Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove $| |y-x| - |z-y| | \leq |x-z|$ How to prove equation of the form $| |y-x| - |z-y| | \leq |x-z| x,y, z \in \Re$ ?
There are 10(?) cases:
(1) $x>y>z$ : $||y-x|-|z-y|| \le x-z$
(2) $x>z>y$ : $|y-x|-z+y \le x-z$
(3) $y>x>z$ : $|y-x-|z-y|| \le x-z$
(4) $y>z>x$ : $|y-x-|z-y||\le |x-z|$
(5) $z>x>y$ : $||y-x|-z+y|\le|x-z|$
(6) $z>y>x$ : $|y-x-z+y|\le|x-z|$
(7) $x=y>z$ : $|x-x|-|z-x|\le|x-z|$
(8) $x=z>y$ : $|y-x|-|x-y| \le |x-x|$
(9) $y=z>x$ : $|z-x|-|z-z| \le |x-z|$
(10)$x=y=z$ : $||x-x|-|x-x||\le |x-x|$
(1) $||y-x|-|z-y||\le x-z \to ||x-y|-|z-y||\le x-z \to |x-y-z+y|\le x-z \\ \to |x-z|\le x-z \to x-z \le x-z$ proved
(2) $|y-x|-z+y\le x-z \to |y-x|+y\le x \to |x-y|+y\le x \to x \le x$ proved
(3) $|y-x-|z-y|| \le x-z \to |y-x-|y-z|| \le x-z \to |y-x-y+z| \le x-z \to \\|-x+z|\le x-z \to |z-x|\le x-z \to |z-x| \le x-z \to |x-z| \le x-z \to \\ x-z\le x-z $ proved
(4) $|y-x-|z-y|| \le |x-z| \to |y-x-|y-z||\le z-x \to |-x+z| \le z-x \to \\ |-x+z| \le z-x \to |z-x| \le z-x \to z-x \le z-x $ proved
(5) $||y-x|-z+y|\le z-x \to |x-y-z+y|\le z-x \to |x-z|\le z-x \to z-x\le z-x$ proved
(6) $|y-x+y-z| \le |x-z| \to |2y-x-z|\le|x-z| \to $ still no idea
(7) $|0-|z-x||\le|x-z| \to |z-x|\le|x-z|$ proved
(8) $|y-x|-|x-y|\le|x-x| \to 0 \le 0$ proved
(9) $|z-x|-|z-z| \le |x-z| \to |z-x|\le|x-z|$ proved
(10)$||x−x|−|x−x||≤|x−x| \to 0\le0$ proved
any ideas how to proceed? what could i do next?
| Recall that by reverse triangle inequality
$$||a|-|b||\le |a-b|$$
indeed
*
*$a,b\ge 0 \implies ||a|-|b||=|a-b|\le |a-b|$
*$a,b\le 0 \implies ||a|-|b||=|-a+b|\le |a-b|$
*$-a=a'\le 0, b\ge 0 \implies ||a|-|b||=|a'-b|\le |-a'-b|=|a'+b|$
*$a\ge 0, b=-b'\le0 \implies ||a|-|b||=|a-b'|\le |a+b'|$
therefore
$$| |x-y| - |z-y| | \leq |(x-y)-(z-y)|=|x-z|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3402725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove a reduction formula for: $\int \frac{x^n dx}{\sqrt{ax^2 + bx + c}}$
For $n > 1\in \Bbb N$, prove that:
$$
J_n = \int \frac{x^n dx}{\sqrt{ax^2 + bx + c}} = \\
{1\over na}\left(x^{n-1}\sqrt{ax^2 + bx + c} - {b\over 2}(2n-1)J_{n-1} - c(n-1)J_{n-2}\right)
$$
I've been working on this for a while without any success. I've first tried to use the fact that:
$$
\int {P_n(x)dx\over \sqrt{ax^2 + bx + c}} = Q_{n-1}(x)\sqrt{ax^2 + bx + c} + \int {\lambda dx \over \sqrt{ax^2 + bx + c}}
$$
where $Q_{n-1}(x)$ is a polymonial of $n-1$ degree at max and coeeficients for $Q_{n-1}(x)$ and $\lambda$ are yet to be determined. Using polynomials from the problem statement we get:
$$
\int \frac{x^n dx}{\sqrt{ax^2 + bx + c}} = Ax^{n-1}\sqrt{ax^2 + bx + c} + \int{\lambda dx \over \sqrt{ax^2 + bx + c}}
$$
Differente both parts of the equality, after which we need to find the coefficients:
$$
\frac{x^n dx}{\sqrt{ax^2 + bx + c}} = {d(Ax^{n-1}\sqrt{ax^2 + bx + c})\over dx} + {\lambda \over \sqrt{ax^2 + bx + c}} = \\
A(n-1)x^{n-2}\sqrt{ax^2 + bx + c} + Ax^{n-1}\frac{2ax + b}{2\sqrt{ax^2 + bx + c}} + {\lambda dx \over \sqrt{ax^2 + bx + c}}$$
Multiplying both sides by $\sqrt{ax^2 + bx + c}$ and skipping some algebraic transformation I was indeed able to get that:
$$
A = {1\over na}
$$
However, the term $\lambda$ appears to be equal to $0$. Which yields:
$$
J_n = {x^{n-1}\over na}\sqrt{ax^2 + bx + c}
$$
And this approach doesn't seem to lead anywhere. I've then tried a different technique. Let $b = 2b_0$, then:
$$
aJ_{n+2} = \int \frac{ax^{n+2}dx}{\sqrt{ax^2 + 2b_0x+c}}\\
2b_0J_{n+1} = \int \frac{2b_0x^{n+1}dx}{\sqrt{ax^2 + 2b_0x+c}}\\
cJ_n = \int \frac{cx^{n}dx}{\sqrt{ax^2 + 2b_0x+c}}
$$
Summing left and right parts we get:
$$
aJ_{n+2} + 2b_0J{n+1} + cJ_n = \\
\int \frac{(ax^2 + 2b_0x + c)x^n dx}{\sqrt{ax^2 + 2b_0x + c}}= \\
\int {x^n\sqrt{ax^2 + 2b_0x + c}} dx
$$
Integration by parts yields:
$$
u = \sqrt{ax^2 + 2b_0x + c}\\
du = {2ax + 2b_0\over 2\sqrt{ax^2 + 2b_0x + c}}dx\\
dv = x^n\\
v = {x^{n+1}\over n+1}\\
\int {x^n\sqrt{ax^2 + 2b_0x + c}} dx = uv - \int vdu = \\
= {x^{n+1}\over n+1} \sqrt{ax^2 + 2b0x + c} - {J_{n+2}\over n+2} - {J_{n+1}\over 2(n+1)}
$$
Which seems to be "the other way round". The question is what technique do I use to prove what's stated in the question section? Hopefully, I didn't make typos in the body. Thank you!
| Late answer but there is an elegant way to derive the reduction formula as follows:
We have
$$
aJ_n + \frac{b}{2}J_{n-1} = \int \frac{x^{n-1}(2ax+b)dx}{2\sqrt{ax^2+bx+c}} = \int x^{n-1} d\left(\sqrt{ax^2+bx+c}\right)
$$
We then proceed by integrating by parts
$$
\int x^{n-1} d\left(\sqrt{ax^2+bx+c}\right) = x^{n-1}\sqrt{ax^2+bx+c} - (n-1)\int x^{n-2}\sqrt{ax^2+bx+c} dx
$$
To calculate $\int x^{n-2}\sqrt{ax^2+bx+c}$, note that
$$
aJ_n + bJ_{n-1} + cJ_{n-2} = \int \frac{x^{n-2}(ax^2+bx+c)}{\sqrt{ax^2+bx+c}} = \int x^{n-2}\sqrt{ax^2+bx+c}
$$
Substituting back
$$
aJ_n + \frac{b}{2}J_{n-1} = x^{n-1}\sqrt{ax^2+bx+c} - (n-1)\left(aJ_n + bJ_{n-1} + cJ_{n-2}\right)
$$
By grouping terms, we easily get the reduction formula. QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3402954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Diophantine equation with quartic polynomial
What are all integral solutions to
$$y^2=x^4+x^3+x^2+x+1$$
The RHS could become
$$(x^2-x+1)(x^2+x+1)+x(x^2+1)$$
or
$$\frac{x^5-1}{x-1}$$
I have no idea how to manipulate the equation into something useful or what the first step should be.
Also, A quartic diophantine equation looks useful, but none of the answers completely solve the question? Thanks!
| If $x \gt 0$,
the only solution is
$x=3, y=11$.
Note:
Wolfy was used extensively
getting this answer.
$y^2
=x^4+x^3+x^2+x+1
$
$y^2
=x^4+x^3+x^2+x+1
\gt x^4
\implies
y > x^2
$.
$(x+1)^4
=x^4+4x^4+,,,
$
so
$y^2
\lt (x+1)^4
$
or
$y < (x+1)^2
$.
$(x^2+x/2+3/8)^2
=x^4 + x^3 + x^2 + (3 x)/8 + 9/64
\lt y^2
$
so
$y > x^2+x/2+3/8$.
$(x^2+x/2+1)^2
=x^4 + x^3 + (9 x^2)/4 + x + 1
\gt y^2
$
so
$y < x^2+x/2+1
$.
If $x = 2n$ then
$4n^2+n+3/8
\lt y
\lt 4n^2+n+1
$,
so there can be no such
integer $y$.
If $x = 2n+1$ then
$x^2+x/2+3/8
=4n^2+4n+1+n+1/2+3/8
=4n^2+5n+15/8
$
and
$x^2+x/2+1
=4n^2+4n+1+n+1/2+1
=4n^2+5n+5/2
$
so
$y
= 4n^2+5n+2
$.
But
$y^2
=16 n^4 + 40 n^3 + 41 n^2 + 20 n + 4
$
and
$x^4+x^3+x^2+x+1
=\dfrac{x^5-1}{x-1}
=\dfrac{(2n+1)^5-1}{2n}
=16 n^4 + 40 n^3 + 40 n^2 + 20 n + 5
$
and the difference is
$n^2-1$
so they are never equal
unless $n = 1$
so $x = 3
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3403537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Please help with this limit of a sequence: $\lim\limits_{n\to\infty} \bigg(n\sqrt[3]{1+\frac1n} - n\sqrt[3]{1+\frac1{n^2}}\bigg)$ I have this limit of a sequence: $$\lim_{n\to\infty} \bigg(n\sqrt[3]{1+\frac1n} - n\sqrt[3]{1+\frac1{n^2}}\bigg).$$
Can I modify this expression to this expression by knowing that fractions goes to $0$?: $\lim n-n$. How to modify this solution to get the limit?
| If $x = (1+1/n)^{\frac{1}{3}}$ and $y = (1+1/n^{2})^{\frac{1}{3}}$, we can multiply and divide the argument by $x^{2}+xy+y^{2}$ to use $x^{3}-y^{3} =(x-y)(x^{2}+xy+y^{2})$.
Note that $$n\bigg{[}\bigg{(}1+\frac{1}{n}\bigg{)}^{\frac{1}{3}}-\bigg{(}1+\frac{1}{n^{2}}\bigg{)}^{\frac{1}{3}}\bigg{]}\frac{\bigg{[}\bigg{(}1+\frac{1}{n}\bigg{)}^{\frac{2}{3}}+\bigg{(}1+\frac{1}{n}\bigg{)}^{\frac{1}{3}}\bigg{(}1+\frac{1}{n^{2}}\bigg{)}^{\frac{1}{3}}+\bigg{(}1+\frac{1}{n^{2}}\bigg{)}^{\frac{2}{3}}\bigg{]}}{\bigg{[}\bigg{(}1+\frac{1}{n}\bigg{)}^{\frac{2}{3}}+\bigg{(}1+\frac{1}{n}\bigg{)}^{\frac{1}{3}}\bigg{(}1+\frac{1}{n^{2}}\bigg{)}^{\frac{1}{3}}+\bigg{(}1+\frac{1}{n^{2}}\bigg{)}^{\frac{2}{3}}\bigg{]}} \\ =\frac{n\bigg{[}\bigg{(}1+\frac{1}{n}\bigg{)}-\bigg{(}1+\frac{1}{n^{2}}\bigg{)}\bigg{]}}{\bigg{(}1+\frac{1}{n}\bigg{)}^{\frac{2}{3}}+\bigg{(}1+\frac{1}{n}\bigg{)}^{\frac{1}{3}}\bigg{(}1+\frac{1}{n^{2}}\bigg{)}^{\frac{1}{3}}+\bigg{(}1+\frac{1}{n^{2}}\bigg{)}^{\frac{2}{3}}} \\ = \frac{1-\frac{1}{n}}{\bigg{(}1+\frac{1}{n}\bigg{)}^{\frac{2}{3}}+\bigg{(}1+\frac{1}{n}\bigg{)}^{\frac{1}{3}}\bigg{(}1+\frac{1}{n^{2}}\bigg{)}^{\frac{1}{3}}+\bigg{(}1+\frac{1}{n^{2}}\bigg{)}^{\frac{2}{3}}}$$
Now, when $n \to \infty$ the denominator goes to 3 and the numerator goes to 1, so your limit is 1/3.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Approximation of Sum of Squares of Sine Given the sum $$\sum_{n=1}^x \sin^2(\frac{33}{n})$$ is there an approximation for this sum? Preferably, it gets closer to the actual sum as $x$ grows larger (like the approximation for the harmonic numbers).
If it helps anyone, these is a closed form for $$\sum_{n=1}^x \sin^2(n)= \frac{1}{4} (2 x - \csc(1) \sin(2 x + 1) + 1)$$
| Let us transform square of sine using $\sin^2 t=(1-\cos 2t)/2$ and using the Taylor expansion of cosine:
\begin{align*}
\sum_{n=1}^x\sin^2\left(\frac{33}n\right)&=
\frac x2-\frac 12\sum_{n=1}^x\cos\left(\frac{66}n\right)
=\frac x2-\frac 12\sum_{n=1}^x\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}\cdot\left(\frac{66}n\right)^{2k}\\
&=\frac x2-\frac 12\sum_{k=0}^\infty\frac{(-4356)^k}{(2k)!}\cdot H_x^{(2k)}
\end{align*}
where $H_x^{(2k)}$ is the generalized harmonic number.
I have got the following additional idea:
\begin{align*}
\sum_{n=1}^x\sin^2\left(\frac{33}n\right)
&=\frac x2-\frac 12\sum_{n=1}^x\cos\left(\frac{66}n\right)
=\frac x2-\frac 12\sum_{n=1}^x\Re\left(\cos\frac 1n+\mathrm i\cdot\sin\frac 1n\right)^{66}\\[12pt]
&=\frac x2-\frac 12\sum_{n=1}^x\Re\left(\sum_{k=0}^{66}\binom{66}{k}\cos^k\frac 1n\cdot\mathrm i^{66-k}\cdot\sin^{66-k}\frac 1n\right)\\[12pt]
&=\frac x2-\frac 12\sum_{n=1}^x\Re\left(\sum_{k=0}^{66}\binom{66}{k}\cos^k\frac 1n\cdot\mathrm i^{66-k}\cdot\sin^{66-k}\frac 1n\right)\\[12pt]
&=\frac x2-\frac 12\sum_{n=1}^x\sum_{\substack{k=0\\\text{$k$ is even}}}^{66}\binom{66}{k}(-1)^{\frac k2+1}\cos^k\frac 1n\cdot\sin^{66-k}\frac 1n\\[12pt]
&=\frac x2-\frac 12\sum_{n=1}^x\sum_{j=0}^{33}\binom{66}{2j}(-1)^{j+1}\cos^{2j}\frac 1n\cdot\sin^{66-2j}\frac 1n\\[12pt]
&\approx\frac x2-\frac 12\sum_{n=1}^x\sum_{j=0}^{33}\binom{66}{2j}(-1)^{j+1}\left(1-\frac 1{2!\cdot n^2}+\frac 1{4!\cdot n^4}-\frac 1{6!\cdot n^6}\right)^{2j}\\[12pt]
&\quad\cdot\left(\frac 1n-\frac 1{3!\cdot n^3}+\frac 1{5!\cdot n^5}-\frac 1{7!\cdot n^7}\right)^{66-2j}
\end{align*}
The more terms of Taylor series of sine and cosine function are used the better approximation you get.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Definite Integral $\int_0^2 \frac{\ln(1+x)}{x^2-x+1}dx$ I was working on this problem, I probably need to abuse symmetry somewhere but can't see how:
$$\int_0^2 \frac{\ln(1+x)}{x^2-x+1}dx$$
| \begin{align*}
I&=\underbrace{\int_0^2\frac{\ln(x+1)}{x^2-x+1}dx}_{x+1\mapsto x}\\
&=\underbrace{\int_1^3\frac{\ln x}{x^2-3x+3}dx}_{x\mapsto 3/x}\\
&=\int_3^1\frac{\ln\left(\frac{3}{x}\right)}{\frac{9}{x^2}-\frac{9}{x}+3}\left(-\frac{3}{x^2}dx\right)\\
&=\int_1^3\frac{\ln 3-\ln x}{x^2-3x+3}dx\\
&=\int_1^3\frac{\ln 3}{x^2-3x+3}dx-I\\
&=\frac{\ln 3}{2}\int_1^3\frac{1}{x^2-3x+3}dx\\
&=\frac{\ln 3}{2}\left[\frac{2}{\sqrt{3}}\arctan\left(\frac{2}{\sqrt{3}}\left(x-\frac{3}{2}\right)\right)\right]_{x=1}^3\\
&=\frac{\ln 3}{\sqrt{3}}\left(\arctan\left(\sqrt{3}\right)-\arctan\left(-\frac{1}{\sqrt{3}}\right)\right)\\
&=\frac{\ln 3}{\sqrt{3}}\left(\frac{\pi}{3}+\frac{\pi}{6}\right)\\
&=\frac{\pi\ln 3}{2\sqrt{3}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim_{x \to \ 0}{\frac{1-\sqrt{\cos x}}{1-\cos\sqrt{x}}}$ without using the l'Hospital rule The task is to evaluate
$$\lim_{x \to \ 0}{\frac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}}.$$
I tried to use some trigonometric identities such as
$$\lim_{x \to \ 0}{\frac{1-\sqrt{\cos\left(x\right)}}{1-\cos\left(\sqrt{x}\right)}}= \lim_{x \to \ 0} \frac{1-
\sqrt{\cos\left(x\right)}}{1-
\cos\left(\sqrt{x}\right)}\cdot\frac{1+\sqrt{\cos\left(x\right)}}{1+\sqrt{\cos\left(x\right)}}$$$$= \lim_{x \to \ 0} \frac{1-
\cos\left(x\right)}{2\sin^{2}\left(\frac{\sqrt{x}}{2}\right)}\cdot\frac{1}{1+\sqrt{\cos\left(x\right)}}$$$$=
\lim_{x \to \ 0} \left(\frac{\sin\left(\frac{\sqrt{x^{2}}}{2}\right)}{\sin\left(\frac{\sqrt{x}}{2}\right)}
\right)^{2}\cdot\frac{1}{1+\sqrt{\cos\left(x\right)}}$$$$=\frac{1}{2}\lim_{x \to \
0}\left(\frac{\sin\left(\frac{\sqrt{x^{2}}}{2}\right)}{\sin\left(\frac{\sqrt{x}}{2}\right)}\right)^{2}$$
and this is where I have a problem.
| Easy with equivalents: as $\sin u\sim_{u\to 0} u$, and equivalence is compatible with multiplication and division, we have
$$\left(\frac{\sin\left(\frac{\sqrt{x^{2}}}{2}\right)}{\sin\left(\frac{\sqrt{x}}{2}\right)}\right)^{\!\!2}\sim_{x\to 0}\left(\frac{\frac{\sqrt{x^{2}}}{2}}{\frac{\sqrt{x}}{2}}\right)^{\!\!2}=\bigl(\sqrt x\bigr)^2=x \qquad(\text{as } x\ge 0)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
what is $\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}- 2\sqrt{x^{2}+x}+x\right)$? $$\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-2\sqrt{x^{2}+x}+x\right)$$$$=\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-1-2\sqrt{x^{2}+x}+1+x\right)$$$$=\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-1\right)+\lim\limits_{x\to\infty}\left(-2x\right)\left(\sqrt{x^{2}+x}+1\right)+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}x\lim\limits_{x\to\infty}\left(x\sqrt{1+\frac{2}{x}}-1\right)+\lim\limits_{x\to\infty}\left(-2x\right)\lim\limits_{x\to\infty}\left(x\sqrt{1+\frac{1}{x}}+1\right)+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}x\left(1\right)+\lim\limits_{x\to\infty}\left(-2x\right)\left(\frac{1}{2}\right)+\lim\limits_{x\to\infty}x^{2}= ∞$$
another way I tried:$$\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-2\sqrt{x^{2}+x}\right)+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-2\sqrt{x^{2}+x}\right)\cdot\frac{\sqrt{x^{2}+2x}+2\sqrt{x^{2}+x}}{\sqrt{x^{2}+2x}+2\sqrt{x^{2}+x}}+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}\frac{x\left(-2x^{2}-2x\right)}{\sqrt{x^{2}+2x}+2\sqrt{x^{2}+x}}+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}\frac{x\left(-2x^{2}-2x\right)}{x\left(\sqrt{1+\frac{2}{x}}+2\sqrt{1+\frac{1}{x}}\right)}+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}\frac{\left(-2x^{2}-2x\right)}{\left(\sqrt{1+\frac{2}{x}}+2\sqrt{1+\frac{1}{x}}\right)}+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}\frac{-2x^{3}-2x}{3}+\lim\limits_{x\to\infty}x^{2}=\lim\limits_{x\to\infty}\frac{x^{3}\left(-2+\frac{3}{x}-\frac{2}{x^{2}}\right)}{3}=-∞$$
and none of them are the answer, why the solutions are not right and can someone use an elementary way to solve the limit?
| As I mentioned in the comments you cant distribute the limit if it does not exists for instance you cant distribute the limit if the limit of one function is infinity as what you did
The idea to get rid of the negative sign make it positive to do that we need to multiply with the conjugate regroup the positive terms together and the negative together then the following
$$\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-2\sqrt{x^{2}+x}+x\right)$$
$$\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}+ x -2\sqrt{x^{2}+x}\right)$$
$$\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-2\sqrt{x^{2}+x}+x\right)\frac{\sqrt{x^{2}+2x}+x+2\sqrt{x^{2}+x}}{\sqrt{x^{2}+2x}+x+2\sqrt{x^{2}+x}}$$
$$\lim_{x\rightarrow \infty} \frac{x\left( x^2 +2x + 2x\sqrt{x^2+2x} + x^2 - 4(x^2+x)\right)}{\sqrt{x^{2}+2x}+x+2\sqrt{x^{2}+x}}$$
$$\lim_{x\rightarrow \infty} \frac{ x\left( 2x\sqrt{x^2+2x}-2x^2 -2x\right)}{\sqrt{x^{2}+2x}+x+2\sqrt{x^{2}+x}}$$
$$\lim_{x\rightarrow \infty} \frac{ 2x^2\left( \sqrt{x^2+2x}-(x +1)\right)}{\sqrt{x^{2}+2x}+x+2\sqrt{x^{2}+x}}$$
Multiply with the conjugate again to get
$$\lim_{x\rightarrow \infty} \frac{2x^2(x^2+2x - (x+1)^2)}{\left(\sqrt{x^2+2x} + x + 2\sqrt{x^2+x}\right)\left(\sqrt{x^2+2x} + (x+1)\right)} $$
Can you continue ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What are the roots of this equation? (Quadratic Equation) $\frac{x}{\sqrt{1-x^2}}=A \Rightarrow \frac{x^2}{1-x^2}=A^2$
$\Rightarrow x^2=A^2(1-x^2)=A^2-A^2x^2$
$\Rightarrow x^2+A^2x^2=A^2$
$\Rightarrow x^2(1+A^2)=A^2$
$\Rightarrow x^2(1+A^2)-A^2=0$
I have tried this;
$\Delta=b^2-4ac=-4(1+A^2)(-A^2)=4(1+A^2)(A^2)$
$x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{\pm2A\sqrt{1+A^2}}{{2(1+A^2)}}$
$x_{1,2}=\frac{\pm A\sqrt{1+A^2}}{1+A^2}$
But I got nothing.
| As you seem to find $x = \frac {A\sqrt{A^2 + 1}}{A^2 + 1}$ to be unsatisfying.
I don't know if you find this any more satisfying.
