Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove that normals for surface of revolution $z=f(\sqrt{x^2+y^2})$ cross axis of revolution $z=f(\sqrt{x^2+y^2})$, $f'\not = 0$
My attempt: considering we have $F(x,y,z)=f(\sqrt{x^2+y^2}) - z$
And using normal equation: $$\frac{x-x_0}{\frac{\partial F}{\partial x}}=\frac{y-y_0}{\frac{\partial F}{\partial y}}=\frac{z-z_... | The normal is not correctly stated. There are several errors. Correcting them,
$u=\sqrt{x^2+y^2}$
$\dfrac{\partial F}{\partial x}=\dfrac{\partial}{\partial x}(f(\sqrt{x^2+y^2})-z)=\dfrac{\partial}{\partial x}f(\sqrt{x^2+y^2})=\dfrac{df(u)}{du}\dfrac{\partial u}{\partial x}=\dfrac{x}{\sqrt{x^2+y^2}}\cdot f'(u)$
$\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that the number $a_{n}=\sqrt{2+\sqrt{2+\sqrt{2+...}}}$, is irrational for every positive integer n. Define $a_{n}=\sqrt{2+\sqrt{2+\sqrt{2+...}}},\quad n=1,2,...$ where the radical contains $n$ twos.
Provide a recursive definition of $a_n$ and then show that the number $a_n$ is irrational for every positive integer... | Using induction we suppose that $a_n$ is irrational. Suppose now that $a_{n+1}$ is rational. Then $a^2_{n+1}=2+a_n$, therefore $a_n$ is rational, contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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An object travelling on $x^2$ I have an object traveling along the curve $y=x^2$. $z$ is the distance from the origin and $dz/{dt}=1$ is the rate it's increasing per unit time. At what rate are my $x$ and $y$'s moving at the point (2,4)? In other words what is $dy/dt$ and $dx/dt$.
I've seen the geometry. Found $z$ as t... | Bu definition,
$ z = \sqrt{\left(x-0\right)^2 +\left(y-0\right)^2 } = \sqrt{x^2 + y^2}$.
First, let us compute $dx/dt:$
Since $y = x^2$ we can rewrite $z$ in the form $ z = z(x,y) = \sqrt{x^2 + x^4}$.
Then by chain rule we have
$$
1 = \frac{d z}{d t} = \frac{d z}{d x} \,\frac{d x}{d t} \implies
\frac{d x}{d t} = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2250641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find a 3 x 3 orthogonal matrix Find a 3x3 orthogonal matrix whose first column is $[\frac1{\sqrt{3}},{- \frac 1{\sqrt{3}}}, \frac1{\sqrt{3}}]$.
I was thinking this has to do with: ${Q}^T•Q=I_{n}$, but I still don't understand how to go about this.
| You need to find an orthonormal basis of $\mathbb{R}^3$ whose first vector is the vector $v_1 = \left( \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)^T$ given to you. This can be done in several ways:
*
*Complete $v_1$ arbitrary to a basis $v_1,v_2,v_3$ of $\mathbb{R}^3$ and perform Gram-Schmidt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $|f'(x)-\frac{\overline{f(x+h)}-\overline{f(x-h)}}{2h}|\le \frac{M_1h^2}{6}+\frac{\epsilon M_2}{h}$ Suppose $f$ is a $C^3$ function on $[x-2h,x+2h]$. Suppose that the computed values $\overline{f(x+h)}$ and $\overline{f(x-h)}$ satisfy $\overline{f(x+h)}=f(x+h)(1+e_1)$ and $\overline{f(x-h)}=f(x-h)(1+e_2)$, wh... | \begin{align*}
\left|f'(x) - \frac{\hat{f}(x+h) - \hat{f}(x-h)}{2h} \right|
&=
\left|f'(x) - \frac{f(x+h)(1+e_1) - f(x-h)(1+e_2)}{2h} \right| \\
&= \left|f'(x) - \frac{f(x+h) - f(x-h)}{2h} -\frac{f(x+h)e_1 - f(x-h)e_2}{2h} \right| \\
&\le
\left|f'(x) - \frac{f(x+h) - f(x-h)}{2h}\right| + \left|\frac{f(x+h)e_1 - f(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
$n^{th}$ derivative of $y=(x \sqrt{1+x^2})^m$ I was trying to find the $n^{th}$ derivative and $y_n(0)$(nth derivative at zero) of the function
$$y=(x \sqrt{1+x^2})^m$$
Here is my approach:
$$y^2=(x^2 (1+x^2)^m$$
$$2yy_1=m(x^2 (1+x^2))^{m-1} \cdot(2x+4x^3)$$
$$2yy_1=m\frac{(x^2 (1+x^2))^{m}}{(x^2 (1+x^2)} \cdot(2x+4x... | You can use the series expansion:
$$
(1+x)^\alpha=\sum_{k=0}^\infty\pmatrix{\alpha \\ k}x^k
$$
where
$$
\pmatrix{\alpha \\ k}=\frac{\alpha\cdot(\alpha-1)\cdot\ldots\cdot(\alpha-k+2)\cdot(\alpha-k+1)}{k\cdot(k-1)\cdot\ldots\cdot2\cdot1}.
$$
Then, if $m$ is a positive integer,
$$
\left(x\sqrt{1+x^2}\right)^m=x^m\sum_{k=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2257800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{5}$ and $abc = 5$, solve for $a^3 + b^3 + c^3$ I recently encountered this question and have been stuck for a while. Any help would be appreciated!
Q: Given that
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{5} \tag{1} \label{eq:1}$$
$$abc = 5 \tag{2} \la... | As DanielV suspects, the question is indeed ill-posed: there are multiple possible outcomes for $a^3 + b^3 + c^3$. First of all notice that if $a$, $b$, and $c$ are positive, then in order for (1) to hold, each of them should be larger than 5. But this results in a contradiction for (2). It follows that at least one sh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Determine a constant that satisifies an inequality Determine the minimum real number $c$ such that $$\sum_ {k=1}^n \frac {a_k}{c^{a_1+a_2+ \dots +a_k}} \le 1,$$ for every positive integer $n$ and positive real numbers $a_1,a_2, \dots, a_n$.
| Let $x>1$ and $f(x)$ be the smallest real number $y$ satisfying that, for any integer $n$ and $n$ numbers $a_1, a_2, \cdots, a_n$, $\sum_ {k=1}^n \frac {a_k}{x^{a_1+a_2+ \dots +a_k}} \le y$. Let's find a closed form expression for $f(x)$.
First of all, $f(x) \ge \lim_{\epsilon \rightarrow 0}\frac{\epsilon}{x^\epsilon-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2259428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Verifying the identity $\frac{1}{\cos x} - \cos x = \sin x \tan x$ $\frac{1}{\cos x} - \cos x = \sin x * \tan x$
I have tried a few things and nothing works
Left Side
$\frac{1}{\cos x} - \cos x$
$1 - (\cos x)^2$
$1 - 1 + \frac{\cos 2x}{2}$
$ \frac{\cos 2x}{2}$
$ \frac{\cos x}{2} \times \frac{\cos}{2}$
... And I am ou... | $$\frac{1}{\cos x} - \cos x = \frac{1- \cos^2 x}{\cos x}$$
Now I think you can finish.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does $\int _0^1 \cfrac{x}{x+\cfrac{x^{2}}{x^2+\cfrac{x^3}{x^3+\cdots}}} \, dx$ exist? $$\int_0^1 \cfrac x {x+\cfrac{x^2}{x^2+\cfrac{x^3}{x^3+\cdots}}} \, dx$$
Does this integral exist?
Any hint will be appreciated.
| $\int _{0}^{1} \frac{x}{x＋\frac{x^{2}}{x^2＋\frac{x^{3}}{x^{3}＋…}}}dx$
Let's call that monstrous integrand $u$. If you notice, $u=\frac{x}{x+xu}=\frac{1}{1+u}$.
$u+u^2=1$, so $u^2+u-1=0$. This means $u=\frac{\sqrt{5}-1}{2}$.
$\int_0^1 \frac{\sqrt{5}-1}{2} \,dx = \boxed{\frac{\sqrt{5}-1}{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2259965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I solve $\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx$ How can I solve the following integral?
$$\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx$$
$$\begin{align}
\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx&=\int \frac{1}{\sqrt{\left(x+\frac72\right)^2-\frac{49}{4}} + 3}\, dx
\end{align}$$
$$u = x+\frac{7}{2}, \quad a = \frac{... | Start from your first result and multiply numerator and denominator with $\sqrt{u^2-a^2}-3$:
$$\int\frac{1\times \color{red}{\sqrt{u^2-a^2}-3}}{\sqrt{u^2-a^2}+3 \times \color{red}{\sqrt{u^2-a^2}-3}}du=\int\frac{\sqrt{u^2-a^2}-3}{u^2-a^2-9}du$$
Now we are going to split this integration into two parts. Each part will be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Angles between lateral faces of any rectangular-based pyramid I was just wondering if anyone had any idea how to solve this problem:
What is the angle between lateral faces of a rectangular-based pyramid with length a, width b, and height h, in terms of a, b and h?
Any responses are much appreciated.
| @Aretino's coordinate/vector approach might be the most accessible, but it's worth mentioning a couple of others.
Given any three-faced "corner", one can determine the dihedral angles from the face angles (or vice-versa) using an appropriate Spherical Law of Cosines. For the problem at hand, with this configuration ..... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving simultaneous equations with complex numbers and moduli Would anyone be able to give me a starting point as to how to approach this? A friend suggested squaring both sides, but I read something that said you can't square both sides of an equation(?).
Complex numbers and simultaneous equations question
If $x$ and... | From the first equation,
$$
\color{blue}{\lvert x + i y \rvert = \sqrt{x^{2}+y^{2}} = 1} \qquad \Rightarrow \qquad \color{blue}{y_{1}=\pm \sqrt{1-x^2}}
$$
From the second equation,
$$
\color{red}{\lvert x -\frac{3}{2} + i y \rvert =\sqrt{\left(x-\frac{3}{2}\right)^2+y^2}}
\qquad \Rightarrow \qquad
\color{red}{y_{2}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\sin(3x)$ is equivalent to $3\sin(x)\cos^2(x)-\sin^3(x)$
The expression $\sin(3x)$ is equivalent to:
A. ...
My book states the right answer is B which is $3\sin(x)\cos^2(x)-\sin^3(x)$.
