Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
If $a,b\in\mathbb R\setminus\{0\}$ and $a+b=4$, prove that $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2\ge12.5$. If $a,b\in\mathbb R\setminus\{0\}$ and $a+b=4$, prove that $$\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge12.5$$
I could expand everything: $$a^2+2+\frac{1}{a^2}+b^2+2+\frac{1}{b^2}\ge12.5$$
Subtract ... | Without loss of generality, we can assume $a\ge b$, so let's write $a=2+x$, $b=2-x$ with $0\le x$. The inequality is clearly satisfied if $a\ge4$, so we need only worry about the range $0\le x\lt2$. Plugging this into the OP's inequality and simplifying like crazy, we find we need only prove
$$f(x)=x^2+{(4+x^2)\over(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/616695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Calculation of $\lambda$, If $x^2+2(a+b+c)\cdot x+3\lambda \cdot (ab+bc+ca) = 0$ has real roots Let $a,b,c$ be the sides of a $\triangle$ where $a\neq b\neq c$ and $\lambda\in \mathbb{R}$. If the roots of the equation
$$x^2+2(a+b+c)\cdot x+3\lambda \cdot (ab+bc+ca) = 0$$ are real , Then which one is Right.
$\bf{Options... | Use that $(a,b,c)$ forms a triangle, hence $$a \leq b+c \\ b \leq c+a \\ c\leq a+b$$ showing $$2(ab+bc+ca) = a(b+c)+b(c+a)+c(a+b) \geq a^2+b^2+c^2.$$
Thus $\lambda \leq \frac{4}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/618036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Taylor expansion of $e^{\cos x}$ I have to find the 5th order Taylor expansion of $e ^{\cos x}$. I know how to do it by computing the derivatives of the function, but the 5th derivative is about a mile long, so I was wondering if there is an easier way to do it.
I'd appreciate any help.
| You can start with the Taylor expansion of $\cos x$ and when you expand
$\exp(\cos x)$, you just throw away terms you know won't affect the final result.
For lower order Taylor expansion, the derivation is actually pretty short and straight forward.
$$\begin{align}
\exp(\cos x)
&= \exp\left(1 - \frac{x^2}{2} + \frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/620065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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evaluation of $\lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} = $ (1) $\displaystyle \lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} = $, where $n\in \mathbb{N}$ and $\lfloor x \rfloor = $ floor function of $x$
(2)$\displaystyle \lim_{x\rightarrow \infty}\l... | Since $x-1\leq \lfloor x\rfloor \leq x$ we have $$ \frac{n\log{x}}{x}\leq \frac{n\log{x}}{\lfloor x\rfloor}\leq \frac{n\log{x}}{x-1}$$ for $x>1$. If the left and right converge to zero as $x\rightarrow \infty$ then the center does by the squeeze lemma. However, the left and right are of the form $\frac{\infty}{\infty... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How prove this stronger than Weitzenbock's inequality:$(ab+bc+ac)(a+b+c)^2\ge 12\sqrt{3}\cdot S\cdot(a^2+b^2+c^2)$ In $\Delta ABC$,$$AB=c,BC=a,AC=b,S_{ABC}=S$$
show that
$$(ab+bc+ac)(a+b+c)^2\ge 12\sqrt{3}\cdot S\cdot(a^2+b^2+c^2)$$
I know this Weitzenböck's_inequality
$$a^2+b^2+c^2\ge 4\sqrt{3}S$$
But my inequality i... | Your inequality is equivalent to $12\sqrt{3}S \le f(a,b,c)$ where
$$f(a,b,c)=\frac{(a+b+c)^2(ab+bc+ac)}{a^2+b^2+c^2}.$$
One link of Hadwiger's inequality (credit to Macavity for the link) says that $12\sqrt{3}S \le g(a,b,c)$ where, defining as in the statement there $Q=(a-b)^2+(b-c)^2+(c-a)^2,$ $g$ is defined as
$$g(a,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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if range of $f(x) = \frac{x^2+ax+b}{x^2+2x+3}$ is $[-5,4]$. Then $a$ and $b$ are If Range of $\displaystyle f(x) = \frac{x^2+ax+b}{x^2+2x+3}$ is $\left[-5,4\; \right]$ for all $\bf{x\in \mathbb{R}}$. Then values of $a$ and $b$.
$\bf{My\; Try}::$ Let $\displaystyle y=f(x) = \frac{x^2+ax+b}{x^2+2x+3} = k$,where $k\in \ma... | Note that $x^2+2x+3\gt 0 $ for any $x\in\mathbb R$.
Hence, we have
$$-5(x^2+2x+3)\le x^2+ax+b\le 4(x^2+2x+3).$$
You can simplify this, and you have two quadratic inequality which will be easy to solve.
You'll have two inequality.
$$6x^2+(a+10)x+b+15\ge0$$
$$3x^2+(8-a)x+12-b\ge0.$$
This leads that each discriminant has ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/621512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Integral: $I=\int\limits_{0}^{1}\dfrac{x^2dx}{\sqrt{3+2x-x^2}}$ Evaluate: $\displaystyle I=\int\limits_{0}^{1}\dfrac{x^2dx}{\sqrt{3+2x-x^2}}$
| As $\displaystyle\frac{x^2}{3+2x-x^2}=\frac{x^2}{2^2-(x-1)^2}$ and we are dealing with definite integral
I would like to use Trigonometric substitution $\displaystyle x-1=2\sin u$ from the very start as we don't need to get back to $x$ again
When $\displaystyle x=0,\sin u=-\frac12\implies u=-\frac\pi6$ and when $\displ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/623754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all integer solutions: $x^4+x^3+x^2+x=y^2$ Find all integer solutions of the following equation:
$$x^4+x^3+x^2+x=y^2$$
| EDIT : The case (1) in my answer has a big mistake. The case (2) is true.
$(x,y)=(0,0)$ is one of the solutions. In the following, suppose that $(x,y)\not=(0,0).$
We have
$$x(x+1)(x^2+1)=y^2.$$
First, we know that $x$ and $x+1$ are coprime, and that $x$ and $x^2+1$ are coprime, too. This leads that there is an integer ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove $ \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1 $ $$
\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1
$$
Wolframalpha shows that it i... | Why not use trig?
for $$\cos\left(\frac{2\pi}{5}\right)+ \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right) + \cos\left(\frac{8\pi}{5}\right) =-1.$$
First of notice that if $a = \frac{2\pi}{5}$ then we have
$$\cos(a) + \cos(2a) + \cos(3a)+ \cos(4a) =-1.$$
Knowing the identity that $$\cos(2a) = 2\cos^2(a)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/625101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 7
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Prove that $\lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4} $. My attempt:
We prove that $$\displaystyle \lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4} $$
It is sufficient to show that for an arbitrary real number $\epsilon\gt0$, there is a $K$
... | Your proof is basically correct, but I would encourage you to practice a bit on articulating exactly what you mean. Where you say
It is sufficient to show that $$\left|
\frac{23n+2}{4n+1} - \frac{23}{4} \right| < \epsilon $$
you mean to say something like
It is sufficient to show that for all $\epsilon\gt0$, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/631462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Evaluating $\prod_{n=2}^\infty {n^3-1\over n^3+1}$
The value of the infinite product $$P = \frac 79 \times \frac{26}{28} \times \frac{63}{65} \times \cdots \times \frac{n^3-1}{n^3+1} \times \cdots$$ is
(A) $1$
(B) $2/3$
(C) $7/3$
(D) none of the above
I wrote first 6 terms and tried to cancel out but did not get any ... | HINT:
Observe that $$n^3-1=(n-1)(n^2+n+1)\text{ and }n^3+1=(n+1)(n^2-n+1)$$
Again, $\displaystyle (n+1)^2-(n+1)+1=\cdots=n^2+n+1$
Just set a few values of $n$ to find the surviving terms
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/634973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Probability that the second roll comes up yellow given the first roll was purple. A bag contains $20$ dice. $5$ of the dice have entirely purple sides, $7$ of the dice have $2$ purple and $4$ yellow sides, and $8$ of the dice have $3$ purple and $3$ yellow sides. If you randomly pick a die, roll it, and observe that th... | The probability of picking the first type of die with all purple sides equals $\Pr(T=1) = \frac{5}{20} = \frac{1}{4}$. The probability of picking the second type of die with 2 purple and 4 yellow sides equals $\Pr(T=2) = \frac{7}{20}$. The probability of picking the third type of die with 3 purple and 3 yellow sides eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/636171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Solve $12x\equiv9\pmod{15}$ Question:
Solve $12x\equiv9\pmod{15}$
My try:
$\gcd(12,15)=3$ so it has at least $3$ solutions.
Now
$15=12\times1+3\\
3=15-12\times 1\\
3=15+2\times(-1)\\
\implies9=15\times3+12\times(-3)\\
\implies12\times(-3)\equiv9\pmod{15}$
So $x\equiv-3\pmod{15}$
Am I correct?
| $\begin{eqnarray}{\bf Hint}\quad 12x\equiv 9\!\!\!\pmod{15} &\iff& 15\mid 12x-9\\ &\iff& \dfrac{12x-9^{\phantom I}}{15} = \dfrac{4x-3}5\in\Bbb Z\\ \\ &\iff& 5\mid 4x-3\\ \\ &\iff& 4x\equiv3\!\!\!\pmod 5 \end{eqnarray}$
Hence $\,{\rm mod}\ 5\!:\,\ 3 \equiv 4x\equiv -x\ \Rightarrow\ x \equiv -3\equiv 2\ $ hence
$$\ x = 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/637041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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(Highschool Pre-calculus) Solving quadratic via completing the square I'm trying to solve the following equation by completing the square:
$x^2 - 6x = 16$
The correct answer is -6,1. This is my attempt:
$x^2 - 6x = 16$
$(x - 3)^2 = 16$
$(x - 3)^2 = 25$
$\sqrt(x -3)^2 = \sqrt(25)$
$x - 3 = \pm5$
$x =\pm5 - 3$
$x = -8,... | I am just going to do the thing first and we will talk about what you did after.
