Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Calculate $ \biggr\lfloor \frac{1}{4^{\frac{1}{3}}} + \frac{1}{5^{\frac{1}{3}}} + ... + \frac{1}{1000000^{\frac{1}{3}}} \biggr\rfloor$ Calculate $ \biggr\lfloor \frac{1}{4^{\frac{1}{3}}} + \frac{1}{5^{\frac{1}{3}}} + \frac{1}{6^{\frac{1}{3}}} + ... + \frac{1}{1000000^{\frac{1}{3}}} \biggr\rfloor$
I am just cl... | Since
$(x^a)' = ax^{a-1}$,
$a\int_n^{n+1} x^{a-1}dx
=(n+1)^a-n^a
$.
Therefore
$\begin{array}\\
v^a-u^a
&=\sum_{n=u}^{v-1}((n+1)^a-n^a)\\
&=\sum_{n=u}^{v-1}a\int_n^{n+1} x^{a-1}dx\\
\end{array}
$
If $a > 1$,
since $x^{a-1}$
is increasing,
$n^{a-1}
\lt \int_n^{n+1} x^{a-1}dx
\lt (n+1)^{a-1}
$.
If $a < 1$,
since $x^{a-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding value of $\lim_{n\rightarrow \infty}\frac{1+\frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n^3}}{\ln(n)}$
Find the value of
$$\lim_{n\rightarrow \infty}\frac{1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n^3}}{\ln(n)}$$
My Try: Using Stolz-Cesaro,
Let $\displaystyle a_{n} = 1+\frac{1}{2}+\frac{1}{3}+\cdots \cdots ... | Let $H_n=\sum_{k=1}^n \frac 1k$. Since $H_n=\log n +O(1)$, $$H_{n^3} = 3\log n +O(1)$$
thus$$\frac{H_{n^3}}{\log n} = 3+O\left(\frac{1}{\log n} \right)=3+o(1)$$
The limit is $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
$\frac {1}{6} + \frac {5}{6.12} + \frac {5. 8}{6.12.18} + \frac {5.8.11}{6.12.18.24} + \cdots.$ How to find the sum
$$\frac {1}{6} + \frac {5}{6.12} + \frac {5. 8}{6.12.18} + \frac {5.8.11}{6.12.18.24} + \cdots.$$
I have tried to make this series in the form $1 + n x + \frac {n (n + 1) x^2}{2!} + \frac {n (n+1) (... | Like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $,
If $\displaystyle S=\frac {1}{6} + \frac {5}{6.12} + \frac {5. 8}{6.12.18} + \frac {5.8.11}{6.12.18.24} + \cdots.$
$$2S=\frac {2}{6} + \frac {2\cdot5}{6.12} + \frac {2\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2913065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving an equation holds for $x \neq 0$ I would like to show that for $x \neq 0$,
$\dfrac{1}{x} = 1 + (1 - x) + (1 - x)^2 + (1 - x)^3/x.$
One way would be to just expand everything, but is there an easier way?
The sum of a geometric series is $1/(1 - y)$ Letting $y = 1 - x,$ we see
$1/x = \sum_{n = 0}^{\infty} (1 - x)... | Notice that the infinite geometric sum is given as following $$a+aq+aq^2+aq^3+\cdots=\dfrac{a}{1-q}$$therefore $$1+(1-x)+(1-x)^2+(1-x)^3+\cdots=\dfrac{1}{1-(1-x)}={1\over x}$$and $$(1-x)^3+(1-x)^4+(1-x)^5+\cdots=\dfrac{(1-x)^3}{x}$$so we have $$1+(1-x)+(1-x)^2+(1-x)^3+\cdots=1+(1-x)+(1-x)^2+(1-x)^3+(1-x)^4+(1-x)^5+\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
The limit of $\frac{n^3-3}{2n^2+n-1}$ I have to find the limit of the sequence above.
Firstly, I tried to multiply out $n^3$, as it has the largest exponent.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} =
\lim_{n\to\infty}\frac{n^3(1-\frac{3}{n^3})}{n^3(\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3})} =
\lim_{n\to\infty}\... | I admire your efforts, another approach is
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} =
\lim_{n\to\infty}\left(\frac12n-\frac14+ \dfrac{\frac34n - \frac{13}{4}}{2n^2+n-1} \right) =\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 2
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Proof by induction, $1/2 + ... + n/2^n < 2$ So I'm having trouble with proving this homework question by induction.
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{n-1}{2^{n-1}} + \frac{n}{2^n} <2
$$
I know how to prove that the series converges to 2 (using things like the ratio method), but actually using induction is ... | Hint:
It's a sum of an arithmetico-geometric progression:
$$S_{n}={\frac {a-(a+nd)\,r^{n}}{1-r}}+{\frac {dr\,(1-r^{n})}{(1-r)^{2}}}$$
Here $d$ is the common difference, $r$ is the common ratio, $n$ denotes the number of terms and $a$ is the first term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Closed form solution for logarithmic inequality We have the following inequality that was very hard to be solved in a closed form. Yet, someone solved it this way, and I can't get to fully understand what is done.
eq: $\sqrt{x-2\sqrt{x-1}} + \sqrt{x+2\sqrt{x-1}} +log_2(x-1)=0$
Solution:
$x\in Df \iff x \geq 1$
We have... | The logic of replacing the square root with an absolute value is the formula $$ \sqrt {x^2} = |x|$$ which is correct.
The problem is that there is a mistake in the solution because $ |\sqrt{x-1} -1|$ is substituted for $ \sqrt{x-\sqrt{x-1}}$ which is not correct.
Note that $$|\sqrt{x-1} -1| =\sqrt {(\sqrt{x-1} -1)^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplifying $\frac{1}{(n+1)!} + \frac{1}{(n+1)(n+1)!} - \frac{1}{n(n!)}$ How can I prove this equality:
$$
\frac{1}{(n+1)!} + \frac{1}{(n+1)(n+1)!} - \frac{1}{n(n!)}= \frac{-1}{n(n+1)(n+1)!}
$$
| We have that
$$\frac{1}{(n+1)!} + \frac{1}{(n+1)(n+1)!} - \frac{1}{n(n!)}=\color{red}1\cdot \frac{1}{(n+1)!} + \color{red}{\frac{1}{n+1}}\cdot \frac{1}{(n+1)!} - \color{red}{\frac{n+1}{n}}\cdot\frac{1}{(n+1)!}=$$
$$=\left( \color{red}{ 1+ \frac{1}{n+1} - \frac{n+1}{n} }\right)\cdot\frac{1}{(n+1)!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculate$\int\limits_{-2}^{0} \frac{x}{\sqrt{e^x+(x+2)^2}}dx$
Calculate $$\int\limits_{-2}^{0} \frac{x}{\sqrt{e^x+(x+2)^2}}dx$$
First I tried the substitution $t=x+2$ and obtained $$\int\limits_{0}^{2} \frac{t-2}{\sqrt{e^{t-2}+t^2}}dt$$ and than I thought to write it as $$\int\limits_{0}^{2} (t-2)\frac{1}{\sqrt{e^{t... | Hint: Let $y = (x+2)e^{-x/2}.$ You should get $I=-2\sinh^{-1}{2}$.
$\displaystyle y = (x+2)e^{-x/2} \implies dy = -\frac{x}{2}e^{-x/2}\,dx$
and $\displaystyle \frac{x\,dx}{\sqrt{e^x+(x+2)^2}} = \frac{xe^{-x/2}\,dx}{\sqrt{1+(x+2)^2e^{-x}}} = \frac{-2\,dy}{\sqrt{1+y^2}}.$
Therefore we have $\displaystyle I = \int_{-2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Solving the equation $\frac{gy}{x} + \frac{2y^2}{x} = \frac{gy}{y^2} + \frac{2x}{y^2}$ for $x$ or $y$ One day, while making and doing math problems, I came across this equation:
$$\frac{gy}{x} + \frac{2y^2}{x} = \frac{gy}{y^2} + \frac{2x}{y^2}$$
After some simple steps, I found $g$, but I couldn't find $x$ or $y$.
He... | Multiplying by $$x\ne 0$$ we get
$$gy+2y^2=\frac{gx}{y}+\frac{2x^2}{y^2}$$
Now you can multiply by $$y^2$$
We get
$$gy^3+2y^4=gxy+2x^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Sort those 3 logarithmic values without using calculator I found this problem interesting, namely we are given three values: $$\log_{1/3}{27}, \log_{1/5}{4}, \log_{1/2}{5}$$
We want to sort those values without using calculator. So I decided to work on them a little bit before comparing anything. Here is what I got:
Fi... | As an alternative and to check recall that
$$\log_b a=\frac{\log_c a}{\log_c b}$$
that is
$$\log_{\frac13}27=\frac{\log_2 27}{\log_2\frac13}=-\frac{\log_2 3^3}{\log_2 3}=-3$$
$$\log_{\frac15}4=\frac{\log_2 4}{\log_2 \frac15}=-\frac{\log_2 2^2}{\log_2 5}=-\frac{2}{\log_2 5}$$
$$\log_{\frac12}5=\frac{\log_2 5}{\log_2 \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Solving the factorial equation $(n + 4)! = 90(n + 2)!$ Solve the equation below:
$(n+4)!
= 90
(n+2)!$
I did this:
$(n+4)(n+3)(n+2)!
= 90
(n+2)!$
$n^2+7n+12+90=0$
$n^2+7n+102=0$
Is there anymore to this?
| Just to be different (and to rip off Mark Bennett's answer):
$ab = k> 0$ with $0< a \le b$ means $a \le \sqrt {k}$ and $b \ge \sqrt {k}$. (and if $0< a < b$ then $a < \sqrt k$ and $b > \sqrt{k}$)
So $(n+3)(n+4) = 90$ means $n+3 \le \sqrt{90} \approx 9.smallchange \le n+4$
As $n+3$ and $n+4$ are consecutive we must have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Induction proof of a series Suppose we have the series
$$f(n) = \sum_{i=1}^n \frac{2i-1}{i^4 - 2 i^3 + 3 i^2 - 2 i + 2}.$$
As a hint it was said that this series "telescopes". I observed the pattern to be $f(n)=\frac{n^2}{n^2 +1}$. I wish to prove this via induction. The base case holds, since $f(1)=\frac{1}{2}$ which... | Note that$$\frac{i^2}{i^2+1}-\frac{(i-1)^2}{(i-1)^2+1}=\frac{2i-1}{i_4-2i^3+3i^2-2i+2}$$and that therefore\begin{align}\sum_{i=1}^n\frac{2i-1}{i^4-2i^3+3i^2-2i+2}&=\sum_{i=1}^n\frac{i^2}{i^2+1}-\frac{(i-1)^2}{(i-1)^2+1}\\&=\frac{n^2}{n^2+1}-\frac{0^2}{0^2+1}\\&=\frac{n^2}{n^2+1}.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2922748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Determine the domain of the inverse of $f(x) = 2x^2 + 8x - 7$, which is a function. Determine the domain where the inverse of $f(x) = 2x^2 + 8x - 7$ is a function.
