Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Closed form for $\prod\limits_{l=1}^\infty \cos\frac{x}{3^l}$ Is there any closed form for the infinite product $\prod_{l=1}^\infty \cos\dfrac{x}{3^l}$? I think it is convergent for any $x\in\mathbb{R}$.
I think there might be one because there is a closed form for $\prod_{l=1}^\infty\cos\dfrac{x}{2^l}$ if I'm not wron... | Let $\displaystyle P=\prod_{i=1}^n\cos\left(\frac x{2^i}\right)$.
$\displaystyle P=\cos\left(\frac x{2}\right)\cos\left(\frac x{4}\right)\cos\left(\frac x{8}\right)\cdots\cos\left(\frac x{2^n}\right)$
$\displaystyle P\sin\left(\frac x{2^n}\right)=\cos\left(\frac x{2}\right)\cos\left(\frac x{4}\right)\cos\left(\frac x{8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Derivatives with different rules I'm having trouble with this one problem that just deals with deriving. I can't seem to figure out how they got their answer. Any help would be appreciated! Thanks!
$
\frac{(x+1)^2}{(x^2+1)^3}
$
The answer is:
$
\frac{-2(x+1)(2x^2-3x-1)}{(x^2+1)^4}
$
I seem to get the wrong answer when ... | Let's use quotient rule:
$$\frac{(x^2+1)^3\frac{d}{dx}(x+1)^2-(x+1)^2\frac{d}{dx}(x^2+1)^3}{{(x^2+1)^3}^2}$$
Do the derivatives out and simplify the denominator:
$$\frac{(x^2+1)^3\cdot 2(x+1)-(x+1)^2\cdot 2x\cdot 3(x^2+1)^2}{(x^2+1)^6}$$
Factor out $2(x+1)(x^2+1)^2$ from the numerator:
$$\frac{2(x+1)(x^2+1)^2((x^2+1)-(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1839196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
A convergent series: $\sum_{n=0}^\infty 3^{n-1}\sin^3\left(\frac{\pi}{3^{n+1}}\right)$ I would like to find the value of:
$$\sum_{n=0}^\infty 3^{n-1}\sin^3\left(\frac{\pi}{3^{n+1}}\right)$$
I could only see that the ratio of two consecutive terms is $\dfrac{1}{27\cos(2\theta)}$.
| $$\sum_{n=0}^\infty 3^{n-1}\sin^3\left(\frac{\pi}{3^{n+1}}\right)$$
By $\sin3\theta =3\sin \theta-4\sin^3\theta$
$$4\sin^3\left(\frac{\pi}{3^{n+1}}\right)=\sin\left(\frac{3\pi}{3^{n+1}}\right)-3\sin\left(\frac{\pi}{3^{n+1}}\right)$$
Then
$$\frac{1}{4}\sum_{n=0}^\infty 3^{n-1}\cdot4\sin^3\left(\frac{\pi}{3^{n+1}}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1840369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluating $\int_0^1 \frac{\arctan x \log x}{1+x}dx$ In order to compute, in an elementary way,
$\displaystyle \int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$
(see Evaluating $\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$)
i need to show, in a simple way, that:
$\displaystyle \int_0^... | Different approach:
start with applying integration by parts
$$I=\int_0^1\frac{\tan^{-1}(x)\ln(x)}{1+x}dx\\=\left|(\operatorname{Li}_2(-x)+\ln(x)\ln(1+x))\tan^{-1}(x)\right|_0^1-\int_0^1\frac{\operatorname{Li}_2(-x)+\ln(x)\ln(1+x)}{1+x^2}dx$$
$$=-\frac{\pi^3}{48}-\int_0^1\frac{\operatorname{Li}_2(-x)}{1+x^2}dx-\color{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 3
} |
Showing that $c_{i}\equiv 0\pmod{p}$ Let the numbers $c_{i}$ be defined by the power series identity $$\frac{1+x+x^{2}+\ldots+x^{p-1}}{(1-x)^{p-1}}= 1+c_{1}x+c_{2}x^{2}+\ldots$$ Show that $c_{i}\equiv 0\pmod{p}$ for all $i\geq 1$.
$\textbf{Solution.}$ First, from the Binomial Theorem we have $$(1-x)^{-n} = 1 + nx + \fr... | In $R:=\mathbb{F}_p[\![x]\!]$, $(1-x)^p=1-x^p$. Because $1-x$ is an invertible element of $R$ (with inverse $1+x+x^2+\ldots$), $$\frac{1-x^p}{(1-x)^p}=1$$ in $R$. Therefore, [...]. The rest is up to you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving $\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2}$ Does anyone have some tips for me how to go about the problem in the image?
$$\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2}$$
I know it's supposed to be simple, but I can't figure out why the solution is:
$90^{\circ}+ 720^{\circ}k$ where $k... | Firstly, denote $x=\frac{\theta}{2}$. Let $t=\tan \frac{x}{2}$. The equation $\sin x + \cos x = \sqrt 2$ became $$\frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2}=\sqrt 2.$$
Therefore $2t + 1 -t^2 = \sqrt 2(1+t^2) \Rightarrow (\sqrt 2+1)t^2 -2t + \sqrt 2 - 1 = 0$. This equation has the solutions $t_1=t_2=\sqrt 2 - 1$. So
$$\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Which of the numbers has the largest number of divisors?
Which of the numbers $1,2,\ldots,1983$ has the largest number of divisors?
Firstly notice that each number less than or equal to $1983$ has at most $4$ different prime divisors since $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 > 1983$. We then analyze each of the case... | If $n=p_1^{e_1}\cdots p_r^{e_r}$, then the number of divisors $d(n)$ of $n$ is given by
$$
d(n)=(e_1+1)(e_2+1)\ldots (e_r+1).
$$
We can have at most four different prime divisors, as you said,because otherwise $n>2\cdot 3\cdot5\cdot 7\cdot 11=2310$. Also, we may assume that we have the smallest primes, i.e., $n=2^{e_1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate $\int \frac{x^2-2}{(x^2+2)^3}dx$ $$\int \frac{x^2-2}{(x^2+2)^3}dx$$
What I did :
Method $(1) $ Re writing $x^2-2 = (x^2+2)-4 $ and partial fractions.
Method $(2) $ Substituting $x^2 = 2\tan^2 \theta $
Is there any other easy methods ?
Some substitution ?
| If
$$I_n=\int\frac{1}{(ax^2+b)^n}dx\quad,\quad n\ge 2$$
$$I_1=\frac{\sqrt{\frac{b}{a}}}{b}\tan^{-1}\left(\sqrt{\frac{a}{b}}x\right)$$
then
$$I_n=\frac{2n-3}{2b(n-1)}I_{n-1}+\frac{x}{2b(n-1)(ax^2+b)^{n-1}}$$
set $a=1$ and $b=2$ apply
$$\frac{x^2-2}{(x^2+2)^3}=\frac{1}{(x^2+2)^2}-\frac{4}{(x^2+2)^3}$$
we have
$$\int \fr... | {
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"url": "https://math.stackexchange.com/questions/1853934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Prove that $\max\{a_i \mid i \in \{0,1,\ldots,1984\}\} = a_{992}$
Consider the expansion
$$\left(1 + x + x^{2} + x^{3} + x^{4}\right)^{496} =
a_{0} + a_{1}x + \cdots + a_{1984}\,\,x^{1984}
$$
Prove that
$\max\left\{a_{i} \mid i \in \left\{0,1,\ldots,1984\right\}\right\} =
a_{992}\ $.
Since $1 + x + x^{2} + x^{3... | As you have noticed, for $(1 + x + x^2 + x^3 + x^4)^n = a_0 + a_1x + \cdots + a_{4n}x^{4n}$, we have
$$
a_i = a_{4n-i}\quad\text{for } 0 \leq i \leq 4n
$$
So it is sufficient to prove that $a_0 < a_1 < \cdots < a_{2n}$. We prove by induction below.
Basis. For $n = 2$, we have
$$
(1 + x + \cdots + x^4)^2 = 1 + 2x + 3x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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On finding the complex number satisfying the given conditions Question:- Find all the complex numbers $z$ for which $\arg\left(\dfrac{3z-6-3i}{2z-8-6i}\right)=\dfrac{\pi}{4}$ and $|z-3+i|=3$
My solution:-
$\begin{equation}
\arg\left(\dfrac{3z-6-3i}{2z-8-6i}\right)=\dfrac{\pi}{4} \implies \arg\left(\dfrac{z-2-i}{z-4-3i... | From the following where $\circ=\pi/4$,
$\qquad\qquad\qquad $
the circle $$(x-4)^2+(y-1)^2=4$$
you've found is correct, but note that we have only the part below the line $y=x-1$ passing through $(2,1)$ and $(4,3)$.
This is because the angle (as measured in counter-clock wise direction) from $\vec{zz_1}$ to $\vec{zz_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Is there a difference between $\sqrt{x+2}+x=0$ and $x^2-x-2=0$ Is there a difference between $\sqrt{x+2}+x=0$ and $x^2-x-2=0$
Solutions are $x=2$ or $x=-1$.
But $x=2$ does not satisfy $\sqrt{x+2}+x=0$.Because $\sqrt{4}+2 \neq0$
So does it mean that they are different ? Why ?
| Ofcourse there lies a difference,
Starting with $ x^2-x-2=0 \implies x^2=x+2 $ , Now follow from here that ,
$ x = \pm \sqrt{x+2} $ , When you take the negative value you get the above result as the solutions are $ x=2 $ or $x=-1 $ , so the negative value satisfies when you have a equation created with the negative sig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Prove: $x^3+y^3\geq \frac{1}{4}(x+y)^3$
Prove: $x^3+y^3\geq \frac{1}{4}(x+y)^3$ for all $x,y$ positive.
Let's look at
$$\begin{split} &(x-y)^2(x+y)\geq 0 \\
\iff &(x-y)(x+y)(x-y)\geq 0\\
\iff& (x-y)(x^2-y^2)\geq 0 \\
\iff &x^3-xy^2-yx^2+y^3\geq 0\\
\iff & 3x^3+3y^3\geq +3xy^2+3yx^2\\
\iff &3x^3+3y^3\geq (x+y)^3 -x^3... | Write $x=m+z$ and $y=m-z$, so that $x+y=2m\gt0$. Then
$$x^3+y^3=(m+z)^3+(m-z)^3=2m^3+6mz^2\ge2m^3={1\over4}(2m)^3={1\over4}(x+y)^3$$
Note that we only need for the midpoint $m$ between $x$ and $y$ to be positive in order for the inequality to hold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Integrate $\int\sin^2x\cos4x\,dx$ I'm having a difficult time solving this integral.
