Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Finding $x^3 + y^3$ when $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $ $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $
$x^3 + y^3 =?$
my answer =
$(3 + \sqrt5)^3 = 47 + 32\sqrt5$
$(3 - \sqrt5)^3 = 47 - 32\sqrt5$
$x^3 + y^3 = \frac{47 + 32\sqrt5}{47 - 32\sqrt5} + \frac{47 - 32\sqrt5}{... | Numerator:
$$(3-\sqrt5)^6+(3+\sqrt5)^6=(14-6\sqrt5)^3+(14+6\sqrt5)^3=8\,(7-3\sqrt5)^3+8(7+3\sqrt5)^3\\
=16\,(7^3+3\cdot7\cdot9\cdot5)=16\cdot7\cdot184.$$
Denominator:
$$(3-\sqrt5)^3(3+\sqrt5)^3=4^3.$$
Ratio:
$$322.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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find minimum of maximum of two functions
I need to find the angle $\theta$ so that: $$\max(\cos^2(\theta),1-\cos^2(45-\theta))$$ is minimized.
OK, so I wrote \begin{align*}f(\theta)&=\max(\cos^2(\theta),1-\cos^2(45-\theta))\\
&=\frac{\cos^2(\theta)+1-\cos^2(45-\theta)+|\cos^2(\theta)-1+\cos^2(45-\theta)|}{2}\end{ali... | A little nudge in the right direction (hopefully!):
Now, using the required formula (which?), we can simplify: \begin{align} 1 - \cos^2(45^\circ - \theta) & = \sin^2(45^\circ - \theta) \\ & = \frac{1 - \sin 2\theta}{2}\end{align}
We are now required to find: $$\min(\max(\frac{1+\cos 2\theta}{2}, \frac{1-\sin 2\theta}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For positive real numbers $a,b,c$ prove that $ a^4 + b^4 + c^4 \ge abc(a+b+c)$ For positive reals $a,b,c$ prove that $$ a^4+b^4+c^4 \ge abc(a+b+c). $$
I tried to pls around trying to reorganize to get AM-GM but i couldn't
Thanks for the help in advance.
| \begin{eqnarray*}
&(x^2-y^2)^2+(y^2-z^2)^2+(z^2-x^2)^2 \\&+2(z^2-xy)^2 +2(x^2-yz)^2+2(y^2-zx)^2\geq 0
\end{eqnarray*}
now divide by $4$ and rearrange.
| {
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"timestamp": "2023-03-29T00:00:00",
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A hard inequality indian olympiad problem If $x,y,z$ are positive real numbers, prove that:
$\left(x+y+z\right)^2\left(yz+xz+xy\right)^2\le 3\left(y^2 + yz + z^2\right)\left(x^2 + xz + z^2\right)\left(x^2 + xy + y^2\right)$.
I have been stuck in it. It is an Indian Olympiad problem. Can you guys help me out, please?
| $\quad3\left(y^2 + yz + z^2\right)\left(x^2 + xz + z^2\right)\left(x^2 + xy + y^2\right)-\left(x+y+z\right)^2\left(yz+xz+xy\right)^2\\=\sum_{sym} \left(2x^4y^2z^0+0.5x^3y^3z^0+0.5x^4y^1z^1-2x^3y^2z^1-x^2y^2z^2\right)\\=2\sum_{sym} \left(x^4y^2z^0-x^3y^2z^1\right)+0.5\sum_{sym} \left(x^3y^3z^0-x^2y^2z^2\right)+0.5\sum_{... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that there are only 3 integer solutions to $2^a+3 ^b=5^c$ I have figured out that $a \pmod{2} = b\pmod{2} =c \pmod{2} $ by making a few modulo tables and using that both sides of the equation must be divisible by $5$, and I have found the three solutions, I'm just at a loss as to how to prove that those are the o... | Let's try out some values of $a$. The tl, dr: we are forced $(a,b,c)=(1,1,1)$ when $a=1$ and to $(a,b,c)=(2,0,1)$ when $a=2$; then by using Pythagorean-triple-like solutions we fi n.v d these constraints eliminate all possibilities for larger $a$ except $(a,b,c)=(4,2,2)$.
$a=0$:
We can't have a power of $3$ and a pow... | {
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"timestamp": "2023-03-29T00:00:00",
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show this inequality $\sum_{cyc}\frac{a^3}{a^2+ab+b^2}\ge\sqrt{\sum a^3}$ let $a,b,c>0$.such $a+b+c=1$ show that
$$\sum_{cyc}\dfrac{a^3}{a^2+ab+b^2}\ge \sqrt{a^3+b^3+c^3}$$
I have show that not stronger inequality:
$$\sum\dfrac{a^3}{a^2+ab+b^2}=\sum_{cyc}\dfrac{a^4}{a^3+a^2b+ab^2}\ge \dfrac{(a^2+b^2+c^2)^2}{\sum (a^3+a... | Also, we can use SOS here.
Indeed, we need to prove that:
$$\sum_{cyc}\frac{a^3}{a^2+ab+b^2}-\frac{a^2+b^2+c^2}{a+b+c}\geq\sqrt{\frac{a^3+b^3+c^3}{a+b+c}}-\frac{a^2+b^2+c^2}{a+b+c}$$ or
$$\frac{(ab+ac+bc)\sum\limits_{cyc}(a^4b^2+a^4c^2-a^3b^2c-a^3c^2b)}{(a+b+c)\prod\limits_{cyc}(a^2+ab+b^2)}\geq$$
$$\geq\frac{(a^3+b^3+... | {
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"timestamp": "2023-03-29T00:00:00",
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Three real roots of $8x^3 – ax^2 + bx – 1 = 0$ in G.P. The equation $8x^3 – ax^2 + bx – 1 = 0$ has three real roots in G.P. If $λ_1 ≤ a ≤ λ_2$, then find ordered pair $(λ_1, λ_2)$.
My approach $f(x)=8x^3 – ax^2 + bx – 1 $
$f'(x)=24x^2 –2ax + b$
For real root
${(4a^2-96b)}>0$
Roots of $f'(x)$ are
$T=\frac{2a+\sqrt{(4a... | Suppose that the 3 roots are $k/r, k, kr$. Their product $k^3$ is equal to $1/8$. Therefore, $k = 1/2$.
Their sum is $a/8 = k(1/r+1+r)$. Hence $1/r+1+r = a/4$.
Finally, $k^2(1/r+1+r)= b/8$, which implies $(1/r+1+r)=b/2$.
If a solution exists, we must have $a =2b$ and $r^2 + (1-b/2) r+1=0$. If this last equation has rea... | {
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If $~\lim_{x\to\infty}\sqrt{64x^2 + ax + 7} - 8x + b = \frac32~ $, find $~ a+b=~?$ If $$~\lim_{x\to\infty}\sqrt{64x^2 + ax + 7} - 8x + b = \frac32~ ,$$ find $~ a+b=~?$
I get $a-16b = 24$,
But how to get $a+b$?
a,b is positive integer
| $$
\sqrt{64x^2 + ax + 7} =\sqrt{(8x)^2 + ax + 7} =\sqrt{(8x)^2 + 2\times(?)\times(8x)+ 7} \\
$$note that $$2\times(?)\times(8x)=ax \to ?=\frac{a}{16}$$so
$$\sqrt{(8x)^2 + 2(\frac{a}{16})(8x)+ 7}=\sqrt{(8x+\frac{a}{16})^2 -(\frac{a}{16})^2+ 7}\\x\to \infty \sqrt{(8x+\frac{a}{16})^2 -(\frac{a}{16})^2+ 7}\sim |8x+\frac{a}... | {
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Can we represents $k$ in partition of consecutive positive integer? Let $a,b$ are non-negative integers.
Let define
$$S(a)=1+2+3+...+a=\sum_{i=0}^{a} i$$
and $$S(b)=1+2+3+...+b=\sum_{i=0}^{b} i$$
Question
Show that each $k\in \mathbb{N}$ and $k \ne 2^t$ $ \forall t\in \mathbb{N} $ can be represent as
$$ k = S(a)-S(b... | You have shown above that it is impossible for any number of the form $2^t$ to be expressed as $S(a)-S(b)$. You have also shown that it is possible for all odd numbers. All you have to do is to extend the same idea for even values.
Case 1: The number is of the form $2^{t-1}m$ where $m \geqslant 2^t+1$ is odd
$$\frac{2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of $f(x, y)$ at $(0, 0)$ I need to show
$$\lim_{(x, y) \rightarrow (0,0)} \frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}} = 0
$$
Not really sure how to go about this without using the epsilon delta definition, which I would prefer not to. Any sort of help is appreciated.
Edit: I do have the inequality:
$$\frac{xy(x^2-y^2)}{... | Use polar coordinates: by letting $x=r\cos(\theta)$ and $y=r\sin(\theta)$, we have that
$$\frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}}= \frac{r^4\sin(2\theta)\cos(2\theta)}{2r^3}=\frac{r\sin(4\theta)}{4}.$$
Therefore
$$\left|\frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}}\right|\leq \frac{r}{4}$$
and the right-hand side goes to $0$ as $r\t... | {
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"timestamp": "2023-03-29T00:00:00",
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a) Find a base for $W_1 \cap W_2$ and b) Find a base for $W_1 + W_2$ Let
$$W_1 = \left\{ \begin{pmatrix}
a & b & c \\
-b & a & b \\
c & b & a \\
\end{pmatrix} :\, a, b, c \in \Bbb R \right\}$$
$$W_2 = \left\{
\begin{pmatrix}
a & 0 & 0 \\
b & 0 & c \\
c & b & a \\
\end{pmatrix} :\, a, b, c \i... | For b), rewrite what you obtained as
$$a \begin{pmatrix}
2& 0& 0\\
0 & 1 & 0 \\
0 & 0& 2 \\
\end{pmatrix}+b \begin{pmatrix}
0 & 1 & 0\\
0 & 0 & 1 \\
0 & 2 & 0 \\
\end{pmatrix}+c \begin{pmatrix}
0 & 0 &1\\
0 & 0 & 1 \\
2 & 0 & 0 \\
\end{pmatrix}$$
Can you say whether ... | {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $(x^2 - 11x + 29)^{(6x^2 + x - 2)}=1$? This question comes from PURE MATHEMATICS 1 for As and A levels. This question is part of exercise 1 and it has 5 Answers : $1/2, - 2/3, 4, 6 $ and $7$ . The first 2 values $1/2$ and $-
2/3 $ I am able to find but the rest I can't. I get this first 2 values by make 1... | It's simple
This equation is true when either $x^2 -11x+29=1$ or $6x^2+x-2 =0$ or $x^2-11x+29=-1 \ \ and \ \ 6x^2+x-2= even$ because $1^k=1$ and $n^0=1$ and $(-1)^{even}=0$
Solve the equations
You have solved second equation so I'll leave that
For the first you will get $$x^2 -11x+28=0$$
$$(x-7)(x-4)=0$$
Solution i... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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how to solve $\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}$? $\displaystyle\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+1}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+n}}\right)^{n}$
left=$\displaystyle\lim_{n\to\infty}{\l... | We can use the sandwich theorem twice for solving the limit
$$\text { Observe that }
\frac{n}{\sqrt{n^2 +n}}\le\alpha_n=\sum_{k=1}^{n}\frac{1}{\sqrt{n^2 +k}}\le\frac{n}{\sqrt{n^2+1}}, \text{ So, } \lim_{n\to\infty}\alpha_n=1.\space
\text { Proceeding as an usual case of } 1^{\infty}, $$
$$ \alpha=\lim_{n\to\inft... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 2
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Factoring $3x^2+4x-4=0$ using the quadratic formula The correct answer is $(3x-2)(x+2)$ but I am getting $(x-\frac{2}{3})(x+2)$, why?
