Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Calculation of $\int_{0}^{\pi}\frac{1}{(5+4\cos x)^2}dx$ Calculation of $\displaystyle \int_{0}^{\pi}\frac{1}{(5+4\cos x)^2}dx$
$\bf{My\; Try}::$ Using $\displaystyle \cos x = \frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$
Let $\displaystyle I = \int_{0}^{\pi}\frac{1}{\left(5+\frac{4-4\tan^2 \frac{x}{2}}{1+\tan^2 \... | After setting
$$
t=\tan\frac{x}{2},
$$
we get
\begin{eqnarray}
I:&=&\int_0^\pi\frac{1}{(5+4\cos x)^2}\,dx=\int_0^\infty\frac{2}{1+t^2}\cdot\left(\frac{1+t^2}{9+t^2}\right)^2\,dt=2\int_0^\infty\frac{1+t^2}{(9+t^2)^2}\,dt\\
&=&2\int_0^\infty\frac{9+t^2-8}{(9+t^2)^2}\,dt=2\int_0^\infty\frac{1}{9+t^2}\,dt-16\int_0^\infty\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/574099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
find limit of $a_n=\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+...+\frac{1}{(2n)^2}$ finding limit of $a_n=\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+...+\frac{1}{(2n)^2}$
I know that i have to use Stolz Cesaro theorem, but the problem is that i need second sequence.
| $$a_n=\frac{1}{n^2}\left( \frac{1}{(1+\frac1n)^2}+\frac{1}{(1+\frac2n)^2}+...+\frac{1}{(1+\frac{n}n)^2} \right)=\frac{1}{n}\left[\frac{1}{n}\left( \frac{1}{(1+\frac1n)^2}+\frac{1}{(1+\frac2n)^2}+...+\frac{1}{(1+\frac{n}n)^2} \right)\right]$$
Now, apply S-C to
$$a_n=\frac{ \left( \frac{1}{(1+\frac1n)^2}+\frac{1}{(1+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/576943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
How prove this $\tan{\frac{2\pi}{13}}+4\sin{\frac{6\pi}{13}}=\sqrt{13+2\sqrt{13}}$ Nice Question:
show that: The follow nice trigonometry
$$\tan{\dfrac{2\pi}{13}}+4\sin{\dfrac{6\pi}{13}}=\sqrt{13+2\sqrt{13}}$$
This problem I have ugly solution, maybe someone have nice methods? Thank you
My ugly solution:
let $$A=... | Straightforward for WA, not so by hand: Let $x = \exp(i \theta)$ be a primitive $13$th root of unity and $y = \tan(\theta) + 4 \sin(3\theta)$. Then $(x, y)$ is a common zero of the polynomials $$i (x^2-1) x^3 + 2i (x^6-1)(x^2+1) + y (x^2+1)x^3 \textrm{ and}\\
x^{12}+x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1.$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/578286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 2,
"answer_id": 1
} |
Proving that $(abc)^2\geq\left(\frac{4\Delta}{\sqrt{3}}\right)^3$, where $a$, $b$, $c$ are the sides, and $\Delta$ the area, of a triangle Let $a$, $b$, $c$ be the sides of $\triangle ABC$.
Prove $$(abc)^2\geq\frac{4\Delta}{\sqrt{3}}$$
where $\Delta$ is the area of the triangle.
(Editor's note: As observed in the... | The actual inequality is $(abc)^2\ge \left(\dfrac{4\Delta}{\sqrt{3}}\right)^3$.I will show this one.
*
*Lemma:
$\displaystyle \sin{\alpha}+\sin{\beta}+\sin{\gamma} \le \frac{3\sqrt{3}}{2}$
when $\alpha$,$\beta$,$\gamma$ are angles of a triangle.Now we see
that $\sin x$ is concave in $(0,\pi)$ so applying Jensen's in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/578876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int\frac{\cot{x}}{1+\sin{x}+\cos{x}} \mathrm dx$ Find this integral:
$$\int\dfrac{\cot{x}}{1+\sin{x}+\cos{x}}\mathrm dx$$
My try: since
$$1+\sin{x}+\cos{x}=2\cos^2{\dfrac{x}{2}}+2\sin{\dfrac{x}{2}}\cos{\dfrac{x}{2}}$$
$$\cot{x}=\dfrac{1-\tan^2{\dfrac{x}{2}}}{2\tan{\dfrac{x}{2}}}$$
so
$$\dfrac{\cot{x}}{1+\sin... | Let
$$I=\int \frac{\cot x}{1+\sin x+\cos x} \operatorname{d}x$$
Substitute $t = \tan\left(\frac{x}{2}\right)$ and $\operatorname{d}t = \frac{1}{2} \sec^2\left(\frac{x}{2}\right) \operatorname{d}x$, and transform the integrand using the substitutions $\sin x = \frac{2 t}{t^2+1}, \cos x = \frac{1-t^2}{t^2+1}$ and $\o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/580331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$ Question is to Evaluate :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$$
what all i could do is :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots=\sum_{n=1}^{\infty} \frac{1}{(n+2)n!}=\sum_{n=1}^{\in... | Since $\displaystyle e^x=\sum_{n=0}^\infty \frac {x^n}{n!}$, then $\displaystyle e^x-1-x=\sum_{n=0}^\infty\frac{x^{n+2}}{(n+2)!}$, and
$$ \frac{e^x-1-x}{x}=\sum_{n=0}^\infty\frac{x^{n+1}}{(n+2)!}, $$
so
$$ \left(\frac{e^x-1-x}{x}\right)'=\sum_{n=0}^\infty\frac{(n+1)x^n}{(n+2)!}. $$
Evaluate at $x=1$, and subtract $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
The indefinite integration of $\frac{1}{\sqrt{n^4-1}}$ I need the indefinite integral:
$$\int\frac{1}{\sqrt{n^4-1}}dn $$
I know it has a relation with the $\tanh^{-1}$ function, but can't find a proper substitution.
| Introduce variables $c, s$ and $\theta$ such that
$$\cos\theta = c = \frac{1}{n}\quad\text{ and }\quad \sin\theta = s = \sqrt{1-c^2} = \frac{\sqrt{n^2-1}}{n}.$$
We have
$$\begin{align}
\int \frac{dn}{\sqrt{n^4-1}}
&= - \int \frac{dc}{c^2\sqrt{\frac{1}{c^4}-1}}
= - \int \frac{dc}{\sqrt{1-c^4}}\\
&= - \frac{1}{\sqrt{2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/583604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Please help me with this inequality $a,b,c > 0$ (no other conditions)
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\sqrt{3\left(a^2+b^2+c^2\right)}$
I tried this:
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{\left (a+b+c\right)^2}{a+b+c}=a+b+c$
then
$a+b+c\geq\sqrt{3\left(a^2+b^2+c^2\right)}$ which is not cor... | if $$a^2\to a,b^2\to b,c^2\to c$$
then $a+b+c=3$,we have
$$\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{c}}+\dfrac{c}{\sqrt{a}}\ge 3$$
so you can see this Inequality. $\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/584881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$\left(\frac1\alpha-\frac1\beta\right)^2$ for $p(x)=x^2+x-2$ If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x)=x^2+x-2$,
then $\left(\frac1\alpha-\frac1\beta\right)^2 is:$
A) $\frac94$
B) $\frac{-9}4$
C) $\frac25$
D) $\frac{-2}5$
This is a homework question from our school's home assignment for c... | $(x+\frac{1}{2})^2-\frac{1}{4} -2 =0 \iff (x+\frac{1}{2})^2=\frac{9}{4} \iff x=1$ or $x=-2$
Since we know that $\alpha$ and $\beta$ are the roots of the polynomial, it must hold that $\alpha =1$ and $\beta=-2$ or $\alpha=-2$ and $\beta=1$
Lets first consider the case in which $\alpha =1$ and $\beta=-2. \Rightarrow (\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/587915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Minimum value of $\left|z^2-z+1\right|+\left|z^2+z+1\right|$ for $z\in \mathbb{C}$ (1) If $\left|z\right| = 1$. Then find minimum value of $\left|z^2+z+4\right|$
(2) If $z\in \mathbb{C}.$ Then minimum value of $\left|z^2-z+1\right|+\left|z^2+z+1\right|.$
$\bf{My\; Try}::$ (1) Given $\left|z\right| = 1\Rightarrow z \bar... | Let $z=x+yi\ (x,y\in\mathbb R)$.
Let us consider the case $|z|=r$ where $r$ is a fixed non-negative real number.
We have
$$\small\begin{align}&\left|z^2-z+1\right|+\left|z^2+z+1\right|
\\\\&=|(x+yi)^2-(x+yi)+1|+|(x+yi)^2+x+yi+1|
\\\\&=\sqrt{(x^2-y^2-x+1)^2+(2xy-y)^2}+\sqrt{(x^2-y^2+x+1)^2+(2xy+y)^2}
\\\\& \stackrel{y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/591264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that Q($\sqrt{2}$, $\sqrt{3}$) is a field Prove that $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \{a+b\sqrt{2} +c\sqrt{3} +d\sqrt{6}\ |\ a,b,c,d \in \mathbb{Q}\}$ is a field.
I am doing the subfield test, but having trouble in showing how to express the inverse in such a form. Anyone can help?
| HINT: $$(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})(a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6})=(a+d\sqrt{6})^2-(b\sqrt{2}+c\sqrt{3})^2:=r+s\sqrt{6}$$ where $r,s\in\mathbb{Q}$. Then $$(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})(a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6})(r-s\sqrt{6})=r^2-6s^2=\alpha$$ and $\alpha$ is rational. Hence you have got the inverse.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/592773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
If $3^n+81$ is a perfect square, then positive integer value $n$ is If $3^n+81$ is a perfect square, Then calculation of a positive integer value of $n$.
$\bf{My\; Try}::$ When $n≤4,$ then easy to know that $3^n+81$ is not a perfect square.
Now let $n=k+4(k∈Z^{+}),$ then $3^{n}+81=81(3^{k}+1).$
So $3^{n}+81$ is a pe... | Note that if $3^n+81 = x^2$, we have
$$(x+9)(x-9) = 3^n$$
Hence, we have
$x+9 = 3^m$ and $x-9 = 3^{n-m}$. Hence, we need two powers of $3$ that differ by $18$, i.e., we need $3^m - 3^{n-m} = 18$.
Now observe that $3^m-3^{m-1} \geq 18$, if $m\geq 3$. Hence, we have limited options for $m$ and $n-m$.
