Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How can $\frac{x^3-4x^2+4x}{x^2-4}$ be both $0$ and "undefined" when $x = 2$? Suppose I have a function defined as $$F(x)= \frac{x^3-4x^2+4x}{x^2-4}$$
Now I want to find the value of $F(2)$. I can do it in 2 ways:
*
*Put $x=2$ and solve the function. It will give:
$$F(2)=\frac{0}{0}$$ which is not defined.
*Solve $... | "Solve $F(x)$" is the wrong terminology. "Simplify $F(x)$" fits better. One solves problems; one solves equations; one evaluates or sometimes simplifies expressions.
$\dfrac 5 0$ is undefined because there is no number $x$ for which $0x = 5.$
But $\dfrac 0 0$ is undefined because there are many numbers $x$ for which $0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2027810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
How can we factorize $a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$? How can we factorize $a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$?
| You just need to do a little rearrangement :
$$a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$$
$$a^2b+a^2c+b^2c+bc^2+ab^2+abc+abc+ac^2$$
$$a^2(b+c)+bc(b+c)+ab(b+c)+ac(b+c)$$
$$(b+c)(a^2+bc+ab+ac)$$
$$(b+c)(a(a+b)+c(a+b))$$
$$(b+c)(a+b)(a+c)$$
$$(a+b)(b+c)(c+a)$$
*And we are done !!!!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2030326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Summation of series with $r^{th}$ term $\frac{1}{\sqrt{a+rx} +\sqrt{a+(r-1)x}} $
Find the summation of series with $r^{th}$ term $t_r$ given by $\displaystyle \frac{1}{\sqrt{a+rx} +\sqrt{a+(r-1)x}} $
Working:
$$t_r=\frac{1}{\sqrt{a+rx} +\sqrt{a+(r-1)x}}$$
$$t_r=\frac{\sqrt{a+rx} -\sqrt{a+(r-1)x}}{a+rx-(a+(r-1)... | Continue from your working $$S_{n} = \sum^{n}_{r=1}t_{r} = \frac{1}{x}\sum^{n}_{r=1}\bigg[\sqrt{a+rx}-\sqrt{a+(r-1)x}\bigg]$$
Now Using Telescopis Sum (expanding summation.)
$$S_{n} = \frac{1}{x}\bigg[\left(\sqrt{a+x}-\sqrt{a}\right)+(\sqrt{a+2x}-\sqrt{a+x})+\cdots \cdots +(\sqrt{a+nx}-\sqrt{a+(n-1)x)}\bigg]$$
So we ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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maclaurin series of $f(x)=xe^{-x}$ I'm not sure how to do: $$f(x)=xe^{-x}$$I did this one so far:
$$ f(x) = x^2 \cos x = x^2 \left( 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} +\dfrac{x^8}{8!} \dots \right) = x^2 - \dfrac{x^4}{2!} + \dfrac{x^6}{4!} - \dfrac{x^8}{6!} +\dfrac{x^{10}}{6!} \dots $$
i know that ... | From
$$e^x= 1 + x+ \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!}+\cdots$$
one gets that
$$
f(x)=e^{-x}= 1 - x+ \dfrac{x^2}{2!} - \dfrac{x^3}{3!} + \dfrac{x^4}{4!}+\cdots
$$ giving
$$
f(x)=xe^{-x}= x - x^2+ \dfrac{x^3}{2!} - \dfrac{x^4}{3!} + \dfrac{x^5}{4!}+\cdots
$$ or
$$
f(x)=xe^{-x}=\sum_{n=0}^\infty(-1)^n \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculation of some limit I'm having problems with limits like that $ \lim_{x → 0} \left(\frac{1- \left(\cos x\right)^{\sin x}} {x^3}\right)$. I need to calculate it in pretty rigorous way.
I can use Taylor and write $\cos x^{\sin x}$ as $\left(1+o\left(x^2\right)\right)^{x + o\left(x^2\right)}$ but I have no idea what... | You can compose polynomial expansions. We'll use expansions at order $3$.
By definition, $\;\cos x^{\sin x}=\mathrm e^{\sin x\ln(\cos x)}$.
Now $\;\cos x=1-\dfrac{x^2}2+o(x^3)$, and $\;\ln(1-u)=-u-\dfrac{u^2}2-\dfrac{u^3}3-\dotsm$, so it will be enough to use the expansion of $\;\ln(1-u)$ at order $1$:
$$\ln(\cos x)=-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2035616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove $2^{1/3} + 2^{2/3}$ is irrational What's the nice 'trick' to showing that the following expression is irrational?
$2^{1/3} + 2^{2/3}$
| Let $x=\sqrt[3]{2}+\sqrt[3]{4}$ then raising to the $3^{rd}$ power and expanding the binomial on the right:
$$x^3 = (\sqrt[3]{2})^3 + 3 \cdot \sqrt[3]{2} \cdot \sqrt[3]{4} \cdot (\sqrt[3]{2}+\sqrt[3]{4}) + (\sqrt[3]{4})^3 = 6 x + 6$$
By the rational root theorem, the equation $x^3-6x-6=0$ can only have divisors of $6$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2036087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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How many three-digit numbers can be generated from 1, 2, 3, 4, 5, 6, 7, 8, 9, such that the digits are in ascending order? Solution for the same is given below
Numbers starting with 12 – 7 numbers
Numbers starting with 13 – 6 numbers;
14 – 5, 15 – 4, 16 – 3, 17 – 2, 18 – 1.
Thus total number of
numbers starting from ... | Here is a trick:
*
*An increasing arrangement must contain unique digits
*There are $\binom{9}{3}$ combinations of $3$ out of $9$ unique digits
*Each combination has exactly $1$ increasing arrangement
*Hence the number of increasing arrangements is $\binom{9}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove or disprove $(a^2+b^2)(b^2+c^2)(c^2+a^2)\leq(a+b)(b+c)(c+a)$ Prove or disprove:
If $a, b, c \in (0, \infty)$ such that $a^2+b^2+c^2=3$ then
$$(a^2+b^2)(b^2+c^2)(c^2+a^2)\leq(a+b)(b+c)(c+a).$$
All my attempts to prove inequality have been unsuccessful. Maybe someone has an idea. Thank you very much!
| It's wrong! Try $c\rightarrow0^+$ and $a=b\rightarrow\sqrt{1.5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $ \alpha_i, i=0,1,2...n-1 $ be the nth roots of unity, the $\sum_{i=0}^{n-1} \frac{\alpha_i}{3- \alpha_i}$ is equal to? If $ \alpha_i, i=0,1,2...n-1 $ be the nth roots of unity, the $\sum_{i=0}^{n-1} \frac{\alpha_i}{3- \alpha_i}$ is equal to?
A) $ \frac{n}{3^n-1} $
B) $ \frac{n-1}{3^n-1} $
C) $ \frac{n+1}{3^n-1} $
... | (For future reference.) Introducing
$$f(z) = \frac{z}{3-z} \frac{nz^{n-1}}{z^n-1}$$
we get
$$S_n = \sum_{q=0}^{n-1}
\frac{\exp(2\pi i q/n)}{3-\exp(2\pi i q/n)}
= \sum_{q=0}^{n-1} \mathrm{Res}_{z=\exp(2\pi i q/n)} f(z).$$
Residues sum to zero so we have
$$S_n + \mathrm{Res}_{z=3} f(z) +
\mathrm{Res}_{z=\infty} f(z) = 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Can someone please explain why this is the first move when solving this integral. The integral which I am solving is $$\int_0^c\sqrt{c^2-x^2}~\mathrm dx$$ and the first thing which they suggest doing is setting $x=(\sqrt{a})/(\sqrt{b}) \sin u$. I understand everything after this but I do not comprehend where and why th... | If you must do this by a method other than recognizing this as the area of one quarter of a circle, then the expression $\sqrt{c^2-x^2}$ suggests using the substitution
\begin{align}
x & = c\sin\theta \\
dx & = c\cos\theta\,d\theta \\[10pt]
\sqrt{c^2 -x^2} & = \sqrt{c^2 - c^2\sin^2\theta} = c\sqrt{1-\sin^2\theta} \\[5p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\int J_5(x) dx= -J_4(x) - \frac {4}{x} J_3(x) - \frac {8}{x^2}J_2(x) +c$
Prove that $$\int J_5(x) dx= -J_4(x) - \frac {4}{x} J_3(x) - \frac
{8}{x^2}J_2(x) +c$$ where $J_n (x)$ is the Bessel function of first kind
and order $n $.
My attempt:
I know the following recurrence relations:
$$\frac {d}{dx}\le... | Following Alex R's comment, I am proceeding as follows:
$$\int J_5(x) dx$$
$$=\int \frac {x^4}{x^4} J_5(x) dx$$
$$=\int x^4 \cdot x^{-4} J_5(x) dx$$
Integrating by parts, we get
$$x^4 \cdot \int x^{-4} J_5(x) dx - \int \left\{\frac {d}{dx}(x^4) \cdot \int x^{-4} J_5(x) dx \right\} dx$$
$$=x^4 \cdot (-1) \cdot x^{-4} J... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Diophantine equation of prime numbers Determine all pairs of prime numbers $(p,q)$ that satisfy the equation
$p^3-q^3=pq^3-1$.
I can easily understand that
$p>q$, $q^3=p^2-p+1$, $q^3\equiv 1 \pmod{p}$, $p^3\equiv -1 \pmod{q}$.
But I don't know how to go on. (I know that p=19 and q=7 works)
| $p^2 - p + 1 = q^3$
$p(p-1) = (q-1)(q^2+q+1)$.
As you noticed, $p>q$, so $q-1$ is coprime with $p$. So $q-1|p-1$, or instead we can write
$k(q-1)=p-1$ and $kp=q^2 + q + 1$ for some integer $k>1$.
Get $p$ from the first equation and plug it to the second one:
$k(kq - k + 1) = q^2 + q + 1$
$q^2 + (1-k^2)q +1-k+k^2=0$
So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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What is wrong with the "proof" for $\ln(2) =\frac{1}{2}\ln(2)$? I have got a question which is as follows:
Is $\ln(2)=\frac{1}{2}\ln(2)$??
The following argument seems suggesting that the answer is yes:
We have the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$
which has a mathematically determined value $... | The series $\sum\limits_{k=1}^n\frac{(-1)^{k+1}}k$ is convergent but not absolutely convergent, i.e. the sum itself has a limit, but $\sum_{k=1}^\infty\left\lvert\frac{(-1)^{k+1}}{k}\right\rvert=+\infty$. Therefore, by Riemann-Dini theorem, for any extended real numbers $\alpha\le \beta\in\Bbb R\cup\{-\infty,\infty\}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
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General solution to $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ is said to have a general solution of $x=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{12}$.
