Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solving cauchy riemann equations, finding all analytic functions I need someone to check my work! I tried doing this as properly as possible, but I have no way to check whether this is correct.
Find $\textit{all}$ analytic functions $f = p(x,y) +iq(x,y) $ such that $p+q = xy$
(I'm using p's and q's since u's and v's lo... | It is of course possible that I overlooked a mistake, but since the result is correct, and I didn't see any error, I'm reasonably convinced that your work is correct.
Here, as in many situations, we can solve the task in an easier way if we apply a certain transformation to the problem. It is advisable to try to look f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1330534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Nesbitt's Inequality for 4 Variables I'm reading Pham Kim Hung's 'Secrets in Inequalities - Volume 1', and I have to say from the first few examples, that it is not a very good book. Definitely not beginner friendly.
Anyway, it is proven by the author, that for four variables $a, b, c$, and $d$, each being a non-negati... | I think, the best way it's C-S:
$$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac}\geq\frac{\left(\sum\limits_{cyc}a\right)^2}{\sum\limits_{cyc}(ab+ac)}=2+\frac{(a-c)^2+(b-d)^2}{\sum\limits_{cyc}(ab+ac)}\geq2.$$
Another way:
Let $(a-c)(b-d)\geq0.$
Thus,
\begin{align}
\sum_{cyc}\frac{a}{b+c}&=\frac{a+d}{b+c}+\frac{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Binomial expansion in descending power For example, find, in ascending powers of $x$, the first three terms in the expansion of $(2+5x)^7$.
So, $(2+5x)^7=2^7+\binom{7}{1}(2^6)(5x)+\binom{7}{2}(2^5)(5x)^2$.
I've no problem to solve this kind of problem.
But now another question wants me to find, in descending powers of... | Exactly the same way, you can just do this with the binomial theorem.
\begin{align*}
\left(2x+\frac{1}{3x}\right)^6 &= \sum_{k=0}^6 {6 \choose k}(2x)^{6-k}\left(\frac{1}{3x}\right)^k \\ &= 2^6x^6+2^5\frac{1}{3}{6 \choose 1} x^4 + 2^4\frac{1}{3^2}{6 \choose 2} x^2+ 2^3\frac{1}{3^3}{6 \choose 3} + \cdots \\ &= 64x^6+ 64... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the sum of first $20$ terms of a sequence Define a sequence $$a_n=\sqrt{1+\left(1-\frac{1}{n}\right)^2}+\sqrt{1+\left(1+\frac{1}{n}\right)^2}$$ for $n \geq 1$. Find $$\sum_{i=1}^{20}\frac{1}{a_i}$$ Some insight on the approach is highly appreciated. What is the general way of solving such problems? Thanks.
| $$\frac{1}{\sqrt{x}+\sqrt{y}}=\frac{\sqrt{x}-\sqrt{y}}{x-y},$$
so by choosing $x=1+\left(1+\frac{1}{n}\right)^2$, $y=1+\left(1-\frac{1}{n}\right)^2$ we have:
$$ \frac{1}{a_n} = \frac{n}{4}\left(\sqrt{1+\left(1+\frac{1}{n}\right)^2}-\sqrt{1+\left(1-\frac{1}{n}\right)^2}\right) $$
or:
$$ \frac{1}{a_n} = \frac{1}{4}\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Two different trigonometric identities giving two different solutions Using two different sum-difference trigonometric identities gives two different results in a task where the choice of identity seemed unimportant. The task goes as following:
Given $\cos 2x =-\frac {63}{65} $ , $\cos y=\frac {7} {\sqrt{130}}$, unde... | Andre already gave the answer, but let me explain the "generalities".
The main problem is the following:
After you reached the point $\sin(x+y) = \sqrt{2}/2$, you concluded that $x+y$ can take both values $\pi/4$ or $3\pi/4$.
You interpreted this to mean
BOTH $x+y = \pi/4$ and $x+y = 3\pi/4$ are solutions. (That ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Roots addition.
$$\sqrt{\frac{a+x^2}{x}-2\sqrt{a}}+\sqrt{\frac{a+x^2}{x}+2\sqrt{a}}=Q $$
One is expected to find $Q$ respecting $a>0$, $x>\sqrt{a}$ .
I'd like to have my solution checked; namely the correct answer is $2\sqrt{x}$ but I simply fail to see where I made a mistake.
And it'd be nice to hear if there is... | Since we have $x\gt \sqrt a\gt 0$, we have
$$x^2\gt a\Rightarrow a-x^2\lt 0.$$
Hence, note that we have
$$\sqrt{\frac{(a-x^2)^2}{x^2}}=\frac{\sqrt{(a-x^2)^2}}{\sqrt{x^2}}=\frac{|a-x^2|}{|x|}=\frac{\color{red}{-}(a-x^2)}{x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
anyone can help me with solving this $x^{x^{3}}=3$? Find the value of $x$ : $x^{x^{3}}=3$
I tied with "log" but I couldn't. any help?
| Assume the domain is $(0,\infty)$. You show the equation $x^{x^3} = 3$ has a unique real solution $x = \sqrt[3]{3}$. Look at the function: $f(x) = x^3\ln x , 0 < x < \infty$. If $0 < x < 1 \Rightarrow x^3\ln x < 0$, so $x^3\ln x < \ln 3$, since $\ln 3 > 0$. Thus: $e^{x^3\ln x} < e^{\ln 3}\Rightarrow x^{x^3} < 3$. And ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1339593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Proving that $\frac{1}{a^2}+\frac{1}{b^2} \geq \frac{8}{(a+b)^2}$ for $a,b>0$ I found something that I'm not quite sure about when trying to prove this inequality.
I've proven that
$$\dfrac{1}{a}+\dfrac{1}{b}\geq \dfrac{4}{a+b}$$
already. My idea now is to replace $a$ with $a^2$ and $b^2$, so we now have
$$\dfrac{1}{... | $\dfrac{1}{a^2}+\dfrac{1}{b^2} \ge \dfrac{2}{ab} \ge \dfrac{8}{(a+b)^2} \iff (a+b)^2 \ge 4ab$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Compute definite integral Question: Compute
$$\int_0^1 \frac{\sqrt{x-x^2}}{x+2}dx.$$
Attempt: I've tried various substitutions with no success. Looked for a possible contour integration by converting this into a rational function of $\sin\theta$ and $\cos \theta$.
| We certainly can use contour integration here. First, through a sub of $x \mapsto x^2$ and a little algebra, we can express the integral as
$$2 \int_0^1 dx \, \sqrt{1-x^2} - 4 \int_0^1 dx \frac{\sqrt{1-x^2}}{2+x^2} = \frac{\pi}{2} - 2 \int_{-1}^1 dx \frac{\sqrt{1-x^2}}{2+x^2}$$
Now consider
$$\oint_C dz \frac{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
Prove that $2^{15}-1$ is divided by $11\cdot31\cdot61$? I have to prove that $2^{15}-1$ is divided by $11\cdot31\cdot61$.
I have proven using congruencies that $2^{15}-1$ is divided by $31$. However we have
$$2^5\equiv 10 \mod{11}$$
$$2^{15}\equiv 10^3=1000\equiv 10 \pmod {11}$$
Therefore
$$2^{15}-1\equiv 9 \pmod{11}.... | You are correct, there is a mistake in the problem. Here are two useful prime factorizations:
$$
\begin{align*}
2^{15}-1&=32767=7\cdot 31\cdot 151\\
2^{15}+1&=32769=3^2\cdot 11\cdot 331
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the indefinite integral $\int \frac{\cos \theta}{ \sqrt{2 - 9 \sin^2 \!\theta}} \mathrm{d}\theta$ I want to evaluate $$\int \dfrac{\cos \theta \, \mathrm{d}\theta}{ \sqrt{2 - 9 \sin^2 \theta}}$$
but I can't seem to get the answer, my working is as below:
| Using the substitution $u = \dfrac{\sqrt{2}}{3} \sin \theta \implies \dfrac{\mathrm{d}u}{\mathrm{d}\theta} = \dfrac{\sqrt{2}}{3} \, \cos \theta$. Our integral can be written as $$\int \dfrac{\cos \theta}{ \sqrt{2 - 9 \sin^2 \theta}} \, \frac{\mathrm{d}\theta}{\mathrm{d}u} \mathrm{d}u = \int \dfrac{\cos \theta}{ \sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find remainder when $777^{777}$ is divided by $16$
Find remainder when $777^{777}$ is divided by $16$.
$777=48\times 16+9$. Then $777\equiv 9 \pmod{16}$.
Also by Fermat's theorem, $777^{16-1}\equiv 1 \pmod{16}$ i.e $777^{15}\equiv 1 \pmod{16}$.
Also $777=51\times 15+4$. Therefore,
$777^{777}=777^{51\times 15+4}={(7... | Fermat's theorem will only work with primes.($16$ is not a prime)
But, it can be solved by the general formula, using Euler's theorem.
Since $gcd(9,16)=1, 9^{\phi(16)} \equiv 1 (mod 16)$. Since $16=2^4, \phi(16)=8$.
Therefore, $9^8 \equiv 1 (mod 16)$, and we have $777^{777} \equiv 9^{777} \equiv 9^9 \equiv 9*9^8 \equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
If $a+b+c+d=1$ then why is the maximum value of $(a+1)(b+1)(c+1)(d+1)$ is ${\left(\frac{5}{4}\right)}^4$? What I know is that for equations of type $x+y=8$, $xy$ attains its maximum value when $x=y$ and this can be proved by either solving the quadratic equation with completing the squares or finding the first and seco... | In light of all the AM-GM answers, here is a variational approach.
Suppose we know that $a+b+c+d=1$, then we also know that for any direction moved $(\delta a,\delta b,\delta c,\delta d)$ that maintains this condition
$$
\begin{align}
0
&=\delta a+\delta b+\delta c+\delta d\\
&=(\delta a,\delta b,\delta c,\delta d)\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
$(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3=24xyz$? The question given is
Show that $(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3=24xyz$.
