ is impossible, it's
$$9+x+9-x-27-9\sqrt[3]{(9+x)(9-x)}=0,$$
which gives $x^2=80$.
I used $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$,
where $a^2+b^2+c^2-ab-ac-bc=0... | {
"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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The extremities of a diameter of a circle.. The extremities of a diameter of a circle have co ordinates $(6,-2)$ and $(6,8)$. What length does the circle intercept along $X$ axis..?
My Attempt:
Let, $(6,-2)\equiv (x_1,y_1)$ and $(6,8)\equiv (x_2,y_2)$ be the end points of diameter of circle.
Now,
$$(x-x_1)(x-x_2)+(y-y... | Continue with your solution -
The lengths of intercepts made by the circle
$$x^2 + y^2 + 2gx + 2fy + c = 0$$
with X and Y axes are $2\sqrt{g^2−c}$ and $2\sqrt{f^2−c}$ respectively.
Or -
Mid point of (6,-2) and (6,8) is (6,3).
Also radius r = $\frac 12 \sqrt{(6-6)^2 + (8+2)^2}$
= $\frac 12 \sqrt {100}$
= 5
Equation ... | {
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"timestamp": "2023-03-29T00:00:00",
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Continuity and integration using Dominated Convergence theorem I have to use the Dominated Convergence Theorem to show that $\lim \limits_{n \to \infty}$ $\int_0^1f_n(x)dx=0$ where $f_n(x)=\frac{n\sqrt{x}}{1+n^2x^2}$.
I did the following:
$$\frac{n\sqrt{x}}{1+n^2x^2} <\frac{n\sqrt{x}}{n^2x^2} = \frac{x^{-\frac{3}{2}}... | Use for example:
$$\frac{n\sqrt x}{1+n^2x^2}=\frac{1}{\sqrt x}\cdot\frac{1}{\frac{1}{nx}+nx}\le \frac{1}{2\sqrt x}$$
since $u+\frac{1}{u}\ge 2$ for $u>0$
And $\int_0^1\frac{1}{2\sqrt x}=1<\infty$
(this is equivalent to AM-GM inequality on the denominator $1+n^2x^2\ge2nx$)
| {
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"source": "stackexchange",
"question_score": "2",
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Determine whether or not $\exp\left(\sum_{n=1}^{\infty}\frac{B(n)}{n(n+1)}\right)$ is a rational number Let $B(n)$ be the number of ones in the base 2 expression for the positive integer n.
Determine whether or not $$\exp\left(\sum_{n=1}^{\infty}\frac{B(n)}{n(n+1)}\right)$$ is a rational number.
Attempt:
I tried to mak... | For $B(n)$ we have the following properties:
\begin{align}
& B(2k) = B(k) & \text{if }n = 2k+1 \\
& B(2k + 1) = B(k) + 1 & \text{if }n = 2k
\end{align}
Hence,
$$S = \sum\limits_{n=1}^{+\infty} \dfrac{B(n)}{n(n+1)} = \sum\limits_{k=0}^{+\infty} \dfrac{B(2k+1)}{(2k+1)(2k+2)} + \sum\limits_{k=0}^{+\infty} \dfrac{B(2k + 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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To find possible values of $(C-A)$ Let $A,B,C$ satisfy $0<A<B<C<2\pi$. If $$\cos(x+A)+\cos(x+B)+\cos(x+C)=0$$ for all x $\in \mathbb{R}$, then i have to find possible values of $(C-A)$.
i don't know how to begin
Thanks
| We have that $$\cos (x+A) + \cos (x+B) + \cos (x + C)=0$$ $$\Rightarrow \cos x (\cos A + \cos B + \cos C) - \sin x (\sin A + \sin B + \sin C)=0$$
We can conclude thus $$\cos A + \cos B + \cos C =0 \tag {1}$$ $$\sin A + \sin B + \sin C =0 \tag {2}$$
Using $(1)$ and $(2)$, we get, $$(\cos A + \cos C)^2 +(\sin A + \sin C)... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Find all real number satisfying $10^x+11^x+12^x = 13^x+14^x$
Find all real number $x$ satisfying $10^x+11^x+12^x = 13^x+14^x$
My Work
Dividing by $13^x$ we get
$$\left( \frac{10}{13} \right)^x +
\left( \frac{11}{13} \right)^x +
\left( \frac{12}{13} \right)^x
= 1 + \left( \frac{14}{13} \right)^x$$
The LHS is a ... | I just solved my problem. A well known fact that
$$a^2 +(a+1)^2 + \cdots + (a+k)^2 = (a+k+1)^2 +(a+k+2)^2+ \cdots (a+2k)^2$$
Satisfies when $a = k(2k+1)$.
So in my question $a = 2(2\cdot2+1)$, So without any thinking $x=2$ is an answer.
This helps suppose the equation was -
$$55^x+56^x+57^x+58^x+59^x+60^x=61^x+62^x+6... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Evaluate the integral $\int \frac{x^2(x-2)}{(x-1)^2}dx$
Find $$\int \frac{x^2(x-2)}{(x-1)^2} dx .$$
My attempt:
$$\int \frac{x^2(x-2)}{(x-1)^2}dx = \int \frac{x^3-2x^2}{x^2-2x+1}dx $$
By applying polynomial division, it follows that
$$\frac{x^3-2x^2}{x^2-2x+1} = x + \frac{-x}{x^2-2x+1}$$
Hence $$\int \frac{x^3-2x^2}{... | The most efficient method is probably to substitute $t := x - 1$, as I see Yves has done in his answer, as this rewrites the integrand as a sum of power functions. In you don't see this, proceeding the standard way goes as follows:
After writing the integrand as a 'proper polynomial fraction', rewrite the denominator i... | {
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"source": "stackexchange",
"question_score": "2",
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Why is $X^4-X^2+1$ reducible over $\Bbb F_5$? I have checked $X^4-X^2+1=0$ and got the solution that the polynom is never equal 0 so it should be irreducibel.
