Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
If $xy$ divides $x^2 + y^2$ show that $x=\pm y$ Problem statement : Let $x,y$ be integers, show that if $xy$ divides $x^2 + y^2$ then $x=\pm y.$
What I have tried:
I can reduce this to the case where $\gcd(x,y)=1$, since if $x$ and $y$ have a common factor, $d$ say, then $d^2$ divides through both $xy$ and $x^2 + y^... | Suppose $xy|x^2+y^2$. Then $xy|x^2+y^2+2xy=(x+y)^2$. But if $\gcd(x,y)=1$, then also
$$
1=\gcd(x,x+y)=\gcd(y,x+y)=\gcd(xy,x+y)=\gcd(xy,(x+y)^2)
$$
from which it then follows that $xy=\pm 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
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Computing a gaussian integral involving both real and imaginary coefficients in a stochastic system I am stuck with the integral:
$$\int_{-\infty}^{\infty} \frac{\exp[-a(x-b)^2]}{1+cx^2} dx$$ where $a,c$ are real and $b$ is purely imaginary.
I tried to solve it by contour integration but the integral along the semicirc... | Since $b$ is purely imaginary, this is a Fourier Transform.
Assuming $a>0$, $c>0$ ,and $\Re(b) = 0$; make the substitution $- \pi s = \Im (b)$ or equivalently $-i\pi s =b$ or $\pi s = ib$
$$\begin{align*}\displaystyle & \int_{-\infty}^{\infty} \frac{\exp\left[-a\left(x-b\right)^2\right]}{1+cx^2} dx\\
\\
&= \int_{-\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$u_{n+1} = a u_n +b u_{n-1} +c$ Is there a closed form formula for the general term of a recurrence sequence satisfying $$u_{n+1} = a u_n +b u_{n-1} +c$$ The sequence I am interested in satisfies $u_{n+1} = 8 u_n - 3 u_{n-1} -4$.
| This is a linear difference equation so can be stated as
$$
u = u_h + u_p
$$
with
$$
u_h(n+1)-au_h(n)-bu_h(n-1) = 0\\
u_p(n+1)-au_p(n)-bu_p(n-1) = c\\
$$
now making $u_h(n) = \phi^n$ and substituting we have
$$
\phi^n\left(\phi-a-b\phi^{-1}\right) = 0
$$
so
$$
\phi = \frac{1}{2} \left(a\pm\sqrt{a^2+4 b}\right)
$$
and
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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With $z\in \mathbb C$ find the maximum value for $\lvert z\rvert$ such that $\lvert z+\frac{1}{z}\rvert=1$
With $z\in \mathbb C$ find the maximum value for |z| such that
$$\left\lvert z+\frac{1}{z}\right\rvert=1.$$
Source: List of problems for math-contest training.
My attempt: it is easy to see that the given c... | Suppose $z + \frac{1}{z} = e^{it}$ for some $t \in \mathbb{R}$. Then solving for $z$ gives
$$z = \frac{e^{it} \pm \sqrt{e^{2it} - 4}}{2} = e^{it} \cdot \frac{1 \pm \sqrt{1 - 4 e^{-2it}}}{2}.$$
Therefore,
$$ |z| = \frac{1}{2} \left| 1 \pm \sqrt{1 - 4 e^{-2it}} \right| \le \frac{1}{2} \left( |1| + |\sqrt{1 - 4 e^{-2it}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\Bigl\{\begin{smallmatrix}x+\frac{3x-y}{x^2+y^2}=3\\y-\frac{x+3y}{x^2+y^2}=0\end{smallmatrix}$ in $\mathbb R$
$$\begin{cases}
x+\dfrac{3x-y}{x^2+y^2}=3 \\
y-\dfrac{x+3y}{x^2+y^2}=0
\end{cases}$$
Solve in the set of real numbers.
The furthest I have got is summing the equations, and I got
$$x^3+(y-3)x^2+(y^... | Note that if $y=0$ then we get $-\frac 1x=0$ on second equation, so the system has no solutions.
So we can assume $y\neq 0$, set $\ x=ty\ $ and substitute.
We get $\begin{cases}t^3y^2+ty^2+3t-1=3t^2y+3y \\t^2y^2+y^2-t-3=0\end{cases}$
And this allow us to isolate $y^2=\dfrac{t+3}{t^2+1}$
Substituting in first equation r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute $\sum\limits_{j = 0}^{m - 1} \left(c_j + 1\right)\ln\left(c_j + 1\right)$ where $c_j = \cos\left(\frac{\pi}{2m}\left(1 + 2j\right) \right)$ As part of solving:
\begin{equation}
I_m = \int_0^1 \ln\left(1 + x^{2m}\right)\:dx.
\end{equation}
where $m \in \mathbb{N}$. I found an unresolved component that I'm unsur... | Here's another, quicker, method (I also don't know if this one works)
Using the same $r_k^{(n)}$ as last time, we apply the $\log\prod_{i}a_i=\sum_i\log a_i$ property to see that
$$\log(1+x^n)=\log\prod_{k=1}^{n}(x-r_k^{(n)})=\sum_{k=1}^{n}\log(x-r_k^{(n)})$$
So
$$I_n=\int_0^1\log(1+x^n)\mathrm dx=\sum_{k=1}^{n}\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
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A Series For the Golden Ratio
Question: Can we show that $$\phi=\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2} $$; where $\phi={1+\sqrt{5} \above 1.5pt 2}$ is the golden ratio ?
Some background and motivation:
Wikipedia only provides one series for the golden ratio - see also the link in the ... | First of all note that
$$\frac1{\sqrt{1-4x}}=\sum_{n=0}^{\infty}\binom{2n}n x^n$$
Lets rewrite your sum as the following
$$\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\frac15\sum_{n=0}^\infty\binom{2n}n\left(\frac1{5^3}\right)^n=\frac15\frac1{\sqrt{1-\left(\frac4{5^3}\right)}}=\frac{\sqrt 5}{11}$$
And therefore you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Find $ \sum_{n=1}^{\infty} \frac{1}{n(n+2)} $ I am doing the telescoping technique with partial sum of this infinite series because I want to find an upper bound of its partial sum.
\begin{equation}
\sum_{n=1}^{\infty} \frac{1}{n(n+2)}
\end{equation}
First I set $ \frac{1}{n(n+2)}= \frac{A}{n}+\frac{B}{n+2}$ and sol... | $$\sum^{\infty}_{n=1}\frac{1}{n(n+2)} = \frac{1}{2}\sum^{\infty}_{n=1}\bigg[\frac{1}{n}-\frac{1}{n+2}\bigg]$$
$$ = \frac{1}{2}\sum^{\infty}_{n=1}\bigg[\bigg(\frac{1}{n}-\frac{1}{n+1}\bigg)+\bigg(\frac{1}{n+1}-\frac{1}{n+2}\bigg)\bigg]$$
$$ = \frac{1}{2}\bigg[\frac{1}{1}+\frac{1}{2}\bigg] = \frac{3}{4}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to divide ${2k^3+3k^2+k-2j^3+3j^2-j}$ with $(k+1-j)$? The question I had was calculating $$\displaystyle\frac{1}{k+1-j}\sum_{i=j}^k i^2$$
Because I didn't know how to do a variable change, I did
$$\frac{1}{k+1-j}\sum_{i=j}^k i^2 = \frac{1}{k+1-j}\left(\sum_{i=1}^k i^2 - \sum_{i=1}^{j-1} i^2\right) = \frac{2k^3+3k^... | Try to write the polynomial in $k$ as a polynomial in $m=k+1$.
$2k^3+3k^2+k=\big[2(k+1)^3-2-6k-6k^2\big]+\big[3(k+1)^2-3-6k\big]+(k+1)-1\\=2(k+1)^3+3(k+1)^2+(k+1)-6(1+2k+k^2)\\=2m^3-3m^2+m$
Therefore, $2m^3-3m^2+m-2j^3+3j^2-j=2(m^3-j^3)-3(m^2-j^2)+m-j\\=(m-j)\big[2(m^2+j^2+mj)-3(m+j)+1\big]$
Divide by $k+1-j=m-j$ and b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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What is the solution for $\int_{0}^{\pi/2}\frac{\cos^2x}{\cos^2x+4\sin^2x}\,dx$ I used the rule :$$\int_{0}^{a}f(x) = \int_{0}^{a}f(a-x)\,dx$$
And got :$$\int_{0}^{\pi/2}\frac{\sin^2x}{\sin^2x+4\cos^2x}\,dx$$
Then, I added both the question and the above integrand, and I got :$$\int_{0}^{\pi/2}\frac{dx}{5}$$
Solving th... | So what you have done is apparently wrong. The equality you used is true, the problemes lies elsewhere.
