Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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how to integrate $\int\frac{1}{x^2-12x+35}dx$? How to integrate following
$$\int\frac{1}{x^2-12x+35}dx?$$
What I did is here:
$$\int\frac{dx}{x^2-12x+35}=\int\frac{dx}{(x-6)^2-1}$$
substitute $x-6=t$, $dx=dt$
$$=\int\frac{dt}{t^2-1}$$
partial fraction decomposition,
$$=\int{1\over 2}\left(\frac{1}{t-1}-\frac{1}{t+1}\ri... | Can you skip the substitution? Certainly!
$\frac {1}{x^2 - 12 x + 35} = \frac {1}{(x-7)(x-5)} = \frac {A}{x-7} + \frac {B}{x-5}$
$A+B = 0\\
-5A - 7B = 1\\
A = \frac {1}{2}, B = -\frac {1}{2}$
$12\int \frac {1}{x-7} - \frac {1}{x-5} \ dx\\
\frac 12 (\ln |x-7| - \ln |x-5|)\\
\frac 12 \ln \left|\frac {x-7}{x-5}\right|$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\lim\limits_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right)$ Evaluate $$\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right)$$
My attempt: $$\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right) = \lim_{x \to 1} \frac{x+2}{x^2+x+1}=1$$
According to the answer key, this limit does not exist... | Using the identity
$A-B=\dfrac{1}{\dfrac{1}{A}}-\dfrac{1}{\dfrac{1}{B}}$ we get $$\dfrac{1}{x-1}-\dfrac{3}{1-x^3}=\dfrac{2-x^3-x}{-x^4+x^3+x-1}$$ so you go from the form $\infty-\infty$ to the form $\dfrac00$ and you can apply L'Hôspital so you have $$\lim_{x \to 1} \frac{1}{x-1}-\frac{3}{1-x^3}=\lim\dfrac{-3x^2-1}{-4x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Let f(x) be a polynomial satisfying $\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}=3,$ $f(2)=5 $ and $f(3)=10$, $f(-1)=2$ and $f(-6)=37$ Let f(x) be a polynomial satisfying $\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}=3,$ $f(2)=5 $ and $f(3)=10$, $f(-1)=2$ and $f(-6)=37$ .
And then there are some question related to this.
In ... | the correct and precise reason for writing f(x)=3(x−2)(x−3)(x+1)(x+6)+(x^2+1) is as follows (I am assuming you have understood why f(x) will be a 4th deg polynomial and how its leading coeff shall be 3 from other answers):
consider g(x)= f(x)-(x^2+1)
clearly, deg(g(x))= 4
now, g(2)= f(2)-(2^2+1)= 0
similarly, g(3)=g(-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
There are only two six-digit integers $N$, each greater than $100,000$. for which $N^2$ has $N$ as its final six digits There are only two six-digit integers $N$, each greater than 100,000 for which $N^2$ has $N$ as its final six digits (or $N^2-N$ is divisible by $10^6$). What are these two numbers?
Is the problem sol... | Yes, we can solve $N^2-N\equiv0\bmod10^6$ with the Chinese remainder theorem
to get $N\equiv0 $ or $1\bmod 2^6=64$ and $N\equiv0$ or $ 1\bmod5^6=15625$.
We don't want the solutions $N\equiv0\bmod2^6$ and $5^6$ or $N\equiv1\bmod2^6 $ and $5^6$.
We want the solutions $N\equiv1\bmod 2^6$ and $N\equiv0\bmod5^6$ and vice ve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Another way to solve $\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$ without the substitution $y=\tan\left(\frac{x}{2}\right)$? Is there another way to solve an integral $$\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$$ without the substitution $y=\tan\left(\frac{x}{2}\right)$?
$\large \int \frac{\sin^3(x)}{1+\cos^2(x)}\ dx$ is easi... | HINT:
$$\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx=\int \frac{(1-\cos^2(x))^2}{1+\cos^2(x)}\ dx$$
$$=\int \frac{(1+\cos^2(x))^2-4(1+\cos^2(x))+4}{1+\cos^2(x)}\ dx$$
$$=\int \left(1+\cos^2(x)-4+\frac{4}{1+\cos^2(x)}\right)\ dx$$
$$=\int \left(\frac{1+\cos(2x)}{2}-3+\frac{4\sec^2x}{\tan^2(x)+2}\right)\ dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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Solving a multiple variable system of congruences using the Chinese Remainder theorem I've tried to solve the following system:
$2x+7y\equiv17\mod35\\3x+8y\equiv18\mod35$
My idea was using the Chinese Remainder theorem, so firstly, I've found that
$3\cdot 5 - 2\cdot 7 = 1$
And that $15$ is $1\mod7$, $\space -14\equiv 1... | Subtracting the first congruence from the second gives
$$ x+y \equiv 1\pmod{35}. $$
Using this in the first congruence gives
$$ 17 \equiv 2(x+y) + 5y \equiv 2+5y \pmod{35}, $$
so that $35 \mid 5(y-3)$, or $7 \mid (y-3)$. From $y \equiv 3\pmod{7}$ and $x+y \equiv 1\pmod{7}$ we get $x \equiv 5\pmod{7}$.
The two congruenc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $\alpha,\beta,\gamma$ are the roots of $x^3+x+1=0$, then find the equation whose roots are: $(\alpha-\beta)^2,(\beta-\gamma)^2,(\gamma-\alpha)^2$ Question:
If $\alpha,\beta,\gamma$ are the roots of the equation, $x^3+x+1=0$, then find the equation whose roots are: $({\alpha}-{\beta})^2,({\beta}-{\gamma})^2,({\gamma... | The standard method:
$$a+b+c=0; ab+bc+ca=1; abc=-1;\\
a^2+b^2+c^2=-2;a^2b^2+b^2c^2+c^2a^2=1;a^4+b^4+c^4=2;\\
a^3=-a-1.$$
First coefficient:
$$(a-b)^2+(b-c)^2+(c-a)^2=\\2(a^2+b^2+c^2)-2(ab+bc+ca)=-6$$
Second coefficient:
$$(a-b)^2(b-c)^2+(b-c)^2(c-a)^2+(c-a)^2(a-b)^2=\\
a^4+b^4+c^4+3(a^2b^2+b^2c^2+c^2a^2)-\\
2(a^3(\unde... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Given $\cos(a) +\cos(b) = 1$, prove that $1 - s^2 - t^2 - 3s^2t^2 = 0$, where $s = \tan(a/2)$ and $t = \tan(b/2)$ Given $\cos(a) + \cos(b) = 1$, prove that $1 - s^2 - t^2 - 3s^2t^2 = 0$, where $s = \tan(a/2)$ and $t = \tan(b/2)$.
I have tried using the identity $\cos(a) = \frac{1-t^2}{1+t^2}$. but manipulating this s... | Alternatively:
$$\cos a=2\cos^2 \frac a2-1; \\
\cos^2 \frac a2=\frac 1{1+\tan^2 \frac a2}=\frac1{1+s^2}\\
\cos a+\cos b=2\cos ^2 \frac a2+2\cos^2 \frac b2-2=\\
\frac2{1+s^2}+\frac2{1+t^2}-2=1.$$
I leave to you to simplify it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How can I show that $A=\begin{pmatrix}a&b\\b&d\end{pmatrix}$ with $b\neq 0$ is diagonalizable?
How can I show that $A=\begin{pmatrix}a&b\\b&d\end{pmatrix}$ with
$b\neq 0$ is diagonalizable?
Sei $b\neq 0$, dann gilt:$$\det(A-\lambda E)=\det \begin{pmatrix} a-\lambda & b \\ b & d-\lambda \end{pmatrix}=(a-\lambda)(d-\la... | *
*The characteristic equation is $(a-\lambda)(d-\lambda)-b^2=\lambda^2-(a+d)\lambda+ad-b^2$;
*The roots a distinct because the discriminant $(a+d)^2-4(ad-b^2)=(a-d)^2+4b^2$ is certainly non-zero.
*$v=(b,d-\lambda)$ solves the system $Av=\lambda v$, for both $\lambda$s (check by direct subtitution).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find range of $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$ Find range of $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$
My attempt :
\begin{align*}
f(x)&=\dfrac{5}{9\cos^2x-6\sin x\cos x+\sin^2x-6\cos^2x}\\
&= \dfrac{5}{(3\cos x+\sin x)^2-6\cos^2x}
\end{align*}
The problem is if I'm going to use
$$-1\leqslant\sin x... | $$f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2 x}=\frac{10}{2\sin^2 x-12 \sin x \cos x+6 \cos^2 x}.$$
$$f(x)=\frac{10}{1-\cos 2x-6 \sin 2x +3(1+\cos 2x)}=\frac{10}{4+2\cos 2x-6 \sin 2x}$$
$$\implies f(x)=\frac{5}{2+\cos 2x-3 \sin 2x}=\frac{5}{2+\sqrt{10}(\cos 2x-3\sin 2x)/\sqrt{10}}$$
$$\implies f(x)=\frac{5}{2+\sqrt{10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A locus problem related to circumcenters and conic sections Given a point $A$, a circle $O$ and conic section $e$, if $BC$ is a moving chord of the circle $O$ tangent to $e$, then prove that
the locus of △$ABC$'s circumcenters $T$ is a conic section.
The question was posted in 纯几何吧 by TelvCohl and remained unsolved for... | Here's something of a brute-force vector proof.
In the figure, $P$ is a point on our given conic, $\bigcirc K$ of radius $r$ is our given circle, and $A$ is our given point. The tangent line at $P$ meets $\bigcirc K$ at $R$ and $R'$, and the circumcenter of $\triangle ARR'$ is $Q$. Points $A'$ and $K'$ are the respecti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Optimizing a quadratic in one variable with parameterized coefficients Recently I was doing a physics problem and I ended up with this quadratic in the middle of the steps:
$$ 0= X \tan \theta - \frac{g}{2} \frac{ X^2 \sec^2 \theta }{ (110)^2 } - 105$$
I want to find $0 < \theta < \frac{\pi}2$ for which I can later ta... | The trick is to write a quadratic in terms of $ \tan \theta $ and not in terms of $X$
$$ 0 = X \tan \theta -\frac{g}{2} \frac{ X^2 (1 +\tan^2 \theta)}{(110)^2} - 105$$
applying the condition that tangent $ D>0$ for the quadratic in terms of $ \tan \theta$(taking g as 10),
I get, this
$$ 1> \frac{20}{(110)^2} ( 105 + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Does a closed solution of this integral exist I saw a problem on Facebook with this integral except it was a definite integral from -2 to +2 and the answer was 0 since the function was odd. I am wondering if a closed solution exists or if this can even be integrated, its not an obvious integral as trig. Substitution wo... | Let $x=2\sin u$ ,
Then $dx=2\cos u~du$
$\therefore\int x^3\cos\dfrac{x}{2}\sqrt{4-x^2}~dx$
$=16\int\sin^3u\cos^2u\cos\sin u~du$
$=16\int\sin^3u(1-\sin^2u)\cos\sin u~du$
$=16\int\sin^3u\cos\sin u~du-16\int\sin^5u\cos\sin u~du$
$=16\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+3}u}{(2n)!}~du-16\int\sum\limits_{n=0}^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Prime dividing $4n^2+1$ congruent to $1\pmod{4}$ Prove that for any integer $n$, any prime $p$ such that $p\mid 4n^2+1$ is congruent to $1\pmod{4}$.