I tried integrating by parts:
$\int\sin^2x\cos4x\,dx$
$u=\sin^2x$, $dv=\cos4x\,dx$
I used the power reducing formula to evaluate $\sin^2x$
$du = 2\sin x\cos x\,dx$, $v=1/4\sin4x$
$uv - \int\ v\,du$
$\dfrac{1}{4}\sin^2x\sin4x - \dfrac{... | Here is the most straight way to solve this problem:
Notice that $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x $
So $\sin^2 x \cos 4x = \sin^2 (1-2\sin^2 2x) = \sin^2 [1-2(1-2\sin^2 x)^2] = -8\sin^6 x +8\sin^4 x -\sin^2 x$
By Reduction Formula: The integral can be easily solved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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Find the probability of getting two sixes in $5$ throws of a die.
In an experiment, a fair die is rolled until two sixes are obtained in succession. What is the probability that the experiment will end in the fifth trial?
My work:
The probability of not getting a $6$ in the first roll is $\frac{5}{6}$
Similarly for ... | Let $X$ denote any value between $1-5$, then the optional sequences are:
*
*$XXX66$
*$X6X66$
*$6XX66$
Calculate the probability of each sequence:
*
*$P(XXX66)=\frac56\cdot\frac56\cdot\frac56\cdot\frac16\cdot\frac16=\frac{5^3}{6^5}$
*$P(X6X66)=\frac56\cdot\frac16\cdot\frac56\cdot\frac16\cdot\frac16=\frac{5^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
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$f(x)$ is a quadratic polynomial with $f(0)\neq 0$ and $f(f(x)+x)=f(x)(x^2+4x-7)$
$f(x)$ is a quadratic polynomial with $f(0) \neq 0$ and $$f(f(x)+x)=f(x)(x^2+4x-7)$$
It is given that the remainder when $f(x)$ is divided by $(x-1)$ is $3$.
Find the remainder when $f(x)$ is divided by$(x-3)$.
My Attempt:
Let $f(x)=ax... | My answer to the question : Quadratic Functional equations.
Should work in the exact same manner, the extra conditions may or may not contradict the solution obtained, but in the case of a contradiction, then such a function does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
A closed form for $1^{2}-2^{2}+3^{2}-4^{2}+ \cdots + (-1)^{n-1}n^{2}$ Please look at this expression:
$$1^{2}-2^{2}+3^{2}-4^{2} + \cdots + (-1)^{n-1} n^{2}$$
I found this expression in a math book. It asks us to find a general formula for calculate it with $n$.
The formula that book suggests is this:
$$-\frac{1}{2}... | Let $S(n)=1^2+2^2+\dots+n^2$ and $T(n)=1^2-2^2+3^2-4^2+\dots+(-1)^{n-1}n^2$.
Suppose first $n=2k$ is even; then
$$
T(n)=T(2k)=
S(n)-2\bigl(2^2+4^2+\dots+(2k)^2\bigr)=
S(n)-8S(k)
$$
Since
$$
S(n)=\frac{1}{3}n\left(n+\frac{1}{2}\right)(n+1)=\frac{n(2n+1)(n+1)}{6}
$$
We have
$$
T(2k)=\frac{2k(4k+1)(2k+1)}{6}-8\frac{k(2k+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve the equation $x^{x+y}=y^{y-x}$ over natural numbers
Solve the equation $x^{x+y}=y^{y-x} \tag 1$ where $x,y \in \mathbb{N}$
$x = 1,y = 1$ is a solution, now suppose $x \ne 1, y \ne 1$.
Obviously $x, y \ne 0$ and $x, y$ have same prime divisors.
Because $x+y \gt y-x$ it follows that $x \mid y$ therefore $y = kx... | From
$x^2=k^{k-1}$,
$k$ must be odd.
Let $k = 2j+1$,
so
$x^2 = (2j+1)^{2j}$
or
$x = (2j+1)^j$.
Then
$y = kx
=(2j+1)(2j+1)^j
= (2j+1)^{j+1}
$.
Check:
$x^{x+y}?y^{y-x}$
$\begin{array}\\
x^{x+y}
&=((2j+1)^j)^{(2j+1)^j+(2j+1)^{j+1}}\\
&=(2j+1)^{j(2j+1)^j(1+(2j+1))}\\
&=(2j+1)^{j(2j+1)^j(2j+2)}\\
&=(2j+1)^{2j(j+1)(2j+1)^j}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Doubt in finding number of integral solutions Problem :
writing $5$ as a sum of at least $2$ positive integers.
Approach :
I am trying to find the coefficient of $x^5$ in the expansion of $(x+x^2+x^3\cdots)^2\cdot(1+x+x^2+x^3+\cdots)^3$ .
which reduces to coefficient of $x^3$ in expansion of $(1-x)^{-5}$ ,which is $$... | Actually you count all the possible permutations that gives $x^5$ in
$\displaystyle \color{red}{\frac{x^2}{(1-x)^2}} \times
\color{blue}{\frac{1}{(1-x)^3}}$,
$\color{red}{(4+1)}+\color{blue}{(0+0+0)}$ counts
$\color{red}{2} \times \color{blue}{1}$ possibilities.
$\color{red}{(3+2)}+\color{blue}{(0+0+0)}$ counts
$\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
integrate $\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx$
$$\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx$$
$$\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx=\int \frac{\sin^2x \cos^2x \sin x }{1+\sin^2x}dx=\int \frac{(1-\cos^2x )\cos^2x \sin x }{1+1-\cos^2x}dx=\int \frac{(1-\cos^2x )\cos^2x \sin x }{2-\cos^2x}dx$$
$u=cosx$
$du=-... | $$\int\frac{\sin^3(x)\cos^2(x)}{1+\sin^2(x)}\space\text{d}x=$$
Use $\sin^2(x)=1-\cos^2(x)$:
$$-\int\frac{\sin(x)(\cos^2(x)-\cos^4(x))}{\cos^2(x)-2}\space\text{d}x=$$
Substitute $u=\cos(x)$ and $\text{d}u=-\sin(x)\space\text{d}x$:
$$\int\frac{u^2-u^4}{u^2-2}\space\text{d}u=$$
Use long division:
$$-\int\left[u^2+1+... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integer solutions to $5m^2-6mn+7n^2 = 1985$
Are there integers $m$ and $n$ such that $$5m^2-6mn+7n^2 = 1985?$$
Taking the equation modulo $3$ gives $n^2-m^2 \equiv 2 \pmod{3}$. Thus, $3 \mid n$ but $3 \nmid m$. How can I use this to find a contradiction?
| Taking modulo $5$ we have
$$n(2n-m)\equiv 0\pmod 5$$
This gives us two possibilities:
*
*If $n$ is a multiple of $5$, write $n=5u$ and $175u^2-30mu+5m^2=1985$, or $35u^2-6mu+m^2=397$. Then
$$m=\frac{6u\pm\sqrt{1588-104u^2}}{2}=3u\pm\sqrt{397-26u^2}$$
*
*If $2n\equiv m\pmod 5$, then $m=5u+2n$ and (after some eas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Prove that $a_n$ is a perfect square
*
*Let $\,\,\,\left(a_{n}\right)_{\ n\ \in\ \mathbb{N}\,\,\,}$ be the sequence of integers defined recursively by
$$
a_{1} = a_{2} = 1\,,\qquad\quad a_{n + 2} = 7a_{n + 1} -a_{n} - 2\quad
\mbox{for}\quad n \geq 1
$$
*Prove that $a_{n}$ is a perfect square for every $n$.
... | For first, we may translate our sequence in order to have some sequence fulfilling $b_{n+2}=7b_{n+1}-b_n$. For such a purpose, it is enough to set $a_n=b_{n}+\frac{2}{5}$, leading to $b_0=b_1=\frac{3}{5}$. Now the characteristic polynomial of the sequence $\{b_n\}_{n\geq 0}$ is
$$ p(x)=x^2-7x+1 $$
with roots given by $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Converting repeating decimal in base b to a fraction the same base The repeating decimal .36666... in base 8 can be written in a fraction in base 8.
I understand simple patterns such as 1/9 in base 10 is .1111.... so 1/7 in base 8 is .1111.
But I'm not too sure how to convert this decimal in this base to the fraction i... | \begin{align}
0.3\bar{6}_8
&= \frac{3}{8} + 6\left(\frac{1}{8^2}+\frac{1}{8^3} + \cdots\right)\\
&= \frac{3}{8} + \frac{6}{8^2}\left(1+\frac{1}{8}+ \frac{1}{8^2} +\cdots\right)\\
&= \frac{3}{8} + \frac{6}{8^2}\frac{1}{1-(1/8)} & \text{geometric series}\\
&= \frac{3}{8} + \frac{3}{28}\\
&= \frac{27}{56}\\
&= \frac{33_8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
How would I solve $n(a) := 3 1 7 2 a 6 3 7 5$ with $9|n(a)$? $n(a) := 3 1 7 2 a 6 3 7 5$
What value needs a to be so that $9|n(a)$?
Is there a way to solve this fast? My initial idea was to write it as a sum like
$5 + 7 * 10 + 3 * 100 + 6 * 1000 + a * 10000 ... $
Because $a + c = b + d (\mod m )$
But this seems a bit t... | A number $n$ is divisible by $9$ if and only if the sum of the digits of $n$ is also divisible by $9$
(if and only if the sum of the digits of the sum of the digits of $n$ is also divisible by $9$) (if and only if the sum of the digits of the sum of the digits of the sum of the digits...)
