Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$\sqrt[5]{2^4\sqrt[3]{16}} = ? $ $$\sqrt[5]{2^4\sqrt[3]{16}} = ? $$
Rewriting this and we have
$$\sqrt[5\cdot 3]{2^{4\cdot 3}4}$$
$$\sqrt[15]{2^{12}2^2}$$
Finally we get
$$\sqrt[15]{2^{12}2^2} = \sqrt[15]{2^{14}} = 2^{\frac{15}{14}}$$
Am I right?
| \begin{align}
\sqrt[5]{2^4 \sqrt[3]{16}}
&=\sqrt[5]{2^4 \sqrt[3]{2^4}}
=\sqrt[5]{2^4 \sqrt[3]{2^3 \cdot 2}}
=\sqrt[5]{2^4 \cdot 2 \sqrt[3]{2}}\\
&=\sqrt[5]{2^5 \sqrt[3]{2}}
=2 \sqrt[5]{\sqrt[3]{2}}
=2 \cdot \sqrt[15]{2}
=2 \cdot 2^{1/15}\\
&=2^{16/15}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2756282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute $\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}$
Compute $\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}$
This question came after an exercise involving finding the $7$th roots of $-1$. The roots were $\operatorname{cis}\frac{\pi}{7},\operatorname{cis}\frac{3\pi}{7},\dots$
This made me wonde... | As $\cos(\pi-x)=-\cos(x)$ this is the same as to compute:
$$\cos \left( \frac{\pi}{7} \right)+\cos \left( \frac{3 \pi}{7} \right)+\cos \left(5 \frac{\pi}{7} \right)$$
But denoting by $\omega_i$ $i\in{0,\ldots 6}$ the $7$th roots of $-1$.
You have:
$$\sum_{i} \omega_i=0$$
and by taking the real part:
$$-1+2\left( \cos \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2757509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Estimation of $\int_0^1 \frac{1}{1+x^4}dx$
Prove that $\dfrac34<\displaystyle\int_0^1 \frac{1}{1+x^4}\,\mathrm dx<\dfrac {9}{10}$.
My working:
We can easily prove that
$$\begin{align}
\frac{1}{1+x^2}&<\frac{1}{1+x^4}<1-x^4+x^8,\forall x\in (0,1) \\
\implies\int_0^1\frac{1}{1+x^2}\,\mathrm dx&<\int_0^1\frac{1}{1+x^4}\... | We can easily prove that
$$\begin{align}
1-x^3&<\frac{1}{1+x^4}<1-\frac{x^4}{2},\forall x\in (0,1) \\
\implies\int_0^1 1-x^3\,\mathrm dx&<\int_0^1\frac{1}{1+x^4}\,\mathrm dx<\int_0^11-\frac{x^4}{2}\,\mathrm dx \\
\implies\frac34&<\int_0^1\frac{1}{1+x^4}\,\mathrm dx<1-\frac {1}{10}=\frac{9}{10}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2757863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A Problem Dealing with putting balls in bin and Expected Value - possible wrong answer Please consider the problem below. Is my answer correct. If is not correct then where did I go wrong?
Problem:
You keep tossing balls into $n$ bins until one of the bins has two balls. For each toss there is a $\frac{1}{n}$ probabili... | Use the pigeonhole principle.
If the number of tosses ($n$) $\geq b+1$ , then at least one bin will contain $2$ balls. If the number of tosses is $0 < n < b$, the probability that each ball will go into a different bin is $$\frac{1 \times 2 \times ...(b-n+1)}{b^n} = \frac{b!}{n!\times b^n}$$
So the probability that a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
$30\mid p_1^4+p_2^4+\cdots+p_{31}^4$. Prove that $p_1=2,p_2=3,p_3=5$ Let $p_1<p_2<\cdots<p_{31}$ be prime numbers such that $30\mid p_1^4+p_2^4+\cdots+p_{31}^4$. Prove that $p_1=2,p_2=3,p_3=5$
I have no clue whatsoever about how to even approach this problem. Any hint is welcome.
| There are $30$ terms in $p_2^4+...+p_{31}^4$ and each of them is $\equiv 1\pmod 2.$ So $2$ divides $p_1^4+30$ so $2$ divides $p_1^4$ so $p_1=2$.
There are $29$ terms in $p_3^4+...+p_{31}^4 $ and each of them is $\equiv 1 \pmod 3 .$ So $3$ divides $2^4+p_2^4 +29$ so $3$ divides $p_2^4$ so $p_2=3.$
There are $28$ ter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2759741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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A question about a palindromic polynomial of even degree In Wikipedia it is stated,
If $p(x)$ is a palindromic polynomial of even degree $2d$, then there is a polynomial $q$ of degree $d$ such that $p(x) = x^d q(x + \frac{1}{x})$.
My question is:
How does one find the polynomial $q$ given $p$.
The specific example ... | Write $q(y) = a_6 y^6 + \cdots + a_1 y + a_0$. Using your example, we have $$\begin{align*}q(x + x^{-1}) =& x^6 - x^5 - x^4 + x^2 + x -2 + x^{-1} + x^{-2} - x^{-4} - x^{-5} + x^{-6} \\
=& a_6 x^6 + a_5 x^5 + (6 a_6 + a_4) x^4 + (5 a_5 + a_3) x^3 + \\&(15 a_6 + 4 a_4 +a_2) x^2 + (10 a_5 + 3 a_3 + a_1) x + \\
&(20a_6 + 6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to solve this complex limits at infinity with trig? Please consider this limit question
$$\lim_{x\rightarrow\infty}\frac{a\sin\left(\frac{a(x+1)}{2x}\right)\cdot \sin\left(\frac{a}{2}\right)}{x\cdot \sin\left(\frac{a}{2x}\right)}$$
How should I solve this? I have no idea where to start please help.
| Trivial with equivalents:
as $x\to\infty$,
*
*$\dfrac{a(x+1)}{2x}\to \dfrac a2$, so $\sin\dfrac{a(x+1)}{2x}\sim_\infty \sin \dfrac a2$,
*$\sin\dfrac{a}{2x}\sim_\infty\dfrac{a}{2x} $,
so that
$$\frac{a\sin \dfrac{a(x+1)}{2x}\,\sin\dfrac{a}{2}}{x\cdot \sin \dfrac{a}{2x}}\sim_\infty\frac{a\sin^2\dfrac a2}{x\,\dfrac a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Probability that if three balls numbered from 1-20 are selected without replacement that at least one will be numbered at least 17 I'm a new user here. I am currently self-learning probability theory (long journey ahead). I was wondering as to why my answer is not correct. I have listed the question below.
Three balls ... |
Find the probability that at least one of three balls selected from an urn containing $20$ balls, numbered from $1$ to $20$, displays a number that is at least $17$.
Method 1: We subtract the probability that none of the three selected balls have numbers larger than $16$ from $1$.
There are
$$\binom{20}{3}$$
ways... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2767202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Evaluate$\int_0^\infty \frac{\log(x)^2}{(1+x)^2}dx$ Trying to evaluate the integral
$$\int_0^\infty \frac{\log(x)^2}{(1+x)^2}dx$$
or equivalently $$\int_{-\infty}^\infty e^x\frac{x^2}{(1+e^x)^2}dx$$
The answer should be $\frac{\pi^2}{3}$
Looks like we can use residue theorem, but the integration path does not seem to b... | We can compute directly.
$$\int_0^{\infty}\frac{\log^2x}{(1+x)^2}dx = \int_{0}^1
\frac{\log^2x}{(1+x)^2}dx + \int_1^{\infty}\frac{\log^2x}{(1+x)^2}dx. $$
With a change of variable $t=\frac{1}{x}$ in the second integral we have
$$\int_{1}^{\infty} \frac{\log^2x}{(1+x)^2} dx = \int_0^1\frac{\log^2t}{(1+t)^2}dt.$$
Then w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2767291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Evaluate $\int \frac{e^{2x}-1}{e^{2x}+1}dx$ using partial fractions
Integrate the function $\frac{e^{2x}-1}{e^{2x}+1}$ using partial fractions.
My Attempt
$$
\int\frac{e^{2x}-1}{e^{2x}+1}dx=\frac{1}{2}\int\frac{2.e^{2x}}{e^{2x}+1}dx-\int\frac{dx}{e^{2x}+1}=\frac{1}{2}\log|e^{2x}+1|-\int\frac{dx}{e^{2x}+1}
$$
Put $t=e... | $$\begin{equation}
\frac{1}{t(t^2+1)}=\frac{A}{t}+\frac{Bt+C}{t^2+1}\implies 1=A(t^2+1)+t(Bt+C)
\end{equation}$$
This is for all t you get that
$$1=t^2(A+B)+Ct+A \quad \color{red}{\forall t}$$
$$A+B=0, A=1, C=0$$
$$(A,B,C)=(1,-1,0)$$
Note that you just integrate $\tanh(x)$
$$\int \frac{e^{2x}-1}{e^{2x}+1} dx= \int \tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2769580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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British Maths Olympiad (BMO) 2004 Round 1 Question 1 alternative approaches? The questions states:
Solve the simultaneous equations (which I respectively label as $
> \ref{1}, \ref{2}, \ref{3}, \ref{4}$)
$$\begin{align} ab + c + d &= 3 \tag{1} \label{1} \\ bc + d + a &= 5
\tag{2} \label{2} \\ cd + a + b &= 2 \tag{3... | Step 1: Obtain $a+c=2$.
Step 2:
Note that
$$ab+bc+cd+ad=(a+c)(b+d)$$
Adding four equations gives
$$(a+c)(b+d)+2(a+c)+2(b+d)=16$$
$$(a+c)(b+d+2)+2(b+d+2)=20$$
$$(a+c+2)(b+d+2)=20$$
With $a+c=2$, we have $b+d=3$.
Step 3:
Further manipulating, 1-2+3-4 gives
$$(a-c)(b+d)=6$$
Thus, $a-c=2$.
Therefore, we have $a=2, c=0$.
St... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2770117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Given that exactly 2 Jacks appear, what is the expected number of Aces that appear? Pick 5 cards from standard 52 cards without replacement. Given that exactly 2 jack cards appear, find the expected number of ace cards that appear.
ATTEMPT
Let $X$ be number of aces chosen and $Y$ number of jacks chosen so we want find ... |
Now, pluggin in into the first equation should give the answer. Is this a correct approach to tackle this problem?
Yes, that would work, although your evaluations are off.
$p_{\small Y}(2) =\mathsf P(Y{=}2) = \left.\binom 4 2\binom {48}{3}\middle/\binom{52}{5}\right.$ is the probability for selecting $2$ from $4$ Jac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2772929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Induction principle: $n^2-7n+12≥0$ for every $n≥3$ How can I prove that $n^2-7n+12≥0$ for every $n≥3$?
I know that for $n=3$ I have $0≥0$ so the inductive Hypothesis is true.