I tried:
$$\sin(x)\cos(2x)+\cos(x)\sin(2x) = \\
\sin(x)(2\cos^2(x)-1)+\cos(x)\cdot2\sin(x)\cos(x) = \\
2\cdot\sin(x)\cos(x... | Notice that \begin{equation}\begin{split}\sin(3x) &= 4\cos^2 x \sin x - \sin x\\&= 3\sin x \cos^2 x +(\sin x \cos^2x -\sin x )\\&=3\sin x \cos^2 x + \sin x\cdot( \cos^2x - 1)\\&= 3\sin x \cos^2 x -\sin^3 x \end{split}\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there a straight forward way to prove that $\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{31} > 3$ Using brute force, it is straight forward to calculate that $\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{31} > 3$
Is there a more elegant way to demonstrate this?
What if I want to find the smallest $n$ such that $\fr... | Claim: $$\sum_{i=1}^{3^n}\frac{1}{i} \geq \frac{7}{6}+\frac{2}{3}n$$
whenever $n\geq2$.
Proof: By Mathematical Induction.
Take $n=3$, then instantly $1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{27}\geq \frac{7}{6}+2 > 3$.
Hope this is not a brutal way and this can be easily generalized.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find cubic function whose graph has horizontal tangents at two points Find the cubic function $y = ax^3 + bx^2 + cx + d$ whose graph has horizontal tangents at the points $(-2,7)$ and $(2,1)$.
I find the derivative and set equation to zero, but then that only gives me one solution. I can't find all solutions for some r... | First, find the derivative of $y$ with respect to $x$:
$$y'=3ax^2+2bx+c$$
And because of the tangent points given, we know that
$$3a(-2)^2+2b(-2)+c=0$$
$$3a(2)^2+2b(2)+c=0$$
And since the graph of the cubic must pass through those points,
$$a(-2)^3+b(-2)^2+c(-2)+d=7$$
$$a(2)^3+b(2)^2+c(2)+d=1$$
When we simplify these e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2266155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Composite Function value Given: $f(f(x)).[1+f(x)] = -f(x)$. Find $f(3)$
Approach to the question:-
$fof(x) + f(x).fof(x) = -f(x)$
$fof(x) + f(x) + f(x).fof(x)+1=1$
$(1+f(x))(1+f(f(x))=1$
This implies that $1+f(x)$ and $1+fof(x)$ are reciprocals.
This gives $f(x)=\frac{1}{x}-1$ or $ f(x)=\frac{-1}{x}-1$ as solutions. (B... | Well $1+f(x)$ and $1+f(f(x))$ are reciprocals of each other and not of $x$ which means $$1+f(x)=\frac1{1+f(f(x))}\\1+f(f(x))=\frac{1}{1+f(x)}$$
Consider this instead since $f(x)=-1$ doesn't satisfy the above condition we can divide by $1+f(x)$ to get
$$f(f(x))=\frac{-f(x)}{1+f(x)}$$
Now putting $f(x)=t$ (this step requ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2267205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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$\cos\gamma+\sin\gamma$ in a 3-5-7 triangle ($\gamma$ is the greatest angle) Clearly $\gamma$ has to be the angle oposite to the side of lenght 7. The law of cosines gives me $$7^2 = 5^2+3^2 - 2\cdot 5\cdot3\cos{\gamma}\Longleftrightarrow\cos{\gamma}=-\frac{7^2-5^2-3^2}{2\cdot5\cdot 3}=-\frac{1}{2}.$$
This value of $\c... | For $0^\circ<\gamma<180^\circ$ – the range of an angle in a non-degenerate triangle – $\sin\gamma$ must be positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2270228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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System of two symmetric equations I'm trying to solve the following system of equations over positive reals, but I've got stuck.
$$
\left\{
\begin{array}{c}
ab + bc + ca = 12 \\
a + b + c + 2 = abc \\
\end{array}
\right.
$$
What I've tried: I've used Vieta's formulas to convert this system of equations to $x^3 + (a... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the our conditions give $v^2=4$ and $w^3=3u+2$.
In another hand,
$$(a-b)^2(a-c)^2(b-c)^2=27(2u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)\geq0,$$
which gives
$$(u-2)^2(2u+5)(6u+13)\leq0$$ or $u=2$, which gives $w^3=8$ and $a$, $b$ and $c$ are roots of the equation
$$x^3-6x^2+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2271917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Problem with square number I have a doubt with this problem :
"Show that there exist an infinite number of integers $n$ such that $$P(n)=\frac{\sum_{k=0}^{n}k^{2}}{n}=\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}$$ is square number."
I find that $P(n)$ is integer if $n$ is prime number.
Sincerely,
| The formula for the sum of the first $n$ square numbers is given by
$$1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$
Meaning that
$$P(n)=\frac{\frac{n(n+1)(2n+1)}{6}}{n}$$
$$P(n)=\frac{(n+1)(2n+1)}{6}$$
So now you just need to find out when
$$\frac{(n+1)(2n+1)}{6}$$
is a perfect square. Now let $a$ and $b$ be some two numbe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show monotonicity Originally, I want to show that
$$
\frac{\sqrt{a \cdot b + \frac{b}{a}x^2}\arctan \left(\frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}\right)}{\sqrt{a \cdot b}\arctan \left(\frac{c}{\sqrt{a \cdot b}}\right)} \geq 1 \ \ \text{for} \ \ x, a,b,c > 0 \ .
$$
To do so, I figured it is sufficient to show tha... | I'll first rearrange and reparametrise the original equality a little to make it easier to work with. Let $\alpha := c/\sqrt{ab}$, $z:= \sqrt{ 1 + (x/a)^2}$. Since $a,b,c >0$, it should be easy to see that the original inequality is equivalent to $$ z \arctan \frac{\alpha}{z} \overset{?}\ge \arctan \alpha$$ for every $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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About $142857$: proof that $\;3 \mid 1^n + 4^n + 2^n + 8^n + 5^n + 7^n$ Most of you already know the flabbergasting properties of cyclic numbers. Among these, $142857$ deserves to be the most famous and admired, being (in base $10$) the smallest and the only one without those spoilsport leading zeros. I personally love... | I'll look at each addend one at a time.
$1^n$:
$$
1^n \equiv 1 \;\pmod 3
$$
$4^n$:
$$\begin{align}
4 &\equiv 1 &\pmod 3\\
4 \cdot 4 &\equiv 4 \equiv 1 &\pmod 3\\
&\;\;\vdots\\
4^n &\equiv 1 &\pmod3
\end{align}$$
$2^n$:
$$\begin{align}
2 &\equiv 2 &\pmod 3\\
2 \cdot 2&\equiv 4 \equiv 1 &\pmod 3\\
2 \cdot 2 \cdot 2 &\equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Multiple of $5$ If two natural nos. $x,y$ are selected at random then the probability that $x^2+y^2$ is a multiple of 5?
I think they should give a finite set for selecting natural nos. otherwise there are infinite ordered pairs of $x$ and $y$.
Is the question incomplete or am I missing some trick?
| Modulo $5$, the residues of $x^2$ take on the values $0, 1, 4, 4, 1$.
So $x^2 + y^2 \equiv 0 \bmod 5$ will be true if both residues are $0$ or if one is $1$ and the other is $4$.
This implies the probability is $\frac{1}{5} \cdot \frac{1}{5} + 2 \cdot \frac{2}{5} \cdot \frac{2}{5} = \frac{9}{25}$
Edit: Let's say we wan... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find tangent equation for a curve which is perpendicular to a line
How do I find the point on the curve $y=1/(2x−1)$ where the tangent line will be perpendicular to the line defined by $x−2y−1=0$?
What I have so far:
The equation of the given line is $y= 0.5x+1$
Since the slope is $-0.5$, the tangent slope must be $... | The line
$x - 2y - 1 = 0 \tag{1}$
may also be written
$y = \dfrac{1}{2}x - \dfrac{1}{2}; \tag{2}$
from this we see its slope is $\frac{1}{2}$. Thus the slope of a line normal or perpendicular to this line is $-2$.
For any value of $x$ other than $\frac{1}{2}$, the slope of the curve
$y(x)= \dfrac{1}{2x - 1} = (2x - 1)... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
} |
Riemann-Stieltjes sum Evaluate $$\lim_{n \to \infty} \sum_{k=1}^n \left[\left(\cos\left(\frac{2\pi(k+1)}n \right) +\cos\left(\frac{2\pi k}n\right) \right) \left(\sin\left(\frac{2\pi(k+1)}n \right) -\sin\left(\frac{2\pi k}n\right) \right) \right]$$
| First, by trigonometric identities involving sums and products, we have
$$\begin{align}
&\left(\cos\left(\frac{2\pi(k+1)}n \right) +\cos\left(\frac{2\pi k}n\right) \right) \left(\sin\left(\frac{2\pi(k+1)}n \right) -\sin\left(\frac{2\pi k}n\right) \right) \\ &=4\cos\left(\frac{2\pi(k+1/2)}n \right)\cos\left(\frac{2\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculate the sum of $\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}$ I would like to calculate the sum for $x\in(-1,1)$ of this: $$\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}$$
So far I managed that
$$\int \frac{x^n}{n}dx = \frac{x^{n+1}}{n(n+1)}, $$
and
$$\sum_{n=1}^\infty \frac{x^n}{n}=-\log(1-x), $$
and
$$ \int -\log(... | When $|x| < 1$:
$$\begin{align*}\sum_{n=1}^\infty \dfrac{x^{n+1}}{n(n+1)} &= \sum_{n=1}^\infty \dfrac{x^{n+1}}{n} - \dfrac{x^{n+1}}{n+1} \\ &= x\sum_{n=1}^\infty \dfrac{x^n}{n} - \sum_{n=1}^\infty \dfrac{x^{n+1}}{n+1} \\ &= x\sum_{n=1}^\infty \dfrac{x^n}{n} - \sum_{n=1}^\infty \dfrac{x^n}{n} + x \\ & = -x\ln(x-1) + \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2290534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
$(a^2+b^2)\cdot (c^2+d^2) = z^2+1$ then $z = ac+bd$ ? I have the following equation :
$$(a^2+b^2)\cdot (c^2+d^2) = z^2+1$$ with $a, b, c, d, z \in \mathbb{N}$
Then how can I prove that : $z = ac+bd$ ?
