Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Asymptotics of $\Sigma x^n/n^p$ I am looking for asymptotics of
$$\sum_{n=1}^N \dfrac{x^n}{n^p}$$
in terms of $N$, where $x>1$ and $p$ is a positive number.
So I am looking for a $O(f(N))$ type result. For example what is the approximate value of
$\sum_{n=1}^N \dfrac{5^n}{n^2}$ for large $N$. One possibility is p... | Just playing around to see what happens.
If $x > 1$
then
$\begin{array}\\
\sum_{n=1}^N \dfrac{x^n}{n^p}
&=x^N\sum_{n=1}^N \dfrac{x^{n-N}}{n^p}\\
&=x^N\sum_{n=0}^{N-1} \dfrac{x^{-n}}{(N-n)^p}\\
&=\dfrac{x^N}{N^p}\sum_{n=0}^{N-1} \dfrac{x^{-n}}{(1-n/N)^p}\\
&=\dfrac{x^N}{N^p}\sum_{n=0}^{N-1} x^{-n}(1-n/N)^{-p}\\
&=\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate $\int^{\frac{\pi}{4}}_{0}\frac{1}{(\sqrt{\sin x})}dx$ $$\int^{\frac{\pi}{4}}_{0}\frac{1}{(\sqrt{\sin x})}dx$$
I tried everything substitution,beta gamma function everything but I am unable to convert this into a standard form so How do I solve this problem.
| $$\int^{\pi}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx = \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx + \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\cos x}} \mathrm dx = 2 \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx$$
Then use the formula
$$B(x,y) = 2\int^{\pi/2}_0 (\sin t)^{2x-1} (\cos t)^{2y-1} d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{c+a}{ca}}$ For $a,b,c$ are positive real number. Prove that
$$\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{c+a}{ca}}$$
Le... | Apply the popular inequality $\sqrt{\dfrac{a^2+b^2}{2}}\ge \dfrac{a+b}{2}$, with $a = \sqrt{2x}, b = \sqrt{2y}$, and do the same for the other two pairs $(b,c)$ and $(c,a)$, and add all $3$ inequalities to get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2203382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
To find factor of a polynomial equation
One of the factors of $4x^2+y^2+14x-7y-4xy+12$ is equal to
*
*$2x-y+4$
*$2x-y-3$
*$2x+y-4$
*$2x-y+3$
Step $1$:
$4x^2+y^2-4xy$ can be simplified as $(2x-y)^2$
Step $2$:
$14x-7y$ can be simplified as $7(2x-y)$
and finally
$(2x-y) (2x-y+7) + 12$
I can able to factor to thi... | You can do $2x-y=k$ and then
$$k^2+7k+12=(k+3)(k+4)$$
and then you get
$$(2x-y+3)(2x-y+4)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Exponential map of Beltrami-Klein model of hyperbolic geometry In the Betrami-Klein model of hyperbolic geometry, geodesics are represented as straight lines. Hence the exponential map of a tangent vector $\mathbf{v}$ at a point $\mathbf{p}$ is $\mathbf{p} + \lambda \mathbf{v}$, where $\lambda$ is a scalar that depends... | I did some calculation and so and if
$P = (p_x,p_y)$ and $Q=(p_x+v_x,p_y+v_y)$ are two points on a chord the ideal points (points on the unit circle are:
$$I_1 = \left ( \frac{v_y \ c + v_x \sqrt{v_x^2 +v_y^2 -c^2}}
{v_x^2 +v_y^2} \ , \
\frac{-v_x \ c + v_y \sqrt{v_x^2 +v_y^2 -c^2} }{v_x^2 +v_y^2} \right) $$
$$I_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is $\displaystyle\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}$ equal to $-\dfrac 12$ or just $-\dfrac{\sin y}{2y}$ Question:
$$\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}=?$$
Here is my try:
\begin{align}\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}&=\lim_{x\rightarrow y}\dfrac{-2 \sin (\frac 12(x+... | Your answer is correct. A simpler approach: let $f(x)= \cos x$. Then
$\dfrac{\cos x-\cos y}{x^2-y^2}=\dfrac{f( x)-f(y)}{x-y}* \dfrac{1}{x+y} \to f'(y)* \dfrac{1}{2y}$ for $x \to y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Limit of $\frac{(\sqrt{x^2+12}-4)}{2-\sqrt{x^3-4}}$ as $x$ goes to 2
$$\lim_{x\to 2}\frac{\sqrt{x^2+12}-4}{2-\sqrt{x^3-4}}$$
I know it must be very trivial, but yet I'm stuck at this problem and other similar problems for quite a long time. I'd be glad if someone showed me a full solution without using L'Hospital and... | Try that
$$\lim_{x \to 2}\frac{\sqrt{x^2+12}-4}{2 - \sqrt{x^3-4}} = \lim_{x \to 2 }\frac{(\sqrt{x^2+12}-4)(\sqrt{x^2+12}+4)(2 + \sqrt{x^3-4})}{(2 - \sqrt{x^3-4})(\sqrt{x^2+12}+4)(2 + \sqrt{x^3-4})} = \lim_{x \to 2}\frac{(x^2-4)(2 + \sqrt{x^3-4})}{(8-x^3)(\sqrt{x^2+12}+4)}$$
Then you can use that the limit of a product ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Is every integer odd $p$ not divisible by $3$ a sum of difference of powers of $2$ and $3$ or vice versa? Does every odd integer $p$ not divisible by $3$ have the form
$2^a+3^b$, $2^a-3^b$, and or $3^b-2^a$ for integers $a$ and $b$?
Please show proofs if possible. Thanks for help.
| No. $53$ is a counterexample.
There are no integers $a,b$ such that $53=2^a+3^b$ since $$2^a\not=53-3^0,53-3^1,53-3^2,53-3^3$$
Suppose that there are integers $a,b$ such that $53=2^a-3^b$.
It is easy to see that $a\ge 2$ and $b\ge 2$.
We have
$$2\equiv (-1)^a-0\pmod 3\implies \text{$a$ is odd}$$
Also, we have
$$1\equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $a^3+a^2+a=9b^3+b^2+b$ and $a,b$ are integers then show $a-b$ is a perfect cube. If $a^3+a^2+a=9b^3+b^2+b$ and $a,b$ are integers then show $a-b$ is a perfect cube.
My attempt:I factorized it like below:
$(a-b)(a^2+b^2+ab+a+b+1)=8b^3=(2b)^3$
I take $gcd(a-b,a^2+b^2+ab+a+b+1)=d$
If $d=1$ then it is clear that $a-b$ i... | Let us consider the Diophantine equation
$$(a-b)(a^2+ab+b^2+a+b+1) = 8b^3.\tag1$$
Let $\ \gcd(a,b) = d.$
$\text{If }\mathbf {d > 1,}\text{ then }a=md,\quad b=nd,\quad \gcd(m,n) = 1,$
$$(m-n)\left((m^2+mn+n^2)d^2 + (m+n)d + 1\right) = 8d^2n^3.\tag2$$
${\text{If }\mathbf{2\mid d},\ \text{then}}$
$$2\mid m-n,\quad 2\nmid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 4,
"answer_id": 3
} |
Trig identities - stuck solving $\tan^2\theta = -\frac 32 \sec\theta$ Solve the equation on the interval $0\leq \theta < 2\pi$
$$\tan^2 \theta = -\frac{3}{2}\sec \theta $$
Here are the steps I have so far:
Identity: $\tan^2 \theta = \sec^2 \theta -1 $
Substitute:
$$\sec^2 \theta -1 = -\frac{3}{2}\sec \theta $$
$$2\sec... | Why factor? It's fine, but in general when faced with something like
$$2\sec^2 \theta +3\sec \theta - 2 = 0$$
you could set $u=\sec\theta$ and then use the quadratic formula.
$$2u^2 +3u - 2 = 0$$
$$u = \frac{-3 \pm \sqrt{9-4(2)(-2)}}{2(2)} = \frac{-3 \pm 5}{4} = \frac12 , -2$$
Then solve for possible $\theta$ from $\se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$1^2$+$2^2$-$2×3^2$+$4^2$+$5^2$-$2×6^2$+$...$+$9997^2$+$9998^2$-$2×9999^2$ ∵ For any positive integer $n$
$n^2-2(n+1)^2+(n+2)^2$
= $n^2-2n^2-4n-2+n^2+4n+4=2$
∵ $1^2$+$2^2$-$2×3^2$+$4^2$+$5^2$-$2×6^2$+$...$+$9997^2$+$9998^2$-$2×9999^2$
= $1^2+(2^2-2×3^2+4^2)+(5^2-2×6^2+7^2)+...+(9995^2-2×9996^2+9997^2)+9998^2-2×9999^2$
... | There would be $((9995-2)/3)+1$ lots of 2.
Following;
$1+2×(((9995-2)/3)+1)+9998^2-2×9999^2$
$2×3332+(9998^2-9999^2)+(1-9999^2)$
$6664+(9998-9999)(9998+9999)+(1-9999)(1+9999)$
$6664-19997-99980000$
= $-99993333$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Closed Form of Series $ \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1} $. Problem:
A closed form for the sum $$ S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1} $$
is $\Large1 - \frac{a^{n+b}}{3^{2^{n+c}}-1},$ where $a, b,$ and $c$ are integers. Find $a+b+c.$
... | HINT:
$$\dfrac1{3^{2^m}-1}-\dfrac1{3^{2^m}+1}=\dfrac2{3^{2^{m+1}}-1}$$
$$\dfrac2{3-1}-S=\dfrac2{3-1}-\dfrac2{3+1}-\left(\dfrac{2^2}{3^2+1}+\dfrac{2^3}{3^4+1}+\cdots+\dfrac{2^n}{3^{2^n}+1}\right)$$
$$=\dfrac{2^2}{3^2-1}-\dfrac{2^2}{3^2+1}-\left(\dfrac{2^3}{3^4+1}+\cdots+\dfrac{2^n}{3^{2^n}+1}\right)=\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving $\sec^3\frac{2\pi}{7}+\sec^3\frac{4\pi}{7}+\sec^3\frac{6\pi}{7}=-88$ and $\sec^2\frac{2\pi}{7}+\sec^2\frac{4\pi}{7}+\sec^2\frac{6\pi}{7}=24$
Prove:
$$\begin{align}
\sec^3\frac{2\pi}{7}+\sec^3\frac{4\pi}{7}+\sec^3\frac{6\pi}{7} &=-88 \tag{1} \\[6pt]
\sec^2\frac{2\pi}{7}+\sec^2\frac{4\pi}{7}+\sec^2\frac{6\pi}{... | Let $\cos\frac{2\pi}{7}=x$, $\cos\frac{4\pi}{7}=y$ and $\cos\frac{4\pi}{7}=z$.