My calculations:
$x = \frac{-4\pm\sqrt{16-(-48)}}{6} = \frac{-4\pm8}{6}$, which gives the factors $(x-\frac{4}{6})(x+\frac{12}{6}) = (x-\frac{2}{3})(x+2)$.
| You are confusing factoring a quadratic with solving a quadratic.
The quadratic $ax^2 + bx + c=0$ will have the same solutions as $x^2 + \frac bax + ca = 0$ (in fact, that was the very first step in developing the quadratic formula) and will have the same as $dx^2 + \frac {db}ax + \frac {dc}a-0$ or any $akx^2 + bkx + c... | {
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"timestamp": "2023-03-29T00:00:00",
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Geometry Ratio in right triangle
In ABC, $\angle ACB=90$. Point O is chosen inside ABC such that [AOB]=[BOC]=[COA]. If OA=23 and OB=11, find $OC^2$.
Using the areas, we can find OD:OE:OF, where D, E, F are the feet of the altitudes from O to BC, AC, and AB, respectively. I have no idea how to proceed
| With $D$ and $E$ as stated, note that $OD=\frac{AC}3$, $OE=\frac{BC}3$, then $AE=\frac{2AC}3$, $BD=\frac{2BC}3$.
Applying Pythagorean theorem to triangles AEO and ODB gives $4AC^2+BC^2=9\times 23^2$ and $AC^2+4BC^2=9\times 11^2$
Adding these gives $5(AC^2+BC^2)=9\times 650$
Replacing $AC^2+BC^2$ with $AB^2$ and rearran... | {
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"url": "https://math.stackexchange.com/questions/3371256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Nice olympiad inequality :$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{3}{7}$ I have this to solve :
Let $x,y,z>0$ such that $x+y+z=3$ then we have :
$$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{3}{7}$$
I try to use Jensen's inequality but the function $f(x)=\fra... | Note that (tangent line trick)
$$\frac{5+2y}{49} - \frac{y^2}{4y^3+3} = \frac{(8y^2+36y+15)(y-1)^2}{49(4y^3+3)}.$$
Thus, we have
$$\frac{y^2}{4y^3+3} \le \frac{5+2y}{49}, \quad \forall y \ge 0.$$
Thus, we have
$$\sum_{\mathrm{cyc}} \frac{xy^2}{4y^3+3} \le \sum_{\mathrm{cyc}} \frac{x(5+2y)}{49}
= \frac{5(x+y+z) + 2(xy+... | {
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How to solve the following ordinary differential equation? Consider the ODE $$y' + x = \sqrt{x^2+y}.$$
I have no idea how to solve this equation. I have tried several ways but neither of them worked. Can anybody help me? Many thanks.
I have tried substituting $u=\sqrt{x^2+y}$, as well as squaring both sides of the ODE.... | This is a heuristic derivation of the complete solution showing a possible motivation chain. Notice that Jack D'Aurizio has given a special solution before.
How can we simplify the ODE?
$$y' + x = \sqrt{x^2+y}\tag{1}$$
The first thing to do would be to get rid the square root.
Hence the set $v^2 = x^2+y$ so that the r.... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can I prove that $4^{n} + 5$ is divisible by $3$. I have trying to prove that $4^{n} + 5$.
I've already proved the base case, so I'm working on the inductive step.
I've done the following:
$4^{n} + 5$
$4^{n+1} + 5$
$4*4^{n} + 5$
But I am unsure where to go from here to prove that it is divisible by 3 since I am un... | You can prove that using induction
We suppose that $4^n + 5 \equiv 3$ is true
Induction base: $n = 1 \Longrightarrow 4^1 + 5 = 9 \equiv 3$ - true.
Induction transition: let $k = n+1 \Longrightarrow 4^k + 5 = 4^{n+1}+5 = 4 \cdot 4^n + 5 = 3\cdot4^n + 4^n + 5$
$3 \cdot 4^n \equiv 3, \ \ 4^n + 5 \equiv 3$
Hence $4^{n+1}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the principal root of $\sqrt{-i}$? Where is the mistake in this solution?
$$\sqrt{-i}=\sqrt {\cos \frac{3 \pi}2+i\sin \frac{3 \pi}2}=\cos \frac{3 \pi}4+i\sin \frac{3 \pi}4=-\frac{\sqrt 2}2+i \frac{\sqrt 2}2$$
WA gives me different result:
$$\frac{\sqrt 2}2-i \frac{\sqrt 2}2$$
Why the principal root must be ... | Solve the quadratic equation
$$x^2=-i=e^{-i\pi/2}\implies x=e^{-i\pi/4+ik\pi},\; k=0,1$$
| {
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Prove the following triangle related inequality
Prove that $\dfrac32\leq \dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} < 2.$ where $a$ ,$b$ ,$c$ are sides of Triangle
I tried using sine rule which makes up the the expression $\sum \dfrac{\sin{A}}{\sin{B}+\sin{C}}$. Now i dont how to find the range of $\sum \dfrac{\sin... | Making $b = x a$ and $c=y a$ the conditions read
$$
\frac 32\le\frac{1}{x+y}+\frac{x}{1+y}+\frac{y}{1+x}\le 2 \ \ \text{s. t.}\ \ x + y\ge 1
$$
so making the lagrangian
$$
L(x,y,\lambda) = \frac{1}{x+y}+\frac{x}{1+y}+\frac{y}{1+x}+\lambda(x+y-1)
$$
We have two stationary points. One at $x^*=1, y^* = 1$ corresponding to... | {
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Calculate the value of $\frac{1}{2}+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+....+\frac{F_n}{2^n}+....$ The Fibonacci sequence starts with 1, 1, 2, 3, 5, 8, 13, ... .(Start from the 3rd term,
each term is the sum of the two previous terms). Let $F_n$ be the $n$th term of this sequence. $S$ is defined as $S=\frac{1}{2}+\fra... | Hint: $$F_n{={\frac {\Phi ^{n}-\Psi ^{n}}{\sqrt {5}}}={\frac {1}{\sqrt {5}}}\left(\left({\frac {1+{\sqrt {5}}}{2}}\right)^{n}-\left({\frac {1-{\sqrt {5}}}{2}}\right)^{n}\right)}$$
$$S = \sum_{n=1}^{\infty}\dfrac{F_n}{2^n}=\dfrac{1}{\sqrt{5}}\sum_{n=1}^{\infty}\left[\left(\dfrac{\Phi}{2}\right)^n- \left(\dfrac{\Psi}{2}\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Ramanujan's Nested Radical By noting Ramanujan's Nested Radical, we have
$3 = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$
On the other hand, we can manipulate the number $4$ by applying the similar principle. Here we have
$\begin{aligned} 4 & = \sqrt{16} \\ & = \sqrt{1+15} \\ & = \sqrt{1+2 \cdot \dfrac{15}{2}} \\ & ... | The generating function is
$$
f(x) = \sqrt{1+(x+1)f(x+1)}\Rightarrow f^2(x) = 1+(x+1)f(x+1)
$$
so making $f(x) = a x + b$ we get $a = 1, b= 2$ or
$$
f(x) = x + 2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Substitution integral with $t=x^3$
$$
\int \frac{3}{t(3+\sqrt[3]{t})} \mathrm{d} t . \text { Substitution } t=x^{3}
$$
$$
\int \frac{3}{x^3(3+\sqrt[3]{x^3})}3x^2
$$
$$
\int \frac{9}{x(3+{x})}
$$
$$
9\int \frac{1}{3x+{x^2}}
$$
Now i'm stuck
my calculation so far
| To finish your calculation:
$$\int\frac{9}{x(3+x)}$$
Take partial fractions
$$=3\int \frac{1}{x+3}dx+2\int\frac{1}{x}dx$$
Let $u=x+3,du=dx$ have
$$=3\int\frac{1}{u}du+2\int\frac{1}{x}dx$$
Then exists $C\in\mathbb{R},s.t.$
$$=3\ln(u)+2\ln(x)+C$$
Substitute $u$ back, we have
$$=3\ln(x+3)+2\ln(x)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, prove the following.
If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, show that $$(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2)=(1-q+s)^2+(p-r)^2.$$
I have tried putting
\... | Your approach ends up proving $(α^2 - 1)(β^2 - 1)(γ^2 - 1)(δ^2 - 1) = (1 - q + s)^2 - (p - r)^2$. This means we will have to do something with that minus sign and to do that we can use the identity $(x + ιy)(x - ιy) = x^2 + y^2$. Putting $P(ι)$, and $P(-ι)$ you can easily solve the problem with the same approach.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3390796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find a such that $ax^{17}+bx^{16}+1$ is divisible by $x^2-x-1$.
Find $a$ such that $ax^{17}+bx^{16}+1$ is divisible by $x^2-x-1$.
I tried taking the roots of the polynomial which are $\frac{1±\sqrt{5}}{2}$
And I got the equation $a(\frac{1±\sqrt{5}}{2})^{17}+b(\frac{1±\sqrt{5}}{2})^{16}+1=0$
Now I don't know what to ... | Hint. If $x=\frac{1±\sqrt{5}}{2}$ then $x^2=x+1$ and
\begin{align}
ax^{17}+bx^{16}+1 &=(ax+b)(x+1)^8+1 \\
&=(ax+b)(x^2+2x+1)^4+1\\
&=(ax+b)(3x+2)^4+1\\
&=(ax+b)(21x+13)^2+1.
\end{align}
Can you take it from here? Note that 3,2, 21,13 are all Fibonacci numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3391666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Can we simply further the expression? We have the set $\left \{z\in \mathbb{C}\mid z\neq 1, \ \left |\frac{z}{z-1}\right |\leq 1\right \}$
Let $z=x+yi$.
We have that \begin{align*}\frac{z}{z-1}&=\frac{x+yi}{(x-1)+yi}=\frac{\left (x+yi\right )\left ((x-1)-yi\right )}{\left ((x-1)+yi\right )\left ((x-1)-yi\right )} \... | It's much easier to rearrange to $|z|\le|z-1|$. Geometrically, the distance of $z$ from $0$ is less than or equal to the distance of $z$ from $1$ in the complex plane, so the complex numbers $z$ satisfying this are those with $\Re(z)\le\frac{1}{2}$, or $x\leq\frac{1}{2}$.
Alternatively, you can write $z=x+iy$ in $|z|\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to calculate $lcm(2,3,4,5,6,7,8,9,10)$ solely by mental arithmetic? Denote $lcm(a,b)$ the lowest common multiple of $a$ and $b.$
Question: How to calculate
$$lcm(2,3,4,5,6,7,8,9,10)$$
solely by mental arithmetic?