I tried Brahmagupta identity but it doesn't seem to work...
| The statement is false. Counter-example:
$$(9^2+7^2)(4^2+1^2) = 2210 = 47^2 + 1
\quad\text{ but }\quad
47 \ne
\begin{cases}
43 &= 9\cdot 4 + 7\cdot 1\\
37 &= 9\cdot 1 + 7\cdot 4
\end{cases}$$
Please note that $47$ is the smallest value of $z$ where $z^2+1$ contains three distinct prime factors of the form $4k+1$. Thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $a,b,c \geq 1$ then prove $\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} < \sqrt{abc+c}$ If $a,b,c \geq 1$, then prove:
$$\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} < \sqrt{abc+c}$$
My try:
$$A=\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} \\ A^2 =(a-1)+(b-1)+(c-1)+2\sqrt{(a-1)(b-1)}+2\sqrt{(a-1)(c-1)}+2\sqrt{(b-1)(c-1)} $$
$$B=\sqrt{abc... | Substitute $a=x^2+1$, $b=y^2+1$ and $c=z^2+1$ for $x,y,z\ge 0$ to get
$$x+y+z\le\sqrt{(z^2+1)\left(\left(x^2+1\right)\left(y^2+1\right)+1\right)}$$
Now after squaring and grouping we obtain
\begin{align*}x^2+y^2+z^2+2xy+2yz+2zx\le(z^2+1)\left((x^2+1)(y^2+1)+1\right) \\
0\le z^2(x^2+1)(y^2+1)-2z(x+y)+x^2y^2-2xy+2 \\
0\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that if $5^n$ begins with $1$, then $2^{n+1}$ also begins with $1$
Prove that if $5^n$ begins with $1$, then $2^{n+1}$ also begins with $1$ where $n$ is a positive integer.
In order for a positive integer $k$ to begin with a $1$, we need $10^m \leq k < 2 \cdot 10^m$ for some positive integer $m$. Thus since $5^... | Note that:
$$2\cdot10^n=5^n\cdot 2^{n+1}$$
If $5^n$ begins with $1$ then
$$10^m<5^n<2\cdot10^m\to10^m< \frac{2\cdot10^n}{2^{n+1}}<2\cdot10^m$$
what give us
$$2^{n+1}< 2\cdot 10^{n-m}\text{ and } 2^{n+1}>10^{n-m}\Leftrightarrow 10^{n-m}<2^{n+1}<2\cdot 10^{n-m}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to figure out either series are absolute or conditional convergence $\sum_1^\infty \cos(\frac{\pi n}{3})(n^{\frac{1}{\sqrt[6]{n+6}}}-1)$ $$\sum_1^\infty \cos\left( \frac{\pi n} 3 \right) \left(n^{\frac 1 {\sqrt[6]{n+6}}}-1\right)$$
I tried to use Dirichlet, but unsuccessfully. Please, give me hints. Thanks a lot.
| \begin{align}
& \Big( \cos \frac{1\pi} 3, \quad \cos\frac{2\pi} 3, \quad \cos\frac{3\pi} 3, \quad \cos\frac{4\pi} 3, \quad \cos \frac{5\pi} 3, \quad \cos\frac{6\pi} 3, \quad \ldots \Big) \\[10pt]
= \frac 1 2 & \Big( \underbrace{1,\ 1,\ 0,\ -1,\ -1,\ 0},\ \underbrace{1,\ 1,\ 0,\ -1,\ -1,\ 0},\ \ldots\ldots\ldots \Big)
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Computing $\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} $
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \space \text{ converges by the alternating series test, but what is its value?} $$
$1 - \sin(x) \space \text{ is an entire function} $, so by Weierstrass Factorization Theorem, it can be written as a product of its zeros, which... | We can write $1-\sin{x}=2\cos^2{(x/2+\pi/4)}$, and I think we can now see the problem: all the roots have to be double roots. They'll be at $(4n + 1)\pi/2$ and $-(4n+3)\pi/2$, as you suggest, but all the factors have to be squared:
$$ 1-\sin{x} = A \prod_{n=0}^{\infty} \left(1-\frac{2x}{(4n+1)\pi}\right)^2 \left(1+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
Calculate $\lim_{x\to a}(2-\frac{x}{a} )^{\tan( \frac{\pi x}{2a})}$
I would like to calculate
$$\lim\limits_{x\to a}\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)},\quad a \in\mathbb{R}^* \,\,\text{fixed} $$
we've
$$\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)}=e^{\tan\left... | Set $y=x-a$, then
$$
\lim_{x\to a}\left(2-\frac{x}{a}\right)^{\tan\left(\frac{\pi x}{2a}\right)}=\lim_{y\to0}\left(1-\frac{y}{a}\right)^{-\cot\left(\frac{\pi y}{2a}\right)}=\\=\lim_{y\to0}\exp
\left(-\frac{\log\left(1-\frac{y}{a}\right)}{\tan\left(\frac{\pi y}{2a}\right)}\right)=\lim_{y\to a}\exp{\left(\frac{y}{a}\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Find $\tan x$ if $x=\arctan(2 \tan^2x)-\frac{1}{2}\arcsin\left(\frac{3\sin2x}{5+4\cos 2x}\right)$ Find $\tan x$ if $$x=\arctan(2 \tan^2x)-\frac{1}{2}\arcsin\left(\frac{3\sin2x}{5+4\cos 2x}\right) \tag{1}$$
First i converted $$\frac{3 \sin 2x}{5+4 \cos 2x}=\frac{6 \tan x}{9+\tan^2 x}$$
So
$$\arcsin\left( \frac{6 \tan ... | Where you have left off,
$$\dfrac{6\tan x}{9-\tan^2x}=\dfrac{2\cdot\dfrac{\tan x}3}{1-\left(\dfrac{\tan x}3\right)^2}$$
Using my answer here, Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,
$$2\arctan\dfrac{\tan x}3=\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proof for sum of product of four consecutive integers I had to prove that
$(1)(2)(3)(4)+\cdots(n)(n+1)(n+2)(n+3)=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}$
This is how I attempted to do the problem:
First I expanded the $n^{th}$ term:$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$.
So the series $(1)(2)(3)(4)+\cdots+(n)(n+1)(n+2)(n+3)$ wil... | There is a mistake here:
$$1^4+2^4+\cdots+n^4=\frac{n(n+1)(2n+1)(3n^2-3n-1)}{30}$$
It should be
$$1^4+2^4+\cdots+n^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$
You approach has no problem, except this minor mistake.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Help with simplification question: $\frac{x}{x-1} - \frac{x}{x+1}$ Write $\frac{x}{x-1} - \frac{x}{x+1}$ as a single fraction in its simplest form.
I got $\frac{x}{x-1}$ but according to the mark scheme this is incorrect. Could someone analyze my working out and see where I went wrong?
Working out:
Combined both fract... | \begin{align}
\frac{x}{x-1}-\frac{x}{x+1} &= \frac{x(x+1)-x(x-1)}{(x-1)(x+1)} \\
&=\frac{x^2+x-(x^2-x)}{x^2-1} \\
&=\frac{x^2+x-x^2+x}{x^2-1} \\
&=\frac{2x}{x^2-1}
\end{align}
Remark: Without the answer scheme, we should know that soemthing went wrong if $$a-b=a$$ if $b \neq 0$ in general.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2301842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to solve the $u'' + u + \sin (x) = 0$ Could someone help me to obtain the general solution to this differential equation?
I need to solve the DE: $$u'' + u + \sin (x) = 0$$
Note: "I got as far as the general solution for $\;u''+u=0,$ but it was the $\;\sin(x)$ term that confused me."
| You are looking at the 2nd order ODE with constant coefficients, which is linear. Hence, general solution comes from solving the homogeneous ODE $$u''+u=0$$ which you can do by substituting $u = e^{rx}$, which yields the equation $r^2+1=0$. You get $r = \pm i$ and hence
$$
u_h(x) = \alpha e^{ix} + \beta e^{-ix} = a \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2302900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{x\to 0} \left\lfloor \frac{\tan 2x}{\sin x} \right\rfloor $
Find the limit
$$\lim_{x\to 0} \left\lfloor \dfrac{\tan 2x}{\sin x} \right\rfloor $$
My try:
$$ \tan 2x =\dfrac{\sin 2x}{\cos 2x}$$
$$\sin 2x =2\cos x\sin x$$
So:
$$\dfrac{\tan 2x}{\sin x}=\dfrac{2\cos x}{\cos 2x}$$
So:
$$\lim_{x\to 0} \left\lflo... | $x\neq0$ and around $0$ we have
$$\frac{2\cos{x}}{\cos2x}>2$$ because it's
$$\frac{2\cos{x}}{2\cos^2x-1}>2$$ or
$$(1-\cos{x})(1+2\cos{x})>0.$$
Thus, $$\lim_{x\rightarrow0}\left[\frac{2\cos{x}}{\cos2x}\right]=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Having trouble integrating $ \ \ \int \frac{x^{3}+3x^{2}+2x+4}{x^{2}(x^{2}+2x+2)} dx $. I'm having trouble integrating $$\int \frac{x^{3}+3x^{2}+2x+4}{x^{2}(x^{2}+2x+2)} dx $$
My approach
*
*$ \frac{x^{3}+3x^{2}+2x+4}{x^{2}(x^{2}+2x+2)} =\frac{Ax+B}{x^{2}}+\frac{Cx+D}{x^{2}+2x+2} $ ,
*or, $ x^{3}+3x^{2}+2x+4=(Ax+B... | $$\int \frac{x^3+3x^2+2x+4}{x^2(x^2+2x+2)}dx=\int \frac{x^3+3x^2+2(x+2)}{x^2(x^2+2x+2)}dx=\int (\frac{2x+3}{x^2+2x+2}-\frac{1}{x}+\frac{2}{x^2})dx=\int \frac{2x+3}{x^2+2x+2}dx-\int\frac{1}{x}dx+2\int\frac{1}{x^2}dx$$
Now solving: $\int \frac{2x+3}{x^2+2x+2}dx$ , write $2x+3$ as $2x+2+1$ and split it, we have:
$$\int \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How prove this inequality with $\sqrt{\sum_{cyc}\frac{a^2}{b^2}}+\sqrt{3}\ge\sqrt{\sum\frac{a}{b}}+\sqrt{\sum\frac{b}{a}}$
Let $a,b,c>0$, show that
$$\sqrt{\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}}+\sqrt{3}\ge\sqrt{\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}}+\sqrt{\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}}$$
I ... | Let $\frac{a}{b}=x$, $\frac{b}{c}=y$ and $\frac{c}{a}=z$.
