Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Show $60 \mid (a^4+59)$ if $\gcd(a,30)=1$
If $\gcd(a,30)=1$ then $60 \mid (a^4+59)$.
If $\gcd(a,30)=1$ then we would be trying to show $a^4\equiv 1 \mod{60}$ or $(a^2+1)(a+1)(a-1)\equiv 0 \mod{60}$. We know $a$ must be odd and so $(a+1)$ and $(a-1)$ are even so we at least have a factor of $4$ in $a^4-1$. Was thinkin... | The question is answered successfully, I would like find the number for which $a^4\equiv1\pmod n$ for all $a$ such that $(a,n)=1$
Using Carmichael Function, If $\lambda(n)=2^m$ where integer $m\ge 1$
$n$ can not have any odd prime factor with exponent $>1$ and for each prime factor $p_i,p_i-1$ must be of the form $2^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/398656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Can anyone show me, how to solve these system of Equations: $$\begin{align*}
x+y+z &= 2 \\
(x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y) &= 1 \\
x^2(y+z)+y^2(z+x)+z^2(x+y) &= -6
\end{align*}$$
Can anyone explain me the solution. I asked it in mathoverflow but god knows why it was closed. Please help and detail the process in step b... | $$x+y+z=2$$
$$x^2+y^2+z^2+2(xy+zx+yz)=4$$
The second equation is $x^2+y^2+z^2+3(xy+zx+zy)=1 \implies zx+yz+yx=-3$
You get $x^2+y^2+z^2=10$
Using the $x+y+z=2$ in the third equation.
$x^2(2-x)+y^2(2-y)+z^2(2-z)=-6 \implies 2(x^2+y^2+z^2)-x^3+z^3+z^3=-6$
A little re-arranging would give
$26 =x^3+y^3+z^3 $
$a^3+b^3+c^3-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/399915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Solving $\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$ Where do I start to solve a equation for x like the one below?
$$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$$
After squaring it, it's too complicated; but there's nothing to factor or to expand?
Ideas?
| Note that
$$ \sqrt{7x-4} - \sqrt{7x-5} = \frac{(7x-4)-(7x-5)}{\sqrt{7x-4} + \sqrt{7x-5}} = \frac{1}{\sqrt{7x-4} + \sqrt{7x-5}} $$
and likewise
$$ \sqrt{4x-1} - \sqrt{4x-2} = \frac{1}{\sqrt{4x-1} + \sqrt{4x-2}}. $$
Plugging these to our equation, we obtain
$$ \sqrt{7x-4} + \sqrt{7x-5} = \sqrt{4x-1} + \sqrt{4x-2}. $$
Thu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/400395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Find the limit without use of L'Hôpital or Taylor series: $\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)$ Find the limit without the use of L'Hôpital's rule or Taylor series
$$\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2x}\right)$$
| Well, you have
$$\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right) = \left(\frac{1}{x} - \frac{1}{\sin x}\right)\left(\frac{1}{x} + \frac{1}{\sin x}\right)$$
$$=\left(\frac{1}{x^2} - \frac{1}{x\sin x}\right)\left(1+ \frac{x}{\sin x}\right)$$
Since the limit of $\left(1+ \frac{x}{\sin x}\right)$ is $2$, your answer will be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/400541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
What is the least nonnegative number $a$ congruent to $3^{340}\pmod{341}$? Find the least nonnegative number $a$ congruent to $3^{340} \pmod{341}$.
What steps should I take to get to the answer?
| As $341=11\cdot31,$
Using Fermat's Little Theorem, $ 3^{10}\equiv1 \pmod {11}$
$\implies 3^{340}=(3^{10})^{34}\equiv1^{34}\equiv1\pmod{11}\ \ \ \ (1)$
Again applying Fermat's Little Theorem, $3^{30}\equiv1\pmod {31}$
$\implies 3^{340}=(3^{30})^{11}\cdot 3^{10}\equiv 1\cdot3^{10}\equiv 3^{10} \pmod {31}$
Now, $3^3\equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/401259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving that $\int_0^{2\pi}\frac{ 1-a\cos\theta}{1-2a\cos\theta + a²} \mathrm d\theta = 0$ for $|a|>1$
Let $|a|>1$. Show that $$\int_0^{2\pi}\frac{ 1-a\cos\theta}{1-2a\cos\theta + a²} \mathrm d\theta = 0$$
I'm also trying to solve this problem. I wasn't sure if I was supposed to start a new thread or contribute to pr... | The standard complex analysis approach is to make the substitution $z = e^{i\theta}$. This will map $[0,2\pi]$ onto the unit circle traversed counter-clockwise once. Using Euler's formulas,
$$ \cos \theta = \frac{e^{i\theta}+e^{-i\theta}}2 = \frac{z+z^{-1}}{2}.$$
We also have $dz = ie^{i\theta}\,d\theta$, i.e.
$$d\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/401588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $5^n - 2^n$ is divisible by $3$ for all nonnegative integers $n$ using mathematical induction Using mathematical induction, prove for all integers n 1 that $5^n - 2^n$ is divisible by 3.
Can someone help me with this?
| This one could be solved by induction but is it much neater if you use that
\[ a^n-b^n = (a-b) \cdot \sum_{i=0}^{n-1} a^i\cdot b^{n-1-i} \]
If you use this one in your problem you see that
\[ 5^n- 2^n = (5-2) \cdot \sum_{i=0}^{n-1} 5^i \cdot 2^{n-1-i}=3 \cdot \sum_{i=0}^{n-1} 5^i \cdot 2^{n-1-i}\]
Which can be dived... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/402731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the fraction $\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}$ when knowing the sums $a+b+c+d$ to $a^4+b^4+c^4+d^4$ How can I solve this question with out find a,b,c,d
$$a+b+c+d=2$$
$$a^2+b^2+c^2+d^2=30$$
$$a^3+b^3+c^3+d^3=44$$
$$a^4+b^4+c^4+d^4=354$$
so :$$\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}=?$$
If the qusetion i... | One way to approach such questions is to view $a, b, c, d$ as roots of a single quartic equation $x^4-s_1x^3+s_2x^2-s_3x+s_4=0$, when we have Vieta's relations
$$s_1=a+b+c+d$$$$s_2=ab+ac+ad+bc+bd+cd=(a+b)(c+d)+ab+cd$$$$s_3=abc+abd+acd+bcd$$$$s_4=abcd$$
We then let $p_k=a^k+b^k+c^k+d^k$ with: $$p_0=4$$$$p_1-s_1=0$$$$p_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/402856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 1
} |
Simple divisibility problem
Suppose $a$ and $b$ are distinct integers and $n\mid a^n-b^n$. Prove that $n\mid\frac{a^n-b^n}{a-b}$.
Here is how I do it: Write $a=b+d$, then $a^n-b^n=(b+d)^n-b^n=\binom{n}{1}b^{n-1}d+\binom{n}{2}b^{n-2}d^2+ \cdots + \binom{n}{n-1}bd^{n-1}+d^n \equiv 0 \pmod{n}$. Since $n| \binom{n}{k}$ f... | We will repeatedly use the trick that if $n=xy$, then $a^n$ is also an $x$th power.
We will approach this by first showing it is true for prime powers.
Base case: $p$ is a prime and $p | a^p - b^p$. Then, we have $a \equiv a^p \equiv b ^p \equiv b \pmod{p}$, and hence $\frac{a^p - b^p}{a-b} \equiv p a^{p-1} \equiv 0 \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/405913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating the time average over energy For more info see the article equations 37
Edit: The $\varepsilon ^3 $ has vanished due to time average. But how to get the 4th order?
Let us define some function for scalar field
$$\phi= \sum_{k=1}^\infty\varepsilon^k \phi_k \tag{1}=\varepsilon^1 \phi_1+\varepsilon^2 \phi_2 +\... | Look,
$$
{\cal{E}} = \frac{1}{2} \left(\partial_t \phi\right)^2
+ \frac{1}{2} \left(\partial_i \phi\right)^2
+U(\phi)$$
Now fill in the expansion
$$\phi= \varepsilon^1 \phi_1+\varepsilon^2 \phi_2 +\varepsilon^3 \phi_3 + \ldots$$
this should give
$${\cal{E}} = \frac{1}{2} \left(\varepsilon^1 \partial_t\phi_1+\varepsi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/407665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
A geometric reason why the square of the focal length of a hyperbola is equal to the sum of the squares of the axes? When I teach conics, I give a simple geometric argument for the Pythagorean relation for ellipses. (By "Pythagorean relation" I mean the fact that the square of the focal length $c$ is the difference of ... | Since the hyperbola is defined as the locus where the difference of the two distances is constant, I don't see any analogous argument. But certainly a limit argument does work: As $(x,y)\to \infty$ (with $x>0$), we have $\left|\frac yx\right|\to\frac ba$, and so
\begin{align*}
\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}&=\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/408954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 5,
"answer_id": 3
} |
Finding another way of doing this integral $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$ Problem :
Integrate : $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$
I have the solution : We can substitute $\sqrt{x}= \cos^2t$ and proceeding further,
I got the the answer which is $-2\sqrt{1-x}+\cos^{-1}\sqrt{x}+\sqrt{x-x^2}+C$
Can we do th... | Write $y = \sqrt{(1-\sqrt{x})/(1+\sqrt{x})}$. Then $y^2 = (1 - \sqrt{x})/(1+\sqrt{x}) = -1 + 2/(1+\sqrt{x}),$
so
$2/(1+y^2) = 1 + \sqrt{x},$
and so
$$x = \Bigl(\dfrac{2}{1+ y^2} - 1\Bigr)^2.$$
Thus $$y\, dx = y \, d\Bigl(\dfrac{2}{1+y^2} - 1\Bigr)^2
= -8 y^2 \Bigl(\dfrac{2}{1+y^2} - 1\Bigr)\dfrac{dy}{(1+y^2)^2}.$$
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/409829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
A Challenging Euler Sum $\sum\limits_{n=1}^\infty \frac{H_n}{\tbinom{2n}{n}}$ Recently, I encountered a strange series involving Harmonic Numbers and Binomial Coefficients both.
