Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Integrate $\int \frac{x}{x^4+4}dx$ I'm having trouble solving the integral
$\int \frac{x}{x^4+4}dx$
I already tried the following but, I'm getting stuck after step 3.
$\int \frac{x}{(x^2)^2+4}dx $
$u = x^2 , du = 2x dx, \frac{du}{2}=xdx$
$\frac{1}{2}\int \frac{1}{u^2+4}du$
| Use $$\frac{x}{x^4+4}=\frac{x}{x^4+4x^2+4-4x^2}=\frac{x}{(x^2-2x+2)(x^2+2x+2)}=$$
$$=\frac{1}{4}\left(\frac{1}{x^2-2x+2}-\frac{1}{x^2+2x+2}\right)=\frac{1}{4}\left(\frac{1}{(x-1)^2+1}-\frac{1}{(x+1)^2+1}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How many positive integer values of n exist such that $\frac{4^n+2^n+1}{n^2+n+1}$ also a positive integer? My question is as follows: find the number of values of the positive integers $n$ such that the fraction below is also a positive integer:
$$\frac{4^n+2^n+1}{n^2+n+1}$$
I have only been able to establish the bases... | Let's prove first that, if $k = 2^s$, then $x^k + x^{k/2} + 1$ is divisible by $x^2+x+1$ over the integers. This is because we can do this single trick, knowing that $(a+b)(a-b) = a^2-b^2$:
$$(x^2+x+1)(x^2-x+1) = (x^2+1)^2 - x^2 = x^4+x^2+1$$
We can apply it again:
$$(x^4+x^2+1)(x^4-x^2+1) = (x^4+1)^2-x^4 = x^8+x^4+1$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the limit $\lim_{x\to 0} x\left(\left[\frac{1}{x}\right] +\left[\frac{2}{x}\right] +\cdots \left[\frac{10}{x}\right] \right)$ Can someone help me finding the following limit
$$ \lim_{x\to 0} x\left(\left\lfloor\frac{1}{x}\right\rfloor +\left\lfloor\frac{2}{x}\right\rfloor +\cdots \left\lfloor\frac{10}{x}\right\rf... | Hint Since $u-1 < \lfloor u \rfloor \leq u$, you have
$$ \left(\frac{1}{x} +\frac{2}{x} +\cdots \frac{10}{x}\right)-10 \leq \left(\left[\frac{1}{x}\right] +\left[\frac{2}{x}\right] +\cdots \left[\frac{10}{x}\right] \right) \leq \left(\frac{1}{x} +\frac{2}{x} +\cdots \frac{10}{x}\right)$$
Therefore,
$$ \frac{55}{x} -10 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3517293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the $n$-th power of a $3{\times}3$ matrix using the Cayley-Hamilton theorem. I need to find $A^n$ of the matrix $A=\begin{pmatrix}
2&0 & 2\\
0& 2 & 1\\
0& 0 & 3
\end{pmatrix}$ using Cayley-Hamilton theorem.
I found the characteristic polynomial $P(A)=(2-A)^2(3-A)$ from which I got $A^3=7A^2-16A+12$. How to c... | Since $p(A)=0$ where $p(x)=(2-x)^2(3-x)$, if we divide $x^n$ by $p(x)$ to get $x^n=p(x)q(x)+r(x)$, then $A^n=p(A)q(A)+r(A)=0q(A)+r(A)=r(A)$, so it suffices to figure out $r$. However, if $n$ is large, actually doing division will not be effective.
Since $p(2)=p(3)=0$, we have that $r(2)=2^n$ and $r(3)=3^n$. However, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3517795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Asymmetric inequality in three variables $\frac{3(a+b)^2(b+c)^2}{4ab^2c} \geq 7+\frac{5(a^2+2b^2+c^2)}{(a+b)(b+c)}$ Consider three real positive variables $a,\ b$ and $c$. Prove that the following inequality holds:
$$\frac{3(a+b)^2(b+c)^2}{4ab^2c} \geq 7+\frac{5(a^2+2b^2+c^2)}{(a+b)(b+c)}$$
My progress: We can prove th... | Making the coordinates transformation
$$
\cases{
\frac{a+b}{a}=x\\
\frac{b+c}{c}=y\\
a b^2 c=z
}
$$
solving for $a,b,c$ we have
$$
\left\{
\begin{array}{rcl}
a & = & \frac{z}{(y-1)^2 \left(\frac{(x-1) z}{(y-1)^3}\right)^{3/4}} \\
b & = & (y-1) \sqrt[4]{\frac{(x-1) z}{(y-1)^3}} \\
c & = & \sqrt[4]{\frac{(x-1) z}{(y-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3519062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solving a limit by two methods with different results I'm considering this limit
$$\lim_{x\to 0}\frac{(1-\cos x^2)\arctan x}{e^{x^2}\sin 2x - 2x + \frac{2}{3}x^3}.$$
My first attempt was using the following equivalent infinitesimals
$$1-\cos x^2 \sim \frac{x^4}{2},\quad \arctan x \sim x, \quad \sin 2x \sim 2x, \quad e^... | Keep in mind that you cannot add equivalents inconsiderately. So you second approach is correct. A simpler counter example is
$$f(x)={\sin{x}-x\over x^2}$$
If I just take equivalents I find $f(x)\to \infty$ obviously incorrect because the limit is $0$ as the Taylor expansions show.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3519645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Maximize $x^2+2y^2$ subject to $y-x^2+1=0$
Maximize $x^2+2y^2$ subject to $y-x^2+1=0$
I tried using Lagrange multiplier method. We have:
$$L(x,y)=x^2+2y^2+\lambda(y-x^2+1)$$
So we have:
$$L_x=2x(1-\lambda)=0$$
$$L_y=4y+\lambda=0$$
One possible solution with $y-x^2+1=0$ is
$x=0$, $y=-1$, $\lambda=4$
But when we calcul... | As in the answer by amd, you need to construct what is called the Bordered Hessian, which is nothing more that the Hessian when $\lambda$ is deliberately included as a new variable. Maybe give it a new name,
$$ h(\lambda, x,y) = x^2 + 2 y^2 + \lambda (y-x^2 +1) $$
The gradient is the triple
$$ h_\lambda = y-x^2 + 1, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3520586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to find:$\lim_{x\to 0}\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{x\arctan x}$
Evalute: $$\lim_{x\to
0}\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{x\arctan x}$$
My attempt:
I used the standard limits from the table:$$\lim_{x\to 0}\frac{\sin x}{x}=1,\;\;\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2... | Alternatively $$\lim_{x\to0}\dfrac{e^{1-\cos^3x}-e^{1-\cos^4x}}{x\arctan x}$$
$$=\lim_{x\to0} e^{1-\cos^4x}\cdot\lim_{x\to0}\dfrac{e^{\cos^4x-\cos^3x}-1}{x\arctan x}$$
$$=-\lim_{x\to0}\cos^3x\cdot\lim_{x\to0} e^{1-\cos^4x}\cdot\lim_{x\to0}\dfrac{e^{(\cos^4x-\cos^3x)}-1}{\cos^4x-\cos^3x}\cdot\lim_{x\to0}\dfrac{1-\cos x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3523158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to prove this symmetric polynomial identity According to this (equation 21 and 26) how to show that
\begin{gather}
\ln(1+\Pi_1t+\Pi_2t^2+\Pi_3t^3+...) =\Pi_1t+\frac{1}{2}(-\Pi^2_1+2\Pi_2)t^2+\frac{1}{3}(\Pi^3_1-3\Pi_1\Pi_2+3\Pi_3)t^3+.. .=\sum_{k=1}^{\infty}\frac{s_k}{k}t^k
\end{gather}
$s_1=\Pi_1 \\ s_2=-\Pi^2_1+... | Given a generating function
$$ f(x) := 1+a_1t+a_2t^2+a_3t^3+\dots \tag{1} $$
whose logarithm is $$ \ln(f(x)) =
\sum_{k=1}^{\infty}\frac{s_k}{k}t^k, \tag{2} $$
differentiate both sides to get
$$ \frac{d}{dx} \ln(f(x)) = f'(x)/f(x) =
\sum_{k=1}^{\infty}s_k t^{k-1}. \tag{3} $$
Multiply both sides by $\,f(x)\,$ to get
$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Solving the trigonometric equation $483\sin\left(\alpha+\frac\pi{3}\right)+16\sqrt3\sin\left(2\alpha+\frac{\pi}{3}\right)+20=0$
I'm trying the value of $\alpha \in [0,\pi]$ that is solution to this trigonometric equation: $$483\sin\left(\alpha+\frac\pi{3}\right)+16\sqrt3\sin\left(2\alpha+\frac{\pi}{3}\right)+20=0$$
I... | Notice that:
$$\sin\left(\alpha+\frac\pi{3}\right) = \sin\left(\alpha\right)\cos\left(\frac\pi{3}\right) +\cos\left(\alpha\right)\sin\left(\frac\pi{3}\right) = \\
= \frac{1}{2}\sin(\alpha) + \frac{\sqrt{3}}{2}\cos(\alpha).$$
Moreover:
$$\sin\left(2\alpha+\frac\pi{3}\right) = \sin\left(\alpha+\left(\alpha+\frac\pi{3}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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value of expression having variables $p,q,r,x,y,z$
If $p,q,r,x,y,z$ are non zero real number such that
$px+qy+rz+\sqrt{(p^2+q^2+r^2)(x^2+y^2+z^2)}=0$
Then $\displaystyle \frac{py}{qx}+\frac{qz}{ry}+\frac{rx}{pz}$ is
what try
$(px+qy+rz)^2=(p^2+q^2+r^2)(x^2+y^2+z^2)$
$p^2x^2+q^2y^2+r^2z^2+2pqxy+2qryz+2prxz=p^2x^2+p^2... | Write $v=(p,q,r)$ and $w=(x,y,z)$. Then the given relation states
$$v\cdot w+|v||w|=0$$
But $v\cdot w=|v||w|\cos\theta$ where $\theta$ is the angle between them, so
$$|v||w|(\cos\theta+1)=0$$
Since none of the scalars are zero, we get $\cos\theta=-1$, so $v=kw$ for some nonzero $k\in\mathbb R$ and
$$\frac{py}{qx}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3532587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given that $x^2 + ax + b > 0$ and $x^2 + (a + np)x + (b + nq) > 0$, prove that $x^2 + (a + mp)x + (b + mq) > 0, m = \overline{1, n - 1}$.
Given that $$\large \left\{ \begin{align} x^2 + ax + b > 0\\ x^2 + (a + np)x + (b + nq) > 0 \end{align} \right., \forall x \in \mathbb R \ (p, q \in \mathbb R \setminus \{0\}, n \in... | Multiply the first inequality by $(n-m)$ and the second by $m$ and add these together
\begin{eqnarray*}
(n-m) (x^2 +ax+b) >0 \\
m(x^2+(a+np)x+(b+np))>0 \\
n(x^2+(a+mp)x+(b+mp))>0.