$\frac {x}{\sqrt {1-x^2}} = \tan (\arcsin x)= A\\
x = \sin (\arctan A)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 3,
"answer_id": 2
} |
Evaluate $\lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)^{n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-(n+1)^2}$ using the definition of $e$
Evaluate $\lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)^{n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-(n+1)^2}.$
I know that $e^x=\lim\limits_{n\to\infty}\left(1+\dfrac{x}{n}\right)^n,$ so I need to somehow convert the limits to this form. I also noticed that $\left(1+\frac{1}{n+1}\right)=\left(\frac{n+2}{n+1}\right)=\left(\frac{n(n+2)}{(n+1)^2}\right)\cdot\left(1+\frac{1}{n}\right).$
Thus, the limit can be rewritten as $$\lim\limits_{n\to\infty}\dfrac{\left(1+\frac{1}{n}\right)^{n^2}}{\left(1+\frac{1}{n}\right)^{(n+1)^2}}\cdot\left(\dfrac{(n+1)^2}{n(n+2)}\right)^{(n+1)^2}\\
=\lim\limits_{n\to\infty}\left(1+\dfrac{1}{n}\right)^{-2n}\cdot\left(1+\dfrac{1}{n}\right)^{-1}\cdot\left(1+\dfrac{1}{n^2+2n}\right)^{n^2+2n}\cdot\left(1+\dfrac{1}{n^2+2n}\right)$$
$$=\dfrac{1}{e^2}\cdot(1)\cdot e\cdot(1)=\dfrac{1}{e}$$
| This is not what I would consider an answer to the question but I decided to provide an in depth asymptotic analysis of the behaviour of this function in the limit as $x\to\infty$. Considering
$$f(x)=\left(1+\frac1x\right)^{x^2}\cdot\left(1+\frac1{x+1}\right)^{-(x+1)^2}$$
then we have
$$\begin{align}
\ln{(f(x))}
&=x^2\ln{\left(1+\frac1x\right)}-(x+1)^2\ln{\left(1+\frac1{x+1}\right)}\\
&=x^2\left(\frac1x-\frac1{2x^2}+\frac1{3x^3}-\frac1{4x^4}+o\left(\frac1{x^4}\right)\right)-(x+1)^2\left(\frac1{x+1}-\frac1{2(x+1)^2}+\frac1{3(x+1)^3}+\frac1{4(x+1)^4}+o\left(\frac1{x^4}\right)\right)\\
&=\left(x-\frac12+\frac1{3x}-\frac1{4x^2}+o\left(\frac1{x^2}\right)\right)-\left((x+1)-\frac1{2}+\frac1{3(x+1)}-\frac1{4(x+1)^2}+o\left(\frac1{x^2}\right)\right)\\
&=-1+\frac1{3x}-\frac1{4x^2}-\frac1{3(x+1)}+\frac1{4(x+1)^2}+o\left(\frac1{x^2}\right)\\
&=-1+\frac1{3x}\left(1-\frac1{1+\frac1x}\right)-\frac1{4x^2}\left(1-\frac1{(1+\frac1x)^2}\right)+o\left(\frac1{x^2}\right)\\
&=-1+\frac1{3x}\left(1-\left(1-\frac1x+\frac1{x^2}+o\left(\frac1{x^2}\right)\right)\right)-\frac1{4x^2}\left(1-\left(1-\frac2x+\frac3{x^2}+o\left(\frac1{x^2}\right)\right)\right)+o\left(\frac1{x^2}\right)\\
&=-1+\frac1{3x}\left(\frac1x-\frac1{x^2}+o\left(\frac1{x^2}\right)\right)-\frac1{4x^2}\left(\frac2x-\frac3{x^2}+o\left(\frac1{x^2}\right)\right)+o\left(\frac1{x^2}\right)\\
&=-1+\frac1{3x^2}+o\left(\frac1{x^2}\right)\\
\end{align}$$
$$\begin{align}
\therefore f(x)
&=\exp{\left(-1+\frac1{3x^2}+o\left(\frac1{x^2}\right)\right)}\\
&=\frac1e\cdot\exp{\left(\frac1{3x^2}+o\left(\frac1{x^2}\right)\right)}\\
&=\frac1e\left(1+\frac1{3x^2}+o\left(\frac1{x^2}\right)+o\left(\frac1{3x^2}+o\left(\frac1{x^2}\right)\right)\right)\\
&=\frac1e\left(1+\frac1{3x^2}+o\left(\frac1{x^2}\right)\right)\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Cubic roots of the equation $x^3-x-2=0$ If $\alpha,\beta,\gamma$ are the cubic roots of the equation $x^3-x-2=0$, then find the value of $\alpha^5+\beta^5+\gamma^5$.
One of the root is real and two roots are imaginary.
I get three equation.
$\alpha+\beta+\gamma=0$;
$\alpha\beta+\beta\gamma+\gamma\alpha=-1$
$\alpha\beta\gamma=2$
But not able to get the requisite result.
| Alternative approach that is very similar to the answer of farruhota.
Suppose that the roots are $a,b,c$.
You know that each value of $a,b,c$ satisfies
$$x^3 = x + 2.$$
From the coefficient of the $x^2$ term, you also know that
$$(a + b + c) = 0.$$
Also, from the coefficient of the $x^1$ term, you know that
$$ab + ac + bc = -1.$$
This allows you to compute $(a^2 + b^2 + c^2)$, as follows:
$$0^2 = (a + b + c)^2 = (a^2 + b^2 + c^2) + 2(ab + ac + bc)$$
$$= (a^2 + b^2 + c^2) + 2(-1) \implies $$
$$(a^2 + b^2 + c^2) = 2.$$
So, now you can compute
$$a^5 + b^5 + c^5$$
$$= (a^3 \times a^2) + (b^3 \times b^2) + (c^3 \times c^2)$$
$$= [(a + 2) \times a^2] + [(b + 2) \times b^2] + [(c + 2) \times c^2]$$
$$= (a^3 + b^3 + c^3) + 2(a^2 + b^2 + c^2)$$
$$= [(a + 2) + (b + 2) + (c + 2)] + 2(2)$$
$$= [0 + 6] + 4 = 10.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3412237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How level sets associated with a quadratic expression can switch from one to two components? So I have two equations:
$$
x^2 + y^2 - z^2 = 1/2
$$
and
$$
x^2 + y^2 - z^2 = -1/2,
$$
which represent single-sheeted hyperboloid, and a two-sheeted hyperboloid, respectively. How do you explain how this different sign has this effect on the level surfaces?
| Note that the expression $x^2+y^2-z^2$ has rotational symmetry about the $z$-axis, which can be seen by introducing polar coordinates:
*
*$x=r\cos t, y= r\sin t \Rightarrow r^2-z^2 = \pm\frac{1}{2}$
This means, any intersection of a plane containing the $z$-axis with $x^2+y^2-z^2 = \pm \frac{1}{2}$ produces the curve $r^2-z^2 = \pm\frac{1}{2}$ with $r\geq 0$.
Factoring shows that we have in both cases a hyperbolic curve:
*
*$(r-z)(r+z) = \pm\frac{1}{2} \Leftrightarrow r+z = \pm\frac{1}{2(r-z)}$
Now, just check the shape of the two curves (for example plotting on an $r$-$z$-coordinate system):
*
*$r^2-z^2 =\frac{1}{2} \Leftrightarrow z = \pm\sqrt{r^2-\frac{1}{2}} \Rightarrow$ only one branch (the seeming two branches are connected at $z =0 \Leftrightarrow r=\frac{1}{\sqrt{2}}$) $\Rightarrow$ one sheet
*$r^2-z^2 =-\frac{1}{2} \Leftrightarrow z = \pm\sqrt{r^2+\frac{1}{2}} \Rightarrow$ two branches $\Rightarrow$ two sheets
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3412550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Verification of the integral: $\int \frac{dx}{a\cos^2x+b\sin(2x)+c\sin^2x}$ I have solved several integrals (of the form below) with given coefficients for $a, b, c$ and then though whether I could generalize that or not. I would like to ask for verification. Problem statement:
Evaluate:
$$
\int \frac{dx}{a\cos^2x+b\sin(2x)+c\sin^2x}
$$
Below is what I got so far. I've started by factoring out $1\over \cos^2x$ to obtain:
$$
I = \int \frac{1}{\cos^2x}\cdot\frac{dx}{a + 2b\tan x + c\tan^2x}
$$
Now substitute $t = \tan x$:
$$
I = \int \frac{dt}{a + 2bt + ct^2}
$$
Now depending on the coefficients, we might have three cases: either there are 2 distinct real roots, both roots are equal, or there are no roots in $\Bbb R$.
If there are no roots, then we could complete the square and consider a standard integral. Suppose:
$$
ct^2 + 2bt + a = c(t-h)^2 + k \\
ct^2 + 2bt + a = ct^2 + 2ch t+ch^2 + k
$$
After solving a linear system of equations one may obtain:
$$
h = {b\over c}\\
k = {ac + b^2\over c}
$$
Hence the integral becomes:
$$
I = {1\over c}\int \frac{dt}{\left(t - {b\over c}\right)^2 + {ac + b^2 \over c^2}}
$$
Which is a standard integral:
$$
\boxed{I = {1\over \sqrt{ac + b^2}}\arctan\left(\frac{c\tan x + b}{\sqrt{ac-b^2}}\right) + Const}
$$
As for the second case, namely when there are two distinct roots. After solving a quadratic equation:
$$
t_{1,2} = \frac{-b \pm \sqrt{b^2 - ac}}{c}
$$
Let:
$$
R_1 = \frac{-b + \sqrt{b^2 - ac}}{c}\\
R_2 = \frac{-b - \sqrt{b^2 - ac}}{c}
$$
So the integral becomes:
$$
I = \int \frac{dt}{(t-R_1)(t-R_2)}
$$
Using partial fraction decomposition:
$$
\frac{1}{ct^2 + 2bt + a} = \frac{A}{t - R_1} + \frac{B}{t-R_2} \\
At - AR_2 + Bt - BR_1 = 1\\
\begin{cases}
A + B = 0\\
-AR_2-BR_1 = 1
\end{cases}\\
A = -B\\
B = {1\over (R_2-R_1)}
$$
Therefore:
$$
I = \int \frac{1}{(R_1 - R_2)(t - R_1)}dt - \int \frac{1}{(R_1-R_2)(t-R_2)}dt
$$
Which yields:
$$
\begin{align}
I &= {1\over R_1 - R_2}\ln|t - R_1| - {1\over R_1 - R_2}\ln|t - R_2| \\
&={1\over R_1 - R_2}\ln|\tan x - R_1| - {1\over R_1 - R_2}\ln|\tan x - R_2| \\
&=\boxed{{1\over R_1 - R_2}\ln\left|{\tan x - R_1\over \tan x - R_2}\right| + Const}
\end{align}
$$
When there are two equal roots, then let the root be denoted by $R_1 = R_2 = R$:
$$
I = \int {dt\over (t - R)^2} = -\frac{1}{t - R}
$$
Or simply:
$$
\boxed{I = -\frac{1}{\tan x - R} = \frac{1}{R-\tan x} + Const}
$$
I'm wondering whether I've done this right, could someone please verify the above? Thank you!
| Taking into account the small corrections in the comments, this is correct for the cases with real roots. The case with no real roots has a subtle error that often gets glossed over in these sorts of problems.
The issue is that when there are no real roots, $[a \cos(x)^2 + b\sin(2x) + c \sin(x)^2]^{-1}$ is continuous everywhere. This means its antiderivative must be everywhere differentiable, but what you have for its antiderivative is not, since $\tan^{-1}$ jumps down by $\pi$ every time its argument switches from $+\infty$ to $-\infty$ (or up if it's switching the other way because $c < 0$). To restore continuity, we need to add (or subtract, if $c < 0$) $\pi$ to it every time $x$ passes an odd multiple of $\pi/2$. This can be done with the floor function, giving
$$
I = \frac{1}{\sqrt{ac - b^2}}\left(\tan^{-1}\left[\frac{c\tan(x)+b}{\sqrt{ac-b^2}}\right] +\pi\left\lfloor\frac{x}{\pi}+\frac{1}{2}\right\rfloor\operatorname{sgn}[c]\right) + Const,
$$
where $\operatorname{sgn}$ is the sign function. As you can see in this plot for the $a=2,\,b=c=1$ case, the extra term restores continuity of the antiderivative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3418024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find the general solution of $\theta$ for which the following quadratic equation is the square of a linear function. Find the general solution of $\theta$ for which the quadratic equation
$$\left(\sin\theta\right)x^2+(2\cos\theta)x+\dfrac{\cos\theta+\sin\theta}{2}$$ is the square of a linear function.
$$D=0$$
$$4\cos^2\theta-2\sin\theta\left(\sin\theta+\cos\theta\right)=0$$
$$2\cos^2\theta-\sin^2\theta-\sin\theta\cos\theta=0$$
$$2\cos^2\theta-2\sin\theta\cos\theta+\sin\theta\cos\theta-\sin^2\theta=0$$
$$2\cos\theta(\cos\theta-\sin\theta)+\sin\theta(\cos\theta-\sin\theta)=0$$
$$(\cos\theta-\sin\theta)(2\cos\theta+\sin\theta)=0$$
$$\tan\theta=1 \text { or } \tan\theta=-2$$
$$\theta=n\pi+\dfrac{\pi}{4} \text { or } \theta=\tan(2\pi-\tan^{-1}(2))$$
$$\theta=n\pi+\dfrac{\pi}{4} \text { or } \theta=n\pi+2\pi-\tan^{-1}(2)$$
$$\theta=n\pi+\dfrac{\pi}{4} \text { or } \theta=\pi(n+2)-\tan^{-1}(2)$$
$$\sin\theta\ne 0$$
$$\theta\ne m\pi \text { where $m \in$ I }$$
$$\theta=n\pi+\dfrac{\pi}{4} \text { can't be integral multiple of $\pi$ as } \theta=\dfrac{\pi(4n+1)}{4}$$
$$\theta=n\pi+\dfrac{\pi}{4} \text { is the valid solution }$$
$$\theta=\pi(n+2)-\tan^{-1}(2) \text { cannot be the integral multiple of $\pi$ as $\tan^{-1}(2)$ is not the integral multiple of $\pi$ } $$
$$\theta=\pi(n+2)-\tan^{-1}(2) \text { is the valid solution }$$
Hence $\theta=\pi(n+2)-\tan^{-1}(2) \text { or } \theta=n\pi+\dfrac{\pi}{4}$
But actual answer is $\theta= 2n \pi+\dfrac{\pi}{4} \text{or } \theta =(2n+1)\pi - \tan^{-1}(2) \text { where $n \in$ I}$
I tried to find out the mistake but didn't get any breakthrough. What am I missing.
| By tangent half angle formulas, by $t=\tan \theta$, we have that
$$4\cos^2\theta-2\sin\theta\left(\sin\theta+\cos\theta\right)=0 \iff 4 \cos^2 \theta-2\sin^2 \theta-2\sin \theta \cos \theta=0$$
$$6\frac{1+\cos (2\theta)}2-\sin (2\theta)-2=0$$
$$3\cos (2\theta)-\sin (2\theta)+1=0$$
$$3\frac{1-t^2}{1+t^2}- \frac{2t}{1+t^2}+1=0$$
$$3-3t^2-2t+1+t^2=0 \iff t^2+t-2=0 \iff (t-1)(t+2)=0$$
and by the original equation we also need $\sin \theta \ge 0$, therefore the solutions are
$$\theta=\frac \pi 4 +2k\pi \quad \lor \quad \theta=\pi -\arctan(2) +2k\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3418655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding the area of a quadrilateral with two points given and other lie on a line (Coordinate Geometry) The question is as follows:
The diagram shows the quadrilateral $ABCD$ in which $A$ is the point $(4, 2)$ and $B$ is the point $(-2, -10)$. The points $C$ and $D$ lie on the line $x = 14$. The diagonal $AC$ is perpendicular to $AB$ and passes through the mid-point, $M$, of the diagonal $BD.$ Find the area of the quadrilateral $ABCD.$
I found $C$ and proceeded to try and find the mid-point, using colinear points. It came to nothing however. Can anyone on this forum help me?
| Since the diagonal $AC$ is perpendicular to the diagonal $AB$ and $AB$ has slope $-\dfrac{4}{3},$ the diagonal $AC$ has slope $\dfrac{3}{4}.$ Since it contains the point $A (4,2),$ it has equation $y-2=\dfrac{3}{4}(x-4).$ So the $y$-coordinate of point $C$ is simply the value of $y$ when $x=14$ is substituted into this equation. This gives that $C$ is the point $(14,9.5).$ Let $D$ have coordinates $(14,a).$ Then the midpoint of $BD$ is $\left(6,5+\dfrac{a}{2}\right).$ Since the diagonal $AC$ passes through this point, we have that $3+\dfrac{a}{2}=\dfrac{3}{2}\Rightarrow a=-3.$
So the area is simply the area of the rectangle with width parallel to the $x$-axis equal to $16$ and with height parallel to the $y$-axis equal to $13$ minus the sum of the areas of the triangles and rectangle that are not included in the quadrilateral. This gives that the area is $16\cdot 13 - \dfrac{1}{2}\cdot16\cdot \dfrac{1}{2}- \dfrac{1}{2}\cdot8\cdot 4-\dfrac{1}{2}\cdot 5\cdot 10-5\cdot 4=\boxed{143.}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Equation of a straight line in the treated segments Problem: $$4x - 7y -2 = 0$$
But the real problem is that in my math book, the result of this is different than my solution.
This is my progress so far:
$$4x - 7y - 2 = 0$$
$$\frac{4x}{2} - \frac{7y}{2} = 1$$
This is result that should be:
$$
\frac{x}{1/2} + \frac{y}{-2/7} = 1
$$
My question is how this is possible?
| Starting from
$$\frac{4x}{2} - \frac{7y}{2} = 1$$
recognize that you can simplify the fraction in the first term since ${4}/{2}={2}/{1}$
$$\frac{2x}{1} - \frac{7y}{2} = 1$$
then observe that $z=\frac{1}{1/z}$ for any $z\neq 0$. Also, $\frac{1}{a/b}=\frac{b}{a}$ if both $a\ne 0$ and $b\ne 0$. This allows you to conclude that
$$\frac{2}{1}=\frac{1}{\frac{1}{2/1}}=\frac{1}{{1}/{2}}$$
$$\frac{7}{2}=\frac{1}{\frac{1}{7/2}}=\frac{1}{{2}/{7}}$$
therefore
$$\frac{2x}{1} - \frac{7y}{2} = \frac{x}{1/2} - \frac{y}{2/7} = \frac{x}{1/2} + \frac{y}{-2/7}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3421336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving $\frac{4}{3} = \frac{3.2}{y}$ This is what I've tried.
$\frac{1}{3.2} \cdot \frac{4}{3} = \frac{3.2}{y} \cdot \frac{1}{3.2}$ to isolate $y$ via product and quotient identity.
When simplified this gives me $\frac{4}{9.6} = y$ so $y = 0.41\overline{6}$
However, the textbook gives an answer of $y=2.4$ and in the example they used the cross-product method which does follow, but what I did should also give me the same answer.
| If you multiply $\dfrac{1}{3.2}$ both side:
$$\dfrac{1}{3.2}\times\dfrac{4}{3}=\dfrac{3.2}{y}\times\dfrac{1}{3.2}\\\dfrac{4}{9.6}=\dfrac{1}{y}\\\dfrac{1}{2.4}=\dfrac{1}{y}\\y=\boxed{2.4}$$
Another way to do this is to use cross method:
$$\dfrac{4}{3}=\dfrac{3.2}{y}\\ 4y=3.2\times 3 \\4y=9.6 \\ y=\boxed{2.4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3421457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Evaluate $\int_0^\frac{1}{16}\int_{y^{\frac{1}{4}}}^\frac{1}{4} \cos{\left(16\pi x^5\right)}dx\,dy$ $\displaystyle\int_0^\frac{1}{16}\int_{y^{\frac{1}{4}}}^\frac{1}{4} \cos{\left(16\pi x^5\right)dx\,dy}$
My attempt:
Reverse the order of integration
Then the integral will be:
$$\displaystyle\int_{x^4}^\frac{1}{16}\int_{0}^\frac{1}{4} \cos{\left(16\pi x^5\right)dx\,dy}$$
Using wolfram, this is what I found:
$$0.0154$$
But the correct answer is $-0.0085$. Where did I go wrong?
| The region of integration has two wings. Thus in reversing the order of integration you are going to have two integrals namely $$\displaystyle\int_0^\frac{1}{16}\int_{y^{\frac{1}{4}}}^\frac{1}{4} \cos{\left(16\pi x^5\right)dx\,dy}=\displaystyle\int_{0}^\frac{1}{4}\int_{0}^{x^4} \cos{\left(16\pi x^5\right)dy\,dx} -$$
$$\displaystyle\int_{\frac {1}{4}}^\frac{1}{2}\int_{x^4}^\frac{1}{16} \cos{\left(16\pi x^5\right)dy\,dx}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3422895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A question from Munkres' Analysis on Manidolds (P. 55, E. 10)
*
*Define $f: \mathbf{R}^{2} \rightarrow \mathbf{R}$ by setting $f(\mathbf{0})=0$, and
$$
f(x, y)=x y\left(x^{2}-y^{2}\right) /\left(x^{2}+y^{2}\right) \text { if }(x, y) \neq 0
$$
i) Show $f$ is of class $C^{1}$ on $\mathbf{R}^{2} .$ [Hint: Show $D_{1} f(x, y)$ equals the product of $y$ and a bounded function, and $D_{2} f(x, y)$ equals the product
of $x \text { and a bounded function. }]$
My Attempt. $D_1 f(x,y)=\frac{d}{d x}\left(\frac{x y\left(x^{2}-y^{2}\right)}{x^{2}+y^{2}}\right)=\frac{y\left(x^{4}+4 x^{2} y^{2}-y^{4}\right)}{\left(x^{2}+y^{2}\right)^{2}}$ and $D_2 f(x,y)=\frac{d}{d y}\left(\frac{x y\left(x^{2}-y^{2}\right)}{x^{2}+y^{2}}\right)=\frac{x\left(-y^{4}-4 x^{2} y^{2}+x^{4}\right)}{\left(x^{2}+y^{2}\right)^{2}}.$
Now, how can I use the hint, may you help? Thanks...
Can we say $\frac{\left(x^{4}+4 x^{2} y^{2}-y^{4}\right)}{\left(x^{2}+y^{2}\right)^{2}}$ is a bounded function in $D_1 f(x,y)$?
| Switch to polar coordinates:
$D_1 f(x,y)=\frac{d}{d x}\left(\frac{x y\left(x^{2}-y^{2}\right)}{x^{2}+y^{2}}\right)=\frac{y\left(x^{4}+4 x^{2} y^{2}-y^{4}\right)}{\left(x^{2}+y^{2}\right)^{2}}=y\left(\frac{r^4(\cos^4 t+4\cos^2 t \sin^2 t-\sin^4t )}{r^4}\right)\le 6y$.
Similarly for $D_2f(x,y)$
| {
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Is $| \lceil \frac{a}{2} \rceil - \lceil \frac{b}{2} \rceil |\geq \lfloor |\frac{a - b}{2}| \rfloor $? Let $a$ and $b$ be integers. Is it true that
$$
\left | \left \lceil \frac{a}{2} \right \rceil - \left \lceil \frac{b}{2} \right \rceil \right |\geq \left \lfloor \left | \frac{a - b}{2} \right |\right \rfloor
$$
Where $\lceil \cdot \rceil$ is the ceiling function, $\lfloor \cdot \rfloor$ the floor function and $|\cdot|$ is the absolute function.
The inequality seems to be true when I check it programatically but I would like to get a proof (or disproof) for this inequality.
| Yes, it is true.
$$
\left | \left \lceil \frac{a}{2} \right \rceil - \left \lceil \frac{b}{2} \right \rceil \right |\geq \left \lfloor \left | \frac{a - b}{2} \right |\right \rfloor
\tag1$$
In the following, $m,n$ are integers.
Case 1 : If $a=2m,b=2n$, then both sides of $(1)$ equal $|m-n|$.
Case 2 : If $a=2m,b=2n+1$, then
$$(1)\iff |m-n-1|\ge \left\lfloor\left |m-n-\frac 12\right|\right\rfloor\tag2$$
If $m-n-\frac 12\ge 0$, then $m-n-1\ge 0$, so$$(2)\iff m-n-1\ge m-n-1$$which is true.
If $m-n-\frac 12\lt 0$, then $m-n-1\lt 0$, so$$(2)\iff -m+n+1\ge -m+n$$which is true.
Case 3 : If $a=2m+1, b=2n$, then
$$(1)\iff |m-n+1|\ge \left\lfloor\left|m-n+\frac 12\right|\right\rfloor\tag3$$
If $m-n+\frac 12\ge 0$, then $m-n+1\ge 0$, so$$(3)\iff m-n+1\ge m-n$$which is true.
If $m-n+\frac 12\lt 0$, then $m-n+1\lt 0$, so$$(3)\iff -m+n-1\ge -m+n-1$$which is true.
Case 4 : If $a=2m+1,b=2n+1$, then both sides of $(1)$ equal $|m-n|$.
| {
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Finding the number of continuous functions \begin{gather*}
No.\ of\ continuous\ function( s) \ \ f:[ 0,1]\rightarrow \ \Re \ \ satisfying\\
\int ^{1}_{0} f( x) dx\ =\ \frac{1}{3} \ +\int ^{1}_{0} f^{2}\left( x^{2}\right) dx\ \ is/are
\end{gather*}
My approach:
I put $x^2$=$t$, giving $2xdx=dt$, but I am not able to find/ proceed further. Can anyone help please?
| You should do the same but inversely:
$$\int ^{1}_{0} f(x) dx = \int ^{1}_{0} f(t^2) 2tdt = \int ^{1}_{0} 2x f(x^2) dx$$
So we have:
$$\int ^{1}_{0} 2x f(x^2) dx = \frac{1}{3} +\int ^{1}_{0} f^{2}\left( x^{2}\right) dx \Longrightarrow \int ^{1}_{0} f^{2}\left( x^{2}\right) dx - \int ^{1}_{0} 2x f(x^2) dx + \frac{1}{3} = 0$$
$$\Longrightarrow 0 = \int ^{1}_{0} \left(f^{2}\left( x^{2}\right) - 2x f(x^2) \right)dx + \frac{1}{3} = \int ^{1}_{0} \left(\left(f\left( x^{2}\right) - x\right)^2 -x^2 \right)dx + \frac{1}{3} = $$
$$ = \int ^{1}_{0} \left(f\left( x^{2}\right) - x\right)^2 dx - \int ^{1}_{0}x^2dx + \frac{1}{3} = \int ^{1}_{0} \left(f\left( x^{2}\right) - x\right)^2 dx - \left[\frac{x^3}{3}\right]_0^1 + \frac{1}{3}$$
$$\Longrightarrow \int ^{1}_{0} \left(f\left( x^{2}\right) - x\right)^2 dx = 0$$
Since left side is a non negative function, it must be zero everywhere:
$$\left(f\left( x^{2}\right) - x\right)^2 = 0 \Longrightarrow f\left( x^{2}\right) - x = 0 \Longrightarrow f\left( x^{2}\right) = x \Longrightarrow f(x) = \sqrt x $$
| {
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Find all $n$ natural numbers such that : $n^{2}-3$ divisible by a perfect square $m>1$ Problem :
Find all $n$ natural numbers such that : $k=n^{2}-3$ divisible by a perfect square $m>1$
I'm going to find the smallest number then find all this numbers
I was tired many time of $n$
$n=2$ so $k=1$ $×$
$n=3$ so $k=6$ $×$
$n=4$ so $k=13$ $×$
$n=5$ so $k=22$ $×$
$n=6$ so $k=33$ $×$
.
