Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Calculating a sum in the coefficient of some generating function I tried to calculate the coefficient of a generating function
\begin{align*}
\frac{1}{n} [u^{n-1}] e^{un} \frac{1}{(1-u)^2}
\end{align*}
and got to
\begin{align*}
\frac{1}{n} [u^{n-1}] e^{un} \frac{1}{(1-u)^2} &= \frac{1}{n} [u^{n-1}] \Bigg(\sum_{k\geq 0}... | One could also work with geometric series instead of using the general binomial theorem:
\begin{align*}
[z^n]g(f(z)) &= \frac{1}{n}[u^{n-1}]\phi(u)^n g'(u) \\
&= \frac{1}{n}[u^{n-1}]e^{nu} \left( \frac{1}{1-u}\right)' \\
&= \frac{1}{n}[u^{n-1}]e^{nu} \left( \sum_{k \geq 0}u^k\right)' \\
&= \frac{1}{n}[u^{n-1}]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Does the pattern of identities for $(x+y+z)^{2n-1}$ continue past $2 n - 1 = 5$? We have the following identities.
\begin{align}
(x+y+z)^3&=x^3+y^3+z^3+3(x+y)(y+z)(z+x)\\
(x+y+z)^5&=x^5+y^5+z^5+5(x+y)(y+z)(z+x)(x^2+y^2+z^2+xy+yz+zx)
\end{align}
I'm wondering is there any pattern like above in the expansion of $(x+y+z)^... | Yes you do keep the factors $x+y,x+z,y+z$. We can prove that, for instance, $x+z$ divides $(x^k+y^k+z^k)-(x+y+z)^k$ for all odd $k$ by observing that the difference goes to zero simultaneously with the claimed factor $x+z$ when we specify $z=-x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proving by induction that $\frac{1}{n} \ge \frac{n!}{n^n}$ I have been trying to prove by induction that for all $n \ge 0, \frac{1}{n} \ge \frac{n!}{n^n} $
Here is my proof so far:
Prove that for $ n = 1, \frac{1}{n} \ge \frac{n!}{n^n} $
$ \frac{1}{1} = 1 $
$ \frac{1!}{1^1} = 1 \le 1 $
Next, assume there exists $ n \... | Looks like a trivial proof holds. $n!$ has $n$ terms, each $\le n$. So $\frac{n!}{n^n}=\frac{1}{n}\times C\le \frac{1}{n}$, since $C=\frac{2}{n}\times...\frac{n}{n}\le 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4444076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Help with $\frac{\sin\theta+\cos\theta}{-\sin\theta+\cos\theta}=?$ Assume that $0<\theta<\frac{\pi}{4}$. If $\sin2\theta=\frac{1}{4}$, then $\frac{\sin\theta+\cos\theta}{-\sin\theta+\cos\theta}=?$
My approach with this problem is to rationalize the denominator in the fraction to get this:
$$
\frac{\sin2\theta+1}{\cos2\... | Alternative approach:
$$\frac{1}{4} = \sin(2\theta) = 2\sin(\theta)\cos(\theta). \tag1 $$
Also, since $0 < \theta < (\pi/4)$, you have that
$$\cos(2\theta) = \sqrt{1 - \left(\frac{1}{4}\right)^2} = \frac{\sqrt{15}}{4}. \tag2 $$
Finally, you have that
$$\cos^2(\theta) - \sin^2(\theta) = \cos(2\theta) = \frac{\sqrt{15}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4444621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
To find the function $g(x)$ in terms of $f(x)$. For each point $(a,b)$ on the graph $y = f(x)$ the point $(3a-1,\frac{b}{2})$ is plotted forming the graph of another function $y = g(x)$. The problem asks us to find the function $g(x)$ in terms of $f(x)$.
Also, we need to describe the transformation.
