Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Examine the continuity of function $f(x)=\frac{2x^2-4x}{|x+1|+|x-3|-2}$ Using the definition of absolute value for
$$|x+1|=\begin{cases}
x+1, & x\ge -1\\
-x-1, & x>-1
\end{cases}$$
and
$$|x-3|=\begin{cases}
x-3, & x\ge 3\\
-x+3, & x>3
\end{cases}$$
and checking cases when $$x<0,x=0,x>0$$
I get this definition of a func... | $$|x+1|=\begin{cases}
x+1& x\ge -1\\
-x-1, & x\color{red}<-1
\end{cases}$$
and
$$|x-3|=\begin{cases}
x-3, & x\ge 3\\
-x+3, & x\color{red}<3
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
integer solution of $(x-y)(x+y)xy=z^2$ By wolfram alpha, integer solution of $(x-y)(x+y)xy=z^2$ is $x=y=z=0$.
How to show that there are not another solutions with $z \neq 0$.
Thanks.
| If $\gcd(a,b) = d$, and $(x,y,z) = (a,b,c)$ is a nontrivial solution, then $(\frac{a}{d}, \frac{b}{d},\frac{c}{d^2})$ is also a solution, where $\gcd(\frac{a}{d},\frac{b}{d}) = 1$. So if there is no solution $(a,b,c)$ having $\gcd(a,b) = 1$, then there are no nontrivial solutions.
Assume $\gcd(a,b) = 1$. We want to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How to simplify a diabolical expression involving radicals A friend and I have been working on this problem for hours - how can the following expression be simplified analytically?
It equals $\frac{1}{2},$ and we have tried the following to no avail:
*
*Substitution of $x = \sqrt{5}$
*Substitution of $x = 2\sqrt{5}... | Hint $\ $ Let $\ a = \sqrt{5+2\sqrt 5},\ b = \sqrt{10+2\sqrt 5}.\,$ Show $\,\color{#c00}{b = (\sqrt5 -1) a},\,$ so scaling the top and bottom of the fraction by $\,8\,$ yields $\ \dfrac{4a+ 2\sqrt 5 a - \color{#c00}{(\sqrt5 -1) a}}{10a + 2\sqrt 5 a}^{\phantom{1^{1^1}}}\!\!\!\!\!\! =\, \dfrac{1}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Derivative of Function with Rational Exponents $f(x)= \sqrt[3]{2x^3-5x^2+x}$ I have a question following:
$$f(x)=\sqrt[3]{2x^3-5x^2+x}$$
Here's what I did,
$$f(x)=\sqrt[3]{2x^3-5x^2+x}
\\ = (2x^3-5x^2+x)^{3\over2}
\\\\f'(x) = {3\over 2}(2x^3-5x^2+x)^{3\over2}(6x^2-10x+1)$$
Did I do this correctly?? Because I have diffe... | No, you didn't. We can write $\sqrt[3]{t}=t^{1/3}$, not $t^{3/2}$. So
$$
f(x)=(2x^3-5x^2+x)^{1/3}
$$
and
$$
f'(x)=\frac{1}{3}(2x^3-5x^2+x)^{-2/3}(6x^2-10x+1)=
\frac{1}{3}\frac{6x^2-10x+1}{\sqrt[3]{(2x^3-5x^2+x)^2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
$x^2-y^2=196$, can we find the value of $x^2+y^2$? $x$ and $y$ are positive integers.
If $x^2-y^2=196$, can we know what the value of $x^2+y^2$ is?
Can anyone explain this to me? Thanks in advance.
| Hint 1: $x^2-y^2 = (x+y)(x-y)$.
Hint 2: Divisors of $196$: $1$, $2$, $4$, $7$, $14$, $28$, $49$, $98$, $196$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find the number of integer solutons to $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 60$. Find the number of integer solutons to $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 60$ if $x_1 \geq2, x_2\geq 5, 2\leq x_3\leq 7, x_4\geq 1, x_5\geq 3, x_6\geq 2$.
Here is my approach: If you let $M = \{\infty x_1, \infty x_2, \infty x_3, \infty... | $$S(x)=\underbrace{(x^2+x^3+\cdots)}_{x_1}\underbrace{(x^5+x^6+\cdots)}_{x_2}\underbrace{(x^2+x^3+\cdots+x^7)}_{x_3}\underbrace{(x+x^2+\cdots)}_{x_4}\\\times\underbrace{(x^3+x^4+\cdots)}_{x_5}\underbrace{{(x^2+x^3+\cdots)}}_{x_6}\\
={x^2\over1-x}{x^5\over1-x}{x^2(1-x^6)\over 1-x}{x\over1-x}{x^3\over1-x}{x^2\over1-x}={x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Easiest way to calculate the determinant of this 4x4 matrix I have this 4x4 matrix:
$$A=
\begin{pmatrix}
2 & 3 & 1 & 0 \\
4 & -2 & 0 & -3\\
8 & -1 & 2 & 1\\
1 & 0 & 3 & 4\\
\end{pmatrix}
$$
I read that it's easy to calculate it by converting the matrix to upper diagonal. ... | As suggested in the comments, Gauss elimination is usually the way to go, and the fastest in this case, too:
$$\det A=
\det\begin{pmatrix}
2 & 3 & 1 & 0 \\
4 & -2 & 0 & -3\\
8 & -1 & 2 & 1\\
1 & 0 & 3 & 4\\
\end{pmatrix}
=
\det\begin{pmatrix}
0 & 3 & -5 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\int \frac { \sin x+\cos x }{ \sin^4 x+\cos^4x}\, dx$? How can one find $$\int \frac { \sin x+\cos x }{ \sin^4 x+\cos^4x}\, dx?$$
| Let
\begin{align}
I &= \int {\frac{{\sin x + \cos x}}{{\sin ^4 x + \cos ^4 x}}dx} \\
&= \int {\frac{{\sin x}}{{\sin ^4 x + \cos ^4 x}}dx} + \int {\frac{{\cos x}}{{\sin ^4 x + \cos ^4 x}}dx} \\
&= \int {\frac{{\sin x}}{{\sin ^4 x + \cos ^4 x}}dx} + \int {\frac{{\sin \left( {{\textstyle{\pi \over 2}} - x} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the number of equations having real roots.
If both $a$ and $b$ belong to the set $\{1,2,3,4\}$ ,
then number of equations of the form $ ax^2+bx+1=0$ having real roots is
$a.)\ 10\\
\color{green}{b.)\ 7}\\
c.)\ 6\\
d.)\ 12\\ $
To solve this I had to make a table and check each of the $16$ cases.
$$\begin{array}... | For real roots,
$b^2\ge{4a}$
Put $a=1$.
$b^2\ge4$
$b$ can be $\{2,3,4\}$.
Now put $a=2$
$b$ can take two values.
Put $a=3$, $b$ can take one value and finally put $a=4$. Here $b$ can take one value.
Hence there are $7$ ordered pairs $(a,b)$ so there are $7$ such equations
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Nesbitt's Inequality for 4 Variables I'm reading Pham Kim Hung's 'Secrets in Inequalities - Volume 1', and I have to say from the first few examples, that it is not a very good book. Definitely not beginner friendly.
Anyway, it is proven by the author, that for four variables $a, b, c$, and $d$, each being a non-negati... | You can indeed use AM-GM, as the book says, to derive the result. Admittedly, it would have been more transparent had the book added one line of derivation as shown below.
For positive $x$ and $y$,
$$\frac{x+y}2\,\frac12\Big(\frac1x+\frac1y\Big)\ge\sqrt{xy}\sqrt{\frac1{xy}}=1,$$
thus
$$\frac1x+\frac1y\ge \frac4{x+y}.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
$1+\sqrt[3]{e^{2a}}\sqrt[5]{e^{b}}\sqrt[15]{e^{2c}} \leq \sqrt[3]{(1+e^{a})^2}\sqrt[5]{1+e^{b}}\sqrt[15]{(1+e^{c})^2}$ The inequality $1+\sqrt[3]{e^{2a}}\sqrt[5]{e^{b}}\sqrt[15]{e^{2c}} \leq \sqrt[3]{(1+e^{a})^2}\sqrt[5]{1+e^{b}}\sqrt[15]{(1+e^{c})^2}$ is true for all $a,b,c\in\mathbb{R}$?
I've tried to use the Bernoul... | Hint: use Holder inequality we have
$$(1+e^a)^{10}\cdot (1+e^b)^3\cdot(1+e^c)^{2}\ge (1+\sqrt[15]{e^{10a}\cdot e^{3b}\cdot e^{2c}})^{15}$$
I add $n=3$ Holder inequality proof,ie:
$$(a^3+b^3+c^3)(p^3+q^3+r^3)(u^3+v^3+w^3)\ge (apu+bqv+crw)^3,a,b,c,p,q,r,u,v,w>0$$
since
$$\sum_{cyc}\left(\dfrac{a^3}{a^3+b^3+c^3}+\dfrac{p^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Limit involving Zeta and Gamma function Can someone help me evaluate this limit?
$$\lim_{x\to +\infty}\frac {\zeta(1+\frac 1x)}{\Gamma(x)}$$
I never came across this kind of limit so I don't even know where to start.
| Hint: Note that for $x>1$, we have $$\zeta \left(1+\frac{1}{x}\right) = \frac{1}{1^{1+\frac{1}{x}}}+\frac{1}{2^{1+\frac{1}{x}}}+\frac{1}{3^{1+\frac{1}{x}}}+\frac{1}{4^{1+\frac{1}{x}}}+\frac{1}{5^{1+\frac{1}{x}}}+\cdots< \\ \frac{1}{1^{1+\frac{1}{x}}}+\frac{1}{2^{1+\frac{1}{x}}}+\frac{1}{2^{1+\frac{1}{x}}}+\frac{1}{4^{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1335165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Why does $\cos(x) + \cos(y) - \cos(x + y) = 0$ look like an ellipse? The solution set of $\cos(x) + \cos(y) - \cos(x + y) = 0$ looks like an ellipse. Is it actually an ellipse, and if so, is there a way of writing down its equation (without any trig functions)?
