Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Asking A Question Why 16 lnx added in the inequality How prove this inequality $a^2+b^2+c^2+8(ab+bc+ac)+3-10(a+b+c)\ge 0$
see last answer why did he add 16lnx but lnx doesnt exist
| This is a well-known trick for some inequalities of the form $f(a) + f(b) + f(c) \ge 0$ under the condition $a, b, c > 0; abc = 1$, assuming that the equality case is $a = b = c = 1$.
The idea is: If we can find a real constant $m$ such that $F(x) := f(x) + m \ln x \ge 0$ for all $x > 0$,
then we have
$0 \le F(a) + F(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4415199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Combinatorial argument why is the following below true? $\frac{12!}{2^6 \cdot 3! \cdot 3!} = {12 \choose 6} \cdot 5^2 \cdot 3^2 \cdot 1^2 $
I'm trying to formulate a combinatorial argument and have singled out a case of the two formulas above hoping to see if anyone can explain why the above is true. If so it may help ... | Suppose we have $12$ people. We will place six people each in two labeled rooms, $A$ and $B$, and then place the people in each room in three pairs.
There are $\binom{12}{6}$ ways of selecting which six of the twelve people will be placed in room $A$. The rest must be placed in room $B$. Line up the six people in ro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4416991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Why is $\lim\limits_{n\to\infty} \prod\limits_{k=1}^n \left(1 + \frac{k+1}{n^2}\right) = \sqrt e$? Mathematica returns these somewhat striking (to me, at any rate) infinite product identities:
$$\lim_{n\to\infty} \prod_{k=1}^n \left(1 + \frac{k+1}{n^2}\right) = \sqrt e$$
and
$$\lim_{n\to\infty} \prod_{k=1}^n \left(1 + ... | Consider any real $p > 0$. Note that for any $x>0$, $$ x \ge \ln(1+x) \ge x - x^2.$$ Indeed, at $x=0$ we have equality, and looking at the derivatives of $x,\ln(1+x),x-x^2$ respectivelly, we get $1,\frac{1}{1+x},1-2x$ respectivelly, and since $1 \ge \frac{1}{1+x} \ge 1-2x$ for any $x>0$ the result follows. Hence $$ \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4417359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Inverse Derivative: Error/Intuition Problem is to find $[f^{-1}(4)]'$ given $f(x)=\frac{x^3+7}{2}$.
Way One: Switch $x$ and $y$ to find inverse function. So, $x=\frac{y^3+7}{2}$. Therefore, $\sqrt[3]{2x-7}=y$ (i.e. our inverse function). So, $f^{-1}(x)=(2x-7)^{\frac{1}{3}}$. So, $\frac{d}{dx}f^{-1}(x)=\frac{2}{3}(2x-7)... | You already know that after switching variables
\begin{align*}
\frac{dy}{dx}=\frac{1}{\frac{3}{2}y^2}\tag{1}
\end{align*}
Since we want to know $\left(f^{-1}(4)\right)^{\prime}$ we obtain again after switching variables from $x=\frac{y^3+7}{2}$ evaluated at $x=4$:
\begin{align*}
4&=\frac{y^3+7}{2}\\
8&=y^3+7\\
y&=1
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4421633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to find the closed form of $\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(p \cos ^{2} x+q \sin ^{2} x+r\right)^{n}}$, where $n\in N$? In my post, I found the integral
$$\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\right)^{2}}=\frac{\pi\left(a^{2}+b^{2}\right)}{4 a^{3} b^{3}}$$
Then I w... | With $a=p+r$ and $b=q+r$
\begin{aligned}I_n=\int_{0}^{\frac{\pi}{2}} \frac{1}{\left(p \cos ^{2} x+q \sin ^{2} x+r\right)^{n}}dx
=\int_{0}^{\frac{\pi}{2}} \frac{1}{\left(a \cos ^{2} x+b \sin ^{2} x\right)^{n}}dx
\end{aligned}
Integrate, instead
\begin{align}
\sum_{n=1}^\infty I_n t^n=&
\int_0^{\frac{\pi}{2}} \sum_{n=1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Is there any other method to compute $\int_{0}^{\frac{\pi}{4}} y \ln (\cos y) d y$? After investigating the integral
$$
\int_{0}^{\frac{\pi}{2}} y \ (\cos y) d y
$$
in the post.
I keep on finding the integral with smaller limit
$$
I:=\int_{0}^{\frac{\pi}{4}} y \ln (\cos y) d y.
$$
As before, I use the Fourier series of... | To avoid infinite series, let
$$S=\int_{0}^{\frac{\pi}{4}} y \ln (2\sin y) d y,\>\>\>
C=\int_{0}^{\frac{\pi}{4}} y \ln (2\cos y) d y
$$
and $K=\int_{0}^{\frac{\pi}{2}} y \ln (\tan y) d y $. Then
\begin{align}
S+C=&\int_{0}^{\frac{\pi}{4}} y \ln (2\sin 2y) d y
\overset{2y\to y}=\frac14 \int_{0}^{\frac{\pi}{2}} y \ln (2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4428738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$ \sum_{n=1}^\infty ( n ( \sum_{k=n}^\infty \frac{1}{k^2})^2 - \frac1n) = \frac32 - \frac12 \zeta(2) + \frac32\zeta(3)$
Prove $$\sum_{n=1}^\infty \left( n \left( \sum_{k=n}^\infty \frac{1}{k^2}\right)^2 - \frac1n\right) = \frac32 - \frac12 \zeta(2) + \frac32\zeta(3)$$
I decided to write the LHS of the equation as
\b... | One way is to reduce the power of summand by summation by parts.
Take
$$a_n=\bigg(\sum_{k=n}^\infty\frac1{k^2}\bigg)^2-\frac1{n^2},\quad b_n=n,\quad B_n=\sum_{k=1}^nb_n=\frac12n(n+1),$$
we wish to find the closed form of
$$S=\sum_{n=1}^\infty a_nb_n.$$
By summation by parts, with some simplifications,
\begin{align*}
S&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4434804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integral $ \int_{-\infty}^\infty \frac{1}{x^2+1} \left( \tan^{-1} x + \tan^{-1}(a-x) \right) dx$ For $a\in\mathbb R$, I want to evaluate the integral
$$ I = \int_{-\infty}^\infty \frac{1}{x^2+1} \left( \tan^{-1} x + \tan^{-1}(a-x) \right) dx.$$
I tried to integration by parts by considering $\left( \tan^{-1}(x) \right)... | Too long for a comment
In fact, the evaluation can be done by means of the complex integration method, and this evaluation is rather straightforward. Just one of the possible contours.
$$\int_{-\infty}^\infty \frac{1}{x^2+1} \left( \tan^{-1} x + \tan^{-1}(a-x) \right) dx=I_1+I_2$$
It is evident that $I_1=0$ (the integr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4435002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\mathbb{Z}_p[x]/(x^2+x+1)$ is not a field I have to prove the following:
Let $p$ be a prime and denote $\mathbb{Z}_p=\mathbb{Z}/p\mathbb{Z}$. Show that $\mathbb{Z}_p[x]/(x^2+x+1)$ is not a field if and only if $p=3$ or $p \equiv 1 \pmod 3$.
First I consider showing that $x^2+x+1$ is reducible, We may write ... | By the quadratic formula, $x^2 + x + 1$ has roots $\frac{-1 \pm \sqrt{-3}}{2}$.
Therefore, we consider when $-3$ has a square root mod $p$. If it does, then $x^2+x+1$ is not prime and $\mathbb{F}_p [x] / (x^2 + x + 1)$ is not a field. Otherwise, $x^2+x+1$ is irreducible and $\mathbb{F}_p [x] / (x^2 + x + 1)$ is a fiel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4435497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The number of integer points on the curve $(7x-1)^2+(7y-1)^2=n$ The number of integral solutions to the equation
$$x^2+y^2=n$$
is defined to be $r_2(n)$ and if $n=2^ap_1^{a_1}\dots p_k^{a_k}q^2$ where $p_i\equiv 1\mod 4$ and $q$ is the product of primes which are $3\mod 4$, then
$$r_2(n)=4(a_1+1)\dots(a_k+1).$$
If we r... | For a prime $p\equiv 1\mod 4$, we have a unique (up to sign) representation as a sum of squares $p=a^2+b^2$ and with the identity
$$p(x^2+y^2)=(ax-by)^2+(bx+ay)^2=(ax+by)^2+(-bx+ay)^2$$
we can define the linear transformations
$$\begin{bmatrix}
x\\
y\end{bmatrix}\mapsto\begin{bmatrix}
a & -b\\
b & a\end{bmatrix}\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4443254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to calculate this limit $ \lim\limits_{x\rightarrow0}\frac{1-\sqrt{1+x^2}\cos{}x}{x^4} $? $ \lim\limits_{x\rightarrow0}\frac{1-\sqrt{1+x^2}\cos{}x}{x^4} $
I tried L'Hopital, but it gets complicated.
I tried also this:
$$ \lim\limits_{x\rightarrow0}\frac{1-(1+x^2)\cos^2{x}}{x^4(1+\sqrt{1+x^2}\cos{x})} $$
$$ \lim\lim... | Another way.-Let $f(x)=\dfrac{1-\sqrt{1+x^2}\cos(x)}{x^4}$. Since $f(x)=f(-x)$ and is well defined for $|x|\gt0$ by prolongement by continuity we can put $f(0)=a$ where $a$ is precisely the asked limit. One has therefore
$$1-\sqrt{1+x^2}\cos(x)=ax^4$$ Now we have
$$1-(1+\frac{x^2}{2}-\frac{x^4}{8}+\cdots)(1-\frac{x^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4444063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find the matrix of a linear transformation from $\Bbb R^{2\times2}$ to $P_2$ Let $T:\Bbb R^{2\times2}\rightarrow P_2$ be a linear transformation, where $\Bbb R^{2\times2}$ is the vector space of all $2\times2$ matrices and $P_2$ is the vector space of all polynomials up to second order.
We consider the basis $B$... | You need to find out how basis vectors transform under $T$:
$$
\begin{split}
T\begin{pmatrix}
1 & 0 \\
0 & 0
\end{pmatrix} &= 2t^2 \\
T\begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix} &= t + t^2 \\
T\begin{pmatrix}
0 & 0 \\
1 & 0
\end{pmatrix} &= 1-t \\
T\begin{pmatrix}
0 & 0 \\
0 & 1
\end{pmatrix} &= 1.