Hence, $xyz=1$ and we need to prove that
$$\sqrt{x^2+y^2+z^2}+\sqrt{3}\geq\sqrt{x+y+z}+\sqrt{xy+xz+yz}.$$
or
$$x^2+y^2+z^2+3+2\sqrt{3(x^2+y^2+z^2)}\geq x+y+z+xy+xz+yz+2\sqrt{(xy+xz+yz)}$$ or
$$x^2+y^2+z^2+3-2(xy+xz+yz)+2\sqrt{3(x^2+y^2+z^2)}-2(x+y+z)+$$
$$+\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there any way to simplify the product of cosines? I recently saw a problem: Estimate the following: $cos(\frac{\pi}{15})cos(\frac{2\pi}{15})\ldots cos(\frac{7\pi}{15})$, and the options were between different consecutive power of ten.
How would I do this, no calculator of course, and is there's any way to shorten an... | $$\prod_{k=1}^7\cos\frac{k\pi}{15}=
\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\left(-\cos\frac{8\pi}{15}\right)\cos\frac{\pi}{5}\cos\frac{\pi}{3}\cos\frac{2\pi}{5}=$$
$$=-\frac{16\sin\frac{\pi}{15}\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{8\pi}{15}}{16\sin\frac{\pi}{15}}\cdot\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the inverse Laplace transform of $\frac{s^2}{s^3 -1}$
Find the inverse Laplace transform of
$$F(s)=\frac{s^2}{s^3 -1}$$
What I've done is to decompose into partial fractions, so I got:
$$\mathcal{L}^{-1}\{\frac{s^2}{s^3 -1}\}=\mathcal{L}^{-1}\{\frac{1}{3(s-1)}+\frac{2s+1}{3(s^2 +s+1)}\}$$
By the properties of... | $s^3-1$ vanishes at the third roots of unity, $1,\omega,\omega^2$. It follows that
$$ \frac{s^2}{s^3-1} = \frac{A}{s-1}+\frac{B}{s-\omega}+\frac{C}{s-\omega^2}\tag{1} $$
and by residues:
$$ A = \text{Res}\left(\frac{s^2}{s^3-1},s=1\right) = \lim_{s\to 1}\frac{s^2}{s^2+s+1}=\frac{1}{3}, $$
$$ B = \text{Res}\left(\frac{s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\frac{3-\tan^2\frac\pi7}{1-\tan^2\frac\pi7}=k\cos\frac\pi7$, find $k$ I need help solving this question:
$$
\text{If }\frac{3-\tan^2\frac\pi7}{1-\tan^2\frac\pi7}=k\cos\frac\pi7\text{, find k.}
$$
I simplified this down to:
$$
\frac{4\cos^2\frac\pi7-1}{2\cos^2\frac\pi7-1}
$$
But am unable to proceed further. The val... | $$k\cos y=\dfrac{3-\tan^2y}{1-\tan^2y}=\dfrac{3\cos^2y-\sin^2y}{\cos^2y-\sin^2y}=\dfrac{4\cos^2y-1}{2\cos^2y-1}$$
$$\iff2k\cos^3y-4\cos^2y-k\cos y+1=0\ \ \ \ (1)$$
Now if $7y=(2n+1)\pi,$
$\cos4y=\cdots=-\cos3y$
Using $\cos2A=2\cos^2A-1,\cos3B=4\cos^3B-3\cos B$
The roots of $$8\cos^4y+4\cos^3y-8\cos^2y-3\cos y+1=0$$ ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Evaluating $\int_0^1 \sqrt{1 + x ^4 } \, d x $ $$
\int_{0}^{1}\sqrt{\,1 + x^{4}\,}\,\,\mathrm{d}x
$$
I used substitution of tanx=z but it was not fruitful. Then i used $ (x-1/x)= z$ and
$(x)^2-1/(x)^2=z $ but no helpful expression was derived.
I also used property $\int_0^a f(a-x)=\int_0^a f(x) $
Please help me out
| For an approximation, you could use a Padé approximant for the integrand. The simplest one would be
$$\frac{3 x^4+4}{x^4+4}=3-\frac{x+2}{x^2+2 x+2}+\frac{x-2}{x^2-2 x+2}$$
$$\int \frac{3 x^4+4}{x^4+4}\,dx=3x+\frac{1}{2} \log \left(\frac{x^2-2 x+2}{x^2+2 x+2}\right)+\tan ^{-1}(1-x)-\tan ^{-1}(1+x)$$ So, using the given ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
How to find expression for this tensor? First consider 1D case. Suppose I would like to compute
\begin{alignat}{2}
m_p = \sum_{x=1}^{N} x^p f(x), \qquad \forall\; p = 1, 2, \dots, P
\end{alignat}
I know it can be expressed as a matrix product as follows
\begin{alignat}{2}
\underbrace{ \begin{bmatrix}
m_1 \\
m_2 ... | People usually stop writing down matrices when dealing with rank-3 tensors and higher. In your case, using the Einstein summation convention,
1D: $m_p = X_{pi} f_i$ with $X_{pi} = i^p$, $f_i = f(i)$
2D: $m_{pq} = X_{pi} Y_{qj} f_{ij}$ with $X_{pi}$ as above, $Y_{qj} = j^q$, $f_{ij} = f(i,j)$.
3D: $m_{pqr} = X_{pi} Y_{q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2315832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum $\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$ How can we find the sum
$$\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$$
WolframAlpha says it's $13/4$ but how to compute it? In general how to approach this kind of sums of rational functions where denominator has distinct roots?
| $$\frac{3n+7}{n(n+1)(n+2)}=\frac{3n+6+1}{n(n+1)(n+2)}=$$
$$=\frac{3}{n(n+1)}+\frac{1}{n(n+1)(n+2)}=\frac{3}{n}-\frac{3}{n+1}+\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)=$$
$$=\frac{3}{n}-\frac{3}{n+1}+\frac{1}{n}-\frac{1}{n+1}-\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)=$$
$$=\frac{7}{2n}-\frac{4}{n+1}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2317136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Given $\triangle ABC$, find point $X$ that minimizes $|\overline{AX}||\overline{BX}|+|\overline{BX}||\overline{CX}|+|\overline{CX}||\overline{AX}|$ Given a $\triangle$$ABC$.Find with proof the point $X$ in the plane of the triangle,for which $$|\overline{AX}||\overline{BX}|+|\overline{BX}||\overline{CX}|+|\overline{CX}... | Let $a = |BC|$, $b = |CA|$, $c = |AB|$, $R_1 = |XA|$, $R_2 = |XB|$, $R_3 = |XC|$.
WOLOG, we will assume $a \ge b \ge c$. We will show that
$$R_1R_2 + R_2R_3 + R_3R_1 \ge \min(ab,bc,ca) = bc \tag{*1}$$
This implies vertex $A$ is always a point that minimize $R_1R_2 + R_2R_3+R_3R_1$.
When $X \in \{ A, B, C \}$, $(*1)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2318783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find $\delta$ such that if $0 < \vert x - 2 \vert < \delta$, then $\vert f(x)-3 \vert < .2$ I'm looking for assistance with the following problem:
Find $\delta$ such that if $0 < \vert x - 2 \vert < \delta$, then $\vert f(x)-3 \vert < .2$ where $f(x)=x^2-1$.
My attempt:
We need $\delta$ such that $\vert f(x)-2 \vert <... | The other solution is of course neat :-)
If $|x^2-4|<0.2$ then we have that $3.8<x^2<4.2$ that is either
*
*$\sqrt{3.8}<x<\sqrt{4.2}$ or
*$-\sqrt{4.2}<x<-\sqrt{3.8}$.
Consider case $1$. From here we get
\begin{align}
-0.05064\sim\sqrt{3.8}-2&<x-2<\sqrt{4.2}-2\sim0.04939
\end{align}
Now if we want $-\delta<x-2<\de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation $\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x=4$ Find $x\in R$ satisfy $$\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x=4$$
The root of equation very bad
My try 1:
Let $\sqrt[3]{x+2}=a;\sqrt[3]{2x-1}=b\Rightarrow a^3-b^3=3-x$
Have: $a+b=4-x$
=>Root of a system of equation bad, too
My try 2:
Use quality $\sqrt[3]{a}\pm \sqrt[3]... | (Too long for a comment.) Building upon this...
Let $\,\sqrt[3]{x+2}=a\,;\;\sqrt[3]{2x-1}=b$
The canonical way to get the polynomial equation in $x$ is to eliminate $a,b$ between:
$$
\begin{cases}
\begin{align}
a^3 &= x+2 \\
b^3 &= 2x-1 \\
a+b +x &= 4
\end{align}
\end{cases}
$$
This is a routine calculation using po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
If $a^3+b^3+c^3=3$ so $2(a^2+b^2+c^2)-(ab+bc+ca) \geq 3$ I was looking at this question, and I derived this inequality from that
Let $a$, $b$ and $c$ be positive numbers such that $a^3+b^3+c^3=3.$ Prove that:
$$2(a^2+b^2+c^2)-(ab+bc+ca) \geq 3$$
I couldn't solve Rozenberg's inequality but I assumed it's true. So if ... | Rozenberg's inequality is solved, find the solution at this link :
If $a^3+b^3+c^3=3$ so $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq\frac{3}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $a \equiv b \pmod{1008}$
Suppose that the positive integers $a$ and $b$ satisfy the equation $$a^b-b^a = 1008.$$ Prove that $a \equiv b \pmod{1008}$.