I have a way to calculate, but it relies on writing on paper.
We can calculate the lcm using re... | Since we have a list of consecutive integers starting from $2$, just pick the largest powers of a prime:
$$\text{lcm}(2,3,4,5,6,7,8,9,10)=\text{lcm}(5,7,8,9)=5\cdot 7\cdot 8 \cdot 9=63\cdot 40=2520.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Which is bigger 5^(√3) and 4^(√5)? How do you know which one is bigger between $5^\sqrt{3}$ and $4^\sqrt{5}$?
For my method I used in $2^\sqrt{3}$ and $3^\sqrt{2}$ I put both numbers in the function $f(x)=x^\sqrt{3}$ so $f(2^\sqrt{3})$ become $2^3$ which is equal to 8 and $f(3^\sqrt{2})$ become $3^\sqrt{6}$ and $3^\sqr... | We make a power of $\sqrt{3}$ both expression:
$\left(5^\sqrt{3}\right)^\sqrt{3}=5^3=125$
$\left(4^\sqrt{5}\right)^\sqrt{3}=4^\sqrt{15}>4^{3.5}=4^3\times 2=128>125$
Therefore, $4^\sqrt{5}>5^\sqrt{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3399681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Primitive instances where $c^3|(a^3+b^3)$ with $a,b,c\in\mathbb{N}$ It is known that by Fermat's Last Theorem there are no solutions to $a^3+b^3=c^3$ for $a,b,c\in\mathbb{N}$. I wondered about how multiplying the $c^3$ by a constant would change this fact.
Accordingly, I have been looking into instances where $c^3|(a^3... | To show that there are infinitely many non-trivial solutions, it suffices to consider solutions of the following form, for $n=0,1,2\dots$
$\qquad a=3$
$\qquad b=24n+5$
$\qquad c=2$
$\qquad d=(3n+1)[9-3(24n+5)+(24n+5)^2]=1728n^3+1080n^2+225n+19$
Note that:
$$a^3+b^3 = (a+b)(a^2-ab+b^2)=(3+24n+5)[9-3(24n+5)+(24n+5)^2]$$
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Equilateral pentagon with increasing/equal angles
Suppose all sides of a convex pentagon ABCDE have the same size, and $\angle A \ge \angle B \ge \angle C \ge \angle D \ge \angle E $. Prove that this pentagon is a regular pentagon.
I know this should be proved using contradiction, but I'm not sure how to reach that p... | WOLOG, we only need to consider the case where the side of the equilateral pentagon is $1$ and the vertices $A,B,C,D,E$ are ordered counterclockiwisely
on circumference of the pentagon.
Let $\alpha,\beta,\gamma,\delta,\epsilon$ be the external angles at $A, B, C, D, E$. We have
$$\angle A \ge \angle B \ge \angle C \ge ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
help solving for delta in epsilon delta proof $$\lim_{x\to 9} \frac{1}{\sqrt{x}} = \frac{1}{3} $$
so the two statements are
$$0<|x-9|<\delta$$
$$\left|\frac{1}{\sqrt{x}}-\frac{1}{3}\right|<\epsilon$$
I've tried to multiply by the conjegate to get
$$\frac{\frac{1}{x}-\frac{1}{9}}{\sqrt{\frac{1}{x}}+\frac{1}{3}}<\epsilo... | We have that
$$\left|\frac{1}{\sqrt{x}}-\frac{1}{3}\right|=\left|\frac{\sqrt{x}-3}{3\sqrt{x}}\right|=\left|\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{3\sqrt{x}(\sqrt{x}+3)}\right|=\left|\frac{x-9}{3\sqrt{x}(\sqrt{x}+3)}\right|$$
then assuming wlog $|x-9|<1 \implies 3\sqrt{x}(\sqrt{x}+3)\ge 3\sqrt{8}(\sqrt{8}+3)$ therefore
$$\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3403270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove that $\lim_{x\to0, y\to0}\frac{x^3+y^3}{x^2+y^2}=0$ Prove that $\lim_{x\to0, y\to0}\frac{x^3+y^3}{x^2+y^2}=0$
My Try
| HINT
Notice that
\begin{align*}
|x|^{3} = |x|\times x^{2} \leq |x|(x^{2} + y^{2}) \Rightarrow \left|\frac{x^{3}}{x^{2} + y^{2}}\right| \leq |x|
\end{align*}
Based on the same procedure, you can prove that
\begin{align*}
|y|^{3} = |y|\times y^{2} \leq |y|(x^{2} + y^{2}) \Rightarrow \left|\frac{y^{3}}{x^{2} + y^{2}}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3406077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
formula for $\left(1\cdot2\cdot...\cdot k\right)+...+\left(n\left(n+1\right)...\left(n+k- 1\right)\right)$ proof if $k$ is a constant then for every $n$ natural numbers we have:
$$\left(1\cdot2\cdot...\cdot
k\right)+\left(2\cdot3\cdot...\cdot\left(k+1\right)\right)+...+\left(n\left(n+1\right)...\left(n+k-
1\right)\rig... | We can prove this by induction. The base case here is when $n=1.$ Clearly,
$n\cdot(n+1)\cdot(n+2)\cdot\dots\cdot(n+k-1)=1\cdot(2)\cdot(3)\cdot\dots\cdot(k)$
and
$\dfrac{(n)\cdot(n+1)\cdot(n+2)\cdot\dots\cdot(n+k)}{1+k}=(1)\cdot(2)\cdot(3)\cdot\dots\cdot(k)=n\cdot(n+1)\cdot(n+2)\cdot\dots\cdot(n+k-1)$
so the base cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find $\lim_{x \to \ 0}{\frac{1-\sqrt{\cos x}}{1-\cos\sqrt{x}}}$ without using the l'Hospital rule The task is to evaluate
$$\lim_{x \to \ 0}{\frac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}}.$$
I tried to use some trigonometric identities such as
$$\lim_{x \to \ 0}{\frac{1-\sqrt{\cos\left(x\right)}}{1-\cos\left(\sqrt{x}\right)... | We have that
$$\frac{1-\sqrt{\cos\left(x\right)}}{1-\cos\left(\sqrt{x}\right)}=
\frac{1-\sqrt{\cos\left(x\right)}}{x}\frac{1+\sqrt{\cos\left(x\right)}}{1+\sqrt{\cos\left(x\right)}}
\frac{x}{1-\cos\left(\sqrt{x}\right)}=$$
$$=x\frac{1-\cos\left(x\right)}{x^2}\frac{1}{1+\sqrt{\cos\left(x\right)}}
\frac{x}{1-\cos\left(\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
If $f(x) = (x-a)^3(x-b)^3$ then what is the nature of the roots of $f^{\prime\prime}(x) = f^\prime (x)$? If we try solving it by finding $f''(x)$ then it is very long and difficult to do, so my teacher suggested a way of doing it, he said find nature of all the roots of $f(x) =f'(x)$, and on finding nature of the roots... | Recall the Leibnitz differentiation formula
$$(fg)^{(n)}=\sum_{k=0}^n f^{(k)} g^{(n-k)}$$
Then
$$\left( (x-a)^3(x-b)^3 \right)'=3(x-a)^2(x-b)^3+3(x-a)^3(x-b)^2=3(x-a)^2(x-b)^3[x-a+x-b]$$
$$\left( (x-a)^3(x-b)^3 \right)''=6(x-a)(x-b)^3+18(x-a)^2(x-b)^2+6(x-a)^3(x-b)=6(x-a)(x-b)[(x-a)^2+3(x-a)(x-b)+(x-b)^2]
$$
Equating... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Show that $\frac{3\> + \>\cos x}{\sin x}$ cannot have any value between $-2\sqrt2$ and $2\sqrt2$ Show that
$$\dfrac{3+\cos x}{\sin x}\quad \forall \quad x\in R $$
cannot have any value between $-2\sqrt{2}$ and $2\sqrt{2}$.
My attempt is as follows:
There can be four cases, either $x$ lies in the first quadrant, secon... | Let the value of this expression at $x=p$ be $q$. Then we have
$${3+\cos p\over\sin p}= q$$
$$\implies\cos p-q\sin p = -3$$
Dividing both LHS and RHS by $\sqrt{q^2+1}$ we have
$$\cos p\cdot {1\over\sqrt{q^2+1}} - \sin p\cdot{q\over\sqrt{q^2+1}} = {-3\over\sqrt{q^2+1}}$$
Let $1\over\sqrt{q^2+1}$ be $\cos r$. So, we hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Coincidence in Diophantine solution parametrization for Pythagorean triples etc Consider a triple of nonnegative integers $(a, b, c)$ such that $c^2 = a^2 + b^2$. This can be viewed as integer triangles with sides $(a, b, c)$ such that $c$ is the side opposite a $90°$ angle. Such triples are well-known as Pythagorean t... | The example I like to show is solving
$$ 2(x^2 + y^2 + z^2) - 113(yz + zx + xy)=0, $$ four "recipes," all made up of binary quadratic forms
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
37 u^2 + 51 uv + 8 v^2 \\
8 u^2 -35 uv -6 v^2 \\
-6 u^2 + 23 uv + 37 v^2
\end{array}
\right)
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve : $\sin\left(\frac{\sqrt{x}}{2}\right)+\cos\left(\frac{\sqrt{x}}{2}\right)=\sqrt{2}\sin\sqrt{x}$ Solve : $\sin\left(\dfrac{\sqrt{x}}{2}\right)+\cos\left(\dfrac{\sqrt{x}}{2}\right)=\sqrt{2}\sin\sqrt{x}$
$$\dfrac{1}{\sqrt{2}}\cdot\sin\left(\dfrac{\sqrt{x}}{2}\right)+\dfrac{1}{\sqrt{2}}\cos\left(\dfrac{\sqrt{x}}{2}\... | First we square both sides to get
$$
\begin{align*}
\left(\sin\left(\frac{\sqrt{x}}{2}\right)+\cos\left(\frac{\sqrt{x}}{2}\right)\right)^2&=\left(\sqrt{2}\sin\left(\sqrt{x}\right)\right)^2,\\
\sin^2\left(\frac{\sqrt{x}}{2}\right)+\cos^2\left(\frac{\sqrt{x}}{2}\right)+2\sin\left(\frac{\sqrt{x}}{2}\right)\cos\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
A general formula to generate functions of power series I have been playing with Maclaurin series lately, I have been able to come across this:
$\dfrac{1}{1+x}=1-x+x^2-x^3+x^4-x^5...$
$\dfrac{1}{(1+x)^2}=1-2x+3x^2-4x^3+5x^4-6x^5+7x^6...$
I found out by accident that:
$\dfrac{1-x}{(1+x)^3}=1-2^2x+3^2x^2-4^2x^3+5^2x^4+6^... | We are looking for polynomials $p_k(x)$ with
\begin{align*}
\frac{p_k(x)}{(1+x)^{k+1}}=\sum_{j=0}^\infty (-1)^j(j+1)^kx^j\qquad\qquad k\geq 0
\end{align*}
We can find $p_k(x)$ as follows:
\begin{align*}
\color{blue}{p_k(x)}&=\left(\sum_{j=0}^\infty(-1)^j(j+1)^kx^j\right)(1+x)^{k+1}\\
&=\left(\sum_{j=0}^\infty(-1)^j(j+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3422095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
how to factorize $a^3 - ab^3 + a^2 + b^2 + 1$ I've got the following question in a book about maths olympiad
factorize $a^3b - ab^3 + a^2 + b^2 + 1$
and the answer to this question is
$a^3b - ab^3 + a^2 + b^2 + 1$$= a^3b - a^2b^2 + ab + a^2b^2 - ab^3 + b^2 - ab + + a^2+ 1$ $= ab(a^2 - ab +1) + b^2 (a^2 - ab + 1) + ... | Another way to guesstimate is to try $b=1$ and $a=1$:
$$a^3b - ab^3 + a^2 + b^2 + 1=a^3+a^2-a+2=(a^2-a+1)(a+2)=(a^2-a+\color{red}1)(a+1+\color{red}1)\\
a^3b - ab^3 + a^2 + b^2 + 1=-b^3+b^2+b+2=(2-b)(b^2+b+1)=(1-b+\color{red}1)(b+b^2+\color{red}1)$$
Hence:
$$a^3b - ab^3 + a^2 + b^2 + 1=(a^2-ab+1)(ab+b^2+1)$$
Note: $\col... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3422724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Determinant of a nondegenerate quadratic form Given a quadratic form $Q$ over a space $V$.