However it is.
My prof gave me that hint:
$X^4-X^2+1=(X^2+aX+b)(X^2+cX+d)$
Unfortunately that did not really help me. How can I solve that equation and where ... | $$x^{4} - x^{2} + 1= x^{4} - 2 x^{2} + 1 + x^{2} = x^{4} - 2 x^{2} + 1 - 4 x^{2} =\\= (x^{2} - 1)^{2} - (2 x)^{2} = (x^{2} - 1 + 2 x) (x^{2} - 1 - 2 x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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$\lim_{z \to \exp(i\pi/3)} \frac{z^3+8}{z^4+4z+16}$ Find $$\lim_{z \to \exp(i \pi/3)} \dfrac{z^3+8}{z^4+4z+16}$$
Note that
$$z=\exp(\pi i/3)=\cos(\pi/3)+i\sin(\pi/3)=\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$$
$$z^2=\exp(2\pi i/3)=\cos(2\pi/3)+i\sin(2\pi/3)=-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$$
$$z^3=\exp(3\pi i/3)=\cos(\pi)+i\... | You know $z^3=(\exp(\pi i/3))^3=\exp(3\pi i/3)=-1$ and $z=\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$ thus with substitution
\begin{eqnarray}
\lim_{z \to \exp(i \pi/3)} \dfrac{z^3+8}{z(z^3+4)+16}
&=&
\lim_{z \to \exp(i \pi/3)} \dfrac{-1+8}{(\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2})(-1+4)+16}\\
&=&
\dfrac{14}{35+3\sqrt{3}i}\\
&=&
\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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finding definite integral involving inverse of cot function finding $\displaystyle \int^{\pi}_{-\frac{\pi}{3}}\bigg[\cot^{-1}\bigg(\frac{1}{2\cos x-1}\bigg)+\cot^{-1}\bigg(\cos x - \frac{1}{2}\bigg)\bigg]dx$
Attempt:
\begin{align}
& \int^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\bigg[\cot^{-1}\bigg(\frac{1}{2\cos x-1}\bigg)+\co... | I see some tricks in here. Ideas, as yet unfinished, I'm out of time now but will try to come back later.
Consider $cot^{-1}(b) = \theta \leftrightarrow cot(\theta) = b$
and then $tan(\theta ) = 1/b$ so $\theta = tan^{-1}(1/b)$
Also $(cos(x) - 1/2 ) = (2 cos(x) -1)/2$
$$f(x) = cot^{-1}(\frac{1}{2cos(x) -1}) + cot^{-1}(... | {
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"source": "stackexchange",
"question_score": "3",
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Solve an integral $\int\frac{\cos^3 x}{\sin^3 x+\cos^3 x}dx$ Solve an integral $$\int\frac{\cos^3 x}{\sin^3 x+\cos^3 x}dx$$
I tried to divide the numerator and denominator by $\cos^4 x$ to get $\sec x$ function but the term ${\sin^3 x}/{\cos^4 x}$ gives $\tan^2 x\sec^2 x\sin x$. How to get rid of $\sin x$ term?
| $\displaystyle\frac{\cos^3x}{\sin^3x+\cos^3x}=\frac{1}{\tan^3x+1}=\frac{\tan^2x+1}{(\tan^2x+1)(\tan^3x+1)}=\frac{1}{\cos^2x(\tan^2x+1)(\tan^3x+1)}$
Now replace $\displaystyle t=\tan x,dt=\frac{dx}{\cos^2x}$ and do partial fractions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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A series involve combination I want find another Idea to find sum of $\left(\begin{array}{c}n+3\\ 3\end{array}\right)$ from $n=1 ,to,n=47$
or $$\sum_{n=1}^{47}\left(\begin{array}{c}n+3\\ 3\end{array}\right)=?$$ I do it first by turn $\left(\begin{array}{c}n+3\\ 3\end{array}\right)$ to $\dfrac{(n+3)(n+2)(n+1)}{3!}=\dfr... | This was my second try :
$$\sum_{n=1}^{47}\left(\begin{array}{c}n+3\\ 3\end{array}\right)=\\
\dfrac16\sum_{n=1}^{47}(n+3)(n+2)(n+1)=\\
\dfrac16.\dfrac14\sum_{n=1}^{47}(\color{red} {n+4-n})(n+3)(n+2)(n+1)=\\
\dfrac{1}{24}\sum_{n=1}^{47}(n+4)(n+3)(n+2)(n+1)-(n+3)(n+2)(n+1)n\\
\\by \space {[f(n)=(n+3)(n+2)(n+1)n]}\\
\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How is a Generator Matrix for a (7, 4) Hamming code created? I see that a generator matrix is created with the following formulae:
$$G = \left[I_{k}|P\right]$$
I do not understand what P is in this case. In my notes, I am told that in a (7, 4) Hamming code situation my
$$G =
\begin{pmatrix} 1 & 0 & 0 & 0 & 1 & 0 & 1 \\... | $G$ represents the mapping from the input to the codewords of the code. You get the codeword vector $c$ by multiplying the input vector $v$ from the left of $G$:
$$c = vG.$$
So, you see that the input vector $\begin{pmatrix} 1 & 0 & 0 & 0 \end{pmatrix}$ maps to the first row of $G$ (if you doubt me, do the calculation)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 1
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Undetermined Coefficients
I tried to change it to $y" + y = (14\sin x-28 \sin^3 x)$. The complementary solution is $C_1\cos x+C_2\sin x$ and the particular solution to $y" + y = 14 \sin x$ is $-7\sin x$. How do you find the particular solution to $y" + y = -28\sin^3 x$ ?