Now for $n+1$ I have $(n+1)^2-7(n+1)+12=n^2-5n+6$ and now I don't know how to go on...
| Use completing the square:
$$n^2-7n+12=\bigg(n-\frac{7}{2}\bigg)^2-\bigg(\frac{7}{2}\bigg)^2+12$$
$$\to\bigg(n-\frac{7}{2}\bigg)^2-\frac{49}{4}+\frac{48}{4}=\bigg(n-\frac{7}{2}\bigg)^2-\frac{1}{4}$$
Note that $k^2 \ge 0$ $\forall k \in \Bbb R$, and for $n\ge 3, \bigg(n-\frac{7}{2}\bigg)^2\ge\frac{1}{4}$, hence $\bigg(n... | {
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Deriving a recurrence equation I've really been stuck on this problem for a while. We have the equation: $s_t = (s_{t-1}/2)+3$. I need to show the steps it would take to show this can be written as: $s_t = 2^{-t}(s_0-6) +6$. I figure it has something to do with telescoping but I'm not sure how this would be done.
| This is just a direct approach. First apply the recursion a couple of times to find a pattern. Then suppose you apply it $i$ times. Then substitute $i=t$ to show what it would look like if you applied it all the way. Bam you're done.
$$\begin{align}
s_t &= \frac{1}{2}s_{t-1}+3 \\
s_t &= \frac{1}{2}\left(\frac{1}{2}... | {
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"timestamp": "2023-03-29T00:00:00",
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If $z^{23}=1$ then evaluate $\sum^{22}_{z=0}\frac{1}{1+z^r+z^{2r}}$
If $z$ is any complex number and $z^{23}=1$ then evaluate $\displaystyle \sum^{22}_{r=0}\frac{1}{1+z^r+z^{2r}}$
Try: From $$z^{23}-1=(z-1)(1+z+z^2+\cdots +z^{22})$$
And our sum $$\sum^{22}_{r=0}\frac{1}{1+z^r+z^{2r}}=\frac{1}{3}+\frac{1}{1+z+z^2}+\f... | Define $f(x)=\frac{1}{1+x+x^2}$. Then we have to calculate
$$
\sum_{k=0}^{22}f(z^k)=\frac{1}{3}+\sum_{k=1}^{22}f(z^k).
$$
Let's work out the polynomial expression for $f$ using $(a+b)(a-b)=a^2-b^2$.
$$
f(s)=\frac{s-1}{s^3-1}=\frac{(s-1)(s^3+1)}{s^6-1}=\frac{(s-1)(s^3+1)(s^6+1)}{s^{12}-1}=\frac{(s-1)(s^3+1)(s^6+1)(s^{12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776254",
"timestamp": "2023-03-29T00:00:00",
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A common tangent line
The graph of $f(x)=x^4+4x^3-16x^2+6x-5$ has a common tangent line at $x=p$ and $x=q$. Compute the product $pq$.
So what I did is I took the derivative and found out that $p^2+3p+q^2+3q+pq=0$. However when I tried to factorize it I didn't find out an obvious solution. Can someone hint me what to ... | The tangent line equation at $x=p$ and $x=q$ is:
$$y=p^4+4p^3-16p^2+6p-5+(4p^3+12p^2-32p+6)(x-p);\\
y=q^4+4q^3-16q^2+6q-5+(4q^3+12q^2-32q+6)(x-q).\\$$
By equating the coefficients, simplifying and denoting $a=p+q, b=pq$ we get:
$$\begin{cases}b=a^2+3a-8\\
3a^3-(6a+8)b+8a^2-16a=0\end{cases}$$
Solving the system we get:
... | {
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"timestamp": "2023-03-29T00:00:00",
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Inequality $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}>1$ Show that $$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}>1,\:\forall n\in\mathbb{N}$$
This is a 9th grade problem.
I was trying to take the greatest numerator, which is the last numerator of the last fraction. But there are only $2n+1$ terms. Right?
After... | Use the method given in this answer by Jack D'Aurizio.
Note: $H_n=1+\frac12+\frac13+\cdots+\frac1n=\sum_{k=1}^n \frac1n$ is called $n$-th harmonic number.
Then:
$$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}=\sum_{k=1}^{2n+1}\frac{1}{n+k}=H_{3n+1}-H_{n}.$$
Consider the sequence: $a_n=H_{3n+1}-H_n$. We will show tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many ways can dominoes cover a $2 \times n$ rectangle? Justify proposed solution. I was able to get that $d_n = d_{n-1}+d_{n-2}$
It isn't finished, because I have to solve this recursive equation.
I read about Binet's formula, but I don't know the steps between this
$$d_n = d_{n-1}+d_{n-2}$$
and this
$$f_n=\frac{1... | Your recurrence relation is correct. The sequence either begins with one vertical tile followed by one of $d_{n - 1}$ admissible sequences of length $n - 1$ or two horizontal tiles followed by one of the $d_{n - 2}$ admissible sequences of length $n - 2$. Hence,
$$d_n = d_{n - 1} + d_{n - 2}$$
The recursion $d_n = d... | {
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"url": "https://math.stackexchange.com/questions/2783655",
"timestamp": "2023-03-29T00:00:00",
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Integration of $ \int_1^2 \frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2} \mathrm dx$ Evaluate $$\int_1^2 \frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2}\mathrm dx$$I believe I should use the partial fractions technique to evaluate this integral, but I am not getting anywhere when I try it.
| \begin{align} \frac {3x^3 + 3x^2 - 5x + 4}{3x^3 -2x^2 + 3x - 2}
& = \frac {(3x^3 - 2x^2 + 3x - 2) + 5x^2 - 8x + 6}{3x^3 -2x^2 + 3x - 2}\\ \\
&\ = 1 + \frac{5x^2 - 8x + 6}{3x^3 -2x^2 + 3x - 2} \end{align}
Factor the denominator
$$3x^3 -2x^2 + 3x - 2 = (3x-2)(x^2 + 1)$$
$$\frac {5x^3 -8x^2 +6}{(3x-2)(x^2+1)} = \frac {A}... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that a sequence does not have a limit How can I show that the sequence $x_n=\frac{(-1)^nn}{n+1}$ does not have a limit by using only the definition of limits?
Attempts:
Let's assume that our sequence has a limit $x$. Then there exists such a number $N$ so that for any $n>N$ $|a_n-a| < \epsilon$. This is also true ... | If
$x_n=\frac{(-1)^nn}{n+1}
$
has a limit $v$,
then,
for any $c > 0$
there is a $n(c)$ such that
$|x_n-v| < c$
for all $n > n(c)$.
In this case,
$x_n$ has two subsequences
that have limits:
$x_{2n} \to 1$
and
$x_{2n+1} \to -1$.
So let's choose a $c$
that is less than
half the distance
between the limits of
these two su... | {
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"url": "https://math.stackexchange.com/questions/2784124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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${1 \over \sqrt{3}} \sum\limits _{r=0}^4 \tan \left( \frac{\pi}{15}+\frac{r\pi}{5} \right)=$? $$\sum _{r=0}^4 \tan \left( \frac{\pi}{15}+\frac{r\pi}{5} \right)=k \sqrt{3}$$. Then evaluation of $k$
solution i try
$$\tan(12^\circ)+\tan(48^\circ)+\tan(84^\circ)+\tan(120^\circ)+\tan(156^\circ)$$
$$=\tan(12^\circ)+\tan(48... | $$S=\tan(12^\circ)+\tan(48^\circ)+\tan(84^\circ)+\tan(120^\circ)+\tan(156^\circ)$$
$$S=\tan(12^\circ)+\tan(48^\circ)+\tan(84^\circ)+\tan(156^\circ)+\sqrt{3}$$
$$S=\frac{\sin(60^\circ)}{\cos(12^\circ)\cos(48^\circ)}+\frac{\sin(240^\circ)}{\cos(84^\circ)\cos(156^\circ)}-\sqrt{3}.$$
$$S=\frac{\sqrt{3}}{\cos (60^\circ)+\co... | {
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"url": "https://math.stackexchange.com/questions/2785821",
"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int\sqrt{1+\sin x}dx$. The integral is:$$\int\sqrt{1+\sin x} dx$$
My first attempt was to multiply by:
$$\frac{\sqrt{1-\sin x}}{\sqrt{1-\sin x}}$$
giving:
$$\int\frac{\sqrt{1-\sin^2(x)}}{\sqrt{1-\sin x}}dx$$
which is:
$$\int\frac{\cos x}{\sqrt{1-\sin x}}dx$$
then make the substitution:
$$u=1-\sin x$$
whic... | $$I = \int \sqrt{1+\sin x}dx$$
Apply substitution $ u=\tan \left(\frac{x}{2}\right)$
(Tangent half-angle substitution)
$$\int \frac{2\sqrt{\left(u+1\right)^2}}{\left(1+u^2\right)\sqrt{u^2+1}}du$$
$$2\int \frac{u+1}{\left(1+u^2\right)\sqrt{u^2+1}}du$$
$$2\left(\underbrace{\int \frac{u}{\left(1+u^2\right)\sqrt{u^2+1}}du... | {
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"url": "https://math.stackexchange.com/questions/2786184",
"timestamp": "2023-03-29T00:00:00",
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How to explain this point transformation? Yesterday we were having a lecture on point coordinates after rotation. The prof. explained that the position of a point after a counterclockwise rotation is obtained from the following formula,
$x=x_0 \operatorname{Cos}(\theta)-y_0 \operatorname{Sin}(\theta), \qquad y=y_0 \ope... | You must use some trig equalities:
$$
\cos\frac{\pi-\theta}{2}=\sin\frac{\theta}{2},\quad
\sin\frac{\pi-\theta}{2}=\cos\frac{\theta}{2},\\
\cos\theta=1-2\sin^2\frac{\theta}{2}=2\cos^2\frac{\theta}{2}-1,\quad
\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}
$$
to obtain:
$$
\begin{align}
&\left(-x_0 \cos\frac{\pi-\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Interpolating $f$ and approximating its derivative We are given $x_0 = 1, x_1 = \frac{4}{3}$ and $x_2 = 2$.
Find a parabola which agrees with function $f(x) = (x + 1)\sin(x)$ in the given points. Afterwards derive a formula for the
approximation of the $f'$ i.e. the derivative in $x_1$. What is the approximation of $f'... | $$
f\left(x \right) = \left(x + 1\right)\sin x
\implies \left\lbrace
\begin{alignedat}{4}
f\left(x_0\right)&= 2\sin \left(1\right) &&\approx 1.68,\qquad &\text{where} \quad x_0&=1 \\
f\left(x_1\right)&= \tfrac{7}{3}\sin \left(\tfrac{4}{3}\right)&\, &\approx 2.27, &\text{where} \quad x_1&=4/3 \\
f\left(x_2\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2788404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Area of square under a curve.
A square having sides parallel to the coordinate axes is inscribed in the region.
{$(x,y):x,y>0:y\le -x^3+3x$}.