To put it simply, we say $Q$ is nondegenerate if for every $x\in V$ there exists at least one $y\in V$ such that $x.y\neq 0$ (where '$.$' is a generalization of the inner product, and $x.y=\frac{1}{2}(Q(x+y)-Q(x)-Q(y)$).
Equiv... | Let's put the focus on the matrix $A$. We have
$$Q(x) = x^T A x.$$
Suppose $Q(x) = 0$. This means $x^T (A x) = 0$, or equivalently, $x$ is orthogonal (under the standard inner product) to $Ax$. Note that this doesn't require that $Ax = 0$ - it only requires that $Ax$ be a vector which lives on the perpendicular subspa... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$| \dfrac{-z+\sqrt{z^2-4}}{2}|\le 1$ Suppose $z=u+iv$, with $v>0$.
$$ m(z)=\dfrac{-z+\sqrt{z^2-4}}{2}$$ satisfies the equation $$m(z)+1/m(z)+z=0 $$ from which we found that $$|m(z)|=|m(z)+z|^{-1} $$
Take the branch of the square root so that the imaginary part of $m(z)$ is postive.
($m(z) $ here is the Stieljes Transfo... | Let $z=a+ib,\sqrt{z^2-r^2}=c+id$ where $a,b,c,d$ are real and $r>0$
$$r^2=z^2-(z^2-r^2)=(a+ib)^2-(c+id)^2=a^2+d^2-b^2-c^2+2i(ab-cd)$$
Equating the imaginary parts, $ab-cd=0\implies\dfrac ac=\dfrac db=k$(say)
$\implies a=ck,d=bk$
Equating the real parts, $$r^2=a^2+d^2-b^2-c^2=(b^2+c^2)(k^2-1)$$
$\implies k^2-1=\dfrac{r... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Compute $\int_0^{\pi/2} x^2\left(\sum_{n=1}^\infty (-1)^{n-1} \cos^n(x)\cos(nx)\right)dx$ How to prove
$$I=\int_0^{\pi/2} x^2\left(\sum_{n=1}^\infty (-1)^{n-1} \cos^n(x)\cos(nx)\right)dx=\frac16\left(\frac{\pi^3}{12}-\pi\operatorname{Li}_2\left(\frac13\right)\right)$$
This problem is proposed by Cornel which can be f... | We can use the following fourier series:$$\frac{1}{a+b\cos t}=\frac{1}{\sqrt{a^2-b^2}}+\frac{2}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos{(nt)},\ a>b$$
Plugging $a=5, b=3$ and $t=2x$ we get:
$$\frac{1}{1+3\cos^2 x}=\frac{2}{5+3\cos(2x)}=\frac{1}{2}+\sum_{n=1}^\infty (-1)^n\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3428840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Calculating the integral $\iint_D\frac{x}{\sqrt{x^2+y}}\,dx\,dy$ I wish to calculate the following integral:
$$\iint_D\frac{x}{\sqrt{x^2+y}} \, dx \, dy$$
where $D$ is the area where $y>0,0<x<1,y<x^2$
I was asked to calculate twice, first integrating by $x$ and then by $y$, and then again the by opposite order.
Here is... | *
*gives you $$\frac{2}{3} \left(2^{3/2}- \frac{2^{3/2}}{2}\right) - \frac{2}{3}$$
It looks different, but can be simplified to $$\frac{2}{3}\left(\frac{2.2^{3/2} - 2^{3/2}}{2}\right) - \frac{2}{3} = \frac{2}{3} \frac{2^{3/2}}{2} - \frac{2}{3} = \frac{2}{3} \sqrt 2 - \frac{2}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int_{0}^{\infty} \frac{x^4}{(1+2x^2)^4} dx$ Evaluate :$$ \int_{0}^{\infty} \frac{x^4}{(1+2x^2)^4} dx$$
using residue theorem. The integral is even and it can be written as :
$$ \frac{1}{2} \int_{-\infty}^{\infty} \frac{x^4}{(1+2x^2)^4} dx$$
The only pole with a positive imaginary part is $ \frac{i}{\sqrt2} $... | You can instead find the laurent series, and pick up the coefficient of the $z^{-1}$ term.
i.e. $(1+2z^2)^{-4} = (1+2(z-\frac{i}{\sqrt2})^2-2(\frac{-1}{2})+4\frac{iz}{\sqrt2})^{-4}= (1+2(z-\frac{i}{\sqrt2})^2+1+2\sqrt2iz)^{-4} = (2\sqrt{2}i(z-\frac{i}{\sqrt2})+2\sqrt2i\frac{i}{\sqrt2}+2+2(z-\frac{i}{\sqrt2})^2)^{-4}=(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find $4x \equiv 7 \pmod{15}$ and $3x \equiv 5 \pmod{16}$ (different exercises) This is how I solved each:
$$4x \equiv 7\pmod {15} \\
4x - 7 \equiv 0\pmod{15} \\
4x-7 = 15k \Leftrightarrow 4x-15k= 7 \\
$$
$$15 = 4*3+3 \\
4=3*1+1\\
3=3*1+0$$
$$1 = 4-3*1 \\
3 = 15-4*3 \\
1 = 4 - (15-4*3)*1 \\
1 = 4-15+4*3 \\
1 = 4*4-15*1... | To correctly pattern-match $\,\color{#c00}x\,$ in the above $\,\rm\color{#0a0}{scaled}\,$ Bezout equations do as follows.
$\qquad\ \begin{align}
\ -7\cdot \color{#90f}{15}&\ +\ 4\cdot 7\cdot 4\,\ =\ \color{#0a0} 7\ \ [=\ \color{#0a0}{7\ {\rm times}} {\rm \ Bezout\ of\ } \gcd(15,4)=1\,]\\[.2em]
\Rightarrow\ \bmod \co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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How to prove $\sum\limits_{k=0}^{N} \frac{(-1)^k {N \choose k}}{(k+1)^2} = \frac{1}{N+1} \sum\limits_{n=1}^{N+1} \frac{1}{n}$ In the process of proving a more complicated relation, I've come across the following equality that I'm having trouble proving:
$$
\sum\limits_{k=0}^{N} \frac{(-1)^k {N \choose k}}{(k+1)^2} = \f... | Introduce the function
$$f(z) = (-1)^n \times n! \times \frac{1}{(1+z)^2}
\prod_{q=0}^n \frac{1}{z-q}.$$
Then we have for $0\le k\le n$
$$\mathrm{Res}_{z=k} f(z)
= (-1)^n \times n! \times \frac{1}{(1+k)^2}
\prod_{q=0}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^{n} \frac{1}{k-q}
\\ = (-1)^n \times n! \times \frac{1}{(1+k)^2}
\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3433059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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How to compute $\lim\limits_{x \to a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}$?
How to compute $\lim\limits_{x \to a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}$?
My process:
\begin{align}
\lim\limits_{x \to a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}... | Similar to @user's answer, starting with $x=a y$, we have
$$A=\frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}=a\frac{ \sqrt[3]{y}-\sqrt{y \left(2-y^3\right)}}{y^{3/4}-1}$$ Now, let $y=1+t$ to get
$$A=a\frac{\sqrt[3]{1+t}-\sqrt{1-2 t-6 t^2-4 t^3-t^4}}{(1+t)^{3/4}-1}$$ Now, using Taylor series arou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3434383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How big is the smallest triangle inscribed in a square A square is divided in 7 areas as show on the figure.
The dots show the corners of the square and the middle points on the edges.
How large a fraction is area $D$ and how do I work it out?
I have tried using trigonometry to calculate the area A. If we say each sid... | Coordinate geometry can also work out area $D$. If the bottom left corner is at the origin, and the square has a side length of $1$, the equations of the three lines are:
$$y = 1-x \tag{1}$$
$$y = \frac{1}{2} + \frac{1}{2}x \tag{2}$$
$$x = \frac{1}{2} \tag{3}$$
Solving $(1) = (2)$ gives the point $\left(\frac{1}{3}, \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3435395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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condition for pair of straight line equation While determining the condition for the pair of straight line equation $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ i.e $\quad$$ax^2+2(hy+g)x+(by^2+2fy+c)=0$
$$x=\frac{-2(hy+g)}{2a}\pm\frac{\sqrt{(hy+g)^2-a(by^2+2fy+c)}}{a}
$$$$x=\frac{-2(hy+g)}{2a}\pm\frac{\sqrt{(h^2-ab)y^2+2(hg-af)y+(g... | If the condition
$$
a x^2 + 2 h x y + b y^2 + 2 g x + 2 f y + c = 0
$$
represents the product of two lines then the set of solutions for this conditions should be one solution point, void or infinite solution points associated to the cases in which we have the two lines intersection, two lines parallel and two lines co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3435685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find x and y such that $xx...x6yy...4=k^2$. Find x and y such that $N=xx...x6yy...4=k^2$, where number of x and y is n.
Solution:
Supose n=1 and N=x6y4 and we must have:
$(x6y)\times 10 +4=k^2$
⇒ $(x6y)\times 10=(k-2)(k+2)$
That is the last digit of (k+2) or (k-2) must be zero, let $k+2 ≡0 \ mod 10$ ⇒ $k ≡8 \mod 10$
L... | This is an interesting problem but so far I only have a partial answer.
$k$ must be of the form $10l+2$ or $10l+8$. Let's look at the second of these possibilities.
$$(10l+8)^2=\frac {10^n-1}{9}10^{n+2}x+\frac {10^n-1}{9}10y +6\times 10^{n+1}+4$$
$$100l^2+160l+64=\left(\frac {10^n-1}{9}\right)\left (10^{n+2}x+10y+540\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factorial Digits Sum
A natural number is a factorion if it is the sum of the factorials of
each of its decimal digits. For example, $145$ is a factorion because
$145 = 1! + 4! + 5!$. Find every $3$-digit number which is a factorion.
My solution:
We can obviously only have factorials $1$–$6$.