In fact when you add the two terms you found :
$$A =\frac{\cos^2x}{\cos^2x+4\sin^2x}$$
$$B =\frac{\sin^2x}{\sin^2x+4\cos^2x}$$
$$\implies A + B = \frac {3 \cos(x)^4 + 3 \sin(x)^4+1}{(3 \sin(x)^2 +1) \cdot (3 \cos(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $x>0$ , $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$. Then find limit
For $x>0$, let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$. Then
$$
\lim_{x\to0^+}x\left(\left\lfloor\frac{1}{x}\right\rfloor
+\left\lfloor\frac{2}{x}\right\rfloor+\dots
+\left\lfloor\f... | $\frac{1}{x} -1< \lfloor \frac{1}{x}\rfloor<\frac{1}{x}$
multiplying by x we get
$\implies$x[$\frac{1}{x} -1]< x\lfloor \frac{1}{x}\rfloor<x[\frac{1}{x}]$
this gives $1-x< \lfloor \frac{1}{x}\rfloor<1$
taking limit x $\to 0^{+}$ by squeez theorem ,we get
$ lim_{x\to 0^{+}} x\lfloor \frac{1}{x}\rfloor=1$
similarly,
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solve linear equation system with Gauss I have the following matrix and have to see if it has solutions depending on $a$. My solution:
$M=
\left[ {\begin{array}{cc}
a & a^2 &| &1 \\
-1 & -1& | & -a \\
1 & a & | & a
\end{array} } \right]
$
My attemp was:
Changing first with third line
$=
\left[ {\begin{ar... | \begin{align}
M=
\left[\begin{array}{cc|c}
a & a^2 & 1 \\
-1 & -1 & -a \\
1 & a & a
\end{array}\right]
&\to
\left[\begin{array}{cc|c}
1 & a & a \\
-1 & -1 & -a \\
a & a^2 & 1
\end{array}\right]
&&R_1\leftrightarrow R_3
\\[4px]&\to
\left[\begin{array}{cc|c}
1 & a & a \\
0 & a-1 & 0 \\
0 & 0 & 1-a^2
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that $\int_a^b (f'(x))^2dx - h\sum_{i=0}^{n-1} \left(\frac{f(x_{i+1})-f(x_i)}{h}\right)^2 = O(h^2) $ Prove that:
$\int_a^b (f'(x))^2dx - h\sum_{i=0}^{n-1} {(\frac{f(x_{i+1})-f(x_i)}{h})}^2 = O(h^2) $
Where $h=\frac{b-a}{n} , x_k=a+kh$ and $f \in C^\infty[a,b]$
First I've tried to estimate the following: $\int_{x_... | We have
\begin{align*}
A(x) &\overset{\text{def}}{=} \int_a^b [f'(x)]^2dx - \sum_{k=0}^{n-1}\left(\frac{f(x_{k+1}) - f(x_k)}{h}\right)^2h \\
&= \sum_{k=0}^{n-1}\left\{\color{red}{\int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} - \color{blue}{\left(\frac{f(x_{k+1}) - f(x_k)}{h}\right)^2h}\right\}
\end{align*}
Now, for $x \in [x_k, x_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3070153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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trigonometric integral with tangent $$\int\biggl( \sqrt[3]{\frac{\sin x+1}{\cos x}}+\sqrt[3]\frac{\sin x-1}{\cos x}\biggr)\frac{1}{\cos^2x}\,dx ,\;x\in\Bigl(-\frac{\pi}{2},\frac{\pi}{2}\Bigr)$$ Can somebody give some tips about how can I evaluate this integral,please? I observed that $\frac{1}{\cos^2x}$ is the derivati... | I will use integration by parts twice. First we write
\begin{align}
I &= \int \left (\sqrt[3]{\frac{\sin x + 1}{\cos x}} + \sqrt[3]{\frac{\sin x - 1}{\cos x}} \right ) \sec^2 x \, dx\\
&= \int \big{(}\sqrt[3]{\tan x + \sec x} + \sqrt[3]{\tan x - \sec x} \big{)} \sec^2 x \, dx.
\end{align}
If we let
$$f(x) = \sqrt[3]{\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3071308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Using the R method for finding all solutions to $\sin(2a) - \cos(2a) = \frac{\sqrt{6}}{2}$. My solution differs from official answer.
How many solutions does
$$\sin(2a) - \cos(2a) = \frac{\sqrt{6}}{2}$$
have between $-90^\circ$ and $90^\circ$?
I used the R method and got
$$2a-45^\circ = \arcsin\left(\frac{\sqrt... | You are right. Here is the image of the function using google.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $ I struggle for a while solving limit of this chain:
$
a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right)
$
I know from WolframAlpha result will be $ \frac{1}{4\sqrt{2}} $, but step-by-step solution is overcomplicated(28 steps). Usually I s... | You may proceed as follows transforming the limit into a derivative of a function at $0$.
First set $n = \frac{1}{t}$ and consider $t \to 0^+$:
\begin{eqnarray*} n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right)
& \stackrel{n=\frac{1}{t}}{=} & \frac{\sqrt{\frac{1}{t^2}+\sqrt{\frac{1}{t^4}+1}}-\sqrt{2}\frac{1}{t}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
$a+b=c+d$ , $a^3+b^3=c^3+d^3$ , prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$
Let $a, b, c, d$ be four numbers such that $a + b = c + d$ and $a^3 + b^3 = c^3 + d^3$. Prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$.
I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I ... | Hint: If $a+b=c+d$ and $ab=cd$ then $\{a,b\}=\{c,d\}$. Use the identity
$$x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)^3-3xy(x+y).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
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Evaluating $\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x)\,dx$
How to prove $$\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x) \, dx = \frac43G + \frac13\pi\ln\left(2+\sqrt3\right),$$where $G$ is Catalan's constant?
I have a premonition that this integral is related to $\Im\operatorname{Li}_2\left(2\pm\sqrt3\right)$.
At... | On the path of Kemono Chen...
\begin{align}J&=\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x)dx\end{align}
Perform the change of variable $y=\operatorname{arcsinh}(2\tan x)$,
\begin{align}J&=\int_0^{+\infty}\frac{2x\cosh x}{4+\sinh^2 x}\,dx\\
&=\int_0^{+\infty}\frac{4x\left(\text{e}^{x}+\text{e}^{-x}\right)}{14+\text{e}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 2
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general solution for double integral of square root of quadratic polynomials I wonder if there is a closed-form solution for double integral of square root of quadratic polynomials. Such as
$$
\int _0 ^1 \int _0 ^1 \sqrt{a \cdot x^2 + b \cdot y^2 + c \cdot x \cdot y + d \cdot x+e\cdot y +1 } \ dx dy
$$
I have tried t... | MAPLE got:
$\int _0^1\int_0^1\sqrt{a\cdot x^2+b\cdot y^2+c\cdot x\cdot y+d\cdot x+e\cdot y +1 }\ dx\ dy$
$$=\frac{1}{8}\left(-4\,ab\ln\left({\frac{c+d+2\,\sqrt{b+e+1}\sqrt{a}}{\sqrt{a}}}\right)+4\,ab\ln\left({\frac{2\,a+c+d+2\,\sqrt{a+b+c+d+e+1}\sqrt{a}}{\sqrt{a}}}\right)-4\,ae\ln\left({\frac{c+d+2\,\sqrt{b+e+1}\sqrt{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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For I the incenter in △ABC, if AB+IC=AC+IB, then △ABC is isosceles.
Let $I$ be the center of the inscribed circle in $\triangle ABC$. Prove that if $$AB+IC=AC+IB$$ then the triangle is isosceles!
| Another way.
In the standard notation we have
$$c+\frac{r}{\sin\frac{\gamma}{2}}=b+\frac{r}{\sin\frac{\beta}{2}}$$ or
$$c+\frac{\frac{2S}{a+b+c}}{\sqrt{\frac{1-\frac{a^2+b^2-c^2}{2ab}}{2}}}=b+\frac{\frac{2S}{a+b+c}}{\sqrt{\frac{1-\frac{a^2+c^2-b^2}{2ac}}{2}}}$$ or
$$b-c=\frac{4S\sqrt{ab}}{(a+b+c)\sqrt{(a+c-b)(b+c-a)}}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The Integral $\int \frac {dx}{(x^2-2ax+b)^n}$ Recently I came across this general integral,
$$\int \frac {dx}{(x^2-2ax+b)^n}$$
Putting $x^2-2ax+b=0$ we have,
$$x = a±\sqrt {a^2-b} = a±\sqrt {∆}$$
Hence the integrand can be written as,
$$
\frac {1}{(x^2-2ax+b)^n}
=
\frac {1}{(x-a-\sqrt ∆)^n(x-a+\sqrt ∆)^n}
$$
Resolving ... | Let $b\neq a^2$,
$$S(n)=\int \frac {dx}{(x^2-2ax+b)^n}$$
With method of undetermined coefficients we find formula
$$S(n)=\frac{Ax+B}{(x^2-2ax+b)^{n-1}}+CS(n-1)$$
We get
$$1=-\left( 2 A n-C-3 A\right) \, {{x}^{2}}-\left( \left( 2 B-2 A a\right) n+\left( 2 C+4 A\right) a-2 B\right) x\\+2 B a n-\left( -C-A\right) b-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Extracting two balls with the same color vs distinct color I am trying to find which has higher probability when extracting from a box without replacement: two balls of the same color, or two balls with different colors.
Let's assume that in the box are $a$ white balls and $b$ black balls.
$$P(\text{distinct color})=P(... | Total number of possible extractions is $\binom{a+b}{2}$
Number of ways to extract:
*
*two white balls: $\binom{a}{2}$
*two black balls: $\binom{b}{2}$
*one of each color: $ab$
Probabilities can be calculated as $$\frac{\binom{a}{2}+\binom{b}{2}}{\binom{a+b}{2}}\text{ for same color and }\frac{ab}{\binom{a+b}{2}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_{0}^{\pi/2}\cos^{2n+1}(x)dx$ How can I compute this Integral for integer $n$ from $0$ to $\pi/2$
$$\int_{0}^{\pi/2}\cos^{2n+1}(x)dx?$$
| Keywords: Wallis integrals.
For $k\geq 0$ integer, define
\begin{align}J_k=\int_0^{\frac{\pi}{2}}\cos^{2k+1} x\,dx\end{align}
Perform integration by parts, ($(\sin x)^\prime=\cos x$ )
\begin{align}J_{k+1}&= \int_0^{\frac{\pi}{2}}\cos^{2k+3} x\,dx\\
&=\Big[\sin x\cos^{2k+2} x\Big]_0^{\frac{\pi}{2}}+(2k+2)\int_0^{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3089328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Evaluate $\sum_ \limits{n=1}^{\infty} \frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1) } $ $$\sum_ \limits{n=1}^{\infty} \frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1) } $$
$$1 \cdot 3 \cdot 5 \cdots (2n+1) = \frac{1 \cdot 2 \cdot 3 \cdots (2n+2)}{2 \cdot 4 \cdot 6 \cdots (2n+2)} = \frac{(2n+2)!}{2^{n+1} \cdot (n+1)!} $$
But the fol... | Observe that
$$
\frac{n}{1\cdot 3\cdot 5\cdots (2n+1)}=\frac{1}{2}\left(\frac{1}{1\cdot 3\cdot 5\cdots (2n-1)}-\frac{1}{1\cdot3\cdot 5\cdots (2n+1)}\right)
$$
Hence
$$\sum_{n=1}^\infty\frac{n}{1\cdot 3\cdot 5\cdots (2n+1)}=\sum_{n=1}^\infty\frac{1}{2}\left(\frac{1}{1\cdot 3\cdot 5\cdots (2n-1)}-\frac{1}{1\cdot 3\cdot 5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Evaluate the path integral of $f(x, y) = y$ over the graph of the semicircle $y = \sqrt{1-x^2}, -1 \leq x \leq 1$ Evaluate the path integral of $f(x, y) = y$ over the graph of the semicircle $y = \sqrt{1-x^2}, -1 \leq x \leq 1$
Solution attempt:
$f(x, y) = y$, along $y = \sqrt{1-x^2}, -1 \leq x \leq 1$
$\vec{r}(t) = (t... | You can also parameterise the curve $x=cost$, $y=sint$
$$r'(t) = <-sint,cost>$$
$$||r'(t)|| = 1$$
$$F = sin t$$
$$I = \int_{0}^{\pi} sint dt = -cost|_{0}^{\pi} = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving irrationality of $\sqrt[3]{3}+\sqrt[3]{9}$ I need to prove $$\sqrt[3]{3}+\sqrt[3]{9}$$
is irrational, I assumed
$$\sqrt[3]{3}+\sqrt[3]{9} = \frac{m}{n}$$
I cubed both sides and got
$$\sqrt[3]{3}+\sqrt[3]{9} = \frac{m^3-12n^2}{9n^3}$$
I tried setting $$\frac{m^3-12n^2}{9n^3} = \frac{m}{n}$$ but that led me nowhe... | Here are two other takes.