*
*So, I started off switching the $y$- (the $f(x)$) and $x$-value, like so:
$x = 2y^2 + 8y - 7$
*Then, I plugged these values into the quadratic formul... | Let $a>0$ and $f(x)=ax^2+bx+c$. There is
$$
\varphi: \left[\frac{-\Delta}{4a},+\infty)\right) \to\left(-\infty, -\frac{b}{2a}\right]
\qquad \mbox{and} \qquad
\psi:\left[ \frac{-\Delta}{4a},+\infty)\right) \to \left[ -\frac{b}{2a}, +\infty,\right)
$$
such that
$$
\begin{matrix}
\varphi ( f(x))=x \qquad f(\varphi(y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2923430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Prove that all $2 \times 2$ orthogonal matrices can be expressed as rotation or reflection Let A be some $2 \times 2$ matrix with real entries. Prove that $A^T$$A$ = $I$ if and only if $A$ is the rotation matrix or the reflection matrix.
My Progress: It can be shown that if $A$ is either the rotation or reflection matr... | With not so many variables running around, we may verify this claim algebraically. Write $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Then $A^TA = \begin{pmatrix} a^2+c^2 & ab+cd \\ ab+cd & b^2+d^2 \end{pmatrix}$. So $\begin{pmatrix} a \\ c\end{pmatrix}$ and $\begin{pmatrix} b \\ d\end{pmatrix}$ are unit vectors... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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What is a fast way to evaluate $\cos(\pi/4 + 2\pi/3)$ and $\cos(\pi/4 + 4\pi/3)$ (maybe using inspection) Given
$\cos(\pi/4 + 2\pi/3)$
and
$\cos(\pi/4 + 4\pi/3)$
I wish to show that these values are given by $(\pm(\sqrt{6}) - \sqrt{2})/4$.
What is a quick way to arrive at these exact solutions (perhaps even by insp... | Use the fact that when angles correspond to equally spaced points around a circle, their sines add up to zero and their cosines do the same. Thus
$\cos(\frac{\pi}{4})+\cos(\frac{\pi}{4}+\frac{2\pi}{3})+\cos(\frac{\pi}{4}+\frac{4\pi}{3})=0$
Put $\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$, then:
$\cos(\frac{\pi}{4}+\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Volume of $x^2+y^2\leq4,\quad z=2+x^2+y^2,\quad z\geq-1$ Find the volume of the solid defined by $$x^2+y^2\leq4,\quad z=2+x^2+y^2,\quad z\geq-1.$$
I found the intersection of surfaces:
$$S\equiv\begin{cases}
x^2+y^2&=4\\
2+x^2+y^2&=z\\
z&=-1
\end{cases}
\equiv
\begin{cases}
x^2+y^2&=4\\
2+4&\neq-1\\
z&=-1.
\end{cases}... | FALSE. Not all three have to intersect; notice that the surfaces $x^2+y^2=4$ and $z=x^2+y^2+2$ intersect, and the surfaces $x^2+y^2=4$ and $z= -1$ intersect, forming a cylinder-like volume with a circular base and a concave paraboloid-shaped "lid."
You may calculate the volume by converting to polar coordinates and con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Implication of equations Just an interesting question I saw online. :)
Is the following statement true or false?
$$\large \dfrac{ab}{c^2} + \dfrac{bc}{a^2} + \dfrac{ca}{b^2} = 3 \iff \dfrac{a + b}{c} + \dfrac{b + c}{a} + \dfrac{c + a}{b} \in \{-3; 6\}$$
$a + b + c \ne 0$
Edit: In case you don't know, $a, b, c$ can be ... | Notice that the equality $\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ca}{b^2}=3$ implies that for $x=ab, y=bc, z=ca$:
$$x^3+y^3+z^3-3xyz=0\leftrightarrow (x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)=0$$.
This in turn implies that either $ab+bc+ca=0 \hspace{0.3cm}(1)$ or $a=b=c \hspace{0.3cm}(2)$.
*
*Case (1):
We easily find by appl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\lim_{x\to-\infty}\sqrt{x^2-5x+1}-x$ results in loss of information Let $f(x) = \sqrt{x^2-5x+1}-x$
Find $\lim_{x\to\infty}f(x)$
$$\lim_{x\to\infty} \sqrt{x^2-5x+1}-x$$
$$\lim_{x\to\infty} \dfrac{x^2-5x+1-x^2}{\sqrt{x^2-5x+1}+x}$$
$$\lim_{x\to\infty} \dfrac{-5x+1}{\sqrt{x^2-5x+1}+x}$$
$$\lim_{x\to\infty} \dfrac... | as far as understanding, you have, in essence,
$$ \left| x - \frac{5}{2} \right| - x . $$
for large positive $x$ you get close to $-5/2,$ but for large negative $x,$ meaning $x$ is negative and $|x|$ is large, you have roughly
$$ 2 |x| + \frac{5}{2} $$
which grows without bound. For example, if $x = -10,$ the ori... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Random walk Catalan sum Starting at $0$ on the number line, you go right $1$ unit with probability $p$ and left $1$ unit with probability $1-p$. What's the probability of ever getting to $n>0$, and how many steps are expected?
Let $q_k$ be the probability of ever getting to $k\ge 0$. $q_k=q_1^k$ since you ever get to $... | We consider the case $p<\frac{1}{2}$.
We obtain
\begin{align*}
\color{blue}{q_1(p)}&=\sum_{k=0}^\infty\frac{1}{k+1}\binom{2k}{2k}p^{k+1}(1-p)^k\\
&=p\sum_{k=0}^\infty\frac{1}{k+1}\binom{2k}{k}\left(p(1-p)\right)^k\\
&=p\cdot\frac{1-\sqrt{1-4p(1-p)}}{2p(1-p)}\tag{1}\\
&=p\cdot\frac{1-(1-2p)}{2(1-p)}\\
&\,\,\color{blu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Why isn't this approach in solving $x^2+x+1=0$ valid? There is this question in which the real roots of the quadratic equation have to be found:
$x^2 + x + 1 = 0$
To approach this problem, one can see that $x \neq 0$ because:
$(0)^2 + (0) + 1 = 0$
$1 \neq 0$
Therefore, it is legal to divide each term by $x$:
$x + 1 ... | You can indeed substitute. First, though, note that $1$ is not a solution to $x = -1 - 1/x$. So, by making that substitution, we are excluding $x = 1$ as a solution to our equation. In a sense, we are looking for a solution of $x^2 +x + 1 = 0$ that is also a solution to $x = -1 - 1/x$.
Here is what we get by substi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2928367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 0
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Generating function: Collecting dollars from children and adults. Here is the problem:
In how many ways can I collect a total of $20$ dollars from $4$ different children and $3$ different adults, if each child can contribute up to $6$ dollars, each adult can give up to $10$ dollars, and each individual gives a nonnegat... | You have made two mistakes in summing the geometric series.
$$1 + x + x^2 + \dots + x^6 = \frac{1-x^\color{red}{7}}{1-x}$$
and
$$1 + x + x^2 + \dots + x^{10} = \frac{1-x^\color{red}{11}}{1-x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solve $\sin^{3}x+\cos^{3}x=1$
Solve for $x\\ \sin^{3}x+\cos^{3}x=1$
$\sin^{3}x+\cos^{3}x=1\\(\sin x+\cos x)(\sin^{2}x-\sin x\cdot\cos x+\cos^{2}x)=1\\(\sin x+\cos x)(1-\sin x\cdot\cos x)=1$
What should I do next?
| Both of $\sin(x),\cos(x)$ must be nonnegative since, if one of them was negative, the equation $\sin^3(x)+\cos^3(x)=1$ would imply that the other one is more than $1$, contradiction.
Thus, we have $0\le \sin(x) \le 1$ and $0\le \cos(x)\le 1$.
If both of $\sin(x),\cos(x)$ are less than $1$, then, since they are both n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2931449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 0
} |
Application of Fermat's little theorem to check divisibility Using Fermat's little theorem to prove: $(i)19\mid 2^{2^{6k+2}}+3$, where $k=0,1,2.....$$(ii)13\mid 2^{70}+3^{70}$My Approach: I couldn't think of how to go with $(i)$ but i tried $(ii)$ to show $2^{70} \equiv 0\pmod {13}$ and $3^{70} \equiv 0\pmod {13}$.Sinc... | In (mod 13): $ 3^{70} = 9^{35} = (-4)^{35} = -(4^{35}) = -(2^{70}) $
So that: $ 2^{70} + 3^{70} = 0 (mod 13) $
The main idea is: $ 3^2 = (-1)2^2 $ (mod 13)
For all odd n: $ 13 $ $|$ $2^{2n} + 3^{2n} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Integrate $\int \sqrt{\frac{1+x}{1-x}}dx$ using $x = \cos(u) $ I have to integrate $\int \sqrt{\frac{1+x}{1-x}}$ using $x = \cos(u)\Rightarrow dx = -\sin(u) du$
My attempts:
$$\int \sqrt{\frac{1+x}{1-x}}dx=\int \sqrt{\frac{(1+x)(1-x)}{(1-x)(1-x)}}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx$$
$$\i... | There are many ways, and a natural temptation is to exploit the fact that $g(z)=\frac{1-z}{1+z}$ is an involution, $g(g(z))=z$. In particular
$$ \int\sqrt{\frac{1+x}{1-x}}\,dx = \int \frac{dx}{\sqrt{g(x)}}\stackrel{x\mapsto g(z)}{=}\int\frac{g'(z)\,dz}{\sqrt{z}}\stackrel{z\mapsto u^2}{=}2\int g'(u^2)\,du =4\int\frac{du... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Locus problem for vertex of equilateral triangle Question
Given an equilateral triangle $PQR$ where $P(1,3)$ is a fixed point and $Q$ is a moving point on the line $x=3.$ Find the locus of $R.$
My attempt
*
*Take $Q$ as $(3,p)$ and $R(h, k).$ Then substitute to the slope of $PQ$ side and $PR$ side to be equal, taki... | I am lazy, so I will pretend $P$ is the origin using shifted coordinates $(x', y')$, and $Q$ is on the line $x' = 2$:
$$\pmatrix{x'\\y'} = \pmatrix{x\\y}-\pmatrix{1\\ 3}$$
Let the coordinates of $Q$ be $(x' = 2,y' = p')$. There are two possible vertices for an equilateral triangle, obtained by rotating $Q$ by $\pm\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Calculate $\int_0^1{x·\lceil1/x\rceil dx}$ I am trying to calculate following integral:
$$\int_0^1{x·\biggl\lceil \frac{1}{x}\biggr\rceil dx}$$
I tried usual change t=1/x but not able to further advance.