Hence, $m=3$ and $n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/593572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to find the minimum of $a+b+\sqrt{a^2+b^2}$ let $a,b>0$, and such
$$\dfrac{2}{a}+\dfrac{1}{b}=1$$
Find this minimum
$$a+b+\sqrt{a^2+b^2}$$
My try: since
$$2b+a=ab$$
so
$$a+b+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+2ab}+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+4b+2a}+\sqrt{a^2+b^2}$$
then I can't
maybe this problem can use AM-GM or Cauchy... | since
$$\dfrac{1}{a}+\dfrac{2}{b}=1$$
then a straight line
$$\dfrac{x}{a}+\dfrac{y}{b}=1$$ cross $P(1,2)$
then
$$a+b+\sqrt{a^2+b^2}=|OA|+|OB|+|AB|$$
In the follow
we have
$$|OC|+|OD|+|CD|\ge|OE|+|OF|+|EF|=|OA|+|OB|+AB|=10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/593703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Triangle and Maxium value Given any triangle ABC with $a \ge b \ge c$ such that $\frac{a^3+b^3+c^3}{\sin^3(A)+\sin^3(B)+\sin^3(C)}=7$, what is the maximum value of $a$?
| Using the Sine Law, i.e.
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k $$
we find that the condition in the question is merely equivalent to $k^3 = 7$.
So $a = k \sin A \le k = \sqrt[3]7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/595633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How find this integral $\int\frac{1}{1+\sqrt{x}+\sqrt{x+1}}dx$ Question:
Find the integral
$$I=\int\dfrac{1}{1+\sqrt{x}+\sqrt{x+1}}dx$$
my solution:
let $\sqrt{x}+\sqrt{x+1}=t\tag{1}$
then
$$t(\sqrt{x+1}-\sqrt{x})=1$$
$$\Longrightarrow \sqrt{x+1}-\sqrt{x}=\dfrac{1}{t}\tag{2}$$
$(1)-(2)$ we have
$$2\sqrt{x}=t-\dfrac{1}{... | Using $\sqrt{x}=u=\tan(\theta)$ and $v=\sin(\theta)$,
$$
\begin{align}
&\int\frac1{1+\sqrt{x}+\sqrt{x+1}}\,\mathrm{d}x\\
&=\int\frac{1+\sqrt{x}-\sqrt{x+1}}{2\sqrt{x}}\,\mathrm{d}x\\
&=\sqrt{x}+\frac x2-\int\frac{\sqrt{x+1}}{2\sqrt{x}}\,\mathrm{d}x\\
&=\sqrt{x}+\frac x2-\int\sqrt{u^2+1}\,\mathrm{d}u\\
&=\sqrt{x}+\frac x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/596467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Point on the graph of $y=\sqrt{4x+13}$ closest to $(5,0)$? Just did this question on an exam earlier today, I'm curious to see if I'm correct.
What point on the graph of $y=\sqrt{4x+13}$ is closest to $(5,0)$?
My answer: $(-1,3)$
| Your method described in the comments seems correct. You start by noting that the distance between $(x,y(x))$ and $(5,0)$ is given by
$$d(x) = \sqrt{(5 - x)^2 + (0 - y(x))^2} = \sqrt{(5 - x)^2 + 4x + 13} = \sqrt{x^2 - 6x + 38}. \tag{1}$$
You want to minimize $d(x)$, so indeed you want to look at the derivative:
$$d'(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/603179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find value of integral: $I=\int_0^{2\pi}\frac{dx}{(2+\cos x)^2}$ Find value of integral: $$I_1=\int_0^{2\pi}\frac{dx}{(2+\cos x)^2}$$ and $$I_2=\int_0^{2\pi}\frac{dx}{(2+\sin x)^2}$$
I don't know how, i need a solution, please
| Let $\displaystyle I = \frac{\sin x}{(2+\cos x)}$
Now Diff. both side w.r. to $x$ , $\displaystyle \frac{dI}{dx} = \frac{d}{dx}\left(\frac{\sin x}{2+\cos x}\right) = \frac{(2+\cos x)\cdot \cos x-\sin x\cdot (-\sin x)}{(2+\cos x)^2}$
$\displaystyle \frac{dI}{dx} = \frac{2\cos x+1}{(2+\cos x)^2}\Rightarrow \frac{dI}{dx} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/604312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
All solutions of $a+b+c=abc$ in natural numbers I was observing some nice examples of equalities containing the numbers $1,2,3$ like $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\pi$ and $\log 1+\log 2+ \log 3=\log (1+2+3)$. I found out this only happens because $1+2+3=1*2*3=6$. I wanted to find other examples in small numbers, b... | If $a=0$ then you require $b+c=0$ and hence $b=c=0$.
Note that you can assume $a\leq b \leq c$. If $a, b, c \geq 2$ then $abc \geq 4c > c + b + a$. Hence at least one of $a,b,c$ is equal to $1$.
Wlog assume $a=1$, and look for solutions to $b+c+1 = bc$. If $b,c\geq 3$ then $bc \geq 3c > b + c + 1$, hence at least one o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/613105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
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if $3^3 2^2 \ | a^2$ then $3^2 2 \ | a $ where a is integer if $3^3 2^2 \ | a^2$ then $3^2 2 \ |a $ where a is integer.
I just cannot see it. please explain this trivial remark.
| Here is a Bezout's Identity approach.
Suppose $9\nmid a$. Then $\gcd(a,9)\mid3$. Thus, there exist $x,y$ so that
$$
ax+9y=3\tag{1}
$$
Then
$$
a^2x^2+27\left(2y-3y^2\right)=9\tag{2}
$$
Thus, $\gcd\!\left(a^2,27\right)\mid9$. Then $27\nmid a^2$. By contraposition, we have
$$
27\mid a^2\implies9\mid a\tag{3}
$$
Suppose ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/615001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Real roots of the equation $1+\sum_{r=1}^{7}\frac{x^{r}}{r} = 0$ The number of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7} = 0$
$\bf{My\; Try}::$ Let $\displaystyle f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}... | Note that
$$f'(x)=\sum_{k=0}^6 x^k=\frac{x^7-1}{x-1}.$$
Now, it is not too hard to see that $f'(x)>0$ for all $x\in\mathbb{R}$ (consider regions $x<-1$, $-1<x<0$, $0<x<1$ and $x>1$). So, there is a real root because the degree of the polynomial is odd, but there is only one because the function is monotonically increas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/617030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
The speed of the top of a sliding ladder
A $5$m ladder is leaning against a wall. If the bottom of the ladder is pulled along the ground away from the wall at a constant rate of $0.4$m/s, how fast will the top of the ladder be moving down the wall when its bottom is $3$m away from the wall.
Is my solution wrong? How ... | First we use Pythagorean Theorem
$$5^2=x^2+y^2 \implies y = \sqrt{25-x^2}.$$
Next we use that we can write $x=0.4t=\frac{2}{5}t$:
$$y=\sqrt{25-\left(\frac{2}{5}t\right)^2} = \sqrt{25-\frac{4}{25}t^2}.$$
Now we calculate the derivative (using the chain rule)
$$ \frac{dy}{dt}= \frac{\frac{d}{dt}\left(25-\frac{4}{25}t^2\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Asymptotics of ${2^n \choose n}$? How can one compute the asymptotics of ${2^n \choose n}$? I know it is bounded below and above by $\left(\frac{2^{n}}{n}\right)^n$ and $\left(\frac{2^{n}e}{n}\right)^n$.
If I plug in Stirling's approximation I get
$$\frac{2^{2^n n+n/2-1/2}}{(2^n-n)^{2^n-n+1/2} n^{n+1/2}\sqrt{\pi}}.$$... | Writing a product as the exponential of the sum of the logarithms is often a fruitful method.
Here we can write
$$\begin{align}
\binom{2^n}{n} &= \prod_{m=1}^n \frac{2^n - (m-1)}{m}\\
&= \frac{2^{n^2}}{n!} \prod_{k=1}^{n-1} \left(1- \frac{k}{2^n}\right)\\
&= \frac{2^{n^2}}{n!} \exp \left(\sum_{k=1}^{n-1} \log \left(1-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How find this $f^{(4)}(0)$ let
$$f(x)=\dfrac{e^x}{1-\sin{x}}$$
Find the value of
$$f^{(4)}(0)=?$$
My try: let
$$\dfrac{e^x}{1-\sin{x}}=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4+\cdots$$
so
$$e^x=(1-\sin{x})(a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4+\cdots)$$
since
$$e^x=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dfrac{x^4}{4}+\cd... | First, note that your $e^x$'s expansion is wrong as Shuchang pointed out.
Why don't you compare the coefficients? You'll get
$$1=a_0$$
$$1=a_1-a_0$$
$$\frac{1}{2!}=a_2-a_1$$
$$\frac{1}{3!}=a_3-a_2+\frac{a_0}{3!}$$
$$\frac{1}{4!}=a_4-a_3+\frac{a_1}{3!}$$
Then, the answer is $4!\cdot a_4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/618601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$a,b$ are roots of $x^2-3cx-8d = 0$ and $c,d$ are roots of $x^2-3ax-8b = 0$. Then $a+b+c+d =$ (1) If $a,b$ are the roots of the equation $x^2-10cx-11d=0$ and $c,d$ are the roots of the equation
$x^2-10ax-11b=0$. Then the value of $\displaystyle \sqrt{\frac{a+b+c+d}{10}}=,$ where $a,b,c,d$ are distinct real numbers.
(2... | The answer for (1) is $11$.
$$abcd=121bd\Rightarrow bd(ac-121)=0\Rightarrow bd=0\ \text{or}\ ac=121.$$
(Note that you have a mistake here too.)
1) The $bd=0$ case : If $b=0$, we have $x(x-10a)=0$. This leads that $c=0$ or $d=0$. This is a contradiction. The $d=0$ case also leads a contradiction.
2) The $ac=121$ case : ... | {
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"url": "https://math.stackexchange.com/questions/619294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to reduce the radical? I was reading a pdf on Cardano's method of solving for the roots of a cubic polynomial, when I noticed an example.
$$x^3+6x-20=0$$
On solving, I got
$$x=\sqrt[3]{10+\sqrt{108}}+\sqrt[3]{10-\sqrt{108}}$$
I went through a calculator and it gave the answer $2$.