*
*If we have a ... | Brainstorming:
As you pointed out. If one of the digits of $N$ is $7$ then $7! > 1000> N$ which is impossible.
If one of the digits is $6$ then $N > 6! = 720$ so one of the digits is at least $7$ which we just showed was impossible.
Also if all three digits, call them $a,b,c$, are all $< 5$ then $a!, b!,c! \le 4! =2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3439887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the non-linear congruence $x^3+2x^2+5x+4\equiv 0\pmod {60}$. Solve the non-linear congruence $x^3+2x^2+5x+4\equiv 0\pmod {60}$.
For this question, I think I might need to use the Chinese remainder Theorem to simplify the problem to several smaller moduli (e.g. possibly $\pmod 4,\pmod 5,\pmod 3$ and simplify the ... | Modulo $4$ we have $x^3+2x^2+x=x(x+1)^2\equiv0$,
which means $x\equiv0$ or $x+1\equiv2$ or $4$.
Modulo $5$ we have $x^3+2x^2+5x+4=(x+1)(x^2+x+4)\equiv(x+1)(x-2)^2\equiv0$,
which means $x\equiv 4$ or $2$.
Modulo $3$ we have $x^3+2x^2+5x+4=(x+1)(x^2+x+4)\equiv(x+1)(x-1)^2\equiv0$,
which means $x\equiv 2$ or $1$.
Can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find all positive integers $x, y, z$ so that $(x+2y)(y+2z)(z+2x)$ is equal to prime power Find all positive integers $x, y, z$ so that $(x+2y)(y+2z)(z+2x)$ is equal to prime power.
My try
Since $x, y, z$ are positive integers it follows that $x+2y, y+2z, z+2x \geq 3$. We have following system of equations:
$$\begin{ali... | Suppose we have a solution $(x, y, z)$. Note that if $d = \gcd(x, y, z) > 1$, then $d$ must also be a power of $p$, and we get a smaller solution $(x', y', z') = (x/d, y/d, z/d)$ with $\gcd(x', y', z') = 1$.
Thus without loss of generality we can assume our solution has $\gcd(x, y, z) = 1$. Now, if $p \neq 3$, then by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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What are the values of the parameters $a,b \in \mathbb{R}$ such that $\lim\limits_{x \to \infty}(\sqrt{x^2+x+1}+\sqrt{x^2+2x+2}-ax-b)=0.$ I have to find the values of $a$ and $b$ (with $a,b \in \mathbb{R}$) such that the following is true:
$$\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-ax-b)=0$$
This is... | Hint: Intuitively, for large $x$, we have $\sqrt{x^2+x+1} = \sqrt{(x+\tfrac{1}{2})^2+\tfrac{3}{4}} \approx x+\tfrac{1}{2}$ and $\sqrt{x^2+2x+2} = \sqrt{(x+1)^2+1} \approx x+1$. But how can we make these observations rigorous?
Using the identity $\sqrt{a}-\sqrt{b} = \dfrac{a-b}{\sqrt{a}+\sqrt{b}}$, we get that$$\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving HM-GM inequality using AM-GM I want to prove
$$\frac{n}{\frac{1}{a_1} + \cdots + \frac{1}{a_n}} ≤ \sqrt[n]{a_1\cdots a_n}$$
Using AM-GM inequality.
My attempt:
Consider arbitrary $x_1,\cdots,x_n$ We know that
$$\sqrt[n]{x_1\cdots x_n} ≤ \frac{x_1 + \cdots + x_n}{n}$$
Replacing all $x$ with $\frac{1}{x}$
$$... | a)Yes, it is correct.
b) A better method would be to take reciprocals in the first step, as both sides are positive.
$$\sqrt[n]{\frac{1}{x_1}\cdots \frac{1}{x_n}} \leq \frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n}$$
$$\iff \sqrt[n]{x_1x_2\cdots x_n}\geq\frac{n}{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Triangle sides $a,b,c$ are in arithmetic progression. Show $\sin^2(A/2)\csc2A$, $\sin^2(B/2)\csc2B$, $\sin^2(C/2)\csc2C$ are in harmonic progression
If sides $a,b,c$ of the triangle ABC are in arithmetic progression, prove that $\sin^2\frac{A}{2}\mathrm{cosec}(2A),\sin^2\frac{B}{2}\mathrm{cosec}(2B),\sin^2\frac{C}{2}\... | The arithmetic progression of sides $b-a = c-b$ leads to
$$\sin B - \sin A = \sin C - \sin B\tag 1$$
With $A+B+C=\pi$,
$$\sin\frac{C}2\sin\frac{B-A}2=\sin\frac{A}2\sin\frac{C-B}2\implies
\frac{\sin\frac{B-A}2}{\sin\frac A2}
= \frac{\sin\frac{C-B}2}{\sin\frac C2}\tag 2$$
Next, let $I_X = {\sin^2\frac{X}{2}\csc2X}$ to e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How many numbers from $1$ to $1000$ can be written as the sum of $4$s and $5$s?
I like the numbers $4$ and $5$. I also like any number that can be added together using $4$s and $5$s. Eg,
$$9 = 4+5 \qquad 40 = 5 + 5 + 5 + 5 + 5 + 5 + 5 +5$$
How many number have this property from 1 to 1000?
Multiples of $4$s and ... | Google frobenius coin problem.
But notice. If $K = 4m + 5n$ is such number then $K+4 = 4(m+1)+5n$ is such a number and all $K + 4a = 4(m+a) + 5n$ will be such numbers.
$12 = 3*4$ and $13=2*4 + 5$ and $14=4+2*5$ and $15=3*5$. So every $12 + 4a$ and $13+4b$ and $14+4b$ and $15+4c$ will be such numbers And that is ever... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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prove: $\nabla \cdot \frac{\vec{r}}{r^3} = 0$ prove:
$\nabla \cdot \frac{\vec{r}}{r^3} = 0$
a few useful identities for proof:
$\vec{r} \cdot \vec{r} = 1$
$\nabla r^n = n r^{n-2} \vec{r}$
$\nabla \cdot (\vec{v} + \vec{w}) = \nabla \cdot \vec{v} + \nabla \cdot \vec{w}$
$\nabla \cdot (A \vec{v}) = (\nabla A) \cdot \vec{... | Solution : $$\nabla \cdot\frac{\vec r}{r^3}$$
$$=(\nabla \cdot \vec r)\frac{1}{r^3} + \vec r \cdot \left( \nabla \frac{1}{r^3} \right)$$
where, $\nabla \cdot(\phi \vec A)=(\nabla \cdot \vec A) \phi + \vec A \cdot (\nabla \phi)$
$$=\frac{3}{r^3}+\vec r \cdot [ (-3)r^{-3-2}\vec r ] $$
where , $ \nabla r^n=nr^{n-2} \vec ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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$k=-\sqrt{3}(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}})$
Let $k$ be equal to:
$$k=-\sqrt{3} \left(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}} \right)$$
I am trying to simplify the expression and express $5+2\sqrt{6}$ and $5-2\sqrt{6}$ as squares. I don't think it's smart to multiply the brackets by $-\sqrt{3}$ at first. Is t... | Squaring gives
\begin{eqnarray*}
k&=& -\sqrt{3} \left(\sqrt{5+ 2\sqrt{6}} + \sqrt{5- 2\sqrt{6}} \right) \\
k^2&=&3 \left(5+ 2\sqrt{6}+5- 2\sqrt{6} +2\sqrt{ (5+ 2\sqrt{6})( 5- 2\sqrt{6}) } \right) \\
k^2&=&3 \left(10 +2\sqrt{ 25- 4 \times{6} } \right) \\
k^2&=& 36 \\
\end{eqnarray*}
Now square root thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456305",
"timestamp": "2023-03-29T00:00:00",
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Evaluate the following integral $\int \frac{2x^2+x\cos x+\sin^{-1}x}{1+\sqrt{1-x^2}}$ I change the integral $\int \frac{2x^2+x\cos x+\sin^{-1}x}{1+\sqrt{1-x^2}}$ into
$$\int \frac{2x^2}{1+\sqrt{1-x^2}} + \int \frac{x\cos x}{1+\sqrt{1-x^2}} + \int \frac{\sin^{-1}x}{1+\sqrt{1-x^2}}$$
and I figure out $\int \frac{2x^2}{... | PARTIAL ANSWER
First solve
\begin{eqnarray}
\mathcal I_1 &=& \int \frac1{1+\sqrt{1-x^2}}dx=\\
&=&\int\frac{1-\sqrt{1-x^2}}{x^2} dx=\\
&=&-\frac1x-\int\frac{\sqrt{1-x^2}}{x^2}dx.
\end{eqnarray}
Using $x=\sin t$ we get
\begin{eqnarray}
\mathcal I_1 &=&-\frac1x - \int \cot^2 t dt=\\
&=&-\frac1x + \cot t+ t+C=\\
&=&-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Mean and variance of first order statistic from a uniform distribution - min(x) So here are the first two moments about the mean and the origin of the first order statistic of a uniform random variable in the interval $[0, 1]$. Maybe someone can derive the moment generating function and post it as a answer.
CDF:
$P\le... | The derivative with respect to $y$ of $1-(1-y)^n$ is $+n(1-y)^{n-1}$ not $-n(1-y)^{n-1}$, and densities should always be non-negative
Step by step:
*
*The derivative of $y^n$ is $+ny^{n-1}$
*The derivative of $(1-y)^n$ is $-n(1-y)^{n-1}$ by the chain rule
*The derivative of $-(1-y)^n$ is $+n(1-y)^{n-1}$
*The de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3460727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}\frac{x^{2}+y^{2}}{x+y}$ I am supposed to evaluate the following limit:
$$\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}\frac{x^{2}+y^{2}}{x+y}$$
My solution:
$$\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}\frac{x^{2... | The answer is wrong, that limit doesn't exist.
$$\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}\frac{x^{2}+y^{2}}{x+y}=\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}\frac{x^{2}+y^{2}+2xy-2xy}{x+y}=\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}(x+y)-2\lim_{\left ( x,y \right )\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3462835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I prove analytically that the golden ratio is less then $\frac{\pi^2}{6}$. Or stated in other terms, prove that
$$\frac{1+\sqrt{5}}{2} < \sum_{n=1}^{\infty}\frac{1}{n^2}$$
| $\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}=\frac12\sum_{n=1}^\infty \frac1{n^2}$ (see here). Now, the inequality is equivalent to $\frac{1+\sqrt5}4<\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}.$
\begin{equation}
\begin{split}
\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}&=\sum_{k=1}^\infty\left(\frac1{(2k-1)^2}-\frac1{(2k)^2}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3464638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove $\sqrt{(a-1)(b-1)}+\sqrt{(a-1)(c-1)}+\sqrt{(b-1)(c-1)}\geq a+b+c+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}$? I am trying to prove this inequality:
$$\sqrt{(a-1)(b-1)}+\sqrt{(a-1)(c-1)}+\sqrt{(b-1)(c-1)}\geq a+b+c+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}$$
if there is condition $a+b+c=1$. $a, b, c$ are positive rational numbers.