Note that $x^2-6x\sim (x-3)^2$. In fact
$$\begin{align}(x-3)^2=x^2-6x+9&=(x^2-6x)+9\\\Rightarrow x^2-6x&=(x-3)^2-9.\end{align}$$
Therefore if
$$x^2-6x=16$$ then
$$\begin{align}(x-3)^2-9&=16
\\\Rightarrow (x-3)^2&=25
\\ \Rightarrow x-3&=\pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/641599",
"timestamp": "2023-03-29T00:00:00",
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How to contruct such a sequence of rational? How to order all rational numbers from $(0,1)$ in a sequence $(x_n)_{n=1}^\infty$ in such a way that $$|x_n-x_k| \geq \frac{1}{(n+1)^2}$$ for $k<n$ ?
| It seems that $\dfrac{1}{(n+1)^2}$ can be improved. If we do the first thing that comes to our mind:
$$
\frac{1}{2},\,\,\,\frac{1}{3},\frac{2}{3},\,\,\,\frac{1}{4},\frac{3}{4},\,\,\,
\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\,\,\,
\frac{1}{6},\frac{5}{6},\,\,\,\frac{1}{7},\frac{2}{7},\frac{3}{7},\frac{4}{7},
\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/642638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Prove $\int \limits_0^b x^3 = \frac{b^4}{4} $ by considering partitions $[0, b]$ in $n$ equal subinvtervals. I was given this question as an exercise in real analysis class. Here is what I came up with. Any help is appreciated!
Prove $\int \limits_0^b x^3$ = $\frac{b^4}{4} $ by considering partitions [0, b] in $n$ equ... | You can consider the Riemann sums. For instance, the left Riemann sum
$$ \sum_{i=0}^{n-1} f(x_i) \Delta x_i = \sum_{i=0}^{n-1} \left(\frac{b}{n}i\right)^3\frac{b}{n} = \frac{b^4}{n^4}\sum_{i=0}^{n-1} i^3 = \frac{b^4}{n^4}\left({n}^{4}/4 - {n}^{3}/2+ {n}^{2}/4\right)\longrightarrow_{n\to \infty} \frac{b^4}{4} $$
To fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/645932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the volume between two surfaces
Find the volume between $z=x^2$ and $z=4-x^2-y^2$
I made the plot and it looks like this:
It seems that the projection over the $xy$-plane is an ellipse, because if $z=x^2$ and $z=4-x^2-y^2$ then $2x^2+y^2=4$ which means that $\displaystyle\frac{x^2}{(\sqrt{2})^2}+\displaystyle\f... | I agree with Fantini that you do not need the $z$ inside the integral. Generally, when you see a lot of squares, especially $x^2+y^2$, then you should be thinking about switching to polar/cylindrical coordinates. This will get the answer for this integral, though since the boundary is an ellipse it is not quite so stra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/650264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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how to find $ \lim\limits_{x \to \infty} \left(\sqrt{x^2 +1} +\sqrt{4x^2 + 1} - \sqrt{9x^2 + 1}\right)$ How can I find this?
$ \lim\limits_{x \to \infty} \left(\sqrt{x^2 +1} +\sqrt{4x^2 + 1} - \sqrt{9x^2 + 1}\right)$
| Here is another tack. If $a > 0$,
$${\sqrt{a^2 x^2 + 1} - ax } = {1\over{\sqrt{a^2 x^2 + 1} + ax }}= O\left
({1\over x}\right). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/651148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Find the value of the expression The expression $ax^2 + bx + 1$ takes the values $1$ and $4$ when $x$ takes the values $2$ and $3$ respectively. How can we find the value of the expression when $x$ takes the value of $4$?
| Let $f(x) = ax^2 + bx + 1$
Plug the values in $f(x)$ and we have:
$$f(2) = a(2)^2 + 2b + 1 = 4a + 2b + 1 = 1$$
$$f(3) = a(3)^2 + 3b + 1 = 9a + 3b + 1 = 4$$
Then just solve the following system of equations to find the values for $a$ and $b$, which is fairly easy:
\begin{cases} 4a + 2b = 0\\ 9a + 3b = 3 \end{cases}
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/652269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that $\frac{e^x + e^{-x}}2 \le e^{x^2/2}$
Prove the following inequality:
$$\dfrac{e^x + e^{-x}}2 \le e^{\frac{x^2}{2}}$$
This should be solved using Taylor series.
I tried expanding the left to the 5th degree and the right site to the 3rd degree, but it didn't help.
Any tips?
| $$\frac{e^x+e^{-x}}{2} = \frac{\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \right) + \left(1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots \right)}{2} = \frac{2 + 2 \frac{x^2}{2!} + 2 \frac{x^4}{4!} + \cdots}{2}$$
$$ = 1 + \frac{x^2}{2} + \frac{x^4}{24} + \frac{x^6}{720}+ \cdots$$
$$e^{\frac{x^2}{2}} = 1 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/654839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 1
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Integrating $\int_0^\infty\frac{1}{1+x^6}dx$ $$I=\int_0^\infty\frac{1}{1+x^6}dx$$
How do I evaluate this?
| Without complex analysis nor special functions:
$$x^6+1=(x^2+1)(x^4-x^2+1)=(x^2+1)(x^2+\sqrt3x +1)(x^2-\sqrt3x+1).$$
($x=\pm i$ are obviously roots and $(x^4-x^2+1)$ is biquadratic)
And now you have the following partial fraction decomposition:
$${1\over x^6+1} =
{Ax+B\over x^2+1} + {Cx+D\over x^2+\sqrt3x +1} + {Ex+F\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/658637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 5
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Sum from $k=1$ to $n$ of $k^3$ $$\sum_{k=1}^n k^3 = \left(\frac{1}{2}n(n+1) \right)^2$$
I want to prove this using induction. I start with $(\frac{n}{2}(n+1))^2 + (n+1)^3$ and rewrite $(n+1)^3$ as $(n+1)(n+1)^2$, then factor out an $(n+1)^2$ from the expression:
$(n+1)^2((\frac{n}{2})^2 + (n+1))$
I'm confused where to ... | $(n+1)^2((\frac{n}{2})^2 + (n+1))=(n+1)^2\frac{n^2+4n+4}{4}=(n+1)^2\frac{(n+2)^2}{2^2}= \left(\frac{1}{2}(n+1)((n+1)+1) \right)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/658956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove trig identity: $\csc \tan - \cos = \frac{\sin^2}{\cos}$ I keep hitting seeming dead-ends.
\begin{align*}
\csc\ x \tan\ x - \cos\ x &= \left(\frac{1}{\sin\ x}\right)\left(\frac{\sin\ x}{\cos\ x}\right) - \cos\ x \\
&= \frac{\sin\ x}{(\sin\ x)(\cos\ x)} - \cos\ x \\
&= \frac{\sin\ x}{(\sin\ x)(\cos\ x)} - \frac{(\... | Your method was also not wrong,
L.H.S=$\frac{sinx}{sinxcosx}$-$\frac{cosxsinxcosx}{sinxcosx}$
= $\frac{sinx-cos^2xsinx}{sinxcosx}$
= $\frac{sinx(1-cos^2)}{sinxcosx}$
= $\frac{1-cos^2}{cosx}$
= $\frac{sin^2}{cosx}$=R.H.S
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/661495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding $\lim_{x \to 0} \frac {a\sin bx -b\sin ax}{x^2 \sin ax}$ witouth L'Hopital, what is my mistake? I was working on this question.
$\lim_{x \to 0} \dfrac {a\sin bx -b\sin ax}{x^2 \sin ax}$
$\lim_{x \to 0} \dfrac {1}{x^2} \cdot \lim_{x \to 0} \dfrac { \frac {1}{abx}}{\frac {1}{abx}} \cdot \dfrac {a\sin bx -b\sin ax... | A good way is to use Taylor expansion. You know what is Taylor series for $sin(x)$. So, develop accordingly the different terms.
For the numerator, you will have $\frac{1}{6} x^3 \left(a^3 b-a b^3\right)+O\left(x^4\right)$ and for the denominator $a x^3+O\left(x^4\right)$. So, the result should be $$\frac{1}{6} \left... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
} |
Prove that $\frac{1}{1*3}+\frac{1}{3*5}+\frac{1}{5*7}+...+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$ Trying to prove that above stated question for $n \geq 1$. A hint given is that you should use $\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1})$. Using this, I think I reduced it to $\frac{1}{2}(\frac{1}{n... | Hint: Multiply and divide by $2$.
What you have done is correct. All you need to observe is terms cancel out in pairs starting from $2nd$ term.
$$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...)$$
So we are left with $1st$ and the last terms.
$$\frac{1}{2}(1-\frac{1}{2n+1})$$
$$\frac{n}{2n... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A formula for a sequence which has three odds and then three evens, alternately We know that triangular numbers are 1, 3, 6, 10, 15, 21, 28, 36... where we have alternate two odd and two even numbers. This sequence has a simple formula $a_n=n(n+1)/2$.
What would be an example of a sequence, described by a similar alge... | How about the sequence
$$a_n=\frac{1+(-1)^{\lfloor n/3\rfloor}}{2},\qquad \begin{array}{c|c|c|c|c|c|c|c|c|c}
n \strut& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\\hline
a_n \strut& 1 & 1 & 1 & 0 & 0 & 0 & 1 & 1 & 1
\end{array}\;\cdots$$
or even simpler, the sequence
$$a_n=\left\lfloor \frac{n}{3}\right\rfloor+1,\qquad \begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/665722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to tackle a recurrence that contains the sum of all previous elements? Say I have the following recurrence:
$$T(n) = n + T\left(\frac{n}{2}\right) + n + T\left(\frac{n}{4}\right) + n + T\left(\frac{n}{8}\right) + \cdots +n + T\left(\frac{n}{n}\right) $$
where $n = 2^k$, $k \in \mathbb{N} $ and $T(1) = 1$.
simplifie... | As an additional comment on this, note that if we define $T(n)$ for all $n$ and not just powers of two with $T(0)=0$ and $T(1)=1$ like this
$$T(n) = \sum_{k=1}^{\lfloor \log_2 n \rfloor} T(\lfloor n/2^k \rfloor)
+ n \lfloor \log_2 n \rfloor,$$
and the binary representation of $n$ is given by
$$n = \sum_{k=0}^{\lfloor \... | {
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"url": "https://math.stackexchange.com/questions/667929",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How prove this $|ON|\le \sqrt{a^2+b^2}$ let ellipse $M:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$,and there two point $A,B$ on $\partial M$,and the point $C\in AB$ ,such $AC=BC$,and the Circle $C$ is directly for the AB circle,for any point $N$ on $\partial C$,
show that
$$|ON|\le\sqrt{a^2+b^2}$$
my try: let
$$A(x_{1},y_{1... | Following Tian's observation and taking
$$A:(a\cos\theta,b\sin\theta),$$
$$B:(a\cos\phi,b\sin\phi)$$
we have to prove
$$4\|OC\|^2+8\|OC\|\|AC\|+4\|AC\|^2\leq 4a^2+4b^2,\tag{1} $$
or:
$$a^2(\cos\theta+\cos\phi)^2+b^2(\sin\theta+\sin\phi)^2+a^2(\cos\theta-\cos\phi)^2+b^2(\sin\theta-\sin\phi)^2+2\sqrt{\left(a^2(\cos\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/668125",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $ It is known that
\begin{align}
\arcsin x + \arcsin y =\begin{cases}
\arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 \le 1 &\text{or} &(x^2+y^2 > 1 &\text{and} &xy< 0);\\
\pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\t... | Hope the graph explains the x,y domain ( |x|< 1, |y|< 1 ) and range.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 1
} |
How can I find all integers $x≠3$ such that $x−3|x^3−3$ How can I find all integers $x≠3$ such that $x−3|x^3−3$?