\end{split}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4444736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding the sum of finite geometric series I'm doing the following summation $\sum_{l=k}^{n}2^l$
$\sum_{l=k}^{n}2^l = 2^k + 2^{k+1} + 2^{k+2} + \ldots+ 2^{n-1} + 2^{n}$
$S_n=a_1\dfrac{1-r^n}{1-r} \therefore S_n=2^k\dfrac{1-(2)^n}{1-2} = 2^{k+n}-2^k$
But my final result seems to be incorrect compared to the one obtaine... | The following summation is
$$\sum_{l=k}^{n}2^l = 2^k + 2^{k+1} + 2^{k+2} + \ldots+ 2^{n-1} + 2^{n}$$
can be written as
$$\sum_{l=0}^{n-k}2^{l+k} = 2^k + 2^{k+1} + 2^{k+2} + \ldots+ 2^{n-1} + 2^{n}$$
by shifting the index $k$-times, which is a series with $n−k+1$ terms
$S_{n-k+1}=a_1\dfrac{1-r^{n-k+1}}{1-r} \therefore... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $\tan^2(x)+\tan(x)=2$ for $0\leq x\leq 2\pi$ I am trying to solve the trigonometric equation
$$\tan^2(x)+\tan(x)=2$$ for $0\leq x\leq 2\pi$. At first glance, I try to rearrange the trigonometric equation into something more manageable such that
\begin{align}
\frac{\sin^2(x)}{\cos^2(x)}+\frac{\sin(x)}{\cos(x)}&=2 ... | Let $u = \tan(x)$
It follows that
$$\tan^2(x) + \tan(x) = 2 \iff u^2 + u - 2 = 0 \iff (u-1)(u + 2) = 0$$
$$\therefore \tan(x) = -2 \; \text{or} \; \tan(x) = 1$$
$$\therefore x = \pi n - \tan^{-1}(2)\; \text{or} \; x = \frac{\pi}{4} + \pi n, \; n \in \mathbb{Z}$$
In the given domain $0 \le x \le 2\pi$, we have
$$x = \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4448886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Difficulty evaluating $\int_0^\infty\frac{1}{(x^3+2)\sqrt{x^2+8}}\,\mathrm{d}x$ In evaluating
$$\int_0^\infty\frac{1}{(x^3+2)\sqrt{x^2+8}}\,\mathrm{d}x$$ I did not have a situation where the polynomial under the square root has lower degree. Neither trigonometric substitutions nor variable changes help, at least how I ... | Substitute $ x =2^{3/2}\tan y$, along with $a=2^{7/6}$
\begin{align}
\int_0^\infty \frac{dx}{(x^3 + 2)\sqrt{x^2+8}} =\frac{1}{2}\int_0^\frac\pi2\frac{\sec y}{1 + (a\tan y)^3} dy
=\frac12 (I_1 + I_2)
\end{align}
where
\begin{align}
I_1=& \int_0^\frac\pi2\frac{\sec y}{1 + a\tan y} dy
=\frac2{\sqrt{1+a^2}}\tanh^{-1} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove $\sum \frac{n^{p+1}}{a_1+2^pa_2+\cdots+n^pa_n}$ is convergent.
Assume $\sum\limits_{n=1}^{\infty} \dfrac{1}{a_n}$ is a convergent positive term series and
$p>0$. Prove $$ \sum_{n=1}^{\infty} \frac{n^{p+1}}{a_1+2^pa_2+\cdots+n^pa_n}$$ is
convergent.
Since
$$a_1+2^pa_2+\cdots+k^pa_k\ge \sqrt[k]{a_1\cdot2^pa_2\cdo... | This is similar to robjohn's solution and it is based on the solution of this page: http://www.math.org.cn/forum.php?mod=viewthread&tid=28918
By Cauchy-Schwarz, we have
$$
\sum_{k=1}^n a_k k^p \sum_{k=1}^n \frac k{a_k} \geq \left(\sum_{k=1}^n k^{\frac{p+1}2}\right)^2\geq \left( \frac{n^{\frac{p+3}2}}{\frac{p+3}2}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4453465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Remarquable identities $f(n) = \frac{a^n}{(a-b)(a-c)} + \frac{b^n}{(b-a)(b-c)} + \frac{c^n}{(c-a)(c-b)}$ Let $n$ be an integer, and
\begin{equation}
f(n) = \frac{a^n}{(a-b)(a-c)} + \frac{b^n}{(b-a)(b-c)} + \frac{c^n}{(c-a)(c-b)}
\end{equation}
\begin{equation}
g(n) = \frac{(bc)^n}{(a-b)(a-c)} + \frac{(ac)^n}{(b-a)(b-c)... | Using the method that @RonaldBlaak used, we see that
$$\begin{align}
\sum_{n\ge0}z^ng(n)&=\sum_{n\ge0}\frac{(bcz)^n}{(a-b)(a-c)}+\sum_{n\ge0}\frac{(acz)^n}{(b-a)(b-c)}+\sum_{n\ge0}\frac{(abz)^n}{(c-a)(c-b)}\\
&=\frac{1}{(a-b)(a-c)(1-bcz)}+\frac{1}{(b-a)(b-c)(1-acz)}+\frac{1}{(c-a)(c-b)(1-abz)}\\
&=\frac{z}{(1-bcz)(1-ac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4453585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Solve: $\sec(2x) \ge\sec(x) , x\in [0,\pi]$\ {$\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}$} Here is the following question:
Solve: $\sec(2x) \ge\sec(x) , x\in [0,\pi]$\
{$\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}$}
Note: This is part b of a question where in part a, I was asked to solve: $\cos(2x)=\cos(x), x\in[0,... | Recall that $\sec(x) = \frac{1}{\cos x}$ and $\cos(2x) = 2 \cos^2(x) - 1$. So, in terms of $\cos(x)$, you have:
$$\frac{1}{2 \cos^2(x) - 1} \ge \frac{1}{\cos(x)}$$
For convenience, let $c = \cos(x)$.
$$\frac{1}{2 c^2 - 1} \ge \frac{1}{c}$$
Now, cross-multiply. Note that because $-1 \le c \le 1$, $2c^2 - 1$ is always ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3$ Prove that if $a,b,c,d$ are positive reals we have:
$$\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3.$$
I think that I have found a equality cas... | Since the inequality is both homogeneous and cyclic, assume that $a = \max(a, b, c, d) = 1$.
It suffices to prove that, for all $b, c, d \in (0, 1]$,
$$\frac{1}{\sqrt{1+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+1}} \le 3.$$
To this end, first, we have
\begin{align*}
\frac{1}{\sqrt{1+b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 1
} |
Find the area of the shaded region. Given, $MT=2$ and $AC=8$. Calculate the area of the shaded region.
$\triangle BMT_(notable)\implies (a, 2a, a\sqrt5)\\
\therefore MT = 2, BT=4\\
\triangle ABF \sim \triangle MBT \implies k = \frac{MT}{BT }=\frac{1}{2} =\frac{AF}{BF}\therefore AF =2.2 = 4, BF = 2.4 = 8 \\
S\triang... |
$MT=2, BT = 4, BM = 2\sqrt5\\
\triangle AGC \sim \triangle MTB\\
\frac{GC}{TB}=\frac{AC}{MB}\\
\frac{GC}{4}=\frac{8}{2\sqrt5}\\
∴GC=\frac{16}{\sqrt5}\\
S_{\triangle MBC} =\frac{1}{2}.MB⋅GC=\frac{1}{2}\cdot 2\sqrt5 \cdot \frac{16}{\sqrt5}=16$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to express a function into powers of $(x-1)$ and $(y-2)$ using Taylor's formula? Use Taylor's formula to express the following in powers of $(x-1)$ and $(y-2)$:
$f(x,y)=x^3 + y^3 + xy^2$
Solution:
$f(1,2)=1 +8 + 4=13$
$f_x (1,2) = 3 + 4=7$
$f_y (1,2) = 12 + 4=16$
$f_{xx} (1,2) = 6$
$f_{yy} (1,2)= 12 + 2=14$
$f_{xy}... | Your approach is fine, although it should be somewhat revised. A way to check your calculation is recalling that when expanding the function at $x_0=1$ and $y_0=2$ we can replace in $f=f(x,y)$:
\begin{align*}
x&=(x-x_0)+x_0=(x-1)+1\\
y&=(y-y_0)+y_0=(y-2)+2
\end{align*}
We obtain
\begin{align*}
f(x,y)&=x^3+y^3+xy^2\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4457413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proof verification that $ \lim_{n\to \infty} \frac{n^2+n-1}{n^2 + 2n +2}=1$ EDIT: I've had some problems uploading this question today as I initially used the mobile verision, hence the quite absurd first proof if you saw it. Here is the full one:
We do this using the epsilon-N definition of the limit of :
$$\forall \v... | HINT
As a suggestion, you do not need to split into two cases. Suppose that $n\geq n_{\varepsilon}$. Then one gets that
\begin{align*}
\left|\frac{n^{2} + n - 1}{n^{2} + 2n + 2} - 1\right| & = \frac{n + 3}{n^{2} + 2n + 2} = \frac{n + 3}{(n + 1)^{2} + 1} \leq \frac{3n + 3}{(n + 1)^{2}} \leq \frac{3}{n} \leq \frac{3}{n_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4458619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why am I getting wrong solution to the system $2x+6y-3z=10$, $5x+2y-1z=12$? Suppose we have this system of equations:
$$\begin{align}
2x+6y-3z&=10\tag{1} \\
5x+2y-1z&=12\tag{2}
\end{align}$$
Why does solving as follows give an incorrect solution?
$$\begin{align}
2x+6y-3z-10&=5x+2y-1z-12 \\
-3x+4y-2z&=-2\tag{3}
\end{ali... | \begin{align}
2x+6y-3z&=10 \tag 1 \\
5x+2y-1z&=12 \tag 2
\end{align}
Matrix form: $AX=b$
$A=\begin{pmatrix} 2 & 6&-3\\5&2&-1\end{pmatrix}$
$X=\begin{pmatrix}x\\y\\z\end{pmatrix}
$
$b=\begin{pmatrix} 10\\12\end{pmatrix}$
Solve : $[A|b]=\begin{pmatrix} 2 & 6&-3&|10\\5&2&-1&|12\end{pmatrix}$
Then you get the solution :
$X... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4458739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 13,
"answer_id": 4
} |
Geometry problem inspired by Babylonian tablets (positive matrix factorization) Consider the following problem: given positive real numbers $a,b,c>0$, find positive real numbers $x,y,z,w >0$ such that
$$x^2+y^2 =a$$
$$z^2+w^2=c$$
and
$$xz+wy=b$$
This has an obvious geometric interpretation: the sum of the areas of two ... | quite a mess. Given positive definite quadratic form $a x^2 \pm 2bxy + c y^2 $ so that $ac>b^2,$ take $r > 0,$ with
$$ 0 < r < \min \left( \frac{b}{a}, \frac{c}{b} \right) $$
Then define $$ p = \frac{c-br}{b-ar} $$
We see that $p > 0.$ It follows from $a r^2 - 2 b r + c > 0$ that $p > r > 0. $
$$
\left(
\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4459925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding a representation of an element of Fibonacci sequence using power series. I need help finding the power series of $f\left(x\right)=\frac{-1}{x^{2}+x-1}$ probably for $x=0$.