What I tried is suppose $a=(y+z-x),\ b=(z+x-y)$ and $c=(x+y-z)$ and then noted that $a+b+c=x+y+z$. So the question statement reduced to $(a+b+c)^3-(a^3+b^3+c^3)$. Then I tried to invoke t... | HINT:
Following your way,
$$(a+b+c)^3-(a^3+b^3+c^3)=3(a+b)(b+c)(c+a)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
$P\left(\limsup \left(X_n=0, X_ {n+1}=1,X_ {n+2}=0 \right)\right)$ "Let $X_1, X_2, ...$ independent random variables where $X_n\sim B(p_n)$ and $p_n = \frac{1}{n}$. Calculate $P\left(\limsup \left(X_n=0, X_ {n+1}=1,X_ {n+2}=0 \right)\right)$"
I suppose that i can use the lemma of Borel-Cantelli, but I don't know how i... | Is that really $X_n = 0$, $X_{n+1} = \color{red}{1}$ and $X_{n+2} = 0$ and not $X_n = 0$, $X_{n+1} = \color{red}{0}$ and $X_{n+2} = 0$?
If so, since the $X_n$'s are independent, the events $X_n = 0$, $X_{n+1} = 1$ and $X_{n+2} = 0$ are independent. Thus, we have
$$P(\{X_n = 0, X_{n+1} = 0, X_{n+2} = 0\}) = P(X_n = 0)P(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{8}{5}\le 2a+b\le 8$ Let $a,b,c,d,e$ be real numbers such that
$$\begin{cases}
a+b+c+d+e=8\\
a^2+b^2+c^2+d^2+e^2=16
\end{cases}$$
Prove that:
$$\dfrac{8}{5}\le 2a+b\le 8$$
| Let we set $(a,b,c,d,e)=\frac{8}{5}\cdot(1,1,1,1,1)+(x_1,x_2,x_3,x_4,x_5)$.
Then we have:
$$\left\{\begin{array}{rcl}x_1+x_2+x_3+x_4+x_5&=&0,\\ x_1^2+x_2^2+x_3^2+x_4^2+x_5^2&=&\frac{16}{5}\end{array}\right.$$
and we have to find the stationary points of $2x_1+x_2$. Lagrange multipliers give:
$$ (2,1,0,0,0) = \lambda(1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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A general method for integration of rational function. $\int\frac {x^3}{1+x^5}$
ATTEMPT:
I did the following substitution:
Let $x=\frac{1}{t}.$
$dx=\frac{-1}{t^2}dt.$
substituting back:
$I=\int\frac{-1}{1+t^5}dt$ which doesn't seems a simpler integration.
Next i substituted $x=p^\frac{2}{5}.$
$dx=\frac{2}{5}\frac{p^... | There is a general closed form for such integrals but it's not elementary or pretty at all.
$$\int \frac{x^m}{1+x^n} \, \mathrm{d}x = \frac{x^{m+1} \, _2F_1\left(1, \frac{m+1}{n};\frac{m+n+1}{n}; -x^n\right)}{m+1} + \mathrm{C}.$$
Where $_2F_1\left(a, b;c; x\right)$ is the hypergeometric function.
With regards to your ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Sum of trigonometric infinite series I am trying to prove that for any $x\geq 1$ we have:
$$ \sum_{m=1}^{\infty} \frac{\sin\frac{(2m-1)\pi}{x}}{\left(\frac{(2m-1)\pi}{x}\right)^3} = \frac{x}{8}(x-1). $$
Could I have some help please? I am thinking that Fourier series could help, but I found nothing until now. Thank you... | Yes. Fourier series can help.
It is equivalent to finding the limiting function of the Fourier series
$\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {(2n - 1)t} \right)}}{{{{(2n - 1)}^3}}}} $ .
Note that$\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {nt} \right)}}{n}} $ converges to $(\pi - t)/2$ for $0 < t < 2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1347535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Integral of $\frac{x^2+1}{(1-x^2)\sqrt{1+x^4}}$ So we have to evaluate $\int\frac{x^2+1}{(1-x^2)\sqrt{1+x^4}}dx$.
My work-
We can write the integrand as $\frac{(x+1)^2-2x}{(1-x)(1+x)\sqrt{1+x^4}}dx$.
So we wish to deduce $\int\frac{(x+1)}{(1-x)\sqrt{1+x^4}}dx-\int\frac{2x}{(1-x^2)\sqrt{1+x^4}}dx$
So lets write it a... | $\bf{My\; Solution::}$ Given $$\displaystyle I = \int\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}dx = \int\frac{1+x^2}{x^2\cdot \left(\frac{1}{x}-x\right)\sqrt{\left(\frac{1}{x}-x\right)^2+\left(\sqrt{2}\right)^2}}dx$$
So $$\displaystyle I = \int\frac{\left(\frac{1}{x^2}+1\right)}{\left(\frac{1}{x}-x\right)\sqrt{\left(\frac{1}{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Looking for help to understand example of Group I am looking for someone to help me to understand what is going on in the following example, from Hersteins "Topics in Algebra".
It says,
Let $G$ be the set of all $2*2$ matrices $$\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}$$ where $a,b,c,d$ are integers modulo $p$,... | $G$ is group because if $x,y \in G$ then $x^{-1} y \in G$ and $G$ is nonabelian because:
$\begin{pmatrix} 1 & 0 \\ 1 & 1 \\ \end{pmatrix} *\begin{pmatrix} 0 & 1 \\ 1 & 1 \\ \end{pmatrix} \neq \begin{pmatrix} 0 & 1 \\ 1 & 1\\ \end{pmatrix} *\begin{pmatrix} 1 & 0 \\ 1 & 1 \\ \end{pmatrix}$
And at last $G$ is finite beca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If det $A = 0$ and $\det B \neq 0$ then show that $abc = -1$ This has been hurting my head for a while now....
If
$$
\det\begin{bmatrix}a&a^2&1+a^3\\b&b^2&1+b^3\\c&c^2&1+c^3\end{bmatrix}=0
$$
And
$$
\det\begin{bmatrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{bmatrix} ≠0
$$
Then show that $abc=-1$.
| First note that,
$$\begin{vmatrix}
a & a^2 & a^3+1 \\
b & b^2 & b^3+1 \\
c & c^2 & c^3+1 \\
\end{vmatrix}
=
\begin{vmatrix}
a & a^2 & a^3 \\
b & b^2 & b^3 \\
c & c^2 & c^3 \\
\end{vmatrix}
+
\begin{vmatrix}
a & a^2 & 1 \\
b & b^2 & 1 \\
c & c^2 & 1 \\
\end{vmatrix}=0$$
Then, $$abc\begin{vmatrix}
1 & a & a^2 \\
1 & b &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Difference quotient of $f(x)= 2-6x+4x^2$ I need to find $f(a), f(a + h)$, and the difference quotient
$$\frac {f(a + h) − f(a)}{h},$$
where $h\neq 0$ and $f(x) = 2-6x+4x^2$.
My work:
$$f(a) = 2-6a+4a^2,\ \ f(a+h) = 2-6(a+h)+4(a+h)^2.$$
I need help on the last one:
$$\frac {f(a+h)-f(a)}{h}.$$
My work:
\begin{split} \... | You need $\frac{f(a+h)-f(a)}{h}$, but you seem to have computed something else not involving $f$. Try this,
\begin{equation}
\frac{f(a+h)-f(a)}{h}=\frac{(2-6(a+h)+4(a+h)^2)-(2-6a+4a^2)}{h} \\
=\frac{2-6a-6h+4a^2+8ah+4h^2-2+6a-4a^2}{h} \\
=\frac{-6h+8ah+4h^2}{h} \\
=-6+8a+4h
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding subgroups of $\mathbb{Z}_{13}^*$
I need to find all nontrivial subgroups of $G:=\mathbb{Z}_{13}^*$ (with multiplication without zero)
My attempt:
$G$ is cyclic so the order of subgroup of $G$ must be $2,3,4,6$
Now to look for $g\in G$ such that $g^2=e,g^3=e,g^4=e,g^6=e$
$$\begin{align}
&12^1=12\mod 13\\
&12... | Since $2$ is a generator of $\Bbb{Z}_{13}^*$ you have that the subgroups are exactly
$$\{ \langle 2^n \rangle : n \mbox{ divides } 12\}$$
i.e.
$$\langle 2 \rangle \\ \langle 2^2 \rangle = \langle 4 \rangle \\ \langle 2^3 \rangle = \langle 8 \rangle \\ \langle 2^4 \rangle = \langle 3 \rangle \\ \langle 2^6 \rangle = \la... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Can someone please check my work?: $\cos^2(x)=1-\sin(x)$ $$\begin{align}\cos^2(x)&=1-\sin(x)\\
1-\sin^2(x)&=1-\sin(x)\\
(1-\sin x)(1+\sin x)&= 1-\sin(x)
\end{align}$$
divide both sides by $1 - \sin(x)$
End up with $1 + \sin(x)$
The answer is supposed to be in radians between $0$ and $2 \pi$.
So I get $1+\sin(x)=0$
$$\s... | There are two correct ways to proceed from
$$(1-\sin x)(1+\sin x) = 1-\sin(x)$$
first way
This will be true if
$1 - \sin x = 0$
$x \in \{90^{\circ}, 270^{\circ}\}$
If not, then we can divide both sides by $1 - \sin x$, getting
$1 + \sin x = 1$
$\sin x = 0$
$x \in \{0^{\circ}, 180^{\circ}\}$
second way
$$(1-\sin x)(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1353326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Solving the equation $\{x\}+\{\frac{1}{x} \}=1$ Solve the equation $$\large\{x\}+\left\{\dfrac{1}{x}\right\}=1$$
given $x\in\mathbb R\setminus \{0\}$ and $\{x\}$ denotes the fractional part of $x$.
| $\{x\}+\{\frac{1}{x}\}=1$
$x+\frac{1}{x}=1+[x]+[\frac{1}{x}]$
$x+\frac{1}{x}$ is an integer.
let $n=x+\frac{1}{x}$
$n=1+[x]+[n-x]=1+[x]+n+[-x]$
$[x]+[-x]=-1$
$x\not\in\mathbb Z$
$n=x+\frac{1}{x}$
$x^2-nx+1=0$
since x is real, discriminant $\geq 0$
$n^2\geq 4$
$|n|\geq 2$
$|n|=2 \iff |x|=1 \implies x\in\mathbb Z$ , cont... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
If $x+y+z=3$, then $\sum_{\text{cyc}}\frac{x^2}{2y^2-y+3}\ge\frac{3}{4}$
Let $x,y,z>0$, be such that $x+y+z=3$. Show that
$$\dfrac{x^2}{2y^2-y+3}+\dfrac{y^2}{2z^2-z+3}+\dfrac{z^2}{2x^2-x+3}\ge\dfrac{3}{4}.$$
I've tried many things but all have failed.
$$\left(\sum_{\text{cyc}}\dfrac{x^2}{2y^2-y+3}\right)\left(\sum... | I would go into homogeneous coordinates,
$$A=3=x+y+z$$
$$X=2x-y-z \quad x=(A+X)/3$$
and cyclically for $Y$ and $Z$. The coordinates $X$, $Y$, $Z$ are orthogonal to $A$, so whatever their values, they respect the constraint to the plane. Additionally, $X+Y+Z=0$.