My Approach:
Considering the equation as
$$
a\cos x+b\sin x=\sqrt{a^2+b^2}\Big(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\... | Hint:
For $(\tan A-\tan45^\circ)\cos x+(\tan A+\tan45^\circ)\sin x=\sec A$
$\cos(x+A+45^\circ)=\cos45^\circ$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$
Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$.
We can rearrange the given equation to $$y^2 = x^2(3y-2)\tag1$$ Thus $3y-2$ must be a perfect square and so $3y-2 = k^2$.
How can we continue?
| We note that $x^2\mid 3x^2y$, so $x^2\mid 2x^2+y^2$, then $x^2\mid y^2$, which implies that $x\mid y$. Let's call $y=kx$, if we replace this into your equation $(1)$ we get $k^2x^2=x^2(3kx-2)$, then $k^2=3kx-2$, which lead us to the quadratic equation $k^2-3xk+2=0$. Using the formula for quadratic equations we deduce t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Distributing $8$ different articles among $7$ boys
Problem Statement:-
Find the number of ways in which $8$ different articles can be distributed among $7$ boys, if each boy is to receive at least one article.
Attempt at a solution:-
First, start by numbering the boys from $1$ to $7$, which can be done in $7!$ ways.... | You have permuted both the boys and the articles, which is incorrect.
I'd prefer to work it out as
[ Choose "lucky" boy ] $\times$ [ Permute articles ]
$= \dbinom71 \times \dfrac{8!}{2!1!1!1!1!1!1!}$
which, of course, could be abbreviated as $ 7\times \dfrac{8!}{2!}$
PS:
On closer scrutiny, I find that the book answe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\vec{AI}.\vec{BH}=0$ First I'm sorry I don't know how to draw the triangle on LaTeX.
We consider $\triangle ABC$ isoscele in $A$. We denote $O$ the midpoint of $[BC]$ and $H$ the orthogonal projection of $O$ on $(AC)$. We also note $I$ the midpoint of $[OH]$.
Is it a good idea to use Apollonius theorem in $... | Since the title invites a vector proof, note that by construction:
$$
\begin{align}
& \overrightarrow{BO}=\overrightarrow{OC} && \text{since } O \text{ is the midpoint of } BC \tag{1} \\
& \overrightarrow{AO} \cdot \overrightarrow{BC}=0 && \text{since } \triangle ABC \text{ is isosceles, so } AO \perp BC\tag{2} \\
& \o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that $\sum\limits_{cyc}(4a^6+5a^5b)\geq\frac{(a+b+c)^6}{27}$ Let $a$, $b$ and $c$ be real numbers. Prove that:
$$4(a^6+b^6+c^6)+5(a^5b+b^5c+c^5a)\geq\frac{(a+b+c)^6}{27}$$
I tried SOS, uvw and more, but without success.
| The following proof is true if they are non-negative reals.
$4(x^3+y^3+z^3)+15xyz\geq (x+y+z)^3$- this directly follows from Schur. So $$4(a^6+b^6+b^6)+5(a^5b+b^5c+c^5a)\geq 4(a^6+b^6+c^6)+15a^2b^2c^2\geq (a^2+b^2+c^2)^3\geq\dfrac{(a+b+c)^6}{27}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Is something wrong with this precalculus question? If $x^{2}-4x+6=0$, then what can be the value of $1-\frac{4}{3x}+\frac{2}{x^{2}}$?
My answer is $1-\frac{4}{3x}+\frac{2}{x^{2}}=\frac{3x^{2}-4x+6}{3x^{2}}=\frac{3x^{2}-x^{2}}{3x^{2}}=\frac{2}{3}$ for $x\neq 0$. But according to the book answer is 2. What is the point i... | $3x^2-4x+6=2x^2+(x^2-4x+6)=2x^2$
You're right. Note that $x$ is complex and not real, but apart from this, the value of the other expression is $2/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How to prove that trace$(ABA^{-1}B^{-1})$=$3$ If $A,B$ are two $3 \times 3$ square matrices and trace(A) is defined as the sum of all diagonal elements. trace$(ABA^{-1}B^{-1})$=$3$
I could easily verify the above for the identity matrix.But I couldn't generalise it.
Please help me in this regard.thanks.
| Counterexample:
$$\begin{align}
A&=\begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix}
&B&=\begin{pmatrix}1&0&0\\1&1&0\\0&0&1\end{pmatrix}
\\A^{-1}&=\begin{pmatrix}1&-1&0\\0&1&0\\0&0&1\end{pmatrix}
&B^{-1}&=\begin{pmatrix}1&0&0\\-1&1&0\\0&0&1\end{pmatrix}
\end{align}$$
results in
$$ABA^{-1}B^{-1}=\begin{pmatrix}3&-1&0\\1&0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Common proof for $(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} $ I'm asking for an alternative (more common?) proof of the following equality, more specifically an alternative proof for the inductive step:
$$(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} (x\neq 1)$$
This is how I proved it... | A Proof with a Bit of Field Theory
Over a field of characteristic $2$, $1+x^{2^r}=(1+x)^{2^r}$ and $1-x=1+x$, so the LHS is $$
\begin{align}
\prod_{i=0}^n\,\left(1+x^{2^i}\right)
&=\prod_{i=0}^n\,(1+x)^{2^i}=(1+x)^{\sum\limits_{i=0}^n\,2^i}=(1+x)^{2^{n+1}-1}
\\
&=\frac{(1+x)^{2^{n+1}}}{1+x}=\frac{1+x^{2^{n+1}}}{1+x}=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
prove $(a^3+1)(b^3+1)(c^3+1)\ge 8$ let $a,b,c\ge 0$ and such $a+b+c=3$ show that
$$(a^3+1)(b^3+1)(c^3+1)\ge 8$$
My research:It seem use Holder inequality,so
$$(a^3+1)(b^3+1)(c^3+1)\ge (abc+1)^3$$
Use AM-GM
$$abc\le\dfrac{(a+b+c)^3}{27}=1$$
what? then I think this method is wrong
| Let $f(x,y,z) = (x^3 +1)(y^3 +1)(z^3 +1)$ subject to the constraint $g(x,y,z) = x + y + z - 3 =0$ and $ x,y,z \ge 0$.
Then using lagrange multipliers we have
*
*$f_x = 3x^2(y^3 +1)(z^3 +1)= g_x= \lambda$
*$f_y = 3y^2(x^3 +1)(z^3 +1)= g_y = \lambda$
*$f_z = 3z^2(x^3 +1)(y^3 +1)=g_z = \lambda$
*$g(x,y,z) = x + y +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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What is the remainder left after dividing $1! + 2! + 3! + ... + 100!$ by $5$? I have this question as a homework.
What is the remainder left after dividing $1! + 2! + 3! + \cdots + 100!$ by $5$?
I tried this: I noticed that every $n! \equiv 0 \pmod{5}$ for every $n\geq 5$.
For $n < 5$:
$$\begin{align*}
1! &\equiv 1 ... | To complement the other answer in less theory-heavy terms, you can view it as:
$$1! + 2! + 3! + 4! + 5! (1 + 6 + [6 \times 7] + \dots + [6 \times \dots \times 100])$$
The final term is clearly divisible by $5$; so we just need to look at $1! + 2! + 3! + 4!$ and its remainder, as you did.
| {
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"url": "https://math.stackexchange.com/questions/2069717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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} |
Choose $a, b$ so that $\cos(x) - \frac{1+ax^2}{1+bx^2}$ would be as infinitely small as possible on ${x \to 0}$ using Taylor polynomial $$\cos(x) - \frac{1+ax^2}{1+bx^2} \text{ on } x \to 0$$
If $\displaystyle \cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \cdots $
Then we should choose $a, b$ in a such way that it's T... | Note you just need to compute the Taylor expansion of
\begin{align}
g(x)&=\cos x+bx^2\cos x-1-ax^2 \\[6px]
&=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}
+bx^2-\frac{bx^4}{2!}+\frac{bx^6}{4!}-1-ax^2+o(x^6)\\[6px]
&=\left(-\frac{1}{2}+b-a\right)x^2+
\left(\frac{1}{24}-\frac{b}{2}\right)x^4+
\left(-\frac{1}{6!}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
If $abc+1=0$, verify that' If $abc+1=0$, prove that:
$\frac {1}{1-a-b^{-1}}+\frac {1}{1-b-c^{-1}} +\frac {1}{1-c-a^{-1}}=1$.
My Attempt:
$abc+1=0$
$abc=-1$.
Now,
$$L.H.S=\frac {1}{1-a-b^{-1}}+\frac {1}{1-b-c^{-1}}+\frac {1}{1-c-a^{-1}}$$
$$=\frac{b}{b-ab-1} + \frac {1}{1-b-c^{-1}} + \frac {c^{-1}}{c^{-1}-1-(ca)^{-1}}$$... | You are on the correct path. Your steps are all correct. Proceeding from there, just note that
$$\frac{b}{b+c^{-1} -1} +\frac{1}{1-b-c^{-1}} +\frac{c^{-1}}{c^{-1}-1+b}$$
$$=\frac{b+c^{-1}}{b+c^{-1} -1} -\frac{1}{b+c^{-1}-1}$$
$$=\frac{b+c^{-1}-1}{b+c^{-1} -1}$$
$$=1$$
Hope this helps you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Short technique of tackling or another method $\int_{0}^{\infty}{2x\over (x^4+2x^2+1)+\sqrt{x^4+2x^2+1}}dx=\ln{2}$ Prove that,
$$I=\int_{0}^{\infty}{2x\over (x^4+2x^2+1)+\sqrt{x^4+2x^2+1}}dx=\ln{2}$$
I try:
$x^4+2x^2+1=(x^2+1)^2$
$$\int_{0}^{\infty}{2x\over (x^2+1)(x^2+2)}dx$$
Let $u=x^2+1$, $du=2xdx$
$$\int_{1}^{\inft... | Set $x=\tan\theta$, we have
$$I=\int_{0}^{\infty}{2x\over (x^2+1)(x^2+2)}dx=\int_{0}^{\frac{\pi}{2}}\frac{2\tan\theta(1+\tan^2\theta)}{(1+\tan^2\theta)(2+\tan^2\theta)}d\theta$$
thus
$$I=\int_{0}^{\frac{\pi}{2}}\frac{\sin2\theta}{1+\cos^2\theta}d\theta=-\ln(1+\cos^2\theta)\Big{|}_{0}^{\frac{\pi}{2}}=\ln(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to compute the dimension of $\Bbb C[x,y]/I$ on $\Bbb C$? The problem is compute the dimension of $\Bbb C[x,y]/I$ over $\Bbb C$ as vector space where $I=\langle(x+2)^2,(x+2)(y+1),(y+1)^3\rangle$. $\Bbb C[x,y]$ is the polynomial ring over $\Bbb C$.