I got a sequence defined as $a_{0}=1,\ a_{1}=1\ \ a_{n}=a_{n-1}+a_{n-2}$ for $n\ge 2$. I found out that the radius of convergence of the ser... | $$\frac{1}{x-\frac{-1+\sqrt{5}}{2}}=\frac{2}{1-\sqrt{5}}\frac{1}{1-x\cdot\frac{2}{\sqrt{5}-1}}$$
Now let’s use rationalising/conjugacy technique: $$\frac{2}{\sqrt{5}-1}=\frac{2(\sqrt{5}+1)}{(\sqrt{5})^2-1^2}=\frac{\sqrt{5}+1}{2}$$
So you get: $$-\frac{1+\sqrt{5}}{2}\cdot\frac{1}{1-x\cdot\frac{1+\sqrt{5}}{2}}$$
Can you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4462473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding the $z$ coordinate of center of mass for $C:= \{ (x,y,z) \in \mathbb{R^3}: \sqrt{x^2+y^2}\leq z \leq 1$ The Problem
Let $C$ be the cone
$$C:= \{ (x,y,z) \in \mathbb{R^3}: \sqrt{x^2+y^2}\leq z \leq 1$$
Assume that $C$ has a constant mass density and find the z coordinate of the center of mass.
The work I have ... | This is a cone of base radius $1$, and height $1$. The center of mass formula is
$ \overline{z} = \displaystyle \dfrac{ \iiint z dV } {\iiint dV} $
The numerator is
$ \displaystyle \iiint z dV = \int_{z=0}^1 (2 \pi z^3)dz = \dfrac{\pi}{2} $
The denominator is
$ \displaystyle \iiint z dV = \int_{z=0}^1 (2 \pi z^2)dz = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Let $a,b,c \in [0,1[$ such that $a+b+c=2$. Prove that $a^3+b^3+c^3+2abc\leq 2$
Let $a,b,c \in [0,1[$ such that $a+b+c=2$. Prove that
$$a^3+b^3+c^3+2abc\leq 2$$
My attempt:
put $a=x^{\frac{2}{3}}$,$b=y^{\frac{2}{3}}$,$c=z^{\frac{2}{3}}$
\begin{align*}
&x^{2}+y^{2}+z^{2}+2x^{\frac{2}{3}}y^{\frac{2}{3}}z^{\frac{2}{3}} \... | New proof:
Using $0 \le (1 - a)(1 - b) = 1 - a - b + ab$, we have $ab \ge a + b - 1$.
We have
\begin{align*}
&a^3+b^3+c^3 + 2abc\\
\le\,& a^2 + b^2 + c^2 + 2abc\\
=\,& (a + b)^2 - 2ab + c^2 + 2abc\\
=\,& (a + b)^2 - 2ab(1 - c) + c^2\\
\le\,& (a + b)^2 - 2(a + b - 1)(1 - c) + c^2 \\
=\,& (2 - c)^2 - 2(1-c)(1-c) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Directional Derivative in $\mathbb{R}^2$
Definition Let $f: \mathbb{R}^n \to \mathbb{R}$ and $u \in \mathbb{R}^n$ be a unit vector. The directional derivative of $f$ in the direction of $u$ is
$$D_uf(j) = \displaystyle\lim_{t \to 0} \frac{f(j+tu) - f(j)}{t}$$
provided that this limit exists.
I am preparing for my su... | Maybe you can recognize that
$Df(\mathbf{x}_0)[\mathbf{u}]
=g'(0)$ with the scalar-valued function
$g(t) = f(\mathbf{x}_0 + t\mathbf{u} )$.
Denote $\mathbf{x}= \mathbf{x}_0 + t\mathbf{u}$.
It is simple to show using chain rule that
$g'(t) = \nabla_\mathbf{x}f(\mathbf{x}_0 + t\mathbf{u}):\mathbf{u}$
from which you can ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Doubt regarding validity of answer in a mod equation question
$$|x^2-2x+2|-|2x^2-5x+2|=|x^2-3x|$$
Find the set of values of $x$.
The answer given is $[0,\frac12]\cup [2,3]$.
What I don't get is how is the solution a range? I'm getting the solution as $0,\frac 12, 2 ,3, \frac 25$. That makes sense to me as the first e... | $$|x^2-2x+2|-|2x^2-5x+2|=|x^2-3x|$$
Since $x^2-2x+2 = x^2-2x+1+1=(x-1)^2 +1 > 0$, $|x^2-2x+2|-|2x^2-5x+2|= x^2-2x+2-|2x^2-5x+2|=|x^2-3x|$
then
$2x^2-5x+2=(2x-1)(x-2)=0$, $x=\frac{1}{2}$, and $x=2$
$|2x^2-5x+2|= 2x^2-5x+2$ when $-\infty \leq x \leq\frac{1}{2}$ or $2 \leq x \leq\infty$
$|2x^2-5x+2|= -(2x^2-5x+2)$ when $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4470083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
find the minimum of $A=\frac{x^3}{3y+1}+\frac{y^3}{3z+1}+\frac{z^3}{3x+1}$ with $x^3+y^3+z^3=3$ With $x,y,z \ge 0, x^3+y^3+z^3=3$: find the minimum of $A=\dfrac{x^3}{3y+1}+\dfrac{y^3}{3z+1}+\dfrac{z^3}{3x+1}$
My attempts: $A=\dfrac{x^6}{3yx^3+x^3}+\dfrac{y^6}{3zy^3+y^3}+\dfrac{z^6}{3xz^3+z^3} \ge \dfrac{(x^3+y^3+z^3)^2... | @SuzuHirose's use of standard Lagrange methods indeed finds an extremum. Here's a color-coded plot of the target function on the cubic surface to show that:
The minima occur when $x = y = z = 1$, etc., as seen in the figure, where blue is small.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to prove $s > \frac{11}{2}t-3t\ln t$ in this problem The problem is:
Given $f(x)=\frac{\ln x}{x}$, line $l$ is the tangent of curve
$y=f(x)$ at $(t,f(t))$, and intersects the curve at another point
$(s,f(s))$ where $s<t$.
(1) Find the range of $t$;
(2) (i) Prove $\ln x\le 1 +\frac{1}{e}(x-e)-\frac{1}{2e^2}(x-e)^2... | For (2)-(ii):
From question (1), we have $t > \mathrm{e}^{3/2}$.
We only need to consider the case when $\frac{11}{2}t - 3t\ln t > 0$, i.e. $t < \mathrm{e}^{11/6}$.
Let
$$G(x) := x F(x) = \ln x - \frac{1-\ln t}{t^2}(x-t)x - \frac{x\ln t}{t}.$$
We have $G(s) = 0$.
We have
$$G'(x) = \frac{(t - x)(t + 2x - 2x\ln t)}{xt^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4475136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find expression of $c_n$, where $c_n = a_n + b_n$ Given the recurrence relation $a_{n+2} = 3a_{n+1} + 6a_n$ and $b_{n+2} = b_{n+1} + b_n$ I am supposed to find an expression of the recurrence relation for $c_n := a_n + b_n$. I tried to find some form of linear dependence to obtain a recurrence relation for $c_n$ but th... | Using that $\,a_{n+2} - 3a_{n+1} - 6a_n = 0\,$ and $\,b_{n+2} = b_{n+1} + b_n\,$:
$$
\require{cancel}
\begin{align}
c_{n+2}-3c_{n+1}-6c_n &= \cancel{(a_{n+2}-3a_{n+1}-6a_n)} + (b_{n+2}-3b_{n+1}-6b_n)
\\ &= -2b_{n+1} - 5 b_n
\end{align}
$$
Then, using that $\,b_{n+2} - b_{n+1} - b_n = 0\,$:
$$
\begin{align}
(c_{n+4} - 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4475812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Estimate the condition number in the second norm of the matrix An Let $A_{n}$ be a matrix of size $n$ for $n \geq 1$, and a structure:
$$
A_{n}=\left[\begin{array}{cccccccc}
\sqrt{21} & 1 & 0 & 0 & \ldots & 0 & 0 & 0 \\
0 & \sqrt{21} & 1 & 0 & \ldots & 0 & 0 & 0 \\
0 & 0 & \sqrt{21} & 1 & \ldots & 0 & 0 & 0 \\
0 & 0 & ... | As you indicated, you essentially need to estimate the eigenvalues of $A^*A$.
Fortunately, in this case it is easy to compute $A^*A$ explicitly.
$$
A^*A=\left[\begin{array}{cccccccc}
22 & \sqrt{21} & 0 & 0 & \ldots & 0 & 0 & 0 \\
\sqrt{21} &22 & \sqrt{21} & 0 & \ldots & 0 & 0 & 0 \\
0 & \sqrt{21} &22 & \sqrt{21} & \ldo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluate $\int \dfrac{dx}{(1+x)\sqrt{1+2x-x^2}}$
$$\int \dfrac{dx}{(1+x)\sqrt{1+2x-x^2}}$$
I completed the square:
$\int \dfrac{dx}{(1+x)\sqrt{2-(x-1)^2}}$
And then substituted $\sqrt 2\sin θ = x-1$ which gives
$\int \dfrac{dθ}{\sqrt2 \sinθ + 2}$
But now I'm stuck. Can someone please help?
| You may continue with $\theta=\frac\pi2+t$
\begin{align}\int \frac{1}{\sqrt2 \sin\theta+ 2}d\theta
= &\int \frac{1}{\sqrt2 \cos t+ 2}dt= \int \frac{1}{2\sqrt2 \cos^2\frac t2+ 2-\sqrt2}dt\\
=&\int \frac{2d(\tan\frac t2)}{(2-\sqrt2)\tan^2\frac t2+(2+\sqrt2)}=\sqrt2\tan^{-1}\frac{\tan\frac t2}{\sqrt2+1}+C
\end{align}
Or, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the maximum of $\frac{abc}{(4a+1)(9a+b)(4b+c)(9c+1)}$,where$a,b,c>0$ $a,b,c>0$, find the maximum of :
$$\frac{abc}{(4a+1)(9a+b)(4b+c)(9c+1)}$$
I try to find the minimum of $\frac{(4a+1)(9a+b)(4b+c)(9c+1)}{abc}=\frac{4a+1}{\sqrt{a}}\cdot\frac{9a+b}{\sqrt{ab}}\cdot\frac{4b+c}{\sqrt{bc}}\cdot\frac{9c+1}{\sqrt{c}}
=\... | Let $Q = \dfrac{abc}{(4a+1)(9a+b)(4b+c)(9c+1)}= \dfrac{abc}{(b+9a)(4a+1)(4b+c)(9c+1)}$. Apply Cauchy-Schwarz inequality twice: $Q \le \dfrac{abc}{(2\sqrt{a}\cdot \sqrt{b}+1\cdot 3\sqrt{a})^2\cdot(2\sqrt{b}\cdot 3\sqrt{c}+\sqrt{c}\cdot 1)^2} = \left(\dfrac{\sqrt{b}}{(2\sqrt{b}+3)(6\sqrt{b}+1)}\right)^2= \left(\dfrac{t}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Calculating complex integral with Residue - where is my fault? \begin{align}\int_0^{2\pi} \frac{\cos(x)}{13+12\cos(x)} dx & = \displaystyle\int_0^{2\pi} \frac{(z+1/z)\frac{1}{2}}{13+12(z+1/z)\frac{1}{2}}\frac{1}{iz} dz \\
& = \cdots \\
&= -i\displaystyle\int_0^{2\pi} \frac{z^2+1}{\left(z+\frac{2}{3}\right)\left(z+\frac... | You forget the residue at $-2/3$, because the change of variable is $z=e^{it}$, so the integral in $z$ is not from $0$ to $2\pi$, is in the unit circle, that includes inside $-2/3$.