If the area of the square is written as $A^{1/3}+B^{1/3}$ square units where $A,B\in \Bbb Z$ and $A>B$, then find
(i)$\sqrt{A-B}$
(ii)Slope of line with x and y intercep... | Define $f(x):=3 x - x^3$. Obviously, the square must have a side length of $f(q)$, and we need to find $q$ that gives the largest square inside $f(x)$.
The line $y=f(q)$ intersects $y=f(x)$ at three points:
$$\left(q,f\left(q\right)\right)\\
\left(\frac{1}{2} \left(-\sqrt{3} \sqrt{4-q^2}-q\right),f\left(\frac{1}{2} \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\int\frac{1}{(\sin x+a\sec x)^2}dx$
Evaluation of $$\int\frac{1}{(\sin x+a\sec x)^2}dx$$
Try: Let $$I=\int\frac{1}{(\sin x+a\sec x)^2}dx=\int\frac{\sec^2 x}{(\tan x+a\sec^2 x)^2}dx$$
put $\tan x=t$ and $dx=\sec^2 tdt$
So $$I=\int\frac{1}{(a+at^2+t)^2}dt$$
Could some help me how to solve above Integral ... | HINT
Let
$$b=\dfrac1{2a},\quad y=t+b,\tag1$$
then
$$I=4b^2\int\dfrac{dt}{(t^2+2bt+1)^2} = 4b^2\int\dfrac{dy}{(y^2+1-b^2)^2} = \dfrac{4b^2}{1-b^2}\int\dfrac{(y^2+1-b^2)-y^2}{(y^2+1-b^2)^2}\,dy,$$
$$I=\dfrac{4b^2}{1-b^2}(I_1+I_2),\tag2$$
where
$$I_1=\int\dfrac{dy}{y^2+1-b^2} = const +
\begin{cases}\dfrac1{\sqrt{1-b^2}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2793823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Partial fraction decomposition trouble I'm trying to do a partial fraction expansion on
$$Y = n \cdot \frac{e^{-pt_{0}}}{(p+n)(p^2+\omega^2)}$$
which gives $A(p^2+\omega^2)+(Bp+C)(p+n) = 1$
which implies
$$A = -B$$
$$Bpn=0\rightarrow B=0\rightarrow A=0$$
which is incorrect. Any help with what I'm missing would be grea... | The condition
$$
A(p^2 + \omega^2) + (B p + C)(p + n ) = 1
$$
leads to the equations
\begin{eqnarray}
A + B &=& 0 \\
B n + C &=& 0 \\
A \omega^2 + C n &=& 1
\end{eqnarray}
whose solution is
\begin{eqnarray}
A &=& \frac{1}{\omega^2 + n^2} = -B \\
C &=& \frac{n}{\omega^2 + n^2}
\end{eqnarray}
So the fraction is
$$
Y =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why is there a pattern to the last digits of square numbers? I was programming and I realized that the last digit of all the integer numbers squared end in $ 0, 1, 4, 5, 6,$ or $ 9 $.
And in addition, the numbers that end in $ 1, 4, 9, 6 $ are repeated twice as many times as the numbers that end in $ 0, 5$
I checked th... | What you are looking at is the residues of squares modulo $10$.
$0^2=\color{red}0\bmod 10\\1^2=\color{blue}1\bmod 10\\2^2=\color{orange}4\bmod 10\\3^2=9\bmod 10\\4^2=\color{green}6\bmod 10\\5^2=\color{brown}5\bmod 10\\6^2=\color{green}6\bmod 10\\7^2=9\bmod 10\\8^2=\color{orange}4\bmod 10\\9^2=\color{blue}1\bmod 10$
As ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 7,
"answer_id": 5
} |
Matrix exponential Let $A,B,C \in \operatorname{Mat}_2(\mathbb{R})$ define the real matrices:
$$A = \begin{pmatrix}
0 & 1 \\
0 & 0 \\
\end{pmatrix},\ B =\begin{pmatrix}
0 & 0 \\
0 & 1 \\
\end{pmatrix},\ C = A+B=\begin{pmatrix}
0 & 1 \\
0 & 1 \\
\end{pmatrix}
$$
Show that
$$\exp(A... | I assume that $e^A = I + A + {1 \over 2!} A^2+ \cdots$.
Note that $A^2 = 0$, hence $e^{At} = I + tA$.
Note that $B^k = B$ for all $k \ge 1$, hence $e^{Bt} = I + B(t + {t^2 \over 2} + \cdots) )= \begin{bmatrix} 1 & 0 \\ 0 & e^t \end{bmatrix}$.
Finally, as Jean-Claude noted, $C^k = C$ for $k \ge 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trigamma identity $\psi_1\left(\frac{11}{12}\right)-\psi_1\left(\frac{5}{12}\right)=4\sqrt 3 \pi^2-80G$ Regarding this integral: Integral $\int_0^1 \frac{\sqrt x \ln x} {x^2 - x+1}dx$ The following conjecture comes: $$\psi_1\left(\frac{11}{12}\right)-\psi_1\left(\frac{5}{12}\right)=4\sqrt 3 \pi^2-80G$$ Where $G$ is Cat... | \begin{align}
\psi_1(\tfrac{11}{12})
-\psi_1(\tfrac{5}{12})
&=4\sqrt 3 \pi^2-80G
\tag{1}\label{1}
.
\end{align}
The constant $G$ - Catalan's constant
has known to appear in relations
\begin{align}
\psi_1(\tfrac14)&=\pi^2+8G
,\\
\psi_1(\tfrac34)&=\pi^2-8G
,
\end{align}
so we can try to start from $\psi_1(\tfrac14)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
${3^n\choose k}$ is divisible by $3$? How can I prove that ${3^n\choose k}$ is divisible by $3$ for all positive integer values of $n$? (where $k$ is any positive integer smaller than $3^n$) Can you use induction? Thanks.
| I just want to demonstrate the relevance of if $p$ is prime and $0<t<p$ then $p$ divides $\displaystyle \binom p t$
Consider $$\begin{split}(x+y)^p
& =\sum_{k=0}^p \binom p k x^{p-l} y^{k} \\
& =x^p+\binom p 1x^{p-1}y+\binom p 2x^{p-2}y^2+\cdots+\binom{p}{p-1}xy^{p-1}+y^p\\
& = x^p + y^p \mod p
\end{split}$$
Hence
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find $x$ if $\cot^{-1}\left(\frac{1}{x}\right)+\cos ^{-1}(-x)+\tan^{-1}(x)=\pi$ if $x \lt 0$ Then Find value of $$\frac{(1-x^2)^{\frac{3}{2}}}{x^2}$$ if
$$ \cot^{-1} \left(\frac{1}{x}\right)+\cos^{-1}(-x)+\tan^{-1}(x)=\pi$$
My try:
Since $x \lt 0$ we have $$ \cot^{-1}\left(\frac{1}{x}\right)=\pi +\tan ^{-1}x$$
Also $$... | A further step to a solution is along the following lines:
Note that if $$cot^{-1}\left(\frac{1}{x}\right)=\theta \Rightarrow{cot(\theta)=\frac{1}{x}=\frac{1}{tan(\theta)}}\Rightarrow{x=tan(\theta)}$$
Therefore
$$cot^{-1}\left(\frac{1}{x}\right)=tan^{-1}\left(x\right)$$
The original problem becomes
$$2tan^{-1}\left(x\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2805697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove using induction that $n^3 − n$ is divisible by 6 whenever $n > 0$. Prove using induction that $n^3 − n$ is divisible by $6$ whenever $n > 0$
My attempt:
Base step: For $n=1$
$1^3 - 1 = 0$.
$0$ which is divisible by $6$.
Thus, $n= 1$ is true.
Assumption step: Let $n=k$
$k^3-k$
Inductive step: $f(k+1)-f(k)$
$(k+1... | In your assumption step, you need to assume the statement is true for $n=k$, i.e. $k^3-k$ is divisible by $6$.
In the induction step, expand and simplify $(k+1)^3-(k+1)$:
$$\begin{aligned}(k+1)^3-(k+1)&=k^3+3k^2+3k+1-k-1\\&=k^3-k+3k^2+3k\\&=(k^3-k)+3k^2+3k\\&=(k^3-k)+3k(k+1)\end{aligned}$$
Note that $k^3-k$ is divisibl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Using the limit definition to find a derivative for $\frac{-5x}{2+\sqrt{x+3}}$ I am trying to find the derivative of $\frac{-5x}{2+\sqrt{x+3}}$ using the limit definition of a derivative.
$$\lim_{h \to 0} \frac {f(x+h)-f(x)}{h}$$
What I did is
$$\lim_{h \to 0} \frac{\dfrac{-5(x+h)}{2+\sqrt{x+h+3}}\dfrac{2-\sqrt{x+h+3}}... | The difference quotient to which we are to apply the operation $\;\lim\limits_{h \rightarrow 0}\;$ is
$$ \frac{f(x+h) - f(x)}{h} \;\; = \;\; \left[\frac{-5(x+h)}{2 + \sqrt{(x+h) + 3\;}\;} \; - \; \frac{-5x}{2 + \sqrt{x+3\;}\;} \right] \; \div \; h $$
$$ = \;\; \left[\frac{-5(x+h)\left(2+\sqrt{x+3\;}\right) \;\; - \;\;... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
solve $L=\int_0^4 \sqrt{1+\frac{9}{4}y^4+\frac{9}{2}y^2}dy$ I need help with this excercise.
Find the arc length of the function $$x=\frac{1}{2}(y^2+2)^{3/2}$$
from $y=0$ to $y=4$.
$$L=\int_a^b \sqrt{1+(\frac{dx}{dy})^2}dy$$
Now, $$x=\frac{1}{2}(y^2+2)^{3/2}$$
$$\frac{dx}{dy}=\frac{3}{2}y\sqrt{y^2+2}$$
Then,
$$L=\int... | We can instead take the parametrization
$y=\sqrt{2}\tan(a), x=\sqrt{2}\sec^3({a})$
$y'^2=2\sec^4(a), x'^2=18\sec^6(a)\tan^2(a)$
We then need to compute
$\int_{0}^{\arctan(2\sqrt{2})} \sqrt{18\sec^6(a)\tan^2(a)+2\sec^4(a)}da$
Making the substiution $u=\tan{a}$ transfomrs the integral to
$\int_{0}^{2\sqrt{2}} \sqrt{18u^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find real part of $\frac{1}{1-e^{i\pi/7}}$ How can you find
$$\operatorname{Re}\left(\frac{1}{1-e^{i\pi/7}}\right).$$
I put it into wolframalpha and got $\frac{1}{2}$, but I have no idea where to begin. I though maybe we could use the fact that $$\frac{1}{z}=\frac{\bar{z}}{|z|^2},$$ where $\bar{z}$ is the conjugate of ... | Well, firstly, since $z\bar{z} = |z|^2$ we know $\frac{1}{z} = \frac{\bar{z}}{|z|^2}$ (1). So let $z = 1 - e^{\frac{i \pi}{7}}$. Then from (1), $$\frac{1}{1 - e^{\frac{i \pi}{7}}} = \frac{\bar{z}}{|z|^2}.$$
Now, by Euler's formula, $e^{\frac{i \pi}{7}} = \cos(\frac{\pi}{7})+i\sin(\frac{\pi}{7}),$ so $z = (1- (\cos(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$1^n-3^n-6^n+8^n$ is divisible by $10$ Prove that $1^n-3^n-6^n+8^n$ is divisible by $10$ for all $n\in\mathbb{N}$
It is divisible by $2$ and $5$ if we rearrange it will it be enough
$(1^n -3^n)$ and $(6^n -8^n)$ is divisible by $2$.