\end{eqnarray*}
Now divide the final inequality by $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given that $2x^2-4xy+6y^2=9$, find the biggest and lowest value of $2x-y$ So, that's the problem. I tried factoring the quadratic equation to get something like $2x-y$ but it doesn't work. Dividing by $y^2$ won't work because the right side is $9$, not $0$. The last idea I have is to say that $2x-y=t => y=2x-t$ and rep... | $$2x^2-4xy+6y^2=9$$ and let $$k=2x-y\Rightarrow y=2x-k$$
$2x^2-4x(2x-k)+6(2x-k)^2=9$
$2x^2-8x^2+4xk+6(4x^2+k^2-4xk)=9$
$-6x^2+4xk+24x^2+6k^2-24xk-9=0$
$$18x^2-20kx+6k^2-9=0$$
For real roots, Discriminant $\geq 0$
$$400k^2-4\cdot 18(6k^2-9)\geq 0$$
$$100k^2-108k^2+162\geq 0$$
$$-8k^2+162\geq 0\Rightarrow k^2-\frac{81}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Show $1+a+b+c\leq 2\lfloor\sqrt{at^2+bt+c}\rfloor$ for all $a,b,c,t\in\Bbb N\cup\{0\}$ with $a\neq 0$, $t\ge 2$ and $a,b,c\le t-1$. Related to a project I'm writing, I came across the problem of showing that
$$1+a+b+c \leq 2\lfloor \sqrt{ at^2+bt+c}\rfloor$$
holds for all nonnegative integers $a,b,c,t$ satisfying $a \... | You're asking to confirm
$$1+a+b+c \leq 2\lfloor \sqrt{ at^2+bt+c}\rfloor \tag{1}\label{eq1A}$$
Since $a,b,c \in \{0,1,\ldots,t-1\}$, each value is at most $t - 1$, so you get
$$1 + a + b + c \le 1 + (t - 1) + (t - 1) + (t - 1) = 3t - 2 \tag{2}\label{eq2A}$$
With $a \ge 4$, you have
$$2\lfloor \sqrt{ at^2+bt+c}\rfloor ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Splitting field of $\alpha=(2+\sqrt{2})^{1/3}$ Given $\alpha=(2+\sqrt{2})^{1/3}$ The problem asks to calculate the minimal polynomial of $\alpha$ over $\mathbb{Q}$ and find it's splitting field. I have the solution and I understood until the point where it shows that the minimal polynomial is $$x^6-4x^3+2=0$$
Now, sinc... | $$x^6-4x^3+2=0$$ $$\Leftrightarrow (x^3-2)^2=2$$
$$\Leftrightarrow x^3-2=\pm \sqrt{2}$$
$$\Leftrightarrow x^3=2\pm \sqrt{2}$$
$$\Leftrightarrow x=\omega^i\sqrt[3]{2+\sqrt{2}},\omega^i\sqrt[3]{2-\sqrt{2}},i=0,1,2,$$ where $\omega$ is a root of $x^2+x+1$, so $\omega^3=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3537038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Proof by induction:$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$ In the very beginning I'm going to refer to similar posts with provided answers:
Induction Inequality Proof with Product Operator $\prod_{k=1}^{n} \frac{(2k-1)}{2k} \leq \frac{1}{\sqrt{3k+1}}$ (answered by Öz... | It is correct.
Anyways you already ended the proof when you stated that
$\frac{4n+3}{4n+5}\prod_{i=1}^n\frac{4i-1}{4i+1}$ $<$ $\sqrt{\frac{3}{4n+7}}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Find the area of the largest rectangle that can be inscribed in the ellipse $ \frac {x^2}{2} + \frac {y^2}{6} = 1$ Find the area of the largest rectangle that can be inscribed in the ellipse $$ \frac {x^2}{2} + \frac {y^2}{6} = 1$$
I know the answer is $x=1$ yields $Area = 4\sqrt{3}$.
I'm able to get $x=1$ by taking t... | Since your $x=1$ is correct, so I will continue from there. First, solve for y.
$$\frac{1^2}{2}+\frac{y^2}{6}=1$$
$$\frac{y^2}{6}=\frac{1}{2} $$
$$y^2=3$$
$$y=\sqrt{3}$$
Then, we double both of those values to find the side lengths. This is because if we didn’t do that, we would be calculating the area of the rectangle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$ A^3 + B^2 = I_n $ and $A^5=A^2$, then $\det(A^2 + B^2 + I_n) \geq 0 $ and $\operatorname{rank}(I_n + AB^2) = \mathrm{rank}(I_n - AB^2) $ Let $A, B$ be square matrices of size $n$, $n \geq 2$, containing real entries.
$\DeclareMathOperator\rank{rank}$
If the following properties take place: $ A^3 + B^2 = I_n $ and... | From the first condition, $B^2=1-A^3$, so that by the second,
$$ 0=A^2-A^5=A^2B^2=AB^2A=B^2A^2.$$
Now
$$(1-AB^2)(1+AB^2)=1-AB^2AB^2=1 $$
so that
$$\operatorname{rank}(1\pm AB^2)=n. $$
Note that
$$ B^4=(1-A^3)^2=1-2A^3+A^6=1-A^3=B^2.$$
Let $U=\operatorname{im} B^2$ and $W=\operatorname{im} (1-B^2)$. Then for $u\in U$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is $a$ in $\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{\sin 6t}{t^3 \cos^2 3t}\right) = -18$
What is $a$ in $\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{\sin 6t}{t^3 \cos^2 3t}\right) = -18$
We have to make this function defined, by change the cosine into sine or tan. So,
$$\lim_{t \to 0} \left(\frac{a}{t^2} - \fr... | Your condition implies that $$\lim_{t\to 0}\frac{at\cos^23t-\sin 6t}{t^3}=-18$$ Adding and subtracting $at$ in numerator we get $$\lim_{t\to 0} \frac{at-\sin 6t}{t^3}-9a\cdot\frac{\sin^23t}{(3t)^2}=-18$$ or $$\lim_{t\to 0}\frac {a-6}{t^2}+\frac{6t-\sin 6t}{t^3}=9(a-2)$$ Using substitution $x=6t$ the second term on left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Smarter way to solve $ \int_0^1\arctan(x^2)\,dx$ I'm trying to solve the following definite integral:
\begin{equation}
\int_0^1 \arctan(x^2)dx
\end{equation}
I tryed first by parial integration, finding:
\begin{equation}
x\arctan(x^2)\Bigl|_0^1-\int_0^1 \dfrac{2x^2}{1+x^4}dx
\end{equation}
Then:
\begin{equation}
\int_0... | First you need to write $x^4+1$ as a product of factors that are irreducible in the ring of real polynomials.$$x^4+1=(x^2+1)^2-(\sqrt 2 x)^2 $$ $$(x^2+1-\sqrt2 x)(x^2+1+\sqrt 2 x)$$ $$=(x^2-\sqrt 2 x+1)(x^2+\sqrt 2 x+1)$$.Then decompose $$\frac{2x^2}{(x^2-\sqrt 2 x+1)(x^2+\sqrt 2 x+1)}$$ into partial fractions as $$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Sum of the series: $\sum_{n=2}^\infty\frac{(-1)^n}{n^2+n-2}$ is? This was a question I confronted in JAM 2016. I tried the following steps:
$$\displaystyle\sum_{n=2}^\infty\frac{(-1)^n}{n^2+n-2}
\implies\sum_{n=2}^\infty\frac{(-1)^n}{(n-1)(n+2)}\implies\sum_{n=2}^\infty(-1)^n\frac{1}{3}\cdot[\frac{1}{(n-1)}-\frac{1}{... | $$S=\sum_{n=2}^\infty\frac{(-1)^n}{n^2+n-2}$$
Change the indice of the sum
$$m=n-1$$
$$
\begin {align}
S=&-\sum_{m=1}^\infty\frac{(-1)^m}{m(m+3)} \\
S=&-\frac 1 3\sum_{m=1}^\infty{(-1)^m}\left ({\frac 1 m -\frac 1 {m+3}} \right ) \\
S=&\frac 1 3 \ln 2 +\frac 1 3 \left ( \sum_{m=1}^\infty{\frac {(-1)^m} {m+3}} \right ) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $a^3+b^3+3abc>c^3$ where a,b,c are triangle sides If $a,b,c$ are triangle sides prove
$a^3+b^3+3abc>c^3$
|
Triangle inequality: If $~a,~ b, ~$and$~ c~$ are the lengths of the sides of the triangle, with no side being greater than $~c~$, then the triangle inequality states that
$~c\leq a+b~$.
Now to prove the given inequality $~a^3+b^3+3abc>c^3~$, we have to use the above property. So we have
$$a^3 + b^3 + 3abc = (a + b)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3545017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $\log_2{\log_x{(x-3y)}}=-1, x\cdot y^{\log_x{y}}=y^{\frac{5}{2}}$ Problem is to solve this system of equation.
$$\log_2{\log_x{(x-3y)}}=-1$$
$$x\cdot y^{\log_x{y}}=y^{\frac{5}{2}}$$
\My attempt:
First equation can be written as $$\log_x{(x-3y)}=\frac{1}{2}$$.
If on the second equation I put $\log_x()$ I get this
... | You have some mistakes when you solve the second one. A $x$ is changed to a $y$. But the approach is correct. First of all $y$ is positive, which means $x$ has to be as well. Then
$$x\cdot y^{\log_x{y}}=y^{\frac{5}{2}}\Rightarrow \log_x \color{red}{x}+\log_x y\cdot \log_x y=\frac{5}{2}\log_x y$$
or
$$1+t^2=\frac{5}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Summation of finite series: Let $f(r)$ be what? Find the sum of the first $n$ terms of
$\displaystyle \frac{1}{1\times4\times7}+\frac{1}{4\times7\times10}+\frac{1}{7\times10\times13}+...$
My working:
Let $\begin{align}\displaystyle\frac{1}{(3x-2)(3x+1)(3x+4)} &\equiv \frac{A}{3x-2}+\frac{B}{3x+1}+\frac{C}{3x+4}\\1&\equ... | Do telescopic summation after doing the partial fractions:
$$T_k=\frac{1}{(3k-2)(3k+1)(3k+4)}$$ $$\implies 18 T_k=\left(\frac{1}{3k-2}-\frac{1}{3k+1}\right)-\left( \frac{1}{3k+1}-\frac{1}{3k+4)}\right)$$
Let $F_k=\frac{1}{3k-2}$
$$S_n=\sum_{k=1}^{n} T_k= \frac{1}{18}(\sum_{k=1}^n [F_k-F_{k+1}]-\sum_{k=1}^n [F(k+1)-F(k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $n^4 \mod 8$ is identically equal to either 0 or 1, $\forall \ n\in \mathbb{N}$. I feel pretty confident about the first half of my solution for this, however I don't like how I used the induction hypothesis on the case for even integers, it feels like it isn't doing anything useful since it is really easy t... | Your idea is okay, but $(2x+2)^4=16 + 64 x + 96 x^2 + 64 x^3 + 16 x^4$,
unlike what you initially put in the question.