$\begin{align}\displaystyle\int_0^{2\pi} \frac{cos(x)}{13+12cos(x)} dx & = \displaystyle\int_\gamma \frac{(z+1/z)\frac{1}{2}}{13+12(z+1/z)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4480941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Computing $D_v^2(f)(x_0)$, if $f(x,y)=x^3+x+y^2-2y+1$, $v=\frac{1}{5}(3,4), x_0=(\frac{1}{3},1)$
Compute $D_v^2(f)(x_0)$, if $f(x,y)=x^3+x+y^2-2y+1$, $v=\frac{1}{5}(3,4), x_0=(\frac{1}{3},1)$ using the formula $D_v^2(f)(x_0)=v^TH_f(x_0)v$.
So, $v^T= \left[
\begin{array}{cc}
\frac{3}{5}\\
\frac{4}{5}
\end{array}
\... | You're almost there, but are misunderstanding the convention. Actually, $$v=\begin{bmatrix} \frac{3}{5} \\ \frac{4}{5}\end{bmatrix}$$ and
$$v^T=\begin{bmatrix} \frac{3}{5} & \frac{4}{5}\end{bmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\underset{z=0}{\text{Res}} \; \frac{(z^6-1)^2}{z^5(2z^4 -5z^2 + 2)}$ . Problem:
Evaluate
$$
\underset{z=0}{\text{Res}} \; \frac{(z^6-1)^2}{z^5(2z^4 -5z^2 + 2)}.
$$
My question:
This is a question from a previous complex analysis qualifying exam that I am trying to work through. I know that the straightforward... | Let $$f(z) = \frac{(z^6-1)^2}{z^5(2z^4 - 5z^2 + 2)}, \quad g(z) = \frac{(z^3-1)^2}{2z^2 - 5z + 2},$$ so that we have $f(z) = g(z^2)/z^5$. So by computing the series expansion of $g$ about $0$, we can obtain the Laruent expansion of $f$, since $g$ has no singularity at $0$.
To this end, observe $$2z^2 - 5z + 2 = (2z-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4487007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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$\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$ is independent of $n$
If $n$ is a positive integer, prove that
$$\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$$
is i... | Here is a variation which is also related to Bill Dubuque's hint. We want to show that
\begin{align*}
f(n)&=\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor\\
&=\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-\left\lfloor\frac{n-17}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4489162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
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Equation of line passing through two given lines (3D) A line from the origin meets the line $x-2\over 1$ = $y-1\over-2$ = $z+1\over1$ And $(x-\frac83)\over2$ = $y+3\over-1$ = $z-1\over1$ at P and Q respectively. If the length $PQ = d$ , then $(d)^2$ is
My attempt :
I assumed general points on the given lines, thus intr... | Let point $P$ be on the first line, then
$P = P_0 + t V $
And let point $Q$ be on the second line, then
$Q = Q_0 + s W $
Since we want the line extending from $P$ to $Q$ to pass through the origin, then it follows that $P$ is on the plane that passes through the origin and contains $Q(s)$. Similarly, $Q$ is on the pla... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4489772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find $f:\mathbb{R} \to \mathbb{R}$ which satisfies $f(x)^2+2f(x)f(y)+y^2=\big(x-f(y)\big)^2+4f(x)y$.
Find $f:\mathbb{R} \to \mathbb{R}$ which satisfies $f(x)^2+2f(x)f(y)+y^2=\big(x-f(y)\big)^2+4f(x)y$.
Let $P(x, y): f(x)^2+2f(x)f(y)+y^2=\big(x-f(y)\big)^2+4f(x)y.$
$P(0, 0): f(0)^2-2f(0)f(0)=\big(0-f(0)\big)^2.$
$\the... | You have shown that $\lvert f(x) \rvert = |x|$
and
$$ f(y)(f(x)+x)=2f(x)y $$
for all $x$ and $y$.
Consider whether $f(1)$ is $1$ or $-1$.
If $f(1) = 1$, then (plugging in $y = 1$)
$$ 1 \cdot (f(x) + x) = 2 f(x) \cdot 1, $$
so $f(x) = x$ for all $x$.
If $f(1) = -1$, then (plugging in $x = y = 1$)
$$ -1 \cdot (-1 + 1) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Check convergence of series I want to check the convergence of the following series $$\sum_{n=0}^{\infty}\left (\frac{3+n^3}{4-n+n^2-3n^3}\right )^n \ \\ \sum_{n=0}^{\infty}\left (\frac{3+n^2}{4+2n^2}\right )^n \\ \sum_{n=1}^{\infty}\left (\sqrt{2n}-\sqrt{n}\right ) $$
I have done the following :
$$\sum_{n=0}^{\infty}... | The limit of the general term is, after your work, $\lim_{n\to\infty}(\sqrt2-1)\sqrt n=\infty$.
Thus since it doesn't go to zero, the series diverges (sometimes called the divergence test).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
show that $0\le \frac a{1+b}+\frac b{1+a} \le 1$ Let $0\le a < b \le 1$
prove that $0\le \frac a{1+b}+\frac b{1+a} \le 1$
This is the answer from book :
Obviously $0\le \frac a{1+b}+\frac b{1+a}$ so :
\begin{align}
&1\ge \frac a{1+b}+\frac b{1+a} \\
\iff & (1+a)(1+b) \ge a(1+a)+b(1+b)\\
\iff & 1-a^2\ge b^2-ab \\
\iff ... | Let $0\le a , b \le 1$
We prove that $$0\le \frac a{1+b}+\frac b{1+a} \le 1$$
Let $$F=\frac a{1+b}+\frac b{1+a}$$
Then $F\ge 0$, note that
$$\frac{a}{1+b} \le \frac{a}{a+b}$$
and
$$\frac{b}{1+a}\le \frac{b}{b+a}$$
Adding the last two we get
$$F\le 1$$
Equality holds when $$a=1=b.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
(Absolute) convergence of $\sum_{k=1}^n (-1)^k \frac{3^k+2}{4^k -6}$ We want to check if the following sequence converges (absolutely):
$$\sum_{k=1}^n (-1)^k \frac{3^k+2}{4^k -6}$$
Since we have $(-1)^k$, we can use the Leibniz criterion.
We have to prove that $a_n$ is a null sequence and that it is monotonically decre... | As Lionel Ricci's comment indicates, rather than trying to use the Leibniz criterion (note that, as described in Leibniz Criterion, this criterion (also called the alternating series test) only indicates the series converges, not that it necessarily converges absolutely), instead use that, for $k \ge 2$, we have $4^k -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Express $|\pi - \frac{23}{7}|$ without the absolute value symbol
Express $\left|\pi - \dfrac{23}{7}\right|$ without the absolute value symbol.
I know I have to check if $\pi - \dfrac{23}{7}$ is greater than (or equal to) zero, but how can I do it analytically (without a calculator)?
I know that $\pi \gt 3=\dfrac{21}{... | I will not use any approximations in this answer.
Consider the definite integral $$\int_0^1\frac{t^4(1-t)^4}{1+t^2}dt.$$ Simply expand the numerator using binomial formula and reduce the numerator in terms of the denominator. I’ll skip a few steps for the sake of brevity: $$\int_0^1\frac{t^4(1-t)^4}{1+t^2}dt=\int_0^1\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
How does one show this complex expression equals a natural number? We have:
$$\left(\frac{10 }{3^{3/2}}i-3\right)^{1/3}+ \frac{7}{3 \left(\frac{10}{3^{3/2}}i-3\right)^{1/3}}=2$$
This comes from solving the cubic equation of $x^3-7x+6=0$ which factors as $(x-2)(x-1)(x+3)=0$
We can simplify the problem into finding just ... | Write your complex numbers in polar. Then you have
$$
\left(-3+i\frac{10}{3\sqrt{3}}\right)^{1/3}=\left[\frac{7\sqrt{7}}{3\sqrt{3}}\exp\left(i\pi-i\sin^{-1}\frac{10}{7\sqrt{7}}\right)\right]^{1/3} = \frac{\sqrt{7}}{\sqrt{3}}\exp\left(\frac{i\pi}{3} - \frac{i}{3}\sin^{-1}\frac{10}{7\sqrt{7}}\right).$$
Now honestly simpl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4497637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Missing solutions from complex cubic root From Cardano's work in the 1600s, we have this famous example of a cubic polynomial equation:
$$x^3-15x-4=0.$$
Plugging the coefficients into the Cardano-Tartaglia formula, we get an expression for the solutions that reduces to:
$$x=\sqrt[3]{2+\sqrt{-121}}+\sqrt[3]{2-\sqrt{-121... |
$x = \sqrt[3]{2+i\sqrt{121}}+\sqrt[3]{2-i\sqrt{121}}$
This is misleading, because a non-zero complex number has $3$ cube roots, and the above makes it look like $x$ could take $9$ values between the different combinations of the two cube roots.
The better way to write the cubic formula is $x = \sqrt[3]{2+i\sqrt{121}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
pseudo C-S inequality? Problem :
For $x,y,z\in\mathbb{R}$,
Find minimum of $$8x^4+27y^4+64z^4$$
where $$x+y+z=\frac{13}{4}$$
I tried to apply C-S inequality but it has little difference,
The form what I know is :
$$(a^4+b^4+c^4)(x^4+y^4+z^4)\ge (ax+by+cz)^4$$
But in this problem, coefficient is form of $()^3$, not a $... | Apply AM-GM inequality repeatedly three times:
$8x^4+\dfrac{243}{2}=8x^4 + \dfrac{81}{2}+\dfrac{81}{2}+\dfrac{81}{2} \ge 4\sqrt[4]{8x^4\cdot \dfrac{81}{2}\cdot \dfrac{81}{2}\cdot \dfrac{81}{2}}=4\sqrt[4]{(27x)^4}= 4|27x|\ge 27(4x)$.