And
$(1^n-6^n)$ and $(8^n-3^n)$ is divisible by $5$.
Hence $\gcd(2,5)$ is $1$ and it ... | Note: $p,q,r,s,t, a ,b, n \in \mathbb{N},$
$f(n):= (8^n-3^n) - (6^n-1^n)= (8-3)p-(6-1)q=5(p-q).$
On the other hand:
$f(n)= (8^n-6^n) -(3^n-1^n)=(8-6)r -(3-1)s=2(r-s)$
Hence:
$f(n)=5(p-q)=2(r-s).$
Euclid's Lemma:
$2$ divides $(p-q)$, i.e.
$p-q=2t$.
Combining:
$f(n)= 5\cdot 2 t.$
Used:
$(a^n-b^n)=$
$(a-b)(a^{n-1}+a^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Proof by induction: is it enough to show 0=0? Let $P_n$ be a proposition, so that $$P_n: 3+11+...+(8n-5)=4n^2-n$$
$P_1:3=4\cdot1^2-1$, $P_2:3+11=4\cdot2^2-2$ etc.
When proving $P_n\to P_{n+1}$, is it enough to show in the second induction step that, by subtracting from both sides we have $$4(n+1)^2-(n+1)=4n^2-n+(8(n+1)... | I would rather
make the use of induction
extremely clear.
First of all,
I would write
$P_n: 3+11+...+(8n-5)=4n^2-n
$
as
$P_n:
\sum_{k=1}^n (8n-5)
=4n^2-n
$.
This makes the start and end terms
of the sum clear
and reduces the chance of error.
Then
I would show the step
from $P_n$
to $P_{n+1}$
like this:
$\begin{array}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is this proof of the following integrals fine $\int_{0}^{1} \frac{\ln(1+x)}{x} dx$? $$\int_{0}^{1} \frac{\ln(1+x)}{x} dx = \int_{0}^{1} \frac1x\cdot (x-x^2/2 + x^3/3 -x^4/4 \ldots)dx = \int_{0}^{1}(1 - x/2 + x^2/3 - x^4/4 \dots)dx = 1-\frac1{2^2} + \frac1{3^3} - \frac{1}{4^4} \dots = \frac{\pi^2}{12}$$
Also,
$$\int_{0... | Let,
$\begin{align}I&=\int_0^1 \frac{\ln x}{1-x}\,dx\\
J&=\int_0^1 \frac{\ln x}{1+x}\,dx\\
\end{align}$
$\begin{align}I-J&=\int_0^1 \frac{2x\ln x}{1-x^2}\,dx\end{align}$
Perform the change of variable $y=x^2$,
$\begin{align}I-J&=\frac{1}{2}\int_0^1 \frac{\ln x}{1-x}\,dx\\
&=\frac{1}{2}I\end{align}$
Therefore,
$J=\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2814357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Inequalities for $\operatorname{rad}(n\text{th pentagonal number})$ and the diagonal of the regular pentagon that represents this polygonal number We denote for integers $n\geq 1$ the square-free kernel, or radical of the natural $n>1$, as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,$$ that is ... | If the $m^{\text{th}}$ pentagonal number is a square, then we have $\DeclareMathOperator{\rad}{rad}$
$$\rad \biggl(\frac{3m^2-m}{2}\biggr) \leqslant \sqrt{\frac{3m^2-m}{2}} \leqslant \sqrt{\frac{3}{2}}\cdot m\,.$$
Since $\sqrt{\frac{3}{2}} < \frac{3}{2} < \frac{1 + \sqrt{5}}{2}$, in this case we will have $(4)$ except ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2814959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the probability of a symmetric coin fall
The prompt: We have 10 coins and 1 of them is non-symmetric (with probability of head equal $\frac{1}{3}$). We toss a randomly selected coin 6 times, and obtain 3 tails. What is the probability that we tossed a symmetrical coin?
The way I went on about solving the prob... | Borrowing your notation, the value we want is: $P(S \mid E) = \frac{P(E \mid S)~P(S)}{P(E)}$, which is Bayes' rule.
Note that:
$$P(E \mid S) = {6 \choose 3} \left( \frac{1}{2} \right)^6 = 0.3125$$
$$P(E \mid NS) = {6 \choose 3} \left( \frac{1}{3} \right)^3 \left(\frac{2}{3} \right)^3=0.219$$
Thus, $P(E) = \sum_i P(E \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Expected length of longest stick The problem is the same as here.
A stick of 1m is divided into three pieces by two random points. Find the average length of the largest segment.
I tried solving it in a different way, and the logic seems fine, however I get a different result to $\frac{11}{18}$.
Here is my solution... | Here is how I would do it.
Lets define $x$ to be the short stick, $y$ to be the medium stick and $z$ to be the long stick.
$x\le y\le z\\
z = 1-x-y\\
x\le y \le \frac {1-x}{2}\\
x\le \frac 13$
$$ \bar z = \frac {\displaystyle\int_0^\frac 13\int_x^{\frac {1-x}{2}} 1-x-y\ dy\ dx}{\displaystyle\int_0^\frac 13\int_x^{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2819667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Maximize $f(x,y)=xy$ subject to $x^2-yx+y^2 = 1$
Use Lagrange multipliers method to find the maximum and minimum values of the function
$$f(x,y)=xy$$
on the curve
$$x^2-yx+y^2=1$$
Attempt:
First I set let $g(x,y)=x^2-xy+y^2-1$ and set $$\nabla f=\lambda\nabla g$$
so
$$(y,x)=\lambda(2x-y,2y-x)$$
then
$$\begin{c... | Geometrical ways:
\begin{align}
x^2-xy+y^2 &= 1 \\
\frac{(x+y)^2}{4}+\frac{3(x-y)^2}{4} &= 1 \tag{1}
\end{align}
With the transformation $(X,Y)=
\left( \dfrac{x+y}{\sqrt{2}}, \dfrac{x-y}{\sqrt{2}} \right)$,
$$\frac{X^2}{2}+\frac{3Y^2}{2}=1 \tag{2} $$
which is an ellipse with semi-major and minor axes $\sqrt{2}$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Why are the solutions of the $3\times 3$ system like that? Consider the problem:$$
\min \quad -x_1^2-4x_1x_2-x_2^2\\ \text{s.t.} \quad x_1^2 + x_2^2 = 1$$
The KKT system is given by\begin{align*}
x_1 (-1 + v) + 2 x_2 &= 0 \tag{1} ,\\ x_2 (-1 + v) + 2 x_1 &= 0 \tag{2},\\ x_1^2 + x_2^2 &= 1 \tag{3}
\end{align*}
The solut... | $(x_1-x_2)^2\geq0$ gives $$x_1x_2\leq\frac{x_1^2+x_2^2}{2}=\frac{1}{2}.$$
The equality occurs for $x_1=x_2$.
Thus,
$$-x_1^2-4x_1x_2-x_2^2=-1-4x_1x_2\geq-1-4\cdot\frac{1}{2}=-3.$$
The equality occurs for
$$(x_1,x_2)\in\left\{\left(\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right),\left(-\frac{1}{\sqrt2},-\frac{1}{\sqrt2}\right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Interesting facts about the numbers 1 - 31 I'm a maths teacher and want to create a calendar for my classroom.
I'm looking to compile some interesting facts about each of the numbers 1 through to 31.
The hard part is they must be at a level where 11-18 year-olds can understand them.
Along the lines of: 2 is the smal... | Very interesting idea- I like it. How about these:
$1$ is the only number with one factor, or the only number which is the factorial of two numbers ($0!=1!=1)$
$2$ is the only even prime
$3$ is the best integer approximation of $\pi$ and $e$.
Every even square is a multiple of $4$.
$5$ is the only odd number for which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the minimum of $f(x)=\frac{4x^2}{\sqrt{x^2-16}}$ without using the derivative?
Find the minimum of function $$f(x)=\frac{4x^2}{\sqrt{x^2-16}}$$ without using the derivative.
In math class we haven't learnt how to solve this kind of problems (optimization) yet. I already know that is solvable using derivat... | Render
$\frac{4x^2}{\sqrt{x^2-16}}=a$
Here, $a$ is the minimum function value.
Square both sides and convert the resulting polynomial relation to the form $P(x)=0$:
$\frac{16x^4}{x^2-16}=a^2$
$x^4-a^2x^2+16a^2=0$
Now if $a$ is to be the minimum function for $f(x)$ value then the polynomial will have a squared factor fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 6
} |
$\frac{a^2} {1+a^2} + \frac{b^2} {1+b^2} + \frac{c^2} {1+c^2} = 2.$ Prove $\frac{a} {1+a^2} + \frac{b} {1+b^2} + \frac{c} {1+c^2} \leq \sqrt{2}.$ $a, b, c ∈ \mathbb{R}+.$
WLOG assume $a \leq b \leq c.$ I tried substitution: $x=\frac{1} {1+a^2}, y=\frac{1} {1+b^2}, z=\frac{1} {1+c^2},$ so $x \geq y \geq z$ and $(1-x)+(1... | Let $a,b,c>0$
$$\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}=2\rightarrow \dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}\leq \sqrt{2}$$
$$\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}=\left( 1-\dfrac{1}{1+a}\right) +\left( 1-\dfrac{1}{1+b}\right) +\left( 1-\dfrac{1}{1+c}\right) =$$
$$=3-\left( \dfrac{1}{1+a}+\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
In $\triangle ABC$, $(b^2-c^2) \cot A+(c^2-a^2) \cot B +(a^2-b^2) \cot C = $?