Take 1
Let $\alpha = \sqrt[3]{3}+\sqrt[3]{9}$. Then $\alpha^3 = 9 \alpha + 12$.
By the rational root theorem, if $\alpha$ is rational, then $\alpha$ is an integer.
Now $1 < \sqrt[3]{3} < 2 $ and $2 < \sqrt[3]{9} < 3 $, and so $3 < \alpha < 5$.
Since $x=4$ is not a root of $x^3 = 9 x + 12$, $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Calculating the determinant of a matrix $n\times n$ I was trying to calculate $detD_{1,1}$ when
$$ D=\begin{pmatrix}0 & -1 & -1 & \ldots & \ldots & -1 & -1 & -1\\
0 & 2 & -1 & \ldots & \ldots & 0 & 0 & 0\\
0 & 0 & 2 & \ldots & \ldots & -1 & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \ldots & \vdots & \vdots & \vdots\... | Move the bottom row of the matrix to the top (permuting the rows cyclically) to find
$$
\begin{vmatrix}0 & -1 & 0 & \ldots & \ldots & 0 & 0 & 4\\
0 & 0 & -1 & \ldots & \ldots & 0 & 0 & 8\\
0 & 0 & 0 & \ldots & \ldots & -1 & 0 & 16\\
\vdots & \vdots & \vdots & \ddots & \ldots & \vdots & \vdots & \vdots\\
\vdots & \vdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simplifying compound fraction not producing answer provided by book I am working on a problem in a textbook(Precalculus Mathematics for Calculus, by James Stewart) and the answer in the back of the book for the problem(1.4 #67) is -xy but I am not getting that. Here is the problem:
\begin{align}
\frac{\frac{x}{y}-\fra... | You were almost there. $\frac{xy(x-y)}{y-x}$(your second last one)$=-xy$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3093144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to change the variables in $\frac{1}{(z+1)^{n+1}}=\frac{(-1)^n}{n!}\frac{d^n}{dz^n}\left( \frac{1}{z+1}\right)$? We know that
$$\frac{1}{(z+1)^{n+1}}=\frac{(-1)^n}{n!}\frac{d^n}{dz^n}\left( \frac{1}{z+1}\right).$$
If we put $z=e^t$ how can we change the variables above equation?
I know that $z\frac{d}{dz}=\frac{d}... | In general, if $f(z)$ is $n$ times differentiable, then we have
$$\left(z\frac{d}{dz}\right)^n\{f(z)\}=\sum_{k=1}^n \begin{Bmatrix} n \\ k \end{Bmatrix}z^k\frac{d^k\,f(z)}{dz^k}$$
where $\begin{Bmatrix} n \\ k \end{Bmatrix}=\frac1{k!}\sum_{j=0}^k (-1)^{k-j}\binom{k}{j}\,j^n$ are the Stirling Numbers of the Second Kind... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simplify $\frac{x^3-x}{x^2+xy+x+y}$ $$\frac{x^3-x}{x^2+xy+x+y}$$
What I did:
$$\frac{x}{xy+x+y}$$
through simplifying the $x$'s.
But it's not right.
What did I do wrong?
| Notice that $$\frac{x^3-x}{x^2+xy+x+y} = \frac{(x+1)(x^2-x)}{(x+1)(x+y)} =\frac{x^2-x}{x+y}$$
P.S.: You cannot simplify the $x$'s without considering all terms of the numerator and the denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3096139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\lim_{n \to \infty} (\frac{n^2-1}{2n^2+3})=\frac{1}{2}$ and $\lim_{n\to\infty} \frac{1}{\ln(n+1)}=0$
Use the definition of the limit of a sequence to prove that $\lim_{n \to \infty} (\frac{n^2-1}{2n^2+3})=\frac{1}{2}$.
We have
$$\begin{align}
\left|\frac{n^2-1}{2n^2+3}-\frac{1}{2}\right| & =\left|\frac{2... | Your answer is correct. However it seems you may have overcomplicated it.
\begin{align}
\lim_{n\to \infty}\left(\frac{n^2-1}{2n^2+3}\right) & = \lim_{n\to \infty}\left(\frac{1-\frac{1}{n^2}}{2+\frac{3}{n^2}}\right)\\
\end{align}
Now use the fact that $\lim_{n\to \infty}\left(\frac{1}{n^k}\right)=0$, where $k$ is any po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3098037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the $f$ of $g$ where $f(x) =x^2+2x+1$ and $g(x)=2x-3$ According to my textbook the $f$ of $g$ where $f(x) =x^2+2x+1$ and $g(x)=2x-3$ is $4x^2-8x+2$, but I get $4x^2-8x+4$.
What am I doing wrong?
My Steps:
$(2x-3)^2 + 2(2x-3)+1$
$(2x-3)(2x-3)+4x-6+1$
$4x^2-12x+9+4x-6+1$
$4x^2-8x+4$
| A simpler approach to your calculation is to factor $f$ first:
$$f(x) = x^2 + 2x + 1 = (x+1)^2.$$
Then, $$f(g(x)) = (g(x)+1)^2 = ((2x-3)+1)^2 = (2x-2)^2 = 2^2(x-1)^2 = 4x^2 - 8x + 4.$$
To check that the book's solution cannot be correct, it suffices to choose $x = 0$, which leads to $g(0) = -3$, and $$f(g(0)) = f(-3) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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A sum of Fibonacci numbers Let $F_n$ be the Fibonacci numbers. I would like to prove this really messy identity:
$$\sum_{n=0}^{\infty}(-1)^n(n+1)^2\frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n+2\choose 1}{2n+2 \choose 2}{2n \choose n}}=F_{k}\tag1$$
which simplifies to:
$$\sum_{n=0}^{\infty}(-1)^n\frac{5n^2... | Here is another answer, in the same spirit as that of @dan_fulea:
Both sides of $$\sum_{n=0}^{\infty}(-1)^n\frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n \choose n}(4n+2)}=F_{k+1}$$ can be considered as sequences defined by the same second order linear recurrence for the index $k$. Then, for them to be equal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3100501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Proving $\left(1+\frac{1}{m}\right)^m < \left(1+\frac{1}{n}\right)^n$ Let $m,n\in \mathbb{N}$. If $m > n$ show that
$$\left(1+\frac{1}{m}\right)^m > \left(1+\frac{1}{n}\right)^n$$
My works: I tried to show if $g(x)=\left(1+\frac{1}{x}\right)^x$ then $g'(x) > 0$.
\begin{align}
g'(x) &= \frac{d e^ { x \ln(1+\frac{1}{x}) ... | $$
\frac{d}{dx} \exp \left( \ln\left(1 + \frac{1}{x}\right) x \right) = \exp \left( \ln\left(1 + \frac{1}{x}\right) x \right) \left( \ln\left(1 + \frac{1}{x}\right) - \frac{1}{x + 1} \right) > 0
$$
for $x > 1$, so that the function is strictly increasing on $(1, \infty)$. Indeed,
$$
\ln\left(1 + \frac{1}{x}\right) = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How do we factor out $x^2 - x -2$ in this expression? Suppose we're given that
$$x^4 - 2x^3 +x-2$$
How do we factor out $x^2 - x -2$ in this expression?
$$(x^4 -x^3- 2x^2)-(x^3-x^2-2x)+(x^2-x-2) = x^2(x^2 -x-2)-x(x^2-x-2)+(x^2-x-2) = (x^2-x-2)(x^2-x+1)$$
This satisfies with what we want to get. However, I do not seem... | An easier way is to note that $x^4-2x^3+x-2$ has the simple roots $x=-1$ and $x=2$, which are obvious candidates by the rational root theorem, so your polynomial is divisible by
$$(x+1)(x-2)=x^2-x-2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3106729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Show, that the sequence $ a_n = \left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{9}\right)\cdot\ldots\cdot\left(1+\frac{1}{3^n}\right) $ converges. I need to show, that the following sequence converges.
I think I can somehow do it by Riemann Integral, but I cannot figure out a way to extract $ \frac{1}{n} $ from it. I a... | If you are ok with using GM-AM (inequality between geometric and arithmetic mean) and with using the well known limit
*
*$\left(1+ \frac{x}{n}\right)^n\stackrel{n\to \infty}{\longrightarrow}e^x$
then you can reason directly as follows:
\begin{eqnarray*} \prod_{k=1}^n \left(1+\frac{1}{3^k}\right)
& \leq & \left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3107650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding all real $x$ such that $1+\sum_{j=1}^n\sin{\frac{j\pi x}{n+1}} = 0$, where $n=18$.
The task is to find all $ x \in \mathbb R $ such that
$$ 1 + \sum_{j=1}^n \sin{\frac{j\pi x}{n + 1}} = 0, \qquad n = 18 $$
What I have tried
Using the following formulas:
$$1. \sin{x} + \sin{y} = 2\sin{\frac{x + y}{2}}\cos{\fra... | Multiply the series by $\frac {\sin \frac {\pi x}{19}}{\sin \frac {\pi x}{19}}$
Then use
$\sin A\sin B = \frac 12 \cos (A-B) - \frac 12 \cos (A+B)$
And the series will telescope.
$1+\sum_\limits{j=1}^{18} \frac {\sin \frac{j\pi}{19}\sin \frac {\pi x}{19}}{\sin\frac {\pi x}{19}}$
$1+\sum_\limits{j=1}^{18} \left(\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3110306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
An algebra question based on sequences and series The following question was asked in JEE Advanced 2014 Paper I:
A pack contains $n$ cards numbered sequentially from $1$ to $n$. Two consecutive cards are removed from the pack and the sum of the numbers on the remaining cards is $1224$. If the number on the smaller ca... | Let the two removed cards be $k$ and $k + 1$.