.
.
I don't know which number must be try it ?
And how I find all this numbers $n=?$
| $$(22+5)^2-3=22^2+220+25-3\equiv -22+22\pmod{121}$$
$$(65-4)^2-3=65^2-520+16-3\equiv-13+13\pmod{169}$$
Finding all solutions requires at least a somewhat advanced theorem, namely the law of quadratic reciprocity.
For example, there are infinitely many $n^2-3$ that are multiple of $13^2$ or $23^2$, but none that is multiple of $17^2$ or $29^2$.
If $p>3$, $n^2-3$ can be a multiple of $p^2$ iff $p\equiv \pm1\pmod {12}$. For $p=2,3$, it is not possible.
| {
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Impossible integral? $ \int \sqrt{1-x^2} \arccos ( \sqrt{1-x^2} ) dx $ Is it possible to solve this integral?
$$ \int \sqrt{1-x^2} \arccos ( \sqrt{1-x^2} ) dx $$
| Take $\cos\theta = \sqrt{1-x^2}$. This yields $-\sin\theta d\theta = -\frac{x}{\sqrt{1-x^2}}dx = -\frac{\sin\theta}{\cos\theta}dx$ so therefore $dx = \cos\theta d\theta$. The integral transforms to
\begin{eqnarray*}
\int \sqrt{1-x^2}\arccos\left (\sqrt{1-x^2}\right )dx & = & \int \cos(\theta)\cdot \theta \cdot \cos\theta d\theta \\
& = & \int \theta \cos^2\theta d\theta \\
& = & \int \frac{\theta}{2}(1+\cos(2\theta)) d\theta \\
& = & \frac{1}{4}\theta^2 + \frac{1}{4}\theta\sin(2\theta) - \frac{1}{4}\int \sin(2\theta)d\theta \\
& = & \frac{1}{4}\theta^2 + \frac{1}{4}\theta\sin(2\theta) + \frac{1}{8}\cos(2\theta) + C \\
& = & \frac{1}{4}\theta^2 + \frac{1}{2}\sin\theta\cos\theta + \frac{1}{8}\cos^2\theta - \frac{1}{8}\sin^2\theta + C \\
& = & \frac{1}{4}\left (\arccos\sqrt{1-x^2}\right )^2 + \frac{1}{2}x\sqrt{1-x^2} -\frac{1}{4}x^2 + \frac{1}{8} +C.
\end{eqnarray*}
| {
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Issue with the following limit $\bigg(2 \sqrt{1 + \frac{1}{n}}\bigg)^n$ Calculate the following limit:
$\bigg(2 \sqrt{1 + \frac{1}{n}}\bigg)^n$
When I calculate it I get to different answers.
First way (Edit: this is where I did the mistake): $$\bigg(2 * \sqrt{1 + \frac{1}{n}}\bigg)^n = \bigg({4 + \frac{4}{n} \bigg)^\frac{n}{2}} = \bigg({4 + \frac{4}{n} \bigg)^{\frac{n}{4} \cdot \frac{4}{n}\cdot\frac{n}{2}}}$$
When we do $\lim_{n \to \infty}\bigg(\bigg({4 + \frac{4}{n} \bigg)^{\frac{n}{4}\cdot \frac{4}{n}\cdot\frac{n}{2}}}\bigg)$ we get $e^2$
Now the second way:
$$\bigg(2 \cdot \sqrt{1 + \frac{1}{n}}\bigg)^n = 2^n\cdot (1 + \frac{1}{n})^{n \cdot \frac{1}{2}}$$
When we do limit out of this we get $2^\infty \cdot \sqrt{e}$ which is of course $\infty$.
Could someone point out the mistake I made?
Edit:
I just realised where my mistake lies! I mistakenly thought that $(4 + \frac{4}{n})^\frac{n}{4} = e$ which is false, actually $(1 + \frac{4}{n})^\frac{n}{4} = e$. The second way of calculating this limit is the correct one!
| Your use of limit for $e^x$ is incorrect.
$$\lim\limits_{n\to\infty}\left(1+\dfrac{4}{n}\right)^n = e^4$$
$$\lim\limits_{n\to\infty}\left(4+\dfrac{4}{n}\right)^n = \infty$$
| {
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Prove $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3} +...+\frac{1}{2n}$
Prove that for all $n ≥ 1$
$$1 -
\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}
= \frac{1}{n+1} + \frac{1}{n+2} +\frac{1}{n+3} +\cdots+\frac{1}{2n}$$
My attempt:
By induction
Base case: $n = 1$
LHS : $1 - \frac{1}{2} = \frac{1}{2}$
RHS : $\frac{1}{1 + 1 } = \frac{1}{2}$
Suppose equality holds for $n$
We need to show that it holds for $n + 1$, i.e
$$1 - \frac{1}{2}+\frac{1}{3}-\frac{1}{4}-+\cdots-\frac{1}{2n+2} = \frac{1}{n+2} + \frac{1}{n+3} +\frac{1}{n+4} +\cdots+\frac{1}{2n+2} $$
We have
$$\begin{align}
1 - \frac{1}{2}+\frac{1}{3}-\frac{1}{4}-\cdots-\frac{1}{2n+2} & = \bigl[1 - \frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots - \frac{1}{2n}\bigr] + \frac{1}{2n+1} - \frac{1}{2n+2} \\
& = \frac{1}{n+1} + \frac{1}{n+2} +\frac{1}{n+3} +\cdots+\frac{1}{2n} + \frac{1}{2n+1} - \frac{1}{2n+2} \\
& =\frac{1}{n+2} +\frac{1}{n+3} +\cdots+\frac{1}{2n} + \frac{1}{2n+1} +\bigl (\frac{1}{n+1} - \frac{1}{2n+2} \bigr)\\
& = \frac{1}{n+2} +\frac{1}{n+3} +\cdots+\frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}
\end{align}$$
As desired. $\Box$
Is it correct?
| Your induction works.
As an alternative,
*
*writing $H_n =1 +\frac12 +\frac13 +\cdots +\frac1{n-1}+\frac1n$
*then $H_{2n} =1 +\frac12 +\frac13 +\cdots +\frac1{2n-1}+\frac1{2n}$
*while $\frac12H_n = \frac12 +\frac14 +\frac16 +\cdots +\frac1{2n-2}+\frac1{2n}$
*so $H_{2n} - \frac12H_n =1 +\frac13 +\frac15 +\cdots +\frac1{2n-3} +\frac1{2n-1}$
*and $H_{2n} - H_n =1 -\frac12 +\frac13 +\cdots +\frac1{2n-1}-\frac1{2n}$
*which is equal to $H_{2n} - H_n = \frac{1}{n+1} +\frac{1}{n+2}+\frac{1}{n+3} +\cdots +\frac1{2n-1}+\frac1{2n}$
This then gives a reasonable approximation, since we know $H_n=\log_e{n}+\gamma+\frac{1}{2n}-\frac{1}{12n^2}+\frac{1}{120n^4}-\cdots$ and so $H_{2n}=\log_e 2+\log_e{n}+\gamma+\frac{1}{4n}-\frac{1}{48n^2}+\frac{1}{1920n^4}-\cdots$ leading to
$H_{2n}-H_n = \log_e 2 - \frac{1}{4n}+\frac{1}{16n^2} -\frac{1}{128n^4} +\cdots$
| {
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Inequality with 5 cyclic variables For postive real numbers $a$,$b$,$c$,$d$ and $e$, prove that
$$\frac{4a^3}{a^2+2b^2+\frac{2b^3}{a}} + \frac{4b^3}{b^2+2c^2+\frac{2c^3}{b}} + \frac{4c^3}{c^2+2d^2+\frac{2d^3}{c}}+ \frac{4d^3}{d^2+2e^2+\frac{2e^3}{d}} + \frac{4e^3}{e^2+2a^2+\frac{2a^3}{e}} \geq $$ $$ \frac{2ab^2+2b^3}{a^2+2b^2 + 2\frac{b^3}{a} } + \frac{2bc^2+2c^3}{b^2+2c^2 + 2\frac{c^3}{b} }+ \frac{2cd^2+2d^3}{c^2+2d^2 + 2\frac{d^3}{c} } + \frac{2de^2+2e^3}{d^2+2e^2 + 2\frac{e^3}{d} } + \frac{2ea^2+2a^3}{e^2+2a^2 + 2\frac{a^3}{e} }$$
Can this be solved directly by $ AM-GM $?
| By AM-GM:
$$\sum_{cyc}\frac{4a^3}{a^2+2b^2+\frac{2b^3}{a}}-\sum_{cyc}\frac{2ab^2+2b^3}{a^2+2b^2+\frac{2b^3}{a}}=2\sum_{cyc}\frac{2a^4-a^2b^2-ab^3}{a^3+2ab^2+2b^3}=$$
$$=2\sum_{cyc}\left(\frac{2a^4-a^2b^2-ab^3}{a^3+2ab^2+2b^3}-(a-b)\right)=2\sum_{cyc}\frac{a^4+a^3b-3a^2b^2-ab^3+2b^4}{a^3+2ab^2+2b^3}=$$
$$=2\sum_{cyc}\frac{a^4-3a^2b^2+2ab^3+a^3b-3ab^3+2b^4}{a^3+2ab^2+2b^3}=2\sum_{cyc}\frac{(a+b)(a^3-3ab^2+2b^3)}{a^3+2ab^2+2b^3}\geq$$
$$\geq2\sum_{cyc}\frac{(a+b)\left(3\sqrt[3]{a^3\cdot\left(b^3\right)^2}-3ab^2\right)}{a^3+2ab^2+2b^3}=0.$$
| {
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The normal to the curve $y=2x^2-x+3$ at the point $P(1,4)$ meets at the curve again at the point $Q$. What is the $x$- coordinate of $Q$? The normal to the curve $y=2x^2-x+3$ at the point $P(1,4)$ meets at the curve again at the point $Q$. What is the $x$- coordinate of $Q$?
Please answer as soon as possible !
$(-4/7), -1, 0$ are wrong answers. Accept them if you find anything!
| The derivative is $4x-1$. Therefore the derivate at that point is 3. From that, the normal slope is $-\frac{1}{3}$ We can form a line from this: $y-4=-\frac{1}{3}(x-1)$ This simplifies to $y=-\frac{x}{3}+\frac{13}{3}$. So we need $$2x^2-x+3=-\frac{x}{3}+\frac{13}{3}$$ From this:$$2x^2-\frac{2x}{3}-\frac{4}{3}$$After multiplying by 3, which preserves the roots:$$6x^2-2x-4$$ Which factors to, $(6x+4)(x-1)$. This makes sense because the one root is at one, the point, and the other is at $x=-\frac{4}{6}=-\frac{2}{3}$. This is your answer.
| {
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finite sum with combinatorics I have two finite sum with combinatorics:$$f(n)=\sum_{k=0}^n (-1)^k 4^{n-k} \binom{2n-k+1}{k}$$ and $$g(n)=\sum_{k=0}^n (-1)^k 4^{n-k}\binom{2n-k}{k}$$
I tried several methods but have no clue how to derive that $f(n)=n+1$ and $g(n)=2n+1$. Can anybody help me?
Any hint is appreciated in advance!
| For the second one, we start as follows:
$$\sum_{k=0}^n (-1)^k 4^{n-k} {2n-k\choose k}
= \sum_{k=0}^n (-1)^k 4^{n-k} {2n-k\choose 2n-2k}
\\ = \sum_{k=0}^n (-1)^k 4^{n-k} [z^{2n-2k}] (1+z)^{2n-k}
\\ = [z^{2n}] (1+z)^{2n}
\sum_{k=0}^n (-1)^k 4^{n-k} z^{2k} (1+z)^{-k}.$$
Now when $k\gt n$ we get zero contribution due to the coefficient
extractor $[z^{2n}]$ and the factor $z^{2k}$, so this enforces the
range of the sum and we may continue with
$$[z^{2n}] (1+z)^{2n}
\sum_{k\ge 0} (-1)^k 4^{n-k} z^{2k} (1+z)^{-k}
\\ = 4^n [z^{2n}] (1+z)^{2n} \frac{1}{1+z^2/(1+z)/4}
\\ = 4^{n+1} [z^{2n}] (1+z)^{2n+1} \frac{1}{4+4z+z^2}
= 4^{n+1} [z^{2n}] (1+z)^{2n+1} \frac{1}{(z+2)^2}.$$
This is
$$4^{n+1} \;\underset{z}{\mathrm{res}}\; \frac{1}{z^{2n+1}}
(1+z)^{2n+1} \frac{1}{(z+2)^2}.$$
We introduce $z/(1+z)=w$ so that $z = w/(1-w)$ and $dz = 1/(1-w)^2 \;
dw,$ to obtain
$$4^{n+1} \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{2n+1}}
\frac{1}{(w/(1-w)+2)^2} \frac{1}{(1-w)^2}
\\ = 4^{n+1} \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{2n+1}}
\frac{1}{(2-w)^2}
\\ = 4^{n+1} [w^{2n}] \frac{1}{(2-w)^2}
= 4^{n} [w^{2n}] \frac{1}{(1-w/2)^2}
= 4^n (2n+1) \frac{1}{2^{2n}} \\ = 2n+1.$$
Remark. This can also be done using the fact that residues sum to
zero, which starting from the residue in $z$ we see that the residue
at infinity is zero, so our sum is
$$- 4^{n+1} \mathrm{Res}_{z=-2} \frac{1}{z^{2n+1}}
(1+z)^{2n+1} \frac{1}{(z+2)^2}
\\ = - 4^{n+1}
\left.\left(\frac{1}{z^{2n+1}}
(1+z)^{2n+1}\right)'\right|_{z=-2}
\\ = - 4^{n+1}
\left.\left(-\frac{2n+1}{z^{2n+2}}
(1+z)^{2n+1}+ \frac{(2n+1)}{z^{2n+1}}
(1+z)^{2n}\right)
\right|_{z=-2}
\\ = (2n+1) \times 4^{n+1}
\left(\frac{(-1)^{2n+1}}{(-2)^{2n+2}}
- \frac{(-1)^{2n}}{(-2)^{2n+1}}\right)
\\ = (2n+1) \times 2^{2n+2}
\left(- \frac{1}{2^{2n+2}}
+ \frac{1}{2^{2n+1}}\right)
= 2n+1.$$
| {
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Determine the maximum Problem: Determine $$\max_{z \in \mathbb{C}, |z|=1} |z^3-z+2|$$
Solution: We have $z = \cos t + i \sin t$, and replacing, we get: $$E=|{{z}^{3}}-z+2|=\sqrt{2\left| 1- \cos 2x+2\cos 3x-2\cos x \right|}.$$
Replacing the formula $\cos 2x, \cos 3x$ with the $\cos x$, and denoting $\cos x=t$, we get
$$E=\sqrt{4\left| 4{{t}^{3}}+{{t}^{2}}-2t+1 \right|},t\in \left[ -1,1 \right].$$
Denoting $f(t)=4{{t}^{3}}+{{t}^{2}}-2t+1$, the maximum on $\left[ -1,1 \right]$ yields
$$\max \left\{ f\left( -\frac{1}{2} \right),f(1) \right\} =\max \left\{ \frac{7}{4},4 \right\}=4.$$
Am I right?
| After
$E^2 = 4\cos(3x)-2\cos(2x)-4\cos x+6$
reducing to $\cos x$ we have
$$
E^2 = 16\cos^3 x-4\cos^2x-16\cos x+8
$$
and now calling $y = \cos x$ the problem becomes
$$
\max_{y\in [-1,1]}\sqrt{16 y^3-4 y^2-16 y+8}
$$
| {
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Find the given limit of $S$?
Find the given limit of $S$ ?
$$S=\lim _{n \rightarrow \infty} \left (\frac{\sin \frac{\pi}{ n+1}} { 1} +\frac{\sin \frac{2\pi}{ n+1}} { 2} + + ....+ \frac{\sin \frac{n\pi}{ n+1}} { n} \right )$$
My attempt :
i construct modified the given series
$\frac{\pi}{ n+1}\left (\frac{\sin \frac{\pi}{ n+1}} { \frac{1\pi}{ n+1}} +\frac{\sin \frac{2\pi}{ n+1}} { \frac{2\pi}{ n+1}} + ... + ....+ \frac{\sin \frac{n\pi}{ n+1}} { \frac{n\pi}{ n+1}} \right )$
After that im not able to proceed further
Any hints/solution will be appreciated
thanks u
| You can write it as a Riemann sum:
$$S_n=\frac{\sin \frac{\pi}{ n+1}} { 1} +\frac{\sin \frac{2\pi}{ n+1}} { 2} + + ....+ \frac{\sin \frac{n\pi}{ n+1}} { n}$$ $$= \sum_{i=1}^n\frac{\sin\frac{i\pi}{ n+1}}{\frac{i}{n+1}}\frac{1}{n+1} \stackrel{n \to \infty}{\longrightarrow}\int_0^1\frac{\sin \pi x}{x}dx = Si(\pi)\approx 1.852$$
| {
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$\lim_{x\to 1} (x^3+2x^2-2)=1$ using the definition to prove that $\lim_{x\to 1} (x^3+2x^2-2)=1$
let $\varepsilon>0$, and $x\in\mathbb{R}$
we must find $\delta>0$ such that if $|x-1|<\delta $ then $|x^3+2x^2-2-1|<\varepsilon$
i do :
$$
|x^3-1+2x^2-2|\leq |x^3-1|+2|x^2-1|=|x^3-1|+2 |(x-1)| |(x+1)|=
$$
$$
|x-1|(|x^2+x+1|+2|x+1|)\leq \delta (|x^2+x+1|+2|x+1|)
$$
how to continue?
| $$|x^3+2x^2-2-1|= |(x-1)(x^2+3x+3)|<\delta |x^2+3x+3|<13\delta <\varepsilon $$
Provided that $$0<\delta < \min \{1,\epsilon/13\}$$
| {
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Find the $n^{th}$ derivative of $y=\frac {x}{x^2+a^2}$ Find the $n^{th}$ derivative of $y=\dfrac {x}{x^2+a^2}$.
My Attempt:
$$y=\dfrac {x}{x^2+a^2}$$
$$y=x.(x^2+a^2)^{-1}$$
Differentiating both sides with respect to $x$
$$y_{1}=x.((-1).(x^2+a^2)^{-2}.2x)+(x^2+a^2)^{-1}$$
$$y_{1}=(-1).(x^2+a^2)^{-2}.2x^2+(x^2+a^2)^{-1}$$
| Hint: $\displaystyle\frac1x\left(\left(\frac xa\right)^2-\left(\frac xa\right)^4+\left(\frac xa\right)^6-\cdots\right)=\frac1x\times\frac{\left(\frac xa\right)^2}{1+\left(\frac xa\right)^2}=\frac x{x^2+a^2}$ and therefore$$\frac x{x^2+a^2}=\frac x{a^2}-\frac{x^3}{a^4}+\frac{x^5}{a^6}-\cdots.$$
| {
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} |
Know that $\tan\left(\alpha-\frac{\pi}{4}\right)=\frac{1}{3}$ calculate $\sin\alpha$
Know that $\tan\left(\alpha-\frac{\pi}{4}\right)=\frac{1}{3}$ calculate
$\sin\alpha$
My proof:
$\tan\left(\alpha-\frac{\pi}{4}\right)=\frac{1}{3}\\
\frac{\sin\left(\alpha-\frac{\pi}{4}\right)}{\cos\left(\alpha-\frac{\pi}{4}\right)}=\frac{1}{3}\\3\sin\left(\alpha-\frac{\pi}{4}\right)=\cos\left(\alpha-\frac{\pi}{4}\right)\\\sin^2\left(\alpha-\frac{\pi}{4}\right)+9\sin^2\left(\alpha-\frac{\pi}{4}\right)=1\\\sin\left(\alpha-\frac{\pi}{4}\right)=\pm\frac{1}{\sqrt{10}}\\
\sin\left(\alpha-\frac{\pi}{4}\right)=\sin\alpha\cos\frac{\pi}{4}-\sin\frac{\pi}{4}\cos\alpha=\frac{\sqrt{2}}{2}\sin\alpha-\frac{\sqrt2}{2}\cos\alpha=\frac{\sqrt2}{2}\left(\sin\alpha-\cos\alpha\right)=\pm\frac{1}{\sqrt{10}}\\\sin\alpha-\cos\alpha=\pm\frac{1}{\sqrt{5}}\\\sin\alpha=\pm\frac{1}{\sqrt{5}}+\cos\alpha$
I have no idea how to determine $\sin\alpha$
| We have $\tan\left(\dfrac\pi4-\alpha\right)=\dfrac{1-\tan\alpha}{1+\tan\alpha}=\dfrac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}$
$$\dfrac{\sin\alpha-\cos\alpha}1=\dfrac{\sin\alpha+\cos\alpha}3=\pm\sqrt{\dfrac{(\sin\alpha-\cos\alpha)^2+(\sin\alpha+\cos\alpha)^2}{1^2+3^2}}$$
$$\dfrac{\sin\alpha-\cos\alpha}1=\dfrac{\sin\alpha+\cos\alpha}3=\pm\sqrt{\dfrac2{1^2+3^2}}=\dfrac{\sin\alpha+\cos\alpha+\sin\alpha-\cos\alpha}{1+3}$$
| {
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Finding a partial derivative with two independent variables and two dependent variables The following problem is from the 7th edition of the book "Calculus and Analytic Geometry" by Thomas and Finney. It is problem 5b in section 16.5. Below is my attempt to solve it. However, I am getting a different answer than the book. Where did I go wrong?
Problem:
Find
$$ \left( \frac{\partial w}{\partial y} \right)_z $$
at the point $(w, x, y, z) = (4, 2, 1, -1)$ if
$$ w= x^2y^2 + yz - z^3 $$
and
$$ x^2 + y^2 + z^2 = 6 .$$
Answer:
The first step in the process is to eliminate the variable $x$ from the first equation.
\begin{align*}
x^2 &= 6 - y^2 - z^2 \\
w &= x^2y^2 + yz - z^3 \\
w &= (6 - y^2 - z^2 )y^2 + yz - z^3 \\
\left( \frac{\partial w}{\partial y} \right)_z &= (6 - y^2 - z^2)(2y) + y^2(-2y) + z \\
\left( \frac{\partial w}{\partial y} \right)_z &= (6 - y^2 - z^2)(2y) - 3y^3 + z \\
\left( \frac{\partial w}{\partial y} \right)_z ( 4, 2, 1, -1 ) &= ( 6 - 1^1 - 1^2)(2(1)) - 3(1^3) - 1 \\
\left( \frac{\partial w}{\partial y} \right)_z ( 4, 2, 1, -1 ) &= ( 4)(2) - 3 - 1 \\
\left( \frac{\partial w}{\partial y} \right)_z ( 4, 2, 1, -1 ) &= 4 \\
\end{align*}
However, the book's answer is $5$.
| Notice that
$$
y^2(-2y)=-2y^3\neq 3y^3.
$$
I think that's where your mistake is.
Indeed, if you consider this, you get
$$
\left(\frac{\partial w}{\partial z} \right)(4,2,1,-1)=(6-1^2-1^2)(2(1))-2(1^3)-1=5,
$$
which the answer in the book.
| {
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How to solve this rational function with logarithm infinite series problem?
Determine whether the series converges absolutely, conditionally, or diverges:
$\sum_{n=5}^\infty \frac{n^3+2n^2+1}{n^5-n^2+\ln n} $
I first thought of using the limit comparison test with a $\frac{1}{n^2}$ series and a $\frac{n^3+2n^2+1}{n^5-n^2+n} $ series. I found both converge. However, I realized that because $\ln n$ is always less than $n$, $\frac{n^3+2n^2+1}{n^5-n^2+\ln n} $ is always bigger than $\frac{n^3+2n^2+1}{n^5-n^2+n} $ so I can't use a series comparison test to finally solve this problem. What should I do to solve this problem?
| Limit comparison test is not necessary here. A simple comparison test will suffice. Since $\ln n>0$ and for every $n>2$, the following holds for $n>5$:
$$\dfrac{n^3+2n^2+1}{n^5-n^2+\ln n}<\dfrac{n^3+2n^2+1}{n^5-n^2}<\dfrac{n^3+2n^2+n}{n^5-n^2}=\dfrac{2n^2+2n+2}{n(n-1)(n^2+n+1)}=\dfrac{2}{n(n-1)}$$
Then you can show that $\displaystyle\sum_{n=5}^{\infty}\dfrac{2}{n(n-1)}$ converges by the definition of infinite series, and thus the original series converges by comparison test.
| {
"language": "en",
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"source": "stackexchange",
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"answer_id": 1
} |
$\int\frac1{(x^2+1)^2}\ dx$ by partial fraction decomposition Is there any possible way to calculate the integral of $\frac{1}{(x^2+1)^2}$ by partial fraction decomposition? I do not know the formulas for the trigonometric method.
Thank you!
| Up to my knowledge no.
I see 3 ways :
*
*a change of variables : $x = \tan t$,
*modify the numerator $\frac{1}{(1+x^2)^2} = \frac{1+x^2-x^2}{(1+x^2)^2} = \frac{1}{(1+x^2)} + \frac{x}{2}\frac{-2x}{(1+x^2)^2} = \arctan'(x) + \frac{x}{2}\left(\frac{1}{1+x^2}\right)'$ and integrate by part the second quantity,
*method by residues theorem.