Thanks!
| If $\frac 1{n} \le x < \frac 1{n-1}$ then $\lceil \frac 1x \rceil = n$. and $\int_{\frac 1{n}}^{\frac 1{n-1}} x\lceil \frac 1x \rceil dx= \int_{\frac 1{n}}^{\frac 1{n-1}} xn dx = n\frac {x^2}2|_{\frac 1n}^{\frac 1{n-1}}= \frac n2(\frac 1{(n-1})^2 -\frac 1{n^2}=\frac n2(\frac {2n-1}{n^2(n-1)^2})=\frac 1{(n-1)^2} - \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2938203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Evaluating $\lim_{n \to \infty}\left(\sqrt{n^2 - n+1}-\left\lfloor\sqrt {n^2 - n+1}\right\rfloor\right)$
How do I evaluate
$$\lim_{n\to\infty}\left(\sqrt{n^2-n+1}-\left\lfloor\sqrt{n^2 - n+1}\right\rfloor\right),n\in\Bbb N$$
Attempt:
I thought of using Squeeze theorem but that could not help.
Secondly, we know th... | Since $n - 1 < \sqrt{n^2 - n + 1} < n$, then\begin{align*}
&\mathrel{\phantom{=}}{} \sqrt{n^2 - n + 1} - [\sqrt{n^2 - n + 1}] = \sqrt{n^2 - n + 1} - (n - 1)\\
&= \frac{n}{\sqrt{n^2 - n + 1} + (n - 1)} → \frac{1}{2}. \quad (n → ∞)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2939373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding series of two functions multiplied with each other Let's say we are given a function defined as ${\frac{\ln(1+t)}{1-t}}$. We want to find the series expansion up to ${t^4}$.
Now we know that we have two function within this larger function which are ${\ln(1+t)}$ and ${\frac{1}{1-t}}$. Now the series and their ... | Note that:
$$\ln(1+t)=\sum\limits_{n=1}^{\infty}(-1)^{n\color{red}{+1}}\frac{\color{red}t^n}{\color{red}n}=t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+O(t^5)\\
\frac{1}{1-t}=\sum\limits_{n=0}^{\infty}\color{red}t^n=1+t+t^2+t^3+O(t^4)\\
$$
Hence:
$$\frac{\ln(1+t)}{1-t}=\frac{1}{1-t}\cdot \ln(1+t)=\\
(1+t+t^2+t^3+O(t^4))... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2940544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplify $\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}$
Simplify
$$\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}$$
Found in a book with tag "Moscow 1982", the stated answer is $1+\sqrt[4]{5}$. Used all tricks that I know but without success. The answer appears to be correct, checked in Wolfram ... | Define $a=\sqrt[4]{5}$. Then $a^4 = 5$.
Consider the expression (under the radical in the denominator): $$
E = 4-3a+2a^2-a^3.\tag{1}
$$
What if we'll multiply it by $1+a$:
$$E(1+a)=(4-3a+2a^2-a^3)(1+a)\\=(4-3a+2a^2-a^3)+(4a-3a^2+2a^3-a^4)\\=4+a-a^2+a^3-a^4 \\= -1+a-a^2+a^3.\tag{2}
$$
Now what if we'll multiply it by $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2942085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Solution to the equation of a polynomial raised to the power of a polynomial. The problem at hand is, find the solutions of $x$ in the following equation:
$$ (x^2−7x+11)^{x^2−7x+6}=1 $$
My friend who gave me this questions, told me that you can find $6$ solutions without needing to graph the equation.
My approach was ... | Take natural logarithm from both sides:
$$\ln (x^2−7x+11)^{x^2−7x+6}=\ln1 \Rightarrow \\
(x^2-7x+6)\cdot \ln |x^2-7x+11|=0 \Rightarrow \\
1) \ x^2-7x+6=0 \Rightarrow x_{1,2}=1,6; \\
2) \ \ln |x^2-7x+11|=0 \Rightarrow |x^2-7x+11|=1 \Rightarrow x^2-7x+11=\pm 1 \Rightarrow \\
x_{3,4,5,6}=2,5,3,4.$$
Note: The found soluti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 5,
"answer_id": 0
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Factorization of $(1+x+x^2+x^3)^2 - x^3$
Factorize : $(1+x+x^2+x^3)^2 - x^3$
I've tried to expand it but the equation will be even more complicated, anyone can give me some hints to solve it without expanding it (or it is necessary to expand it)?
| $(1+x+x^2+x^3)^2-x^3 = (1+x+x^2)^2 + 2x^3(1+x+x^2) + x^3(x-1)(1+x+x^2) = (1+x+x^2)(1+x+x^2+x^3+x^4)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Divisors of $\left(p^2+1\right)^2$ congruent to $1 \bmod p$, where $p$ is prime
Let $p>3$ be a prime number. How to prove that $\left(p^2+1\right)^2$ has no divisors congruent to $1 \bmod p$, except the trivial ones $1$, $p^2+1$, and $\left(p^2+1\right)^2$?
When $p=3$, you also have $p+1$ as a divisor of $\left(p^2... | Assume there are factors of $\def\num{(p^2+1)^2}\num$ which are congruent to $1$ modulo $p$. Then both the factor and the quotient are $\equiv1\pmod p$ as $\num\equiv1\pmod p$, so we can write $(np+1)(mp+1)=\num$.
Then
$$
(nmp+n+m)p+1=p^4+2p^2+1,
$$
so $nmp+n+m=p^3+2p$. This shows that $p$ divides $n+m$.
Without loss o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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RMO practice problem inequality Let $a_n$ & $b_n$ be two sequences such that $a_0$ , $b_0$ > 0 and $a_{n+1}$ = $a_n$ + $\frac{1}{2b_n}$ & $b_{n+1}$ = $b_n$ + $\frac{1}{2a_n}$ $\forall$ n $\geq$ 0. Then prove that $$max(a_{2018},b_{2018}) > 44.$$
Anyone Please help me with this question.. How to approach this?
| Without loss of generality, suppose that $a_k>b_k$ for some k. Then $a_{k+1}=a_k+\frac{1}{2b_k}>a_k+\frac{1}{2a_k}$. Therefore it suffices to prove that $c_{2018}>44$, where $c_0$ is arbitrary positive number and $c_{n+1}=c_n+\frac{1}{2c_k}$.
Claim. $c_n\ge\sqrt{n+1}$ for all $n>0$.
We prove the claim by mathematical i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For what values of k does this system of equations have a unique / infinite / no solutions? My system of equations is:
\begin{cases}
x + 5y- 6z = 2 \\
kx + y - z = 3 \\
5x - ky + 3z = 7
\end{cases}
So the augmented matrix is:
$$ \left[
\begin{array}{ccc|c}
1&5&-6&2\\
k&1&-1&3\\
5&-k&3&7
\end{array}
\right] $$
I r... | From my comments above, I always find the determinant first which gives us
$$\det \begin{bmatrix} 1&5&-6 \\ k&1&-1 \\ 5&-k&3 \end{bmatrix} = 2 (k-2) (3 k-2)$$
This tells us we may have to account for $$k = 2, k = \dfrac{2}{3}$$
The RREF is given by the steps
*
*Swap $R_1$ and $R_2$
*Set $R_2 \longleftarrow R_2 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2947062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Partitioning a number into set of coprimes such that their product is maximum and sum of partition is the number itself Example 1)
Let the number be 7. Then we have a set {3,4}, so the product of the numbers is 10 and the numbers are mutually coprime.
Example 2)
for n=12, coprimes set = {3,4,5}, where product is 60 whi... | The product is given by the Landau function, shown in OEIS A000793 and beginning
$$ 1, 2, 3, 4, 6, 6, 12, 15, 20, 30, 30, 60, 60, 84, 105, 140, 210,\\ 210, 420, 420, 420, 420, 840, 840, 1260, 1260, 1540, 2310$$
To get the set you can just factor these numbers. The set always consists of powers of distinct primes. The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2947190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Image and Kernel of a linear transformation from a matrix to a polynomial Let the following be the linear transformation for $T$ from a $2$ x $2$ matrix to a second degree polynomial.
$T\begin{pmatrix}a&b\\ c&d\end{pmatrix}=\left(a-c\right)x^2+c\left(x-1\right)+b$
Find a basis for $Im(T)$ and $Ker(T)$.
So this is what ... | Use a systematic approach. Let
$$
E_1=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},\quad
E_2=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},\quad
E_3=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix},\quad
E_4=\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}
$$
and let $p_1=1$, $p_2=x$, $p_3=x^2$. Then
\begin{align}
TE_1&=p_3\\
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2952211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Factor $10^6-1$ completely I know kind of a very elementary method to factor this number. Consider the following:
$$10^6-1 = (10^3-1)(10^3+1)=9 \times 11 \times (10^2+10+1)(10^2-10+1) = 9 \times 11 \times 111\times 91$$
I would then factor each number individually.