In $\triangle ABC$,
$$(b^2-c^2) \cot A+(c^2-a^2) \cot B +(a^2-b^2) \cot C = \text{?}$$
I tried solving this by cosine rule but it is becoming too long. Any short step solution for this?
| Using Proof of the sine rule
and Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
$$(b^2-c^2)\cot A$$
$$=4R^2(\sin^2B-\sin^2C)\cot A$$
$$=4R^2\sin(B+C)\sin(B-C)\cdot\dfrac{-\cos(B+C)}{\sin(B+C)}$$
$$=-4R^2\sin(B-C)\cos(B+C)=2R^2(\sin2C-\sin2B)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove by induction that $n^4-4n^2$ is divisible by 3 for all integers $n\geq1$. For the induction case, we should show that $(n+1)^4-4(n+1)^2$ is also divisible by 3 assuming that 3 divides $n^4-4n^2$. So,
$$ \begin{align} (n+1)^4-4(n+1)&=(n^4+4n^3+6n^2+4n+1)-4(n^2+2n+1)
\\ &=n^4+4n^3+2n^2-4n-3
\\ &=n^4+2n^2+(-6n^2+6n^... | Given that $6n^2$ is divisible by $3$, it's all a mater of showing that $4n^3-4n$ is divisible by $3$, and we can show that by showing that $n^3-n$ is divisible by $3$
And we know the later is true as follows:
if $n$ itself is divisble by $3$, then obviously $n^3-n$ is divisible by $3$. So, the only other options are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2828422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
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Distance of point $P$ from an ellipse If $ \frac {x^2}{a^2} + \frac {y^2}{b^2} = r^2$ is an ellipse, with the parameterization $x(θ)≔r(a \cos θ,b \sin θ ),$ I have to find the value of $θ$ giving the minimum distance from $P(p,q)$ (not on the ellipse) to the ellipse is given by a quartic in $t= \tan( \frac {θ}{2}).$... | One can normalize the length to $r=1$. The distance from $P$ to a point of the ellipse can be written as
\begin{equation}
d(\theta)=\sqrt{(p-a\cos \theta)^2+(q-b\sin\theta)^2}
\end{equation}
It is minimal if
\begin{equation}
a(p-a\cos\theta)\sin\theta-b(q-b\sin\theta)\cos\theta=0
\end{equation}
with $u=\tan\theta/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The number of incongruent solutions of $x^4-14x^2+36\equiv 0\pmod{2019}$ I was asked to find the number of the incongruent solutions of $x^4-14x^2+36\equiv 0\pmod{2019}$, and the fact $2019=673\times 3$ is given.
My attempt: Given congruence is equivalent to
$$
\begin{cases}
x^4-14x^2+36\equiv 0\pmod{673}\\
x^4-14x^2+3... | I found a solution without finding the square root of 13 modulo 673 directly. $x^4-14x^2+36\equiv 0\pmod{673}$ is equivalent to
$$
x^4-12x^2+36\equiv 2x^2\pmod{673}.
$$
Also, there is an integer $r$ such that $r^2\equiv 2\pmod{673}$, since $\left(\dfrac{2}{673}\right)=1$. Thus given congruence is factored to
$$
(x^2-rx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Finding the shaded area in a triangle Here is the diagram:
I only know that the middle segment is a median of the big triangle. But nothing else.
| Let $h$ be the height of the triangles. Then the area of the large triangle is $$\Delta_{\text{ large}} = \frac{1}{2}\times 6\times \sqrt{h^2+36} \times \sin a$$ and the area of the white triangle is $$\Delta_{\text{ white}} =\frac{1}{2} \times\sqrt{h^2+1}\times\sqrt{h^2+9}\times\sin a$$
But since the large triangle h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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"answer_id": 0
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What tools one should use for inequalities? If $a,b,c>0$ prove that:
$$\frac{1}{a+4b+4c}+\frac{1}{4a+b+4c}+\frac{1}{4a+4b+c}\leq \frac{1}{3\sqrt[3]{abc}}.$$
My first try was the following:
$$\sum_{cyc}\frac{1}{a+4b+4c}\leq\sum_{cyc}\frac{1}{\sqrt[3]{16abc}}=\frac{1}{\sqrt[3]{16abc}}$$
But $\frac{1}{\sqrt[3]{16abc}}\geq... | This is sort of an ugly proof which uses Maclaurin inequalities.
Define constants $A,B,C, \alpha,\beta,\gamma$ through following polynomial:
$$P(\lambda) = (\lambda-a)(\lambda-b)(\lambda-c)
= \lambda^3 - A\lambda^2 + B\lambda - C
= \lambda^3 - 3\alpha\lambda^2 + 3\beta^2\lambda - \gamma^3$$
By Vieta's formulas, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2837548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$
Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$
Is there a way to solve this rather than just declaring a matrix $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ and then trying to solve a system of cubic equations? My attempt... | Presumably $A$ is real. Observe that $A^3-I$ is nonzero but nilpotent (because $(A^3-I)^2=0$). Therefore $A^3$ and in turn $A$ are not diagonalisable. Hence $A$ has repeated eigenvalues. So, if $A$ is real, its eigenvalues must be real (or else the trace of $A$ would become non-real) and equal to $1$ (because $A^3-I$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
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Solution of polynomial Find the set of values of $k$ for which the equation $x^4+kx^3+11x^2+kx+1=0$ has four distinct positive root.
Attempt:
$x^4+kx^3+11x^2+kx+1=0$
$x^2+kx+11+{k\over x}+{1\over x^2}=0$
$x^2 + {1\over x^2} +k(x+{1\over x})+11=0$
$(x + {1\over x})^2 +k(x+{1\over x})+13=0$
I don't know how to proceed a... | By your work let $f(u)=u^2+ku+13,$ where $u=x+\frac{1}{x}.$
Hence, $|u|>2$ and we need to solve the following system:
$$f(2)>0,$$
$$\frac{-k}{2}>2$$ and
$$k^2-52>0$$ or
$$f(-2)>0,$$
$$\frac{-k}{2}<-2$$ and
$$k^2-52>0,$$ which gives
$$-8.5<k<-\sqrt{52}$$ or
$$\sqrt{52}<k<8.5.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Evaluating indefinite integrals.
If $$\int\sqrt{\dfrac{x}{a^3-x^3}}dx=\dfrac{d}{b}{\sin^{-1}} \bigg(\dfrac{x^\frac{3}{2}}{a^\frac{3}{2}}\bigg)+C$$ where $b,d$ are relatively prime find $b+d$.
My solution:
$$\displaystyle\int\sqrt{\dfrac{x}{a^3-x^3}}dx=\int\dfrac{1}{x}\sqrt{\dfrac{x^3}{a^3-x^3}}dx$$
Then let $x^3=t\im... | Hint: As suggested in the comments, differentiate both sides of the equation.
Edit: If you do this, you obtain $$\left(\frac{x}{a^3-x^3}\right)^{1/2}=\frac db \frac {1}{\sqrt{1-[\left(\frac xa\right)^{3/2}]^2}}\cdot \left(\frac1 a\right)^{3/2}\frac32 x^{1/2}.$$ Hopefully, you can rearrange this to obtain $$\frac{x^{1/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The logic in radical symplification I'm having troubles while studying radicals, namely with converting expressions with the form $$\sqrt{a+b\sqrt{c}}$$ to $$a+b\sqrt{c}$$ and vice versa. When I'm dealing with these kind of problem, I usually add and subtract $\sqrt{c}^2$. In example: $$\sqrt{27+10\sqrt{2}}$$ I factor ... | For a different approach, let $\alpha = \sqrt{113+8\sqrt{7}}$.
Then $\alpha$ is a root of $$(x^2-113)^2-8^27=(x^2 - 2 x - 111) (x^2 + 2 x - 111)$$ whose roots are $\pm 1 \pm 4 \sqrt 7$. We just have find the correct signs. Since $\alpha$ is positive, $\alpha= \pm 1 + 4 \sqrt 7$. Now $(1 + 4 \sqrt 7)^2=113+8\sqrt{7}$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Range of $f(x)= \frac{\tan{x}}{\tan{3x}} $
Prove that for the values of $x$ where the following $f(x)$ is defined, $f(x)$ does not lie between $\frac{1}{3}$ and $3$. $$f(x)=\frac{\tan{x}}{\tan{3x}}$$
My Attempt:
I wrote down, $$\tan{3x}=\frac{3\tan{x}-\tan^3{x}}{1-3\tan^2{x}}$$
This reduced $f(x)$ to,
$$f(x)=\frac{1-... | \begin{align*}
f(x)&=\frac{1-3\tan^2{x}}{3-\tan^2{x}}\\
&=\frac{1-3\frac{\sin^2{x}}{\cos^2{x}}}{3-\frac{\sin^2{x}}{\cos^2{x}}}\\
&=\frac{\cos^2{x}-3\sin^2{x}}{3\cos^2{x}-\sin^2{x}}\\
&=\frac{\cos^2{x}-3(1-\cos^2{x})}{3\cos^2{x}-(1-\cos^2{x})}\\
&=\frac{4\cos^2{x}-3}{4\cos^2{x}-1}\\
&=\frac{4\cos^2{x}-1}{4\cos^2{x}-1}-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find the minimum value of $\frac{a+b+c}{b-a}$ Let $f(x)=ax^2+bx+c$ where $(a<b)$ and $f(x)\geq 0$ $\forall x\in R$.
Find the minimum value of $$\frac{a+b+c}{b-a}$$
If $f(x)\geq 0$ $\forall x\in R$ then $b>a>0$ and $b^2-4ac\leq 0$ implying that $c>0$. After this not able to find way out.
| As you noted, we have $b > a > 0$, and $b^2-4ac\le 0$, hence $c\ge {\large{\frac{b^2}{4a}}}$.
Letting $t=b-a$, we have $t > 0$, and $b=a+t$.
\begin{align*}
\text{Then}\;\;&\frac{a+b+c}{b-a}\\[4pt]
&\ge \frac{a+b+\frac{b^2}{4a}}{b-a}\\[4pt]
&=\frac{(2a+b)^2}{4a(b-a)}\\[4pt]
&=\frac{(3a+t)^2}{4at}\\[4pt]
&=\frac{9a^2+6a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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What is the number of connected components of solutions to $y^3 +3xy^2 - x^3 = 1$
What is the number of connected components of solutions to $y^3 +3xy^2 - x^3 = 1$
Attempt
$x^3 - 3y^2x + 1 - y^3 = 0$
$p := -3y^2$, $q := 1 - y^3$
The discriminant is
$Q = (p/3)^3 + (q/2)^2 = (-y^2)^3 + (1 - y^3)^2 / 8 = -y^6 + 1/8 -... | There are three disjoint components to the solution curves.
Consider the disjoint regions formed by the following half lines:
$1)$ $$ \{ (x,0): x\ge 0\}$$
$2)$ $$ \{(x,-x): x\le 0\}$$
$3)$ $$ \{(0,y): y\le 0 \}$$
Each component is contained in one and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2845452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove by induction $\sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}{4}$ In this problem I had to use the method of mathematical induction in order to solve $$\sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}{4}$$
After proving that $P_1$ is true, and writing out $P_k$ I proceeded to solve for $P_{k+1}$. I wrote it out, using $P_k$ to substitu... | We have that
$$\sum_{i=1}^{n+1} i^3=\frac{n^2(n+1)^2}{4}+\frac{4}{4}(n+1)^3=\frac{(n+1)^2(n^2+4n+4)}{4}=\frac{(n+1)^2(n+2)^2}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove the equation $\left(2x^2+1\right)\left(2y^2+1\right)=4z^2+1$ has no solution in the positive integers Prove the equation
$$\left(2x^2+1\right)\left(2y^2+1\right)=4z^2+1$$
has no solution in the positive integers
My work:
1) I have the usually problem
$$\left(nx^2+1\right)\left(my^2+1\right)=(m+n)z^2+1$$
in the p... | The question is concerned only with $x,y,z\ge 1$. $x=y$ does not give a solution (see comments). WLOG assume $x>y$. Look at the equation $\mod x$.