What motivates this is the following example. The solution... | @Mr Spock: You can go from $\cos(x)+\cos(y)=\cos(x+y)$ to a rational function (you ask about it!) via identities $$\tan (\frac x2)=t$$$$\tan (\frac y2)=s$$ $$\tan (\frac {x+y}{2})=\frac{t+s}{1-ts}$$ from which you have
$$\cos (x)=\frac{1-t^2}{1+t^2}$$ $$\cos (y)=\frac{1-s^2}{1+s^2}$$
Since $$\tan (\frac{x+y}{2})=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 9,
"answer_id": 4
} |
Find the sum of first $20$ terms of a sequence Define a sequence $$a_n=\sqrt{1+\left(1-\frac{1}{n}\right)^2}+\sqrt{1+\left(1+\frac{1}{n}\right)^2}$$ for $n \geq 1$. Find $$\sum_{i=1}^{20}\frac{1}{a_i}$$ Some insight on the approach is highly appreciated. What is the general way of solving such problems? Thanks.
| First you calculate $\frac{1}{a_i}$ and write it a bit differently:
$\frac{1}{a_n} =
\frac{\sqrt{1+(1-\frac{1}{n})^2}-\sqrt{1+(1+\frac{1}{n})^2}}{(1+(1-\frac{1}{n})^2)-(1+(1+\frac{1}{n})^2)}
=\frac{\sqrt{1+(1-\frac{1}{n})^2}-\sqrt{1+(1+\frac{1}{n})^2}}{-\frac{4}{n}}
=\frac{1}{4} \left ( \sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Two different trigonometric identities giving two different solutions Using two different sum-difference trigonometric identities gives two different results in a task where the choice of identity seemed unimportant. The task goes as following:
Given $\cos 2x =-\frac {63}{65} $ , $\cos y=\frac {7} {\sqrt{130}}$, unde... | Simply: Solving for an angle using its sine value is ambiguous. (This is one reason that the Law of Cosines is preferred over the Law of Sines for determining angles from sides.)
Here's an analogous situation:
$$\begin{align}x^2 + 2 x = 35 \quad(1)\\
x^2 - 2 x = 15 \quad(2)
\end{align}$$
Adding $(1)+(2)$, we see that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Partial fraction of $\frac 1{x^6+1}$ Can someone please help me find the partial fraction of $$1\over{x^6+1}$$
?
I know the general method of how to find the partial fraction of functions but this seems a special case to me..
| We have
$$\begin{align}x^6+1&=(x^2)^3+1\\&=(x^2+1)(x^4-x^2+1)\\&=(x^2+1)(x^4+2x^2+1-3x^2)\\&=(x^2+1)((x^2+1)^2-(\sqrt 3\ x)^2)\\&=(x^2+1)(x^2-\sqrt 3\ x+1)(x^2+\sqrt 3\ x+1)\end{align}$$
Now setting $$\frac{1}{x^6+1}=\frac{a}{x^2+1}+\frac{bx^2+c}{x^4-x^2+1}$$
will give you $a=\frac 13,b=-\frac 13,c=\frac 23.$
Then, set... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
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Roots addition.
$$\sqrt{\frac{a+x^2}{x}-2\sqrt{a}}+\sqrt{\frac{a+x^2}{x}+2\sqrt{a}}=Q $$
One is expected to find $Q$ respecting $a>0$, $x>\sqrt{a}$ .
I'd like to have my solution checked; namely the correct answer is $2\sqrt{x}$ but I simply fail to see where I made a mistake.
And it'd be nice to hear if there is... | Set $\sqrt{a}=b$ so the first summand is
$$
\sqrt{\frac{b^2+x^2}{x}-2b}=
\sqrt{\frac{(x-b)^2}{x}}=\frac{x-b}{\sqrt{x}}
$$
because, by assumption, $x>b$ (and so also positive).
Similarly
$$
\sqrt{\frac{b^2+x^2}{x}+2b}=
\sqrt{\frac{(x+b)^2}{x}}=\frac{x+b}{\sqrt{x}}
$$
Thus your expression is
$$
Q=\frac{x-b}{\sqrt{x}}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Proving that $\frac{1}{a^2}+\frac{1}{b^2} \geq \frac{8}{(a+b)^2}$ for $a,b>0$ I found something that I'm not quite sure about when trying to prove this inequality.
I've proven that
$$\dfrac{1}{a}+\dfrac{1}{b}\geq \dfrac{4}{a+b}$$
already. My idea now is to replace $a$ with $a^2$ and $b^2$, so we now have
$$\dfrac{1}{... | The problem is simple enough that you can just be mechanical about it
$$
\frac{1}{a^2}+\frac{1}{b^2}\geq\frac{8}{(a+b)^2}\iff(a^2+b^2)(a+b)^2-8a^2b^2\geq0.
$$
The claim now follows because
$$
(a^2+b^2)(a+b)^2-8a^2b^2=(a-b)^2(a^2+4ab+b^2)\geq0,\quad a,b>0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
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Evaluate the indefinite integral $\int \frac{\cos \theta}{ \sqrt{2 - 9 \sin^2 \!\theta}} \mathrm{d}\theta$ I want to evaluate $$\int \dfrac{\cos \theta \, \mathrm{d}\theta}{ \sqrt{2 - 9 \sin^2 \theta}}$$
but I can't seem to get the answer, my working is as below:
| Hint
$$\int\dfrac{\cos{\theta}}{\sqrt{2-9\sin^2{\theta}}}d\theta=\int\dfrac{d\sin{\theta}}{\sqrt{2-9\sin^2{\theta}}}$$
and
$$\int\dfrac{1}{\sqrt{2-9t^2}}dt=\dfrac{1}{\sqrt{2}}\cdot\int\dfrac{1}{\sqrt{1-\left(\frac{3}{\sqrt{2}}t\right)^2}}dt=\dfrac{1}{3}\cdot\int\dfrac{1}{\sqrt{1-\left(\frac{3}{\sqrt{2}}t\right)^2}}d\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find remainder when $777^{777}$ is divided by $16$
Find remainder when $777^{777}$ is divided by $16$.
$777=48\times 16+9$. Then $777\equiv 9 \pmod{16}$.
Also by Fermat's theorem, $777^{16-1}\equiv 1 \pmod{16}$ i.e $777^{15}\equiv 1 \pmod{16}$.
Also $777=51\times 15+4$. Therefore,
$777^{777}=777^{51\times 15+4}={(7... | Alternative to other answers: Note that $9^2=81\equiv 1\pmod{16}$, so $777^{777}\equiv 9^{777}=9\cdot(9^2)^{388}\equiv 9\pmod{16}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Expand $\binom{xy}{n}$ in terms of $\binom{x}{k}$'s and $\binom{y}{k}$'s Motivated by this question, I want to find a complete set of relations for the ring of integer-valued polynomials, where the generators are the polynomials $\binom{x}{n}$ for $n\in \mathbb{N}$. The best way to do this is would be to describe how t... | There's another formula for the expression in equation (3) on the bottom of page 183 of my paper here: http://www.sciencedirect.com/science/article/pii/S0022404905002161
The formula is ${{xy} \choose n} = \underset{l_1 + 2 l_2 + \cdots + nl_n = n}{\sum} {{\sum_{i=1}^n l_i} \choose {l_1, l_2, \ldots, l_n}} {y \choose ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\frac{8}{5}\le 2a+b\le 8$ Let $a,b,c,d,e$ be real numbers such that
$$\begin{cases}
a+b+c+d+e=8\\
a^2+b^2+c^2+d^2+e^2=16
\end{cases}$$
Prove that:
$$\dfrac{8}{5}\le 2a+b\le 8$$
| Use Cauchy-Schwarz inequality we have
$$3(c^2+d^2+e^2)\ge (c+d+e)^2\Longrightarrow 48-3a^2-3b^2\ge (8-a-b)^2$$
then let $2a+b=t$, you have
$$8a^2+(8-7t)a+2t^2-8t+8\le 0$$
$$\Longrightarrow \Delta_{a}=(8-7t)^2-32(2t^2-8t+8)\ge 0\Longrightarrow \dfrac{8}{5}\le t\le 8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A general method for integration of rational function. $\int\frac {x^3}{1+x^5}$
ATTEMPT:
I did the following substitution:
Let $x=\frac{1}{t}.$
$dx=\frac{-1}{t^2}dt.$
substituting back:
$I=\int\frac{-1}{1+t^5}dt$ which doesn't seems a simpler integration.
Next i substituted $x=p^\frac{2}{5}.$
$dx=\frac{2}{5}\frac{p^... | What you already made is $$I=\int\frac {x^3}{1+x^5}dx=-\int\frac {dt}{1+t^5}$$ Now $$1+t^5=\prod_{i=1}^5 (t-\alpha_i)$$ where $\alpha_i=-(-1)^{\frac i5}$ which means that, using partial fraction decomposition, $$\frac {1}{1+t^5}=\sum_{i=1}^5 \frac{\beta_i}{t-\alpha_i}$$ $$\int\frac {dt}{1+t^5}=\sum_{i=1}^5 \beta_i \log... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Substitutions that transform Fermat Equations to Elliptic Curves I was reading Chapter 1 of Elliptic Curves - Number Theory and Cryptography by Lawrence C Washington. He was considering Fermat equations $$a^4+b^4=c^4\text{ and }a^3+b^3=c^3.$$ For the 1st equation, let $$x = 2\frac{b^2+c^2}{a^2},\quad y=\frac{4b(b^2+c^2... | @Zilin J.:I have a method that sends all binomial addition $a+b=c$ (so, for example, $a^4+b^4=c^4$) to an elliptical curve, noted $V_A$, of the shape $$X^3+Y^3=AZ^3$$, where $A$ is a cube-free number, and this curve is birational equivalent to a Wierstrass one, the curve $$y^2z=x^3-432A^2z^2$$ (one can obviously work w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How to solve the integral $\int\tan^{3}x \sec^{3/2}x\; dx$? How to solve the following indefinite integral
$$\int \tan^{3}x \sec^{3/2}x \; dx$$
to get the solution in the form of
$$\large\frac{2}{7}\sec^{7/2}x - \frac{2}{3}\sec^{3/2}x +c$$
I tried taking
$$u = \sec^{2}x \implies du = \tan x \; dx$$
$$\large \int \ta... | While the OP is interested in understanding the source of an error from a specific substitution, I thought that it would be of some interest to show another way to evaluate this indefinite integral.
To that end, we can simply write
$$\tan^3 x\sec^{3/2}x=(1-\cos^2x)\sin x\cos^{-9/2}x$$
so that we have
$$\begin{align}
\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1347207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Number of elements in discrete $n$-dimensional simplex such that $x_1 \leq \ldots \leq x_n$ For positive integers $n,d$, how many elements are there in the set $S = \{(x_1,\ldots,x_n) \in \mathbb{Z}^n\ |\ 0 \leq x_1 \leq \ldots \leq x_n \wedge \sum_i x_i = d \}$?