Taking the equation modulo $1008$ we get $a^b \equiv b^a \pmod{1008}$, but I didn't see how to use this to get that $a \equiv b \pmod{1008}$.
| We have $1008 = 2^4 \cdot 3^2 \cdot 7$. First note that $a \equiv b \pmod{2}$ since $1008$ is even. Note that $2 \nmid a,b$ since if $2 \mid a,b$ then $a,b \leq 5$. Therefore, $a^b \equiv a \pmod{8}$ and $b^a \equiv b \pmod{8}$, so $a \equiv b \pmod{8}$. Then since $a^4 \equiv 1 \pmod{16}$, we have $a^b \equiv a^a \pmo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2322466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f (x)$
If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f(x)$.
$f'(x)$ is the first derivative of $f (x)$.
I have no idea about this question, please help me.
| We can use an integrating factor, $u(x)$:
$$
\begin{align}
v(x)(f(x)+f'(x))
&=\frac{\mathrm{d}}{\mathrm{d}x}(u(x)\,f(x))\\
&=u'(x)\,f(x)+u(x)\,f'(x)\tag{1}
\end{align}
$$
If we set $\frac{u'(x)}{u(x)}=1$, we see that $(1)$ is satisfied. Thus, we get that we can use $u(x)=v(x)=e^x$. That is,
$$
\begin{align}
\frac{\math... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 5
} |
Prove that each divisor of $5^{(4n+1)}+5^{(3n+1)}+1 =$ $1 \pmod {10}$ Prove that if $d$ divides $5^{(4n+1)}+5^{(3n+1)}+1$ for any $n$, then $d =$ $1 \pmod {10}$.
A similar statement can be proven, where $d$ divides $(3^{(2n+1)}+1)/4$, for any $n$, then $d =$ $1 \pmod {6}$ by first showing that if $-3$ is a quadratic r... | Each divisor of $5^{4n+1}+5^{3n+1}+1$ is odd, hence it is enough to show that any odd prime divisor of $5^{4n+1}+5^{3n+1}+1$ is $\equiv 1\pmod{5}$. Now we may consider the splitting field of $f(x)=5x^4+5x^3+1$ over $\mathbb{F}_p$ for any prime $p>5$. The complex roots of $f(x)$ are given by
$$ \frac{1}{4}\left(-1\color... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
probability or rolling 1 1 2 with six dice What is the probability of rolling at least two ones and at least one two (order not important) with six 6-sided die?
Equivalently, to win a game, I need two 1s and a 2, (1,1,2). What is the probability of my rolling at least this with 6 dice.
There are $6^6=46656$ possible r... | Approach via inclusion-exclusion and De Morgan's laws.
Let $A$ be the event that you roll at least two $1$'s. Let $B$ be the event that you roll at least one $2$.
We are attempting to calculate then $Pr(A\cap B)$
$Pr(A\cap B)=1-Pr((A\cap B)^c)=1-Pr(A^c\cup B^c) = 1-Pr(A^c)-Pr(B^c)+Pr(A^c\cap B^c)$
We calculate each te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the Nth derivative using Maclaurin series The prompt is to find the 8th derivative of the function f(x) defined as,
$$f(x) = \frac{3}{1+x-2x^2}$$
$$f^{(8)}(0) = ?$$
To find the maclaurin series, I proceeded by finding the derivatives of the function at 0 as follows,
$$f^{(1)}(x) = \frac{ -3 + 12x }{ (1 + x -2x^2)... | Suppose we want to do this without expanding in partial fractions. Then we can just perform a long division of the fraction to obtain the coefficient of $x^8$. But a faster division algorithm is Newton-Raphson division, $n$ steps will yield the coefficients up to $x^{2^n -1}$ This method can be generalized
such that i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Complex Integration - $\int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx$
Exercise :
Show that :
$$\int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx = \frac{4\pi}{5}\sin\bigg(\frac{2\pi}{5}\bigg)$$
Attempt :
$$\int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx = \int_{-\infty}^{\infty} \frac{1}{x^4 + x^3 + x^2 + x + 1}dx $$
$$x^4 + ... | Brute force also helps:
$$\int\limits_{-\infty}^{+\infty}\frac{x-1}{x^5-1}dx=\int\limits_{-\infty}^{+\infty}\frac{1}{x^4+x^3+x^2+x+1}dx=$$
$$=\int\limits_{-\infty}^{+\infty}\frac{1}{x^2\left(x^2+\frac{1}{x^2}+x+\frac{1}{x}+1\right)}dx=\int\limits_{-\infty}^{+\infty}\frac{1}{x^2\left(\left(x+\frac{1}{x}\right)^2+x+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Prove that $2\cdot \cos \frac{72}{2}\cdot \cos \frac{24}{2}+2\cdot \sin \frac{96}{2}\cdot \sin \frac{72}{2}=0.5$. Additional data added To solve it I have tried some options, where in one of them I applied product to sum formulas, which seemed to be very helpful, but didn't get the answer.
Used these formulae:
$$\co... | Using the formulae and supposing that you work on degrees :
$$ \cos(a+b)+\cos(a−b)=2\cos(a)\cos(b)$$
$$−\cos(a+b)+\cos(a−b)=2\sin(a)\sin(b) $$
we apply them and transform your given equation :
$$2\cos \frac{72^\circ}{2} \cos \frac{24^\circ}{2}+2\sin \frac{96^\circ}{2} \sin \frac{72^\circ}{2}= \cos\bigg(\frac{72^\circ}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\sum_{cyc}(1-x)^2\ge \sum_{cyc}\frac{z^2(1-x^2)(1-y^2)}{(xy+z)^2}$
Let $x,y,z>0$. Show that
$$\sum_{cyc}(1-x)^2\ge \sum_{cyc}\dfrac{z^2(1-x^2)(1-y^2)}{(xy+z)^2}.$$
Source: In a blog entry posted in 2013, it was said that this problem was proposed by Dongyi Wei (full marks both at 50th IMO 2009 and 49th IM... | The idea to prove this inequality is to use the well-know Ky-Fan inequality :
Before that we make the following substitution:
$x=\frac{1}{a}$
$y=\frac{1}{b}$
$z=\frac{1}{c}$
We get the following inequality :
$$\sum_{cyc}\frac{(1-a)^2}{a^2}\geq \sum_{cyc} \frac{(1-a^2)(1-b^2)}{(ab+c)^2}$$
Now we have the Ky-fan inequali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Prove: If n is a perfect square, then n+3 is not a perfect square Use either direct proof, proof by contrapositive, or proof by contradiction.
Using proof by contradiction method
Assume n is a perfect square and n+3 is a perfect square (proof by
contradiction)
There exists integers x and y such that $n = x^2$ and $n+3... | First, as mentioned by others, $1$ is a perfect square and $1+3$ is a perfect square. So you need to prove "If $N > 1$ is a perfect square, then $n+3$ is not a perfect square.
Let $n = m^2$ where $m$ and $n$ are positive integers and $n > 1$. Then we must have $m > 1$. Since $m$ is an integer, we can say that $m \ge 2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Log with $\sqrt x$ base I'd like to know how this simplification happened:
$$\frac{1}{2}\log _{\sqrt{2}}\left(x-2\right)=\log _2\left(x-2\right)$$
$$
\begin{array}{l}
\color{red}{2 \log _{2} x+\log _{\frac{1}{2}}(1-\sqrt{x})=\frac{1}{2} \log _{\sqrt{2}}(x-2 \sqrt{x}+2) \quad } \color{blue}{0<x<1} \\
\Leftrightarrow 2... | $$\log_a(x) = \frac {\log_b(x)} {\log_b(a)}$$ for any $b$.
therefore: $\frac 1 2 \log_{\sqrt 2}(x-2) = \frac {log_2(x - 2)} {2 \log_2(\sqrt 2)} = \frac {log_2(x - 2)} {2 \times \frac 1 2} = \log_2(x-2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
} |
Smith Normal Form of this polynomial non square matrix in $\mathbb Q[t]$
Given $$A=\begin{bmatrix}t^3-t^2&t^2-t\\2t^3&t^4-t\\t&2t^3-2t^2\end{bmatrix} \in M_{3\times 2}(\mathbb Q[t])$$
I'm looking for the matrices $S,P,Q$ such that $S=QAP$, where $Q$ are the row and $P$ are the column transformations and $S$ is its S... | Since $-4t^5+5t^4-t=-t(t-1)(4t^3-t^2-t-1)$ and $-2t^5+4t^4-2t^3+t^2-t=-t(t-1)(2t^3-2t^2-1)$, you only need to find an integer multiple of the GCD of $4t^3-t^2-t-1$ and $2t^3-2t^2-1$ using Euclidean algorithm. This should be straightforward:
$$
\begin{align}
4t^3-t^2-t-1 &= 2(2t^3-2t^2-1) + (3t^2-t+1)\\
9(2t^3-2t^2-1) &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
show this inequality $f(\sqrt{x_{1}x_{2}})\ge \sqrt{f(x_{1})\cdot f(x_{2})}$ let $$f(x)=\dfrac{1}{3^x}-\dfrac{1}{4^x},x\ge 1$$
For any $x_{1},x_{2}\ge 1$,show that
$$f(\sqrt{x_{1}x_{2}})\ge \sqrt{f(x_{1})\cdot f(x_{2})}$$
or
$$\left(\dfrac{1}{3^{\sqrt{x_{1}x_{2}}}}-\dfrac{1}{4^{\sqrt{x_{1}x_{2}}}}\right)^2 \ge \left(\d... | Let $$g(x)=\ln\left(\left(\frac{1}{3}\right)^x-\left(\frac{1}{4}\right)^x\right).$$
Thus, $$g'(x)=\frac{3^x\ln4-4^x\ln3}{4^x-3^x}<0$$
because $$3^x\ln4-4^x\ln3<0\Leftrightarrow\left(\frac{4}{3}\right)^x\geq\frac{\ln4}{\ln3},$$
for which it's enough to prove that
$$\frac{4}{3}\geq\frac{\ln4}{\ln3},$$
which is $3^4>4^3.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2338165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Let $z_0 = \left(a,b\right)$ find $z$ such that $z^2 = z_0$ where $z_0 , z$ are complex numbers. Let $z_0 = \left(a,b\right)$ find $z$ such that $z^2 = z_0$ where $z_0 , z$ are complex numbers.