Alternatively, you could argue that if $2|n$ then $8|n^3|n^4$, so $n^4\equiv0\pmod8$,
and if $2\nmid n$ then $2|n-1, n+1, $ and $n^2+1$, so $8|n^4-1$; i.e., $n^4\equiv1\pmod8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Given a sequence $a_n =\frac{n^2 - 3n + (-1)^n}{3n^2 - 7n + 5}$, prove that $\lim_{n \to \infty} a_{n} = g$. The given Sequence is $a_n = \frac{n^2 - 3n + (-1)^n}{3n^2 - 7n + 5}$.
I showed that the Sequence $a_n$ converges towards a Value $g = \frac{1}{3}$.
How do I determine for each $\epsilon > 0$ an $n_0$ so that: $... | This follows directly from the definition of a Limit of a sequence. if $a_n$ converges to $g$, then $lim_{n\to \infty} a_n =g$, such that $\forall \epsilon >0 \exists n_0 \in \mathbb{N} \text{ such that } \forall n\ge n_0 : |a_n -g|<\epsilon$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find a limit with sqrt $\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$ $$\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$$
I don't know how to rewrite or rationalize in order to find the limit.
| To write the complete solution, as per the hint in the comments I gave:
$$
\begin{aligned}
\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right) &= \lim_{x \to \infty}x^2\left(\sqrt{(x^2+3)^2} - \sqrt{x^4 + 6x^2}\right)\\
&=\lim_{x \to \infty}x^2\left(\frac{(x^2+3)^2-x^4-6x^2}{\sqrt{(x^2+3)^2} + \sqrt{x^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
} |
Find $\lim_{x \to \infty} x^3 \left ( \sin\frac{1}{x + 2} - 2 \sin\frac{1}{x + 1} + \sin\frac{1}{x} \right )$ I have the following limit to find:
$$\lim\limits_{x \to \infty} x^3 \bigg ( \sin\dfrac{1}{x + 2} - 2 \sin\dfrac{1}{x + 1} + \sin\dfrac{1}{x} \bigg )$$
What approah should I use? Since it's an $\infty \cdot 0$ ... | Let $t=\frac1x$. Then,
$$\lim_{x \to \infty} x^3 \left ( \sin\frac{1}{x + 2} - 2 \sin\frac{1}{x + 1} + \sin\frac{1}{x} \right )
=\lim_{t \to 0} \frac1{t^3} \left ( \sin\frac{t}{1 + 2t} - 2 \sin\frac{t}{1+t } + \sin t \right )$$
Use $\frac 1{1+a} = 1-a+a^2+O(a^3)$ to expand,
$$\sin\frac{t}{1 + 2t} - 2 \sin\frac{t}{1+t }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Closed form of integral $\int_0^1 \frac{1-\frac{1}{\sqrt[5]{x}}}{1-x} \, dx$ Originally this integral was interesting for me as it represents the specific harmonic number $H_{-\frac{1}{5}}$ and also because Mathematica returned the wrong value $0$ for the integral. I posted the probem here https://mathematica.stackexch... | Note
$$I=\int_0^1 \frac{1-\frac{1}{\sqrt[5]{x}}}{1-x} dx
\overset{ x=t^5}=\int_0^1 \frac{-5t^3}{t^4+t^3+t^2+t+1}dt$$
$$=\sqrt5\int_0^1\left( \frac{2\phi_-t+1}{t^2+2\phi_-t+1}
- \frac{2\phi_+t+1}{t^2+2\phi_+t+1}\right)dt = \sqrt5 J(\phi_-) - \sqrt5 J(\phi_+)$$
where $\phi_{+}=\frac{1+\sqrt5}4= \cos \frac\pi5$, $\phi_{-}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the biggest value of $|a-b|+|b-c|+|c-a|$ when $a^2+b^2+c^2=1$ Given that $a^2 + b^2 + c^2 = 1$
How can I prove that the maximum value of |a - b| + |b - c| + |c - a| is $2\sqrt{2}$ ?
In extends, given that $a_1^2 + a_2^2 + a_3^2 + ... + a_{2020}^2 = 1$ find the maximum value of $|a_1 - a_2| + |a_2 - a_3| +|a_3 - a_... | Without loss of generality, assume that $a\geq b\geq c$. Then:
$$|a-b|+|b-c|+|c-a| = 2(a-c)$$
And we have:
$$
\begin{aligned}
(a-c)^2&=2(a^2+c^2)-(a+c)^2\\
&\leq 2(a^2+c^2) \\
&\leq 2(a^2+b^2+c^2)\\
&=2
\end{aligned}$$
Therefore $2(a-c)\leq 2\sqrt{2}$. Equality occurs when $(a,b,c) = \left(\frac{1}{\sqrt{2}}, 0,-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How many complex numbers $z$ are there such that $|z|=1$ and $z^{5040} - z^{720}$ is a real number? My attempt:
If $z^{5040} - z^{720}$ is real, that means that their imaginary parts are equal.
$e^{5040i\theta} - e^{720i\theta} = k$ , where $k$ is a real number
$\sin{5040\theta} = \sin{720\theta} $
Let $u=720\theta$
$... | Solve $\sin{7u} = \sin{u} $ as follows,
$$ \sin{7u} - \sin{u} = 2\cos 4u \sin 3u=0$$
$\sin4u = 0$ yields $ u = \frac {\pi n}4$ and $\cos3u=0$ yields $u=\frac\pi6+ \frac {\pi n}3$. Then,
$$z= e^{ \frac { i2\pi n}{8\cdot 720}}, \>\>\>\>\>
z= e^{ \frac{i\pi}{6\cdot 720 }+i\frac { i2\pi n}{6\cdot 720}}$$
Thus, number of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the relative maximum and relative minimum of the function $f(x) = 49x + \frac{4}{x}$ Find the relative maximum and relative minimum of the function $f(x) = 49x + \frac{4}{x}=49x+4x^{-1}$
Solution:
Step 1: Find the values of $x$ where $f'(x)=0$ and $f'(x)$ DNE.
$f'(x)=49-4x^{-2}=49-\frac{4}{x^2}$
We can see that $... | Rewrite as
\begin{eqnarray*}
49x+\frac{4}{x} = \left( 7 \sqrt{x} - \frac{2}{\sqrt{x}} \right)^2 +28.
\end{eqnarray*}
From this it easy to see that a minimum of $28$ occurs at $x=2/7$.
Also
\begin{eqnarray*}
49x+\frac{4}{x} = \left( 7 \sqrt{x} + \frac{2}{\sqrt{x}} \right)^2 -28.
\end{eqnarray*}
similarly a maximum of $-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3558946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find $\int_{\frac{1}{4}}^4\frac{(x+1)\arctan x}{x\sqrt{x^2+1}} dx$ I have to find this integral:
$$\int_{\frac{1}{4}}^4\frac{(x+1)\arctan x}{x\sqrt{x^2+1}} dx$$
My attempt was to split the integral:
$$\int_{\frac{1}{4}}^4\frac{(x+1)\arctan x}{x\sqrt{x^2+1}}=\int_{\frac{1}{4}}^4\frac{x\arctan x+1}{x\sqrt{x^2+1}}+\int_{\... | Hint. Let $t=1/x$ then
$$\begin{align}I&:=\int_{\frac{1}{4}}^4\frac{(x+1)\arctan(x)}{x\sqrt{x^2+1}} dx=
\int^{\frac{1}{4}}_4\frac{(1/t+1)\arctan(1/t)}{(1/t)\sqrt{1/t^2+1}} \frac{-dt}{t^2}\\
&=\int_{\frac{1}{4}}^4\frac{(t+1)\arctan(1/t)}{t\sqrt{t^2+1}} \,dt
=\frac{\pi}{2}\int_{\frac{1}{4}}^4\frac{t+1}{t\sqrt{t^2+1}} \,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show that the restriction of $f$ to any straight line through $B$ has a local minimum in $B$ I have the following problem:
We have the function: $f(x,y)=(x^2+y^2-6x)\cdot (\frac{1}{8}y^2-x)$
We also have the point $B=(0,0)$.
Show that the restriction of $f$ to any straight line through $B$ has a local minimum in $B$. ... | Straight lines through $(0, 0)$ take the form
$$y = mx$$
for some $m \in \Bbb{R}$, when not vertical. When vertical, the equation becomes
$$x = 0.$$
This means two cases. For the first case, substitute $(x, y) = (x, mx)$ into $f(x, y)$:
\begin{align*}
f(x, mx) &= (x^2 + m^2 x^2 - 6x)\cdot\left(\frac{1}{8}m^2x^2 - x\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Rolling at Least 2 2's on 3 fair dice Say we roll three 6-sided dice.
What is the probability that at least two of the faces are a 2?
The sample space here is 216.
Rolling a 2 on one dice is 1/6.
Rolling two 2s is 1/6 * 1/6 = 1/36
The third dice is at least two 2s so the probability it isn't a 2 is 5/6 and the prob... | The following rolls are all possible:
$$2,2,1 \\ 2,1,2 \\ 1,2,2 \\ \vdots \\ 2,2,6 \\ 2,6,2 \\ 6,2,2$$
The probability of getting exactly two $2$'s followed by a non-$2$: $$\dfrac{1}{6}\cdot \dfrac{1}{6}\cdot \dfrac{5}{6}$$
The probability of getting a $2$, a non-$2$, then a $2$:
$$\dfrac{1}{6}\cdot \dfrac{5}{6}\cdot \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3569381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$\sum_{n=1}^\infty (n!)^{-\frac{1}{n}}$ I have to study the character of this series
$$\sum_{n=1}^\infty (n!)^{-\frac{1}{n}}$$
and I tried with the ratio test:
$ \frac{a_{n+1}}{a_n}= \frac{((n+1)!)^{-\frac{1}{n+1}}}{(n!)^{-\frac{1}{n}}}=
\frac{(n!)^{\frac{1}{n}}}{((n+1)!)^{\frac{1}{n+1}}}=
\frac{(n!)^{\frac{1}{n}}}{((... | $\frac{1}{n}=\frac{1}{n^{n*\frac{1}{n}}}<\frac{1}{n!^{\frac{1}{n}}}$
Applying the comparison test and knowing that the series $\sum_{n=1}^\infty \frac{1}{n}$ diverges, also our series diverges
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the general solution to $\csc \theta + \sec \theta = 1$
Find the general solution to $$\csc\theta + \sec\theta =1$$
This is how I solved.