I am using this as a step to prove that there are infinitely many primes that are congruent to $1\pmod{4}$, but I want to prove this using this method.
| Suppose that $p$ is a prime dividing $4n^2+1$.
Then, if we define $x=2n$:
$x^2 \equiv -1 \pmod{p}$
$\left(x^2\right)^{\frac{p-1}{2}} \equiv (-1)^{\frac{p-1}{2}} \pmod{p} $
$x^{p-1} \equiv (-1)^{\frac{p-1}{2}} \pmod{p}$
$(-1)^{\frac{p-1}{2}} \equiv 1 \pmod{p}\ $ by Fermat's theorem
$(-1)^{\frac{p-1}{2}} = 1$
And, s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\tan^{-1}\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\frac{\pi}{4}+\frac 12 \cos^{-1}x^2$ Let the above expression be equal to $\phi$
$$\frac{\tan \phi +1}{\tan \phi-1}=\sqrt{\frac{1+x^2}{1-x^2}}$$
$$\frac{1+\tan^2\phi +2\tan \phi}{1+\tan^2 \phi-2\tan \phi}=\frac{1+x^2}{1-x^2}$$
$$\frac{1+\t... | Your mistake
$\sin y=a\stackrel{\text{to}}{\longrightarrow}\sin^{-1}(\sin y)=\begin{cases}2n\pi+y&y\in\text{I, IV quadrant}\\(2n-1)\pi-y&y\in\text{II, III quadrant}\end{cases}=\sin^{-1}a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
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Exercises 6 and 7 in 1.1 (page 4) of Topology and Groupoids, by Ronald Brown I am not sure which keywords to use to search for previously asked questions dealing with these problems. I did search for "Topology and Groupoids" but didn't find anything relevant.
Problems:
*Let $C$ be a neighborhood of $c \in \mathbb{R}... | Let $ n$ enough great such that
$$0<a-\frac 1n<x<a+\frac 1n$$
$$0<b-\frac 1n<y<b+\frac 1n$$
then
$$0<ab -\frac 1n(a+b)+\frac{1}{n^2}<xy<ab+\frac 1n(a+b)+\frac{1}{n^2}$$
and
$$|ab-xy|<\frac{|a+b|}{n} +\frac{1}{n^2}$$
Since $$\lim_{n\to+\infty}\frac{|a+b|}{n}+\frac{1}{n^2}=0$$
we can choose $ n$ such that
$$\frac{|a+b|}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761635",
"timestamp": "2023-03-29T00:00:00",
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How to evaluate $\int x^5 (1+x^2)^{\frac{2}{3}}\ dx$? I am trying to evaluate $\int x^5 (1+x^2)^{\frac{2}{3}} dx$
This is apparently a binomial integral of the form $\int x^m (a+bx^k)^ndx$. Therefore, we can use Euler's substitutions in order to evaluate it. Since $\dfrac{m+1}{k} = \dfrac{5+1}{2} = 3 \in \mathbb{Z}$ w... | Alternatively, denote $x^2=t\Rightarrow 2xdx=dt$:
$$\int x^5 (1+x^2)^{\frac{2}{3}}dx=\frac12\int t^2 (1+t)^{\frac{2}{3}}dt=\\
\frac12\int (t^2-1+1)(1+t)^{\frac23}dt=\int(t-1)(1+t)^\frac53dt+\int(1+t)^\frac23dt=\\
\frac12\int(t+1-2)(1+t)^{\frac53}dt+\frac12\int(1+t)^{\frac23}dt=\\
\frac12\int(1+t)^{\frac83}dt-\int (1+t)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Check and comment my proof of $a+b \geq 2 \sqrt{ab}$ I want to prove
$$
a+b \geq 2 \sqrt{ab}, \quad \text{for} \quad a, b>0 \tag 1
$$
First question:
Isn't this wrong? Shouldn't it be
$$
a+b \geq 2 \sqrt{ab}, \quad \text{for} \quad a, b\geq0 \tag 2
$$
Or
$$
a+b > 2 \sqrt{ab}, \quad \text{for} \quad a, b>0 \tag 3
$$
... | No, it's not wrong. It is simply not as strong as it could be. Your (2) is also right, but that doesn't make (1) wrong.
However, your (3) is wrong: if $a=b$, then $a+b=2\sqrt{ab}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to evaluate $\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx$? I am trying to evaluate
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx \quad (1)$$
The typical way to confront this kind of integrals are the conjugates i.e:
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx = $$... | Substitute $x$ with $\cos{2\theta}$.
The above integral will simplify to:
$-2\int \frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} \sin{2\theta} \mathrm d\theta $.
Then you can use integration by parts.
Note that: $-\mathrm d(\cos\theta - \sin\theta) = (\cos\theta + \sin\theta) \mathrm d\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Show that $\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N\frac{1}{N+n}=\int\limits_1^2 \frac{dx}{x}=\ln(2)$ Show that:
$$\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N\frac{1}{N+n}=\int\limits_1^2 \frac{dx}{x}=\ln(2)$$
My attempt:
We build a Riemann sum with:
$1=x_0<x_1<...<x_{N-1}<x_N=2$
$x_n:=\frac{n}{N}+1... | Without Rieman sums.
$$S_N=\sum\limits_{n=1}^N\frac{1}{N+n}=H_{2 N}-H_N$$ Using the asymptotics of harmonic numbers
$$S_N=\log (2)-\frac{1}{4 N}+\frac{1}{16 N^2}+O\left(\frac{1}{N^4}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3766146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
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Find the area between $f(x) = x^2+3x+7 $ and $g(x) = xe^{x^3+4}$ for $x \in [3,5]$.
Calculate the area between the two functions, $f(x)$, $g(x)$, for $x \in [3,5]$.
$$f(x)=x^2+3x+7$$
$$g(x)=xe^{x^3+4}$$
To determine the area between the functions I used the formula $A= \int_a^b|f(x)-g(x)|dx$. Therefore, I have:
\begi... | $$\int{x\,e^{x^3+4}}\,dx=-\frac{e^4 x^2 \Gamma \left(\frac{2}{3},-x^3\right)}{3
\left(-x^3\right)^{2/3}}=-\frac{1}{3} e^4 x^2 E_{\frac{1}{3}}\left(-x^3\right)$$
$$\int_3 ^5 {x\,e^{x^3+4}}\,dx=\frac{e^4}{3} \left(9 E_{\frac{1}{3}}(-27)-25 E_{\frac{1}{3}}(-125)\right)$$ Since the arguments ar quite large, for an eval... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3767518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding the Last Eigenvalue for a Matrix $K$ is a $3 \times 3$ real symmetric matrix such that $K = K^3$. Furthermore, we are given that:
\begin{align*}
K(1, 1, 1) \ \ & = \ \ (0, 0, 0) \\
K(1, 2, -3) \ \ & = \ \ (1, 2, -3)
\end{align*}
So we know that $0, 1$ are two of the eigenvalues of $K$. What can I do to ascert... | Let $w = (1,1,1)\times (1,2,-3) = (-5,4,1)$ and consider
$$Kw = w, \quad Kw = 0, \quad Kw = -w.$$
In all three cases $K$ is symmetric, $K = K^3$ and we get that the last eigenvalue is $1,0$ and $-1$ respectively. The respective matrices of $K$ in basis $\{(1,1,1), (1,2,-3),(-5,4,1)\}$ are
$$\begin{bmatrix} 0 & 0 & 0 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that if $a$ and $b$ have the same sign then $|a + b| = |a| + |b| $, and if $a$ and $b$ have opposite signs then $|a+b| < |a| + |b|$ I'm considering the different cases for $a$ and $b$
Case 1) $ a\geq 0$ and $ b\geq 0$
Given both terms are positive, $ a + b \geq 0 $
$$
|a+b| = a + b = |a| + |b|\\
$$
Case 2) $ a< 0$... | Dealing with absolute values, the general strategy is: if you can avoid the cases method, do it.
Here , we have a typical example of this strategy: as both expressions are non-negative numbers, we compare them by comparing their squares, using that $|x|^2=x^2\,$:
$$|a+b|^2=a^2+2ab+b^2,\enspace \text{ and }\enspace (|a|... | {
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"answer_count": 4,
"answer_id": 1
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$f\left( x \right) = {x^3} + x$, then $\int\limits_1^2 {f\left( x \right)dx} + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx} $
If $f\left( x \right) = {x^3} + x$, then
$$\int\limits_1^2 {f\left( x \right)dx} + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx} $$
is________.
My approach is as follows:
$$g = {f^... | $$f(x)=x^3+x$$
Let $$K=\int_{1}^{2} f(x) dx+2\int_{1}^{5} f^{-1}(2x) dx=I+J$$
$$ I=\int_{1}^{2} (x^3+x) dx=\frac{21}{4}$$
$$J=2\int_{1}^{5} f^{-1}(2x)dx=\int_{2}^{10} f^{-1}(z) dz$$
Let $f^{-1}(z)=t \implies z=f(t) \implies dz=f'(t) dt$, then
$$J=\int_{1}^{2} t f'(t) dt=\int_{1}^{2}t(3t^2+1)dt=\frac{51}{4}.$$
Finally $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Given ellipse of axes $a$ and $b$, find axes of tangential and concentric ellipse at angle $t$ Let’s say I have an ellipse with horizontal axis $a$ and vertical axis $b$, centered at $(0,0)$.