With this, the sum is rewritten:
$$\sum \frac{(A+X)^2}{2(A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Determining $\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx$
$$\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx$$
Attempt:
Simplification of the root factor:
$$\sqrt{x^4+2x^3-x^2+2x+1}=\frac{1}{x}\sqrt{\left(x^2+\frac{1}{x^2}\right)+2\left(x+\frac{1}{x}\right)-1}.$$
Arranging the rest of the factors as:
$... | Maybe it's not the most rapid way, but it seems work. We have, taking $u=t+1
$ $$\int\frac{\sqrt{t^{2}+2t-3}}{t+2}dt=\int\frac{\sqrt{u^{2}-4}}{u+1}du
$$ and taking $u=2\sec\left(v\right)
$ we have $$=4\int\frac{\tan^{2}\left(v\right)\sec\left(v\right)}{2\sec\left(v\right)+1}dv=4\int\frac{\tan^{2}\left(v\right)}{\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Existence of $\sqrt{-1}$ in $5$-adics, show resulting sum is convergent. I know that to prove the existence of a square root of $-1$ in $\mathbb{Z}_5$, I can just plug $x = -5$ and $a = 1/2$ into the Taylor expansion$$(1 + x)^a = \sum_{n=0}^\infty \binom{a}{n} x^n.$$However, I am confused on how I would make sure that ... | The convergence should be in $\mathbb{Z}_5$, that works differently than in $\mathbb{R}$.
I try first an approximation
$$2^2 = 4 = -1 + 5$$
Say I have $a$ with $a^2 = -1$. Then
$$\left(\frac{a}{2}\right)^2 = \frac{-1}{-1+5}= \frac{1}{1-5}$$
So take
$$\frac{a}{2} = (1-5)^{-1/2}$$
We have the binomial formula
$$(1-t)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Finding the intersection point between two lines using a matrix I'm trying to find the intersection point (if any) of two lines. Long story short, here are the parametric equations:
For the first line:
$$x = 3 + 4\lambda_1\\
y = 4 + \lambda_1\\
z = 1$$
For the second line:
$$x = -1 + 12\lambda_2\\
y = 7 + 6\lambda_2\\
... | There is an error in your initial matrix: the second row should be [0 3 -4] instead of [0 3 4] because the equation is $3\lambda_2=-4$.
I row reduced the matrix with that last row and got the same values: $\lambda_1=-5$ and $\lambda_2=\frac{-4}{3}$
Also note that you only really need two rows because you have two unkno... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I evaluate this integral :$\int \frac{\sqrt{-x^2-x+2}}{x^2}dx$? Is there someone who can show me how to evaluate this integral:
$$\int \frac{\sqrt{-x^2-x+2}}{x^2}dx.$$
I have tried many changes of variables but I haven't succeeded yet. Thank you for any help.
| $\begin{gathered}
\int {\frac{{\sqrt { - {x^2} - x + 2} }}{{{x^2}}}} \,dx \hfill \\
= - \frac{{\sqrt { - {x^2} - x + 2} }}{x} - \int {\frac{{\left( { - 2x - 1} \right)}}{{2\sqrt { - {x^2} - x + 2} }}} \frac{{ - 1}}{x}\,dx\quad \quad \left( {{\text{Integrating by parts}}} \right) \hfill \\
= - \frac{{\sqrt { -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving the integral series $\int _0^1\left(1-x^2\right)^n\,dx=\frac{2}{3}\cdot \frac{4}{5}\cdot\ldots\cdot \frac{2n}{2n+1}$
We have the series $\left(I_n\right)_{n\ge 1\:}$ where $$I_n=\int _0^1\left(1-x^2\right)^n\,dx.$$
Prove that $$I_n=\frac{2}{3}\cdot \frac{4}{5}\cdot\ldots\cdot \frac{2n}{2n+1}.$$
I tried to int... | Here is a way you could do it without induction, using the Beta function and the Gamma function. Let $\sqrt{t}=x$, so that
$$
I_n=\int_0^1(1-t)^n\frac{t^{1/2}}{2}dt.
$$
This now looks like the Beta function, so we have
$$
I_n=\frac{\Gamma(1/2)\Gamma(n+1)}{2\Gamma ((n+1)+1/2)}.
$$
Using the fact that $\Gamma(1/2)=\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 0
} |
How to solve a system of logarithmic equations? I need to create a function with the following properties:
$$f(1)=1$$
$$f(65)=75$$
$$f(100)=100$$
Additionally, the function needs to grow logarithmically. So that gives three equations:
$$A \cdot \ln(B \cdot 1 + C) = 1$$
$$A \cdot \ln(B \cdot 65 + C) = 75$$
$$A \cdot \ln... | This is not an answer, just what I tried using Count Iblis's hint.
Let $$f(x) = A\ln(x+B)+ C$$
You can always write the function with the coefficient of $x$ being $1$, so we assume that it is.
\begin{align*}
f(1)&=A\ln(1+B)+ C=1\\
f(65)&=A\ln(65+B)+ C=75\\
f(100)&=A\ln(100+B)+ C=100
\end{align*}
We use Count Iblis's ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Examining the convergence of $\int_{0}^{1}\left(\left\lceil \frac{1}{x} \right\rceil-\left\lfloor \frac{1}{x} \right\rfloor\right) \, dx$ Okay so I'm trying to determine whether $\int_{0}^{1}\left(\left\lceil \frac{1}{x} \right\rceil-\left\lfloor \frac{1}{x} \right\rfloor\right) \, dx$ converges and if so, to what valu... | Yes, you are correct. But it is much faster if you say that $\lceil \frac{1}{x} \rceil-\lfloor \frac{1}{x} \rfloor$ is $1$ except in a set of measure zero. If you don't know what the meausre of a set is, just say that the set is countable. Then
$$\int_0^1\left(\left\lceil \frac{1}{x}\right \rceil-\left\lfloor \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Solve $2^{a+3}=4^{a+2}-48,\ a\in \mathbb{R}$
Solve $2^{a+3}=4^{a+2}-48,\ a\in \mathbb{R}$
I tried to simplify it ,
$2^{a+3}=4^{a+2}-48\\
2^{a+3}=2^{2(a+2)}-2^4\cdot 3\\
2^{2a}-2^{a-1}- 3=0\\
$
I don't know how to go from here.
This question is from chapter quadratic equations, so i think there must be hidden quadrati... | $$2^{a+3}=4^{a+2}-48\Longleftrightarrow$$
$$48+2^{a+3}-4^{a+2}-48=0\Longleftrightarrow$$
$$-8\left(2^a-2\right)\left(3+2^{a+1}\right)=0\Longleftrightarrow$$
$$\left(2^a-2\right)\left(3+2^{a+1}\right)=0\Longleftrightarrow$$
$$\left(2^a-2\right)=0 \vee \left(3+2^{a+1}\right)=0\Longleftrightarrow$$
$$2^a=2 \vee 3+2^{a+1}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Infinite number of ways to write $1=\frac{1}{n}+\frac{1}{a_1}+\cdots+\frac{1}{a_k}$ How can I show that there is an infinite number of ways in which $1$ can be written in the form $$1=\frac{1}{n}+\frac{1}{a_1}+\cdots+\frac{1}{a_k},$$ where $n>1$ is an integer (this number is fixed), each $a_i$ is an integer and $n<a_1<... | Hint: For any integer $a_k > 1$, we have $\dfrac{1}{a_k} = \dfrac{1}{a_k+1} + \dfrac{1}{a_k^2+a_k}$, where $a_k+1 < a_k^2+a_k$.
So, if $1 = \dfrac{1}{n}+\dfrac{1}{a_1}+\cdots+\dfrac{1}{a_{k-1}}+\dfrac{1}{a_k}$ for some integers $n < a_1 < \cdots < a_{k-1} < a_k$, then we also have $1 = \dfrac{1}{n}+\dfrac{1}{a_1}+\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Is the substitution of standard angles while proving the equality of trigonometric formulas allowed? Here is a problem that my class 10 maths teacher gave me:
Prove that $\sec^4\theta$ - $\sec^2\theta$ = $\tan^4\theta$ + $\tan^2\theta$
She expected me to use trigonometric identities to prove such equality, but I inst... | LHS:
$$
\sec^4\theta - \sec^2\theta = \sec^2\theta(\sec^2\theta - 1) = \frac{1}{\cos^2\theta}\frac{1-\cos^2\theta}{\cos^2\theta}=\frac{\sin^2\theta}{\cos^4\theta}
$$
RHS:
$$
\tan^4\theta + \tan^2\theta = \tan^2\theta (1 + \tan^2\theta) = \frac{\sin^2\theta}{\cos^2\theta}\frac{\sin^2\theta + \cos^2\theta}{\cos^2\theta}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Formulae for sequences Given that for $1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{n^2(n+1)^2}{4}$
deduce that $(n+1)^3 + (n+2)^3 +\cdots+ (2n)^3 = \frac{n^2(3n+1)(5n+3)}{4}$
So far:
the sequence $(n+1)^3 + (n+2)^3 +\cdots+ (2n)^3$ gives $2^3 + 3^3 + 4^3 +\cdots,$ when n=1.