I tried expand $x^3−3$ as a sum but I couldn't find a way after that.
| I am sitting with this exact problem right now. I used a more brute approach to solving the problem but I think that I am missing some solutions. Could someone take a look at my solution?
Consider $$\dfrac{x^{3}-3}{x-3}=\cdots =1-\dfrac{x(x+1)(x-1)}{x-3}.$$ Therefore $$x-3\vert x^{3}-3 \Leftrightarrow x-3\vert x~~or~~x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/672854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving trig equation with two ways gives two different answers It is given that $x$ satisfies
$$ 3\sin x + 4\cos x = 5 $$
We have to find the value(s) of $\tan (x/2) $. My approach is this:
$$\begin{align} &3\sin x + 4\cos x = 5 \\
&\implies \dfrac 3 5\sin x + \dfrac 4 5 \cos x = 1 \\
&\implies \cos ( x - \arccos(\dfr... | There is a sign error. The solution of $ \dfrac{2y}{1-y^2} = \dfrac 3 4 $ is $ y = -3, \dfrac 1 3 $.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How can you find the cubed roots of $i$? I am trying to figure out what the three possibilities of $z$ are such that
$$
z^3=i
$$
but I am stuck on how to proceed. I tried algebraically but ran into rather tedious polynomials. Could you solve this geometrically? Any help would be greatly appreciated.
| You can solve this geometrically if you know polar coordinates.
In polar coordinates, multiplication goes $(r_1, \theta_1) \cdot (r_2, \theta_2) = (r_1 \cdot r_2, \theta_1 + \theta_2)$, so cubing goes $(r, \theta)^3 = (r^3, 3\theta)$. The cube roots of $(r, \theta)$ are $\left(\sqrt[3]{r}, \frac{\theta}{3}\right)$, $\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/674621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 1
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Find the value of $\lim_{x \to - \infty} \left( \sqrt{x^2 + 2x} - \sqrt{x^2 - 2x} \right)$ I am stuck on this. I would like the algebraic explanation or trick(s) that shows that the equation below has limit of $-2$ (per the book). The wmaxima code of the equation below.
$$
\lim_{x \to - \infty} \left( \sqrt{x^2 + 2x} ... | The direct approach of just factoring an $|x|$ from each piece is not fruitful: It leads to
$$ |x| \Big(\sqrt{1 + 2/x} - \sqrt{1 - 2/x}\Big)$$
The first term grows, and the second term tends to $0$, so there's a balance between them.
Multiply top and bottom by the conjugate to find that
\begin{align*}
\sqrt{x^2 + 2x} ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Incongruent solutions to $7x \equiv 3$ (mod $15$) I'm supposed to find all the incongruent solutions to the congruency $7x \equiv 3$ (mod $15$)
\begin{align*}
7x &\equiv 3 \mod{15} \\
7x - 3 &= 15k \hspace{1in} (k \in \mathbb{Z}) \\
7x &= 15k+3\\
x &= \dfrac{15k+3}{7}\\
\end{align*}
Since $x$ must be an integer, we mus... | Let us take your equation $7x=15k+3$ and multiply both sides by $13$.
We get $91x=195k+39$. From where $(6\cdot 15+1)x=15\cdot 13k+2\cdot 15+9$, or what is the same $x=15(13k-6x+2)+9$. Since $k$ was certain integer we could say that $K:=13k-6x+2$ is certain integer and get $x=15K+9$.
So, all solutions leave the same re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/675867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Trigonometric Series Proof I am posed with the following question:
Prove that for even powers of $\sin$:
$$ \int_0^{\pi/2} \sin^{2n}(x) dx = \dfrac{1 \cdot 3 \cdot 5\cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n} \times \dfrac{\pi}{2} $$
Here is my work so far:
*
*Proof by induction
$P(1) \Rightarrow n = 2 \Rightarrow... | This proof makes use of the gamma and beta functions. We have the basic identity $$B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)} $$
Note that $$B(x, y) = \int_0^1 t^x(1 - t)^y \,dt $$
by definition. Put $x = \sin^2\theta$. Thus we have $$\int_{0}^{\pi/2}\sin^{2m + 1}\theta\cos^{2n + 1}\theta \,\,d\theta = \frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/680340",
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"source": "stackexchange",
"question_score": "1",
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An inequality for sides of a triangle Let $ a, b, c $ be sides of a triangle and $ ab+bc+ca=1 $. Show
$$(a+1)(b+1)(c+1)<4 $$
I tried Ravi substitution and got a close bound, but don't know how to make it all the way to $4 $.
I am looking for a non-calculus solution (no Lagrange multipliers).
Do you know how to do it?
| WOLOG assume that $a\geq b\geq c>0$. The constraints the problem imposes are $c=\frac{1-ab}{a+b}$, $b+c>a$. Equivalently, we have $\frac{1-ab}{a+b}+b > a\geq b$ and $b\geq \frac{1-ab}{a+b}$. These inequalities yield $$0\leq a^2-b^2<1-ab<1\quad (1)$$ and $$ab\geq \frac{1-b^2}{2}.\quad(2)$$
Note that $(1)$ implies $0<b\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/681433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Trying to get a bound on the tail of the series for $\zeta(2)$ $$\frac{\pi^2}{6} = \zeta(2) = \sum_{k=1}^\infty \frac{1}{k^2}$$ I hope we agree.
Now how do I get a grip on the tail end $\sum_{k \geq N} \frac{1}{k^2}$ which is the tail end which goes to zero?
I want to show that $\sqrt{x}\cdot \mathrm{tailend}$ is bound... | The tail is
$$\frac{1}{N^2}+\frac{1}{(N+1)^2}+\frac{1}{(N+2)^2}+\cdots.$$
For $N\gt 1$ this is less than
$$\frac{1}{(N-1)(N)}+ \frac{1}{(N)(N+1)}+ \frac{1}{(N+1)(N+2)}+\cdots.$$
Note that $\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{t+1}$. Using this we find that the tail is less than
$$\left(\frac{1}{N-1}-\frac{1}{N}\righ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Relations and Combinatorics exercise Be $A=\{1,2,3,\ldots,10\}$ Determine how many equivalence relations can be defined in $A$ with exactly two equivalence classes.
Determine how many equivalence relations can be defined in $A$ with exactly three equivalence classes.
What I did for two is:
$${10\choose1}+{10\choose2}+\... | If the chosen subset is $\{1,2,3\}$, then it's the SAME equivalence relation as if the chosen subset is $\{4,5,6,7,8,9,10\}$. Therefore when you add $\dbinom{10}{3}+\dbinom{10}{7}$, you're counting each of those equivalence relations twice.
Similarly with $\dbinom{10}{1}+\dbinom{10}{9}$ and with $\dbinom{10}{2}+\dbino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/688831",
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"source": "stackexchange",
"question_score": "2",
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Show that $(x^2-yz)^3+(y^2-zx)^3+(z^2-xy)^3-3(x^2-yz)(y^2-zx)(z^2-xy)$ is a perfect square and find its square root. Show that $(x^2-yz)^3+(y^2-zx)^3+(z^2-xy)^3-3(x^2-yz)(y^2-zx)(z^2-xy)$ is a perfect square and find its square root.
My work:
Let, $x^2-yz=a,y^2-zx=b,z^2-xy=c$. So, we can have,
$a^3+b^3+c^3-3abc=(a+b+... | Use the formula $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=\frac{1}{2}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$$ and then notice that $a-b=x^2-y^2+zx-yz=(x-y)(x+y+z)$ This will lead you to get the answer as $(x^3+y^3+z^3-3xyz)^2$ as pointed out by Ewan Delanoy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/691078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Finding J-invariant of Legendre form of Elliptic Curve PROBLEM: Put the Legendre equation $y^2 = x(x − 1)(x − λ)$ into Weierstrass
form and use this to show that the j-invariant is
j = $2^8\frac{(λ2 − λ + 1)^3}{λ^2(λ − 1)^2}$ .
Recall:
Weierstrass equation form: E: y^2 = x^3 + Ax +B
and
J(E) = 1728$\frac{4A^3}{4A^3... | In general, if a curve is given by $Y^2=X^3+AX^2+BX+C$, a change of variables $Y=y$ and $X=x-A/3$ will provide a model for the curve of the form $y^2=x^3+A'x+B'$.
The reason is that
$$X^3+AX^2+\cdots = (x-A/3)^3 + A(x-A/3)^2+\cdots$$
and the coefficient in $x^2$ is given by $3(-A/3)+A=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/692309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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To find the measure of the given angle
In triangle $ABC$ angle $\widehat C=60°$. $(AD)$ and $(BE)$ are perpendicular on $(BC)$ and $(AC)$ respectively. $M$ is the midpoint of $[AB]$. How to find the measure of angle $\widehat{EMD}$ in degrees?.
| Here is a heavyweight, straightforward solution.