I'm hoping that the order constraints on the $x_1,\ldots... | Introduce change of variable
$$\begin{cases}
x_1 &= y_1\\
x_2 &= y_1 + y_2\\
x_3 &= y_1 + y_2 + y_3\\
&\;\vdots\\
x_n &= y_1 + y_2 + y_3 + \cdots + y_n
\end{cases}$$
We have
$$\begin{array}{c}
0 \le x_1 \le x_2 \le x_n\\
\text{ and }\\
x_1 + x_2 + \cdots + x_n = d
\end{array}
\quad\iff\quad
\begin{array}{c}
y_1, y_2, \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral of $\frac{x^2+1}{(1-x^2)\sqrt{1+x^4}}$ So we have to evaluate $\int\frac{x^2+1}{(1-x^2)\sqrt{1+x^4}}dx$.
My work-
We can write the integrand as $\frac{(x+1)^2-2x}{(1-x)(1+x)\sqrt{1+x^4}}dx$.
So we wish to deduce $\int\frac{(x+1)}{(1-x)\sqrt{1+x^4}}dx-\int\frac{2x}{(1-x^2)\sqrt{1+x^4}}dx$
So lets write it a... | My method :
Am guiding you in different way which is too easy to understand
Make it a try ...
First step : remove $x^2$ from numerator and $x$ from first factor and $x^2$ from under square root function
Then it becomes $\int\frac{1+\frac1{x^2}}{(\frac1{x} -x)\sqrt{x^2+\frac{1}{x^2}}}$
Next
Why don't you take $\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show that $p \in \left[\frac{4^m}{2\sqrt{m}},\frac{4^m}{\sqrt{2m+1}}\right]$
If the number of ways in which $m$ identical apples can be put in $2m$ boxes, so that no box contains more than one apple, is $p$, prove that $$p \in \left[\frac{4^m}{2\sqrt{m}},\frac{4^m}{\sqrt{2m+1}}\right]$$
I did this as follows :
L... | Here is another rather elementary answer.
Let $a_m=\frac{1}{4m}\binom{2m}{m}$. We show the following is valid
\begin{align*}
\frac{1}{2\sqrt{m}}\leq a_m \leq \frac{1}{\sqrt{2m+1}}\qquad\qquad m\geq 1\tag{1}
\end{align*}
We start similarly to @CuriousGuest
\begin{align*}
a_m&=\frac{1}{4^m}\binom{2m}{m}=\frac{(2m)!}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Straight line is tangent to the curve. The straight line $y=mx+1$ is tangent to the curve $x^2+y^2-2x+4y=0$. Find the possible values of $m$.
My attempt
Substitute the $y=mx+1$ into the equation $x^2+y^2-2x+4y=0$. $$x^2+(mx+1)^2-2x+4(mx+1)=0$$ $$x^2+m^2x^2+2mx+1-2x+4mx+4=0$$ $$(1+m^2)x^2+6mx-2x+5=0$$ $$(1+m^2)x^2+(6m-... | The general equation of a tangent to a circle $X^2 + Y^2 = a^2$ is given by $Y = mX \pm a\sqrt{1 + m^2}$.
For the purpose of your question, $X = x - 1$, $Y = y + 2$, $a = \sqrt{5}$. Also,
$$
y = mx + 1 \\
\implies y + 2 = m(x - 1) + (m + 3) \\
\implies Y = mX + (m + 3).
$$
Hence,
$$
(m + 3) = \sqrt{5} \sqrt{1 + m^2}.
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $x+y+z=3$, then $\sum_{\text{cyc}}\frac{x^2}{2y^2-y+3}\ge\frac{3}{4}$
Let $x,y,z>0$, be such that $x+y+z=3$. Show that
$$\dfrac{x^2}{2y^2-y+3}+\dfrac{y^2}{2z^2-z+3}+\dfrac{z^2}{2x^2-x+3}\ge\dfrac{3}{4}.$$
I've tried many things but all have failed.
$$\left(\sum_{\text{cyc}}\dfrac{x^2}{2y^2-y+3}\right)\left(\sum... | By C-S $\sum\limits_{cyc}\frac{x^2}{2y^2-y+3}\geq\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(2x^2y^2-x^2y+3x^2)}$. Thus, it remains to prove that $\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(2x^2y^2-x^2y+3x^2)}\geq\frac{3}{4}$, which is $\sum\limits_{cyc}(3x^4-x^3y-2x^3z+x^2y^2-x^2yz)\geq0$, which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Ratio of $\frac{\sin x_1 }{\sin x_2}$ where $f(x_1)=f(x_2)$ for a trigonometric sum $f(x)$ If $f(x) = \cos(x+a_1)+\frac12\cos(x+a_2)+\frac14\cos(x+a_3)+\cdots+\frac1{2^{n-1}}\cos(x+a_n)$, where $a_1, a_2, ... a_n$ are some constants and $f(x_1)-f(x_2)=0$, where $x_2 \neq m\pi$, find
$$\frac{\sin x_1}{\sin x_2}+\frac{\l... | Use the addition formula for cosine
$\cos(A+B) = \cos A \cos B - \sin A \sin B$
to write
$f(x) = \cos{x} \displaystyle\sum\limits_{i=1}^{n}{\dfrac{2}{2^{n-k}}\cos a_k} - \sin{x} \displaystyle\sum\limits_{i=1}^{n}{\dfrac{2}{2^{n-k}}\sin a_k}$
Regard the summations as fixed since you are given $a_1,a_2,\dots,a_n$
Let $\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Proving $\sin 20^\circ \, \sin 40^\circ \, \sin 60^\circ \, \sin 80^\circ =\frac{3}{16}$ The problem is to prove that
$$\sin 20^\circ \, \sin 40^\circ \, \sin 60^\circ \, \sin 80^\circ =\frac{3}{16}$$
All my attempts were to get them in $\sin (2A)$ form after eliminating $\sin 60^\circ$ in both sides. Unfortunately, al... | HINT:
Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
$$\sin(60^\circ-x)\cdot\sin x\sin(60^\circ+x)=\sin x[\sin^2120^\circ-\sin^2x]=\dfrac{3\sin x-4\sin^3x}4$$
$$\implies4\sin(60^\circ-x)\cdot\sin x\sin(60^\circ+x)=\sin3x$$
Set $x=20^\circ$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How find $\sum_{0 \leq k: \leq 100 \ gcd \left( k, 100 \right) =1} f \left( \frac{k}{100} \right)$ if $f \left( x \right) = \frac{9^x}{3+9^x}$ ? How find of $\sum_{0 \leq k: \leq 100 \ gcd \left( k, 100 \right) =1} f \left( \frac{k}{100} \right)$ if $f \left( x \right) = \frac{9^x}{3+9^x}$ ?
| We can prove $(k,100)=(100-k,100)$
Now prove $f(x)+f(1-x)=1$
Set $x=\dfrac k{100}$
Now $\phi(100)=\phi(25)\cdot\phi(4)=40$
If $\displaystyle S=\sum_{0\le k\le 100;\left( k, 100 \right) =1} f \left( \frac k{100} \right),$
$\displaystyle S+S=\sum_{0\le k\le 100;\left( k, 100 \right) =1}\left[f \left( \frac k{100} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proof - for all integers $y$, there is integer $x$ so that $x^3 + x = y$ For all integers $y$, there is an integer $x$ so that $$x^3 + x = y.$$
This is what I have done so far:
Proof: Suppose $y$ is some integer. We want to prove that $$x^3 + x = y$$ for some integer $x$.
I am thinking this is false. For example, if $y... | This is indeed not true. Two proofs:
Using basic calculus:
Note that $x^3+x$ is strictly increasing. (Its derivative is $3x^2+1$ and squares are positive)
Now note that $1^3+1=2$ and $0^3+0=0$. Therefore the $x$ such that $x^3+x=1$ must statistify $0<x<1$, so its not an integer.
Using number theory only:
Note that $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Converting a matrix from one base base to another. I have this basis $B = ((1,0,1),(0,1,-1),(1,-1,0))$
That is represented by:
$$[T]_B = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 2 \\ 1 & 1 & 2 \end{pmatrix}$$
I want to convert this representing matrix into standard basis, but I somehow misunderstood it since I have mistake... | First you find $ [i]_{E^B} $ and $ [i]_{B^E} $, where $i$ is the identity map. Now I assume you are calling $ E $ as the standard basis vectors. To find $ [i]_{B^E} $, the $j^{th} $ column is given by the linear combination of $E_j$ in terms of the vectors from $B$. This is easy because this is already how $B$ is writt... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Minimizing a summation? I have absolutely no idea how to approach this problem. I've been looking through notes, and I think I missed this when my professor discussed this in class.
$$
\text{Consider the data}\\
i\: x_i\: y_i\\
1\:2\:1\\
2\:3\:2\\
3\:3\:3\\
4\:4\:6\\
5\:5\:5\\
\text{As discussed in class, compute the l... | Let our best fit line be described as $mx + b$
We would hope that we could solve for $m$ and $b$ without any contradictions using the system of equations:
$\begin{cases} 2m+b = 1\\
3m + b = 2\\
3m + b = 3\\
4m + b = 6\\
5m + b = 5\end{cases}$
Which can be expressed as the following matrix equation:
$\begin{bmatrix} 2 &... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int \frac{(x+1)}{(x^2+2)^2}dx$ This is the question I want to ask
Integrate $$\int \frac{(x+1)}{(x^2+2)^2}\,dx.$$
I tried it using algebric manipulation
Integrate $x/(x^2+2)^2+1/(x^2+2)^2$.
Then the latter part would not be solved.
| We split the integral into two components as
$$\int\frac{x+1}{(x^2+2)^2}dx=\int\frac{x}{(x^2+2)^2}dx+\int\frac{1}{(x^2+2)^2}dx\tag 1$$
The first integral on the right-hand side of $(1)$ is trivial to evaluate using the substitution $x^2=u\implies x\,dx\to \frac12 du$. Then,
$$\begin{align}
\int\frac{x}{(x^2+2)^2}dx&... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
The converges of $ \sqrt 2 +\sqrt { 2-\sqrt 2} +\sqrt { 2-\sqrt { 2+\sqrt 2} } + \cdots =$ I would like to know wheather this series converge or diverge?
$\sqrt 2 +\sqrt { 2-\sqrt 2} +\sqrt { 2-\sqrt { 2+\sqrt 2} } +\sqrt { 2-\sqrt { 2+\sqrt { 2+\sqrt 2} } } +\cdots$
My work:
multiplyIing both demominator and nume... | First, the sum we want can be rewritten as $\displaystyle\;\sum_{n=0}^\infty \sqrt{\epsilon_n}\;$
where
$$\epsilon_n = 2 - \overbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}^{n\; \text{terms}}$$
Notice for $n \ge 0$,
$$\epsilon_{n+1} = 2 - \sqrt{4 - \epsilon_n} = \frac{\epsilon_n}{2+\sqrt{4-\epsilon_n}}$$
It is easy to see... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve $z^3 + \overline z = 0$ I need to solve this:
$$z^3 + \overline z = 0$$
how should I manage the 0?