Since $a$ goes to ... | Let $$X=3a-1\;\text{ and }\; g(X)=\frac b2$$
with
$$b=f(a) $$
or
$$2g(X)=f(\frac{X+1}{3})$$
So
$$g(X)=\frac 12f(\frac{X+1}{3})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3$ Prove that if $a,b,c,d$ are positive reals we have:
$$\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3.$$
I think that I have found a equality cas... | Again partial hint :
Using the formula :
$$\cos\left(\operatorname{arccot}\left(\frac{a}{b}\right)\right)-\sqrt{\frac{a^{2}}{a^{2}+b^{2}}}=0$$
We use Karamata's inequality with $-\cos(x)$:
$$\frac{a}{b}\leq \frac{b}{c}\leq \frac{c}{d}\leq \frac{d}{a}$$
And $$3=4\cos\left(\operatorname{arccot}\left(\frac{3}{\sqrt{7}}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 4
} |
Calculate the integral by Riemann $\int_{0}^{1}\sqrt{x}\,dx$ Calculate the integral by Riemann $\displaystyle \int _{0}^{1}\sqrt{x} \, dx$
$$\displaystyle a_{k} =0+k\cdot \frac{1}{n} =\frac{k}{n}$$
$$\displaystyle \Delta x=\frac{1-0}{n} =\frac{1}{n}$$
\begin{align*}
\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left(\sqrt{... | Here is a direct way to look at this if you would like to stick with a Riemann sum that uses equally spaced subintervals. The strategy is, in small doses, to replace the irrational $\sqrt{k}$ values with nearby integers.
$$\begin{align}
\lim_{n\to\infty}\frac{1}{n\sqrt{n}}\sum_{k=1}^n\sqrt{k}
&=\lim_{n\to\infty}\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4457035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
A number theory question that is probably wrong Prove that $a^3-b^3 = 2011$ has no integer solutions.
I think the question is wrong as
$a^3-b^3 = 2011$
$(a-b)(a^2+ab+b^2) = 2011$
As $2011$ is prime so the only factors $2011$ has are $1$ and $2011$ itself.
So if $a-b = 1$ and $a^2+ab+b^2 = 2011$ , then we can say that
$... | The possible values or $x^3\pmod 9$ are $\{0,1,-1\}$ only.
So their difference can only take the values $\{0,1,2,-1,-2\}$, yet $2011\equiv 4\pmod 9$ so it is not reachable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4457570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$f(x)=x^{3}-3 x+a$. Given that it has $3$ integer roots $\alpha, \beta, \gamma$. Find all possible values of $a$. So, I tried vieta and with some algebra, I got to the point $0=\alpha^{2}+\beta^{2}+\gamma^{2}+2(\alpha \gamma+\gamma \beta+\alpha \beta)$ where $(\alpha \gamma+\gamma \beta+\alpha \beta)=a$. I don't know h... | $$f(x) = (x - \alpha)(x - \beta)(x - \gamma) = x^3 - 3x + a$$
$$f(x) = x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \alpha\gamma + \beta\gamma)x - \alpha\beta\gamma = x^3 - 3x + a$$
Matching up the coefficients gives you the system of equations:
$$\alpha + \beta + \gamma = 0$$
$$\alpha\beta + \alpha\gamma + \bet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4459391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How do you finish the equations for $S^2 + \frac{7}{4}S= \frac{1}{2}$? Complete the Square:
$$S^2 + \frac{7}{4}S = \frac{1}{2}$$
$$S^2 + \frac{7}{4}S \_\_\_\_\_\_\_ = \frac{1}{2}$$
$$S^2 + \frac{7}{4}S + \frac{49}{64} = \frac{81}{64}$$
THIS is where I get stuck. I know that the Square Root of $\frac{81}{64}$ is $\frac{... | I'm thinking that with the algebra/precalculus tag, the equation you're asking about is $$s^2+\frac {7}{4} s = \frac {1}{2}$$
Completing the square is correct; when you have
$$s^2 + \frac {7}{4}s + \frac {49}{64} = \frac {1}{2} + \frac {49}{64}$$ you can factor the left hand side; then you'll have $$\left(s + \frac {7}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4462165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the maximum and minimum value of this function? Find the maximum and minimum values of the $f(z) = \vert 1 + z\vert + \vert 1-z+z^2\vert$ on the $\{z\in \mathbb{C} \vert$ $\vert z \vert =1, Re(z) \geq 0\}$ (Here the $\mathbb{C}$ is the set of
complex numbers.)