Here is my attempt.
taking absolute values on both sides, we obtain
$\left|z\right|^2 = \left|z_0\right|^2$
we let $z = u +v... | It would be better to write the complex number $z_0$ in rectangular form:
$$z_0 = a + bi = \sqrt{a^2 + b^2}\cdot\bigg({\frac{a}{\sqrt{a^2 + b^2}} + \frac{b}{\sqrt{a^2 + b^2}}i}\bigg)$$
Then write the complex number $z_0$ in polar form:
$$z_0 = {r_0}\exp(i\arg{z_0})$$
where $r_0 = \sqrt{a^2 + b^2}$ and $\arg{z_0} = \arc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Fourier sine series of $x^2$ When I am trying to compute $$\frac{2}{l}\int_{0}^{l}x^2\sin\left(\frac{n\pi x}{l}\right)dx$$
I let $u = x^2$, $du = 2x dx$, $dv = \sin\left(\frac{n\pi x}{l}\right)$, $v = -\frac{l}{n \pi}\cos\left(\frac{n\pi x}{l}\right)$. Thus applying integration by parts I get $$-\frac{2l^2}{n\pi}\cos(n... | The integral in question is as follows.
\begin{align}
\int x^2 \, \sin\left(\frac{n \pi x}{l}\right) \, dx &= - \frac{l \, x^2}{n \pi} \cos\left(\frac{n \pi x}{l}\right) + \frac{2 l}{n \pi} \, \int x \, \cos\left(\frac{n\pi x}{l} \right) \, dx \\
&= \left( 2 \, \left(\frac{l}{n \pi}\right)^3 - \frac{l x^2}{n \pi} \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I solve $\int_0^\infty \frac{x^2}{x^4+1} \, dx$?
$$\int_0^\infty \frac{x^2}{x^4+1} \; dx $$
All I know this integral must be solved with beta function, but how do I come to the form $$\beta (x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\;dt \text{ ?}$$
| $$\begin{eqnarray*}I=\int_0^{+\infty}\frac{x^2\,dx}{x^4+1}\stackrel{\text{parity}}{=} \frac{1}{2} \int_{-\infty}^{+\infty}\frac{dx}{x^2+\frac{1}{x^2}}&=&\frac{1}{2} \int_{-\infty}^{+\infty}\frac{dx}{\left(x-\frac{1}{x}\right)^2+2} \\[8pt]
(\text{by Glasser's Master Theorem})&=&\int_0^{+\infty}\frac{dx}{x^2+2}=\color{re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Prove $ \left|x-1\right|<1$ implies $ \left|x^2-4x+3\right|<3$. Question:
Let $ x$ be a real number. Prove that if $ \left|x-1\right|<1$ then $ \left|x^2-4x+3\right|<3$.
We can write $ \left|x^2-4x+3\right|<3 $ as $ |x-3||x-1| < 3$. If I start with $ \left|x-1\right|<1$, how can I show that $ |x-3| < 3$ ?
| $|x-1|<1\implies |x^2-4x+3|=|(x-1)^2+2(1-x)|\leq $ $\leq |(x-1)^2|+|2(1-x)|<1^2+2\cdot 1=3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
If $a,b$ are roots of $3x^2+2x+1$ then find the value of an expression It is given that $a,b$ are roots of $3x^2+2x+1$ then find the value of:
$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3$$
I thought to proceed in this manner:
We know $a+b=\frac{-2}{3}$ and $ab=\frac{1}{3}$. Using this I tried to c... | Plug $x=\frac{1-y}{1+y}$ in the given equation. We get:
$$\frac{3 (1-y)^2}{(y+1)^2}+\frac{2 (1-y)}{y+1}+1=0$$
Expanding and collecting, we have:
$$y^2-2 y+3=0$$
whose solutions are $$y_1=\frac{1-a}{1+a};\;y_2=\frac{1-b}{1+b}$$
We also know that sum of roots is $s=y_1+y_2=2$ and product is $p=y_1y_2=3$.
The sum of cubes... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Issue with Baby Rudin Example 7.3 Example 7.3 of Baby Rudin states that the sum
\begin{align}
\sum_{n=0}^{\infty}\frac{x^2}{(1 + x^2)^n}
\end{align}
is, for $x \neq 0$, a convergent geometric series with sum $1 + x^2$. This confuses me. As far as I know, a geometric series is a series of the form
\begin{align}
\sum_{... | You should write the first few terms to understand its behavior:
$$\sum_{n=0}^{\infty}\frac{x^2}{(1 + x^2)^n}=x^2+\frac{x^2}{1+x^2}+\frac{x^2}{(1+x^2)^2}+\frac{x^2}{(1+x^2)^3}+\cdots=$$
$$x^2\left(1+\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+\frac{1}{(1+x^2)^3}+\cdots\right)\stackrel{x\ne 0}=$$
$$x^2\cdot \frac{1}{1-\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Are these two distinct series representations for $\frac{1}{1-x^2}$ correct? Let $f(x)=\frac{1}{1-x^2}$.
With a $u=x^2$ substitution, $f(x)$ becomes $g(u)=\frac{1}{1-u}$, which being the limit of a geometric series has the form $$g(u)=\sum_{k=0}^\infty u^k$$
Plugging the substitution back in yields a power series repr... | We have
\begin{align}
\sum_{k \ge 0} \frac{x^k}{1 + x} &= \sum_{k \ge 0} \sum_{\ell \ge k} (-1)^{\ell-k}x^\ell \\
&= \sum_{\ell \ge 0} \sum_{k \le \ell} (-1)^{\ell-k}x^\ell \\
&= \sum_{\ell \ge 0} x^\ell \sum_{k \le \ell} (-1)^{\ell-k}
\end{align}
where
$$ \sum_{k \le \ell} (-1)^{\ell-k} = \begin{cases} 0 & \text{if $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2349503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Squares and Rationalization I was solving this question, and I'm hitting a wall.
If ${1\over{\sqrt{2011+\sqrt{2011^2-1}}}}=\sqrt{m}-\sqrt{n}$, where $m$ and $n$ are positive integers, what is the value of $m+n$?
I tried to solve this question with two approaches:
${\sqrt{m}-\sqrt{n}}={1\over{\sqrt{2011+\sqrt{2011^2... | HINT:
$$(a+\sqrt{a^2-1})(a-\sqrt{a^2-1})=a^2-(a^2-1)=?$$
$$\implies\sqrt{a+\sqrt{a^2-1}}\cdot\sqrt{a-\sqrt{a^2-1}}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Confirming my solution to the expansion of $z/(z-1)(z+3)$ I would just like a confirmation of my work for the following problem.
Represent the function $$ \frac{z}{(z-1)(z+3)} $$ as a Laurent Series around $z=0$ when $1<|z|<3$.
Using partial fraction decomposition to simplify the problem a bit,
$$ \frac{1}{(z-1)(z+... | The function
\begin{align*}
f(z)&=\frac{z}{(z-1)(z+3)}\\
&=\frac{1}{4(z-1)}+\frac{3}{4(z+3)}\\
\end{align*}
has two simple poles at $z=1$ and $z=-3$.
Since we want to find a Laurent expansion with center $0$, we look at the poles $1$ and $-3$ and see they determine three regions.
\begin{align*}
|z|<1,\qquad\quad
1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Integration by trig substitution - why is my answer wrong? Attempting integral:
$$-\int \frac{dx}{\sqrt{x^2-9}}$$
Let $x = 3\ sec\ \theta$ so that under the square root we have:
$$\sqrt{9\ sec^2\ \theta - 9}$$
$$\sqrt{9(sec^2\ \theta - 1)}$$
$$\frac{dx}{d\theta} =3\ sec\ \theta\ tan\ \theta$$
The $1/3$ and the $3$ canc... | HINT:
If $x=3\sec\theta$
$$\tan^2\theta=\left(\dfrac x3\right)^2-1=\dfrac{x^2-9}9$$
$|\tan\theta|=\dfrac{\sqrt{x^2-9}}3$
$\sec\theta+|\tan\theta|=\dfrac{x+\sqrt{x^2-9}}3$
$\implies\ln(\sec\theta+|\tan\theta|)=\ln(x+\sqrt{x^2-9})-\ln3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Multiplication of a continuous function with non continuous $$f(x) = \begin{cases} e^x e^{-1} & ,0 \leq x\leq 1
\\ \lfloor x \rfloor+ 1 & ,1< x < 3 \end{cases}$$
and $$g(x) = x^2 -ax + b $$ such that $f(x)g(x)$ is continuous in $[0,3)$ then find the ordered pair $(a,b)$
| You need to make it continuous at $x=1$. That is
\begin{align}
\lim_{x \to 1^-}f(x)g(x) &= \lim_{x \to 1^+}f(x)g(x) \\
e^{-2}(1-a+b) &= 2(1-a+b) \\
(e^{-2}-2)(1-a+b) &= 0 \\
(1-a+b) &= 0 \\
b &= a-1 \\
\hline
g(x) &= x^2 + ax +(a-1)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to find the value of this expression Suppose I have a general quadratic equation $ ax^2 + bx + c = 0 $
And I have $ \alpha , \beta $ as roots, then what is the simplest way to find $ (a\alpha + b)^{-2} + (a\beta + b)^{-2} $ ?
I know the formula for finding the sum and product of roots, but that doesn't helped me i... | Alternative hint: derive the equation whose roots are $y_1=(a\alpha + b)^{-2}$ and $y_2 = (a\beta + b)^{-2}\,$.
Let $y=(ax+b)^{-2}\,$, then $\;\displaystyle ax+b=\pm \frac{1}{\sqrt{y}} \;\;\iff\;\; x = \frac{-b \pm \frac{1}{\sqrt{y}}}{a} = \frac{-b\sqrt{y}\pm1}{a\sqrt{y}}\,$.
Substituting back into the original equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Small doubt on this trigonometry question If $\alpha, \beta$ are complementary angles such that $b \sin\alpha = a $, then find the value of $\sin \alpha \cos \beta - \cos \alpha \sin \beta$.