I want to compute $a’$ and $b’$ of a smaller ellipse centered at $(0,0)$, with the axes rotated by some angle $t$, tangent to th... | Let the ratio $r= \frac{a’}a=\frac{b’}b$. Then, the inscribed ellipse with the tilt angle $t$ is
$$\frac{(x\cos t+y\sin t)^2}{r^2a^2}+ \frac{(-x\sin t+y\cos t)^2}{r^2b^2}=1\tag 1
$$
Also, the inscribed ellipse can be expressed as
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}-k\left(
\frac{x\sin \theta}{a}-\frac{y\cos \theta}{b}... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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EGMO 2014/P3 : Prove that there exist infinitely many positive integers $n$ such that $\omega(n) = k$ and $d(n)$ does not divide $d(a^2+b^2)$ We denote the number of positive divisors of a positive integer $m$ by $d(m)$ and the number of distinct prime divisors of $m$ by $\omega(m)$. Let $k$ be a positive integer. Prov... | I couldn't do this solution without @Raheel 's hint. It was all about $5 \pmod 6$ ! Also, I will be really grateful if someone proof reads it?Thanks in advance.
Case 1 : For $k=1$,we can take $n=2^{p-1}$, where p is an odd prime. Let's say for contradiction, there exist some $a$ and $b$ such that $a+b=n$ and $d(n)|d(... | {
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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If $a_n=100a_{n-1}+134$, find least value of n for which $a_n$ is divisible by $99$
Let $a_{1}=24$ and form the sequence $a_{n}, n \geq 2$ by $a_{n}=100 a_{n-1}+134 .$ The first few terms are
$$
24,2534,253534,25353534, \ldots
$$
What is the least value of $n$ for which $a_{n}$ is divisible by $99 ?$
We have to find.... | Claim 1: a number is divisible by $9$ if and only if the sum of its digits (written in base $10$) is $0\pmod 9$.
Proof: if $n=\sum_{k=0}^md_k10^k$, then since $10^k\equiv 1\pmod 9$, we have $n\equiv \sum_{k=0}^m d_k\pmod 9$. Thus $9$ divides $n$ if and only if $9$ divides $\sum_{k=0}^md_k$.
Claim 2: a number is divisib... | {
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"source": "stackexchange",
"question_score": "2",
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Proving a solution of a Bernoulli type equation Prove that
\begin{equation}
y(x) = \sqrt{\dfrac{3x}{2x + 3c}}
\end{equation}
is a solution of
\begin{equation}
\dfrac{dy}{dx} + \dfrac{y}{2x} = -\frac{y^3}{3x}
\end{equation}
All the math to resolve this differential equation is already done. The exercise simply asks to p... | Your work seems fine and we obtain
$$\dfrac{dy}{dx} =\dfrac{y}{2x} - \dfrac{y^3}{3x}$$
which should be the correct differential equation.
I don't understand why you have added the term $P(x)y$ in the last step.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a^2 + b^2 + c^2 = 1$, what is the the minimum value of $\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}$?
Suppose that $a^2 + b^2 + c^2 = 1$ for real positive numbers $a$, $b$, $c$. Find the minimum possible value of $\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}$.
So far I've got a minimum of $\sqrt {3}$. Can anyo... | We have
$$\left(\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}\right)^2 = \frac {a^2b^2}{c^2} + \frac {b^2c^2}{a^2} + \frac {c^2a^2}{b^2}+2(a^2+b^2+c^2).$$
Using the AM-GM inequality, we get
$$\frac {a^2b^2}{c^2} + \frac {b^2c^2}{a^2} + \frac {c^2a^2}{b^2} = \frac{1}{2} \sum \left(\frac {a^2b^2}{c^2} + \frac {b^2c^2}{a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Orthocenter, Circumcenter, and Circumradius In triangle $ABC,$ let $a = BC,$ $b = AC,$ and $c = AB$ be the sides of the triangle. Let $H$ be the orthocenter, and let $O$ and $R$ denote the circumcenter and circumradius, respectively. Express $HO^2$ in terms of $a,$ $b,$ $c,$ and $R.$
I know what orthocenter, circumcent... | In the standard notation we obtain: $$OH^2=\vec{OH}\cdot\vec{OH}=\left(\vec{OA}+\vec{OB}+\vec{OC}\right)^2=$$
$$=3R^2+2\sum_{cyc}\vec{OA}\cdot\vec{OB}=3R^2+2R^2\sum_{cyc}\cos\measuredangle AOB=$$
$$=3R^2+2R^2\sum_{cyc}\cos2\gamma=3R^2+2R^2\sum_{cyc}\left(2\left(\frac{a^2+b^2-c^2}{2ab}\right)^2-1\right)=$$
$$=R^2\sum_{c... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving that $\int_{-1}^{1} \frac{\{x^3\}(x^4+1)}{(x^6+1)} dx=\frac{\pi}{3}$, where $\{.\}$ is positive fractional part Here, $\{-3.4\}=0.6$.
The said integral can be solved using $\{z\}+\{-z\}=1$, if $z$ is a non-zero real number;
after using the property that $$\int_{-a}^{a} f(x) dx= \int_{0}^{a} [ f(x)+f(-x)] dx$$
... | $$I=\int\dfrac{x^4+1}{x^6+1}\,dx=\int\dfrac{x^4+1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\,dx=\int\left(\dfrac{x^2+1}{3\left(x^4-x^2+1\right)}+\dfrac{2}{3\left(x^2+1\right)}\right)dx$$
$$\int\dfrac{x^2+1}{x^4-x^2+1}\,dx=\int\dfrac{x^2+1}{\left(x^2-\sqrt{3}x+1\right)\left(x^2+\sqrt{3}x+1\right)}\,dx=\int\left(\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778395",
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"source": "stackexchange",
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How to prove that $\frac{\cos(x)-\cos(2x)}{\sin(x)+\sin(2x)} = \frac{1-\cos(x)}{\sin(x)}$ in a simpler way. EDIT: Preferably a LHS = RHS proof, where you work on one side only then yield the other side.
My way is as follows:
Prove: $\frac{\cos(x)-\cos(2x)}{\sin(x)+\sin(2x)} = \frac{1-\cos(x)}{\sin(x)}$
I use the fact... | Using compound angle formula, we have
$$
\begin{aligned}
\frac{\cos x-\cos 2 x}{\sin x+\sin 2 x} & =\frac{2 \sin \frac{3 x}{2} \sin \frac{x}{2}}{2 \sin \frac{3 x}{2} \cos \frac{x}{2}} \\
& =\frac{2 \sin ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \\
& =\frac{1-\cos x}{\sin x}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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Square equal to sum of three squares For which integers $n$ there exists integers $0\le a,b,c < n$ such that $n^2=a^2+b^2+c^2$?
I made the following observations:
*
*For $n=1$ and $n=0$ those integers doesn't exist.
*If $n$ is a power of 2 those integers doesn't exist. Let $n=2^m$ with $m>0$ the smallest power of 2... | If we want $a^2+b^2+c^2=n^2$, then $a^2+b^2=n^2-c^2=(n-c)(n+c)$.
So a way to generate solutions is to choose $a$ and $b$ and then try to find $n$ and $c$ that work.
Example: $a=10$, $b=11$. So $a^2+b^2=100+121=221$.
Now $221=13*17=(15-2)(15+2)$. So a solution should be $10^2+11^2+2^2=15^2$.
You could also write $221=2... | {
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"source": "stackexchange",
"question_score": "6",
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How to find $a$, $b$, $c$ such that $P(x)=ax^3+bx^2+cx$ and $P\left(x\right)-P\left(x-1\right)=x^2$ I'm trying to find $a$, $b$ and $c$ such that $P(x)=ax^3+bx^2+cx$ and $P\left(x\right)-P\left(x-1\right)=x^2$.
After expanding the binomial in $P(x-1)$, I end up getting
$3ax^2-3ax+2bx+a-b=x^2$. What next? Using $3a = 1$... | Alternatively. We see that $p(0)=0$.
If $x=1$ we get $$1 = p(1)-p(0) = a+b+c \implies \boxed{a+b+c=1}$$
If $x=2$ we get $$4 = p(2)-p(1) = 7a+3b+c \implies \boxed{7a+3b+c=4}$$
If $x=-1$ we get $$1 = p(0)-p(-1) = a-b+c \implies \boxed{a-b+c=1}$$
Now solve this system...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the determinant of a $5\times 5$ matrix Let
$$A = \left[\begin{array}{rrrrr}6 & 2 & 2 & 2 & 2 \\ 2 & 6 & 2 & 2 & 2 \\ 2 & 2 & 6 & 2 & 2 \\2 & 2 & 2 & 6 & 2 \\ 2 & 2 & 2 & 2 & 6\end{array}\right] \in {M}_{5}(\mathbb{R})$$
Which of following options is $\det(A)$ ?
*
*$4^4 \times 14$
*$4^3 \times 14$
*$4^2 \... | $$\begin{vmatrix}6 & 2 & 2 & 2 & 2 \\ 2 & 6 & 2 & 2 & 2 \\ 2 & 2 & 6 & 2 & 2 \\2 & 2 & 2 & 6 & 2 \\ 2 & 2 & 2 & 2 & 6\end{vmatrix}=\begin{vmatrix}4 & 0 & 0 & 0 & -4 \\ 0 & 4 & 0 & 0 & -4 \\ 0 & 0 & 4 & 0 & -4 \\0 & 0 & 0 & 4 & -4 \\ 2 & 2 & 2 & 2 &\ \ 6\end{vmatrix}=\begin{vmatrix}4 & 0 & 0 & 0 & -4 \\ 0 & 4 & 0 & 0 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A Problem on Inverse Trigonometric Functions and Differential Equation.
$f(x)=\left(\sin \left(\tan ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right)^{2}-1,\ |x|>1$
If $\displaystyle\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} x}\left(\sin ^{-1}(f(x))\right)$ and
$y(\sqrt{3})=\frac{\pi... | $$\dfrac{d}{dx}(\sin^{-1}f)= \dfrac{1}{\sqrt{1-f^2}}\cdot \dfrac{df}{dx}=P\cdot Q \;;$$
$$P=\dfrac{1}{\sqrt{1-(2x/(1+x^2))^2}} = \dfrac{1+x^2}{1-x^2} $$
Differentiating by Quotient rule
$$Q=\dfrac{2}{(1+x)^2}; \quad PQ=\dfrac{2}{1-x^2};\; $$
EDIT1:
Stopped too early by mistake. Continuing,
$$\dfrac{dy}{dx}= \dfrac{1}{... | {
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On proving $a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$. This problem was a "warm-up" problem by the author. Note: $a, b, c$ are non-negative numbers.
$$a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$$
I tried to remove the $2$ from ${b+c\over 2}$ and got this-
$$ 4(a^3+b^3+c^3-3abc) \geq (b+c-2a)^3 $$
... | Because $$a^3+b^3+c^3-3abc-2\left(\frac{b+c}{2}-a\right)^3$$
$$=\frac{3}{4}\Big[4a^3-4(b+c)a^2+2(b^2+c^2)a+b^3-b^2c-bc^2+c^3\Big]$$
$$=\frac{3}{4}\Big[a(2a-b-c)^2+a(b-c)^2+(b+c)(b-c)^2\Big]\geqslant 0.$$
The equality occurs for $$a(2a-b-c)^2=a(b-c)^2=(b+c)(b-c)^2=0.$$
*
*$a=0$.