I have tried in this way.
For $f(x,y)$ in $\Bbb C[x,y]$ it has the fo... | A basis of $C[x,y]$ is $\{(x+2)^m(y+1)^n: m,n\ge 0\}$.
In $C[x,y]/I$, $(x+2)^2$ and $(y+1)^3$ are identified with $1$, hence, just because of them the basis reduces to
$$
(x+2)^m(y+1)^n, \quad m=0,1,\,\,n=0,1,2.
$$
But as $(x+2)(y+1)$ is also identified with $1$, then we are reduced to
$$
1, x+2, y+1, (y+2)^2.
$$
Hen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Why do I first need to bring $-4x$ into the numerator in $\lim_{x\to \infty} 4x^2/(x-2) - 4x$ I tried solving the question in the title as follows:
$$\lim_{x\to \infty} \frac{4x^2}{x-2} - 4x \to 4x - 4x \to 0$$
However, apparently that first step ($\to 4x - 4x$) was wrong, and I should first have brought the second $4x... | Hint
You can use long division first and get $$\frac{4x^2}{x-2}=4 x+8+\frac{16}{x}+\cdots$$
The other way could be $$\frac{4x^2}{x-2}=4\frac{x^2}{x-2}=4\frac{x^2-4x+4+4x-4}{x-2}=4\frac{(x-2)^2+4(x-1)}{x-2}=4\left(x-2+4\frac{x-1}{x-2} \right)=4\left(x-2+4\frac{x-2+1}{x-2} \right)=4\left(x-2+4\left(1+\frac{1}{x-2}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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How to differentiate $y=\ln(x+\sqrt{1+x^2})$? I'm trying to differentiate the equation below but I fear there must have been an error made. I can't seem to reconcile to the correct answer. The problem comes from James Stewart's Calculus Early Transcendentals, 7th Ed., Page 223, Exercise 25.
Please differentiate $y=\ln(... | Let $u=x+\sqrt{1+x^2}$. Then
$$u'=1+\left[\frac{1}{2}(1+x^2)^{-\frac{1}{2}}\cdot 2x\right]=1+\frac{x}{\sqrt{1+x^2}}=\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}.$$
Thus,
$$y'=\frac{1}{u}\cdot u'=\frac{1}{x+\sqrt{1+x^2}}\cdot \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}=\frac{1}{\sqrt{1+x^2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2073046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Find coefficient of $x^n$ in $(1+x+2x^2+3x^3+.....+nx^n)^2$ Find coefficient of $x^n$ in
$(1+x+2x^2+3x^3+.....+nx^n)^2$
My attempt:Let $S=1+x+2x^2+3x^3+...+nx^n$
$xS=x+x^2+2x^3+3x^4+...+nx^{n+1}$
$(1-x)S=1+x+x^2+x^3+....+x^n-nx^{n+1}-x=\frac{1-x^{n+1}}{1-x}-nx^{n+1}-x$
$S=\frac{1}{(1-x)^2}-\frac{x}{1-x}=\frac{1-x+x^2... | Concluding my attempt
Let $S=1+x+2x^2+3x^3+...+nx^n$
$xS=x+x^2+2x^3+3x^4+...+nx^{n+1}$
$(1-x)S=1+x+x^2+x^3+....+x^n-nx^{n+1}-x=\frac{1-x^{n+1}}{1-x}-nx^{n+1}-x$
$S=\frac{1}{(1-x)^2}-\frac{x}{1-x}=\frac{1-x+x^2}{(1-x)^2}$. (Ignoring terms which have powers of x greater than $x^n$)
So one can say that
coefficient of $x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Choosing one type of ball without replacement. Suppose I have $9$ balls, among which $3$ are green and $6$ are red. What is the probability that a ball randomly chosen is green?
It is $\dfrac{3}{9}=\dfrac{1}{3}$.
If three balls are randomly chosen without replacement, then what is the probability that the three balls ... | The probability that the first ball selected is green is $P_1 = \frac {3}{9} $. After selecting a green balls, there are now $2$ green and a total of $8$ balls. Now the probability of selecting a green balls is $P_2 = \frac {2}{8} $. Now there are a total of $7$ balls and a green ball. Now the probability is $P_3=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Show that:$\sum\limits_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}={1\over 12}(1-\ln{2})$ Show that
$$\sum_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}={1\over 12}(1-\ln{2})$$
My try:
We split into partial decomposition
$$n={A\over 2n-1}+{B\over 2n+1}+{C\over 4n-1}+{D\over 4n+1}$$
Setting $n={1\over 2}$, ${-1\over2}$ we have... | If we consider
$$ f(x)=\frac{x}{(4x^2-1)(16x^2-1)} $$
we may compute its partial fraction decomposition through the residue theorem:
$$ f(x) = \frac{1}{24}\left(\frac{1}{x-\tfrac{1}{2}}+\frac{1}{x+\tfrac{1}{2}}\right)-\frac{1}{24}\left(\frac{1}{x-\tfrac{1}{4}}+\frac{1}{x+\tfrac{1}{4}}\right)$$
and that leads to:
$$\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
From complex rotation matrix to real matrix Let's consider $R_{n}$ to be an $n \times n$ real rotation matrix. $R_{n}$ can be diagonalized with a unitary matrix $U_{n}$ to $D_{n}$, which is composed of blocs of
$ \left( \begin{array}{ccc}
e^{j\theta} & 0 \\
0 & e^{-j\theta} \end{array} \right) $ and eventually ones.... | Notice that $$\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix} = \begin{pmatrix}\tfrac{1}{\sqrt{2}} & -\tfrac{j}{\sqrt{2}} \\ -\tfrac{j}{\sqrt{2}} & \tfrac{1}{\sqrt{2}}\end{pmatrix} \begin{pmatrix}e^{j\theta} & 0\\ 0 & e^{-j\theta}\end{pmatrix} \begin{pmatrix}\tfrac{1}{\sqrt{2}} & \tfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Angles in triangles are in AP, find:$\dfrac{a}{c}\sin 2A+\dfrac{c}{a}\sin 2C$
If angles of a $\triangle{ABC}$ are in A.P. then find the value of : $$\dfrac{a}{c}\sin 2A+\dfrac{c}{a}\sin 2C$$ where $a,b,c$ are sides and $A,B,C$ are angles.
My attempts:
Let $\angle A =x-d , \angle B=x, \angle C=x+d\ \implies \angle B=... | I think there is a typo in your question. It should be to find the value of $\frac{a}{c}\sin 2C + \frac{c}{a}\sin 2A$.
Assuming the case it is so, then the problem simplifies to: $$\frac{\sin A}{\sin C}(2\sin C\cos C) + \frac{\sin C}{\sin A}(2\sin A\cos A) = 2\sin A\cos C + 2\sin C\cos A = 2\sin (A+C) = 2\sin 2B = 2\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Can I express $x^7y+xy+x+1$ by a repeated use of operation $xy+x+y+1$? A certain calculator can only give the result of $xy+x+y+1$ for any two real numbers $x$ and $y$.
How to use this calculator to calculate $x^7y+xy+x+1$ for any given $x$ and $y$?
When $x$ and $y$ are equal, it will give $(x+1)^2$. But I cannot pr... | Follow these steps:
*
*Give $(x,x)$ to the calculator to get $x^2 + 2x + 1$ and subtract $2x + 1$ from this. You can even get $2x$ from the calculator by giving in $(x,1)$ to get $2x + 2$.
*Now, using prev result, give $(x^2, x^2)$ to get $x^4+ 2x^2 + 1$. Use previous result to obtain $x^4$.
*Using previous result... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $AC: CB$ in $\triangle XYZ$ Problem:
In $\triangle XYZ$, $XY = 4$, $YZ = 7$, and $XZ = 9$. Let $M$ be the midpoint of $\overline{XZ}$, and let $A$ be the point on $\overline{XZ}$ such that $\overline{YA}$ bisects angle $XYZ$. Let $B$ be the point on $\overline{YZ}$ such that $\overline{YA} \perp \overline{AB}$. Le... | (Wrong!!! This is the case for $YB\perp AB.$)
Following your attempts, we get $\frac{11}{3}\times\frac{AC}{CB}\times\frac{YB}{7} = 1.$ Hence it remains to find the length of $YB.$ Let $K$ be the point on $YZ$ such that $XK\perp YZ.$ Then we can use the area of the triangle to find $XK$:
$$\sqrt{10(10-9)(10-7)(10-4)}=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Is it possible to find the sum of the infinite series $1/p + 2/p^2 + 3/p^3 + \cdots + n/(p^n)+\cdots$, where $p>1$? Is it possible to find the sum of the series:
$$\frac{1}{p} + \frac{2}{p^2} +\frac{3}{p^3} +\dots+\frac{n}{p^n}\dots$$
Does this series converge? ($p$ is finite number greater than $1$)
| Let:
$
S = {1\over p} + {2\over p^2} + {3 \over p^3} + \cdots \\
\implies {S\over p} = {1\over p^2} + {2\over p^3} + {3\over p^4} + \cdots.
$
Hence,
$
S\left ( 1 - {1\over p} \right ) = {1\over p} + {1\over p^2} + {1\over p^3} + \cdots \\
S\left ( 1 - {1\over p} \right ) = \frac{1}{p}\frac{1}{1-{1\over p}} = \frac{1}{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find the number of three elements sets of positive integers $\{a,b,c\}$ such that $a\times b\times c=2310$. Find the number of three elements sets of positive integers $\{a,b,c\}$ such that $a\times b\times c=2310$.
My attempt:
Now, $2310=2\times 3\times 5\times 7\times 11=a\times b\times c$
$a=2^{x_{1}}3^{y_{1}}5^{z_{... | I got a combinatorial solution.