$27y^4 + 81 = 27y^4+ 27+27+27 \ge 4\sqrt[4]{27y^4\cdot 27\cdot 27\cdot 27}=4|27y| \ge 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4499469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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How to show $\sum_{k=1}^{80} {1 \over \frac{k}{81} - \frac{1}{2} - \frac{\iota}{2}} + {1 \over \frac{k}{81} - \frac{1}{2} +\frac{\iota}{2}} = 0$? I came across this sum while doing the homework my teacher gave us on series. It was originally $$a_k = \frac{k}{81}, S = \sum_{k=1}^{80} {a_k^2 \over 1+2a_k^2-2a_k}$$
I then... | Notice that :
$$S = \sum_{k = 1}^{80} \dfrac{k^2}{81^2 + 2 k^2 - 2 \times 81 \times k} = \sum_{k = 1}^{80} \dfrac{k^2}{(81 - k)^2 + k^2}$$
and :
$$\sum_{k = 1}^{80} \dfrac{k^2}{(81 - k)^2 + k^2} = \sum_{k = 1}^{80} \dfrac{(81 - k)^2}{k^2 + (81 - k)^2}$$
which allows us to deduce that :
$$2 S = \sum_{k = 1}^{80} \dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4501816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Radius of convergence opposite result I am trying to find the radius of convergence of the series $\sum_{n=1}^\infty{\frac{(n!)^3 \cdot x^n}{(3n)!}}$. I know that it should be $|x|<27$, but i get the exactly the opposite result ($27<|x|$). I am using the ratio test to determine it (e.g. $\rho = \lim_{n\to\infty} \left|... | There shouldn't be any $x$'s. Then you will get $r=\lim\dfrac{ \mid a_n\mid}{\mid a_{n+1}\mid}=27$.
That's you use the $a_n$ in $$\sum a_nx^n$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\sum_{r=0}^{1010} \binom{1010}r \sum_{k=2r+1}^{2021}\binom{2021}k$ I need to find the summation
$$S=\sum_{r=0}^{1010} \binom{1010}r \sum_{k=2r+1}^{2021}\binom{2021}k$$
I tried various things like replacing $k$ by $2021-k$ and trying to add the 2 summations to a pattern but was unable to find a solution. For... | For $N=1010$ or others
\begin{equation}
S_N\equiv \sum_{r=0}^N\binom{N}{r}\sum_{k=2r+1}^{2N+1}\binom{2N+1}{k}
=
\end{equation}
\begin{equation}
\sum_{r=0}^N\binom{N}{r}\sum_{k=2r+1}^{2N+1}\binom{2N+1}{k}
+
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2r}\binom{2N+1}{k}
-
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2r}\binom{2N+1}{k}
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Parameterizing the parabola $9x^2 +y^2-6xy+4x-4y+1=0$
Find parametrization of curve given by equation:
$$9x^2 +y^2-6xy+4x-4y+1=0$$
My attempt:
Let's notice that \begin{split} 9x^2 +y^2-6xy+4x-4y+1=0 & \iff (3x)^2 -6xy + y^2 +4x -4y +1=0\\
& \iff (3x-y)^2 +6x -2y + 1 -2x -2y=0\\
& \iff (3x-y)^2 +2(3x -y) + 1^2 -2x -... | One possible parametrization can be obtained by setting $u = 3x-y+1$ and $v = 2(x + y)$, so that
$$
\begin{cases}
u = 3x - y + 1 \\
v = 2(x + y)
\end{cases} \Longleftrightarrow
\begin{cases}
x = \frac{1}{8} \left(2u + v - 2 \right)\\
y = \frac{1}{4} \left(-2u + 3v + 2 \right)
\end{cases}
$$
and $9x^2 +y^2-6xy+4x-4y+1=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Spivak: How do we divide $1$ by $1+t^2$, to obtain $\frac{1}{1+t^2}=1-t^2+t^4-t^6+...+(-1)^nt^{2n}+\frac{(-1)^{n+1}t^{2n+2}}{1+t^2}$? In Chapter 20, "Approximation by Polynomial Functions" in Spivak's Calculus, there is the following snippet on page $420$
The equation
$$\arctan{x}=\int_0^x \frac{1}{1+t^2}dt$$
suggests... | I haven't read the book but if the author has already introduced polynomial long division or synthetic division, then either one can be used to evaluate even "weird-looking" fractions like
$$\frac 1{1+t^2}$$
For division of polynomials where the denominator has a degree greater than one, then we neglect the highest ord... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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If $\frac{x}{y}+\frac{y}{x}\ge2$ and $\sum_{cyc}\frac{x}{y+z}\ge\frac32$, can we say $\sum_{cyc}\frac{w}{x+y+z}\ge\frac43$? We know that $$\frac{x}{y}+\frac{y}{x}\ge2$$ and $$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}\ge\frac32$$ Can we say that
$$\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{x+w+z}+\frac{z}{x+y+w}\ge\frac43... | This answer is a generalization of the method used in the first answer given above.
Problem:
Prove that:
$$\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\ge\frac{n}{n-1}$$
Proof:
Using AM-GM inequality we have,
$$\begin{align}&\sum_{1\le i<j\le n} (x_i^2+x_j^2)=(n-1)\sum_{1\le i\le n} x_i^2\ge 2\left(\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Find the radius of the circle cutting sides of equilateral triangle Consider the figure below. The triangle is equilateral. Find the radius of the circle.
My try:
I have dropped perpendiculars from the center of the circle to respective chords naming them $x,y,z$ and named the other line segments as $m,n,p,q,s,t$ as s... | If you set up a Cartesian coordinate system with its origin at the center of the circle and $BC$ as a horizontal line beneath it, then the coordinates of the chord endpoints work out to:
$$G = \left( - \frac{5}{2}, - \sqrt{R^2 - \frac{25}{4}} \right)$$
$$T = \left( \frac{5}{2}, - \sqrt{R^2 - \frac{25}{4}} \right)$$
$$N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
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How do I make a change of variable for $\;\lim_{x\to 0}\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}$? I can't use l'hopital, so change of variable is the only way.
$$\lim_{x\to 0}\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}$$
| We have: $\dfrac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}=\dfrac{\dfrac{x^2}{\sqrt{x^2+p^2}+p}}{\dfrac{x^2}{\sqrt{x^2+q^2}+q}}=\dfrac{\sqrt{x^2+q^2}+q}{\sqrt{x^2+p^2+p}}\to \dfrac{2q}{2p} = \dfrac{q}{p}$ as $x \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Show that $4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ}) = 1$. The problem asks us to show that the following equation holds true. $$4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ}) = 1$$
This equation has been verified on my calculator.
Perhaps some basic trigonometric formulas will be enough to solve th... | From DeMoivre's Theorem:
$$(\cos(12°) + i \sin(12°))^5 = \cos(60°) + i\sin(60²) = \frac{1}{2} + i\frac{\sqrt{3}}{2}$$
For brevity, let $c = \cos(12°)$ and $s = \sin(12°)$.
$$(c + is)^5 = \frac{1}{2} + i\frac{\sqrt{3}}{2}$$
$$c^5 + 5isc^4 - 10s^2c^3 - 10is^3c^2 + 5s^4c + is^5 = \frac{1}{2} + i\frac{\sqrt{3}}{2}$$
Splitt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Show for $0 < A < B, \int_{-\infty}^\infty \frac{A^2 - A + B^2 - B + (A + B - 2AB)\cos(v) + v^2}{(A^2-2AB\cos(v)+B^2+v^2)^2}dv > 0$. The integral in question doesn't have a closed form solution as far as I know, but it does converge. The denominator is clearly positive, and the numerator of the integrand is always posi... | If $A + B - 2AB \le 0$, then
$$\mathrm{NUM} \ge A^2 - A + B^2 - B + (A + B - 2AB) = (A - B)^2 > 0.$$
If $A + B - 2AB > 0$ and $A + B \ge 2$, then
$$\mathrm{NUM}
\ge A^2 - A + B^2 - B - (A + B - 2AB) = (A + B)(A + B - 2) \ge 0 .$$
If $A + B - 2AB > 0$ and $A + B < 2$, using $\cos v \ge 1 - v^2/2$, then
\begin{align*}
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimize $f(a,b,c) = \dfrac{1}{a²+b²+c²} +\dfrac{1}{9abc}$ subject to $a+b+c=1$ and $a,b,c>0$
Minimize $f(a,b,c) = \dfrac{1}{a²+b²+c²} +\dfrac{1}{9abc}$ subject to
$a+b+c=1$ and $a,b,c>0$
My attempt:
Since $f(a,b,c) \ge 0$, then we can minimizing the addends individually. Furthermore, minimizing fractions is the same... | Let's try this with AM-GMs. First note that as $(x+y+z)^2\geqslant 3(xy+yz+zx)$,
$$(ab+bc+ca)^2 \geqslant 3abc(a+b+c)=3abc $$
$\begin{align*}
\implies 9abc(a^2+b^2+c^2) & \leqslant 3(a^2+b^2+c^2)(ab+bc+ca)^2 \\
&\leqslant 3\left(\frac{a^2+b^2+c^2+2(ab+bc+ca)}3 \right)^3 \\
&=\frac19(a+b+c)^6=\frac19
\end{align*}$
Hen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Limit of $u_{n}=\frac{1}{n !} \sum_{k=0}^{n} k !$ I encountered a sequence $(u_n)_{n \in \mathbb{N}} $ defined as
$$ u_{n}=\frac{1}{n !} \sum_{k=0}^{n} k ! $$
And I wonder what is the limit. It seems to be 1 but even Wolfram Alpha cannot figure it out.
My first idea was to write $$ \frac{k!}{n!} = \frac{1}{(k+1)...(n-1... | Write it as
$$\frac{1}{n!} \sum_{k=0}^n k! = \frac{n!}{n!} + \frac{(n-1)!}{n!}+...+\frac{1!}{n!} = 1+\frac{1}{n}+\frac{1}{n(n-1)}+...+\frac{1}{n!}$$
Now this is clearly greater or equal to $1$. On the other hand the last $(n-1)$ terms (all except the first two) are smaller than $\frac{1}{n(n-1)}$ so we have an upper bo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4515299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Argue on elementary grounds that the average value of $x/y$ exceeds 1 if x and y are chosen randomly between 1 and 2 Textbook problem:
Two numbers, first $x$ and then $y$, are chosen at random between $1$ and $2$. What is the average value of the quotient $\frac{x}{y}$? Can you argue on elementary grounds that the ans... | We can argue as follows. The probability distribution is symmetric w.r.t. interchanging $x$ and $y$. This means that we can compute the expectation value by restricting $x$ to be larger than or equal to $y$ and computing the expectation value of $f(x,y) = \frac{1}{2}\left(\frac{x}{y} + \frac{y}{x}\right)$ over this mod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4515396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
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$\frac{1}{\log_2(x-2)^2}+\frac{1}{\log_2(x+2)^2}=\frac{5}{12} $ Find all x's
It is given that $$\frac{1}{\log_2(x-2)^2}+\frac{1}{\log_2(x+2)^2}=\frac{5}{12} $$ Find all possible values of $x$.