What would be your guess?
Ans: $$c_1 \sin(x\sqr... | Using double-angle formula, you can rewrite $\sin(x)\cos(2x)$ as
$$ \sin(x)\cos(2x) = \frac{1}{2}[\sin(3x) - \sin(x)].$$
Indeed,
\begin{align*}
\sin(3x) = \sin(2x + x) & = \sin(2x)\cos(x) + \cos(2x)\sin(x) \\
\sin(x) = \sin(2x-x) & = \sin(2x)\cos(x) - \cos(2x)\sin(x).
\end{align*}
Thus, this gives
$$ y'' + y = 7[\sin(... | {
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"timestamp": "2023-03-29T00:00:00",
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Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$
Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$.
I expanded the brackets and applied AM-GM on all of the eight terms to get :
$$\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big) \geq 3\... | Another way.
We need to prove that $$27\prod_{cyc}((a+b+c)^2+6a^2)\geq125(a+b+c)^6$$ or
$$\sum_{sym}\left(8a^6+15a^5b+159a^4b^2+160a^3b^3-\frac{249}{2}a^4bc-336a^3b^2c+\frac{237}{2}a^2b^2c^2\right)\geq0$$ or
$$168\sum_{sym}(a^3b^2-2a^3b^2c+a^2b^2c^2)+$$
$$+\sum_{sym}\left(8a^6+15a^5b+159a^4b^2-8a^3b^3-\frac{249}{2}a^4b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Given the positive numbers $a, b, c$. Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le \frac{3}{2}$ Given the positive numbers $a, b, c$ satisfy $a+b+c\le \sqrt{3}$. Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le \frac{3}{2}$
My Try (Edited from C... | Note that $f(x)=\frac{x}{\sqrt{x^2+1}}$ is a concave function for $x>0$. This follows as $$f''(x)=-\dfrac{3x}{(x^2+1)^{\frac{5}{2}}}<0$$ Now use Jensen's inequality. Note that $$f(a)+f(b)+f(c) \le 3 f\left(\frac{a+b+c}{3}\right)=3f\left(\frac{\sqrt{3}}{3}\right)=\frac{3}{2}$$
As $a+b+c=\sqrt{3}$.
The maximum is achiev... | {
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"source": "stackexchange",
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"answer_id": 0
} |
Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Let $\omega$ be a complex number such that $\omega^7 = 1$ and $\omega \neq 1$. Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Then $\alpha$ and $\beta$ are roots of the quadratic
[x^2 + p... | $$q=\alpha\beta=3+\sum_{j=1}^6w^j=3+w\frac{w^6-1}{w-1}=3+\frac{1-w}{w-1}=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $\left|z^3 + {1 \over z^3}\right| \le 2$ then $\left|z + {1 \over z}\right| \le 2$
$\displaystyle \left|z^3 + {1 \over z^3}\right| \le 2$ prove that $\displaystyle \left|z + {1 \over z}\right| \le 2$
$$\left|z^3 + {1 \over z^3}\right| = \left(z^3 + {1 \over z^3}\right)\left(\overline z^3 + {1 \over \overline{z}^3... | Let $a = \lvert z + \frac 1z \rvert$. Then
$$
a^3 = \left\lvert \left( z + \frac 1z \right)^3 \right\rvert =
\left\lvert z^3 + 3z + \frac 3z + \frac{1}{z^3} \right\rvert
\le \left\lvert z^3 + \frac{1}{z^3} \right\rvert
+ 3 \left\lvert z + \frac{1}{z}\right\rvert \le 2 + 3a
$$
so that
$$
0 \ge a^3 - 3a - 2 = (a-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How can we show that $\sum_{n=0}^{\infty}{2n^2-n+1\over 4n^2-1}\cdot{1\over n!}=0?$ Consider
$$\sum_{n=0}^{\infty}{2n^2-n+1\over 4n^2-1}\cdot{1\over n!}=S\tag1$$
How does one show that $S=\color{red}0?$
An attempt:
$${2n^2-n+1\over 4n^2-1}={1\over 2}+{3-2n\over 2(4n^2-1)}={1\over 2}+{1\over 2(2n-1)}-{1\over (2n+1)}... | This is as simple as $1-2-3$.
$(1)$
Note that
$$\begin{align}
\sum_{n=0}^\infty \frac{1}{(2n-1)n!}&=-1+\sum_{n=0}^\infty\frac{1}{(2n+1)(n+1)!}\\\\
&=-1+\sum_{n=0}^\infty \left(\frac{2}{2n+1}-\frac{1}{n+1}\right)\frac{1}{n!}\\\\
&=-1+\sum_{n=0}^\infty \left(\frac{2}{2n+1}\right)\frac1{n!}-\sum_{n=0}^\infty\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove with the $\epsilon - \delta$ proof that $\lim_{(x, y) \rightarrow (a, a)}\frac{x^3 - y^3}{x^2 - y^2} = \frac{3a}{2}$ So that is the question. Here is my partial work:
Let $x = s + a$ and $y = t + a$.