According to Mathematica:
$$\displaystyle \sum_{n=1}^\infty \frac{H_n}{\binom{2n}{n}} = -\frac{2\sqrt{3} \pi}{27}(\log (3)-2)+\frac{2}{27} \le... | Note: there is a minus sign missing in front of the 1st term on the left side of your evaluation.
Put the integral into the form
$$I=\int_0^{1}\frac{\left(y^2-y+1\right)\ln\left(y^2-y+1\right)-2\ln\left(y^2-y+1\right)-2\left(y^2-y+1\right)+2}{\left(y^2-y+1\right)^2}dy.$$
Making the change of variable $y=\frac12+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/409909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 2,
"answer_id": 0
} |
Integrals using Arctangens We want to find $\displaystyle \int\dfrac{12}{16x^2 +1}$
I rewrote it to the form $ 3 \cdot \dfrac{1}{u^2 + 1} \cdot u' $ where $u=4x$.
I found out that the correction sheet does the same thing, but their next step leaves my puzzled:
$$ F(x) = 3 \arctan (4x) + C$$
Where did the $u' = 4$ go to... | As you noted, if $u = 4x$, then $du = 4dx$, yielding
$$\begin{split}
\int \frac{12dx}{16x^2+1}
&= 3 \int \frac{4dx}{(4x)^2+1} \\
&= 3 \int \frac{u'dx}{(u)^2+1} \\
&= 3 \int \frac{du}{u^2+1} \\
&= 3 \arctan(u) + C \\
&= 3 \arctan(4x)+C
\end{split}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/410961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Limit with roots I have to evaluate the following limit:
$$ \lim_{x\to 1}\dfrac{\sqrt{x+1}+\sqrt{x^2-1}-\sqrt{x^3+1}}{\sqrt{x-1}+\sqrt{x^2+1}-\sqrt{x^4+1}} . $$
I rationalized both the numerator and the denominator two times, and still got nowhere. Also I tried change of variable and it didn't work.
Any help is gratefu... | To ease typing, we temporarily manipulate the top and bottom separately.
The top is $\sqrt{(x-1)(x+1)}+\left(\sqrt{x+1}-\sqrt{x^3+1}\right)$ (we changed the order). "Rationalize" the part in parentheses. The top becomes
$$\sqrt{(x-1)(x+1)}-\frac{x(x-1)(x+1)}{\sqrt{x+1}+\sqrt{x^3+1}}.\tag{1}$$
Do the same rationalizi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/411676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
no. of solution of the equation $[x]^2+a[x]+b = 0$ is If $a$ and $b$ are odd integer. Then the no. of solution of the equation $[x]^2+a[x]+b = 0$ is
where $[x] = $ greatest Integer function
My Try:: Let $[x] = y$. Then equation become $y^2+ay+b = 0$
Now If given equation has real Roots, Then $\displaystyle y = \frac{... | If $y := \lfloor x \rfloor$ is even, $y^2$ is even, $ay$ is even, $b$ is odd, hence $y^2 + ay + b$ is odd, so $y^2 + ay+b \ne 0$. If on the other hand, $y$ is odd, $y^2$ and $ay$ are odd also, as is $b$, so $y^2 + ay+ b$ is odd again. Hence $y^2 + ay + b$ is an odd integer for every integer $y$. So the equation does n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/413655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solve this $\;2\sqrt[4]{\frac{x^4}{3} + 4} = 1 + \sqrt {\frac{3}{2}} |y| \ldots$ Solve
$\begin{cases}
2\sqrt[4]{\frac{x^4}{3} + 4} = 1 + \sqrt {\frac{3}{2}} |y|\\[8pt]
2\sqrt[4]{\frac{y^4}{3} + 4} = 1 + \sqrt {\frac{3}{2}} |x|
\end{cases}$
| Following Ron Gordon's suggestion, and noting that the if $(x,y)$ is a solution, then $(\pm x, \pm y)$ are all solutions, we need to solve:
$$2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}}x$$
Both sides to the fourth power:
$$\frac{16x^4}{3}+64=1+2\sqrt6x+9x^2+3\sqrt6x^3+\frac{9x^4}{4}\\$$
It turns out that this expre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/413971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of $\lim\limits_{n\to \infty}\left(\frac{1^\frac{1}{3}+2^\frac{1}{3}+3^\frac{1}{3}+\dots+n^\frac{1}{3}}{n\cdot n^\frac{1}{3}} \right)$ I want to evaluate this limit.
$$\lim_{n\to \infty}\left(\frac{1^\frac{1}{3}+2^\frac{1}{3}+3^\frac{1}{3}+\dots+n^\frac{1}{3}}{n\cdot n^\frac{1}{3}} \right)$$
What I did is:
set $f... | Hint: An integral over $[0,\infty)$ does not look right to me, it seems better to observe that with your $f$ as above
\begin{align*}
\frac{1^{1/3} + \cdots + n^{1/3}}{n \cdot n^{1/3}} &= \frac{f(1/n) + f(2/n) + \cdots + f(1)}{n}\\
&= \sum_{i=1}^n f(i/n) \cdot \frac 1n
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/414807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find the product of the following determinants (involving logarithms with different bases) Find the product of the following determinants:
$$\begin{vmatrix}
\log_3512 & \log_43 \\
\log_38 & \log_49
\end{vmatrix} * \begin{vmatrix}
\log_23 & \log_83 \\
\log_34 & \log_34
\end{vmatrix}$... | HINT:
Putting $\log_32=a,$
$\log_3512=9\log_32=9a, \log_38=3\log_32=3a,$
$\log_43=\frac{\log 3}{\log 4}=\frac1{\frac{\log 4}{\log 3}}=\frac1{\log_34}=\frac1{2\log_32}=\frac1{2a}$ and so on
$$\begin{vmatrix}
\log_3512 & \log_43 \\
\log_38 & \log_49
\end{vmatrix} =\begin{vmatrix}
9a & \frac1{2a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/415246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Finding the Equation of Parabola
Write the equation in the form $y=a(x-h)^{2}+k$ with zeros -4 and 8, and an optimal value of 18.
I'm not sure what "optimal value" means first of all- I think it means that the maximum value has a y-value of 18. What I've done so far:
$y=a(x+4)(x-8)$. Then to calculate the x-value of ... | Use factored form: $y=a(x+s)(x+t)$
$y=a(x+4)(x-8)$
'Optimal Value'= $y$ value of $(x,y)$
$18$ = $y$ value of $(x,y)$
$x$ value of $(x,y)= (s+t)/2
= (-4+8)/2
= 4/2
= 2$
Therefore $(x,y) = (2,18)$
Substitute $(2,18)$ into $y=a(x+4)(x-8)$
$18=a(2+4)(2-8)$
$18=a(6)(-6)$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/416931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
trouble expanding taylor series about a point other than zero using geometric series I'm trying to understand how to use a Taylor series expansion to correctly expand a population growth function about a point other than zero using the geometric series.
For expansion about $t=0$, I get:
$$
\begin{align}
p(t) &= 2 ln (t... | You can expand $2\log(t+4)$ around $t=0$ and then plug in $t = x-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the range of $ f(x)=9^x - 3^x+1$
Problem:-Find range of function $ f(x)=9^x - 3^x+1$, here the domain of $f$ is $\mathbb R$.
Solution: $ f(x)=9^x - 3^x+1$. Let $f(x)=y$. Then
$$ \begin{split}y&=9^x - 3^x+1\\
y&=3^{2x} - 3^x+1
\end{split}$$
Let $3^x= u$. Then $ y=u^2 - u+1$, so
$$ u^2 - u+1-y=0.$$
Am I do... | Looks good. To finish it off, we complete the square and notice that:
$$
y = u^2 - u + 1 = \left(u^2-u+\dfrac{1}{4}\right)+\dfrac{3}{4} = \left(u-\dfrac{1}{2}\right)^2+\dfrac{3}{4} \ge 0+\dfrac{3}{4}=\dfrac{3}{4}
$$
since the square of any real number must be nonnegative. Hence, the range is:
$$
\{y\in \mathbb{R} \mid ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/419271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find a minimal polynomial (field theory) I was asked to find a minimal polynomial of $$\alpha = \frac{3\sqrt{5} - 2\sqrt{7} + \sqrt{35}}{1 - \sqrt{5} + \sqrt{7}}$$ over Q.
I'm not able to find it without the help of WolframAlpha, which says that the minimal polynomial of $\alpha$ is $$19x^4 - 156x^3 - 280x^2 + 2... | $$\alpha-\alpha\sqrt{5}+\alpha\sqrt{7}=3\sqrt{5}-2\sqrt{7}+\sqrt{35},$$
$$\alpha-\sqrt{35}=(\alpha+3)\sqrt{5}-(\alpha+2)\sqrt{7},$$
$$(\alpha-\sqrt{35})^2=[(\alpha+3)\sqrt{5}-(\alpha+2)\sqrt{7}]^2,$$
$$\alpha^2+35-2\alpha\sqrt{35}=5(\alpha+3)^2+7(\alpha+2)^2-2\sqrt{35}(\alpha+2)(\alpha+3),$$
$$2\sqrt{35}(\alpha^2+4\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/422233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 2
} |
Induction to prove that $1^2+ 3^2 + 5^2+ \cdots +(2n + 1)^2=\frac{(n+1)(2n+1)(2n + 3)}{3}$ Have I started this right? I know I have to add $(k+1)$ but why?
Use mathematical induction to prove that $$1^2+ 3^2 + 5^2+ \cdots +(2n + 1)^2=\frac{(n+1)(2n+1)(2n + 3)}{3}.$$
BASIS STEP: $P(0)$ is true since $$\begin{align}(2(... | You need to assume that it holds for some $k\ge0,$ not some $k>0$, for otherwise the induction doesn't go through (and you've only proved it for $k=0$).