$1 + 2 + .... n = \frac {n(n+1)}2,$ and removing $k$ and $k+1$ and adding them up we get $\frac {n(n+1)}2 - k -(k+1) =\frac {n(n+1)}2 - 2k-1 = 1224,$
so $k = \frac {n(n+1)}4 - \frac {1225}2,$ which means $\frac {n(n+1)}2$ is odd and that
$\frac {n(n+1)}2 > 1225$.
So $n(n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3110449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the particular solution of the differential equation $\sqrt{x} + \sqrt{y}y' =0$, with $y(1) = 9$.
I get:
$\sqrt{y}dy = -\sqrt{x}dx \Rightarrow \frac{2}{3}y^{\frac{3}{2}} = -\frac{2}{3}x^{\frac{3}{2}}+C$.
What should I do from here?
| Multiply through by $3/2$ and rename $\frac32 C = c$
$$ y^{3/2} = c - x^{3/2} $$
The initial condition $y(1)=9$ gives $c = 28$. Solving for $y$ gives
$$ y^{3/2} = 28 - x^{3/2} \implies y = (28-x^{3/2})^{2/3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding limit of a (Laurent?) series I've been practicing series for my upcoming Calculus 1 exam, and I've stumbled upon this one:
$1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + ... + \frac{1}{1 + 2 + 3 + ... + n}$
The task is to find the limit.
| As $1+2+\ldots +n = \frac{n}{2}(n+1)$ you are looking for the sum of the series
$S = \sum_{n=1}^{\infty} \frac{1}{\frac{k}{2}(k+1)} = 2 \sum_{k=1}^{\infty} \frac{1}{k(k+1)}$.
Nos separating you get that $S_n = 2\sum_{k=1}^{n} \frac{1}{k} - \frac{1}{k+1} = 2(1 - \frac{1}{n+1})$. Now taking limit when $n \rightarrow \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3116293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Why is $\frac{\frac{a}{b} - \frac{b}{a}}{\frac{a+b}{ab}}$ = a - b? I'm given the complex rational expression
$$\frac{\frac{a}{b} - \frac{b}{a}}{\frac{a+b}{ab}}$$
and asked to simplify. The solution provided is $a - b$; however I get $\frac{a^2 - b^2}{a+b}$.
My working:
Numerator first:
$$\frac{a}{b} - \frac{b}{a}$$
Le... | $$a^2-b^2=(a-b)(a+b)$$ If $a+b\ne 0$, you can divide both the numerator and denominator by it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3119058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Suppose $a, b, c \in I$ such that greatest common divisor of $x^2 + ax + b$ and $x^2 + bx + c$ is $(x + 1)$ and the least common multiple...
Suppose $a, b, c \in I$ such that greatest common divisor of $x^2 + ax + b$ and $x^2 + bx + c$ is $(x + 1)$ and the least common multiple of $x^2 + ax + b$ and $x^2 + bx + c$ is ... | Using GCD $\times$ LCM = Product, we get
$$(x+1)(x^3 - 4x^2 + x + 6) = (x^2+ax+b)(x^2+bx+c)$$
Now it remains to equate coefficients on both sides to get $(a, b, c) = (-1, -2, -3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3120909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
u-sub vs trig-sub are giving different answers for $\int\frac{x+4}{x^2+2x+5}dx$ When I complete the square in the denominator and solve using u-sub, I can get the right answer:
$$\int\frac{x+4}{x^2+2x+5}dx$$
$$\int\frac{x+4}{x^2+2x+1+4}dx$$
$$\int\frac{(x+1)+4}{(x+1)^2+4}dx$$
$$u=x+1$$
$$\int\frac{u+3}{u^2+4}du$$
$$I_1... | You plugged in the wrong expression for $\cos \theta$, it should be $2/\sqrt{u^2+4}$. Notice that the Pythagorean theorem tells you that the length of the hypotenuse should be $\sqrt{x^2+2x+5}$.
Note the minus sign outside of the logarithm. $$-\ln\left|\dfrac{2}{\sqrt{u^2+4}} \right|=\ln\left|\dfrac{\sqrt{u^2+4}}{2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3122242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Undo this transformation I have two variables $x$, $y$ and calculate the following:
$a = \frac{x}{\sqrt{x^2+y^2}}$, $b = \frac{y}{\sqrt{x^2+y^2}}$
Using $a$ and $b$ is there a way I can derive my original $x$ and $y$?
| By squaring we get $$a^2=\frac{x^2}{x^2+y^2},b^2=\frac{y^2}{x^2+y^2}$$ and we get
$$a^2+b^2=\frac{x^2+y^2}{x^2+y^2}=1$$
and you can not compute $x$ or $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Computing $\int_{-a}^a \int_{-b}^b \frac 1{(x^2 + y^2 +c^2)^{3/2}}\, dxdy$. The question is exactly as in the title:
$$\int_{-a}^a \int_{-b}^b \frac 1{(x^2 + y^2 +c^2)^{3/2}}\, dxdy$$
It's been so long since the last time I tried to calculate something like this. I first thought about polar coordinates but that does... | Let
$$ F(c)=\int_{-a}^a\int_{-b}^b \frac {1}{(x^2 + y^2 +c^2)^{1/2}}dxdy $$
and then
$$ F'(c)=-c\int_{-a}^a\int_{-b}^b \frac {1}{(x^2 + y^2 +c^2)^{3/2}}dxdy $$
or
$$ \int_{-a}^a\int_{-b}^b \frac {1}{(x^2 + y^2 +c^2)^{3/2}}dxdy=-\frac{F'(c)}{c}. $$
Since
$$ \int\frac {1}{(x^2 + c^2)^{1/2}}dx=\ln(x+\sqrt{x^2+c^2})$$
one ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3125280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the $n^{th}$ term derivative of $f(x) = (x^2-x-1)(\ln(1-x))$? I have the first three terms but am struggling with finding the $n^{th}$ term derivative of the function. Here is my work:
$$\\$$
$$f(x) = (x^2-x-1)(\ln(1-x)) $$
$$f'(x) = (2x-1)(\ln(1-x))-\left(\dfrac{x^2-x-1}{1-x}\right)$$
$$f''(x) = \dfrac{3x^2-5x... | Use the following.
$$f'''(x)=\frac{2x^2-5x+1}{(x-1)^3}=\frac{2x^2-4x+2-x+1-2}{(x-1)^3}=\frac{2}{x-1}-\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3}.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
finding the mean and variance descriptive statistics Suppose the following data are obtained by recording $X$, the number of customers that arrive at an automatic banking machine during $15$ successive one-minute
time intervals.
Q) Record the mean and variance.
mean is
$u_{X} = \sum_{x=1}^{15} x f_{X}(x) = 1.67$
usin... | Your method finds the population variance where
$$\begin{align*}
\mathsf{Var}(X)
&=\mathsf E(X^2)-\mathsf E(X)^2\\\\
&=0^2\left(\frac{4}{15}\right)+1^2\left(\frac{3}{15}\right)+2^2\left(\frac{4}{15}\right)+3^2\left(\frac{2}{15}\right)+4^2\left(\frac{2}{15}\right)-\left(\frac{25}{15}\right)^2\\\\
&=4.6-\left(\frac{25}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3127041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Factor $x^5-x+15$ It's possible to factor $x^5-x+15$. WolframAlpha gives the answer of: $$(x^2+x+3)(x^3-x^2-2x+5)$$
According to the wikipedia article on quintic functions, the general form $x^5-x+a$ is factorable only when $a=±15$, $±22440$, or $±2759640$.
Question: How would one factor such an expression? For me, it ... | One can factor $$f(x) := x^5 - x \pm 15$$ manually (over $\Bbb Z$) without too much fuss.
First, if $f$ has a linear factor, it has a rational root and (because $f$ is monic) any rational root must be an integer. But $f(0) \equiv f(1) \equiv 1 \pmod 2$, so $f$ has no root modulo $2$ and hence no integer root and hence ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3128335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
In equilateral ABC , Proving : $3(a^4 + b^4 + c^4 + s^4 ) = (a^2 + b^2 + c^2 + s^2 )^2$ My question is that:
In equilateral $\triangle ABC$, $AB= s$
There is a point P in same plane , the distance to three vertices are $a, b, c $
then
$$ 3(a^4 + b^4 + c^4 + s^4 ) = (a^2 + b^2 + c^2 + s^2 )^2 $$
Is this equation true?... | This is not a solution, although the equality seems to be true. One way to approach the problem might be to work with the points themselves as opposed to distances.
We can put your triangle on a coordinate plane and say that Point A is at $(-\frac{s}{2} , 0)$, Point B is at $(\frac{s}{2}, 0)$, and Point C is at $(0, \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3129491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Finding the vertex, axis, focus, directrix, and latus rectum of the parabola $\sqrt{x/a}+\sqrt{y/b}=1$
Find the vertex, axis, focus, directrix, and length of latus rectum of the parabola
$$\displaystyle \sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1$$
Try: Curve
$$\sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1$$
represents a ... | The given equation $\displaystyle \sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1$
can be simplified as
$$\displaystyle \sqrt{\frac{x}{a}} = 1 - \sqrt{\frac{y}{b}}$$
Squarring both sides,
$$\displaystyle \frac{x}{a}= 1 - 2\sqrt{\frac{y}{b}} + \frac{y}{b}$$
$$\displaystyle \frac{x}{a}-1+ \frac{y}{b}=- 2\sqrt{\frac{y}{b}}$$
Squ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3130185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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What's maximum value of $x (1-x^2)$ for $0 < x <1$? Since we are being taught about AM-GM inequality, I decided to use the method however I am getting two different answers by slightly different methods.
Method 1
\begin{equation}
v=x (1-x^2)$
\implies v^2=x^2 (1-x^2)^2
\end{equation}
Using the AM-GM-inequality we obtai... | Notice the A.M/G.M theorem says:
$a_1 + a_2 + .. + a_n \ge n\sqrt[n]{a_1a_2...a_n}$. And that $a_1 + a_2 + .. + a_n = n\sqrt[n]{a_1a_2...a_n}$ if and ONLY IF $a_1 = a_2 = .... = a_n$.
Your calculations for the inequalities were correct. But you ignored the requirements for equality.
So by your calculations.
$v \le ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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What would be the best way to memorize the 10 by 10 multiplication table? Hear me out before you start downvoting please. I have a learning disability so no matter how hard I try I can’t memorize the table. Please give some tips/hints on how to memorize the table. Thanks in advance.
| Without more context on why you're having difficulty, we can't specifically tailor an answer to you.