The question itself is straightforward, but while solving for $\cos \alpha $ and $ \sin \beta$ , which sign should I take? Or sho... | Maybe the form of the answer : $X - 1$ should make you think about the following transformation
\begin{align*}\sin \alpha \cos \beta - \cos \alpha \sin \beta &= 2 \sin \alpha \cos \beta - (\sin \alpha \cos \beta + \cos \alpha \sin \beta) \\
& = 2 \sin \alpha \cos \big( \frac{\pi}{2} - \alpha \big) - \sin(\alpha+\beta) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Express the differential equation as power series
My attempt: $y=\sum^{\infty}_{n=0}a_n x^n \rightarrow y'=\sum^{\infty}_{n=0}na_n x^{n-1}\rightarrow y''=\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2}\\
(2+x^2)y"+x^2y'+3y=(2+x^2)\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2}+x^2\sum^{\infty}_{n=0}na_n x^{n-1}+3\sum^{\infty}_{n=0}a_n x^n\... | Note:
$$(2+x^2)y''+x^2y'+3y=$$
$$(2+x^2)(2a_2+3\cdot 2a_3x+4\cdot 3a_4x^2+5\cdot 4a_5x^3+\cdots)+$$
$$x^2(a_1+2a_2x+3a_3x^2+\cdots)+$$
$$3(a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+\cdots)=$$
$$4a_2+3a_0+(2(3\cdot 2)a_3+3a_1)x+$$
$$2(4\cdot 3)a_4+a_1+2a_2+3a_2)x^2+$$
$$(2(5\cdot 4)a_5+2a_2+3\cdot 2a_3+3a_3)x^3+$$
$$(2(6\cdot 5)a_6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $y = \frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Prove, using an algebraic method,that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Hence, determine the minimum and maximum points $y=\frac{2x}{x^2 +1}$ .
What I tried:
Firstly, I thought of using partial fractions but since $x^2 +1 =... | We need to prove that$$\left|\frac{2x}{x^2+1}\right|\leq1$$ or
$$x^2-2|x|+1\geq0$$ or
$$\left(|x|-1\right)^2\geq0.$$
The equality occurs for $|x|=1$, which gives:
$(1,1)$ is a maximum point and $(-1-1)$ is a minimum point.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 9,
"answer_id": 0
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Math subject GRE Exam 9768 Q.16 The question is in the following picture:
And the answer is (D).
At first I agreed with this answer when I added the given three vectors component wise,but when I added the forth component in the three vectors I understood that it is impossible that we can get 5, so my opinion change... | The question is the same as asking for what $m$ the following equation has a solution:
$$
\begin{pmatrix}
0&0&1\\
1&0&1\\
1&0&2\\
1&1&0
\end{pmatrix}
\begin{pmatrix}
x\\y\\z
\end{pmatrix}=
\begin{pmatrix}
1\\2\\m\\5
\end{pmatrix}\tag{1}
$$
Working on column vectors might make some observation easier. Note that (1) is e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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Integral $\int \frac{x^2}{x^4+x^2+1}\ dx$ I have taken $(x^4+x^2+1)=u$ to differentiate it w.r.t. $x$ to get
$$\int \frac{x}{2u(2x^2+1)}\ du$$
| $$((x^2+1)+x)((x^2+1) -x) = x^4 + 2x^2 + 1 - x^2 = x^4 + x^2 + 1$$
Then use partial fraction method.
Edit: More hints,
$$((x^2+1)+x) + ((x^2+1) -x) = 2(x^2+1)$$
$$\int\frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Explain why the graph of $y=\frac{4x}{x^2+1}$ and $y=2\sin(2\arctan x)$ are the same. Explain why the graph of $y=\frac{4x}{x^2+1}$ and $y=2\sin(2\arctan x)$ are the same.
The first equation is of the form of Newton's Serpentine. When you graph the second equation it appears to overlap the first equation.
I'm not sure ... | We use the identity
$$\sin(2\theta) = \frac{2\tan\theta}{1 + \tan^2\theta}$$
Let $\arctan x = \theta$. Then $x = \tan\theta, -\frac{\pi}{2} < x < \frac{\pi}{2}$. Hence,
$$2\sin(2\arctan x) = 2\sin(2\theta) = 2 \cdot \frac{2\tan\theta}{1 + \tan^2\theta} = 2 \cdot \frac{2x}{1 + x^2} = \frac{4x}{1 + x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2367735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove this trigonometric inequality about the angles of $\triangle ABC$
In $\Delta ABC$ show that
$$\cos{\frac{A}{2}}+\cos\frac{B}{2}+\cos\frac{C}{2}\ge \frac{\sqrt{3}}{2} \left(\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\frac{A-B}{2}\right)$$
since
$$\frac{\sqrt{3}}{2}\left(\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\df... | Denote the difference $\text{LHS}-\text{RHS}$ by $S(A,B,C)$. We need to show that $S(A,B,C)\ge0$. We begin with a simple
Proposition. For $\alpha\in[0,\pi/2]$ we have the following inequality
$$2\cos\frac{\pi-\alpha}4-\sqrt3\cos\frac{\pi-3\alpha}4<\sqrt3$$
Proof. $\frac{\pi-3\alpha}4\in[-\frac\pi8,\frac\pi4]$, thu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2368630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Which area is larger, the blue area, or the white area? In the square below, two semicircles are overlapping in a symmetrical pattern. Which is greater: the area shaded blue or the area shaded white?
My Solution
Let the length of each side of the square be $2r$.
The area of the square is $4r^2$.
The two semi-circles h... | If you just divide the lower left part of the blue area and move each part $90$ degree up to join them to the main blue part, then the area of the new blue shape will be equal to the white part.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
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What is the largest 3-digit number which has all 3 digits different and is equal to 37 times the sum of it's digits?
What is the largest 3-digit number which has all 3 digits different and is equal to 37 times the sum of it's digits?
-Question 28, Junior Division, AMC 2016
I found the solution here, page 51, however ... | "What is the largest 3-digit number...."
Let $N = abc$ have three digits. $N = 100a + 10b + c$. That's what the writing numbers of digits mean. $593 = 500 + 90 +3$ and $abc = 100a + 10b + c$.
" which has all 3 digits different"
So $a \ne b; a\ne c; b \ne c$.
" and is equal to 37 times the sum of it's digits?"
So $N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2371949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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What is the probability that Cathy wins?
Alice, Bob and Cathy take turns (in that order) in rolling a six sided
die. If Alice ever rolls a 1, 2 or 3 she wins. If Bob rolls a 4 or a 5
he wins, and Cathy wins if she rolls a 6. They continue playing until
a player wins. What is the probability (as a fraction) that ... | What you are leaving out is the possibility that nobody wins the first round. There is a $\frac{1}{18}$ chance that Cathy wins AND does so in the first round. Her overall odds will be $\sum_{n=1}^{\infty}(\frac{1}{2})^n(\frac{2}{3})^n(\frac{5}{6})^{n-1}(\frac{1}{6})=\frac{6}{5}\frac{1}{6}\sum_{n=1}^{\infty}(\frac{5}{18... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2374181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} =\sqrt \frac{x}{2}$ The value of $x$ (considering only the positive root) satisfying the equation $$\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} = \sqrt{\frac{x}{2}}$$ is?
Please, ca... | Let
$ a = \sqrt{5+ 2\sqrt{6}}, \ \ b= \sqrt{5- 2\sqrt{6}}.$
$$ \frac{ a +b}{a-b} = \sqrt{\frac{x}{2}}$$
$$ \frac{(a+b)(a+b)}{(a-b)(a+b)} = \frac{(a+b)^2}{a^2 - b^2} = \sqrt{\frac{x}{2}}$$
$$ \frac{\left(\sqrt{5+ 2\sqrt{6}} + \sqrt{5-2\sqrt{6}}\right)^2}{\left(\sqrt{5+2\sqrt{6}}\right)^2-\left(\sqrt{5-2\sqrt{6}})\right... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Difficult Homogeneous Differential Equation Solve the differential equation:
$$\frac{dy}{dx}=\frac{\sqrt{x^2+3xy+4y^2}}{x+2y}$$
I tried to solve it by putting $t=x+2y$ but that lead to a very complicated integral. The hint given is that equation is reducible to homogeneous form.
| The integral that Noé AC derived is expressible in terms of elementary functions although it is highly messy. Integration of any rational function $R(u,\sqrt{P_2(u)})$, where $P_2$ is a quadratic, can be reduced to the integration of a rational function.
The first step is the reduction to a rational function of $\sin\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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bisector theorem , cosine rule or other method? A fixed point $A(a,0)$, where $a>0$ and a straight line $l$ is given $x = -1$, $B$ is a moving point on the second quadrant and lies on the straight line $l$, the angle bisector of $\angle BOA$ intersects $AB$ at point $C$.
(a) Find the locus of point $C$ and state the ra... | Let $B=(-1,t)$, $C=(x,y)$.
Then
$$\frac{y}{x-a}=\frac{t}{-1-a}$$
Angle AOB can be obtained to be
$$\cos AOB=\frac{-a}{a\sqrt{1+t^2}}=\frac{-1}{\sqrt{1+t^2}}$$
Using
$$\cos^2 AOC = \frac{1+\cos AOB}{2}$$
we have
$$\frac{y^2}{x^2}=\tan^2 AOC = \sec^2 AOC - 1=\frac{2}{1-\frac{1}{\sqrt{1+t^2}}}-1=\frac{2\sqrt{1+t^2}}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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using Parseval's identity to estimate the value of $\Sigma_{n=1}^\infty \frac{1}{(2n-1)^2(2n+1)^2}$ There is a problem with two parts; The first part is asking to find Fourier series for $f(x)=|\sin(x)|$ on $[-\pi,\pi]$. And the second part wants to estimate the following using Parseval's identity:
$$
\sum_{n=1}^\infty... | I guess I did it!
I want to know the value of:
$$
\sum_{n=1}^\infty \frac{1}{(4n^2-1)^2}=\sum_{n=1}^\infty \frac{1}{((2n)^2-1)^2}
$$
So let $m=2n$, then $\sum_{n=1}^\infty \frac{1}{(m^2-1)^2}$.