$2310=2 \times 3 \times 5 \times 7 \times 11$
$\dfrac{\dbinom{5}{1}\dbinom{4}{1}\dbinom{3}{3}}{2!}=10\\
\dfrac{\dbinom{5}{2}\dbinom{3}{1}\dbinom{2}{2}}{2!}=15\\
\dbinom{5}{3}\dbinom{2}{2}=10\\
\dbinom{5}{4}\dbinom{1}{1}=5$
Total: $40$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
Relation between inverse tangent and inverse secant I've been working on the following integral
$$\int\frac{\sqrt{x^2-9}}{x^3}\,dx,$$
where the assumption is that $x\ge3$. I used the trigonometric substitution $x=3\sec\theta$,which means that $0\le\theta<\pi/2$. Then, $dx=3\sec\theta\tan\theta\,dx$, and after a large n... | Hint. Applying the chain rule, for $x>3$, one may check that
$$
\left(\frac16\sec^{-1}\frac{x}{3}\right)'=\frac{1}{2x\sqrt{9-x^2}}
$$$$
\left(-\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}\right)'=\frac{1}{2x\sqrt{9-x^2}}
$$ since
$$
\lim_{x \to 3^+}\left(\frac16\sec^{-1}\frac{x}{3}\right)=0,\quad \lim_{x \to 3^+}\left(-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Identifying the digits of $37 \cdot aaaa\ldots a$. With a calculator, I have noticed that the integer $37$ multiplied with some particular numbers yields numbers with some structures.
For instance, let $aaaa\ldots a$ be a natural number of $n$ identical digits. Then, $ 37 \cdot aaaa\ldots a$ is a number with $n+1$ or ... | Nice observation!
This follows because $aaa\cdots a = a \cdot 111 \cdots 1$ and
$37 \cdot 11 = 407$
$37 \cdot 111 = 4107$
$37 \cdot 1111 = 41107$
$\cdots$
Indeed, let $u_n = 111 \cdots 1$ ($n$ ones). Then $u_n = \dfrac{10^n-1}{9}$ and
$$
\begin{align}
37 \cdot 111\cdots 1 \quad (n \text{ ones}) &=
37u_n\\&= 36u_n + u_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Integral of $2x^2 \sec^2{x} \tan{x}$ I've been trying for a while to find $\int{( 2x^2 \sec^2{x} \tan{x} )} dx$, using integration by parts.
I always end up getting a more complicated integral in the second part of the equation.
For example:
$$ \int{( 2x^2 \sec^2{x} \tan{x} )} dx =
\\ 2x^2 \tan^2x - \int{\tan{x} \cdot ... | Use integration by parts with $u=x^2$ and $v=\tan^2(x)$. Then, we have
$$\int 2x^2 \tan(x)\sec^2(x)\,dx=x^2\tan^2(x)-2\int x\tan^2(x)\,dx$$
Continue with a subsequent integration by parts with $u=x$ and $v=\tan(x)-x$ to obtain
$$\begin{align}
\int 2x^2 \tan(x)\sec^2(x)\,dx&=x^2\tan^2(x)-2\left(x\tan(x)-x^2-\int(\tan(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Decompose $X^8 + X^7 + X^6 + X^4 + 1$ over $\mathbb{Z}_2$ Is there a clever way to decompose $f(X) = X^8 + X^7 + X^6 + X^4 + 1$ into irreducible factors over $\mathbb{Z}_2$? Or to see that it is irreducible itself?
This is what I have thought about so far: $f(0) \ne 0$ and $f(1) \ne 0$, so $f$ does not have linear fact... | Note that $(x^8+x^7+x^6+x^4+1)(x+1)=x^9 + x^6 + x^5 + x^4 + x + 1$.
Thus, if $\alpha^5=1$, then $\alpha$ is a root of the latter polynomial. That is, the latter polynomial is divisible by $x^5+1=(x^4 + x^3 + x^2 + x + 1)(x+1)$. But this means that
$x^4 + x^3 + x^2 + x + 1$ is a factor of your original polynomial. Now... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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show that $a_{n+874}=a_{n}$,if such $a_{n+2}=\left\lceil \frac{4}{3}a_{n+1}-a_{n}+0.5\right\rceil$
Let the sequence $\{a_{n}\}$ be such that $a_{1}=1, a_{2}=100$, and $$a_{n+2}=\left\lceil \dfrac{4}{3}a_{n+1}-a_{n}+0.5\right\rceil$$
Prove that the sequence $\{a_{n}\}$ is periodic.
I have used a computer and found t... | N.B. This is not complete answer, but just extrapolation of the clue provided by @SangchulLee.
First of all we can rewrite the recurrence as:
$$a_{n+2}=\frac{4a_{n+1}}{3}-a_{n}+v_{n+1}$$
where
$$v_{n+1}=\begin{cases}
1 & a_{n+1} = 0 \bmod 3\\
\frac{2}{3} & a_{n+1} = 1 \bmod 3 \\
\frac{4}{3} & a_{n+1} = 2 \bmod 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
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Prove that $3 \mid (a+b+c+d)$
Given $a,b,c,d \in \mathbb{Z}$ satisfying $a^3+b^3 = 2(c^3-8d^3)$, prove that $3 \mid (a+b+c+d)$.
I first factorized $a^3+b^3$ to get $a^3+b^3 = (a+b)(a^2-ab+b^2)$. I wasn't sure how to use the right-hand side to get $a+b+c+d$. How can we prove that $3 \mid (a+b+c+d)$?
| $x^3\equiv x\bmod3$, which you can prove by trial and error in this case. This should reduce it down to $a+b\equiv2c-d\bmod3$. Adding $c+d$ to both sides yields $a+b+c+d\equiv3c\equiv0\bmod3$, which means $3|(a+b+c+d)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Hint in integration $\int\frac{x^{2}}{\left(x\cos x-\sin x \right )\left( x\sin x+\cos x \right )}\,\mathrm{d}x$ In the following integration
$$\int \frac{x^{2}}{\left ( x\cos x-\sin x \right )\left ( x\sin x+\cos x \right )}\, \mathrm{d}x$$
I tried alot. But does not get any proper start.
Can anybody provide me a hin... | Hint. One may observe the following trick
$$
\begin{align}
\frac{x^2}{(x \cos x-\sin x)(x\sin x+\cos x)}&=\frac{x\cos x(x \cos x-\sin x)+x\sin x(x\sin x+\cos x)}{(x \cos x-\sin x)(x\sin x+\cos x)}
\\\\&=\frac{x\cos x}{x\sin x+\cos x}+\frac{x \sin x}{x \cos x-\sin x}
\\\\&=\frac{(x\sin x+\cos x)'}{(x\sin x+\cos x)}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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There is a single pair $(a,b)$ such that $x^2+2(1+a)x+(3a^2+4ab+4b^2+2)=0$. For this pair, what is $a+b$?
There is a single pair $(a,b)$ such that $x^2+2(1+a)x+(3a^2+4ab+4b^2+2)=0$. For this pair, what is $a+b$?
I know how to find rational roots now because of my last question asked, but what about real roots? What d... | $$0=x^2+2(1+a)x+(3a^2+4ab+4b^2+2) =\\(x+1+a)^2+(3a^2+4ab+4b^2+2-1-2a-a^2)
=\\(x+1+a)^2+(2a^2+4ab+4b^2+1-2a)
=\\(x+1+a)^2+(a^2+4ab+4b^2)+(a^2-2a+1) \\
=(x+1+a)^2+(a+2b)^2+(a-1)^2
$$
Since
$$(x+1+a)^2+(a+2b)^2+(a-1)^2=0$$
you get
$$x+1+a=0\\
a+2b=0\\
a-1=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to integrate $\int\sqrt{\frac{4-x}{4+x}}$? Let
$$g(x)=\sqrt{\dfrac{4-x}{4+x}}.$$
I would like to find the primitive of $g(x)$, say $G(x)$.
I did the following: first the domain of $g(x)$ is $D_g=(-4, 4]$. Second, we have
\begin{align}
G(x)=\int g(x)dx &=\int\sqrt{\dfrac{4-x}{4+x}}dx\\
&=\int\sqrt{... | Substitute $x=4\frac{1-y^2}{1+y^2}$ so that $y=\sqrt{\frac{4-x}{4+x}}$:
$$
\begin{align}
\int\sqrt{\frac{4-x}{4+x}}\,\mathrm{d}x
&=4\int y\,\mathrm{d}\frac{1-y^2}{1+y^2}\\
&=-16\int\frac{y^2}{\left(1+y^2\right)^2}\,\mathrm{d}y\\
&=-16\int\frac{\tan^2(\theta)}{\sec^2(\theta)}\,\mathrm{d}\theta\\
&=-16\int\sin^2(\theta)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Let $f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ Then the value of $ \int^{3/4}_{1/4}f(f(x))\mathrm dx$ If $\displaystyle f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ then the value of $\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))\mathrm dx$
$\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}(f(x))^3-1.5(f(x))^2+f(x)+... | Apply kings rule, I=int(f(f(1-x))) =integral( -x3 +1.5x2 -x +0.75)
Add the two 2I = int(1)
I= 0.25.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A simple proof by induction $P(n) = \frac{6^{2n} - 3^n}{11} \in \mathbb{N}$ Verify by induction that $P(n) = \frac{6^{2n} - 3^n}{11} \in \mathbb{N} \quad \forall n \ge 1 \in \mathbb{N}$
Basis: $P(1) \Rightarrow \frac{33}{11} \in \mathbb{N}$.
Induction: if the statement holds for some $n$ $\Rightarrow$ holds for $n+1$
I... | We have $P(n)=\frac{6^{2n}-3^n}{11} \in \mathbb{N}$.
You have proven that the base case is true.
Now, assume true for $n=k$.
$$6^{2k}-3^k=11p \tag{1}$$
Where $p \in \mathbb{N}$.
True for $n=k+1$.
$6^{2(k+1)}-3^{k+1}=6^{2k+2}-3^{k+1}=36 \cdot 6^{2k}-3 \cdot 3^k \tag{2}$
Now, here comes the trick.
We can substitute from ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Solve the system of equations ... Solve the system of equations :
(EDIT : The problem does not say anything about the nature of $x$ and $y$ (integer, natural number ,..etc.) )
$4xy + 4(x^2 + y^2) + {\frac {3} { (x+y) ^ 2 } } = \frac {85} {3} $
$2x + {\frac {1} {x+y}} = \frac {13} {3}$
I do not know how to approach the... | Less elegant than Anurag A's answer and using brute force.
Considering the equations $$4xy + 4(x^2 + y^2) + {\frac {3} { (x+y) ^ 2 } } = \frac {85} {3}\tag 1$$
$$2x + {\frac {1} {x+y}} = \frac {13} {3}\tag 2$$ extract $y$ from $(2)$; this gives $$y=\frac{-6 x^2+13 x-3}{6 x-13}\tag 3$$ Replace $(3)$ in $(1)$ an simplify... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Positive pairs of integral values satisfying $2xy − 4x^2 +12x − 5y = 11$ The number of positive pairs of integral values of $(x, y)$ that solves
$2xy − 4x^2 +12x − 5y = 11$ is?
I rearranged it to $(2x-5)(y+1-2x)=6$, which took quite a bit of time.