"Memorizing" the times table isn't the most fruitful endeavor, but instead you should try and memorize the patterns that pop up. For example, $$\begin{matrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\2 & & & & 10 & & ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Image of a region bounded by lines $x-y <2$ and $x+y>2$, under mapping $w=1/z$ I got $z = x+iy$ and $w=u+iv=\frac1 z = \frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}$
$u(x,y) = \frac{x}{x^2+y^2}$ and $v(x,y) = -\frac{y}{x^2+y^2}$
also $x(u,v) = \frac{u}{u^2+v^2}$ and $\;y(u,v) = -\frac{v}{u^2+v^2}$
$x-y<2 \implies \dfrac{u+v}{u^... | Answering my own question. I would like input/correction form any complex-analysis expert out there.
$\frac{u+v}{u^2+v^2} <2$ when compared to the standard equation of circle:
$x^2+y^2+\frac{B}{A}x+\frac{C}{A}y+D = 0$, gives a region in uv- plane which is outside the circle centered at (0.25,0.25) and radius=$\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Units in this ring Consider the ring $ R = \mathbb{Z}[\frac{1 + \sqrt{-15}}{2}]$, then what are the units of this ring?
My try;
Let $\alpha \in R$. Then $\alpha = a + b(\frac{1 + \sqrt{-15}}{2})$, for some $a,b \in \mathbb{Z}$.
Now, $\alpha$ is a unity iff $N(\alpha) = 1$ iff $(a + b(\frac{1 + \sqrt{-15}}{2}))(a - b(... | In general $\alpha$ is a unit in $R$ if and only if $N(\alpha)$ is a unit in $\Bbb{Z}$. You haven't considered the case $N(\alpha)=-1$.
Second, the norm of $\alpha=a+b\left(\tfrac{1 + \sqrt{-15}}{2}\right)$ is not equal to
$$N(\alpha)=\left(a + b\left(\frac{1 + \sqrt{-15}}{2}\right)\right)\left(a - b\left(\frac{1 + \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating indefinite integrals of the form $\int \frac{x^2 \,dx}{a x^5 + b}$
Evaluate the indefinite integral
$$\int \frac{x^2 \,dx}{a x^5 + b},$$
for real parameters $a, b \neq 0$.
No apparent substitutions simplify the expression (if the exponent of $x$ in the denominator were an integral multiple of $3$, the form... | Old post, but the original author bumped it so...
We want to scale the problem so we can concentrate on
$$\int\frac{x^2dx}{x^5+1}=\sum_{n=0}^4\int\frac{A_ndx}{x-x_n}=\sum_{n=0}^4A_n\ln(x-x_n)+C_1$$
Here $x_n^5=-1=e^{i(2n+1)\pi}$ so $x_n=e^{\frac{i(2n+1)\pi}5}$,
$$A_n=\lim_{x\rightarrow x_n}\frac{(x-x_n)x^2}{x^5+1}=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3135455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Given three a-triangle-sidelengths $a,b,c$. Prove that $3\left((a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\right)\geqq b(a+b-c)(a-c)(c-b)$ . If you are interested in IMO 1983 please see: Given three a-triangle-sidelengths $a,b,c$. Prove that:
$$3\left ( a^{2}b(a- b)+ b^{2}c(b- c)+ c^{2}a(c- a) \right )\geqq b(a+ b- c)(a- c)(c... | The task is homogenius. Let WLOG
$$a+b+c=2,\quad a,b,c \in(0,1),\tag1$$
$$f(a,b,c) = 3\big(a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\big) - b(a+b-c)(a-c)(c-b).\tag2$$
Using of the substiutions
$$a=x,\quad b= 1-xy,\quad c=1-x+xy,\quad x,y\in(0,1),\tag3$$
provides the conditions $(1)$ and allows to get
$\quad f\big(x,1-xy,1-x+xy\big... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
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Prove that the number $7^n+1$ is divisible by $8$ if $n$ is odd. In the case where $n$ is even, give the remainder of the division of $7^n+1$ by $8$.
Prove that the number $7^n+1$ is divisible by $8$ if $n$ is odd. In the case where $n$ is even, give the remainder of the division of $7^n+1$ by $8$.
*
*If $n$ is od... | Note that
$\forall n \in \Bbb N, \; 7^n + 1 = (8 - 1)^n + 1 = \displaystyle \sum_0^n \dfrac{n!}{k!(n - k)!}8^{n - k}(-1)^k + 1 = \sum_0^{n - 1} \dfrac{n!}{k!(n - k)!}8^{n - k}(-1)^k + (-1)^n + 1 , \tag 1$
via the binomial theorem; when $n$ is odd this yields
$7^n + 1 = \displaystyle 8\sum_0^{n - 1} \dfrac{n!}{k!(n - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integrate $\int \frac{dx}{(x^2-1)^{\frac{3}{2}}}$ via trig substitution $x = a\sec\theta, dx = \sec\theta \tan\theta$
$\int \frac{dx}{(x^2-1)^{\frac{3}{2}}}$ =
$ \int \frac{dx}{(\tan^2\theta)^{\frac{3}{2}}}$ =
$ \int \frac{dx}{\tan^{\frac{7}{2}}\theta}$ =
$\int \frac{\sec\theta \tan\theta}{\tan\theta ^{\frac{7}{2}}}$... | $$\int \frac{dx}{(x^2-1)^{\frac{3}{2}}} =
\int \frac{\tan\theta\sec\theta d\theta}{\underbrace{(\tan^2\theta)^{\frac{3}{2}}}_{\tan^3\theta}} =
\int \frac{\sec\theta d\theta}{\tan^2\theta}=
-\frac1{\sin \theta}=-\frac{x}{\sqrt{x^2-1}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How do find the fourth derivative of $e^{x^2}$ So using chain rule for the first derivative:
$$f' = 2x\cdot e^{x^2}$$
then using product rule for the $$f'' = 2x(2xe^{x^2}) + e^{x^2}\cdot 2$$
Is there an easy way to get the third? Does the second deriative simplify to:
$$4x^2e^{x^2} + 2e^{x^2} = (4x^2 + 2)(e^{x^2})$$
Ca... | Use Maclaurin expansion: $y=e^{x^2}=\sum_{n=0}^{\infty} \frac{x^{2n}}{n!}$:
$$\begin{align}y'&=\sum_{n=1}^{\infty} \frac{2nx^{2n-1}}{n!}\\
y'' &=\sum_{n=1}^{\infty} \frac{2n(2n-1)x^{2n-2}}{n!}\\
y''' &=\sum_{n=2}^{\infty} \frac{2n(2n-1)(2n-2)x^{2n-3}}{n!}\\
y^{(4)}&=\sum_{n=2}^{\infty} \frac{2n(2n-1)(2n-2)(2n-3)x^{2n-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Trig Subsitution When There's No Square Root I would say I'm rather good at doing trig substitution when there is a square root, but when there isn't one, I'm lost.
I'm currently trying to solve the following question:
$$Ar \int_a^\infty \frac{dx}{(r^2+x^2)^{(3/2)}}$$
Anyway, so far, I have that:
$$x = r\tan \theta$$
$... | You are doing $(r\sec\theta)^3=r^6\sec^6\theta$. Oops! ;-)
There's a slicker way to do it.
Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes
$$
\frac{A}{r}\int_{a/r}^{\infty}\frac{1}{(1+u^2)^{3/2}}\,du
$$
Now let's concentrate on the antiderivative
$$
\int\frac{1}{(1+u^2)^{3/2}}\,du=
\int\frac{1+u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3142908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Picture proof that the area of a right triangle is $xy$ I stumbled on the following result by accident:
Let $A, B, C$ be the vertices of a right triangle, with opposite side lengths $a, b, c$ respectively, where $\angle C = 90^\circ$ and $a^2 + b^2 = c^2$.
Draw the incircle, and let $x, y, z$ be the length of the tang... | Here is an answer to @6005's comment "Maybe there is a way to also show $(x-z)(y-z)=z^2$ pictorally".
Referring to the pic below, it's immediate from the construction that, comparing it with the cut up area, A, of the triangle, the shaded area, $S$ equals $A-2z^2$. Then, $W = (x-z)(y-z) = 2A -S -A = 2z^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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$\lim\limits_{n\to\infty} \prod\limits_{k=1}^{n} \left( 1 + \tan{\frac{k}{n^2}} \right) $ I want to calculate $$\lim\limits_{n\to\infty} \prod_{k=1}^{n} \left( 1 + \tan{\frac{k}{n^2}} \right) $$
Taking logarithms, it's enough to find
$$\lim\limits_{n\to\infty} \sum_{k=1}^{n} \log\left( 1 + \tan{\frac{k}{n^2}} \right).$... | Too long for a comment.
As I wrote in comments, composition of Taylor series is not only good for the limit but also allows quick and reasonable approximation of the partial product
$$P_n= \prod_{k=1}^{n} \left( 1 + \tan \left(\frac{k}{n^2}\right) \right)$$ Doing what you did taking logarithms and Taylor expansion, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find the number of roots of the equation, $x^3 + x^2 +2x +\sin x = 0$ in $[-2\pi , 2\pi]$.
Find the number of roots of the equation,
$$x^3 + x^2 +2x +\sin x = 0$$
in $[-2\pi , 2\pi]$.
What I have tried:
$$x^3 + x^2 +2x = -\sin x$$
$$x^2 +x +2 = \frac{-\sin x }{x}$$
$$(x + \frac{1}{2})^2 + \frac{7}{4} = \frac{-\sin ... | $\forall x~(f'(x)=3x^2+2x+2+\cos x \gt 0)$ (the polynomial term is always at least $5/3$), so note that $f(-2 \pi) \lt 0, f(2 \pi) \gt 0$ and conclude that there is exactly $1$ root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Find minimum of $4(a^3 + b^3 + c^3) + 15abc$ subject to $a + b + c = 2$
$a$, $b$ and $c$ are three sides of a triangle such that $a + b + c = 2$. Calculate the minimum value of $$\large P = 4(a^3 + b^3 + c^3) + 15abc$$
Every task asking for finding the minimum value of an expression containing the product of all of t... | First, using the identity
$$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca),$$
we have
$$a^3 + b^3 + c^3 - 3abc = 8 - 6(ab + bc + ca). \tag{1}$$
Second, using three degree Schur
$$a^3 + b^3 + c^3 + 3abc \ge ab(a + b) + bc(b + c) + ca(c + a),$$
we have
$$a^3 + b^3 + c^3 + 6abc
\ge ab(a + b) + bc(b ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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Prove that $\left\lfloor \frac {n+3}{2} \right\rfloor=\left\lceil \frac {n+2}{2}\right\rceil$ Prove that $$\left\lfloor \frac {n+3}{2} \right\rfloor=\left\lceil \frac {n+2}{2}\right\rceil$$
I have tried to solve this on my own, and I want to check my solution.