My question is how can we prove
... | As $\displaystyle10+\sqrt{108}=10+6\sqrt3,$
we can write $\displaystyle10+6\sqrt3=(a+b\sqrt3)^3$ where $a,b$ are real rationals
$\displaystyle\implies10+6\sqrt3=a^3+2a^2\cdot b\sqrt3+3a(b\sqrt3)^2+(b\sqrt3)^3$
$\displaystyle\implies10+6\sqrt3= a^3+9ab^2+3(a^2b+b^3)\sqrt3$
Comparing the rational & the irrational parts ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Continuous $f$ satisfying $f(2x)=f(x-1/4)+f(x+1/4)$ on $(-1/2,1/2)$ What are the continuous functions $f\colon (-\frac{1}{2},\frac{1}{2}) \to \mathbb{C}$ that satisfy the following functional equation, and how are they derived?
$$f(2x)=f(x-\frac{1}{4})+f(x+\frac{1}{4})\,\,\,\,\,\,\,\,\text{for }x \in (-\frac{1}{4},\fra... | If $f$ is as in the problem statement, then, from the functional equation and continuity of $f$ at $x=0$, we see that
$$\lim_{x \ \to -\frac{1}{4}} f(2x) - f(x-\frac{1}{4}) = f(0).$$
Now, suppose $f_0$ is any continuous function on $(-\frac{1}{2},0]$ such that
\begin{align*} \lim_{x \ \to -\frac{1}{4}} f_0(2x) - f_... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Integration of $\int \frac{dx}{a+f^2(x)}$ I want to solve a integral of the form:
$$
\int \frac{dx}{a+f^2(x)}
$$
in my particular case I got
$$
\int \frac{dx}{5+\cos^2(x)}
$$
in my case I followed this process:
$$
\int \frac{dx}{5+\cos^2(x)} \\
let \ t = tg(\frac{x}{2}) =>
dx = \frac{2dt}{1+t^2}\\
\int \frac{dx}{5+\... | Let $f$ be the function such that for all real number $x$, $f(x) = \frac{1}{5+\cos^{2}(x)}$. The primitive you want to compute is
$$ \int f(x) \, dx $$
Since $f(x) \, dx$ is invariant under the change of variables $x \rightarrow \pi+x$, the change of variables $t=\tan(x)$ might lead to more simple computations. Let's... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove: $xy+yz+zx\leq \frac{16}{3}$ For $x,y,z\in R$ and $x^2+xy+y^2=1$; $y^2+yz+z^2=16$
Prove: $xy+yz+zx\leq \frac{16}{3}$
| $16=[(\frac x2+y)^2+\frac 34x^2][(\frac z2+y)^2+\frac 34z^2]$
Using the Cauchy-Schwarz inequality, we get
$16\ge \frac 34[(\frac x2+y)z+x(y+\frac z2)]^2$
$\Rightarrow xy+yz+zx\le \frac 8{\sqrt 3}<\frac {16}3$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How find this sum $I_n=\sum_{k=0}^{n}\frac{H_{k+1}H_{n-k+1}}{k+2}$ $$I_n=\sum_{k=0}^{n}\dfrac{H_{k+1}H_{n-k+1}}{k+2}$$
where $$H_{n}=1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}$$
my try:since
$$I_n=\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{2}+\dfrac{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right)}{... | Maybe I have got the definitive trick. I recall my previous $(5)$:
$$I_n=\frac{3}{2}\sum_{t=3}^{n+3}\frac{H_{t-1}^2-H_{t-1}^{(2)}}{t}.\tag{5}$$
Partial summation gives (I set $H_{0}^{(j)}=0$ for consistency):
$$\sum_{n=1}^{m}\frac{H_{n-1}^{(2)}}{n}=H_m H_{m-1}^{(2)}-\sum_{n=1}^{m-1}\frac{H_n}{n^2}=H_m H_m^{(2)}-\sum_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/627936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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$Z^2-YZ-Y^2+X^2+2XY$ is an irreducible polynomial How to show that $Z^2-YZ-Y^2+X^2+2XY$ is an irreducible polynomial in $\Bbb{C}[X,Y,Z]$?
| Suppose that $Z^2-YZ-Y^2+X^2+2XY$ is reducible. Since this is a degree two polynomial in $Z$ with coefficients in $\Bbb C[X,Y]$ it splits into a product of two polynomials of degree one in $Z$, that is, $Z^2-YZ-Y^2+X^2+2XY=(Z+f(X,Y))(Z+g(X,Y))$. We get $f(X,Y)+g(X,Y)=-Y$ and $f(X,Y)g(X,Y)=-Y^2+X^2+2XY$. Now plug in the... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the sum of $\sum_{n=1}^\infty \frac{x^{n-1}}{3^nn}$ - What is wrong with my solution? I have to find the sum of the following power series:
NOTE: please assume that x is in the convergence domain.
$$\sum_{n=1}^\infty \frac{x^{n-1}}{3^nn}$$
My solution:
*
*$$S(x) = \sum_{n=1}^\infty \frac{x^{n-1}}{3^nn} = \sum_{n... | Your mistake is in step 4, where you claim
$$\frac{\frac{x}{3}}{1-\frac{x}{3}} = \frac{x}{3}\left(1 - \frac{3}{x}\right).$$
The correct value is
$$\frac{\frac{x}{3}}{1-\frac{x}{3}} = \frac{x}{3-x},$$
which then leads to
$$\begin{gather}
(x\cdot S(x))' = \frac{1}{3-x}\\
x\cdot S(x) = -\log (3-x) + C\\
S(x) = -\frac{\log... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/630520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the limit $\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$
Find the following limit: $$\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$$
Well I tried to do the $\exp\left(\frac{ \ln\frac{\sin x}{x}}{x^2}\right)$ then apply LHR but I seem to get to endless dervivations...
... | $\require{cancel}$
$$\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$$
$$f(x):=\left(\frac{\sin x}x\right)^{1/{x^2}}$$
$$\ln f(x)= \dfrac{1}{x^2}\ln \left( \dfrac{\sin x}{x}-1+1\right)$$
$$\ln f(x)=\dfrac{\ln \left( \dfrac{\sin x}{x}-1+1\right)}{\dfrac{\sin x}{x}-1}\times \dfrac{\dfrac{\sin x}{x}-1}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Taylor's series when x goes to infinity Let $f(x) = \frac {x^3}{(x+1)^2}$. Find constants a, b, c, so that $f(x) = ax + b + \frac cx + o(\frac 1x)$ as $x$ goes to $\pm \infty$. So i know that i can't take Taylor series as $x$ goes to infinity. So i am assuming i have to make some kind of substitution. I tried making $... | I would start off here by using long division to simplify things a bit:
$$
\frac{x^3}{(x+1)^2}=\frac{x^3}{x^2+2x+1}=\cdots=x-2+\frac{3x+2}{x^2+2x+1}.
$$
Heuristically speaking, it is clear that as $x\rightarrow\infty$, that last bit is $\frac{3}{x}+o(\frac{1}{x})$. So, this would suggested that
$$
\frac{x^3}{(x+1)^2}=... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrate $\int \frac{(1+x^2)dx}{(1−x^2)\sqrt{1+x^4}}$ Integrate $\displaystyle \int \frac{(1+x^2)dx}{(1−x^2)\sqrt{1+x^4}}$
I don't know how to do this one. I need some suggestions. Thank you!
| These types of integrals are best evaluated using a substitution of the form $$u = x - x^{-1}, \quad du = 1 + x^{-2} \, dx.$$ Note that $u^2 = x^2 + x^{-2} - 2$, and we can write the integrand as $$\frac{1+x^2}{(1-x^2)\sqrt{x^4+1}} dx = \frac{x^2(1+x^{-2}) \, dx}{x^2(x^{-1}-x)\sqrt{x^2+x^{-2}}} = -\frac{du}{u\sqrt{u^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the number of roots of equation $z^5 -12z^2+14=0$ that lie in the region {$z \in \Bbb C : 2 \leq |z|< \frac {5}{2}$}
Problem: Find the number of roots of equation $z^5 -12z^2+14=0$ that lie in the region
{$z \in \Bbb C : 2 \leq |z|< \frac {5}{2}$}
Solution :
In $f(z)=z^5 -12z^2+14=0$, there is $2$ change in si... | The idea is to use Rouché's theorem.
On the outer circle $\lvert z\rvert = \frac{5}{2}$, we have
$$\lvert z\rvert ^5 = \frac{3125}{32} = 97 + \frac{21}{32} > 89 = 75 + 14 = 12\lvert z\rvert^2+14 \geqslant \lvert -12z^2 + 14\rvert,$$
so by Rouché's theorem, $f(z) = z^5 - 12z^2 + 14$ has as many zeros in the disk $\lvert... | {
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"url": "https://math.stackexchange.com/questions/635867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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I'd like to get explain about complex roots If $x^6+1=0$ so $x^6=-1$, then we have to find the roots at $\mathbb{C}$.
I saw that the roots are $$\Large{e^{(\frac{\pi}{6}+\frac{2k\pi}{6})i}}\;\small{k=0,1,2,3,4,5}$$
this what I understand. maybe I wrong...
My question is why we are putting the $\frac{\pi}{6}$?
Thank you... | As you said, $x^6 = -1$. We want to find all complex solutions. You may have been exposed to complex numbers as things of the form $a + bi$, where $a,b$ are real. However, there is another form you can write them in: $re^{i\theta}$, where $r,\theta$ are real, and $r \ge 0$.