Then $d(x, y) = \sqrt{s^2 + t^2}$. If $\sqrt{s^2 + t^2} < \delta$, then $s^2 + t^2 < \delta^2$.
So then
\begin{eqn... | Here is one approach: if $a=0,\ $ the result is almost immediate. Otherwise, assume $a\neq 0,\ $ set $x-a=\delta \cos t, y-a=\delta \sin t,\ $
take $0<\delta <1$ so small that $|2a+\delta (\sin t+\cos t)|>\frac{|a|}{2},\ $
Then,
$\left | \frac{\left ( (a+\delta \cos t)^{2}+((a+\delta \cos t))(a+\delta \sin t)+(a+\del... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
An integral of rational function with third power of cosine hyperbolic function Prove
$$\int_{-\infty}^{\infty}\frac{1}{(5 \pi^2 + 8 \pi x + 16x^2)
}\frac{\cosh\left(x+\frac{\pi}{4} \right)}{\cosh^3(x)}dx =
\frac{2}{\pi^3}\left(\pi \cosh\left(\frac{\pi}{4} \right)-4\sinh\left(
\frac{\pi}{4}\right) \right)$$
Attem... | Brevan Ellefsen answer was an inspiration for me to solve it using Contour integration,
Consider
$$f(z) = \frac{\sinh(z)}{z \sinh^3(z-\pi/4)}$$
If we integrate around a contour of height $\pi$ and strech it to infinity we get
By taking $T \to \infty $
$$\color{red}{\int^{i\pi/2+\infty}_{-i\pi/2+\infty}f(x)\,dx}+\col... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Solve $\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9}$ without using L'Hôpital's rule $\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9}$
To do this I tried 2 approaches:
1: If $\lim_{x \rightarrow 0} \frac{\ln(x+1)}{x} = 1$, $\lim_{x\to1}\frac{\ln(x)}{x-1}=1$ and $\lim_{x\to0}\frac{\ln(x+1)}x=\lim_{u\to1}\fra... | Hint: Let $x=u+3$ and use log rules.
$$\frac{\ln\sqrt{x-2}}{x^2-9}=\frac{\frac12\ln(x-2)}{(x+3)(x-3)}=\frac1{2(u+6)}\frac{\ln(u+1)}u\to\frac1{12}$$
To the first and second try, you assumed that just because the logarithm goes to 1 and the denominator goes to 0 that the limit is 1, but that is only true if the stuff ins... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find the coefficient of $x^{29}$ in the given polynomial. The polynomial is :
$$
\left(x-\frac{1}{1\cdot3}\right) \left(x-\frac{2}{1\cdot3\cdot5}\right) \left(x-\frac{3}{1\cdot3\cdot5\cdot7}\right) \cdots \left(x-\frac{30}{1\cdot3\cdot5\cdots61}\right)
$$
What I've done so far : The given polynomial is an expression of... | HINT:$$\underbrace{1.3,1.3.5,...,1.3.5....61}_{30-terms} $$so
$$\underbrace{(x-\frac{1}{1.3})(x-\frac{2}{1.3.5})(x-\frac{3}{1.3.5.7})......(x-\frac{30}{1.3.5.....61})}_{30-terms }$$we have
,to obtain $x^{29}$
we have
when you want
$x^{29}$ from $(x+a_1)(x+a_2)(x+a_3)...(x+a_{30})$ you have to find this
$$(x+0)(x+0)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2151030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to show that if $9n^2=a^2+b^2$, $a$ and $b$ are multiples of $3$ To be honest, I don't know where to start with this problem:
Let $n\in \mathbb{N}$. Prove that if $9n^2$ is the sum of two perfect squares $(a^2,b^2)$, then $a$ and $b$ are multiples of $3$.
| To clarify on vadim's answer after concluding that $$\cases{0^2=0\equiv 0\mod3 \\1^2=1\equiv 1\mod3 \\2^2=4\equiv 1\mod3 }$$
we can make a table of all combinations:
$$\begin{array}{|cc|cc|}\hline a(mod3)&b(mod3)&a^2(mod3)&b^2(mod3)\\\hline
0&0&0&0\\0&1&0&1\\0&2&0&1\\1&0&1&0\\1&1&1&1\\1&2&1&1\\2&0&1&0\\2&1&1&1\\2&2&1&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2153527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Proving Saalschutz Theorem I saw this in a pdf, and I'm wondering
Questions:
*
*How do you prove Saalschutz Theorem:
$$_3F_2\left[\begin{array}{c,c}-x,-y,-z\\n+1,-x-y-z-n\end{array}\right]=\dfrac {\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(z+x+n+1)}{\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(x+y+z+n+1)}\tag... |
Prove$$_3F_2\left[\begin{array}{c,c}-x,-y,-z\\n+1,-x-y-z-n\end{array}\right]=\dfrac {\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(z+x+n+1)}{\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(x+y+z+n+1)}$$
Proof: Begin with the identity$$(1-z)^{a+b-c}\space_2F_1(a,b;c;z)=_2F_1(c-a,b-a;c;z)\tag1$$This can be easily prov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2155025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Minimize $P=5\left(x^2+y^2\right)+2z^2$ For $\left(x+y\right)\left(x+z\right)\left(y+z\right)=144$, minimize $$P=5\left(x^2+y^2\right)+2z^2$$
I have no idea. Can you make a few suggestions?