We have,
\begin{align}
\csc\theta + \sec\theta &=1\\
\frac1{\sin\theta} + \frac1{\cos\theta}& =1\\
\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta} &=1\\
(\sin\theta + \cos... | Your squaring of the equation $$\cos x+\sin x=\cos x\>\sin x\tag{1}$$ has introduced spurious solutions. In fact the value ${1\over2}\arcsin\bigl(2-2\sqrt{2}\bigr)\approx-0.488147$ does not solve the given problem.
Drawing the graphs of $x\mapsto \cos x+\sin x$ and $x\mapsto\cos x\>\sin x$ shows a symmetry with respec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3577196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Maximum value of $\frac{a}{1+bc} + \frac b{1+ac} + \frac{c}{1+ab}$ given $a^2 + b^2 + c^2 = 1$ Given that the constraint of $a, b, c$, for which $a, b, c$ are non-negative real numbers, is $a^2+b^2+c^2=1,$ find the maximum value of $$\frac{a}{(1+bc)}+\frac b{(1+ac)}+\frac{c}{(1+ab)}.$$
For this question I have tried us... | For non-negative variables we obtain: $$\sum_{cyc}\frac{a}{1+bc}\leq\sum_{cyc}\frac{a\sqrt2}{a+b+c}=\sqrt2$$
because $$2(1+bc)^2\geq(a+b+c)^2$$ it's
$$2+4bc+2b^2c^2\geq1+2(ab+ac+bc)$$ or
$$2b^2c^2+(b+c-a)^2\geq0.$$
The equality occurs fot $c=0$ and $a=b=\frac{1}{\sqrt2},$ which says that we got a maximal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate Variance with Binomial Expansion IQ is normally distributed with mean 100 and standard deviation 15. An IQ of 130 or above is considered gifted, and 150 and above is considered genius. There are 350 million people living in the United States.
Let $X$ be the IQ of a person randomly selected from the population... | You have
\begin{align}
15^2 & = \operatorname{Var}(X) = \operatorname E\big((X-100)^2\big) \\
& = \operatorname E(X^2) - 200 \operatorname E(X) +100^2 \\
& = \operatorname E(X^2) - 100^2
\end{align}
Therefore
$$
\operatorname E(X^2 ) = 100^2 + 15^2 = 32\,500.
$$
\begin{align}
\operatorname{Var}(X^2) & = \operatorname E... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3583326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Integral of $\int \sin^2x\cos^4xdx$ $$\int \sin^2x\cos^4xdx$$
I tried
$$I = \int (1-\cos^2x)\cos^4xdx = \int \frac{\sec^2x-1}{\sec^6x}dx = \int \frac{\tan^2x}{\sec^6x}dx$$
Take $\tan x = t \implies \sec^2xdx = dt$
$$I = \int \frac{t^2}{(t^2+1)^4}dt$$
And I could not proceed further from here.
| You can calculate it directly without substitution as follows:
$$\sin^2x\cos^4x=(\sin x\cos x)^2\cos^2x =\frac 14\sin^22x\cdot \frac 12(1+\cos 2x)$$
$$= \frac 18 \cdot\frac 12(1-\cos 4x) +\frac 18\sin^2 2x\cos 2x$$
Hence,
$$\int \sin^2x\cos^4x\; dx = \frac x{16}-\frac{\sin 4x}{64} + \frac{\sin^3 2x}{48}+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3583796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
If $A, B, C$ are angles of $\Delta ABC$ and $\sin (A-\pi /4) \sin (B-\pi/4) \sin (C-\pi/4)=\frac{1}{2\sqrt 2}$...
If $A, B, C$ are angles of $\Delta ABC$ and $\sin (A-\pi /4) \sin (B-\pi/4) \sin (C-\pi/4)=\frac{1}{2\sqrt 2}$, then prove that $\sum \tan A \tan B=\sum \tan A$
Solving the given equation, we get$$(\sin A... | Hint:
Both
\begin{align}
\sin (A-\pi /4) \sin (B-\pi/4) \sin (C-\pi/4)=\frac{1}{2\sqrt 2}
\tag{1}\label{1}
\end{align}
and
\begin{align}
\tan A\tan B+\tan B\tan C+\tan C\tan A
&=
\tan A+\tan B+\tan C
,
\tag{2}\label{2}
\end{align}
expressed in terms of semiperimeter $\rho$,
inradius $r$ and circumradius $R$ of the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
On A Splitting Equation of an Egyptian fraction to Egyptian fractions such that all produced fractions have odd denominators. My question is: Do we have an splitting equation where we can produce fractions with odd denominators?
To split an Egyptian fraction to Egyptian fractions, we can use the splitting equation belo... | A general solution is for every positive positive integer $\ n\ $ :
*
*If $n$ is odd , then $$\frac{1}{3n+2}+\frac{1}{6n+3}+\frac{1}{18n^2+21n+6}=\frac{1}{2n+1}$$ is a solution with odd denominators
*If $n$ is even , then $$\frac{1}{3n+3}+\frac{1}{6n+3}+\frac{1}{6n^2+9n+3}=\frac{1}{2n+1}$$ is a solution with odd de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3585135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Trig substitution for $\sqrt{9-x^2}$ I have an integral that trig substitution could be used to simplify.
$$ \int\frac{x^3dx}{\sqrt{9-x^2}} $$
The first step is where I'm not certain I have it correct. I know that, say, $\sin \theta = \sqrt{1-cos^2 \theta}$, but is it correct in this case $3\sin \theta = \sqrt{9 - (3\c... | You can solve without the trigonometric substitution, decomposing as
$$\frac{x^3}{\sqrt{9-x^2}}=9\frac{x}{\sqrt{9-x^2}}-x\sqrt{9-x^2}$$
and the antiderivatives are immediate (by $u=x^2$):
$$-9\sqrt{9-x^2}-\frac13(9-x^2)^{3/2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Proving $a^2 + b^2 + c^2 \geqslant ab + bc + ca$
Let $a, b, c \in \mathbb{R}$. Show that $a^2 + b^2 + c^2 \geqslant ab + bc + ca$.
My reasoning went as follows and I would like to know if it's correct.
$a^2 + b^2 + c^2 \geqslant ab + bc + ca$
$\Leftrightarrow a^2 + b^2 + c^2 -ab - bc - ca \geqslant 0$
$\Leftrightarro... | Add up the inequalities below
$$a^2 + b^2 \ge 2ab, \>\>\>\>\>b^2 + c^2 \ge 2bc, \>\>\>\>\>c^2 + a^2 \ge 2ca$$
to arrive at,
$$a^2 + b^2 + c^2 \geqslant ab + bc + ca$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Evaluating $\int _0^1\frac{\ln ^2\left(x^2+1\right)}{x+1}\:dx$ using real methods I've stumbled upon that interesting integral here, the OP managed to transform the integral into something more approachable using contour integration and proved that
$$\int _0^1\frac{\ln ^2\left(x^2+1\right)}{x+1}\:dx\:=\:-\pi G+\frac{5}... | Integrate by parts
$$\int _0^1\frac{\ln ^2(x^2+1)}{x+1}~dx
=\ln^32 -4I \tag1$$
where $I$ is given below, along with its twin $J$
$$I=\int_0^1\frac{x\ln(1+x)\ln(x^2+1)}{1+x^2}~dx,\>\>\>\>\>
J=\int_0^1\frac{x\ln(1-x)\ln(x^2+1)}{1+x^2}~dx $$
Then, evaluate their sum and difference as follows
\begin{align}
J+I=& \int_0^1\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $\frac1{a(1+b)}+\frac1{b(1+c)}+\frac1{c(1+a)}\ge\frac3{1+abc}$ I tried doing it with CS-Engel to get $$
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geq \frac{9}{a+b+c+ a b+b c+a c}
$$
I thought that maybe proof that $$
\frac{1}{a+b+c+a b+b c+a c} \geq \frac{1}{3(1+a b c)}
$$ or $$
3+3 a b c \geq a+b+... | For positive variables by AM-GM we obtain:
$$\sum_{cyc}\frac{1}{a(1+b)}=\frac{1}{1+abc}\sum_{cyc}\frac{1+abc}{a(1+b)}=\frac{1}{1+abc}\left(\sum_{cyc}\frac{1+abc}{a(1+b)}+1-1\right)=$$
$$=\frac{1}{1+abc}\sum_{cyc}\frac{1+a+ab+abc}{a(1+b)}-\frac{3}{1+abc}=$$
$$=\frac{1}{1+abc}\sum_{cyc}\frac{1+a+ab(1+c)}{a(1+b)}-\frac{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$2^{\sin(x) + \cos(y)} = 1$ , $16^{\sin^2(x) + \cos^2(y)} = 4$ (system of equations) My progress so far:
$2^{\sin(x) + \cos(y)} = 1$
$16^{\sin^2(x) + \cos^2(y)} = 4$
||||||||||||||||||||
$\sin(x) + \cos(y) = 0$
$\sin^2(x) + \cos^2(y) = \frac12$
I think I'm on the right track, but I'm not sure how to continue. Thanks i... | Suppose $\sin x + \cos y = 0$. Then
$$\sin ^2x + \cos ^2y = (\sin x + \cos y) ^2 -2\sin x\cos y = -2\sin x\cos y. $$
Thus we have left to consider
$$\begin{cases} \sin x + \cos y = 0 \\ \sin x\cos y = -\frac{1}{4} \end{cases} $$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Unable to prove an assertion with induction I need to prove:
$ \displaystyle \sum_{k=1}^n\frac{1}{(5k + 1) (5k + 6)} = \frac{1}{30} - \frac{1}{5(5n + 6)} $ with mathematical induction for all $n \in \mathbb{N}$.
After successfully proving it for n = 1, I try to prove it in the Induction-Step for n + 1:
$ \displaystyle... | You started fine. Note that to prove by induction a sequence of equalities of the type$$a_1+a_2+\cdots+a_n=b_n$$is the same thing as proving that $a_1=b_1$ and that, for each $n\in\mathbb N$, $a_{n+1}=b_{n+1}-b_n$. So, you should prove that\begin{align}\frac1{5\bigl(5(n+1)+1\bigr)\bigl(5(n+1)+6\bigr)}&=\frac1{30}-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Compute $\left[\begin{smallmatrix}1-a & a \\ b & 1-b\end{smallmatrix}\right]^n$ Compute $\begin{bmatrix}1-a & a \\ b & 1-b\end{bmatrix}^n$, where the power of $n\in\mathbb N$ denotes multiplying the matrix by itself $n$ times; $a,b\in[0,1]$.