So it can be $2*3$ , $3*2$, $6*1$ or $1*6$ which gives us 2 possible pos... | First of all, $6=\pm1\cdot\pm6=\pm1\cdot\pm6=\pm2\cdot\pm3=\pm3\cdot\pm2=\pm6\cdot\pm1$
We've got the following equations:
$$\begin{cases} 2x = 5+a\\-2x+y = -1+b\end{cases}$$
We can rearrange it to:
$$\begin{cases} x = 2.5+\frac{a}{2}\\y = 4+a+b\end{cases}$$
We want $x$ and $y$ to be integer, so $a$ must be an odd numb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Any solution for $\iiint\frac{x^2+2y^2}{x^2+4y^2+z^2}\,dv$ I tried to solve this triple integral but couldn't integrate the result.
$$\iiint\frac{x^2+2y^2}{x^2+4y^2+z^2}\,dv$$ and the surface to integrate in is $$x^2+y^2+z^2\le1$$
Is there any way to transform the integral into polar coordinates?
| Notice that:
$$\iiint \frac{x^2+2y^2}{x^2+4y^2+z^2} \mbox{d}v =
\iiint \frac{x^2+4y^2+z^2-2y^2-z^2}{x^2+4y^2+z^2} \mbox{d}v
= \iiint 1-\frac{z^2+2y^2}{x^2+4y^2+z^2} \mbox{d}v $$
But by symmetry $x \leftrightarrow z$, we have:
$$\iiint \frac{x^2+2y^2}{x^2+4y^2+z^2} \mbox{d}v
= \iiint \frac{z^2+2y^2}{x^2+4y^2+z^2} \mbo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Definite integral of $\int_0^{1}x^{7}\sqrt{\frac{1+x^{2}}{1-x^{2}}}dx$. When i integrate $\int_{0}^{1}x^{7}\sqrt{\frac{1+x^{2}}{1-x^{2}}}dx$, putting
$t^{2}=1-x^{2}$ then integration $\int{(1-t^{2})}^{3}\sqrt{t^{2}+2}dt$, but i don't understand how to proceed . please someone help me. Thank you.
| Substitute $\text{u}=\frac{1}{1-x^2}$:
$$\mathcal{I}=\int x^7\cdot\sqrt{\frac{1+x^2}{1-x^2}}\space\text{d}x=\frac{1}{2}\int\frac{\left(\text{u}-1\right)^3\cdot\sqrt{2\text{u}-1}}{\text{u}^5}\space\text{d}\text{u}$$
Now, substitute $\text{v}=\sqrt{2\text{u}-1}$:
$$\mathcal{I}=2\int\frac{\text{v}^2\cdot\left(\text{v}^6-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of type $\frac{0}{0}$: $\lim_{x\to0} \frac{\sin(x^n)-\sin^n(x)}{x^{n+2}}$ $$\lim_{x\to 0}\ \frac{\sin\left(x^n\right)-\sin^n\left(x\right)}{x^{n+2}} ~~~~~~~ \mbox{for} ~~~ n \geq 2$$
It's a multiple choice problem and the answer is $\dfrac{n}{6}$.
I tried it for $n=2$ and i got the answer $\dfrac{2}{6}$ which fi... | Write it
$$\frac{\sin (x^n) - (\sin x)^n}{x^{n+2}} = \underbrace{x^{2n-2}}_{\to 0}\underbrace{\frac{\sin (x^n) - x^n}{x^{3n}}}_{\to - \frac{1}{6}} + \frac{x^n - (\sin x)^n}{x^{n+2}},$$
and then
$$\frac{x^n - (\sin x)^n}{x^{n+2}} = \frac{x-\sin x}{x^3}\sum_{k = 0}^{n-1} \frac{x^{n-1-k}(\sin x)^k}{x^{n-1}}$$
using $a^n -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help me to calculate $\int_{0}^{\pi/4}\sqrt{1+\tan x}\,\mathrm dx$ How to calculate $$\int_{0}^{\pi/4}\sqrt{1+\tan x}\,\mathrm dx$$
My attempt:
Let
$$I=\int_{0}^{\pi/4}\sqrt{1+\tan x}\,\mathrm dx$$
substitute $\sqrt{1+\tan x}=t$,then
$$I=\int_{1}^{\sqrt{2}}\frac{2t^{2}}{t^{4}-2t^{2}+2}\,\mathrm dt=\int_{1}^{\sqrt{2}}\f... | Hint:
\begin{align*}
I&=\int_{1}^{\sqrt{2}}\frac{2t^{2}}{t^{4}-2t^{2}+2}\,\mathrm{d}t\\
&=\int_{1}^{\sqrt{2}}\frac{\sqrt{2}+t^{2}+\left ( t^{2}-\sqrt{2} \right )}{t^{4}-2t^{2}+2}\,\mathrm{d}t \\
&=\int_{1}^{\sqrt{2}}\frac{\displaystyle\frac{\sqrt{2}}{t^{2}}+1}{t^{2}-2+\displaystyle\frac{2}{t^{2}}}\,\mathrm{d}t+\int_{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Two fair dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on different numbers? $F =$ The probability that the dice land on different numbers
$F = {(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,5),... | The mistake in your work is that while the question asks us to find that the numbers on the dice are different, in writing the sample space for event $E $, you have also included $(6,6) $ where the numbers are not different. Correct that and we will get, $$P=\frac{\frac{10}{36}}{\frac {5}{6}} =\frac {1}{3} $$ Hope it h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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If $q^k n^2$ is an odd perfect number with Euler prime $q$, does $I(n^2) \geq 5/3$ imply $k=1$? Let $\sigma(x)$ denote the sum of the divisors of $x \in \mathbb{N}$. Denote the abundancy index of $y \in \mathbb{N}$ by $I(y)=\sigma(y)/y$.
If $\sigma(N)=2N$, then $N$ is said to be perfect.
Euler proved that every odd pe... | We consider three cases:
Case 1 $I(n^2) = 5/3$
This implies that
$$2 - I(n^2) = \frac{2n^2 - \sigma(n^2)}{n^2} = 2 - \frac{5}{3} = \frac{1}{3},$$
which implies that $(2n^2 - \sigma(n^2)) \mid n^2$, since
$$\frac{n^2}{2n^2 - \sigma(n^2)} = 3$$
then holds. This means that $n^2$ is deficient-perfect, which is true if and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How do you compute the $\gcd(1+n+n^2,1+n+n^2+s+2ns+s^2)$ I would like to prove the following claim which I think is true:
Claim: Let $n,$ $m$ and $s$ be positive numbers. Fix $s$, then for every positive number $n$ the $\gcd(1+n+n^2,1+n+s+n^2+2ns+s^2)$ will be equal to a divisor of $1+5s^2+s^4.$
For example for every... | Take $n=1$ and $s=3$. Then
$$\displaylines{
1+n+n^2=3\cr
1+n+n^2+s+2ns+s^2=21\cr
\gcd(3,21)=3\cr
1+5s^2+s^4=127\cr
3\ \hbox{is not a factor of}\ 127\ .\cr}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to compute the sum $\sum_{n=0}^\infty \tfrac{n^2}{2^n}$?
How to find this sum :
$\sum_{n=0}^\infty \dfrac{n^2}{2^n}$
$\sum_{n=0}^\infty \dfrac{n^2}{2^n}=\dfrac{1}{2}+\dfrac{4}{4}+\dfrac{9}{8}+\dfrac{16}{16}+\dfrac{25}{32}+\dfrac{36}{64}+\dfrac{49}{128}+\dots$
Now $\sum_{n=0}^\infty \dfrac{n}{2^n}\leqslant \sum_{n... | Interesting... let's look at it.
$S=\frac{1}{2}+\frac{4}{4}+\frac{9}{8}+...$
$\frac{S}{2}=\frac{1}{4}+\frac{4}{8}+\frac{9}{16}+...$
$\frac{S}{2}=\frac{1}{2}+\frac{3}{4}+\frac{5}{8}+...$
$\frac{S}{4}=\frac{1}{4}+\frac{3}{8}+\frac{5}{16}+...$
$\frac{S}{4}=\frac{1}{2}+\frac{2}{4}+\frac{2}{8}+...=\frac{1}{2}+1=\frac{3}{2}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help for series calculation $\sum_{n\ge1} \frac{1}{4n^3-n}$ I want to find the following series.
$$\sum_{n\ge1} \frac{1}{4n^3-n}$$
So, fisrt of all, patial fraction : $$\frac{1}{4n^3-n}=\frac{1}{2n+1}+\frac{1}{2n-1}-\frac{1}{n}.$$
Next, I consider the geometric series $$\sum_{n\ge1} x^{2n-2}+x^{2n}-x^{n-1}=\frac{1}{1-... | Following a comment of mine above:
*
*without checking too carefully, the issue in your step most likely resides in the "I integrate both sides of (1) from 0 to 1." You then integrate the LHS termwise -- why can you do that? The interval of convergence of your power series is $1$, so it is not obvious you can swap $... | {
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"url": "https://math.stackexchange.com/questions/2106470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What is the Laurent expansion of $f(z)=\frac{1}{(z-1)(z-2)(z-3)}$ for $1<|z|<2$ Find the Laurent expansion for the function $f(z)$ in the domains $i),ii)$
$f(z)=\frac{1}{(z-1)(z-2)(z-3)}$
$i)$ $0<|z|<1$
$ii)$ $1<|z|<2$
What I did for the first:
$f(z)=\frac{1}{(z-1)(z-2)(z-3)}=\frac{-1}{(z-2)}+\frac{1}{2(z-3)}-\frac{1}{... | 1. Laurent expasions of $\; f_1(z)=\dfrac{1}{z-1}$
$$\frac{1}{z-1}=\begin{cases} -\dfrac{1}{1-z}=-\displaystyle\sum_{n=0}^{+\infty}z^n,\quad |z|<1,\quad (L_1) \\\dfrac{1}{z}\dfrac{1}{1-\dfrac{1}{z}}= \dfrac{1}{z}\displaystyle\sum_{n=0}^{+\infty}\left(\frac{1}{z}\right)^n=\sum_{n=0}^{+\infty}\frac{1}{z^{n+1}},\quad |z|>... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Derivative of arcsin In my assignment I need to analyze the function
$f(x)=\arcsin \frac{1-x^2}{1+x^2}$
And so I need to do the first derivative and my result is:
$-\dfrac{4x}{\left(x^2+1\right)^2\sqrt{1-\frac{\left(1-x^2\right)^2}{\left(x^2+1\right)^2}}}$
But in the solution of this assignment it says
$f'(x)=-\frac... | Use that $(1+x^2)^2\sqrt{1-\frac{(1-x^2)^2}{(1+x^2)^2}}=(1+x^2)\sqrt{(1+x^2)^2-\frac{(1+x^2)^2(1-x^2)^2}{(1+x^2)^2}}=(1+x^2)\sqrt{((1+x^2)+(1-x^2))((1+x^2)-(1-x^2))}=(1+x^2)\sqrt{2\cdot2x^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2109428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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Find all possible positive integer $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $
Find all possible positive integers $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $.