$1\cdot (2y^2+1)\equiv 4z^2+1\mod x$.
Let $z\equiv a\mod x$. Then $2y^2+1=4(nx+a)^2+1$.
Subtracting $1$ from each side and dividing by $2$ we get $y^2=2(nx+a)^2\Rightarrow ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
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Find min of $P = \dfrac{1}{(a-b)^2} + \dfrac{1}{(b-c)^2} + \dfrac{1}{(c-a)^2}$ Let $a, b, c \in \mathbb{R}^+$ such that $a^2 + b^2 + c^2 = 3$. Find the minimum value of $P = \dfrac{1}{(a-b)^2} + \dfrac{1}{(b-c)^2} + \dfrac{1}{(c-a)^2}$?
In fact, we have that $P \geq \dfrac{4}{ab +bc +ca}$. However, the equality can n... | Let $D = \{ (a,b,c) \in (0,\infty)^3 : a^2 + b^2 + c^2 = 3 \}$ and $\bar{D}$ be its closure. Instead of looking for a minimum of $P = \frac{1}{(a-b)^2} + \frac{1}{(b-c)^2} + \frac{1}{(c-a)^2}$ over $D$, we need to look at the minumum over the larger $\bar{D}$.
This is because $P$ doesn't achieve any local minimum ove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2848219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Is $\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=1$?
Evaluate $$\lim_{x\to-\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}$$
My attempt:
$$\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=\lim_{x\to-\infty}\frac{5x}{-\sqrt{4x^2}}=\frac{-5}{2}$$ According to the answer key, it actually equals $1$.
Thanks in advance.... | Note that $\sqrt {4x^2} = 2|x|$ and in this case $|x|=-x$, thus $$\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=\lim_{x\to-\infty}\frac{5x}{3x-\sqrt{4x^2}}= \lim_{x\to-\infty} \frac{5x}{3x+2x }=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Let $V_m(r)=\operatorname{vol}B(0,r)$ and prove $V_{n+1}(r) = 2V_n(1)\int_0^r(r^2-t^2)^{n/2}\,\mathrm dt$
Let $V_{m}(r)$ the volume of the ball with center in origin and radius $r$ in $\mathbb{R}^{m}$. Prove the inductive relation $V_{n+1}(r) = 2V_{n}(1)\int_{0}^{r}(r^{2}-t^{2})^{n/2}\,\mathrm dt$ and conclude that
... | $\def\peq{\mathrel{\phantom{=}}{}}\def\d{\mathrm{d}}$For any $n \geqslant 1$ and $(x_1, \cdots, x_{n + 1}) \in \mathbb{R}^{n + 1}$, note that\begin{align*}
&\mathrel{\phantom{\Longleftrightarrow}} (x_1, \cdots, x_{n + 1}) \in B_n(0, r)\\
&\Longleftrightarrow \sum_{k = 1}^{n + 1} x_k^2 < r^2 \Longleftrightarrow -r < x_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$ Prove that
$$ \min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$
Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$
Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4 $.
I used Cauchy-Schwarz to get $\q... | Notice
$$\begin{align}
a + b + \frac1a + \frac1b
= &\; \left(a + \frac{1}{2a} + \frac{1}{2a}\right)
+ \left(b + \frac{1}{2b} + \frac{1}{2b}\right)\\
\\
\color{blue}{\rm AM \ge \rm GM \rightarrow\quad}
\ge &\; 3\left[\left(\frac{1}{4a}\right)^{1/3} + \left(\frac{1}{4b}\right)^{1/3}\right]\\
\color{blue}{\rm AM \ge \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 7
} |
$\tan(a) = 3/4$ and $\tan (b) = 5/12$, what is $\cos(a+b)$ It is known that
$$\tan(a) = \frac{3}{4}, \:\:\: \tan(b) = \frac{5}{12} $$
with $a,b < \frac{\pi}{2}$.
What is $\cos(a+b)$?
Attempt :
$$ \cos(a+b) = \cos(a) \cos(b) - \sin(a) \sin(b) $$
And we can write $\tan(a) = \sin(a)/\cos(a) = 0.3/0.4 $
and $ \sin(b)/\co... | $$ \tan(a+b)= \frac {\tan a +\tan b}{1-\tan a \tan b} = \frac {3/4 +5/12}{1-(3/4)(5/12)}= \frac {56}{33}$$
$$ \sec^2(a+b)=1+\tan^2(a+b)=\frac {4225}{1089}=(\frac {65}{33})^2$$
$$\cos(a+b)= 33/65$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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} |
Positive integer solutions to $a^3 + b^3 = c^4$ Let $n$ and $m$ be positive integers. We know that
$n^3 + m^3 = n^3 + m^3$
Multiply both sides by $(n^3 + m^3)^3$; on the LHS you distribute, and on the RHS you use power addition rule;
$(n^4 + nm^3)^3 + (m^4 + mn^3)^3 = (n^3 + m^3)^4$
So we get an infinite array of solut... | $$793^3+854^3=183^4$$ and $$183$$ is not the sum of two cubes.
| {
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"question_score": "5",
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Verifying the limit of the expressions I have calculated the limit of the following expressions as follows:
$$\lim_{x \rightarrow 0} \frac{x \times \frac{2}{x}}{5x + 3 + e^{-2x}} = \lim_{x \rightarrow 0} \frac{2x}{5x^2 + 3x + xe^{-2x}} = \lim_{x \rightarrow 0} \frac{2}{10x + 3 + e^{-2x} -2xe^{-2x}} (\text{using Le Chat... | As suggested by Bernard in the above comment, more simply we have
$$\lim_{x \rightarrow 0} \frac{\color{red}x \times \frac{2}{\color{red}x}}{5x + 3 + e^{-2x}} = \lim_{x \rightarrow 0} \frac{2}{5x + 3 + e^{-2x}} \stackrel{\text{by continuity}}= \frac2{3+1}=\frac12$$
and similarly for the second we have
$$\lim_{x \righta... | {
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Prove that $\sum_{j=0}^{\infty}(-1)^j\left(\frac{1}{5+6j}+\frac{1}{1+6j} \right) = \frac{\pi}{3}$ I was trying to solve the integral $$\int_{0}^{\infty}\frac{x^4}{1+x^6}dx$$
using series. Now I'm stuck at the series below.
How to prove that
$$\sum_{j=0}^{\infty}(-1)^j\left(\frac{1}{5+6j}+\frac{1}{1+6j} \right) = \frac... | Denote
$$
S = \sum_{j=0}^{\infty}(-1)^j \left( \frac{1}{6j+1} + \dfrac{1}{6j+5}\right).
$$
And denote
$$
T= \sum_{j=0}^{\infty} (-1)^j \dfrac{1}{2j+1}.
$$
According to Leibniz formula for $\pi$, $\;T=\dfrac{\pi}{4}$.
Then
$$S - \dfrac{1}{3}T = \\
S - \dfrac{1}{3}\sum_{j=0}^{\infty} (-1)^j \dfrac{1}{2j+1} = \\ S-\sum_{... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Double fractional part integral Let $\{\}$ denote the fractional part, does the following integral have a closed form ?
$$\int_{0}^{1}\int_{0}^{1}\bigg\{\frac{1}{x\,y}\bigg\}^2dx\,dy$$
| This is an incomplete answer that only addresses the 1-dimensional case.
We split the integral into continuous pieces:
$$\begin{align}
\int_0^1 \left\{ \frac{1}{x} \right\}^2 \mathrm{d}x
&= \sum_{n=1}^{\infty} \int_{\frac{1}{n+1}}^{\frac{1}{n}} \left\{ \frac{1}{x} \right\}^2 \mathrm{d}x \\\\
&= \sum_{n=1}^{\infty} \int... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove: $\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$
Prove: $$\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$$
ok, what I saw instantly is that:
$$\sin\frac{\pi}{20}+\sin\frac{3\pi}{20}=2\sin\frac{2\pi}{20}\cos\frac{\pi... | Since $\surd2\sin\frac14\pi=\surd2\cos\frac14\pi=1$, we may write the difference as$$\surd2\left(\cos\frac\pi4\cos\frac\pi{20}+\sin\frac\pi4\sin\frac\pi{20}\right)-\surd2\left(\cos\frac\pi4\cos\frac{3\pi}{20}-\sin\frac\pi4\sin\frac{3\pi}{20}\right)-\frac{\surd2}2$$$$=\frac{\surd2}2\left(2\cos\frac\pi5-2\cos\frac{2\pi}5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858498",
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"source": "stackexchange",
"question_score": "6",
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"answer_id": 5
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Prove the limit below by using epsilon-delta definition Prove $$\lim_{z\rightarrow4+3i}\overline{z}^2 =(4-3i)^2$$ by using epsilon-delta definition.
My working:
Assume $|z-(4-3i)|<1 \implies |z|<6$
\begin{align}
|\overline{z}^2-(4-3i)^2| &=|[\overline{z}-(4-3i)][\overline{z}+(4-3i)]|\\
&=|\overline{z}-(4-3i)||\overline... | Careful, you can assume $|z - (4+3i)| < \delta$, not $|\overline{z} - (4+3i)| < \delta$.
If we pick $0 < \delta < -5+\sqrt{25+\varepsilon}$ we have
\begin{align}
\left|\overline{z}^2 - (4-3i)^2\right| &= |\overline{z} - (4-3i)|\cdot |\overline{z} + (4-3i)|\\
&= \left|\overline{z - (4+3i)}\right|\cdot \left|\overline{z ... | {
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Verifying that $\left[\int_a^b(x-p)\,dx\right]^2\le(b-a)\int_a^b(x-p)^2\,dx$ without Cauchy-Schwarz
For real numbers $a,b,p$, verify that $$\left[\int_a^b(x-p)\,dx\right]^2\le(b-a)\int_a^b(x-p)^2\,dx\tag{1}$$ without using the Cauchy-Schwarz inequality.