Edit:
I considered using induction, computed the desired matrix:
$$\begin{bm... | Let $ n $ be a positive integer, $ a,b\in\left[0,1\right] $ such that $ a+b\neq 0 \cdot $
Denoting $ J=\left(\begin{matrix}-a&a\\b&-b\end{matrix}\right) $, observe that : $ \left(\begin{matrix}1-a&a\\b&1-b\end{matrix}\right)=I_{2}+J \cdot $
We have that: \begin{aligned} J^{2}=\left(\begin{matrix}-a&a\\b&-b\end{matrix}\... | {
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"timestamp": "2023-03-29T00:00:00",
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If $p$ and $q=2p−1$ are primes, and $N=pq,$ then $N$ is pseudo-prime for all possible bases $b$ that are quadratic residues modulo $2p−1.$ ... Being $n \in \mathbb Z,$ $n$ is said to be pseudo-prime with respect to the base $b$ if it is compound and also verifies the congruence $b^{n−1}\equiv 1 \pmod n$ where $n|b^{n-... | We want to show that if $p$ and $q:=2p-1$ are prime, then $N:=pq$ satisfies the property that if $$a \equiv u^2 \pmod{q},$$
and $p,q\nmid a$,
then
$$a^{N-1}\equiv 1 \pmod{N}.$$
One solution is to use the Chinese remainder theorem. In this context,
it says that it suffices to show both
$$a^{N-1}\equiv 1 \pmod{p}$$
an... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For non-negative reals $x,y$ such that $x+y\le 4$ prove that $y(x-3)(y-3) \le 3(4-y)$ For non-negative reals $x,y$ such that $x+y\le 4$ prove that $y(x-3)(y-3) \le 3(4-y)$
ATTEMPT
We transform the equation into $xy^2+12y-3xy-3y^2\le 12$
I noticed that for $x=0, y=2$ equality is achieved, but I am lost here.
| You want to show that if $x$ and $y$ are nonnegative real numbers with $x+y\leq4$ then
$$f(x,y)=y(x-3)(y-3)-3(4-y),$$
is negative. Taking the derivative with respect to $x$ we get
$$\frac{\partial f}{\partial x}=y(y-3),$$
which is nonzero if $y\neq0$ and $y\neq3$. The derivative is everywhere positive for $y>3$ and so ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1+\sin^4 x}{\rm d}x$ Someone gives a solution as follows:
\begin{align*} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1+\sin^4 x}{\rm d}x&=2\int_0^{\frac{\pi}{2}}\frac{\sin^2 x}{1+\sin^4 x}{\rm d}x\\ &=2\int_0^{\frac{\pi}{2}}\frac{\cos^2 x}{1+\cos... | The red-colored line is obtained with the substitution $t = \frac1u$ as follows
\begin{align}
\int_0^{\infty} \frac{2}{(1+t^2)^2+1} dt
& =\int_{0} ^{\infty} \frac{2}{t^4+2t^2+2}dt\\
&=\int_{0}^{\infty} \frac{1}{\frac1{2u^2}+1+{u^2}} du\\
& =\frac{1}{2}\int_{-\infty}^{+\infty}\frac{1}{\left(u-\frac{1}{u\sqrt{2}}\right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $A$ and $B$ have non-negative eigen values then can we conclude that $A+B$ has non-negative eigen values? Let $A$ and $B$ be $n\times n$ matrices whose all eigen values are non-negative real numbers. What can I say about the signs of the eigen values of $A+B$ ? First of all is it possible that the eigen values of $A... | Take $A=\begin{pmatrix} 1& 2\\0&1\end{pmatrix}$, $B=\begin{pmatrix} 1& 0\\2&1\end{pmatrix}$. Then both have $1$ as unique eigenvalue but $A+B$ has zero as eigen value.
Take $C=\begin{pmatrix} 1& 1\\0&1\end{pmatrix}$, $D=\begin{pmatrix} 1& 0\\-1&1\end{pmatrix}$. Then both have $1$ as unique eigen value but $C+D$ has no ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3601299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that: $\left|\int\limits_a^b\frac{\sin x}{x}dx\right|\le \frac{2}{a}$ Prove that: $$\left|\int\limits_a^b\frac{\sin x}{x}dx\right|\le \frac{2}{a}$$
$0<a<b$
I'm not asking for a complete solution to this, just a hint or a push in the right direction. I've been struggling with this and haven't really made any progr... | \begin{align*}
I &= \int_{a}^b \dfrac{\sin x}{x}dx \\
&= \dfrac{-\cos x}{x}|_{a}^b+ \int_{a}^b\dfrac{\cos x}{x^2}dx \\
&= \dfrac{\cos a}{a} - \dfrac{\cos b}{b}+ J \\
&\le \dfrac{1}{a}+\dfrac{1}{b}+ \int_{a}^b \dfrac{1}{x^2}dx \\
&= \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{a}-\dfrac{1}{b} \\
&= \dfrac{2}{a} \tex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Fermat's Little Theorem: Proving an integer exists which satisfies the following criteria mod $5$ This question has been tantalizing me for a day now:
Given integers $a,b,c,d$ such that $d \neq 0$ (mod $5$) and $m$ an integer for which $am^3 + bm^2 +cm+d \equiv 0$ (mod $5$), prove there exists an integer $n$ for which... | $m\not\equiv0\bmod5$, since otherwise $d\equiv0\bmod5$, contradicting the given facts. Hence, since $5$ is a prime number, $m$ has an inverse modulo $5$ – let $n$ be this inverse. Multiplying $am^3+bm^2+cm+d\equiv0\bmod5$ by $n^3$, we can easily verify that $n$ satisfies the final congruence.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to use prove this $p^4\equiv p\pmod {13}$
Let a prime number $p$, and $n$ a positive integer such $$p\mid n^4+n^3+2n^2-4n+3.$$
Show that $$p^4\equiv p\pmod {13}.$$
A friend of mine suggested that I might be able to use the results problem.
| My take on this problem is to solve the quartic equation using the quartic formula. To use this formula, the $x^3$ term needs to be eliminated, which can be done by substituting $y=x-\frac {1}{4}$. The result is, after quite a bit of algebra, is
$y^4+\frac{13}{8} y^2 - \frac{39}{8}y -\frac{1053}{256}=0$
To solve this, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
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Show that $1+2^n+2^{2n}$ is divisible by $7$, when $n$ is not a multiple of $3$ Problem taken from a paper on mathematical induction by Gerardo Con Diaz. Although it doesn't look like anything special, I have spent a considerable amount of time trying to crack this, with no luck.
Show that $1+2^n+2^{2n}$ is divisible b... | A bit longer:
Assume $\tau(n)=\top,n\equiv 1\pmod{3}$
$$1+2^n+2^{2n}=7k\implies 2^{2n}=7k-2^n-1,\;k\in\mathbb Z$$
$\tau(n+1)=\top\implies\;n+1\equiv 2\pmod{3}$
$$1+2^{n+1}+2^{2(n+1)}=1+2\cdot 2^{n}+4\cdot2^{2n}=1+2\cdot2^n+4(7k-2^n-1)\\=28k-2\cdot2^n-3=28k-(2^{n+1}+3)$$
$$7\mid2^{n+1}+3\implies7\mid 2\cdot2^n-4=2(2^n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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This inequality maybe is a form of conditional Chebyshev's inequality let $a_{i},b_{i}>0$, show that
$$\sum_{i=1}^{n}a_{i}b_{i}\ge \dfrac{2}{n+\sqrt{\sum_{i=1}^{n}\dfrac{b_{i}}{a_{i}}\sum_{i=1}^{n}\dfrac{a_{i}}{b_{i}}}}\sum_{i=1}^{n}a_{i}\sum_{i=1}^{n}b_{i}\tag{1}$$
I try:since
$$\sum_{i=1}^{n}\dfrac{b_{i}}{a_{i}}\sum... | Proof: We have the following identity:
\begin{align}
&\sum a_i b_i - \frac{2}{n + \sqrt{\sum \tfrac{b_i}{a_i} \sum \tfrac{a_i}{b_i}}} \sum a_i \sum b_i\\
=\ & \sum \frac{b_i}{a_i}
\left(a_i - \frac{\frac{a_i}{b_i}\sqrt{\sum \tfrac{b_i}{a_i} \sum \tfrac{a_i}{b_i}} + \sum \frac{a_i}{b_i}}
{\left(n+\sqrt{\sum \tfrac{b_i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3618210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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On proving a sequence converges by induction. I would like to prove by induction that given a sequence of positive real numbers $\{x_k \}_{k \in \mathbb{N}}$ such that
$$x_{k+1} \le \left(1 - \frac{2c \ell}{k} + \left(\frac{c}{k} \right)^2 M_1 L^2 \right) x_{k} + \frac{c^2}{k^2} M $$
where $0<c < \max\{ 1/ \sqrt{M... | This part of the answer is preliminary and is intended to show the problems with the proposed approach.
Put $A= M_1L^2$ and $D=\ell^2-A$. Assume that $M\ge 0$ and $D\le 0$. It follows that $P(k)=1 - \frac{2c \ell}{k} + \left(\frac{c}{k} \right)^2 A\ge 0$ for each $k$.
In the appoach from the question in order to can... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to integrate with a function in the limit $\int ^a _0 (\int_0 ^ \sqrt {a^2 - x^2} (a^2 - x^2 - y^2)dy)dx = 303$
I'm trying to solve the above question where the integral has a function as the limit.
Seeing the similarity between the limit and the variable integration, I substituted the $\sqrt {a^2 - x^2}$ with $u$... | Hint. You do the inner integral first. That is, integrate with respect to $y$ alone, treating $x$ as constant. When you put in the limits, you get an expression that depends on $x$ alone. Then integrate this and substitute the constant limits -- this is the outer integral.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\int_{0}^{\frac{\pi}{2}}\text{erf}(\sqrt{a}\cos(x))\text{erf}(\sqrt{a}\sin(x))\sin(2x)dx=\frac{e^{-a}-1+a}{a}$ How to prove that for $a>0$:
*
*$\int_{0}^{\frac{\pi}{2}}\text{erf}(\sqrt{a}\cos(x))\text{erf}(\sqrt{a}\sin(x))\sin(2x)dx=\frac{e^{-a}-1+a}{a}$
*$\int_{0}^{\frac{\pi}{2}}\text{erf}\ ^2(\sqrt{a}\cos... | As noted by @Zacky the first result can be adapted from this answer from @DrZafarAhmedDSc. Using the series expansion
\begin{equation}
\operatorname{erf}(z)= e^{-z^2} \sum_{n=0}^{\infty} \frac{z^{2n+1}}{\Gamma(n+3/2)}.