The brackets in the formula for the second sequenc... | You can use the following relation :
$$\sum_{k=n+1}^{2n}k^3=\sum_{k=1}^{\color{red}{2n}}k^3-\sum_{k=1}^{n}k^3$$
(To get the first sum on RHS, you only need to replace the $n$ in the given equation with $\color{red}{2n}$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding $4$ variables using $3$. if I have:
$ x=\dfrac{a-.5b-.5c+.25d}{a+b+c+d}$
$ y=\dfrac{\dfrac{b\sqrt{3}}{2}+\dfrac{c\sqrt{3}}{2}+\dfrac{d\sqrt{3}}{4}}{a+b+c+d}$
$ z=a+b+c+2d $
Then how do I get back to:
$ a= $ , $ b= $ , $ c= $ , and $ d= $ ?
When there is a divide by zero error, $ x=0 $ and $ y=0 $.
When
... | Usually you can't :
The system can be rewritten as
$$x(z-d) - \frac{d}{4} = a - \frac{b}{2} - \frac{c}{2}$$
$$y(z-d) - \frac{d\sqrt{3}}{4} = \frac{b\sqrt{3} }{2} - \frac{c\sqrt{3}}{2}$$
$$ z-2d = a+b+c$$
This can be seen as a linear system with unknown $a$,$b$ and $c$, and it's not hard to show that that this system h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Partial derivative of $f(x,y) = (x/y) \cos (1/y)$ So I'm not really sure whether I'm correct as several people are saying some of my syntax is wrong, where others are saying I have a wrong answer. I have checked my answer using Wolfram Alpha and it appears to be correct; could anyone please confirm/clarify?
Calculate t... | $$\frac { \partial f\left( x,y \right) }{ \partial y } =\frac { \partial \left( \frac { x }{ y } \cos { \frac { 1 }{ y } } \right) }{ \partial y } =x\frac { \partial \left( \frac { 1 }{ y } \right) }{ \partial y } \cos { \frac { 1 }{ y } +\frac { x }{ y } \frac { \partial \left( \cos { \frac { 1 }{ y } } \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Mathematical Induction proof for a cubic equation. If $ x^3 = x +1$, prove by induction that $ x^{3n} = a_{n}x + b_n + \frac {c_n}{x}$, where $a_1=1, b_1=1, c_1=0$ and
$a_n = a_{n-1} + b_{n-1}, b_n = a_{n-1} + b_{n-1} + c_{n-1}, c_n = a_{n-1} + c_{n-1}$ for $n=2,3,\dots $
For $n=1$ we have $x_3 = a_1x + b_1 + \fra... | The basic idea is that $x^3=x+1$ implies
$$x^2=1+{1\over x}\quad\text{and}\quad x^4=x^2+x=1+{1\over x}+x$$
So if $x^{3k}=a_kx+b_k+{c_k\over x}$ then
$$\begin{align}
x^{3(k+1)}&=x^3\cdot x^{3k}\\
&=x^3(a_kx+b_k+{c_k\over x})\\
&=a_kx^4+b_kx^3+c_kx^2\\
\end{align}$$
Now replace the $x^4$, $x^3$, and $x^2$ in the last li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
$\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n^2+0}}+\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+2n}}+\cdots+\frac{1}{\sqrt{n^2+(n-1)n}}$
Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n^2+0}}+\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+2n}}+\frac{1}{\sqrt{n^2+3n}}+\cdots+\frac{1}{\sqrt{n^2+(n-1)n}}$
... | It can be shown that, for $\alpha\in(-1,0]$ and $N\in\mathbb{N}$, $$\frac{(N+1)^{\alpha+1}-1}{\alpha+1}\leq\sum_{k=1}^N\,k^\alpha\leq\frac{N^{\alpha+1}}{\alpha+1}\,.$$ Define $S_N$ for each $N\in\mathbb{N}$ to be $$\sum_{k=1}^N\,\frac{1}{\sqrt{k}}=\sum_{k=1}^N\,k^{-\frac{1}{2}}\,,$$ we have $$2(\sqrt{N+1}-1)\leq S_N \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Sum of the series $\sum\limits_{n=0}^\infty \frac{1}{(3n+1)^3}$ The following result matches very good numerically:
$$\sum_{n=0}^\infty \frac{1}{(3n+1)^3}=\frac{13}{27}\zeta(3)+\frac{2\pi^3}{81\sqrt{3}}.$$
Though I'm not sure how to approach this. How can we prove it? Also, is it possible to find closed form for $$\su... | Let $$S_1=\sum_{n=0}^{\infty} \frac{1}{(3n+1)^3} ,S_2=\sum_{n=0}^{\infty} \frac{1}{(3n+2)^3}.$$
It's easily seen that $$S_1+S_2+\sum_{n=1}^{\infty}\frac{1}{(3n)^3}=\zeta(3).$$
That is, $$S_1+S_2=\zeta(3)-\frac{1}{27}\zeta(3)=\frac{26}{27}\zeta(3)\tag{1}$$
However, $$S_1-S_2=\sum_{n=0}^{\infty} \frac{1}{(3n+1)^3}-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric equation with sine and cosine So the equation is $3\cos ^2t + 5\sin t = 1$
Now I have simplified this to $$3(1-\sin ^2t) + 5\sin t -1 = 0$$
which leads to $$-3\sin ^2t + 5\sin t + 2 = 0$$
Then I get $$-3t^2 + 5 t +2 = 0$$
Is this the correct way to go with this equation then use $t = t/2 \pm \sqrt {(t/2)^... | We have, $$3\cos ^2t + 5\sin t = 1$$ $$\implies 3(1-\sin ^2t) + 5\sin t = 1$$ $$\implies 3\sin ^2t+ 5\sin t-2=0$$ Factorizing the expression, we get $$ (3\sin t-1)(\sin t+2)=0$$ $$\text{if}\ 3\sin t-1=0 \implies \sin t=\frac{1}{3}$$$$\implies \color{blue}{t=2n\pi+\sin^{-1}\left(\frac{1}{3}\right)}$$ $$\text{Or} \ \colo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solve $10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$ How to solve the following equation?
$$10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$$
My attempt:
$$ 10x^4 - (7x^2+1)(x^2+x+1)=0$$
Thats all i can
Update
Tried to open brakets and simplify:
$$(7x^2+1)(x^2+x+1) = 7x^4+7x^3+7x^2+x^2+x+1=7x^4+7x^3+8x^2+1 $$
$$10x^4 - (7x^2+1)(x^2+x+1)= 3x^4... | Set $A=x^2,B=x^2+x+1$.
Then,
$$\begin{align}10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2&=10A^2-7AB+B^2\\&=(2A-B)(5A-B)\\&=(2x^2-(x^2+x+1))(5x^2-(x^2+x+1))\\&=(x^2-x-1)(4x^2-x-1)\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
creative method to obtain range of newton function ?! I am searching for more proof that the range of $y=\frac{x}{x^2+1}$ is $ \frac{-1}{2}\leq y \leq \frac{+1}{2}$
these are my tries :
domain is $\mathbb{R}$
first : $$\quad{y=\frac{x}{x^2+1}\\yx^2+y=x \rightarrow x^2y-x+y=0 \overset{\Delta \geq 0 }{\rightarrow} ... | by $AM-GM$ we have $\frac{x^2+1}{2}\geq |x|$, from here we obtain $\frac{|x|}{x^2+1}\le \frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Am I getting the right answer for the integral $I_n= \int_0^1 \frac{x^n}{\sqrt {x^3+1}}\, dx$?
Let $I_n= \int_0^1 \dfrac{x^n}{\sqrt {x^3+1}}\, dx$. Show that $(2n-1)I_n+2(n-2)I_{n-3}=2 \sqrt 2$ for all $n \ge 3$. Then compute $I_8$.
I get an answer for $I_8={{2 \sqrt 2} \over 135}(25-16 \sqrt 2)$, could somebody plea... | Perform the integration in steps:
$$
\begin{array}{rcl}
I_8
&=& \displaystyle \int_0^1 \frac{x^8}{\sqrt{x^3 + 1}} dx
= \displaystyle \int_0^1 x^6 \frac{x^2}{\sqrt{x^3 + 1}} dx\\
&=& \displaystyle
\left[ \frac{2}{3} x^6 \sqrt{x^3+1} \right]_0^1
- \int_0^1 4 x^3 x^2 \sqrt{x^3+1} dx\\
&=& \displaystyle
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
A question about differential function If $f(x)=f'(x^{2})+2x$, then $f(1)=?$ and $f''(1)=?$
Sorry. I am going to check the original problem, and then i will update.
| Consider the equation $f(x) = f'(x^2) + 2x$. It can be seen that $f(1) = 4$, $f'(1) = 2$ and $f^{(n+2)}(1) = 0$ for $n \geq 0$. The demonstration of how these values are obtained is as follows.
Since $f(x)$ is differentiable, which is evident from the differential equation, then consider the function in a series expan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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For two vectors $a$ and $b$, why does $\cos(θ)$ equal the dot product of $a$ and $b$ divided by the product of the vectors' magnitudes? While watching a video about dot products (https://www.youtube.com/watch?v=WDdR5s0C4cY), the following formula is presented for finding the angle between two vectors:
For vectors $a$,... | There are several derivations of this online. Here's where you can start.
Define two vectors $\textbf{a}$ and $\textbf{b}$. Then $ \textbf{a} - \textbf{b}$ is the vector that connects their endpoints and makes a triangle.
Therefore, we have a triangle with side lengths $|\textbf{a}|$, $|\textbf{b}|$, and $|\textbf{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Prove the relation for cos inverse Prove the relation $\cos^{-1}x_0=\dfrac{\sqrt {1-x^2_0}}{x_1\cdot x_2\cdot x_3\cdots \text{ ad inf.}}$ where the successive quantities $x_r$ are connected by the relation $x_{r+1}=\sqrt{\frac{1}{2}(1+x_r)}$
My attempt:
$$x_1=\sqrt{\frac{1}{2}(1+x_0)}$$
$$x_2=\sqrt{\frac{1}{2}(1+x_1)}$... | Let $\cos(\theta)=x_0$, so that $\sqrt{1-x_0^2}=\sin(\theta)$.