Let me show you my solution.
Put $z = e^{i\theta}$... | For your given function, your answer is correct. The maximum is attained when $\theta = 0$, giving $z = 1$, hence $$f(1) = |1+1| + |1^2-1+1| = 3.$$ This is clearly greater than $\sqrt{3}$ claimed by the note. The minimum is attained when $\theta \in \pm \pi/3$, giving $z \in \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$, he... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4462316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A Clarkson-type inequality $|a+b+c|^q + \sum\limits_{\mathrm{cyc}} |a + b - c|^q \ge \sum\limits_{\mathrm{cyc}} (|a + b|^q + |a - b|^q)$ Show that, for $a,b,c \in \mathbb{R}$ and $q \geq 2$:
$$ |a+b|^{q} + |a-b|^{q} + |a+c|^q + |a-c|^q + |b+c|^q + |b-c|^q \\\leq \\|a+b+c|^q + |a+b-c|^q + |a-b+c|^q + |a-b-c|^q $$
When $... | WLOG, assume that $a\ge b \ge c \ge 0$.
Fact 1: It holds that
\begin{align*}
&(a+b-c)^q + (a-b+c)^q + |b+c-a|^q \\
\ge\,& a^q + b^q + c^q
+ (a - b)^q + (b - c)^q + (a - c)^q.
\end{align*}
(The proof is given at the end.)
Fact 2: It holds that
$$(a + b + c)^q + a^q + b^q + c^q
\ge (a + b)^q + (b + c)^q + (c + a)^q.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If $A=\begin{bmatrix}2&1&1\\0&1&0\\1&1&2\end{bmatrix}$, find $A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$
If $A=\begin{bmatrix}2&1&1\\0&1&0\\1&1&2\end{bmatrix}$, find $A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$
Attempt $1:$ $A=I+B$, where $B=\begin{bmatrix}1&1&1\\0&0&0\\1&1&1\end{bmatrix}$
$B^n$ comes out to be $2^{n-1}B$
Aft... | Following your attempt 1 and using
$$\begin{cases}
A-I &= B\\
B^2 &= 2B
\end{cases}$$ you get that
$A^2-4A + 3I=0$, i.e. that $q(A)=0$ where $q(x) = x^2-4x +3$. You want to get $p(A)$ where $p(x)=x^8-5x^7+7x^6-3x^5+x^4-5x^3+8x^2-2x+1$.
Performing the long division of $p$ by $q$ you get
$$p(x) = (x^6-x^5+x^2-x+1)q(x) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4472508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How can we know how many times a number is divisible by 2? For example 36:
36/2=18
18/2=9
So for 36 it is divisible by 2 2 times.
For 2ⁿ×(2k-1) it is divisible by 2 n times.
But how do I know for 3k-2?
Is there a formula to find how many times a number is divisible by 2?