| By AM-GM
$1728=(3x+3y)(2x+2z)(2y+2z)\leq \left(\frac{5x+5y+4z}{3}\right)^3$
$\Rightarrow 5x+5y+4z\geq 36$
By Cauchy-Schwarz we have:
$18P=(5x^2+5y^2+2z^2)(5+5+8)\geq (5x+5y+4z)^2\geq 36^2$
$$\Rightarrow P\geq 72$$
Hence $P_{\min}=72\Leftrightarrow (x,y,z)=(2,2,4)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2156334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
A self inverse function $f(x)=\frac{ax+1}{x-b}$
If $$f(x)=\frac{ax+1}{x-b} \forall x \in\mathbb{R}-b,ab\neq1,a\neq1$$ is a self inverse function such that $$\frac{f(4)}{4}=\frac{f(12)}{12}={f\left(\frac{1+b}{1-a}\right)}$$The question is to find out $a$ and $b$
For a self inverse function $f(f(x))=x$.So I tried to p... | Suppose $b\neq 0$. Then $f(0)=-1/b$ and if $f$ is self inverse
$$0=f(-1/b)=\frac{a/b-1}{b+1/b}=\frac{a-b}{b^2+1}.$$
It follows that $a=b$. Then
$$f(f(x))=\frac{a\frac{ax+1}{x-a}+1}{\frac{ax+1}{x-a}-a}=\frac{a^2x+a+(x-a)}{ax+1-a(x-a)}=\frac{a^2x+x}{1+a^2}=x$$
and $f$ is self-inverse. It suffices now to check the remaini... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Which of the following numbers is greater? Which of the following numbers is greater?
Without using a calculator and logarithm.
$$7^{55} ,5^{72}$$
My try
$$A=\frac{7^{55} }{5^{55}×5^{17}}=\frac{ 7^{55}}{5^{55}}×\frac{1}{5^{17}}= \left(\frac{7}{5}\right)^{55} \left(\frac{1}{5}\right)^{17}$$
What now?
| Consider that
$$
\frac{7^{55}}{5^{72}} = 7\left(\frac{7^3}{5^4}\right)^{18}
$$
Now, hand-calculation is easy enough on the bracketed term: $7^3=7\times49=343$ and $5^4=25^2=625$. As such, the bracketed term is just a little larger than $1/2$, and certainly less than $1/\!\sqrt{2}$. Therefore,
$$
7\left(\frac{7^3}{5^4}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 2
} |
Which of the following $u=\sqrt{c+1} - \sqrt{c}$ and $v=\sqrt{c} - \sqrt{c-1}$ is greater? If
$$u=\sqrt{c+1} - \sqrt{c}$$
and
$$v=\sqrt{c} - \sqrt{c-1}$$
then, which among $u$ and $v$ is greater?
Please help without substituting values.
| $u=\frac{(\sqrt{c+1}-\sqrt{c})(\sqrt{c+1}+\sqrt{c})}{\sqrt{c}+\sqrt{c+1}} = \frac{1}{\sqrt{c}+\sqrt{c+1}}$
By analogy.
$v=\frac{1}{\sqrt{c}+\sqrt{c-1}}$.
So,$v>u$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$. Find $(a+b+c)$ Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$.
Find $(a+b+c)$.
I computed the whole product ;If $(a+b+c)=x\implies (1+x)(1+\frac{bc+ca+ab}{abc})=16$. Unable to view how to p... | We know :$a>0 \to a+\frac1a \geq 2$ and now;
$$(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\\
1+1+1+1+a+b+c+\frac1a+\frac1b+\frac1c+\frac ab+\frac ac +\frac ba+\frac bc +\frac ca+\frac cb=\\
4+(a+\frac 1a)+(b+\frac 1b)+(c+\frac 1c)+(\frac ba+\frac ab)+(\frac ac+\frac ca)+(\frac cb+\frac bc)\geq 4+3(2)+3(2)\\ l.h.s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2163597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the limits without L'Hôpital:$\lim_{ x \to 0 }\frac{x-\sin x}{x-\tan x}=? $ Find the limits without L'Hôpital's rule
$$\lim_{ x \to 0 }\frac{x-\sin x}{x-\tan x}=? $$
My Try:
$$\lim_{ x \to 0 }\frac{\sin(\pi-x)-\sin x}{\tan(\pi+x)-\tan x}=?\\\lim_{ x \to 0 }\frac{2\sin(\frac{\pi}{2}-x)\cos(\frac{\pi}{2})}{\frac{\s... | You can write the Sin and the Tan as exponential series, and then you get
$$x-\sin(x)=\frac{x^3}{6}-\frac{x^5}{120}+\frac{x^7}{5040}-\frac{x^9}{362880}+O\left(x^{11}\right
)$$
and
$$-\frac12 (x-\tan(x))=\frac{x^3}{6}+\frac{x^5}{15}+\frac{17 x^7}{630}+\frac{31
x^9}{2835}+O\left(x^{11}\right)$$
If you divide the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Laplace transform of Bessel function of order zero I'm trying to prove that the Laplace transform of the function
$$
J_0(a\sqrt{x^2+2bx})
$$
is
$$
\frac{1}{\sqrt{p^2+a^2}} \exp\left\{bp- b\sqrt{p^2+a^2} \right\}
$$
as asserted in the eqworld. This formula can also be found in the book "Tables of Integral Transforms" pa... | Here is the complete answer the my first question. Let $J_0$ denote the Bessel function of order zero. We want to show that
\begin{equation}
\color{red}{\mathcal{L}(J_0(a\sqrt{x^2+2bx}))} = \frac{1}{\sqrt{p^2+a^2}} \exp\left\{bp- b\sqrt{p^2+a^2} \right\}
\end{equation}
with $a$ and $b$ arbitrary real numbers.