E.g. $18324$ has sum of dig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum to infinity of trignometry inverse: $\sum_{r=1}^\infty\arctan \left(\frac{4}{r^2+3} \right)$ If we have to find the value of the following (1)
$$
\sum_{r=1}^\infty\arctan \left(\frac{4}{r^2+3} \right)
$$
I know that
$$
\arctan \left(\frac{4}{r^2+3} \right)=\arctan \left(\frac{r+1}2 \right)-\arctan \left(\frac{r-1}2... | One may observe that summing
$$
u_{r+1}-u_{r-1}
$$ may be simplified as a telescoping sum:
$$
\sum_{r=1}^N\left(u_{r+1}-u_{r-1}\right)=\sum_{r=1}^N\left(u_{r+1}-u_{r}\right)+\sum_{r=1}^N\left(u_{r}-u_{r-1}\right)=u_{N+1}+u_N-u_1-u_0.
$$
From the identity
$$
\arctan \left(\frac{4}{r^2+3} \right)=\arctan \left(\frac{r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Integral $\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$ $$\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$$
$$\tan x + \cot x + \csc x + \sec x=\frac{\sin x + 1}{\cos x} +\frac{\cos x + 1}{\sin x} $$
$$= \frac{\sin x +\cos x +1}{\sin x \cos x}$$
$$t= \tan {\frac{x}{2}}$$
On solving ,
$$\frac{1}{\tan x + \cot x +... | $$\begin{aligned}\int \frac{d x}{\tan x+\cot x+\csc x+\sec x}&=\int \frac{\cos x \sin x}{1+\cos x+\sin x} d x \\&= -\frac{1}{2} \int \frac{1-(\cos x+\sin x)^2}{1+\cos x+\sin x} d x \\& = -\frac{1}{2} \int(1-\cos x-\sin x) d x\\&= \frac{1}{2}(-x+\sin x-\cos x)+C\end{aligned} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Proving that $\lim_{x\to 2}\frac{x^2-5x}{x^2+2}=-1$ using the $\epsilon$-$\delta$ definition of a limit My attempt:
$$
\left|\frac{x^2-5x}{x^2+2}+1\right|<\left|\frac{x^2-5x}{x^2+2}\right|<
\left|\frac{x^2-5x}{x^2}\right|<\frac{1}{x^2}|x^2-5x|,$$
using the restriction $|x-2|<2$, so $0<x<4$, thus
$$\frac{1}{x^2}|x^2-5x... | Let $\epsilon>0$ be given. We have to find a number $\delta>0$ such that $\left|\frac{x^2-5x}{x^2+2}+1\right|<\epsilon$ whenever $|x-2|<\delta$. But, as Andre notes,
$$
\left|\frac{x^2-5x}{x^2+2}+1\right|=\left|\frac{2x^2-5x+2}{x^2+2}\right|=|x-2|\left|\frac{2x-1}{x^2+1}\right|.
$$
We find a positive constant $C$ such ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show that $a_n$ is decreasing
$a_1 = 2, a_{n+1} = \frac{1}{3 - a_n}$ for $n \ge 2$. Show $a_n$ is decreasing.
First we need to show $a_n > 0$ for all $n$.
$a_2 = 1/2$ and $a_3 = 2/5$ and $a_4 = 5/13$
One way we can do this is by showing $3- a_n > 0$. Thus suppose it holds for $n$ then we need to show $\frac{3(3 - a_... | Let us go for the overkill, i.e. to find an explicit formula for $a_n$. We may set $a_n=\frac{p_n}{q_n}$, with $p_1=2,q_1=1$, then deduce from $a_{n+1}=\frac{1}{3-a_n}$ the recurrence relation:
$$ w_{n+1}=\begin{pmatrix}p_{n+1} \\ q_{n+1} \end{pmatrix} = \begin{pmatrix}0 & 1 \\ -1 & 3 \end{pmatrix} \begin{pmatrix}p_{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Quadratic Inequality in terms of variable $x$
Find the values of $a$ for which the inequality $x^2+ax+a^2+6a<0\;\forall x \in (1,2)$
$\bf{My\; Try::}$ We can Write Equation as $$x^2+ax+\frac{a^2}{4}+\frac{3a^2}{4}+6a<0$$
So $$\left(x+\frac{a}{2}\right)^2+\frac{3a^2+24a}{4}<0$$
Now how can i solve after that, Help req... | Rewrite the inequality as $$(2a+x+6)^2< (6-x)(3x+6)$$
The right hand side is positive for $1<x<2$, so we get
$$-\sqrt{(6-x)(3x+6)}<2a+x+6<\sqrt{(6-x)(3x+6)}$$
and
$$-x-6-\sqrt{(6-x)(3x+6)}<2a<-x-6+\sqrt{(6-x)(3x+6)}$$
$$\frac{-x-6-\sqrt{(6-x)(3x+6)}}{2}<a<\frac{-x-6+\sqrt{(6-x)(3x+6)}}{2}$$
The maximum of the right ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Show that $3^n+4^n+\dots+(n+2)^n=(n+3)^n$ has no answers for $n\ge 6$. Considering
$$3^n+4^n+\dots+(n+2)^n=(n+3)^n$$
Clearly $n=2$ and $n=3$ are solutions of this equation and this equality does not hold for $n=4$ and $n=5$.
How can I show this equation has no solutions for $n>5$.
Thanks.
| Set $f(x)=\left(1-\frac{1}{x+3}\right)^x=\left(\frac{x+2}{x+3}\right)^x$ where $x\ge 6$. We have
$$f'(x)=\left(\frac{x+2}{x+3}\right)^x\left(\ln\left(\frac{x+2}{x+3}\right)+\frac{x}{x^2+5x+6}\right)$$
$f'(x)<0$ for $x\ge 6$. Hence $f(x)\le f(6)=\left(\frac 89\right)^6<\frac 12$, i.e
$$\left(1-\frac{1}{x+3}\right)^x< ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Can anyone tell me the formula for this vector field? 3Blue1Brown's video on curl
$$V=\bigg[ \begin{array}{ccc}
P\left(x,y\right) \\
Q\left(x,y\right) \end{array} \bigg] $$
*
*$P$ gives you the x component at all points in space.
*$Q$ gives you the y component at any point in space.
What is the explicit formula fo... | Working through Andrew D. Hwang's answer above:
$$F\left(x,y\right)=\left(\frac{2(-y)}{1+\sqrt{x^2+y^2}},\frac{2(x)}{1+\sqrt{x^2+y^2}}\right)$$
$$G\left(x,y\right)=\left(\frac{-2(y)}{1+\sqrt{(x-6)^2+(y)^2}}+\frac{-2(y)}{1+\sqrt{(x+6)^2+(y)^2}}-\frac{-2(y-6)}{1+\sqrt{(x)^2+(y-6)^2}}-\frac{-2(y+6)}{1+\sqrt{(x)^2+(y+6)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Integration by means of partial fraction decomposition I'm trying to solve this indefinite integral by means of partial fraction decomposition:
$\int\dfrac{x+1}{\left(x^2+4x+5\right)^2}\ dx$.
The denominator has complex (but not real) roots because $\Delta<0$; so, according with my calculus book, i try to decompose the... | Here is an easy way
$$\int { \frac { x+1 }{ \left( x^{ 2 }+4x+5 \right) ^{ 2 } } } dx=\frac { 1 }{ 2 } \int { \frac { 2x+4-2 }{ \left( x^{ 2 }+4x+5 \right) ^{ 2 } } } dx=\\ =\frac { 1 }{ 2 } \int { \frac { 2x+4 }{ \left( x^{ 2 }+4x+5 \right) ^{ 2 } } dx-\int { \frac { dx }{ \left( x^{ 2 }+4x+5 \right) ^{ 2 } } } } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Show that the matrices' linear transformations are similar. Let be $T: M_2(\mathbb{C}) \longrightarrow M_2(\mathbb{C})$ the following linear transformation $$ T\left( \begin{array}{cc}
x & y\\
z & w\\
\end{array} \right) = \left( \begin{array}{cc}
0 & x\\
z-w & 0\\
\end{array} \right) $$
a) Determine the matrix $T$ con... | Your answer for the first one is correct. I didn't check the second one fully but I think you made some mistakes. For instance, writing $B = (b_1,\ldots,b_4)$, we have $$T(b_1) = T\begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix}0&1\\-1&0\end{pmatrix} \neq \begin{pmatrix}1&0\\0&1\end{pmatrix} - \begin{pmatrix}1&0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the value $\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$ The value of $$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$$is equal to
*
*$\frac{\pi}{6}$
*$\frac{\pi}{4}$
*$\frac{\pi}{3}$
*$\frac... | Hint the second bracket is $\tan(\frac{\pi}{3}-\tan^{-1}\sqrt{2})$ now can you proceed further
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
how can I prove this $\sum_{k=1}^{n}\frac{1}{4k^{2}-2k}=\sum_{k=n+1}^{2n}\frac{1}{k}$ How can I prove the following equation?:
$$s=\sum_{k=1}^{n}\frac{1}{4k^{2}-2k}=\sum_{k=n+1}^{2n}\frac{1}{k}$$
Simplifying both terms of the equation:
$$\sum_{k=1}^{n}\frac{1}{4k^{2}-2k} = \sum_{k=1}^{n}\frac{1}{(2k-1)2k} =(\sum_{k=1}^... | Let's do a cute induction on $n$:
For $n = 1$, we have $\frac{1}{4 - 2} = \frac{1}{2}$, so the claim holds.
Assume $n > 1$ and the claim holds for $n - 1$. Then $$\sum_{k = 1}^n \frac{1}{4k^2 - 2k} = \frac{1}{4n^2 - 2n} + \sum_{k = 1}^{n - 1} \frac{1}{4k^2 - 2k} = \frac{1}{4n^2 - 2n} + \sum_{k = n}^{2(n -1)} \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$\tan^{-1}x$, $\tan^{-1}y$, $\tan^{-1}z$ are in arithmetic progression, as are $x$, $y$, $z$. Show ...
$\tan^{-1}x, \tan^{-1}y, \tan^{-1}z $ are in arithmetic progression, as are $x$, $y$, $z$. (We assume $y \ne 0,1,-1$.) Show:
*
*$x$, $y$, $z$ are in geometric progression
*$x$, $y$, $z$ are in harmonic progression... | (Not an answer so much as an elaborate comment ...)
For fun, I made a "trigonograph" to see why we might expect the elements to be equal (item (3)). Here, the arithmetic progressions are $\alpha$, $\gamma$, $\beta$ and $\tan\alpha$, $\tan\delta$, $\tan\beta$. (Also, I take all values to be positive, and assume $\alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Integrate $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$$
$$f(x) = (\tan \ x)^{\frac{2}{3}}, \ f'(x) = \frac{2}{3} \cdot (\tan \ x)^{-\frac{1}{3}} \cdot \sec^2x$$
$$\therefore \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\fra... |
$$
\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{2}{3}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx =
\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{f'(x)}{f(x)}dx$$
This step is wrong since
$$f(x) = (\tan \ x)^{\frac{2}{3}}, \ f'(x) = \frac{2}{3} \cdot (\tan \ x)^{-\frac{1}{3}} \cdot \sec^2x dx$$
Then $$\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find $\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $ Find $$\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $$
Any hints please?