Thus, $(b+c)(b-c)^2=0,$ which gives $b... | {
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"source": "stackexchange",
"question_score": "1",
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Given $\triangle ABC$ can we construct point $O$ such that $AO\times BC=BO\times AC=CO\times AB$? Given a $\triangle ABC$, is it possible to construct, with compass and straightedge, a point $O$ such that
$$AO\cdot BC=BO\cdot AC=CO\cdot AB$$
Does that point exist?
| These points are known
in ETC
as 1st and 2nd isodynamic points,
the triangle centers $X_{15}$ and $X_{16}$.
Isodynamic point:
In Euclidean geometry, the isodynamic points of a triangle are points
associated with the triangle, with the properties that ... the
distances from the isodynamic point to the triangle vertice... | {
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Express the value $z$ below in polar form, and the value $w$ in the form $a+bi$. I have been having a lot of issues on determining how to work through problems of the sort and I would be very grateful if somebody could provide me with a guided/ explained answer to enable me to understand how to work through these.
Give... | Just do it. If $z = a + bi$ than $|z| =\sqrt{a^2 + b^2}$ and
$z = \sqrt{a^2+ b^2}(\frac a{a^2 + b^2} + \frac b{a^2 + b^2} i)$. Let $r= \sqrt{a^2 + b^2} = |z|$.
Now $\frac a{a^2 + b^2}=x$ and $\frac b{a^2 + b^2}=y$ are two numbers such that $x^2 + y^2 = 1$. That means that there must be some angle $\theta$ so that $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $x, y, z$ be positive distinct integers. Prove that $(x+y+z)(xy+yz+zx-2)\ge9xyz$ I just came across the following question, in a book, which has as its topic contest-math:
Let $x, y, z$ be positive distinct integers. Prove that $(x+y+z)(xy+yz+zx-2)\ge9xyz$
I solved the question, in the following way:
Without loss o... | Let $x<y<z$, $y=x+1+a$ and $z=x+2+a+b$,
where $a$ and $b$ are non-negative integers.
Thus, we need to prove that
$$(x+y+z)(xy+xz+yz)-9xyz\geq2(x+y+z)$$ or $$\sum_{cyc}z(x-y)^2\geq2(x+y+z)$$ or
$$(x+2+a+b)(1+a)^2+(x+1+a)(2+a+b)^2+x(b+1)^2\geq2(3x+3+2a+b).$$
Can you end it now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove $\frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}$? Give $a,b,c>0$. Prove that: $$\dfrac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}.$$
My direction: (we have the equation if and only if $a=b=c$)
$a^{n+1}+a^nb+a^nc \ge 3a^n\sqrt[3]{abc}$
$b^{n+1}+b^na+b^nc \ge 3b^n\sqrt[3]{abc}$
$c... | Because for natural $n$ by AM-GM we obtain: $$\begin{aligned} \sum_{\text{cyc}}a^{n+1}&=\frac{1}{3(n+1)}\sum_{\text{cyc}}\left((3n+1)a^{n+1}+b^{n+1}+c^{n+1}\right) \\
&\geq \sum_{\text{cyc}}\sqrt[3n+3]{a^{(3n+1)(n+1)}b^{n+1}c^{b+1}} \\
&=\sum_{\text{cyc}}a^{n+\frac{1}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}} \\
&=\sqrt[3]{ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Partial reciprocal sum How can I show that $$\sum_{k=1}^{n}\frac{1}{n+k}\leq\frac{3}{4}$$ for every integer $n \geq 1$?
I tried induction, estimates with logarithms and trying to bound the sum focusing on the larger terms or things like $\frac{1}{n+1}+\frac{1}{n+2}\leq\frac{2}{n+1}$ but nothing seems to work. Do you ha... | Here is an elementary proof. Let $$ u_n = \sum_{k=1}^n \frac{1}{n+k}$$
For all $n \geq 1$, we have
$$u_{n+1}-u_n = \sum_{k=1}^{n+1} \frac{1}{n+1+k} - \sum_{k=1}^n \frac{1}{n+k}$$ $$= \sum_{k=2}^{n+2} \frac{1}{n+k} - \sum_{k=1}^n \frac{1}{n+k} = \frac{1}{2n+2} + \frac{1}{2n+1} - \frac{1}{n+1}$$ $$ = \frac{1}{2n+1}-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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For $a>1$, show that $\frac{1}{1+x}-\frac{1}{1+ax} \leq \frac{\sqrt{a}-1}{\sqrt{a}+1}$, $x \geq 1$ I'm self-learning the analysis I "by Herbert Amann" and got stuck in this problem. It's in Chapter IV Taylor's theorem.
For $a>1$ and $x\geq 1$ show that $$\frac{1}{1+x}-\frac{1}{1+ax} \leq \frac{\sqrt{a}-1}{\sqrt{a}+1}$... | We need to prove that $$\frac{ax-x}{(1+ax)(1+x)}\leq\frac{\sqrt{a}-1}{\sqrt{a}+1}$$ or
$$\frac{x(\sqrt{a}+1)}{(1+ax)(x+1)}\leq\frac{1}{\sqrt{a}+1},$$ which is true by C-S:
$$\frac{x(\sqrt{a}+1)}{(1+ax)(x+1)}\leq\frac{x(\sqrt{a}+1)}{(\sqrt{x}+\sqrt{ax})^2}=\frac{1}{\sqrt{a}+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3789967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x$, $y$ and $z$ satisfy $xy=1$, $yz=2$ and $xz=3$, what is the value of $x^2+y^2+z^2$? Express your answer as a common fraction.
If $x$, $y$ and $z$ satisfy $xy=1$, $yz=2$ and $xz=3$, what is the value of $x^2+y^2+z^2$? Express your answer as a common fraction.
I tried going along the path of computing $(x+y+z)... | $$\text{Compilation of all *Hints*}$$
We know that $$(x+y+z)=xyz(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx})$$
From here we see that $$(x+y+z)=xyz(\frac{11}{6})$$
Also, $(xyz)^2=6$ , so it would be nice to say that
$$(x+y+z)^2=6\big(\frac{11}{6}\big)^2$$
Then write the expansion for $$(x+y+z)^2$$ and keep substituting till... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Why $8^{\frac{1}{3}}$ is $1$, $\frac{2\pi}{3}$, and $\frac{4\pi}{3}$ The question is:
Use DeMoivre’s theorem to find $8^{\frac{1}{3}}$. Express your answer in complex form.
Select one:
a. 2
b. 2, 2 cis (2$\pi$/3), 2 cis (4$\pi$/3)
c. 2, 2 cis ($\pi$/3)
d. 2 cis ($\pi$/3), 2 cis ($\pi$/3)
e. None of these
I think that ... | We could look at it like this:
$$8^{\frac13}=2.1^{\frac13}=2\cdot \text{CiS}\left(\frac{2k\pi}{n}\right)$$
Now for different values of $k$, we have different answers: (here $n$ is $3$)
$$k=1\implies 8^{\frac13}=2\cdot\text{CiS} \left(\frac{2\pi}{3}\right)$$
$$k=2\implies8^{\frac13}=2\cdot\text{CiS}\left(\frac{4\pi}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Finding $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ this limit according to wolframalpha is equal to $0$.
So this is my work thus far
$\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}... | Hint:
WLOG $x=\tan y\implies y\to\dfrac\pi2$
$$\dfrac{3\tan y+2\tan^3y-2\sec^3y}3$$
$$=\dfrac{3\sin y\cos^2y+2\sin^3y-2}{3\cos^3y}$$
The numerator $$=3(1-\sin^2y)\sin y+2\sin^3y-2=\cdots=(1-\sin y)^2(2\sin y+1)$$
Finally use $$\dfrac{1-\sin y}{\cos y}=\dfrac{\cos y}{1+\sin y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
Directional derivative and gradient of a differentiable function Let $u = f(x,y,z)$ be a differentiable function in $\mathbb{R^3}$
Given the function satisfies: $f(x,y,x^2 + y^2) = 2x+y$ for all $x,y$
And the directional derivative of the point $(0,2,4)$ in the direction: $(-2,1,2)$ is equal to $-\frac{5}{3}$.
Calculat... | For simplicity, I will denote $\nabla f(0,2,4) = (f_x(0,2,4), f_y(0,2,4) , f_z(0,2,4)) = (a,b,c)$
You need to use the chain rule on this known fact $f(x,y,x^2+y^2) = 2x+y$:
$$\frac{ \partial }{\partial x}f(x,y,x^2+y^2) = f_x(x,y,x^2+y^2) \cdot \frac{\partial}{\partial x}x + f_y(x,y,x^2+y^2) \cdot \frac{\partial}{\parti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solving $\sqrt[3]{x+1} - \sqrt[3]{x-1} = \sqrt[3]{x^2-1}$ for real $x$
Solve the equation in the Real number system:
$$\sqrt[3]{x+1} - \sqrt[3]{x-1} = \sqrt[3]{x^2-1}$$
I have attempted using $(A-B)^3 = A^3 - B^3 - 3.A.B.(A-B)$ with $A = \sqrt[3]{x+1}$ , $B = \sqrt[3]{x-1}$ and $(A-B) = \sqrt[3]{x^2-1}$, however I en... | Put $a=\sqrt[3]{x+1}$ and $b= - \sqrt[3]{x-1}$.
Then $a+b=-ab$, hence $a$ satisfy $a^2+pa+p=0$ where $p=ab$ from which
$$a=\frac{-p\pm\sqrt{p^2-4p}}2$$
Consequently,
\begin{align}
x
&=a^3-1\\
&=-pa(a+1)-1\\
&=-p(-pa-p+a)-1\\
&=(p-1)\frac{-(p^2-2p-2)\pm p\sqrt{p^2-4p}}2
\end{align}
On the other hand, we have $a^3+b^3=2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find the argument of $z = {\left( {2 + i} \right)^{3i}}$ $z = {\left( {2 + i} \right)^{3i}}$
My approach is as follow
$z = {\left( {2 + i} \right)^{3i}} = {\left( {{{\left( {2 + i} \right)}^3}} \right)^i} = {\left( {8 + {i^3} + 12i - 6} \right)^i} = {\left( {2 + 11i} \right)^i}$
$\ln z = i\ln \left( {2 + 11i} \right)$
... | $$Z=(2+i)^{3i} \implies \ln Z=3i \ln (2+i)=3 i[\ln \sqrt{5}+i \tan^{-1}\frac{1}{2}]$$
$$\implies \ln Z= -3\tan^{-1}(1/2)+3i\ln \sqrt{5} \implies Z=e^{-3\cot^{-1}2}~~ e^{3i \ln \sqrt{5}}$$
$$\implies \arg(Z)=3\ln\sqrt{5}=2.414<\pi$$
AS this lies in $(-\pi, \pi]$, it is also the principal value of the argument.