| For $k = 1,2,3,\ldots$ and so $3k-2 = 1,4,7,\ldots$, the highest power of $2$ dividing $3k-2$ are:
$$1,4,1,2,1,16,1,2,1,4,1,2,1,8,1,\underbrace2_{k=16},\ldots$$
This looks like some translated or transformed version of the highest power of $2$ dividing $k$ (https://oeis.org/A006519), but instead term $4^n=2^{2n}$ first... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
problem canceling in multiplication of fractions now that i have a grasp on formatting mathematical fractions and whatnot i present you with my latest confusion
$$3\frac15 \times 1\frac23 \times 2\frac 34$$
converted into improper fractions
$$\frac{16}5 \times \frac 53 \times \frac{11}4$$
this much i understand. the fo... | We know that $\dfrac{a}{b}\times\dfrac{c}{d}=\dfrac{a\times c}{b\times d}$ and that we can commute $\dfrac{a}{b}\times\dfrac{c}{d}=\dfrac{c}{b}\times\dfrac{a}{d}=\dfrac{c}{d}\times\dfrac{a}{b}$ and so on, of course in the nominator and denominatior. Then, in your exercise:
$$\frac{16}5 \times \frac 53 \times \frac{11}4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Why are the two calculations of $ \int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x $ give two distinct answers? One of the calculations:
$$
\begin{aligned}
\int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x &=\int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{1}{\sqrt{x}} \mathrm{~d} x \\
&=2 \int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \ma... | They are both correct. Let $y=2 \arcsin \sqrt x$. Then $x=(\sin (\frac y 2))^{2}$. So $2x-1=2(\sin (\frac y 2))^{2}-1=\sin (y+\frac {\pi} 2)$. So $y=\sin (2x-1)-\frac {\pi} 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the integer solution: $a+b+c=3d$, $\: a^{2} + b^{2} + c^{2}= 4d^{2}-2d+1$ Find the integer solutions:
$$a+b+c=3d$$
$$a^{2} + b^{2} + c^{2}= 4d^{2}-2d+1$$
Attempt:
Notice that $a=b=c=d=1$ is a solution.
Other facts:
Notice that $a^{2} + b^{2} + c^{2} > 0$, so $4d^{2}-2d+1>0$. Notice that $4d^{2}-2d+1$ has negati... | Your equation has no non-zero solutions.
Lemma. The Diophantine equation $3x^2+y^2=2z^2$ has no non-zero solutions in integers. Proof. Let $x_0$ is smallest positive integer such that there are integers $y_0$ and $z_0$ such that $3x_0^2+y_0^2=2z_0^2$. If $y_0 \not \equiv 0 \pmod 3$ then $z_0 \not \equiv 0 \pmod 3$. Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
What is the solution to this non-linear second order differential equation? I'm trying to solve the following non-linear second order differential equation:
$$\tag{1}
\frac{d\, }{dx} \Bigl( \frac{1}{y^2} \, \frac{dy}{dx} \Bigr) = -\, \frac{2}{y^3},
$$
where $y(x)$ is an unknown function on the real axis. I already kno... | Solve
\begin{align*}
u^{\prime \prime}&=2 u^{3}
\end{align*}
Multiplying both sides by $u^{\prime}$ gives
\begin{align*}
u^{\prime} u^{\prime \prime}&=2 u^{3} u^{\prime}
\end{align*}
Integrating both sides w.r.t. $x$ gives
\begin{align*}
\int{u^{\prime} u^{\prime \prime}\, \mathrm{d}x} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding Solutions to a Symmetric Non-Linear Equation (for Some Cases Beside $x = y = z$)
Find all values $x,y,z$ (whether real or complex) such that :
$2x-2y+z^{-1} = 2022^{-1}\\ 2z-2x+y^{-1}=2022^{-1}\\ 2y-2z + x^{-1} = 2022^{-1}$
I know that for case when $x=y=z$ i can easily find : $x=y=z = 2022$ (which seems to... | To simplify the calculations, doing the same as @Dietrich Burde did in his answer, let $a=\frac 1{2022}$ and you end with the cubic
$$12 a x^3-36 x^2-3 a^3 x+a^2=0$$
which has three real roots since
$$\Delta=1296 a^2 \left(a^4+12\right)^2 \quad > 0~~\forall a$$ Then the three roots are given by
$$x_k=\frac 1 a +\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Probability of exactly $7$ unreplaced draws from $\{1,...,10\}$ to get $1,2,3$? A box contains tickets numbered $1$ to $10$. Tickets are drawn at random without replacement until the numbers $1, 2$ and $3$ are drawn. Find the probability that exactly $7$ draws are required.