Now, given the assumption that $$1^2+ 3^2 + 5^2+ \cdots +(2k + 1)^2=\frac{(k+1)(2k+1)(2k + 3)}{3},$$ you need to add $\bigl(2(k+1)+1\bigr)^2$ to both sides, and show t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/422359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Indefinite integration : $\int \frac{1+x-x^2}{\sqrt{(1-x^2)^3}}$ Problem :
Solve : $\int \frac{1+x-x^2}{\sqrt{(1-x^2)^3}}$
I tried :
$\frac{1-x^2}{\sqrt{(1-x^2)^3}} + \frac{x}{\sqrt{(1-x^2)^3}}$
But it's not working....Please guide how to proceed
| you had a good guees! see
$$\int \frac{1+x-x^2}{(1-x^2)^{\frac{3}{2}}}dx = \int \frac{1-x^2}{(1-x^2)^{\frac{3}{2}}}+\frac{x}{(1-x^2)^{\frac{3}{2}}}dx
=\int (1-x^2)^{\frac{-1}{2}}dx+\int\frac{x}{(1-x^2)^{\frac{3}{2}}}dx$$
note that $[(1-x^2)^{\frac{-1}{2}}] '= -\frac{1}{2}(1-x^2)^\frac{-3}{2}(-2x)= \frac{x}{(1-x)^{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/423286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
What is the $8^{th}$ term of $\left(3x-\frac{y}{2}\right)^{10}$? What is the $8^{th}$ term of $\left(3x-\frac{y}{2}\right)^{10}$?
My solution: I'am not sure if I'am correct :)
$^{10}C_r (3x)^{10-r} \left(-\frac{y}{2}\right)^r$
where $r= 7$ since we start at $r=0$
$^{10}C_7 (3x)^3 \left(-\frac{y}{2}\right)^7$
$\left(-12... | $$(3x-(1/2)y)^{10}$$ general term is $$T_k=\binom{n}{k}a^{n-k}b^k$$ in our case we have
$$a=3x,b=-\frac{1}{2}y,n=10,k=7$$
so
$$T_7=\binom{10}{7}(3x)^3(-\frac{1}{2}y)^7=-\frac{405}{16}x^3y^7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/425594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Differentiation and Rate of Change Im slaving over this one
using rate of change and Differentiation
I tried the chain rule but that is not the route as it got passed back
Question: find rate of change v if
V = x^3 * y^2
where
x=cos3t and y =sin3t
which gives
v = cos^3(3t) * sin^2(3t)
This is what i have so far
dv/dx... | Since $x=\cos 3t$ and $y = \sin 3t$, by chain rule we have:
$$
\dfrac{dx}{dt} = -3 \sin 3t \qquad \text{and} \qquad \dfrac{dy}{dt} = 3 \cos 3t
$$
Hence, by product rule, we have:
$$ \begin{align*}
\dfrac{dV}{dt}
&= \left(3x^2 \dfrac{dx}{dt}\right)(y^2) + (x^3)\left(2y \dfrac{dy}{dt}\right) \\
&= (3\cos^2 3t) (-3 \sin 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/426025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Explanation of an example of linear transformation This is an example from a text (Linear Algebra, Freidberg). I am trying to follow along, and I feel like I should know this from vector calc but I am missing something silly.
The example is:
Define the Linear transformation $T:P_2(R)\rightarrow M_{2x2}(R)$ by
$$T(\... | Your transformation is from the space of polynomials of degree at most 2 to the space of $2\times 2$ matrices with real coefficients.
They tell you that the polynomials $1$, $x$, and $x^2$ are a basis for $P_2(\mathbb{R})$; the next three matrices are just the results of applying your transformation to these three poly... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/427137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving for $a,b,c,d$ where $a^2 + b^2 + c^2 + d^2 = 630^2$ How could one solve for $a,b,c,d$ where:
$$a^2 + b^2 + c^2 + d^2 = 630^2,\ a>b>c>d$$
$a,b,c,d$ squared is equal to the square of $630$, and $a$ is larger than $b$, and so forth.
$a,b,c,d$ is also of such form that:
$$n = x + (x+1) + (x+2) + (x+3), \ x\in\mathb... | Since $a, b, c, d$ are all of the form $4x+6$ for some $x \in \mathbb{N}$, we see that $a,b,c,d \equiv 2 \pmod 4$. It follows that $a^2+b^2+c^2+d^2 \equiv 4+4+4+4 \equiv 0 \pmod 8$. However, $630^2 \equiv 4 \pmod 8$. Therefore there is no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/427742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Factor $x^4 - 11x^2y^2 + y^4$ This is an exercise from Schaum's Outline of Precalculus. It doesn't give a worked solution, just the answer.
The question is:
Factor $x^4 - 11x^2y^2 + y^4$
The answer is:
$(x^2 - 3xy -y^2)(x^2 + 3xy - y^2)$
My question is:
How did the textbook get this?
I tried the following methods (exa... | A start: Divide the original by $y^4$, and set $z=x/y$. We are trying to factor $z^4-11z^2+1$ as a product of two quadratics. Without loss of generality we may assume they both begin with $z^2$. So we are looking for a factorization of type $(z^2+az+b)(z^2+cz+d)$. By looking at the coefficient of $z^3$ in the product, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/430602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 0
} |
Express $\cos 6\theta $ in terms of $\cos \theta$ I think I'm supposed to use the chebyshev polynomials, as in $$ \cos n \theta = T_n(x) = \cos(n \arccos x)$$
But no idea what now?
| Since $$\cos 2a =2\cos ^{2}a -1, \qquad\sin 2a
=2\sin a\cos a,$$ $$\cos (a+b)=\cos a\cos b-\sin a\sin b,$$
and
$$
\begin{eqnarray*}
\cos 3\theta &=&\cos (2\theta +\theta ) \\
&=&\cos 2\theta \cos \theta -\sin 2\theta \sin \theta \\
&=&( 2\cos ^{2}\theta -1) \cos \theta -2\sin ^{2}\theta \cos
\theta \\
&=&( 2\cos ^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/432254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
If $A = \tan6^{\circ} \tan42^{\circ},~~B = \cot 66^{\circ} \cot78^{\circ}$ find the relation between $A$ and $B$ My trigonometric problem is:
If $A = \tan6^{\circ} \tan42^{\circ}$ B = cot$66^{\circ} \cot78^{\circ}$ find the relation between $A$ and $B$.
Working :
$$B = \cot 66^{\circ} \cot78^{\circ} = 1- \frac{\tan... | $$A=Tan(6^\circ)\,Tan(42^\circ)=Tan(2.3^\circ)Tan(45^\circ-3^\circ)=\frac{2Tan(3^\circ)}{1-Tan^2(3^\circ)}\frac{1-Tan(3^\circ)}{1+Tan(3^\circ)}=\frac{2Tan(3^\circ)}{\left(1+Tan(3^\circ)\right)^2}$$
$\:$
$$ B=Tan(24^\circ)\,Tan(12^\circ)=Tan(2.12^\circ)Tan(12^\circ)=\frac{2Tan^2(12^\circ)}{1-Tan^2(12^\circ)} \implies$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/432322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How can one calculate this diferential equation $y\ dy = 4x(y^2+1)^2\ dx \text{ and } y(0) = 1$
I'm trying:
$\dfrac{y\ dy}{ (y^2+1)^2 }= 4x\ dx$
but I can't figure out what to do now.
| Good start, you separated the variables. Now integrate.
For the integral of the left-hand side, make the substitution $u=y^2+1$. You should arrive at
$$\int \frac{1}{2} \cdot\frac{1}{u^2} \,du,$$
which is easy.
The integral of the right-hand side is $2x^2+C$. Use the initial condition to find $C$.
Added detail: Let $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/432460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Is this function bounded? Next question about integral $\int_{\partial M} \frac{1}{||y-x||} n_y \cdot \nabla_y \frac1{||y-x||} dS_y$.
Let $\partial M$ be $C^2$ closed surface in $\mathbb{R}^3$, $M$ is open. Show that
$$
f(x) = \frac{\int_{\partial M} \left| \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} \r... | Why not thinking it the other way around? Integrals like this are typical of electromagnetic theory.
Let
\begin{multline}
\iint_M \mbox{div}\left(\frac{1}{\|y-x\|}\nabla_y \frac{1}{\|y - x\|}\right) dy = \\
\iint_M \left\{\frac{1}{\|y-x\|}\Delta_y \frac{1}{\|y-x\|} + \left\|\nabla_y \frac{1}{\|y-x\|}\right\|^2\right\}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/435472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Covergence of $\frac{1}{4} + \frac{1\cdot 9}{4 \cdot 16} + \frac{1\cdot9\cdot25}{4\cdot16\cdot36} + \dotsb$ I am investigating the convergence of the following series: $$\frac{1}{4} + \frac{1\cdot 9}{4 \cdot 16} + \frac{1\cdot9\cdot25}{4\cdot16\cdot36} + \frac{1\cdot9\cdot25\cdot36}{4\cdot16\cdot36\cdot64} + \dotsb$$
T... | $$2\cdot 4\cdots (2n)=2^n n!,\quad 1\cdot 3\cdots (2n-1)=\frac{(2n)!}{2^nn!}$$
By Stirling:
$$\frac{1^2\cdot 3^2\cdots (2n-1)^2}{2^2\cdot 4^2\cdots (2n)^2}=\frac{1}{16^n}\left(\frac{(2n)!}{n!^2}\right)^2\sim \frac{1}{16^n}\left[\frac{\left(\frac{2n}{e}\right)^{2n}\sqrt{4\pi n}}{\left(\frac{n}{e}\right)^{2n}2 \pi n}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/435675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Calculate sum of squares of first $n$ odd numbers Is there an analytical expression for the summation $$1^2+3^2+5^2+\cdots+(2n-1)^2,$$ and how do you derive it?
| One can give a combinatorial version of the argument below. But it takes some time to tell the right story. So for now we settle for the magic math approach. We have
$$\binom{2k+1}{3}-\binom{2k-1}{3}=\frac{(2k+1)(2k)(2k-1)-(2k-1)(2k-2)(2k-3)}{3!}.\tag{1}$$
Note that the above equation even holds when $k=1$, if we use ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/437835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
Volume using disk method $y = 2\sqrt{x} \quad y=x$ $y = 2\sqrt{x}$
$y=x$
about $x=-2$
I know that $y=x$ is on the outside and they meet at 4.
I need these in terms of y since I rotate about y.