Putting: $AB=c$, $AC=b$ and $BC=a$.
Let us see $\triangle DMB$ first:
*
*$BM=\frac{c}{2}$
*$BD=c\cos B$ [why?]
So, applying the cosine rule on $DMB$, we have $DM=\sqrt{(\frac{c}{2})^2+c^2\cos ^2B-c^2\cos ^2B}=\frac{c}{2}$. Similarly, doing this on $\triangle MAE$, ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove the trigonometric identity $\sin^4{x} = \frac{3-4\cos{2x}+\cos{4x}}{8}$ I need to show the steps to prove this identity:
$$\sin^4{x} = \frac{3-4\cos{2x}+\cos{4x}}{8}$$
I know that $\cos{2x}=\cos^2{x}-\sin^2{x}$. From there I do not know what to do.
The solution should look like:
$$\sin^4{x}=sin^4{x}$$
I need to p... | Let $ z = \cos(\theta) +i \sin(\theta)$
We can then find through De Moive's formula and the fact that $\cos -\theta = \cos \theta$ and $\sin -\theta = - \sin \theta$:
$ z - \frac1z \equiv 2i\sin\theta$
Thus: ($i^4=1$)
$ (z-\frac1z)^4 \equiv 16\sin^4\theta$
Expanding the RHS:
$ (z-\frac1z)^4 \equiv z^4 - 4z^2 + 6 - 4z^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determinant algebra If $A$ and $B$ are $4 \times 4$ matrices with $\det(A) = −2$, $\det(B) = 3$, what is $\det(A+B)$?
At first I approached the problem that $\det(A+B) = \det(A) + \det(B)$ but this general rule would not hold true, so I do not know how to approach the problem from here.
| The number $\det(A+B) $ is not determined by the determinants of $A$ and $B$. For instance if
$$
A=\begin{bmatrix} 1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&-2\end{bmatrix}, \ \
B=\begin{bmatrix} -1&0&0&0\\0&-1&0&0\\0&0&1&0\\0&0&0&3\end{bmatrix}
$$
then $A$ and $B$ are as required, with $\det A+B=0$. While if$$
A=\begin{bmatri... | {
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"url": "https://math.stackexchange.com/questions/697259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Help with a simple derivative I am trying to solve $\dfrac {6} {\sqrt {x^3+6}}$
and so far I made it to $6(x^3+6)^{-\frac 1 2}$
then I continued and now I have $(x^3+6)^{- \frac 3 2} * 3x^2$
and I cannot figure out what how to find the constant that should be before the parenthesis.
| $\dfrac{d}{dx} \left(\dfrac{6}{\sqrt{x^3+6}}\right) = 6 \dfrac{d}{dx}\left(\dfrac{1}{\sqrt{x^3+6}}\right) = 6\cdot -\dfrac{1}{2}\left(x^3+6\right)^{\frac{-3}{2}}\cdot 3x^2 = -9x^2\left(x^3+6\right)^{\frac{-3}{2}}$ via chain rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/699607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to determine the eigenvectors of this matrix? I have some problems to determine the eigenvectors of a given matrix:
The matrix is:
$$
A = \left( \begin{array}{ccc}
1 & 0 &0 \\
0 & 1 & 1 \\
0 & 0 & 2
\end{array} \right)
$$
I calculated the eigenvalues first and got $$ \lambda_1 = 1, \lambda_2 = 2, \lambda_3 = 1$$
T... | Every linear combination of $EV_{1}=\pmatrix{1\\0\\0}$ and $EV_3=\pmatrix{0\\1\\0}$ is a eigenvector with eigenvalue $1$.
$EV_{1,3} = span\{\left( \begin{array}{ccc}
1 \\
0 \\
0
\end{array} \right),
\left( \begin{array}{ccc}
0 \\
1 \\
0
\end{array} \right),
\left( \begin{array}{ccc}
1 \\
1 \\
0
\end{array} \right)\}$
i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/701673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Prove $\frac {1} {1+x^2} = \sum^{n-1}_{j=0} (-1)^j x^{2j} + (-1)^n \frac {x^{2n}} {1+x^2}, x\in \mathbb R, n \in \mathbb N$ by induction Prove $\frac {1} {1+x^2} = \sum^{n-1}_{j=0} (-1)^j x^{2j} + (-1)^n \frac {x^{2n}} {1+x^2}, x\in \mathbb R, n \in \mathbb N$ by induction.
$n = 1$: $\sum^{(1)-1}_{j=0} (-1)^j x^{2j} + ... | The equation in question is
$\dfrac{1}{1 + x^2} = \sum_{j = 0}^{n - 1} (-1)^j x^{2j} + (-1)^n \dfrac{x^{2n}}{1 + x^2} \tag{1}$
which, upon multiplication by $1 + x^2$ is seen to be equivalent to
$1 = (1 + x^2) \sum_{j = 0}^{n - 1} (-1)^j x^{2j} + (-1)^n x^{2n} \tag{2}$
or
$1 - (-1)^n x^{2n} = (1 + x^2) \sum_{j = 1}^{n ... | {
"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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Where is the mistake in this limit calculation? I got this limit:
$$\lim_{x\to1}\frac{\sqrt[3]{x}-1}{\sqrt[4]{x}-1} \implies \lim_{x\to1}\frac{\frac{x-1}{\sqrt[3]{x²}+\sqrt[3]{x}+1}}{\sqrt[4]{x}-1} \implies \lim_{x\to1}\frac{x-1}{\sqrt[3]{x²}+\sqrt[3]{x}+1}*\frac{1}{\sqrt[4]{x}-1}*\frac{\sqrt[4]{x}+1}{\sqrt[4]{x}+1} \i... | If fractional exponents are difficult to work with, you can also try a variable substitution: let $x = u^{12}$, so that $\sqrt[3]{x} = u^4$ and $\sqrt[4]{x} = u^3$. Then we have $$\frac{\sqrt[3]{x} - 1}{\sqrt[4]{x} - 1} = \frac{u^4 - 1}{u^3 - 1} = \frac{(u-1)(u+1)(u^2+1)}{(u-1)(u^2+u+1)} = \frac{(u+1)(u^2+1)}{u^2+u+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/703251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
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Why doesn't this calculation work? I want to find some closed form for $\gcd(x^3+1,3x^2 + 3x + 1)$ but get $7$ which is not always true.
| First
$$
3(x^3+1)-(3x^2+3x+1)(x-1)=2x+4
$$
and
$$
2(3x^2+3x+1)-3(2x+4)(x-1)=14
$$
Thus, we have
$$
(3x^2-6x+5)(3x^2+3x+1)-9(x-1)(x^3+1)=14
$$
and thus, $(x^3+1,3x^2+3x+1)\mid14$. Since $3x^2+3x+1=6\binom{x+1}{2}+1$, it is always odd. Thus, we can improve the statement to
$$
(x^3+1,3x^2+3x+1)\mid7
$$
If we look mod $7$,... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Solve inequality: $\frac{2x}{x^2-2x+5} + \frac{3x}{x^2+2x+5} \leq \frac{7}{8}$ Rational method to solve $\frac{2x}{x^2-2x+5} + \frac{3x}{x^2+2x+5} \leq \frac{7}{8}$ inequality?
I tried to lead fractions to a common denominator, but I think that this way is wrong, because I had fourth-degree polynomial in the numerator... | I think I'm found nice solving method. If $x\neq 0$:
$$\frac{2x}{x^2-2x+5} + \frac{3x}{x^2+2x+5} \leq \frac{7}{8} \equiv \frac{2}{x-2+\frac{5}{x}} + \frac{3}{x+2+\frac{5}{x}} \leq \frac{7}{8}$$
Let $t=x+\frac{5}{x}$.
$$\frac{2}{t-2} + \frac{3}{t+2} \leq \frac{7}{8}$$
$$\frac{(t-6)(7t+2)}{(t-2)(t+2)} \geq 0$$
$t \in (-... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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linear algebra exercise problem, please help me with this linear combination To express each equation of this system $$x_1-x_3=0 \qquad x_2+x_3=0$$ as a linear combination of this system $$-x_1+x_2+4x_3=0 \qquad x_1+3x_2+8x_3=0 \qquad \frac{1}{2}x_1+x_2+\frac{5}{2}x_3=0$$
but I found $\begin{pmatrix}-a+b+\frac{c}{2}\\ ... | let's call the equations in the basis set as follows:
$$\begin{align}
E1 &: -x_1+x_2+4x_3 &=0 \\
E2 &: x_1+3x_2+8x_3 &=0 \\
E3 &: \frac{1}{2}x_1+x_2+\frac{5}{2}x_3 &=0
\end{align}$$
Now, we want to find linear combinations of these three equations that give each of the following equations:
$$\begin{align}
R1 &: x_1 -... | {
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"question_score": "1",
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Two Dimensional Delta Epsilon Proof I was dawdling in some 2D delta epsilon examples, and I was wondering how to prove that the limit of $x^2+2xy+y^2$ has limit 3 as $(x,y)\rightarrow(1,1)$, using epsilon delta.
| Using Cauchy-Schwarz inequality twice:
$(x^2 + 2xy + y^2 - 4)^2 = ((x+y)^2 - 2^2)^2 = (((x-1) + (y-1))((x-1) + (y-1) + 4))^2$ <=
$2*((x-1)^2 + (y-1)^2)*18((x-1)^2 + (y-1)^2 + 1) = S$ .
So let epsilon e > 0 be given, we need to choose delta = d > 0 so that the square-root of S is less than epsilon e. First we want:
$... | {
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"answer_count": 2,
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Order of operations when using evaluation bar Suppose we have the function
\begin{align*}
f(x) = \sin(x)
\end{align*}
with first derivative
\begin{align*}
\frac{d}{dx}f(x) = \cos(x).
\end{align*}
If we evaluate $f'(x)$ at $x=0$, the result depends on whether you evaluate $f(0)$ or differentiate $f(x)$ first.
\begin{ali... | *
*These are not equivalent indeed. $$\displaystyle \frac{d}{dx}\left(f(x)\mid_{x = 0}\right)$$will always be 0.