To integrate by parts, use : $\int u'v = uv - \int uv'$ with $u = \frac{1}{1+x^2}$ and $v=\frac{x}{2}$, thus
$$
\begin{array}{rcl}
\displaystyle\int\frac{1}{(1+x^2)^2}\mathrm{d}x &=& \displaystyle\int \arctan'(x) + \frac{x}{2}\left(\frac{1}{1+x^2}\right)'\mathrm{d}x\\
& =& \displaystyle \arctan(x) + \frac{1}{2}\frac{x}{1+x^2} - \frac{1}{2}\int \frac{1}{1+x^2}\mathrm{d}x\\
&=& \displaystyle\frac{1}{2} \left(\arctan(x) + \frac{x}{1+x^2}\right) + \mathsf{cte}
\end{array}
$$
| {
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} |
What is wrong with my series expansion of $\cos(\sin(x))$ I need to expand the Maclaurin series for $f(x)=\cos(\sin(x))$
I take the first derivative of this function and I obtain $f'(x)=\sin(\sin(x))\cdot -\cos(x)$
I then assume that the series of $f(x)=\cos(\sin(x))$ takes the form:
$b_0+b_1x+b_2x^2+b_3x^3+b_4x^4+b_5x^5+b_6x^6+b_7x^7+O(x^8)$
The series for $-\cos(x)=-1+\dfrac{x^2}{2}-\dfrac{x^4}{24}+\dfrac{x^6}{720}-\dfrac{x^8}{40320}+\dfrac{x^{10}}{3628800}-\dfrac{x^{12}}{479001600}$
I let $a_0$ through $a_n$ denotes the coefficients of $-cos(x)$, so I have:
$a_0=-1, a_1=0, a_2=\dfrac{1}{2}, a_3=0, a_4=-\dfrac{1}{24},...$
The series for $f'(x)=\sin(\sin(x))\cdot -\cos(x)=b_1+2b_2x+3b_3x^2+4b_4x^3+5b_5x^4+6b_6x^5+7b_7x^6+8b_8x^7$
This means the differentiate the series of $f(x)=\cos(\sin(x))$
The Cauchy product of two power series is defined as:
$A= a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+...$
$B= b_0+b_1x+b_2x^2+b_3x^3+b_4x^4+...$
$A\cdot B=a_0b_0+(a_0b_1+a_1b_0)x+(a_0b_2+a_1b1_1+a_2b_0)x^2...$
I equate the coefficients of $f'(x)=\sin(\sin(x))\cdot -\cos(x)$
It is here that I am lost, this method works well for series expansion of $e^{cos(x)}$ and $e^{sin(x)}$, but it doesn't seem to work here. I don't know the series expansion of $\sin(\sin(x))$
Is there any better method than employing directly the Taylor formula for $x=0$. An elegant way of expanding this functions rather than brute force calculation.
| Rather than calculating derivatives. Substitute
\begin{eqnarray*}
\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\cdots
\end{eqnarray*}
into
\begin{eqnarray*}
\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\cdots.
\end{eqnarray*}
This gives
\begin{eqnarray*}
\cos(\sin(x))=1 &-& \frac{1}{2} \left( x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\cdots \right)^2 \\
& +& \frac{1}{24}\left( x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\cdots \right)^4 \\
&-& \frac{1}{720} \left( x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\cdots \right)^6+\cdots.
\end{eqnarray*}
and expand upto the order you require
\begin{eqnarray*}
\cos(\sin(x))=1 -\frac{x^2}{2} +\frac{5x^4}{24}+\cdots.
\end{eqnarray*}
| {
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If $\triangle$ is the area of triangle with side lengths $a,b,c$, then show that $\triangle \le\dfrac{1}{4}\cdot\sqrt{(a+b+c)abc}$ If $\triangle$ is the area of triangle with side lengths $a,b,c$, then show that $\triangle \le\dfrac{1}{4}\cdot\sqrt{(a+b+c)abc}$. Also show that equality occurs in the above inequality when $a=b=c$
My attempt is as follows:-
$$\triangle=\sqrt{s(s-a)(s-b)(s-c)}$$
$$\triangle=\dfrac{1}{4}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$$
$$b+c-a>0,a+c-b>0,a+b-c>0$$
So we can apply $A.M\ge G.M$
$$\dfrac{b+c-a+a+c-b+a+b-c}{3}\ge\left((b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{3}$$
$$\dfrac{a+b+c}{3}\ge\left((b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{3}\tag{1}$$
As $a>0,b>0,c>0$, we can also say
$$\dfrac{a+b+c}{3}\ge(abc)^\frac{1}{3}\tag{2}$$
But we can't say from this that $(abc)^\frac{1}{3}\ge\left((b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{3}$
So I tried $G.M\ge H.M$ for $a,b,c$
$$(abc)^\frac{1}{3}\ge\dfrac{3abc}{ab+bc+ca}$$
$$\dfrac{ab+bc+ca}{3}\ge(abc)^\frac{2}{3}\tag{3}$$
Seems like it didn't produce anything useful, so I tried $G.M\ge H.M$ for $b+c-a,a+c-b,a+b-c$
$$\left((b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{3}\ge\dfrac{3(b+c-a)(a+c-b)(a+b-c)}{((b+c-a)(a+c-b)+(a+c-b)(a+b-c)+(b+c-a)(a+b-c))}$$
$$\left((b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{3}\ge\dfrac{3(b+c-a)(a+c-b)(a+b-c)}{2ab+2bc+2ca-a^2-b^2-c^2}\tag{4}$$
But again it didn't produce anything useful, so I tried something else.
Applying $A.M\ge G.M$ for $a+b+c,a+b-c,b+c-a,a+c-b$
$$\dfrac{2(a+b+c)}{4}\ge\left((a+b+c)(b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{4}$$
$$\dfrac{a+b+c}{2}\ge\left((a+b+c)(b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{4}$$
Squaring both sides
$$\dfrac{(a+b+c)^2}{4}\ge\left((a+b+c)(b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{2}$$
By cachy's inequality:-
$$(1^2+1^2+1^2)(a^2+b^2+c^2)\ge(a+b+c)^2$$
$$3(a^2+b^2+c^2)\ge(a+b+c)^2$$
$$\dfrac{3(a^2+b^2+c^2)}{4}\ge\dfrac{(a+b+c)^2}{4}\tag{5}$$
Hence
$$\dfrac{3(a^2+b^2+c^2)}{4}\ge\left((a+b+c)(b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{2}\tag{6}$$
Dividing by $4$
$$\dfrac{3(a^2+b^2+c^2)}{16}\ge\triangle$$
So I tried all these approaches, but unfortunately didn't arrive at the required result. What can we do here? Please help me in this.
| $$\triangle=rs=\dfrac{abc}{4R}$$ where $2s=a+b+c$ and $R,r$ are the in-radius & circum-radius respectively
$$\sqrt{(a+b+c)abc}=\sqrt{\dfrac{2\triangle}r\cdot4\cdot\triangle\cdot R}=4\triangle\sqrt{\dfrac R{2r}}$$
Now use
True or False: The circumradius of a triangle is twice its inradius if and only if the triangle is equilateral.
or
Why is the inradius of any triangle at most half its circumradius?
| {
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"answer_id": 0
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Simple proof of Inverse Discrete Fourier Transformation (IDFT)? Discrete Fourier Transformation (DFT) is defined by:
$X_k = \sum_{n=0}^{N-1} x_n \exp(\frac{-2 \pi i k n}{N}) ; 0\leq k \leq N-1$
And Inverse Discrete Fourier Transformation (IDFT) is defined by:
$x_n = \frac{1}{N} \sum_{k=0}^{N-1} X_k \exp(\frac{2 \pi i k n}{N})$
How can I proof IDFT using DFT?
| Observe
\begin{align}
\frac{1}{N}\sum^{N-1}_{k=0} X_k \exp\left(\frac{2\pi i kn}{N} \right) =&\ \frac{1}{N}\sum^{N-1}_{k=0} \left(\sum^{N-1}_{m=0}x_m\exp\left(-\frac{2\pi i km}{N} \right)\right) \exp\left(\frac{2\pi i kn}{N} \right) \\
=&\ \frac{1}{N}\sum^{N-1}_{m=0}\sum^{N-1}_{k=0} x_m \exp\left(\frac{2\pi i k(n-m)}{N} \right)\\
=&\ \sum^{N-1}_{m=0}x_m\left(\frac{1}{N}\sum^{N-1}_{k=0} \exp\left(\frac{2\pi i k(n-m)}{N}\right)\right)\\
=& \sum^{N-1}_{m=0} x_m \delta_{n, m} = x_n.
\end{align}
Note that I have used the fact that
\begin{align}
\frac{1}{N}\sum^{N-1}_{k=0} \exp\left(\frac{2\pi i k(n-m)}{N}\right)= \frac{1}{N}\frac{1-\exp(2\pi i(n-m))}{1-\exp(\frac{2\pi i(n-m)}{N})} =0
\end{align}
if $n\ne m$. However, if $n=m$ then
\begin{align}
\frac{1}{N}\sum^{N-1}_{k=0} \exp\left(\frac{2\pi i k(n-m)}{N}\right)=\frac{1}{N}\sum^{N-1}_{k=0} 1 = \frac{N}{N} =1.
\end{align}
Another way is to consider the DFT matrix. Let $e(x) = \exp(\frac{2\pi i x}{N})$ then
\begin{align}
F = \frac{1}{\sqrt{N}}
\begin{bmatrix}
e(1\cdot 1) & e(1\cdot 2) & \cdots & e(1\cdot (N-1))\\
e(2\cdot 1) & e(2\cdot 2) & \cdots & e(2\cdot (N-1))\\
\vdots & \vdots & \ddots & \vdots\\
\vdots & \vdots & \ddots & \vdots\\
e((N-1)\cdot 1) & \cdots & \cdots & e((N-1)\cdot (N-1))
\end{bmatrix}.
\end{align}
Let us check that the columns of $F$ are orthonormal. Observe
\begin{align}
(\text{column m})^H(\text{column n}) = \frac{1}{N}\sum_{\ell=0}^{N-1}e(-\ell \cdot m)e(\ell \cdot n) = \frac{1}{N}\sum_{\ell=0}^{N-1}e(\ell \cdot (n-m)) = \delta_{n, m}
\end{align}
which means the columns are orthogonal. It's also clear that each column has norm 1. So $F$ is a unitary matrix, which means
\begin{align}
F^{-1} = F^H =
\frac{1}{\sqrt{N}}
\begin{bmatrix}
e(-1\cdot 1) & e(-2\cdot 1) & \cdots & e(-(N-1)\cdot 1)\\
e(-1\cdot 2) & e(-2\cdot 2) & \cdots & e(-(N-1)\cdot 2)\\
\vdots & \vdots & \ddots & \vdots\\
\vdots & \vdots & \ddots & \vdots\\
e(-1\cdot(N-1)) & \cdots & \cdots & e(-(N-1)\cdot (N-1))
\end{bmatrix}.
\end{align}
| {
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Prove that the minimum values of $x^2+y^2+z^2$ is $27$ with given condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$. Question: Prove that the minimum values of $x^2+y^2+z^2$ is $27$, where $x,y,z$ are positive real variables satisfying the condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$.
From AM$\ge$ GM, we have $\left( \dfrac{x^2+y^2+z^2}{3}\right)^3\ge (xyz)^2=(xy+yz+zx)^2$. Is it possible to show thae result from this relation?
| By Holder,
$$\left(x^2+y^2+z^2 \right)\cdot \left(\frac1x+\frac1y+\frac1z \right)^2\geqslant (1+1+1)^3$$
As $x=y=z=3$ achieves this, we have a minimum.
| {
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Prove an inequality : $\sum_{cyc}\frac{a^3}{abu+b^2v}\geq \frac{a+b+c}{u+v}$ without Jensen's inequality I'm interested in the following problem :
Let $a,b,c>0$ be the variables and $u,v>0$ be constant then we have :
$$\sum_{cyc}\frac{a^3}{abu+b^2v}\geq \frac{a+b+c}{u+v}$$
Rewrrting the inequality, we have :
$$\sum_{cyc}\frac{a}{a+b+c}\frac{1}{\frac{b}{a}u+\frac{b^2}{a^2}v}\geq \frac{1}{u+v}$$
As the function :
$$f(x)=\frac{1}{xu+x^2v}$$
Is convex (with the condition of positivity) we can apply Jensen's inequality and then we have:
$$\sum_{cyc}\frac{a}{a+b+c}f\Big(\frac{b}{a}\Big)\geq f\Big(\frac{a\frac{b}{a}+b\frac{c}{b}+c\frac{a}{c}}{a+b+c}\Big)=f(1)=\frac{1}{u+v}$$
Done !
My question is have you an alternative proof wich doesn't use Jensen's inequality ?
Thanks for sharing your time and knwoledge .
| By C-S
$$\sum_{cyc}\frac{a^3}{uab+vb^2}=\sum_{cyc}\frac{a^4}{ua^2b+vb^2a}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(ua^2b+va^2c)}.$$
Thus, it's enough to prove that
$$(u+v)\sum_{cyc}(a^4+2a^2b^2)\geq(a+b+c)\sum\limits_{cyc}(ua^2b+va^2c)$$ or
$$\sum_{cyc}((u+v)a^4-ua^3b-va^3c)+(u+v)\sum_{cyc}(a^2b^2-a^2bc)\geq0,$$ which is true by Rearrangement and Muirhead.
| {
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What's the coefficent of $x^{20}$ in $(x^3+x^4+x^5+...)^5$? Only hint is needed. I know about Binomial/Multinomial expansion but I got stuck on this series; it doesn't look like anything I've solved before. I already searched for any hint/formula and couldn't find one, any help is appreciated.
| Can be written as
$x^{15}(1-x)^{-5}$
Use binomial expansion for $(1-x)^{-5}$
That is
$$x^{15}(1+5x+ \frac{5 \cdot 6}{2!}x^2 + \frac{5 \cdot 6 \cdot 7}{3!}x^3 + \frac{5 \cdot 6 \cdot 7 \cdot 8}{4!}x^4 + \frac{5 \cdot 6 \cdot 7 \cdot 8 \cdot 9}{5!}x^5 + \cdots)$$
Coefficient of $x^{20}$ is
$x^{15}×$ coefficient of $x^5$ .
Which is $$\frac{5 \cdot 6 \cdot 7 \cdot 8 \cdot 9}{5!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculating $\int_0^1\frac{1}{1+x}\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)dx$
How to prove in a simpe way that
$$\int_0^1\frac{1}{1+x}\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)dx=\frac{13}{8}\ln2\zeta(2)-\frac{33}{32}\zeta(3)\ ?$$
where $\operatorname{Li}_2$ is the dilogarithm function.
If we integrate by parts, the integral boils down to
$$\int_0^1\left(\frac{1}{x}-\frac{2x}{1+x^2}\right)\left[2\ln(1-x)\ln(1+x)-\ln(1+x)\ln(1+x^2)\right]dx$$
and that seems long and complicated, any other ideas?
| $$I:=\int_0^1\frac{1}{1+x}\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)dx\overset{\large x\to\frac{1-x}{1+x}}=\int_0^1 \frac{1}{1+x}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx$$
$$J:=\int_0^1 \frac{1}{1-x}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx$$
$$I=\frac12\left((I+J)+(I-J)\right)=\boxed{\int_0^1 \frac{1}{1+x}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx=\frac{13}{8}\zeta(2)\ln 2-\frac{33}{32}\zeta(3)}$$
$$J=\frac12\left((I+J)-(I-J)\right)=\boxed{\int_0^1 \frac{1}{1-x}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx=\frac58\zeta(2)\ln 2+\frac{19}{32}\zeta(3)}$$
$$I-J=-2\int_0^1\frac{x}{1-x^2}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx\overset{\large \frac{1-x^2}{1+x^2}\to x}=\color{blue}{\int_0^1 \frac{\operatorname{Li}_2(x)}{1+x}dx}-\color{red}{\int_0^1 \frac{\operatorname{Li}_2(x)}{x}dx}$$
$$\overset{\color{blue}{IBP}}=\color{blue}{\zeta(2)\ln 2+\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}dx}-\color{red}{\sum_{n=1}^\infty \frac{1}{n^2}\int_0^1 x^{n-1}dx}=\boxed{\zeta(2)\ln 2-\frac{13}{8}\zeta(3)}$$
See here for the first integral.
$$I+J=2\int_0^1 \frac{1}{1-x^2}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx\overset{\large \frac{1-x^2}{1+x^2}\to x}=\int_0^1 \frac{\operatorname{Li}_2(x)}{x\sqrt{1-x^2}}dx$$
Now we're going to use the following result:
$$\int_0^1 \frac{\ln t}{t-\frac{1}{x}}dt=- \sum_{n=1}^\infty x^n\int_0^1 t^{n-1}\ln t\,dt=-\sum_{n=1}^\infty x^n \left(\frac{1}{n}\right)'=\sum_{n=1}^\infty \frac{x^n}{n^2}=\operatorname{Li}_2(x)$$
$$\Rightarrow I+J=- \int_0^1\int_0^1 \frac{\ln t}{(1-tx)\sqrt{1-x^2}}dx dt\overset{x\to \sin x}=- \int_0^1\ln t\int_0^\frac{\pi}{2} \frac{1}{1-t\sin x}dxdt$$
$$=-\int_0^1\ln t \ \frac{\frac{\pi}{2}+\arcsin t}{\sqrt{1-t^2}} dt\overset{t=\sin x}=-\int_0^\frac{\pi}{2}\left(\frac{\pi}{2}+x\right)\ln (\sin x) dx=\boxed{\frac94\zeta(2)\ln 2 -\frac7{16}\zeta(3)}$$
See here and here for the above integrals.
Also to prove a leftover integral from above we will consider: $$f(a)=\int_0^\pi \frac{dx}{1+\sin a\sin x}\overset{\tan\frac{x}{2}=y}=\int_0^\infty \frac{dy}{1+y^2+2\sin a y}$$
$$=\int_0^\infty \frac{dy}{(\sin a+y)^2+\cos^2 a}=\frac{1}{\cos a}\arctan\left(\frac{\sin a+y}{\cos a}\right)\bigg|_0^\infty=\frac{\frac{\pi}{2}-a}{\cos a}$$
$$\Rightarrow \int_0^\frac{\pi}{2}\frac{1}{1-t\sin x}dx=\frac12 f(-\arcsin t)=\frac{\frac{\pi}{2}+\arcsin t}{\sqrt{1-t^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3460434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$\frac{1}{5}\big((4 + \sqrt{15})^{2n} + (4 - \sqrt{15})^{2n} + 8\big)$ is the sum of 3 consecutive squares
Prove that for every positive integer $n$, the number
$$
\large \frac{(4 + \sqrt{15})^{2n} + (4 - \sqrt{15})^{2n} + 8}{5}
$$
can be expressed as a sum of squares of three consecutive integers.
Attempt. Since
$$
(m - 1)^2 + m^2 + (m + 1)^2 = 3m^2 + 2, \forall m \in \mathbb Z^+, m \ge 1,
$$
it suffices to show that
$$
(4 + \sqrt{15})^{2n} + (4 - \sqrt{15})^{2n} = 15m^2 + 2, \,\,\text{for some}\,\,\, m \in \mathbb Z^+,
$$
$$\implies \sum_{p = 1}^{n - 1}[(4 + \sqrt{15})^{2p} + (4 - \sqrt{15})^{2p}] = 15m^2, \forall m \in \mathbb Z^+, m \ge 1$$
But this couldn't use any mathematical induction, since $m$ hasn't been known to be on a sequence for $n = 1, 2, 3, 4, \cdots$
| Let $m=\dfrac{(4+\sqrt{15})^n-(4-\sqrt{15})^n}{\sqrt{15}}$.
$m$ is an integer, because, in the binomial expansions in the numerator,
terms with even powers of $\sqrt{15}$ cancel, leaving a numerator that is an integer times $\sqrt{15}$.
Furthermore, $m^2=\dfrac{(4+\sqrt{15})^{2n}+(4-\sqrt{15})^{2n}-2}{15},$ so
$(m-1)^2+m^2+(m+1)^2=3m^2+2=\dfrac{(4+\sqrt{15})^{2n}+(4-\sqrt{15})^{2n}-2}{5}+2,$ as desired.
| {
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"url": "https://math.stackexchange.com/questions/3461117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is it true that, for any Pythagorean triple $4ab > c^2$?
Is it true that, for any Pythagorean triple $4ab > c^2$?
So this came up in a proof I was working on and it seems experimentally correct from what I've tried and I would imagine the proof is similar to proving,
$$ab < \frac{c^2}{2}$$
The idea I have for this approach is,
$$4ab > c^2$$
$$4ab > a^2 + b^2$$
(Then maybe something with the triangle inequality?)
So is this statement true (it seems to be), and how can I prove it?
A counter-example would also be acceptable.
| If $a^2 + b^2 = c^2$ then
$c^2 < 4ab$ is the same thing as claiming $a^2 + b^2 < 4ab$
This is true if and only if $a^2 - 2ab + b^2 < 2ab$ which is to say
$(a-b)^2 < 2ab$
There shouldn't be any reason that should be true in general. If let $m = $ average/midpoint of $a,b$, that is, $m = \frac {a+b}e$ and $e = $ the radius of the interval $a$ to $b$, that is $e = |m-a| =|b-m| = \frac{|a-b|}2$ then we have
$(a-b)^2 < 2ab \implies$
$4e^2 < 2(m-e)(m+e) = 2(m^2 - e^2)\implies$
$3e^2 < m^2$
Which utterly need not be true!
Of course finding a pythagorean triplet where $a= m\pm e$ and $b=m\mp e$ are integers so that $a^2 + b^2 = 2(m^2 + e^2)$ is not merely and integer but a perfect square may put a wrinkle in finding a counterexample.
But not an insurmountable wrinkle.
We need, for a counter-example, $3e^2 \ge m^2$ which just require $e$ be significantly large, which is utterly irrelevent (it would seem) to $2(m^2 + e^2)$ being a perfect square.
... But to find a counter example:
.....
Back to square 1: We can find a Pythogorian triple $a^2 + b^2 = c^2$ by letting $b$ be any odd integer and if $b^2 = 2a+1$ so $c^2 = (a+1)^2$ ... retrosolving by letting $a = \frac {b^2-1}2$.
So we want a counter example of $c^2 = a^2 + b^2 = a^2 + 2a + 1 = (a+1)^2 \ge 4ab$ or substitutint $a = \frac {b^2-1}2$, we want a case where
$(\frac {b^2+1}2)^2 \ge 4\frac{b^2 -1}2b$.
Surely we can find a counter example.
$(b^2 + 1)^2 \ge 8(b^2-1)b$
$b^4 + 2b^2 + 1 \ge 8b^3 -8b$
$b^4 -8b^3 +2b^2 - 8b + 1 \ge 0$
yeah that can be solved...
(Obviously as $b\to \infty$ the $b^4-8b^3 + 2b^2 - 8b + 1\to \infty$ so there is some $K$ where $b > K$ will always have $b^4 -8b^3 +2b^2 - 8b + 1\ge 0$.)
Taking a sledgehammer and letting $b > 8$, say, $b=9$ and $a = \frac {b^2 -1}2=40$ so $40^2 + 9^2 = 41^2> 4*9*40=36*40$
Yeah... it's not true.
(Actually our sledgehammer was fairly precise. $b^4 -8b^3 +2b^2 - 8b + 1$ has only two real solutions; one between $0$ and $1$ (far closer to $0$ than to $1$) and the other between $7$ and $8$. So of primmitive triplets where $b$ is odd and $b^2 = 2a+1$. $a = 40, b=9, c=41$ is the smallest primitive triplet (and thus smallest of all triplets) where $c^2 > 4ab$.)
( so $c^2 < 4ab \iff \frac {c}{\gcd(a,b,c)} < 41$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\int_0^\infty \frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x$ using Feynman's trick We are expected to solve an integral similar to $\int_0^\infty \frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x$ using contour integration, but I was wondering whether it would be possible to use the so-called Feynman's trick, i.e. differentiating under the integral sign. I tried using $$F(t) = \int_0^\infty \frac{\sqrt{x}e^{-t\sqrt{x}}}{x^2+2x+5}\mathrm{d}x$$
so that
$$F^\prime(t) = \int_0^\infty \frac{-xe^{-t\sqrt{x}}}{x^2+2x+5}\mathrm{d}x$$
Sadly, this does not work. So I'm looking for any starting point to approach this problem. I appreciate any help.
| Well, you could solve this problem using Differentiation Under the Integral Sign, however I think that it wouldn't be an easy task and probably would end up in a tricky differential equation. Instead of it, I offer you a solution that just requires some substitutions.
$$I=\int_{0}^{\infty}{\frac{\sqrt x}{x^2+2x+5}dx}\overbrace{=}^{x\rightarrow\sqrt{5t}}5^{\frac{3}{4}}\int_{0}^{\infty}{\frac{\sqrt t}{{5\ t}^2+2\sqrt5t+5}dt}$$
Let's make some rearrangements:
$$I=\color{red}{\frac{2}{\sqrt[4]{5}}\int_0^{\infty}\frac{\frac{1}{2}\frac{1}{\sqrt t}}{\left(\sqrt t-\frac{1}{\sqrt t}\right)^2+\frac{10+2\sqrt5}{5}}dt}\overbrace{=}^{t\rightarrow \frac{1}{t}}\color{blue}{\frac{2}{\sqrt[4]{5}}\int_0^{\infty}\frac{\frac{1}{2}\frac{1}{t\sqrt t}}{\left(\sqrt t-\frac{1}{\sqrt t}\right)^2+\frac{10+2\sqrt5}{5}}dt}$$
Summing the integrals red and blue:
$$2I={\frac{2}{\sqrt[4]{5}}\int_0^{\infty}\frac{\frac{1}{2}\frac{1}{\sqrt t}+\frac{1}{2}\frac{1}{t\sqrt t}}{\left(\sqrt t-\frac{1}{\sqrt t}\right)^2+\frac{10+2\sqrt5}{5}}dt}\overbrace{=}^{\sqrt t-\frac{1}{\sqrt t}=u}\frac{2}{\sqrt[4]{5}}\int_{-\infty}^{\infty}\frac{du}{u^2+\frac{10+2\sqrt5}{5}}$$
$$2I=\frac{2}{\sqrt[4]{5}}\sqrt{\frac{5}{2\left(5+\sqrt5\right)}}\left[\arctan{\left(u\sqrt{\frac{5}{2\left(5+\sqrt5\right)}}\right)}\right]_{-\infty}^\infty$$
Hence:
$$I=\frac{\pi}{\sqrt[4]{5}}\sqrt{\frac{5}{2\left(5+\sqrt5\right)}}=\frac{\pi}{2\sqrt{\phi}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Increasing numerator o decreasing denominator? Which fraction is bigger? How can i demonstrate it?
$$\frac{a+1}{b}$$ or $$\frac{a}{b-1}$$
| If $1<b<a+1$ then
\begin{eqnarray*}
ab+b-a-1 < ab \\
(a+1)(b-1) < ab \\
\frac{a+1}{b} < \frac{a}{b-1}.