From known trigonometry, we have
$$x_{r+1}=\cos\left(\frac{\arccos(x_r)}2\right),$$
and by recurrence
$$x_r=\cos\left(\frac{\arccos(x_0)}{2^r}\right)=\cos\left(\frac{\theta}{2^r}\right).$$
Hence the denominator is the infinite product
$$\prod_1^\infty\cos\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find all values that solve the equation For which values a, the equation
$$ a\sin{x}+(a+1)\sin^2{\frac{x}{2}} + (a-1)\cos^2{\frac{x}{2}} =1 $$
has a solution?
My idea: I think it's possible to factorize equation or reduce equation to the form like: $a(\sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}}) =1 $
Let's go:
$$ 2a\sin{... | $$ a\sin{x}+(a+1)\sin^2{\frac{x}{2}} + (a-1)\cos^2{\frac{x}{2}} =1 $$
$$ a\sin{x}+a\left(\sin^2{\frac{x}{2}}+\cos^2{\frac{x}{2}}\right)-\left(\cos^2{\frac{x}{2}}-\sin^2{\frac{x}{2}}\right)=1 $$ $$ a\sin x+a-\cos x=1 $$
$$ a\sin x-\cos x=(1-a) $$ $$ \frac{a}{\sqrt{1+a^2}}\sin x-\frac{1}{\sqrt{1+a^2}}\cos x=\frac{(1-a)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Parallelogram ABCD There's a parallelogram $ABCD$. I'm given point $A(3,12)$ and point $B(-1,5)$. Given the equations of the lines $BC$ and $AC$ are $y=8x+13$ and $y=3x+3$ respectively.
How to find the coordinates of the point of intersection between the diagonals $BD$ and $AC$? And the coordinates of $D$?
I've no i... | Let the coordinated of $D$ be $(a, b)$. Since $AD\parallel BC$ hence the lines $AD$ & $BC: y=8x+13$ have equal gradients $$\frac{b-12}{a-3}=8$$ $$b-12=8a-24$$ $$\implies b=8a-12\tag 1$$ Now, solving $BC: y=8x+13$ & $AC: y=3x+3$ , we get coordinates of the vertex $C(-2, -3)$
Since $CD\parallel AB$ hence the lines $CD$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Multiplicating inequalities I have two inequalities: $|x|\leq\sqrt{x^2+y^2}$ and $|y|\leq\sqrt{x^2+y^2}, \forall x,y \in \Bbb R$, can I multiply these inequalities to get $|xy|\leq x^2+y^2$?
If yes, what is the justification? If not, why?
| You can multiply these inequalities, because both sides of both inequalities are nonegative. Here is an explaination:
$$|x| \leq \sqrt{x^2+y^2}$$
$|y| \geq 0$, so you can multiply both sides by $|y|$:
$$|xy| \leq |y|\sqrt{x^2+y^2}$$
But we know that $|y| \leq \sqrt{x^2+y^2}$ and $0 \leq \sqrt{x^2+y^2}$, so:
$$|y|\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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Finding all the triangles $ABC$ satisfying $\sum \frac{a^{2}\cos\frac{B-C}{2}}{\sin\frac{A}{2}}=2(a^2+b^2+c^2)$ $\triangle ABC$ has $BC=a, CA=b, AB=c$ and satisfies
$$\dfrac{a^{2}\cos\dfrac{B-C}{2}}{\sin\dfrac{A}{2}}+\dfrac{b^{2}\cos\dfrac{C-A}{2}}{\sin\dfrac{B}{2}}+\dfrac{c^{2}\cos\dfrac{A-B}{2}}{\sin\dfrac{C}{2}}=2(a... | Converting all the angles in sides the given condition will look like
$$\sum \frac{a^{2}\cos\frac{B-C}{2}}{\sin\frac{A}{2}}\frac{2sin(\frac{B+C}{2})}{2cos(\frac{A}{2})}=2(a^2+b^2+c^2)$$ $$\sum \frac{a^{2}(sin{B}+sinC)}{\sin{A}}=2(a^2+b^2+c^2)$$$$\sum \frac{a^{2}(\frac{b}{R}+\frac{c}{R})}{\frac{a}{R}}=2(a^2+b^2+c^2)$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do you solve for θ in the equation $\tan \frac{\theta}{5} + \sqrt{3} = 0$ $$\tan \frac{\theta}{5} + \sqrt{3} = 0$$
Alright so the $\frac{\theta}{5}$ is confusing me.
Would it be wrong to do
\begin{eqnarray}
\tan \frac{\theta}{5}&=&-\sqrt{3}\\
\frac{\theta}{5}&=&\tan^{-1}(-\sqrt{3})\\
\theta&=& 5\tan^{-1}(-\sqrt... | hint: Add $n\pi$ to your answer !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Find the maximum value of $72\int\limits_{0}^{y}\sqrt{x^4+(y-y^2)^2}dx$ Find the maximum value of $72\int\limits_{0}^{y}\sqrt{x^4+(y-y^2)^2}dx $ for $y\in[0,1].$
I tried to differentiate the given function by using DUIS leibnitz rule but the calculations are messy and I tried to solve directly by integrating it but th... | Let
$$
I(y) = \int_0^y \sqrt{x^4 + (y-y^2)^2}dx
$$
Let's find $y$ such that $dI/dy=0$. By Leibniz integral rule we have
$$
\frac{dI}{dy} = \int_0^y \frac{\partial}{\partial y}\sqrt{x^4 + (y-y^2)^2}dx + \sqrt{y^4 + (y-y^2)^2}
$$
But $dI/dy>0$ for $y>0$ ($dI/dy=0$ at $y=0$). So, maximum is reached for $y=1$:
$$
I(1)=\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
help with trigonometric equations How do I solve this?
$$\cos3x=\cos^2x-3\sin^2x$$
| $$ \cos(3x)=4\cos^3(x)-3\cos(x),\quad 4\cos^2(x)-3=\cos^2 x-3\sin^2 x $$
hence by setting $z=\cos x$ we have to solve:
$$ 4z^3-4z^2-3z+3 = (z-1)(4z^2-3) = 0 $$
so $\cos x=1$ or $\cos x=\frac{\sqrt{3}}{2}$, from which $x=2k\pi$ or $x=\pm\frac{\pi}{6}+2k\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Orthocenter of triangle $DEF$ is same as the circumcenter of triangle $ABC$ $D,E,F$ are mid points of the sides of the triangle $ABC$,then prove that the orthocenter of triangle DEF is same as the circumcenter of triangle ABC.
I cannnot figure out what coordinates to suppose for A,B,C.I tried taking $(x_1,y_1),(x_2,y_2... | Let's consider a $\triangle ABC$ with side BC coinciding with the x-axis such that vertex B is at origin $(0, 0)$ & vertex C is at $(a, 0)$ then vertex A will be at $\left(\frac{a\tan C}{\tan B+\tan C}, \frac{a\tan B\tan C}{\tan B+\tan C}\right)$. The mid-points $D, E$ & $F$ of the sides $BC$, $AC$ & $AB$ respectively ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Explanation of inverse trig euation solution The equation is:
$\arctan 3 + 2\arctan2 = \pi + arccot 3$
They go on and assign $\arctan 3 = \theta$ and $\arctan2 = \phi$. Therefore, $\frac{\pi}{4}< \theta < \frac{\pi}{2}$, same for $\phi$. Which I follow.
Then they use $\tan(A+B)$ formula to show that that $\tan(\theta+2... | First of all, $\arctan x+\text{arccot}x=\dfrac\pi2$ (Proof)
So, the problem reduces to $\arctan 2+\arctan3=\dfrac{3\pi}4$
Now from this or Ex$\#5$ of Page $\#276$ of this,
$$\tan^{-1}x+\tan^{-1}y=\begin{cases} \tan^{-1}\dfrac{x+y}{1-xy} &\mbox{if } xy<1 \\\pi+ \tan^{-1}\dfrac{x+y}{1-xy} & \mbox{if } xy> 1. \end{cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Show that $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ is rational using the rational zeros theorem What I've done so far:
Let $$r = \sqrt{4+2\sqrt{3}}-\sqrt{3}.$$
Thus, $$r^2 = 2\sqrt{3}-2\sqrt{3}\sqrt{4+2\sqrt{3}}+7$$
and $$r^4=52\sqrt{3}-28\sqrt{3}\sqrt{4+2\sqrt{3}}-24\sqrt{4+2\sqrt{3}}+109.$$
I did this because in a similar exam... | HINT: use that $$4+2\sqrt{3}=(1+\sqrt{3})^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
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Evaluate the integral $\int_0^{\frac{\pi}{2}} \frac{1}{a + \sin^2 \theta}$ using the residue theorem Can someone show me how to compute this integral using the residue theorem:
$$\int_0^{\frac{\pi}{2}} \frac{1}{a + \sin^2 \theta}d\theta$$
| Notice,
$$\int_0^{\frac{\pi}{2}} \frac{1}{a + \sin^2 \theta}d\theta$$
$$=\int_0^{\frac{\pi}{2}} \frac{\sec^2\theta}{a\sec^2\theta + \sec^2\theta\sin^2 \theta}d\theta$$
$$=\int_0^{\frac{\pi}{2}} \frac{\sec^2\theta}{a+a\tan^2\theta + \tan^2 \theta}d\theta$$
$$=\int_0^{\frac{\pi}{2}} \frac{\sec^2\theta}{a+(a+1)\tan^2\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Method of partial fractions when denumerator cannot be factorized? Suppose I'm given an expression: \begin{align*} P(x) = \frac{1+x}{1-2x-x^2}. \end{align*} The denumerator cannot be readily factorized, so I found the zeros, which are \begin{align*} \lambda_1 = -1 - \sqrt{2} \qquad \lambda_2 = -1 + \sqrt{2}. \end{align... | Notice, we have $$\frac{x+1}{1-2x-x^2}$$ $$=\frac{x+1}{-(x^2+2x-1)}$$ $$=\frac{-(x+1)}{(x^2+2x+1)-2}$$ $$=\frac{-x-1}{(x+1)^2-(\sqrt2)^2}$$ $$=\frac{-x-1}{(x+1-\sqrt2)(x+1+\sqrt2)}$$ Now, let
$$\frac{-x-1}{(x+1-\sqrt2)(x+1+\sqrt2)}=\frac{A}{x+1-\sqrt2}+\frac{B}{x+1+\sqrt2}$$
By comparing the corresponding coefficien... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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$\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ $\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ has value equal to
$(A)0\hspace{1cm}(B)\frac{3}{4}\hspace{1cm}(C)\frac{3}{4}\hspace{1cm}(D)2 $
I tried to solve this question by putting $x-\frac{1}{x}=t$ and limits have changed to $\frac{-3}{2}$ to $\frac... | The fact that the limits are reciprocals of each other gives you your first clue. Splitting the integral up about the geometric mean $\sqrt{2\cdot\frac 12} = 1$, we have
$$I = \int_{1/2}^1 \frac{1}{x}\sin (x-1/x) \ dx + \int_1^2 \frac{1}{x}\sin (x-1/x) \ dx$$
Now change variables in the first of those two integrals by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Let $(a, b, c)$ be a Pythagorean triple. Prove that $\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2$ is greater than 8 and never an integer.