The geometric interpretation of $a + bi$ is r... | {
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"url": "https://math.stackexchange.com/questions/636014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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different results for the solution of bessel function with exponential I have this integral
$$
\int_0^\infty e^{-\alpha x}K_1(\beta \sqrt{x}) \, dx.
$$
for $\Re[\alpha] >0$, and $\Re[\beta]>0$
According to the (Table of Integrals, Series, and Products, Seventh Edition), equation 6.614.5 the solution of the above integ... | There is a special relation between Tricomi's U and the Bessel K functions, see e.g. http://functions.wolfram.com/07.33.03.0006.01:
$$U(a, 2 a - 1, z) = \frac{e^{\frac{z}{2}} z^{\frac{3}{2} - a}}{2(a - 1) \sqrt{\pi}}\left(K_{a - \frac{1}{2}} \left(\frac{z}{2}\right) - K_{a - \frac{3}{2}}\left(\frac{z}{2}\right)\right)$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Calculating Bernoulli Numbers from $\sum\limits_{n=0}^\infty\frac{B_nx^n}{n!}=\frac x{e^x-1}$ How is the Bernoulli numbers? For example, found that in internet
$$\sum_{n=0}^\infty\frac{B_nx^n}{n!}=\frac x{e^x-1}$$
but if I want to find $B_2$ then
$$B_0+B_1x+\frac{B_2x^2}{2}+\sum_{n=3}^\infty\frac{B_nx^n}{n!}=\frac x{e... | Note that
$$\frac{e^z-1}z=\frac1 z\sum_{n=1}^\infty\frac1{n!}z^n=\sum_{n=1}^\infty\frac1{n!}z^{n-1}=\sum_{n=0}^\infty\frac1{(n+1)!}z^n$$
and we can use Mertens’ multiplication theorem to get
$$1=\left(\sum_{n=0}^\infty\frac{B_n}{n!}z^n\right)\left(\sum_{n=0}^\infty\frac1{(n+1)!}z^n\right)=\sum_{n=0}^\infty\sum_{k=0}^n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/642625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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How to factorize $2x^2+5x+3$? I'm doing pre-calculus course at coursera.org and I'm in trouble with this solution $$2x^2 +5x +3 = (2x+3)(x+1)$$
By trial, using ac-method I got stuck:
$$
ac = (2)(3) = 6\\
6 + ? = 5 \Rightarrow~ ? = 5 - 6 = -1
$$
Then,
$$2x^2+6x-x+3 = 2x(x+3)-x+3$$
At this point I could not get the answe... | In general
$$ax^2+bx+c=a(x-x_1)(x-x_2)$$where $x_1,x_2$ are the roots of equation
$$ax^2+bx+c=0$$
in case $a=2,b=5,c=3$ and roots are
$$x_{1,2}=\frac{-5\pm\sqrt{5^2-4\cdot2\cdot3}}{2\cdot2}=\frac{-5\pm1}{4},x_1=-1,x_2=-\frac{3}{2}$$ and put these above we get
$$2x^2+5x+3=2(x-(-1))(x-(-\frac{3}{2}))=$$
$$=2(x+1)(x+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/644975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify the expression : $\tan(\theta) +2\tan(2\theta) +2^2\tan(2^2 \theta) +\cdots +2^{14} \tan(2^{14}\theta) +2^{15} \cot(2^{15} \theta)$ How to simplify the expression:
$\tan(\theta) +2\tan(2\theta) +2^2\tan(2^2 \theta) +\ldots +2^{14} \tan(2^{14}\theta) +2^{15} \cot(2^{15} \theta)$
I am not getting any clue how ... | HINT:
Use $$\cot A-\tan A=\frac{\cos^2A-\sin^2A}{\cos A\sin A}=2\cot2A$$ repeatedly
So, we have $$\cot\theta-\tan\theta=2\cot2 \theta$$
and $$2(\cot2\theta-\tan2\theta)=2(2\cot2^2\theta)$$
$$2^2(\cot2^2\theta-\tan2^2\theta)=2^2(2\cot2^3\theta)$$
and so on
Now add the relations.
Reference : Double-Angle Formulas
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\frac3x - \frac4y = 1$ and $\frac7x + \frac2y = \frac{11}{12}$ How can we solve the following simultaneous equations:
$$\frac3x - \frac4y = 1$$
$$\frac7x + \frac2y = \frac{11}{12}$$
| $\frac{3}{x}-\frac{4}{y}=1$ ---(1)
$\frac{7}{x}+\frac{2}{y}=\frac{11}{12}$ ---(2)
2(2): $\frac{14}{x}+\frac{4}{y}=\frac{11}{6}$ ---(3)
(1)+(3): $\frac{17}{x}=\frac{17}{6}$
Hence $x=6$, and $y=-8$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that the triangle which satisfy the inequality $\frac{\sin^2 A+\sin^2 B+\sin^2 C}{\cos^2 A+\cos^2 B+\cos^2 C}=2$ Show that the triangle which satisfy the inequality $\dfrac{\sin^2 A+\sin^2 B+\sin^2 C}{\cos^2 A+\cos^2 B+\cos^2 C}=2$ is right angled.
My work:
$\sin^2 A+\sin^2 B+\sin^2 C=2(\cos^2 A+\cos^2 B+\cos^2 ... | The last equation gives: cos(2A) + cos(2B) + cos(2C) = - 1 => 2cos(A+B)cos(A-B) + 2(cos(C))^2 = 0 ==> -2cos(C)cos(A-B) + 2(cos(C))^2 = 0 ==> 2cos(C)(-cos(A-B) + cos(C)) = 0. So cos(C) = 0 gives C = pi/2 or cos(A-B) = cos(C). So A-B = C or -C and this means A = B + C or B = A + C so 2A = pi so A =pi/2 or 2B = pi so B = ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Are those rings fields? Let $f(x) = x^4+x^2+1 \in Z_{2}$ and $A = Z_{2}/f(x)$
*
*Is $A$ a field?
Let $f(x) = x^4-x^2+1 \in Z_{7}$ and $B = Z_{7}/f(x)$
*
*Is $B$ a field?
I know that a ring $A = Z_{n}/f(x)$ is a field $\Leftrightarrow$ $f(x)$ has no roots in $Z_{n}$ . But this happens only for 2nd-degree and 3... | Good luck allowed this argument in answer to the second question.
In characteristic $7$, we get $(a+b)^6=a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6$. In $\mathbb F_7[x]$, $(x^2+1)^6=x^{12}-x^{10}+x^8-x^6+x^4-x^2+1$, which is clearly congruent to $1$ modulo your polynomial $x^4-x^2+1$. Thus, in $B$, $x^2+1$ is a sixth root ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/647764",
"timestamp": "2023-03-29T00:00:00",
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How prove this $f(n)\le f(n+1)$ where $f(n)=\sum_{k=1}^{n}\frac{n}{n^2+k^2}$ let $$f(n)=\sum_{k=1}^{n}\dfrac{n}{n^2+k^2}$$
prove or disprove
$$f(n)\le f(n+1)$$
this inequality is found when I deal this follow limit:
$$\lim_{n\to\infty}f(n)=\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{1+(k/n)^2}=\int_{0}^{1}\... | This answer is essentially the same approach as vesszabo's; I include it just to give another perspective. Set $g(x)=1/(1+x^2)$. Consider the trapezoidal approximation
$$\int_0^1 g(x) dx \approx \frac{1}{n} \left( \frac{g(0)+g(1/n)}{2} + \frac{g(1/n)+g(2/n)}{n} + \cdots + \frac{g((n-1)/n)+ g(1)}{2} \right)$$
$$=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/648367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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On solving $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}}$ How do we show that there is only one solution to,$$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}}$$
I guess it is only $x=2$.
Please help.
| Let $\;f(x) = \sqrt{2+x}\;$ and $\;g(x) = \sqrt[3]{6+x}$, they are strictly increasing function in $x$ when $x \ge -2$.
Since $(x+2)^3 - (x+6)^2 = (x-2)(x^2 + 7x + 14)$ and $x^2 + 7x + 14 > 0$ for all $x$,
we have
$$\begin{cases} f(x) > g(x) > 2,& x > 2\\f(x) = g(x) = 2,& x = 2\\f(x) < g(x) < 2, & x <2\end{cases}$$
So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/649570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Taylor Series Expansion of $\frac{1}{1+x^2}$ about $x=a$ Let $$f(x)=\frac{1}{1+x^2}$$ Consider its Taylor series expansion about a point $a\in \mathbb{R}$. What is the radius of convergence of this series??
About $x=0$ we could expand it like
$$(1+x^2)^{-1}= 1-x^2+(-1)(-2) \frac{x^2}{2!}+(-1)(-2)(-3)\frac{x^3}{3!}+\do... | Here is a pretty simple way to get the actual Taylor series centered at $a$:
Using partial fractions:
$\dfrac{1}{z^2+1} = \dfrac{i}{2}\left(\dfrac{1}{z+i} - \dfrac{1}{z-i}\right)$.
Now, we can expand each of the the two using the geometric series:
$\displaystyle\dfrac{1}{z+i} = \dfrac{1}{(z-a)+(a+i)} =\dfrac1{a+i}\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/649677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Formula for sides of a triangle where the Perimeter equals to the Area I was wondering if there is a formula that could generate the values of the sides of a triangle where his area equals to his perimeter. I only found that if the triangle is equilateral then
$$l=\frac{12}{√3}$$
where $l$ is the side of the triangle.... | Given any triangle $T$ there is a triangle $T'$ similar to $T$ such that the area of $T'$ is the same as the perimeter of $T'$.
Suppose the perimeter of $T$ is $P$ and the area of $T$ is $A$. Then dilate $T$ by a factor of $\frac{P}{A}$ to produce the triangle $T'$. (That is, multiply all the lengths by $\frac{P}{A}$.... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding all positive integers $x,y,z$ that satisfy $3^x - 5^y = z^2$
Find all positive integers $x,y,z$ that satisfy:
$$3^x - 5^y = z^2.$$
I think that $(x,y,z)= (2,1,2)$ will be the only solution. But how to prove that?
| I will give a brief review to Yiyuan's answer, and then complete the proof. We have:
$$3^x-5^y=z^2$$
Working $\pmod4$, we have $(-1)^x-1\equiv 0,1 \pmod4$, so $x$ has to be even. Substituting $x=2k$ and $y=a+b$:
$$(3^k-z)(3^k+z)=5^y\implies 3^k-z=5^a, 3^k+z=5^b\implies 2\times3^k=5^a+5^b$$
If $a,b\ge1$, then the RHS is... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given $x+y$ and $x\cdot y$, what is $x^3+ y^3$ ? I have been looking at an assortment of high school number sense tests and I noticed a reoccurring problem that states what x+y is and what $x\cdot y$ is then asks for $x^3+ y^3$. I want to know how to work these problems. I have a couple of examples.
$x+y=5$ and $x\cdot... | Its simple. Use the identity
${x^3 + y^3 =(x+y)(x^2 +y^2 -xy) }$
In the above identity;
make modifications and bring it to this form :
${x^3+y^3 = (x+y)((x+y)^2 -3xy) }$
Now putting the values of ${x+y}$ and ${xy}$ in the above equations, you should get the correct answer.
:)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 3
} |
Parabola $\sqrt {x}+\sqrt {y}=1 $ How do I prove that the equation $\sqrt {x}+\sqrt {y}=1 $ is part of parabola.