I don't know how to start with. Any hint or full solution will be helpful.
| Let $d$ be a positive integer such that $d$ divides both $3^{n-1}+5^{n-1}$ and $3^{n}+5^{n}$.
then d divides $5\cdot (3^{n-1}+5^{n-1})-( 3^{n}+5^{n} )= 3^{n-1}\cdot 2$, and
$3\cdot(3^{n-1}+5^{n-1})-( 3^{n}+5^{n} )= -5^{n-1}\cdot 2$
Thus, as $\gcd(5,3)=1$, it follows that $d$ is either $1$ ou $2$.
Since $3^{n-1}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2109523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Explain why $(a−b)^2 = a^2 −b^2$ if and only if $b = 0$ or $b = a$. This is a question out of "Precalculus: A Prelude to Calculus" second edition by Sheldon Axler. on page 19 problem number 54.
The problem is Explain why $(a−b)^2 = a^2 −b^2 $ if and only if $b = 0$ or $b = a$.
So I started by expanding $(a−b)^2$ to $(a... | we have $$(a-b)^2=a^2-b^2$$ we know that $$(a-b)^2=a^2+b^2-2ab$$ and if we use the first equation we get
$$2b^2-2ab=0$$ thus we get $$b=0$$ or $$b=a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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How can I find $\sum_{cyc}\sin x\sin y$ $x$, $y$ & $z$ are real number such that
$$\frac{\sin{x}+\sin{y}+\sin{z}}{\sin{(x+y+z)}}=\frac{\cos{x}+\cos{y}+\cos{z}}{\cos{(x+y+z)}}=2$$
find the value
$$\sum_{cyc}\sin x\sin y$$
All help would be appreciated.
| I used the Blue's beautiful idea.
Let $e^{ix}=a$, $e^{iy}=b$ and $e^{iz}=c$.
Hence, $\sin{x}=\frac{a-\frac{1}{a}}{2i}=\frac{a^2-1}{2ai}$, $\cos{x}=\frac{a^2+1}{2a}$, $\sin{y}=\frac{b^2-1}{2bi}$, $\sin{x}=\frac{c^2-1}{2ci}$ and $\cos{x}=\frac{c^2+1}{2c}$.
Thus, $\sum\limits_{cyc}\sin{x}=2\sin(x+y+z)$ gives $\sum\limits_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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The ratio of their $n$-th term. The sum of $n$ terms of two arithmetic series are in the ratio of $(7n+ 1) : (4n+ 27)$. We have to find the ratio of their $n$-th term.
�
I tried to find the ratio by using the formula of summation of A.P.
But it becomes too long due to many variables that is $a_1,a_2,d_1,d_2$
| Let $a_1$, $a_2$ be the first terms and $d_1$, $d_2$ the common differences of the given APs. Then, their sum of $n$ terms are given by: $S_{n}=\frac{n}{2}\left[2 a_{1}+(n-1) d_{1}\right]$ and $S_{n}^{\prime}=\frac{n}{2}\left[2 a_{2}+(n-1) d_{2}\right]$
On dividing, we get:
$$
\frac{2 a_{1}+(n-1) d_{1}}{2 a_{2}+(n-1) d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Analytic Proof of alternating series Having some problems figuring out the following proof. If we have a series $a_n$ that is both nonnegative and decreasing. Then if we consider the corresponding alternating series $$\sum (-1)^{n+1}a_n$$ I need to prove that the sequence of odd partial sums $$s_{2n+1}= \sum _{k=1}^{2n... | Let
$A_n
=\sum_{k=1}^n (-1)^{k+1}a_k
$.
Then
$\begin{array}\\
A_{n+2}
&=\sum_{k=1}^{n+2} (-1)^{k+1}a_k\\
&=\sum_{k=1}^{n} (-1)^{k+1}a_k+(-1)^{n+2}a_{n+1}+(-1)^{n+3}a_{n+2}\\
&=A_n+(-1)^{n+2}(a_{n+1}-a_{n+2})\\
\end{array}
$
Putting $2n$
for $n$,
$\begin{array}\\
A_{2n+2}
&=A_{2n}+(-1)^{2n+2}(a_{2n+1}-a_{2n+2})\\
&=A_{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2115554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $N = \frac{{n_1}(n_1+1)}{2}\cdot{d_1}$, is it possible to have $N = \frac{{n_2}(n_2+1)}{2}\cdot{d_2}$? If $$N = \frac{{n_1}(n_1+1)}{2}\cdot{d_1},$$
is it possible to have $$N = \frac{{n_2}(n_2+1)}{2}\cdot{d_2}?$$
Here, $d_i, n_i \in \mathbb{N}$, $d_i > 1$ and $n_1 \neq n_2$.
That is,
Question 1
If a number can be r... | Let $\dfrac{n_1(n_1+1)}{n_2(n_2+1)} = \dfrac{d_2}{d_1}$
It follows that
$\dfrac{{n_1}(n_1+1)}{2}\cdot{d_1} = \dfrac{{n_2}(n_2+1)}{2}\cdot{d_2}$
Example $T(6) = 21$ and $T(8)=36$. So $\dfrac{T(6)}{T(8)} = \dfrac{7}{12}$
Hence
$\dfrac{{6}(6+1)}{2}\cdot{12} = \dfrac{{8}(8+1)}{2}\cdot{7}$
If $n_1$ and $d_1$ are given {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $a^4-a^3+a^2+2$ when $a^2+2=2a$ Find the value of $a^4-a^3+a^2+2$ when $a^2+2=2a$
My Attempt,
$$a^4-a^3+a^2+2=a^4-a^3+2a$$
$$=a(a^3-a^2+2)$$
What's next?
| Hint $\ $ We seek the value of $\,f(a)\,$ given that $\, g(a) = a^2-2a+2 = 0.\,$ By the Polynomial Division Algorithm we can write $\ f = q\,g + r\,$ so $\, g=0\,\Rightarrow\, f = r = f\bmod g.\,$ So it suffices to compute the remainder $\, r = (f\bmod g).\,$ Doing so easily yields $\,f = r = 0.$
Because $\, g = 0\,$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Angle between two lines explanation I gave two lines $q_1 = 2x - y + 2 = 0$ and $q_2 = x + 2y - 3 = 0$.
They have vectors $n_1 = (2,1)$ and $n_2 = (1,2)$.
When I have to find angle between them I must apply that formula:
$$\cos \theta = \frac{2\cdot 1 + (-1)\cdot2}{\sqrt{4 + 1}\sqrt{1 + 4}} = 0 \Rightarrow \theta = ... | Let $\theta$ be an angle.
$\cos \theta = \frac{2.1 + (-1).2}{\sqrt 5. \sqrt 5}$
$\cos \theta = \frac{2 - 2}{\sqrt 5. \sqrt 5}$
$\cos \theta = \frac{0}{5}$
$\cos \theta = 0$
$\cos \theta = \cos 90°$
$\theta = 90°$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Integral of product of two error complementary functions (erfc) Could you please help me to show that the integral
$$
\int_0^{\infty} \mathrm{erfc}(ax) \, \mathrm{erfc}(bx)\, \mathrm{d}x
$$
is equal to
$$
\frac{1}{ab\sqrt{\pi}} (a+b-\sqrt{a^2+b^2}),
$$
where
$$
\mathrm{erfc}(y)=\frac{2}{\sqrt{\pi}} \int_y^{\infty} \... | A multiple integral approach. Since
$$
\int_{0}^{\infty}x^2 e^{-\alpha x^2}\mathrm{d}x \overset{\alpha x^2 \to x}{=} \frac{1}{2\alpha^{\frac{3}{2}} }\int_{0}^{\infty} x^{\frac{3}{2}-1}e^{-x} \mathrm{d}x = \frac{1}{2\alpha^{\frac{3}{2}} } \Gamma\left(\frac{3}{2}\right) = \frac{\sqrt{\pi}}{4\alpha^{\frac{3}{2}} }
$$
We... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find a formula for $\cos(5x)$ in terms of $\sin(x)$ and $\cos(x)$ I was asked to find a formula for $\cos(5x)$ in terms of $\sin(x)$ and $\cos(x)$.
I tried to use the formula $\cos(5x) + i\sin(5x) = (\cos(x)+i\sin(x))^5
$ and what I get is
$16i\sin^5(x) - 20i\sin^3(x) + 5i\sin(x) + \cos(x) + 16 \sin^4(x) \cos(x) -... | Your way is right. But I think it should be
$$(\cos x + i\sin x)^5 = \cos^5 x + 5i\cos^4x\sin x - 10\cos^3x\sin^2 x - 10i\cos^2x\sin^3x + 5\cos x\sin^4x +i\sin^5x$$
So, you have
$$\cos 5x = \cos^5x-10\cos^3x\sin^2x+5\cos x\sin^4x$$
$$\sin 5x = \sin^5x-10\cos^2x\sin^3x+5\cos^4 x\sin x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2123388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How do I solve $(x+1)^5 +36x+36 = 13(x+1)^3$? I tried $$(x+1)^5 + 36 x + 36 = 13 (x +1)^3\\
(x+1)^5 + 36(x+1) = 13 (x +1)^3\\
(x+1)^4 +36 = 13 (x+1)^2
$$
But, don't understand how to solve further. Can somebody show step by step please. Thanks!
| It's a simple problem. Well
$$(x+1)^5+36x+36=13(x+1)^3$$
$$\implies (x+1)^5–13(x+1)^3+36(x+1) = 0$$
Let $y=x+1$
$$\implies y^5-13y^3+36y = 0$$
$$\implies y(y^4-13y^2+36)= 0$$
$$\implies y(y^2-9)(y^2-4)= 0$$
$$\implies y = 0,\pm 2,\pm 3$$
$$\implies x+1 = 0,\pm 2,\pm 3$$
$$\implies x = -4, -3, -1, 1, 2$$
and we’re done.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Prove the $\frac {2r+5}{r+2}$ is always a better approximation of $\sqrt {5}$ than $r$.
Prove the $\frac {2r+5}{r+2}$ is always a better approximation of $\sqrt {5}$ than $r$.