I know that a complex number is in this form: z = a + ib so:
$$z^3 = \rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace$$
$$\overline z = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace$$
but how about... | Given
$$
z^3 = - \bar{z}.
$$
First we note that
$$
|z^3| = |-\bar{z}| \Longrightarrow |z|^3 = |z|.
$$
Therefore
$$
z = 0 \vee z = \exp(\zeta \pi \mathbf{i}).
$$
The case $z \ne 0$
We obtain
$$
\exp(3 \zeta \pi \mathbf{i}) = \exp( \pi \mathbf{i} - \zeta \pi \mathbf{i}).
$$
Whence
$$
3 \zeta = 2 k + 1 - \zeta \Longrighta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 8
} |
Solve $\log_3(3x + 2) = \log_9(4x + 5)$ for $x$
Solve for $x$
$$
\log_3(3x + 2) = \log_9(4x + 5)
$$
I changed the bases of the logs
$$
\frac {\log_{10}(3x + 2)} {\log_{10}(3)} = \frac {\log_{10}(4x + 5)} {\log_{10}(9)}
$$
Now I'm stuck, I don't know how to eliminate the logs.
On WolframAlpha I've seen that $\dfrac {\... | Notice, formula $$\color{blue}{log_{b^n}(a)=\frac{1}{n}log_{b}(a)}$$ Now, we have $$log_{3}(3x+2)=log_{9}(4x+5)$$ $$\implies log_{3}(3x+2)=log_{3^2}(4x+5)$$
$$\implies log_{3}(3x+2)=\frac{1}{2}log_{3}(4x+5)$$ $$\implies 2 log_{3}(3x+2)=log_{3}(4x+5)$$ $$\implies log_{3}(3x+2)^2=log_{3}(4x+5)$$ $$\implies (3x+2)^2=4x+5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
First principle derivative of a square root and conjugates I'm trying to find the derivative of the equation: $$g(x)=\sqrt {x+2}-3x^2$$.
I can find the solution just fine using the power rule but am finding trouble with First Principles.
Essentially, I understand getting as far as $$\displaystyle\lim_{h\to 0}\frac{\sqr... | I am taking from the step you got stuck $$\lim_{h\to 0}\frac{\sqrt {x+h+2}-3x^2-6xh-3h^2 -\sqrt {x+2}+3x^2}{h}$$ $$=\lim_{h\to 0}\frac{\sqrt {x+2}\left(1+\frac{h}{x+2}\right)^{1/2}-\sqrt {x+2}-6xh-3h^2 }{h}$$ Use binomial expansion $(1+x)^n=1+nx+ \ldots $ where, $|x|<1$ $$=\lim_{h\to 0}\frac{\sqrt {x+2}\left(1+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving that the sum of the first $2n$ terms of the series $1^2 - 3^2 + 5^2 - \cdots$ is $-8n^2$ by induction
Use mathematical induction to prove the following for the first $2n$ terms of the series $$1^2 - 3^2 + 5^2 - 7^2 + \cdots = -8n^2.$$
As we have odd numbers that are squared we could use $n = 2k-1$.
But the $2... | $1^2−3^2+5^2−7^2+⋯=−8n^2$
Last term of this series is $(4n-3)^2-(4n-1)^2$
$\Rightarrow P(n):1^2−3^2+5^2−7^2+⋯+(4n-3)^2-(4n-1)^2=−8n^2$
For n=1
$\Rightarrow P(1):1^2−3^2=−8(1)^2$
$\Rightarrow P(1):-8=−8$ which is true
Let P(k)be true,
$P(k):1^2−3^2+5^2−7^2+⋯+(4k-3)^2-(4k-1)^2=−8k^2$.......(1)
We have to prove P(k+1) as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Given the differential equation, how to solve the y function with x as the independent variable? $y\frac{dy}{dx} = x(y^4 + 2y^2 + 1)$
$y = 1$ when $x = 4$
I tired to integrate by substitution, but it doesn't seem to work out.
| To solve this, you can use separation of variables:
$$ y \frac{dy}{dx} = x(y^4+2y^2+1)$$
$$ \frac{y}{y^4+2y^2+1}dy = x\ dx$$
$$ \frac{y}{(y^2+1)^2}= x \ dx$$
Using the substitution $u = y^2+1$, $\frac{du}{dy}= 2y $
$$ \frac{1}{2}\int u^{-2} du= \int x \ dx $$
$$ \frac{-1}{2(y^2+1)} = \frac{x^2}{2} + c$$
where c is a co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Proving $\frac1{4ab}\left(\frac{(b+1)^{b+1}}{b^b}\right)^a<\binom{a(b+1)}a<\left(\frac{(b+1)^{b+1}}{b^b}\right)^a$ Let $a\in\mathbb N$, and $b\in\mathbb R, b\geq 1$. How to prove that $$\frac{1}{4ab}\left(\frac{(b+1)^{b+1}}{b^b}\right)^a<\binom{a(b+1)}{a}<\left(\frac{(b+1)^{b+1}}{b^b}\right)^{a}?$$
| $\def\e{\mathrm{e}}$Lemma 1: If $$f(x) = \left( 1 + \dfrac{1}{x} \right)^x,\ g(x) = \left( 1 + \dfrac{1}{x} \right)^{x + 1}, \quad x \geqslant 1$$
then $f$ is strictly increasing but $g$ is strictly decreasing, and$$
\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} g(x) = \e.
$$
(This is a well-known result.)
Lemma 2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Differentiate the Function: $y=\sqrt{x^x}$ $y=\sqrt{x^x}$
How do I convert this into a form that is workable and what indicates that I should do so?
Anyway, I tried this method of logging both sides of the equation but I don't know if I am right.
$\ln\ y=\sqrt{x} \ln\ x$
$\frac{dy}{dx}\cdot \frac{1}{y}=\sqrt{x}\ \frac... | That way could work though you made some mistakes, but an easier way shifts the square root to a fractional exponent.
$$\begin{align}
y&=\sqrt{x^x} \\[2ex]
&= \left(x^x\right)^{1/2} \\[2ex]
&= x^{x/2} \\[2ex]
\ln y&= \ln x^{x/2} \\[2ex]
&= \frac x2\ln x \\[2ex]
\frac{dy}{dx}\frac 1y &=\frac 12\ln x+\frac x2\frac 1x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Mathematical Induction proof for a cubic equation. If $ x^3 = x +1$, prove by induction that $ x^{3n} = a_{n}x + b_n + \frac {c_n}{x}$, where $a_1=1, b_1=1, c_1=0$ and
$a_n = a_{n-1} + b_{n-1}, b_n = a_{n-1} + b_{n-1} + c_{n-1}, c_n = a_{n-1} + c_{n-1}$ for $n=2,3,\dots $
For $n=1$ we have $x_3 = a_1x + b_1 + \fra... | You should put "$=$" between things ALREADY KNOWN to be equal, not with things you are trying to prove to be equal. You should not start a chain of equalities with the very equality you're trying to prove.
Thus
\begin{align}
x^{3(k+1)} & = x^3 x^{3k} & & \text{(This known, by routine algebra.)} \\[10pt]
& = x^3 ( a_kx... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Trigonometric equation with sine and cosine So the equation is $3\cos ^2t + 5\sin t = 1$
Now I have simplified this to $$3(1-\sin ^2t) + 5\sin t -1 = 0$$
which leads to $$-3\sin ^2t + 5\sin t + 2 = 0$$
Then I get $$-3t^2 + 5 t +2 = 0$$
Is this the correct way to go with this equation then use $t = t/2 \pm \sqrt {(t/2)^... | you have $$-3\sin(t)^2+5\sin(t)+2=0$$ let $$\sin(t)=u$$ then we have $$-3u^2+5u+2=0$$ divided by $-3$ gives $$u^2-\frac{5}{3}u-\frac{2}{3}=0$$
solving this equation we obtain
$$u_{1,2}=\frac{5}{6}\pm\sqrt{\frac{25}{36}+\frac{24}{36}}$$
from here you will come to the result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Asymptotic expansion of integral with hyperbolic functions Consider the integral given by
$$f(r)=\int_{0}^{\tanh(r)} \arccos\left(\frac{\sigma}{\sinh(r)\sqrt{1-\sigma^2}}\right)\cdot \frac{1}{\sqrt{\sigma^2+a^2}}d\sigma,$$
where $a>0$. I am wondering how one can derive the asymptotic power series for $r\downarrow 0$?
... | Consider $\;\sigma:=\tanh(u)\,$ then :
\begin{align}
f(r)&:=\int_{0}^{\tanh(r)} \arccos\left(\frac{\sigma}{\sinh(r)\sqrt{1-\sigma^2}}\right)\cdot \frac{1}{\sqrt{\sigma^2+a^2}}\,d\sigma\\
&=\int_{0}^{r} \arccos\left(\frac{\sinh(u)}{\sinh(r)}\right) \frac{1}{\sqrt{\tanh(u)^2+a^2}}\,\frac{du}{\cosh(u)^2}\\
\end{align}
Set... | {
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Determine if this series $ \sum_{n=0}^\infty\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}$ converges Determine if the following series converges:
$$
\sum_{n=0}^\infty\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}.
$$
(http://i.stack.imgur.com/qWiuy.png)
I don't know how to start.
| Use the limit comparison test.
Let $\displaystyle a_n=\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}$ and $\displaystyle b_n=\frac{n^6}{n^7}=\frac{1}{n}$.
Since
$$\begin{align}\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}&=\lim_{n\to\infty}\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}\cdot\frac{n}{1}\\
&=\lim_{n\to\infty}\frac{n^7\left... | {
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Solving an exponential equation that includes division and multiplication The question is simplify the expression $\left(\dfrac{a^2}{27}\right)^{1/3}\left(\dfrac{64}a\right)^{2/3}$
1: Multiply on both sides equals $\dfrac{a^{2/3}}{27^{1/3}}\cdot \dfrac{64^{2/3}}{a^{2/3}}$
Does this give me $\dfrac{a^{2/3}}{3} \cdot \d... | Assuming that $a\neq 0$; then
$$\Big(\frac{64}{a}\Big)^{\frac{2}{3}}=\Big(\frac{4^6}{a^2}\Big)^{\frac{1}{3}};$$
then $$\Big(\frac{a^2}{3^3}\Big)^{\frac{1}{3}}\Big(\frac{16^3}{a^2}\Big)^{\frac{1}{3}}=\frac{16}{3}.$$
So your answer is correct. Clearly if $a=0$ this expression is not define.
| {
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Solving two diophantine equations. Find at least one 5-tuple of positive integers which satisfy the following two equations
$$a^2-d^2=3(b^2-c^2)$$
$$e^2-b^2=3(d^2-c^2)$$
such that no three of the 5 positive integers $a, b, c, d, e$ form an Arithmetic progression and $a<b<c<d<e$.