Here we h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$x+y+z=xyz$ and $x,y,z>0$. Proof that $\frac{x}{1+x^2}+\frac{2y}{1+y^2}+\frac{3z}{1+z^2}=\frac{xyz(5x+4y+3z)}{(x+y)(y+z)(z+x)}$ $x+y+z=xyz$ and $x,y,z>0$.
Proof that $$\frac{x}{1+x^2}+\frac{2y}{1+y^2}+\frac{3z}{1+z^2}=\frac{xyz(5x+4y+3z)}{(x+y)(y+z)(z+x)}$$
So far, I have managed to reduce $(x+y)(y+z)(z+x)$:
$$(x+y)(y+... | Though tedious, this is a purely mechanical problem:
Expanding the denominators, the problem becomes equivalent to the polynomial
$$
(x + y) (y + z) (z + x)(x(1 + z^2)(1 + y^2) + 2y(1 + z^2)(1 + x^2) + 3z(1 + y^2)(1 + x^2)) \\ -(1 + z^2)(1 + y^2)(1 + x^2)(x y z (5 x + 4 y + 3 z))
$$
being 0 given $xyz=x+y+z$.
Expand th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding triangle of maximal area given two vertices and condition on internal bi-sector
$A=(2,5), B=(5,11)$ and a point $P$ moves such that internal bi-sector of $\angle APB$ passes through $(4,9).$ The maximum area of $ \triangle APB\;$ is __?
My attempt:
I found that $(4,9)$ lies on the line segment of $AB.$ Then I... |
Since the point $D=(4,9)$ belongs to $AB$,
it is a foot of the bisector of $\angle ACB$,
and a known expression for it is
\begin{align}
D&=\frac{aA+bB}{a+b}
\tag{1}\label{1}
.
\end{align}
With known coordinates we have
\begin{align}
(4,9)&=
\left(\frac{2a+5b}{a+b},\, \frac{5a+11b}{a+b} \right)
\tag{2}\label{2}
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $a$, $b$, $c$ are the roots of $x^3-6x^2+3x+1=0$, find all possible values of $a^2b+b^2c+c^2a$
Let $a$, $b$, $c$ be the roots of
$$x^3 - 6x^2 + 3x + 1 = 0$$ Find all possible values of $a^2 b + b^2 c + c^2 a$. Express all the possible values, in commas.
I've already tried to bash out all the roots, Vieta's Formula... | Let $p = a + b + c$, $q = ab + bc + ca$ and $r = abc$.
As pointed out by @Donald Splutterwit, we may let $A = a^2b + b^2c + c^2a$ and $B = ab^2 + bc^2 + ca^2$,
then $A + B$ and $AB$ are both symmetric which can both be expressed in terms of polynomials of $p, q, r$.
Indeed, we have
\begin{align}
A + B &= a^2b + b^2c +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $\alpha\in(0^\circ;90^\circ)$ simplify $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha.\cos^2\alpha+\cos^4\alpha$ For $\alpha\in(0^\circ;90^\circ)$ simplify $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha\cdot\cos^2\alpha+\cos^4\alpha.$
My try: $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha\cdot\cos^2\alpha+\cos^4\alpha=\sin^2\alpha+\dfr... | Express everything in terms of $ \cos \alpha=c $
$$ (1-c^2) + (1-c^2)/c^2 + (1-c^2) c^2 + c^4$$
$$= \dfrac{c^2-c^4+1-c^2+c^4-c^6+c^6}{c^2} = \dfrac{1}{c^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Is it possible to solve this identity by "inspection"? I was asked to prove the following identity (starting from the left-hand side):
$$(a+b)³(a⁵+b⁵)+5ab(a+b)²(a⁴+b⁴)+15a²b²(a+b)(a³+b³)+35a³b³(a²+b²)+70a⁴b⁴=(a+b)^8.$$
I'm trying to solve it by a sort of "inspection", but I haven't made it yet. Of course I could try to... | $$(a+b)^3(a^5+b^5)+5ab(a+b)^2(a^4+b^4)+15a^2b^2(a+b)(a^3+b^3)+35a^3b^3(a^2+b^2)+70a^4b^4=(a+b)^8$$
Note that $a+b|(a+b)^3$, so except the last two terms in the LHS, every term is divisible by $(a+b)^2$, in fact you can take $35a^3b^3$ common from the last two terms to have $$35a³b³(a²+b²)+70a⁴b⁴=35a^3b^3(a+b)^2$$ so th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Inverse of a matrix and matrix relation Let $A$ be the matrix $$ A= \begin{pmatrix}
2 & -1 & -1\\
0 & -2 & -1 \\
0 & 3 & 2 \\
\end{pmatrix} $$ I am trying to find $A^{-1}$ as a relation of $I_{3}, A$ and $A^{2}$ and also to prove that $A^{2006}-2A^{2005}=A^{2}-2A$. For the first one I noticed that $$A^{n}= \begin{pma... | Since the characteristic polynomial of $A$ is $-\lambda ^3+2\lambda^2+\lambda-2$, you know, by the Cayley-Hamilton theorem, that $-A^3+2A^2+A-2\operatorname{Id}_3=0$. Therefore,$$A^{-1}=-\frac12A^2+A+\frac12\operatorname{Id}_3.$$On the other hand,\begin{align}A^{2006}-2A^{2005}&=A^{2003}(A^3-2A^2)\\&=A^{2003}(A-2\opera... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3808412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Verify my proof that for any $n>1$, if $n^n+1$ is prime, then $n=2^{2^k}$ for some integer $k$. I am solving a problem and I am respectfully asking someone to critique my work and offer suggestions on formatting or point out any glaring logical errors. Here is the problem:
Prove that for any $n>1$, if $n^n+1$ is prime... | Your proof is correct in its ideas, but contains a few unclear and even false statements. I'll list a few points of improvement:
...if $n$ is odd, then $n^n$ is odd and thus $n^n+1$ is even, so $n$ must be even.
How does it follow that $n$ must be even? If you are concluding this because you have reached a contradict... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Distance between $\left\{(x,y,z):\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\right\}$ and $\big\{(x,y,z): x+y+z=h\big\}$ Calculate the distance between $$M=\left\{(x,y,z) \in \mathbb{R}^{3}: \frac{x^2}{a^2}+ \frac{y^2}{b^2}+\frac{z^2}{c^2}=1\right\}$$ and $$H=\big\{(x,y,z) \in \mathbb{R}^{3}: x+y+z=h\big\}$$ wher... | I'm assuming when you say the distance between $M$ and $H$ you mean $Dist(M, H) = \inf\{||x-y|| \ |\ x\in M, y\in H\}$. Additionally, I'll assume that $M$ and $H$ don't intersect.
Let $\tilde{M} = \{(x,y,z) \in \mathbb{R}^{3}: \frac{x^2}{a^2}+ \frac{y^2}{b^2}+\frac{z^2}{c^2} \leq 1\}$ and notice that this is convex set... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Distribution of $Z=\left\{\begin{matrix} X+Y & \operatorname{if} & X+Y<1\\ X+Y-1 & \operatorname{if} & X+Y>1 \end{matrix}\right.$ Let $X\perp Y$ be two random variables with uniform distribution on $[0,1]$. How is it possible that $F_Z(z)=z$?
Initially I wrote $F_Z(z)=\mathbb{P}(Z\leq z)=\mathbb{P}(X+Y\leq z,X+Y<1)+\ma... | For $z\in\left(0,1\right)$ we find:
$$\begin{aligned}P\left(Z\leq z\right) & =P\left(Z\leq z,X+Y<1\right)+P\left(Z\leq z,X+Y>1\right)\\
& =P\left(X+Y\leq z,X+Y<1\right)+P\left(X+Y-1\leq z,X+Y>1\right)\\
& =P\left(X+Y<z\right)+P\left(1<X+Y<1+z\right)\\
& =\frac{1}{2}z^{2}+\left[\frac{1}{2}-\frac{1}{2}\left(1-z\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3811639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to show that $ C := A^3-3A^2+2A = 0$?
Let $$A = \begin{pmatrix} -1 & 1 & 2 \\ 0 & 2 & 0 \\ -1 & 1 & 2 \end{pmatrix}$$
Let $C:= A^3-3A^2+2A $.
Show that $C=0$.
I know that $A$ is diagonalisable with $\operatorname{spec}(A)=\{0,1,2\}$.
I have no clue how to approach that problem. Any advice?
| Method 1: you can solve it by direct computation:
$$A^2 = \begin{pmatrix} -1 & 3 & 2 \\ 0 & 4 & 0 \\ -1 & 3 & 2 \end{pmatrix}$$
and
$$A^3 = \begin{pmatrix} -1 & 7 & 2 \\ 0 & 8 & 0 \\ -1 & 7 & 2 \end{pmatrix}.$$
You can actually argue that
$$A^n = \begin{pmatrix} -1 & \sum_{k=0}^{n-1} 2^k & 2 \\ 0 & 2^n & 0 \\ -1 & \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3813075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Is my Integral Solution Correct? Compute $\int \frac{n^x}{n^{2x} + 8n^x + 12} \, dx$, where $n$ is a positive real number.
UPDATED SOLUTION:
First, set $n^x = u.$ We have $du = n^x\log(n)\, dx.$
Subbing in $u$ and $\frac{du}{u\log(n)},$ we have $\frac{1}{\log n}\int \frac{1}{u^{2} + 8u + 12} \, du.$
Using partial fract... | Alternatively, you could have used the roots of the quadratic to write
$$\frac1{u^2+8u+12}=\frac1{(u+6)(u+2)}=\frac A{u+6}+\frac B{u+2}$$
You determine $A$ and $B$ using $$(A+B)u=0$$ and $$2A+6B=1$$
So $A=-B$ and $4B=1$. Therefore $$\frac1{u^2+8u+12}=\frac 14\left(\frac 1{u+2}-\frac 1{u+6}\right)$$
You will get the sam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3813616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Complex Roots in a Complex Equation
Problem: The equation $$\frac{x}{x+1} + \frac{x}{x+2} = kx$$ has exactly two complex roots. Find all possible complex values for $k.$
Progress: I was online today, and I saw this problem. Heres what I tried: $$(x+1)x+(x+2)x=kx(x+1)(x+2),$$ and I immediately got stuck :(. I'm quite... | Good!