Could'nt think of any approach till now...
| When I have a rational function I usually start by making a part of the denominator "appear" in the numerator, and since we're dealing with trigonometric function, the following identity is quite handy:
$$\frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}=\frac{x^2\cos^2x+x^2\sin^2x}{(x \cos x - \sin x )(x \sin x + \co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Inequality $a+b+c > abc$, prove $a^2+b^2+c^2 > abc$ Given, $$a+b+c > abc$$
Prove $$a^2+b^2+c^2 > abc$$
I tried to square the first one but it wouldn't work.
| $$a^2+b^2+c^2\geq ab+bc+ac$$ holds, since it is equivalent with $$(a-b)^2+(b-c)^2+(c-a)^2\geq 0.$$ Also $$ab+bc+ac\geq\sqrt{3abc(a+b+c)}.$$ Previous inequality holds, since it is equivalent with $$a^2b^2+b^2c^2+c^2a^2\geq abc(a+b+c),$$
which is equivalent with $$(x-y)^2+(y-z)^2+(z-x)^2\geq 0$$ for $x=bc$, $y=ac$, $z=a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the maximum of the value $x+y+z$ if such condition $n(x+y+z)=xyz$ Let $n$ be give postive intger, and $x\ge y\ge z$ are postive integers,such
$$n(x+y+z)=xyz$$
Find the $(x+y+z)_{\max}$
I have see this problem only answer is $(n+1)(n+2)$,iff $x=n(n+2),y=n+1,z=1$
and How do it?
| SUMMARY: it really is true that
$$(x+y+z)_{\max} = n^2 + 3n + 2,$$
this occurring only when $z=1, y=n+1, x=n^2 + 2n $
This works. I will leave $n=1,2,3$ as exercises for the reader. We take $n \geq 4.$ We are given $x \geq y \geq z \geq 1,$ with
$$ xyz = nx + ny + nz, $$
$$ x y z^2 = nxz + nyz + n z^2, $$
$$ z^2 xy -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Evaluate $\int_{3}^{\infty}\frac{dx}{x^2-x-2}$ $$\int_{3}^{\infty}\frac{dx}{x^2-x-2}=\int_{3}^{\infty}\frac{dx}{(x-2)(x+1)}$$
$$\frac{1}{(x-2)(x+1)}=\frac{A}{(x-2)}+\frac{B}{(x+1)}$$
$$1=Ax+A+Bx-2B$$
$$1=(A+B)x+A-2B$$
$A+B=0\iff A=-B$
$-3B=1$
$B=-\frac{1}{3}$, $A=\frac{1}{3}$
$$\int_{3}^{\infty}\frac{dx}{(x-2)(x+1)}=\f... | $\frac{\ln(4)}{3} = \frac{\ln(2^2)}{3} = \frac{2\ln(2)}{3}$
And to handle the first two terms, combine the logs:
$\frac{1}{3}\ln(t-2) - \frac{1}{3}\ln(t+1) = \frac{1}{3}\ln(\frac{t-2}{t-1})$
Then taking the limit as $t \rightarrow \infty$ will make it $\frac{1}{3}\ln(1) = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve $\int_{0}^{1}\frac{1}{1+x^6} dx$ Let $$x^3 = \tan y\ \ \text{ so that }\ x^2 = \tan^{2/3}y$$
$$3x^2dx = \sec^2(y)dy$$
$$\int_{0}^{1}\frac{1}{1+x^6}dx = \int_{1}^{\pi/4}\frac{1}{1+\tan^2y}\cdot \frac{\sec^2y}{3\tan^{2/3}y}dy = \frac{1}{3}\int_{1}^{\pi/4} \cot^{2/3}y\ dy$$
How should I proceed after this?
EDITED: C... | You can take the long way using partial fractions, which gives you:
$$\underbrace{\frac{1}{3(x^2+1)}}_{f_1(x)}+\underbrace{\frac{\sqrt 3 x+2}{6(x^2+\sqrt3 x+1)}}_{f_2(x)}+\underbrace{\frac{3\sqrt3x-6}{18(-x^2+\sqrt3x-1)}}_{f_3(x)}$$
$$\int_0^1f_1(x)dx=\frac 13\int_0^1\frac 1{1+x^2}dx=\left[\arctan(x)\right]_0^1$$
$$\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Laurent series of $f(z) = \frac{z^2+1}{\sin(z)}$ How does one find the first four terms of the Laurent series of $f(z) = \frac{z^2+1}{\sin(z)}$? My approach was this:
$(z^2+1) = f(z)\sin(z) = \left(\sum\limits_{n=-\infty}^\infty c_n z^n \right)\sin(z)=\left(\sum\limits_{n=-\infty}^\infty c_n z^n \right)\left(z-\frac{z^... | $$\frac{z^2+1}{\sin z}=\biggl( z+\frac 1z\biggr)\frac z{\sin z},$$
so all you have to find is the expansion of
$$\frac z{\sin z}=\frac 1{1-\cfrac{z^2}6+\cfrac{z^4}{120}+\cfrac{z^6}{5040}+o(z^7)}$$
It can be obtained with a division by increasing powers of $x$:
$$\begin{array}{rrrrr}
&&1&{}+\dfrac{z^2}6&{}+\dfrac{7z^4}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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General formula for two sequences Please help me to find general 'analytical' formula fot these two sequences $\{x_n\}$, $n=1,2,3,\ldots$.
I. $0, 1, 2, 2, 3, 6, 6, 7, 14, 14, \ldots$
This meens the following.
$x_{1}=0$, $x_{2}=x_{1}+1$, $x_{3}=2x_{2}$, $x_{4}=x_{3}$, $\ldots$, $x_{3k-1}=x_{3k-2}+1$, $x_{3k}=2x_{3k-1}$,... | For case II), shift the index to start from $0$ i.e. put $n=m+1$.
Then you have a function of period $6$:
$$
f\left( m \right) = \left\{ \begin{gathered}
1\quad \left| {\;\bmod \left( {m,6} \right) = 0} \right. \hfill \\
0\quad \left| {\;\bmod \left( {m,6} \right) \ne 0} \right. \hfill \\
\end{gathered} \right.
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to find a characteristic polynomial of 5x5 matrix From an example that I am looking at, the characteristic polynomial of
$\begin{bmatrix}
-1&4&0&0&0 \\
0&3&0&0&0 \\
0&-4&-1&0&0 \\
3&-8&-4&2&1 \\
1&5&4&1&4
\end{bmatrix}$
is $(x-3)^3(x+1)^2$. I understand how to find the characteristic polynomial of 2x2 and 3x3 ma... | Directly using elementary operations that don't change the value of the determinant. Develop by the second row twice:
$$\begin{vmatrix}
x+1&-4&0&0&0 \\
0&x-3&0&0&0 \\
0&4&x+1&0&0 \\
-3&8&4&x-2&-1 \\
-1&-5&-4&-1&x-4
\end{vmatrix}=(x-3)\begin{vmatrix}
x+1&0&0&0 \\
0&x+1&0&0 \\
-3&4&x-2&-1 \\
-1&-4&-1&x-4
\end{vmatrix}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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On convergence of $\cos(2\pi \alpha n!)$
Let $x\in \mathbb R$
Prove that there exists some $\alpha_x$ such that the sequence $\cos(2\pi \alpha_x n!)$ converges to $\cos(x)$
I've been stumped with this problem for a while. I've been looking for some $\alpha$ such that $$\forall n, 2\pi \alpha n! = x + 2\pi k_n + \epsi... | For any $x \in \mathbb{R}$, take a $y \in [0,\pi]$ such that $\cos x = \cos y$.
Define
$$\alpha = \sum_{n=1}^\infty \frac{1}{n!}\left\lfloor \frac{y}{2\pi} n\right\rfloor$$
For any $N > 0$, we have
$$\alpha N! =
\underbrace{\sum_{n=1}^N \frac{N!}{n!}\left\lfloor \frac{y}{2\pi} n\right\rfloor}_{\in \mathbb{Z}}
+ \frac{... | {
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How would you show that the series $\sum_{n=1}^\infty \frac{(2n)!}{4^n (n!)^2}$ diverges? How would you show that the series $$\sum_{n=1}^\infty \frac{(2n)!}{4^n (n!)^2}$$ diverges? Wolfram Alpha says it diverges "by comparison", but I'd like to know to what you would compare it? I've tried some basic things to no avai... | Here is my attempt:
$a_n=\dfrac{(2n)!}{4^n(n!)^2}$
First i expanded factorial terms:
$a_n=\dfrac{2n(2n-1)(2n-2)\cdots 3\cdot 2\cdot 1}{4^n n(n-1)(n-2)\cdots 2\cdot 1\cdot n(n-1)(n-2)\cdots 2\cdot 1}$
At the numerator, rearrange the $n$ even factors and the $n$ odd factors (not necessary but useful for my description)
$... | {
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Definite integral and limit $\lim_{n \rightarrow \infty} (n((n+1)I_{n}-\frac{\pi}{4})$ Given $I_{n} = \int_{0}^{1} x^{n} \arctan(x)dx $
Calculate:
$\lim_{n \rightarrow \infty} (n((n+1)I_{n}-\frac{\pi}{4})$
| This is probably a too complex answer.
Considering $$J_n=\int x^n \tan^{-1}(x)\,dx=\frac{x^{n+1} \left((n+2) \tan ^{-1}(x)-x \,
_2F_1\left(1,\frac{n}{2}+1;\frac{n}{2}+2;-x^2\right)\right)}{(n+1) (n+2)}$$ where appears the hypergeometric function, $$I_n=\int_0^1 x^n \tan^{-1}(x)\,dx=\frac{H_{\frac{n-2}{4}}-H_{\frac{n... | {
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"answer_count": 4,
"answer_id": 2
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Gauss circle problem : a simple asymptotic estimation.
Find the number of integer points lying in or inside a circle of radius $n\in \mathbb N$ centered at the origin.