Hence,
$$x+y+z=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{2\sin\frac{\pi}{7}}=$$
$$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The quotient of the euclidean division $\text{Let } n,\ a_1,\ a_2, \ldots, \ a_n \text{ be natural numbers, such that:} $
$$n = q_1a_1 + r_1, 0\le r_1<a_1$$
$$q_1 = q_2a_2 + r_2, 0\le r_2<a_2$$
$$\vdots$$
$$q_{k-1} = q_ka_k + r_k, 0\le r_k<a_k$$
I have to prove that
$$\begin{cases}
\hfill q_k = \left\lfloor\frac{n}{\p... |
Proof $\mathbb{1}:$
$\boxed{\left\lfloor \frac{n}{\prod_{i=1}^{k} a_i} \right\rfloor = \left\lfloor\frac{\left\lfloor\frac{n}{\prod_{i=1}^{k-1} a_i}\right\rfloor}{a_k}\right\rfloor}$
Note that $\bigg\lfloor\frac{[x]}{m}\bigg\rfloor=\bigg\lfloor\frac{x}{m}\bigg\rfloor$ when $m$ is positive integer as in this case. Wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2219398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Compute the mean for the $f(x)=\dfrac {3}{2}\sqrt {x-1},\;1 \leq x \leq 2$ using probability distribution. How I compute the mean for the $$f(x)=\dfrac {3}{2}\sqrt {x-1},\;1 \leq x \leq 2$$ using probability distribution.
My attempt$$\mu_x=E(X)=\int_{-\infty}^{\infty}xf(x)dx\\= \int\limits_{1}^{2}\,x\cdot\dfrac {3}{2}... | Now, use integration by parts formula$$\displaystyle\int u\; dv=u\;v-\int v\;du\\\text{Let }u=x,\;\;dv=\sqrt{x-1} \; dx\\du=dx,\;\;v=\dfrac{2}{3}(x-1)^{3/2}\\ = \dfrac{3}{2} \left[x\cdot\dfrac{2}{3}(x-1)^{3/2} - \int \dfrac{2}{3}(x-1)^{3/2} \,dx\right]^2_1\\ =\left[ \dfrac{3}{2} \left[x\cdot\dfrac{2}{3}(x-1)^{3/2} - \d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2220406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Matrix inversion question with different dimensions
Let $A=\left[\begin{matrix}0&0&0\\-1&0&0\\0&-1&0\end{matrix}\right]$ and $B=\left[\begin{matrix}4&1\\5&2\\6&3\end{matrix}\right]$ verify the matrix equation $X=AX+B$
*
*What are the dimensions of $X$?
*Find the matrix $X$. Is the matrix you found the unique solut... | We know $X=\left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]$ so we can say
\begin{align}X&=AX+B\\
\left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]&=\left[\begin{matrix}0&0&0\\-1&0&0\\0&-1&0\end{matrix}\right]\left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]+\left[\begin{matrix}4&1\\5&2\\6&3\end{matrix}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2220990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Integrate: $\int (x^2+a^2)^{-3/2} \cdot dx$
Integrate: $\int (x^2+a^2)^{-3/2} \cdot dx$
My Approach:
$\int (x^2+a^2)^{-3/2} \cdot dx$
$\int (x^2+a^2)^{-3/2} \cdot d(a^2+x^2)\cdot \frac{dx}{d(x^2+a^2)}$
But this doesn't give the right answer. I showed this to my friend and he said $d(x^2+a^2)$ is not possib... | Following your idea we get $du=2xdx$ so :
$$I=\int (x^2+a^2)^{-3/2} dx=\int\frac{2xdx}{2x(x^2+a^2)^{3/2}}=\int \frac{du}{2\sqrt{u^4-a^2u^3}}=\frac{1}{2}\int\frac{du}{u^{3/2}\sqrt{u-a^2}} $$
Let $v=\frac{1}{u-a^2}$, we get $dv=-\frac{1}{(u-a^2)^2}$ :
$$I=-\frac{1}{2}{\displaystyle\int}\dfrac{1}{\left(a^2v+1\right)^\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Given $f(n)$ how to find the maximum value of $f(n^5)$? A friend of mine recently asked me this question.
For a positive integer $n$, a function $f(n)$ is defined as:
$f(n)=$ sum of digits in $n$.
given $f(n)=5$ find the maximum value of $f(n^5)$.
I tried solving this problem by putting random values, but my friend gav... | Consider the terms generated by $(a+b+c+d+e)^5$. There will be $5$ of the form $a^5$ which will have a coeffiecient of $1$ & each will therefore give a contribution of $1$, there will be $20$ of the form $a^4 b$ which will have a coeffiecient of $5$ & will therefore give a contribution of $5$, etc ... terms whose coef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Sum of the serie $\sum_{n=1}^{\infty}\frac{1}{4n^3-n}$ I am trying to calculate the sum of this infinite series
$$\sum_{n=1}^{\infty}\frac{1}{4n^3-n}.$$
I only know that
$$\frac{1}{4n^3-n}=-\frac{1}{n}+ \frac{1}{2n+1} +\frac{1}{2n-1}.$$
Can you help me, please?
thanks.
| $$\sum_{n=1}^{N}\frac{1}{4n^3-n}=-\sum_{n=1}^{N}\frac{1}{n}+\sum_{n=1}^{N} \left(\frac{1}{2n+1} +\frac{1}{2n-1}\right)\\=-\sum_{n=1}^{N}\frac{1}{n}+2\sum_{n=0}^{N} \frac{1}{2n+1}-1-\frac{1}{2N+1}\\= -\sum_{n=1}^{N}\frac{1}{n}+2\sum_{n=1}^{2N+1}\frac{1}{n}-2\sum_{n=1}^{N}\frac{1}{2n}-1-\frac{1}{2N+1}$$
Now use $\sum_{n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
How I find the angle 'a'.
Please let me know if you can prove the angle a is 150°.
| Let $\triangle ABC$ satisfy $\angle BAC=96^\circ$, $\angle ABC = \angle ACB=42^\circ$. Let $CK$ be a line with K on $AB$ such that $\angle BCK=18^\circ$ and $A,K$ are on the same side of $BC$. Let $F$ be the point on $BC$ such that $\angle AFC=60^\circ$. Let $AF$ meet $KC$ at $D$. Our goal is to prove that $\angle DBC=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2228763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
$\int_0^1 \frac{1}{1-t} ( \frac{1}{\sqrt{1-4 x}}-\frac{1}{\sqrt{1-4 t x}}) \, dt=\frac{2}{\sqrt{1-4x}}\log (\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}})$ How to prove
$$\int_0^1 \frac{1}{1-t} \bigg( \frac{1}{\sqrt{1-4 x}}-\frac{1}{\sqrt{1-4 t
x}}\bigg) \, dt=\frac{2}{\sqrt{1-4x}}\log \bigg(\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}... | Hint 1:
$$\sqrt{1-4xt}-\sqrt{1-4x}={4x(1-t)\over\sqrt{1-4xt}+\sqrt{1-4x}}$$
which suggests the change of variable $u=4xt$.
Hint 2: Letting $1-u=w^2$ (and abbreviating $\sqrt{1-4x}$ to $a$), we have
$$\int{du\over a\sqrt{1-u}(\sqrt{1-u}+a)}={1\over a}\int{-2wdw\over w(w+a)}$$
Now use partial fractions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2229419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove by Induction $1\cdot 2+2\cdot 5+3\cdot 8+4\cdot 11+...+ n(3n-1) = n^2(n+1)$ Prove by induction that the following equality holds true for all n that's an element of a natural number.
$$1\cdot 2+2\cdot 5+3\cdot 8+4\cdot 11+...+ n(3n-1) = n^2(n+1)$$
My work:
Base Case: $n = 1$
l.s = 2
r.s = 2
True
Induction Hypothe... | The claim is as follows:
$$\sum_{k=1}^{n}k(3k-1)=n^2(n+1)$$
So consider the base case, that is, when $n=1$
Then:
$$\sum_{k=1}^{1}k(3k-1)=1(3(1)-1)=2 \ \checkmark$$
Assume the claim holds for the $n^{th}$ case, that is:
$$\sum_{k=1}^{n}k(3k-1)=n^2(n+1)$$
Then show it holds for the $(n+1)^{th}$ case:
$$\sum_{k=1}^{n+1}k(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2229745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find area of triangle given one angle Question: In triangle ABC, all three side lengths are integers and angle A = 60 degrees. The point D lies on the side AC so that BC = BD and DC = 2. Find the smallest possible area of the triangle ABC.
I know that because BC=BD, one of the inner triangles is isosceles. However, I d... | The first condition is already fairly restrictive. Since $\angle A = 60^\circ$, $\cos \angle A = \frac12$, so we have $$a^2 = b^2 + c^2 - bc$$ by the Law of Cosines, which is nontrivial to satisfy over the integers. (We take $a,b,c$ to be the lengths of the sides opposite $A,B,C$.)
The existence of point $D$ gives us a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2229965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Limit of $S_n = \frac{1}{2n + 1} + \frac{1}{2n + 3} + ... + \frac{1}{4n - 1}$ when $n\to\infty$ $$S_n = \frac{1}{2n + 1} + \frac{1}{2n + 3} + ... + \frac{1}{4n - 1}$$
the task is to find $$\lim_{n \to \infty} S_n$$
I've tried different ways, but all I could do is to make an estimation that the limit is somewhere betwee... | $\displaystyle \sum\limits_{k=2n+1}^{4n-1} \frac{1}{k} - \sum\limits_{k=n+1}^{2n-1} \frac{1}{2k} \approx (\ln 4 – \ln 2) - \frac{1}{2}(\ln 2 – \ln 1) = \frac{\ln 2}{2} $
Explanation (with suitable variables $a,b,n,q,u,v$, e.g. $a\in\mathbb{N}$, $n\to\infty$):
$\displaystyle \prod\limits_{k=1}^{an} (1+\frac{x}{k}) \ap... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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How to explain for my daughter that $\frac {2}{3}$ is greater than $\frac {3}{5}$? I was really upset while I was trying to explain for my daughter that $\frac 23$ is greater than $\frac 35$ and she always claimed that $(3$ is greater than $2$ and $5$ is greater than $3)$ then $\frac 35$ must be greater than $\frac 23$... | In a first step you can observe with the child that
$$
\frac{2}{3} = 1-\frac{1}{3}
\quad \text{ and } \quad
\frac{3}{5} = 1 - \frac{2}{5}.
$$
In words, you are looking at what is remaining of the pie when you have taken either one third or two fifths.