My attempt:rotation in 45 degrees brings the equation to $ -2a^2=1-2\sqrt {2}b $ when $ x= \frac {a-b} {\sqrt {2} } $ and $ y= \frac {a+b} {\sqrt {2} } $. It is a parabola, why is it only part of it? (also ... | Rearranging, we get $\sqrt{y} = 1 - \sqrt{x}$, which becomes $y = 1 - 2\sqrt{x} + x$ when we square both sides. Rearranging again and squaring both sides, we get $(y-x-1)^2 = y^2 + x^2 + 1 - 2xy - 2y + 2x = 4x$.
Generally, when there is an equation of the form $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, if $B^2 - 4AC = 0$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/655104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $\sqrt{3\cos^2 x - \sin 2x} = - \sin x$ Please, can you suggest something for solving this equation: I have to find the solutions included in interval $\left[3\pi/2, 2\pi\right]$:
$$\sqrt{3\cos^2 x - \sin 2x} = - \sin x$$
This is what I did:
$$\begin{array}{crcl}
\Longrightarrow & 3\cos^2 x - \sin 2x &=& \sin^... | from your last step
1−sin2x+2cos2x=0
=>(sin^2x+cos^2x)-2sinxcosx+2(cos^x-sin^2x)=0
=>3cos^2x-2sinxcosx-sin^2x=0
=>3cos^2x-3sinxcosx+sinxcosx-sin^2x=0
=>3cosx(cosx-sinx)+sinx(cosx-sinx)=0
=>(cosx-sinx)(3cosx+sinx)=0
=>cosx-sinx=0 or, 3cosx+sinx=0
=>tanx=1, or tanx=-3
=>x=π/4, or x=2π/3
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Can I use this formula with pseudo determinants instead of usual determinants? Let $A$ be a matrix with $A^+$ Moore-Penrose inverse. Let also $Det()$ denote the pseudo-determinant of a matrix.
Does the formula (which assumes the existence of $A^{-1}$)
$$ det\left( \begin{array}{cc}
A & B \\
C & D \end{array} \right)... | The classical formula
${\rm Det}(R) = {\rm Det}( \left[ \begin{array}{cc} A & B \\ C & D \end{array} \right] )
= {\rm Det}(D) {\rm Det}(A-BD^+C)$ for block matrices
does not hold for pseudo determinants ${\rm Det}$ and Moore pseudo inverse $D^+$, as already
$$ \left[ \begin{array}{cc} A & B \\ C & D \end{array} \right]... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Supremum calculation Calculate $\sup(\sum_{k=n+1}^{\infty}\frac{|x_{k}|^{2}}{4^{k} })$, where $x=(x_{1},x_{2},....)$ is a member of $l_{2}$ and the supremum is take over all $x$ with $||x||= 1$.
My intuition says the answer is $\frac{1}{4^{n+1}}$.
| Your intuition is correct. It follows from that
$\sum_{k=n+1}^\infty\frac{|x_k|^2}{4^k} \leq \frac{1}{4^{n+1}}\sum_{k=n+1}^\infty|x_k|^2\leq \frac{1}{4^{n+1}}\sum_{k=1}^\infty|x_k|^2=\frac{1}{4^{n+1}}\|x\|_{l_2}^2=\frac{1}{4^{n+1}}$
thus the supremum is at most $\frac{1}{4^{n+1}}$. We also have that this value is obtai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/659442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
parametric solution for the sum of three square Is there a parametric integer solution for $x,y,z,t$ when the sum of three square is equal to a square, i.e,
$$x^2+y^2+z^2=t^2$$?
| EDIT: I knew i had written up Jones and Pall 1939 before, but it was on a question with different intent: Every integer vector in $\mathbb R^n$ with integer length is part of an orthogonal basis of $\mathbb R^n$
J-P 1939: as long as $t$ is odd, find all solutions to
$$ a^2 + b^2 + c^2 + d^2 = t, $$
including permuting... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/660143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $\frac{a^3}{a^6 + 1}$ given $a$ is a root of a quadratic equation
If $a$ is a root of the equation $x^2 - 3x + 1 = 0$, then find the value of $\frac{a^3}{a^6 + 1}$.
So, I figured we can use the quadratic formula, and formed the following equation:
$$a=\frac{-(-3)+\sqrt{9-4}}{2(1)}\implies a=\frac{3+\sqrt5}2$$
Bu... | So
$$a^2 + 1 = 3a$$
and this gives:
$$\frac{a}{a^2 + 1} = \frac{1}{3},$$
and
$$(a^2 + 1)^2 = 9a^2 \implies a^4 + 1 = 7a^2.$$
So
$$\frac{a^2}{a^4 - a^2 + 1} = \frac{a^2}{6a^2} = \frac{1}{6}.$$ And finally
$$\frac{a^3}{a^6 + 1} = \frac{a}{a^2 +1}\cdot \frac{a^2}{a^4 - a^2 + 1} = \frac{1}{3}\cdot \frac{1}{6} = \frac{1}{18... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/660488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability problem with binomial/multinomial distribution Mary knows the answers to $20$ of the $25$ multiple choice questions on the Psychology $101$ exam, but she has skipped several of the lectures, she must take random guesses for the other five. Assuming each question has four answers, what is the probability she... | We need to make some assumptions. We will assume the $5$ questions she does not know the answers to are equally likely to be any $5$ of the $25$ questions.
We want the probability she gets exactly $2$ of the last $5$ questions wrong. This can happen in various ways. The last $5$ questions may contain $2$ questions she... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/661661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The equation $3^n+4^m=5^k$ in positive integer numbers Please help me to prove that the equation $3^n + 4^m = 5^k$
where $n$, $m$, $k$ are positive integer numbers has only the solution $n=m=k=2$.
I know how to prove it for $n=m=k$.
If $3^x + 4^x = 5^x$ then $(3/4)^x + 1 = (5/4)^x$,
and this equation has at most one so... | Reducing the equation mod $3$ shows that $k$ must be even, and reducing mod $4$ shows the same for $n$, so let's rewrite the whole thing as
$$3^{2n}+2^{2m}=5^{2k}$$
and show that $(n,m,k)=(1,2,1)$ is the only solution in positive integers.
Now
$$2^{2m}=5^{2k}-3^{2n}=(5^k-3^n)(5^k+3^n)$$
implies $5^k-3^n=2^a$ and $5^k+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/662121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How prove $\measuredangle CDE=2\measuredangle ABE$ In rectangular $ABCD$,and $E\in AC$,such
$$BE=\sqrt{2}\cdot AE$$
show that
$$\measuredangle CDE=2\measuredangle ABE$$
My try: let $$AB=a,AD=b,\dfrac{AE}{AC}=k,$$
then
$$AE=k\sqrt{a^2+b^2},BE=k\sqrt{2(a^2+b^2)}$$
I know have this nice relsut
$$AE^2+EC^2=BE^2+ED^2$$
th... | I solve this problem in another way. Let us establish a coordinate system which $A$ is original point and $AB$ is y-axis and $AD$ is x-axis. And $AB=a$, $AD=b$.
In this frame, we have: $A(0,0)$, $B(0,a)$, $C(b,a)$, $D(b,0)$. For the sake of simplicity, we define $k:=\frac{b}{a}$ and equation of $AC$ can be written as $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/663182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $|x + 1| + |x - 3| = 6$, solve for $x$ $|x + 1| + |x - 3| = 6$. Solve for X.
So I know when you have a problem like this: |x| = 6, you solve by doing x=6 and x=-6. That doesn't help us much in the above example.
You can also solve problems of this fashion by negative the variable portion of the equation. For ex:
$|... | $|x+1| = 6 - |x-3|$
$x + 1 = 6 - |x-3|$ or $x+1 = -6 + |x-3|$
For the first equation: $|x-3| = 5 - x$ so
$x-3 = 5 -x$ or $ x-3 = x-5$. Only the first part is valid, it gives you the solution $x=4$.
Second equation: $x+7 = |x-3|$. So
$x+7 = x-3$ or $-x - 7 = x-3$. Only the second equation is valid, that gives you $x=-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/665171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$p \equiv 5 \mod8\Rightarrow p=(2x+y)^{2}+4y^{2}$ If $p \equiv 5 \mod8$ , then $p=(2x+y)^{2}+4y^{2}$,for some x and y integers.
Thanks
Here is my approach:
I know $p \equiv 5 \mod8\Rightarrow $
$p \equiv 1 \mod4\Rightarrow $
$n^{2}+m^{2}=p\equiv 5 \mod8 \Rightarrow$
from $t^{2}\equiv 0,1~or ~4 \mod8$ we get $n^{2} \eq... | $p \equiv 5 \mod8\Rightarrow $
$p \equiv 1 \mod4\Rightarrow $
$n^{2}+m^{2}=p\equiv 5 \mod8 \Rightarrow$
from $t^{2}\equiv 0,1~or ~4 \mod8$ we get $n^{2} \equiv 1 \mod8$ and $m^{2} \equiv 4 \mod8 \Rightarrow$
n is odd and $ m=2y$ where y is odd $\Rightarrow$
n-y is even $\Rightarrow$
n-y=2x and m=2y $\Rightarrow$
$p=(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/668092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How prove this $\sqrt{a^2(\cos{x}+\cos{y})^2+b^2(\sin{x}+\sin{y})^2}+\sqrt{a^2(\cos{x}-\cos{y})^2+b^2(\sin{x}-\sin{y})^2}\le 2\sqrt{a^2+b^2}$ let $a,b>0$ is give numbers,for any $x,y\in R$,show that
$$\sqrt{a^2(\cos{x}+\cos{y})^2+b^2(\sin{x}+\sin{y})^2}+\sqrt{a^2(\cos{x}-\cos{y})^2+b^2(\sin{x}-\sin{y})^2}\le 2\sqrt{a^2... | Let us make use of the trigonometric identities for the sum of angles i.e.
$\cos(a \pm b) = \cos a \cos b \mp \sin a \sin b$
$\sin(a \pm b) = \sin a \cos b \pm \cos a \sin b$
Thus
$\cos x+ \cos y = 2\cos (\frac{x+y}{2})\cos (\frac{x-y}{2})$
$\cos x- \cos y = -2\sin (\frac{x+y}{2})\sin (\frac{x-y}{2})$
$\sin x+ \sin y ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Find a formula for $\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$ I need to find a clear formula (without summation) for the following sum:
$$\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$$
Well, the first few elements look like this:
$1,1,1,2,2,2,2,2,3,3,3,...$
In general, we have $(2^2-1^2)$ $1$'s, $(3^2-2^2)$ $2$'s e... | I am very bad in the area of discrete mathematics (as well in other) but I have been fascinated by the problem set in your post.