Denoting ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\cos 20^{\circ} + \cos 100^{\circ} + \cos {140^{\circ}} = 0$ Assume $A = \cos 20^{\circ} + \cos 100^{\circ} + \cos 140^{\circ}$ . Prove that value of $A$ is zero.
My try : $A = 2\cos 60^{\circ} \cos 40^{\circ} + \cos 140^{\circ}$ and I'm stuck here
| $\cos 20^{\circ}+\cos 100^{\circ}+\cos 140^{\circ}=\\
=\cos 20^{\circ}+\cos (120^{\circ}-20^{\circ})+\cos (120^{\circ}+20^{\circ})=\\
=\cos 20^{\circ}+\cos 120^{\circ}\cos 20^{\circ}+\sin 120^{\circ}\sin 20^{\circ} +\cos 120^{\circ}\cos 20^{\circ}-\sin 120^{\circ}\sin 20^{\circ} =\\
=\cos 20^{\circ}+2\cos 120^{\circ}\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2168706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What is the problem of difference between instruction's answer and my answer? Here is the question and its answer from instructor.
But I noticed that my solution is slightly different,
$$\int \sin^{4}x\,dx = \int (\frac{1-\cos{2x}}{2})^{2}\,dx = \frac{1}{4}\int{1-2\cos{2x} + \cos^{2}{2x}}\,dx = $$
$$ \frac{1}{4}\int{1... | You have $\dfrac{1}{4}\cdot \left(x+\dfrac{1}{2}x\right) = \dfrac{1}{4}\cdot \dfrac{3}{2}x = \dfrac{3}{8}x$.
The answer on the board has
$$\dfrac{1}{2}x- \dfrac{1}{8}x = \dfrac{3}{8}x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2169547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I add the terms in the binomial expansion of $(100+2)^6$? So, I stumbled upon the following question.
Using binomial theorem compute $102^6$.
Now, I broke the number into 100+2.
Then, applying binomial theorem
$\binom {6} {0}$$100^6(1)$+$\binom {6} {1}$$100^5(2)$+....
I stumbled upon this step. How did they add... | The long way:
$$
\begin{align}
& 10^{12} + 12 \cdot 10^{10} + 6 \cdot 10^9 + 16 \cdot 10^7 + 24 \cdot 10^5 +192 \cdot 10^2 + 64 \\
=\; & 10^{12} + (10+2) 10^{10} + 6 \cdot 10^9 + (10+6) 10^7 + (20+4) 10^5 +(100+90 +2) 10^2 + 60+4 \\
=\; & \color{red}1\cdot10^{12} + \color{red}1 \cdot 10^{11} + \color{red}2\cdot 10^{10}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Does this vector lie in the vector subspace $U$?
Given are the vectors $v_{1}=\begin{pmatrix} 1\\ 3\\ 5
\end{pmatrix}, v_{2}=\begin{pmatrix} 4\\ 5\\ 6 \end{pmatrix},
v_{3}=\begin{pmatrix} 6\\ 4\\ 2 \end{pmatrix}$ from
$\mathbb{R}^{3}$.
The vector subspace $U = \text{span}\left\{v_{1},v_{2},v_{3}\right\}$
Does ... | The complete matrix of the system, with Gaussian elimination:
\begin{align}
\begin{bmatrix}
1 & 4 & 6 & 2 \\
3 & 5 & 4 & 4 \\
5 & 6 & 2 & 5
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 4 & 6 & 2 \\
0 & -7 & -14 & -2 \\
0 & -14 & -28 & -5
\end{bmatrix}
&& \begin{aligned}R_2&\gets R_2-3R_1 \\ R_3&\gets R_3-5R_1\end{aligned}
\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Determine: $S = \frac{2^2}{2}{n \choose 1} + \frac{2^3}{3}{n \choose 2} + \frac{2^4}{4}{n \choose 3} + \cdots + \frac{2^{n+1}}{n+1}{n \choose n}$ We are given two hints: consider $(n+1)S$; and use the Binomial Theorem. But we are not to use calculus.
My consideration of $(n+1)S$ goes like this:
\begin{align*}
\sum\... | Observe that
$$\frac1{k+1}{n\choose k}=\frac{n!}{(k+1)\cdot k!\cdot(n-k)!}=\frac{n!}{(k+1)!\cdot(n-k)!}=\frac{1}{n+1}\cdot{n+1\choose k+1}$$
Then
\begin{align*}\frac{2^2}{2}{n\choose 1}+\frac{2^3}{3}{n\choose 2}+\ldots+\frac{2^{n+1}}{n+1}{n\choose n}&=\frac1{n+1}\left[\sum_{k=1}^n2^{k+1}{n+1\choose k+1}\right]\\
&=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to simplify ${(1+2i)}^6$? How to simplify ${(1+2i)}^6$ using De Moivre's formula?