Please help.
| Here is a nice context to this seemingly random system. If there is a $5$-tuple $a,b,c,d,e$ such that,
$$\begin{aligned}
a^2-d^2&=3(b^2-c^2)\\[1.5mm]
e^2-b^2&=3(d^2-c^2)
\end{aligned}\tag1$$
then for $k = 2,4$,
$$(a + c)^k + (-a + c)^k + (2 d)^k = (c + e)^k + (c - e)^k + (2 b)^k\tag2$$
and for $k = 1,2,3,5$,
$$(a + 2 c... | {
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Prove the relation for cos inverse Prove the relation $\cos^{-1}x_0=\dfrac{\sqrt {1-x^2_0}}{x_1\cdot x_2\cdot x_3\cdots \text{ ad inf.}}$ where the successive quantities $x_r$ are connected by the relation $x_{r+1}=\sqrt{\frac{1}{2}(1+x_r)}$
My attempt:
$$x_1=\sqrt{\frac{1}{2}(1+x_0)}$$
$$x_2=\sqrt{\frac{1}{2}(1+x_1)}$... | Let $x_0:=\cos y$. Want to prove that $y = RHS$. Also we know, that in case of convergence, $x_r \to 1$ because $\bar x = \sqrt{\frac12(1+\bar x)} \Rightarrow \bar x^2 - \frac12 \bar x - \frac12 = 0$ and $\bar x \ge 0$. The roots of the quadratic are $1$ and $-\frac12$.
We now need to show that
$$y = \frac{\sqrt{1-\c... | {
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Why does $\sqrt{6} + \sqrt{10} + \sqrt{15}$ have four conjugates? I am having trouble understanding how algebraic number $\sqrt{6} + \sqrt{10} + \sqrt{15}$ has four conjugates.
Minimal polynomial is $x^4-62 x^2-240 x-239$ according to Wolfram Alpha.
Factorized:
$$\left(x-2\sqrt{15 (4-\sqrt{15})}-8\sqrt{4-\sqrt{15}}-\sq... | The conjugates of an algebraic number are (by definition) the roots of its minimal polynomial. The number of (distinct) roots of an irreducible polynomial over the rationals is equal to its degree, that is four.
Thus once you know the minimal polynomial "it is clear."
There is some wiggling room as one might or might ... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that the angle between $OP$ and the normal to the curve at $P$ satisfies the following I'm struggling to answer the following question below
I've already worked out the gradient to the curve at $P$, but I'm having difficulty answering the second part of the question. MY attempt is as follows:
I focus on $\triangl... | Notice, gradient of the curve: $x^2+y^2+2axy=1$ $$\frac{d}{dx}(x^2+y^2+2axy)=\frac{d}{dx}(1)$$ $$2x+2y\frac{dy}{dx}+2ax\frac{dy}{dx}+2ay=0$$ $$2(ax+y)\frac{dy}{dx}=-2(x+ay)$$ $$\frac{dy}{dx}=\frac{-2(x+ay)}{2(ax+y)}$$ $$\implies \color{blue}{\text{slope of tangent}\,, \frac{dy}{dx}=\frac{-(x+ay)}{ax+y}}$$
Above is th... | {
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Calculate derivative of integral I tried to calculate the derivative of this integral:
$$\int_{2}^{3+\sqrt{r}} (3 + \sqrt{r}-c) \frac{1}{2}\,{\rm d}c $$
First I took the anti-derivative of the integral:
$$\frac{1}{2}\left(\frac{-c^2}{2}+c\sqrt{r}+3c\right)$$
Then I evaluated the integral:
$$-\frac{1}{2}\left(\frac{3+\... | $$\int_{2}^{3+\sqrt{r}} \left(\frac{1}{2}\left(3+\sqrt{r}-c\right)\right)dc=$$
$$\frac{1}{2}\cdot\int_{2}^{3+\sqrt{r}} \left(3+\sqrt{r}-c\right)dc=$$
$$\frac{1}{2}\left(\int_{2}^{3+\sqrt{r}} (3)dc+\int_{2}^{3+\sqrt{r}} (\sqrt{r}) dc-\int_{2}^{3+\sqrt{r}} (c)dc\right)=$$
$$\frac{1}{2}\left(\left[3c\right]_{2}^{3+\sqrt{r... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Trigonometric substitution and triangles I'm learning trigonometric substitutions and am having a bit of trouble understanding the intuition behind the conversions (why do most use secant?). If you could explain the conversions geometrically using a triangle, that would be very helpful. For example, if we have $$\int \... | It's not that it's wrong to take on the integral the way you propose -- it's just not awfully convenient. If we implicitly differentiate your substitution equation, we obtain
$$ \cos \theta \ d\theta \ = \ \frac{x \ \cdot \frac{1}{2} \ (x^2 - 4)^{-1/2} \ \cdot 2x \ - \ (x^2 - 4)^{1/2} \ \cdot \ 1}{x^2} \ dx $$
$$ = \ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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New idea to solve this equation I was teaching $\left \lfloor x \right \rfloor$ function properties and equation . I solved this equation in my class . My works are show below. Some students ask me for new Idea...,and now I am looking for various method to solve this (like this) equation .$$\left \lfloor x\right \rf... | Note that for any integer $n$, $\;\lfloor x+n\rfloor=\lfloor x\rfloor+n $,and that $$\lfloor 2x\rfloor=\begin{cases}2\lfloor x\rfloor&\text{if}\enspace 0\le x-\lfloor x\rfloor<\dfrac12,\\2\lfloor x\rfloor+1&\text{if}\enspace \dfrac12\le x-\lfloor x\rfloor<1.\end{cases} $$
The given equation implies $\;1=\bigl\lfloor\lf... | {
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Find, in terms of $s$,the coordinates of the point where this normal cuts the curve again.
a) Find the equation of the normal at the point $(2s,\frac{2}{s})$ to the curve whose parametric equations are $x=2s,y=\frac{2}{s}$
b) Find, in terms of $s$,the coordinates of the point where this normal cuts the curve again.
I... | The parametric equation of the curve is $x=2s$ & $y=\frac{2}{s}$
Now, the equation of the curve in cartesian coordinates is obtained by eliminating $s$ as follows $$y=\frac{2}{\frac{x}{2}}=\frac{4}{x}$$ $$\implies \color{blue}{xy=4}$$ Above, equation represents a rectangular hyperbola.
Now, differentiating the equatio... | {
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$A+B+C=2149$, Find $A$ In the following form of odd numbers
If the numbers
taken from the form where $A+B+C=2149$
Find $A$
any help will be appreciate it, thanks.
| $A$ is going to be the $k$th entry in the $n$th row, where $1 \leq k \leq n$. That will make $B$ the $k$th entry in the $n+1$st row, and $C$ the $k+1$st entry in the $n+1$st row.
So the first question is, can we write a formula for $A$ in terms of $k$ and $n$? If we consider the first entry in the $n$th row, there are ... | {
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How do you solve for θ in the equation $\tan \frac{\theta}{5} + \sqrt{3} = 0$ $$\tan \frac{\theta}{5} + \sqrt{3} = 0$$
Alright so the $\frac{\theta}{5}$ is confusing me.
Would it be wrong to do
\begin{eqnarray}
\tan \frac{\theta}{5}&=&-\sqrt{3}\\
\frac{\theta}{5}&=&\tan^{-1}(-\sqrt{3})\\
\theta&=& 5\tan^{-1}(-\sqrt... | Notice, $$\tan \frac{\theta}{5}+\sqrt 3=0$$ $$\tan \frac{\theta}{5}=-\sqrt 3=-\tan \frac{\pi}{3}$$ Writing the general solution for $\theta$ as follows $$\frac{\theta}{5}=n\pi-\frac{\pi}{3}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\theta=5n\pi-\frac{5\pi}{3}}}$$
Where, $n$ is any integer
| {
"language": "en",
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Closed form for an infinite product The following fascinating formula appears in the paper "On gamma quotients and infinite products" by M.Chamberland and A.Straub (see page 9):
$$\prod_{n=2}^\infty\left(1-\frac{1}{n(n-1)}\right)=-\frac1\pi\cos\left(\frac{\sqrt5}{2}\pi\right).$$
Unfortunately, it's given without any re... | This formula is an application of the theorem $1.1$ for $n=2$ from your Chamberland & Straub paper (when $a+b=c+d$) :
$$\tag{1}\prod_{k\ge 0}\frac {(k+a)(k+b)}{(k+c)(k+d)}=\frac {\Gamma(c)\Gamma(d)}{\Gamma(a)\Gamma(b)}$$
Observe first that : $\quad\displaystyle n^2-n-1=\left(n-\frac{\sqrt{5}+1}2\right)\left(n+\frac{\sq... | {
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Show that $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ is rational using the rational zeros theorem What I've done so far:
Let $$r = \sqrt{4+2\sqrt{3}}-\sqrt{3}.$$
Thus, $$r^2 = 2\sqrt{3}-2\sqrt{3}\sqrt{4+2\sqrt{3}}+7$$
and $$r^4=52\sqrt{3}-28\sqrt{3}\sqrt{4+2\sqrt{3}}-24\sqrt{4+2\sqrt{3}}+109.$$
I did this because in a similar exam... |
is there another approach to these type of problems
Since
$$4=1+3,3=1\times 3$$
we can have
$$4+2\sqrt 3=1+3+2\sqrt{1\times 3}=(1+\sqrt 3)^2$$
| {
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How to find eigenvalues of matrix $\begin{bmatrix} 3& a+1\\a+1&3 \end{bmatrix}$ I want to find the eigenvalues of the following matrix
$\begin{bmatrix} 3& a+1\\a+1&3 \end{bmatrix}$ expressed in $a$ using $\begin{bmatrix} \lambda - 3& a+1\\a+1&\lambda-3 \end{bmatrix}$.