Then, you can show that cutting a third into two pieces gives ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2231366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "67",
"answer_count": 29,
"answer_id": 25
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Evaluate $\int_0^{\infty} \frac{1}{1+x^6} dx$ using Contour Integral My attempt:
I am considering a contour C consisting of upper half circle $|z|=R$ and the real axis from $-R$ to $R$ and finding $\int_C \dfrac{1}{z^6+1}dz$ around it.
First, I find the poles of this function, which are given by $z^6+1=0$
The poles are... | You have miscalculated the residues at $\dfrac{\pm\sqrt{3} + i}{2}$. Note that if $\zeta$ is a simple zero of $f$, then the residue of $\dfrac{1}{f}$ at $\zeta$ is $\dfrac{1}{f'(\zeta)}$, so here for $f(z) = z^6 + 1$, we have the residues
$$\frac{1}{f'(\zeta)} = \frac{1}{6 \zeta^5} = \frac{\zeta}{6\zeta^6} = - \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Analyzing quadratic forms $x^2-3xy+y^2$, $2xy+yz-3xz$ and $x^2+y^2+2xy-xt+2yt$. Positive, negative or indefinite?
By means of successive coordinate changes, write each one of the
quadratic forms below as a sum of terms of the type $\pm u^2$ and
decide wich ones are positive, negative or indefinite:
$$A(x,y) = x^2-... | There is an algorithm for solving $P^T A P = D,$ where $A$ is some given symmetric matrix, $D$ is diagonal, while $\det P = 1.$ See reference for linear algebra books that teach reverse Hermite method for symmetric matrices
In this case we get
$$
P =
\left(
\begin{array}{rr}
1 & \frac{3}{2} \\
0 & 1
\end{array}
\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2234781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Complex numbers involving roots of unity Let $z\in \mathbb{C}$ and $n\in \mathbb{N}, n \geq 1$. Solve the following equation: $$z+z^2+\dots+z^n=n|z|^n$$
Obviously, $(z,n)=(0,n)$ and $(1,n)$ are solutions, $\forall n \geq 1$. Considering $z \neq 0,1$ I tried to use $$z+z^2+\dots+z^n=z(1+z+\dots+z^{n-1})=z(z-\epsilon)(z-... | Expanding on the work of @WB-man by looking at the solutions graphically.
$n=2$
$$
f(z) = z+ z^2 -2 \left| z\right|^{2}
$$
Find the roots of the real and imaginary components. Solve
$$
\begin{align}
\text{Re } f &= x - x^2 - 3 y^2 = 0, \\
\text{Im } f &= y + 2 x y2 = 0, \\
\end{align}
$$
The solution for the real comp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $x^2 - 3x + 2$ is a factor of $x^4 - px^2 +q$, then find the value of $p$ and $q$ If $x^2 - 3x +2$ is a factor of $x^4-px^2+q$ then find the value of $p$ and $q$.
My attempt:
$$x^2-3x+2$$
$$x^2-2x-x+2$$
$$x(x-2)-1(x-2)$$
$$(x-1)(x-2)$$
How do I proceed further?
P.S: Edit after Deepak's comment!
| EDIT1
Factors of $ (x^2 - 3 x + 2) $ are $(x-1),(x-2)$ and each factor produces a zero remainder. Each factor zero remainder requires that
$$ 1-p+q = 0, \, 16-4p+q = 0 $$
Solving
$$ p=5, q=4 $$
and original expression was
$$ x^4 - 5 x^2 + 4 =(x-1)(x+1)(x-2)(x+2)$$
BTW the answer is same even if problem is set in an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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$\int \arcsin(3x - 4x^3)dx$
$$\int \arcsin(3x - 4x^3)dx$$
I think the best idea here is to use integration by parts and remove the arcsine from the integral.
$$\int \arcsin(3x - 4x^3)dx = x\arcsin(3x - 4x^4) - \int {3x(1-4x^2) \over \sqrt{1- (3x - 4x^3)^2}} dx$$
Now to integrate $\displaystyle \int {3x(1-4x^2) \over ... | Another approach:
Let, $x=\sin t\implies dx=\cos t\ dt$
\begin{align*}
\int\arcsin\left(3x-4x^3\right)\ dx&=\int\arcsin\left(3\sin t-4\sin^3t\right)\cos t\ dt\\
&=\int\arcsin(\sin3t)\cos t\ dt\\
&=\int3t\cos t\ dt\\
&=3\left[t\int\cos t\ dt-\int\left\{\dfrac{d}{dt}(t)\cdot \left(\int\cos t\ dt\right)\right\}\right]\\
&... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Find the Saddle Points of $f(x,y) = x^3 +3xy^2-3x^2-3y^2+4$ I started by finding the partial derivatives :
$$ f_x(x,y) = 3x^2+3y^2-6x,$$
$$f_y(x,y) = 6xy -6y$$
but I cant seem to find the critical points from here.
| To find critical points, we solve:
$$\begin{cases}
3x^2 + 3y^2 - 6x &=& 0 \\
6xy - 6y &=& 0
\end{cases}$$
From the second equation, we obtain $x=1$ or $y=0$.
Putting $x=1$ to the first equation gives $y=\pm1$.
Putting $y=0$ to the first equation gives $x=2$ or $x=0$.
By the Second partial derivative test:
$$H(x,y) = \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\frac{4^{2x}+4^x+1}{2^{2x}+2^x+1}=13$ for $x$
Solve $$\frac{4^{2x}+4^x+1}{2^{2x}+2^x+1}=13$$ for $x$
Any help is appreciated. I'm entering a challenge and can't reach the solution.
| Alternative approach: notice that
$$t^4+t^2+1 = (t^2+1)^2-t^2 = (t^2+t+1)(t^2-t+1)$$
so
$$\frac{t^4+t^2+1}{t^2+t+1} = t^2-t+1$$
so either
$$t^2+t+1=0$$
or
$$t^2-t+1=13$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Solve : $x+y=2010$, $(\sqrt[3]{x}+\sqrt[3]{y})(\sqrt[3]{x}+11)(\sqrt[3]{y}+11)=2942$
Given$$x+y=2010,\\(\sqrt[3]{x}+\sqrt[3]{y})(\sqrt[3]{x}+11)(\sqrt[3]{y}+11)=2942,$$
find the value of $\sqrt[3]{x}+\sqrt[3]{y}$.
My attempts:
Let $u=\sqrt[3]{x} , v=\sqrt[3]{y}$.
$(u+v)(u^2-uv+v^2)=2010 \tag{1}$
$(u+v)(uv+11u+11v+1... | From what you've done, we have that
$$(1)\iff 2010=(u+v)^3-3uv(u+v)\tag3$$
and that
$$(2)\iff uv(u+v)=2942-11(u+v)^2-121(u+v)\tag4$$
From $(3)(4)$, we have
$$2010=(u+v)^3-3(2942-11(u+v)^2-121(u+v)),$$
i.e.
$$t^3+33t^2+363t-10836=0\tag5$$
where $t=u+v$.
Now checking if a divisor of $10836=2^2\times 3^2\times 7\times 43$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Simple inequality $\left|\frac{3x+1}{x-2}\right|<1$
$$\left|\frac{3x+1}{x-2}\right|<1$$
$$-1<\frac{3x+1}{x-2}<1$$
$$-1-\frac{1}{x-2}<\frac{3x}{x-2}<1-\frac{1}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2} \text{ , }x \neq 2$$
$${-x+1}<{3x}<{x-3} \text{ , ... | Square to remove the absolute value: if $x\neq 2$,
$$\left|\frac{3x+1}{x-2}\right|<1\iff (3x+1)^2<(x-2)^2\iff8x^2+10x-3<0.$$
Now this quadratic polynomial has reduced discriminant $\Delta'=5^2+24=49$, whence the roots $-3/4$ and $1/4$, and it is negative between the roots, so the solutions are
$$-\frac32<x<\frac14.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Root of the quintic $x^5 − 5x^4 + 30x^3 − 50x^2 + 55x − 21=0$ What is real root of the quintic $x^5 − 5x^4 + 30x^3 − 50x^2 + 55x − 21=0$?
Some remarks:
*
*I saw this quintic in wikipedia
*Real root is given $x=1+{\sqrt[ {5}]{2}}-\left({\sqrt[ {5}]{2}}\right)^{2}+\left({\sqrt[ {5}]{2}}\right)^{3}-\left({\sqrt[ {5}]{... | First off, I don't know of a systematic way to solve it. But suppose one has the magic inspiration to effect the substitution $x=y+2\,$, then the equation in $y$ turns out to be: $$y^5 + 5 y^4 + 30 y^3 + 90 y^2 + 135 y + 81 = 0$$
At this point, the equation became easy to solve. The ratios between symmetric coefficient... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Finding maximum value of $x^2y$ given following condition.
Find the maximum value of $x^2y$ given the constraint that : $$x+y+\sqrt{2x^2+2xy+3y^2} =k ~\text{(constant)}~ \text{and} ~~ x,y \in \mathbb{R^+}$$ (Answer is in terms of '$k$')
I tried using AM-GM , $$\frac x2+\frac x2+y \ge 3\Bigg(\frac{x^2y}{4}\Bigg)^{\fra... | I think you mean that the variables are positives.
By the given we obtain:
$$2x^2+2xy+3y^2=(k-x-y)^2$$ or
$$x^2+2y^2+2k(x+y)=k^2$$ or
$$(x+k)^2+2\left(y+\frac{k}{2}\right)^2=\frac{5k^2}{2}$$ or
$$2(x+k)^2+(2y+k)^2=5k^2.$$
Now, by AM-GM and Holder we obtain:
$$5k^2=2(x+k)^2+(2y+k)^2\geq3\sqrt[3]{\left((x+k)^2(2y+k)\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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User remainder theorem to estimate $ \sum^{\infty}_{k=1} \frac{12(-1)^{k+1}}{k^2} $ within $\frac{2}{5}$ I am trying to use the alternate series test to estimate:
$ \sum^{\infty}_{k=1} \frac{12(-1)^{k+1}}{k^2} $
The question asks:
By the alternating series test, which of the following is known to be an estimate of L ... | In order to estimate
$$ \eta(2) = \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2} $$
we may quote Euler. He noticed (in his third or fourth proof of the Basel problem, $\approx 1741$) that we may simply exploit the trigonometric integral
$$ \int_{0}^{1}\frac{\arcsin(x)}{\sqrt{1-x^2}}\,dx = \frac{1}{2}\arcsin^2(1) = \frac{\pi^2}{8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of $6(M+m)$ Let $q$, $r$ are unit vectors such that $\vec p = \vec q \times \vec p + \vec r$. $M$ is the maximum and $m$ is the minimum value of $[\vec p \vec q \vec r]$, then we have to find the value of $6(M+m)$.