\end{equation}
we have
\begin{align}
\mathcal{I}_1&=\int_{0}^{\frac{\pi}{2}}\operatorname{erf}(\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3651510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that sequence $ x_{(n+1)}= \frac {a}{1+x_n}$ is convergent to positive root of $x^2+x-a=0,$where $a >0$ and $x_1 >0$ Prove that sequence $$ x_{(n+1)}= \frac {a}{1+x_n}$$ is convergent to positive root of $x^2+x-a=0,$where $a >0$ and $x_1 >0$
we have $$x_{(n+2)}-x_n=\frac{-a(x_{(n+1)}-x_{(n-1)})}{(1+x_{(n-1)})(1+x... | Yes, the behavior is opposite indeed, but, as a continuation of your work, let's look at
$$x_{n+2}=\frac{a}{1+x_{n+1}}=\frac{a}{1+\frac{a}{1+x_n}}=\frac{a(x_n+1)}{1+x_n+a}$$
or $x_{n+2}=f(x_n)$ where $f(x)=\frac{a(x+1)}{1+x+a}$. $f(x)$ is ascending because $f'(x)=\left(\frac{a}{1+x+a}\right)^2$. It is also worth menti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3651794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find Maclaurin series for $x^2\ln(1+2x)$ We recently started learning about Maclaurin series. I feel confused a little, can anyone help me with this?
$$f(x) =x^2\ln(1+2x),\;\vert x\vert\lt\frac{1}{2}$$
| One way to assess your problem is to it the following way by using the following equation: \begin{equation}f(x)=f(0)+f'(0)\frac{x}{1!}+f''(0)\frac{x^2}{2!}+f'''(0)\frac{x^3}{3!}+\dots\end{equation}
So when assessing the problems like one has to realize that $x^2$ is a polynomial form, and $\ln(1+2x)$. So the natural lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Find a closed form to the solution of $\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-x}}}}}=x$ Hi I try to solve the following nested radical :
$$\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-x}}}}}=x$$
Miraculously the related polynomials is a quintic .More precisely :
$$ x^5 - x^4 - 4 x^3 + 3 x^2 + 3 x - 1=0$$
I know that we can... | One of the solution that can be obtained easily is by substituting $x = 2\cos\theta$
steps as follows
$\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-x}}}} = x$ will become $\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-2\cos\theta}}}} = 2\cos\theta$
Now by applying Half angle cosine formula we can simplify as follows
$\sqrt{2+\sqrt{2+\sqrt{2-2\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3658357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I approach this inequality? Let $a, b$ and $c$ be three non-zero positive numbers. Show that:
$$\sqrt{\frac{2a}{a + b}} + \sqrt{\frac{2b}{b + c}} + \sqrt{\frac{2c}{a + c}} \leq 3$$
I know the triangular inequality would help here, but I don't know how to approach it.
I started by $a+b≥a$ then that gives $\frac{... | By C-S $$\sum_{cyc}\sqrt{\frac{2a}{a+b}}\leq\sqrt{2\sum_{cyc}\frac{a}{(a+b)(a+c)}\sum_{cyc}(a+c)}=$$
$$=\sqrt{\frac{8(ab+ac+bc)(a+b+c)}{\prod\limits_{cyc}(a+b)}}\leq3,$$ where the last inequality it's just $$\sum_{cyc}c(a-b)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3659032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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$m(\angle BOD)=60^{\circ}$ iff $k=\sqrt{3}$. Let $\triangle {ABC}$ cu $m(\angle C)=90^{\circ}$ and $D\in [BC], E\in [AC]$ s.t. $\frac{BD}{AC}=\frac{AE}{CD}=k$.
If $BE\cap AD=\{O\}$ show that $m(\angle BOD)=60^{\circ}$ iff $k=\sqrt{3}$.
I tried to prove it with trigonometry, with tangent, but there are a lot of computa... |
Let us prove that $k=\sqrt 3\implies \angle BOD=\angle AOE=60^\circ.$
First, construct lines perpendicular to lines $AC$ and $BC$ at points $E$ and $D$. Their intersection is point $F$. Also construct line $OG$ perpendicular to $AC$ at point $G$ and line $OH$ perpendicular to line $BC$ at point $H$.
To simplify calcu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving a first order differential equation by finding an integrating factor : Edit Problem:
Solve the following differential equations by first finding an integrating factor.
$$ (y^2(x+1) + y ) \, dx + ( 2xy + 1 ) \, dy = 0 $$
Answer:
\begin{align*}
M_y &= 2(x+1)y + 1 = 2xy + 2y + 1 \\
N_x &= 2y \\
\frac{1}{N(x,y)} \l... | $$(y^2(x+1) + y ) \, dx + ( 2xy + 1 ) \, dy = 0$$
Rearrange some terms:
$$y^2xdx+y^2dx + y dx + xdy^2 + dy = 0$$
$$y^2xdx+(y^2dx +xdy^2)+ y dx + dy = 0$$
$$y^2xdx+dxy^2+ y dx + dy = 0$$
Multiply by $e^x$:
$$y^2xde^x+e^xdxy^2+ y de^x + e^x dy = 0$$
$$dxy^2e^x+ de^xy= 0$$
Integrate:
$$xy^2e^x+ e^xy= C$$
Note ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3663339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to improve $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$ I have proved this inequality $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$.
Using $\left|\sin(nx)\right|\leq n\left|\sin(x)\right|$ on $[0,\frac{\pi}{2n}]$ and $\... | We have the elementary estimate
$$1 \le \frac{z^4}{\sin^4 z} \le 1 + z^2 \varepsilon$$
where
$$\varepsilon= \frac{\pi^2}{4} - \frac{4}{\pi^2}.$$
Let $z = (y/n)$ and multiply both sides by $\sin^4 y/y^4$. Then for $y \in [0,n \pi/2]$,
one has:
$$ \frac{\sin^4 y}{y^4}
\le
\left(\frac{\sin(y)/n}{\sin (y/n)}\right)^4
\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3664974",
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"source": "stackexchange",
"question_score": "6",
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Calculating the limit of $\lim\limits_{n \to \infty} (\frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + ...+\frac{1}{2n})$ Hello everyone how can I calculate the limit of:
$\lim\limits_{n \to \infty} (\frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + ...+\frac{1}{2n})$
| Hint:
Factor out $n$ in all denominators:
$$\frac{1}{n+1} + \frac{1}{n+2} + …+\frac{1}{2n}=\frac1n\biggl(\frac1{1+\frac1n}+\frac1{1+\frac2n}\dots+\frac1{1+\frac nn}\biggr)$$
and recognise a Riemann sum.
| {
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Any $(x, y, z)$ can satisfy the $5x^2+2y^2+6z^2-6xy-2xz+2yz<0$? Please tell me whether there any $(x, y, z)$
which can satisfy the $5x^2+2y^2+6z^2-6xy-2xz+2yz<0$ ?
No process or just solve it by calculator are both fine.
Thank you.
| The expression $$5x^2+2y^2+6z^2-6xy-2xz+2yz$$ may be written as $$3(x-y)^2+(x-z)^2+(y+z)^2+x^2-2y^2+4z^2$$ Notice that the only term with a negative sign is $-2y^2$, all others are positive. Setting $x=z=0$ we have $$3(x-y)^2+(x-z)^2+(y+z)^2+x^2-2y^2+4z^2\geq2y^2\geq 0$$
Hence, the expression is never negative.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do we prove this continued fraction for the quotient of gamma functions Given complex numbers $a=x+iy$, $b=m+in$ and a gamma function $\Gamma(z)$ with $x\gt0$ and $m\gt0$, it is conjectured that the following continued fraction holds
$$\frac{\displaystyle4\Gamma\left(\frac{2a+3}{4}\right)\Gamma\left(\frac{2b+3}{4}\... | This can be deduced from Gauss' continued fraction
$$\frac{_2F_1(a+1,b;c+1;z)}{_2F_1(a,b;c;z)}=\cfrac{c}{c+\cfrac{(a-c)bz}{c+1+\cfrac{(b-c-1)(a+1)z}{c+2+\cfrac{(a-c-1)(b+1)z}{c+3+\cfrac{(b-c-2)(a+2)z}{c+4+\ddots}}}}}$$
which evaluates the continued fraction in $(1a)$ as
$$\frac{(2a+1)(2b+1)}{a+b+2}\frac{_2 F_1\left(b+\... | {
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"source": "stackexchange",
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Polynomial transformations and Vieta's formulas
Let $f(x)$ be a monic, cubic polynomial with $f(0)=-2$ and $f(1)=−5$. If the sum of all solutions to $f(x+1)=0$ and to $f\big(\frac1x\big)=0$ are the same, what is $f(2)$?
From $f(0)$ I got that $f(x)=x^3+ax^2+bx-2$ and from $f(1)=-5$ that $a+b = -4$ however I'm not sur... | When $f(x+1)=0$, we have,
$$(x+1)^3+a(x+1)^2+b(x+1)-2=0$$
Whose sum of roots can be obtained by "negative of coefficient of $x^2$ upon coefficient of $x^3$", which comes out to be $-\frac{3+a}{1}$.
Similarly, when $f(\frac 1x)=0$, we have,
$$\frac {1}{x^3}+a\frac{1}{x^2}+b\frac{1}{x}-2=0$$
Since $x\neq 0$, we can rearr... | {
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how is this solved? $\lim_{x→0} \frac{1}{x^2}\left(\left(\tan(x+\frac{\pi}{4})\right)^{\frac{1}{x}}−e^2\right)$ I know how to solve the trigonometric part of the limit, i.e $\left(\tan(x+\frac{\pi}{4})\right)^{1/x}$, which I think is $e^2$, I do not however, know how to carry forward the question with that.
Also, I am ... | We use Taylor series around $0$:
$$
\tan \left( {x + \frac{\pi }{4}} \right) = 1 + 2x + 2x^2 + \frac{8}{3}x^3 + \cdots .
$$
Hence, by using Taylor series for the logarithm and the exponential function,
\begin{align*}
\left( {\tan \left( {x + \frac{\pi }{4}} \right)} \right)^{\frac{1}{x}} & = \exp \left( {\frac{1}{x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $x^2+y^2-\sqrt{2}xy < 4$ provided $1\le x,y \le \frac{5}{2}$. If possible I would like an elegant solution to the following problem:
Let $x,y\in\mathbb{R}$ such that $1\le x \le \frac{5}{2}$ and $1\le y \le \frac{5}{2}$. Prove that
$x^2+y^2-\sqrt{2}xy < 4$
I'm aware that you can use Lagrange multipliers bu... | Let $f(x,y)=x^2+y^2-\sqrt2xy.$
Thus, $f$ is a convex function of $x$ and $f$ is a convex function of $y$.
Id est, $$\max{f}=\max\{f(1,1),f(1,2,5),f(2.5,2.5)\}=f(1,2.5)=7.25-2.5\sqrt2<4.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Formula for number of permutations that have $1$ flanked by an even integer on each side. Suppose we have the sequence $(1,2,3,...,n)$ permuted. If $n$ is an even integer greater than 3, find a formula for the number of permutations that have $1$ flanked by an even integer on each side.