*For that reason, this notation always means the first form.
| {
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How prove this $\frac{\cos{x}-\sin{y}}{\sin{x}-\cos{y}}=\frac{1-2\cos{x}}{1-2\sin{x}}$ let $\dfrac{\pi}{2}<x<\dfrac{3\pi}{2},0<y<\dfrac{\pi}{2}$,and such
$$\dfrac{1-\sin{x}}{1-\cos{x}}=\dfrac{1-\sin{y}}{1-\cos{y}}$$
show that
$$\dfrac{\cos{x}-\sin{y}}{\sin{x}-\cos{y}}=\dfrac{1-2\cos{x}}{1-2\sin{x}}$$
my idea:
$$\Longle... | I'm sure that there is at least one sublimer solution than this as this is not how the problem came along , but its legitimate.
If we cross multiply the relation to be proved, we find
$$\sin x+\sin y-(\cos x+\cos y)=-2\cos(x+y)$$
Using Prosthaphaeresis & Double Angle Formulas,
$$2\cos\frac{x-y}2\left(\sin\frac{x+y}2-\... | {
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"source": "stackexchange",
"question_score": "2",
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Prove that $2^x \cdot 3^y - 5^z \cdot 7^w = 1$ has no solutions Prove that $$2^x \cdot 3^y - 5^z \cdot 7^w = 1$$ has no solutions in $\mathbb{Z}^+$, if $y\ge 3$.
| Here's a proof without Størmer's theorem, relying purely on (a lot of) modular arithmetic:
Observation 1: $z$ is odd.
Reducing mod $3$ shows that
$$1=2^x\cdot3^y-5^z\cdot7^w\equiv-(-1)^z\pmod{3}.$$
Observation 2: $x=2$ and $y\equiv2\pmod{12}$.
Reducing mod $8$ shows that if $x\geq3$ then
$$1=2^x\cdot3^y-5^z\cdot7^w\eq... | {
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"source": "stackexchange",
"question_score": "3",
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Solve congruence system $\ x\equiv m_i-1 \pmod{m_i}\,$ for $\,i = 1,\ldots, k$ Find natural number $x$ so that
$$\begin{align}x&\equiv 9\pmod{10}\\ x&\equiv8\pmod9\\ &\ \ \vdots\\ x&\equiv 1\pmod2\end{align}$$
| \begin{align}
x &\equiv 9 \pmod{10} \\
x &\equiv 8 \pmod 9 \\
x &\equiv 7 \pmod 8 \\
x &\equiv 6 \pmod 7 \\
x &\equiv 5 \pmod 6 \\
x &\equiv 4 \pmod 5 \\
x &\equiv 3 \pmod 4 \\
x &\equiv 2 \pmod 3 \\
x &\equiv 1 \pmod 2 \\
\end{align}
Is equivalent to
\begin{align}
x &\equiv -1 \pmod{10} \... | {
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"source": "stackexchange",
"question_score": "1",
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Prove that $\sqrt[3]{\sec\frac{2\pi }{7}} + \sqrt[3]{\sec\frac{4\pi }{7}} + \sqrt[3]{\sec\frac{8\pi }{7}} = \sqrt[3]{8-6\sqrt[3]{7}}$ Prove that:
$$\sqrt[3]{\sec\frac{2\pi }{7}} + \sqrt[3]{\sec\frac{4\pi }{7}} + \sqrt[3]{\sec\frac{8\pi }{7}} = \sqrt[3]{8-6\sqrt[3]{7}}$$
Thank you!
Avdiu...
| $$
\sec\frac{2\pi}{7},\qquad
\sec\frac{4\pi}{7},\qquad
\sec\frac{8\pi}{7}
$$
are the three zeros of the polynomial $z^3+4z^2-4z-8$ .
The cube-roots of those zeros are the zeros of the polynomial $y^9+4y^6-4y^3-8$ . Using as a hint the right-hand side above, we can factor this polynomial to get a polynomial of degree... | {
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"source": "stackexchange",
"question_score": "2",
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What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression using sums $\sum$ and double sums $\sum$$\sum$?
| The sum can be derived via combinatorial argument.
Count the number of quintuples $(x_1,x_2,x_3, x_4,x_5)$ with the property $x_5> \max\limits_{i\in S}$ from the set $S=\{1,2,...,n,n+1\}$,
then as $x_5$ varies from $2$ to $n+1$, the positions $(x_1,x_2,x_3, x_4)$
the positions can be filled in $\sum\limits_{i=1}^n i^4... | {
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"source": "stackexchange",
"question_score": "5",
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Solving the functional equation $f\big(4f(x)-3x\big)=x$ Find all functions $f:[0,+\infty)\to [0,+\infty)$ such that $f(x)\geq \frac{3x}{4}$ and
$$f\big(4f(x)-3x\big)=x,\forall x\in[0,+\infty)$$
| Note that $4f(x) - 3x\geq 0$ for all $x\geq 0$.
Given a lower bound of the form $f(x) \geq a x$ with $a>0$ for $x \geq 0$. Let $x > 0$ and set $y = 4f(x) - 3x \geq 0$. Then
$$f(y) = x \geq a y = a(4f(x) - 3x) = 4af(x) - 3 ax,$$
so $f(x) \leq \frac{(3a + 1)x}{4a}$.
Similarly, given an upper bound of the form $f(x)\leq... | {
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"source": "stackexchange",
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Definite integral by u-substitution (1/u^2), u given $$\int_{-3}^0 \frac{-8x}{(2x^2+3)^2}dx; u=2x^2+3$$
I need help solving this integral -- I'm completely bewildered. I've attempted it many times already and I don't know what I am doing wrong in my work, and I seem to be having the same problem with other definite int... | Since the derivative of $2x^2 + 3$ is almost the numerator, we set $u = 2x^2 + 3$. Then $du = 4x dx$ or $dx = du/(4x)$. Then take the integral so,
$$-\int \frac{8x}{(2x^2 + 3)^2}dx = -\int \frac{2}{u^2} du = \frac{2}{u} + C = \frac{2}{2x^3 + 3} + C.$$
Therefore
$$-\int_{-3}^0 \frac{8x}{(2x^2 + 3)^2}dx = 2\left[\frac{1}... | {
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Rationalization of $\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$
Question:
$$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$$ equals:
My approach:
I tried to rationalize the denominator by multiplying it by $\frac{\sqrt{2}-\sqrt{3}-\sqrt{5}}{\sqrt{2}-\sqrt{3}-\sqrt{5}}$. And got the result to be (after a long calculat... | $$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\cdot\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$$
$$=\frac{2\sqrt{6}\cdot(\sqrt{2}+\sqrt{3}-\sqrt{5})}{(\sqrt{2}+\sqrt{3})^2-5}=\frac{2\sqrt{6}\cdot(\sqrt{2}+\sqrt{3}-\sqrt{5})}{5+2\sqrt{6}-5}$$
$$=\frac{2\sqrt... | {
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"source": "stackexchange",
"question_score": "12",
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How to prove/show that the sequence $a_n=\frac{1}{\sqrt{n^2+1}+n}$ is decreasing? How to prove/show that the sequence $a_n=\frac{1}{\sqrt{n^2+1}+n}$ is decreasing?
My idea:
*
*$n^2<(n+1)^2 /+1$
*$n^2+1<(n+1)^2+1/ \sqrt{}$
*$\sqrt{n^2+1}<\sqrt{(n+1)^2+1}/+n$
*$\sqrt{n^2+1}+n<\sqrt{(n+1)^2+1}+n$
And now I'm stuck... | You have to prove $$\frac{1}{\sqrt{n^2+1}+n}>\frac{1}{\sqrt{(n+1)^2+1}+n+1}$$
i.e., $$\sqrt{n^2+1}+n< \sqrt{(n+1)^2+1}+n+1$$
i.e., $$\sqrt{n^2+1}< \sqrt{(n+1)^2+1}+1$$
i.e., $$n^2+1< (n+1)^2+1+1+2(\sqrt{(n+1)^2+1})$$
i.e., $$n^2< (n+1)^2+1+2(\sqrt{(n+1)^2+1})$$
i.e., $$0 <2n+1+1+2(\sqrt{(n+1)^2+1})$$
i.e., $$0 <n+(\sqr... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Proving an inequality using induction Use induction to prove the following:
$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2^n}\geq1+\frac{n}{2}$
What would the base case be? Would it still be $n=0$
so $\frac{1}{1}+\frac{1}{2}\geq 1+\frac{0}{2}$, which holds true.
then how would you prove for $n$ and $n+1$ to pro... | Base case for $n = 0$: $\frac{1}{1} \geq 1$
Inductive step: Assume $\displaystyle \sum_{k = 1}^{2^n} \frac{1}{k} \geq 1 + \frac{n}{2}$
Then, after substituting the inductive step, what we want to prove is $\displaystyle \sum_{k = 2^n + 1}^{2^{n+1}} \frac{1}{k} \geq \frac{1}{2}$
This is clear to show by a comparison:
$\... | {
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"source": "stackexchange",
"question_score": "2",
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How to evaluate $\int_0^1\frac{1+x^4}{1+x^6}\,dx$ $$\int_0^1\frac{1+x^4}{1+x^6}\,dx$$
Can anyone help me solve the question? I am struggling with this.
| $$ \begin{align} \int_{0}^{1} \frac{x^{4}+1}{x^{6}+1} \ dx &= \int_{0}^{1} \frac{x^{4}+1+x^{2}-x^{2}}{x^{6}+1} \ dx \\ &= \int_{0}^{1} \frac{x^{4}-x^{2}+1}{x^{6}+1} \ dx + \int_{0}^{1}\frac{x^{2}}{x^{6}+1} \ dx \\ &= \int_{0}^{1} \frac{x^{4}-x^{2}+1}{(x^{2}+1)(x^{4}-x^{2}+1)} \ dx + \int_{0}^{1}\frac{x^{2}}{(x^{3})^{... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Equation with floor function How would one solve an equation with a floor function in it:
$$a - (2x + 1)\left\lfloor{\frac {a - 2x(x + 1)}{2x + 1}}\right\rfloor - 2x(x + 1) = 0$$
$a$ is a given and can be treated as a natural number, and all $x$ other than integers can be discarded. At least one non-trivial solution wo... | Rearranging we get,
$$\frac{a - 2x(x + 1)}{(2x + 1) } =\left\lfloor{\frac {a - 2x(x + 1)}{2x + 1}}\right\rfloor $$
This means,$\frac{a - 2x(x + 1)}{(2x + 1) }$ is an integer.
$$\frac{a - 2x(x + 1)}{(2x + 1) }=k$$
$$a=2xk+k+2x^2+2x$$
$$2x^2+2(k+1)x+k-a=0$$
Applying the quadratic formula,
$$x = \frac{-2(k+1)\pm\sqrt{4... | {
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How to find characteristic polynomial of this matrix?