But the $a$-term makes it difficult for me to find ... | \begin{align}
\left| \begin{matrix} \lambda -3 & a+1 \\
a+1 &\lambda-3 \end{matrix} \right| & = (\lambda-3)(\lambda-3) - (a+1)(a+1) \\
&= \lambda^2-6\lambda+9-(a^2+2a+1) \\
&= \lambda^2-6\lambda+9-a^2-2a-1 \\
&= \lambda^2-6\lambda-a^2-2a+8 \\
&= \lambda^2-6\lambda-(a^2+2a-8)\\
&= \lambda^2-6\lambda-(a+4)(a-2) \\
&= (\l... | {
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"url": "https://math.stackexchange.com/questions/1388263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrating Factor by Inspection $(x^3+xy^2+y)dx + (y^3+xy^2+x)dy=0$ $$(x^3+xy^2+y) \hspace{.1cm} dx + (y^3+xy^2+x)\hspace{.1cm} dy=0$$ So I tried to solve this problem but can't figure out my integrating factor all I can see here is if I distribute first I can get a $y \hspace{.1cm} dx + x dy$ so that would be $d(xy)$... | Hint:
$(x^3+xy^2+y)~dx+(y^3+xy^2+x)~dy=0$
$\dfrac{dx}{dy}=-\dfrac{y^3+xy^2+x}{x^3+xy^2+y}$
With reference to http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=180:
Let $u=\dfrac{x}{y}$... | {
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Evaluate the integral $\int_0^{\frac{\pi}{2}} \frac{1}{a + \sin^2 \theta}$ using the residue theorem Can someone show me how to compute this integral using the residue theorem:
$$\int_0^{\frac{\pi}{2}} \frac{1}{a + \sin^2 \theta}d\theta$$
| We can simplify the problem by (i) exploiting the even symmetry of the integrand, (ii) using the identity
$$\sin^2 \theta =\frac{1-\cos 2\theta}{2}$$
and (iii) enforcing the substitution $2\theta \to \theta$. We can write then write the integral of interest $I(a)$ as
$$I(a)=\frac12 \int_0^{2\pi}\frac{1}{(2a+1)-\cos \... | {
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How can I prove remainder side of this inequality? Let $x$, $y$ be two positive numbers such that $x^4+y^4=x^2+y^2$. Prove that
$$1\leqslant x+y\leqslant 2.$$
With $x+y\leqslant 2$. I tried
We have $$x^4+y^4\geqslant \dfrac{(x^2+y^2)^2}{2}.$$
Therefore, $$x^2+y^2\geqslant \dfrac{(x^2+y^2)^2}{2}$$
or
$$(x^2+y^2)^2-2(x^... | let $x+y=u,xy=v^2 \implies u^2 \ge 4v^2 \\ x^2+y^2=u^2-2v^2,x^4+y^4=(x^2+y^2)^2-2(xy)^2 =(u^2-2v^2)^2-2v^4=u^4-4u^2v^2+2v^4 \implies \\
u^4-(4v^2+1)u^2+2v^4+2v^2=0 \\ u^2=\dfrac{4v^2+1+ \sqrt{8v^4+1}}{2}$
$u^2=\dfrac{4v^2+1-\sqrt{8v^4+1}}{2} \le 2v^2$
$v^2\ge 0 \implies
u^2=\dfrac{4v^2+1+ \sqrt{8v^4+1}}{2} \ge \... | {
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If 2 roots of the equation $(p-1)(x^2+x+1)^2-(p+1)(x^4+x^2+1)$ are real and distinct and $f(x)=\frac{1-x}{1+x}$... Question: If 2 roots of the equation $(p-1)(x^2+x+1)^2-(p+1)(x^4+x^2+1)$ are real and distinct and $f(x)=\frac{1-x}{1+x}$, then $f(f(x))+f(f(\frac{1}{x})) = ?$
(a)p
(b)2p
(c)-p
(d)-2p
Attempt: the give eq... | $\bf{My\; Solution::}$ Given
$$(p-1)(x^2+x+1)^2-(p+1)(x^4+x^2+1) = 0$$
$$\displaystyle (p-1)(x^2+x+1)^2=(p+1)(x^4+x^2+1)$$
We can write it as $$\displaystyle \frac{p+1}{p-1}=\frac{(x^2+x+1)^2}{(x^4+x^2+1)}=\frac{(x^2+x+1)^2}{(x^2-x+1)(x^2+x+1)}=\frac{x^2+x+1}{x^2-x+1}$$
Now We can Write it as $$\displaystyle \frac{\lef... | {
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Method of partial fractions when denumerator cannot be factorized? Suppose I'm given an expression: \begin{align*} P(x) = \frac{1+x}{1-2x-x^2}. \end{align*} The denumerator cannot be readily factorized, so I found the zeros, which are \begin{align*} \lambda_1 = -1 - \sqrt{2} \qquad \lambda_2 = -1 + \sqrt{2}. \end{align... | $$\frac{1+x}{1-2x-x^2}=\frac{-1-x}{x^2+2x-1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$n^{th}$ derivative of $y=x^2\cos x$ I am stuck with Leibniz formula
$$D^{n}y = \sum_{k=0}^{n} \binom{n}{k} \, x^{(2k)}\cos^{(n-k)}x$$
Could someone show how to do it?
| I would use the fact that:
$$\cos^{(n)}(x)=\cos\Bigl(x+\frac{n\pi}2\Bigr)$$
and that $$(x^2)'=2x,\quad (x^2)''=2,\quad (x^2)^{(n)}=0\enspace\text{si}\enspace n>2.$$
Thus Leibniz's rule gives:
$$\bigl(x^2\cos x\bigr)^{(n)} =x^2\cos^{(n)}(x)+2nx\cos^{(n-1)}(x)+n(n-1)\cos^{(n-2)}(x)$$
Also, note that $\;\cos^{(n-2)}(x)=-\... | {
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Prove that $\left (1+\frac ba \right )\left (1+\frac ac \right )\left ( 1+ \frac cb\right )\ge 4+3\sqrt2$ Given $a,b,c>0$ and $a^2\ge b^2+c^2$. Prove that $$\left (1+\frac ba \right )\left (1+\frac ac \right )\left ( 1+ \frac cb\right )\ge 4+3\sqrt2$$
This is my try:
I expanded the LHS, and I have to show that $\disp... | We have to show $\displaystyle\frac a b +\frac a c +\frac b c +\frac b a +\frac c a +\frac c b \geq 2+3\sqrt2\quad\color{red}{(1)}$
$(1)\iff \displaystyle \left ( b+c \right )\left ( \frac{a}{bc}+\frac{1}a \right )+\left (\frac b c +\frac c b \right ) \ge2+3\sqrt2$
We know that $\displaystyle {b+c\ge 2\sqrt{bc},\quad... | {
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"answer_id": 0
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Solve trigonometric equation $ 3 \cos x + 2\sin x=1 $ Solve trigonometric equation: $$ 3 \cos x + 2\sin x=1 $$
I tried to substitue $\cos x = \dfrac{1-t^2}{1+t^2}, \sin x = \dfrac{2t}{1+t^2}$. Yet with no results.
| Let $u = \cos x$ and $v = \sin x$. Thus, $3u + 2v = 1$ and $u^2 + v^2 = 1$. Isolating one of the variables , we have
$$
u^2 + \dfrac{(1 - 3u)^2}{4} = 1 \quad \Rightarrow \quad 13u^2 - 6u - 3 = 0
$$
and so on...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Solve the binomial equation Solve the binomial equation $$z^4 = -8$$
Below is the steps i have done
1: I have taken |-8| that is 8 and then done 8^(1/4) which is 2^(1/4).
2: Since $z=r(cos\alpha+isin\alpha)$ leads me to
$r^4(cos4\alpha+isin4\alpha)=-8(cos\pi/2+isin\pi/2)$
Divide by 4 since the z term is raised by fo... | Notice, $$z^4=-8$$
$$z^4=8i^2$$ $$z=\sqrt[4]{8i^2}$$
$$z=(2)^{3/4}\sqrt{i}$$ $$z=(2)^{3/4}\sqrt{\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}}$$
$$=(2)^{3/4}\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)^{1/2}$$
$$=(2)^{3/4}\left(\cos\left(2k\pi+\frac{\pi}{2}\right)+i\sin\left(2k\pi+\frac{\pi}{2}\right)\right)^{1/2}$$
$$=(2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Sum of an infinite series. I have found a series whose $n$-th term is denoted by $\dfrac{n}{a^n}$. Here $a$ is a constant. I tried to find the sum but failed. Is there any formula for its infinite sum or just an approximation?
| notice, we have $$T_n=\frac{n}{a^n}$$ $$S_n=\sum_{n=1}^{n} T_n=\sum_{n=1}^{n}\frac{n}{a^n}$$ $$S_n=\frac{1}{a}+\frac{2}{a^2}+\frac{3}{a^3}+\frac{4}{a^4}+\ldots +\frac{n}{a^n}\tag 1$$ Multiplying by $\frac{1}{a}$ & rewriting as follows
$$\frac{1}{a}S_n= \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{a^2}+\frac{2}{a^3}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Area between the curve and x-axis Find the area between the curve $y={2\over x-1}-1$ and the $x$-axis over the interval $[2,4]$
Would this solve the question $\int_{2}^4( {2\over x-1}-1)dx$ ?
So...
$[2\ln(x-1)-x+c]_2^4$ would be incorrect because the curve would be below the x-axis for $3$ to $4$
| As I mentioned in the comment, you should divide the interval $[2, 4]$ to two parts,
\begin{align*}S = &\int_2^3\left(\frac{2}{x - 1} - 1\right) dx \;{\color{red} -}\int_3^4\left(\frac{2}{x - 1} - 1\right)dx \\
= & \left[2\log(3 - 1) - 2\log(2 - 1) - 1\right] - \left[2\log(4 - 1) - 2\log(3 - 1) - 1\right] \\
= & 4\log ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve trigonometric inequality $\cos x \geq \sin^2 x - \cos^2 x $ Solve trigonometric inequality $$\cos x \geq \sin^2 x - \cos^2 x $$
My incorrect solution:
$$\cos^2 x-\sin^2 x \geq -\cos x $$
$$\cos 2x \geq \cos (\pi - x) $$
which means:
$$ 2x \geq -(\pi + x)$$
$$ x \geq -\pi $$
Which is wrong.
And
$$ 2x \leq 2\p... | $$0\le\cos x+\cos2x=2\cos\dfrac{3x}2\cos\dfrac x2=2\cos^2\dfrac x2\left(4\cos^2\dfrac x2-3\right)$$
$$\implies0\le4\cos^2\dfrac x2-3=2(1+\cos x)-3=2\cos x-1\iff\cos x\ge\dfrac12$$
$2n\pi-\dfrac\pi3\le x\le2n\pi+\dfrac\pi3$ where $n$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Advanced Algebra Manipulation/Inequality Proof: $\frac{4x^3(x^2+y^2)-2x(x^4+y^4)}{(x^2+y^2)^2} \leq 6|x|$ I need to show that
$$\frac{4x^3(x^2+y^2)-2x(x^4+y^4)}{(x^2+y^2)^2} \leq 6|x|$$
by starting with the left side of the inequality and working from there.