Let $(a, b, c)$ be a Pythagorean triple, i.e. a triplet of positive integers with $a^2 + b^2 = c^2$.
a) Prove that $$\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 > 8$$
b) Prove that there ... | For now, I prove you part (a) with a different approach. Dividing by $c^2=a^2+b^2$, taking the square root, and multiplying by $-1/2$, the inequality can be rewritten as
$$
\frac{2}{\frac{1}{a}+\frac{1}{b}}<\sqrt{\frac{a^2+b^2}{2}},
$$
which is simply HM-QM. Equality cannot hold because it cannot be the case that $a=b$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Solve trigonometric equation $ 3 \cos x + 2\sin x=1 $ Solve trigonometric equation: $$ 3 \cos x + 2\sin x=1 $$
I tried to substitue $\cos x = \dfrac{1-t^2}{1+t^2}, \sin x = \dfrac{2t}{1+t^2}$. Yet with no results.
|
$$ 3 \cos x + 2\sin x=1 $$
$y:=\tan\big(\frac x 2\big)$ then $\sin(x)=\frac{2y}{y^2+1}$ and $\cos(x)=\frac{1-y^2}{y^2+1}$
$$-1+\frac{3}{y^2+1}+\frac{4y}{y^2+1}-\frac{3y^2}{y^2+1}=0$$
$$\frac{2y^2-2y-1}{y^2+1}=0$$
$$2y^2-2y-1=0$$
$$y^2-y=\frac 1 2$$
Add $\frac 1 4$ to both sides:
$$y^2-y+\frac 1 4=\frac 3 4$$
$$\bigg(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Why I am getting different answer? I have just started learning single variable calculus. I'm confused in a problem from sometime. I didn't get why my answer is different from the book.
$$
\require{cancel}
\begin{align}
&\int\sin x \sin 2x \sin 3x\,dx\\
&=\int\sin x\;\,2\sin x\cos x \left(3\sin x - 4\sin^3 x\right)\,dx... | $\bf{My\; Solution::}$ Let $\displaystyle \int \sin x\cdot \sin 2x \cdot \sin 3x dx$
Using the formula
$\bullet 2\sin A \cdot \sin B = \cos (A-B)-\cos (A+B)$
$\bullet 2\cos A \cdot \sin B = \sin (A+B)-\sin (A-B)$
So $$\displaystyle I = \frac{1}{2}\int \left[2\sin 3x \cdot \sin x\right]\cdot \sin xdx = \frac{1}{2}\int ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find the least positive integer $n$ so that $\left ( 1-\frac{1}{s_{1}} \right ) \cdots \left ( 1-\frac{1}{s_{n}} \right )=\frac{51}{2010}$
Find the least positive integer $n$ for which there exists a set $\left \{ s_{1}, s_{2},....,s_{n} \right \}$ consisting of $n$ distinct positive integers such that $$\left ( 1-\f... | Suppose that for some $n$ there exist the desired numbers; we may assume that $s_{1}< s_{2}< ...< s_{n}$. Surely $s_{1}> 1$ since otherwise $1-\frac{1}{s_{1}}=0 $. So we have $2\leq s_{1}\leq s_{2}-1\leq s_3 - 2\leq ....\leq s_{n} - (n-1)$, hence $s_{i}\geq i+1$ for each $i=1,...,n$. Therefore $$\frac{51}{2010}=\left (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Finding equation of straight line that is tangent to $y = 2^x$. Problem: Find an equation of the straight line that is tangent to $y= 2^x$ and that passes through the point $(1,0)$.
Attempt: Let $(a, 2^a)$ be the point of tangency. Now we have that $y' = 2^x \ln(2)$, which evaluated at the tangency point becomes $y' = ... | To show the answers are the same, we need to show that if $a=\frac{\ln 2+1}{\ln 2}$ then $2^a=2e$.
Note that $a=1+\frac{1}{\ln 2}$, so
$$2^a=2\cdot 2^{1/\ln 2}=2\cdot (e^{\ln 2})^{1/\ln 2}=2e.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Inverse Trigonometric Function: Find the Exact Value of $\sin^{-1}\left(\sin\left(\frac{7\pi}{3}\right)\right)$ $$\arcsin\left(\sin\left(\frac{7\pi}{3}\right)\right)$$
I cannot use this formula, correct? $f(f^{-1}(x))=x$
The answer in the book is $\frac{\pi}{3}$
How do I approach solving a problem such as this?
The in... | You have to consider the restricted ranges of the inverse trig functions. Since $\sin\left(\frac{7\pi}{3}\right)=\frac{\sqrt{3}}{2}$, you want $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}$, since the range of inverse sine is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the other end of the Diameter For a National Board Exam Review
A circle has it center at $(3,-2)$ and one end of a diameter at $(7,2)$. Find the other end of the diameter.
Answer is $(-1,6)$
$$m=\frac{y^2-y^1}{x^2-x^1}=\frac{2-(-2) }{7-3}$$
$$=(3-7)^2+(-2-2)^2=r^2=32$$
$$r =\sqrt{32}$$
Plugin $(7,2)$ into
$$y... | Here's another approach:
The center of a circle is the midpoint of any diameter. If the coordinates of the unknown point are $(a,b)$, using the midpoint formula:
$\frac{a+7}{2}=3$, and...(see if you can work out the equation for $b$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sum of (arithmetic?) infinite series How the heck do I find the sum of a series like $\sum\limits_{n=3}^\infty\frac{5}{36n^{2}-9}$? I can't seem to convert this to a geometric series and I don't have a finite number of partial sums, so I'm stumped.
| $$\frac{5}{9}\sum \frac{1}{4x^2-1}$$
$$=\frac{5}{18}\sum \frac{1}{2x-1}-\frac{1}{2x+1}$$
$$=\frac{5}{18}[ \frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}...]$$
$$=\frac{5}{18}[\frac{1}{5}]$$
$$=\frac{1}{18}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Evaluation of $\int\frac{1}{1-\tan^2 x}dx$ Evaluation of $\displaystyle \int\frac{1}{1-\tan^2 x}dx$
$\bf{My\; Try::}$ We can write $$\displaystyle \int\frac{1}{(1-\tan x)\cdot (1+\tan x)}dx = \frac{1}{2}\int\frac{(1+\tan x)+(1-\tan x)}{(1-\tan x)\cdot (1+\tan x)}dx$$
So We get $$\displaystyle = \frac{1}{2}\int\frac{1}{... | Perhaps this might be a bit shorter
\begin{align*}
\frac{1}{1-\tan^2 x} & = \frac{\cos^2x}{\cos^2x-\sin^2x}\\
& = \frac{1+\cos2x}{2\cos2x}\\
& = \frac{1}{2}\sec 2x+\frac{1}{2}.
\end{align*}
Now you just have to integrate $\sec 2x$. This is a routine problem (multiply numerator and denominator by $\sec 2x + \tan 2x$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Value of $\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}$ Here is the question:
$$\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}} = \;?$... | The hint
$$
\frac{ \sqrt{1+\sin(x)}+\sqrt{1+\cos(x)} }{ \sqrt{1-\sin(x)}+\sqrt{1-\cos(x)} }=1+\sqrt2 \quad \text{for } x\in [0,\pi/2]
$$
is spot on.
We have
$$
\begin{align}
\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{99}}
&=
\sqrt{10}\left(\sqrt{1+\sqrt{0.01}}+\sqrt{1+\sqrt{0.99}}\right)
\\&=
\sqrt{10}\left(\sqrt{1+\sin(t)}+\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 4
} |
Prove that every positive integer less than or equal to the square root of a is a near factor of a In many computer languages, the division operation ignores remainders. Let's denote this by the operation $//$, so for instance $13//3 = 4$. If for some $b$, $a//b = c$ then we say that $c$ is a near factor of $a$. Thus, ... | Disclaimer: This is not quite a complete solution. It proves that every $\boldsymbol{k}$ with $\boldsymbol{k+1 \le \sqrt{a}}$ is a near factor.
Let $m$ be the smallest integer for which $m(m+1) > a$.
Consider the (not necessarily strictly) decreasing sequence
$$
a // m, a // (m+1), a // (m+2), \ldots, a // (a-1), a // ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculus Integral from Partial Fractions When you have an irreducible quadratic factor repeated you can get integrals that look like $\int \dfrac{dx}{(x^2+a)^m}$, where $m>1$, integer, and $a>0$. What is the best way to integrate this function? Is there more than one way?
| The recurrence relation derived by Jack D'Aurizio in his answer above can actually be solved explicitly for $I_m$. First let me quote it below, but changing the recurrence index to $n$:
$$
I_{n+1} = \frac{2n+1}{2n}\,I_{n} + f_n(t)
\qquad (n \ge 1) \qquad (1)
$$
where
$$
\begin{align}
I_0 &= t \\[0.05in]
I_1 &= \ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Is $(x^2 + 1) / (x^2-5x+6)$ divisible? I'm learning single variable calculus right now and at current about integration with partial fraction. I'm stuck in a problem from few hours given in my book. The question is to integrate $$\frac{x^2 + 1}{x^2-5x+6}.$$
I know it is improper rational function and to make it proper ... | Given $\displaystyle \int\frac{x^2+1}{x^2-5x+6}dx = \int\frac{(x^2-5x+6)+(5x-5)}{x^2-5x+6}dx = \int 1dx+5\int\frac{(x-1)}{(x-2)(x-3)}dx$
Now Using Partial fraction Method::
$\displaystyle \frac{x-1}{(x-2)\cdot (x-3)} = \frac{A}{(x-2)}+\frac{B}{(x-3)}$
So we get $x-1 = A(x-3)+B(x-2) = (A+B)x+(-3A-2B)$
After Camparing th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Confusing question on exponents and algebra. I was going through some sample papers of math, and I found this question which I cant solve:
If $abc=1$, find $1/(1+a+b^{-1})+1/(1+b+c^{-1})+1/(1+c+a^{-1})$.