Let, $A=\begin{bmatrix} 4&0&1&0\\1&1&1&0\\0&1&1&0 \\0&0&0&4 \end{bmatrix}$. Knowing that $4$ is one of its eigenvalues, find the characteristic polynomial of $A$.
Well if $4$ is an eigenvalues of $A$, one should have $|A-4I_{4}|=0$ .
And so,
$\begin{vmatrix} 0&0&1... | ...and how to find the charateristic polynomial of the original matrix, $A$
$$\begin{align}\mathrm{det}(A - \lambda \mathrm{I}) &= 0 \tag{1}\\
\begin{vmatrix}(4-\lambda)&0&1&0 \\
1&(1-\lambda)&1&0\\
0&1&(1-\lambda)&0\\
0&0&0&(4-\lambda) \end{vmatrix} &= 0 \tag{2}\\
(4-\lambda)\begin{vmatrix}(4-\lambda)&0&1\\
1&(1-\lamb... | {
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How to write $x=2\cos(3t) y=3\sin(2t)$ in rectangular coordinates? How would I write the following in terms of $x$ and $y$? I think I use the inverse $\cos$ or $\sin$?
$$x=2\cos(3t)\,, \quad y=3\sin(2t)$$
| We have $\cos(3t)=\frac{x}{2}$. Using the identity $\cos 2u=2\cos^2(u)-1$, we get
$$\cos(6t)=2\left(\frac{x}{2}\right)^2-1.\tag{1}$$
We also have $\sin(2t)=\frac{y}{3}$. Using the not so well-known identity $\sin(3u)=3\sin u-4\sin^3 u$, we get
$$\sin(6t)=3\left(\frac{y}{3}\right)-4\left(\frac{y}{3}\right)^3.\tag{2}$$
... | {
"language": "en",
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"question_score": "3",
"answer_count": 3,
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Prove $\int\limits_{0}^{\pi/2}\frac{dx}{1+\sin^2{(\tan{x})}}=\frac{\pi}{2\sqrt{2}}\bigl(\frac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\bigr)$
Prove the following integral
$$I=\int\limits_{0}^{\frac{\pi}{2}}\dfrac{dx}{1+\sin^2{(\tan{x})}}=\dfrac{\pi}{2\sqrt{2}}\left(\dfrac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\right)$$
This in... | I hope nobody cares that i exhume this question, but i found it interesting that it is possible to obtain this integral by a relativly straightforward contour integration method.
Observe that,following the question opener and using parity, that we can rewrite the integral as
$$
\frac{1}{2}\int^{\infty}_{-\infty}\frac{1... | {
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Find the least value of x which when divided by 3 leaves remainder 1, ... A number when divided by 3 gives a remainder of 1; when divided by 4, gives a remainder of 2; when divided by 5, gives a remainder of 3; and when divided by 6, gives a remainder of 4. Find the smallest such number.
How to solve this question in 1... | So we need $\displaystyle x=3a+1=4b+2=5c+3=6d+4$
which can also the written as $\displaystyle x=3(a+1)-2=4(b+1)-2=5(c+1)-2=6(d+1)-2$
So, we need to find $x$ such the remainder $=-2$ for the divisors $3,4,5,6$
Now, the smallest number which is divisible by $3,4,5,6$ is lcm$(3,4,5,6)=60$
So, $60m-2$ (where $m$ is an inte... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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cyclic subgroup elements I'm having hard time finding elements of the cyclic subgroup $\langle a\rangle$ in $S_{10}$, where $a = (1\ 3\ 8\ 2\ 5\ 10)(4\ 7\ 6\ 9)$
This is my attempt:
\begin{align}
a^2 &= (1\ 8\ 5\ 10)(4\ 6\ 9) \\
a^3 &= (1\ 3\ 5\ 10)(4\ 7\ 9\ 6) \\
a^4 &= (1\ 5\ 10)(4\ 9\ 7) \\
a^5 &= (1\ 3\ 8\ 2\ 10)(7... | Your calculations look wrong. Keep in mind that $a = (1\ 3\ 8\ 2\ 5\ 10)(4\ 7\ 6\ 9)$ is the map $a:\{1,2,\dots,10\}\to\{1,2,\dots,10\}$ given by
$$
a : 1\mapsto 3, 3\mapsto 8, 8 \mapsto 2, 2\mapsto5, 5\mapsto 10, 10\mapsto 1, 4\mapsto 7, 7\mapsto 6, 6 \mapsto 9, 9\mapsto 4.
$$
Applying this map twice yields
$$
a^2 : 1... | {
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"source": "stackexchange",
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How to calculate the diagonal matrix The question says: find the eigenvalues and corresponding eigenvectors of the matrix $A$. This I could do. But then it says: hence find a non-singular matrix $P$ and a diagonal matrix $D$ such that $A + A^2 + A^3 = PDP^{-1}$ , where
$$
A =\begin{pmatrix}
6 & 4 & 1 \... | You can find $Q$ such that $A=QDQ^{-1}$ with $D$ diagonal, right? Then show that $$A+A^2+A^3=Q(D+D^2+D^3)Q^{-1}$$
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Proof by induction that $n^3 + (n + 1)^3 + (n + 2)^3$ is a multiple of $9$. Please mark/grade. What do you think about my first induction proof? Please mark/grade.
Theorem
The sum of the cubes of three consecutive natural numbers is a multiple of 9.
Proof
First, introducing a predicate $P$ over $\mathbb{N}$, we rephra... | Formulation, base case, inductive hypothesis, inductive step, it all looks good. :)
One might also conclude with a clarifying statement about what has been done - that the hypothesis is true for all $n \in \Bbb N$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Taylor series of $\sqrt{1+x}$ using sigma notation I want help in writing Taylor series of $\sqrt{1+x}$ using sigma notation
I got till
$1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5x^4}{128}+\ldots$ and so on.
But I don't know what will come in sigma notation.
| If $f(x) = \sqrt{1+x} = (1+x)^\frac{1}{2}$, then $f'(x) = \frac{1}{2}(1+x)^{-\frac{1}{2}}$ and $f''(x) = \frac{1}{2}\frac{-1}{2}(1+x)^{-\frac{3}{2}}$. In general, $$
\frac{d}{dx}(1 + x)^{\frac{-n}{2}} = \frac{-n}{2}(1 + x)^{\frac{-n-2}{2}} \text{,}
$$
and therefore the $n$-th derivative of $f(x) = \sqrt{1+x}$ is (for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/732540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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Cartesian to polar coordinates - Integration $$\iint_R \frac{1}{1+x^2+y^2} \,dA$$
$$R=\left\{(r,\theta):1\le r\le 2,0\le \theta \le \pi\right\}$$
limits of outer integral are $0$ to $\pi$ and inner integral are $1$ to $2$. I wanted to confirm if i did the problem right.
My answeR: $(1/2)\ln(5/2)\pi$
| Let $r=\sqrt{x^2+y^2}$ and $dA=dxdy =rdrd\theta$, then
$$\int \int_{R}\frac{dA}{1+x^2+y^2}=\int_{\theta=0}^{\theta=\pi}\int_{r=1}^{r=2}\frac{r drd\theta}{1+r^2}= $$
$$=\int_{\theta =0}^{\theta =\pi}\left(\frac{1}{2} \int_{r=1}^{r=2}\frac{2r dr}{1+r^2}\right)d\theta=\int_{\theta=0}^{\theta=\pi}d\theta \cdot \left.\frac{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Demonstrate that $\displaystyle \frac{(2n - 2)!!}{(2n - 3)!!} \simeq 1.7 \sqrt{n}$ As in the title, I know that
$\displaystyle \frac{(2n - 2)!!}{(2n - 3)!!} = \frac{(2n - 2)(2n - 4)\cdots 4 \cdot 2}{(2n - 3)(2n - 5) \cdots 3 \cdot 1} \simeq 1.7 \sqrt{n}$
Could you give some hint to prove it?
(should I look the series e... | Use the Stirling formula:
$$\frac{(2n - 2)(2n - 4)\cdots 4 \cdot 2}{(2n - 3)(2n - 5) \cdots 3 \cdot 1}
=\frac{(2n - 2)^2(2n - 4)^2\cdots 4^2 \cdot 2^2}
{(2n - 2)!}
\\=
2^{2n-2}
\frac{(n - 1)^2(n - 2)^2\cdots 2^2 \cdot 1^2}
{(2n - 2)!}
\\=2^{2n-2}
\frac{((n - 1)!)^2}{(2n - 2)!}
\sim 2^{2n-2}\frac{(n-1)^{2n-2}e^{-2n+2}2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/738107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find the $a_{10}+a_{2014}$ if $a_{n+1}=\frac{8}{5}a_{n}+\frac{6}{5}\sqrt{4^n-a^2_{n}}$
The sequence $\{a_{n}\}$ satisfies $a_{0}=1$,and
$$a_{n+1}=\dfrac{8}{5}a_{n}+\dfrac{6}{5}\sqrt{4^n-a^2_{n}},n\ge 0$$
Find the $a_{10}+a_{2014}$.