My steps:
Set $x=\left\lfloor \frac {n+3}{2}\right\rfloor $... | You can pull integers out of floor/ceiling, so that setting $n=2k$ or $n=2k+1$ and pulling $k+1$
$$\left\lfloor \frac {2k+3}{2}\right \rfloor=\left\lceil \frac {2k+2}{2}\right\rceil$$ is equivalent to
$$\left\lfloor \frac {1}{2}\right \rfloor=\left\lceil \frac {0}{2}\right\rceil,$$
and
$$\left\lfloor \frac {2k+1+3}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3154243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the surface area of two solids of revolution I have a couple of questions that are similar in nature:
1) I am trying to find the surface area of this when I rotate it around the x-axis. I have $y = \sqrt{5-x}$ when $3 \leq x \leq 5$
Say $dy/dx = \frac{1}{2}(5-x)^{\frac{-1}{2}}$. Then $(dy/dx)^2 = \frac{1}{4}(5... | For the first question, notice:
$$\begin{align}
\sqrt{5-x} * \sqrt{1 + \frac{1}{4} (5-x)^{-1}} &= \sqrt{(5-x)\left( 1 + \frac{1}{4} (5-x)^{-1} \right)} \\
&= \sqrt{(5-x)+(1/4)} \\
&= \sqrt{-x+ \frac{21}{4}}
\end{align}$$
Thus the integral becomes
$$SA = 2\pi \int_3^5 \sqrt{-x+ \frac{21}{4}} dx$$
A simple $u$-substitut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3156270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Solving $f(yf(x)+x/y)=xyf(x^2+y^2)$ over the reals
Find all functions $f:\mathbb{R}\to \mathbb{R}$ such that $f(1)=1$ and for all real numbers $x$ and $y$ with $y \neq 0$, $$f\Bigg (yf(x)+\frac{x}{y}\Bigg)=xyf(x^2+y^2)$$
This seems quite hard. $f(x)=\begin{cases}\frac{1}{x}, x\neq 0 \\ 0, x=0 \end{cases}$ works by i... | Let $f:\textbf{R}\rightarrow\textbf{R}$, such that $f(1)=1$ and
$$
f\left(yf(x)+\frac{x}{y}\right)=xyf(x^2+y^2)\textrm{, }\forall (x,y)\in\textbf{R}\times\textbf{R}^{*}\tag 1
$$
Set $x=0$, $y=1$ in (1), then easily $f(0)=0$ (the case $f(0)\neq 0$ is trivial). For $y=1$ in (1) we get
$$
xf(x^2+1)=f(f(x)+x)\textrm{, }x\n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3156382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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Is it possible to express accuracy knowing sensitivity and specificity?
I know the value of the sensitivity Se and the value of the specificity of Sp, they are equal to 78.65 and 90.00, respectively. I know nothing but this. Can I somehow of the equations, which in the photo express the value of the accuracy Ac?
| Consider
$$X = \frac{A}{A+B}, \ Y = \frac{C}{C+D}, \ Z = \frac{A+C}{A+B+C+D}$$
Then $X$ and $Y$ don't determine $Z$. Counterexample:
$$A=1, B = 2, C = 3, D = 9 \Rightarrow X = \frac{1}{3}, Y = \frac{1}{4}, Z = \frac{4}{15}$$
$$A=3, B = 6, C = 1, D = 3 \Rightarrow X = \frac{1}{3}, Y = \frac{1}{4}, Z = \frac{4}{13}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the following system of equations - (3)
Solve the following system of equations:
$$\large
\left\{
\begin{align*}
3x^2 + xy - 4x + 2y - 2 = 0\\
x(x + 1) + y(y + 1) = 4
\end{align*}
\right.
$$
I tried writing the first equation as $(x - 2)(3x + y - 10) = -18$, but it didn't help.
| The resultant of $3\,{x}^{2}-xy-4\,x+2\,y-2$ and $
x \left( x+1 \right) +y \left( y+1 \right) -4$ with respect to $y$ is
$$ 10\,{x}^{4}-24\,{x}^{3}-10\,{x}^{2}+42\,x-8$$
which is irreducible over the rationals. Its roots can be written in terms of
radicals, but they are far from pleasant. There are two real roots,
$... | {
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"url": "https://math.stackexchange.com/questions/3158340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Calculate the minimum value of $\left(\frac{n + p}{mx} + \frac{p + m}{ny} + \frac{m + n}{pz}\right) \cdot \sqrt{yz + zx + xy}$. Going back a few more years and you can find more and more interesting problems over the years as time turns back. I am still surprised at how easy this competition has become. Then I come acr... | Let $a = mx, b = ny, c = pz$. We have that $\dfrac{n + p}{mx} + \dfrac{p + m}{ny} + \dfrac{m + n}{pz}$
$$ = \frac{n + p}{a} + \frac{p + m}{b} + \frac{m + n}{c} = \left(\frac{1}{b} + \frac{1}{c}\right) \cdot m + \left(\frac{1}{c} + \frac{1}{a}\right) \cdot n + \left(\frac{1}{a} + \frac{1}{b}\right) \cdot p$$
$$ = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3158987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Reasoning about inequalities involving floor functions I am working on the beginning of an inductive argument and I wanted to confirm that my base case is sound.
Let $f(x) = \lfloor x\rfloor - \left\lfloor\dfrac{x}{2}\right\rfloor$ where is $x$ is a positive real.
Let $u,v$ be positive integers such that $u \ge 7$ and ... | I have gone through all of your steps fairly carefully and didn't see any mistakes. As for being clear, it took me a bit of time to figure out some of your lines in $(3)$ as you sometimes did several steps in one line. However, this is a minor point. Also, I've seen quite a few math proofs which are considerably harder... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3159319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Reaching upon $9=1$ while solving $x$ for $3\tan{(x-15^{\circ})}=\tan{(x+15^{\circ})}$
$x$ for $3\tan{(x-15^{\circ})}=\tan{(x+15^{\circ})}$
Substituting $y=x+45^{\circ}$, we get
$$3\tan{(y-60^{\circ})}=\tan{(y-30^{\circ})}$$
$$3\frac{\tan y - \sqrt3}{1+\sqrt3\tan y}=\frac{\tan y - 1/\sqrt3}{1+1/\sqrt3\cdot\tan y}$$ ... | Another way to avoid confusion:
$$\dfrac31=\dfrac{\sin(x+15^\circ)\cos(x-15^\circ)}{\cos(x+15^\circ)\sin(x-15^\circ)}$$
Apply Componendo et Dividendo
$$\dfrac{3+1}{3-1}=\dfrac{\sin2x}{\sin30^\circ}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Consecutive integers divisible by prime powers There's a similar question on here already but I don't think the answers are applicable in this case.
Question: Find the smallest three consecutive integers for which the first integer is divisible by the square of a prime; the second integer by the cube of a prime; and th... | Hmm...
Well, by CRT $n \equiv 1 \pmod {p^2}$
$n \equiv 0 \pmod {q^3}$
$n \equiv-1 \pmod {r^4}$ will have unique solutions for any the primes $p, q, r$.
If I blanketly choose $r=2; q=3;p=5$ I will get.
$n\equiv 1 \pmod {25}$ and $n\equiv 0 \pmod 27$. $n= 27j = 1 + 25k = 1-2k + 27k$. If $k=14$ we have $j= 13$. This gi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solve the equation $\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$.
Solve the equation: $$\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$$
Here's what I've done.
Let $\sqrt{x + 2} = a$ and $\sqrt{3 - x} = b$
$\implies
\left\{
\begin{align}
a^2 + b^2 &= 5\\
a^2 - b^2 &= 2x - 1
\end{align}
\right.$.
We have that $a - b = (x -... | $$\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9 \implies (\sqrt{x + 2} - 2) - (\sqrt{3 - x} - 1) = x^2 - 6x + 8$$
$$ \implies (x - 2)\left(\frac{1}{\sqrt{x + 2} + 2} + \frac{1}{\sqrt{3 - x} + 1}\right) = (x - 2)(x - 4)$$
$$\implies \left[ \begin{align*} x - 2 &= 0\\ \frac{1}{\sqrt{x + 2} + 2} + \frac{1}{\sqrt{3 - x} + 1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Where does this proof of convergence fail? Given the series, $$\sum_{n=1}^{\infty} (-1)^{n}\frac{n}{n+1}$$
I know we can immediately conclude that it is obviously divergent by the divergence test. But I want to know where exactly am I going wrong in the following 'proof' as I have just started learning about convergenc... | Let
$s_m
=\sum_{n=1}^{} (-1)^{n}\frac{n}{n+1}
$.
Then
$\begin{array}\\
s_{2m}
&=\sum_{n=1}^{2m} (-1)^{n}\frac{n}{n+1}\\
&=\sum_{n=1}^{m} ((-1)^{2n-1}\frac{2n-1}{2n-1+1}+(-1)^{2n}\frac{2n}{2n+1})\\
&=\sum_{n=1}^{m} (-\frac{2n-1}{2n}+\frac{2n}{2n+1})\\
&=\sum_{n=1}^{m} \frac{-(2n-1)(2n+1)+4n^2}{2n(2n+1)}\\
&=\sum_{n=1}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3163595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the mean slope of $f(x) = 2x^3-6x^2-90x+6$ on the interval $[-5,8]$ I need to find the mean slope of $f(x) = 2x^3-6x^2-90x+6$ on the interval $[-5,8]$. I am getting $\frac{518}{13}$ but this is wrong. Here is my work.
Using the mean value theorem $f'(c)=\frac{f(b) - f(a)}{b - a}$
$f(b) = 2(-5)^3-6(-5)^2-90(-5)+6 =... | Your error is a simple mistake in computing $f(8)$. To be precise, you write that
$$2(8)^3-6(8)^2-90(8)+6 = 1024-384-66,$$
where indeed $2\times8^3=1024$ and $6\times8^2=384$. You seem to have computed $9\times8$ instead of $90\times8$.