Should $\vec p$ be zero? As in $\vec p = \vec q \times \vec p + \vec r$, $p$ cannot be e... |
$$\mathbf{p}=\mathbf{q} \times \mathbf{p}+\mathbf{r} \tag{1}$$
$\mathbf{p} \cdot \mathbf{p}$ gives
\begin{align}
p^2 &= \mathbf{p} \cdot \mathbf{r} \\
p^2 &= p\cos \theta
\end{align}
$$p=\cos \theta \tag{2}$$
$\mathbf{p} \cdot \mathbf{q}$ gives
$$\mathbf{p} \cdot \mathbf{q}=\mathbf{q} \cdot \mathbf{r}=\cos \ph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is $1+x+x^2+x^3...$? What is the difference between these two series?
$$
\begin{align}
1+x+x^2+x^3+...+x^n+\mathcal O(x^{n+1})&=\frac{1}{1-x}\\
\\
1+r+r^2+r^3+...+r^{n-1}&=\frac{r^n-1}{r-1}\\
\end{align}
$$
I can't wrap my head around it; they both start with $1+x+x^2+x^3...$? Of course the first is the maclaurin... | The second (finite) series is a particularization of the first (with a shifted index):
$$
1 + x + x^{2} + x^{3} + \cdots + x^{n} + O(x^{n+1})= \frac{1}{1 - x}
$$
means: There exists constant $C > 0$ and $\delta > 0$ such that if $|x| < \delta$, then
$$
\left|\frac{1}{1 - x} - (1 + x + x^{2} + x^{3} + \cdots + x^{n})\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Simplify sum over trigonometric functions I need to prove that assuming $N>2$,
$$\frac{1}{N}\sum_{k=1}^{N}\sin\left(\frac{2k}{N}\pi\right)\sin\left(\frac{2\pi}{N}(k-x)\right) = \frac{1}{2}\cos\left(\frac{2\pi}{N} x\right).$$
It should be achievable solely by using trigonometric identities but I am stuck and can't find ... | $$(\sin \frac{2\pi k}{n})(\sin \frac{2\pi (k-x)}{n})=\frac{1}{2}(\cos\frac{2\pi x}{n}-\cos\frac{2\pi (2k-x)}{n})$$
Using $$\sin A \sin B=\frac{1}{2} . \cos (A-B) - \frac{1}{2}.\cos (A+B)$$
Now, $$\frac{1}{n} \sum^n_{k=1} .\frac{1}{2}\cos\frac{2\pi x}{n}-\frac{1}{2}.\frac{1}{n} \sum^n_{k=1}\cos\frac{2\pi (2k-x)}{n}$$
Fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int_C\frac{z+1}{z^2-2z}dz$, where $C$ is the circle $|z|=3$
Evaluate the contour integral $\int_C\frac{z+1}{z^2-2z}dz$ using Cauchy's residue theorem, where $C$ is the circle $|z|=3$.
I see that the function has 2 singularities, at 0 and 2, so I need to find the residue of each. By examining the Laurent ... | You can do this without calculating any residues. For any $R>2,$ Cauchy's theorem shows your integral equals
$$\int_{|z|=R} \frac{z+1}{z^2-2z}\, dz = \int_0^{2\pi} \frac{(Re^{it} + 1)iRe^{it}}{R^2e^{2it} - 2Re^{it}}\, dt.$$
Do a little work to see this equals
$$\tag 1 i\int_0^{2\pi} \frac{1 + e^{-it}/R}{1 - 2e^{-it}/R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determine the unique vector x in the row space for A, for which Ax = b Given the Vector $b = (9, -6, 3)^T$
$A =\begin{pmatrix}
1 & 2 & 0 & 1\\
-1 & -1 & -1 & 0 \\
1 & 0 & 2 &-1
\end{pmatrix}$
I know that the row space is $ \begin{pmatrix} 1 & 0 & 2& -1\end{pmatrix}$ , $ \begin{pmatrix... | Let us take a vector $\ x=(x_1,x_2,x_3,x_4) $
For us to solve the given problem we need to solve the the system $\ Ax=b$
In our case$\ A=\pmatrix{1&2&0&1\\-1&-1&-1&0\\1&0&2&-1}$ and $\ b=\pmatrix{9\\-6\\3}$
$\ Ax=\pmatrix{1&2&0&1\\-1&-1&-1&0\\1&0&2&-1}*\pmatrix{x_1\\x_2\\x_3\\x_4}=\pmatrix{9\\-6\\3}$
After solving this... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2251271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find roots of $f = x^4 - 2x^3 + x^2 + 4x +a \in \Bbb R[x]$ knowing the real part of one of them I have an algebra exam and I've taken this exercise from some old tests.
Let $f = x^4 - 2x^3 + x^2 + 4x +a \in \Bbb R[x]$. Find all $a \in \Bbb R$ for which $f$ has a root $z \in \Bbb C - \Bbb R$ such that the real part of ... | The sum of the two other (possibly complex) roots of $f$ is $0$ so we have two options:
*
*The two other roots are pure-imaginary and conjugate. In this case, $f$ is divisible by $x^2 + u^2$ (and the roots are $\pm iu$) for some $u \in \mathbb{R}$.
*The two other roots are real. In this case, $f$ is divisible by $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find the number b such that the line y=b divides the region bounded by y=0 and y=4-x^2 into two regions with equal area. Find the number b such that the line y=b divides the region bounded by y=0 and y=4-x^2 into two regions with equal area.
I know how to graph it out but I have no idea what to do after that.
| The area between $y=4-x^2$ and $y=0$ is $\int_{-2}^{2}(4-x^2)\,dx = \frac{32}{3}$.
We have to find some constant $b\in(0,4)$ such that the area between $y=4-x^2$ and $y=b$ equals $\frac{16}{3}$. That leads to the equation
$$ \int_{-\sqrt{4-b}}^{\sqrt{4-b}}(4-b-x^2)\,dx = \frac{16}{3} $$
and ultimately to $b=\color{red... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the nature of the triangle Consider a triangle $ABC$ for which $\tan \frac{A}{2} = \frac{a}{b + c}$, where $a = BC, \: b = AC, \:c = AB$.
Then, the triangle is:
A) equilateral
B) right triangle with $A = \frac{\pi}{2}$
C) right triangle with $B = \frac{\pi}{2}$ or $C = \frac{\pi}{2}$
D) acute
E) obtuse
All I cou... | Option C) is correct by trig simplification.
By Sine Rule $a$ is proportional to $\sin A $ with circumcircle diameter as constant of proportionality.
Using shorthand for $ ( s= \sin, c=\cos )$
$$ \tan A/2 = \dfrac{s_{A/2}}{c_{A/2}}= \dfrac{s_A}{s_B+s_C}= \dfrac{2 s_{A/2}c_{A/2}}{2 s_{(B+C)/2}c_{(B-C)/2}} \tag1 $$
sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2253784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Solving the Diophantine Equation $ax^2 + bx + c = dy^2 + ey + f$? As a fun topic for personal research, I am trying to determine everything I can about Pythagorean triples for which all three integers are one less than a perfect square (please don't say anything about this if there is a known solution, I want to try to... | Legendre gave a general solution to the equation
$$
ax^2+bxy+cy^2+dx+ey+f=0.
$$
Your equation/system can certainly be put in this form, and then the parameterization given by Legendre would provide you with the answers you seek.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2254658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Integrate $\int_{0}^{\pi}\frac{\sin(x)}{(\sin^{2} (x) + k\cos^{2}(x))^{1/2}}\,dx$ I am faced with a dilemma. I have the following integral:
$$I(g,b) = \int_{0}^{\pi} \frac{\sin(\theta)}{(\sin^2(\theta) + \frac{g}{b}\cos^2(\theta))^{1/2}} d\theta = \int_{0}^{\pi} \frac{\sin(\theta)}{(1 - (1-\frac{g}{b})\cos^2(\theta))^... | Case 1: $ \ $ Let $k=g/b>1$. The indefinite integral is
$$
\int{\sin t\over\sqrt{\sin^2 t + k \cos^2t}} dt = -\frac{\log(\sqrt{2(k - 1)} \cos t + \sqrt{(k - 1) \cos2t + k + 1})}{\sqrt{k - 1}} + C
$$
$$
= -{\sinh^{-1}(\sqrt{k-1} \cos t)\over\sqrt{k-1}} + C_1.
$$
The definite integral from $0$ to $\pi$ is
$$
\int_0^{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2254834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value of $\alpha^6+\beta^6$ is
If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value
of $\alpha^6+\beta^6$ is
I know that, here, $\alpha\beta=4$ and $\alpha + \beta = 2$ and use that result to find $\alpha^2 + \beta^2$ using the expansion... | Hint: Here $\alpha^2=2\alpha -4$ and same for $\beta$, as they are the roots of the equation $x^2-2x+4=0$.
Now calculate the value of ${\alpha}^6$ and ${\beta}^6$ in terms of $\alpha^2 and \beta^2$. Then substitute their value again.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Certain properties of numbers such that $n \mid 2^n+1$ A natural number $n>1$ is called good if$$n \mid 2^n+1.$$ For example, $n=3$ is good, as $3 \mid 2^3+1=9$. Prove that if $N_1$ and $N_2$ are good, then:
*
*$\mathrm{lcm}(N_1,N_2)$ and $\gcd(N_1,N_2)$ are good,
*$N_1\cdot N_2$ is good.
This seems pretty diffi... | You can prove "good" ness pretty easily without knowing much more about $N_1 $ and $N_2$, but I thought I'd share the properties that make numbers "good", and thus you can see that your problem becomes trivial.
We are going to go case by case where we assume something about $n$, take it to its conclusion, and then sew ... | {
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"url": "https://math.stackexchange.com/questions/2258139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the coefficient of $x^2y^2z^4$ in the expansion of $(x-2y+z)^8$. I need help solving this question. How do I start?