I'm not sure how to start this,... | Since $1$ must be flanked on each side by an even integer, it can't be at either end. Thus, $1$ must be in one of the $n - 2$ spots in between the end points. Then there are $\frac{n}{2}$ possible even integers that can go before it and, once that is chosen, there are $\frac{n}{2} - 1$ remaining even integers that can ... | {
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Finding $\lim_{n\to\infty} \left(\frac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}$
Find the following limit without using the L'Hopital rule:
$$\lim_{n\to\infty} \left(\dfrac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}$$
Answer: $e^{-1}$
My attempt: Since the limit is of the form $1^{\infty}$, I decided to use the... | $ \lim_{n \to \infty} l = -1 $, not $ 0 $. Indeed,
\begin{align}
2n + 3 - \sqrt{1 + \frac{1}{n}}(2n + 3) + \frac{1}{n}
&= \left(1 - \sqrt{1 + \frac{1}{n}}\right)(2n + 3) + \frac{1}{n} \\
&= \frac{(1 - \sqrt{1 + 1/n})(1 + \sqrt{1 + 1/n})}{1 + \sqrt{1 + 1/n}} \cdot (2n + 3) + \frac{1}{n} \\
&= \frac{-1/n}{1 + \sqrt{1 + 1... | {
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"source": "stackexchange",
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Prove that a power series satifies that $x^2y''-4xy'+(x^2+6)y=0$ Consider the power series
$$
\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}z^{2n+3}
$$
I have shown that f is differentiable and that
$$
f'(x) = \sum_{n=0}^\infty \frac{(-1)^n(2n+3)}{(2n+1)!}x^{2n+2}
$$
and
$$
f''(x) = \sum_{n=0}^\infty \frac{(-1)^n(2n+3)(2n+2)... | It is plain that the power series is convergent for all $z$ because the factorial denominator grows much faster than any power of $z$. Convergent power series may be differentiated term by term.
It is then a question - as you have started - of working it through.
Let $S = z^2 f''(z) - 4zf'(z) + (z^2 + 6) f(z)$, so th... | {
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If $(1 + \operatorname{cis}x)(1 + \operatorname{cis} 2x) = a + bi$, prove that $a^2 + b^2 = (4\cos x \cos\frac x2)^2$ Could anyone help me with this? I'm stuck.
If $$(1 + \operatorname{cis}x)(1 + \operatorname{cis} 2x) = a + bi$$ prove that $$a^2 + b^2 = \left(4\cos x \cos\frac x2\right)^2$$
For reference, $\operator... | Note
\begin{align}
a^2+b^2 &= |a + bi|^2 \\
&= |(1 + \operatorname{cis}x)(1 + \operatorname{cis} 2x) |^2\\
&= |1 + e^{i x}|^2|1 + e^{i 2x}|^2\\
&= |e^{-\frac x2}+ e^{-\frac x2}|^2 |e^{-i x}+ e^{ ix}|^2\\
&= |2\cos\frac x2|^2 |2\cos x|^2\\
&= (4\cos x \cos\frac x2)^2
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof of the binomial theorem by induction clarification This is an exercise from Spivak's "Calculus".
3.d. Prove the "binomial theorem": If $a$ and $b$ are any numbers and $n$ is a natural number, then
\begin{align} (a+b)^n &= a^n + {n \choose 1}a^{n-1}b + {n \choose 2}a^{n-2}b^2 + \dots + {n \choose n-1}ab^{n-1} + b... | It is Pascal's rule:
$${n\choose j}+{n\choose j-1}={n+1\choose j}$$
So we have
\begin{align}
(a+b)^{n+1}&=(a+b)(a+b)^n\nonumber \\
&=(a+b)\sum_{j=0}^{n}{n\choose j}a^{n-j}b^j\nonumber \\
&=\sum_{j=0}^{n}{n\choose j}a^{n-j+1}b^j +\sum_{j=0}^{n}{n\choose j}a^{n-j}b^{j+1}\nonumber \\
&={n\choose 0}a^{n+1}+\sum_{j=1}^n{n\c... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluation of $\int \:\frac{1}{\sqrt[3]{\left(1+x\right)^2\left(1-x\right)^4}}dx$ using trig substitution Recently I came accross this integral :
$$ \int \:\frac{1}{\sqrt[3]{\left(1+x\right)^2\left(1-x\right)^4}}dx $$
I would evaluate it like this, first start by the substitution:
$$ x=\cos(2u) $$
$$ dx=-2\sin(2u)du $$... | Your result is correct, which can be obtained alternatively by substituting $t= \frac{1+x}{1-x}$ to arrive at
$$ \int \:\frac{dx}{\sqrt[3]{\left(1+x\right)^2\left(1-x\right)^4}}
=\frac12\int t^{-2/3}dt = \frac32 t^{1/3}+C=\frac32 \sqrt[3]\frac{1+x}{1-x} +C$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Technique for simplifying, e.g. $\sqrt{ 8 - 4\sqrt{3}}$ to $\sqrt{6} - \sqrt{2}$ How to find the square root of an irrational expression, to simplify that root. e.g.:
$$
\sqrt{ 8 - 4\sqrt{3} } = \sqrt{6} - \sqrt{2}
$$
Easy to verify:
\begin{align}
(\sqrt{6} - \sqrt{2})^2 = 6 - 2\sqrt{12} +2
= 8 - 4 \sqrt{3}
\end{alig... | I'm assuming you're looking for a formula to de-nest $\sqrt{a+b\sqrt{c}}$.
To derive such a formula, equate this to $x+y\sqrt{c}$.
$a+b\sqrt{c}=(x+y\sqrt{c})^2=x^2+cy^2+2xy\sqrt{c}$
$a+b\sqrt{c}=x^2+cy^2+2xy\sqrt{c}$
On matching up coefficients, we get:
$
\begin{cases}
x^2+cy^2=a\\
2xy=b
\end{cases}
$
Substitute $y=\fr... | {
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"source": "stackexchange",
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Why does Stolz- Cesaro fail to evaluate the limit of $\dfrac{n + n^2 + n^3 + n^4 + \ldots + n^n}{1^n + 2^n + 3^n + 4^n + \ldots +n^n}$, I need to find the limit of the sequence
$\dfrac{n + n^2 + n^3 + n^4 + \ldots + n^n}{1^n + 2^n + 3^n + 4^n + \ldots +n^n}$,
My strategy is to use Stolz's Cesaro theorem for this seque... | If you approximate the sums with the corresponding integrals it converges to 0
$$
L = \frac{\int_{1}^{n}n^x dx}{\int_{1}^{n} x^n dx} = \frac{n^{n+1} + n^n }{n^{n+1}\log n - \log n} = 0
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For angles $A$ and $B$ in a triangle, is $\cos\frac B2-\cos \frac A2=\cos B-\cos A$ enough to conclude that $A=B$? Brief enquiry:
$$\cos\frac B2-\cos \frac A2=\cos B-\cos A$$
Optionally
$$\sqrt\frac{1+\cos B}{2}-\cos B=\sqrt\frac{1+\cos A}{2}-\cos A$$
Is above equality sufficient to prove that it implies $A=B$?
Deta... | After more thinking, I realized that the answer is in fact "Yes".
As before, let $a=\cos\frac{A}{2},b=\cos\frac{B}{2}, 0<a,b<1$
The identity becomes:
$2a^2-a=2b^2-b$
which can be rewritten as:
$(a-b)\left(a+b-\frac{1}{2}\right)=0$
So either $a=b$, which gives $A=B$, as $\cos$ is $1:1$ on $\left[0,\frac{\pi}{2}\right]$
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Parabola transformation Find the real affine change of coordinates that maps the parabola in the $xy$-plane to the parabola in the $uv$-plane
$$4x^2 + 4xy + y^2 - y + 1 = 0$$
$$4u^2 + v = 0$$
My attempt:
Since there is an $xy$ term, we know that there is a rotation. Thus suppose there is a $x'y'$ coordinate system befo... | To obtain affine change of coordinates simply observe that
$4x^2+4xy+y^2-y+1=(2x+y)^2+(1-y)=4\left(x+\frac{y}{2}\right)^2+(1-y)$. So take \begin{align*}u&=x+\frac{y}{2}\\v&=1-y\end{align*}
This gives us
$$\begin{bmatrix}u\\v\end{bmatrix}=\begin{bmatrix}1&\frac{1}{2}\\0&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}+\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3700024",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Inequality involving AM-GM but its wierd Let a, b, c be positive real numbers. Prove that
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+ \frac{3\cdot\sqrt[3]{abc}}{a+b+c} \geq 4$
Ohk now i know using AM-GM that
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq 3 \cdot \sqrt[3] {\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a}} = 3 \c... | From
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant \frac{(a+b+c)^2}{ab+bc+ca},$$
and $ \sqrt[3]{abc} \geqslant \frac{3abc}{ab+bc+ca},$ we need to prove
$$\frac{(a+b+c)^2}{ab+bc+ca} + \frac{9abc}{(ab+bc+ca)(a+b+c)} \geqslant 4,$$
equivalent to
$$ (a+b+c)^2 + \frac{9abc}{a+b+c} \geqslant 4(ab+bc+ca),$$
or
$$a^2+b^2+c^2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What's the sum function of $\sum_{n=1}^{\infty} \frac{x^n(x-1)}{n(x^{2n+1}-1)}$? Notice that
\begin{align*}
\frac{x^n(x-1)}{n(x^{2n+1}-1)}&=\frac{1}{n}\cdot\frac{x^n}{\sum_{k=0}^{2n}x^k}=\frac{1}{n\sum_{k=0}^{2n}x^{k-n}}=\frac{1}{n\sum_{k=-n}^{n}x^{k}}
\end{align*}
This will help?
| What about
\begin{align*}
(1-x)\sum_{n=1}^\infty \frac{1}{n} \frac{x^n}{1-x^{2n+1}} = (1-x) \sum_{m=0}^\infty \sum_{n=1}^\infty \frac{1}{n} x^{n + (2n +1)m} = (x-1) \sum_{m=0}^\infty x^m \ln(1-x^{2m+1}),
\end{align*}
where we used the series representation of the logarithm.