My idea: since
$$5a_{n+1}-8a_{n}=6\sqrt{4^n-a^2_{n}},n\ge 0$$
so
$$25a^2_{n+1}-8... | We have $a_0=1, a_1=\frac{8}{5}, a_2=4, a_3=\frac{32}{5}$. This leads us to see the pattern $$a_{2n}=2^{2n}, a_{2n+1}=\frac{2^{2n+3}}{5}$$ This is easily proven by induction. Thus $a_{10}+a_{2014}=2^{10}+2^{2014}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integrating partial fractions I have
$\int{\frac{2x+1}{x^2+4x+4}}dx$
Factorising the denominator I have
$\int{\frac{2x+1}{(x+2)(x+2)}}dx$
From there I split the top term into two parts to make it easier to integrate
$\int{\frac{2x+1}{(x+2)(x+2)}}dx$ = $\int{\frac{A}{(x+2)}+\frac{B}{(x+2)}}dx$
=$\int{\frac{A(x+2)}{(x+2)... | You want to try a split like
$$\frac{2 x+1}{(x+2)^2} = \frac{A}{x+2} + \frac{B x}{(x+2)^2}$$
Then $A+B=2$ and $2 A=1$. The decomposition is then
$$\frac{2 x+1}{(x+2)^2} = \frac12 \frac1{x+2} + \frac{3}{2} \frac{x}{(x+2)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/741861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Optimize function on $x^2 + y^2 + z^2 \leq 1$ Optimize $f(x,y,z) = xyz + xy$ on $\mathbb{D} = \{ (x,y,z) \in \mathbb{R^3} : x,y,z \geq 0 \wedge x^2 + y^2 + z^2 \leq 1 \}$. The equation $\nabla f(x,y,z) = (0,0,0)$ yields $x = 0, y = 0, z \geq 0 $ and we can evaluate $f(0,0,z) = 0$.
Now studying the function on the boun... | $f=xy(z+1) \le \dfrac{x^2+y^2}{2}(z+1)=\dfrac{(z+1)(k-z^2)}{2},k\le1$
$f'_z=\dfrac{k-z^2-2z^2-2z}{2}=0,z=\dfrac{-2 \pm \sqrt{4+12k}}{6}=\dfrac{-1 \pm \sqrt{1+3k}}{3}$
for max: $z=\dfrac{-1 + \sqrt{1+3k}}{3},f_{max}=\dfrac{(2+\sqrt{1+3k})(6k-1+2\sqrt{1+3k})}{2*27}\le \dfrac{32}{2*27}$ ,
when $k=1$ get max becasue $6k-1+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Area bounded by two circles $x^2 + y^2 = 1, x^2 + (y-1)^2 = 1$ Consider the area enclosed by two circles: $x^2 + y^2 = 1, x^2 + (y-1)^2 = 1$
Calculate this area using double integrals:
I think I have determined the region to be $D = \{(x,y)| 0 \leq y \leq 1, \sqrt{1-y^2} \leq x \leq \sqrt{1 - (1-y)^2}\}$
Now I can't se... | Solve the system: $x^2 + y^2 = 1 = x^2 + (y - 1)^2$ we have: $x =\dfrac{\sqrt{3}}{2}$ and $y = \dfrac{1}{2}$, and $x = -\dfrac{\sqrt{3}}{2}$ and $y = \dfrac{1}{2}$. So the set up for the integral is:$\int_\frac{-\sqrt{3}}{2}^{\frac{\sqrt{3}}{2}} \int_{1 -\sqrt{1 - x^2}}^{\sqrt{1 - x^2}} 1dydx$ which can be easily compu... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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AM-GM inequality On the wikipedia page on
"Nesbit's inequality", the fifth proof ends as follows:
$$ \frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}\geq 6$$
which is true, by AM-GM inequality.
I am wondering if the inequality is obvious / immediate from just looking at it and how you see this immediately without resorting t... | $$ \frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}\geq 6$$ is equivalent to
$$ \frac{x^2 z+z^2x + y^2 z+z^2 y + x^2 y + y^2 x}{6}\geq xyz$$
which is true by the AM-GM inequality:
$$ \frac{x^2 z+z^2x + y^2 z+z^2 y + x^2 y + y^2 x}{6}\geq \sqrt[6]{x^6y^6z^6}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Best strategy for rolling $20$-sided and $10$-sided dice There is a $20$-sided (face value of $1$-$20$) die and a $10$-sided (face value of $1$-$10$) dice. $A$ and $B$ roll the $20$ and $10$-sided dice, respectively. Both of them can roll their dice twice. They may choose to stop after the first roll or may continue to... | $B$ should reroll anything $5$ or below as he is likely to improve, while with a $6$ he is more likely to get worse. He then has $\frac 1{20}$ chance of getting each number $1$ to $5$ and $\frac 3{20}$ chance of any number $6$ to $10$.
$A$ can then compute the chance that each number wins. Any number $11$ through $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Convergence of $ \sum_{n=1}^{\infty} (\frac{n^2+1}{n^2+n+1})^{n^2}$
Find if the following series converge: $$\displaystyle \sum_{n=1}^{\infty} \left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}$$
What I did:
$$a_n=\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}$$
$$b_n=\frac {1}{(n+1)^{n^2}}$$
$$\forall n\ge 1 : a_n < b_n \ \Rightar... | This is likely what I'd have a Calculus student do. Use the root test:
$$\lim_{n \rightarrow \infty} \left| \left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}\right|^\frac{1}{n} = \lim_{n \rightarrow \infty} \left( \frac{n^2+1}{n^2+n+1}\right)^n$$
$$\begin{array} \\
\lim_{n \rightarrow \infty} \ln \left(\left( \frac{n^2+1}{n^2+n... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding correlation coefficient Two dice are thrown. $X$ denotes number on first die and $Y$ denotes maximum of the numbers on the two dice. Compute the correlation coefficient.
| We outline a basically computational approach, leaving details to you. The mean of $X$ is easily found to be $\frac{7}{2}$. For the variance of $X$, we first find $E(X^2)$, which is
$$\frac{1}{6}(1^2+2^2+3^2+4^2+5^2+6^2).$$
Thus $\text{Var}(X)=\frac{91}{6}-\left(\frac{7}{2}\right)^2=\frac{35}{12}$.
Next we find the mea... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I solve this definite integral: $\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$? $$\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$$
I have already solved the indefinite integral by transforming $\sin^{4}x + \cos^{4}x$ as follows:
$\sin^{4}x + \cos^{4}x = (\sin^{2}x + \cos^{2}x)^{2} - 2\cdot\sin^{2}x\cdot\cos^{2... | Let $z=e^{i x}$; then $dx = -i dz/z$ and the integral is equal to
$$-i 8 \oint_{|z|=1} dz \frac{z^3}{z^8 + 6 z^4+1}$$
By the residue theorem, the integral is then equal to $i 2 \pi$ times the sum of the residues at each pole inside the unit circle. The residue at each pole $z_k$ is equal to
$$-i 8 \frac{z_k^3}{8 z_k^7... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the solutions of: $\sin x+\cos x=\sin^2 x+0.5\sin{2x}$ Find the solutions of: $\sin x+\cos x=\sin ^2 x+0.5\sin{(2x)}$
How can I find the solutions ?
| Your equation, as you wrote it, is not true for all $x \in \mathbb{R}$, nor even for most of them. Consider $x=0$, in which case the left hand side evaluates to $1$ and the right hand side equals $0$.
We can still find what values of $x$, if any, satisfy your equation.
Recall that
$$\sin (2x) = 2\sin x \cos x$$
Subs... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that $(x + 1)^{(2n + 1)} + x^{(n + 2)}$ can be divided by $x^2 + x + 1$ without remainder I am in my pre-academic year. We recently studied the Remainder sentence (at least that's what I think it translates) which states that any polynomial can be written as $P = Q\cdot L + R$
I am unable to solve the following:
... | Hint: try to check if complex roots of $x^2+x+1$ are also roots for $(x+1)^{2n+1}+x^{n+2}$.
Also there is another way to prove it. Let see that $((x+1)^2-x)\cdot (x+1)^{2n-1}$ also divided by $x^2+x+1$. So we just need to prove that $x\cdot (x+1)^{2n-1}+x^{n+2}$ is divided by $x^2+x+1$. In this way we can come to prove... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove $x^3$ is strictly increasing I am trying to use $f(x)=x^3$ as a counterexample to the following statement.
If $f(x)$ is strictly increasing over $[a,b]$ then for any $x\in (a,b), f'(x)>0$.
But how can I show that $f(x)=x^3$ is strictly increasing?
| Consider $a^3-b^3$, where $a\gt b$. We want to show this is positive.
We have
$$a^3-b^3=(a-b)(a^2+ab+b^2).$$
Note that $a^2+ab+b^2$ is positive unless $a=b=0$. There are many ways to do this. For example,
$$a^2+ab+b^2=\frac{1}{2}\left(a^2+b^2+(a+b)^2\right).$$
More conventionally, complete the square. We have $4(a^2+a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Quadratic equations and inequalities $\sqrt{4n+1}<\sqrt{n} + \sqrt{n+1}<\sqrt{4n+2}$ and $[\sqrt{n}+\sqrt{n+1}] = [\sqrt{4n+1}]$ For every positive integer $n$, prove that
$$\sqrt{4n+1}<\sqrt{n} + \sqrt{n+1}<\sqrt{4n+2}$$
Hence or otherwise, prove that $[\sqrt{n}+\sqrt{n+1}] = [\sqrt{4n+1}]$, where $[x]... | Clearly $\lfloor\sqrt{4n+1}\rfloor\le\lfloor\sqrt n+\sqrt{n+1}\rfloor\le \lfloor\sqrt{4n+2}\rfloor$. Thus the claim could only be wrong if $\lfloor\sqrt{4n+1}\rfloor< \lfloor\sqrt{4n+2}\rfloor$, i.e. if there exist an ineger $m$ with $\sqrt {4n+1}<m\le\sqrt {4n+2}$, equivalently: with $4n+1<m^2\le 4n+2$. But we cannot ... | {
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"timestamp": "2023-03-29T00:00:00",
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How to prove uniform convergence for sequences $f_n = x^n(1-x), f_n = \frac {n^3x} {1+n^4x^2}$ and $ f_n = \sqrt {n} xe^{-nx^2}$ on $[0,1]$
How to prove uniform convergence for sequences $f_n = x^n(1-x), f_n = \frac {n^3x} {1+n^4x^2}$ and $ f_n = \sqrt {n} xe^{-nx^2}$ on $[0,1]$
I've already shown that the following ... | I will show the 1st function $f_n(x) = x^n\cdot (1 - x)$ to converge uniformly on $[0, 1]$. $f_n'(x) = nx^{n-1} - (n+1)x^n = 0 \iff x = 0, \dfrac{n}{n+1}$. So $\displaystyle \sup_{x \in [0, 1]} |f_n(x) - 0| = f_n\left(\frac{n}{n+1}\right) = \dfrac{1}{n+1}\cdot \dfrac{1}{(1 + \frac{1}{n})^n} \to 0$ as $n \to \infty$.
F... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For which $a$ is a matrix $A$ diagonalizable? Say I have a matrix $A_a$ with
$$A_a:= \left(\begin{array}{c}
2 & a+1 & 0 \\
-a & -3a & -a \\
a & 3a+2 & a+2
\end{array}\right)$$
I was wondering if there was an easy way to determine for which $a$ the matrix would be diagonalizable.
I tried to determine its ei... | Given:
$$A_a:= \left(\begin{array}{c}
2 & a+1 & 0 \\
-a & -3a & -a \\
a & 3a+2 & a+2
\end{array}\right)$$
We find the eigenvalues using $|A_a - \lambda I| = 0$, yielding:
$$-a^2 \lambda +2 a^2-2 a \lambda ^2+7 a \lambda -6 a-\lambda ^3+4 \lambda ^2-4 \lambda = 0 \implies (\lambda-2)(\lambda - (1-a-\sqrt{a+... | {
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Find all integer solutions of $x^4 + 2x^3 + 2x^2 + 2x +5 = y^2$ Find all integer solutions to $x^4 + 2x^3 + 2x^2 + 2x +5 = y^2$.
I'm in a dead end. I've transformed the expression in the following state:
$(x^2+1)(x+1)^2 = y^2 -4$
I couldn't see anyway in which I could work with this expression in this state, so I co... | Hint: $(x^2+x)^2 < y^2 < (x^2+x+1)^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Last digits, numbers Can anyone please help me?
1) Find the last digit of $7^{12345}$
2) Find the last 2 digits of $3^{3^{2014}}$.
Attempt: 1)
By just setting the powers of $7$ we have $7^1 = 7$, $7^2=49$, $7^3=343$, $7^4 = 2401$, $7^5 = 16807$, $7^6 = 117649$, $\dots$
After the power of $4$, the last digits will rep... | First of all, finding last $n$ digits in Base $10$ essentially finding modulo $\displaystyle10^n$
Using Carmichael function, $\displaystyle3^{\lambda(100)}\equiv1\pmod{100}$ as $(3,100)=1$
$$\implies3^{3^{2014}}\equiv3^{3^{2014}\pmod{\lambda(100)}}\pmod{100}$$
Now $\displaystyle\lambda(100)=20,$ so we need to find $\di... | {
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$-5| 2+4x | = -32(x+3/4)- | x | + 1$ This was my attempt:
$$-5| 2+4x | = -32\left(x+\frac34\right)- | x | + 1\\
\implies|2+4x|=\frac{-32x-24- | x | + 1}{-5}\\
\implies2+4x=\pm \frac{-32x+-24- x + 1}{-5}\\
\implies4x=\pm \frac{-33x+-23 }{-5}-2\\
\implies-20x=\pm (-33x-23)+10\\
\implies-20x= -33x-13\text{ or} -20x=33x+3... | You made a mistake between the second and the third line:
$$|2+4x|=\frac{-32x-24- | x | + 1}{-5}\\
2+4x=\pm \frac{-32x-24- \color{red}{x} + 1}{-5}\\$$ instead it should have been $$2+4x=\pm \frac{-32x-24- \color{red}{|x|} + 1}{-5}.\\$$
Going on you get
$$-10-20x=\pm(-32x-24- |x| + 1)$$
Now you should treat the 2 case... | {
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Is this a valid method of finding magnitude of complex fraction If I have a complex fraction $\dfrac{a+bi}{c+di}$ and I want the magnitude, then will it be $\left|\dfrac{a+bi}{c+di}\right|=\dfrac{|a+bi|}{|c+di|}$?
Scratch that ... I just found the answer on another page; however, I'm still unclear why it's true?
| $\frac{a+bi}{c+di} = \frac{a+bi}{c+di} * \frac{c-di}{c-di} = i (\frac{b c}{c^2+d^2}-\frac{a d}{c^2+d^2})+\frac{a c}{c^2+d^2}+\frac{b d}{c^2+d^2}$. At this point, you should be able to get the magnitude easily. Yes, it'll be cumbersome computation wise, but that should be it.
Suppose $e = \frac{b c}{c^2+d^2}-\frac{a d}... | {
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solve this problem of trigonometry. It is given :
$$\sin(A-B)/\sin B = \sin(A + Y)/\sin (Y)$$
We have to prove $$\cot B - \cot Y = \cot(A + Y) + \cot(A - B).$$
Please help me solving this. I have tried to solve this by analyzing $$\cot B - \cot Y$$
From the given equation,we get $$\frac{\sin A \cos B - \cos A \sin B... | OK, you started with:
$$
{\sin(A−B)\over \sin B}={\sin(A+Y)\over\sin Y}
$$
Which led you to:
$$
\cot B−\cot Y=2\cot A
$$
Let $U = A - B$, which imples $B = A - U$.
Let $V = -(A+Y)$, which implies $Y = -(A+V)$.
Substitute these values into the first equation to get:
$$
{\sin U \over \sin (A - U)} = {\sin (-V) \over \sin... | {
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why $\int \frac{1}{1+\sin(x)+\cos(x)}dx = \ln\left | \tan(\frac{x}{2})+1 \right |+const.$? This is how I solve $$\int \frac{1}{1+\sin(x)+\cos(x)}dx$$, but I got the wrong answer, and the correct answer is $$\ln\left | \tan(\frac{x}{2})+1 \right |+\text{const}.$$
How to solve this？
| \begin{align}
\sin x &= \frac{2\tan(x/2)}{1 + \tan(x/2)} \\
\cos x &= \frac{1-\tan^2x}{1+ \tan^2x} \\
1+\sin x+\cos x &= \frac{2 + 2\tan(x/2)}{1+\tan(x/2)} \\
I &= \int\frac{dx}{1+\sin x+\cos x} \\
&= \int\frac{\sec(x/2)}{2+\tan(x/2)} dx
\end{align}
Let $z=\tan(x/2)$, then $dz=\frac12\sec(x/2)dx$ and
\begin{align}
I &=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/769114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the eigen values and vectors of the matrix Find the eigenvalues and eigenvectors of the matrix $A$:
$$A = \begin{bmatrix}-2 & 2 & -3\\2 & 1 & -6\\-1 & -2 & 0\\\end{bmatrix}.$$
$$A - \lambda I = \begin{bmatrix}-2-\lambda & 2 & -3\\2 & 1-\lambda & -6\\-1 & -2 & -\lambda\\\end{bmatrix}\\
\det(A-\lambda I)=(-2-\lambda... | This is easier to proceed if we permute the matrix:
$$
\pmatrix{2& -7& -3\\
-4& 2& -6\\
-2&-1&-5} \rightarrow \pmatrix{2& -7& -3\\
0& -12& -12\\
0&-8&-8}
$$
This clearly leads to the equation:
$$
-12x - 12z = 0 \rightarrow z = -x \\
\text{using the first equation, }2y - 7x - 3z = 0 \rightarrow y = \frac{7x + 3z}{2} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/769783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I solve this square root problem? I need to solve the following problem:
$$\frac{\sqrt{7+\sqrt{5}}}{\sqrt{7-\sqrt{5}}}=\,?$$
| $\frac{7+\sqrt{5}}{7-\sqrt{5}}=\frac{7+\sqrt{5}}{7-\sqrt{5}}\cdot\frac{7+\sqrt{5}}{7+\sqrt{5}}=\frac{(7+\sqrt{5})^{2}}{49-5}=\frac{(7+\sqrt{5})^{2}}{44}$
Then you are taking the root of the quotient (note for $a,b >0$ $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$) and you get
$\frac{7+\sqrt{5}}{\sqrt{44}}=\frac{7+\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/770259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Why does $\int^{ab}_{a} \frac{1}{x} dx = \int^{b}_{1} \frac{1}{t} dt$? I can't understand how the integral having limits from $a$ to $ab$ in Step 1 is equivalent to the integral having limits from $1$ to $b$. I'm a beginner here. Please explain in detail.
\begin{align*}
\ln(ab) = \int^{ab}_{1} \frac{1}{x} dx &= \int^{... | Your are asking why
$$ \int_{a}^{ab} \frac{1}{x} dx = \int_{1}^b \frac{1}{t} dt $$
Well, it follows by substituting $x = at \implies dx = a dt $.
Now, the limits in the first integral are $x =a$ to $x=ab$. Hence, if $x =a $, then $ t = \frac{a}{a} = 1 $. and if $ x = ab $ , then $ t = \frac{ab}{a} = b$ which are your ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/771015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
A 3-minute algebra problem I have just taken advance math test level 2 and there are several problems that have been bugging me. This is the first question:
If $x,y>0$, then determine the value of $x$ that satisfies the system
of equations: \begin{align} x^2+y^2-xy&=3\\ x^2-y^2+\sqrt{6}y&=3\\
\end{align}
I can answ... | Substracting second equation to the first you obtain $$2y^2-xy-\sqrt{6}y=0$$
Then, factoring $$y(2y-x-\sqrt{6})=0$$
Thus $2y-x-\sqrt{6}=0$ cause $x,y$ are greater than $0$. So $y=\frac{x+\sqrt{6}}{2}$, putting this expression into the second equation you have
\begin{align}
x^2-\left(\frac{x+\sqrt{6}}{2}\right)^2+\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/771551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Multivariable Chain Rule Does anyone know how to use the multivariable chain rule to solve the following problem?
If $G(x^2+y^2, x+y)=(7x+3y, x+5\,y)$, knowing that $G:\mathbb{R}^2\rightarrow \mathbb{R}^2$, find $(G^{-1})'(24,8)$
| Let $f:\mathbb{R}^2\longrightarrow\mathbb{R}^2$ defined by $(x,y)\mapsto(7x+3y,x+5y)$, observe
$$G^{-1}\circ f(x,y)=(x^2+y^2,x+y)$$
applying Chain Rule it follows
\begin{align}
DG^{-1}[f(x,y)]Df(x,y)&=\begin{bmatrix}2x&2y\\1&1\end{bmatrix}\\
DG^{-1}[f(x,y)]\begin{bmatrix}7&3\\1&5\end{bmatrix}&=\begin{bmatrix}2x&2y\\1&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/771751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"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.