SOURCE : Inequalities (Page Number 4 ; Question Number 207)
I tried a lot of approaches, but without success.
I rewrote $\frac {2r+5}{r+2}$ as... | Rather than doing manipulations with $\sqrt{5}$ it is better to use just the rationals. We can see that $$\left(\frac{2r+5}{r+2}\right)^{2}-5=\frac{4r^{2}+20r+25-5r^{2}-20r-20}{(r+2)^{2}}=\frac{5-r^{2}}{(r+2)^{2}}$$ Since $r>0$ it follows that $(2r+5)/(r+2)$ is a better approximation to $\sqrt{5}$ than $r$, but in a di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How is the infinite sum of a series calculated by symbolic math? I wonder how Wolfram can solve this series and provide the solution symbolically:
$$\sum_{k=1}^\infty\frac 1{(2k-1)^4}$$
In this particular case I know how to use a Fourier series on a triangle function to get the result by employing Parseval's theorem, b... |
I do not know how symbolic programs evaluate series in closed form. But I thought it might be instructive to see one methodology to solve the problem of interest. It is to that end we now proceed.
Note that $\zeta(4)=\sum_{n=1}^\infty\frac{1}{n^4}$. Next, we can write $\zeta(4)$ as
$$\begin{align}
\zeta(4)&=\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find all conditions for $x$ that the equation $1\pm 2 \pm 3 \pm 4 \pm \dots \pm n=x$ has a solution. Find all conditions for $x$ so that the equation $1\pm 2 \pm 3 \pm 4 \pm \dots \pm 1395=x$ has a solution.
My attempt: $x$ cannot be odd because the left hand side is always even then we have $x=2k(k \in \mathbb{N})$ al... | The solution for $n$ as in the title is as follows:
Let $p = (n-3)(n+4)/2$ then the solutions for x are $-p-4, -p, -p+2, \ldots,p-2, p, p +2, p+4$ where from the term $-p$ on we go to the term $p$ each time adding $2$. So only the first and the last two terms are irregular. We can verify:
*
*for $n = 4$ we have $-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 7
} |
What is $x$, if $3^x+3^{-x}=1$? I came across a really brain-racking problem.
Determine $x$, such that $3^x+3^{-x}=1$.
This is how I tried solving it:
$$3^x+\frac{1}{3^x}=1$$
$$3^{2x}+1=3^x$$
$$3^{2x}-3^x=-1$$
Let $A=3^x$.
$$A^2-A+1=0$$
$$\frac{-1±\sqrt{1^2-4\cdot1\cdot1}}{2\cdot1}=0$$
$$\frac{-1±\sqrt{-3}}{2}=0$$
I... | Note that the function $f(x)=3^x+3^{-x}$ has first derivative $f'(x)=3^x-3^{-x}$ and second derivative $f''(x)=3^x+3^{-x}$. Since $f''(x)>0$ always, this is a convex function that attains a global minimum wherever $f'(x)=0$, so wherever $3^x=3^{-x}$. This occurs at $x=0$, in which the function takes value $f(0)=3^0+3^0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Conditional probability and calculation if $P(F|E)=3P(F|E^c)$
$P(E)=0.3$
$P(F)=0.4$
Then $P(E|F)=??$
How do i use the conditional probability equation that has been given in the question
| From the first equality you wrote,
$$P(F|E) = \frac{P(F\cap E)}{P(E)} = \frac{P(F\cap E)}{.3} = 3P(F|E^{c}) = 3\frac{P(F\cap E^{c})}{P(E^{c})} = 3\frac{P(F\cap E^{c})}{.7} $$
We can rearrange some terms to get
$$P(F\cap E) = \frac{3*.3}{.7} P(F\cap E^{c}) = \frac{9}{7}P(F\cap E^{c})$$
Now, since $P(F) = P(F\cap E) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find that solution satisfying $ϕ(1) = 3ϕ(0)$ for $y' + 5y = 2$. Find that solution satisfying $ϕ(1) = 3ϕ(0)$ for the following second order linear ordinary differential equation:
$y' + 5y = 2$
I found the solution to be $ϕ(x) = \frac{2}{5} + ce^{5x}$.
Now how do I find a particular solution satisfying $ϕ(1) = 3ϕ(0)... | Firstly it should be $$y'=2-5y\\ \frac { dy }{ 2-5y } =dx\\ \int { \frac { dy }{ 2-5y } } =\int { dx } \\ \int { \frac { d\left( 5y-2 \right) }{ 5y-2 } =-5\int { dx } } \\ \ln { \left| 5y-2 \right| =-5x+C } \\ 5y-2={ Ce }^{ -5x }\\$$
$$ \phi \left( x \right) =\frac { 2 }{ 5 } +{ Ce }^{ -5x }$$
then we are given $ϕ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
A closed form for $\sum _{j=0}^{\infty } -\frac{\zeta (-j)}{\Gamma (j)}$ Is there a closed form for
$$\sum _{j=0}^{\infty } -\frac{\zeta (-j)}{\Gamma (j)}$$
where $\zeta (-j)$ Zeta function and $\Gamma (j)$ Gamma function.
I tried everything, but I still can not solve it. Any Ideas?
| The reflection formula for the Riemann zeta function gives
$$
\begin{align}
\zeta(-k)
%&=\zeta(k+1)\frac{\Gamma\!\left(\frac{k+1}2\right)}{\pi^{\frac{k+1}2}}\frac{\pi^{-\frac{k}2}}{\Gamma\!\left(-\frac{k}2\right)}\frac{\Gamma\!\left(1+\frac{k}2\right)}{\Gamma\!\left(1+\frac{k}2\right)}\\
%&=\zeta(k+1)\frac{\Gamma\!\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Antiderivative of a product of a rational function and a square root. Let $0<x<1$ and $p$ be a non-negative integer.
Consider a following definite integral:
\begin{equation}
{\mathbb I}_p(x):=\int\limits_1^x \frac{(1-v^2)^p}{v^{2p+2}} \cdot \sqrt{\frac{1-v}{1+v}} dv
\end{equation}
We have computed it in a following way... | Let us complete the calculation given above by Felix Marin. Clearly that result above is correct however it can be simplified further and what's more important expressed through elementary functions. We demonstrate here how is this done. Firstly we use the series expansion of the incomplete beta function. We have:
\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to do this question that talks about dependency of x Let $x > 0$. Prove that the value of the following expression doesn't depend on x
$$\int_{0}^{x} \frac{1}{1+t^2} dt + \int_{0}^{\frac{1}{x}} \frac{1}{1+t^2}dt$$
Attempt:
Left: f'(x) = $\frac{1}{1+x^2}$
Right: f'(x) = $\frac{1}{1+(\frac{1}{x})^2} -\frac{1}{x^2}$
... | To answer the question as-is after having been heavily edited, let for $x \gt 0\,$:
$$f(x) = \int_{0}^{x} \frac{1}{1+t^2} dt + \int_{0}^{\frac{1}{x}} \frac{1}{1+t^2}dt$$
Then, using the Leibniz integral rule:
$$
f'(x) = \frac{1}{1+x^2} \,-\, \frac{1}{x^2} \cdot \frac{1}{1+ \cfrac{1}{x^2}} = \frac{1}{1+x^2} \,-\, \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the maximum and minimum value of $\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$ Find the maximum and minimum value of $$\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$$.
i simplified and reach to expression as follows :
$5 + 2\sin(x)[\sin(x)-3\cos(x)]$. How do i go from here?
Thanks
| The expression can be simplified to $$\cos^2 x + \sin^2 x +2\sin^2 x +2 -3\sin 2x $$ $$=3 -3\sin 2x + 2\frac {1-\cos 2x}{2} $$ $$=4 - 3\sin 2x - \cos 2x $$
Now we know that $$-\sqrt{a^2 +b^2} \leq a\sin \alpha + b \cos \alpha \leq \sqrt {a^2 +b^2} $$
Hope you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find minimal value of $abc$ if the quadratic equation $ax^2-bx+c = 0$ has two roots in $(0,1)$ If $$ ax^2-bx+c = 0 $$ has two distinct real roots in (0,1) where a, b, c are natural numbers then find the minimum value of product abc ?
| Since $a,b,c $ are positive the roots are trivially greater than 0.
What remains is to solve the inequality:
$\frac{b + \sqrt{b^2-4ac}}{2a} <1$
This reduces to $ a+c>b$
But the roots being real and distinct we have $b^2 >4ac$
Combining both we have :
$a^2 + c^2 + 2ac > b^2 > 4ac$
$b^2 > 4ac$ tells us $b> 2$ (why?)
$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the integral $\int \frac{x^2(x-2)}{(x-1)^2}dx$
Find $$\int \frac{x^2(x-2)}{(x-1)^2} dx .$$
My attempt:
$$\int \frac{x^2(x-2)}{(x-1)^2}dx = \int \frac{x^3-2x^2}{x^2-2x+1}dx $$
By applying polynomial division, it follows that
$$\frac{x^3-2x^2}{x^2-2x+1} = x + \frac{-x}{x^2-2x+1}$$
Hence $$\int \frac{x^3-2x^2}{... | $$\int\frac{-x}{x^2-2x+1} dx=\int\frac{1-x}{(1-x)^2} dx+\int\frac{-1}{(1-x)^2} dx=-\ln(1-x)+\frac{1}{1-x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
Bessel to integral form Can someone tell me how the Bessel function be this form.
$$\frac{2}{π}\int_0^1 \frac{\cos (xt)}
{\sqrt{1-t^2}} dt = J_0(x)$$
| $\newcommand{\dif}{\mathrm d}$
$$\begin{align*} \int_0^1 \frac{\cos (xt)}{\sqrt{1-t^2}} \dif t
&= \int_0^1 \sum_{j=0}^\infty \frac{(-1)^j (xt)^{2j}}{(2j)!\sqrt{1-t^2}} \dif t \\
&= \sum_{j=0}^\infty \frac{(-1)^j x^{2j}}{(2j)!} \int_0^1 \frac{t^{2j} \dif t}{\sqrt{1-t^2}}
\end{align*}$$
Observe that (this part needs some... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Divergence of $\vec{f} = \frac{1}{r^2} \hat{r}$ Why am I getting zero divergence of function $\vec{f} = \frac{1}{r^2} \hat{r}$, where $r$ is the distance from the origin and $\hat{r}$ is the unit vector in the radial direction. The divergence of this function over a sphere of radius $R$, which includes the origin.