What I did: $$\frac{1}{\log_2(x-2)}+\frac{1}{\log_2(x+2)}=\frac{5}{6} $$ $$\log_{(x-2)}2 +\log_{(x+2)}2 =\frac{5}{6}$$
What i... | Ah you meant $\log_2((x-2)^2)$ not $(\log_2(x-2))^2$? That wasn't clear to me. Are you sure that its +- 6? Wolframalpha gives completely different answers, no matter whether I use $\log_2((x-2)^2)$ or $(\log_2(x-2))^2$.
This might help you a bit, but don't expect really nice answers unless you've made a mistake in the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Simplifying $\sqrt{a^{4 n}} \cdot \frac{3}{a^{n}} \cdot a^{1+n} \cdot(2 a)^{3 n-1}$ as far as possible, using only positive integers So far here is my steps but I am unsure of how to do the final step
$$\sqrt{a^{4 n}} \cdot \frac{3}{a^{n}} \cdot a^{1+n} \cdot(2 a)^{3 n-1}$$
$$\left(a^{4 n}\right)^{\frac{1}{2}} \cdot \f... | Just to clarify the comments, I have a full worked solution provided.
\begin{align}
&\quad\sqrt{a^{4n}} \cdot \frac{3}{a^{n}} \cdot a^{1+n} \cdot(2 a)^{3 n-1}\\
&=\left(a^{4n}\right)^{\frac{1}{2}}\cdot3\cdot\frac{1}{a^n}\cdot a^{1+n}\cdot2^{3n-1}\cdot a^{3n-1}\\
&=a^{\frac{4n}{2}}\cdot3\cdot a^{-n}\cdot a^{n+1}\cdot2^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How can we simplify the limit of the angle at a vertex of a traingle as the triangle gets arbitrarily small? (the limit involves the law of cosines) Let $f$ be a mapping $\mathbb{R} \implies \mathbb{R}$
Let $\forall x, r \in \mathbb{R}$ let $T(x, r)$ be the triangle whose three vertices are the following points:
*
*$... | Define $A = \frac{f(x) - f(x - r)}{r}$ and $B = \frac{f(x + r) - f(x)}{r}$. If $f$ is differentiable, then $A$ and $B$ both converge to $f'(x)$ as $r \rightarrow 0$. But this generalization allows for piecewise functions that "turn the corner" at $(x, f(x))$. Anyhow, if we make the substitutions
$$f(x - r) = f(x) - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Convergence of $1-\frac{1 \times 3}{3!} + \frac{1 \times 3 \times 5}{5!} +....+ \frac{1 \times 3 \times 5 \times ... \times(2n-1)}{(2n-1)!} - ...$ I was asked to prove whether
$$1-\frac{1 \times 3}{3!} + \frac{1 \times 3 \times 5}{5!}- \frac{1 \times 3 \times 5 \ \times 7}{7!} +....+ \frac{1 \times 3 \times 5 \times ..... | Even easier, we can explicitly evaluate the sum. As you noticed, if for $n \ge 0$ $$a_n = \frac{(-1)^n}{(2n+1)!} \prod_{k=1}^n (2k+1),$$ then $$\begin{align}
a_n &= \frac{(-1)^n}{(2n+1)!} \prod_{k=1}^n (2k)(2k+1) \left(\prod_{k=1}^n (2k) \right)^{-1} \\
&= \frac{(-1)^n}{(2n+1)!} (2n+1)! \left( 2^n \prod_{k=1}^n k \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Limit of $(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$ Find $a,b$ of:
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$$
I can't use L'hopital, I tried multiplying by the conjugate, and solving it,
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-(ax+b))\cdot \frac{\sqrt{4x^2+2x+1}+(ax+b)}{\sqrt{4x^2+2x+1}+(ax+b)}$$
$$=\li... | It should be a question to find a and b.
We need to eliminate $x^2$ term, otherwise limit is zero, so $a=\pm2$
$$=\lim_{x \to \infty}\frac{4x^2+2x+1-(ax+b)^2}{\sqrt{4x^2+2x+1}+(ax+b)} = \lim_{x \to \infty}\frac{x\left(2-2ab+\frac{1-b^2}{x}\right)}{x\left(\sqrt{\frac{4x^2+2x+1}{x^2}}+a+\frac{b}{x}\right)}=\frac{2-2ab}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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For $0For $0<t\leq 1$ show
$$ \text{ln}(t)\geq \frac{t-1}{2t+2}\left(1+\sqrt{\frac{2t^2+5t+2}{t}}\right)$$
On a side note, I've been taking this summer class on inequalities. The professor told us that his main source are all sorts of math olympiads. Given that I'm gonna take Calc 3 next semester and Real Analysis aft... | Fact 1: If $x \in (0, 1]$, then
$$\ln x \ge \frac{(x - 1)(1 + x^{1/3})}{x + x^{1/3}}.$$
(See [1], page 272. This inequality is due to Karamata.)
By Fact 1, it suffices to prove that
$$\frac{(t - 1)(1 + t^{1/3})}{t + t^{1/3}}
\ge \frac{t-1}{2t+2}\left(1+\sqrt{\frac{2t^2+5t+2}{t}}\right)$$
or
$$\frac{1 + t^{1/3}}{t + t^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
finding integral of $ \int \frac{1}{\sin x + \sqrt{3} \cos x}\ dx $ In $ \int\frac{1}{\sin x + \sqrt{3} \cos x}\ dx $, If I multiply and divide by $1/2$ I get $$ \int\frac{1/2}{\frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x}\ dx ,$$
then I can write $ 1/2=\sin (\frac{\pi}{6}) $ and $ \sqrt{3}/2=\cos (\frac{\pi}{6}) $ a... | Both of your methods have a mistake in them. In your first method, you should have
$$\frac{\sin\left(\frac{\pi}{6}\right)}{\sin\left(\frac{\pi}{6}\right)\sin\left(x\right)+\cos\left(\frac{\pi}{6}\right)\cos\left(x\right)} = \frac{\sin\left(\frac{\pi}{6}\right)}{\cos\left(\frac{\pi}{6}-x\right)}.$$
Similarly, your secon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
In how many ways can the ice hockey team be formed?
A ice hockey team consists of 1 goaltender, 2 defencemen and 3 forwards. The coach has available to him: 3 goaltenders, 7 defencemen, 10 forwards and 4 players that can play both position of a defenceman and a forward. In how many ways can the team be formed?
If th... | Given the explanation here
Combination problem - picking a basketball team with restrictions one could expect two solutions, 50916 and 42273.
\begin{array}{c|c}
play & rest & \\
\hline
1 + g.G & 1 + G + \frac {G^2}{2!} + \frac {G^3}{3!} \\
1 + d.D + d^2\frac {D^2}{2!} & 1 + D + \frac {D^2}{2!} + \cdots + \frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4533892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove $\frac{1+xy+xz}{(1+y+z)^2} + \frac{1+zy+zx}{(1+y+x)^2} + \frac{1+yx+yz}{(1+x+z)^2} \ge 1$ for $x,y,z > 0 $
Let $x, y, z > 0$. Prove that
$$\frac{1+xy+xz}{(1+y+z)^2} + \frac{1+zy+zx}{(1+y+x)^2} + \frac{1+yx+yz}{(1+x+z)^2} \ge 1.$$
I encountered this problem today and thought that it was straightforward, and to b... | By the Cauchy-Schwarz inequality,
$$
(1+xy+xz)(1+y/x+z/x)
\\
=\left(1,\sqrt{xy},\sqrt{xz}\right)^2
\left(1,\sqrt{ y/x},\sqrt{z/x}\right)^2\geq(1+y+z)^2
$$
which we can rearrange to
$$
{1+xy+xz\over(1+y+z)^2}\geq{x\over x+y+z}.
$$
The rest is smooth.
Credit
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4538156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to differentiate $y=\sqrt{\frac{1-x}{1+x}}$? It is an example question from "Calculus Made Easy" by Silvanus Thompson (page 68-69). He gets to the answer $$\frac{dy}{dx}=-\frac{1}{(1+x)\sqrt{1-x^2}}$$ The differentiation bit of the question is straightforward, but I'm having trouble simplifying it to get the exact ... | Continuing from your second-last step:
$$\frac{dy}{dx}=-\frac{(1+x)^\frac12}{2(1+x)\sqrt{1-x}}-\frac{(1-x)^\frac12}{2(1+x)\sqrt{1+x}}$$
$$=-\frac{\sqrt{1+x}^2}{2(1+x)\sqrt{1-x}\sqrt{1+x}}-\frac{\sqrt{1-x}^2}{2(1+x)\sqrt{1+x}\sqrt{1-x}}$$
$$=\frac{-(1+x)-(1-x)}{2(1+x)\sqrt{1-x^2}}=\frac{-1}{(1+x)\sqrt{1-x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4542228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
How to solve this quadratic system of equations? The equations are:
$$ \begin{aligned} b + d &= c^2 - 6 \\ b - d &= -\frac{1}{c} \\ b d &= 6 \end{aligned} $$
and I want integer solutions for this.
I tried using various methods such as using $(a+b)^2-(a-b)^2=4ab$ or trying to solve for $b+d$ using the last two equations... | If $b, c, d$ are integers,
then $c$ can only be
$1$ or $-1$.
In either case,
$b+d=-5$.
From the last,
$(b,d)$ is one of
$(1,6), (2,3),
(3,2), (6,1)$,
or these replaced by their negatives.
The only ones that work are
$(-2,-3)$ and $(-3,-2)$.
If $c=1$, then $b-d=-1$ so
$b=-3$, $d=-2$.
If $c=-1$, then $b-d=1$ so
$b=-2$, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4543408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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How to use little-o notation to solve $\lim\limits_{x\to 0} \frac{\sin{(x^2)}-\sin^2{(x)}}{\sin{(x^2)}\sin^2{(x)}}$ with Taylor polynomials? Consider the limit
$$\lim\limits_{x\to 0} \frac{\sin{(x^2)}-\sin^2{(x)}}{\sin{(x^2)}\sin^2{(x)}}\tag{1}$$
Let me preface my question by saying that I am studying the Chapter of Sp... | *
*$o(x^4)$ means "any function $f(x)$ such that $\frac{f(x)}{x^4}\to0$" (here: when $x\to0$). So $o(x^4)-o(x^4)$ means "any difference of any two such functions" (which may be $\ne$), and such a difference is again some $o(x^4)$.
*My lazy writing was only to imitate the style of Spivak's solution you reported in you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4544105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find numbers divisible by 6 Find the number of all $n$, $1 \leq n \leq 25$ such that $n^2+15n+122$ is divisible by 6.