Apart from this, your work is entirely correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3169874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the remainder when the polynomial $1+x^2+x^4+x^6+....+x^{22}$ is divided by $1+x+x^2+x^3+...+x^{11}$ Find the remainder when the polynomial $$1+x^2+x^4+x^6+....+x^{22}$$ is divided by $$1+x+x^2+x^3+...+x^{11}$$
$1+x^2+x^4+x^6+....+x^{22}=\frac{x^{24}-1}{x^2-1}$
$1+x+x^2+x^3+...+x^{11}=\frac{x^{12}-1}{x-1}$
Now$$\... | Upon applying: $\ fg\bmod fh\, =\, f(g\bmod h) =\, $ mod Dstributive Law with $\,z = x^2$
$\begin{align}
&\ \,1+\cdots+z^{11}\,\bmod\, (1+\cdots+z^{5})(x+1)\\[.3em]
=\ &(1+\cdots+z^{5})\big(1+\color{#c00}{z^6}\,\bmod x+1 \big)\\[.3em]
=\ &(1+\cdots+z^{5})\big( 1 + \color{#c00}1\big)\, \ {\rm by}\ \color{#c00}{z^6}\eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Fourier Series for $\cos^3(x)$ Consider the familiar trigonometric identity: $\cos^3(x) = \frac{3}{4} \cos(x) + \frac{1}{4} \cos(3x)$
Show that the identity above can be interpreted as Fourier series expansion.
so we know that cos is periodic between $\pi$ and $-\pi$ and $\cos$ is an even function, therefore, $\cos^3$ ... | Recall: from a familiar trigonometric identity,
$$\cos^3(x) = \cos(x)\color{blue}{\cos^2(x)} = \cos(x)\color{blue}{(1 - \sin^2(x))}$$
Thus,
$$\int \cos^3(x)dx = \int (1-\sin^2(x))\cos(x)dx$$
Make the $u$-substitution $u = \sin(x), du = \cos(x)dx$ and you should be able to find the result easily. You should get
$$\int \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3172485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Solving for coefficients in polynomial I have the following equation:
$$8\eta_c-6x\eta_c+\frac{x^3}{2}\eta_c=2.5\eta_c+C_{01}x^0+C_{11}x+C_{21}x^2+C_{31}x^3$$
If I solve for these coefficients, I get:
$$C_{01}=5.5\eta_c$$
$$C_{11}=-6\eta_c $$
$$C_{21}=0$$
$$C_{31}=1/2\eta_c$$
If I change my equation from (x) to (x-1) l... | Rearranging the original equation gives
$$(C_{01}-5.5\eta_c)+(C_{11}+6\eta_c)x+C_{21}x^2+(C_{31}-\frac{\eta_c}{2})x^3=0$$
Now if we replace $x$ with any function $f(x)$ for example $f(x)=x-1$ we get
$$(C_{01}-5.5\eta_c)+(C_{11}+6\eta_c)f(x)+C_{21}f^2(x)+(C_{31}-\frac{\eta_c}{2})f^3(x)=0$$
Now this equation is always tr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3173356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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A nice Nesbitt inequality from a strange inequality Given $a,\,b,\,c> 0$$,$ prove that$:$
$$\frac{a}{b+ c}+ \frac{b}{c+ a}+ \frac{c}{a+ b}+ \frac{63}{5}\left [ \frac{2\,c^{\,2}}{(\,a+ b\,)^{\,2}}- \frac{c}{a+ b} \right ]\geqq 0$$
See$:$ $\lceil$ https://artofproblemsolving.com/community/c6h354642p1923888 $\rfloor$
The ... | We have: $$f\left(a,b,c\right)=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{63}{5}\left[\frac{2c^2}{(a+b\, )^2}-\frac{c}{a+b}\right]\ge 0$$
Let $t=\dfrac{a+b}2>0$. We hope that $f(a,b,c)\ge f(t;t;c)$. Indeed,
$$f(a,b,c)-f\left(t,t,c\right)=\frac{\left(a-b\right)^2\left(a+b+c\right)}{\left(a+c\right)\left(b+c\right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A probability question: Poor Alex
Alex remembers all but the last digit of his friend's telephone number. He decides to choose the last digit at random in an attempt to reach him. Given that, Alex has only enough money to make two phone calls, the probability that he dials the right number before running out of money ... | Alex tries the first number. The probability to get it wrong is $9/10$. If he does, he chooses the second number. The probability that he get the second number wrong is $8/9$ (assuming that he does not try to dial the same number expecting a different result). Then the total probability that he gets the number wrong is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solutions of sin(x) = cos(x) I know that the solutions to the equation $\sin(x) = \cos(x)$ are :
$ x= \frac{\pi}{4}$ (45°) ; $ x= \frac{5 \pi}{4}$ (225°)
However when I try to solve it algebraically I get the following :
$$ \sin x = \cos x$$
$$ \sin^2 x = \cos^2 x$$
$$ \sin^2 = 1 - \sin^2 x$$
$$ 2\sin^2 x = 1$$
$$... | $\cos x=\sin x=\cos\left(\frac{\pi}{2}-x\right)$ thus $x=\frac{\pi}{2}-x \bmod [2\pi]$ or $x=-\frac{\pi}{2}+x \bmod [2\pi]$. The first equation has $\pi/4$ as solution and the second does not provide a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3175461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation... | Fixed Point Iteration
This is not efficient but it's another valid way to solve the system. Treat the system as a matrix equation and rearrange to get $\begin{bmatrix} x\\ y\end{bmatrix}$ on the left hand side.
Define
$f\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} (36-2y)/3 \\ (64-5x)/4\end{bmatrix}$
Start with ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 14,
"answer_id": 9
} |
Determinant with trigonometric functions
If
$$
\begin{vmatrix}
\sin 2x & \cos^2x & \cos 4x \\
\cos^2x & \cos2x & \sin^2x \\
\cos^4x & \sin^2x & \sin 2x \\
\end{vmatrix} = a_0 + a_1\sin x + a_2\sin^2x +\cdots+ a_n \sin^n x
$$
Then what is the value of $a_0$?
How do I solve this?
Thank you so much!!
| As noted in the comments, let $x=0$. Then we have
$$
\begin{vmatrix}
\sin 2x & \cos^2x & \cos 4x \\
\cos^2x & \cos2x & \sin^2x \\
\cos^4x & \sin^2x & \sin 2x \\
\end{vmatrix} = a_0 + a_1\sin x + a_2\sin^2x +\cdots+ a_n \sin^n x \\
\iff
\begin{vmatrix}
0 & 1 & 1 \\
1 & 1 & 0 \\
1 & 0 & 0 \\
\end{vmatrix} = a_0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Quickest way to find $a^5+b^5+c^5$ given that $a+b+c=1$, $a^2+b^2+c^2=2$ and $a^3+b^3+c^3=3$
$$\text{If}\ \cases{a+b+c=1 \\ a^2+b^2+c^2=2 \\a^3+b^3+c^3=3}
\text{then}\ a^5+b^5+c^5= \ ?$$
A YouTuber solved this problem recently and, though he spent some time explaining it, took over 40 minutes to solve it.
Like the v... | This answer is almost in the same spirit as @Quang Hoang's, but I hope this answer will add something. Let $$
P(z) = (z-a)(z-b)(z-c)=z^3-\sigma_1 z^2+\sigma_2 z-\sigma_3
$$ where $\sigma_1=a+b+c$, $\sigma_2=ab+bc+ca$ and $\sigma_3=abc$ by Vieta's formula. Note that for $z\in \{a,b,c\}$,
$$
z^{n+3} =\sigma_1 z^{n+2}-\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
Numerical Methods: calculate $b/a$ without division
Calculate $b/a$ in a calculator that only adds, subtracts and multiplies.
This problem is in the textbook for my numerical methods class. Obviously you can calculate it by
$$ \frac{1}{a} + \frac{1}{a} +\frac{1}{a} + ... \frac{1}{a} + = \frac{b}{a} $$
So the difficul... | Do you know how long division by hand works? Subtract the largest multiple of $a$ from $b$.
That will be the quotient (value before the decimal point). Let the difference (reminder) be $d$. Then put a decimal point and consider $10 \cdot d$. Repeat the procedure to as many decimal values needed or until you get a repea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3186607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find the third expression of Taylor's for $f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}$ around $x=0$
Find the third expression of Taylor's for $f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}$ around $x=0$
My try:Let $y=x+1$. Then: $$f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}=\frac{1}{(2y-1)(4-\frac{1}{y^2})^2}=-\frac{1}{2}\cdot\frac{1}{... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3188040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the median in a triangle with trigonometry In a triangle $ABC$, $AB=7$, $AC=4$ and $\angle CAB=50º$. Let $M$ be the midpoint of $BC$. Determinate $AM$.
My try
I applied law of cosines $3$ times, first to find $BC$, then I let $\angle BCA=\alpha$ and $\angle ABC=130-\alpha$, and applied law of cosines in triangle $... | The Law of Cosines tells us that
$$CB^2=7^2+4^2-2\cdot7\cdot 4\cdot\cos50=65-56\cos50$$
Finally, in virtue of Apollonius’s theorem, we obtain that
\begin{align*}
AM^2&=-\frac{BC^2}{4}+\frac{AC^2}{2}+\frac{AB^2}{2}\\
&=\frac{56\cos50-65+2\cdot 4^2+2\cdot 7^2}{4}\\
&=\frac{56\cos50+65}{4}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3189380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving ${\lim\limits_ {n\to\infty}}\frac{6n^3+5n-1}{2n^3+2n+8} = 3$ I'm trying to show that $\exists \,\varepsilon >0\mid\forall n>N\in\mathbb{N}$ such that:
$$\left|\frac{6n^3+5n-1}{2n^3+2n+8}-3\right| < \varepsilon$$
Let's take $\varepsilon = 1/2$:
$$\left|\frac{6n^3+5n-1}{2n^3+2n+8}-3\right| = \left|\frac{6n^3+5n-1... | Don't worry about $\epsilon$.
Instead,
try to get the difference
in a simple form
by assuming $n$
is as large as you need.
Then getting $n$
is much simpler.