I am a little confused with this question. My... | Probability $1,2,3$ are in the first $7$ draws is the probability none in last $3$ are $1,2,3$: $\frac{7 \choose 3}{10 \choose 3}$
Probability $1,2,3$ are in the first $6$ draws is the probability none in last $4$ are $1,2,3$: $\frac{7 \choose 4}{10 \choose 4}$
Probability exactly $7$ draws are required is the differen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
Properties of Continued fractions of $\sqrt n$ I got to know from this Brilliant article (no pun intended) that the continued fraction expansion expression of $\sqrt n$ is of the form $[a_0,\overline{a_1,a_2,\dots ,a_k}]$ for integers $a_i$ where
*
*$a_0=\lfloor\sqrt n\rfloor$ and $a_k=2a_0$
*$a_1,a_2\dots ,a_{k-1}$... | here I give an example fo converting the square root to continuous fraction:
We want to find $\sqrt{41}$. The greatest perfect square less than $41$ is $36=6^2$ , so we may write:
$\sqrt{41}\approx 6 +\frac 1 x$
We have:
$\frac 1 x=\sqrt {41}-6\Rightarrow x=\frac 1{\sqrt{41}-6}\space\space\space\space(1)$
Multiplying n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4487826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding a closed form of $f(n) = \frac{n-4}{n}f(n-1)+1\ ,\ n \geq 5$ where $f(4) = 1$. I'm trying to find the closed form of
$f(n) = \frac{n-4}{n}f(n-1)+1\ ,\ n \geq 5$ where $f(4) = 1$.
Empirically, it turns out the answer is simply $f(n)=1+\frac{n-4}{5}\ ,\ n \geq 4$, but I'm having a hard time getting there. I've tr... | Alt. hint: write it as $\lambda_nf(n) = \lambda_{n-1}f(n-1) + \mu_n$ where $\lambda_n=\mu_n=n(n-1)(n-2)(n-3)$.
[ EDIT ] After telescoping $\,\lambda_nf(n) = \lambda_4 f(4) + \sum_{k=5}^n \mu_k\,$ where $\,\lambda_4 = 24\,$, $\,f(4)=1\,$, and the sum of falling factorials can be calculated as shown at $\sum r(r+1)(r+2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4488489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Find the maximum value of $M$ with given constraints.
Let $a, b \in \mathbb{R}$ such that $n^{4a-\log_5{n^2}} \leq 25^{40-b^2}$ for all positive real number $n$. Find the maximum value of $M=a^2+b^2+a-3b$.
What I have tried:
Taking the logarithm base 5 of both sides from the inequality, we get $(4a-2\log_5{n})\log_5n... |
$x^2+2ax-b^2+40
\geq0$
You dropped a sign: should be
$x^2-2ax-b^2+40
\geq0.$
This won't matter to the rest of your analysis, which is good. You reached the point of realizing that
you must have
$$a^2 + b^2 = 40 \implies $$
$$ -\sqrt{40} \leq a \leq \sqrt{40} ~~\text{and}~~
-\sqrt{40} \leq b \leq \sqrt{40}. \tag1 $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4488705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve the differential equation $L^2 y=0$ where $L=x^2\frac{d^2}{dx^2}+x\frac{d}{dx}-(x^2+1)\text{id}$ Consider the linear differential equation defined by
$$
L^2y=0,\quad \text{where}\ L:=x^2\frac{d}{dx^2}+x\frac{d}{dx}-(x^2+1)\text{id}. \tag{1}$$
Here $\text{id}$ denotes the identity operator, and $L^2$ denotes the o... | Per a c.a.s. the general solution is
\begin{split}y(x) = A I_1(x) + B K_1(x) + C \left[-I_1(x) \int_{x_0}^x \frac{K_1(t)^2 \,dt}{t} + K_1(x) \int_{x_0}^x \frac{I_1(x) K_1(x)}{x}\,dt \right] \qquad \qquad \\+ D \left[\frac{1}{2} K_1(x) \left(I_0(x)^2 - I_1(x)^2\right) - I_1(x) \int_{x_0}^x \frac{I_1(x) K_1(x)}{x}\,dt\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4489188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
why does $1/((x-1)(x^2 - 1)) = (x^4-1)/(x-1) \cdot (x^8 - 1)/(x^2 - 1) \cdot (x^{16} - 1)/(x^4 - 1) \cdots$
Why does $1/((x-1)(x^2 - 1)) = (x^4-1)/(x-1) \cdot (x^8 - 1)/(x^2 - 1) \cdot (x^{16} - 1)/(x^4 - 1) \cdots$?