$x = y$
$x = \frac{y^2}{4}$
$$\pi \int_0^4 (y - (-2))^2 - \left(\frac{y^2}{4} - (-2)\right)^2 dy$$
$$\pi \int_0^4 (y + 2)^2 - \... | As pointed out in the other answer, your expansion of the subtracted term is off:
$$\pi \cdot \int_0^4 (y + 2)^2 -\left (\frac{y^2}{4} + 2\right)^2 dy$$
So far, so good. Your set up looks fantastic.
But note that expanding the subtracted square, we get:
$$\pi \cdot \int_0^4 y^2 + 4y +4 - \left(\frac{y^4}{16} +4\cdot\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/438271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that, if $0 < x < 1$, then $(1+\frac{x}{n})^n < \frac1{1-x}$ More fully,
if $n\ge 2$ is an integer
and
$0 < x < 1$,
prove that
$(1+\frac{x}{n})^n < \frac1{1-x}$.
In addition,
if $c > 1$ and
$0 < x \le \frac{c-1}{c}$,
prove that
$(1+\frac{x}{n})^n < 1+cx$.
Proofs by elementary means
(no calculus or limits)
are pa... | \begin{align}
\left ( 1+\frac{x}{n} \right )^n
&=1+n\cdot \frac{x}{n}+\frac{n(n-1)}{2}\frac{x^2}{n^2}+\cdots+\frac{x^n}{n^n}\\
&<1+x+x^2+\cdots\\
&=\frac{1}{1-x}
\end{align}
for $n\geq 2$, $0<x<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/438927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Number of terms in the expansion of $(1+x^3+x^5) ^n$? How can we find the number of terms in the expansion of $(1+x^3+x^5)^n$ ?
| Note that
$$(1+x^3+x^5)^n = \sum_{k=0}^n \binom{n}{k} x^{3 k} (1+ x^2)^k$$
There are $n+1$ terms. For the $k$th term, there are $k+1$ terms. Therefore, there are
$$\sum_{k=0}^n (k+1) = \frac12 (n+1)(n+2)$$
terms before combining. However, there are many terms that will combine (i.e., like powers). To get an accurate... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/439065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving the multiple integral $\int \int_D \sqrt{x^2+y^2+3}dx dy$, D:{$1\le x^2+y^2\le 4, y\le \sqrt3x, y\ge \frac x{\sqrt 3}$} $\iint_D \sqrt{x^2+y^2+3}\, dx dy$, $D=\left\{(x,y) \in \mathbb{R}^2 \mid 1 \leq x^2+y^2\le 4, y\le \sqrt3x, y\ge \frac x{\sqrt 3} \right\}$.
So I've started by drawing two circles, One with $... | I think you've done almost everything right, but $\;1\le r\le\frac6{\sqrt{10}}=\frac{3\sqrt2}{\sqrt 5}\;$ ! , so
$$\int\limits_1^2r\sqrt{r^2+3}\,dr=\left.\frac12\frac23(r^2+3)^{3/2}\right|_1^2=\frac13\left(7^{3/2}-8\right)$$
and together with
$$\int\limits_{\pi/6}^{\pi/3}d\theta=\frac{\pi}6\;,\;\;\text{we get:}$$
$$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/439595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $z^3=z+\overline{z}$ I have been trying to solve an equation $z^3=z+\overline{z}$, where $\overline{z}=a-bi$ if $z=a+bi$. But I cant find any clues on how to move forward on that one. Please help.
| Writing out real and imaginary parts of $z$ and separating real and imaginary parts yields
$$
x^3+3ix^2y-3xy^2-iy^3=2x
$$
therefore,
$$
x^3-3xy^2=2x\implies x=0\quad\text{or}\quad x^2-3y^2=2
$$
and
$$
3x^2y-y^3=0\implies y=0\quad\text{or}\quad3x^2=y^2
$$
If $y=0$, then $x^3=2x\implies x\in\{0,\sqrt2,-\sqrt2\}$.
If $x=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/440000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 4
} |
system of differential linear equations $y'=\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}y$ find the solution to the problem $y'=\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}y, y(0)=\begin{pmatrix}4\\0\end{pmatrix}$
I know i have to find the eigenvalues and eigenvectors of the matrix $A=\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}... | Let
$$I_2 = \begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\quad\text{ and }\quad J_2 = \begin{pmatrix}0 & 1\\0 & 0\end{pmatrix}$$
Notice $I_2 J_2 = J_2 I_2$, $I_2^2 = I_2$ and $J_2^2 = 0_2$, we have:
$$e^{At} = e^{I_2t + J_2t} = e^{I_2t}e^{J_2t} = (I_2 + I_2t + I_2^2\frac{t^2}{2!} + \cdots)( I_2 + J_2t+ J_2^2\frac{t^2}{2}+\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/441121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find all values of a for which the equation $x^4 +(a-1)x^3 +x^2 +(a-1)x+1=0$ possesses at least two distinct negative roots Find all values of a for which the equation $$x^4 +(a-1)x^3 +x^2 +(a-1)x+1=0 $$ possesses at least two distinct negative roots.
I am able to prove that all roots would be negative .How to proceed ... | I decided to transform my comment into an answer.
As $x=0$ is not a root, divide the equation by $x^2$ and solve for $x+\frac{1}{x}$ as
RGB has pointed out.
You will get two solutions $y_1$ and $y_2$ for $x+\frac{1}{x}$:
$$x+\frac{1}{x}=\frac{-(a-1)-\sqrt{(a-1)^2+4}}{2}=y_1$$
and
$$x+\frac{1}{x}=\frac{-(a-1)+\sqrt{(a-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/444286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Integral $ \dfrac { \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx} { \int_0^{\pi/2} (\sin x)^{\sqrt 2 - 1} dx} $ I have this difficult integral to solve.
$$ \dfrac { \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx} { \int_0^{\pi/2} (\sin x)^{\sqrt 2 - 1} dx} $$
Now my approach is this: split $(\sin x)^{\sqrt 2 + 1}$ and $(\sin x)^... | Just like what @O.L pointed out, you can use Beta function to express the solution. In fact, you can change variables to get
\begin{eqnarray*}
&&\frac{\int_0^{\pi/2}\sin^{\sqrt{2}+1}xdx}{\int_0^{\pi/2}\sin^{\sqrt{2}-1}xdx}=\frac{\int_0^{1}u^{\sqrt{2}+1}(1-u^2)^{-1/2}du}{\int_0^{1}u^{\sqrt{2}-1}(1-u^2)^{-1/2}du}\\
&=&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/445808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 0
} |
If $\alpha +\beta = \dfrac{\pi}{4}$ prove that $(1 + \tan\alpha)(1 + \tan\beta) = 2$
If $\alpha +\beta = \dfrac{\pi}{4}$ prove that $(1 + \tan\alpha)(1 + \tan\beta) = 2$
I have had a few ideas about this:
If $\alpha +\beta = \dfrac{\pi}{4}$ then $\tan(\alpha +\beta) = \tan(\dfrac{\pi}{4}) = 1$
We also know that $\tan... | You've reached here :
$$1 = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}
$$
Let's continue it in this way:
$$\large\begin{align}
\Rightarrow & \tan\alpha + \tan\beta+\tan\alpha\tan\beta = 1\\
\Rightarrow &\tan\alpha(1+\tan\beta) + \tan\beta = 1\\
\Rightarrow &\tan\alpha(1+\tan\beta) + 1+ \tan\beta = 2\\
\Righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/446402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Find $F_{n}$ in : $F_{n} +2F_{n-1} + ... + (n+1)\cdot F_{0} = 3^{n}$ I'm stuck with the question for a while : Find in $$F_{n} +2F_{n-1} + ... + (n+1)\cdot F_{0} = 3^{n}$$
the element $F_{n}$ .
Placing $n-1$ instead on $n$ results in :
$$F_{n-1} +2F_{n-2} + ... + (n-1+1)\cdot F_{0} = 3^{n-1}$$
$$ F_{n-1} +2F_{n-2} + .... | The condition
$$F_{n} +2F_{n-1} + ... + (n+1)\cdot F_{0} = 3^{n}$$
is equivalent to the equality of coefficients of $z^n$ in below expression:
$$( F_0 + F_1 z + F_2 z^2 + \cdots )(1 + 2z + 3z^2 + \cdots ) = 1 + 3z + 3^2 z^2 + \cdots$$
Notice $\frac{1}{(1-z)^2} = 1 + 2z + 3z + \cdots$ and $\frac{1}{1-3z} = 1 + 3z + 3^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/448516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Show that $\frac {\sin(3x)}{ \sin x} + \frac {\cos(3x)}{ \cos x} = 4\cos(2x)$ Show that
$$\frac{\sin(3x)}{\sin x} + \frac{\cos(3x)}{\cos x} = 4\cos(2x).$$
|
$$\frac {\sin3x} {\sin x} = 3-4\sin^2x$$
as $\sin3x = 3\sin x-4\sin^3x$
$$\frac {\cos3x} {\cos x} = 4\cos^2x - 3$$
as $\cos3x = 4\cos^3x -3\cos x$
adding the two above we get $4(\cos^2x - \sin^2x)=4\cos2x.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/448673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Properties of Triangle - Trigo Problem : In $\triangle $ABC prove that $a\cos(C+\theta) +c\cos(A-\theta) = b\cos\theta$ Problem :
In $\triangle $ABC prove that $a\cos(C+\theta) +\cos(A-\theta) = b\cos\theta$
My approach :
Using $\cos(A+B) =\cos A\cos B -\sin A\sin B and \cos(A-B) = \cos A\cos B +\sin A\sin B$, we get... | Law of sines states that
$$\frac{\sin{A}}{a} = \frac{\sin{C}}{c}$$
which means that $c \sin{A} - a \sin{C} = 0$. The desired result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/449596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Initial Value, First Order Differential Equation: Weird natural log separation Solve the initial value first order differential equation problem:
$y' = \displaystyle\frac{y^5}{x(1+y^4)},\ y(1) = 1$
\begin{align}
\frac{1+y^4}{y^5}dy &= \frac 1x dx\\
\left(\frac 1{y^5} + \frac 1y\right)dy &= \frac 1x dx\\
-\frac 1 {4y^4}... | You got nearly the conclusion. Let's elevate to power $-4$ after your equation :
\begin{align}
e^{\ln y - \frac 1 {4y^4}} &= e^{\text{ln} x + C_1}\\
e^{-4\ln y + \frac 1 {y^4}} &= e^{-4\,\text{ln} x - 4\;C_1}\\
\frac 1{y^4}\;e^{\frac 1 {y^4}} &= \frac {C_2}{x^4}\\
\end{align}
So that from the definition of the LambertW... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/449758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to find the sum of the series $1+\frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}$? How to find the sum of the following series?