Hints from the textbook said to work from these inequality "t... | Interestingly, this inequality is not tight. In fact, we can show that
$$\left|\frac{4x^3(x^2+y^2)-2x(x^4+y^4))}{(x^2+y^2)^2}\right|\le 2|x|$$
To that end, we write
$$\begin{align}
\frac{4x^3(x^2+y^2)-2x(x^4+y^4))}{(x^2+y^2)^2}&=\frac{2x}{(x^2+y^2)^2}\left(2x^2(x^2+y^2)-(x^4+y^4)\right)\\\\
&=\frac{2x}{(x^2+y^2)^2}\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
nice classical nonhomogeneous inequality Let $a,b,c$ be positive reals and $abc=1$. Prove that $$10(a^4+b^4+c^4)+21\ge 17(a^3+b^3+c^3).$$
I have found a solution using MV and I'm wondering if there is a nice solution.
| This is just a solution, not necessarily a nice one. I started by simple rearrangement, and I got
$$ 10(a^4+b^4+c^4 - 3abc) \geq 17(a^3+b^3+c^3 - 3abc)$$ but I couldn't think of any well-known inequality that can prove this. So I decided to do straight calculus:
Let $g(a,b,c) = 1-abc$ and $f(a,b,c)=10*(a^4+b^4+c^4)-17... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Evaluate the integration : $\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$ $$\int{\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx}$$
$$\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx=\int \frac{(1-\sin x)(2-\sin x)}{\sqrt{(1-\sin x)(2-\sin x)(1+\sin x)(2+\sin x)}}dx$$
I am stu... | Hint:
$$\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$$
$$=\int\frac{\cos x\sqrt{4-\sin^2 x}}{(1+\sin x)(2+\sin x)}\,dx$$
( multiplying numerator & denominator by $(1+\sin x)(2+\sin x)$ under square root sign.)
Now put , $\sin x=z$.
Expand Hint :
Then ,
$$=\int\frac{1}{1+z}\sqrt{\frac{2-z}{2+z}}\,dz$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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What is the residue obtained from this integral? Consider the following integral in two complex variables $z_1$ and $z_2$:
$$\frac{1}{(2\pi i)^2}\oint_{{|z_1|=\epsilon}\atop{|z_2|=\epsilon}}dz_1 dz_2\frac{1}{z_1 z_2\left(1+a+\frac{z_1}{z_1+z_2}\right)\left(1+b+\frac{z_2}{z_1+z_2}\right)}$$
If I perform the contour inte... | (EDITED)
The poles at $z_1 = 0$ and $z_2=0$ are not the only ones inside the circles $|z|=\epsilon$. Note that
$$ \dfrac{1}{z_1 z_2 (1+a+z_1/(z_1 + z_2)(1+b+z_2/(z_1+z_2))} = \dfrac{(z_1 + z_2)^2}{z_1 z_2 ((a+2) z_1 + (a+1) z_2)((b+1) z_1 + (b+2) z_2}$$
so (for fixed $z_2$ on the circle $|z|=\epsilon$) this has poles... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the other end of the Diameter For a National Board Exam Review
A circle has it center at $(3,-2)$ and one end of a diameter at $(7,2)$. Find the other end of the diameter.
Answer is $(-1,6)$
$$m=\frac{y^2-y^1}{x^2-x^1}=\frac{2-(-2) }{7-3}$$
$$=(3-7)^2+(-2-2)^2=r^2=32$$
$$r =\sqrt{32}$$
Plugin $(7,2)$ into
$$y... | Let $(a, b)$ be other end of the diameter then the center $(3, -2)$ is the mid-point of line joining the ends points of diameter $(a, b)$ & $(7, 2)$
Hence the coordinates of the center $(3, -2)$ are given as $$\left(\frac{a+7}{2}, \frac{b+2}{2}\right)\equiv(3, -2)$$ By comparing the corresponding coordinates, we get $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Sum of (arithmetic?) infinite series How the heck do I find the sum of a series like $\sum\limits_{n=3}^\infty\frac{5}{36n^{2}-9}$? I can't seem to convert this to a geometric series and I don't have a finite number of partial sums, so I'm stumped.
| Notice, we have $$\sum_{n=3}^{\infty}\frac{5}{36n^2-9}$$ $$=\frac{5}{9}\sum_{n=3}^{\infty}\frac{1}{4n^2-1}$$
$$=\frac{5}{9}\sum_{n=3}^{\infty}T_n$$
$$\sum_{n=3}^{\infty}\frac{5}{36n^2-9}=\frac{5}{9}\lim_{n\to \infty}\sum_{n=3}^{n}T_n\tag 1$$
Where, $T_n$ is nth term of the series
given as
$$T_n=\frac{1}{4n^2-1}= \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Evaluation of $\int\frac{1}{1-\tan^2 x}dx$ Evaluation of $\displaystyle \int\frac{1}{1-\tan^2 x}dx$
$\bf{My\; Try::}$ We can write $$\displaystyle \int\frac{1}{(1-\tan x)\cdot (1+\tan x)}dx = \frac{1}{2}\int\frac{(1+\tan x)+(1-\tan x)}{(1-\tan x)\cdot (1+\tan x)}dx$$
So We get $$\displaystyle = \frac{1}{2}\int\frac{1}{... | We can also put $\tan\left(x\right)=u
$ to get $$\int\frac{1}{1-\tan^{2}\left(x\right)}dx=\int\frac{1}{\left(1-u^{2}\right)\left(u^{2}+1\right)}du=$$ $$=\int\frac{1}{2\left(u^{2}+1\right)}du+\int\frac{1}{4\left(u+1\right)}du-\int\frac{1}{4\left(u-1\right)}du.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Value of $\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}$ Here is the question:
$$\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}} = \;?$... | Let $f(n)=\frac{\sum_{k=1}^{n} \sqrt{\sqrt{n+1}+\sqrt{k}}}{\sum_{k=1}^{n} \sqrt{\sqrt{n+1}-\sqrt{k}}}$, and the original problem is to calculate $f(99)$.
Now we claim that $f(n)=\sqrt{2}+1$ for $\forall n>0$.
$f(n)=\frac{\sqrt{\sqrt{2}+1}}{\sqrt{\sqrt{2}-1}} \cdot \frac{\sum_{k=1}^{n} \sqrt{(\sqrt{n+1}+\sqrt{k})(\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 3
} |
If $3x^2 -2x+7=0$ then $\left(x-\frac{1}{3}\right)^2 =$? If $\ 3x^{2}-2x+7=0$ then $$\left(x-\frac{1}{3}\right)^2 =\text{?} $$
I am so confused. It is a self taught algebra book.
The answer is: $ \large -\frac{20}{9}$ but I don't know how it was derived.
Please explain.
Thanks for everyone who commented! I understand ... | Starting from, $$3x^2-2x+7=0$$ Dividing by 3:$$x^2-\frac{2x}{3}+\frac{7}{3}=0$$ Completing the square: $$\left(x-\frac 13\right)^2-\frac{1}{9}+\frac{7}{3}=0$$ $$\left(x-\frac 13\right)^2=\frac{1}{9}-\frac{7}{3}$$ $$\left(x-\frac 13\right)^2=\frac{1-21}{9}$$
$$\left(x-\frac 13\right)^2=-\frac{20}{9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 8,
"answer_id": 6
} |
Sigma Notations I have troubles understanding the sigma notation. If for example we have $c_i$ as $$c_i=\frac {x_i-x}{\sum(x_i-x)^2}$$ $$\sum c_i=\sum\frac{x_i-x}{\sum(x_i-x)^2}$$ Do we distribute the sigma to both top and bottom? But then the bottom would have double sigmas which dont make sense?
| Your notation is confusing.
Assuming $i$ goes from $0$ to $n$ and $c_i$ is actually
$$c_i=\frac{x_i-x}{\sum^n_{k=0}(x_k-x)^2}.$$
Then, we have
\begin{align}
\sum_{i=0}^n c_i
&= c_0 + c_1 + \dots + c_n\\
&=\frac{x_0-x}{\sum^n_{k=0}(x_k-x)^2} + \frac{x_1-x}{\sum^n_{k=0}(x_k-x)^2} + \dots +\frac{x_n-x}{\sum^n_{k=0}(x_k-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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percentage of integers such that $n^4 \pmod{16} \equiv 1$? How do I find the percentage of numbers $n$ in the list $1^4, 2^4, ... 1000^4$ such that $n \pmod{16} \equiv 1$? I know that for any $x$, if $x \pmod{16} \equiv 1$, then $x^n \pmod{16} \equiv 1$, so I know for sure that there are at least around $\lfloor \frac{... | We have
$$
\begin{align}
(2n)^4
&=16n^4\\
&\equiv0&\pmod{16}
\end{align}
$$
and
$$
\begin{align}
(2n+1)^4
&=16n^4+32n^3+24n^2+8n+1\\
&\equiv16\frac{n(n+1)}2+1&\pmod{16}\\
&\equiv1&\pmod{16}
\end{align}
$$
Therefore, $8$ out of $16$ equivalence classes mod $16$ satisfy $n^4\equiv1\pmod{16}$. So, I'd say $50\%$ of the eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is $(x^2 + 1) / (x^2-5x+6)$ divisible? I'm learning single variable calculus right now and at current about integration with partial fraction. I'm stuck in a problem from few hours given in my book. The question is to integrate $$\frac{x^2 + 1}{x^2-5x+6}.$$
I know it is improper rational function and to make it proper ... | You can use long division. Another way is as follows:
\begin{align*}
\frac{x^2+1}{x^2-5x+6}&=\frac{(x^2-5x+6)+(5x-5)}{x^2-5x+6}\\
&=\frac{x^2-5x+6}{x^2-5x+6}+\frac{5x-5}{x^2-5x+6}\\
&=1+\frac{5x-5}{x^2-5x+6}
\end{align*}
As next step notice that $$x^2-5x+6=(x-2)(x-3)$$
and expand in partial fractions:
\begin{align*}
\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
How to solve equation: $ \frac{x+1}{9x^3-4x} + \frac{3x}{27x^3 - 12x + 8} - \frac{1}{(3x-2)^2}=0 $ How to solve this equation?
$$
\frac{x+1}{9x^3-4x} + \frac{3x}{27x^3 - 12x + 8} - \frac{1}{(3x-2)^2}=0
$$
I try
$$
\frac{81x^5 - 81x^4 - 90 x^3 + 36 x^2 + 16x -16}{x(3x-2)^2(3x+2)(27x^3 - 12x + 8)}=0
$$
And then
$$
81x^... | This is really a very unpleasant problem.