Attempt at solution
*The term that contains $x^2y^2z^4$ is ${8 \choose 4} (-2y)^2(z)$ ?
*
| We can also iteratively apply the binomial theorem in order to find the coefficients. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$.
We obtain
\begin{align*}
[x^2y^2z^4](x-2y+z)^8&=[x^2y^2z^4]\sum_{j=0}^8\binom{8}{j}x^j(-2y+z)^{8-j}\tag{1}\\
&=[y^2z^4]\binom{8}{2}(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the value of $a$ and $b$ for the curve $y=ax^2+\frac{b}{x}$ given $\frac{dy}{dx}=-5$ at $(2,-2)$. The curve $y= ax^2 + \frac{b}{x} $ has a gradient of $-5$ at the point $(2,-2)$. Find the value of $a$ and $b$.
These are my workings -
$$y= ax^2 + \frac{b}{x} \tag{1}$$
Sub $x=2 , y=-2$
$$8a+b=-4 \tag{2}$$
$$... | You can easily obtain simultaneous equations from both $y$ and $\frac{dy}{dx}$. You are generally on the right track (You've found an equation using $y=ax^2+\frac{b}{x}$). However, for the derivative on equation $(3)$, you seem to have forgotten to substitute $x=2$. If you had substituted it, you would have gotten the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2259482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the inequality $\sqrt { x + 2} - \sqrt { x + 3} < \sqrt { 2} - \sqrt { 3}$ Solve for $x$ real the inequality $$\sqrt { x + 2} - \sqrt { x + 3} < \sqrt { 2} - \sqrt { 3}.$$ Obviously $x\ge-2$. After that I tried to square the whole inequality, which led me to $x < - \frac { 18} { 4\sqrt { 6} - 5}$. Now, the answer... | If we differentiate, $\frac d{dx}(\sqrt{x+2}-\sqrt{x+3})=\frac{1}{2}\Big(\frac1{\sqrt{x+2}}-\frac1{\sqrt{x+3}}\Big)>0$. So, provided $x\geq-2$, the LHS is continuous and increasing with $x$ and it will be equal to the RHS when $x=0$. Therefore it is less than the RHS when $-2\leq x<0$.
To see it is increasing without c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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How to calculate $14^{2017^{2017}} \mod 60$? $14^{2017^{2017}} \mod 60$
So, I know that I should begin with decomposing $60$ to the prime factors, which are: $ 3, 2^2, 5$, now I should calculate $14^{2017^{2017}} mod$ all three of these prime factors.
My question is, what is the easiest way of calculating $14^{2017^{20... | Sketch:
Since $14\equiv 2\pmod 3$,
$$14^{2017^{2017}}\equiv 2^{2017^{2017}}\pmod 3.$$
Now, we know that the powers of $2$ satisfy
\begin{align*}
2^1=2&\equiv 2\pmod 3\\
2^2=4&\equiv 1\pmod 3\\
2^3=8&\equiv 2\pmod 3\\
2^4=16&\equiv 1\pmod 3\\
2^5=32&\equiv 2\pmod 3
\end{align*}
and so on.
In other words, since $2^2\equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given $x,y,z >0$ and $xy^2z^3 = 108 $, what is the minimum value of $x+y+z$? Given $x,y,z >0$ and $xy^2z^3 = 108 $, what is the minimum value of $x+y+z$ ?
This is a homework problem, so if someone could just give me an outline/hint of the method used to solve this it would be much appreciated.
Thanks
| To prove the solution is $x=1,y=2,z=3$ we can write the constrained problem as
$min f(y,z)$ where $f(y,z) = \frac{108}{y^2 z^3} + y + z$.
Solving for the minimum of $f(y,z)$ we get:
$$
\frac{\partial f}{\partial y} = -2 \frac{108}{y^3 z^3} +1 = 0
$$
$$
(yz)^3 = 216
$$
$$
yz = 6
$$
$$
\frac{\partial f}{\partial z} = -3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim_{n \to \infty} M(n)/n$ where $M(n)$ is the largest integer satisfying a binomial inequality Let $M (n) $ be the largest integer such that ${m \choose n-1}>{m-1 \choose n} $, then what is the value of $\lim _{n \to \infty} \frac {M (n)}{n} $?
Attempt: After solving and simplifying the given condition, I get ... | Note that from the condition we get the ratio of the two binomials to be $\frac{mn}{(m-n+1)(m-n)}$ so our condition simplifies to $(m-n+1)(m-n) - mn <0$. That is $m^2 - 3mn + m - n + n^2< 0$. (note that this is different to your condition). We can simplify this to $m^2 - m(3n-1) + n^2 - n < 0$. This quadratic has roots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the determinant represents a straight line. Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$, then show that the equation
represents a straight line.
Expansion of determinant is tedious. I can see that matrix is symmetric but not able to use this fact. Could someone help me with this?
| \begin{eqnarray*}
Det\left[
\begin{array}{ccc}
ax-by-c&bx+ay&cx+a \\
bx+ay&-ax+by-c&cy+b\\
cx+a&cy+b&-ax-by+c \\
\end{array}
\right]=(x^2+y^2+1)(a^2+b^2+c^2)(ax+by+c).
\end{eqnarray*}
The first two factors are positive so $\color{red}{ax+by+c=0}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove by induction that $2^n + 4^n \leq 5^n$ I'm trying to prove by induction that $2^n + 4^n \leq 5^n$. Through some value plugging I've established that the induction must start from $n = 2$ because $2^2 + 4^2 \leq 5^2 \equiv 20 \leq 25$; for $n = 1$ it doesn't hold since $2 + 4 \geq 5$.
Now I assume that $2^k + 4^... | It can be done much easier: assume $2^k+4^k\le 5^k$, and see
\begin{align}
5^{k+1}&=5\cdot5^k\\
&\ge 5\cdot 2^k+5\cdot4^k\\
&\ge2\cdot 2^k+4\cdot4^k\\
&=2^{k+1}+4^{k+1}\\
\end{align}
Completing the induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2266291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Solving the differential equation $y'' + y = 8\cos(x) \cos(2x)$ I'm solving the differential equation $y'' + y = 8\cos(x) \cos(2x)$
I started to find the homogeneous solution:
We search the zeros of the associated polynomial:
$$r^2 + 1 = 0$$
This yields $r = i$ or $r = -i$
Hence, the homogeneous solution is:
$$y_h = e^... | The C.F.$=c_1\cos x+c_2\sin x$,
By inverse operator rule: When
\begin{align*}
P.I.&=\dfrac{1}{f(D^2)}\cos(ax+b)\ \text{ put }D^2=-a^2,\text{ when }[f(-a^2)\neq0]\\
&=x\dfrac{1}{f'(D^2)}\cos(ax+b)\ \text{ put }D^2=-a^2,\text{ when }[f(-a^2)=0,\ f'(-a^2)\neq0]\\
&=x^2\dfrac{1}{f''(D^2)}\cos(ax+b)\ \text{ put }D^2=-a^2... | {
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"url": "https://math.stackexchange.com/questions/2266905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Prove $\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$ For real numbers $a, b, c, d >0$ I have to prove that $$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$$ always holds. I tried using the mean inequalities for multiple numbers but I always get in trouble with the d... | Use this inequality:
$\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab$ equivalent $\frac{1}{4}(a+b+c+d) \geq \sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$ Now just prove$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq \frac{1}{4}(a+b+c+d)$ equivalent $4(a^2+b^2+c^2+d^2)\geq (a+b+c+d)^2$ equivalent $3(a^2+b^2+c^2+d^2)\geq 2(ab+ac+ad+ bc + bd ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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The value of infinite product At a first look I observed that every term of this product is less than 1 so the limit will also be less than 1 but clueless about how to find limit .
$$P=\frac{7}{9}\cdot\frac{26}{28}\cdot\frac{63}{65}\cdots\frac{n^3-1}{n^3+1}\cdots.$$
| Note that
$$\ln\left(\frac{n^3-1}{n^3+1}\right)=\ln\left(\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}\right)\\=\ln(n-1)-\ln(n+1)+\ln((n+1)^2-(n+1)+1)-\ln(n^2-n+1).$$
Hence, by taking the logarithm of the partial product we obtain a telescoping sum,
$$\ln\left(\prod_{n=2}^N \frac{n^3-1}{n^3+1} \right)=\sum_{n=2}^N \ln\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Halving a vector and halving an angle
An object is subjected to two equal forces along two different directions. If the magnitude of one of them is halved, the angle which the new resultant makes with the other component force is also halved. What is the angle between the forces?
Well I'm not sure whether the vector... | Let's set
$$
\left\{
\begin{array}{ll}
\vec{p} &= p \cos{\alpha} \hat{i} + p \sin{\alpha}\hat{j} \\
\vec{p^*} &=p \cos{\alpha} \hat{i} - p \sin{\alpha}\hat{j} \\
\end{array}
\right.
$$
$\Rightarrow \vec{p} + \vec{p^*} = 2p \cos{\alpha}\hat{i}$. But
$\frac{1}{2}\vec{p} + \vec{p^*} = \frac{3}{2}p \cos{\alpha} \hat{i} - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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If $t=\tan{\frac{x}{2}}$, then $\cos{x}$ can be expressed as... If $t=\tan{\frac{x}{2}}$, then $\cos{x}$ can be expressed as
a) $\frac{1+t^2}{1-t^2}$
b) $\frac{2t}{1+t^2}$
c) $\frac{1-t^2}{1+t^2}$
d) $\frac{2t}{1-t^2}$
Attempt: I tried using the half angle formula but it just leaves me with an expression in terms of $... | Note that $$\sin^{2}\left(\frac{x}{2}\right) = \frac{1-\cos x}{2},\quad \cos^{2}\left(\frac{x}{2}\right) = \frac{1+\cos x}{2},$$
so $$\tan^{2}\left(\frac{x}{2}\right) =\frac{1-\cos x}{1+\cos x}.$$
Setting $$t = \tan \left(\frac{x}{2}\right)$$
we get
\begin{align*}
t^{2} &= \frac{1-\cos x}{1+\cos x}\\ \implies t^{2} + t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Solve for xy in terms of a and b $$x^2 + xy + y^2 = a$$
$$x + y = b$$
$xy = ?$
I tried this and did this:
$xy = a - x^2 - y^2$
$xy = a - (x^2 + y^2)$
$xy = a - (x + y)(x - y)$
$xy = a - b(x - y)$
At this point I can't think of anything to do to represent the $x - y$ part in terms of $a$ and $b$.