Yes, multiplying both sides by $(x+ 1)(x+ 2)$ "reduces" the equation to $(x+ 1)x+ (x+ 2)x= kx(x+ 1)(x+ 2)$. Now, I would do the indicated multiplications to get $x^2+ x+ x^2+ 2x= kX*(x^2+ 3x+ 2)= kx^3+ 3kx^2+ 2kx$ which can be written as $kx^3+ (3k- 1)x^2+ (2k- 1)x= 0$. The left side can be factored as $x(kx^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3814402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Relationship between constants so that the center of curvature of the helix is contained in the cylinder I am studying the topic of torsion and curvature and I ran into the following problem:
Consider the helix $\alpha(t)=(a\sin t,a\cos t,bt)$. Get the relationship between constants
so that the center of curvature of t... | Since the principal normal is horizontal, you will need $1/\kappa<2|a|$, so that seems to require $|b|<|a|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3815653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$, where $x+y+z=0$ Question :
Let $x,y,z$ be reals satisfying $x+y+z=0$. Prove the following inequality:$$6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$$
My Attempts :
It’s obvious that $x,y,z$ are either one negative and two positive numbers or one positive and two negative num... | The difference
$$
(x^2+y^2+z^2)^3-6(x^3+y^3+z^3)^2
$$
for $x+y+z=0$ can be written
$$
2 (x-y)^2 (y-z)^2 (z-x)^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
If $a, b, c, d>0$ and $abcd=1$ prove that an inequality holds true If $a, b, c, d>0$ and $abcd=1$ prove that:
$$\frac{a+b+c+d}{4}\ge\frac{1}{a^3+b+c+d}+\frac{1}{a+b^3+c+d}+\frac{1}{a+b+c^3+d}+\frac{1}{a+b+c+d^3}$$
I attempted to solve it in the following way:
$$\begin{equation}\frac{1}{a^3+b+c+d}+\frac{1}{a+b^3+c+d}+\f... | This is just a thought about the possibility to continue the proof. The inequality:
$$ \frac{a+b+c+d}{4} \ge \frac{\sqrt{bcd}+\sqrt{acd}+\sqrt{abd}+\sqrt{abc}}{4}$$
cannot hold for all $a,b,c,d>0$ with $abcd=1$. In fact, for $N>1$ set $a=b=c=N$, $d=1/N^3$. Then $LHS < N$ and $ RHS > N^{3/2}/4$. Moreover, if $N>16$ then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3819122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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If $a^2+b^2-ab=c^2$ for positive $a$, $b$, $c$, then show that $(a-c)(b-c)\leq0$
Let $a$, $b$, $c$ be positive numbers. If $a^2+b^2-ab=c^2$. Show that
$$(a-c)(b-c)\leq0$$
I have managed to get the equation to $(a-b)^2=c^2-ab$, but I haven't been able to make any progress.
Can someone help me?
| WLOG, we assume that $a \le b$ and we want to show that
$$a \le \sqrt{a^2+b^2-ab}\le b$$
which is equivalent to
$$a^2\le a^2+b^2-ab \le b^2$$
The first inequality reduces to $0 \le b(b-a)$ which is clearly true.
The second inequality is $a^2-ab=a(a-b) \le 0$ which is true again.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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Prove $\lim\limits_{x^2 + y^2 \to +\infty} x^2 -2xy + 2y^2 = +\infty$
Prove that $$\lim_{x^2 + y^2 \to +\infty} x^2 -2xy + 2y^2 = +\infty$$
My attempt:
$$x^2 + 2y^2 = x^2+y^2 + y^2 \implies \lim_{x^2 + y^2 \to +\infty}x^2 +2y^2 = +\infty$$
Then, from Cauchy-Schwarz:
$$x^2 + 2y^2 \geq 2\sqrt2xy \geq 2xy $$
Thus,
$$x... | Note: If $xy<0$, then $2\sqrt{2}xy \leq 2xy$. So the Cauchy-Schwartz argument may not hold.
However, $x^2-2xy+y^2 =(x-y)^2 \geq 0$. So as $x^2+y^2\rightarrow\infty$, $x^2-2xy+y^2+y^2\rightarrow\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3822564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Prove that $a^3 + b^3 + 3abc > c^3$ Suppose $a,b,c$ are the sides of a non-degenerate triangle. Prove that $a^3 + b^3 + 3abc > c^3.$
I was thinking that this inequality looked suspiciously like $a^3 + b^3 + c^3 - 3abc,$ which factors as $(a + b + c) (a^2 - a b + b^2 - a c - b c + c^2).$ However, I have little to no id... | By your hint we need to prove that: $$(a+b-c)(a^2+b^2+c^2-ab+ac+bc)>0.$$
Can you end it now?
$$a^2+b^2+c^2-ab+ac+bc=b(b+c-a)+a^2+c^2+ac>0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Minimize $(x+y)(y+z)(z+x)$ given $xyz(x+y+z) = 1$ $x,y,z$ are positive reals and I am given $xyz(x+y+z) = 1$. Need to minimize $(x+y)(y+z)(z+x)$. Here is my approach.
Using AM-GM inequality
$$ (x+y) \geqslant 2 \sqrt{xy} $$
$$ (y+z) \geqslant 2 \sqrt{yz} $$
$$ (z+x) \geqslant 2 \sqrt{zx} $$
So, we have
$$ (x+y)(y+z)(z... | $(x+y)(y+z)(z+x)=(z+x)(y(x+y+z)+xz)=(\frac{1}{zx}+zx)(x+z)$
now we can use $$\frac{1}{zx}+zx\ge 4{(\frac{1}{27{(xz)}^2})}^{1/4}$$
(HINT:$\frac{1}{zx}=\frac{1}{3zx}+\frac{1}{3zx}+\frac{1}{3zx}$)
also we can use $$x+z\ge 2\sqrt{xz}$$
Multiplying we get $$(x+y)(y+z)(z+x)\ge \frac{8}{3^{3/4}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3829530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability that atleast $3$ from $5$ dice show the same face I think I have solved this problem correctly, but my solution does not seem to match the one in the textbook - Problem IV.2 from Feller's introduction to probability theory & applications page 101. Could you help with where I've possible gone wrong?
Note. Th... | In case you are interested in alternative approaches, here is a slightly different solution, also using the formula for the realization of at least $m$ among $N$ events.
Let's see if we can solve a simpler problem: What is the probability that at least $3$ of $5$ dice roll a $1$? Here $A_i$ is the event that die $i$ ro... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve system of equations using inverse matrix? System of equations is the following:
$$x + 4y + 2z = 10$$
$$4x - 3y+0z = 6$$
$$2x + 2y + 2z = 14$$
Here is my solution:
$$det(A) = 1 *(-3 * 2 - 0 * 2) -4 * (4 * 2 - 0 * 2) + 2 * (4 * 2 - (-3) * 2)$$
$$= -6 -32 + 28$$
$$= -10$$
$$
+\begin{pmatrix}
-3 & 0 \\... | There are mistakes in the last 3 determinants.
$+\begin{vmatrix}4 & 2 \\-3 & 0\end{vmatrix} = 0 + 6 = 6$
$-\begin{vmatrix}1 & 2 \\4 & 0\end{vmatrix} = -(0-8)=+8$
$+\begin{vmatrix}1 & 4 \\4 & -3\end{vmatrix} = -3 - 16 = -19$
Try using these values.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If complex number $a, b, c, d,$ and $|a|=|b|=|c|=|d|=1$, why $|a(c+d)|+|b(c-d)|\leq 2\sqrt{2}$? If we have 4 complex number $a, b, c, d,$ and $|a|=|b|=|c|=|d|=1$, So, how to prove that $|a(c+d)|+|b(c-d)|\leq 2\sqrt{2}$？
I try to separate $|a(c+d)|+|b(c-d)|$ to $|a||(c+d)|+|b||(c-d)|$ than I get $|(c+d)|+|(c-d)|$.... | Let $c=\operatorname{cis}\alpha$ and $d=\operatorname{cis}\beta$.
Thus, by C-S we obtain: $$|a(c+d)|+|b(c-d)|=|c+d|+|c-d|=\sqrt{2+2\cos(\alpha-\beta)}+\sqrt{2-2\cos(\alpha-\beta)}=$$
$$=2|\cos\frac{\alpha-\beta}{2}|+2|\sin\frac{\alpha-\beta}{2}|\leq2\sqrt{(1+1)\left(\cos^2\frac{\alpha-\beta}{2}+\sin^2\frac{\alpha-\beta... | {
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"source": "stackexchange",
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Getting a cubic equation from 3 equations Here is the problem:
Three numbers have a sum of $5$ and the sum of their squares is $29$.
If the product of the three numbers is $−10$, what are the three
numbers? Express your answer in simplest radical form.
I used Vieta's formulas to get $x^3-5x^2+12x+10=0$
What should I ... | As @WhatsUp noted in the comments, the equation should have been $x^3-5x^2-2x+10=0$, and so you can find by inspection or by the Rational Root Theorem that $x=5$ is a root. Divide and get $x^2-2=0$, and so your answers should be $x=5, \sqrt{2}, -\sqrt{2}$.
Edit: Another (essentially equivalent, but more detailed) metho... | {
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Determine convergence of the sequence $x_0=1 , x_{n+1}=x_n (1+ 2^{-(n+1)})$ I want to check if the following sequence converges:
$$x_0=1 , x_{n+1}=x_n \left(1+ \frac{1}{2^{n+1}}\right)$$
I proved the sequence is increasing :
$\cfrac{x_{n+1}}{x_n}=1+ \cfrac{1}{2^{n+1}} \gt 1$
Now I should prove it is bounded above. le... | First show that $x_n \leq \sqrt{2}^{n+1}$ through mathematical induction. Then,
$0 \leq x_{n+1} - x_n = x_n \frac{1}{2^{n+1}} \leq \frac{1}{\sqrt{2}^{n+1}}$. You can easily check that given sequence is Cauchy sequence.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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How to integrate this area? I need to find the following area between the shapes made by $r=3+2\sin{t}$ and $r=2$:
My problem is if I need to integrate with any function or not. First, to check the angle I just did $2=r=3+2\sin{t} \Rightarrow \sin{t} = -\frac{1}{2} \Rightarrow$ the angle goes from $\frac{7\pi}{6}$ and... | Consider the following diagram. Here $dθ$ is the infinitesimally small angle such that we the variation of $r$ with $θ$ can be neglected.