The problem asks for all $(a,b)\in \mathbb Z^2$ such that $a^2+b^2\leq n^2$. Looking at such points lying strictly in the North-East quadrant, pick an... | Pick a square with unit side length centered at every lattice point inside the region $x^2+y^2=n^2$, i.e. the circle with radius $n$ centered at the origin. Those squares entirely cover the circle with radius $n-\frac{1}{\sqrt{2}}$, but any square lies inside a circle with radius $n+\frac{1}{\sqrt{2}}$, hence the numbe... | {
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How to solve this determinant
Question Statment:-
Show that
\begin{align*}
\begin{vmatrix}
(a+b)^2 & ca & bc \\
ca & (b+c)^2 & ab \\
bc & ab & (c+a)^2 \\
\end{vmatrix}
=2abc(a+b+c)^3
\end{align*}
My Attempt:-
$$\begin{aligned}
&\begin{vmatrix}
\\(a+b)^2 & ca & bc \\
\\ca & (b+c)^2 & ab \\
\\bc & ab & (c+a)^2 ... | One can use factor theorem to get a simpler solution. If we put $a=0$, we get
\begin{align*}
\begin{vmatrix}
(a+b)^2 & ca & bc \\
ca & (b+c)^2 & ab \\
bc & ab & (c+a)^2 \\
\end{vmatrix}
=\begin{vmatrix}
b^2 & 0 & bc \\
0 & (b+c)^2 & 0 \\
bc & 0 & c^2 \\
\end{vmatrix} = 0
\end{align*}
Hence $a$ is a factor. Similar... | {
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How can we prove this inequality? Let $a$ , $b$ and $c$ be positive real numbers and $a+b+c=1$ How can we show this inequality?
$$a^2+b^2+c^2+2\sqrt{3abc}\le 1$$
Thanks.
| Set $a=x^2$ , $b=y^2$ and $c=z^2$ and define
$$f(x,y,z)=x^4+y^4+z^4+2\sqrt{3}xyz-1$$
and
$$g(x,y,z)=x^2+y^2+z^2-1=0$$
By application of Lagrange method, we have
$$\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$$
thus
\begin{cases}
4x^3+2\sqrt{3}yz=2\lambda x\\
4y^3+2\sqrt{3}xz=2\lambda y\\
4z^3+2\sqrt{3}xy=2\lambda z\\
\end{... | {
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Inequalities in integration We are given the value of $S_n$ below and we have to find if $S_n$ is greater or smaller than $\dfrac{\pi}{3\sqrt{3}}$.
$$S_n=\sum\limits_{k=1}^n \frac{n}{n^2+kn+k^2}$$
$$S_n<\dfrac{\pi}{3\sqrt{3}} \quad \text{or}\quad S_n>\dfrac{\pi}{3\sqrt{3}}$$
I tried it as follows:
$$S_n=\frac{1}{n}\sum... |
Look at this picture. The red curve is the function $\displaystyle f(x)=\frac{1}{x^2+x+1}$
The red curve is decreasing on $[0,1]$, since $\displaystyle f'(x)=-\frac{2x+1}{\left(x^2+x+1\right)^2}<0$ for $0\leq x \leq 1$.
The gray shaded area is the value of $\displaystyle\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\left(\frac{k}... | {
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"question_score": "1",
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Smallest $n$ such that $H_{n,2^{-1}} \geq n$ where $H_{n,2^{-1}}$ are generalized Harmonic numbers Consider the generalized harmonic numbers evaluated at $2^{-1}$ $$H_{n,2^{-1}} = 1+ {1 \above 1.5pt \sqrt{2}}+ \ldots+ {1 \above 1.5pt \sqrt{n}}$$ The table below lists some initial values:
$$\begin{array}{nc|cccccc}
n &1... | Note that
$$H_{n,2^{-1}}=\int_1^{n+1}\frac{dt}{\sqrt{\lfloor t\rfloor}}$$
So
$$\int_1^{n+1}\frac{dt}{\sqrt{t}}\le H_{n,2^{-1}}\le1+\int_2^{n+1}\frac{dt}{\sqrt{t-1}}$$
That is,
$$2(\sqrt{n+1}-1)\le H_{n,2^{-1}}\le 1+2(\sqrt{n}-1)$$
For example, $1998 \le H_{999,999,\,2^{-1}} < 1999$. This means that $\mathfrak a(1998)\l... | {
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$5$ divides one of $x,y,z$ in $x^2+y^2=z^2$? 1) If $x$, $y$, and $z$ are positive integers for which $gcd(x,y,z) =1$ and $x^2+y^2=z^2$, show that $3$$∣$$xy$.
2) Now again if $x$, $y$, and $z$ are positive integers for which $gcd(x,y,z) =1$ and $x^2+y^2=z^2$, can a similar result to the one in (part 1) be said mod $5$? ... | For the second part since $5$ is prime, if $5 \mid xy \implies 5 \mid x$ or $5\mid y$. For the first part, we have: $x = 2mn, y = m^2-n^2=(m-n)(m+n)$. Thus if $3 \nmid x$, then $3 \nmid m, 3 \nmid n$. Thus if $m = n \pmod 3$, then $m-n = 0 \pmod 3$, otherwise say $m = 1\pmod 3, n = 2\pmod 3 \implies m+n = 0 \pmod 3$.
| {
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Prove that $ABC$ is right-angled Prove that if $\cos^2{A} + \cos^2{B} + \cos^2{C} = 1$, then $ABC$ is right-angled.
I only found that $\sin^2{A} + \sin^2{B} + \sin^2{C} = 2$, but I have no idea what to do next.
Thank you in advance for your answers!
| As you found that $\sin^2{A} + \sin^2{B} + \sin^2{C} = 2$ note that $\sin^2 \theta + \cos^2 \theta =1$
Then $1-\cos^2 A + 1- \cos^2 B + 1 - \cos^2 C = 2$
$ 3 -(\cos^2 A + \cos^2 B + \cos^2 C)= 2$
$3 - 2 = \cos^2 A + \cos^2 B + \cos^2 C$
$\cos^2 A + \cos^2 B + \cos^2 C =1$
| {
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Prove that if $a+b+c=1$ then $\sum\limits_{cyc}\frac{1}{\sqrt{a^2+b^2}}\le\frac{9\sqrt{2}}{2}$ Let $a,b,c>0$,and such $a+b+c=1$,prove or disprove
$$\dfrac{1}{\sqrt{a^2+b^2}}+\dfrac{1}{\sqrt{b^2+c^2}}+\dfrac{1}{\sqrt{c^2+a^2}}\le\dfrac{9\sqrt{2}}{2}\tag{1}$$
My try:since
$$\sqrt{a^2+b^2}\ge\dfrac{\sqrt{2}}{2}(a+b)$$
it ... | It's wrong. Try $a=b\rightarrow0^+$
| {
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Mathematical induction problem: $\frac12\cdot \frac34\cdots\frac{2n-1}{2n}<\frac1{\sqrt{2n}}$ This is a problem that I tried to solve and didn't come up with any ideas
.?$$\frac{1}{2}\cdot \frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n}}.$$
All I get is $\frac{1}{\sqrt{2n}}\cdot\frac{2n+1}{2n+2}<\frac{1}{\sqrt{2n+2... | Prove by induction the stronger inequality: for any integer $n\geq 1$,
$$\frac{1}{2}\cdot \frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}.$$
The basic step is true: $1/2<1/\sqrt{3}$.
Then in the inductive step you will obtain the inequality
$$\frac{1}{\sqrt{2n+1}}\cdot\frac{2n+1}{2n+2}<\frac{1}{\sqrt{2n+3}}$$
t... | {
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Find coefficients of polynomial, knowing its roots are consecutive integers In the function $f(x)= x^3-15x^2+ax+b$ the graph has $3$ consecutive points where it crosses the x-axis.
These $3$ points are consecutive integers. Find $a$ and $b$ for this is you know that $a$ and $b$ are real numbers.
How do I start to fin... | The brute force method that doesn't require any formulas or theorems would be like so:
You know that there are $3$ consecutive zeroes $(n-1)$, $n$, $(n+1)$. It's important to pick them like so as opposed to $n$, $(n+1)$, $(n+2)$ because this will save you a lot of algebra.
$0=(n-1)^3-15(n-1)^2+a(n-1)+b $
$0=n^3-15n^2+a... | {
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How do I factorize a polynomial $ax^2 + bx + c$ where $a \neq 1$ How do I factorize a polynomial $ax^2 + bx + c$ where $a \neq 1$?
E.g. I know that $6x^2 + 5x + 1$ will factor to $(3x + 1)(2x + 1)$, but is there a recipe or an algorithm that will allow me to factorize this?
| The roots of a quadratic polynomial $ax^2+bx+c$ are $x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$. Then the factorization is just $a(x-x_1)(x-x_2)$.
In your example, $a=6,b=5,c=1$, so $x_1=\frac{-5+\sqrt{5^2-4\cdot 6\cdot1}}{2\cdot 6}=-\frac13$ and $x_2=\frac{-5-\sqrt{5^2-4\cdot 6\cdot1}}{2\... | {
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"source": "stackexchange",
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Find all integer solutions to the equation $\frac{xy}{z}+\frac{xz}{y}+\frac{zy}{x}=15$ I was going through one of my Mathematics books and I came to this problem:
Find all integer solutions to the equation: $\frac{xy}{z}+\frac{xz}{y}+\frac{zy}{x}=15$
I tried a few things to begin with but none went in the right directi... | Since obviously $xyz\ne 0$, you have $$(xy)^2+(xz)^2+(yz)^2=15xyz\iff (y^2+z^2)x^2-(15yz)x+(yz)^2=0$$ hence the discriminant must be non-negative.
$$(15yz)^2-4(y^2+z^2)(yz)^2\ge 0$$ It follows $$y^2+z^2\le 56$$ This involves only the set $\mathcal S$ of the first seven squares
$$\mathcal S=\{1,4,9,16,25,36,49\}$$ Sear... | {
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How do I get the closed form of this recurrence using generating functions? Recurrence: $T(n) = n + nT(n-1)$ and $T(0) = 0$.
What I tried:
Let $G(x) = \sum_{n=0}^{\infty} T(n) x^n$ so that $xG'(x) = \sum_{n=0}^{\infty} nT(n)x^n$
Solving:
\begin{align}
G(x) &= \sum_{n=0}^{\infty} T(n) x^n
\\&= 0x^0 + \sum_{n=1}^{\infty... | $$G(x)-xG'(x)=\frac{x}{(1-x)^2}$$
Now use the fact that
$$\left(\frac{G(x)}{x} \right)'=\frac{G'(x) x -G(x)}{x^2}$$
to reduce your equation to
$$\left(\frac{G(x)}{x} \right)'=-\frac{1}{x (1-x)^2}$$
Integrate and you are done.