Is there a faster method? The great hint is that this ... | \begin{align}
10^6 - 1 &= 1000000-1\\
&= 999999 \\
&= 3^2 \times 111111 \\
&= 3^2 \times 111\times 1001 \\
&= 3^2 \times 111 \times (1100 - 99)\\
&= 3^2 \times 111 \times 11 \times (100-9)\\
&= 3^2 \times 111 \times 11 \times 91 \\
&= 3^2 \times (3 \times 37)\times 11 \times 7 \times 13
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2952786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Not sure how to solve $\lim_\limits{x\to0}{{\sqrt{x^2+1}-1}\over\sqrt{x^2+16}-4}$ So I got this problem:
Determine the following limit value:
$$\lim_\limits{x\to0}{{\sqrt{x^2+1}-1}\over\sqrt{x^2+16}-4}$$
What I tried is:
$\large{\lim_\limits{x\to0}{{\sqrt{x^2+1}-1}\over\sqrt{x^2+16}-4}\cdot{\sqrt{x^2+16}+4\over\sq... | Now, you can make the following:
$$\frac{(\sqrt{x^2+1}-1)(\sqrt{x^2+16}+4)}{x^2+16-16}=\frac{(x^2+1-1)(\sqrt{x^2+16}+4)}{x^2(\sqrt{x^2+1}+1)}\rightarrow\frac{4+4}{1+1}=4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2952887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
sharp bounds for $(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{k}{n})$ Is there any good bounds or estimation of $(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{k}{n})$ and $(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{k}{n})$, $1 < k < n$?
What I actually want is $\sum_{1\le k \le n}(1+\frac{1}{n})(1+\frac{2}{n})\cdots(... | Classic products, whose results are $\left({\pm 1 \over n} \right)^k P(1\pm n,k)$ where $P$ is the Pochammer symbol.
And what you "really" want is:
$$-\frac{2^{2 n+1} e^{-n} n^{n+1} \Gamma \left(\frac{1}{2} (2 n+3)\right) \Gamma (-2 n-1,-n)}{\sqrt{\pi }}+e^{-n} (-n)^{n+1} \Gamma
(-n,-n)-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2953515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $\frac {1}{3+x^2+y^2} + \frac {1}{3+y^2+z^2} +\frac{1} {3+x^2+z^2}\leq \frac {3}{5} . $ Let $x, y, z>0$ s.t. $x+y+z=3$.
Show that $$\frac {1}{3+x^2+y^2} + \frac {1}{3+y^2+z^2} +\frac{1 } {3+x^2+z^2}\leq \frac {3}{5}\ . $$
My idea: $$3 + x^2 + y^2 \geq 1 + 2x+ 2y=7-2z $$
I notice that $f (t)=\frac {1}{7-2t} $... | The second inequality.
By C-S we obtain:
$$\sum_{cyc}\frac{a}{a^2+bc}=\sum_{cyc}\frac{a^2(b+c)^2}{(a^3+abc)(b+c)^2}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}a(b+c)\right)^2}{\sum\limits_{cyc}(a^3+abc)(b+c)^2}=\frac{4(ab+ac+bc)^2}{\sum\limits_{cyc}(a^3+abc)(b+c)^2}.$$
Thus, it's enough to prove that
$$8(ab+ac+bc)^2(a^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2954303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Closed-form expression for infinite series related to a Gaussian Consider the following infinite series, where $x$ is indeterminate and $r$ is held constant:
$\displaystyle 1 + \frac{x}{r} + \frac{x^2}{r^2} + \frac{x^3}{r^3} + ...$
It is relatively easy to see that the above, for $\frac{x}{r} < 1$, converges to
$\displ... | Your original series is
$f(x, r)
=\sum_{n=0}^{\infty} \dfrac{x^n}{r^{(n^n)}}
$.
This is not the same as
$g(x, r)
=\sum_{n=0}^{\infty} \dfrac{x^n}{(r^n)^n}
=\sum_{n=0}^{\infty} \dfrac{x^n}{r^{n^2}}
$.
Also,
you went from
$\displaystyle 1 + \frac{x}{e} + \frac{x^2}{e^4} + \frac{x^3}{e^9} + ...
$
to
$\displaystyle 1 + x \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Which number can I erase? All positive integers greater than $2$ are written on a board. First we erase number $3$ and $5$.
With 4 positive integers $a,b,c,d$ satisfying $a+b=c+d$, if $ab$ is erased, then $cd$ can be erased, otherwise $cd$ cannot be erased.
For example, $3=3 \times 1$, $3+1=4=2+2$ , then $2 \times 2... | Somewhat a long comment (not a complete answer).
The first thing that I would suggest is that you should work through some examples and see what happens. I'll help with some of the initial steps.
To think about what you're erasing, let $n$ be a number that has been erased. Then, for each pair of factors $ab$ so that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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How big does $r$ need to be, to ensure that $x^4+y^4 >r^2$ for all $x^2+y^2 =r^2$? Consider the circle $x^2+y^2 = r^2$ for some fixed $r>0$. How big does $r$ need to be, to ensure that $$x^4+y^4 >r^4$$ for all $x^2+y^2 =r^2$? I know that $x^2<x^4$ whenever $|x| >1$, but I'm not sure how to use that here.
| By C-S $$(1^2+1^2)(x^4+y^4)\geq(x^2+y^2)^2.$$
Thus, $$x^4+y^4\geq\frac{1}{2}r^4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Expected value and standard deviation of a pmf function
In a family, the probability mass function of the number of people $x$ who have contracted the flu is given by $$P(x) = Kx \qquad x\in\{0,1,\ldots,N\}$$ where $N$ is the number of people in the family.
(a) If nine people are expected to have flu in the family, ca... | This is not a binomial function, more like a triangular one. We cannot immediately derive $N$ from $K$ or the other way round, but we can derive two equations first. For the expected value:
$$1\cdot K+2\cdot2K+\dots+N\cdot NK=9$$
$$K(1^2+2^2+\dots+N^2)=9$$
For the pmf summing to 1:
$$K(1+2+\dots+N)=1$$
Thus we have
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2960347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Induction proof: $\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +...+\frac{n}{2^n}$ $<2$ Prove by induction the following.
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +\dots+\frac{n}{2^n}<2.$$
Caveat: The $<$ will be hard to work with directly. Instead, the equation above can be written in the form,
$$\frac{1}{2^1}... | Let the sum of the $n$ first terms be $S_n$. We have the recurrence
$$S_{n+1}=S_n+\frac{n+1}{2^{n+1}}$$ or
$$2^{n+1}S_{n+1}=2\cdot2^nS_n+n+1.$$
This hints the change of variable that leads to
$$R_{n+1}=2R_n+n+1.$$
This is a linear recurrence which we will solve a usual:
*
*homogeneous part, $R_{n+1}=2R_n$, so that $... | {
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"timestamp": "2023-03-29T00:00:00",
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Find $P(x,y,z)=x^n+y^n+z^n-\prod\limits_{k=0}^{n-1}(x+\omega_n^ky+\omega_n^{-k}z)$, where $\omega_n$ denotes a primitive $n$th root of unity
Find $P(x,y,z)=x^n+y^n+z^n-\prod\limits_{k=0}^{n-1}(x+\omega_n^ky+\omega_n^{-k}z)$, where $\omega_n$ denotes a primitive $n$th root of unity.
I have manually multiplied the term... | We need to find a closed form expression of
$$
\prod_{k=0}^{n-1}(x+\omega^ky+\omega^{-k}z),
$$ where $\omega=e^{\frac{2\pi i}{n}}$ is the primitive $n$-th root of unity. Now let us consider the polynomial
$$
F(r) = r^n\prod_{k=0}^{n-1}(x+r\omega^ky+r^{-1}\omega^{-k}z)=\prod_{k=0}^{n-1}(xr+r^2\omega^ky+\omega^{-k}z)
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2962964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Proving that $\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2}=\frac{G}{\sqrt 3} -\frac{\pi^2}{24}$ Trying to show using a different approach that $\int_0^1 \frac{\sqrt x \ln x}{x^2-x+1}dx =\frac{\pi^2\sqrt 3}{9}-\frac{8}{3}G\, $ I have stumbled upon this series: $$\sum_{n=1}^\infty \frac{\sin\left... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "17",
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Successive differentiation BSc 2
Question : if $y^3 + x^3 - 3axy = 0$, show that $$y'' = -\frac{(2a^3 xy)}{(y^2-ax)^3}$$
How I did
I differentiated the whole expression twice (as shown below)
*
*$3y^2y'+3x^2-3x^2-3ay-3axy'=0$
*$6yy''+6x-3yy'-3ay'-3axy''=0$
Then I kept $y''$ on the left and transferred everything ... | I got $$3y^2y'+3x^2-3ay-3axy'=0$$ or
$$y'=\frac{ay-x^2}{y^2-ax}$$
so
$$y''=\frac{(ay'-2x)(y^2-ax)-(ay-x^2)(2yy'-a)}{(y^2-ax)^2}$$
$$y''=\frac{y'(-a^2x-ay^2+2xy)-2xy^2+ax^2+a^2y}{(y^2-ax)^2}$$
$$y''=\frac{(ay-x^2)(-ax^2-ay^2+2x^2y)-(2xy^2-ax^2-a^2y)(y^2-ax)}{(y^2-ax)^3}$$
Expanding all i got
$$y''=\frac{-2a^3xy+6ax^2y^... | {
"language": "en",
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"source": "stackexchange",
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Proving that if $x^4 + 5x + 1 < 27$ then $x < 2$ I need help trying to figure out how I can prove a statement like this. So looking at this I can conclude that this statement is of the form $P \Rightarrow Q$
$$
P: x^4 + 5x + 1 < 27
$$
$$
Q: x < 2
$$
I wanted to try and prove this by contrapositive , so this state would... | If $x\ge 2$ is it true that $x^4 \ge 16$?
If $x \ge 2$ is it true that $5x \ge 10$?
So if $x \ge 2$ is it true that $x^4 + 5x + 1 \ge 16 + 10 + 1 = 27$?
...
It could get tedious to go to axioms and prove that if $x\ge 2 > 0$ then $x^4 \ge 2x^3 \ge 4x^2 \ge 8x \ge 16$ via the axiom: if $a > 0$ and $m < n$ then $am < an$... | {
"language": "en",
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"source": "stackexchange",
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A basic theorem of prime numbers. Checking My Proof
*
*Let $n$ be a positive integer. Prove that $n^5 -1$ is prime if and only if $n=2$.
$(\Rightarrow)$ Assume $n^5-1$ is prime. We will show $n=2$. If $n=3$, then clear $n^5-1$ is not prime.
Claim. If $n>2$, then $n^5-1$ is not prime.
Proof of the claim. We will do ... | Your proof is fine but really long and complicated and hard to read.
Simpler to just say:
$n^5 -1 = (n-1)(n^4 + n^3 + n^2 + n + 1)$.