Clearly, we need to find the summation (integral) of the orange areas :
Now consider the following diagrams:
Our Orange area = Blue Area - Green Area
Also, the figures can be app... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In the Euclidean Ring ${\Bbb Z}_3[x]$, what is the remainder of the euclidean division of $x^4+2x^3+x^2+2x+2$ by $x^2+2$? So far I've $(x^2+2)(x^2+2x-1)+4-2x$ but the options for the remainder are:
$2x+1$,
$2x$,
$x+1$, or
$x$.
| Hint:
$$-2=1\pmod 3\;,\;\;4=1\pmod 3\;$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve system of modular equivalence with parameter Given system of modular equivalences. Find the smallest natural parameter $a$ that system has solutions
$$\begin{cases}
x \equiv a \mod 140\\
x \equiv 3 \mod 91\\
x \equiv 2a \mod 39
\end{cases}$$
Of course, solution will be appreciated but if you don't want to ... | If the system is consistent then it has solutions.
Since $140=4\cdot 5\cdot 7$ then the first system is equivalent to $$x \equiv a \pmod {4, 5, 7}$$
and similarly for the other congruences we have $x \equiv 3 \pmod{7, 13}$ and $x\equiv 2a \pmod{3,13}$.
From the first we have $x\equiv a \pmod{7}$ and from the second $x\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to prove $\sum_{k=0}^{n}2^{2k}\binom{2n}{2k}=\frac{9^{n}+1}{2}$ by using mathematical induction? My question is how to prove $\sum_{k=0}^{n}2^{2k}\binom{2n}{2k}=\frac{9^{n}+1}{2}$ by using mathematical induction?
I can finish this question by considering the binomial $(1-x)^{2n}$. First, prove that
$$\binom{2n}{0}+... | Use Pascal’s identity twice to get the identity
$$\binom{n}k=\binom{n-2}{k-2}+2\binom{n-2}{k-1}+\binom{n-2}k\,,$$
and then use that at the start of the induction:
$$\begin{align*}
\sum_{k=0}^{n+1}2^{2k}\binom{2n+2}{2k}&=\sum_{k=0}^{n+1}2^{2k}\left(\binom{2n}{2k-2}+2\binom{2n}{2k-1}+\binom{2n}{2k}\right)\\
&=\sum_{k=1}^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that $ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$ By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that
$$ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$$
Here's what I've done so far (starting from after expansion):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x... | Hint Use $$a^3+b^3=(a+b)\left({(a+b)}^2-3ab \right)$$
with
$a=\cos^2 x, b=\sin^2 x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
The identity $\arctan(x) + \arctan(x^3) = \arctan(2x+\sqrt{3}) + \arctan(2x-\sqrt{3})$ I came to this identity while doing some indefinite integrals.
$\arctan(x) + \arctan(x^3) = \arctan(2x+\sqrt{3}) + \arctan(2x-\sqrt{3})$
Seems weird to me, no idea why it's correct but it is.
I wonder if there's some geometric or tri... | Hint:
\begin{eqnarray*}
\tan^{-1}(A)+\tan^{-1}(B)=\tan^{-1} \left( \frac{A+B}{1-AB} \right).
\end{eqnarray*}
$$\tan^{-1}(x)+\tan^{-1}(x^3)=\tan^{-1} \left( \frac{x(1+x^2)}{1-x^4}\right)=\tan^{-1} \left( \frac{x}{1-x^2} \right).$$
\begin{eqnarray*}\tan^{-1}(2x+\sqrt{3})+\tan^{-1}(2x-\sqrt{3})&=&\tan^{-1} \left( \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3843570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x+y+z=xyz$, prove $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
If $x+y+z=xyz$, prove
$\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
given that $x^2~,~y^2~,~z^2\ne1$
I came across... | Let $z=0$.
Thus, $x+y=0$ and
$$\sum_{cyc}\frac{2x}{1-x^2}=\frac{2x}{1-x^2}-\frac{2x}{1-x^2}=0=\prod_{cyc}\frac{2x}{1-x^2}.$$
Now, let $xyz\neq0.$
Thus, we need to prove that:
$$\sum_{cyc}\frac{1}{\frac{2x}{1-x^2}\cdot\frac{2y}{1-y^2}}=1.$$
Indeed, $$\sum_{cyc}\frac{1}{\frac{2x}{1-x^2}\cdot\frac{2y}{1-y^2}}=\sum_{cyc}\f... | {
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$f(m, n) = f(m-1, n) + f(m, n-1) + f(m-1, n-1)$ show that $\sum_{n=0}^{\infty} f(n, n) x^n = \frac{1}{\sqrt{1 - 6x + x^2}}$ Problem Statement: Let $f(m, 0) = f(0, n) = 1$ and $f(m, n) = f(m-1, n) + f(m, n-1) + f(m-1, n-1)$ for $m, n > 0$. Show that
$$\sum_{n=0}^{\infty} f(n, n) x^n = \frac{1}{\sqrt{1 - 6x + x^2}}$$
I w... | Let
$$G(x,t)=F(xt,xt^{-1}).$$
Then
$$G(x,t)=\sum_{i,j}f(i,j)x^{i+j}t^{i-j}
=\sum_{r,s}f((r+s)/2,(r-s)/2)x^rt^s$$
so the terms with $t^0$ therein add to $\sum_r f(r,r)x^{2r}$, essentially what you're
after.
But
$$G(x,t)=\frac{1}{1-x^2-xt-xt^{-1}}$$
and
\begin{align}
1-x^2-xt-xt^{-1}&=
-xt^{-1}\left(t-\frac{1-x^2}{2x}+\f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Representing $\cos(\frac{π}{11})$ as cyclic infinite nested square roots of $2$ How can we represent $2\cos(\frac{\pi}{11})$ and $2\cos(\frac{\pi}{13})$ as cyclic infinite nested square roots of 2
I have partially answered below for $2\cos(\frac{\pi}{11})$ which I derived it accidentally.
Currently I have figured out ... | For your solution
$ 2 \cos \left(\frac{\pi }{11}\right)=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-2 \cos \left(\frac{\pi }{11}\right)}}}}} $
I have add it (with your name) to Page 3 of my Notebook:
http://eslpower.org/Notebook.htm
I found the same result independently on July,2021.
Here is my method:
Since $(2^5+1) \bmod... | {
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"timestamp": "2023-03-29T00:00:00",
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Role of coprimality in proof of Fermat's Little Theorem This is the start of the proof for FLT:
I was curious -- I know that all the elements of S are unique because gcd(a,p) = 1, but I was wondering --
What would be an example in which the elements were not unique?
Also, in the next step:
what happened to the a? it ... |
I was curious -- I know that all the elements of S are unique because gcd(a,p) = 1, but I was wondering -- What would be an example in which the elements were not unique?
Well, one where $p$ isn't prime and if $\gcd(a,p) \ne 1$ then you'd have $\frac p{\gcd(a,p)}$ distinct terms and the repeated $a$ times (minus a fi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Understand how to evaluate $\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ We are given this limit to evaluate:
$$\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
In this example, if we try substitution it will lead to an indeterminate form $\frac{0}{0}$. So, in order to evaluate this limit, we can multiply this exp... | On the 4th row of the second calculation, you arrive at the following expression:
$$
\frac{2-x}{2-x}
$$
This expression is actually $\frac{0}{0}$ (undefined) at $x=2$. Equating it to $1$ - which is what you did is the same as removing the singularity (the point of where the function is not defined) in the expression an... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the equation of the circle I've been experiencing a difficulties in answering this. I hope someone will help me in solving this
Find the equation of the circle through the points $(2,8),(7,3)$ and $(-2,0)$.
|
The question can be restated as
Given triangle $ABC$ with coordinates of vertices $A=(-2,0)$,
$B=(7,3)$, $C=(2,8)$, find the equation of its circumscribed circle.
First, find the squares of the side lengths of $\triangle ABC$:
\begin{align}
a^2&=50
,\quad
b^2= 80
,\quad
c^2=90
\tag{1}\label{1}
.
\end{align}
Second, ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to arrive at correct parametric form In question 2 of this quiz, the resulting parametric vector form is $x = x_3 \begin{align}
\begin{bmatrix}
1 \\
1 \\
1 \\
\end{bmatrix}
\end{align}$. How do they arrive at $x = x_3 \begin{align}
\begin{bmatrix}
1 \\
... | If $x=(x_1,x_2,x_3)^T$ is a solution, then you already commented that you know that $x_1=x_2=x_3$. But then
$$
\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}= \begin{pmatrix} x_3 \\ x_3 \\ x_3\end{pmatrix}= x_3 \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The equation $x^4-x^3-1=0$ has roots $\alpha,\beta,\gamma,\delta$. Find $\alpha^6+\beta^6+\gamma^6+\delta^6$ The equation $x^4-x^3-1=0$ has roots $\alpha,\beta,\gamma,\delta$.
By using the substitution $y=x^3$, or by any other method, find the exact value of
$\alpha^6+\beta^6+\gamma^6+\delta^6$
This is a problem from F... | Define the polynomial function $p$ by
$$p(x) = x^4-x^3-1$$
Also define the function $q$ by
$q(x)=p\left(x^{1/3}\right)$. Then,
$$q(t)=t^{4/3}-t-1$$
By definition of $q$, we have that $\alpha^3,\beta^3,\gamma^3,\delta^3$ are roots of the equation $q(t)=0$.
$$q(t)=0\iff t^4 - (t+1)^3=0$$
The polynomial $t^4-(t+1)^3$ is a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Range of $f(z)=|1+z|+|1-z+z^2|$ when $ |z|=1$
Find the Range of $f(z)=|1+z|+|1-z+z^2|$ when $z$ is a complex $ |z|=1$
I am able to get some weaker bounds using triangle inequality.
$f(z)< 1+|z|+1+|z|+|z^2|=5$
also $f(z)>|1+z+1-z+z^2|=|z^2+2|>||z^2|-2|=1$,but
these are too weak!.as i have checked with WA
Is there an ... | Note that $1-z+z^2=(1+r)(1+r^2)$ where $r$ is the cube root of unity.
Also, since $|z|=1$, $z\bar z=1$
Now rewrite the given expression as $√ (1+z)(1+\bar z) + √(1-z+z^2)(1-\bar z +\bar z^2)$
$\Rightarrow √ 1+z+\bar z+1 + √(z+r)(z+r^2)(\bar z+r)(\bar z+r^2)$
$\Rightarrow √(2+2\cos\theta) + √(z\bar z+zr+\bar zr+ r^2)(z\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Proving that $\sum_{k=0}^{k=n} \binom{2n}{k} \cdot k = 2^{2n -1} \cdot n$ I have to show that $\sum_{k=0}^{k=n} \binom{2n}{k} \cdot k = 2^{2n -1} \cdot n$.