Any help would be appre... | $x^2 + xy + y^2 = a$ ....(1)
$x + y = b$
Squaring both sides,
$x^2 + y^2 + 2xy = b^2$
$x^2 + y^2 = b^2 - 2xy$
Put value of $x^2 + y^2$ in equation (1),
$b^2 - 2xy + xy = a$
$b^2 - xy = a$
$xy = b^2 - a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Identify the plane defined by $|z-2i| = 2|z+3|$ I tried:
$$|z-2i| = 2|z+3| \Leftrightarrow \\
|x+yi-2i|=2|x+yi+2|\Leftrightarrow \\
\sqrt{x^2+(y-2)^2}=\sqrt{4((x+2)^2+y^2)} \Leftrightarrow \\
\sqrt{x^2+y^2-4y+4} = \sqrt{4x^2+24x+36+4y^2} \Leftrightarrow \\
x^2+y^2-4y+4 = 4x^2+24x+36+4y^2 \Leftrightarrow \\
y^2-4y-4y^2=... | Naive geometric solution:
The locus of points whose distances are a fixed ratio (not equality) from two given points is a circle. Call the points $A$ and $B$. The circle's center lies on the (extended) line $AB$.
If you want all the points twice as far from $A$ as from $B$, then the two points where its circumference i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Teach me how to transform integrals Trig for $\int _{0}^{\pi }\sin ^{2}\left( \psi \right) \sin \left( m\psi \right) d\psi $ How to transform the integral
$$\int _{0}^{\pi }\sin ^{2}\left( \psi \right) \sin \left( m\psi \right) d\psi $$
to
$$\int _{0}^{\pi }\left( \dfrac {1} {2}-\dfrac {1} {2}\cos 2\psi \right) \sin ... | The general approach
Considering trigonometric integrals there are a myriad of different techniques to solve them. Sometimes integration by parts is the best option, other times a clever substitution or a trigonometric identity saves the day. Sadly, the only way to know which one to use is to experiment and solve as ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Integral inequality with two increasing functions Let $f,g:[0,1]\rightarrow\mathbb{R}^+$ be increasing functions such that $f\leq g$. Is there a constant $c>0$ (independent of $f,g$) for which there exists some $r\geq 0$ (possibly dependent on $f,g$) such that
$$\int_{x:f(x)\leq r\leq g(x)}g(x)dx+\int_{f(x)\geq r}f(x)d... | $$
\newcommand {\diff} {\mathrm d}
\newcommand{\spc}[1] {\quad #1 \quad}
$$
For convenience, rewrite the inequality as $\max\limits_r \frac A B \ge c$, where
$$A = \int\limits_{f(x) \leq r \leq g(x)} g(x)\; \diff x + \int\limits_{f(x) \geq r} f(x)\; \diff x, \quad B = \int\limits_0^1g(x)\; \diff x$$
For $f(x) = \frac 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find the point on the graph of $f (x) = \sqrt x$ closest to the point $(3,0)$ $(x,y)=(x,\sqrt{x})$
$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
$=\sqrt{x^4-5x^2+9}$
$g(x)=x^4-5x^2+9$ , $g'(x)=4x^3-10x=0$ $: x=0, x=+-\sqrt{10}/2$
The question is: Should I put those x-values in g(x) or the orginal graph,$
y=\sqrt{x}$. To me, it... | You want to minimize the function
$$
f(x)=\|(x,\sqrt{x})-(3,0)\|^2=\|(x-3,\sqrt{x})\|^2=(x-3)^2+(\sqrt{x})^2=x^2-5x+9
$$
The function $f$ reaches its minimum value at
$$
h=-\dfrac{-5}{2(1)}=\dfrac{5}{2}
$$
and the minimum value is
$$
k=f(h)=\dfrac{11}{4}
$$
Hence the point of $y=\sqrt{x}$ closest to the point $(3,0)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find all the points on the cone $z^2=x^2+4y^2$ that are the closest to the point $(0,0,c)$
From all the points on the cone $z^2=x^2+4y^2$ find the closest to the point $(0,0,c)$. State explicitly the minimal distance. $c$ is a constant.
Lagrange multipliers can be used here.
Let the constraint function $g=x^2+4y^2-z^... | Minimizing the squared distance,
$$\begin{array}{ll} \text{minimize} & x^2 + y^2 + (z-c)^2\\ \text{subject to} & z^2 = x^2 + 4 y^2\end{array}$$
Let the Lagrangian be
$$\mathcal L (x,y,z,\lambda) := x^2 + y^2 + (z-c)^2 + \lambda (x^2 + 4 y^2 - z^2)$$
Taking the partial derivatives and finding where they vanish, we obtai... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Taylor expansion of composite functions Suppose one has a function of the form: $$f(x)=\sqrt{1+a\cos^{2}(x)}$$ where $a$ is some parameter.
Firstly, what is the correct method for Taylor expanding such a composite function? Should one just repeatedly use the chain rule $(f\circ g)'(x)=f'(g(x))g'(x)$ to determine the c... | $$f(x)=\sqrt{1+a - a\sin^{2}(x)} = \sqrt{1+a}\sqrt{1 - {a\over a+1}\sin^2x}$$
$$ = \sqrt{1+a}\left(1 - {1\over2}{a\over a+1}\sin^2x + {1\over2!}{1\over2}\left({1\over2}-1\right)\left(-{a\over a+1}\right)^2\sin^4x\right) +O(x^6)$$
$$ = \sqrt{1+a}\left(1 - {1\over4}{a\over a+1}(1-\cos2x) - {1\over32}{a^2\over (a+1)^2}(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the locus of $|z-2i|=3|z+3|$ I got as far as
$$|z-2i|=3|z+3| \Leftrightarrow \\
(\ldots) \Leftrightarrow \\
x^2-y^2+4y-4=9x^2+54x+81-9y^2 \Leftrightarrow \\
x^2-9x^2-y^2+9y^2+4y-4-54x-81=0\Leftrightarrow \\
-8x^2+8y^2+4y-85-54x=0 \Leftrightarrow \\
8y^2+4y-8x^2-54x-85=0\Leftrightarrow \\
y^2+\frac{1}{2}y-x^2-\fr... | i have $$|x+iy-2i|=3|x+3+iy|$$ from here we get
$$\sqrt{x^2+(y-2)^2}=3\sqrt{(x+3)^2+y^2}$$
can you finish now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $ABCD$ be a square with sidelength 1 and $AKL$ is an equilateral triangle where $K$ lies on BC and L lies on CD. What is the area for AKL? I tried to use the Pythagorean Theorem to find one side of $AKL$, but there isn't enough information. What am I missing?
| Let $a$ be the sidelength of the triangle $AKL$. Then $K$ and $L$ are the intersection points between the circle $\omega$ centered in $A$ with radius $a$ and $BC$ and $CD$ respectively. Since both the circle $\omega$ and the square $ABCD$ are symmetric with respect to the line $AC$ then $|BK| = |DL|$.
Let $b = |BK|$. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Solve a system of linear equations subject to Boolean constraints Say I have an equation on the form
$$c_1^{\top} = c_2^{\top} X$$
where $c_1, c_2 \in \mathbb N^n$ and $X \in \{0,1\}^{n \times n}$ is a square matrix. How do I solve for $X$?
| (let's call $c_1 = a$ and $c_2 = b$ for the rest of the question)
Let's say $x \in \mathbb{R}^{n\times x}$ (with $x_{ij}\in \{0,1\}$) and $a,b\in\mathbb{R}^n$. This makes:
$$ \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix}^T = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix}^T \begin{bmatrix} x_{11} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that surfaces $x + 2y – lnz + 4 = 0$ and $x^2 - xy – 8x + z + 5 = 0$ are tangent at $(2,-3,1)$. $x + 2y – ln(z) + 4 = 0$
$x^2 - xy – 8x + z + 5 = 0$
$\nabla [1,2,\frac{-1}{z}] $
$\nabla [2x-y-8,-x,1] $
$\nabla(P0) [1,2,-1] $
$\nabla(P0) [-1,-2,1] $
I've stuck at this point and i don't know what to do next.
| Two surfaces $S_1$ and $S_2$ are tangent at a point $P$ if and only if
$P \in S_1 \cap S_2, \tag{1}$
i.e., $P$ lies in each of $S_1$, $S_2$; and
$T_PS_1 = T_PS_2, \tag{2}$
that is, the tanget planes to $S_1$ and $S_2$ at $P$ are the same. Taking $S_1$ to be the surface
$f_1(x, y, z) = x + 2y - \ln z + 4 = 0, \tag{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating the Limit with Series Expansion Problem
Evaluate the following using series expansion.
$$ \lim_{x \rightarrow 0} \frac{\cosh{x}-\cosh{2x}}{x \cdot \sinh{x} }$$
I really don't know what I should do. If I expand $\cosh$ and $\sinh$ normally, (they'd look something like $\sum_{n = 0}^{\infty}\frac{x^{2n}}{(2n)... | As other answers said, you are correct and the limit is easy to find using this way.
However, the same procedure can be used to see how is approached the limit. Just use one extra term in the expansions $$\cosh(x)=1+\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ $$\cosh(2x)=1+2 x^2+\frac{2 x^4}{3}+O\left(x^6\right)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Factorise $x^5+x+1$
Factorise $$x^5+x+1$$
I'm being taught that one method to factorise this expression which is $=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1$
$=x^3(x^2+x+1)-x^2(x^2+x+1)+x^2+x+1$
=$(x^3-x^2+1)(x^2+x+1)$
My question:
Is there another method to factorise this as this solution it seems impossible to invent it?
| Alternatively note that $$\begin{align}x^5 + x + 1 &= x^5 - x^2 + x^2 + x + 1\\ & =x^2(x^3-1) + \color{red}{x^2+x+1} \\ & = x^2(x-1)\color{red}{(x^2+x+1)} + \color{red}{x^2+x+1} \\& =\color{blue}{(x^3-x^2+1)}\color{red}{(x^2+x+1)} \end{align}$$
where we used the well known identity $x^3 - 1 = (x-1)(x^2+x+1)$ in the th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2293493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Why is $x^2+x$ the same as writing $(x+0.5)^2-0.25$? I find it extremely weird that a translation vertically by a variable should cause the graph to move both vertically and horizontally. Also, why 0.5 and 0.25?
| Consider a more general case: ($a > 0$)
\begin{align}
a \, x^2 + b \, x &= (\sqrt{a})^2 x^2 + 2 \cdot \frac{b}{2 \, \sqrt{a}} \cdot \sqrt{a} x + \frac{b^2}{4 \, a} - \frac{b^2}{4 \, a} \\
&= \left(\sqrt{a} x + \frac{b}{2 \, \sqrt{a}}\right)^2 - \frac{b^2}{4 \, a}.