I think the equality holds as formal power series. Expanding the LHS, one should get $(1+x+x^2 +\cd... | "But doing cancellation for the RHS gives that it equals $(x^2+1)/(x−1)\cdot(x^4+1)(x^8+1)\cdots$, which clearly contradicts the equality."
In fact $(1 + x^2)(1 + x^4)(1 + x^8)(1 + x^{16}) \cdots = 1 + x^2 + x^4 + x^6 + x^8 + x^{10} + \ldots = \frac{1}{1 - x^2}$ -- the first equality is just the generating-functional v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
For a triangle, prove that $\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}=\frac{1}{r^2} $
Context: I made this question and am contemplating about submitting it as a contest question. The contest is not a large one; its scope does not comprise even our whole class. Its just a friendly, small contes... | By the result $1$ found in my post, $$\displaystyle x yz=r^{2}(x+y+z),\tag*{(*)} $$
where $r$ is the in-radius of the triangle ABC of sides $a,b$ and $c$ and $a=y+z, b=z+x $ and $c=x+y$.
Dividing the equation (*) by $xyzr^2$ yields the required equation
$$
\frac{1}{y z}+\frac{1}{x z}+\frac{1}{x y}=\frac{1}{r^{2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\ln^3(\sin(x))$ using Fourier series
Show that $$\int_0^{\frac\pi2} \ln^3(\sin(x)) \, dx = -\frac\pi2 \ln^3(2) - \frac{\pi^3}8 \ln(2) - \frac{3\pi}4 \zeta(3)$$
I have seen a method for this elsewhere, but I would specifically like to reproduce this result in the same way I've computed the similar integrals... | $$\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n^2-H_n^{(2)}}{n+1}\cos(2(n+1)x)\\
=\Re\sum_{n=1}^\infty \frac{H_n^2-H_n^{(2)}}{n+1}(-e^{2ix})^{n+1}$$
$$=\Re\left\{-\frac13 \ln^3(1+e^{2ix})\right\}$$
$$=x^2\ln(2\cos x)-\frac13\ln^3(2\cos x)\tag{*}$$
where we used the generating function
$$\sum_{n=1}^\infty \frac{H_n^2-H_n^{(2)}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
How do I handle equations like $f(x)-x = f'(x)$? How do I handle equations like $f(x)-x = f'(x)$?
Integrating both sides gets me nowhere... I have been trying to guess an example, but with no avail. I cannot think of any other approaches ...
Edit: ok i noticed $f(x) =x+1$ ... can someone guide be to how i can solve for... | I know that the below solution doesn't work in many situations, but I like to share its idea.
Suppose $f(x) = \sum_{n \geq 0}a_nx^n$ is the power series of $f$. Because $f(x) - x = f'(x)$, term by term differentiating of the power series gives us
\begin{align*}
\sum_{n \geq 0}a_nx^n - x = \sum_{n \geq 1}na_nx^{n-1} \im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Finding Equation of tangent to the curve $xy^2=25$ at point $(5,5)$
Find equation of the tangent to the curve $xy^2=25$ at point $(5,5)$
My Working
I tried this sum using two different approaches. But I'm getting two different answers. I wonder if any is correct. Can you please guide me on this
First approach;
use pr... | You did the differentiation right.
$$\frac{dy}{dx} = - \frac{y}{2x} = - \frac{25}{x^2y}$$
Note that if $xy^2 = 25$, then $- \frac{25}{2x^2y} = - \frac{25y}{2x^2y^2} = - \frac{25y}{2x(xy^2)} = - \frac{25y}{2x(25)} = - \frac{y}{2x}$, so the two expressions are equivalent.