... | We assume throughout that $a,b,c$ are positive. Using the condition $a+b+c = 1$, we may write $$\sqrt{(a-1)(b-1)} + \sqrt{(b-1)(c-1)} + \sqrt{(a-1)(c-1)} = \sqrt{(b+c)(a+c)} + \sqrt{(a+c)(a+b)} + \sqrt{(b+c)(a+b)}$$ Note that $$(b+c)(a+c) = c^2 + ab + c(a+b) \ge c^2 + ab + 2c\sqrt{ab} = (c+\sqrt{ab})^2$$ using the two-... | {
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"timestamp": "2023-03-29T00:00:00",
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Maximum value of $\frac{x}{y+1}+\frac{y}{x+1}$ while $0\leq x,y \leq 1$ What is the maximum value of $\frac{x}{y+1}+\frac{y}{x+1}$ while $0\leq x,y \leq 1$?
Wolfram Alpha plots this expression on a 3d graph, but I want to solve it algebraicly, by modifying the expression
My Attempts
1) add and substract 2 at the equa... | When $ 0\le x,y \le 1$, then
$$\frac{x}{y+1} \le \frac{x}{x+y}$$
$$\frac{y}{x+1} \le \frac{y}{x+y}$$
Adding them we have
$$\frac{x}{y+1}+\frac{y}{x+1} \le \frac{x+y}{x+y}=1.$$
Equality holds when $x=0$ and $y=1$ or $x=1$ and $y=0$ and the maximum of $1$ is attained.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3475979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $I_m=\int \frac{x^m}{\sqrt{ax^2+c}}\, dx$, show that $amI_m=x^{m-1}\sqrt{ax^2+c}-(m-1)c I_{m-2}$. Problem: If $I_m=\int \dfrac{x^m}{\sqrt{ax^2+c}}\, dx$, show that $amI_m=x^{m-1}\sqrt{ax^2+c}-(m-1)c I_{m-2}$.
Observe
\begin{align*}
I_m= & \int \dfrac{x^m}{\sqrt{ax^2+c}}\, dx\\
= & \dfrac{1}{a}\int x^{m-1}d(\s... | Your third and fourth steps are wrong. The correct steps are
$$I_m = \frac{1}{a}\Bigg[x^{m-1}\sqrt{ax^2+c}-\int(m-1)x^{m-2}\sqrt{ax^2+c}\ dx\Bigg]$$
$$=\frac{1}{a}\Bigg[x^{m-1}\sqrt{ax^2+c}-(m-1)\int \frac{x^{m-2}(ax^2+c)}{\sqrt{ax^2+c}}dx\Bigg]$$
$$=\frac{1}{a}\Bigg[x^{m-1}\sqrt{ax^2+c}-a(m-1)I_m-c(m-1)I_{m-2}\Bigg]$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving by induction that strictly lower triangular matrix is nilpotent
Let matrix $A \in \mathbb{K}^{n \times n}$ have the property $a_{ij} = 0$ for $1 \leq i \leq j \leq n$. Show that $A^n = 0$.
Proof by induction:
Base Case:
for $n=2:
A=
\left[ {\begin{array}{cc}
0 & 0 \\
a_{21} & 0 \\
\end{array} }... | If you try to deduce the $n=3$ case from $n=2$ that you describe, this happens:
$$
\begin{pmatrix}
0 & 0 & 0 \\
a_{21} & 0 & 0 \\
a_{31} & a_{32} & 0
\end{pmatrix}^{3} =
\begin{pmatrix}
0 & 0 & 0 \\
a_{21} & 0 & 0 \\
a_{31} & a_{32} & 0
\end{pmatrix}^{2}
\begin{pmatrix}
0 & 0 & 0 \\
a_{21} & 0 & 0 \\
a_{31} & a_{32} &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$ydx - (4x^2y + x)dy = 0$, where is my mistake? $ydx - (4x^2y + x)dy = 0$
I found two ways to solve this equation and they give two different answers, but I couldnt find where I did mistake.
Solution 1:
Write $x$ as a function of $y$
$y\frac{dx}{dy} - (4x^2y + x) = 0$
$x' - \frac{x}{y} - 4x^2 = 0$
It is a Bernoulli equ... | You made a sign mistake here since $z=\frac 1x \implies z'=-\frac {x'}{x^2}$
$$z' - \frac{z}{y} = 4$$
It should be
$$z' + \frac{z}{y} =- 4 \implies z' =- \frac{z}{y} \implies z=\frac {C(y)}{y}$$
$$\implies C(y)=-2y^2+K$$
For the first one. The Bernouilli equation gives the same answer:
$$yx'-4x^2y-x=0$$
$$x'-\frac x ... | {
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"source": "stackexchange",
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Why should it be $\sqrt[3]{6+x}=x$? Find all the real solutions to:
$$x^3-\sqrt[3]{6+\sqrt[3]{x+6}}=6$$
Can you confirm the following solution? I don't understand line 3. Why should it be $\sqrt[3]{6+x}=x$?
Thank you.
$$
\begin{align}
x^3-\sqrt[3]{6+\sqrt[3]{x+6}} &= 6 \\
x^3 &= 6+ \sqrt[3]{6+\sqrt[3]{x+6}} \\
x &= ... | Begin as in OP $\begin{align}
x^3-\sqrt[3]{6+\sqrt[3]{x+6}} &= 6 \tag{1}\\
x^3 &= 6+ \sqrt[3]{6+\sqrt[3]{x+6}} \tag{2}\\
x &= \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{x+6}}} \tag{3}
\end{align}$
Setting $t=\sqrt[3]{6+x}$, the equation $(3)$ rewrites
$$t^3-6=\sqrt[3]{6+\sqrt[3]{6+t}},$$ equivalent to $(1).$ Hence $t=x.$
EDIT
The... | {
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"source": "stackexchange",
"question_score": "18",
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Can it be shown, $n^4+(n+d)^4+(n+2d)^4\ne z^4$? We know, $n^4+(n+d)^4= z^4$ has no solution in positive integers $n,d,z$.
Can it be shown, $n^4+(n+d)^4+(n+2d)^4= z^4$ has no solution in positive integers $n,d,z$?
I am check upto $1\le n, d, z\le 150$ without finding a counter example.
PARI/GP
for(n=1,150,for(d=1,150,fo... | First, assume there's at least one solution and then let $m = n + d$, with $m$ being the smallest positive integer which works. The equation then becomes
$$\begin{equation}\begin{aligned}
z^4 & = (m - d)^4 + m^4 + (m + d)^4 \\
& = m^4 - 4md^3 + 6m^2d^2 - 4m^3d + d^4 + m^4 + \\
& \; \; \; \; m^4 + 4md^3 + 6m^2d^2 + 4m^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3481061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find value of $a_1$ such that $a_{101}=5075$ Let $\{a_n\}$ be a sequence of real numbers where $$a_{n+1}=n^2-a_n,\, n=\{1,2,3,...\}$$
Find value of $a_1$ such that $a_{101}=5075$.
I have
$$a_2=1^2-a_1$$
$$a_3=2^2-a_2=2^2-1^2+a_1$$
$$a_4=3^2-2^2+1^2-a_1$$
$$a_5=4^2-3^2+2^2-1^2+a_1$$
$$\vdots$$
$$a_{101}=100^2-99^2+98^... | In the most simple way:
$$A_{n+1}+A_n=n^2~~~(1)$$ is a non-homogeneous recurrence equation. Its homogeneous part
$$A_{n+1}+A_n=0 \implies A_{n+1}=-A_n \implies A_n= (-1)^n S~~~(2)$$ In the RHS of (1) being $n^2)$, we can take $A_n =P n^2+Q n+R$; inserting this in (1), we get
$2P=1,P+Q=0,P+Q+2R=0 \implies P=1/2, Q=1/2, ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the maximum of a sum Given $x + y + z = 0, x + 1 > 0, y + 1 > 0, z + 4 > 0$, find the maximum of $$Q=\frac{x}{x+1} + \frac{y}{y+1} + \frac{z}{z+4}$$
The answer is $\frac{1}{3}$
I've tried inverting each of the fractions then using AM-GM on them, but it doesn't give me the correct answer
| By C-S
$$\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+4}=3+\frac{x}{x+1}-1+\frac{y}{y+1}-1+\frac{z}{z+4}-1=$$
$$=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{4}{z+4}\right)=$$
$$=3-\frac{1}{6}(x+1+y+1+z+4)\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{4}{z+4}\right)\leq$$
$$\leq3-\frac{1}{6}(1+1+2)^2=\frac{1}{3}.$$
The equality occurs... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many ways are there to arrange students and teachers in two lines under the given conditions? There are $24$ children and $4$ teachers. There are $2$ pairs of twins within the 24 children. We want to arrange everyone in two lines. In each line, there will be $12$ children, their order of appearance (or just order... | Consider the line containing a particular teacher.
There are $\begin{pmatrix}20\\10\\\end{pmatrix}$ ways of choosing the children in this line and, for each of these choices, $\begin{pmatrix}2\\1\\\end{pmatrix}$ ways of choosing the pair of twins and $\begin{pmatrix}3\\1\\\end{pmatrix}$ ways of choosing the fellow tea... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find min and max of $A = \sqrt{x-2} + 2\sqrt{x+1} + 2019 -x$
Find minimum and maximum of $$A = \sqrt{x-2} + 2\sqrt{x+1} + 2019 -x$$
How to solve this problem without using derivatives?
Michael Rozenberg's edit:
Because with a derivative it's not so easy:
$$A'(x)=\frac{1}{2\sqrt{x-2}}+\frac{1}{\sqrt{x+1}}-1=\frac{1... | By C-S and AM-GM we obtain:
$$\sqrt{x-2}+2\sqrt{x+1}-x+2019=\sqrt{x-2}+4\sqrt{\frac{x+1}{4}}-x+2019\leq$$
$$\leq\sqrt{(1+4)\left(x-2+4\cdot\frac{x+1}{4}\right)}-x+2019=\sqrt{5(2x-1)}-x+2019\leq$$
$$\leq\frac{5+2x-1}{2}-x+2019=2021.$$
The equality occurs for $x=3,$ which says that we got a maximal value.
The minimum doe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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let $f(x)=x+\frac{1}{x} \ \ x \geq 1$ and $g(x)=x^2+4x-6$ the find Min of $g(f(x))=?$
Let $f(x)=x+\frac{1}{x}$ for all $x \geq 1$ and $g(x)=x^2+4x-6$. Find minimum of $g\circ f$.
My try:
The domain of $g\circ f$ is $[1,+\infty)$ and we have
$$g(f(x))=\left(x+\frac{1}{x}\right)^2+4\left(x+\frac{1}{x}\right)-6.$$
At $x... | 1)AM-GM:
$x>0$:
$x+1/x \ge 2 \sqrt{x(1/x)}=2$;
$f(1)=2=\min ${$f(x)| x \ge 1$}.
2) $g(x)=x^2+4x-6 =$
$(x+2)^2-10$ is increasing for $x \ge -2$.