\end{eqnarray*}
Similarly if $b>a+1>0$
\begin{eqnarray*}
\frac{a+1}{b} > \frac{a}{b-1}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For which integers $n$ are there solutions to the equation $x^2-y^2=n$
For which integers $n$ are there solutions to the equation $x^2-y^2=n$
I believe $n$ must be divisible by $3$, since for any integers $x,y$ mod $3$. $x^2=3k+1$ and $y^2=3n+1$, thus $x^2-y^2=3(n-r)$.
Then I would have $n=0$ with $x,y=0$.
I also found that $n$ if it is even must be divisible by a power of $2$ greater then $1$. Since if $n$ is even then one of $(x+y),(x-y)$ is even, so both are even. Then $4\mid n$.
I'm not sure if there are more restrictions, or how to determine if there are no more. Assuming these are done correctly.
| $(m+1)^2-m^2=2m+1$ gives us all positive odd $n$, eg $1=1^2-0^2,3=2^2-1^2$.
$(m+2)^2-m^2=4(m+1)$ gives us all positive multiples of 4.
$m^2-m^2=0$ and swapping $x,y$ gives us negative odd numbers and negative multiples of 4.
Any square is 0 or 1 mod 4. So we cannot have $x^2-y^2=2\bmod4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472214",
"timestamp": "2023-03-29T00:00:00",
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Locus $\{e^{i \theta} + be^{-i \theta} \ : \theta \in [0, 2 \pi] \}$ for fixed $b \in \mathbb{C}$
What is the locus of $\{e^{i \theta} + be^{-i \theta} \ : \theta \in [0, 2 \pi] \}$ for fixed $b \in \mathbb{C}$?
Write $ w := u +iv = z+b \bar z$, where $z$ traverses the unit circle exactly once.
After substituting $Re(z)= \cos \theta, Im(z) = \sin \theta$ and some algebra I get,
$ (1): \ v^2 + \frac{(1-b)^2 u^2}{(1+b)^2} = 1-b ^2$ for $ b \neq 1,-1$ and $(2): \ w=2Rez, (3): \ w= -2Imz$ for $ b=1$ and $b=-1$ respectively.
If it weren't for the fact that $b$ is complex I would say that $(1)$ is just a conic section. Can someone help me finish this off/ tell me where I've gone wrong? Many thanks!
| Writing $b = p + iq$ for $p, q \in \mathbb{R}$, we can expand the expression as
\begin{align*}
e^{i\theta} + be^{-i\theta}
&= (\cos\theta + i\sin\theta) + (p+iq)(\cos\theta-i\sin\theta) \\
&= ((1+p)\cos\theta + q\sin\theta) + i((1-p)\sin\theta + q\cos\theta).
\end{align*}
So if we write $x + iy = e^{i\theta} + be^{-i\theta}$ for real $x, y$, we are led to the system of linear equations
$$ \begin{pmatrix} x \\ y \end{pmatrix}
= \begin{pmatrix} 1+p & q \\ q & 1-p \end{pmatrix} \begin{pmatrix} \cos\theta \\ \sin\theta \end{pmatrix}. \tag{*}$$
If you are familiar with spectral theorem, this allows to deduce that the locus is a possibly degenerate ellipse whose major and minor axes are determined by the eigenvalue/eigenvector pairs of the 2 by 2 matrix appearing in $\text{(*)}$.
Otherwise, we can still tackle this problem by more elementary means. We have two cases:
*
*If $|b| \neq 1$ so that the 2 by 2 matrix appearing in $\text{(*)}$ is invertible, then solving $\text{(*)}$ to obtian
$$ \begin{pmatrix} \cos\theta \\ \sin\theta \end{pmatrix} = \frac{1}{1-|b|^2} \begin{pmatrix} 1-p & -q \\ -q & 1+p \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} $$
and then plugging this to the relation $\cos^2\theta + \sin^2\theta = 1$, we end up with
$$ ((1-p)x - qy)^2 + (-qx + (1+p)y)^2 = (1-|b|^2)^2. $$
This is the equation of the locus, and it is indeed an ellipse.
*If $|b| = 1$, then write $b = e^{i\alpha}$ and note that
$$ e^{i\theta} + be^{-i\theta}
= e^{i\alpha/2} \bigl( e^{i(\theta-\alpha/2)} + e^{-i(\theta-\alpha/2)} \bigr)
= 2 e^{i\alpha/2} \cos(\theta-\alpha/2). $$
So the locus is the line segment $[-2, 2]$ rotated around $0$ by $\alpha/2$ radian.
| {
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"url": "https://math.stackexchange.com/questions/3473671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $2^x≥x^2$ I could not understand this proof. How are we able to divide both sides of the inequality with different things and claim that the inequality stays the same? Plus, is this an induction proof?
If $x \in \mathbb{Z}$ and $x \ge 4$ then $2^x \ge x^2$ holds.
When $x = 4$ we have $2^4 = 16 \ge 16 = 4^2$.
We $x > 4$ we have $\frac{2^{x + 1}}{2^x} = 2$ and
$$
\frac{(x + 1)^2}{x^2}
= \left(1 + \frac{1}{x}\right)^2
\le \left(\frac{5x}{4}\right)^2
$$
since $2 \le 1.5625$
| The inductive hypothesis is the line $2^x\ge x^2$ for $x\ge4$. Essentially what these steps suggest is the following:$$(x+1)^2\le(1.25x)^2\le2x^2\le2\cdot2^x=2^{x+1}$$which completes the induction proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3473888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Maximum value of $\frac{x}{y+1}+\frac{y}{x+1}$ while $0\leq x,y \leq 1$ What is the maximum value of $\frac{x}{y+1}+\frac{y}{x+1}$ while $0\leq x,y \leq 1$?
Wolfram Alpha plots this expression on a 3d graph, but I want to solve it algebraicly, by modifying the expression
My Attempts
1) add and substract 2 at the equation and we get $\frac{x+y+1}{y+1}+\frac{x+y+1}{x+1}$ and the numerator is same
=>failed
2) use AM-GM or Cauchy-Schwarz inequality
=>also failed
| Since
$$
\delta\left(\frac{x}{y+1}+\frac{y}{x+1}\right)
=\left(\frac1{y+1}-\frac{y}{(x+1)^2}\right)\delta x
+\left(\frac1{x+1}-\frac{x}{(y+1)^2}\right)\delta y
$$
to get an interior critical point, we need
$$
\frac1{y+1}=\frac{y}{(x+1)^2}\quad\text{and}\quad\frac1{x+1}=\frac{x}{(y+1)^2}
$$
which have only the singular solution $x=y=-1$, which, even if allowed, are outside of $[0,1]\times[0,1]$. Therefore, we are searching for boundary critical points.
Due to the symmetry, we need only consider $x=0$ and $x=1$.
For $x=0$, we get $\frac{x}{y+1}+\frac{y}{x+1}=y$ and the maximum is at $y=1$.
For $x=1$, we get $\frac{x}{y+1}+\frac{y}{x+1}=\frac1{y+1}+\frac y2$. Then, the critical point is $y=\sqrt2-1$, which gives $\frac{x}{y+1}+\frac{y}{x+1}=\sqrt2-\frac12\lt1$. The endpoints $y\in\{0,1\}$ both give $\frac{x}{y+1}+\frac{y}{x+1}=1$.
Thus, the maximum is $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find limit of $\frac{|x|^3 y^2+|x|y^4}{(x^2+y^2)^2}$ I have to show that $\lim_{{x\choose y}\to { 0 \choose 0}} \frac{|x|^3 y^2+|x|y^4}{(x^2+y^2)^2} = 0$. But I cannot figure out the trick you need to find an upper estimation which goes to $0$. Do you have any hints?
EDIT:
I think I got it:
$\lim_{{x\choose y}\to { 0 \choose 0}} \frac{|x|^3 y^2+|x|y^4}{(x^2+y^2)^2} = \lim_{{x\choose y}\to { 0 \choose 0}} \frac{|x|^3 y^2}{(x^4+2x^2 y^2 +y^4)} +\lim_{{x\choose y}\to { 0 \choose 0}} \frac{|x|y^4}{(x^4+2x^2 y^2 +y^4)}\leq \lim_{{x\choose y}\to { 0 \choose 0}} \frac{|x|^3 y^2}{x^2 y^2}+\lim_{{x\choose y}\to { 0 \choose 0}} \frac{|x|y^4}{y^4}=\lim_{{x\choose y}\to { 0 \choose 0}} |x|+ \lim_{{x\choose y}\to { 0 \choose 0}} |x|=0.$
What do you think?
| Hint. We have that
$$\frac{|x|^3 y^2+|x|y^4}{(x^2+y^2)^2}=|y|\cdot\frac{|x||y|}{x^2+y^2}. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What's the size of an angle in a triangle with sides $\sin(x), \cos(x),$ and $\tan(x)$?
Imagining a scalene triangle with sides $\sin(x), \cos(x)$ and $\tan(x)$, how would you find angle $x$ if it was between $\cos(x)$ and $\sin(x)$ when $0<x<\frac{\pi}{2}$?
I tried using the law of cosines but it lead nowhere and honestly haven't gone very far.
$$\cos(x)=\frac{\sin^2(x)+\cos^2(x)-\tan^2(x)}{2\sin(x)\cos(x)}\\
\cos(x)=\frac{1-\tan^2(x)}{\sin(2x)}$$
| Starting from @automaticallyGenerated's answer, facing a quintic polynomial, tou will need a numerical method.
Let us consider that we look for the zero of function
$$f(x)=\sin^2(x)- \cos^2(x)+2\cos^4(x)\sin(x)$$ and use Newton method starting with $x_0=0$; this will provide the following iterates
$$\left(
\begin{array}{cc}
n & x_n \\
0 & 0.000000 \\
1 & 0.500000 \\
2 & 0.480810 \\
3 & 0.481195
\end{array}
\right)$$
Edit
Graphing or using inspection, we can notice that the solution is close to $\frac \pi 6$. So, making a Taylor expansion around this point, we have
$$f(x)=\frac{1}{16}+t-\frac{392 }{507}t^2+\frac{7552 }{19773}t^3+\frac{504320
}{771147}t^4+O\left(t^5\right)$$ where $t=\frac{13\sqrt{3}}{16} \left(x-\frac{\pi }{6}\right)$.
Now, using series reversion (using $f(x)=y$), we have
$$x=\frac{\pi }{6}+u+\frac{49 }{26 \sqrt{3}}u^2+\frac{817}{507} u^3+\frac{22975 }{26364
\sqrt{3}}u^4+O\left(u^5\right)$$ where $u=\frac{16 y-1}{13 \sqrt{3}}$.
Make $y=0$ to get the approximation
$$x \sim \frac{\pi }{6}-\frac{497738471}{6776839836 \sqrt{3}}=0.481194$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove that $a + b \geq 4k$ Let $a, b$ and $k$ be positive integers with $k> 1$ such that
$lcm (a, b) + gcd (a, b) = k (a + b)$.
Prove that $a + b \geq 4k$
Solution:
Let $a=da_0$, $b=db_0$, thus $da_0b_0+d=kd(a_0+b_0) \implies k=\frac{a_0b_0+1}{a_0+b_0}$.
Assuming to contrary, say $k > \frac{a+b}4 \implies 4a_0b_0+4 > d(a_0^2+b_0^2+2a_0b_0) > a_0^2+b_0^2+2a_0b_0 \implies 4 > (a_0-b_0)^2$.
Thus $|a_0-b_0| < 2$. Considering $a_0 \geq b_0$, we get $a_0$ as one of $b_0,b_0+1$.
As $k$ is an integer, $\frac{a_0b_0+1}{a_0+b_0}$ is an integer $\implies a_0+b_0 \mid a_0b_0+1$.
Case 1: $a_0=b_0$: $2a_0 \mid a_0^2+1 \implies \mid 2a_0^2+2-2a_0^2 \implies \mid 2$. Thus $a_0=b_0=1$, which is a contradiction as $k>1$.
Case 2: $a_0=b_0+1$: $2b_0+1 \mid b_0^2+b_0+1 \implies \mid 2b_0^2+2b_0+2 -b_0(2b_0+1) \implies \mid b_0+2 \implies \mid 2b_0+4-2b_0-1 \implies \mid 3$. Thus $b_0=1 \implies a_0=2$, which is a contradiction, as $k>1$.
How to prove it without contradiction? An elegant method
| As you noted, it's enough to consider $a$ and $b$ coprime. Then, the problem is to show the implication
\begin{align*}
ab + 1 \stackrel{(1)}{=} k(a+b) \Longrightarrow a+b \stackrel{(2)}{\geqslant} 4k.
\end{align*}
Geometrically, (1) says that $(a,b)$ is a lattice point on the graph of the function
\begin{align*}
f_k(x) = \frac{kx-1}{x-k},
\end{align*}
and (2) says that $(a,b)$ lies above the line $\ell_k : x+y = 4k$. In other words, the problem is to show that no primitive lattice point $(a,b)$ on the graph of $f_k$ can lie (strictly) below $\ell_k$.
By doing a straightforward computation, one finds that the intersections of $\ell_k$ and the graph of $f_k$ occur at $x = 2k \pm 1$, and that $\ell_k$ is strictly below the (relevant part of the) graph for $x$ outside the interval $\left[ 2k-1, 2k+1 \right]$. Hence the only possibility for a lattice point $(a,b)$ on the graph of $f_k$ to lie below $\ell_k$ is that its $x$-coordinate is $2k$. But then its $y$-coordinate fails to be an integer, precisely because we require $k \geqslant 2$. In detail,
\begin{align*}
f_k(2k) = \frac{2k^2 - 1}{k} = 2k - \frac{1}{k}.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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How to integrate $\int_{C} \frac{z^2}{(z^4 +1)}$ where $C$ is $|z + i| = 1$ I am attempting to do this using Cauchy's integral theorem and formula. However I am unable to conclude if a singularity exists at all for me to apply any of those two techniques or any other theorem.
| Using partial fraction decomposition
$$\frac{z^2}{z^4+1}=
\frac{1}{2(z^2+i)}+\frac{1}{2(z^2-i)}=\\
\color{red}{\frac{1}{4\cdot\frac{1-i}{\sqrt{2}}\cdot\left(z-\frac{1-i}{\sqrt{2}}\right)}}-
\frac{1}{4\cdot\frac{1-i}{\sqrt{2}}\cdot\left(z+\frac{1-i}{\sqrt{2}}\right)}+
\frac{1}{4\cdot\frac{1+i}{\sqrt{2}}\cdot\left(z-\frac{1+i}{\sqrt{2}}\right)}-
\color{red}{\frac{1}{4\cdot\frac{1+i}{\sqrt{2}}\cdot\left(z+\frac{1+i}{\sqrt{2}}\right)}}$$
Given $\left|\frac{1-i}{\sqrt{2}}+i\right|<1$ and $\left|-\frac{1+i}{\sqrt{2}}+i\right|<1$, only the values in red fall inside $|z+i|=1$ circle.
Using Cauchy's integral formula
$$f(a)=\frac{1}{2\pi i} \int\limits_{\gamma}\frac{f(z)}{(z-a)}dz$$
where $f(z)=1$
$$\int\limits_{|z+i|=1}\frac{z^2}{z^4+1}dz=
\int\limits_{|z+i|=1}\color{red}{\frac{1}{4\cdot\frac{1-i}{\sqrt{2}}\cdot\left(z-\frac{1-i}{\sqrt{2}}\right)}}dz-\int\limits_{|z+i|=1}\color{red}{\frac{1}{4\cdot\frac{1+i}{\sqrt{2}}\cdot\left(z+\frac{1+i}{\sqrt{2}}\right)}}dz=\\
\frac{2\pi i}{4\cdot\frac{1-i}{\sqrt{2}}}-\frac{2\pi i}{4\cdot\frac{1+i}{\sqrt{2}}}=
\frac{\pi i}{\sqrt{2}\cdot(1-i)}-\frac{\pi i}{\sqrt{2}\cdot(1+i)}=\\
\frac{\pi}{\sqrt{2}}\left(\frac{i}{1-i}-\frac{i}{1+i}\right)=-\frac{\pi}{\sqrt{2}}$$
| {
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$a_n = a_1 + (n-1)d$, $b_n=\frac{\sqrt[n]{a_1 \cdot a_2 ... \cdot a_n}}{\frac{a_1+a_2...+ a_n}{n}}$ Prove $\lim_{n \to \infty}b_n$ exists, and find it
Let $a_n = a_1 + (n-1)d$ and $b_n=\frac{\sqrt[n]{a_1 \cdot a_2 \cdot\ldots \cdot a_n}}{\frac{a_1+a_2+\ldots+ a_n}{n}}$
Prove that $\lim_{n \to \infty}b_n$ exists and find it for $d>0$ and $a_1>0$.
My attempt:
Because an arithmetic average is bigger than a geometric one, then $0\le b_n \le 1$
The sum of $a_n: S_n = \frac{n(2a_1 +(n-1)d)}{2}$ (arithmetic sequence)
Then $b_n = \frac{\sqrt[n]{a_1 \cdot a_2 \cdot... \cdot a_n}}{\frac{\frac{n(2a_1 +(n-1)d)}{2}}{n}} = \frac{2\sqrt[n]{a_1 \cdot a_2 \cdot... \cdot a_n}}{2a_1 +(n-1)d}$
I tried to prove that $b_n \ge b_{n+1}$, but couldn't (then I would have been able to say that $\lim_{n \to \infty}b_n$ exists.)
Any hints would be appreciated. Thanks!
| This is a sketch.
Set $A_n:=\prod^n_{k=1}\frac{a+b(k-1)}{a+\frac{b}{2}(n-1)}$. Then
\begin{align*}
\frac{A_{n+1}}{A_n}&= \frac{a+bn}{a+\frac{b}{2}n}\left(\frac{a+\frac{b}{2}(n-1)}{a+\frac{b}{2}n}\right)^n \\
&= \frac{a+bn}{a+\frac{b}{2}n}\left(1-\frac{b}{2a+bn}\right)^n\rightarrow2 e^{-1}
\end{align*}
The convergence of $A_{n+1}/A_n$ implies the converge of $\sqrt[n]{A_n}$ and they have the same limit.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why combination with repetition doesn't use the same formula as unitary coefficient equation Unitary coefficient equation uses the formula
$$ x_1 + x_2 + \ldots + x_k = m $$
$$ C(m + k - 1, k - 1) $$
(considering $x$ can be $0$, i.e solutions in positive integers)
But for combination with repetition, my book says that the formula is $C(m + k - 1, k)$ (note it is not $k-1$) and says that it is equivalent to proposing the problem as:
$$x_1 + x_2 + \ldots + x_k = m$$
The weird thing is that in the example for combination with repetition is exactly the way of a equation with unitary coefficients.
Example: How many different sets of three coins can be made if each coin can be 10, 25, 50 cents or 1 dollar.
It says that this can be thought as $x_1 + x_2 + x_3 + x_4 = 3$
In this case is $C(6, 3)$ which seems to correlate with the first theorem.
But another example says "In a library there 20 math books each one with unlimited quantity. How many elections of 10 books can be done if repetitions are allowed?"
And puts as answer $C(29, 10)$ instead of $C(29, 9)$
As my thinking is that it should be $C(20 + 10 - 1, 10 - 1)$
I am a bit confused.
Idiot-proof answer is much appreciated.
| The number of solutions of the equation
$$x_1 + x_2 + x_3 + \ldots + x_k = m$$
in the nonnegative (not positive) integers is
$$\binom{m + k - 1}{k - 1} = \binom{m + k - 1}{m}$$
The author of your book seems to be using the formula
$$\binom{m + k - 1}{m}$$
in which case $\binom{m + k - 1}{k}$ is a rather unfortunate typographical error.
How many different sets of three coins if each coin can be 10 cents, 25 cents, 50 cents, or one dollar?
There are four different types of coins, so $k = 4$. A total of three coins are selected, so $m = 3$. Thus, we obtain the equation
$$x_1 + x_2 + x_3 + x_4 = 3$$
Our formula yields
$$\binom{m + k - 1}{k - 1} = \binom{3 + 4 - 1}{4 - 1} = \binom{6}{3} = \binom{3 + 4 - 1}{3} = \binom{m + k - 1}{m}$$
In a library, there are $20$ books, each one with unlimited quantity. How many selections of $10$ books can be done if repetitions are allowed?
In this case, there are $20$ types of books, so $k = 20$. A total of $10$ books are selected, so $m = 10$. Our formula yields
$$\binom{m + k - 1}{k - 1} = \binom{10 + 20 - 1}{20 - 1} = \binom{29}{19} = \binom{29}{20} = \binom{10 + 20 - 1}{20} = \binom{m + k - 1}{m}$$
where
$$\binom{29}{19} = \binom{29}{10}$$
since
$$\binom{n}{j} = \binom{n}{n - j}$$
Your confusion stems from what appears to be a typographical error in the text and, in the second problem, from distinguishing between $k$ and $m$.
| {
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$x$, $y$, $z$ are positive integers such that $x + y \mid xy$ and $y + z \mid yz$. Prove that $\gcd(x, y, z) > 1$.
$x$, $y$, $z$ are positive integers such that $x + y \mid xy$ and $y + z \mid yz$. Prove that $\gcd(x, y, z) > 1$.
Let $\gcd(x, y) = m$ and $\gcd(y, z) = n$, we have that $$\gcd(m, n) > 1 \implies \gcd(x, y, z) > 1$$
Suppose that $\gcd(m, n) = 1$ and $$x = mx', y = mny', z = nz' (\gcd(x', y', z') = 1)$$
We have that $$\left. \begin{align} \frac{xy}{x + y} = \frac{mx' \cdot mny'}{mx' + mny'} = \frac{mnx'y'}{x' + ny'}\\ \frac{yz}{y + z} = \frac{mny' \cdot nz'}{mny' + nz'} = \frac{mny'z'}{my' + z'} \end{align} \right\} \in \mathbb N \implies \frac{mnx'y'z'}{\gcd(x' + ny', my' + z')} \in \mathbb N$$
Then I was stuck.
| If $x+y\mid xy$, then we have $xy=k(x+y)$ for some positive integer $k$.
Rewrite it as $(x-k)(y-k)=k^2$. Let $d=\gcd(x-k,y-k)$. We then have $x-k= a^2d, y-k=b^2d,k=dab$ for some coprime positive integers $a,b$. That is, $x=a(a+b)d$ and $y=b(a+b)d$.
Similarly, we have $y=p(p+q)s$ and $z=q(p+q)s$ for some positive integers $p,q,s$ with $p,q$ coprime.
Suppose $\gcd(x,y,z)=1$. Since $a+b$ divides $x,y$, it must be prime to $z$, and therefore prime to $(p+q)s$. But it divides $y=p(p+q)s$, so it must divide $p$. In particular, we have $p\geq a+b$.
Similarly, $p+q$ must divide $b$, and hence $b\geq p+ q$.
This is a contradiction.
| {
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Compute the $PV\int_0^{\frac{π}{3}}\frac{\cos (4x)}{\cos (3x)}dx$ Problem :
Evaluate the closed form of : $PV\displaystyle\int_0^{\frac{π}{3}}\frac{\cos (4x)}{\cos (3x)}dx$
Wolfram alpha give me :
$I=PV\displaystyle\int_0^{\frac{π}{3}}\frac{\cos (4x)}{\cos (3x)}dx=\sqrt{3}-\coth^{-1}{\sqrt{3}}$
But i can't get by my try as following :
$\cos (4x)=8\cos^{4} x-8\cos^{2} x+1$
And
$\cos (3x)=4\cos^{3} x-3\cos x$
And I know that
$PV\displaystyle\int_0^{\frac{π}{3}}\frac{1}{\cos (3x)}dx=0$
So I need to find
$J=PV\displaystyle\int_0^{\frac{π}{3}}\frac{\cos^{2} x}{\cos (3x)}dx=PV\displaystyle\int_0^{\frac{π}{3}}\frac{\sin^{2} x}{\cos (3x)}dx$
Now take $y=\cos x$
$J=\displaystyle\int_{\frac{1}{2}}^{1}\frac{\sqrt{1-x^2}}{4x^{3}-3x}dx$
From here I don't know how I complete
| Continue with $PV\displaystyle\int_0^{\frac{π}{3}}\frac{dx}{\cos (3x)}=0$ to rewrite the integral as,
$$I =PV \int_0^{\frac{π}{3}}\frac{8(\cos^{2} x-1)\cos^2 x}{4\cos^3 x -3 \cos x}dx
=PV \int_0^{\frac{π}{3}}\frac{-4\sin^{2}x\> d(2\sin x)}{1-4\sin^2 x }dx$$
Substitute $t = 2\sin x$,
$$I =PV \int_0^{{\sqrt3}}\left(1-\frac1{1-t^2}\right)dt
=\sqrt3-\int_0^{1-\epsilon}\frac{dt}{1-t^2}
-\int_{1+\epsilon}^{{\sqrt3}}\frac{dt}{1-t^2}$$
$$=\sqrt3 - \tanh^{-1}t|_0^{1-\epsilon}-\coth^{-1}t|_{1+\epsilon}^{{\sqrt3}}
=\sqrt3 - \coth^{-1}\sqrt3$$
where $(\tanh^{-1}t)'=\frac1{1-t^2}\> \text{for}\> t^2<1$ and $(\coth^{-1}t)'=\frac1{1-t^2}\> \text{for}\> t^2>1$ are used.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the smallest integer $n$ greater than $1$ such that the last $3$ digits of $n^2$ are the same as the last $3$ digits of $n$. Find the smallest integer $n$ greater than $1$ such that the last $3$ digits of $n^2$ are the same as the last $3$ digits of $n$.