Using your calculations:
$\begin{array}\\
\left|\dfrac{6n^3+5n-1}{2n^3+2n+8}-3\right|
&= \left|\dfrac{6n^3+5n-1 - 6n^3 -6n -24}{2n^3+2n+8}\right|\\
&= \left|\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3190400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Find $\lim_{ n \to \infty} \frac{e^{-n}}{\sqrt{n}} \sum_{k=0}^\infty \frac{\sqrt{k+ n }}{k!} (n+a)^k$ I am trying to find the following limit:
\begin{align}
\lim_{ n \to \infty} \frac{e^{-n}}{\sqrt{n}} \sum_{k=0}^\infty \frac{\sqrt{k+ n }}{k!} (n+a)^k
\end{align}
for some fixed $a>0$.
Things that tired.
We can com... | Defining $p(k, \lambda) = \frac{\lambda^k}{k!}e^{-\lambda}$, then
\begin{align*}
f(k, n) = \sqrt{1 + \frac{k}{n}}e^a \frac{(n+a)^k}{k!}e^{-(n+a)} = \sqrt{1 + \frac{k}{n}}e^ap(k, n+a)
\end{align*}
So the summation is equivalent to, for a random variable $X \sim \text{Pois}(n+a)$,
\begin{align*}
e^{a}\mathbb{E}\left[\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3190518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Show with induction that $\sum_{k=1}^{n} \frac{k^{2}}{2^{k}} = 6 - \frac{n^2+4n+6}{2^{n}}$ Show with induction that
$\sum_{k=1}^{n} \frac{k^{2}}{2^{k}} = 6 - \frac{n^2+4n+6}{2^{n}}$
n = 1
$LHS = \frac{1}{2}$
$RHS = 6 - \frac{1+4+6}{2} = \frac{1}{2}$
n = p
$LHS_{p} = \frac{1^{2}}{2^{1}} + \frac{2^{2}}{2^{2}} + \frac{3^{... | Hint: The left-hand side is given by $$\frac{(k+1)^2}{2^{k+1}}-\frac{k^2-4k+6}{2^k}=-2^{-k-1} \left(k^2-10 k+11\right)$$
and the right-hand side: $$-{\frac {{k}^{2}-2\,k+3}{{2}^{k+1}}}$$ so your formula is not true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3191370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proof for range of average We can prove that average of two numbers $a,b$ where $a<b$ will be between $a$ and $b$ as follows
$a < b$ $a + a < a + b$a < $\dfrac{a + b}{2 }$
$a < b$ $a + b < b + b$ $\dfrac{a + b}{2 }< b$
Thus $a < \dfrac{a + b}{2 }< b$
Similarly, is it true always and can be proved, that average of th... | Yes, it is always true.
From what you've already got, we have
$$a<b<\frac{b+c}{2}<c\implies a<\frac{b+c}{2}$$
$$\implies 2a<b+c\implies 3a<a+b+c\implies a<\frac{a+b+c}{3}$$
Also, we have
$$b-a>c-b\implies a+c\lt 2b\implies a+b+c\lt 3b\implies \frac{a+b+c}{3}\lt b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3194481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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A trigonometry question from STEP examination Show that if at least one of the four angles A ± B ± C is a multiple of π, then
$$\sin^4A + \sin^4 B + \sin^4 C − 2 \sin^2 B \sin^2 C − 2 \sin^2 C \sin^2 A
− 2 \sin^2 A \sin^2 B + 4 \sin^2 A \sin^2 B \sin^2 C = 0$$
I want to start with proving $\sin(A+B+C)$ or $(\sin(A)+\si... | Hint:
First of all writing $\sin A=a$ etc.,
$$a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=(a^2+b^2-c^2)^2-(2ab)^2$$
$$=(a+b+c)(a+b-c)(a-b+c)(a-b-c)$$
Now if $A+B+C=\pi$
by this $\sin A+\sin B+\sin C=4\cos\dfrac A2\cos\dfrac B2\cos\dfrac C2$
and by this $\sin A+\sin B-\sin C=4\sin\dfrac A2\sin\dfrac B2\cos\dfrac C2$
Use $\sin2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3195456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Finding the determinant of a tridiagonal matrix
$$\begin{vmatrix}x&1&0&0&⋯\\-n&x-2&2&0&⋯\\0&-(n-1)&x-4&3&⋯\\⋮&⋱&⋱&⋱&⋮\\0&⋯&-2&x-2(n-1)&n\\0&0&⋯&-1&x-2n\end{vmatrix}_{(n+1)×(n+1)}$$Find the value of the above determinant.
This problem comes from an advanced algebra book. I want to solve it with elementary transformati... | Edit. Call your matrix $A$ and let $L$ be the zero-indexed Pascal matrix defined by
$$
l_{ij}=\begin{cases}
\binom{n-j}{i}&\text{ when }\ 0\le i\le n-j\le n,\\
0&\text{ otherwise}.
\end{cases}
$$
It is known that $(L^{-1})_{ij}=(-1)^{i+j}l_{ij}$. E.g. when $n=5$,
$$
L=\pmatrix{1&0&0&0&0&0\\ 5&1&0&0&0&0\\ 10&4&1&0&0&0\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3196991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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How to show that $ab+bc+ca\le \frac34$ Let $a,b$ and $c$ be positive real numbers such that $(a+b)(b+c)(c+a) = 1$ , hen show that $$ab+bc+ca\le \frac34$$
I believe I need to use AM-GM inequality and use the fact $(a+b)(b+c)(c+a) = 1$ Using AM-GM in $a+b,b+c$ & $c+a$, I get $a+b+c\ge \frac32$. Any hint will be thankful.... | Consider the polynomial $$
\begin{align}p(x) &= (x-a)(x-b)(x-c)\\
&= x^3 - Ax^2 + Bx - C\end{align}\quad\text{ where }\quad
\begin{cases} A = a + b + c \\B = ab+bc+ca \\C = abc\end{cases}$$
Notice
$$(a+b)(b+c)(c+a) = (A-c)(A-a)(A-b) = p(A) = BA - C$$
We have
$$(a+b)(b+c)(c+a) = 1 \iff BA - C = 1 \iff B = \frac{1+C}{A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Finding floor of reciprocal sum
Evaluation of
$$\bigg \lfloor \frac{1}{\sqrt[3]{1}}+\frac{1}{\sqrt[3]{2^2}}+\frac{1}{\sqrt[3]{3^2}}+\cdots +\frac{1}{\sqrt[3]{(1000)^2}}\bigg\rfloor$$
Where $\lfloor x\rfloor $ is the floor of $x$
Try: It seems like we can solve it using Telescopic sums and that the sum lies between $2... | Note that $\displaystyle \sum_{k=1}^{1000}\frac{3}{\sqrt[3]{(k+1)^2}+\sqrt[3]{k(k+1)}+\sqrt[3]{k^2}}<\sum_{k=1}^{1000}\frac{1}{\sqrt[3]{k^2}}<1+\sum_{k=2}^{1000}\frac{3}{\sqrt[3]{(k-1)^2}+\sqrt[3]{k(k-1)}+\sqrt[3]{k^2}}$.
$\displaystyle \sum_{k=1}^{1000}\frac{3}{\sqrt[3]{(k+1)^2}+\sqrt[3]{k(k+1)}+\sqrt[3]{k^2}}=\sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
|x²-2x| + |x-4| > |x²-3x+4| , How do I solve for all real x? How do I solve this for all real x?
|x²-2x| + |x-4| > |x²-3x+4|
Looking at the question it is clear that it states |a| + |b| > |a-b|. How to proceed?
| First, note that $x^2-3x+4$ is always positive because discriminant $D=(-3)^2-4\times4<0$.
Case 1: $x\in(-\infty, 0]$
In this range: $|x^2-2x|=x^2-2x$, $|x-4|=-(x-4)$ and the inequality becomes:
$$x^2-2x-(x-4)>x^2-3x+4$$
$$x^2-3x+4>x^2-3x+4$$
...which is never true. So the ineqality has no solution in the given range.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3206431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
One of $a^5b-ab^5,b^5c-bc^5,c^5a-ca^5$ is divisible by $8$ if $a,b,c∈\mathbb{Z}^+$ Given that $a,b,c$ are there distinct positive integers, prove that among $a^5b-ab^5,b^5c-bc^5,c^5a-ca^5$, there is at least one that is divisible by $8$.
I have no clue what to do and cannot even create cases as these are three random w... | The three numbers are cyclic permutations of $ab(a^4-b^4)$. Therefore, at least one of the products must be formed from either $2$ even numbers or $2$ odd numbers.
If $a$ and $b$ are both even, then so is $(a^4-b^4)$, so $ab(a^4-b^4)$ is the product of three even numbers, which must be divisible by $8$.
If $a$ and $b$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3206597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Optimization problem in a single variable. $$f(x) = x^3 - 3x $$
Find the maximum value of the function $f(x)$ on the set of real numbers satisfying
$$x^4 + 36 \le 13x^2$$
I have only learned how to solve such question when the constraints are in an equality rather than an inequality through the Lagrangian method. Gen... | $$\begin{align}
x^4+36 &\le 13x^2 \\
x^4-13x^2+36 &\le 0\\
x^4-9x^2-4x^2+36 &\le 0\\
x^2(x^2-9)-4(x^2-9) &\le 0\\
(x^2-2^2)(x^2-3^2) &\le 0\\
(x+3)(x+2)(x-2)(x-3) &\le 0\\
\end{align}$$
This inequality is only satisfied when $2 \le x \le 3$ or $-3 \le x \le -2$.
$f(x)=x^3-3x=x(x^2-3)$ so $f(x)$ is an odd function. A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$a_n=(1-\frac{1}{n})a_{n-1}+\frac{1}{n}a_{n-2}$, $\lim_{n\to \infty}a_n$ is Given $a_1,a_2,n\in \mathbb N$
$$a_n=(1-\frac{1}{n})a_{n-1}+\frac{1}{n}a_{n-2}$$
Then $\lim_{n\to \infty}a_n$ is
(A) $2(a_2-a_1)+a_1e^{-1}$
(B) $2(a_1-a_2)e^{-1}+a_2$
(C) $2(a_1-a_2)e^{-1}+a_1$
(D) $2(a_2-a_1)e^{-1}+a_1$
My attempt,
$a_1,a_2\in... | Let $f(x) = \sum_{n\geq 1} a_n x^n$. Then
\begin{align*}
(1-x) f'(x)
&= a_1 + (2a_2 - a_1) x + x \sum_{n \geq 1} ((n+2) a_{n+2} - (n+1) a_{n+1}) x^n \\
&= a_1 + (2a_2 - a_1) x + x \sum_{n \geq 1} a_n x^n \\
&= a_1 + (2a_2 - a_1) x + x f(x).
\end{align*}
This gives a first-order linear ODE, and solving this together wit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"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.