Edit To address the updated question:
$$ G'(x) -\frac{1-x}{x^2}G(x) =-\frac{1}{x(1-x)^2} $$... | {
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Why is solution to inequality $\sqrt{1 - x} - \sqrt{x} > \frac{1}{\sqrt{3}}$ equal to interval $[0, \frac{3 - \sqrt{5}}{6})$? Given inequality $\sqrt{1 - x} - \sqrt{x} > \frac{1}{\sqrt{3}}$ we can easily determine, that it's domain is $D = [0, 1]$. Because each term is real, we can take square of the inequality, which ... | Well, there is this but I'm not sure how to generalize it.
$\sqrt{1-x} - \sqrt{x} > 0$ so $1-x > x$ so $x < \frac 12$ and $x \in [0,1/2)$.
When you square but sides of $\sqrt{1-x} - \sqrt{x} > 1/\sqrt{3}$ to get $\frac{1}{3} > \sqrt{x}\sqrt{1 -x}$ you are extraneously adding the possibility that $\sqrt{x} - \sqrt{1-x} ... | {
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Let a, b, c, x and y be integers. Why are there no solutions for $\sqrt {a^2 + b^2 + c^2} = x$, where $x=2^y$ or $x=5 \times 2^y$ Edit - An addition condition: $a$, $b$, and $c$ do not equal $0$.
I'm really digging into 3D vectors and their properties. I've decided to look and see which combinations of three integers w... | It appears you do not permit any of $a,b,c$ to be zero. You can write it as $a^2+b^2+c^2=x^2$. In that case, it comes from the fact that the squares $\bmod 8$ are $0,1,4$ To have $2^{2y}$ be the sum of three squares you would have to have all of the squares be $0 \bmod 8$ or one be $0 \bmod 8$ and two of them be $4 ... | {
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What are some easy but beautiful patterns in Pascal's Triangle? Pascal's triangle has wide applications in Mathematics.I have seen that the most important applications relate to the binomial coefficients and combinatorics.
Are there some other beautiful interesting patterns in Pascal's triangle that can be found by sel... |
There is a nice hexagonal property in the Pascal triangle:
\begin{array}{ccccccccccccccccc}
&&&&&&&&1\\
&&&&&&&1&&1\\
&&&&&&1&&2&&1\\
&&&&&1&&3&&3&&1\\
&&&&1&&4&&6&&4&&1\\
&&&1&&\mathbf{\color{blue}{5}}&&\mathbf{\color{blue}{10}}&&10&&5&&1\\
&&1&&\mathbf{\color{blue}{6}}&&\mathbf{\color{red}{15}}&&\mathbf{\color{... | {
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"source": "stackexchange",
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Indeterminate Limit using Second Fundamental Theorem of Calculus I am trying to find $$\lim_{x\to 0}\frac{\int_0^x(x-t)\sin(t^2) \, dt}{\ln(1+x^4)}$$It seems that I am meant to use the 2nd Fundamental Theorem of Calculus to solve this, but I have never used it on an integral that has $x$ in it. Do I approach it any dif... | $$\frac{\int_{0}^{x}(x-t)\sin(t^2)\,dt}{\log(1+x^4)}$$ is of the form $\frac uv=\frac 00$ if $x=0$. So, let us use L'Hospital rule considering $\frac {u'}{v'}$.
$$u=\int_{0}^{x}(x-t)\sin(t^2)\,dt \implies u'=\frac d {dx} \int_{0}^{x}(x-t)\sin(t^2)\,dt$$ The fundamental theorem of calculus gives $$\frac d {dx} \int_{0}^... | {
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"source": "stackexchange",
"question_score": "3",
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simplify $\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$ simplify $$\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$$
1.$\frac{3}{2}$
2.$\frac{\sqrt[3]{65}}{4}$
3.$\sqrt[3]{2}$
4.$1$
I equal it to $\sqrt[3]{a}+\sqrt[3]{b}$ but I cant find $a$ and $b$
| Let $X=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$ From $(A+B)^3=A^3+B^3+3AB(A+B)$ we get the equation $$X^3-3ABX-(A^3+B^3)=0$$
We have $(AB)^3=({5+2\sqrt{13}})({5-2\sqrt{13}})=-27\Rightarrow {AB}=-3$ and $A^3+B^3=10$. Hence our equation $$X^3+9X-10=0\iff (X-1)(X^2+X+10)=0$$
Thus $$X=1$$ It is the $4$ the asked simp... | {
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Prove that every number of the sequence $49,4489,444889,\ldots$ is a perfect square
Prove that every number of the sequence $49,4489,444889,\ldots$ is a perfect square.
We can observe that $49=7^2, 4489=67^2, 444889=667^2, \ldots$
I have tried expanding terms of the sequence, and to express it as a whole square. But... | We can prove it as follows:
\begin{align*}
\underbrace{666\ldots6}_{n}\,7^2
&= \left[7 + 6(10^1 + 10^2 + \cdots + 10^n) \right]^2 \\
&= \left[7 + 6 \frac{10^{n+1} - 10}{9}\right]^2 \\
&= \left[ \frac{2 \cdot 10^{n+1} + 1}{3}\right]^2 \\
&= \frac{4 \cdot 10^{2n+2} + 4 \cdot 10^{n+1} + 1}{9} \\
&= 4 \cdot \frac{10^{2n+2}... | {
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"source": "stackexchange",
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If the chord $x+y=b$ of the curve ... If the chord $x+y=b$ of the curve $x^2+y^2-2ax-4a^2=0$ subtends a right angle at the origin, prove that: $b(b-a)=4a^2$
My Approach.
Given,
Equation of the chord,
$$x+y=b$$
$$\frac {x+y}{b}=1$$
Now,
Equation of the curve,
$$x^2+y^2-2ax-4a^2=0$$
$$x^2+y^2-2ax=4a^2$$
$$(b-y)^2+(b-x)^... | The combined equation of the lines joining the origin to the end points of the chord can be obtained by "homogenising" the equation of the curve. This is
\begin{align*}
x^2+y^2 - 2ax\left(\frac{x+y}{b}\right) - 4a^2\left(\frac{x+y}{b}\right)^2 = 0
\end{align*}
These lines are perpendicular if the sum of the coefficie... | {
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} |
Find derivative of $\frac{x}{\sqrt{x^2 +2}}$ using definition I recognise that the derivative of $\frac{x}{\sqrt{x^2 +2}}$ is given by the expression $\lim_{h\to 0}\frac{\frac{x+h}{\sqrt{(x+h)^2}+2}-\frac{x}{\sqrt{x^2 +2}}}{h}$. But I have trouble proceeding from here. Any kind soul, please help!
| we have $$\frac{1}{h}\frac{(x+h)\sqrt{x^2+2}-x\sqrt{(x+h)^2+2}}{\sqrt{(x+h)^2+2}\,\sqrt{x^2+2}}$$
and now multiply numerator and denominator by $$(x+h)\sqrt{(x^2+2)}+x\sqrt{((x+h)^2+2)}$$
can you proceed?
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the smallest positive integer that satisfies the system of congruences $N \equiv 2 \pmod{11}, N \equiv 3 \pmod{17}. $ Find the smallest positive integer that satisfies the system of congruences
\begin{align*}
N &\equiv 2 \pmod{11}, \\
N &\equiv 3 \pmod{17}.
\end{align*}
The only way I know to solve this problem is... | How it can be done using 'Euclid's algorithm': $N\equiv 3$ mod 17 can be written as N= 17i+ 3. Since 17= 11+ 6 is equivalent to 6 mod 11, we can write $N\equiv 2$ (mod 11) as $N\equiv 17i+ 3\equiv 6i+ 3= 2$ (mod 11) which is the sae as $6i= 2- 3= -1= 10$ (mod 11) or, dividing by 2, $3i\equiv 5$ (mod 11).
That is the... | {
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"timestamp": "2023-03-29T00:00:00",
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Symmetric polynomials with Vieta's and Newton's theorems
Let $ x_{1}, x_{2}, x_{3}$ be the solutions of the equation $ x^3 -3x^2 + x - 1 = 0.$
Determine the values of $$\frac{1}{{x_{1}x_{2}}} + \frac{1}{{x_{2}x_{3}}} + \frac{1}{{x_{3}x_{1}}}$$
and also
$$ x_{1}^{3}+x_{2}^{3}+x_{3}^{3}$$
| Given $(x-x_1)(x-x_2)(x-x_3)=x^3-3x^2+x-1$, the elementary symmetric polynomials of $x_1,x_2,x_3$ are given by Vieta's formulas:
$$ e_1=x_1+x_2+x_3=3,\quad e_2=x_1 x_2+x_1 x_3+x_2 x_3 = 1,\quad e_3=x_1 x_2 x_3 = 1 \tag{1}$$
hence
$$ \frac{1}{x_1 x_2}+\frac{1}{x_1 x_3}+\frac{1}{x_2 x_3} = \frac{e_1}{e_3} = \color{red}{3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Does $A^TMA = MA^2$ where $A$ is symmetric and $M$ is not necesarily symmetric Let $A, M \in \mathbb{R}^{2\times 2}$, $A$ is symmetric
Does $A^TMA = MA^2$ hold regardless of whether $M$ is symmetric?
I have tested some simple examples and have not found a violation. Does a counter example exist?
| Another counterexample:
$$
\begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix}
\begin{pmatrix}a & b \\ c & d\end{pmatrix}
\begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix}
=\begin{pmatrix}4a & 2b \\ 2c & d\end{pmatrix},
$$
but
$$
\begin{pmatrix}a & b \\ c & d\end{pmatrix}
\begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix}
\begin{pmatrix}2 & 0... | {
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Help on a very interesting integral: $\int_0^1\frac{x^5(1-x)^4}{1+x^3}\,dx$ I've been starting to work on problems with difficult integrals, and I came to
one of my problems where I end up having to integrate
$$\int_0^1 \frac{x^5(1-x)^4}{1+x^3} dx.$$
Wolfram seems to time out on integrals like these, so I decided to ta... | Using the binomial theorem, expand $(1-x)^4$ in the numerator:
$$
I = \int_0^1 \frac{x^5(1-x)^4}{1+x^3}\,dx = \int_0^1 \frac{x^5(1 - 4x + 6x^2 - 4x^3 + x^4)}{1+x^3}\,dx \\
= \int_0^1 \frac{x^5 - 4x^6 + 6x^7 - 4x^8 + x^9}{1+x^3}\,dx.
$$
Using polynomial long division,
$$
I = \int_0^1 \left(x^6 - 4x^5 + 6x^4 - 5x^3 + 5x^... | {
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Existence of a quadrilateral with side lengths $a,b,c,d$ where $a+b+c>d$ and $d$ is largest length I tried to prove triangle inequality, given side lengths $a, b, c$ if $a + b \gt c$, then a triangle always exists, by placing the side length $c$ on the $x$-axis and proving that intersection points of two circles of rad... | Assume $a < b < c < d$.