So far I've got $n^2 = 1000k + n$ which means $n^2 ≡ n \mod 1000$. I don't know how to proceed since 1000 seems to be a bit high.
| In a practical way, the number $N$ is, $\forall p>=0$ $$p\times1000+(100a+10b+c)$$
We can ignore the prefix $p\times1000$ as it doesn't affect the $3$ last digits of $N^2$
We want $c$ and the last digit of $c^2$ to be equal
$$c\equiv c^2\pmod{10}$$
That only works for
$$\bbox[7px,border:1px solid gray]{c\in\{0,1,5,6\}}$$
For $b$, we want the second digit of $(10b+c)^2$ to be equal to $b$ (note: $\lfloor x\rfloor$ denotes the integer part of $x$)
$$b\equiv \lfloor\frac{(10b+c)^2}{10}\rfloor\pmod{10}$$
Plugging our $c$ values, $b$ can only be $0$ if $c=0$, also $0$ if $c=1$, $2$ if $c=5$ and $7$ if $c=6$
$$\bbox[7px,border:1px solid gray]{c=0\implies b = 0}\\
\bbox[7px,border:1px solid gray]{c=1\implies b = 0}\\
\bbox[7px,border:1px solid gray]{c=5\implies b = 2}\\
\bbox[7px,border:1px solid gray]{c=6\implies b = 7}\\$$
Not so many valid candidates after all.
Finally, for $a$, we want the third digit of $(100a+10b+c)^2$ to be equal to $a$
$$a\equiv \lfloor\frac{(100a+10b+c)^2}{100}\rfloor\pmod{10}$$
We have only $4$ cases to check since $b$ and $c$ values are strongly correlated. Injecting our $(b,c)$ values,we get
$$\bbox[7px,border:1px solid gray]{c=b=0\implies a = 0}\\
\bbox[7px,border:1px solid gray]{c=1,b=0\implies a = 0}\\
\bbox[7px,border:1px solid gray]{c=5,b=2\implies a = 6}\\
\bbox[7px,border:1px solid gray]{c=6,b=7\implies a = 3}\\$$
Meaning the last $3$ digits of $N$ can be any of
$$000, 001, 625, 376$$
Since we want the smaller valid number $>1$, the answer is
$$\bbox[8px,border:1px solid blue]{376}\\$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find minimum value of $P = \frac{y + z}{x}$
Let $x,y,z > 0$ such that $3x(x+y+z)=yz$ . Find the minimum value of $$P=\frac{y+z}{x}$$
$$3x(x+y+z)=yz\Leftrightarrow 3(x+y)(x+z)=4yz$$
Or $$3(\frac{x}{y}+1)(\frac{x}{z}+1)=4$$
Let $a=x/y, b=x/z ( a,b>0)$
We have: $$4=3(a+1)(b+1)\ge3 \cdot 2 \sqrt a \cdot 2 \sqrt b=12 \sqrt {ab} \rightarrow \sqrt {ab}\le \frac 1 3$$
And $$P=\frac{y+z}{x}=\frac{1}{a}+\frac{1}{b}\ge \frac{2}{\sqrt {ab}}\ge 6$$
But by WA $Min_P=6+4\sqrt 3$. Is my solution is wrong?
| Use Lagrange multipliers. Call the function you want to minimize $f$ and the constraint $g$. From the condition on the partial derivatives on $y$ and $z$ of $f - \Lambda g$ you can deduce that $y=z$. Finding $x$ with respect to $y$ gives you $x=\frac{\sqrt{12}-3}{3}y$, and this leads to the solution.
You'd have to check this is actually a minimum (in principle this is only a critical point), but you should be able to verify that by yourself.
| {
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"timestamp": "2023-03-29T00:00:00",
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$x^{4\log x} = \frac{x^{12}}{{10}^8} $ all sum of x? ${10}^8 = \frac{x^{12}}{x^{4\log x}}$
${10}^8 = x^{12 - 4\log x}$
$x > 0$
$x \neq 1$
by guessing, x = 10 is true. but how sum of all x?
| As you wrote: $x\ne1\;\;\&\;\;x>0$
Another way:
$\log_xx^{4\log{x}}=\log_x{\frac{x^{12}}{10^8}}$
$4\log x\cdot\log_xx=\log_x{x^{12}}-\log_x{10^8}$
$4\log x=12-8\log_x10\Bigg/:4$
$\log{x}=3-2\log_x10$
$\log{x}=\frac{\log_xx}{\log_x10}=\frac{1}{\log_x10}$
$\frac{1}{\log_x10}=3-2\log_x10\Bigg/\cdot\log_x10$
$2\log^2_x10-3\log_x10+1=0$
$2\log^2_x10-2\log_x10-\log_x10+1=0$
$2\log_x10(\log_x10-1)-(\log_x10-1)=0$
$(\log_x10-1)(2\log_x10-1)=0$
$\log_x10=1\implies x_1=10$
$\log_x10^2=1\implies x_2=100$
$x_1+x_2=110$
| {
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"url": "https://math.stackexchange.com/questions/3485672",
"timestamp": "2023-03-29T00:00:00",
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If $a=\sin\theta+\cos\theta$ and $b=\cos\theta-\sin\theta$, express $\sum_k\sin^k\theta(\sin^k\theta-\cos^k\theta)$ in terms of $a$ and $b$
If $a=\sin\theta+\cos\theta$ and $b=\cos\theta-\sin\theta$, then express
$$\sin \theta(\sin \theta-\cos \theta)+\sin^2\theta(\sin^2\theta-\cos^2\theta)+\sin^3\theta(\sin^3\theta-\cos^3\theta) +\cdots$$
in terms of $a$ and $b$.
My observation :
$\underbrace{\sin \theta(\sin \theta-\cos \theta)}_1 +$$\underbrace{\sin^2\theta(\sin^2\theta-\cos^2\theta)}_2 +\underbrace{\sin^3\theta(\sin^3\theta-\cos^3\theta)}_3 +\cdots\\$
$i.)$$\;\sin \theta(\sin \theta-\cos \theta) =\sin \theta (-b) =\bbox[4px, yellow] {-b\sin \theta} $ $\\$
$ii.) \;\sin^2 \theta(\sin^2 \theta-\cos^2 \theta)=\sin^2 \theta(\sin\theta-\cos
\theta)(\sin \theta + \cos\theta)=\bbox[yellow, 4px]{-ab\sin^2\theta}
\\$
$\mathrm {Now} \\$
$iii.)$ $\sin^3\theta(\sin^3\theta-\cos^3\theta)=\sin^3\theta (\sin\theta-\cos\theta)(\sin^2\theta +\sin\theta \cos\theta+\cos^2\theta) \\$
$=\sin^3\theta(-b)\left(1+\boxed {\sin\theta\cos\theta} \rightarrow\left\{\frac{a^2-b^2} {4} \right\} \right) \\$
$=\underline{\bbox[yellow, 4px] {-b\sin^3\theta\left(1+\frac{a^2-b^2}{4}\right)}}$
Can anyone suggest a way out?
| Subtracting b from a we have $a-b=\sin(x)+\cos(x)-(\cos(x)-\sin(x))$ thus $\frac{a-b}{2}=\sin(x)$
| {
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"url": "https://math.stackexchange.com/questions/3486158",
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Show that $r^{2} = 3$ from a geometric series
The sum of the first two terms of geometric sequence $a_n=a_0r^n$ is $7$, and the sum of the first six terms is $91$.
Show that $r^{2} = 3$.
How do we go about this?
Here's what I have tried -
$\displaystyle \Rightarrow \frac{\dfrac{a(r^{6}-1)}{r-1}}{\dfrac{a(r^{2}-1)}{r-1}} = \frac{91}{7}$
$\displaystyle \Rightarrow \frac{(r^{6}-1)}{(r^{2}-1)} = \frac{91}{7}$
$\Rightarrow\;???$
Any help would be highly appreciated.
| Note that $$r^6-1=(r^3+1)(r^3-1).$$
From @MMM's answer, we can observe that for all $a$ and $b$, $$a^3-b^3=(a-b)(a^2+ab+b^2). $$ Let $a=r$ and $b=\pm 1$. $$\therefore r^6-1=(r+1)(r^2-r+1)\cdot (r-1)(r^2+r+1).$$ Since $r^2-1=(r+1)(r-1)$, it follows then that $$(r^2-r+1)(r^2+r+1)=13.$$
Can you take it from here? :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Proof by Induction $2\cdot 7^n + 3\cdot 5^n - 5$ is multiple of $24$ Proof by induction that $2\cdot 7^n + 3\cdot 5^n - 5$ is a multiple of $24$.
I tried solving but got stuck.
Show that it is true for n=1
$$2\cdot 7^1 + 3\cdot 5^1 - 5 = 14 + 15 - 5 = 24$$
Assume it true for $n = k$
$$2\cdot 7^k + 3\cdot 5^k - 5 = 24g$$
Show it is true for $n= k + 1$
$$2\cdot 7^{k+1} + 3\cdot 5^{k+1} - 5$$
is a multiple of 24
$$(2\cdot 7)(2\cdot 7^k) + (3\cdot 5^k)(3\cdot 5) - 5$$
$$2\cdot7(24g + 5)3\cdot 5 - 5$$
I'm stuck and don't know how to proceed
| Hint:
$24 $ divides $2\cdot7^n+3\cdot5^n-5$ for $n=0$ and $1,$
and $2\cdot7^{n+2}+3\cdot5^{n+2}-5=49(2\cdot7^n+3\cdot5^n-5)-24(3\cdot5^n-10).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Evaluate $\int \cos 2\theta \ln\left(\frac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)d\theta$ $$\int \cos 2\theta \ln\left(\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)d\theta$$
My attempt is as follows:-
$$\ln\left(\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)=t\tag{1}$$
$$\dfrac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\cdot\dfrac{\left(\cos\theta-\sin\theta\right)^2-(-\sin\theta-\cos\theta)(\cos\theta+\sin\theta)}{(\cos\theta-\sin\theta)^2}=\dfrac{dt}{d\theta}$$
$$\dfrac{2}{\cos2\theta}=\dfrac{dt}{d\theta}$$
Let's calculate $\cos2\theta$ from equation $1$
$$\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}=e^t$$
$$\dfrac{1+\tan\theta}{1-\tan\theta}=e^t$$
Applying componendo and dividendo
$$\dfrac{2}{2\tan\theta}=\dfrac{e^t+1}{e^t-1}$$
$$\dfrac{e^t-1}{e^t+1}=\tan\theta$$
$$\cos2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}$$
$$\cos2\theta=\dfrac{(e^t+1)^2-(e^t-1)^2}{(e^t+1)^2+(e^t-1)^2}$$
$$\cos2\theta=\dfrac{4e^t}{2(e^{2t}+1)}$$
$$\cos2\theta=\dfrac{2e^t}{e^{2t}+1}\tag{2}$$
So integral will be
$$\dfrac{1}{2}\cdot\int \left(\dfrac{2e^t}{e^{2t}+1}\right)^2dt$$
$$\dfrac{1}{2}\cdot\int \dfrac{4e^{2t}}{(1+e^{2t})^2}$$
$$e^{2t}+1=y$$
$$2e^{2t}=\dfrac{dy}{dt}$$
$$2e^{2t}dt=dy
$$\int \dfrac{dy}{y^2}$$
$$-\dfrac{1}{y}+C$$
$$-\dfrac{1}{1+e^{2t}}+C$$
$$-\dfrac{1}{1+e^{\ln\left(\dfrac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right)^2}}+C$$
$$-\dfrac{1}{1+\dfrac{1+\sin2\theta}{1-\sin2\theta}}+C$$
$$-\dfrac{1-\sin2\theta}{2}+C$$
$$\dfrac{\sin2\theta}{2}+C'$$
And this should be actually wrong because if we differentiate the result, it will give $\cos2\theta$, but integrand is $\cos 2\theta \ln\left(\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)$
What am I missing here, checked multiple times, but not able to get the mistake. Any directions?
| With integration by parts you get
$$
\begin{align}
&\int \cos 2\theta \ln\left(\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right)\mathrm{d}\theta\\
&=\frac 12\sin2\theta\ln\left(\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right)-\frac 12\int \sin2\theta\left(\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\right)\left(\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right)^{'}\mathrm{d}\theta \\
&=\frac 12\sin2\theta\ln\left(\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right)-\int\tan2\theta\,\mathrm{d}\theta
\end{align}$$
so you just need to use
$$\int \tan x\,\mathrm{d}x=-\ln|\cos x|+C $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Given a fair coin, what is the mean of the number of tails before we toss a head?
Question: Given a fair coin, what is the mean of the number of tails before we toss a head?
Let $N$ be the number of tails before we toss a head. Then
$$E(N) = \sum_{n=1}^\infty n P(N=n).$$
Since
$$P(N=n) = \left(\frac{1}{2}\right)^{n-1}.$$
Then by using geometric series, we have
\begin{align*}
E(N) & = \sum_{n=1}^\infty n P(N=n) \\
& = \sum_{n=1}^\infty n \left(\frac{1}{2}\right)^{n-1} \\
& = \frac{1}{(1-\frac{1}{2})^2} \\
& = 4.
\end{align*}
Are my calculations correct?
| $P(N=n)=\frac{1}{2}^{n+1}$ not $P(N=n)=\frac{1}{2}^{n-1}$.
$P(N=n)$ is the probability that we get n tails and then one head so $P(N=n)=\frac{1}{2}^{n}\frac{1}{2}=\frac{1}{2}^{n+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Inverse trigonometry: How to find $x$ When $\sin^{-1}\left(\frac{2x}{1+x²}\right)=2\tan^{-1}(x)$? I thought that it was for all $x\in\mathbb{R}$ but it was incorrect, so please help!
| Use
Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
OR
Why is $\arctan\frac{x+y}{1-xy} = \arctan x +\arctan y$?
$$2\arctan x=\begin{cases} \arctan\dfrac{2x}{1-x^2} &\mbox{if } x^2<1\\ \pi+\arctan\dfrac{2x}{1-x^2} & \mbox{if } x^2>1\\\text{sign}(x)\cdot\dfrac\pi2 & \mbox{if } x^2=1\end{cases} $$
Finally if $\arctan\dfrac{2x}{1-x^2}=y,\tan y=\dfrac{2x}{1-x^2},-\dfrac\pi2<y<\dfrac\pi2$
$\sec y=+\sqrt{1+\left(\dfrac{2x}{1-x^2}\right)^2}=\dfrac{1+x^2}{|1-x^2|}$
$\sin y=\dfrac{\tan y}{\sec y}=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int \dfrac{x^2}{(x\cos x-\sin x)(x\sin x+\cos x)}dx$ $$\int \dfrac{x^2}{(x\cos x-\sin x)(x\sin x+\cos x)}dx$$
My attempts is as follows:-
Integrating by parts:-
$$\dfrac{1}{(x\cos x-\sin x)(x\sin x+\cos x)}\cdot\dfrac{x^3}{3}-\dfrac{1}{3}\int\dfrac{x^3\left(-x\sin x(x\sin x+\cos x)+x\cos x(x\cos x-\sin x)\right)}{(x\cos x-\sin x)^2(x\sin x+\cos x)^2}dx$$
$$\dfrac{1}{(x\cos x-\sin x)(x\sin x+\cos x)}\cdot\dfrac{x^3}{3}-\dfrac{1}{3}\int\dfrac{x^3\left(-x^2\sin^2x-x\sin x\cos x-x\cos x\sin x+x^2\cos^2x\right)}{(x\cos x-\sin x)^2(x\sin x+\cos x)^2}dx$$
$$\dfrac{1}{(x\cos x-\sin x)(x\sin x+\cos x)}\cdot\dfrac{x^3}{3}-\dfrac{1}{3}\int\dfrac{x^3\left(x^2\cos2x-x\sin 2x\right)}{(x^2\cos x\sin x+x\cos^2x-x\sin^2x-\sin x\cos x)^2}dx$$
$$\dfrac{1}{(x\cos x-\sin x)(x\sin x+\cos x)}\cdot\dfrac{x^3}{3}-\dfrac{1}{3}\int\dfrac{x^3\left(x^2\cos2x-x\sin 2x\right)}{(\cos x\sin x(x^2-1)+x\cos 2x)^2}dx$$
I thought numerator and denominator in the second would be quite similar, but its not, so I was not able to proceed from here.
Any directions?
| It's just $$\ln\left|\frac{x\sin{x}+\cos{x}}{x\cos{x}-\sin{x}}\right|+C,$$ which we can get by the definition of the integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\lim_{n\to\infty}\{\frac{3}{n}\sum_{k=1}^n[1+8\sin^2(\frac{k\pi}{n})]^{-1}\}^{2^n}$ How to calculate the limit below?
$$\lim_{n\to\infty}\{\frac{3}{n}\sum_{k=1}^n[1+8\sin^2(\frac{k\pi}{n})]^{-1}\}^{2^n}$$
Since I used Riemann integration to work out that $$\lim_{n\to\infty}[\frac{3}{n}\sum_{k=1}^n(1+8\sin^2(\frac{k\pi}{n}))^{-1}]=1$$, I've been trying to express $$\frac{3}{n}\sum_{k=1}^n(1+8\sin^2(\frac{k\pi}{n}))^{-1}=(1+\frac{C}{2^n}+o(\frac{1}{2^n})),\quad n\to+\infty$$
Can anyone render me some hints?
| From what it seems$$\frac{3 }{n}\sum _{k=1}^n \frac{1}{1+8 \sin ^2\left(\frac{\pi k}{n}\right)}=\frac{2^n+1}{2^n-1}=1+\frac 2 {2^n}+ \cdots$$
So, now we consider
$$S=\left(\frac{2^n+1}{2^n-1}\right)^{2^n}$$ Let $x=2^n$ making
$$S=\left(\frac{x+1}{x-1}\right)^{x}\implies \log(S)=x \log\left(\frac{x+1}{x-1}\right)=x \log\left(1+\frac{2}{x-1}\right)$$ So, by Taylor
$$\log(S)=2+\frac{2}{3 x^2}+\frac{2}{5 x^4}+O\left(\frac{1}{x^6}\right)$$
$$S=e^{\log(S)}=e^2+\frac{2 e^2}{3 x^2}+\frac{28 e^2}{45
x^4}+O\left(\frac{1}{x^6}\right)\qquad \text{with}\qquad x=2^n$$
Trying for $n=4$, the exact result is
$$\frac{48661191875666868481}{6568408355712890625}\approx 7.40836885$$ while the above truncated formula gives
$$\frac{739207 }{737280}e^2 \approx 7.40836859$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$(x+y)\frac{dy}{dx} -(4x+y)=0, y(1)=2$ Hi I am stuck solving this problem:
'Employ a change of variables $z=\frac{y}{x}$ to solve the differential equation: $$(x+y)\frac{dy}{dx} -(4x+y)=0, y(1)=2$$
Enter the result $y(3)$.
I have done the substitution and then integrated by I cannot get an explicit expression for y, which leaves me confused as to how I can find $y(3)$.
Please help!
| Taking $z = \frac{y}{x}$ yields $y=xz$, and hence $\frac{dy}{dx} = x\frac{dz}{dx}+z$.
From this, we find by substitution
\begin{align*}
(x+xz)(x\frac{dz}{dx}+z) - (4x+xz) &= 0 \\
\text{ divide by } x, \text{ and simplify}\\
x \frac{dz}{dx} +z +zx \frac{dz}{dx} +z^2 -4-z &= 0 \\
\frac{dz}{dx}x(1+z) &= 4-z^2 \\
\frac{dz(1+z)}{4-z^2} &= \frac{dx}{x} \\
dz \left( \frac{3}{4(2-z)} - \frac{1}{4(2+z)} \right) &= \frac{dx}{x}
\end{align*}
Solving this differential equation, we get for some constant $K$,
$$\frac{1}{|(z+2)(z-2)^3|} = Kx^4.$$
Substituting $z = \frac{y}{x}$ again, we get after simplifying
$$ 1 = K|y^2-4x^2|(y-2x)^2 $$
Using the information that $y(1)=2$, we should be able to find $K$, but we get the following.
\begin{align*}
1 &= K|2^2-4 \cdot 1^2|(2-2 \cdot 1)^2 \\
1 &= 0
\end{align*}
Are you sure you copied the condition $y(1)=2$ correctly? There is definetily some mistake in your formulation of the problem anyway.
The next step would be to solve the polynomial equation
$$1 = K|y^2-4 \cdot 3^2|(y-2 \cdot 3)^2$$
Finding $y$ that fits this will then be the solution to your problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Two rectangles: The $1$st has twice the perimeter of the $2$nd and the $2$nd has twice the area of the $1$st. How can this be solved using just algebra, where the first rectangle has sides $a$ and $b$, and the second rectangle has sides $c$ and $d$?
These are the two equations that follow:
$$a + b = 2(c + d)$$
and
$$2ab = cd$$
| Given $a,b,c,d>0$, the condition is $$\begin{cases}c+d=\frac{a+b}2\\ cd=2ab\end{cases}$$
Let's adopt the convention $c\le d$, so that the system is satisfied if and only if $$\begin{cases}c=\frac{a+b}4-\frac12\sqrt{\left(\frac{a+b}2\right)^2-8ab}\\ d=\frac{a+b}4+\frac12\sqrt{\left(\frac{a+b}2\right)^2-8ab}\end{cases}\\ \begin{cases}c=\frac{a+b-\sqrt{a^2-30ab+b^2}}4\\ d=\frac{a+b+\sqrt{a^2-30ab+b^2}}4\end{cases}$$
Notice that, under the hypothesis, both those numbers are strictly positive, provided that the square root exists, i.e. provided that $a^2-30ab+b^2\ge 0$. Under the condition $a,b>0$ this is the case if and only if $\frac ab\le 15-4\sqrt{14}$ or $\frac ab\ge15+4\sqrt{14}$.
Therefore the set of solutions $(c,d)$ with $c\le d$ is parametrised by the region of plane $\begin{cases} \frac ab\ge 15+4\sqrt{14}\\ b>0 \end{cases}\lor\begin{cases} \frac ab\le 15-4\sqrt{14}\\ b>0 \end{cases}$ via the function $$(a,b)\mapsto \left(\frac{a+b-\sqrt{a^2-30ab+b^2}}4,\frac{a+b+\sqrt{a^2-30ab+b^2}}4\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
on $p^n+q^n=(p+q)^k$ Evaluate the condition for integer solutions for $n$.
I found this question in an old book:
Find smallest number of form $2^n+3^n$ divisible by $625$. This solution is from that book; n must be odd and we may write:
$$2^n+3^n$$ $$=2^n+(-1)^n(2-5)^n$$ $$=2^n+(-1)^n\cdot 2^n-(-1)^n\cdot2^{n-1}\cdot5\cdot n+(-1)^n\cdot2^{n-2}\cdot5^2\frac{n(n-1)}{2}-(-1)^n\cdot2^{n-3}\cdot5^3\frac{n(n-1)(n-2)}{6}+625 N$$
$$2^n+3^n=5n\big[2^{n-1}-(n-1)2^{n-2}.5+\frac{(n-1)(n-2)}{3}2^{n-2}.5^2\big]+625 N;\ n\geq 4$$
The value inside box bracket is not divisible by $5$, so $n$ must be divisible by $125$ if $2^n+3^n$ must be divisible by $625$.
Similar reasoning can be used for any primes $p$ and $q$ such that:
$p^n+q^n≡0 \ mod (p+q)^k$
The condition is
$n=(p+q)^{k-1}$.
Now we try to apply Euler $\phi$ function:
$\phi(625)=625\big(1-\frac{1}{5}\big)=500$
$2^{500}≡1 \mod 625$
$3^{500}≡ 1 \ mod 625$
$3^{500}-2^{500}≡0 \mod 625$
$(3^{125}-2^{125})(3^{125}+2^{125})(3^{250}+2^{250})≡0 \mod 625$
Only $3^{125}+2^{125}$ can be divisible by 625. But $3^{125}+2^{125}$ can be factorized more and more.Suppose we can not use first method because p and q are too large, then how can we be sure the smaller factors are not divisible by $625$? is $n=125$ the smallest number?
| A GENERAL PROOF
If $p$ and $q$ are distinct primes then $p,q,p,p+q$ are pairwise coprime.
We require $n$ to be odd and then $$\frac{p^n+q^n}{p+q}=p^{n-1}-qp^{n-2}+ ... +q^{n}\equiv np^{n-1}\pmod {p+q}$$
Therefore $p^n+q^n$ is divisible by $(p+q)^{k}$ if and only if $n$ is divisible by $(p+q)^{k-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a$, $b$, $c$, $d$, $e$ are positive integers such that $abcde=a+b+c+d+e$, then what is the maximum value of $e$?
If $a$, $b$, $c$, $d$, $e$ are positive integers such that $abcde=a+b+c+d+e$, then what is the maximum value of $e$?
The answer is $5$.
What I tried:
Let $a\leq b \leq c \leq d \leq e$. Then
$$abcde \leq 5e \quad\Rightarrow\quad abcd \leq 5$$
For the maximum value of $e$, here $a=b=c=1$ and $d=5$.
So $$e=\frac{a+b+c+d}{abcd-1}=\frac{8}{4}=2$$
But this is not the answer.
How do I solve it? Help me, please.
| It's impossible that $a=b=c=d=1.$
Thus, $abcd\geq2$ and
$$e=\frac{a+b+c+d}{abcd-1}\leq5.$$
The equality occurs for $(a,b,c,d,e)=(1,1,1,2,5).$
For a proof of the last inequality we can use the following reasoning.
Let $a=1+x$, $b=1+y$, $c=1+z$ and $d=2+t$.