I am sure that Daniel Fisher's comment and Sami Ben Romdhane's answer are very useful; however, I have not been able to finish the work.
So, what I used is computer simulation and data re... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 1
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Determine value $\lim_{n\to +\infty} v_n$ Let $(u_n)$ sequence satisfy: $$\left\{\begin{matrix}u_1=3\\u_{n+1}=\frac{1}{5}\left(u_n^2+u_n+4\right),\: n=1, \: 2, ...\end{matrix}\right.$$
Set $v_n=\sum_{k=1}^n \frac{1}{u_k+3}$. Determine value $\lim_{n\to +\infty} v_n$
I think have to find a and b satisfy to $\frac{1}{u_k... | Let us expand
$\large a(\frac{1}{u_k+b}-\frac{1}{u_{k+1}+b})$
Based on the given sequence recursion we obtain
$\large a(\frac{1}{u_k+b}-\frac{5}{u_k^2+u_k+4+5b})=a\frac{u_k^2+u_k+4+5b-5u_k-5b}{(u_k+b)(u_k^2+u_k+4+5b)}=a\frac{u_k^2-4u_k+4}{(u_k+b)(u_k^2+u_k+4+5b)}$
Simplifying further we have
$\large a\frac{(u_k-2)^2}{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Scratch work for delta-epsilon proof for $\lim_{x \to 13} \sqrt{x-4} = 3$
Prove $\lim_{x \to 13} \sqrt{x-4} = 3$.
We need to show for all $E> 0$ there exists $D > 0$ such that
if $0 < |x - 13| < D$, then $|\sqrt{x-4} - 3| < E$. Let me write D for delta and E for epsilon please.
Scratch-work here: $\color{darkred}{\t... | If you found the algebra a bit hard to decipher, maybe this will help.
Assume that $|x-13|<1$ just means that we assume that the distance between $x$ and 13 is less than one. $x$ is the input to your function, it's distance from the number 13 is less than 1. If you want to compute your limit, i.e. what happens to $f(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/676556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
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If $a+b+c = 7\;\;,a^2+b^2+c^2 = 23$, then $a^3+b^3+c^3=$ If $a,b,c\in \mathbb{R}$ and $a+b+c = 7\;\;,a^2+b^2+c^2 = 23$ and $\displaystyle \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} = 31$. Then $a^3+b^3+c^3 = $
$\bf{My\; Trial\; Solution::}$ Given $a^2+b^2+c^2 = 23$ and
$a+b+c = 7\Rightarrow (a+b+c)^2 = 49\Rightarrow (a... | Let $p(x) = (x-a)(x-b)(x-c)$. Then the following identity holds:
$$ a^{3} + b^{3} + c^{3} = (a+b+c)^{3} - 3p(a+b+c). $$
Since we know that $p(x) = x^{3}- 7 x^{2} + 13 x + \frac{621}{31}$, we now have the answer.
| {
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"url": "https://math.stackexchange.com/questions/677184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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How to use $x + \frac{1}{x} = 7$ to compute $x^2 + \frac{1}{x^2}$. I am not sure how to approach this question:
You know that $x + \frac{1}{x} = 7$. Compute $x^2 + \frac{1}{x^2}$.
I have tried adding $x + \frac{1}{x}$ to get $\frac{x^2 +1}{x}$ but can't see if this was useful or not.
I need help in getting started.
| Notice $\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2$.
| {
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"url": "https://math.stackexchange.com/questions/679402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating a trigonometric limit What is the limit as $x$ approaches $0$ of: $$\frac{\sqrt{1+\tan x} - \sqrt{1+\sin x}}{1+x-\cos x}?$$
We cannot use L'Hôpital's rule or anything advanced like Taylor series. I reduced it to, by considering the numerator's conjugate:
$$\frac{1}{2}\lim_{x \to 0}\frac{\tan x - \sin x}{1+x... | The title of the question is a bit intimidating as this limit involves very basic trigonometric manipulation and standard limits like $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$. Here goes the simple solution $$\begin{aligned}L &= \lim_{x \to 0}\frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{1 + x - \cos x}\\
&= \lim_{x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/681774",
"timestamp": "2023-03-29T00:00:00",
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Prove that $\sqrt{8}=1+\dfrac34+\dfrac{3\cdot5}{4\cdot8}+\dfrac{3\cdot5\cdot7}{4\cdot8\cdot12}+\ldots$ Prove that $\sqrt{8}=1+\dfrac34+\dfrac{3\cdot5}{4\cdot8}+\dfrac{3\cdot5\cdot7}{4\cdot8\cdot12}+\ldots$
My work:
$\sqrt8=\bigg(1-\dfrac12\bigg)^{-\frac32}$
Now, I suppose there is some "binomial expansion with rati... | The binomial series is "just" the Taylor series of $(1+x)^{\alpha}$ at $x=0$.
Start deriving $f(x)=(1+x)^{\alpha}$ by $x$ and you get $\alpha(1+x)^{\alpha-1}$, then $\alpha(\alpha-1)(1+x)^{\alpha-2}$ for the first derivative etc.
The $n$th derivative is
$$\frac{d^n}{dx^n} (1+x)^{\alpha} = \alpha(\alpha-1)\cdots(\alpha... | {
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"url": "https://math.stackexchange.com/questions/681949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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An equation, where the solution does not exist, but on solving the equation we got a solution. why this is happening? The solution of the equation
$\sqrt{(x+1)} -\sqrt{(x-1)}= \sqrt{(4x-1)}$
is $\frac{5}{4}$,but when we put $x=\frac{5}{4}$ in the given equation, then it does not satisfy the equation.
Actually, if we ta... | Write the equation as
$$
\sqrt{x+1}=\sqrt{x-1}+\sqrt{4x-1}
$$
Then you must have
\begin{cases}
x+1\ge0\\
x-1\ge0\\
4x-1\ge0
\end{cases}
which boils down to $x\ge1$. Now square, you're sure not to add spurious solutions, because both sides represent non negative numbers:
$$
x+1=x-1+4x-1+2\sqrt{(x-1)(4x-1)}
$$
or
$$
-4x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/685687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 2
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Find the equation of a circle containing three given points Problem:
Determine the equation of the circle that passes through three points, $J(-3, 2)$, $K(4, 1)$, and $L(6, 5)$.
I thought of using systems like so:
$$\left\{ \begin{array}{rcl} (x+3)^2 + (y-2)^2 = r^2 \\ (x-4)^2 + (y-1)^2 = r^2 \\ (x-6)^2 + (y-5)^2 = r... | Use the equation $x^2+y^2+2gx+2fy+c=0$. Substitute those values, you will get three equations involving g,f,c. Substituting the values back to the equation will be your equation. The answer will be $x^2+y^2-2x-10y+1=0$ or $(x-1)^2+(y-5)^2=25$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/685777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Evaluating $\iint_D 6x-2x^2y^2+6y\,dx$$dy$
Evaluate $\iint_D 6x-2x^2y^2+6y\,dx dy$ where $D$ is the rectangle given by $-2 \leq x \leq 3$ and $-2 \leq y \leq 1$.
I've done this problem two ways. The first time I got $-630$ and then the second time I got $\frac{-1030}{9}$. Lon capa said that they were both wrong . Thi... | $$ \int_{-2}^1 \int_{-2}^3 6x-2x^2y^2+6y\ dxdy= \int_{-2}^1 \left(6\int_{-2}^3 x\ dx-2y^2\int_{-2}^3 x^2\ dx+6y\int_{-2}^3 dx\right)dy $$
$$ = \int_{-2}^1 \left(6\frac{x^2}{2}\Bigg|_{-2}^3 -2y^2\frac{x^3}{3}\Bigg|_{-2}^3 +6yx\Bigg|_{-2}^3 \right)dy = \int_{-2}^1 \left(3(9-4) -\frac{2}{3}y^2(27+8) +6y(3+2) \right)dy $$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/686561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Maclaurin series of $\frac{1}{1+\sin x}$ Find the terms through degree four of the Maclaurin series of $f(x)$.
$$f(x) = \frac{1}{1+\sin x}$$
My work:
The Maclaurin series for $\sin x$ up to degree $4$ is $x - \frac{x^3}{6} + \frac{x^5}{120}$
The Maclaurin series for $\frac{1}{1+x}$ up to degree $4$ is $1 - x + x^2 - x^... | You did it right and nothing has to be added to the answers you received.
I do not know if you were obliged to use these steps since there is a direct way of doing this expansion applying the basic rules, that is to say that the Taylor series of $f(x)$ built at $x=0$ just write (up to the fourth degree)
$$f(0)+x f'(0)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/692406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove, that two equations are equivalent EDIT: Missed something very important! Sorry! We have $x^4+1=2(2x-1)^{1/4}$ not $x^4+1=2\sqrt{2x-1}$. One friend of mine told me that the equation $x^4+1=2(2x-1)^{1/4}$, where $x\geq \frac{1}{2}$
is equivalent to $$x^4+1=2x$$. How did he obtain this? The equation $x^4+1=2(2x-1)^... | This is false. Suppose the equivalence as given: $x^4 + 1 = 2\sqrt{2x-1}$ exactly when $x^4 + 1 = 2x$ for $x \ge 1/2$. In particular, this means that if we assume $x^4 + 1 = 2x$, then
$$x^4 + 1 = 2\sqrt{2x-1} = 2\sqrt{(x^4 +1)-1} = 2x^2,$$
so $x^4 - 2x^2 + 1 = (x^2 -1)^2 = 0$ and $x = 1$.