Note that this is obviously an immediate consequence of C-S, and... | Do the substitution $t=x-p$, so the inequality becomes
$$
\left(\int_{a-p}^{b-p} t\,dt\right)^{\!2}\le (b-a)\int_{a-p}^{b-p} t^2\,dt
$$
Setting $a-p=A$ and $b-p=B$, this is the same as proving that
$$
\left(\int_{A}^{B} t\,dt\right)^{\!2}\le (B-A)\int_{A}^{B} t^2\,dt
$$
that is,
$$
\frac{(B^2-A^2)^2}{4}\le (B-A)\frac{B... | {
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"source": "stackexchange",
"question_score": "3",
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The probability density function of a random variable is given by $f_X(x) = \frac{3}{208}x^2$ Find $E(X)$ The probability density function of a random variable is given by $$f_X(x) = \frac{3}{208}x^2$$ whenever $2 < x < 6$. $0$ otherwise
$(a)$ Compute $E(X)$
$(b)$ Compute $E[X(X+1)]$
My attempt
$(a)$
$$E(X) = \int_{2}... | The probability density function of a random variable is given by $$f_X(x) = \frac{3}{208}x^2$$ whenever $2 < x < 6$. $0$ otherwise
$(a)$ Compute $E(X)$
$(b)$ Compute $E[X(X+1)]$
$(a)$
$$E(X) = \int_{2}^{6} x\frac{3}{208}x^2dx = \frac{3}{208}\cdot\frac{1}{4}(x^4)\bigg|_{2}^{6} = 60/13$$
$(b)$
$$E(X(X+1)) = E(X^2 + X)... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Evaluating $\int _0^\infty\,\frac{\ln x}{(1+x^2)^2}\,\text{d}x$ using the Residue Theorem To evaluate this integral
$$\int\limits_0^\infty\frac{\ln x}{(1+x^2)^2} \text{d}x\,,$$
Let us consider the following function.
$$f(z)=\frac{\ln^2 z}{(1+z^2)^2}=\frac{\ln^2 z}{(z+i)^2(z-i)^2}\,.$$
The integral over $C_R$ and $C_\ep... | Let $f(z):=\dfrac{\big(\ln(z)\big)^2}{\left(1+z^2\right)^2}$ for $z\in\mathbb{C}\setminus\mathbb{R}_{\leq 0}$ (i.e., the chosen branch cut of $\ln(z)$ is the negative real line). For a real number $\epsilon$ such that $0<\epsilon<1$, define $\gamma_\epsilon$ to be the contour
$$\left[\epsilon,\frac{1}{\epsilon}\right]... | {
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"source": "stackexchange",
"question_score": "3",
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Given a matrix $A$ find $A^n$. $A=$$
\begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix} $
Find $A^n$.
My input:
$A^2= \begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix} = \begin{bmatrix}
1 & 4\\
0 & 1
\end{bmatrix} $
$A^3 ... | You can use this too
$$A=\begin{pmatrix}1&2\\0&1\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}+\begin{pmatrix}0&2\\0&0\end{pmatrix}$$
$$A=I_2+B$$
And B is a nilpotent matrix $\implies B^2=0$
$$A^n=(I_2+B)^n$$
Use binomial theorem
| {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Evaluate $\int_0^{\frac{\pi}{2}}\frac{\sin x\cos x}{\sin^4x+\cos^4x}dx$
Evaluate
$$
\int_0^{\frac{\pi}{2}}\frac{\sin(x)\cos(x)}{\sin^4(x)+\cos^4(x)}dx
$$
I used the substitution $\sin x =t$, then I got the integral as $$\int_0^1 \frac{t}{2t^4-2t^2+1}dt $$
After that I don't know how to proceed. Please help me with... | Perform the change of variable $y=\sin^2 x$,
$\begin{align}J&=\int_0^{\frac{\pi}{2}}\frac{\sin(x)\cos(x)}{\sin^4(x)+\cos^4(x)}dx\\
&=\frac{1}{2}\int_0^1 \frac{1}{x^2+(1-x)^2}\,dx\\
&=\frac{1}{2}\int_0^1 \frac{1}{2x^2-2x+1}\,dx\\
&=\int_0^1 \frac{1}{(2x-1)^2+1}\,dx\\
\end{align}$
Perform the change of variable $y=2x-1$,... | {
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"question_score": "2",
"answer_count": 5,
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The product of three consecutive integers is ...? Odd? Divisible by $4$? by $5$? by $6$? by $12$? If i have the product of three consecutive integers:
$n(n+1)(n+2)$, so the result is:
$A)$ Odd
$B)$ Divisible by $4$
$C)$ Divisible by $5$
$D)$ Divisible by $6$
$E)$ Divisible by $12$
My thought was:
$i)$ If we have three ... | To prove your statement $i)$ you can use the technique of induction.
First consider $a=1$ therefore the product becomes
$$1\cdot(1+1)\cdot(1+2)=1\cdot2\cdot3$$
which is clearly divisible by $3$ since one of the factors is $3$ itself. Next, by setting $a=n$ and $a=n+1$, we get
$$\begin{align}
n(n+1)(n+2)&\equiv0\mod(3)... | {
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Evaluating the integral $\int \frac1{(2+3\cos x)^2}\mathrm dx$ Please someone give me an idea to evaluate
this: $$\int \frac1{(2+3\cos x)^2}\mathrm dx$$
I don't even know how to start cause even multiplying an dividing by $\cos^2x$ does not work, so help me here.
| Integrating by parts,
$$\int\dfrac{dx}{(a+b\cos x)^2}=\int\dfrac{\sin x}{(a+b\cos x)^2}\cdot\dfrac1{\sin x}dx$$
$$=\dfrac1{\sin x}\int\dfrac{\sin x}{(a+b\cos x)^2}dx-\left(\dfrac{d(1/\sin x)}{dx}\cdot\int\dfrac{\sin x}{(a+b\cos x)^2}dx\right)dx$$
$$=\dfrac1{b\sin x(a+b\cos x)}+\int\dfrac{\cos x}{b(1-\cos^2x)(a+b\cos x)... | {
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"source": "stackexchange",
"question_score": "4",
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Rows of a Matrix is divisible by 19, show that its Determinant is also divisible by 19 I came across the following problem while self studying:
Let
\begin{equation}
A =
\begin{bmatrix}
2 & 1 & 3 & 7 & 5\\
3 & 8 & 7 & 9 & 8\\
3 & 4 & 1 & 6 & 2\\
4 & 0 & 2 & 2 & 3\\
7 & 9 & 1 & 5 & 4\\
\end{bmatrix}
\end{equation}
Use ... | Note that $$10^{10} \det(A) = 10^{4+3+2+1+0} \det(A) = \det \begin{pmatrix} 2 \cdot 10^{4} & 1\cdot 10^3 & 3\cdot 10^2 & 7\cdot 10 & 5 \\
3\cdot 10^{4} & 8 \cdot 10^3 & 7\cdot 10^2 & 9\cdot 10 & 8 \\
3\cdot 10^{4} & 4\cdot 10^3 & 1\cdot 10^2 & 6\cdot 10 & 2 \\
4\cdot 10^{4} & 0 \cdot 10^3 & 2 \cdot 10^2 & 2 \cdot 10 &... | {
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How to show that $\cos(\sin^{-1}(x))$ is $\sqrt{1-x^2}$?
How to show that $\cos(\sin^{-1}(x))$ is $\sqrt{1-x^2}$?
I remember having to draw a triangle, but I'm not sure anymore.
| Given $\cos(\sin^{-1}x)$
Let $\sin^{-1}x=\theta$
$$\implies\sin\theta=x\\\sin^2\theta=x^2\\1-\cos^2\theta=x^2\\\cos^2\theta=1-x^2\\\cos\theta=\pm\sqrt{1-x^2}\\\theta=\cos^{-1}\pm\sqrt{1-x^2}$$
Now, plug in $\theta=\cos^{-1}\pm\sqrt{1-x^2}$ in $\cos(\sin^{-1}x)$ and we get$$\pm\sqrt{1-x^2}$$But note that $\sin^{-1}x$ is... | {
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Show that:$\sum_{i=1}^n x_{i} \cdot \sum_{i=1}^n \frac{1}{x_{i}} \geq n^2 $ Show that:$$\sum_{i=1}^n x_{i} \cdot \sum_{i=1}^n \frac{1}{x_{i}} \geq n^2 $$
The following hints are also given: $$\left(\frac{x}{y} + \frac{y}{x} \geq 2 \right) \land x,y \gt 0$$
Base Case: For n = 2
$$\left(1+2\right) \cdot \left(\frac{1}{1}... | Alternative way :
It's a direct consequence of the inequality of Chebyshev .
| {
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Integral with inverse trigonometric function How do I integrate $$\int \cot^{-1}\sqrt{x^2+x+1}\ dx$$
I don't understand how to proceed?
I did try the question using integration by parts, but it didn’t help.
| A solution without using hyperbolic functions
Integration by parts gives
$$x\cot^{-1}\left( \sqrt{x^2+x+1} \right)+\frac{1}{2}\int \frac{x(2x+1)}{\sqrt{x^2+x+1}(x^2+x+2)}\,dx$$
i can split the integrated on the right as
$$
\begin{align}
& =2\int{\frac{1}{\sqrt{{{x}^{2}}+x+1}}dx}-\int{\frac{1+2x}{2\sqrt{{{x}^{2}}+x+1... | {
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Let $m$ be the largest real root of the equation $\frac3{x-3} + \frac5{x-5}+\frac{17}{x-17}+\frac{19}{x-19} =x^2 - 11x -4$ find $m$
Let $m$ be the largest real root of the equation $$\frac3{x-3} + \frac5{x-5}+\frac{17}{x-17}+\frac{19}{x-19} =x^2 - 11x -4.$$ Find $m$.
do we literally add all the fractions or do we do ... | We need to solve
$$\frac3{x-3}+1+ \frac5{x-5}+1+\frac{17}{x-17}+1+\frac{19}{x-19}+1 =x^2 - 11x $$ or
$$x\left(\frac{1}{x-3} + \frac{1}{x-5}+\frac{1}{x-17}+\frac{1}{x-19}\right) =x^2 - 11x$$ or
$$2x(x-11)\left(\frac{1}{(x-3)(x-19)}+\frac{1}{(x-5)(x-17)}\right)=x(x-11),$$ which gives $x_1=0$, $x_2=11$ or
$$\frac{1}{x^2-2... | {
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How to prove $1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} <\sqrt{n}\ .\left(2n-1\right)^{1/4} $
Prove that $$1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} <\sqrt{n}\ .\Bigl(2n-1\Bigr)^{\frac{1}{4}} $$
My Approach :
I tried by applying Tchebychev's Inequality for two same sets of numbers;
$$1 , \frac{1}{\sqrt... | Your inequality should be not strong.
We can prove that for all natural $n\geq1$ the following inequality holds:
$$\sum_{k=1}^n\frac{1}{\sqrt{k}}\leq\sqrt{n}\sqrt[4]{2n-1}.$$
Indeed,
$$\sum_{k=1}^n\frac{1}{\sqrt{k}}\leq1+\int\limits_1^{n}\frac{1}{\sqrt x}dx=2\sqrt{n}-1.$$
Thus, it's enough to prove that
$$2\sqrt{n}-1... | {
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Prove $\lim_{(x,y) \to (0,0)} \frac{\exp(xy)\cdot xy\cdot(x^2-y^2)}{x^2+y^2} =0$. I want to show that
$$\lim_{(x,y) \to (0,0)} \frac{\exp(xy)\cdot xy\cdot(x^2-y^2)}{x^2+y^2} =0.$$
Is it valid to do it like this:
$$\lim_{(x,y) \to (0,0)} \left|\frac{\exp(xy)\cdot xy\cdot(x^2-y^2)}{x^2+y^2}\right| \leq \lim_{(x,y) \to ... | We have that $e^{xy}\to 1$ and by polar coordinates
$$\left|\frac{xy\cdot(x^2-y^2)}{x^2+y^2}\right|=r^2|(\cos \theta\sin \theta)(\cos^2\theta-\sin^2\theta)|\le r^2\to 0$$
| {
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"source": "stackexchange",
"question_score": "1",
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Compare $\arcsin (1)$ and $\tan (1)$
Which one is greater: $\arcsin (1)$ or $\tan (1)$?