$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{n}$$
This is a harmonic progression. So, is the following formula correct?
$\frac{(number ~of ~terms)^2}{sum~ of~ al... | That formula is not correct to sum the first few terms of the harmonic series. Trying it with even the first three would mean that $$\frac{3^2}{1+2+3} = \frac{9}{6} = 1.5 \neq \frac{1}{1} + \frac{1}{2} + \frac{1}{3}$$.
The harmonic series actually diverges, so the sum of the series as we let $n$ get large doesn't exist... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/451558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 5,
"answer_id": 0
} |
The first two moments of $\int_0^1 B_s^2 \, ds$ I was trying to solve the following problem from Continuous Martingales and Brownian Motion by Daniel Revuz and Marc Yor, but got my solution back as the answer for variance was wrong. I have already recomputed it several times and spelled out everything explicitly withou... | It turned out that my solution was actually correct.
In addition, here is a little empirical proof in MATLAB:
m = 10000; % number of paths
n = 10000; % number of integration segments
dt = 1 / n; % length of one segment
B = cumsum(normrnd(0, sqrt(dt), m, n), 2); % draw m paths of B
X = dt * sum(B.^2, 2); % integrate
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/453303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Show that $\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha = \sin4\alpha\cos\alpha - \cos4\alpha\sin\alpha$
Show that $\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha = \sin4\alpha\cos\alpha - \cos4\alpha\sin\alpha$
I know that $\sin2\alpha = 2\sin\alpha\cos\alpha$
so
$$\sin2\alpha\cos\alpha=2\sin\alpha\cos^2\alpha$$
and $\... | Just remember that $\sin(\alpha+\beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ and $\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$ and apply the r.h.s. of each of these equalities to your example.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/453442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
show that $\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}}=\frac {(2n)!\pi}{2^{2n}(n!)^2}$ show that:
$$\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}}=\frac {(2n)!\pi}{2^{2n}(n!)^2}$$
where $n=0,1,2,3,\ldots$.
is there any help?
thanks for all
| Let $I_k=\int_{-\infty}^\infty\frac{dx}{(x^2+1)^{k+1}}$. An easy integration shows that $I_1=\frac{\pi}{2}$.
Evaluate $I_k$ using integration by parts, letting $du=dx$ and $v=\frac{1}{(x^2+1)^{k+1}}$. We get $u=x$ and $dv=-(k+1)(2x)\frac{1}{(x^2+1)^{k+2}}$. Thus
$$I_k=\left.\frac{x}{(x^2+1)^{k+1}}\right|_{-\infty}^\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/453783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Quadratic Diophantine Equation in Four Variables Consider the equation: $$d^2 = 6 + a^2 - 3b^2 + 3c^2$$ where $a, b, c, d$ are integers.
Is it necessarily the case that $a$ and $b$ have the same parity and that $c$ and $d$ have the opposite parity to that of $a$ and $b$? If so, why?
Example 1: $a = 11, b = 5, c = 12... | Every square is congruent to either $0$ or $1$ $\pmod 4$, hence consider the equation modulo $4$ to get $$\{0,1\} \equiv 2 + \{0,1\} - \{0,3\} + \{0,3\} \pmod{4}$$ where $\{m,n\}$ denotes both possibilities.
Now if $a$ and $b$ are both even we get $a^2 \equiv 0 \pmod{4}$ and $3b^2 \equiv 0 \pmod{4}$, hence $$\{0,1\} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/454033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to get value of $a^2b^2+b^2c^2+c^2a^2$ If $a+b+c=0$ and $a^2+b^2+c^2=36$, what is the value of $a^2b^2+b^2c^2+c^2a^2$?
| $$(a+b+c)=0\implies (a+b+c)^2=0 $$
$$\implies a^2+b^2+c^2+2(ab+bc+ca)=0$$
$$\implies 36+2(ab+bc+ca)=0$$
$$\implies (ab+bc+ca)=-18$$
$$\implies (ab+bc+ca)^2=324$$
just solve this whole square you can find answer
$$\implies a^2b^2+b^2c^2+c^2a^2+2(ab^2c+bc^2a+a^2bc)=324$$
$$\implies a^2b^2+b^2c^2+c^2a^2+2abc(b+c+a)=324$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/454771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve that ode I'm trying solve the following differential equation:
$$
x\left(\frac{dx}{dy}\right)^2+y\frac{dx}{dy}=x
$$
I tried to rewrite it this way:
$$
y(x)=x\frac{dy}{dx}+f\left(\frac{dy}{dx}\right)
$$
| Let $v = \frac{dx}{dy}$ then we face $xv^2+yv=x$ this gives $v^2+\frac{y}{x}v-1=0$ solve this quadratic equation for $v$ to obtain:
$$ v = \frac{-y/x \pm \sqrt{y^2/x^2+4}}{2} $$
Hence,
$$ \frac{dx}{dy} = \frac{-y/x \pm \sqrt{y^2/x^2+4}}{2} $$
Suppose $z = y/x$ then $xz=y$ and $z\frac{dx}{dy}+x\frac{dz}{dy}=1$,
$$ \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/456171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
$\|A-B\|^2 = ?$ We know that if $x,y \in \mathbb{R}$
\begin{equation}
(x-y)^2 = x^2 -2xy + y^2
\end{equation}
If $x,y$ are vectors in $\mathbb{R}^n$ we have
\begin{equation}
|x-y|^2=|x|^2 - 2 \ x \cdot y +|y|^2.
\end{equation}
where $x\cdot y$ is the usual scalar product. We know that there are several ways to define $... | If the norm is induced by some inner product $\langle\cdot, \cdot\rangle$, it must obey the parallelogram law:
\begin{align*}
\|x+y\|^2 + \|x-y\|^2 &= \|x\|^2 + 2\langle x, y\rangle + \|y\|^2 + \|x\|^2 - 2\langle x, y\rangle + \|y\|^2\\
&= 2\|x\|^2 + 2\|y\|^2
\end{align*}
A first indication that something is wrong is t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/457462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How many isosceles triangles with total side length $100$ are there? Let the sum of the three sides of a triangle be $100,$ and all the sides are positive integers length, how many possible isosceles triangles are there?
| x=2*a+100-2*a if (100-2*a)
2*34+32=100
2*35+30=100
2*36+28=100
2*37+26=100
2*38+24=100
2*39+22=100
2*40+20=100
2*41+18=100
2*42+16=100
2*43+14=100
2*44+12=100
2*45+10=100
2*46+8=100
2*47+6=100
2*48+4=100
2*49+2=100
number of solution: 16
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/458456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Sum of $1/(x+1) +2x/(x+1)(x+2) +3x^2/(x+1)(x+2)(x+3)+\ldots$ till n terms Sum of $$\frac{1}{(x+1)}+\frac{2x}{(x+1)(x+2)}+\frac{3x^2}{(x+1)(x+2)(x+3)}+\ldots$$ till n terms ...
I have no clue how to solve this ... I tried multiplying quantities ... such as $(x+2)-(x+1)$
| We have
$$ \frac{kx^{k-1}}{(x+1)(x+2)\cdots(x+k-1)(x+k)}\\= \frac{x^{k-1}}{(x+1)(x+2)\cdots(x+k-1)}- \frac{x^{k}}{(x+1)(x+2)\cdots(x+k)},\, k\geq2$$
so by telescoping
$$\sum_{k=1}^{n} \frac{kx^{k-1}}{(x+1)(x+2)\cdots(x+k-1)(x+k)}=\frac{1}{x+1}+\frac{x}{x+1}-\frac{x^{n}}{(x+1)(x+2)\cdots(x+n-1)(x+n)}\\=1-\frac{x^{n}}{(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/459851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
How to compute $\prod\limits^{\infty}_{n=2} \frac{n^3-1}{n^3+1}$
How to compute
$$\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}\ ?$$
My Working :
$$\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}= 1 - \prod^{\infty}_{n=2}\frac{2}{n^3+1}
= 1-0 = 1$$
Is it correct
| HINT:
If $$t_n=\frac{n^3-1}{n^3+1}=\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}$$
$$t_{n+1}=\frac{n\{(n+1)^2+n+1+1\}}{(n+2)\{(n+1)^2-(n+1)-1\}}=\frac{n\{(n+1)^2+n+1+1\}}{(n+2)(n^2+n+1)}$$
and $$t_{n-1}=\frac{(n-2)\{(n-1)^2+n-1+1\}}{n\{(n-1)^2-(n-1)-1\}}=\frac{(n-2)(n^2-n+1)}{n\{(n-1)^2-(n-1)-1\}}$$
Alternatively, let $\displ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/462082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Double Integral Question on unit square Hints on solving following double integral will be appreciated.