Suppose that WLOG, $a+b > d$. Then , construct a triangle with lengths $a,b,d$, and at the vertex joining $b$ and $d$, make a separation and attach the length $c$ to make a quadrilateral. Similarly for the other pairwise sums. Hence, we will assume that each pairwise sum is less than $d$.
Let $\... | {
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Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$. Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$.
So $(a^2+b^2)(c^2+d^2) = a^2c^2+a^2d^2+b^2c^2+b^2d^2$
and $(ac+bd)^2 = a^2c^2+2acbd+b^2d^2$
So the problem is reduced to proving that $a^2d^2+b^2c^2\ge2acbd$ but I am not sure ... | Hint: use Cauchy-Schwarz in $\mathbb{R}^2$ on the vectors $(a,b)$ and $(c,d)$. This technique should provide a one-line proof of the desired result.
More directly, from what you've already computed, you can observe that
$$
a^2d^2 - 2abcd + b^2c^2 = (ad - bc)^2 \geq 0
\text{,}
$$
so $a^2d^2 + b^2c^2 \geq 2abcd$.
| {
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In any triangle $ ABC $ $\frac {sin(B-C)}{sin (B+C)}=\frac {b^2-c^2}{a^2} $. In any triangle $ ABC $ prove that $\frac {sin(B-C)}{sin (B+C)}=\frac {b^2-c^2}{a^2} $.
Please help. Thanks in advance.
|
In the above configuration, $R \sin(B-C) = OJ_A = M_A H_A$ and $R\sin(B+C)=R\sin(A)= BM_A$. Moreover, by the Pythagorean theorem we have $H_A B^2-H_A C^2=AB^2-AC^2=c^2-b^2$, hence
$ b^2-c^2 = (H_A B+H_A C)(H_A B-H_A C) $ and
$ H_A B - H_A C = \frac{c^2-b^2}{a}.$
It follows that
$$ BH_A = \frac{1}{2}\left(\frac{c^2-b^2... | {
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Skip a number in a summation $$\sum_{n=1}^{10} n^2$$
Returns:
$1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100$
and I would like it to return:
$1 + 9 + 25 + 49 + 81 + 121 + 169 + 225 + 289 + 361$
How will I go about this and, more importantly, how does it work?
| You are looking for $1^2 + 3^2 +5^2 +7^2 +9^2 +11^2+13^2+15^2+17^2 +19^2$
so whats wrong with
$$\sum_{n=0}^9(2n+1)^2$$
or
$$\sum_{n=1}^{10}(2n-1)^2$$
It's $2n$ because we are going up in twos
| {
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How to integrate $ \int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}} $? I am having a little problem with my maths homework. The problem is as follows:
\begin{equation}
\int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}}
\end{equation}
I tried to do the following but got stuck halfway:
Let $\ \ x \ = asin\theta, \ he... | Assuming $a>0$ and applying the substitutions $x=az$, $z=\sin\theta$:
$$ I(a)=\int_{0}^{a}\frac{dx}{x+\sqrt{a^2-x^2}}=\int_{0}^{1}\frac{dz}{z+\sqrt{1-z^2}}=\int_{0}^{\pi/2}\frac{\cos\theta}{\sin\theta+\cos\theta}\,d\theta $$
but due to the substitution $\theta=\frac{\pi}{2}-\varphi$ we also have $I(a)=\int_{0}^{\pi/2}\... | {
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Prove that the function $(12-6x+x^2) e^x - (12+6x+x^2)$ is positive on $(0, \infty)$
Prove that the function $(12-6x+x^2) e^x - (12+6x+x^2)$ is positive on $(0, \infty)$.
I am unable to prove this.
| Let $f(x)=x+\ln(x^2-6x+12)-\ln(x^2+6x+12)$.
Since $f'(x)=1+\frac{2x-6}{x^2-6x+12}-\frac{2x+6}{x^2+6x+12}=\frac{x^4}{(x^2-6x+12)(x^2+6x+12)}\geq0$,
we obtain that $f(x)\geq f(0)=0$.
Done!
| {
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For how many positive integers $a$ is $a^4−3a^2+9$ a prime number? I understand that there are many posts on the problems similar to mine. I have tried my best, but still get different answers from the answer sheet. Can anyone help me? Also is there a simple way to find $a$?
For how many positive integers $a$ is $a^4-... | Here is a fundamental fact about prime numbers: IF: a prime number $p$ factors as $p = xy$, where $x$ and $y$ are integers, THEN: $x = \pm 1$ or $y = \pm 1$.
You have already correctly noticed that
$$
a^4-3a^2+9=(a^2+3+3a)(a^2+3-3a).
$$
Assuming that $a^4 - 3a^2 + 9$ is prime, it follows by the above result that there ... | {
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Easier way to calculate the derivative of $\ln(\frac{x}{\sqrt{x^2+1}})$? For the function $f$ given by
$$
\large \mathbb{R^+} \to \mathbb{R} \quad x \mapsto \ln \left (\frac{x}{\sqrt{x^2+1}} \right)
$$
I had to find $f'$ and $f''$.
Below, I have calculated them.
But, isn't there a better and more convenient way to do t... | Hint
$$\ln(\frac{x}{\sqrt{x^2+1}})=\ln x-\frac{1}{2}\ln(x^2+1)$$
| {
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$\lim_{n\rightarrow\infty} n^2C_n$ for $C_n=\int_{\frac{1}{n+1}}^{\frac{1}{n}}\frac{\tan^{-1}nx}{\sin^{-1}nx}dx$ equals? Let $$C_n=\int_{\frac{1}{n+1}}^{\frac{1}{n}}\frac{\tan^{-1}nx}{\sin^{-1}nx}dx$$ then $\lim_{n\rightarrow\infty} n^2C_n$ equals?
I am having trouble in finding the integral.Wolfram alpha too doesnt gi... | By the MVT for integrals, there is $\xi_n\in[\frac{1}{n+1},\frac1n]$ such that
$$ \int_\frac{1}{n+1}^\frac1n\frac{\tan^{-1}(nx)}{\sin^{-1}(nx)}dx=\frac{\tan^{-1}(n\xi_n)}{\sin^{-1}(n\xi_n)}\frac{1}{n(n+1)}.$$
Note that both $\sin^{-1}x$ and $\tan^{-1}x$ are increasing in $[0,1]$ and hence
$$\frac{\tan^{-1}(n\cdot\frac1... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove the inequality $\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$ Let $a,b,c>0; a+b+c=1$. Prove the inequality
$$\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$$
My work so far:
I tried AM-GM and used fact $a+b+c=1$.
| An homogenization and a full expanding give:
$$\sum\limits_{cyc}(a^6b^2+2a^5b^3+a^5c^3+2a^4b^4+a^6bc-2a^5b^2c+4a^5c^2b+$$
$$+a^4b^3c+4a^4c^3b-7a^4b^2c^2-7a^3b^3c^2)\geq0$$
which is obvious by AM-GM.
| {
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How to find the square roots of $z = 5-12i$ I am asked the following question:
Find the square roots of $z = 5-12i$
I know that this problem can be easily solved by doing the following:
$$
z_k^2 = 5-12i\\
(a+bi)^2 = 5-12i\\
(a^2-b^2) + i(2ab) = 5-12i\\
\\
\begin{cases}
a^2 - b^2 = 5\\
2ab = -12
\end{cases} \quad \Rig... | Your method should work fine in both cases (though the trigonometry is a bit more complicated), but I point out that your first method can still work on the square roots of $2i$: We see that
$$
a^2-b^2 = 0
$$
so $a = \pm b$, and then, secondly, $2ab = 2$. This gives us $a = b = \pm 1$ as the solutions, so the square r... | {
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Easier way to discover the area of a right triangle In the following right triangle: $y-x=5$ and the altitude to the hypotenuse is 12. Calculate its area.
I've managed to discover its area using the following method, but it ends up with a 4th degree equation to solve. Is there an easier way to solve the problem?
$ha=x... | Slightly easier: Write the unknown area as $A:=xy/2$. From $12 \sqrt{x^2+y^2} = xy$ deduce $$x^2+y^2=(A/6)^2.$$ Substitute this into $x^2-2xy+y^2=25$. This gets you a quadratic for $A$:
$$
\left(\frac A6\right)^2-4A-25=0
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $3^1+3^2+3^3+3^4+\cdots+3^n=\frac{3^{n+1}-3}{2}$ (suggestions for improvement on basic induction style/proof) Please see my proof to the following proposition below. From a standpoint of critiquing does this seem like a great execution or would someone feel that this proof was overly cumbersome? Specifically... | Your proof is good!
Here is another nice way to prove it:
$$S=3^1+3^2+3^3+...+3^n\\3S=3(3^1+3^2+3^3+...+3^n)\\3S=3^2+3^3+3^4+...+3^{n+1}\\3S-S=3^{n+1}[-3^n+3^n]-...-[3^2+3^2]-3^1\\2S=3^{n+1}-3\\S=\frac{3^{n+1}-3}{2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determine $ax^4 + by^4$ for system of equations I found the following recreational problem without further specification for $a,b$.
Let $x,y$ be real numbers s.t.
$a + b = 6$,
$ax + by = 10$,
$ax^2 + by^2 = 24$,
$ax^3 + by^3 = 62$.
Determine $ax^4 + by^4$.
I am new to problem solving exercises like this and therefo... | $$ \color{red}{3 \cdot 62 - 24 = 162}. $$
$$ 3 \cdot 10 - 6 = 24. $$
$$ 3 \cdot 24 - 10 = 62. $$
This is the observation of @Ross,
$$ (xy-1)(x-y)^2 = 0. $$ Ross points out in comment (below) that an ingredient was $44=abxy(x-y)^2,$ so that
$$ a \neq 0, b \neq 0, x \neq 0, y \neq 0, x \neq y. $$ Finally,
$$ xy ... | {
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About the primitives of $\frac1{a^2 + (2x)^2}$ So I was to find the integral $\int^{\frac{3}{2}}_0\frac{1}{9+4x^2}dx$
I noticed that the denominator is equal to $3^2+(2x)^2$ and thought I could use the integral $\int\frac{1}{a^2+x^2}dx = \frac{1}{a}arctan(\frac{x}{a}) + C$.