But noting the polynomial $p(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/695004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Partial fractions Im working on some partial fraction calculus 2 homework problems and I am having some issues with one problem. The question reads:
Evaluate the integral $\displaystyle \int\dfrac{2-(5x^2+x)\;dx}{(x-1)(x+1)^2}$
I approached it by splitting it into the form
$$\dfrac{2-(5x^2+x)}{(x-1)(x+1)^2} = \dfrac ... | Your answer is very close except the constant multiplied by $\ln|x + 1|$. First, look at
$$\dfrac{2 - x - 5x^2}{(x - 1)(x + 1)^2} = \dfrac{A}{x - 1} + \dfrac{B}{x + 1} + \dfrac{C}{(x + 1)^2}$$
Multiply both sides by $(x - 1)(x + 1)^2$ to get
$$2 - x - 5x^2 = A(x + 1)^2 + B(x - 1)(x + 1) + C(x - 1)$$
If $x = 1$, then $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/696662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Different solutions of $x+y+z=10$ where $x$, $y$, $z$ are all positive integers and $x, y, z \leq 10$ The number of solutions to the equation $x+y+z=10$ where $x,y,z$ are positive integers, is given by ${k−1 \choose n−1}$, where in this case $k=10,n=3$, giving us ${9 \choose 2} = 36$
Now we have
$x + y + z = 10$ wit... | Following up on Trismegistos' idea of using generating functions: note that limiting the values to 10 really makes no difference, each variable can at most take the value 8 (two at 1, other one is 8). But if there is no limit, we can just write:
\begin{align}
[z^{10}] (z + z^2 + \cdots)^3
&= [z^{10}] z^3 (1 + z + z^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/700216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 6
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Number theory: Odd number and powers of 2 Is is true that for any odd natural number $x > 2$, there exists a positive natural number $y$, such that $x^y = 2^n+1$ or $x^y=2^n-1$ where $n$ is a also natural $> 0$. This cannot be solved by simple group theory methods, since we demand that $x^y$ be exactly $2^n+1$ or $2^n-... | This is to show without the Catalan conjecture/theorem that $11^y\pm 1$ cannot be a power of $2$. Begin with the polynomial $p=x^2-x-1$ whose value at $x=4$ is $11.$ Note that the last two terms in the expansion of $p^y$ are $yx+1$ if $y$ is even, and $-yx-1$ if $y$ is odd. This means that if $y$ is even then $p^y+1$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/702901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Functional equation $xf(y)+yf(x)=f(x+y)^2-f\left(x^2\right)-f\left(y^2\right)$ Here is a nice problem:
Let $f:\mathbb R\to\mathbb R$ be a function, $\mathbb R$ is the set of real numbers, satisfying the following properties: $f(1)$ is an integer and
$$xf(y)+yf(x)=f(x+y)^2-f\left(x^2\right)-f\left(y^2\right)\text,$$
fo... | You can show that the only functions $ f : \mathbb R \to \mathbb R $ satisfying
$$ x f ( y ) + y f ( x ) = f ( x + y ) ^ 2 - f \left( x ^ 2 \right) - f \left( y ^ 2 \right) \tag 0 \label 0 $$
are the constant zero function and the identity function. It's easy to check that those in fact are solutions. We try to prove t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/703309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is this fraction even possible to put into partial fractions? I'm to integrate $\int\frac{x}{x^2+6x+13}dx$
But I'm finding it impossible to do anything with it.
Since $x^2+6x+13$ is an irreducible quadratic factor, and the only factor it means that the partial fraction should be in the form...
$\frac{Ax+B}{x^2+6x+13}$ ... | Actually, it is a simple partial fraction and your mistake is
$$x = (Ax+B)\color{red}{(x^2+6x+13)}$$ the red part should be deleted and so
$$x = (Ax+B)\implies A=1,B=0$$
For the integration, you have
$$\frac{1}{2}\int\frac{(2x+6)-6}{x^2+6x+13}dx\\\frac{1}{2}\int\frac{(2x+6)}{x^2+6x+13}-\frac{6}{(x+3)^2+4}dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/705326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Prove that $\frac{x^2+1}{x^2(1-x)}>8$ for $x\in(0,1)$
Show that $\displaystyle \frac{x^2+1}{x^2(1-x)}>8$ for $x\in(0,1)$
I thought about derivative but I think it's too complicated, do you have any ideas?
Progress
*
*instead show $x^2+1 > 8x^2(1-x) = 8x^2 - 8x^3$. -- k.stm
*I have $(2x+1)(4x^2-2x+1)>7x^2$ am I on ... | We have, $\displaystyle \frac{x^2+1}{x^2(1-x)} = \displaystyle \frac{x+\frac{1}{x}}{x(1-x)}= \displaystyle \frac{2 + (\sqrt x - \frac{1}{\sqrt x})^2}{\frac{1}{4} - (x-\frac{1}{2})^2} > 8$.
(inequality is strict since $x\in(0,1)$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/706697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Question from Spivak's Calculus. Hint makes no sense. I'm currently working through Spivak. This question has me a little bit tied up in knots. I was able to answer it, but I can't make heads or tails of the hint.
Here's the question:
Find out when $(x+y)^5=x^5+y^5$. Hint: from the assumption $(x+y)^5=x^5+y^5$ you shou... | $$(x+y)^3 = x^3+3x^2y+3xy^2+y^3.$$
Suppose $x^3 + 2x^2y+2xy^2+y^3 = 0$. Add to both sides $x^2y+xy^2$:
$$x^3+2x^2y+2xy^2+y^3 + \color{red}{x^2y+xy^2} = \color{red}{x^2y+xy^2} \\
x^3+3x^2y+3xy^2+y^3 = x^2y+xy^2 \\
(x+y)^3 = xy(x+y).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/707114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that $\frac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\sin^2\theta$ The question is:
Prove that: $$\dfrac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\sin^2\theta$$
My proof is shown below. If anyone has an alternate proof please, please post it. Thanks!
| Proof:
$$\dfrac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\dfrac{\left(\dfrac{1}{\cos\theta}\cdot\sin\theta\right)}{\tan\theta+\cot\theta}$$
$$=\dfrac{\left(\dfrac{\sin\theta}{\cos\theta}\right)}{\left(\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}\right)}$$
$$=\dfrac{\left(\dfrac{\sin\theta}{\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/709134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Asymptotic behavior of $\sum_{k=1}^n \binom{n}{k} \left(\frac{ck}{n}\right)^k$ I am looking to show that
$$\lim_{n \rightarrow \infty}\frac{1}{e^n}\sum_{k=1}^n \binom{n}{k} \left(\frac{ck}{n}\right)^k = 0. $$
In my application, $c = (e+1)/2 \approx 1.85914\ldots$. I have been looking all over the place, but I can't se... | To estimate the sum , we first consider even $n$ and note the following:
*
*$c = \frac{(1+e)}{2} \approx 1.85914\dots < e$, and therefore, the last term of the sum for $k=n$ can be ignored as $\binom{n}{n}\left(\frac{c}{e}\right)^n \rightarrow 0$ for $n \rightarrow \infty$.
*$\binom{n}{k} = \binom{n}{n-k}$, and t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Linear Algebra question on diagonlization Please check my work My first question is that is a basis for each eigenspace the same thing as a corresponding eigenvector for an eigenspace?
Could someone tell me if im doing this correctly?
I have the matrix $A=\begin{bmatrix}
4 & 3 & 3\\
-9 & -8 & -9\\
3 & 3 & 4\\
\end{bmat... | The negative sign as no bearing on your answer because you're interested in the roots of the polynomial and $-(\lambda-1)^2(\lambda+2)=0\iff (\lambda-1)^2(\lambda+2)=0$.
You can check what you did yourself by computing $P^{-1}AP$.
In fact
$$\begin{align}P^{-1}AP &=\begin{bmatrix} -1& -1 &-1\\ -3 & -2 & -3\\ 1 & 1 & 2\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/711486",
"timestamp": "2023-03-29T00:00:00",
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How prove this inequality $\sum_{k=2}^{49}\frac{1}{k^2}\ge\frac{9}{10}\ln{2}$? show that
$$\sum_{k=2}^{49}\dfrac{1}{k^2}\ge\dfrac{9}{10}\ln{2}$$
my idea: since
$$\dfrac{1}{k^2}\ge\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$$
so
$$\sum_{k=2}^{49}\dfrac{1}{k^2}\ge\sum_{k=2}^{49}\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right)=... | $$\sum_{k=2}^{49} \dfrac{1}{k^2} = \sum_{k=1}^\infty \dfrac{1}{k^2} - 1 - \sum_{k=50}^\infty \dfrac{1}{k^2} = \dfrac{\pi^2}{6} - 1 - \sum_{k=50}^\infty \dfrac{1}{k^2}$$.
The function $f(x) = 1/x^2$ is strictly decreasing on $[49, +\infty)$ so the right hand Riemann sum $\sum_{k=50}^\infty \dfrac{1}{k^2}$ is less than $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/712132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\cdots$ converges. Consider the series:
$1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\cdots$
Is it convergent?
I believe I need to find a way to split the terms into additive and subtraction terms, however I'... | \begin{align}
\dots &= \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots \right) - \left(\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \dots \right) \\
&= \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots \right)- \frac{1}{2} \, \left(1 - \frac{1}{2} + \frac{1}{3} - \dots \right) \\
&= \tan^{-1}(1) - \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/712413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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There is 10 people that pick-up random number between 1 to 20 There is 10 people that pick-up random number between $1$ to $20$. More then one person can pick up same number (i.e. the pick-ups are independent).
What is the probability that the minimum number of all the people is grater then $8$?
My answer is $\frac{12^... | Your answer $\dfrac{12^{10}}{20^{10}}$ to the first question is right: all $10$ people must pick numbers from among the $12$ numbers greater than $8$, the probability of which is $\left(\dfrac{12}{20}\right)^{10}$.
For the second question (minimum exactly $8$), your second answer $\left(\dfrac{13}{20}\right)^{10} - \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/712963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How would you determine the transformation matrix? Suppose there exists a linear transformation $T$ where $T: \mathbb{R^3} \to \mathbb{R^5}$ and $T(\textbf{x}) = \text{A} \textbf{x}$. Given
$$ \text{A} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1\\ 2\\ 1 \\3 \end{pmatrix} ,
\quad
\text{A} \begin{... | Let $v_1=(1,1,1)^T, v_2=(1,2,3)^T$ and $v_3=(0,1,1)^T$ then we verify easily that $B=(v_1,v_2,v_3)$ is a basis of $\Bbb R^3$. Let $B_c$ the canonical basis and $P$ the change matrix from $B$ to $B_c$. Now if $\textbf{x}$ is a vector of $\Bbb R^3$ such that $X=(x,y,z)^T$ is its coordinates in $B_c$ then $PX=(\alpha,\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/713329",
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Find constants $a$ and $b$ such that $ \lim_{x \to 0} \frac{1}{bx-\sin x} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}}=1$
Find constants $a$ and $b$ such that $$ \lim_{x \to 0} \frac{1}{bx-\sin x} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}}=1$$
First,$a$ should be positive to make sure the limit is meaningful as $x \to 0^-$ .