Please help me with this.... I have spent almost 3 hours on this question...
Thanks for the help.
| $$\frac 1{1+a+b^{-1}}=\frac b{b+ab+1}=\frac b{1+b+c^{-1}}$$
The last equation follows because, $ab=1/c$.
In a similar manner,
$$\frac 1{1+c+a^{-1}}=\frac a{a+ac+1}=\frac a{a+b^{-1}+1}=\frac{ab}{1+b+c^{1}}$$
So, putting this altogether gives
$$\begin{align}
& \frac 1{1+a+b^{-1}}+\frac 1{1+b+c^{-1}}+\frac 1{1+c+a^{-1}} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve equation: $ \frac{x+1}{9x^3-4x} + \frac{3x}{27x^3 - 12x + 8} - \frac{1}{(3x-2)^2}=0 $ How to solve this equation?
$$
\frac{x+1}{9x^3-4x} + \frac{3x}{27x^3 - 12x + 8} - \frac{1}{(3x-2)^2}=0
$$
I try
$$
\frac{81x^5 - 81x^4 - 90 x^3 + 36 x^2 + 16x -16}{x(3x-2)^2(3x+2)(27x^3 - 12x + 8)}=0
$$
And then
$$
81x^... | the equation $$81 x^5-81 x^4-90 x^3+36 x^2+16 x-16=0$$ can only be solved by a numerical method, e.g. the Newton method
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
What base system satisfies the equation For the equation:
$$5x^2 - 50x +125 = 0$$
$x=5$ and $x=8$ are solutions.
This is in another base, what are the steps required to find out what base it is in?
Thanks
| Let $b$ be the base of the coefficients in the quadratic. Then, $(50)_b = 5b$ and $(125)_b = b^2+2b+5$.
Hence, the quadratic in base-$10$ is $5x^2-5bx+(b^2+2b+5) = 0$.
Since $x = 5$ is a solution, we have $125 - 25b + (b^2+2b+5) = 0$, i.e. $b^2-23b+130 = 0$.
Since $x = 8$ is a solution, we have $320 - 40b + (b^2+2b+5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx$ $$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=\frac{u}{v}$$ where $u$ and $v$ are in their lowest form. Find the value of $\dfrac{1000u}{v}$
$$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^2(x^2-1)+1}}dx$$ I put $x^2... | Given $$\displaystyle I = \int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx = \int_{1}^{2}\frac{x^2-1}{x^3\cdot x^2\sqrt{2-2x^{-2}+x^{-4}}}dx$$
$$\displaystyle = \int_{1}^{2}\frac{x^{-3}-x^{-5}}{\sqrt{2-2x^{-2}+x^{-4}}}dx$$
Now Let $2-2x^{-2}+x^{-4} = t^2\;,$ Then $\displaystyle \left(x^{-3}-x^{-5}\right)dx = \frac{2t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Prove that the ratio of the areas of the triangles $A'B'C'$ and $ABC$ is $2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$ If the bisectors of the angles of a triangle $ABC$ meet the opposite sides in $A',B',C'$,prove that the ratio of the areas of the triangles $A'B'C'$ and $ABC$ is $2\sin \frac{A}{2}\sin \frac{B}{2... | Let me try. One has $$\frac{B'A}{B'C} = \frac{BA}{BC} \Rightarrow \frac{B'A}{AC} = \frac{BA}{BA+BC},$$
$$\frac{C'A}{C'B} = \frac{CA}{CB} \Rightarrow \frac{C'A}{AB} = \frac{CA}{CA+CB}$$.
So, one has $$\frac{S_{AB'C'}}{S_{ABC}} = \frac{AB'.AC'}{AC.AC} = \frac{BA.CA}{(BA+BC)(CA+CB)}.$$
Similarly, one has
$$\frac{S_{A'BC'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Where does this sequence $\sqrt{7}$,$\sqrt{7+ \sqrt{7}}$,$\sqrt{7+\sqrt{7+\sqrt{7}}}$,.... converge? The given sequence is $\sqrt{7}$,$\sqrt{7+ \sqrt{7}}$,$\sqrt{7+\sqrt{7+\sqrt{7}}}$,.....and so on.
the sequence is increasing so to converge must be bounded above.Now looks like they would not excee... | Here's a quick proof that the sequence converges.
The sequence is given by the recursive formula $a_0 = 0$, and $a_{n+1} = \sqrt{7 + a_n}$.
Using the method suggested by the previous poster, let $L$ be the positive solution to the equation $x^2 = x + 7$.
We can prove that $a_n < L$ for all $L$ by induction. It is clea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Find the values of 'a' in a $4\times 4$ matrix(A) when the determinant is less than 2012 The matrix is $A \ =\begin{pmatrix}
7 & 1 & 3 & -2\\
-2 & 1 & -12 & -1 \\
1 & 16 & -4 & a \\
2 & 4 & 2 & 2 \\
\end{pmatrix}$
Where $\... | $A \ =\begin{pmatrix}
7 & 1 & 3 & -2\\
-2 & 1 & -12 & -1 \\
1 & 16 & -4 & a \\
2 & 4 & 2 & 2 \\
\end{pmatrix}\cong \begin{pmatrix}
13 & 1 & 5 & -5\\
-5 & 1 & -25 & -3 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sum\limits_{cyc}\sqrt{\frac{a+b}{c}}\ge2\sum\limits_{cyc}\sqrt{\frac{c}{a+b}}$ Let $a,b,c$ be positive numbers. Then we need to prove
$\sqrt{\frac{a+b}{c}}+\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}\ge2\left(\sqrt{\frac{c}{a+b}}+\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}\right).$
I have an idea to set $x=\fr... | $$\sum_{cyc}\sqrt{\frac{a+b}{c}}-2\sum_{cyc}\sqrt{\frac{c}{a+b}}=\sum_{cyc}\frac{a+b-2c}{\sqrt{c(a+b)}}=$$
$$=\sum_{cyc}\frac{b-c-(c-a)}{\sqrt{c(a+b)}}=\sum_{cyc}(a-b)\left(\frac{1}{\sqrt{b(a+c)}}-\frac{1}{\sqrt{a(b+c)}}\right)=$$
$$=\sum_{cyc}\frac{(a-b)^2c}{\sqrt{ab(a+c)(b+c)}\left(\sqrt{a(b+c)}+\sqrt{b(a+c)}\right)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the min and max distance from origin of the curve $\vert z+\frac{1}{z}\vert=a$ $z$ is a complex number, by the way.
I've tried a lot of things and always end up with a huge algebraic mess and I've wondered if anyone of you has any idea on how to approach this problem.
One of the most "satisfactory" things I've don... | Short Solution:
By the Triangle Inequality,
$$a=\left|z+\frac{1}{z}\right|\geq \left||z|-\frac{1}{|z|}\right|\,,$$
giving
$$-a\leq |z|-\frac{1}{|z|}\leq +a\,,$$
which means
$$\frac{\sqrt{a^2+4}-a}{2} \leq |z| \leq \frac{\sqrt{a^2+4}+a}{2}\,.$$
Since $z=\pm \frac{\sqrt{a^2+4}+a}{2}$ and $z=\pm\frac{\sqrt{a^2+4}-a}{2}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Understanding degrees of freedom in relation to rank for $\sum_{i=1}^{n}(y_i-\bar{y})^2$ So I'm looking at this website which states:
One of the questions an instrutor [sic] dreads most from a mathematically unsophisticated audience is, "What exactly is degrees of freedom?" It's not that there's no answer. The mathema... | The sum of all the $n$ columns is $(0,0,0,\ldots,0)^T$ and so surely the column rank cannot be $n$ as you claim it to be?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $ a + b + c \mid a^2 + b^2 + c^2$ then $ a + b + c \mid a^n + b^n + c^n$ for infinitely many $n$
Let $ a,b,c$ positive integer such that $ a + b + c \mid a^2 + b^2 + c^2$.
Show that $ a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer $ n$.
(problem composed by Laurentiu Panaitopol)
So far no i... | There's one more solution (it isn't mine). One can even prove that $(a + b + c) \mid (a^n + b^n + c^n)$ for all $n=3k+1$ and $n=3k+2$. It's enough to prove that
$a + b + c \mid a^n + b^n + c^n$ => $a + b + c \mid a^{n+3} + b^{n+3} + c^{n+3}$.
The proof is here:
https://vk.com/doc104505692_416031961?hash=3acf5149ebfb... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find $ab+cd$ given that $a^2+b^2=c^2+d^2=1$ and $ac+bd=0$? It is given that $a^2+b^2=c^2+d^2=1 $
And it is also given that $ac+bd=0$
What then is the value of $ab+cd$ ?
| If $b=0,ac=0\implies c=0\implies ab+cd=0$
Else $ac+bd=0\iff ac=-bd\iff\dfrac ab=\dfrac{-d}c=k$(say)
$\implies a=bk, d=-ck$
If $a^2+b^2=c^2+d^2,$ not necessarily $=1$
$b^2(1+k^2)=c^2(1+k^2)\implies b^2=c^2$ if $1+k^2\ne0$
Now $ab+cd=(bk)b+c(-ck)=k^2(b^2-c^2)=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 4
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How to find the value of $(a+b+c)(a+b+d)(a+c+d)(b+c+d)$ from the following equation? I have a question about polynomial.
Given a polynomial:
$$x^4-7x^3+3x^2-21x+1=0$$
Given too that the roots of this polynomial are $a, b, c,$ and $d$.
Find the value of $(a+b+c)(a+b+d)(a+c+d)(b+c+d)$?