My attempt. We know that:
\begin{align*}
n^2+15n+122 & \equiv n^2+3n+2 \pmod{6}
\end{align*}
But $n^2+3n+2=(n+1)(n+2)$, then $n^2+15n+122 \equiv (n+1)(n+2) \pmod{6}$, now we have
\begin{align*}
n(n^2+15... | You were almost there with your first line. You have $n^2+3n+2=(n+1)(n+2)\text{ mod 6}$
$(n+1)$ and $(n+2)$ are two consecutive numbers so one of them is even. That gives you that this polynomial is divisible by $2$ for all $n$.
If $n$ is either congruent to $1$ or $2$ mod $3$ then $(n+2)$ or $(n+1)$, respectively, is ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to simplify $\ln |x-1| = 2\ln |\frac {y}{x}-1| - 3\ln |\frac{y}{x} - 2| + C; C = const$ to $(y-2x)^3 = C(y-x-1)^2$? How to simplify $\ln |x-1| = 2\ln |\frac {y}{x}-1| - 3\ln |\frac{y}{x} - 2| + C; C = const$ to $(y-2x)^3 = C(y-x-1)^2$?
I'm trying to solve $(2x - 4y + 6)dx + (x+y-3)dy = 0$. The two lines intersect a... | You first set, correctly, $u=\fracμτ=\frac{y-2}{x-1}$ but then in the transformation back you used $u=\frac yx$. This is not compatible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4547277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding solutions of $x$ for $(\sin x -1)(\sqrt{2} \cos x +1)=0$ in the given interval $0 \le x \le 2\pi$
Finding solutions of $x$ for $(\sin x -1)(\sqrt{2} \cos x +1)=0$ in the given interval $0 \le x \le 2\pi$
Firstly, $\sin x - 1=0 \implies x= \frac{\pi}{2}$
Next, $(\sqrt{2} \cos x +1)=0 \implies x= \frac{3\pi}{4}... | Your overlooked the fact that there are two values of $x \in [0, 2\pi]$ such that $\sqrt{2}\cos x + 1 = 0$.
As we can see from the sine graph, there is only one value of $x \in [0, 2\pi]$ such that $\sin x = 1$.
A number $x$ is a solution of the equation $\sqrt{2}\cos x + 1 = 0 \iff \cos x = -\frac{1}{\sqrt{2}} = -\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4547800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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find a closed form formula for $\sum_{k=1}^n \frac{1}{x_{2k}^2 - x_{2k-1}^2}$
Let $\{x\} = x-\lfloor x\rfloor$ be the fractional part of $x$. Order the (real) solutions to $\sqrt{\lfloor x\rfloor \lfloor x^3\rfloor} + \sqrt{\{x\}\{x^3\}} = x^2$ with $x\ge 1$ from smallest to largest by $x_1,x_2,\cdots$. Provide a clos... | Using Cauchy-Bunyakovsky-Schwarz inequality, we have
$\frac{\{x\}}{\lfloor x\rfloor}
= \frac{\{x^3\}}{\lfloor x^3\rfloor}$
or
$$\lfloor x^3\rfloor = x^2 \lfloor x\rfloor. \tag{1}$$
Clearly, $x=1, 2, \cdots$ are solutions of (1).
Let $k \in \mathbb{Z}_{>0}$.
Consider the solutions of (1) in $(k, k + 1)$ i.e. $k < x < k ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$ by partial fractions
Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$
I know that we can use substitution to make the expression simplier before solving it, but I am trying to solve this by using partial fractions only.
$\frac{x^2}{(x-3)(x+2)^2} = \frac{A}{x-3} + \frac{B}{... | There is another way of finding $A$, $B$ and $C$ which is more efficient than forming and solving simultaneous equations.
Starting with the identity $$x^2=A(x+2)^2+B(x-3)(x+2)+C(x-3)$$
Choose values of $x$ to make the brackets zero.
So, putting $x=2$ gives $$4=C(-5)\implies C=-\frac45$$
Putting $x=3$ gives $$9=A(25)\im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The equation of the line parallel to $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ intersecting the lines $9x+y+z+4=0 =5x+y+3z$ & $x+2y-3z-3=0=2x-5y+3z+3$ The equation of the line parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ intersecting the lines $9x + y + z + 4 = 0 = 5x + y + 3z$ & $x + 2y - 3z - 3 = 0 = 2x - 5y ... | Answer D since none of the lines in A, B, C is parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}.$ Indeed, solving the pairs of associated homogeneous equations, we find that:
*
*(A) is directed by the vector $(-1,1,3)$
*(B) by $(1,1,-3)$
*(C) by $(1,0,-1).$
If the question was more open, I would have solved it ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4558855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$S_{n}$ is the sum of every third element in the $n$th row of the Pascal triangle, beginning on the left with the second element. Find $S_{100}$. Problem:
Let $S_{n}$ be the sum of every third element in the $n$th row of the Pascal triangle, beginning on the left with the second element. Find the value of $S_{100}$.
... | Let $\omega =\exp(2\pi i/3)$ be the primitive cube root of unity. Because $$\frac{1+\omega^k+\omega^{2k}}{3}=\begin{cases}1&\text{if $3 \mid k$}\\0&\text{otherwise}\end{cases}$$
we have $$\sum_k a_{3k} = \sum_k a_k \frac{1+\omega^k+\omega^{2k}}{3}.$$
Now take $$a_k=\binom{n}{k+1}$$ and apply the binomial theorem to ea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4559818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Whose probability of winning is higher? There are $10$ blue marbles and $5$ red marbles in a black bag. Two players $A$ and $B$ take turns taking non-refundable (without replacement) $1$ marble each time (A takes first). The game ends when someone gets a red ball and that person loses. Whose probability of winning... | You can calculate the winning probability of the person to move recursively as a function of the number of blue balls in the bag, since as long as the game is still going, the number of red balls will always be $\ 5\ $. If $\ w_n\ $ is the probability that the person to draw will win when there are $\ n\ $ blue balls ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Number of tangential quadrilaterals with distinct sidelengths chosen in set $\{1,2,\ldots, 8\}$
There are $8$ line segments of length $1,2,3,\ldots,8$ $\textrm{units}$. How many quadrilaterals can be made from these line segments such that circles can be inscribed in the quadrilaterals made$?$
I applied Pitot theore... | A good way to proceed is to take the twenty-eight possible pairs of side lengths and classify them according to the sum of the lengths:
$\{1,2\}$
$\{1,3\}$
$\{1,4\},\{2,3\}$
$\{1,5\},\{2,4\}$
$\{1,6\},\{2,5\},\{3,4\}$
$\{1,7\},\{2,6\},\{3,5\}$
$\{1,8\},\{2,7\},\{3,6\},\{4,5\}$
$\{2,8\},\{3,7\},\{4,6\}$
$\{3,8\},\{4,7\}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4561145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Evaluate $\int_0^1 \cos^{-1} x\ dx$ by first finding $\frac{d}{dx}(x\cos^{-1} x)$ Question
Evaluate $$\int_0^1 \cos^{-1} x\ dx$$ by first finding the value of $$\frac{d}{dx}(x\cos^{-1} x).$$
My Working
As the question said to evaluate $$\frac{d}{dx}(x\cos^{-1} x),$$ I used the product rule to differentiate. We first ha... | After seeing the helpful comments, I can start using $u$-substitution for the solution. Let $u=1-x^2$, then
\begin{align}
\frac{du}{dx}&=-2x\\
du&=-2x\ dx\\
\int_0^1\frac{x}{\sqrt{1-x^2}}&=-\frac{1}{2}\int^1_0\frac{-2x}{\sqrt{1-x^2}}\ dx\\
&=-\frac{1}{2}\int^1_0 \frac{1}{\sqrt{u}}\ du\\
&=-\frac{1}{2}\int^1_0 u^{-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4563516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Prove that $\sum_{y=0}^{p-1}\left(\frac{y}{p}\right)\left(\frac{y+d}{p}\right)=-1$ The question is from The number of solutions of $ax^2+by^2\equiv 1\pmod{p}$ is $ p-\left(\frac{-ab}{p}\right)$
And I wonder how to prove the following equation although someone gives the trick:
\begin{split}
\sum_{y=0}^{p-1}\left(\frac{y... | Here is my answer:
We can replace $\left(\frac{y}{p}\right)$ with $\left(\frac{y^*}{p}\right)$ such that:
\begin{split}
\left(\frac{y}{p}\right) = \left(\frac{y^*}{p}\right).
\end{split}
We can easily verify it by using primitive root.
When $y = 0$ there is no inverse, so we ignore the first term of $\sum_{y=0}^{p-1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4564096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If integers $s$, $x$, $y$, $z$ are such that $s=5xy+5yz+5xz=2x^2+2y^2+2z^2$, then $10s$ is a perfect square I'm having a problem with a question:
Consider integer $s$ and integers $x,y,z$ such that
$$s = 5xy+5yz+5xz \quad\text{and}\quad s= 2x^{2} + 2y^{2} + 2z^{2}$$ Prove that $10s$ is a perfect square.
I'm pretty su... | $(xy+yz+xz)=\frac 15 s$ and $x^2 + y^2 + z^2 = \frac 12s$
Now $(x^2 + y^2 + z^2) + 2(xy + yz+xy)= (x+y+z)^2$
so $\frac 12 s + \frac 25 s = \frac 9{10}s=(x+y+z)^2$
So $10s = \frac {100}9(x+y+z)^2 = (\frac {10}3(x+y+z))^2$.
Dang.
So close!
But if we can prove $x+y+z$ is divisible by $3$ we will be okay. Can we?
Well, i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4567737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Variable substitution with 3 variables I'm trying to determine the following triple integral
$$
\iiint_{K}\left[\left(x - a\right)^{2} +
\left(y - b\right)^{2} + \left(z - c\right)^2\right]{\rm d}x\,{\rm d}y\,{\rm d}z,\quad
K = \left\{\left(x,y,z\right): x^{2} + y^{2} + z^{2}
\leq 1\right\}.
$$
I tried the following va... | By using your change of coordinates and the given limits you are not evaluating the given integral, but $\iiint_K (x^2+y^2+z^2) dxdydz$.
The correct limits are different and not so easy to find.
On the other hand, by expanding the squares, we split the integral into three terms,
$$\begin{align}
I&=\iiint_K ((x-a)^2+(y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4569572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Finding $x+y$, given $xy= 1$, $x^2+y^2=5$, $x^3+y^3=8$ This problem is from a math competition, but I think is wrong:
Find the value of $x+y$ if:
$$\begin{align}
xy &= 1 \\
x^2 + y^2 &= 5 \\
x^3+y^3 &= 8
\end{align}$$
Solution (I think is wrong):
$x^3 + y^3 = (x + y)(x^2-xy+y^2) = (x+y)(5-1) = 4(x+y)$
So we have:
$... | We can also take a geometric interpretation of the problem. The curves corresponding to the equations
$$ xy \ = \ 1 \ \ \ , \ \ \ x^2 \ + \ y^2 \ = \ 5 \ \ \ , \ \ \ x^3 \ + \ y^3 \ = \ 8 $$
all have symmetry about the line $ \ y \ = \ x \ $ (only the first two, however, has symmetry about the origin). This su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4571660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Find for what values of a and b in R the limit exists (No De L'Hopital) I was given this exercise in my math course at university.