Thus, $x$, $y$, $z$ and $t$ are non-negatives and we need to prove that:
$$5(x+1)(y+1)(z+1)(t+2)\geq5+x+y+z+t+5,$$which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve in $\mathbb{R^{3}}$ : $\begin{cases}x+y+xy=19\\y+z+yz=11\\x+z+xz=14\end{cases}$ Solve in $\mathbb{R^{3}}$ the following system :
$$\begin{cases}x+y+xy=19\\y+z+yz=11\\x+z+xz=14\end{cases}$$
My try but I think not complete :
The system equivalent :
$$\begin{cases}(x+1)(y+1)=20\\(y+1)(z+1)=12\\(x+1)(z+1)=15\end{cases}$$
Then from first equation we have :
$$x+1=\frac{20}{y+1}$$
So third equation give :
$$z+1=3\frac{y+1}{4}$$
Second equation $\implies $
$(y+1)^{2}=16$ then $y=3,-5$
This mean $x=\frac{17}{3}$ or $x=-6$ also $z=3$ or $z=-3$
I'm correct or no ??
| Alternatively, take logs in
$$\begin{cases}(x+1)(y+1)=20\\(y+1)(z+1)=12\\(x+1)(z+1)=15\end{cases}. $$
Let $l_x=\ln(|x+1|)$, $l_y=\ln(|y+1|)$, and $l_z=\ln(|z+1|)$ and happily solve
$$\begin{array}{ccc|c}
1 & 1 & 0 & \ln(20)\\
0 & 1 & 1 & \ln(12)\\
1 & 0 & 1 & \ln(15)
\end{array}
$$
in just two steps for $l_x$, $l_y$, and $l_z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Nature of the critical point $(0,0)$ of the function $f(x,y)=x^6-2x^2y-x^4y+2y^2$ Consider the function $$f(x,y)=x^6-2x^2y-x^4y+2y^2.$$ The point $(0,0)$ is a critical point. Observe,
\begin{align*}
f_x & = 6x^5-4xy-4x^3y, f_x(0,0)=0\\
f_y & = 2x^2-x^4+4y. f_y(0,0)=0\\
f_{xx} & = 30x^4-4y-12x^2y, f_{xx}(0,0)=0\\
f_{xy} & = 4x-4x^3, f_{xy}(0,0)=0\\
f_{yy} & = 4, f_{yy}=4
\end{align*}
So, in order to determine the nature of the above critical point, we need to check the Hessian at $(0,0)$ which is $0$ and hence the test is inconclusive. $$ H(x,y)= \det \begin{pmatrix} f_{xx} & f_{xy}\\ f_{yx} & f_{yy} \end{pmatrix}=\det \begin{pmatrix} 0 & 0 \\ 0 & 4 \end{pmatrix}=0$$So, I tried to see the function on slices like $y=0$ and $y=x$ but nothing worked. So please suggest me how do I find the nature of the critical point in this case?
| With $g(x) = x^2$ we have
$$
f(g(x),y) = g(x)^3-2g(x)y-g(x)^2y +2y^2
$$
or
$$
f(g,y)=g^3-2g y-g^2y+2y^2
$$
and the hessian of $f(g,y)$ is
$$
H(g,y) = \left(
\begin{array}{cc}
6 g-2 y & -2 g-2 \\
-2 g-2 & 4 \\
\end{array}
\right)
$$
and also
$$
H(0,0) = \left(
\begin{array}{cc}
0 & -2 \\
-2 & 4 \\
\end{array}
\right)
$$
with eigenvalues
$$
\left\{2 \left(1+\sqrt{2}\right),2 \left(1-\sqrt{2}\right)\right\}
$$
characterizing a saddle point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find determinant of an almost diagonal matrix except first row and column $$\begin{vmatrix}
0 & 1 & 1 & & 1\\
1 & a_1 & & & \\
1 & & a_2 & & \\
\vdots & & & \ddots & \\
1 & & & & a_n
\end{vmatrix}$$
If I suppose $a_1,\dots ,a_n\ne0$, I know how to do it, multiply the $i$-th column by $-1/a_i$ and add to the first column. Doing that I get the correct solution $ -a_1a_2 ...a_n(1/a_1+...+1/a_n)$.
How to avoid the zero case ?
| If $a_k = 0$, then we row expand on $(k+1)^\text{th}$ row and we get:
$$\left|
\begin{matrix}
0 & 1 & 1 & \cdots & 1 & 1 & 1 &\cdots & 1 \\
1 & a_1 & 0 & \cdots & 0 & 0 & 0&\cdots & 0 \\
1 & 0 & a_2 & \cdots & 0 & 0 & 0&\cdots & 0 \\
\vdots & & & \ddots & & & & & \vdots \\
1 & 0 & 0 & \cdots & a_{k-1} & 0 & 0 &\cdots & 0 \\
1 & 0 & 0 & \cdots & 0 & a_k & 0 &\cdots & 0 \\
1 & 0 & 0 & \cdots & 0 & 0 & a_{k+1} &\cdots & 0 \\
\vdots & & & \ddots & & & & & \vdots \\
1 & 0 & 0 & \cdots & 0 & 0 & 0 &\cdots & a_n \\
\end{matrix}\right|
=
(-1)^{k + 1}\left|
\begin{matrix}
1 & 1 & \cdots & 1 & 1 & 1 &\cdots & 1 \\
a_1 & 0 & \cdots & 0 & 0 & 0&\cdots & 0 \\
0 & a_2 & \cdots & 0 & 0 & 0&\cdots & 0 \\
\vdots & & \ddots & & & & & \vdots \\
0 & 0 & \cdots & a_{k-1} & 0 & 0 &\cdots & 0 \\
0 & 0 & \cdots & 0 & 0 & a_{k+1} &\cdots & 0 \\
\vdots & & \ddots & & & & & \vdots \\
0 & 0 & \cdots & 0 & 0 & 0 &\cdots & a_n \\
\end{matrix}\right|
$$
Row expand the matrix on the right (starting the bottom row), and you should get the determinant being $(-1)^{k+1}(-1)^ca_1 \cdots a_{k-1}a_{k+1} \cdots a_n$ for some integer $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Total number of distinct $x\in[0,1]$ for which $\int_{0}^{x}\frac{t^2}{1+t^4}dt=2x-1$
Find total number of distinct $x\in[0,1]$ for which $$\int_{0}^{x}\frac{t^2}{1+t^4}dt=2x-1$$
My multiple attempts are as follows:-
Attempt $1$:
$$t^2=\tan\theta$$
$$2t\dfrac{dt}{d\theta}=\sec^2\theta$$
$$dt=\dfrac{\sec^2\theta}{2\sqrt{\tan\theta}}d\theta$$
$$\int_{0}^{\tan^{-1}x^2}\dfrac{\tan\theta}{1+\tan^2\theta}\dfrac{\sec^2\theta}{2\sqrt{\tan\theta}}d\theta=2x-1$$
$$\int_{0}^{\tan^{-1}x^2}\sqrt{\tan\theta}d\theta=4x-2$$
$$\int_{0}^{\tan^{-1}x^2}\dfrac{\sqrt{\sin\theta}}{\sqrt{\cos\theta}}=4x-2$$
$$\int_{0}^{\tan^{-1}x^2}\dfrac{\sin\theta}{\sqrt{\sin\theta}\sqrt{\cos\theta}}=4x-2$$
$$\cos\theta=y$$
$$\int_{1}^{\frac{1}{\sqrt{1+x^4}}}-\dfrac{dy}{\sqrt{y}\sqrt{\sqrt{1-y^2}}}=4x-2$$
Now it is unsolvable from here.
Attempt $2$:
It seems like this integral is unsolvable, so let's apply Riemann sum
$$\lim_{h\to0}h(f(0)+f(h)+f(2h)+f(3h)\cdots\cdots f(nh))=2x-1$$
$$nh=x$$
$$\lim_{h\to0}h\sum_{r=0}^{n}f(rh)=2x-1$$
$$\lim_{h\to0}h\sum_{r=0}^{n}\dfrac{r^2h^2}{1+r^4h^4}=2x-1$$
If we put the limit $h\rightarrow 0$, then $0\cdot(0+0+0\cdots\cdots)$ where inside the bracket we have indeterminate form as $n\rightarrow \infty$
Now I was not getting how to proceed further.
| Hint: Notice, that:
$$\int\frac{x^2dx}{x^4+1}=
\frac{1}{2}\int\frac{2x}{x^4+1}xdx$$
Now knowing, that $\frac{d}{dx}x=1$ and $\frac{d}{dx}\arctan{x^2}=\frac{2x}{x^4+1}$, we can solve it by parts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the probability for having real roots with the polynomial $x^2+ax+b$ Choose two real numbers $a$ and $b$ satisfying $1<a<3$ and $-1<b<1$ randomly.
Find the probability that the polynomial $x^2+ax+b$ has two real roots.
NB: uniform distribution
| Let $\left[x^{2}>4y\right]$ denote the function with arguments $x,y$
that takes value $1$ if $x^{2}>4y$ and takes value $0$ otherwise.
Then:
$$\begin{aligned}P\left(a^{2}>4b\right) & =\mathbb{E}\left[a^{2}>4b\right]\\
& =\frac{1}{4}\int_{1}^{3}\int_{-1}^{1}\left[x^{2}>4y\right]dydx\\
& =\frac{1}{4}\int_{1}^{3}\int_{-1}^{\min\left(\frac{1}{4}x^{2},1\right)}dydx\\
& =\frac{1}{4}\int_{1}^{3}\min\left(\frac{1}{4}x^{2},1\right)+1dx\\
& =\frac{1}{4}\int_{1}^{2}\left(\frac{1}{4}x^{2}+1\right)dx+\frac{1}{4}\int_{2}^{3}2dx\\
& =\frac{1}{4}\left[\frac{1}{12}x^{3}+x\right]_{1}^{2}+\frac{1}{2}\\
& =\frac{19}{48}+\frac{24}{48}=\frac{43}{48}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all real solution for k: $2\sin^2(x)+6\cos^2\left(\frac x4\right)=5-2k$ I have been stuck at this problem for some time now. I'd really apprechiate your help. Thanks.
$$2\sin^2(x)+6\cos^2(\frac x4)=5-2k$$
| Too long for a comment but I cannot resist when I see an equation !
As @lhf answered, we are looking for the maximum of
$$f(x)=2 \sin ^2(x)+6 \cos ^2\left(\frac{x}{4}\right)$$ so for the zero's of
$$f'(x)=2 \sin (2 x)-\frac{3}{2} \sin \left(\frac{x}{2}\right)$$
Let $x=2 \sin ^{-1}(t)$ to get as a new equation
$$8 t \left(1-2 t^2\right) \sqrt{1-t^2}-\frac{3 t}{2}=0$$ Discarding the trivial $t=0$ and squaring
$$1024 t^6-2048 t^4+1280 t^2-247=0$$ which is a cubic in $t^2$ and, using $z=t^2$, the solutions are given by
$$z_k=\frac{2}{3}+\frac{1}{3} \cos \left(\frac{2 \pi k}{3}-\frac{1}{3} \cos
^{-1}\left(\frac{13}{256}\right)\right) \qquad \text{with} \qquad k=0,1,2$$ Back to $t$ and $x$, the final solution is
$$x_*=2 \sin ^{-1}\left(\sqrt{\frac{2}{3}-\frac{1}{3} \sin \left(\frac{\pi }{6}+\frac{1}{3} \cos^{-1}\left(\frac{13}{256}\right)\right)}\right)\approx 1.33019$$ and then
$$f(x_*)=4+\sqrt{3\left(1+\cos \left(\frac{\pi }{6}+\frac{1}{3} \sin ^{-1}\left(\frac{13}{256}\right)\right) \right) }+$$ $$\cos \left(2 \sin ^{-1}\left(-\frac 13+\frac 23\sin \left(\frac{\pi }{6}+\frac{1}{3} \cos ^{-1}\left(\frac{13}{256}\right)\right)\right)\right)\approx 7.24701$$
Edit
This is exactly the same result as the elegant one provided by @bjorn93 (which, I must confess, I did not pay attention when I started working the problem).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $f(x)=x^{3} +5x +3$ Prove that $f(x)$ is uniformly continuous on any closed interval. Let $$f(x)=x^{3} +5x +3$$
Prove that $f(x)$ is uniformly continuous on any closed interval.
My Attempt :
Let $[a,b]$ any closed interval such that $a<b$
I need to prove that for every $\epsilon > 0 $ exists $\delta > 0 $ such that if $x,y \in [a,b] , |x-y|<\delta \implies |f(x)-f(y)|<\epsilon$
$|(x^{3} +5x +3)-(y^{3} +5y +3)| = |5(x-y)+(x^3 - y^3)|=|(x-y)(x^2 + xy +y^2 +5)|$
now I need to find when $(x^2 + xy +y^2 +5) \leq$ something with $a or b$
so what i did is :
$4a^2\leq (x+y)^2 \leq 4b^2 \rightarrow 4a^2-xy+5 \leq (x^2 + xy +y^2 +5) \leq 4b^2-xy+5$
my question is can I assume the $4b^2-xy+5 \leq 4b^2+5$ to continue the question ?
| No, you cannot. That inequality may well be false.
But you can do this: take $M\in(0,\infty)$ such that $\lvert a\rvert,\lvert b\rvert\leqslant M$. Then, if $x,y\in[a,b]$,$$\lvert x^2+xy+y^2+y\rvert\leqslant M^2+M^2+M^2+5=3M^2+5.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Given $x \in \mathbb R$ such that $8x^3 - 4x^2 - 4x + 1 = 0$. Find all rationals $p, q, r$ such that $px + q(2x^2 - 1) + r(4x^3 - 3x) = -4$.
Given $x \in \mathbb R$ such that $8x^3 - 4x^2 - 4x + 1 = 0$. Find all rationals $p, q, r$ such that $$\large px + q(2x^2 - 1) + r(4x^3 - 3x) = -4$$
We have that $$8x^3 - 4x^2 - 4x + 1 = 0 \iff (-1) + 2x - 2 \cdot (2x^2 - 1) + 2 \cdot (4x^3 - 3x) = 0$$ and $$px + q(2x^2 - 1) + r(4x^3 - 3x) = -4 \iff (-4) \cdot (-1) + px + q(2x^2 - 1) + r(4x^3 - 3x) = 0$$
This implies that $(p + 8)x + (q - 8)(2x^2 - 1) + (r + 8)(4x^3 - 3x) = 0$.
One solution is that $p + 8 = q - 8 = r + 8 = 0 \implies (p, q, r) = (-8, 8, -8)$
I suspect to there be other solutions of $(p, q, r)$, for example, $$x = \frac{1}{6} \cdot \left[1 - \sqrt[3]{\frac{7}{2}(3\sqrt3i + 1)} + \sqrt[3]{\frac{7}{2}(3\sqrt3i - 1)}\right]$$, according to WolframAlpha, is a solution to the equation $8x^3 - 4x^2 - 4x + 1 = 0$.
Right... So WolframAlpha is untrustworthy, according to Michael Rozenberg, the solutions to $8x^3 - 4x^2 - 4x + 1 = 0$ are $\cos\dfrac{\pi}{7}, \cos\dfrac{3\pi}{7}, \cos\dfrac{5\pi}{7}$, which is correct when plugged into the original equation.
| $8x^3 - 4x^2 - 4x + 1 = 0$ and $4rx^3+2qx^2+(p-3r)x+(4-q)=0$.
Case 1
If the original cubic is irreducible over the rationals then the coefficients of the various powers of $x$ in the two cubics must be in proportion. (Otherwise, one or more roots would satisfy a rational polynomial of degree less than $3$ obtained as a linear combination of the two cubics.) Then
$$4r=8t, 2q=-4t,p-3=-4t,4-q=t.$$
This gives the solution $p=-8,q=8,r=-8.$
Case 2
Otherwise the cubic has a rational root $\frac{a}{b}$ with $(a,b)=1$ and so
$$8a^3-4a^2b-4ab^2+b^3=0$$ Then $a$ divides $b^3$ and so $a=1$. Then $b$ would have to be $\pm2$ but these do not satisfy the equation and so there are no further solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3503008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determining by hand whether a big matrix is positive definite Consider the following matrix:
$$A = \begin{pmatrix}
3 & -1 & 0 & 0 & 0 & -1 \\
-1 & 3 & -1 & 0 & -1 & 0 \\
0 & -1 & 3 & -1 & 0 & 0 \\
0 & 0 & -1 & 3 & -1 & 0 \\
0 & -1 & 0 & -1 & 3 & -1 \\
-1 & 0 & 0 & 0 & -1 & 3
\end{pmatrix}$$
It was part of an exam question, where one needed to prove that it is symmetric positive definite in order to prove that both the Gauss-Seidel and Jacob method converge. I was not able to solve this by hand.
How can one approach this in the most efficient way in a time-sensitive setting? I personally tried to just calculate the eigenvalues, but failed because of the $6 \times 6$ determinant. Another approach would be to compute the determinants of the principal minors, but this also would be a lot of work for this small of a part of an exam.
Thanks!
| To show that $A$ is positive semidefinite, it suffices to note that $A$ is symmetric and diagonally dominant with non-negative diagonal entries. From there, we could row-reduce to verify that $A$ is invertible.
Another approach: we can write $A$ as a sum
$$
A = \begin{pmatrix}
2 & -1 & 0 & 0 & 0 & -1 \\
-1 & 2 & -1 & 0 & 0 & 0 \\
0 & -1 & 2 & -1 & 0 & 0 \\
0 & 0 & -1 & 2 & -1 & 0 \\
0 & 0 & 0 & -1 & 2 & -1 \\
-1 & 0 & 0 & 0 & -1 & 2
\end{pmatrix} +
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & -1 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 \\
0 & -1 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 1
\end{pmatrix}.
$$
It is easy to compute the eigenvalues of both matrices to verify that they are positive semidefinite (the first matrix in particular is circulant). It follows that $A$ is the sum of positive semidefinite matrices and is therefore positive semidefinite.
In fact, we could also use this decomposition to see that $A$ is positive definite. For positive semidefinite matrices $B,C$, we have
$$
\ker(B + C) \subset \ker(B) \cap \ker(C).
$$
In this case, the two matrices being added have distinct $1$-dimensional kernels so that $\ker(B) \cap \ker(C) = \{0\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral with partial fractions I'm trying to calculate the following integral:
$$I=\int \frac{\arctan(x)}{x^4}dx$$
My steps so far are:
Per partes:
$$\frac{-\arctan(x)}{3x^3}+\int{\frac{1}{1+x^2} \frac{1}{3x^3}}dx
=\frac{-\arctan(x)}{3x^3}+\frac{1}{3}\int{\frac{1}{1+x^2} \frac{1}{x^3}}dx$$
and now I want to do partial fractions. However, with this integral, I fail to do partial fractions. Could you help me?
Thanks
| A rational function $P(x)/Q(x)$ can be rewritten using Partial Fraction Decomposition:
$$
\frac{P(x)}{Q(x)} = \frac{A_1}{a\,x + b} + \dots + \frac{A_2\,x + B_2}{a\,x^2 + b\,x + c} + \dots
$$
where for each factor of $Q(x)$ of the form $(a\,x + b)^m$ introduce terms:
$$
\frac{A_1}{a\,x + b} + \frac{A_2}{(a\,x + b)^2} + \dots + \frac{A_m}{(a\,x + b)^m}
$$
and for each factor of $Q(x)$ of the form $\left(a\,x^2 + b\,x + c\right)^m$ introduce terms:
$$
\frac{A_1\,x + B_1}{a\,x^2 + b\,x + c} + \frac{A_2\,x + B_2}{\left(a\,x^2 + b\,x + c\right)^2} + \dots + \frac{A_m\,x + B_m}{\left(a\,x^2 + b\,x + c\right)^m}\,.
$$
In light of all this, you have:
$$
\frac{1}{x^3\left(x^2+1\right)} = \frac{A_1}{x} + \frac{A_2}{x^2} + \frac{A_3}{x^3} + \frac{A_4\,x + B_4}{x^2 + 1}
$$
i.e.
$$
\frac{1}{x^3\left(x^2+1\right)} = \frac{\left(A_1 + A_4\right)x^4 + \left(A_2 + B_4\right)x^3 + \left(A_1 + A_3\right)x^2 + A_2\,x + A_3}{x^3\left(x^2+1\right)}
$$
which turns out to be an identity if and only if:
$$
\begin{cases}
A_1 + A_4 = 0 \\
A_2 + B_4 = 0 \\
A_1 + A_3 = 0 \\
A_2 = 0 \\
A_3 = 1
\end{cases}
\; \; \; \; \; \;
\Leftrightarrow
\; \; \; \; \; \;
\begin{cases}
A_1 = -1 \\
A_2 = 0 \\
A_3 = 1 \\
A_4 = 1 \\
B_4 = 0
\end{cases}
$$
from which what you want:
$$
\frac{1}{x^3\left(x^2+1\right)} = -\frac{1}{x} + \frac{1}{x^3} + \frac{x}{x^2+1}\,.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Simplify trigonometric expression $\frac{1 - \cos A + \cos B - \cos(A + B)}{1 + \cos A - \cos B - \cos(A + B)}$
Simplify
$$\frac{1 - \cos A + \cos B - \cos(A + B)}{1 + \cos A - \cos B - \cos(A + B)}$$
How can I simplify this trigonometric expression?
Can you explain it if possible?
| Apply the identities
$$\cos A -\cos B =-2\sin\frac{A-B}2\sin\frac{A+B}2,\>\>\>\>\>\cos(A + B) = 1-2\sin^2\frac{A + B}2$$
to simplify
$$\frac{1 - \cos A + \cos B - \cos(A + B)}{1 + \cos A - \cos B - \cos(A + B)}$$
$$=\frac{1 + 2\sin\frac{A-B}2\sin\frac{A+B}2- (1-2\sin^2\frac{A + B}2)}{1 -2\sin\frac{A-B}2\sin\frac{A+B}2 - (1-2\sin^2\frac{A + B}2)}$$
$$=\frac{2\sin\frac{A+B}2(\sin\frac{A-B}2+\sin\frac{A+B}2)}{2\sin\frac{A+B}2(-\sin\frac{A-B}2+\sin\frac{A+B}2)}=\frac{2\sin \frac A2\cos\frac B2}{2\cos\frac A2\sin\frac B2}=\frac{\tan \frac A2}{\tan \frac B2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral of $\frac{1}{(ax^2+bx+c)}$ from $-\infty$ to $\infty$ Can someone spot my error? I am trying to find a general solution for
$$\int_{-\infty}^\infty \frac {1}{ax^2 + bx + c} \, dx $$ using complex analysis.
I substitute a complex variable, $z$, for $x$ then find the roots and take only the root above the real axis
$$z = \frac{-b}{2a} + \frac{\sqrt{4ac - b^2}}{2a}i$$
This tells me that:
$$\int_{-\infty}^\infty \frac 1 {ax^2 + bx + c} \, dx = \int \frac {dz} {\left( z + \frac{b+\sqrt{4ac-b^2}i}{2a} \right) \left(z+\frac{b-\sqrt{4ac-b^2}i}{2a} \right)}$$
Finding the residue at $\ z = \displaystyle\frac {-b}{2a} + \frac{\sqrt {4ac - b^2}}{2a}i$
$$R\left(\frac {-b}{2a} + \frac{\sqrt{4ac - b^2}}{2a}i\right)
= \frac{1}{\left(\frac{-b+i\sqrt{4ac-b^2}}{2a}+\frac{-b+i\sqrt{4ac-b^2}}{2a}\right)} = \frac{1}{\frac{2i\sqrt{4ac-b^2}}{2a}} = \frac{a}{i\sqrt{4ac-b^2}}$$
$$\int_{-\infty}^\infty \frac {1}{(ax^2 + bx + c)}dx= {2\pi i} R = \frac{2a\pi}{\sqrt {4ac-b^2}}$$
but checking some example on wolfram alpha, I can see that the actual value is $\frac{2\pi}{\sqrt {4ac-b^2}}$. My question is where does that $a$ go? What mistake did I make?
Thanks!
| You had
$$
\left( z + \frac{b+\sqrt{4ac-b^2}i}{2a} \right) \left(z+\frac{b-\sqrt{4ac-b^2}i}{2a} \right)
$$
where you needed
$$
a \left( z + \frac{b+\sqrt{4ac-b^2}i}{2a} \right) \left(z+\frac{b-\sqrt{4ac-b^2}i}{2a} \right).
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the Long Term behavior of a stochastic matrix. I am struggling to find a formula for $P_{11}^n$ of a $3$ x $3$ stochastic matrix. I am comfortable with handling the 2x2 case as it usually just involves converting the system into a recurrence relation between $P_{11}^{n+1}$ and $P_{11}^n$. However I am unsure of what to do in the 3x3 case. The specific matrix I am having a problem with is
\begin{bmatrix}
0 & 2/3 & 1/3 \\
1/3 & 0 & 2/3 \\
2/3 & 1/3 & 0
\end{bmatrix}
| Here $P$ is a doubly stochastic matrix - both the rows and the columns sum to one. This implies that $P$ has the unique stationary distribution $\pi$ which is the uniform distribution. We can verify this by computing
$$
\begin{bmatrix}\frac13&\frac13&\frac13\end{bmatrix}\begin{bmatrix}
0 & 2/3 & 1/3 \\
1/3 & 0 & 2/3 \\
2/3 & 1/3 & 0
\end{bmatrix}
= \begin{bmatrix}\frac13&\frac13&\frac13\end{bmatrix}.
$$
However, the $n$-step probabilities do not have a nice form:
$$
P^n = \left(
\begin{array}{ccc}
\frac{1}{3} \left(2\ 3^{-\frac{n}{2}} \cos \left(\frac{5 n \pi }{6}\right)+1\right) & \frac{1}{6} \left(2\ 3^{-\frac{n}{2}} \left(\sqrt{3} \sin \left(\frac{5 n \pi }{6}\right)-\cos \left(\frac{5 n \pi }{6}\right)\right)+2\right) & \frac{1}{6} \left(2-2\ 3^{-\frac{n}{2}} \left(\cos \left(\frac{5 n \pi }{6}\right)+\sin \left(\frac{5 n \pi }{6}\right) \sqrt{3}\right)\right) \\
\frac{1}{6} \left(2-2\ 3^{-\frac{n}{2}} \left(\cos \left(\frac{5 n \pi }{6}\right)+\sin \left(\frac{5 n \pi }{6}\right) \sqrt{3}\right)\right) & \frac{1}{3} \left(2\ 3^{-\frac{n}{2}} \cos \left(\frac{5 n \pi }{6}\right)+1\right) & \frac{1}{6} \left(2\ 3^{-\frac{n}{2}} \left(\sqrt{3} \sin \left(\frac{5 n \pi }{6}\right)-\cos \left(\frac{5 n \pi }{6}\right)\right)+2\right) \\
\frac{1}{6} \left(2\ 3^{-\frac{n}{2}} \left(\sqrt{3} \sin \left(\frac{5 n \pi }{6}\right)-\cos \left(\frac{5 n \pi }{6}\right)\right)+2\right) & \frac{1}{6} \left(2-2\ 3^{-\frac{n}{2}} \left(\cos \left(\frac{5 n \pi }{6}\right)+\sin \left(\frac{5 n \pi }{6}\right) \sqrt{3}\right)\right) & \frac{1}{3} \left(2\ 3^{-\frac{n}{2}} \cos \left(\frac{5 n \pi }{6}\right)+1\right) \\
\end{array}
\right).