If $n-1$ and $n^4 + n^3 + n^2 + n + 1$ are non trivial factors (not equal to $1$ or $n^5 -1$) then this is not prime.
So the only way for $n^5 -1$ to be prime is if either $n-1 = 1$ or $n... | {
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"source": "stackexchange",
"question_score": "2",
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The rth term in $(1+x)^{1/x} = e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ Let
$$(1+x)^{\frac {1}{x}} = e.G(x)$$
Taking logarithm on both sides,
$$\frac {1}{x} \log {(1+x)} = 1 + \log {G(x)}$$
Putting in the Taylor expansion for $\log {(1+x)}$ we have,
$$\frac {1}{x}(x - \frac {1}{2}x^2 + \frac... | Let $G(x)=(1+x)^{1/x}$
$\ln G(x)=\dfrac{\ln(1+x)}x=1-\dfrac x2+\dfrac{x^2}3-\dfrac{x^3}4+\cdots$
$$G(x)=e\cdot e^{-x/2}\cdot e^{-x^2/3}\cdot e^{-x^3/4}\cdots$$ ignoring terms containing $x^4$
$$=e\left(1-\dfrac{x}2+\dfrac{\left(-\dfrac x2\right)^2}{2!}+\dfrac{\left(-\dfrac x3\right)^3}{3!}+\cdots\right)\left(1-\dfrac{x... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Finding the limit of the sequence $(1, \frac{1}{2+1}, \frac{1}{2+\frac{1}{1+1}},\cdots)$ I am struggling to find the limit of the sequence:
$$1, \cfrac{1}{2+1}, \cfrac{1}{2+\cfrac{1}{1+1}}, \cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2+1}}}, \cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+1}}}},\cdots$$
Normally, for these types... | Based on the first five terms I got the following:
$a_1=1$, $\hspace {0,2cm}$ $a_2=\frac{1}{3}$, $\hspace {0,2cm}$
$a_3=\frac{1}{2+\frac{1}{1+a_1}}$
$a_4=\frac{1}{2+\frac{1}{1+a_2}}$
$a_5=\frac{1}{2+\frac{1}{1+a_3}}$
So the recurrence relationship $\hspace {0,2cm}$ $a_n=\frac{1}{2+\frac{1}{1+a_{n-2}}}=\frac{1+a_{n-2}}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove that $ -2 + x + (2+x)e^{-x}>0 \quad \forall x>0$ I need to prove that
$ -2 + x + (2+x)e^{-x}>0 \quad \forall x>0$
If I define $f(x) = -2 + x + (2+x)e^{-x}$ and plot it I can see it's a monotonously growing function, and f(0)=0. Then $f(x)>0$ if $x>0$.
However I can't find the way to prove this.
Ideally I would... | This approach requires an extra condition $x\lt2.$
\begin{align}\ln\left(1+\dfrac{x}{2}\right)-\ln\left(1-\dfrac{x}{2}\right)&=\left[\dfrac{x}{2}-\dfrac{1}{2}\left(\dfrac{x}{2}\right)^2+\dfrac{1}{3}\left(\dfrac{x}{2}\right)^3+\cdots\right]-\left[-\dfrac{x}{2}-\dfrac{1}{2}\left(\dfrac{x}{2}\right)^2-\dfrac{1}{3}\left(\d... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim\limits_{n \to \infty}\frac{n}{\sqrt[n]{n!}}$. Solution
Notice that
$$(\forall x \in \mathbb{R})~~e^x=1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots.$$
Let $x=n$ where $n\in \mathbb{N_+}$. Then we obtain
$$e^n=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\cdots>\frac{n^n}{n!}.$$
Thus, we obtain
$$e>\frac{n}{... | Using Riemann sums:
$$ \log\frac{\sqrt[n]{n!}}{n} = \frac{1}{n}\sum_{k=1}^{n}\log\left(\tfrac{k}{n}\right)\to \int_{0}^{1}\log(x)\,dx = -1$$
hence $\frac{n}{\sqrt[n]{n!}}$ converges to $e$.
| {
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"source": "stackexchange",
"question_score": "9",
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Calculate $\lim_{k \to \infty}(1+2^{k+1})^{(2^{k-2})-2}\cdot\frac{(2^k-1)^2}{(2^k-1)!!}$ I would like to calculate this limit. I tried to look for known limits which can be substituted to parts of this limit but couldn't find information about the double factorial.
$$\lim\limits_{k \to \infty}(1+2^{k+1})^{(2^{k-2})-2}\... | HINT
Let indicate $n=2^k-1\to \infty$ then we have
$$(1+2^{k+1})^{(2^{k-2})-2}\cdot\frac{(2^k-1)^2}{(2^k-1)!!}=(2n+3)^{\frac{n+1}4-2}\cdot\frac{n^2}{n!!}=(2n+3)^{\frac{n-7}4}\cdot\frac{n^2}{2^nn!}$$
then use Stirling approximation or by Root test
$$\sqrt[n]{(2n+3)^{\frac{n-7}4}\cdot\frac{n^2}{2^nn!}}
\sim \sqrt[n]{\fra... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Trigonometric Identities: Given $\tan(2a)=2$ and $\frac{3\pi}{2}Given $\tan(2a)=2$ and $\frac{3\pi}{2}<a<2\pi$ find value of $\tan(a)$
I first found the values of $\cos(2a)$ and $\sin(2a)$ and then used the half angle formula.
$$\tan(a)=\tan\frac{2a}{2}=\frac{1-\cos(2a)}{\sin(2a)}
\implies\frac{5}{2\sqrt 5}\left(1-\fra... | You made two mistakes, but your method does work.
If $\tan(2a)=2$, then either
Case 1: $0< 2a +2 n \pi<\pi/2$, $\cos(2a)=1/\sqrt{5}$, and $\sin(2a)= 2/\sqrt{5}$ (first quadrant), or
Case 2: $\pi<2a+2 n \pi<3 \pi/2$, $\cos(2a)=-1/\sqrt{5}$, and $\sin(2a)= -2/\sqrt{5}$ (third quadrant)
for some integer $n$.
The prob... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove following inequality Prove that $(\frac{2a}{b+c})^\frac{2}{3}+(\frac{2b}{a+c})^\frac{2}{3}+(\frac{2c}{a+b})^\frac{2}{3} ≥ 3$
What I tried was to use AM-GM for the left side of this inequality, what I got was
$3(\frac{8abc}{(a+b)(b+c)(c+a)})^\frac{2}{9}≥ 3$ and $(\frac{8abc}{(a+b)(b+c)(c+a)})^\frac{2}{9}≥ 1$ or ju... | By AM-GM
$$\sum_{cyc}\left(\frac{2a}{b+c}\right)^{\frac{2}{3}}=\sum_{cyc}\frac{1}{\sqrt[3]{\left(\frac{b+c}{2a}\right)^2\cdot1}}\geq\sum_{cyc}\frac{1}{\frac{\frac{b+c}{2a}+\frac{b+c}{2a}+1}{3}}=\sum_{cyc}\frac{3a}{a+b+c}=3.$$
| {
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"url": "https://math.stackexchange.com/questions/2974588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve the equation $2x^3+x^2-7x-6=0$ given that the difference of two roots is $3$ Q:Solve the equation $2x^3+x^2-7x-6=0$ given that the difference of two roots is $3$.My book solve it leting the roots of the equation be $\alpha,\alpha+3,\beta$ then find the equation whose roots are $\alpha-3,\alpha,\beta-3$.And i know... | Here is one more approach. First, we do not compute anything, just make sentences. One of the paired roots is $a$, say, the other one is $a\pm 3$, and there is one more root, such that the sum of all three is (Vieta) a rational number, well $-1/2$, but it is not so important, it is rational, this is important. So all t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2976543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find bases B3 and B2 for $\Bbb R^3$ and $\Bbb R^2$ given a linear transformation and its matrix A linear transformation T is defined by
T: $\Bbb R^3$ $\rightarrow$ $\Bbb R^2$ $\Rightarrow$ T$\begin{pmatrix}x\\y\\z\end{pmatrix}$ = $\begin{pmatrix}2x+y\\y+2z\end{pmatrix}$
Find bases $\mathscr B_3'$ and $\mathscr B_2'$ fo... | I do not think your approach is correct.
We can simplify the answer by taking the ordered bases
$$\begin{align}
\mathscr B_3' & = \left\{
\varepsilon_1 = \begin{pmatrix}1\\0\\0\end{pmatrix},
\varepsilon_2 = \begin{pmatrix}0\\1\\0\end{pmatrix},
\varepsilon_3 = \begin{pmatrix}0\\0\\1\end{pmatrix}\right\},\\\\
\mathscr B_... | {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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complex number with high powers The question:
$\frac{(-1-\sqrt{3}i)^{73}}{2^{73}}$
I really do not even know where to begin. Am I suppose to expand $(-1-\sqrt{3}i)$ 73 times?
| Ok, what I got was:
$(-\frac{1}{2}-\frac{\sqrt{3}}{2}i)^{73}$
$|z| = \sqrt{(-1/2)^2+\sqrt{3}/2} = \sqrt{1} = 1$
tan$^{-1}$ ($-.5/ \sqrt{3}/2) = \frac{-2}{3}2\pi$
$z^{73} = 1($cos$(\frac{-2}{3}2\pi)+i$sin$(\frac{-2}{3}2\pi)$)
$z = 1^{73}($cos$(\frac{-2}{3}2\pi*73)+i$sin$(\frac{-2}{3}2\pi8*73)$)
= $-.5 + \frac{\sqrt{3... | {
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"question_score": "3",
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"answer_id": 4
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If $\tan(A-B)=1$ and $\sec(A+B)=2/\sqrt{3}$, find the minimum positive value of $B$
If $\tan(A-B)=1$ and $\sec(A+B)=2/\sqrt{3}$, find the minimum positive value of $B$
I am using $\tan(A-B)=1$, so $A-B=n\pi+\pi/4$ and $A+B = 2n\pi\pm\pi/6$. Solving these I am getting $B =7\pi/24$ and $A =37\pi/24$.