From the first part and Parseval's identity,
$$
\frac{2}{\pi}\int_0^\pi \sin^2(x)dx=\frac{16}{2\pi^2}\sum_{m=1}^\infty \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}$
Let $a$, $b$ and $c$ be real numbers such that $a + b + c = 0$ and define:
$$P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}.$$
What is the value of $P$?
This question came in the regional maths olympiad. I tried AM-GM a... | plugging $c=-a-b$ in your term, we get $${\frac {{a}^{2}}{2\,{a}^{2}+b \left( -a-b \right) }}+{\frac {{b}^{2}}{
\left( -a-b \right) a+2\,{b}^{2}}}+{\frac { \left( -a-b \right) ^{2}
}{ab+2\, \left( -a-b \right) ^{2}}}
$$
simplifying this we get $1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Floor functions of powers increase What is the smallest real number $x$ such that $\lfloor x^n\rfloor<\lfloor x^{n+1}\rfloor$ for all positive integers $n$?
In particular, we have $\lfloor x\rfloor<\lfloor x^2\rfloor<\lfloor x^3\rfloor<\dots$. For the first inequality to hold it must be that $x\geq\sqrt{2}$. But if $x=... | Let $x=\sqrt[3]{3}$. First, you've already directly verified that $\lfloor x\rfloor<\lfloor x^2\rfloor<\lfloor x^3\rfloor$, as they are equal $1$, $2$, and $3$, respectively. Now, for all $n\ge3$:
$$x^{n+1}-x^n=x^n(x-1)=\left(\sqrt[3]{3}\right)^n\left(\sqrt[3]{3}-1\right)>3\cdot0.4>1,$$
so the numbers grow by more than... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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show that $7\mid p^3-p$ if $p$ is a prime divisor of $n^3+n^2-2n-1$
Let $p$ be a prime number, and $n$ a positive integer such
$$p\mid n^3+n^2-2n-1, \quad n\ge 2.$$
Show that $$7\mid p^3-p.$$
It maybe can use Fermat's little theorem?
| An alternative way is to consider the prime divisors of
$a^3+a^2b-2ab^2-b^3$, and reduce the numbers by applying Thue's lemma.
Call a prime $p$ "bad" if $p\not\equiv0,\pm1\pmod7$, and there exist some co-prime integers $a,b$ such that $p$ divides $a^3+a^2b-2ab^2-b^3$. Suppose that there at least one bad prime, and take... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 1
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Finding third row of orthogonal matrix?
Find a $3\times3$ orthogonal matrix whose first two rows are $\Big[\frac{1}{3},\frac{2}{3},\frac{2}{3}\Big]$ and $\left[0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right]$.
I tried two approaches.
One, finding vector cross product of given two rows.
Second, assuming third row as $... | One approach that would be to find the cross product.
$(\frac 13, \frac 23, \frac 23)\times (0, \frac 1{\sqrt 2},-\frac 1 {\sqrt2}) = (-\frac 4{3\sqrt 2},\frac {1}{3\sqrt2},\frac {1}{3\sqrt2})$
Another would be to say by inspection any vector that is orthogonal to the second vector must have the from $(a,b,b)$
Choose a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Inequality $(xy+z)^2+(yz+x)^2+(zx+y)^2\geq\sqrt{2}(x+y)(y+z)(z+x)$ Let $x,y,z\geq 0$. Prove that
$$(xy+z)^2+(yz+x)^2+(zx+y)^2\geq\sqrt{2}(x+y)(y+z)(z+x).$$
Expanding gives
$$\sum x^2y^2+\sum x^2+6xyz\geq \sqrt{2}\sum x^2y+2\sqrt{2}xyz.$$
If we use AM-GM on the left-hand side, we get
$$\frac{1}{2}(x^2y^2+x^2)\geq x^2y$$... | Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, our inequality it's
$$\sum_{cyc}(x^2y^2+2xyz+x^2)\geq\sqrt2\sum_{cyc}\left(x^2y+x^2z+\frac{2}{3}xyz\right)$$ or
$$9v^4-6uw^3+6w^3+9u^2-6v^2\geq\sqrt2(9uv^2-w^3),$$
which is a linear inequality of $w^3$,
which says that it's enough to prove our inequality for an extr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2386218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating limits Evaluate the limit without L’Hôpital rule:
$$
\lim_{x \to 0}\frac{\sin^2{x}+2\ln\left(\cos{x}\right)}{x^4}
$$
My work is:
\begin{align}
L&=\lim_{x \to 0}\frac{\sin^2{x}-x^2}{x^4}+\lim_{x \to 0} \frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\lim_{x \to 0}\frac{\sin{x}-x}{x^3}
\lim_{x \to 0}\frac{\sin{x... | With Taylor expansion
$$\sin (x)=x-\frac {x^3}{6}+x^4\epsilon (x) $$
$$\sin^2 (x)=x^2-\frac {x^4}{3}+x^5\epsilon (x) $$
$$\cos (x)-1=-\frac {x^2}{2}+\frac {x^4}{24}+x^5\epsilon (x) $$
$$\ln (X+1)=X-\frac {X^2}{2}+X^2\epsilon (X) $$
$$\ln (\cos (x))=\ln \Bigl(\cos (x)-1+1\Bigr) $$
$$=-\frac {x^2}{2}+\frac {x^4}{24}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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Show that the roots of the equation $x^2-4abx+(a^2+2b^2)^2=0$ are imaginary. Show that the roots of the equation $x^2-4abx+(a^2+2b^2)^2=0$ are imaginary.
My Attempt:
$$x^2-4abx+(a^2+2b^2)^2=0$$
Comparing above equation with $Ax^2+Bx+C=0$, we get
\begin{align}
A&=1 \\
B&=-4ab \\
C&=(a^2+2b^2)^2
\end{align}
Now,
\begin{a... | If $a=b=0$ and the root is $0$.
However, if $(a,b) \neq 0$, then
the discriminant is equal to $-4(a^4+4b^4)<0$, hence the roots are imaginary.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solution of first order nonlinear differential equation without inverse function The following differential equation is part of a path integral problem in quantum mechanics:
$$(\partial_{r} y)^2 = y^4 (-1+p y -q^2 y^2)^{3/2}$$
with $p>0$ and $q>0$ and $p,q \in \mathbb{R}$
Additionally it is required, that
$$\lim_{r\to ... | I have an example for a simplification:
$$\int y^{-2} (-1+p y -q^2 y^2)^{-3/4} dy = \int y^{-2} \left(1-\frac{p^2}{4 q^2}+\left(q y-\frac{p}{2 q}\right)^2\right)^{-{3/4}} dy $$
now we substitute
$x = q y-\frac{p}{2 q}$ with $dy = \frac{1}{q} dx$ which leads to
$$\int y^{-2} (-1+p y -q^2 y^2)^{-3/4} dy = q \int \left( x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The Sum of 5 Consecutive Integers is 505, what is the Third number in this Sequence? So I've never covered this and I'm a bit confused.
$$x + (x+1) + (x+2) + (x+3) + (x+4)$$
So $5x + 10 = 505$ which equals $x = 99$. So isn't the third integer $2$? Which would mean that it equals to $1 + 2 + (99) = 102$? Why is the corr... | In overly pedantic detail, you might think of the problem this way:
We have five consecutive integers that add to 505. Let the five numbers be $a$, $b$, $c$, $d$, and $e$. Since they are consecutive integers, we may assume that they are ordered $a < b < c < d < e$, from which it follows that
\begin{align}
b &= a+1 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding the number of divisors of $N=2^7\times3^5\times5^3$ Find the number of divisors of $N=2^7\times3^5\times5^3$ which are of the form $4t+1$, where $t$ is a natural number.
I understood the fact, that $5$ can be written in the form of $4t+1$. Also, every even power of $3$ can be written in the form of $4t+1.$
for ... | We multiply instead of adding because we're counting the number of ways to mix and match elements from a set of $4$ with elements from a set of $3$. Watch:
$5^03^0, 5^03^2,5^03^4$
That's $3$.
$5^13^0, 5^13^2,5^13^4$
That's another $3$.
$5^23^0, 5^23^2,5^23^4$
Another $3$.
$5^33^0, 5^33^2,5^33^4$
Another $3$.
That's $3$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Isosceles triangle inscribed in an ellipse.
Find the maximum area of an isosceles triangle, which is inscribed inside an ellipse $\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$ with its (unique) vertex lying at one of the ends of the major axis of the ellipse.
The three vertices of the triangle would be $(a,0), (x,y), ... | Everything is correct until you use the value of $x$ to calculate the area.
You should have $$A^2=(-\frac{3a}{2})^2b^2(1-\frac 14)$$ which will give you the correct answer $$A=\frac{3\sqrt{3}}{4}ab$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solve the equation $\sqrt{x^3+1}\left(4x-1\right)=2x^3+x^2+1$ Solve the equation $$\sqrt{x^3+1}\left(4x-1\right)=2x^3+x^2+1$$
$$\Leftrightarrow \sqrt{x^3+1}=\frac{2x^3+x^2+1}{4x-1}$$
$$\Leftrightarrow \sqrt{x^3+1}-\left(x+1\right)=\frac{2x^3+x^2+1}{4x-1}-\left(x+1\right)$$
$$\Leftrightarrow \frac{x^3+1-\left(x+1\right... | after squaring rearranging and factorizing we get
$$-(x-2) x (x+1) \left(4 x^3-8 x^2+9 x-4\right)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $x+y+z+w=29$ where x, y and z are real numbers greater than 2, then find the maximum possible value of $(x-1)(y+3)(z-1)(w-2)$
If $x+y+z+w=29$ where x, y and z are real numbers greater than 2, then find the maximum possible value of $(x-1)(y+3)(z-1)(w-2)$.
$(x-1)+(y+3)+(z-1)+(w-2)=x+y+z+w-1=28$
Now $x-1=y+3=z-1=w-... | $x=8, y=4, z=8, w=9$ is the solution.
edit:
your solution is also correct. I don't know why are you multiplying $6∗10∗6∗5$?
$x−1=y+3=z−1=w−2=7$ means 7*7*7*7=2401
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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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.