$$\int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \text{d}y \ \text{d}x$$
| Since the integrand has a non-integrable singularity at the origin, we must evaluate it by iterated integration. So let's consider the inner integral first:
$$\begin{align}
\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}\, dy &= \frac{1}{x}\int_0^{1/x} \frac{1-t^2}{(1+t^2)^2}\, dt \qquad\qquad\qquad\qquad\qquad (y = x\cdot t)\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/467663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Closed-form Geometric series of both increasing and decreasing variables? This question comes from the formula
$$x^n - a^n = (x-a)(x^{n-1}a^0 + x^{n-2}a^1 + .... + x^1a^{n-2} + x^0a^{n-1})$$
which can be verified by summing the second factor as a geometric series. My question is, how do you express the second factor... | Observe that the series has common ratio $r = \frac{a}{x}$. Hence:
\begin{align*}
(x-a)(x^{n-1}a^0 + x^{n-2}a^1 + \ldots + x^1a^{n-2} + x^0a^{n-1})
&= (x-a) \cdot \dfrac{x^{n-1}(1-(\frac{a}{x})^n)}{1-\frac{a}{x}} \\
&= (x-a) \cdot \dfrac{x}{x} \cdot \dfrac{x^{n-1}(1-\frac{a^n}{x^n})}{1-\frac{a}{x}} \\
&= (x-a) \cdot \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/468684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Series $\frac{k^2}{k!}$ with $ k=1$ to $\infty$ A practice Math Subject GRE asked me to compute
$\sum_{k=1}^\infty \frac{k^2}{k!}$. The sum is equal to $2e$, but I wasn't able to figure this out using Maclarin series or discrete PDFs. What's the most elementary way to solve this? Thank you.
| $e^x=1+x+\dfrac{x^2}{2!}+\cdots+\dfrac{x^k}{k!}+\cdots$
Differentiating both sides,
$e^x=0+1+\dfrac{2x}{2!}+\cdots+\dfrac{kx^{k-1}}{k!}+\cdots$
Multiplying both sides by $x$,
$xe^x=x+\dfrac{2x^2}{2!}+\cdots+\dfrac{kx^k}{k!}+\cdots$
Again differentiating both sides,
$(x+1)e^x=1+\dfrac{2^2x}{2!}+\cdots+\dfrac{k^2x^{k-1}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/470912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\lceil(\sqrt{3}+1)^{2n}\rceil$ is divisible by $2^{n+1}$. Let $n$ be a positive integer. Prove that $\lceil(\sqrt{3}+1)^{2n}\rceil$ is divisible by $2^{n+1}$.
I tried rewriting $\lceil(\sqrt{3}+1)^{2n}\rceil$ as $m*2^{n+1}$ for some m, but couldn't get anywhere.
| Let $(1+\sqrt{3})^{2n}=a_n+b_n\sqrt{3}$. Then $(1-\sqrt{3})^{2n}=a_n-b_n\sqrt{3}$.
Thus $(1+\sqrt{3})^{2n}+(1-\sqrt{3})^{2n}$ is the integer $2a_n$. Since $|1-\sqrt{3}|\lt 1$, we have
$$2a_n=\left\lceil(\sqrt{3}+1)^{2n}\right\rceil.$$
Note that
$$(1+\sqrt{3})^{2n+2}=(1+\sqrt{3})^{2n}(4+2\sqrt{3})=(a_n+b_n\sqrt{3})(4+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/472482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Solving $x^4-2 x^2-x+1 = 0$ for $x$ How could one solve the following, giving the answer as a closed form, not as an estimation:
$$\text{Solve }x^4-2 x^2-x+1 = 0\text{ for } x$$
Where $x$ is $\text{real.}$
I found this one particularly hard. Any help will be very much appreciated.
| As others have mentioned, there are no rational solutions but you can find the exact solution through the formula for a quartic function (though why one would need it is another question). The fact that the coefficient of $x^3$ is zero makes things a bit easier, but not by much.
For: $$a x^4 + c x^2 + d x + e=0$$
The ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/473000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Find the area bounded by the curves $y=1-\cos(2x)$, $y = 0$, for $0\leq x \leq 2 \pi$ Find the area bounded by the curve $y=1-\cos(2x)$, the $x$ axis, and the lines $x=0$ and $x=2\pi$
Can someone please show me how this question is completed. This is causing much confusion.
Thanks everyone!
| First, we plot the graph, and look at the region to be integrated. Below, compliments of Wolfram Alpha, we have the graph of $\color{blue}{y = 1 - \cos 2\pi}$, above the line $\color{purple}{y = 0}$, from $x = 0$ to $x = 2\pi$:
From here, it's relatively straight forward: our bounds of integration are from $x = 0$ to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/473500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to show that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ indirectly? I found this amazingly beautiful identity here. How to prove that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ without directly multiplying the factors? (I've already verified it that way). ... | Expanding on Potato's answer:
First lets make $a^3 + b^3 + c^3 -3abc\ $ a polynomial in $a$ so we get $a^3 - a3bc + b^3 + c^3$. This has $3$ roots so lets express it as $(a+p)(a+q)(a+r)$. But also know: $$a^3 + b^3 + c^3 -3abc = (a+b+c)(a^2 + b^2 + c^2 -ab -bc - ac)$$
Therefore lets say $p=b+c$. So our polynomial is no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/475354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 6,
"answer_id": 4
} |
How to prove this inequality for $a_{n}<5$ Let $a_{1}=1,a_{2}=2$ and be related by the recurrence
$$a_{n+2}=\dfrac{a^2_{n+1}(1+a_{n})+(1+2a_{n})a_{n+1}-a^2_{n}}{a_{n+1}+a^2_{n}+a_{n}+1}$$
for $n\in N.$
How can I show that
$a_{n}<5,n\in N$?
my idea: I think we have
$$a_{n+2}=pa_{n+1}+qa_{n}$$
because we have
$$a_{n+2}a... | Let us start with @Did's simplified form of the recurrence relation (with typos corrected): $$b_{n+2}=\dfrac{b_nb_{n+1}^2}{b_n^2-b_n+b_{n+1}}\quad (n=1,2,...)\quad\text{with}\quad b_1=2\text{ and }b_2=3.$$Write $c_n:=\dfrac {b_{n+1}+b_n^2}{b_nb_{n+1}}\;(n=1,2,...).$ Then substituting for $b_{n+2}$ in the corresponding ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/477133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Finding minima and maxima of $\sin^2x \cos^2x$ I am trying to find the turning points of $\sin^2x \cos^2x$ in the range $0<x<\pi$.
So far I have:
$\begin{align} f'(x) & =\cos^2x \cdot 2\sin x \cdot \cos x + \sin^2x \cdot 2\cos x \cdot -\sin x \\
& = 2 \cos^3x \sin x - 2\sin^3x \cos x \\
& = 2 \sin x \cos x (\cos^2x - \... | if you use the Double angle formulae
$$\sin^2(x)\cos^2(x)=(1-\cos2x)/2 \cdot (1+\cos2x)/2 = (1-\cos^2(2x))/4 = \sin^2(2x)/4 = (1-\cos4x)/8$$
so minimum is $1/8$ at $x=45\circ$ and maximum is 1/4 at $x=90\circ$
remember $\sin^2(x)=(1-\cos2x)/2$ and $\cos^2(x)=(1+\cos2x)/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/477636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
How to calculate the coordinates of a triangle's orthocentre? How does one calculate the coordinates of a triangle's orthocentre?
I was surfing through the net and got this formula:
$$x-\rm{coordinate}= \frac{x_1\tan A+x_2\tan B+x_3\tan C}{\tan A+\tan B+\tan C}$$
$$y-\rm{coordinate}= \frac{y_1\tan A+y_2\tan B+y_3\tan ... |
Here $A\equiv (x_1,y_1)$,$B\equiv (x_2,y_2)$,$C\equiv (x_3,y_3)$. I'll use the usual notation for $a,b,c,R$ and $A,B,C$.
Using simple trigonometry, $BP=c\cos B$, $PC=b\cos C\implies\dfrac{BP}{PC}=\dfrac{c\cos B}{b\cos C}$ $$\implies P_x= \dfrac{x_2b\cos C+x_3c\cos B}{b\cos C+c\cos B}= \dfrac{x_2b\cos C+x_3c\cos B}{a}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/478069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Sum of the series $1+\frac{1\cdot 3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$ Decide if the sum of the series
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+ \cdots$$
is: (i) $\infty$, (ii) $1$, (iii) $2$, (iv) $4$.
| The series can be rewritten as
$$
\begin{align}
&1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\dots\\
&8\left(\frac1{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\dots\right)\\
&=8\sum_{k=1}^\infty\frac{(2k-1)!!}{(2k+2)!!}\tag{1}
\end{align}
$$
This is reminiscent of the series ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/479610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 1
} |
A problem on limit It seems
$\lim_{n \rightarrow \infty} \sum_{r=1}^{n-1}\binom{n}{r-1}
\sum_{j=0}^{n-r+1} (-1)^j \binom{n-r+1}{j} \frac{1}{r+j+1}$
is $\frac{1}{2}$. Here's a plot of the function for $n \leq 300$:
But I can not prove this. Any hints?
| Note that $\int_{0}^{1}x^{r+j}dx=\frac{1}{r+j+1}$.
So the sum becomes:
$\sum_{r=1}^{n-1}\binom{n}{r-1}\sum_{j=0}^{n-r+1}\binom{n-r+1}{j}(-1)^{j}\int_{0}^{1}x^{r+j}dx$
$=\int_{0}^{1}\sum_{r=1}^{n-1}\binom{n}{r-1}\sum_{j=0}^{n-r+1}\binom{n-r+1}{j}(-1)^{j}x^{r+j}dx$
$=\int_{0}^{1}\sum_{r=1}^{n-1}\binom{n}{r-1}x^{r}\big(\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/480160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Showing Whether a Sequence is Bounded Above or Not I am trying to solve the following problem about a sequence:
Consider the sequence ${a_n}$ where $a_n = 1 + \frac{1}{1 \cdot 3} + \frac {1}{1 \cdot 3 \cdot 5} + \frac {1}{1 \cdot 3 \cdot 5 \cdot 7} + ... + \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)}.$ Decide whether ${a... | This may be informal, but just showing that the sequence is monotone decreasing would mean the first term is the upper bound for the sequence.