The book I am refe... | You correctly have
$$
A=B+\frac{\pi}{4}+n\pi
$$
Therefore $A+B=2B+\pi/4+n\pi$. Hence
$$
\cos\left(2B+\frac{\pi}{4}+n\pi\right)=\frac{\sqrt{3}}{2}=\cos\frac{\pi}{6}
$$
Therefore either
$$
2B+\frac{\pi}{4}+n\pi=\frac{\pi}{6}+2m\pi
$$
or
$$
2B+\frac{\pi}{4}+n\pi=-\frac{\pi}{6}+2m\pi
$$
In the first case
$$
2B=-\frac{\pi}{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} \binom{n}{k} = 1+\frac{1}{2}+...+\frac{1}{n}$
How can I prove that $$\sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} \binom{n}{k} = 1+\frac{1}{2}+...+\frac{1}{n}.$$
I tried an induction but couldn't prove that $$\sum_{k=1}^{n+1} \frac{(-1)^{k-1}}{k} \binom{n+1}{k} = \frac{1}{n+1}+\... | In order to verify the inductive step, recall that $\binom{n+1}{k}=\binom{n}{k-1}+\binom{n}{k}$. Therefore
$$\begin{align}\sum_{k=1}^{n+1} \frac{(-1)^{k-1}}{k} \binom{n+1}{k}&=
\sum_{k=1}^{n+1} \frac{(-1)^{k-1}}{k} \binom{n}{k-1}+
\sum_{k=1}^{n+1} \frac{(-1)^{k-1}}{k} \binom{n}{k}\\
&=
\frac{1}{n+1}\sum_{k=1}^{n+1} (-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983060",
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"source": "stackexchange",
"question_score": "2",
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Calculate the limit (Squeeze Theorem?) I have to calculate the limit of this formula as $n\to \infty$.
$$a_n = \frac{1}{\sqrt{n}}\bigl(\frac{1}{\sqrt{n+1}}+\cdots+\frac{1}{\sqrt{2n}}\bigl)$$
I tried the Squeeze Theorem, but I get something like this:
$$\frac{1}{\sqrt{2}}\leftarrow\frac{n}{\sqrt{2n^2}}\le\frac{1}{\sqrt{... | As an alternative by Stolz-Cesaro
$$\frac{b_n}{c_n} = \frac{\frac{1}{\sqrt{n+1}}+\cdots+\frac{1}{\sqrt{2n}}}{\sqrt n}$$
$$\frac{b_{n+1}-b_n}{c_{n+1}-c_n} = \frac{\frac{1}{\sqrt{2n+2}}+\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt n}$$
and
$$\frac{\frac{1}{\sqrt{2n+2}}+\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}$ without L'Hopital I have the following limit question, where different indices of roots appear in the numerator and the denominator
$$\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}.$$
As we not allowed to use L'Hopital, I want to learn how we can pro... | Make the change: $x-2=t^4$. Then:
$$\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}=\lim_{t\to1}\frac{\sqrt[3]{t^4+7}-2}{t-1}=\\
\lim_{t\to1}\frac{\sqrt[3]{t^4+7}-2}{t-1}\cdot \frac{\sqrt[3]{(t^4+7)^2}+2\sqrt[3]{t^4+7}+2^2}{\sqrt[3]{(t^4+7)^2}+2\sqrt[3]{t^4+7}+2^2}=\\
\lim_{t\to1} \frac{(t^4+7)-2^3}{(t-1)(\sqrt[3]{... | {
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"source": "stackexchange",
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Show that $2 < (1+\frac{1}{n})^{n}< 3$ without using log or binommial coefficient $2 <(1+\frac{1}{n})^{n}< 3$
Is it possible to show the inequality without using binommial coefficients thus only by induction? The leftinequality can be shown using bernoulli inequality.
| Let $a_n=\left(1+\frac{1}{n}\right)^n$. For any $n>1$ the inequality $a_n>2$ is trivial. We have $a_{n+1}>a_n$ by AM-GM:
$$\sqrt[n+1]{1\cdot a_n} = \text{GM}\left(1,1+\tfrac{1}{n},\ldots,1+\tfrac{1}{n}\right) \stackrel{\text{AM-GM}}{<} \tfrac{1}{n+1}\left[1+n\cdot\left(1+\tfrac{1}{n}\right)\right] = 1+\tfrac{1}{n+1}.$$... | {
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"source": "stackexchange",
"question_score": "1",
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What am I doing wrong solving this system of equations? $$\begin{cases}
2x_1+5x_2-8x_3=8\\
4x_1+3x_2-9x_3=9\\
2x_1+3x_2-5x_3=7\\
x_1+8x_2-7x_3=12
\end{cases}$$
From my elementary row operations, I get that it has no solution. (Row operations are to be read from top to bottom.)
$$\left[\begin{array}{ccc|c}
2 & 5 & -8 & ... | Note 1: Referring to 5xum's answer, in case you don't know the final answer, any solution will do, e.g.:
$$(x_1,x_2,x_3)=(0,0,-1);(0,0,-1);(1,0,-1);(12,0,0) \ \ \text{(respectively)}$$
Note 2: In step $3$, you can reduce column $3$ instead of column $2$:
$$\left[
\begin{array}{ccc|c}
1&8&-7&12\\
0&2&-3&1\\
0&-3&1... | {
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Evaluating $\sum_{(a,b,c)\in T}\frac{2^a}{3^b 5^c}$, for $T$ the set of all positive integer triples $(a,b,c)$ forming a triangle
Let $T$ be the set of all triples $(a,b,c)$ of positive integers for which there exist triangles with side lengths $a$, $b$, $c$. Express
$$\sum\limits_{(a,b,c)\,\in\,T}\;\frac{2^a}{3^b ... | The main difficulties of this problem is how to handle the triangle inequalities:
$$a+b > c, b+c > a, c+a > b$$
The beauty of Ravi substitution (mentioned by @maveric in comment)
$$a = v + w, b = u + w, c = u + v$$
is under Ravi substitution,
$a, b, c$ satisfies the triangle inequalities if and only if $u,v,w$ are po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
If $\sin x+\sin^2x+\sin^3x=1$, then find $\cos^6x-4\cos^4x+8\cos^2x$
If $\sin x+\sin^2x+\sin^3x=1$, then find $$\cos^6x-4\cos^4x+8\cos^2x$$
My Attempt
\begin{align}
\cos^2x&=\sin x+\sin^3x=\sin x\cdot\big(1+\sin^2x\big)\\
\text{ANS}&=\sin^3x\cdot\big(1+\sin^2x\big)^3-4\sin^2x\cdot\big(1+\sin^2x\big)^2+8\sin x\cdot\bi... | Let us denote $y = \sin x$. The relation we have for $y$ is then $y + y^2 + y^3 = 1$, or also if we multiply by $y-1$, we get $y^4 = 2y - 1$. The idea is simply to write the expression in $\cos^2 x$ given in terms of $y$, and use the relation to simplify it. We have
\begin{align*}
\cos^6x-4\cos^4x+8\cos^2x
&= (1 - y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
A rank identity Let $A,B$ be $n\times n$ matrix, $A^2=A$, $B^2=B$, how to show $\operatorname{rank}(A-B)=\operatorname{rank}(A-AB)+\operatorname{rank}(B-AB)$.
It seems impossible to reduce $\left(\begin{array}{cc}
A-AB&0\\
0&B-AB
\end{array}\right)$ to be $\left(\begin{array}{cc}
A-B&0\\
0&0
\end{array}\right).$
| The mistake I made is that when doing row operation matrix has to be multiplied on the left, when column operation it's on the right.
It seems like
$$\begin{align*}
\pmatrix{
A-AB & 0\\
0 & B-AB
}&\to\pmatrix{
A-AB & (A-AB)+-(B-AB)\\
0 & B-AB
}\\
&\to\pmatrix{
A-AB & A-B\\
0 & B-AB
}\\
&\to\pmatrix{
A(A-B) & A-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Does the integral $\int_{0}^{\infty} \frac{x^3\,\cos{(x^2-x)}}{1+x^2}$ diverge Does the integral
$$J:=\int_{0}^{\infty} \frac{x^3\,\cos{(x^2-x)}}{1+x^2}dx $$
diverge ?
If we integrate by parts we find
$$J=\lim_{a\rightarrow +\infty} \frac{a^3}{(1+a^2)(2a-1)}\cos{(a^2-a)} -\\
\int_{0}^{\infty}\sin{(x^2-x)}
\left( \frac... | The integral is divergent because the limit does not exist.
Let $f(a)=\frac{a^3\,\cos{(a^2-a)}}{(1+a^2)(2a-1)}$.
Take the sequence $a_{n}=:\frac{1+\sqrt{1+8\pi n}}{2}$
that diverges to infinity and satisfies $a_{n}^2-a_{n}=2n\pi$.
Then $f(a_{n})\rightarrow \frac{1}{2}$.
Now, take the sequence $b_{n}:=\frac{1+\sqrt{1+4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Recurrence Relation - # of binary strings with given property Let $a_n$ be the number of binary strings of length $n$ with the property that each entry is adjacent to at least one entry of the same type.
ex: $11000111$ is a valid string but $11011000$ is not valid
$\textbf{(a) Find $a_1,a_2,a_3,a_4,a_5,a_6,a_7$}$
If s... | Using $z$ for ones and $w$ for zeros we get the generating function
$$F(z, w) = (1+z^2+z^3+\cdots)
\times \sum_{q\ge 0} (w^2+w^3+\cdots)^q (z^2+z^3+\cdots)^q
\\ \times (1+w^2+w^3+\cdots).$$
This is
$$\left(1+\frac{z^2}{1-z}\right)
\times \sum_{q\ge 0} \frac{w^{2q} z^{2q}}{(1-w)^q (1-z)^q}
\\ \times \left(1+\frac{w^2}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2996975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Additional methods for integral reduction formula So I have successfully found a reduction formula for
$$I_{m,n}=\int\frac{dx}{\sin^m(ax)\cos^n(ax)}$$
Went as follows:
$$\int\frac{dx}{\sin^m(ax)\cos^n(ax)}=\int\csc^m(ax)\sec^n(ax)dx=\int\csc^m(ax)\sec^{n-2}(ax)\sec^2(ax)dx\\
\begin{vmatrix}u=\csc^m(ax)\sec^{n-2}(ax)\\... | This is a known integral:
$$\frac{\sin ^{1-m}(a x) \sin ^2(a x)^{\frac{m-1}{2}} \cos ^{1-n}(a x) \,
_2F_1\left(\frac{m+1}{2},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(a x)\right)}{a (n-1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Combinatorics. A box contains 20 balls numbered $1,2,3,...,20$.