I have found that $r=\sqrt 5$ and $\tan x=2$ but I can't find the exact value of $x$.
| With $\tan(\theta)=2$ and the angle addition formula for tangent:
$$
\begin{align}
\tan(6\theta)
&=\tan(3\theta+3\theta)\\
&=\frac{2\tan(3\theta)}{1-\tan^2(3\theta)}
=\frac{2\tan(\theta+2\theta)}{1-\tan^2(\theta+2\theta)}\\
&=\frac{2\frac{2+\tan(2\theta)}{1-2\tan(2\theta)}}{1-\left(\frac{2+\tan(2\theta)}{1-2\tan(2\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Proof by induction the divisiblity Proof by induction, that
$$x_n=10^{(3n+2)} + 4(-1)^n\text{ is divisible by 52, when n}\in N $$
for now I did it like that:
$$\text{for } n=0:$$
$$10^2+4=104$$
$$104/2=52$$
$$\text{Let's assume that:}$$
$$x_n=10^{(3n+2)} + 4(-1)^n=52k$$
$$\text {so else}$$
$$4(-1)^n=52k-10^{3n+2}$$
... | Assume that for $n\in\Bbb N$, $$10^{(3n+2)} + 4(-1)^n\text{ is divisible by 52.} $$ This means that $$\frac{10^{(3n+2)} + 4(-1)^n}{52}\in\Bbb Z.$$ Now,
$$\begin{align}
\frac{10^{[3(n+1)+2]} + 4(-1)^{n+1}}{52}&=\frac{10^{(3n+2)}10^3 + 4(-1)^n(-1)}{52}\\
&=\frac{10^{(3n+2)}10^3 + 4(-1)^n(10^3-1001)}{52}\\
&=\frac{10^3\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\sum\limits_{cyc}\frac{a}{a^2+ab+b^2+3}\leq\frac{1}{2}$
Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$\frac{a}{a^2+ab+b^2+3}+\frac{b}{b^2+bc+c^2+3}+\frac{c}{c^2+ca+a^2+3}\leq\frac{1}{2}$$
I think this inequality is very interesting because most of the contest's inequalities are homogeneous... | I think I get something :
Lemma
Let $f:\mathbb{R} \mapsto\mathbb{R^+}$ a real convex function for all $a,b,c,\alpha,\beta,\gamma$ positive real numbers then :
$$\frac{f(a)}{f(a)^2+f(a)f(b)+f(b)^2+3}+\frac{f(b)}{f(b)^2+f(b)f(c)+f(c)^2+3}+\frac{f(c)}{f(c)^2+f(c)f(a)+f(a)^2+3}\leq\frac{f(a)+\alpha f(b)}{\alpha(f(a)^2+f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2178346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Asymptotic expansion of cosine integral Can anybody help with this problem to find the full asymptotic expansion of
$\int_1^\infty \frac{\cos(xt)\, d t}{t}$ from Bender & Orzsag). Does it work by Taylor expansion?
| In this case, integration by parts is the method of choice. You can show easily that (for $n>0$)
$$\int_1^\infty\!dt\, \frac{\cos(x t)}{t^n} = -\frac{\sin(x)}{x} + \frac{n}{x} \int_1^\infty\!dt\, \frac{\sin(x t)}{t^{n+1}} .$$
Similarly
$$\int_1^\infty\!dt\, \frac{\sin(x t)}{t^n} = \frac{\cos(x)}{x} - \frac{n}{x}\int_1^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2178805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve $\int_{0}^{\infty}\frac{\ln(2x)}{4+x^2}dx$ by contour integration I'm a little stuck with this one. I've found the singularities to be at $\pm 2i$ and $0$ (branch point). So far, using a branch cut at $2\pi$ I've found that
$$\int_{0}^{\infty}\frac{\ln(2r)}{4+r^2}dr+\int_{0}^{\infty}\frac{\ln(2r)+2\pi i}{4+r^2}dr... |
METHODOLOGY $1$: COMPLEX ANALYSIS
As already established in the answer left by @ZaidAlyafeai, the classical approach begins by analyzing the contour integral
$$I=\oint_C \frac{\log^2(2z)}{z^2+4}\,dz$$
where $C$ is the "keyhole" contour with the keyhole taken along the positive real axis. In that case $0\le \arg(z)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Multivariable Limit of Fraction: $\lim_{x,y \to (0,0)}\frac{x^4+y^4}{x^4 +(x+\sqrt[3]{y})^2+y^4}$ I am supposed to find the following limit, which seems to be zero, but I am having trouble proving it.
$$\lim_{x,y \to (0,0)}\frac{x^4+y^4}{x^4 +(x+\sqrt[3]{y})^2+y^4}$$
| One has, as $(x,y) \to (0,0)$, with $x=-\sqrt[3]{y}$,
$$
\frac{x^4+y^4}{x^4 +(x+\sqrt[3]{y})^2+y^4}=\frac{x^4+y^4}{x^4 +0+y^4} \to1
$$ and one has, as $(x,y) \to (0,0)$, with $x=0$,
$$
\frac{x^4+y^4}{x^4 +(x+\sqrt[3]{y})^2+y^4}=\frac{y^4}{y^{2/3}+y^4}=\frac{1}{\frac1{y^{10/3}}+1} \to 0
$$ the limit can't exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can we define a 'rectangular coordinate' on a curved surface? We use rectangular coordinate on a flat plane, so can we use it in a curved surface, like the axis is somehow bent?