\end{align}
In this example the values of $(a,b)$ are $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2293553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
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The sum of the coefficients of $P(x)$ of 4th degree given that $P(x)\equiv x-1 \pmod{2x^2+4}\equiv x-1 \pmod{x^2-x-2}\equiv 141 \pmod{x+2}$ Let there be a polynomial $P(x)$ of 4th degree. The remainder of $P(x)$ divided by both $2x^2+4$ and $x^2-x-2$ is equal to $x-1$. So:
$$P(x)\equiv x-1 \pmod{2x^2+4}\equiv x-1 \pmod... | Hint 1:$$P\left( x \right) =a\left( 2{ x }^{ 2 }+4 \right) \left( { x }^{ 2 }-x-2 \right) +x-1\\ P(x)\equiv 141{ \quad \left( mod\quad x+2 \right) \Rightarrow }P\left( -2 \right) =141\\ P\left( -2 \right) =48a-3=141\\ 48a=144\\ a=3\\ P\left( x \right) =3\left( 2{ x }^{ 2 }+4 \right) \left( { x }^{ 2 }-x-2 \right) +x-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Determine exact values using angle sum and difference identity (trig) Ok so we have just learnt the basic angle sum and difference identities:
$$\begin{array}{l}\cos \left( {A \pm B} \right) = \cos A\cos B \mp \sin A\sin B\\\sin \left( {A \pm B} \right) = \sin A\cos B \pm \cos A\sin B\\\tan \left( {A \pm B} \right) = \... | Here you go,
$$\begin{align}\tan \left( {{{45}^0} - {{30}^0}} \right) &= \frac{{\tan {{45}^0} - \tan {{30}^0}}}{{1 + \tan {{45}^0}\tan {{30}^0}}}\\\\
&= \frac{{1 - \frac{1}{{\sqrt 3 }}}}{{1 + \left( 1 \right)\left( {\frac{1}{{\sqrt 3 }}} \right)}}\\\\
&= \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}\\ \\
&= \frac{{\sqrt 3 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is the limit to infinity of this function with summation? I need someone with knowledge of limits to infinity and limits of summations to please work out the following:
$$ \lim_{i\to \infty} \frac{2^{i} + \sum_{j=0}^{i-1} (-1)^j2^j}{2^{i+1}} $$
For context, I want to determine if the following sequence approaches ... | Your sequence does not converge. Firstly $\sum_{j=0}^{i-1} (-1)^j 2^j$ is sum of geometric series with quotient -2 and is equal to $\frac{1}{3} (1-(-2)^i)$.
We have
$$ \lim_{i\to \infty} \frac{2^{i} + \sum_{j=0}^{i-1} (-1)^j2^j}{2^{i+1}} =
\lim_{i\to \infty} \frac{2^{i}}{2^{i+1}} + \lim_{i\to \infty} \frac{ 1-(-2)^{i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Express $\frac {a^2+(a+b)^2}{a(a+b)}$ in terms of $x$ and $y$. Given that ${ab}=x$ and ${a+b}=y$. Express $\dfrac{a^2+(a+b)^2}{a(a+b)}$ in terms of $x$ and $y$.
I try many way still cannot succeed. Please use some elementary way to help me to solve this. As I am not familiar with MathJax, so if exists any mistake pl... | As suggested in the comments, $a$ and $b$ are the roots of $t^2-yt+x=0$. We get this from: $$a=\frac xb\implies\frac xb+b=y\implies b^2-yb+x=0$$ and similarly for $a$ leads to the same equation.
Solving this gives $$t=\frac{y\pm\sqrt{y^2-4x}}{2}$$
Wlog, let $a$ be the root with the $+$ sign and $b$ be the root with th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the power series representation of the function I have to find $a_n$ so, my first attempt was to do partial fraction, but with no real solution. Any recommendation will be appreciate $$\frac{1}{(2x-3)(x^2-x+1)}=\sum_{n=0}^\infty a_n (x-1)^n$$
| Note that the geometric series is valid also in the complex field
$$
\frac{1}
{{1 - z}} = \sum\limits_{0\, \leqslant \,n} {z^{\,n} } \quad \left| \begin{gathered}
\;z \in \;\mathbb{C}\; \hfill \\
\,\left| z \right| < 1 \hfill \\
\end{gathered} \right.
$$
and that the partial fraction decomposition can be operated... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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every integer can be represented in the form $x^2+y^2-z^2$ Every integer can be represented in the form $x^2+y^2-z^2$ and show that $6$ actually requires all three terms.
I put
*
*$z=y+1$
*$x=n^2+3$
*$y=3n^2+4+(n^4-n)/2$
what does it mean that $6$ actually requires all three terms?
| Every odd number is of the form $2n+1=(n+1)^2-n^2=(n+1)^2-n^2+0^2$ and every even integer is of the form $2n=(n+1)^2-n^2-1^2$.
Now the second part:
Let, $6=x^2+y^2$ i.e $z=0$.. Now, $x,y\ne1$ as if one of them is one the other one will be $\sqrt{5}$. $x,y\ne2$ as if one of them is $2$ other one will be $\sqrt{2}$. We d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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How to prove that $\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$? Is there a trick to show that
$$\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$$
is true ?
| Note that
\begin{align*}
(\sqrt{-3-2i}+\sqrt{-3+2i})^2 &= (-3-2i) + 2\sqrt{(-3-2i)(-3+2i)} + (-3+2i) \\
&= -6 + 2\sqrt{13}\\
&= 2(\sqrt{13} - 3).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
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Evaluate series $\sum\limits_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$
Determine the value of
$$\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$$
or $$\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots$$
for $x\in\mathbb{R}$.
The answer is $\dfrac{x}{1-x}$ for $x\in(0,1)$. To prove this, notice
$$\f... | $$\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}
$$
$$
\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}.(1-x^{2^{n-1}})}{(1-x^{2^n}).(1-x^{2^{n-1}})}
$$
$$
\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}-x^{2^{n}}}{(1-x^{2^n}).(1-x^{2^{n-1}})}
$$
$$
\sum_{n=1}^{\infty}\frac{-(1-x^{2^{n-1}})+1-x^{2^{n}}}{(1-x^{2^n}).(1-x^{2^{n-1}})}
$$
$$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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Find matrices $P$ and $Q$ to check if matrices $A$ and $B$ are equivalent( $ Q^{-1}AP = B$ )? $ \begin{align}&A=\begin{pmatrix}
6 & -3 & -6 & -12 & 9 & 9 & 0 & 3 & -6 & 9 \\
12 & 4 & 8 & -4 & 8 & -12 & 4 & 4 & 8 & -4\\
\end{pmatrix} \\\\
& B = \begin{pmatrix}
4 & -4 & 0 & -8 & -12 & ... | If you only want to know if $A$ and $B$ are equivalent or not, you only need to show whether they have the same rank. A short explanation of this is that applying invertible square matrix on the left is just applying multiple elementary row operations, and applying invertible square matrix on the right is just applying... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Definite integral of $\frac{1}{\sqrt{\tan x}}$ i would like to ask you for help to evaluate $$\int_{0}^{\pi/2}\frac{1}{\sqrt{\tan x}}dx$$
I didn't find an appropriate substitution.
Thanks for help.
| Hint
As in the previous answer, let
$$z=\sqrt{\tan(x)}\implies x=\tan ^{-1}\left(z^2\right)\implies dx=\frac{2 z}{z^4+1}\,dz$$ This makes $$\int\frac{dx}{\sqrt{\tan x}}=2\int \frac{dz}{z^4+1}$$ Now $$z^4+1=(z^2+1)^2-2z^2=(z^2+\sqrt 2 z+1)(z^2-\sqrt 2 z+1)$$ Now, using partial fraction decomposition $$\frac 1{z^4+1}=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2318042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find closed formula and limit for $a_1 =1$, $2a_{n+1}a_n = 4a_n + 3a_{n+1}$ Tui a sequence $(a_n)$ defined for all natural numbers given by
$$a_1 =1, 2a_{n+1}a_n = 4a_n + 3a_{n+1}, \forall n \geq 1$$
Find the closed formula for the sequence and hence find the limit.
Here, what I have done:
$$2a_{n+1}a_n = 4a_n + 3a_{... | If you have proved that the sequence has a limit $l$, then $l$ can be calculated by the following:
$$\lim_{n\to \infty}2a_{n+1}a_n = \lim_{n\to \infty}(4a_n + 3a_{n+1}) $$
and hence
$$2l^2=4l+3l$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Differentiate $y = x^{\sqrt{x}}$ (Simplification)
I understand the gist of the problem. Just the simplification confused me.
How did $\sqrt{x}/x$ turn into $1/\sqrt{x}$?
Why does that circled 1 turn into a 2?
| (1) $\displaystyle \sqrt{x}\cdot\frac{1}{x}=\sqrt{x}\cdot\frac{1}{(\sqrt{x})^2}=\frac{1}{\sqrt{x}}$.