3) $g(f(x))=(f(x)+2)^2-10 \ge $
$(2+2)^2-10=6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Po-Shen Loh's new way of solving quadratic equation Quadratic equation, $ax^2+bx+c=0$ and its solution is quadratic equation, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Now setting $a=1$ then we have $x^2+bx+c=0$
$$x=\frac{-b\pm \sqrt{b^2-4c}}{2}$$ rewrite as
$$x=-\frac{b}{2}\pm \sqrt{\left(\frac{b}{2}\right)^2-c}$$
In thi... | Suppose that $A$ is the arithmetic mean of the roots of the quadratic $P$, and $R$ is the geometric mean [where $P(x) = ax^2+bx+c = a(x-r_1)(x-r_2)$], then we have this equation:
$$\begin{align}
P(x) &= 0\\
ax^2+bx+c &= 0\\
x^2+\frac{b}{a}x+\frac{c}{a} &= 0\\
x^2+\beta x+\gamma &= 0\\
\end{align}$$
But we can also look... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 8,
"answer_id": 6
} |
Evaluate $\int \frac{\sqrt{\sin^4x+\cos^4x}}{\sin^3x\cos x}dx$ $$\int \dfrac{\sqrt{\sin^4x+\cos^4x}}{\sin^3x\cos x}$$
My multiple attempts are as follows:-
Attempt $1$:
$$\int \dfrac{\sqrt{1-2\sin^2x\cos^2x}\cdot\sin x}{\sin^4x\cos x}dx$$
$$\cos x=t$$
$$-\sin x=\dfrac{dt}{dx}$$
$$\int -\dfrac{\sqrt{1-2t^2(1-t^2)}}{t(1-... | Claim. The "real answer" is incorrect.
Proof. Let $\csc(x)=\frac1{\sin(x)}$ and $\sec(x)=\frac1{\cos(x)}$. Then, for the given solution at points where $\frac{\csc(x)-1}{\csc(x)+1}>0$, $$\frac{\mathrm d}{\mathrm dx}\bbox[10px,#ffd]{\frac{1}{4} \big(-2\csc (x)-\log (\csc (x)-1)+\log (\csc (x)+1)\big)}=\frac{\csc(x)}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Why does this pattern occur: $123456789 \times 8 + 9 = 987654321$ I came across the following:
$\begin{align} 1 \times 8 + 1 &= 9 \\ 12 \times 8 + 2 & = 98 \\ 123 \times 8 + 3 & = 987 \\ 1234 \times 8 + 4 & = 9876 \\ 12345 \times 8 + 5 & = 98765 \\ 123456 \times 8 + 6 & = 987654 \\ 1234567 \times 8 + 7 & = 9876543 \\ 1... | This is something I noticed but I'm still thinking about if it means anything:
$$\boxed{1\cdot8+1=9}\\\downarrow$$
$$10\cdot8+10=90$$
$$10\cdot8+18=98$$
$$(10+2)\cdot8+2=98$$
$$\boxed{12\cdot8+2=98}\\\downarrow$$
$$120\cdot8+20=980$$
$$120\cdot8+27=987$$
$$(120+3)\cdot8+3=987$$
$$\boxed{123\cdot8+3=987}\\\downarrow$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 3
} |
If $a_k:=\tan( \sqrt{2} + \frac{k\pi}{2011})$, then evaluate $ \frac{a_1+a_2+\cdots+a_{2011}}{a_1a_2~\cdots~a_{2011}}$
Set $$
a_k = \tan{\left( \sqrt{2} + \frac{ k \pi }{2011} \right)}\quad\text{for}~k=1,2,3,\ldots, 2011
$$
Find the value of
$$
\frac{a_1 + a_2 + \ldots + a_{2011}}{a_1 a_2 \ldots a_{2011}}
$$
C... | Using Sum of tangent functions where arguments are in specific arithmetic series,
$$\tan(2m+1)x=\dfrac{\binom{2m+1}1 t-\binom{2m+1}3t^3+\cdots+(-1)^m\binom{2m+1}{2m+1}t^{2m+1}}{\binom{2m+1}0-\binom{2m+1}2t^2+\cdots++(-1)^m\binom{2m+1}{2m}t^{2m}}$$
So, if $\tan(2m+1)x=\tan y$
$(2m+1)x=k\pi+y$ where $k$ is any integer
$x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\arctan x_1 \cdot \arctan x_2$, where $x_1$ and $x_2$ are roots of $x^2 - 2\sqrt{2}x + 1 = 0$. I am told that $x_1$ and $x_2$ are the roots of the following equation:
$$ x^2 - 2\sqrt{2}x + 1 = 0 $$
And I have to find the following:
$$\hspace{6cm} \arctan x_1 + \arctan x_2 \hspace{5cm} (1)$$
$$\hspace{6.cm} \arcta... | The roots of the equation are $ x_1 = \sqrt{2} + 1$ and $ x_2 =\sqrt{2} - 1 $. So we can explicitly work out $ \arctan(x_1) \, \cdot \, \arctan(x_2) = \frac{\pi}{8} \, \cdot \, \frac{3\pi}{8} = \frac{3\pi^2}{64} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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$4$ by $4$ Determinant with variables I was given this $4$ by $4$ determinant
$$
\begin{vmatrix}
x & a & a & a \\
a & x & a & a \\
a & a & x & a \\
a & a & a & x
\end{vmatrix}
= 0 $$
Clearly one of the answers is $x=a,$ how do i find the other answer?
I've tried splitting it into $4$ "$3$ by $3$" determinants but t... | Subtracting the 2nd row from the first, the 3rd from the second, the 4th from the third, tou obtain the determinant
$$\begin{vmatrix}
x-a&a-x&0&0\\0&x-a&a-x&0\\0&0&x-a&a-x\\ a&a&a&x
\end{vmatrix}=(x-a)^3\begin{vmatrix}
1 & -1 & 0 & 0\\ 0 & 1 & -1 & 0\\ 0 & 0 & 1 & -1\\ a&a&a&x
\end{vmatrix}$$
Then, adding successively... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to factorize $X^5-1$ into irreductible factors? I have to factorize the polynomial $P(X)=X^5-1$ into irreducible factors in $\mathbb{C}$ and in $\mathbb{R}$, this factorisation happens with the $5$th roots of the unity.
In $\mathbb{C}[X]$ we have $P(X)=\prod_{k=0}^4 (X-e^\tfrac{2ki\pi}{5})$.
In $\mathbb{R}[X]$ the... | There are various ways to render separate values of $\cos(2\pi/5)$ and $\cos(4\pi/5)$ given the equation
$1+2\cos(2\pi/5)+2\cos(4\pi/5)=0$
I demonstrate one approach. Separate out the factor of $2$ from the last two terms and apply the trigonometric sum-product relations:
$0=1+2(\cos(2\pi/5)+\cos(4\pi/5))=1+4\cos(2\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496594",
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"source": "stackexchange",
"question_score": "1",
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How to derive this series
The function $\dfrac1{1-x}$, equal to $$1 + x + x^2 + x^3 + \cdots,$$ can also be developed according to the series $$1 + \frac{x}{1 + x} + \frac{1\cdot2\cdot x^2}{(1 + x)(1 + 2x)} + \frac{1\cdot2\cdot3\cdot x^3}{(1 + x)(1 + 2x)(1 + 3x)} + \cdots $$ when $x$ is positive and smaller than $1$.
... | Proof of the Formula
Below, we show inductively
$$
\frac1{1-x}=\sum_{k=0}^{n-1}\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}+\frac{n!\,x^n}{\prod_{j=1}^n(1+jx)}\frac{1+nx}{1-x}\tag1
$$
where the empty sum is $0$ and the empty product is $1$.
Gautschi's Inequality says that
$$
\begin{align}
\frac{n!\,x^n}{\prod_{j=1}^n(1+jx)}
&=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Calculate the maximum value of $\lfloor x\lfloor x \rfloor \rfloor + \lfloor y\lfloor y \rfloor \rfloor$.
Given negatives $x$ and $y$ such that $\left(\dfrac{x}{2} - 1\right)^2 + \left(\dfrac{y}{2} - 1\right)^2 \le \dfrac{125}{2}$. Calculate the maximum value of $$\large \lfloor x \lfloor x \rfloor \rfloor + \lfloor y... | Well, we have the following problem:
$$\left(\frac{x}{\Delta}+\alpha\right)^2+\left(\frac{\text{y}}{\Delta}+\alpha\right)^2\le\frac{\rho}{\Delta}\tag1$$
Where all the constants are real numbers.
Now, if we multiply the LHS and the RHS with $\Delta^2$ we get:
$$\Delta^2\cdot\left(\left(\frac{x}{\Delta}+\alpha\right)^2+\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Without using characteristic equation, how can I show genenal formula of this sequence? I know that, solution of the sequence $x_n$ so that
$$x_{n+2}-2 x_{n+1}= 2 x_{n+1} - 4 x_{n}$$
is $x_n = (A+ Bn)2^n$. I want to show this without using characteristic equation. I tried
\begin{align*}
x_{n+2}-2 x_{n+1} &= 2 x_{n... | Use induction.
We are given a sequence $x_1,x_2,\dots$ satisfying the recursion
$$
x_{n+2}-2 x_{n+1}= 2 x_{n+1} - 4 x_{n},
$$
and we want to show that for some $A,B$ we have $x_n = (A + Bn)2^n$ for all $n$.
Select $A,B$ such that $x_n = (A + Bn)2^n$ holds for $n = 1,2$ (or note that such an $A,B$ necessarily exist)... | {
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"source": "stackexchange",
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Area of the curve $y=\sin\theta \cos^2\theta$; $x=\sin^2\theta \cos\theta$ bounded by tangent parallel to axes Area bounded by tangents of the curve given by $y=\sin\theta \cos^2\theta$; $x=\sin^2\theta \cos\theta$, which are parallel to co-ordinate axes(other than axes) is
(1)$\frac{4}{27}$
(2) $\frac{27}{4}$
(3)$\fr... | No need to eliminate the parameter. Tangents to $(x(t),y(t))$ have slope $\frac{\mathrm dy}{\mathrm dx}$. By the chain rule,
$$\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dy}{\mathrm d\theta}\frac{\mathrm d\theta}{\mathrm dx}=\frac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}=\frac{\cos^3\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the first three terms of the Laurent series near $z=0$
Let $$f(z) = \frac{1-e^z}{\sin^2z}$$
Find the first three terms of the Laurent series near $z=0$ and then evaluate $$I =\oint_{|z| = 1}f(z)dz$$
If we write $\sin^2 z = \frac{1-\cos2z}{2}$ and then try to use Taylor expansions of $e^z$ and $\cos z$, we s... | $$e^z=1+z+\frac{z^2}2+\frac{z^3}6+\ldots\,,\;\sin z=z-\frac{z^3}6+\frac{z^5}{120}-\ldots\implies$$
$$\implies\sin^2z=z^2-\frac13z^4+\frac{2z^4}{45}\ldots\implies$$
$$\frac1{\sin^2z}=\frac1{z^2\left(1-\frac13z^2+\ldots\right)}=\frac1{z^2}\left(1+\frac{z^2}3+\frac{z^4}9+\ldots\right)\implies$$
$$\frac{1-e^z}{\sin^2z}=\le... | {
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"url": "https://math.stackexchange.com/questions/3504053",
"timestamp": "2023-03-29T00:00:00",
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Number of solutions via generating function How to find generating function for number of solutions
$x_1+x_2+x_3=n$ in set of positive integers such that $x_1 \ge x_2 \ge x_3$ and $x_1<x_2+x_3$.
| Hint: First rewrite $x_2 = x_3 + d_{2,3}$, with $d_{2,3} \geq 0$ and $x_1 = x_2 + d_{1,2} = x_3 + d_{2,3} + d_{1,2}$ with $d_{1,2} \geq 0$. In terms of $d_{1,2} \geq 0$, $d_{2,3} \geq 0$, $x_3 \geq 1$, we have arranged for all three variables to be (so far) independent.