We may rewrite the sequence in the following way $a_{n+1}=\frac{a_n}{2n+1}$. We can then see whether the sequence is infact monotone decreasing
$a_n-a_{n+1}=a_n-\frac{a_n}{2n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/480266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Idea of the Proof : Existence of a & b so that (Any integer greater than 8) = 3a + 5b
Claim: Prove that for every integer $n \geq 8$, there exist nonnegative integers $a$ and $b$ such that $n = 3a + 5b.$
Proclaimed solution : Let $n ∈ \mathbb{Z}$ with $n ≥ 8.$ $\text{ Then } n = 3q, \text{ where } q ≥ 3, \quad \text{ ... | $3a=3\cdot a+0\cdot5$ which needs $a\ge0\implies 3a\ge 0$
$3a+1=3(a-3)+2\cdot5$ which needs $a\ge3\implies 3a+1\ge 10$ so disallows $1,4,7$
$3a+2=3(a-1)+5\cdot5$ which needs $a\ge1\implies 3a+2\ge 5$ so disallows $2$
Iterating with $5$
$5b=5\cdot b+0\cdot3$ which needs $b\ge0\implies 5b\ge 0$
$5b+1=5(b-1)+2\cdot3$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/481240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
$g(x+y)+g(x)g(y)=g(xy)+g(x)+g(y)$ for all $x,y$. Find all functions $g:\mathbb{R}\to\mathbb{R}$ with $g(x+y)+g(x)g(y)=g(xy)+g(x)+g(y)$ for all $x,y$.
I think the solutions are $0, 2, x$. If $g(x)$ is not identically $2$, then $g(0)=0$. I'm trying to show if $g$ is not constant, then $g(1)=1$. I have $g(x+1)=(2-g(1))g(x... | Here's a much less murky solution with strong influence from Calvin Lin's excellent answer.
$$
g(x + y) + g(x)g(y) = g(xy) + g(x) + g(y) \tag{0}
$$
First of all, plugging in $y = 1$ gives
$$ g(x + 1) + g(x)g(1) = 2g(x) + g(1)$$
For readability let $g(1) = k$ and we have
$$
g(x + 1) = (2-k)g(x) + k \tag{1}
$$
Plugging i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/481673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 6,
"answer_id": 3
} |
Compute $ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$ Compute $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$$
Of course we have $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \left( \frac{x}{1+x^2} \right) ^2 dx = \lim_{ A \to \inft... | This integral may be evaluated using the residue theorem. The integrand has poles at $x=\pm i$; if we close with a contour in the upper half plane, then we need only worry about the pole at $x=i$. The integral is therefore
$$i 2 \pi \left [\frac{d}{dx} \frac{x^2}{(x+i)^2} \right ]_{x=i} = i 2 \pi\left [\frac{i 2 x}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 4
} |
Rolling ellipses I'm struggling to prove the following.
Set one ellipse in contact with a congruent one so that the minor axis of one is aligned with the major axis of the other. Now roll one round the other. The locus of the centre of the rolling ellipse is a circle centre the centre of the other, radius a + b.
Is the... | The locus is not a circle.
Counterexample.
Consider ellipses with semi-axes $a=2,b=1$. Let equation of static
ellipse is
$$
\dfrac{x^2}{2^2} + \dfrac{y^2}{1^2} = 1.
$$
Here 3 steps are shown:
Suppose the locus is a circle (with radius $r = a+b =3$).
Then must be an instant/moment (see $2$nd image), when ellipses are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
How to show $\sqrt{4+2\sqrt{3}}-\sqrt{3} = 1$ I start with $x=\sqrt{4+2\sqrt{3}}-\sqrt{3}$, then
$\begin{align*}
x +\sqrt{3} &= \sqrt{4+2\sqrt{3}}\\
(x +\sqrt{3})^2 &= (\sqrt{4+2\sqrt{3}})^2\\
x^2 + (2\sqrt{3})x + 3 &= 4+ 2\sqrt{3}\\
x^2 + (2\sqrt{3})x - 1 - 2\sqrt{3} &= 0
\end{align*}$
So I have shown that there is so... | Here’s an approach that doesn’t require you to spot a not-terribly-obvious factorization. Multiplying by the conjugate to get rid of some of the square roots is a fairly natural thing to do, and
$$\left(\sqrt{4+2\sqrt3}-\sqrt3\right)\left(\sqrt{4+2\sqrt3}+\sqrt3\right)=1+2\sqrt3\;.$$
Thus, the desired result holds if a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Solving a system of equations using modular arithmetic modulo 5 Give the solution to the following system of equations using modular arithmetic modulo 5:
$4x + 3y = 0 \pmod{5}$
$2x + y \equiv 3 \pmod{5}$
I multiplied $2x + y \equiv 3 \pmod 5$ by $-2$, getting $-4x - 2y \equiv -6 \pmod{5}$.
$-6 \pmod{5} \equiv 4 \... | There's also a "cheats" method available here. There are only $25$ possible values of $(x,y) \in (\mathbb{Z}_5)^2$. We can just check them one-by-one, and see which ones work.
We could do this by hand, or on a computer. In GAP:
for x in [0..4] do
for y in [0..4] do
if((4*x+3*y) mod 5=0 and (2*x+y) mod 5=3) the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/484467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Prove that a group generated by two elements of order $2$, $x$ and $y$, is isomorphic to $D_{2n}$, where $n = |xy|.$ I am completely stuck at the question
Let $G$ be a finite group and let $x$ and $y$ be distinct elements of order 2 in $G$ that generate $G$. Prove that $G \cong D_{2n}$, where $n = |xy|.$
I have prove... | Personally, I like the following presentation for $D_{2n}$:
$$\langle x,y\mid x^n=y^2=(xy)^2=1\rangle$$ Now assume that the following presentation is given:
$$G=\langle x,y\mid x^2=y^2=(xy)^n=1\rangle$$ So we have:
$$
\begin{align*}
G=\langle x,y&\mid x^2=y^2=(xy)^n=1\rangle\\
\cong
\langle x,y,a&\mid x^2=y^2=(xy)^n=1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/487128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Factorizing $f(x) = 2x^4 +x^3 -x^2 +8x-4$ I tried plugging in values of $x$ until I found $f(x)=0$. I got that $(x+2)$ is a factor. Is this the only way to find factors of this polynomial?
Then I used polynomial division to get $f(x) = (x+2)(2x^3 -3x^2+5x-2)$. However I tried different values of x but still couldn't ge... | You might want to take some of the guessing out of finding rational roots by making use of the rational root theorem. From Wikipedia:
In algebra, the rational root theorem (or rational root test) states a constraint on rational solutions (or roots) of the polynomial equation
$$a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0 = 0$$
wi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/487324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
find this limit $\lim_{x\to0^{+}}\frac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$ find the limit.
$$\lim_{x\to0^{+}}\dfrac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$$
my try:
$$\tan{x}=x+\dfrac{1}{3}x^3+o(x^3),\sin{x}=x-\dfrac{1}{6}x^3+o(x^3)$$
so
$$\tan{(\tan{x})}=\tan{x}+\dfrac{1}{3}(\tan{x})^3+o(\t... | Yours is easily the simplest $direct$ method.
My first idea for the limit is that that really looks like a difference quotient for the derivative of $\tan$. Distilling the core idea, I make the following conjecture:
Let $f(x)$ be continuously differentiable at $a$. Then,
$$ \lim_{\substack{(x,y) \to (a,a) \\ x \neq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/490470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Right triangle with equal permeter and height - how to find side lengths? The question:
Suppose there is a right triangle with sides $a$ and $b$ and
hypotenuse $c$. Its perimeter is the same as its area, and $b = 6$.
What are its side lengths?
I just cannot figure out how to do this! The second sentence isn't par... | If $a=6$, then area is equal to $(6b)/2=3b$. If that is equal to the perimeter, then $a+b+c=3b$ so $a+c=2b$ so $6+c=2b$. You also know that $a^2+b^2=c^2$ so $36+b^2=c^2$.
From the $6+c=2b$ you get $c=2b-6$, and so $36+b^2=(2b-6)^2$. Expand and you get a quadratic on $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/491082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Fermat primes relation to $2^n+1$ If $2^n+1$ is prime how to show that $n$ must be a power of $2$
Being at elementary level I am at a loss what to begin with?
| Suppose that $n$ is odd. Then $$x^n=(x+1-1)^n=(-1)^n= -1\mod x+1$$ so that $x+1\mid x^n+1$. Thus if $2^n+1$ is not composite, it must be the case $n=2^u$. If you want to be explicit $$\begin{align}
\frac{{{x^n} + 1}}{{x + 1}} &= \frac{{{{( - x)}^n} - 1}}{{( - x) - 1}} \cr
&= \frac{{{u^n} - 1}}{{u - 1}} \cr
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/491830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Pre calculus fraction simplify question Simplify:
$$\frac{\frac{16x^4}{81} - y^4}{\frac{2x}{3} + y}$$
Wolfram alpha confirms the answer from the answer sheet: Wolframalpha answer
| $$\frac{\frac{16x^4}{81} - y^4}{\frac{2x}{3} + y}=\frac{16x^{4}-3^4y^4}{81}\cdot \frac{3}{2x+3y}=\frac{(4x^2-9y^2)(4x^2+9y^2)}{81}\cdot \frac{3}{2x+3y}=\frac{(2x-3y)(2x+3y)(4x^2+3y^2)}{81}\cdot \frac{3}{2x+3y}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/491975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Different Ways of Integrating $3\sin x\cos x$ I am asking this question for my son who is in (equivalent) twelfth
grade and I failed to answer his query.
When he tries to integrate $3\sin x\cos x$, he finds that
this can be done in at least following three ways.
And these three ways do not produce equivalent results.
O... | You're forgetting that an indefinite integral must include a constant of integration; for any chosen constant $C$, we have that
$$\frac{d}{dx}\left(-\frac{3}{4}\cos(2x)+C\right)=3\sin(x)\cos(x),$$
and that is precisely the relationship captured by the statement that
$$\int 3\sin(x)\cos(x)\,dx=-\frac{3}{4}\cos(2x)+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/495159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Partial Derivates Question How would I go about taking the partial derivative of:
$\frac{a+b^2}{a^2b}$ with respect to $a$ or $b$?
Thanks!
| The partial derivative of $\frac{a+b^2}{a^2b}$ with respect to $a$ is computed by treating $b$ as a constant.
Thus, $\dfrac{\partial u}{\partial a}[\frac{a+b^2}{a^2b}]$ = $\frac{a^2b(1) - (a+b^2)(2ab)}{a^4b^2}$ = $\frac{a^2b - 2a^2b - 2ab^3}{a^4b^2}$ = $\frac{-a-2b^2}{a^3b}$.