How to find without using graph or calculator?
I tried using $\sin(\tan1)\leq1$, but how do I eliminate the possibility of an equality without using the calculator?
| $\arcsin(1)=\frac{\pi}{2}$ while from the Weierstrass product for the cosine function we have
$$ \tan(1) = \sum_{n\geq 0}\frac{8}{(2n+1)^2 \pi^2-4 }=2\sum_{n\geq 0}\left[\frac{1}{(2n+1)\pi-2}-\frac{1}{(2n+1)\pi+2}\right]$$
such that an effective integral representation of $\tan(1)$ through an almost-Gaussian integral i... | {
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"question_score": "10",
"answer_count": 7,
"answer_id": 1
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What does unique factorization exactly mean here? Given $a$ and $b$ are integers:
$$ 3\cdot 2 \cdot 5^2 \cdot 7^a = 2 \cdot 3^2 \cdot 7 \cdot 5^b $$
The answer is as follows:
Since both sides of this equation are integers and have unique factorizations, it follows that $a = 1$ and $b = 2$ is the only solution.
What d... | Every positive integer is either prime or composite or the number one.
If the number is composite then it can be factored as a product of smaller numbers. Ex: $120 = 12\cdot10$ or we could say $120 = 30\cdot4$ or we could say $120 = 6\cdot10\cdot2$.
If we factor a number into smaller composites and then factored thos... | {
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Number of Non negative integer solutions of $3a+2b+c+d=19$ Find Number of Non negative integer solutions of $3a+2b+c+d=19$
My attempt:
we have $$2b+c+d=19-3a$$ Required solutions is coefficient of $t^{19-3a}$ in
$$( 1-t^2)^{-1}(1-t)^{-1}(1-t)^{-1}=\frac{1}{(1-t)^3(1+t)}$$
By partial fractions we get
$$( 1-t^2)^{-1}(1... | HINT.-Checking directly is not hard.
$$3a+2b=19$$ has only three solutions $(1,8),(3,5),(5,2)$.
When $a=6$,$$2b+c+d=1$$ has two solutions $(0,0,1),(0,1,0)$.
When $a=5$, $$2b+c+d=4$$ has five solutions from which one has to be discarded $(1,1,1)(2,0,0),(0,1,3),(0,2,2),(0,3,1)$ we have till now
$$(1,8,0,0),(3,5,0,0),(5,2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ then $\rho\ge-1/2$ This may be trivial but I am not able to prove that if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ for $a,b,c\in\mathbb{R}$ then $\rho\ge-1/2$. Can anybody help me please?
Thanks!
| You can show that the inequality is true for all $a,b,c\in\mathbb{R}$ if and only if $-\dfrac{1}{2}\leq \rho \leq 1$. Indeed, setting $a$, $b$, and $c$ to be $1$ gives
$$3+6\rho\geq 0\,,\text{ or equivalently }\rho\geq -\frac12\,.$$
Taking $(a,b,c)$ to be $(1,0,-1)$ leads to
$$2-2\rho\geq 0\,,\text{ whence }\rho\leq 1... | {
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order of the splitting field of $x^5 +x^4 +1 $? what is the order of the splitting field of
$x^5 +x^4 +1 = (x^2 +x +1)( x^3 +x+1)$ over $\mathbb{Z_2}$
i thinks it will $6$ because $2.3 = 6$
Pliz help me...
| Yes, it is everything ok. This is an answer, not a comment, in order to insert the following confirmation using computer assistance, here sage:
sage: F = GF(2)
sage: R.<x> = PolynomialRing(F)
sage: f = x^5+x^4+1
sage: f.factor()
(x^2 + x + 1) * (x^3 + x + 1)
sage: K.<a> = f.splitting_field()
sage: K
Finite Field in a o... | {
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Double series convergent to $2\zeta(4)$? Using a computer I found the double sum
$$S(n)= \sum_{j=1}^n\sum_{k=1}^n \frac{j^2 + jk + k^2}{j^2(j+k)^2k^2}$$
has values
$$S(10) \quad\quad= 1.881427206538142 \\ S(1000) \quad= 2.161366028875634 \\S(100000) = 2.164613524212465\\$$
As a guess I compared with fractions $\pi^p/q... | So before I start, I've never even attempted to evaluate a double sum before, so there could very well have been an easier way.
$$\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{j^2+jk+k^2}{j^2k^2(j+k)^2} = \sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{k^2(j+k)^2} +\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{jk(j+k)^2} + \sum_... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 0
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Find the maximum of the $| \left( w + 2 \right) ^3 \left( w - 3 \right)^2|$ with $|w|=1$ Let $w \in \mathbb{C}$, and $\left | w \right | = 1$. Find the maximum of the function $| \left( w + 2 \right) ^3 \left( w - 3 \right)^2|$
Since $$|(w+2)^3(w-3)^2|=|w^5-15w^3-10w^2+60w+72|$$
Let $w=\cos x+i \sin x$. Then we have... | Calling
$$
z_1=\rho_1 e^{i\phi_1} = \sqrt{(\cos\phi+2)^2+\sin^2\phi} = \sqrt{5+4\cos\phi}\\
z_2=\rho_2 e^{i\phi_2} = \sqrt{(\cos\phi-2)^2+\sin^2\phi} = \sqrt{10-6\cos\phi}
$$
so
$$
|z_1^3z_2^2| = \rho_1^3\rho_2^2 = \left(5+4\cos\phi\right)^{\frac 32}|10-6\cos\phi|
$$
now
$$
\frac{d}{d\phi}\left( \left(5+4\cos\phi\ri... | {
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For positive number $a,b$, when $a,b$ satisfies $2a^2 +7ab+3b^2=7$, what is maximum value of $a+{\sqrt{3ab}}+2b$ Question is
For positive numbers $a,b$ such that $2a^2 +7ab+3b^2 = 7$, what is the maximum value of $a+{\sqrt{3ab}}+2b$?
I use AM-GM Inequality to do this
$$(2a+b)(a+3b)=7\text{ and }2a+b=\frac{7}{a+3b}.$... | Let $a＝x^2,b＝3y^2$ and $t＝\frac{x}{y}$,only need to find the maxium of this
$$\frac{7(t^2+3t+6)^2}{2t^4+21t^2+2}$$
and ...
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Differential equation: $\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$
The solution of $\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$ is given by:
a) $(x+2)^4 (1+\frac{2y}{x})= ke^{\frac{2y}{x}}$
b) $(x+2)^4 (1+ 2\frac{(y-2)}{x+2})= ke^{\frac{2(y-2)}{x+2}}$
c) $(x+2)^3 (1+ 2\frac{(y-2)}{x+2})= ke^{\frac{2(y-2)}{x+2}}$
d... | Let $X=x+2$ and $Y=y-2$, so the given DE is equivalent to
$$\dfrac{\mathrm d Y}{\mathrm d X}=\frac{(X+Y)^2}{X Y}$$
last DE is homogeneous, so can be transformed into a separable DE by making $Y =uX$ as follows
$$\dfrac{\mathrm d Y}{\mathrm d X}=\frac{(X+Y)^2}{X Y}\qquad\iff\qquad u+X\frac{\mathrm d u}{\mathrm dX}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883253",
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Why isn't this true for $x<0$?
Prove that
$$\lfloor{x}\rfloor=\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2}{2^2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+
\ldots$$
$x\geq{0}$
I asked this question here yesterday. Although I was able to come up with a proof, the second part was lef... | For $x=-1$ the lest side is $-1$ and the right side is $0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find the auxiliary circle of a non-standard ellipse?
Given the equation of conic C is $5x^2 + 6xy + 5y^2 = 8$, find the equation S of its auxiliary circle?
Now, I know that the equation C is an ellipse.
Since $\Delta = 5*5(-8) - (-8)(3^2) \neq 0$
And,
$ab - h^2 = 5*5 - 3^2 > 0$
Which is the condition for an el... | Since the conic is centre origin, we may use polar coordinates:
\begin{align}
8 &= 5r^2\cos^2 \theta+6r^2\cos \theta \sin \theta+5r^2\sin^2 \theta \\
r^2 &= \frac{8}{5+6\cos \theta \sin \theta} \\
&= \frac{8}{5+3\sin 2\theta}
\end{align}
Now,
$$\frac{8}{5+3} \le r^2 \le \frac{8}{5-3} \implies 1 \le r^2 \le 4 \\$$... | {
"language": "en",
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How to solve $\sin^2(x)+\sin2x+2\cos^2(x)=0$ How do you solve $\sin^2(x)+\sin2x+2\cos^2(x)=0$?
I have been able to rewrite it as $(\sin(x)+\cos(x))^2+\cos^2(x)=0$. Not obviously useful, I think
| $\sin^2(x) + \cos^2(x) = 1$, so we have:
$$\sin 2x + \cos^2 x + 1 = 0$$
Now, $\cos^2 x ≥ 0$, and $\sin 2x ≥ -1$, so the only solutions are when $\cos^2 x = 0$ and $\sin 2x = -1$.
When $\cos^2 x = 0$, $x = -\frac{\pi}{2} + 2\pi n$ or $x = \frac{\pi}{2} + 2\pi n$. However, $\sin \left( 2(-\frac{\pi}{2}) \right) = 0$ and ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
How to find all the $z$ that satisfy $(1+i)z^4=(1-i)|z|^2$? Would you please help me solve this? I need all the $z$ that satisfy the equality
$$(1+i)z^4=(1-i)|z|^2.$$
I tried doing this:
$$
\begin{aligned}
(1+i)z^4&= (1-i)z\overline z\\
(1+i)z^4 -(1-i)z\overline z &= 0\\
z[(1+i)z^3-(1-i)\overline z]&= 0
\end{aligned}
$... | $$(1+i)z^4=(1-i)|z|^2\iff iz^4=z\bar z\iff\begin{cases}z=0\\iz^3=\bar z\end{cases}$$ $$iz^3=\bar z\iff i(a+ib)^3=(a-ib)\iff {ia^3 - 3 a^2 b -i 3 a b^2 + b^3}=(a-ib)\\\iff b^3- 3 a^2 b +i(a^3 - 3 a b^2)=a-ib\iff\begin{cases}b^3- 3 a^2 b=a\\a^3 - 3 a b^2=-b\end{cases}$$
We can also notice that $|b^3- 3 a^2 b +i(a^3 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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.