(2) $\displaystyle \frac{1}{\sqrt{x}}+\frac{\ln x}{2\sqrt{x}}=\frac{2}{2}\cdot\frac{1}{\sqrt{x}}+\frac{\ln x}{2\sqrt{x}}=\frac{2}{2\sqrt{x}}+\frac{\ln x}{2\sqrt{x}}=\frac{2+\ln x}{2\sqrt{x}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Show that $2222^{5555} + 5555^{2222}$ is divisible by 7 (without modular arithmetic) I tried using the following approach:
$$x=2222^{5555}+5555^{2222} = (2222^5)^{1111}+(5555^2)^{1111}$$
Now we know $(x^n+y^n)$ is divisible by $(x+y)$ for odd natural number $n$. So,
$$x=(2222^5+5555^2)k,\ k\in N$$
$$x=(1111^2)(32\cdot1... | We have
\begin{align}
2222^5 + 5555^2&=2222^5 +5555^5-5555^5+ 5555^2\\
&=2222^5 +5555^5-5555^2( 5555^3-1)\\
\end{align}
$2222^5 +5555^5$ is divisible by $2222+5555=7(1111)$ as $5$ is odd.
$$5555^3-1=(7\times793+4)^3-1=7p+4^3-1=7p+63$$
is divisible by $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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$8\sin x\cos x-\sqrt{6} \sin x- \sqrt{6} \cos x+1=0$, Solve for $x$
Solve for $x:0\leq x \leq \frac{\pi}{2}$
$$8\sin x\cos x-\sqrt{6} \sin x- \sqrt{6} \cos x+1=0$$
My attempt,
I changed it into $$1+4 \sin 2x-2\sqrt{3} \sin (x+\frac{\pi}{4})=0$$
| This is a variation of Michael Rozenberg's answer,
$$\cos x+\sin x=\sqrt2\cos\left(\dfrac\pi4-x\right)$$
Set $\dfrac\pi4-x=y\iff x=?$
$2\sin x\cos x=\sin2x=\sin2\left(\dfrac\pi4-y\right)=\cos2y=2\cos^2y-1$
So, we have $$0=4(2\cos^2y-1)-2\sqrt3\cos y+1=8\cos^2y-2\sqrt3\cos y-3$$ which is a quadratic equation in $\cos y$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving a relation related to quadratic equation Question:If $α$ and $β$ be the roots of $ax^2+2bx+c=0$ and $α+δ$, $β+δ$ be those of $Ax^2+2Bx+C=0$, prove that, $\frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}$.
My Attempt: Finding the sum of roots and product of roots for both the equations we get,
$α+β=\frac{-2b}{a}$
$αβ=\... | WLOG, $A=a=1$ (otherwise you can divide the trinomials by their leading coefficient).
Then
$$(x+\delta)^2+2B(x+\delta)+C=x^2+2(\delta+B)x+\delta^2+2B\delta+C=x^2+2bx+c,$$ and
$$b^2-c=(\delta+B)^2-(\delta^2+2B\delta+C)=B^2-C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
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Is it true: $\sum_{n=1}^\infty \frac{a_n}{b_n}$&$\sum_{n=1}^\infty \frac{b_n}{c_n}$ are irrational=>$\sum_{n=1}^\infty \frac{a_n}{c_n}$ is irrational Assume that the:
$$\sum_{n=1}^\infty \frac{a_n}{b_n}$$ and $$\sum_{n=1}^\infty \frac{b_n}{c_n}$$
are convergent and irrational, then if
$$\sum_{n=1}^\infty \frac{a_n}{c_... | Another example with all terms positive is
$$ a_n = 2n+1, \qquad b_n = n(n+1)(2n-1), \qquad c_n = n(n+1)(n+2)(n+3)(2n+1). $$
Then we can prove that
$$ \sum_{n=1}^{\infty} \frac{a_n}{b_n} = \frac{1}{3} + \frac{8}{3}\log 2, \qquad \sum_{n=1}^{\infty} \frac{b_n}{c_n} = \frac{9}{15} - \frac{8}{15}\log 2, \qquad \sum_{n=1}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Maximal and minimal value of a function $4x^2-4xy+y^2$ at the border $x^2+y^2=25$. So my first thought was to find the critical points of the $4x^2-4xy+y^2$ and I calculated.
$$F_{x}=8x-4y,\quad F_{y}=2y-4x$$
and I found out that critical point are satisfying $x=2y$.
Then I put then to equation $x^2+y^2=25$.
And I calc... | $x =u + 2v\\
y = 2u - v$
$4x^2 -4xy + y^2 = 4(u^2 + 4uv + 4v^2) - 4(2u^2 -3uv - 2v^2) + (4u^2 - 4uv + v^2) = 25 v^2$
$x^2 + y^2 = (u^2 + 4uv + 4v^2) + (4u^2 - 4uv + v^2) = 5u^2 + 5v^2 = 25$
Minimize $25v^2$ constrained by: $5u^2 + 5v^2 = 25$
$v = 0, u = \pm \sqrt 5\\
(x,y) = (\sqrt 5,2\sqrt 5), (-\sqrt 5, -2\sqrt5)$
M... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
To compute $\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2|^2 dz$ where $\mathcal{C}$ is the unit circle in $\mathbb{C}$
Let $\mathcal{C}$ denote the unit circle in $\mathbb{C}$ centred at the origin taken anticlockwise. Compute the value of the integrtal $$\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2|^2 dz$$
I don't recon w... | We want the integral
$$\frac{1}{2\pi i}\int_C |1+z+z^2|^2dz=\frac{1}{2\pi}\int_0^{2\pi}|1+z+z^2|^2z~d\theta$$
since $z=e^{i\theta}$.
Now consider the absolute value portion,
$$
\begin{align}
|1+z+z^2|^2
&={(1+z+z^2)(1+z+z^2)^*}\\
&={(1+z+z^2)(1+z^{-1}+z^{-2})}\\
&={(1+z+z^2)(1+z^{-1}+z^{-2})\frac{z^2}{z^2}}\\
&={\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
The function $f(x)=(1-\frac{\sqrt{21-4a-a^2}}{a+1})x^3+5x+\sqrt7$ is increasing at every point of its domain. Find the set of values of $a$ for which the function $f(x)=(1-\frac{\sqrt{21-4a-a^2}}{a+1})x^3+5x+\sqrt7$ is increasing at every point of its domain.
For $f(x)$ to be increasing,$f'(x)=3(1-\frac{\sqrt{21-4a-a^... | The domain gives $a^2+4a-21\leq0$ and $a\neq-1$, which gives $-7\leq a\leq3$ and $a\neq-1$.
Now, $1-\frac{\sqrt{21-4a-a^2}}{a+1}=0$ for $a=2$ and use the intervals method.
Maybe the mistake was in solving of the equation:
$$a+1=\sqrt{21-4a-a^2}.$$
We have $a>-1$ and $21-4a-a^2\geq0$, which gives $-1<a\leq3$.
Now, aft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $(x^2-4x+7)^3+p(x^2-4x+7)^2+q(x^2-4x+7)+r=0$ has no real roots, Then $p+2q+r$ is
If $f(x)=x^3+px^2+qx+r$ has three distinct non negative integer roots and
$(x^2-4x+7)^3+p(x^2-4x+7)^2+q(x^2-4x+7)+r=0$ has no real roots,
Then $p+2q+r$ is
$\bf{Attempt:}$ Assuming $f(x) = x^3+px^2+qx+r = (x-a)(x-b)(x-c)$,
where $a,b,... | $$
f(x^2-4x+7)=((x-2)^2+(3-a))((x-2)^2+(3-b))((x-2)^2+(3-c))
$$
has no real roots so $3-a$, $3-b$, and $3-c$ are at least $1$. Hence
$$
\{a,b,c\}=\{0,1,2\}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\arctan(\tan(-\frac{3\cdot \pi}{5}))+\operatorname{arccot}(\cot(-\frac{3\cdot \pi}{5}))=?$ We know that $$\DeclareMathOperator{\arccot}{arccot}\arctan(-x)=-\arctan(x)$$ $$\arccot(-x)=\pi-\arccot(x)$$ $$\cot(-x)=-\cot(x)$$ $$\tan(-x)=-\tan(x)$$So I have figured out that $$\arctan(\tan(-\frac{3\cdot \pi}{5}))=\arctan(-\... | The function $\arctan(\tan x)$ returns a “normalized” $x$, that is the unique angle $x'\in(-\pi/2,\pi/2)$ so that $\tan x=\tan x'$.
Since $-3\pi/5<-\pi/2$, but $2\pi/5=-3\pi/5+\pi\in(-\pi/2,\pi/2)$, you have
$$
\arctan\Bigl(\tan\Bigl(-\frac{3\pi}{5}\Bigr)\Bigr)=\frac{2\pi}{5}
$$
Do similarly for the arccotangent, but w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Convergent, non-geometric, infinite sum as a fraction? How would you go about finding the sum of an infinite series that is known to be convergent, but is not geometric?
The particular example I am interested in is as follows: $\sum_{j=0}^{\infty}\left [(6j+1)(\frac{1}{2})^{6j+2}\right ]$.
The only catch to this is t... | We have
\begin{eqnarray*}
\sum_{j=0}^{\infty} x^j =\frac{1}{1-x}.
\end{eqnarray*}
Differentiate this
\begin{eqnarray*}
\sum_{j=0}^{\infty} j x^{j-1} =\frac{1}{(1-x)^2}.
\end{eqnarray*}
Your sum can be rewitten as
\begin{eqnarray*}
\frac{6}{256}\sum_{j=0}^{\infty} j \left( \frac{1}{64} \right)^{j-1} +\frac{1}{4}\sum_{j... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\sum_{n=1}^{\infty}(-1)^{n-1}\left({\beta(n)\over n}-\ln{n+1\over n}\right)=\ln\sqrt{2\over \pi}\cdot{2\over \Gamma^2\left({3\over 4}\right)}?$
$$\sum_{n=1}^{\infty}(-1)^{n-1}\left({\beta(n)\over n}-\ln{n+1\over n}\right)=\ln\left(\sqrt{2\over \pi}\cdot{2\over \Gamma^2\left({3\over 4}\right)}\right)\tag1$$
Where $\b... | Let us consider the two sums separately:
$$
S_1 = \sum_{n=1}^\infty (-1)^{n-1}\frac{\beta(n)}{n} \\
S_2 = \sum_{n=1}^\infty (-1)^{n-1}\ln\frac{n+1}{n}
$$
For $S_1$, let us first consider the related sum, for which $S_1$ is the limiting value $S(1)$:
$$
S(x) = \sum_{n=1}^\infty (-1)^{n-1}\frac{\beta(n)x^n}{n} \\
S'(x) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
How do I solve $\int\frac{dx}{\sin x+\cos x-1}$? Please help me find the following indefinite integral:
$$\int\dfrac{dx}{\sin x+\cos x-1}$$
I have tried many different substitutions to no avail. Any help is appreciated.
| Another way, using trigonometric identities
$$\sin{2x} = 2 \cdot \sin{x} \cdot \cos{x}$$
$$1 - \cos{x} = 2 \cdot \sin^2\frac{x}{2}$$
is as follows:
$$\int\dfrac{dx}{\sin x+\cos x-1}$$
$$= \int\dfrac{dx}{\sin x-(1 - \cos x)}$$
$$= \int\dfrac{dx}{2 \cdot \sin\frac{x}{2} \cdot \cos\frac{x}{2}-(2\cdot sin^2\frac{x}{2})}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2338996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
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.