If yes, is there any application? Also, can we generalize this to higher dimension?
| Preliminary
\begin{align*}
\mathbf{x}(u,v)
&= \begin{pmatrix} x(u,v) \\ y(u,v) \\ z(u,v) \end{pmatrix} \\
\mathbf{x}_u &= \frac{\partial \mathbf{x}}{\partial u} \\
\mathbf{x}_v &= \frac{\partial \mathbf{x}}{\partial v} \\
\mathbf{N} &=
\frac{\mathbf{x}_u \times \mathbf{x}_v}{|\mathbf{x}_u \times \mathbf{x}_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integration by Substitution I have the following problem: $\int (x+2)(2x-3)^6dx$
What I did was allow $u = 2x-3$
$dx= \frac{1}{2}du$
Since I am integrating w.r.t u I decided to let $x=\frac{u+3}{2}$
my new equation is $\int ( \frac{u+3}{2}+2)(u^6)\frac{1}{2}du$
which I then simplify to $\int ( \frac{u+7}{4})(u^6)du$
I ... | You did all the work here (the hard part).... To finish it off, note that
$$\begin{align} \int \left( \frac{u+7}{4}\right)(u^6)\,du &= \frac 14\int(u+7)(u^6)\,du \\ \\ &= \frac 14\int (u^7 + 7u^6)\,du \\ \\ &= \frac 14\left(\frac{u^8}{8} + u^7\right) + C \\ \\ &= \frac 14 u^7\left(\frac u8 + 1\right)\end{align}$$
All t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Testing for the Convergence/Divergence of a series Given only that the series $\sum a_n$ converges, either prove that the series $\sum b_n$ converges or give a counterexample, when we define $b_n$ by,
i ) $\frac{a_n}{n}$
ii) $a_n \sin(n ) $
iii) $n^{\frac{1}{n}} a_n$
Is there any general approach to such questions?
| Counterexample (ii):
The series
$$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty\frac{\sin n}{n},$$
converges by the Dirichlet test. Using $\cos 2n = 1 - 2 \sin^2 n$ we have
$$\sum_{n=1}^m a_n \sin n = \sum_{n=1}^m\frac{\sin^2 n}{n} = \sum_{n=1}^m\frac{1}{2n}- \sum_{n=1}^m\frac{\cos 2n}{2n}$$
The first sum on the RHS dive... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2190204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Why remainders of division give base expression? $214 = 3 · 71 + 1$
$71 = 3 · 23 + 2$
$ 23 = 3 · 7 + 2 $
$7 = 3 · 2 + 1 $
$2 = 3 · 0 + 2$
As a result, to obtain a base $3$ expansion of $214$, we take the remainders of divisions and we get that $(214)_{10} = (21221)_3$.
Question: why do the remainders give base $3$ expa... | By definition of the base $3$ numeration,
$$|abcde|_3=a\cdot3^4+b\cdot3^3+c\cdot3^2+d\cdot3+e.$$
Divide by $3$ to get the remainder $e$ and the quotient
$$a\cdot3^3+b\cdot3^2+c\cdot3+d.$$
Divide by $3$ to get the remainder $d$ and the quotient
$$a\cdot3^2+b\cdot3+c.$$
Divide by $3$ to get the remainder $c$ and the qu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Write $(-\sqrt{3} + i)^{11}$ as $x+iy$, where $x,y \in \mathbb{R}$ Write $(-\sqrt{3} + i)^{11}$ as $x+iy$, where $x,y \in \mathbb{R}$
I was wondering if there was a pattern or method of calculating this more efficiently. I feel like i am not really using complex analysis.
Let $x = -\sqrt{3} + i$
$\Rightarrow x^2 = 2 - ... | Given that $x=-\sqrt{3}+i$ we have that $\vert x\vert=2$ and $\arg(x)=\frac{2\pi}{3}$ so in exponential form
$$ x=2\exp\left(\frac{2\pi}{3}i\right)$$
Thus
\begin{eqnarray}
(-\sqrt{3}+i)^{11}&=&2^{11}\exp\left(\frac{22\pi}{3}i\right)\\
&=&2048\exp\left(\frac{4\pi}{3}i\right)\\
&=&2048\left(-\frac{\sqrt{3}}{2}-\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\sum\limits_{cyc}(4x^6+5x^5y)\geq\frac{\left(\sum\limits_{cyc}(x^2+xy)\right)^3}{8}$
Let $x$, $y$ and $z$ be real numbers. Prove that:
$$4(x^6+y^6+z^6)+5(x^5y+y^5z+z^5x)\geq\frac{(x^2+y^2+z^2+xy+xz+yz)^3}{8}$$
I think cyclic homogeneous polynomial sixth degree inequalities with three variables are inte... | 6*(8*(4*(x^6+y^6+z^6)+5*(x^5*y+y^5*z+z^5*x))-(x^2+y^2+z^2+x*y+y*z+z*x)^3)-(75*((x-y)^6+(x-z)^6+(z-y)^6)+432*(y*z*(z-y)^4+y*x*(x-y)^4+x*z*(z-x)^4)+567*(y^2*z^2*(z-y)^2+y^2*x^2*(x-y)^2+x^2*z^2*(z-x)^2)+36*(x^2(x^2-y*z)^2+y^2*(y^2-x*z)^2+z^2*(z^2-y*x)^2)+120*(x*y*(x^2-y*z)^2+y*z*(y^2-x*z)^2+z*x*(z^2-y*x)^2)+120*(y^3*z*(y-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.