What I know is that $\sum_{k=0}^{k=n} \binom{n}{k} \cdot k = 2^{n -1} \cdot n$.
How do I proceed from there?
| \begin{align}
\sum_{k=0}^n k\binom{2n}{k}
&= \sum_{k=1}^n k\binom{2n}{k} \\
&= \sum_{k=1}^n 2n\binom{2n-1}{k-1} \\
&= n \sum_{k=1}^n \left(\binom{2n-1}{k-1} + \binom{2n-1}{k-1}\right) \\
&= n \sum_{k=1}^n \left(\binom{2n-1}{k-1} + \binom{2n-1}{2n-k}\right) \\
&= n \sum_{j=0}^{2n-1} \binom{2n-1}{j} \\
&= n \cdot 2^{2n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3866083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Find asymptotic behavior of $a_n$ We are given $a_1=1$ and $a_{n+1}=\dfrac 1 {\sqrt n \ a_n}$.
At first I found through continuous analogue that $a_n=2\sqrt{\sqrt n}$.
Then I got the first inequality : $a_{n+1}^2 -a_1 ^2 >2 \left(1+ \dots +\dfrac 1 {\sqrt n} \right)$.
Then I obtained $a_n^2 > \dfrac {2n}{\sqrt n}$.
For... | It follows by induction that for any $n\geq 2$,
$$
a_{2n} = \frac{{\sqrt {2n - 2} \cdots \sqrt 2 }}{{\sqrt {2n - 1} \sqrt {2n - 3} \cdots \sqrt 3 }} = \sqrt {\frac{{(2n - 2)!!}}{{(2n - 1)!!}}}
$$
and
$$
a_{2n + 1} = \frac{{\sqrt {2n - 1} \cdots \sqrt 3 }}{{\sqrt {2n} \sqrt {2n - 2} \cdots \sqrt 2 }} = \sqrt {\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
solving for $x$:$\cos^2 3x+\frac{\cos^2 x}{4}=\cos 3x\cos^4 x$
solve for x:$$\cos^2 3x+\frac{\cos^2 x}{4}=\cos 3x\cos^4 x$$
My attempt:
completing square: $${(\cos 3x-\frac{\cos x}{2})}^2=\cos 3x\cos x (1-\cos^3 x)$$ or
$$\cos x\cos 3x\ge 0$$
also by some basic identities :$$1+\cos 6x+\frac{1+\cos 2x}{4}=(\cos 4x+\c... | $$\color{blue}{\cos^2{3x}} - (\cos^4{x})\color{blue}{\cos{3x}} + \dfrac{\cos^2{x}}{4}=0$$
$$ \color{blue}{\cos{3x}} = \dfrac{\cos^4{x} \pm \sqrt{\cos^8{x} - \cos^2{x}}}{2}$$
Use $\cos^8{x} - \cos^2{x} \le 0$.
Similarly,
$$\cos{3x}\color{red}{\cos^4{x}} - \dfrac{\color{red}{\cos^2{x}}}{4} - \cos^2{3x}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3871800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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I have a problem from Karush-Kuhn-Tucker condition maximize: $xy-x$ and s.t. $x^2+y^2 \leq 9$ and $x \ge 0$
Then, I tried to find points using lagrange method. I used 4 cases and they are $(1)$ $\lambda_1 =0$ and $\lambda_2 = 0$
forrect?
| Maximize $xy - x$, given constraints: $x^2 + y^2 \leq 9, x \ge 0$
Taking $xy - x = \lambda (x^2 + y^2 -9)$
Taking derivative wrt $x$ and $y$, we get -
$y-1 = 2 \lambda x$ ...(i)
$x = 2 \lambda y$ ...(ii)
From (ii), $\lambda = \frac{x}{2y}$ and substituting in (i)
we get $y(y-1) = x^2 \implies y(y-1) + y^2 \leq 9$
That ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3874809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the roots of the polynomial $x^3-2$. Find the roots of the polynomial $x^3-2$.
If $\alpha$ be the root of this polynomial i.e., $\alpha^3=2$,
then $(\zeta \alpha)^3=2$, where $\zeta$ is $3^{rd}$ root of unity.
Hence the solutions of this equations are
$$\zeta{2}^\frac{1}{3}$$, $\zeta$ a $p^{th}$ root of unity
And... | Forward
i) Euler's formula: $e^{i\theta}=\cos\theta + i\sin\theta$
ii) De Moivre's formula: $\left(\cos x+i\sin x\right)^n=\cos (nx) + i\sin (nx)$
Theorem
If $z\ne 0$, and if $n$ is a positive integer, then there are exactly $n$ distinct complex numbers $z_0,z_1,\dots,z_{n-1}$ such that $z_k^n=z$, $k=0,\dots,n-1$. Thes... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3876285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Solving ODE, check my answer. I'm trying to solve this ODE but I'm not sure with my answer. Anyone please check my answer. If my work is not true, tell my mistake. Thanks.
Solve the ODE $$\left(x^2+2xy\right)\dfrac{dy}{dx}=y^2-2xy.$$
Solution.
\begin{alignat}{2}
&& \left(x^2+2xy\right)\dfrac{dy}{dx}&=y^2-2xy\nonumbe... | Since that $$-\frac{5}{3}\ln(u+3)-\frac{1}{3}\ln(u)=\ln(x)+c \overbrace{\implies}^{u=\frac{y}{x}}\boxed{-\frac{5}{3}\ln\left(\frac{y}{x}+3 \right)-\frac{1}{3}\ln \left(\frac{y}{x} \right)=\ln(x)+c}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Differential equation with integrating factor
Hey I am supposed to solve the following differential equation:
$(1-x^{2}y)dx+x^{2}(y-x)dy=0$
I found integrating factor:
$\varphi (x)=-\frac{2}{x}$
So I multiply my original equation and I got:
$\left ( \frac{1}{x^{2}} -y\right )dx+\left ( y-x \right )dy=0$
But then I tr... | $(1-x^2y)~dx+x^2(y-x)~dy=0$
$x^2(y-x)~dy=(x^2y-1)~dx$
$(y-x)\dfrac{dy}{dx}=y-\dfrac{1}{x^2}$
Let $u=y-x$ ,
Then $y=u+x$
$\dfrac{dy}{dx}=\dfrac{du}{dx}+1$
$\therefore u\left(\dfrac{du}{dx}+1\right)=u+x-\dfrac{1}{x^2}$
$u\dfrac{du}{dx}+u=u+x-\dfrac{1}{x^2}$
$u\dfrac{du}{dx}=x-\dfrac{1}{x^2}$
$u~du=\left(x-\dfrac{1}{x^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3880485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Simplify $\frac{\sin t - \cos t}{\sin t + \cos t}$ $$\frac{\sin t - \cos t}{\sin t + \cos t}$$
I am to simplify this into $$\tan(t - \frac{\pi}{4})$$
I'm not sure how to carry on, though. Multiplying numerator and denominator by denominator (or numerator) doesn't get me anywhere.
| Step-by-step
$$ \dfrac{\sin t - \cos t}{\sin t + \cos t}$$
$$= \dfrac{\dfrac{\sin t}{\cos t} - \dfrac{\cos t}{\cos t}}{\dfrac{\sin t}{\cos t} + \dfrac{\cos t}{\cos t}}$$
$$= \dfrac{\tan t - 1}{\tan t + 1}$$
$$= \dfrac{\tan t - 1}{1 + \tan t \color{blue}{\cdot 1}}$$
Substitute $\color{blue}{\tan \frac{\pi}{4} = 1}$
$$= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3883650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the max. and min. points of $g(x,y,z) = x^2 + y^2 + z^2$ The problem
Let $X$ a submanifold of $\mathbb{R}^3$ defined by $$ X = \left\{ (x,y,z) \mid (x^2 + y^2 + z^2 - 5)^2 = 16(1-z^2) \right\} $$
Let $\Phi(x,y,z) = (x^2 + y^2 + z^2 - 5)^2 - 16(1-z^2) $. Let $f$ a function defined by
$$ f : \mathbb{R}^3 \rightarrow... | Your working is correct -
$\left( \begin{array}{c}
2x\\ 2y\\ 2z
\end{array} \right) = \lambda \left( \begin{array}{c}
4x(x^2 + y^2 + z^2 - 5) \\
4y(x^2 + y^2 + z^2 - 5) \\
4z(x^2 + y^2 + z^2 + 3) \\
\end{array} \right)$
Also, $(x^2 + y^2 + z^2 - 5)^2 = 16(1-z^2)$.
From above system of equations, please note the possibl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3888140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
prove thatt $\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$
prove that $$\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$$ where $x,y,z$ are positives such that $xy+yz+xz=1$
By Holder;$$\left(\sum_{cyc} \frac{1}{{(x+y)}^2} \right){\left(\sum yz+zx \right)}^2\ge {\sum \left(z^{2/3} \right)}^{3}$$.
Hence it suffices to prove $$\sum {\le... | This is the solution by Vo Quoc Ba Can.
Need to prove:
$$(xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\geqq \frac{9}{4}$$
Wlog, assume $x\geq y \geq z,$ we have the following lemma:
$$\sum\limits_{cyc} \frac{1}{(x+y)^2}\geqq \frac{1}{4xy}+\frac{2}{(x+z)(y+z)}\quad \quad(1)$$
$$\Leftrighta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is it valid to apply L'hopital rule to evaluate the limit? $$\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}$$
In the book I am reading, the limit evaluated in this way:
$$\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}\times \frac{\sqrt{x^2+x+1}+1}{\sqrt{x^2+x+1}+1}=\lim_{x\to 0^+}\frac{2\tan x(1-\... | Yes since the expression is in the form $\frac 0 0$ we are allowed to use l'Hospital to obtain
$$\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}=\lim_{x\to 0^+}\frac{2 \sec^2 x - 2 \cos x}{\frac{2 x + 1}{2 \sqrt{x^2 + x + 1}}}=\frac{0}{\frac12}=0$$
Note that often use l'Hospital's rule is esplicitely not all... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3890051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Stronger than Nesbitt's inequality using convexity and functions Hi it's a refinement of Nesbitt's inequality and for that, we introduce the function :
$$f(x)=\frac{x}{a+b}+\frac{b}{x+a}+\frac{a}{b+x}$$
With $a,b,x>0$
Due to homogeneity we assume $a+b=1$ and we introduce the function :
$$g(a)=\frac{a}{1-a+x}$$
Showing... | There are very many refinements of the Nesbitt's inequality.
For example. For positives $a$, $b$ and $c$ we have:
1.$$\frac{a}{\sqrt[3]{4(b^3+c^3)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2};$$
2.$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{(a+b+c)^2}{2(ab+ac+bc)};$$
3.$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3890210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"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.