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Dividing a rope into three random pieces. Expected longest length A rope of 1m is divided into three pieces by two random points. Find the average length of the largest segment.
This question has been answered here:
Average length of the longest segment
Someone had a solution:
let the cuts be at $X, Y$, with $Y \gt X$:
Image of cut positions
Then each piece is equally likely to be the longest, and the expected length of the longest piece doesn't depend on which piece we choose. Then we can calculate $\mathop{\mathbb{E}}(X|X \text{ is the longest piece} )$.
We have the three inequalities:
$$X \gt Y-X \implies Y < 2X$$
$$X \gt 1-Y \implies Y > 1-X$$
and, from our setup,
$$Y \gt X$$
These can be represented by the following diagram:
Diagram of inequalities
Then the area satisfying our inequalities is the two triangles A and B. So we wish to find the expected value of $X$ within this area.
The expected value of $X$ in A is $\bar{X}_A = \frac{1}{2}-\frac{1}{3}(\frac{1}{2}-\frac{1}{3}) = \frac{8}{18}$.
The expected value of $X$ in B is $\bar{X}_B = \frac{1}{2}+\frac{1}{3}(\frac{1}{2}) = \frac{4}{6} = \frac{12}{18}$
The area of A is $A_A = \frac{1}{2} \times \frac{1}{2}\times (\frac{1}{2}-\frac{1}{3}) = \frac{1}{24}$.
The area of B is $A_B = \frac{1}{2} \times \frac{1}{2}\times \frac{1}{2} = \frac{1}{8} = \frac{3}{24} = 3 A_A$.
So $\mathop{\mathbb{E}}(X|X \text{ is the longest piece} ) = \frac{\tfrac{8}{18} + 3\left(\tfrac{12}{18}\right)}{4} = \frac{11}{18}$
I understand everything but I got lost when he said calculate expected value of $X$ within triangles. How do you exactly find the expected value of $X$ over the triangle $A$ and $B$? How did he come up with $\mathop{\mathbb{E}}(X|X \text{ is the longest piece} )$?
| $X$ is a linear function on each triangle. The average value of a linear function on a triangle is the average of its values at the three vertices.
| {
"language": "en",
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Proving a five digit number is divisible by 3 if the sum of its digits is divisible by 3 I'm trying to construct a direct proof to show that a five-digit number is divisible by 3 if the sum of its five digits is divisible by 3. What I was thinking of doing was expanding the five-digit number, but then I get stuck in figuring out how to bring in congruence modulo n and the definition of divides into my proof. Any help is appreciated in making a clear and concise proof. Here's what I have so far:
Suppose that you have a five-digit number $n$ that is written $abcde$. Then,
\begin{align*}
n&=10^4a+10^3b+10^2c+10d+e\\
&=(9999+1)a+(999+1)b+(99+1)c+(9+1)d+e\\
&=(9999a+999b+99c+9d)+(a+b+c+d+e)\\
&=3(3333a+333b+33c+3d)+(a+b+c+d+e)\\
\end{align*}
| Using congruence:
Since $3 \equiv 0 \pmod 3$, from your working, we have $n \equiv a+b+c+d+e
\pmod{3}$
Using direct division.
$$n-(a+b+c+d+e) = 3(3333a+333b+33c+3d)$$
Hence $n-(a+b+c+d+e)=3\alpha$ is a multiple of $3$. If $n$ is a multiple of $3$, then we can write $n = 3\beta$, and we have $a+b+c+d+e=3(\beta-\alpha)$. Similarly if $a+b+c+d+e$ is a multiple of $3$.
In general, if $n = \sum_{i=0}^d n_i \cdot 10^{i}$, then since $10 \equiv 1 \pmod{3}$, we have $n \equiv \sum_{i=0}^d n_i$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determining whether the polynomials span $P_2$
Determine whether the following polynomials span $\mathbb P_2$
$p_1 = 1+x+2x^2$
$p_2 = 3+x$
$p_3 = 5-x+4x^2$
$p_4 = -2- 2x +2x^2$
Attempt:
I created the following augmented matrix:
$$
\begin{bmatrix}
1 & 3 & 5 & -2 & a \\
x & x & -x & -2x&bx\\
2x^2 & 0 & 4x^2 & 2x^2 & cx^2\\
\end{bmatrix}
$$
which reduces to:
$$
\begin{bmatrix}
1 & 3 & 5 & -2 & a \\
0 & 1 & 3 & 0 &(a-b)/2\\
0 & 0 & 1 & 1/2 & (c+a-3b)/2\\
\end{bmatrix}
$$
Now, I believe that the system is consistent and has infinitely many solutions so the polynomials span $\mathbb{P_2}$.
Is this approach correct?
| If $P_2$ is the vector space of real polynomials of order smaller or equal to 2, then you can think of the given polynomials as vectors in $\mathbb {R^3}$ as;
$p_1=(1,1,2)$, $p_2=(3,1,0)$, $p_3=(5,-1,4)$, $p_4=(-2,-2,2)$ (Why?). Then check the rank of the following matrix: \begin{pmatrix}
1 & 1 & 2\\
3 & 1 & 0\\
5 & -1 & 4\\
-2 & -2 & 2
\end{pmatrix}
If the rank is equal to 3, then the answer is affirmative.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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How do you rewrite a determinant of a matrix into a polynomial by induction?
$$\det\begin{bmatrix} {x} & {0} & {\cdots} & {\cdots} & {0} & {a_{1}} \\ {-1} & {x} & {0} & {\cdots} & {0} & {a_{2}} \\ {\ddots} & {\ddots} & {\ddots} & {\ddots} & {\vdots} & {\vdots} \\ {\cdots} & {0} & {-1} & {x} & {0} & {a_{n-3}} \\{\cdots} & {\cdots} & {0} & {-1} & {x} & {a_{n-2}} \\{\cdots} & {\cdots} & {\cdots} & {0} & {-1} & {a_{n-1}+x} \end{bmatrix}=a_1+a_2x+\cdots+a_{n-1}x^{n-2}+x^{n-1}$$
How can I replace the determinant on the left by induction to get $a_1+a_2x+\cdots+a_{n-1}x^{n-2}+x^{n-1}$? I need this to determine the determinant of the companion matrix and I don't understand how this step is done. Thanks in advance!
| We wish to prove the following by induction:
$$
\det
\begin{pmatrix}
t & 0 & \cdots & 0 & a_0 \\
-1 & t & \cdots & 0 & a_1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -1 & t+a_{n-1} \end{pmatrix}
=a_0+a_1t+\cdots+a_{n-1}t^{n-1}+t^n
$$
When $n=1$ we have that
$$
\det
\begin{pmatrix}
a_0+t
\end{pmatrix}
=a_o+t
$$
Now supposing that we've proved the claim holds for $n$, we will try to prove the claim holds for $n+1$. Starting with
$$
\det
\begin{pmatrix}
t & 0 & \cdots & 0 & a_0 \\
-1 & t & \cdots & 0 & a_1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -1 & t+a_n \end{pmatrix}
$$
we do co-factor expansion on the first row we get
$$
t\cdot\det
\begin{pmatrix}
t & 0 & \cdots & 0 & a_1 \\
-1 & t & \cdots & 0 & a_2 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -1 & t+a_n
\end{pmatrix}
+(-1)^n\cdot a_0\cdot\det
\begin{pmatrix}
-1 & t & \cdots & 0 \\
0 & -1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & -1
\end{pmatrix}
$$
The determinant on the right is $(-1)^n$, and from our induction hypothesis, the determinant of the left is $a_1+a_2t+\cdots+a_nt^{n-1}+t^n$. Hence we have that
$$
\begin{align*}
\det
\begin{pmatrix}
t & 0 & \cdots & 0 & a_0 \\
-1 & t & \cdots & 0 & a_1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -1 & t+a_n \end{pmatrix}
&= t\cdot(a_1+a_2t+\cdots+a_nt^{n-1}+t^n)+(-1)^n\cdot a_0\cdot(-1)^n\\
&= a_1t+\cdots+a_nt^n+t^{n+1}+a_o \\
&= a_0+a_1t+\cdots+a_nt^n+t^{n+1}
\end{align*}
$$
This shows the claim holds for $n+1$ assuming it held for $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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When $\frac {a^3-b^3}{a^2-b^2}$ is an integer? If $\frac {a^3-b^3}{a^2-b^2}$ is an integer, then supposing $a-b \ne 0$ we have that also$\frac {a^2+ab+b^2}{a+b}$ is an integer.
For which $a, b\in\mathbb Z$, the fraction $\frac {a^2+ab+b^2}{a+b}$ is an integer?
| Assuming $a\neq b$, we want that $\frac{a^2+ab+b^2}{a+b}=(a+b)-\frac{ab}{a+b}$ is an integer. The solutions of
$$ \frac{ab}{a+b} = k, $$
assuming $a+b\neq 0$, are the solutions of
$$ ab-ka-kb = 0, $$
i.e. the solutions of
$$ (a-k)(b-k) = k^2, $$
which depend on the couples of divisor/complementary divisor of $k^2$. In general, for any $d\mid k^2$ we have the solution $$a=d+k,\qquad b=\frac{k^2}{d}+k.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3523454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does the following function admit a maximum? I should determine if a maximum value for the function exist and, if it is so, the maximum points.
$$f(x, y, z) = (1 + x^
2
)e^{-z}$$
over the region:
$$D = \{(x, y, z) ∈ R^
3
|x = y^
2 + z^
2
, x^2 − x + y^
2 + z^
2 ≤ 2\}$$
The set D is clearly closed but I am not sure how to prove that it is compact here. Thank you..
| Your region $D$ consists of the intersection of the paraboloid $x=y^2+z^2$ and the interior of the sphere $(x-\tfrac12)^2+y^2+z^2=\tfrac94$ (after completing the square). So $D$ is closed and bounded, and thus compact by Heine-Borel.
As for the maxima/minima, since the roles of $x$ and $z$ are independent (and there is no $y$ in the formula, you need to maximize/minimize on $x$ and $z$ independently.
For max: you want $z=0$ and maximize $1+x^2$. In the paraboloid we have $x=y^2+z^2$, so the formula for the sphere because $x^2-x+x\leq 2$, so $x^2\leq 2$. Thus $1+x^2$ maximizes as $1+2=3$ when $x=\pm\sqrt2$. So the maximum is $3$, achieved at points, $(\pm\sqrt2,\pm2^{1/4},0)$ (the values for $y$ coming from $\pm\sqrt2=y^2+0^2=y^2$.
For the minimum you want $x=0$ to minimize $1+x^2$, and $z$ as big as possible; but from $0=y^2+z^2$ you get $y=z=0$, so the minimum is $1$ at $(0,0,0)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that the equation:$x^{2}+x=3+\ln(x+2) $ has only one real solution at $[0,\infty ) $ My attempt:
$x^{2}+x-(3+\ln(x+2))=0$
$x_{1,2}=\dfrac{-1\pm \sqrt{1+12+4\ln(x+2)}}{2} \Rightarrow 13+4\ln(x+2)>0 $
$ \Rightarrow 4\ln(x+2)>-13 $
$\ln(x+2)>-\frac{13}{4} $
$x+2>e^{-\frac{13}{4}} $
$x>e^{-\frac{13}{4}}+2 $
What did I do wrong?
Edit: Thank you all for the help:)
| A simple proof is that $x^2+x$ and $3+\ln (x+2)$ are convex and concave functions respectively on $[0,\infty)$ and both strictly increasing. Also$$x^2+x|_{x=10}<3+\ln(x+2)|_{x=0}$$and $$x^2+x|_{x=0}>3+\ln(x+2)|_{x=10}$$hence aa unique root exists within $(0,10)$ and the proof is complete.
| {
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Let $G=\{A \in GL(3, \mathbb R) \mid A\vec x = \vec x\}$. Suppose that $\vec x = [1\ 0\ 0]^T$. Describe all the orthogonal matrices in $G$ Let $\vec x$ be a non-zero vector in $\mathbb R^3$ and let $G=\{A \in GL(3, \mathbb R) \mid A\vec x = \vec x\}$. Suppose that $\vec x = [1\ 0\ 0]^T$. Describe all the matrices in $G$, as defined above, that are also orthogonal matrices.
This is my attempt:
$$
\begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i \\
\end{pmatrix}\begin{pmatrix}
1\\
0\\
0\\
\end{pmatrix}=\begin{pmatrix}
a\\
d\\
g\\
\end{pmatrix}=\begin{pmatrix}
1\\
0\\
0\\
\end{pmatrix}
$$
Hence $a=1, d=0, g=0$
Then we need to find values or restrictions of $b, c, e, f, h, i$ such that $$
\begin{pmatrix}
1 & b & c \\
0 & e & f \\
0 & h & i \\
\end{pmatrix}\begin{pmatrix}
1 & 0 & 0 \\
b & e & h \\
c & f & i \\
\end{pmatrix}=\begin{pmatrix}
1 & 0 & 0 \\
b & e & h \\
c & f & i \\
\end{pmatrix}\begin{pmatrix}
1 & b & c \\
0 & e & f \\
0 & h & i \\
\end{pmatrix}=I_3
$$
Here
$$
\begin{pmatrix}
1 & b & c \\
0 & e & f \\
0 & h & i \\
\end{pmatrix}\begin{pmatrix}
1 & 0 & 0 \\
b & e & h \\
c & f & i \\
\end{pmatrix}= \begin{pmatrix}
b^2+c^2+1 & cf+be & ic+bh \\
cf+be & f^2+e^2 & if+he \\
ic+bh & if+he & h^2-1 \\
\end{pmatrix}
$$
Where we can determine $h^2-1=1, h^2=2$
Then$$
\begin{pmatrix}
1 & 0 & 0 \\
b & e & h \\
c & f & i \\
\end{pmatrix}\begin{pmatrix}
1 & b & c \\
0 & e & f \\
0 & h & i \\
\end{pmatrix}= \begin{pmatrix}
1 & b & c \\
b & b^2+h^2+e^2 & bc+ih+fe \\
c & bc+ih+fe & c^2+f^2-1 \\
\end{pmatrix}
$$
Where we find that $b=c=0, b^2+h^2+e^2=1$, but then $e^2$ has to be $-1$, which contradicts to such matrix should be $\in GL(3, \mathbb R)$
Finally conclude that such matrix does not exist. Am I correct?
| Since $A^TA=I$ we have $A^Tx = x$ so the subspace spanned by $x$ is $A$ and $A^T$ invariant, so $A$ has the form
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & a & b \\ 0 & c & d \end{bmatrix}$.
Since $A$ is orthogonal, we see that the submatrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is must be orthogonal and since it is $\mathbb{R}^2 \to \mathbb{R}^2$ we see that it must be a rotation (proper or improper).
| {
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Why can't I do $u$ substitution on $\int \cos^4x dx$? I tried to rewrite the integral as $$\int (\cos^2x)(\cos^2x)dx$$ which in turn I used the trig identity of $\cos^2x = 1 - \sin^2x$ and therefore changed $$\int (\cos^2x)(\cos^2x)dx = \int (1 - \sin^2x)(1 - \sin^2x)dx = \int ( \sin^4x - \sin^2x + 1) dx$$
I then tried to do u substitution where $u=\sin x$ and $du=\cos x dx$
But my answer was completely wrong...I ended up getting $$\frac{\sin^5x}{5}+\sin x-\frac{2\cos^3x}{3}+K, \quad K\in \mathbb R$$
Am I doing $u$ substitution wrong or is there a better way to solve this integral?
| Your substitution is not correct. Here are some steps for solving given integral:
$$\cos^4x = (\cos^2x)^2, \quad \cos^2x = \frac{1+\cos2x}{2}.$$
Thus, we can write:
$$\cos^4x = \frac{1 + 2\cos2x + \cos^22x}{4}=\frac{1}{4}+\frac{1}{2}\cos2x+\frac{1}{4}\cos^22x.$$
Now, we will apply once again a formula for square of cosines:
$$\cos^4x = \frac{1}{4}+\frac{1}{2}\cos2x+\frac{1}{4}(\frac{1+\cos4x}{2})=\frac{1}{4}+\frac{1}{2}\cos2x+\frac{1}{8}+\frac{1}{8}\cos4x.$$
The solution for given integral is now easy to obrain...
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the series representation of $\ln\left(\frac{1+x}{1-x}\right)$
Given that $\frac{1}{1-x}=\sum^{\infty}_{n=0}x^n$, what is the series representation of $\ln\left(\frac{1+x}{1-x}\right)$?
Differentiating $\ln\left(\frac{1+x}{1-x}\right)$ results in: $\frac2{(1-x)^2}$. This means that $\ln\left(\frac{1+x}{1-x}\right) = 2\int\frac1{(1-x)^2}\ dx$
Can I now say that $\ln\left(\frac{1+x}{1-x}\right)=2\left(\sum^{\infty}_{n=0}\int x^n\right)^2$ ? This would get: $2\sum^{\infty}_{n=0}\frac{x^{2(n+1)}}{(n+1)^2}$.
I'm assuming the answer is wrong because the answer key did not agree with me. Did I mess up somewhere in this problem?
The answer key only doubles the $x$ in this step: $\ln\left(\frac{1+x}{1-x}\right) = 2\int\frac1{(1-x)^2}\ dx$ so instead of $2\int\frac1{(1-x)^2}\ dx$ they get just $2\int\frac1{1-x^2}\ dx$. Why is this correct? (or is it incorrect?)
| Integrate the given equation
\begin{eqnarray*}
\ln\left( \frac{1}{1-x}\right) =\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}. \\
\end{eqnarray*}
Now (with the given equation, again) sub $ x \rightarrow -x$ and integrate
\begin{eqnarray*}
\ln( 1+x) =\sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1}. \\
\end{eqnarray*}
Add these and note that the even powers cancel, thus we have
\begin{eqnarray*}
\ln\left( \frac{1+x}{1-x}\right) = 2 \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}. \\
\end{eqnarray*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluate limit of the form $0^{\infty}$ $$\lim_{n\to\infty}\left(\frac{n^4-3\cdot n^3-n^2+2\cdot n-1}{n^5+n^4-n^3-3\cdot n^2-3\cdot n+1}\right)^{\frac{6\cdot n^5-2\cdot n^4-2\cdot n^3+n^2-2\cdot n}{9\cdot n^4-2\cdot n^3+n^2+3\cdot n}}$$
It is $$\lim_{n\to \infty}\left(\frac{1}{n}+o\left(\frac{1}{n^2}\right)\right)^{(n+o(1))}$$
How can I bring it to a form that I can compute?
| This is one of the many cases where composition of Taylor series is useful.
We have $y_n=A_n^{B_n}$ where
$$A_n=\frac{n^4-3 n^3-n^2+2 n-1}{n^5+n^4-n^3-3 n^2-3 n+1}\quad \text{and} \quad B_n=\frac{6 n^4-2 n^3-2 n^2+n-2}{9 n^3-2 n^2+n+3}$$
We shall work with
$$\log(y_n)=B_n \log(A_n)$$
Using the long division, we have that
$$A_n=\frac{1}{n}-\frac{4}{n^2}+\frac{4}{n^3}+O\left(\frac{1}{n^4}\right)$$
$$\log(A_n)=- \log(n)-\frac{4}{n}-\frac{4}{n^2}+O\left(\frac{1}{n^3}\right)$$
$$B_n=\frac{2 n}{3}-\frac{2}{27}-\frac{76}{243 n}-\frac{377}{2187
n^2}+O\left(\frac{1}{n^3}\right)$$
$$\log(y_n)=-\frac{2}{3} n \log (n)+\frac{2 }{27}\log (n)-\frac{8}{3}+\frac{4 (19 \log (n)-144)}{243 n}+O\left(\frac{1}{n^2}\right)$$ Now,using the expansion to $O\left(\frac{1}{n}\right)$
$$y_n=e^{\log(y_n)}={e^{-8/3}} n^{2/27}n^{-2 n/3}=n^{\frac{2(1-9 n)}{27} }e^{-8/3}$$ as @Gary wrote in comments.
For $n=100$, the relative error is $0.97$% and for $n=1000$, it is $0.02$%.
Edit
We can make it more general for the case where
$$A_n=\frac{a}{n}+\frac{b}{n^2}+O\left(\frac{1}{n^3}\right)\quad \text{and} \quad B_n=c n+d+O\left(\frac{1}{n}\right)$$ and get
$$y_n=e^{\frac{b c}{a}} \left(\frac{a}{n}\right)^{c n+d}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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some special points of $A\sin(\omega x+\varphi)+B$ consider
$$f(x)=A\sin(\omega x+\phi)+B$$
$A>0,\omega>0,|\phi|<\frac{\pi}2$
if we have $f(1)=2,f(2)=\frac12,f(3)=-1,f(4)=2$
how to directly find one solution?
My way is to notice $(2,1/2)$ is the mid point of $(1,2)$ and $(3,-1)$ , guess $B=\frac 12$, then get $\omega=\frac{2\pi}3$, then get $\phi=-\frac{\pi}3$,finally
$$
f(x)=\frac{1}{2}+\sqrt{3} \sin \left(\frac{2 \pi }{3}x-\frac{\pi }{3}\right)
$$
But if without noticing the mid point by accident, how to discovery it?
| Note $f(1)=f(4)$, or
$$\sin(4\omega +\varphi)-\sin(\omega +\varphi)=2\cos\frac{5\omega+2\phi}2\sin\frac{3\omega}2=0$$
which yields $\sin\frac{3\omega}2=0$, or
$$\omega = \frac{2\pi}3$$
Then,
$$f(1) = 2 =A\sin \left(\frac{2\pi}3 + \phi\right) + B\tag 1$$
$$f(2) = \frac12 =A\sin \left(\frac{4\pi}3 + \phi\right) + B\tag 2$$
Take (2) - (1) to get $-\frac32 = A \sin\phi$ and plug it into $f(3) = -1 =A\sin\phi + B$ to get
$$B=\frac12$$
Then, take (2) + (1) to get $ A\cos\phi = \frac{\sqrt3}2$. Along with $-\frac32 = A \sin\phi$ to have $\tan\phi = -\sqrt3$, which yields
$$\phi = -\frac\pi 3,\>\>\>\>\>A=\sqrt3$$
| {
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Find the point, if any, the graph of $f(x) = \sqrt{8x^2+x-3}$ has a horizontal tangent line Section 2.5 #14
Find the point, if any, the graph of $f(x) = \sqrt{8x^2+x-3}$ has a horizontal tangent line.
Okay, so having a horizontal tangent line at a point on the graph means that the slope of that tangent line is zero. The derivative of a function is another function that tells us the slope of the tangent line at any given point on the graph of the original function.
Thus, to find where the graph of $f(x) = \sqrt{8x^2+x-3}$ has a horizontal tangent line, we need to take the derivative, set it equal to zero, and solve for $x$. This will give us the $x$-coordinate of where the graph of $f(x)$ has a horizontal tangent line. To find the corresponding $y$ value, we plug the $x$ value that we found into the original equation.
In this problem, when we plug the $x$ value we find into the original equation, we get an imaginary number, which means that no point on the graph of $f(x)$ has a horizontal tangent line, and thus our answer is DNE, does not exist. Let's go through the motions!!!
$f(x) = \sqrt{8x^2+x-3}=(8x^2+x-3)^{1/2}$
$f'(x) = \frac{d}{dx}(8x^2+x-3)^{1/2}$
Time do the chain rule!!!
$$\begin{align}
f'(x) &= \frac{(8x^2+x-3)^{-1/2}}{2}\frac{d}{dx}(8x^2+x-3)\\
&= \frac{(8x^2+x-3)^{-1/2}}{2}(16x+1)\\
&= \frac{(16x+1)}{2(8x^2+x-3)^{1/2}}
\end{align}$$
Alright, we have our derivative. We want to find horizontal tangent lines, so we set this equal to zero and solve for $x$
$$0 = \frac{(16x+1)}{2(8x^2+x-3)^{1/2}}$$
multiplying both sides of the equation by $2(8x^2+x-3)^{1/2}$ we get
$0 = (16x+1)$
And thus $x = \frac{-1}{16}$
Now, we plug this value into the original equation to get the corresponding $y$ value, because remember, we are looking for a point on the graph where the horizontal line is tangent, so our answer will be in $(x,y)$ format, is it exists, (which in this case, it won't)..
$f(\frac{-1}{16}) = \sqrt{8(\frac{-1}{16})^2+\frac{-1}{16}-3}$
But $8(\frac{-1}{16})^2+\frac{-1}{16}-3<0$, so taking its square root will give us an imaginary number. Thus the answer is DNE
| Shortcut:
$x\mapsto \sqrt x$ has non-zero derivative everywhere, hence $f(x)=\sqrt{8x^2+x-3}$ having a horizontal tangent is the same as $g(x)=8x^2+x-3$ having a horizontal tangent for the same $x$, and $g(x) \geqslant 0$ for that $x$ so that we can take the square root.
$g'(x)=16x+1$ which is zero for $x_0=-1/16$.
$g(x_0) = \tfrac8{16^2}-\tfrac1{16}-3 = \tfrac1{32}-\tfrac1{16}-3 = -\tfrac1{32}-3 < 0$ hence no horizontal tangent.
The first is intuitive, but you also see it from
$$\big(u(v(x))\big)' = v'(x)\cdot u'(v(x))$$
If $u'$ is non-zero everywhere, then the only way that the product can be zero is when $v'$ is zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3537941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
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