A box contains $20$ balls numbered $1,2,3,...,20$. If 3 balls are randomly taken from the box, without replacement, what is probability that one of them is the average of the other two?
Which one of these is the correct answer?
*
*7/45
*3/38
*6/155
... | Refer to the table:
$$\begin{array}{c|c|c}
\text{Middle (average) number}&\text{Sets}&\text{Number of sets}\\
\hline
2&\{1,2,3\}&1\\
3&\{1,3,5\},\{2,3,4\}&2\\
4&\{1,4,7\},\{2,4,6\},\{3,4,5\}&3\\
\vdots&\vdots&\vdots\\
9&\{1,9,17\},\{2,9,16\},\cdots,\{8,9,10\}&8\\
10&\{1,10,19\},\{2,10,18\},\cdots,\{9,10,11\}&9\\
11&\{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2998227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the general solution to the ODE $x\frac{dy}{dx}=y-\frac{1}{y}$ I have been working through an ODE finding the general solution and following the modulus through the equation has left me with four general solutions, as shown below. Online ODE solvers, however, have only calculated the two (positive and negative), o... | The equation in question is $$xy'(x)=y(x)-\frac1{y(x)},$$ and you are looking for functions $y:\mathbb{R}\setminus\{0\}\rightarrow\mathbb{R}$ which satisfy the above equation and the inequality $-1\lt{y(x)}\lt1,$ as well as the implicit $y(x)\neq0.$ Given these restrictions, we have that $$\frac{y(x)}{y(x)^2-1}y'(x)=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2999612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Locus problem on circle and parabola
Circles are drawn through the vertex of a parabola $y^2 = 4ax$ to cut the parabola orthogonally at the other point. Find the locus of the centers of the circles.
What I did:
The tangent must surely be the diameter of the circle. Also, the line through the vertex perpendicular to t... | We are given
\begin{gather}
\tag{1}\label{eq:1}
x = \frac{at^4}{2(t^2+2)}, \\
\tag{2}\label{eq:2}
y = \frac{3at^3+4at}{2(t^2+2)}.
\end{gather}
Combining \eqref{eq:1} and \eqref{eq:2},
$$
t\frac{y}{a} - \frac{x}{a} = \frac{2t^4 + 4t^2}{2(t^2+2)} = t^2,
$$
whence
\begin{equation}
\tag{3}\label{eq:3}
\left(t^2 + \frac{x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$ Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$
My try:
By Lagrange Multiplier method we have
$$L(x,y,z,\lambda, \mu)=(x^2+6y^2+4z^2)+\lambda(x+2y+z-4)+\mu(2x^2+y^2-16)$$
For $$L_x=0$$ we get
$$2x+\lambda+4\mu x=0 \tag{1}$$
For $$L_y... | Hint:
Let $u=x^2+6y^2+4(4-2y-x)^2$
$\iff5x^2-16(2-y)x+22y^2-16y+16-u=0$
As $x$ is real, the discriminant will be $\ge0$
$$\implies256(y-2)^2\ge20(22y^2-16y+16-u)$$
$$\implies64(y-2)^2\ge5(22y^2-16y+16-u)$$
$$\implies5u\ge46y^2-176y+176=46\left(y-\dfrac{44}{23}\right)^2+176-46\left(\dfrac{44}{23}\right)^2$$
Now $y^2=16-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Is integration by substitution always a reverse of the chain rule? To integrate $\int x^3\sin(x^2+1)dx$, I took the following approach:
\begin{align*}
\begin{split}
\int x^3\sin(x^2+1)dx&=\int x^3\sin(u)\cdot\frac{1}{2x}du\\
&=\frac{1}{2}\int x^2\sin(u)du\\
&=\frac{1}{2}\int(u-1)\sin(u)du\\
&=\frac{1}{2}\int u\sin(u)-\... | The integrand is
\begin{align}
x^3 \sin(x^2+1) &= \frac{x^2}{2} \sin(x^2+1)(2x)\\
&= \frac{(x^2 +1 - 1)}{2} \sin(x^2+1) (2x) \\
&= f(x^2+1) (2x)
\end{align}
where $f$ is the function defined by
$$
f(u) = \frac{(u-1)}{2} \sin(u).
$$
If $F$ is an antiderivative of $f$, then $F(x^2+1)$ is an antiderivative of the integran... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate and Simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$ I am trying to evaluate and simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$.
I am getting $\frac{11}{10}$ but the answer is $\frac{3-4\sqrt{3}}{10}$
My Process:
$\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$
$\cos[\cos^{-1}(\frac{3}{5})] + \co... | The $\cos $ function doesn't behaves in linear way. In the 2nd step in your calculation you have treated it like a linear function. Just use the formula: $$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Solve $\int\frac{2x-3}{(x^2+x+1)^2}dx$ $\int\frac{2x-3}{(x^2+x+1)^2}dx$
$\int\frac{2x-3}{(x^2+x+1)^2}dx=\int\frac{2x+1}{(x^2+x+1)^2}dx-\int\frac{4}{(x^2+x+1)^2}dx$
First integral is easily integrable but substituting $x^2+x+1=t$ but i cannot integrate the second integral.
| Hint:
As $x^2+x+1=\dfrac{(2x+1)^2+3}4,$ set $2x+1=\sqrt3\tan t$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why is my solution incorrect for solving these quadratic equations? $$\frac2x -\frac5{\sqrt{x}}=1 \qquad \qquad 10)\ \frac3n -\frac7{\sqrt{n}} -6=0$$
I have these two problems. For the first one I create a dummy variable,
$y = \sqrt x$ then $y^2 = x$.
Substituting this in the first equation, I get:
$\displaystyle \fr... | You are doing your algebra correctly, the only problem is when you have a positive and a negative value for a square root you have to ignore the negative one.
I checked the answer $$ x= \frac {58-10 \sqrt {33}}{4}$$ for your first problem and it does work nicely.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3012370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
Points $A$ and $B$ lie on the parabola $y=2x^2+4x-2,$such that the origin is the mid point of the line segment $AB$ Points $A$ and $B$ lie on the parabola $y=2x^2+4x-2,$such that the origin is the mid point of the line segment $AB$.Find the length of the line segment $AB$
$y=2(x^2+2x-1)=2(x+1)^2-4\implies (y+4)=2(x+2)... | Let the points be $A(x_1,2x_1^2+4x_1-2)$ and $B(x_2,2x_2^2+4x_2-2)$. Then:
$$\begin{cases}\frac12(x_1+x_2)=0\\ \frac12(2x_1^2+4x_1-2+2x_2^2+4x_2-2)=0\end{cases} \Rightarrow (x_1,x_2)=(\pm 1,\mp 1).$$
Hence: $A(1,4)$, $B(-1,-4)$ and $AB=\sqrt{(-1-1)^2+(-4-4)^2}=\sqrt{68}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3014268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Smallest value of $a^2 + b^2 + c^2+ d^2$, given values for $(a+b)(c+d)$, $(a+c)(b+d)$, and $(a+d)(b+c)$
If $a$, $b$, $c$, $d$ belong to $\mathbb{R}$, and
$$(a+b)(c+d)=143 \qquad (a+c)(b+d)=150 \qquad (a+d)(b+c)=169$$
Find the smallest possible value of
$$a^2 + b^2 + c^2+ d^2$$
I thought of adding $7$ to the fi... | Let $\lambda = a + b + c + d$. When you sum over the 3 equations.
$$(a+b)(c+d)=143,\quad (a+c)(b+d)=150,\quad (a+d)(b+c)=169\tag{*1}$$
LHS sums to
$$2(ab+ac+ad+bc+bd+cd) = (a+b+c+d)^2 - (a^2+b^2+c^2+d^2)$$
while RHS sums to $462$. This leads to
$$a^2 + b^2 + c^2 + d^2 = \lambda^2 - 462$$
To minimize $a^2+b^2+c^2+d^2$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3014786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Modulo and probability How can I prove that 4 modulo 5 is 4?
My though is floor of (4 / 5) is 0 then the remaining is = to the modulo.Am I right?
| You proved it what your proved $\equiv \pmod n$ was an equivalence relationship.
$4 \equiv 4 \pmod 5$ because, being an equivalence relationship, equivalence modulo $n$ is reflexive. i.e. for all $a$, $a \equiv a \pmod n$.
Of course we had to prove equivalence modulo $n$ was an equivalence relationship in the first pl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3016331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solve differential equation $\frac{dx}{dt}=a-b x-cx(1-x)=cx^2-x(b+c)+a$ I want to solve the following first- order nonlinear ordinary differential equation:
$\frac{dx}{dt}=a-b x-cx(1-x)=cx^2-x(b+c)+a$
where a,b and c are constants. I rewrote the equation:
$\leftrightarrow 1=\frac{1}{cx^2-x(b+c)+a}\frac{dx}{dt}\\
\leftr... | The main step is converting the fraction
$$\frac{1}{cx^2+x(b+c)+a}$$
into the form, expected from the integral tables:
$$\int\frac{1}{t^2+q^2}dt=\frac{1}{q}\arctan \frac{t}{q}$$
You take out the extra $c$, complete the square and change variables:
$$\frac{1}{c}\frac{1}{\color{red}{(x+\frac{b+c}{2c})}^2-(\frac{b+c}{2c})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3017166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
limit of $\sqrt{x^6}$ as $x$ approaches $-\infty$ I need to solve this limit:
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}}$$
The answer is $-3$, but I got 3 instead. This is my process:
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}} =
\lim_{x \to - \infty}{\frac {\sqrt{x^6(9-\frac {5}{x^2})}... | You are forgetting that $\sqrt{x^6}=|x^3|$ and $\frac{|x^3|}{x^3}=-1$ when $x$ is negative
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3017300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
About proof: $\cot^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac x2$ I have the following question:
Prove that: $$ \cot^{-1}\Biggl(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\Biggl) = \frac x2, \ x \in \biggl(0, \frac \pi4\biggl) $$
The s... | We see that
$$\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}$$
$$= \sqrt{(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}})^2}$$
$$= \sqrt{\frac{1+\sin x +1-\sin x + 2\cdot\sqrt{1-\sin^2 x}}{1+\sin x +1-\sin x -2\cdot\sqrt{1-\sin^2 x}}}$$
$$= \sqrt{\frac{1+\cos x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3021289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
How to prove this algebra question? If $x=y^2+z^2$, $y=x^2+z^2,$ and $z=x^2+y^2$ then show that $$\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=1$$
| hint
$x-y=y^2-x^2=(y-x)(y+x)$. So either $x=y$ or $x+y+1=0$. Do the same with other pairs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3021574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that integral of $ 1-\cos\left(\frac{1}{x^2}\right) $ is finite I need to prove that
$$ \int_0^{\infty} \left(1 - \cos\left(\frac{1}{x^2}\right)\right) dx < \infty $$
My attempt:
$$ \forall x\in[0,\infty] \hspace{1cm} 0 < 1 - \cos\left(\frac{1}{x^2}\right) < 2 \tag{1}. $$
Using L'Hôpital's rule I can show that... | $\int_0^\infty(1-\cos (1/x^2))dx= \int_0^1(1-\cos (1/x^2))dx+\int_1^\infty(1-\cos (1/x^2))dx\leq 2+\int_1^\infty2\sin^2 \left(\frac{1}{2x^2}\right)dx\leq 2+\int_1^\infty2\sin \left(\frac{1}{2x^2}\right)dx\leq2+\int_1^\infty2 \frac{1}{2x^2}dx \text{ (as } \sin x\leq x \text{ forall } x>0) =2+\int_1^\infty \frac{1}{x^2}d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3022672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Trying to prove an equation I would like to receive some help about the next problem.