The partial derivative of $\frac{a+b^2}{a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/497970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Determinant of matrix obtained by commuting matrices The Question is to prove that :
For Commuting $n\times n$ matrices $A,B,C,D$ over a field $F$,
Determinant of $\left(\begin{array}{cccc}
A & B \\
C & D \\
\end{array} \right)$ is given by $\det(AD-BC)$
I have no idea how to proceed for this except at the case of $n... | Here is a method that works if $F = \mathbb{C}$. Let $\mathcal{F} = \{A, B, C, D\}$ and let $P\in M_{n}(\mathbb{C})$ such that $P^{-1}XP$ is upper triangular for each $X\in \mathcal{F}$. Then
\begin{equation*}
\det\begin{pmatrix}
A & B\\
C & D
\end{pmatrix} = \det\begin{pmatrix}
P^{-1} & 0\\
0 & P^{-1}
\end{pmatrix}\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/498336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Improving bound on $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}}$ An old challenge problem I saw asked to prove that $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < 3$. A simple calculation shows the actual value seems to be around $2.8$, which is pretty close to $3$ but leaves a gap. Can someone find $C < 3$ and prove $\sqrt{2 \sqrt{3 \sq... | Iif $a(n)=\sqrt{ 2\sqrt {3\sqrt {4...\sqrt{n}}}}$,
$$a(n)^{2}= 2\sqrt {3\sqrt {4...\sqrt{n}}} \tag{1}$$
$$\dfrac{1}{2^{2}}.a(n)^{2}= 3\sqrt {4...\sqrt{n}} \tag{2}$$
$$\dfrac{1}{3^{2}}.\dfrac{1}{2^{2^{2}}}.a(n)^{2^{2}}= 4\sqrt{...\sqrt{n}} \tag{3}$$
$$\vdots$$
$$\prod_{i=0}^{n-2}\Big(\dfrac{1}{(n-i)^{2^{i}}}\Bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/498774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 4
} |
Increasing and bounded sequence proof Prove that the sequence $a_n= 1+ \frac 12+ \frac 13+\cdots+ \frac 1n-\ln(n)$ is increasing and bounded above. Conclude that it’s convergent.
This what I got so far
Proof:
Part 1: Proving $a_n$ is increasing by induction.
Base:
$a_1=1$
$a_2=1+\frac 12= \frac 32$
$a_1≤a_2$
So the b... | @Patrick's answer shows that the sequence is decreasing. However, we can show boundedness and covergence in a single step since the terms of the sequence are non-negative. I only provide a skeleton solution, which will need fleshing out with a limit argument.
We can begin by rewriting the sequence as:
\begin{equation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/499733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
How many ways to form $7$-digit numbers from $\{1,2,\ldots,9\}$ in which $4$ is not the next of $5$? I have tried in several cases:
*
*the digits don't contain five only, so the number of possible ways is $8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2$
*the digits don't contain four only, so the number of possible ways is ... | The correct results for the first three cases you enumerated are:
*
*The number contains $4$ but not $5$: we obtain $7\cdot\binom{7}{6}\cdot 6!$ different numbers (the $7$ accounts for the position of the $4$, the binomial for choosing $6$ other digits out of the $7$ possible and the $6!$ for ordering them).
*The n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/501215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How many weak compositions of $n$ with $k$ parts up to reflection? The number of weak compositions of $n$ with $k$ parts is given by:
$$ \binom{n+k -1}{k-1}. $$
Consider the following identification: two compositions $(s_1 + \cdots + s_k), (t_1 + \cdots + t_k)$ are the same if $t_i = s_{k-i}$ for $i = 1,\ldots,k$. In ... | This is a simple application of Polya Counting. Let us recall for a moment that the generating function of weak compositions of $n$ into $k$ parts is
$$Z(E_k)\left(\frac{1}{1-z}\right)$$
where $$Z(E_k) = a_1^k$$
is the cycle index of the identity group so that we get the OGF
$$\left(\frac{1}{1-z}\right)^k.$$
Now when w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/502724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is there an equation that will graph a line segment? Is there an equation that, given the two points of a line segment, will result, when graphed for x on a real graph, in a line segment?
| A very simple way to do this would be to add and then subtract the square roots of an expression to define the domain. For example:
Given $(3,34)$ and $ (5, 52)$, first find the equation for the line that goes between the two points:
$(52-34)/(5-3)x + b = y$
$y = 9x + b$
$34 = 9(3) + b$
$b = 7$
Therefore, the equation ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/502968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 5
} |
Prove $\sqrt{s_n+1} = \frac{1}{2}(1+\sqrt{5})$ This is to prove how the limit of $s_n$ converges to $\frac{1}{2}(1+\sqrt{5})$.
Assume: $s_1 = 1$; for $n \geq 1$, $s_{n+1} = \sqrt{s_n + 1}$.
How to prove this converges to $\frac{1}{2}(1+\sqrt{5})$?
| Consider applying the Contraction Mapping Theorem. Just check that the conditions for the theorem are satisfied (I'll leave those details to you).
We have $s_{n+1} = \sqrt{s_n + 1}$. Take $F: x \to \sqrt{x +1}$. By the Contraction Mapping Theorem we can conclude that $F$ has a unique fixed point $s$.
This gives us:
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/504427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Division by fractions Why is it that to divide by a fraction you need to multiply by the reciprocal of that fraction?
For instance:
$\eqalign{
& 1 \div \frac{1}{3} \cr
& = 1 \times 3 \cr
& = 3 \cr} $
Thank you.
| Note that
$$1 \div \frac{1}{3} = \frac{1}{\frac{1}{3}} = \frac{1}{\frac{1}{3}}\times 1 = \frac{1}{\frac{1}{3}}\times \frac{3}{3} = \frac{1\times 3}{\frac{1}{3}\times 3} = \frac{3}{1} = 3.$$
More generally, you can use the above manipulations to show $\dfrac{a}{b}\div\dfrac{c}{d} = \dfrac{ad}{bc}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/504732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving a system of three linear equations with three unknowns Is my working correct or am I completely wrong? Have I missed anything out? Any feedback is appreciated.
Question: Consider the following system of equations
$2x + 2y + z = 2$
$−x + 2y − z = −5$
$x − 3y + 2z = 8$
Form an augmented matrix, then reduce this m... | You can reduce your matrix further. remember that you can multiply and/or divide each row so you end up obtaining
$$
\begin{pmatrix}
1 & 0 & 0 & 1\\
0 & 1 & 0 & -1\\
0 & 0 & 1 & 2\\
\end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/504800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
fourier expansion of $\coth$ and justifying an identity The problem:
Justify the following equalities:
$$\cot x = i\coth (ix) = i \sum^\infty_{n=-\infty} \frac{ix}{(ix)^2+(n\pi)^2}=\sum^\infty_{n=-\infty}\frac{x}{x^2+(n\pi)^2}$$
I am trying to figure out how to start this. When I insert the Euler identity of $
\coth$... | Adding a bit of diversity to this discussion we suppose that we seek to prove that
$$\coth x = \sum_{n=-\infty}^\infty \frac{x}{x^2+\pi^2 n^2}.$$
Rather than use the standard technique (absolutely nothing wrong with it) of applying a circular contour to
$$ f(z) = \pi \cot(\pi z) \frac{x}{x^2+\pi^2 z^2}$$
we use Mellin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/505210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Common linear and Quadratic factors (i) If $ax^5+bx^2+c$ has a factor of the form $x^2+px+1$ prove that:
$(a^2-c^2)(a^2-c^2+bc)=(ab)^{2}$
(ii)In this case prove that:
$ax^5+bx^2+c$ and $cx^5+bx^3+a$ have a common quadratic factor
| Proving part (ii) independently is less tedious than part (i). Here it is:-
Let $f(x) = ax^5 + bx^2 + c$
Also $f(x) = (x^2 + px + 1)Q_3 (x)$; where $Q_3 (x)$ is a quotient polynomial of degree 3…….(1)
Let $g(x) = x^2 + px + 1$ such that g(x) = 0 has m and n as roots.
From product of roots, $n = 1/m$.
$f(n) = f(1/m) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/505689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
integrating $\int \frac{dt}{(t+2)^2(t+1)}$ I'm practicing to solve a whole, and I am not able to solve this one, could you help me? $$\int \frac{dt}{(t+2)^2(t+1)}$$I tried $$\frac{1}{(t+2)^2(t+1)}=\frac{A}{(t+2)^2}+\frac{B}{(t+2)}+\frac{C}{(t+1)}\\1=A(t+1)+B(t+2)(t+1)+C(t+2)^2\\t=-2\Longrightarrow 1=-A\Longrightarrow \... | $$\left(\frac1{t+2}+\log\frac{t+1}{t+2}\right)'=-\frac1{(t+2)^2}+\frac{t+2}{t+1}\frac1{(t+2)^2}=$$
$$=-\frac1{(t+2)^2}+\frac1{(t+1)(t+2)}=\frac{-t-1+t+2}{(t+1)(t+2)^2}=\frac1{(t+1)(t+2)^2}\;\color\red\checkmark$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/506030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find asymptotic of recurrence sequence Given a sequence $x_1=\frac{1}{2}$, $x_{n+1}=x_n-x_n^2$. It's easy to see that it limits to $0$.
The question is: is there exists an $\alpha$ such, that $\lim\limits_{n\to\infty}n^\alpha x_n\neq0$.
I tried to find explicit formula for $x_n$, but did not succeed. I stucked at that ... | Let $y_n = \frac{1}{x_n} - n$, we have
$$y_{n+1} = \frac{1}{x_{n+1}}-(n+1) = \frac{1}{x_n} + \frac{1}{1-x_n} - (n+1) = y_n + \frac{x_n}{1-x_n}$$
This implies for $n > 1$,
$$y_n = y_1 + \sum_{k=1}^{n-1} \frac{x_k}{1-x_k}
\quad\iff\quad
\frac{1}{x_n} = n + 1 + \sum_{k=1}^{n-1}\frac{x_k}{1-x_k}
\tag{*1}$$
Since $0 \le x_k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/506723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.