Now we need to implement $x_1 < x_2 + x_3$, or,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3505691",
"timestamp": "2023-03-29T00:00:00",
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If $\frac{\sin(b+\theta)\sin a}{\sin b}=\frac{\sin(c+\theta)\sin b}{\sin c}=\frac{\sin(a+\theta)\sin c}{\sin a}$, for arbitrary $\theta$, then $a=b=c$
If $$\frac{\sin(b+\theta)\sin a}{\sin b}= \frac{\sin(c+\theta)\sin b}{\sin c}= \frac{\sin(a+\theta)\sin c}{\sin a}$$ for some arbitrary constant $\theta$, then prove $a... | First, let's combine the two equations into one and eliminate the $\theta$. Take the first equation, expand $\sin(b+\theta)$ and $\sin(c+\theta)$.
$$(\sin \theta \cot b + \cos \theta)\sin a = (\sin \theta \cot c + \cos \theta)\sin b$$
$$\sin \theta (\cot b \sin a - \cot c \sin b) = -\cos \theta (\sin a - \sin b)$$
Do t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving differential equation by separating variables Solve the following differential equation:
$$\frac yx \frac {dy}{dx}=\sqrt {1+x^2+y^2+x^2y^2}$$
I have tried like this:
$$\frac yx \frac{dy}{dx}=\sqrt {1+x^2+y^2(1+x^2)}\\
\implies \frac yx \frac {dy}{dx}=\sqrt {(1+x^2)(1+y^2)}\\
\implies \int \frac {ydy}{\sqrt {1+y... | Hint:
Rewrite as below and integrate.
$$\left(\sqrt{1+y^2}\right)^{-1/2}d(y^2)-\left(\sqrt{1+x^2}\right)^{1/2}d(x^2)=0.$$
Mind the asymmetry in the exponents.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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solution of fuctional equation: $f(g(x))=x^2$. For any $x \in (1,+\infty)$, $f(g(x))=x^2$, $g(f(x))=x^4$.
How to find the $f(x)$， $g(x)$ statisfying above functional equation.
Suppose inverse function existed.
The problem can be simplified to $f^2(x)=f(x^4)$.
How to find $f(x)$?
| Solving
$$
f(x)^2 = f(x^4) .
\tag{1}$$
We do not require that $f$ be continuous!
Let $f$ be arbitrary $[2,2^4) \to (1,+\infty)$. Extend recursively:
For $x \in [2^4,2^{4^2})$, let $f(x) = f(\sqrt[4]{x}\;)^2$. This means that $f$ satisfies $(1)$ for $x \in [2,2^{4})$.
For $x \in [2^{4^2}, 2^{4^3})$, let $f(x) = f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508678",
"timestamp": "2023-03-29T00:00:00",
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Find $a, b$ so polynomial is divisible
Find values $a,b\in\Bbb{R}$ so the polynomial $$P(x)=6x^4-7x^3+ax^2+3x+2$$
is divisible by the polynomial
$$Q(x)=x^2-x+b$$
So what I know and how I do these problems most of the time, is since:
$$P(x)=Q(x)D(x)$$
I would know that by plugging the roots of $Q(x)$ in $P(x)$ sho... | Let $D(x) = 6x^2+cx+d$ (to fix the coefficient of $x^4$). Then multiply out:
$$Q(x)D(x) = (x^2-x+b)(6x^2+cx+d) = 6x^4+(c-6)x^3+(6b+d-c)x^2+(bc-d)x+bd = 6x^4-7x^3+ax^2+3x+2$$
Equate coefficients:
$$c-6=-7 \Longrightarrow c=-1$$
$$6x^4-7x^3+(6b+d+1)x^2-(b+d)x+bd = 6x^4-7x^3+ax^2+3x+2$$
$$bd = 2 \Longrightarrow d=\dfrac{2... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the volume cut off from the sphere $x^2+y^2+z^2=a^2$ by the cylinder $x^2+y^2=ax$
Find the volume cut off from the sphere $x^2+y^2+z^2=a^2$ by the cylinder $x^2+y^2=ax$
Attempt: The projection of the Cylinder ( denoted $D$) on the $xy$ plane is a circle which has the equation: $x^2+y^2=ax ~~~\equiv~~~(x-a/2)^2+y... | $V=4\displaystyle\int_{0}^{\pi/2} \displaystyle\int_{0}^{a\cos\theta}r\sqrt{{a^2}-r^2}\ drd\theta$
$=\dfrac{4a^3}{3}\displaystyle\int_{0}^{\pi/2} (1-\sin^3\theta)d\theta$
$=\dfrac{4a^3}{3}\left(\dfrac{\pi}{2}-\dfrac{2}{3}\right)=\dfrac{2a^3}{3}\left({\pi}-\dfrac{4}{3}\right)$
note: the factor of $4$ accounts for symmen... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the next $x \in \mathbb{Z}^+$ such that $x^2 \equiv 17 \pmod{2^{n+1}}$ Only parts 4. and 5. are relevant, but for completion's sake, I'll just include the whole problem.
Let $N$ be any positive and let $S(N)$ be the set of remainders when the square numbers $0,1,4,9,\dots$ are divided by $N$.
*
*Prove that a... | If $\lambda$ is even, we're obviously done. If not, then notice that $x^2 \equiv 17 + 2^n \pmod{2^{n+1}}$, and so $\left( x - 2^{n-1} \right)^2 \equiv x^2 - 2^n + 2^{2n-2} \equiv \left( 17 + 2^n \right) - 2^n + 0 \equiv 17 \pmod{2^{n+1}}$. This is not too different conceptually from the standard proof that the multipli... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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On the Diophantine equation $m^2 - p^k = 4z$, where $z \in \mathbb{N}$ and $p$ is a prime satisfying $p \equiv k \equiv 1 \pmod 4$ The title says it all:
Do there always exist $m, p, k \in \mathbb{N}$ such that $m^2 - p^k = 4z$, $z \in \mathbb{N}$ and $p$ is prime satisfying $p \equiv k \equiv 1 \pmod 4$?
MY ATTEMPT ... | This is a partial answer.
Your idea works for $z=2^{2n}$.
Claim : For $z=2^{2n}$, if $2^{n+2}+1$ is prime, then $(m,p,k)=(2^{n+1}+1,2^{n+2}+1,1)$. If $2^{n+2}+1$ is not prime, then there is no solution.
Proof :
For $z=2^{2n}$, we have
$$m^2 - p^k = 2^{2n+2}\iff p^k=(m-2^{n+1})(m+2^{n+1})$$
So, there is a non-negative ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve inequality using AM-GM / Cauchy-Schwarz Given positive real numbers $a$, $b$, $c$ satisfying $a^2+b^2+c^2=3$. Prove that
$$\dfrac{1}{a^2+7}+\dfrac{1}{b^2+7}+\dfrac{1}{c^2+7} \leq \dfrac{1}{4} \left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right).$$
I have tried using $a+b\leq \sqrt{2\left(a^2+b^2\right)}=\sqr... | Let $a+b+c=3u.$
Thus, $u\leq1$ and by C-S $$\sum_{cyc}\frac{1}{a^2+7}=\sum_{cyc}\frac{b^2+c^2+6}{(a^2+1+1+5)(1+b^2+c^2+5)}\leq$$
$$\leq\sum_{cyc}\frac{b^2+c^2+6}{(a+b+c+5)^2}=\frac{24}{(a+b+c+5)^2}=\frac{24}{(3u+5)^2}.$$
In another hand, by C-S again:
$$\sum_{cyc}\frac{1}{a+b}=\frac{1}{2(a+b+c)}\sum_{cyc}(a+b)\sum_{cyc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516730",
"timestamp": "2023-03-29T00:00:00",
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Cyclic Inequality $(x^8+1)(x^4+1)(y^8+1)(y^4+1)(z^8+1)(z^4+1)≥((x^2+1)(y^2+1)(z^2+1))^2$ I have been challenged to prove the following:
If $xy+yz+zx=3$ for positive $x,y,z$, then $(x^8+1)(x^4+1)(y^8+1)(y^4+1)(z^8+1)(z^4+1)≥((x^2+1)(y^2+1)(z^2+1))^2$.
Initial attempts at AM-GM are not very helpful for me, I don't think.... | Note: This answer is very similar with Michael Rozenberg's, maybe a little more direct, but I think it's worth posting.
Using Holder's inequality:
$$8(x^8+1)=(1^4+1^4)(1^4+1^4)(1^4+1^4)[(x^2)^4+1] \geq (x^2+1)^4$$
and from Cauchy-Schwarz:
$$2(x^4+1) = (1^2+1^2)((x^2)^2+1) \geq (x^2+1)^2$$
Multiplying with similar inequ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find volume of body given by $x^2+y^2+z^2 \leq a^2$ and $x^2+y^2 \geq a(a-2z)$, $a>0$. Is this answer right? What are mistakes in my solution? It appears to be an upside-down 'bowl', cut out of a sphere by a paraboloid. Moving to cylindrical coordinates,
\begin{align}
4 \int_0^{\pi/2}\ d\phi \int_{0}^{a}\ r \ dr \int... | *
*Your setup is the canonical one, with the exception of taking $0\leq \phi\leq \pi/2$ and then multiplying by $4$: you could have just integrated from $0$ to $2\pi$.
*You are assuming that $a>0$, which is probably the case but you don't say.
*From the first to the second line: instead of $\sqrt{a^2-r^2}-\frac{a^... | {
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"url": "https://math.stackexchange.com/questions/3522668",
"timestamp": "2023-03-29T00:00:00",
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Trisected sides of a scalene triangle Scalene triangle $\bigtriangleup ABC$ has area 45. Points $P_1$ and $P_2$ are located on side $AB$ such that $AP_1 = P_1P_2 = BP_2$. Additionally, the points $Q_1$ and $Q_2$ are located on side $AC$ such that $AQ_1 = Q_1Q_2 = CQ_2$. The area of the intersection of triangles $BQ_1Q_... |
Let [.] denote areas and I = [ABC] = 45. Observe that the intersection area [DEGF] is equal to
$$[DEGF] = [DBC] - [EBC] - [FBC] + [GBC]\tag 1$$
Evaluate [EBC], one of the four triangle areas on the RHS, as follows.
$$\frac{[EBC]}{[P_2BC]}=\frac{EC}{P_2C}
= \frac{[Q_1BC]}{[Q_1P_2BC]} = \frac{[Q_1BC]}{I - [AP_2Q_1]}
... | {
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"url": "https://math.stackexchange.com/questions/3523552",
"timestamp": "2023-03-29T00:00:00",
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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.