The problem:
I'm trying to prove the next equation:
$$\sum_{k = 0}^{n} \frac{(-1)^{-k}}{k!(n - k)!} = 0 \quad, n = 1, 2, ...$$
My work until now:
$$\sum_{k = 0}^{n} \frac{(-1)^{-k}}{k!(n - k)!} = \sum_{k = 0}^{n} \frac{1}{(-1)^{k}k!(n... | $\sum_{k=0}^n \frac{(-1)^{-k}}{k!(n-k)!}=1/(n!)\sum_{k=0}^n \frac{n!(-1)^{k}}{k!(n-k)!}=\frac{1}{n!}\sum_{k=0}^n(-1)^{k}{n\choose k}=\frac{1}{n!}(1-1)^n=0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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For $0 \leq a,b \leq 1$, why does $aFor $0 \leq a,b \leq 1$, why does $a<b$ imply $\frac a{1-a }< \frac b{1-b}$?
Do we need to compute derivatives to validate this inequality? Or talk about increasing/decreasing functions?
| Rearrange the inequality as follows:
\begin{align}&\frac{a}{1-a}<\frac{b}{1-b}\\&\frac{a}{a-1}>\frac{b}{b-1}\\&\frac{a-1+1}{a-1}>\frac{b-1+1}{b-1}\\&1+\frac{1}{a-1}>1+\frac{1}{b-1}\\&\frac{1}{a-1}>\frac{1}{b-1}\end{align}
From $a<b$ it follows that $a-1<b-1<0$ and the inequality is thus satisfied.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Expression for $\cos^{-1}x\pm\cos^{-1}y$ As mentioned in Proof for the formula of sum of arcsine functions $\arcsin x+\arcsin y$ for $\sin^{-1}x+\sin^{-1}y$
$$
\sin^{-1}x+\sin^{-1}y=
\begin{cases}
\sin^{-1}( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \;\;;x^2+y^2 \le 1 \\\
\pi - \sin^{-1}( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \;\;;x^2+y... | Alternatively, to get rid of all those cases, we can write:
$$\sin^{-1}x\pm\sin^{-1}y=\text{atan2}(x\sqrt{1-y^2}\pm y\sqrt{1-x^2},\ \sqrt{1-x^2}\sqrt{1-y^2}\mp xy)$$
and:
$$\cos^{-1}x\pm\cos^{-1}y=\text{atan2}(y\sqrt{1-x^2}\pm x\sqrt{1-y^2},\ xy\mp\sqrt{1-x^2}\sqrt{1-y^2})$$
See atan2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3027787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $a, b, c \in \mathbb{Z^+}$ and $a^2+b^2=c^2$ then ${1\over2}(c-a)(c-b)$ is a perfect square.
Prove that if $a, b, c \in \mathbb{Z^+}$ and $a^2+b^2=c^2$ then ${1\over2}(c-a)(c-b)$ is a perfect square.
I have tried to solve this question and did pretty well until I reached the end, so I was wondering if I... | Multiply by conjugate:
$${ab\over a+b+c}={ab\over a+b+c}\cdot \frac{a+b-c}{a+b-c}=\frac{ab(a+b-c)}{2ab}=\frac{a+b-c}{2}\in \mathbb Z^+,$$
because:
$$a+b>c$$
and there are two cases for $a^2+b^2=c^2$: 1) $a,b,c$ are even; 2) one is even, the other two are odd. And for each case, $a+b-c$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3029557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
How to solve for unknown matrix? How can I solve this?
$$ \begin{bmatrix}
1 & 1 \\
1 & 2 \\
\end{bmatrix}
X +
\begin{bmatrix}
2 & -1\\
-1 & 1\\
\end{bmatrix}
X
\begin{bmatrix}
1 & 5 \\
1 & 2 \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 1\\
1 & 1\\
\end{bmatrix}
$$
I know there's similar question like:
Solve... | If we can't see the trick suggested in the comments, by $X=\begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix}$ we obtain
$$ \begin{bmatrix}
1 & 1 \\
1 & 2 \\
\end{bmatrix}X= \begin{bmatrix}
a+b & a+2b \\
c+d & c+2d \\
\end{bmatrix}$$
$$
\begin{bmatrix}
2 & -1\\
-1 & 1\\
\end{bmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find limit $\lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)$ $\displaystyle \lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}}\displaystyle -\sqrt{2x^{4}}\right)$
$\displaystyle \lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^... | Another approach is using substitution $x=\cot t$:
\begin{align}
\lim_{x\to\infty}\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}}-\sqrt{2x^{4}}\right)
&=\lim_{t\to0}\dfrac{\sqrt{\cos^2t+\cos t}-\sqrt{2\cos^2t}}{\sin t} \\
&=\lim_{t\to0}\dfrac{\cos t(1-\cos t)}{\sin t(\sqrt{\cos^2t+\cos t}+\sqrt{2\cos^2t})} \\
&=\lim_{t\to0}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solutions to $a,\ b,\ c,\ \frac{a}{b}+\frac{b}{c}+\frac{c}{a},\ \frac{b}{a} + \frac{c}{b} + \frac{a}{c} \in \mathbb{Z}$ I came across a puzzle in a Maths Calendar I own. Most of them I can do fairly easily, but this one has me stumped, and I was hoping for a hint or solution. The question is:
What are the solutions to
... | Suppose that $\displaystyle a,b,c,\frac{a}{b}+\frac{b}{c}+\frac{c}{a},\frac{a}{c}+\frac{b}{a}+\frac{c}{b} \in \mathbb Z$.
Consider polynomial
$$P(x)=\left(x-\frac{a}{b}\right)\left(x-\frac{b}{c}\right)\left(x-\frac{c}{a}\right) = x^3-\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)x^2+\left(\frac{a}{c}+\frac{b}{a}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3033895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Minimizing distance from a point to a parabola Problem: The point on the curve $x^2 + 2y = 0$ that is nearest the point $\left(0, -\frac{1}{2}\right)$ occurs at what value of y?
Using the distance formula, I get my primary equation: $L^2 = (x-0)^2 + \left(y-\left(-\frac{1}{2}\right)\right)^2$
However, when using the se... | If $x=2t, y=-2t^2$
We need to minimize $(2t-0)^2+(2t^2+1/2)^2=4t^4+6t^2+1/4=\dfrac{(4t^2+3)^2-8}4\ge\dfrac{0+9-8}4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving this equation: $3^{\log_{4}x+\frac{1}{2}}+3^{\log_{4}x-\frac{1}{2}}=\sqrt{x}$
Solve this equation:
$$3^{\log_{4}x+\frac{1}{2}}+3^{\log_{4}x-\frac{1}{2}}=\sqrt{x}\qquad (1)$$
I tried to make both sides of the equation have a same base and I started:
$$(1)\Leftrightarrow 3^{\log_{4}x}.\sqrt{3}+ \frac{3^{\lo... | Let's generalise a bit, with parameters say $a, b \in (0, \infty)\setminus \{1\}$ subject to $a^2 \neq b$ and let's try to solve the equation:
$$a^{\log_{b}x+\frac{1}{2}}+a^{\log_{b}x-\frac{1}{2}}=\sqrt{x}$$
Notice that the left-hand side can be rewritten as
$$a^{\log_{b}x}(\sqrt{a}+\frac{1}{\sqrt{a}})=b^{\log_{b}a\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Show that point does not belong to a plane Problem
Show that point $\textbf{q}$ does not belong to plane defined by these 3 points:
$$ \textbf{p}_1 = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}, \textbf{p}_2=\begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}, \textbf{p}_3 = \begin{bmatrix} 2 \\ 4 \\ 0 \end{bmatrix} $$
Point $\... | Define ${\bf u} = {\bf p}_2 - {\bf p}_1$ and ${\bf v} = {\bf p}_3 - {\bf p}_1$, a normal vector to the plane is
$$
\hat{\bf n} = \frac{{\bf u}\times {\bf v}}{|{\bf u}\times {\bf v}|} = \frac{1}{\sqrt{19}}\pmatrix{3 \\ -1 \\ -3}
$$
A point ${\bf x}$ belongs to the plane if
$$
\hat{\bf n}\cdot({\bf x} - {\bf p}_1) = 0
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3037467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Expressing $\int_0^{\pi/2}\frac{\sin 2x}{x+1}\,dx$ using $A = \int_0^\pi\frac{\cos x}{(x+2)^2}\,dx$ If $$A = \int_0^\pi\frac{\cos x}{(x+2)^2}\,dx,$$
then find the value of $$\int_0^{\pi/2}\frac{\sin 2x}{x+1}\,dx$$ in terms of $A$.
| As pointed out in the comments, the substitution $x \mapsto 2x $ gives
$$ \int_0^{\pi/2} \frac{\sin 2x}{2x+2}2dx = \int_0^\pi \frac{\sin x}{x+2}dx $$
Integration by parts gives
$$ \int_0^\pi \frac{\sin x}{x+2}dx = -\frac{\cos x}{x+2}\Bigg\vert_0^\pi - \int_0^\pi \frac{\cos x}{(x+2)^2}dx = \frac{1}{\pi+2} + \frac12 - A ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3037875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"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.