Hence, $[\frac{1}{3}arctan(\frac{2z}{3})]^{3/... | The same reason why $\sin (2x)$ is not an anti-derivative (= primitive function) of $\cos (2x)$, although $\sin (x)$ is an anti-derivative of $\cos (x)$.
To see what happens, you can use the substitution method for integrals (let $t = kx$ etc).
Or the other way around: because the chain rule for derivatives would produ... | {
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definite integral of a trigonometric function with log $$I=\int_0^{\frac\pi 2} \log(1+\tan x)\,dx$$
I attempted it by substituting $(\frac\pi 2-x)$ as $x$ but we get $\log(1+\cot x)$ not sure how to proceed... and on adding $$2I=\log[(1+\tan x)(1+\cot x)]$$
| Another solution using Clausen function
Start with $$ I=\int \limits^{\frac{\pi }{2} }_{0}\ln\left( 1+\tan \left( x\right) \right) dx=\overbrace{\int \limits^{\frac{\pi }{2} }_{0}\ln\left( \sin \left( x\right) +\cos \left( x\right) \right) dx} \limits^{I_{1}}-\overbrace{\int \limits^{\frac{\pi }{2} }_{0}\ln\left( \cos ... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$ Can someone please see the work I have so far for the following proof and provide guidance on my inductive step?
Prove that if $m,n\in\mathbb{N}$, then $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$
Base Case. Let ... | A combinatorial proof is also possible. I have $m+n+1$ white balls numbered $0$ through $m+n$. I’m going to paint $m+2$ of them red, then choose any of the red balls except the one with the highest number and put a gold star on it, and I want to know how many different outcomes are possible.
Suppose that the highest-nu... | {
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Simplify the following $(1\cdot2)+(2\cdot3)+(3\cdot4)+\dots+(n\cdot(n+1))$ Simplify the following $(1\cdot2)+(2\cdot3)+(3\cdot4)+\dots+(n\cdot(n+1))$.
How do i do this by using these identities.
$C^r_r + C^{r+1}_r + C^{r+2}_r + ....+C^{n}_r = C^{n+1}_{r+1}$
Or
$C^r_0 + C^{r+1}_1 + C^{r+2}_2 + ....+C^{r+k}_k = C^{r+k+1}... | Hint:
$$(1\times2)+(2\times3)+(3\times4)+\dots(n\times(n+1)) =\sum_{k=1}^nk(k+1) \\
\begin{align}
& =2\sum_{k=1}^n\frac{k(k+1)}2 \\
& =2\sum_{k=1}^nC_2^{k+1} \\
\end{align}$$
| {
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If $A < B < C < D$ where $A, B, C, D \in \mathbb{N}$, does the following series of inequalities hold? PROBLEM STATEMENT
If $A < B < C < D$ where $A, B, C, D \in \mathbb{N}$, does the following series of inequalities hold?
$$\frac{A}{D}<\frac{A}{C}<\frac{A}{B}<\frac{B}{D}<\frac{B}{C}<\frac{C}{D}<1$$
$$1<\frac{D}{C}... | You are correct. Consider the following examples:
$$A=1,B=3,C=4,D=5$$
In which $$\frac{D}{C}<\frac{C}{B},\frac{D}{B}<\frac{B}{A}$$
Also:
$$A=2,B=3,C=4,D=6$$
yields $$\frac{D}{C}>\frac{C}{B}, \frac{D}{B}>\frac{B}{A}$$
| {
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Expressing $\sqrt[3]{7+5\sqrt{2}}$ in the form $x+y\sqrt{2}$ Express $\sqrt[3]{(7+5\sqrt{2})}$ in the form $x+y\sqrt{2}$ with $x$ and $y$ rational numbers.
I.e. Show that it is $1+\sqrt{2}$.
| If it simplifies, then $7+5\sqrt 2$ is a cube $(a+b \sqrt 2)^3$, in the ring of integers of $\Bbb Q(\sqrt 2)$, which is $\Bbb Z[\sqrt 2]$, so $a$ and $b$ must be integers (sometimes you can only deduce that $2a,a+b,2b$ are integers but it's still very good)
Moreover, you have $2a = (7+5\sqrt 2)^\frac 13 + (7-5\sqrt 2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $f$ if $f'(x)=\dfrac{x^2-1}{x}$ knowing that $f(1) = \dfrac{1}{2}$ and $f(-1) = 0$ I am asked the following problem:
Find $f$ if $$f'(x)=\frac{x^2-1}{x}$$
I am not sure about my solution, which I will describe below:
My solution:
The first thing that I've done is separate the terms of $f'(x)$
\begin{align*}
f'(x... | Looks fine! However a little typo when calculate $f(-1)=0$ not $-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1941023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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} |
Solving the equation $(x^2-3x+1)^2=4x^2-12x+9$ I need to solve the following equation:
$$(x^2-3x+1)^2=4x^2-12x+9.$$
I think I need to bring everything to one side but I don't know anything else.
| First you need to factor before bring the terms into one side. $$(x^2-3x+1)^2=4x^2-12x+9$$ $$(x^2-3x+1)^2=(2x-3)^2$$ Now you subtract the right hand side, where it becomes $$(x^2-3x+1)^2-(2x-3)^2=0$$ You can factor by using differences of squares where $a^2-b^2=(a+b)(a-b)$.
Now finish this off on your own:)
If you rea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
Product of digits of a 7 digit number is $2^43^4$ Find the number of seven digit numbers whose product of digits is $2^43^4$.
One method is to list out all possible sets of seven digits that give this product, and then find number of permutations for each case. But there are too many cases that way.
Is there a shorter ... | I propose to count these numbers with respect to the number of digits 6.
We have five cases: ZERO, ONE, TWO, THREE, FOUR. In the following formulas we place the $6$s then the powers of $2$ (i. e. $2,4,8$) and finally the powers of $3$ (i. e. $3,9$).
If there are ZERO digits 6 then we have
$$
\binom{7}{4}\left[\binom{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the global minimum of a function of two variables without derivatives If $f:(\mathbb{R}^{+*})^2\to\mathbb{R}$ is defined by $$f(x,y)=\sqrt{x+y}\cdot\Big(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\Big),$$
how would you prove that $f$ has a global minimum without using all the differential calculus tools ? I am asked to... | $$f(x,y) = \sqrt{x+y} \cdot \left( \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} \right) = \sqrt{x+y} \frac{\sqrt{x} + \sqrt{y}}{\sqrt{xy}}$$
Squaring we get
$$f(x,y)^2 = (x+y) \frac{x + y + 2 \sqrt{xy}}{xy} = \frac{(x+y)^2}{xy} + 2 \frac{x+y}{\sqrt{xy}}$$
Now AM-GM implies that the first summand and second summand are bot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1945661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$1\cdot3\cdot5 \cdots (p - 2) = (-1)^{m+k+1} \pmod p$, and $2\cdot4\cdot6\cdots(p - 1) = ( -1)^{m +k} \pmod p$. Prove that if $p$ is a prime having the form $4k + 3$, and if $m$ is the number of quadratic residues less than $\frac p2$, then we have
$$1\cdot3\cdot5 \cdots (p - 2) = (-1)^{m+k+1} \pmod p, \text{ and } 2\c... | Let $\phi: \mathbb{Z}_p \to \{-1,1\}$, where $\phi(n) = 1$ is a homomorphism given by $1$ if $n$ is a quadratic residue modulo $p$ and $-1$ if not.
Now consider $1 \cdot 3 \cdots (p-2)$. Now switch every number $x$ bigger than $\frac p2$ to $(p-x)$. Eventually as you will make $\frac{p-3}{4} = k$ changes we have that:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate : $\int _0^{\infty } \sin (ax^2)\cos \left(2bx\right)dx$ How can I solve
$$\int_{0}^{\infty} \sin \left (ax^2 \right) \cos \left (2bx \right)dx$$
Thanks!
| Concerning the antiderivative, start writing$$\sin \left (ax^2 \right) \cos \left (2bx \right)=\frac12\left(\sin(ax^2+2bx)+\sin(ax^2-2bx)\right)$$ Now $$ax^2+2bx=\left(\sqrt a x +\frac{b}{\sqrt a}\right)^2-\frac{b^2}a$$ $$ax^2-2bx=\left(\sqrt a x -\frac{b}{\sqrt a}\right)^2-\frac{b^2}a$$ Now, consider $$I=\int \sin(ax... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1947642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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$\sum_{n=1}^\infty\frac1{n^6}=\frac{\pi^6}{945}$ by Fourier series of $x^2$
Prove that
$$\sum_{n=1}^\infty\frac1{n^6}=\frac{\pi^6}{945}$$ by the Fourier series of $x^2$.
By Parseval's identity, I can only show $\sum_{n=1}^\infty\frac1{n^4}=\frac{\pi^4}{90}$. Could you please give me some hints?
| Considering $f(x)=x^3$ By Parseval identity we can prove that $\sum_{n=1}^{\infty}\frac{1}{n^6}=\frac{\pi^6}{945}$.
Then $$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx=0,$$$$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx=0$$ and \begin{align}b_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx\\&=(-1)^{n+1}\frac{2\pi^2}{n}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Proving $\frac{\sin x}{x} + \frac{x^2}{4} >1$, for $x \in [0,\frac{\pi}{2}]$
Prove that, for every $x \in \left(0,\frac{\pi}{2}\right)$,
$$\frac{\sin x}{x} + \frac{x^2}{4} >1$$
I have tried using differentiation to prove that the left-hand side is strictly increasing on the interval, but no success. Please, I need... | Fix $x \in (0, \pi/2)$. Using Taylor's theorem
\begin{align}
\sin x = x-\frac{1}{3!}x^3+\frac{\cos(\xi)}{5!}x^5
\end{align}
where $\xi \in (0, x)$.
Hence it follows
\begin{align}
\frac{\sin x}{x}+\frac{1}{4}x^2-1 = \left(\frac{1}{4}-\frac{1}{6}\right)x^2+\frac{\cos(\xi)}{5!}x^4>0
\end{align}
since $\cos x$ is non-negat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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