Then I ch... | Using the L'Hôpital's rule we have
$$ \lim_{x \to 0} \frac{1}{bx-\sin x} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}}=\lim_{x \to 0}\frac{x^2}{(b-\cos x) \sqrt{a+x}}=\lim_{x \to 0} \frac{1}{\frac{(b-\cos x)}{x^2} \sqrt{a+x}}=1\\\iff (b=1)\land\left(\frac{\sqrt a}{2}=1\right)\iff (b=1)\land(a=4)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/715835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Safe prime mod 24 Given a safe-prime $p = 2q + 1$ where $q$ is also a prime and $p \gt 7$, I've read in a crypto.se answer that either $p \equiv 11 \pmod {24}$ or $p \equiv 23 \pmod {24}$.
I understand the proofs of why $p^2 \equiv 1 \pmod {24}$, and $p \equiv 1 \pmod 6$ or $p \equiv 5 \pmod 6$ for any prime $p$, and I... | A prime must be $\{1,5\} \text{ mod } 6$, so $\{1,5,7,11\} \text{ mod } 12$, so $\{1,5,7,11,13,17,19,23\} \text{ mod } 24$. This is true for $q$. Double each of these and add 1:
$1 \rightarrow 3 \not \in \{1,5,7,11,13,17,19,23\}$
$5 \rightarrow 11 \in \{1,5,7,11,13,17,19,23\}$
$7 \rightarrow 15 \not \in \{1,5,7,11,13... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/718097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the limit $\lim\limits_{x\to0+}\left(\frac{3^x+5^x}{2}\right)^{\frac1x}$ Evaluate
$$
\displaystyle\lim_{x\to0+}\left(\frac{3^x+5^x}{2}\right)^{\displaystyle\frac{1}{x}}
$$
And actually I have my answer and just need someone to verify this for me since I haven't done something like this for a long time.
First... | If the limit $L=\lim_{n\rightarrow a} f(x)^{g(x)}$,
is of the form $1^{\infty}$ maning $\lim_{x\rightarrow a}f(x)=1, \lim_{x\rightarrow a} g(x) = \infty.$ Then $L=\exp[\lim_{x\rightarrow a}(f(x)-1)g(x)]$. So in this case
$$L= \exp \left [\lim_{x\rightarrow 0} \frac{1}{x} \left( \frac{3^x + 5^x}{2}-1 \right)\right]=
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/720185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the determinant without row expansion Show that the determinant of the matrix
\begin{bmatrix} 1& a& a^3\\
1& b& b^3\\
1& c& c^3\end{bmatrix}
is $(a-b)(b-c)(c-a)(a+b+c)$ without expanding.
I was able to get out $(a-b)(b-c)(c-a)$ but couldn't complete.
| \begin{bmatrix} 1& a& a^2&a^3\\
1& b& b^2 &b^3\\
1& c& c^2 &c^3\\
1& X& X^2 &X^3
\end{bmatrix}is the well known Vandermonde determinant. When expend with respect to the last line, this is a polynomial
whose $X^2$ coefficient is the opposite of the result.
Now, expending
$$
(c-a)(b-a)(c-b)(X-c)(X-b)(X-a) \\= (c-a)(b-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/720589",
"timestamp": "2023-03-29T00:00:00",
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Does there exist bivariate polynomials $p$ and $q$ such that $p(x,y)^2 = q(x, y)^2 ( x^2 + y^2)$? Does there exist bivariate polynomials $p$ and $q$ such that $p(x,y)^2 = q(x, y)^2 ( x^2 + y^2)$ for all real $x$ and $y$?
| The answer is no.
If $p(x,y)^2 = q(x, y)^2 ( x^2 + y^2)$ for all real $x$ and $y$, then $p(X,Y)^2 = q(X,Y)^2 (X^2 + Y^2)$ in $\mathbb R[X,Y]$.
Since $X^2+Y^2$ is irreducible (hence prime) in $\mathbb R[X,Y]$ we get $p(X,Y)=(X^2+Y^2)p_1(X,Y)$, and therefore $(X^2+Y^2)p_1(X,Y)^2=q(X,Y)^2$. Using the same argument we hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/720773",
"timestamp": "2023-03-29T00:00:00",
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Given six numbers $x,y,z,a,b,c$ which satisfy the following relations. Express $x+y+z$ in terms of $a,b,c$ Given six numbers $x,y,z,a,b,c$ which satisfy the following relations
$y^2+yz+z^2=a^2$
$z^2+zx+x^2=b^2$
$x^2+xy+y^2=c^2$
Express $x+y+z$ in terms of $a,b,c$
My attempt:
$\dfrac{(y^3-z^3)}{(y-z)}=a^2,\dfrac{(z^3-... | I'm using $p$, $q$, $r$ for $x$, $y$, $z$, because GeoGebra doesn't like the latter for labels. Also, I'm assuming $p$, $q$, $r$ non-negative. (I believe that minor tweaks to the argument would allow for negative values.)
Arranging edges of length $p$, $q$, $r$ on symmetric rays about point $O$ gives $\triangle ABC$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/721928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int\frac{\sqrt {25 - x^2}}{ x^4}$ I'm pretty sure the method used is trig substitution. But I'm having trouble setting up and solving the problem.
| Subsititute $\theta=\sin^{-1} \dfrac{x}5\implies \dfrac{d\theta}{dx}=\dfrac{1}{\sqrt{25-x^2}}$, so we get,
\begin{align}
\\\\\\&\int\frac{\sqrt{25-x^2}}{x^4}dx
\\=&\int\frac{25-x^2}{x^4}\dfrac{1}{\sqrt{25-x^2}}d\theta
\\=&\int \dfrac{25-25\sin^2 \theta}{625\sin^4\theta}d\theta
\\=&\dfrac{1}{25}\int\dfrac{\cos^2 \theta}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/722692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove $\frac{\sin(a)}{\sin(b)}<\frac{a}{b}<\frac{\tan(a)}{\tan(b)}$ for $0not sure how to approach the following $\frac{\sin(a)}{\sin(b)}<\frac{a}{b}<\frac{\tan(a)}{\tan(b)}$ for $0<b<a<\pi/2$. Hints would be appreciated!
| Note that the function $\frac{\sin x}x$ is decreasing:
$$\left(\frac{\sin x}x\right)'=\frac{x\cos x-\sin x}{x^2}=\frac{\cos x}{x^2}(x-\tan x)<0,$$
because $\cos x>0$ and $\tan x>x$ on $(0,\frac\pi2)$.
So for $\frac\pi2>a>b>0$ we have $\frac{\sin a}a<\frac{\sin b}b\Longrightarrow\frac{\sin a}{\sin b}<\frac ab$.
Addition... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/724012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find a value $c$ such that $\left\|\begin{pmatrix} x^2 - y^2\\2xy \end{pmatrix}\right\| \leq |c|\left\|\begin{pmatrix} x\\y \end{pmatrix}\right\|$ This is part of a larger problem where I am trying to find the derivative of a vector valued function. I feel like I'm missing something simple. NOTE: $c$ can be a function ... | We compute
$$(x^2 - y^2)^2 + (2xy)^2 = x^{4} - 2x^{2}y^{2}+ y^{4} + 4x^{2}y^{2} = x^{4} + 2x^2 y^2 + y^4 = (x^2 + y^2)^2$$
Hence
$$\sqrt{(x^2 - y^2)^2 + (2xy)^2} = \sqrt{(x^2 + y^2)^2} = x^2 + y^2$$
We are trying to find $c$ for which
$$ (x^2 + y^2) \leq c \sqrt{x^2 + y^2}$$
If we let $z = x^2 + y^2$ we obtain
$$ z \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/724778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Factorising a complex polynomial over $\mathbb{C}$ I'm given $f(z)=z^6-1$ to factorise over $\mathbb{C}$.
My working is as follows up to the point I don't understand:
$f(-1)=0$ and $f(1)=0$
So $(z+1)$ and $(z-1)$ are factors
$(z+1)(z-1)=z^2-1$
$(z^2-1)(z^4+pz^3+qz^2+rz+s)=z^6-1$
$z^6+pz^5+qz^4-z^4+rz^3-pz^3+... | You're answer is right; what you're missing is that you can actually simplify those square roots by actually computing their real and imaginary parts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/725943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the number of series Find the number of series $(a_1,..., a_{2n})$ that have terms from ${\{0,...9\}}$ so that:
$$
11|\sum_{i=1}^{n}a_i-\sum_{i=n+1}^{2n}a_i
$$
(this is not a homework)
There is a similar problem (Problem 6) in the IMO 1995.
| For a fixed $n$, let $ A_r $ be the number of sequences of $n$ terms, whose sum is $r \pmod{11}$. Then, we want the value of $ \sum A_r ^2 $.
As pointed out by ABC, the generating function $f(x) = \left ( 1 + x + x^2 + \ldots + x^9 \right) ^n $ gives us the number of sequences of $n$ terms, with a total sum of $r$. We ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/726157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the area of the surface obtained by rotating the curve of parametric equations Rotate about the $x$ axis
$x = 2t-2/3t^3$
$y = 2t^2$
$0 \leq t \leq 1$
I did the integral of $\sqrt{(2-2t^2)^2+(4t)^2}$ and got $(2x(x^2+3))/3$ and then I did the integral of $2\pi 2t^2 ((2x(x^2+3))/3)$ and I get $22\pi/9$ but apparentl... | The surface area integral for a figure revolved about the $ \ x-$ axis is
$$ S \ = \ 2 \pi \ \int \ y \ \ ds \ , $$
which, for a parametric curve, will be
$$ S \ = \ 2 \pi \ \int \ y(t) \ \sqrt{ \ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dx}{dt} \right)^2} \ \ dt \ \ = \ \ 2 \pi \ \int \ y(t) \ \sqrt{ \ \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/728617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Group-Isomorphism problem I want to find an group-isomorphism $$ \psi : (\mathbb{Z}/8\mathbb{Z},+) \longrightarrow \mathbb{F}_9^\times $$
which should be used to multiply elements in $\mathbb{F}_9$ or to find the inverse element in an easy way.
| As Seth and Pedro indicated the existence of such an isomorphism follows from (and is equivalent to) the cyclicity of the multiplicative group $\Bbb{F}_9^*$. To exhibit an explicit isomorphism you need to specify a construction of $\Bbb{F}_9$ and find a generator of the multiplicative group (aka a primitive element).
E... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/730809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
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Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.