The question is to find, without using De l'Hopitals and other methods which may use derivates and similars, for what values of $a$ and $b$ in $\mathbb{R}$ the following statement is true.
... | Assuming the limit exists:
\begin{equation}
\lim_{x \to 0^+} \dfrac{\dfrac{\sin x}{x} -ax - b}{x} = \lim_{x \to 0^+} \dfrac{\sin x -ax^2 - bx}{x^2}=\lim_{x \to 0^+} \left(\dfrac{\sin x- bx}{x^2}-a\right)
\end{equation}
Now consider:
\begin{equation}
\lim_{x \to 0^+} \dfrac{\sin x- bx}{x^2}
\end{equation}
If $b\neq1$:
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4576690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Determining probability of a draw in simple coin flipping game Consider a 2-player game where each player flips a coin every round. If a player flips heads, they get a point for that round. Either both players get a point, only one, or none. The game ends when either of the players get to 10 points. If the game ends wi... | Here is a semi-solution, for general $ k $. I try and compute for $ n = 3 $. Unfortunately, it looks like I have a different value. Of course, there may be some calculation mistake. For $ k = 10$, you will need to do some calculations to finish it off.
The number of coin tosses until you get $ k $ heads follow a nega... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4577755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Are there any other decent methods to evaluate $\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x?$ We first split the integrand into 3 parts as
\begin{aligned}
\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x &= \underbrace{\int_0^1 \frac{\ln \left(1+x^2\right)}{1+x^2} d x}_J+\underbrace{\int_0^1 \frac{\ln (1+x)}{1+x... | Substitute $x=\tan t$. Then
$$1-x^4=4x^2\frac{1-x^2}{1+x^2}\left(\frac{1+x^2}{2x}\right)^2
=4\tan^2t\ \frac{\cos 2t}{\sin^2 2t}
$$
and
\begin{aligned}
&\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x \\
=& \int_0^{\pi/4}\ln8+2\ln \tan t +\ln (2\cos 2t)
-2\ln (2\sin 2t) \ d t\\
=&\ \frac{3\pi}4\ln2 - 2G
\end{aligned}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Linear approximation of $\sqrt[7]{e}$ I need to find a linear approximation $\sqrt[7]{e}$. I know that for $x_0 = a + \Delta x$
$$
f(x_0) \approx f(a)+f'(a)\cdot \Delta x
$$
Thus my inital function is $f(x) = \sqrt[7]{x}$ and $e = 1 + (e - 1)$, so
$$
f(e) \approx \sqrt[7]{1} + \frac{1}{7}1^{\frac{-6}{7}}\cdot(e - 1) = ... | $[k,k]$ Padé approximants $P_k$ of $e^{\frac{1}{n}}$ are not bad
$$P_1=\frac{2 n+1}{2 n-1}\qquad P_2=\frac{12 n^2+6 n+1}{12 n^2-6 n+1} \qquad P_3=\frac{120 n^3+60 n^2+12 n+1}{120 n^3-60 n^2+12 n-1}$$ and so on are more then decent. For $n=7$ they generate
$$\left(
\begin{array}{ccc}
k & P_k & \log_{10} \left(\left| P_k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4583585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that the equation $X^2=J_n$has no solution if $n\geq2$, if $J_n$ is the jordan block of size $n$ with zeros on diagonal. More clearly:
$$
J_n=
\begin{pmatrix}
0 & 1 & 0 &\cdots & 0 & 0 \\
0 & 0 & 1 &\cdots & 0 & 0 \\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0 & 0 & 0 &\... | If $X^2$ has rank $n-1$, then $n-1\ge\operatorname{rank}(X)\ge\operatorname{rank}(X^2)=n-1$. Therefore $\dim XV=\dim X^2V=n-1$, where $V=\mathbb F^n$. However, as $X^2V=X(XV)\subseteq X(V)=XV$, we must have $X^2V=XV$. It follows that $X^kV=XV\ne0$ for every integer $k>0$, which is a contradiction to the assumption that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4587913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show that scalene triangle $\triangle ABC$ is a right-triangle if $\sin(A)\cos(A)=\sin(B)\cos(B)$ As the title suggests, this is a college entrance exam practice problem from Japan, it is as follows:
Given a scalene triangle $\triangle ABC$, prove that it is a right triangle if $\sin(A)\cos(A)=\sin(B)\cos(B)$
I found... | Your method is correct.
If you can use trigonometric identities, proceed as follows:
$2\sin(A)\cos(A)=2\sin(B)\cos(B)$
$\sin(2A)=\sin(2B)$
Thus $2A$ and $2B$ are either equal or supplementary, so correspondingly $A$ and $B$ are equal or complementary. But equality of angles implies the triangle is isosceles, a contradi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4588868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int_0^1\arcsin^2(\frac{\sqrt{-x}}{2}) (\log^3 x) (\frac{8}{1+x}+\frac{1}{x}) \, dx$ Here is an interesting integral, which is equivalent to the title
$$\tag{1}\int_0^1 \log ^2\left(\sqrt{\frac{x}{4}+1}-\sqrt{\frac{x}{4}}\right) (\log ^3x) \left(\frac{8}{1+x}+\frac{1}{x}\right) \, dx = \frac{5 \pi ^6}{1134}-\... | Too long for comment: Let
$$I_1 = \int_0^1\log^2\left(\sqrt{\frac x4+1}+\sqrt{\frac x4}\right) \log^3(x) \, \frac{dx}x \\
I_2 = \int_0^1 \log^2\left(\sqrt{\frac x4+1}+\sqrt{\frac x4}\right) \log^3(x) \, \frac{dx}{x+1}$$
so the desired integral is $I=I_1+8I_2$.
Taking a cue from Ali Shadhar's solution to a related probl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4592985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 1,
"answer_id": 0
} |
Prove that $1^3 + 2^3 + ... + 2004^3$ is divisible by $2005$. This is a first year high school problem. There's no series or any other more "advanced" math involved. There should just be a way to factor $2005$ out of the sum.
Judging by similar problems I assume that it is, in general, true that, for a natural number $... | Here's maybe a more 9th-grade-friendly approach, assuming you know how to factorise a sum of two cubes:
$\begin{eqnarray} S & = & 1^3 + 2^3 + \ldots + 2003^3 + 2004^3 \\
& = & (1^3 + 2004^3) + (2^3 + 2003^3) + \ldots + (1002^3 + 1003^3) \\
& = & (1 + 2004)(1^2 + 1 \times 2004 + 2004^2) + (2 + 2003)(2^2 + 2 \times 2003 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4593128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ I would like to solve the equation $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ analytically, without using Wolfram Alpha. I have tried several substitutions, including $x=\cos(t)$, $\sqrt{1-x^2}=t$, but they all fail and eventually result in a quartic polynomial which i... | Observe $\sqrt {1-x^2}≥0$ implies that, $x≥0.$
Thus, under the restriction $0≤x≤\frac {\sqrt 2}{2}$ we have:
$$\begin{align}x\sqrt{1-x^2}&=\sqrt{1-x^2}-x\\
\left(x\sqrt {1-x^2}\right)^2&=1-2x\sqrt {1-x^2}\end{align}$$
Letting $x\sqrt {1-x^2}=u$, we have:
$$\begin{align}&u^2=1-2u\\
\implies &u^2+2u-1=0\end{align}$$
Thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4593991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Integration of Gauss ${}_{2}F_{1}$ hypergeometric function The indefinite integral representation of Gauss hypergeometric function is
$$\int {{}_2{F_1}\left( {a,b;c;z} \right)dz = \frac{{c - 1}}{{\left( {a - 1} \right)\left( {b - 1} \right)}}} {}_2{F_1}\left( {a - 1,b - 1;c - 1;z} \right)$$
However I don't understand h... | Consider the integral in the form
$$ I = \int_{-\infty}^{\infty} (1 + p \, t^2) \, {}_{2}F_{1}\left(a, b; c; \frac{q^2 \, t^2}{4 \, \alpha \, (1 + p \, t^2)} \right) \, dt. $$
Converting the ${}_{2}F_{1}$ into its series form then
$$ I = \sum_{n=0}^{\infty} \frac{(a)_{n} \, (b)_{n}}{n! \, (c)_{n}} \, \left(\frac{q^2}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
In $\triangle ABC$, if $AC=4$ and $BC=5$, and $\cos(A-B)=\frac{7}{8}$, find $\cos(C)$ The problem is as the title suggests, in the given figure below, the goal is to find the Cosine of $\angle C$. I tried multiple ways of approaching this, such as with the Law of Sines, area formula, etc but none of them seemed to lead... | Here is another way to solve for $\cos(C)$, by area calculations, done in 2 ways.
From Bob Dobb' answer, we have:
$\displaystyle \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R = \sqrt{\frac{128}{5}}$
$\displaystyle Δ = \frac{a\,b\,c}{4\,R} \qquad \qquad \quad
→ Δ^2 = \frac{125}{32}\,c^2$
Using my own trian... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Computing Denominator of Bayes theorem - independently vs conditionally Lets say that you have n independent variables (assume n is even) $x_1$ to $x_n$,
Let $X = [x_1, x_2, x_3, ...., x_n]$
We have another variable $y$ that can only take two values $0$ and $1$ such that
$P(y=0)=P(y=1)=0.5$,
Given that $P(x_i=1|y=1) = ... | You're asking which of two options you should pick, but I think both of those ideas will be important here!
For convenience I'll assume $n$ is even. If it's odd you need to handle the end of the products carefully since $X$ will have more $1$s than $0$s.
First, for any $k$ we have
$$
\begin{align}
P(x_k = 0) &= P(x_k =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove $\sum_{n=1}^{+\infty}\dfrac{1}{(n+1)\sqrt[e]{n}}$ converges to a number $I can prove that $\sum_{n=1}^{+\infty}\dfrac{1}{(n+1)\sqrt[e]{n}}$ converges, but don't know how to transform the $\sqrt[e]{n}$ item to prove it less than $e$, could someone help/hint me?
| We use $\int x^{-1-1/e} dx = -ex^{-1/e} + C$.
$$\sum_{n=1}^{\infty} \frac{(1+1/n)^{1/e}}{(n+1)^{1+1/e}} = \frac{1}{2} + \frac{1}{3 \times 2^{1/e}} + \frac{1}{4 \times 3^{1/e}} + \sum_{n=4}^{\infty} \frac{(1+1/n)^{1/e}}{(n+1)^{1+1/e}} $$
$$ \le \frac{1}{2} + \frac{1}{3 \times 2^{1/e}} + \frac{1}{4 \times 3^{1/e}} + (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.