If you consider the function $$f(x)=81 x^5-81 x^4-90 x^3+36 x^2+16 x-16$$ its derivative $$f'(x)=405 x^4-324 x^3-270 x^2+72 x+16$$ shows four roots (which are supposed to be expressed with radicals - see here) but they are so complex that I shall not put the solution here. Thei... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Proving $\binom{n}{m}+2\binom{n-1}{m}+......+(n-m+1)\binom{m}{m} = \binom{n+2}{m+2}$
For $m,n\in\mathbb{N},\;n\geq m$, prove the following:
$$
\tag{i}\binom{n}{m}+\binom{n-1}{m}+\binom{n-2}{m}+......+\binom{m}{m} = \binom{n+1}{m+1}
$$
$$
\tag{ii}\binom{n}{m}+2\binom{n-1}{m}+3\binom{n-2}{m}+......+(n-m+1)\binom{m}{m}... | Maybe a bit late now but here is my answer:
Consider counting the number of possible bit-strings of length $n+i$:
$$ \underbrace{111...\overbrace{\textbf{1}}^{i^{\text{th}}\ 1}...111}_{m+i\ 1\text{s}}\ 000...000 $$
So, here $n$ is the number of digits to the right of the $i^\text{th}\ 1$.
The number of bit strings is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\sum\limits_{cyc}\sqrt{\frac{a+b}{c}}\ge2\sum\limits_{cyc}\sqrt{\frac{c}{a+b}}$ Let $a,b,c$ be positive numbers. Then we need to prove
$\sqrt{\frac{a+b}{c}}+\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}\ge2\left(\sqrt{\frac{c}{a+b}}+\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}\right).$
I have an idea to set $x=\fr... | It is a consequence of Chebychev's inequality:
$$
\sum_{cyc}\sqrt{\frac{a+b}{c}}≥2\sum_{cyc}\sqrt{\frac{c}{a+b}}\iff\sum_{cyc}\frac{a+b-2c}{\sqrt{c(a+b)}}≥0
$$
Since the $a+b-2c$ and $\frac{1}{\sqrt{c(a+b)}}$ are ordered in the same way, we can apply Chebychev's inequality to obtain:
$$
\sum_{cyc}\frac{a+b-2c}{\sqrt{c(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Understanding degrees of freedom in relation to rank for $\sum_{i=1}^{n}(y_i-\bar{y})^2$ So I'm looking at this website which states:
One of the questions an instrutor [sic] dreads most from a mathematically unsophisticated audience is, "What exactly is degrees of freedom?" It's not that there's no answer. The mathema... | Take en example: the matrix
$$
A=\left[\matrix{1 & 1\\2 & 2}\right].
$$
Applying your method: the first column
$$
\left[\matrix{1 \\ 2}\right]=\left[\matrix{1 \\ 1}\right]+\left[\matrix{0 \\ 1}\right],
$$
and the second column
$$
\left[\matrix{1 \\ 2}\right]=2\left[\matrix{1 \\ 1}\right]-\left[\matrix{1 \\ 0}\right],
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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If $ a + b + c \mid a^2 + b^2 + c^2$ then $ a + b + c \mid a^n + b^n + c^n$ for infinitely many $n$
Let $ a,b,c$ positive integer such that $ a + b + c \mid a^2 + b^2 + c^2$.
Show that $ a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer $ n$.
(problem composed by Laurentiu Panaitopol)
So far no i... | It seems that there's a partial solution.
Suppose that $\mathrm{gcd}(a,a+b+c)=\mathrm{gcd}(b,a+b+c)=\mathrm{gcd}(c,a+b+c)=1$.
Then for $n=k\cdot \phi(a+b+c)+1 \, (k=1,2, \ldots )$, where $\phi$ is Euler's function, we have:
$$
(a^n+b^n+c^n)-(a^2+b^2+c^2)=a^2 (a^{n-1}-1) + b^2 (b^{n-1}-1) +
c^2 (c^{n-1}-1),
$$
where... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 0
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How to find $ab+cd$ given that $a^2+b^2=c^2+d^2=1$ and $ac+bd=0$? It is given that $a^2+b^2=c^2+d^2=1 $
And it is also given that $ac+bd=0$
What then is the value of $ab+cd$ ?
| I assume $a,b \in \mathbb{R}$. Since $a^2+b^2 = 1$, we have $-1 \leq a \leq 1$ and likewise $-1 \leq b \leq 1$. Let us take $a = \cos(\alpha)$ and $b = \sin(\alpha)$ without loss of generality. Similarly, $c = \cos(\beta)$ and $d = \sin(\beta)$.
We have $ac + bd = \sin(\alpha) \sin(\beta) + \cos(\alpha) \cos(\beta) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 3
} |
How to find the value of $(a+b+c)(a+b+d)(a+c+d)(b+c+d)$ from the following equation? I have a question about polynomial.
Given a polynomial:
$$x^4-7x^3+3x^2-21x+1=0$$
Given too that the roots of this polynomial are $a, b, c,$ and $d$.
Find the value of $(a+b+c)(a+b+d)(a+c+d)(b+c+d)$?
My attempt:
It seems I need to app... | HINT: Let $p(x)$ be the polynomial. Note that
$$(a+b+c)(a+b+d)(a+c+d)(b+c+d)=(7-d)(7-c)(7-b)(7-a)=p(?)\;.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Trying to solve $\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$ The equation is
$$\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$$
I solve it thus:
$$
\begin{cases}
2\cos^2(x)-\sqrt3=2\sin^2(x) \\
-\sqrt2 \sin(x)\ge 0 \iff \sin(x)\le 0
\end{cases}
$$
The first equation boils down to
$$2\cos^2(x)=2(1-\cos^2(x))+\sqrt3$$
$$4c... | Using $\cos2y=1-2\sin^2y=2\cos^2y-1$ on
$$2\sin^2x=2\cos^2x-\sqrt3$$
$$\iff1-\cos2x=1+\cos2x-\sqrt3\iff\cos2x=\dfrac{\sqrt3}2=\cos30^\circ$$
$$2x=360^\circ n\pm30^\circ\iff x=180^\circ n\pm15^\circ$$ where $n$ in any integer
Case$\#1:$ $+\implies x=180^\circ n+15^\circ$
But as $\sin x<0,x$ lies in the third or in the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Inequality with Four Numbers I am trying to prove the following inequality for real numbers $a,b,c,d$ all of which are greater than $1$
$8(abcd+1) > (a+1)(b+1)(c+1)(d+1)$
I tried the following approaches :
Used the AM-GM inequality
Tried to form a polynomial whose roots are a,b,c,d
Tried the trigonometric substituti... | $(a-1)(b-1)>0$ as $a,b>1 \implies ab+1>a+b $, or $2(ab+1)>ab+1+a+b=(a+1)(b+1)$.
Similarly $(c+1)(d+1)<2(cd+1)$.
Now $ab,cd>1$ then by similar method $(ab+1)(cd+1)<2(abcd+1)$ hence $4(ab+1)(cd+1)<8(abcd+1) $ therefore $(a+1)(b+1)(c+1)(d+1)<8(abcd+1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show that the function is continuous To show that the function $f: \mathbb{R}^2 \rightarrow\mathbb{R}$ with $f=\left\{\begin{matrix}
\frac{x^3-y^3}{x^2+y^2} & , (x,y) \neq (0,0)\\
0 & , (x,y)=(0,0)
\end{matrix}\right.$ is continuous on $(0,0)$ we have to show that $|f(x,y)-f(x_0,y_0)| \leq L ||(x,y)-(x_0,y_0)||$ so ... | Continue by observing that for any $\varepsilon > 0$, all values in the origin-less open disc of radius $\delta = \varepsilon/(2\sqrt{2})$ have absolute value $< \varepsilon$; this, in conjunction with $f(0, 0) = 0$, establishes the continuity of $f$ at $(0, 0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
A way to calculate e? Define three sequences:
The first sequence is $$n^n: 1,\ 4,\ 27,\ 256,\ 3125,\ 46656, \ldots$$
The second sequence is that of the ratios between adjacent members of the first series, or $$\frac{(n+1)^{n+1}}{n^n}: 4,\ \frac{27}4,\ \frac{256}{27}, \ \frac{3125}{256},\ \frac{46656}{3125},\ldots.$$
Th... | This is a continuation of @Mark Bennet's answer in which I will show, why the limit is indeed $e$.
We will concentrate on the term $T_n=(n+1)\left(\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n\right)$.
Firstly, we make a few modifications:
$$
T_n=(n+1)\left(\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 6,
"answer_id": 1
} |
Does the limit $\lim\limits_{x\to0}\left(\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}\right)$ exist? Does the limit: $$\lim\limits_{x\to0}\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}$$ exist?
| You may use the Taylor expansion, as $u \to 0$,
$$
\frac1{1-u}=1+u+O(u^2)
$$ and the Taylor expansion, as $x \to 0$,
$$
\arctan x=x-\frac{x^3}3+O(x^5),
$$ giving
$$
\begin{align}
\frac1{x\arctan x}&=\frac1{x(x-\frac{x^3}3+O(x^5))}\\\\
&=\frac1{x^2}\frac1{(1-\frac{x^2}3+O(x^4))}\\\\
&=\frac1{x^2}(1+\frac{x^2}3+O(x^4))\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Trigonometric equation $\sin v = -1/\sqrt{2}$ I'm trying to solve the following: $$\sin(v) = -\frac{1}{\sqrt{2}}$$
My attempt:
$$-\sin(v) = \frac{1}{\sqrt{2}}$$
$$\sin(-v) = \frac{1}{\sqrt{2}}$$
$$v_1 = -\frac{\pi}{4} - 2\pi n $$
$$v_2 = -\pi + \frac{\pi}{4} - 2\pi n = \frac{-3\pi}{4} - 2\pi n$$
$v_1 $ is correct, but... | Observe that $$-\frac{3\pi}{4} + 2\pi = \frac{5\pi}{4}$$ If you substitute $-m - 1$ for $n$ in the expression $$-\frac{3\pi}{4} - 2\pi n, n \in \mathbb{Z}$$ you obtain $$-\frac{3\pi}{4} - 2\pi(-m - 1) = -\frac{3\pi}{4} + 2\pi m + 2\pi = \frac{5\pi}{4} + 2\pi m$$
where $m = -n - 1 \in \mathbb{Z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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.