$$\n... | $\hat{r} = (x,y,z)/\sqrt{x^2 + y^2 + z^2}$
$$
F(x,y,z) = \frac{1}{(x^2 + y^2 + z^2)^{3/2}} (x,y,z) \\
\frac{\partial}{\partial x} F = \frac{\partial}{\partial x} \frac{x}{(x^2 + y^2 + z^2)^{3/2}} = \frac{-2x^2 + y^2 + z^2}{(x^2 + y^2 + z^2)^{5/2}} \\
\frac{\partial}{\partial y} F = \frac{\partial}{\partial y} \frac{y}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Difficult integral with trigonometric substitution: $\int\frac{\sqrt{1-x}}{\sqrt x}dx$ This is the integral I am talking about
$$\int\frac{\sqrt{1-x}}{\sqrt x}dx$$
As you can tell I tried substitution $u = \sqrt x$, and from there I went to $u = \sin{\theta}$ like here.
$$\newcommand{\dd}{\; \mathrm{d}}
\int \frac{\... | $$
I:=\int \frac{\sqrt{1-x}}{\sqrt{x}} d x
$$
Since $0 <x<1$, therefore let $x=\sin ^{2}\theta$, then $d x=2 \sin \theta \cos \theta d\theta.$
$$\begin{aligned}
I &=\int \frac{\sqrt{\cos ^{2} \theta}}{\sqrt{\sin ^{2} \theta}} \cdot 2 \sin \theta \cos \theta d \theta =\int 2 \cos ^{2} \theta d\theta =\int(1+\cos 2 \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2140318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Does $\sqrt{a+b}\leq \sqrt{a}+\sqrt{b}$? In the case of $$\sqrt{(x_n-\ell_1)+(y_n-\ell_2)}\leq \sqrt{(x_n-\ell_1)^2} + \sqrt{(y_n-\ell_2)^2} = |x_n-\ell_1|+|y_n-\ell_2|$$
it is true, if we take the rise the two sides in the power of $2$ we get:
\begin{align}
& (x_n-\ell_1)+(y_n-\ell_2)\leq \left( \sqrt{(x_n-\ell_1)^2}+... | Quite simply, the answer to your question from the title is "yes", if we assume that $a$ and $b$ are non-negative real numbers.
Since both of the sides are positive, we can square the entire equation, and we get a trivial inequality: $$ \sqrt{a+b} \leq \sqrt{a} + \sqrt{b} \iff a+b \leq a+b+2\sqrt{ab} \iff 0 \leq 2\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Particle starts at $(0,-3)$ and moves clockwise around origin on graph $x^2+y^2=9$, find parametric equation Question
particle starts at $(0,-3)$ and moves clockwise around origin on graph $x^2+y^2=9$, revolve in $9$ seconds find parametric equation in term of $t$.
What I've done so far:
I first thought that the grap... | Let's start with the value of $\theta$, in radians, as a function of time. Because the particle is travelling at a constant velocity, revolving in $9$ seconds, $\dfrac{d\theta}{dt} = -\dfrac{2\pi}{9}$. The minus sign is because the particle is travelling clockwise. Then
$\theta(t) = -\dfrac{2\pi}{9}t + \theta_0$. Sinc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
To prove that prove that $cos^8 \theta sec^6 \alpha , \frac{1}{2 } ,sin^8 \theta cosec^6 \alpha $ are in A.P If $\cos^4 \theta \sec^2 \alpha , \frac{1}{2 } ,\sin^4 \theta \csc^2 \alpha $ are in A.P ,
then prove that $\cos^8 \theta \sec^6 \alpha , \frac{1}{2 } ,\sin^8 \theta \csc^6 \alpha $ are in A.P
Now i have reach... | We have $\cos^4\theta\sec^2\alpha+\sin^4\theta\csc^2\alpha=1$
Set $\cos^2\theta=a,\sin^4\theta=(1-a)^2$ to solve for $a$ to find $a=\cos^2\alpha$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2142174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$ $\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$
My try:
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}=$
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}\times\frac{\sqrt{1+\sqrt{x+2}+\sqrt3}}{\sqrt{1+\sqrt{x+2}+\sqrt3}}=$
$\lim_{x\to2}\frac{\sqrt{x+2\sq... | Following StackTD's comment
let $\sqrt{1+\sqrt{x+2}}-\sqrt3=y\implies y\to0$
and $1+\sqrt{x+2}=(y+\sqrt3)^2=y(y+2\sqrt3)+3$
$$x+2=(y^2+2\sqrt3y+2)^2=y^2(y+2\sqrt3)^2+4y(y+2\sqrt3)+4$$
$$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}}-\sqrt3}{x-2}=\lim_{y\to0}\dfrac y{y^2(y+2\sqrt3)^2+4y(y+2\sqrt3)}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2142540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
differential linear equation of order one $(2xy+x^2+x^4)\,dx-(1+x^2)\,dy=0$ I have no idea how to solve it. Should be linear equation of order one since I am passing through this chapter, but I can't put into the form of $$y'+P(x)y=Q(x)$$
Here is the equation: $$(2xy+x^2+x^4)\,dx-(1+x^2)\,dy=0$$
It is not exact since p... | $$(2xy+x^2+x^4)\,dx−(1+x^2)\,dy=0\\
(1+x^2) \frac {dy}{dx} = (2xy+x^2+x^4)\\
y' - \frac {2x}{1+x^2} y = x^2$$
Integrating factor
$$e^{\int \frac {-2x}{1+x^2}} = e^{-\ln(1+x^2)} = \frac {1} {(1+x^2)}$$
$$\frac {1}{1+x^2}y = \int \frac {x^2}{1+x^2}\\
\frac {1}{1+x^2}y = x - \arctan x + C\\
y = x^3 + x - (1+x^2)\arctan x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving for varible with use of ln Looking for some help with trying to solve for $x$ in the following equation,
$$0=5-re^{-x^2}$$
The solution is,
$$x^2=\ln\frac{r}{5}$$
| \begin{align*}
\implies 5 = re^{-x^2} \iff \frac{5}{r} = e^{-x^{2}} \iff \ln\frac{5}{r} = -x^{2} \iff -\ln\frac{5}{r} = x^2 \iff \ln \frac{r}{5} = x^{2}
\end{align*}
Note: $\ln a^{n} = n\ln a$. In this case, $n=-1$, i.e. $\ln a^{-1} = -\ln a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Related Rates: Angular Velocity The figure shows a rotating wheel with radius 40 cm and a connecting rod AP with length 1.2 m. The pin P slides back and forth along the x-axis as the wheel rotates counterclockwise at a rate of 360 revolutions per minute.
*
*Find the angular velocity of the connecting rod, $\frac{d\... | We have
$$
\sin(\theta) = \frac{s}{r} \\
\sin(\alpha) = \frac{s}{l}
$$
such that equating on $s$ we get
$$
r \sin(\theta) = l \sin(\alpha) \iff \\
\alpha = \arcsin\left( \frac{r}{l} \sin(\theta) \right)
$$
Then the time derivative is
$$
\dot{\alpha}
= \frac{1}{\sqrt{1-\left( \frac{r}{l} \sin(\theta)\right)^2}}
\frac{r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Linear Equations system The system
$$\begin{cases}x-y+3z=-5\\5x+2y-6z=\alpha \\2x-y+\alpha z = -6 \end{cases}$$
for which $\alpha$ values the linear equation system:
*
*has no solution
*has one solution
*has more than one solution
I started to do Gauss elimination on it, but i have no idea what i am looking fo... | Gaussian elimination:
\begin{align}
\left[\begin{array}{ccc|c}
1 & -1 & 3 & -5 \\
5 & 2 & -6 & \alpha \\
2 & -1 & \alpha & -6
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & -1 & 3 & -5 \\
0 & 7 & -21 & \alpha+25 \\
0 & 1 & \alpha-6 & 4
\end{array}\right]
&&\begin{aligned} R_2&\gets R_2-5R_1 \\ R_3&\gets R_3-2R_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Without using calculus, how can I find the formula for the area of the shaded region in the attached figure in terms of a and b? Note that the center of the larger circle is on the edge of the smaller.
|
If we consider the centers of both the circles on y axis, origin as the center of the bigger circle, then let the two circles intersect at $A=(x_1,y_1), B=(x_2,y_2)$
Equation of the bigger circle is $x^2+y^2=b^2...(I)$
Equation of the smaller circle $(x-h)^2+(y-k)^2=a^2$, here $h=0, k=-a$
$\therefore x^2+(y+a)^2=a^2.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determine all possible real values of h so that the equation has 4 real roots The equation is as following: $$x(x+1)(x+h)(x+1+h)=h^2$$
I did the multiplication and got the expression: $$x^4+2(h+1)x^3+(h+1)^2x^2+hx+h=0$$
From now on I do not know how to proceed.
| At $h = 0$ you have double roots at $x=-1$ and $x = 0$
$x(x+1)(x+h)(x+1+h)=h^2\\
u = x+ \frac h2$
$(u-\frac h2)(u+\frac h2)(u+1-\frac h2)(u+1+\frac h2)=h^2\\
(u^2-\frac {h^2}4)((u+1)^2-\frac {h^2}{4})=h^2$
$(u^2-\frac {h^2}4)((u+1)^2-\frac {h^2}{4})$ has its maximum at $u = -\frac 12$
If the max $(u^2-\frac {h^2}4)((u+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the remainder of $\frac{3^{11}-1}{2}$ divided by $9$ I have the following question:
Find the remainder of $$\frac{3^{11}-1}{2}$$ divided by $9$
I tried to reformat the question:
$$\frac{3^{11} -1 } {2} \times \frac{1}{9} = \frac{3^{11}-1}{18}$$
Since $3^2 = 9$
$$\frac{3^2(3^9) -1}{3^2 \times2}$$
I don't know whe... | We have that $$\frac {3^{11}-1}{2} = \frac {3-1}{2}(3^{10} + 3^9 + \cdots + 3^2+3+1) \tag {1} $$
We know that $3^2 =9 \equiv 0 \mod 9$ and similarly $3^3 =3 (3^2) \equiv 0 \mod 9$ and so on. Thus, $(1)$ reduces to, $$1 (0 + 0 + 0 \cdots 0 + 3 +1 ) \equiv 4 \mod 9$$
Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
About the formulas to determine the integral ideal with specific norm (1) For $\mathbb{Q}(\sqrt{10})$, the ring of integers is $\mathbb{Z}[\sqrt{10}]$, and to determine the integral ideal with norm $3$, we can use the formula below:
By $X^2 − 10\equiv (X + 1)(X − 1)\text{ mod 3}$, then we get the conjugate ideals with ... | This answer, despite its length, is not a complete answer either, but I think it'll give you some insights. Your question feels like another one I've answered before, although I think the earlier question was more concerned with numbers than with ideals.
Regardless, I feel that the excessive early emphasis on defining ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
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