My attempt:
It seems I need to app... | Let me try. You have $$(a+b+c)(a+b+d)(a+c+d)(b+c+d) = (7-a)(7-b)(7-c)(7-d) = 7^4 - 7^3(a+b+c+d) + 7^2(ab+ac+ad+bc+bd+cd)-7(abc+abd+bcd+acd) + abcd$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Closed-form of $\operatorname{Li}_2\left(1 \pm i\sqrt{3}\right)$ I've found the following identity while I was going through a quite difficult path.
$$
\Re\operatorname{Li}_2\left(1 \pm i\sqrt{3}\right) = \frac{\pi^2}{24} -\frac{1}{2}\ln^2 2 - \frac{1}{4}\operatorname{Li}_2\left(\tfrac{1}{4}\right),$$
where $\operato... | Here is a solution only using dilogarithm identities:
$$ \operatorname{Li}_2(z) + \operatorname{Li}_2(1-z) = \zeta(2) - \log z \log(1-z). \tag{1} $$
$$ \operatorname{Li}_2(1-z) + \operatorname{Li}_2(1-\tfrac{1}{z}) = -\tfrac{1}{2}\log^2 z. \tag{2} $$
$$ \operatorname{Li}_2(z) + \operatorname{Li}_2(-z) = \tfrac{1}{... | {
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"url": "https://math.stackexchange.com/questions/1410276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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For Which Value The Matrix is Diagonalizable?
For which values of $a$ the matrix $\left(\begin{array}{ccc} 2 & 0 & 0 \\ 2 & 2 & a \\ 2 & 2 & 2 \end{array}\right)$ is diagonalizable:
*
*above $\mathbb{R}$
*above $\mathbb{C}$
We need to look at the characteristic polynomial which is $(x-2)^3-2a(x-2)=x^3-6x^2+2x(6-a)... | $x^3-6x^2+2x(6-a)+4(a-2)$. Clearly it shows that $x=2$ is a zero of the polynomial. Now ,
$x^3-6x^2+2x(6-a)+4(a-2)=0$
$\implies (x-2)(x^2-4x+4-2a)=0$
$\implies x=2 , x=\frac{4\pm\sqrt{16-4(4-2a)}}{2}=2\pm\sqrt{2a}.$
If $a=0$ then , all roots are $2$ and then the matrix is NOT diagonalizable(check!)
If $a\not=2$ then ... | {
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"url": "https://math.stackexchange.com/questions/1410351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to compute fraction sums? For example, $$\sum\limits_{k=1}^{n}\frac{1}{(2k-1)(2k+1)}=\frac{n}{2n+1}$$
Is there an easier way to evaluate fraction sums (without using partial sums)?
| A general approach which works quite well is partial fraction decomposition:
$$
\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}\left(\frac{1}{(2k-1)}-\frac{1}{(2k+1)}\right)
$$
After this, you can use the technique of telescoping.
In your example, this yields:
\begin{align}
& \sum_{k=1}^{n}{\frac{1}{(2k-1)(2k+1)}} = \sum_{k=1}^{n}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Trying to solve $\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$ The equation is
$$\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$$
I solve it thus:
$$
\begin{cases}
2\cos^2(x)-\sqrt3=2\sin^2(x) \\
-\sqrt2 \sin(x)\ge 0 \iff \sin(x)\le 0
\end{cases}
$$
The first equation boils down to
$$2\cos^2(x)=2(1-\cos^2(x))+\sqrt3$$
$$4c... | Take advantage of the fact that $\cos^2{x}=1-\sin^2{x}$. Then solve for $\sin{x}$, then solve for $x$. $\frac{\pi}{12}=15^o$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Limits using definite integration $F(k)$ = $$ \lim_{n\to \infty}{\frac{1^k + 2^k +...+n^k}{(1^2 + 2^2 +...+n^2)*(1^3 + 2^3 +...+n^3)}} $$
I need help in finding $F(5)$ and $F(6)$.
I tried converting it into summation form and using the progression formulas of $n^2$ and $n^3$ but it was of no use.
| Theorem: Define $S_r(n)=1^r+2^2+...+n^r$, $$\frac{n^{r+1}}{r+1}\leq S_r(n)\leq\frac{(n+1)^{r+1}}{r+1}$$ for any positive $r\geq0$
Then $$\frac{12n^{k+1}}{(k+1)(n+1)^3(n+1)^4}\leq{\frac{1^k + 2^k +...+n^k}{(1^2 + 2^2 +...+n^2)\cdot(1^3 + 2^3 +...+n^3)}}\leq \frac{12(n+1)^{k+1}}{(k+1)n^3n^4}$$
So, $$F(6)=\frac{12}{7}$$ a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that the function is continuous To show that the function $f: \mathbb{R}^2 \rightarrow\mathbb{R}$ with $f=\left\{\begin{matrix}
\frac{x^3-y^3}{x^2+y^2} & , (x,y) \neq (0,0)\\
0 & , (x,y)=(0,0)
\end{matrix}\right.$ is continuous on $(0,0)$ we have to show that $|f(x,y)-f(x_0,y_0)| \leq L ||(x,y)-(x_0,y_0)||$ so ... | You can take also another approach. Since $x^3-y^3=(x-y)(x^2+y^2+xy)$ you need show that $\frac{xy(x-y)}{x^2+y^2}$ goes to $0$ when $(x,y)\to 0$. So, call $x=r\cos \theta$, $y=\sin \theta$ and $r^2=x^2+y^2$. We have $\frac{xy(x-y)}{x^2+y^2}=\frac{r^3(\cos^2 \theta \sin \theta-\cos \theta \sin^2\theta)}{r^2}$ and this g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trouble understanding inequality proved using AM-GM inequality I am studying this proof from Secrets in Inequalities Vol 1 using the AM-GM inequality to prove this question from the 1998 IMO Shortlist. However, I'm lost on the very first line of the solution.
Let $x,y,z$ be positive real numbers such that $xyz =1$. P... | I hope one of the $3$ parts below would help you understand the proof above.
Part 1: the AM-GM ineq is: $\dfrac{x^3}{(1+y)(1+z)}+\dfrac{1+y}{8}+\dfrac{1+z}{8} \geq 3\sqrt[3]{\dfrac{x^3(1+y)(1+z)}{(1+y)(1+z)\cdot 8\cdot 8}} = \dfrac{3x}{4}$.
Part 2: $\sum_{cyclic} \left(\dfrac{1+y}{8}+\dfrac{1+z}{8}\right) = 2\sum_{cycl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Does the limit $\lim\limits_{x\to0}\left(\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}\right)$ exist? Does the limit: $$\lim\limits_{x\to0}\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}$$ exist?
| \begin{align}
& \lim_{x\to0} \left( \frac 1 {x \arctan x} - \frac 1 {x^2} \right) = \lim_{x\to0} \frac {x - \arctan x} {x^2 \arctan x} \\[10pt]
= {} & \lim_{x\to0} \frac{1 - \frac 1 {1+x^2}}{\frac{x^2}{1+x^2} + 2x \arctan x } \qquad \text{(L'Hopital)} \\[10pt]
= {} & \lim_{x\to0} \frac{x^2}{x^2 + 2x(1+x^2) \arctan x} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
trying to solve $\sqrt{\cos(x)-2\cos(2x)}+\sqrt{2}\cos(2x)=0$ The equation is
$$\sqrt{\cos(x)-2\cos(2x)}+\sqrt{2}\cos(2x)=0$$
The system is
$$
\begin{cases}
\cos(x)-2\cos(2x)=2\cos^2(2x) \\
-\sqrt{2}\cos(2x)\ge 0 \iff \cos(2x)\le 0
\end{cases}
$$
The equation:
$$\cos(x)-2(2\cos^2(x)-1)=2(2\cos^2(x)-1)^2$$
$$\cos(x)-4\c... | Hint: You can factorise: $$8c^4-4c^2-c=c(2c+1)(4c^2-2c-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find all integers $m,n$ for which $m^2+n^2$ is a square and $\sqrt{\frac{2m^2+2}{n^2+1}}$ is rational This is a repost of my old question here. The question is as follows:
Find all integers m and n, such that $m^2 + n^2$ is a square and $\sqrt{\frac{2(m^2+1)}{n^2+1}}$ is rational.
I have made no progress on this pro... | Here is my amateur attempt to your problem.
The number is rational iff there exist integers $\,p, q\in \mathbb N\,$ such that $\displaystyle\,\sqrt{\frac{k\left(m^2+1\right)}{n^2+1}}=\frac{p}{q}.\,$
$\,m^2 + n^2\,$ is square $\iff$ there exist an integer $\,r\in \mathbb N\,$ such that $\,m^2 + n^2 = r^2\,$
Thus we get... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Nonseparable differential equations How can I solve the equation
$\frac{dy}{dx} =\frac{x^2-y}{x-y^2}$?
I've tried few substitutions such as $y=xv$ and $y=x/v$ but all to no avail!
Please, help.
| $$\frac { dy }{ dx } =\frac { x^{ 2 }-y }{ x-y^{ 2 } } \\ \quad \quad \quad \quad $$ $$\Downarrow \\$$$$ \left( x^{ 2 }-y \right) dx-\left( x-y^{ 2 } \right) dy=0\\ \quad \quad \quad \quad \quad $$$$\Downarrow \\ \left( { x }^{ 2 }dx+{ y }^{ 2 }dy \right) -\left( ydx+xdy \right) =0\\ \quad \quad \quad \quad \quad \quad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1415892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving that $\frac{1}{2}<\frac{2}{3}<\frac{3}{4}<$...$<\frac{n-1}{n}$ In an attempt to find a pattern, I did this:
Let a,b,c,d be non-zero consecutive numbers. Then we have:
$a=a$
$b=a+1$
$c=a+2$
$d=a+3$
This implies:
$\frac{a}{b}=\frac{a}{a+1}$
$\frac{b}{c}=\frac{a+1}{a+2}$
$\frac{c}{d}=\frac{a+2}{a+3}$
I don't know ... | Here's an idea. In general you want to show that
$$\frac{a}{a+1} < \frac{a+1}{a+2}.$$
What happens if you multiply $\frac{a}{a+1}$ by $\frac{(a+1)^2}{a+2}$? What do you know about the number $\frac{(a+1)^2}{a+2}$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
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