But I am not sure, if it is possible to give ... | {
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Prove $(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$ For $a,b,c>0; abc=1.$ Prove$:$ $$(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$$
My proof by SOS is ugly and hard if without computer$:$
$$\left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{3}-9\,abc \left( {a}^{3}+{b} ^{3}+{c}^{3} \right)$$
$$=\frac{1}{8}\, \left( b-c \right) ^{6}+{\frac {117... | By applying Vacs's ineq we obtain $$(a^2+b^2+c^2)^3\ge 3(ab^3+bc^3+ca^3)(a^2+b^2+c^2)$$
So it's suffice to prove $$(ab^3+bc^3+ca^3)(a^2+b^2+c^2)\ge 3abc(a^3+b^3+c^3)$$
$$\Leftrightarrow \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge \frac{3(a^3+b^3+c^3)}{a^2+b^2+c^2}$$
$$\Leftrightarrow (a-b)^2\left(\frac{a^2+c^2}{b}-a\r... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Limitations of Arithmetic-Geometric Inequality? I have to find the range of function $f(x) = x+(1/x) +1$, where $x$ is positive. Now I did it with two ways which we can see below, in equations $(1)$ and $(2)$, by using the AM-GM inequality.
$$\frac 1 3 \left( x + \frac 1 x + 1 \right) \ge \sqrt[3]{x \cdot \frac 1 x \c... | look, this isn't a "limitation" of the AM-GM inequality. for example:
$$\frac{1+2}{2} \leq \sqrt{2}$$
and
$$\frac{1+1+1}{3} \leq \sqrt[3]{1}$$
as you can see in the first example the geometric mean isn't certain to be the biggest the amount could be, in your example, the minimum is the first one, but the second one sti... | {
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Tried to apply the ratio test to determine the convergence interval, but get the limit as a constant. So here is the series: $\Sigma_{n=1}^{\infty} \frac{x^{2n}}{1+x^{4n}}$
$$\left| \frac{x^{2(n+1)} (1+x^{4n})}{x^{2n}(1+x^{4(n+1)})}\right| = \left|\frac{x(1+x^{4n})}{1+x^{4n+4}} \right| \overset{\text{ n } \rightarrow \... | Let's proceed from your work. Corrections are highlighted in red. We have
$$h(x) = \left| \frac{x^{2(n+1)} (1+x^{4n})}{x^{2n}(1+x^{4(n+1)})}\right| = \left|\frac{x^{\color{red}{2}}(1+x^{4n})}{1+x^{4n+4}} \right|.$$ At this point, we need to consider separate cases. The reason is becuase for $n \ge 1$, the behavior ... | {
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$p^2+1=q^2+r^2$. Strange phenomenon of primes Problem:
Find prime solutions to the equation
$p^2+1=q^2+r^2$
I welcome you to post your own solutions as well
I have found a strange solution which I can't understand why it works(or what's the math behind it.) Here it is through examples
Put $r=17$(prime)
Now $17^2-1=16\t... | Let $r=193$, a prime. Then $$r^2-1=9313^2-9311^2=3107^2-3101^2=323^2-259^2$$ are the only expressions of $r^2-1$ as a difference of two squares, but $9313=67\times139$, and $3101$ and $259$ are both multiples of $7$. Thus, $p^2+1=q^2+r^2$ is impossible in primes $p,q$ for this prime value of $r$.
| {
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Find solution set of $\dfrac{8^x+27^x}{12^x+18^x}=\dfrac{14}{12}$ What I've done is factoring it.
$$\dfrac{2^{3x}+3^{3x}}{2^{2x}\cdot 3^{x}+3^{2x}\cdot{2^{x}}}=\dfrac{7}{2\cdot 3}$$
This looks like it can be factored more but it doesn't work from my attempts.
| Well you've reached a good point:
$$\dfrac{2^{3x}+3^{3x}}{2^{2x}\cdot 3^{x}+3^{2x}\cdot{2^{x}}}$$
$$=\dfrac{2^{3x}+3^{3x}}{2^{x}\cdot 3^{x}(2^x+3^{x})}$$
We can let $2^x=a$ and $3^x=b$ and get
$$\frac{a^3+b^3}{ab(a+b)}=\frac{a^2-ab+b^2}{ab}=\frac{a}{b}-1+\frac{b}{a}$$
Now let $z=\frac{a}{b}$, we get
$$z-1+\frac{1}{z}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Multiplying $P(x) = (x-1)(x-2) \dots (x-50)$ and $Q(x)=(x+1)(x+2) \cdots(x+50)$
Let $P(x) = (x-1)(x-2) \dots (x-50)$ and $Q(x)=(x+1)(x+2) \cdots(x+50).$
If $P(x)Q(x) = a_{100}x^{100} + a_{99}x^{99} + \dots + a_{1}x^{1} + a_0$, compute $a_{100} - a_{99} - a_{98} - a_{97}.$
I've been quite stuck with this one. If I mul... | Clearly $a_{99}=a_{97}=0$, since only even degree terms appear. As you've noticed, $a_{100}=1$; so really we just need to find $a_{98}$. But this is just negative $1$ times the sum of squares up to $50$, i.e. $-\sum_{k=1}^{50} k^2 =- \frac{50\cdot 51\cdot 101}{6}=-42925$. So the answer is $42926$.
Edit: it seems I was ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the number of functions which satisfy given conditions If f :{1,2,3,4} → {1,2,3,4} and y = f(x) be a function defined such that |f(α) – α| ≤ 1, for α ∈ {1,2,3,4} and m be the number of such functions, then m/5 is equal
| Okay.
$|f(a) - a| \le 1$ so $-1 \le f(a)-a < 1$ so $a-1 \le f(a) \le a+1$.
so $f(a)4$ may be one of three values $f(a)$ could be $a-1$ if $a-1$ is in range; that is if $a-1\ge 1$ or $a \ge 2$.
Or $f(a)$ could be $a$.
Or $f(a)$ could be $a+1$ if $a+1$ is in range; that is if $a+1 \le 4$ of $a \le 3$.
So
For $a = 1$ ther... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\int_0^\infty \frac{1}{1+x^4}dx$ using the Residue Theorem I'm trying to evaluate the integral
$$\int_0^\infty \frac{1}{1+x^4}dx $$
using the Residue Theorem.
My approach:
Let's consider
$$\oint_\Gamma f$$
with $f(z)=\frac{1}{1+z^4}$ and $\Gamma = \Gamma_1 + \Gamma_2$, where:
*
*$\Gamma_1:[-R,R]\rightarrow \mathbb{... | The poles lying inside the contour are $\alpha_1=i\pi/4, \alpha_2=i3\pi/4$.
$Res[f(z),\alpha_1]=\frac{1}{4z^3}\rvert_{\alpha_1}=\frac{1}{4}e^{-3i\pi/4}$
$Res[f(z),\alpha_2]=\frac{1}{4z^3}\rvert_{\alpha_2}=\frac{1}{4}e^{-9i\pi/4}$
$ \text{After using the estimation principle to squeeze the integral on upper part to zero... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to Prove $\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$ Question:- Prove that
$\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$
On factoring the denominator we get,
$\int_{0}^{\infty}\frac {1}{(x^4+x^2+1)(x^4-x^2+1)}dx$
Partial fraction of the integrand contains big terms with their long... | Let $(a,b,c,d)$ to be the complex roots of $x^8+x^4+1=0$. So, after partial fraction decomposition, the integrand write
$$\frac{1}{(a-b) (a-c) (a-d) \left(x^2-a\right)}+\frac{1}{(b-a) (b-c) (b-d) \left(x^2-b\right)}+$$
$$\frac{1}{(c-a) (c-b) (c-d) \left(x^2-c\right)}+\frac{1}{(d-a) (d-b) (d-c) \left(x^2-d\right)}$$
Thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3728939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluating the indefinite Harmonic number integral $\int \frac{1-t^n}{1-t} dt$ It is well-known that we can represent a Harmonic number as the following integral:
$$H_n = \int_0^1 \frac{1-t^n}{1-t} dt$$
The derivation of this integral doesn't need you to derive the indefinite integral first, so now I'm wondering what ... | The integrand is (in the complex field) a meromorphic function, with one pole coinciding with one of the zeros, and therefore removable,
leaving a polynomial defined over the whole complex field. So it is its integral.
A) polynomial form
Let's define
$$
\eqalign{
& I_{\,n} (x,a) = \int_{t\, = \,a}^{\;x} {{{1 - t^{\,n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3729781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Integrate $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ How can I evaluate this integral $$\int\dfrac{2\cos x-\sin x}{3\sin x+5\cos x } \mathop{dx}$$
I tried using the half angle formula
$$\sin x=\dfrac{2\tan\dfrac x2}{1+\tan^2\dfrac x2}, \cos x=\dfrac{1-\tan^2\dfrac x2}{1+\tan^2\dfrac x2}$$
substituted and simplif... | Rewrite numerator: $2\cos x-\sin x=\frac{7}{34}(3\sin x+5\cos x)+\frac{11}{34}\frac{d}{dx}(3\sin x+5\cos x)$
$$\int\dfrac{2\cos x-\sin x}{3\sin x+5\cos x }$$
$$=\int\dfrac{\frac{7}{34}\left(3\sin x+5\cos x\right)+\frac{11}{34}\left(3\cos x-5\sin x\right)}{3\sin x+5\cos x }dx$$
$$=\frac{7}{34}\int\ dx+\frac{11}{34}\int ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Solve for $y$ in $\frac{dy}{dx}-\frac{3y}{2x+1}=3x^2$ I saw a challenge problem on social media by a friend, solve for $y$ in $$\frac{dy}{dx}-\frac{3y}{2x+1}=3x^2$$
I think this is an integration factor ODE
$$\frac{1}{{(2x+1)}^{\frac{3}{2}}} \cdot \frac{dy}{dx}-\frac{3y}{{(2x+1)}^{\frac{5}{2}}}=\frac{3x^2}{{(2x+1)}^{\f... | This ODE is linear. Consider the homogeneous
$$
y_h' -3\frac{y_h}{2x+1}=0
$$
This ODE is separable with solution
$$
y_h = C_0(2x+1)^{\frac 32}
$$
now assuming for the particular $y_p = C_0(x)(2x+1)^{\frac 32}$ after substitution in the complete ODE we obtain
$$
C_0'(x) = \frac{3x^2}{(2x+1)^{\frac 32}}
$$
thus
$$
C_0(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Letting $k= n+5$ we get that $1^3+2^3+3^3 +\dots+ (k-5)^3 \equiv 1... | Let $$S:=\sum_{k=1}^{n} k^{3}=\left[\frac{n}{2}(n+1)\right]^{2}$$
Then $$\begin{aligned} 4S &=[n(n+1)]^{2} \equiv[-5(-5+1)]^{2}\equiv 400 \quad (\operatorname{mod} n+5) \end{aligned}$$
Given that $S \equiv 17 \bmod (n+5)$, we have
$$
\begin{array}{l}\quad 4(17) \equiv 400 \quad (\bmod n+5) \\
\Rightarrow n+5 \mid 400... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3737447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Limit of a fraction involving square and third roots. $$\lim_{n\to\infty} \frac{\sqrt{n^2+1} - \sqrt{n^2+n}}{\sqrt[3]{n^3+1} - \sqrt[3]{n^3+n^2+1}}$$
Is it a good idea to substitute the numerator and denominator using that $a-b=\frac{a^2-b^2}{a+b}$ and $a-b=\frac{a^3-b^3}{a^2+ab+b^2}$, since I’ll end up having to multi... | Yes, you can make that approach work. It's a little tedious and messy looking, but here's what you get:
$${\sqrt{n^2+1}-\sqrt{n^2+n}\over\sqrt[3]{n^3+1}-\sqrt[3]{n^3+n^2+1}}\\={(n^2+1)-(n^2+n)\over(n^3+1)-(n^3+n^2+1)}\cdot{(\sqrt[3]{n^3+1})^2+\sqrt[3]{n^3+1}\sqrt[3]{n^3+n^2+1}+(\sqrt[3]{n^3+n^2+1})^2\over\sqrt{n^2+1}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.