The problem is that the question is ill-formed: I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4499995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the integral $I=\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx$ To evaluate the integral
$$I=\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx$$
I define $t=-1+\sqrt{\frac{4}{x}-3}$, then we have $x=\frac{4}{(t+1)^2+3}$ and
$$dx=\frac{-8(t+1)}{(t^2+2t+4)^2}dt$$
So we get
$$\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx=8\... | $ \text {Let } y=\sqrt{-1+\sqrt{\frac{4}{x}-3}}\textrm{ then ,}$
$ \displaystyle \begin{aligned}I&=16 \int_{0}^{\infty} \frac{y^{2}\left(y^{2}+1\right) d y}{\left(y^{4}+2 y^{2}+4\right)^{2}}\\&=4\left[3 \underbrace{\int_{0}^{\infty} \frac{y^{2}\left(y^{2}+2\right)}{\left(y^{4}+2 y^{2}+4\right)^{2}} d y}_{J}+\underbrace... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
$\lim\limits_{x\to 0} \frac{x^2\cos x - 6 \ln(1+x^2) +5x^2}{(e^{\sqrt[4]{1+4x^3+8x^4}}-e)\arcsin(x)}$ without L'Hosptial rule. I have a belief that it is possible to evaluate any limit without using L'Hospital rule or series expansions. However the limit $$\lim\limits_{x\to 0} \frac{x^2\cos x - 6 \ln(1+x^2) +5x^2}{(e^{... | With the hint of eyeballfrog, I was able to arrive at the following solution.
We write the nuemreator as $x^2(\cos x - 1) - 6(\ln(1+x^2)-x^2)$ then dividing top and bottom by $x^4$ the limit becomes $$\lim\limits_{x\to 0} \frac{\frac{\cos x - 1}{x^2}-\frac{6}{x^4}(\ln(1+x^2)-x^2)}{\frac{\arcsin x}{x} \cdot \frac{e^{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim_{x \to \infty} \sqrt{x^2 + 2x} - x$. Why doesn't $x^2$ overpower $2x$ when $x\to\infty$, so that the answer is $0$ (instead of $1$)?
Evaluate $$\lim_{x \to \infty} \sqrt{x^2 + 2x} - x$$
Hello!
I was solving this problem, and here is my approach:
Inside the square root $x^2$ overpowers $2x$ as $x \to... | $\lim_{x \to \infty} \sqrt{x^2+2x}-x =$
$\lim_{x \to \infty} (\sqrt{x^2+2x}-x) \cdot \frac{\sqrt{x^2+2x}+x}{\sqrt{x^2+2x}+x} =$
$\lim_{x \to \infty} \frac{(\sqrt{x^2+2x}-x)\cdot(\sqrt{x^2+2x}+x)}{\sqrt{x^2+2x}+x} =$
$\lim_{x \to \infty} \frac{|x^2+2x|-x^2}{\sqrt{x^2+2x}+x} =$
$\lim_{x \to \infty} \frac{x^2+2x-x^2}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Calculate the sum of $\left\lfloor \sqrt{k} \right\rfloor$ I'm trying to calculate for $n\in \mathbb{N}$ the following sum :
$\sum_{k=1}^{n^2}\left\lfloor \sqrt{k} \right\rfloor$.
I tried putting in the first terms, which gave me
$\sum_{k=1}^{n^2}\left\lfloor \sqrt{k} \right\rfloor=(1+2+3+\cdots+n)+\left\lfloor \sqrt{2... | The sum can be calculated as follows:
$$
\sum_{k=1}^{n^2} \lfloor \sqrt k \rfloor = 1 + 1 + 1 + \overbrace{2 + .. + 2}^{5 \space \text{terms}}+ \overbrace{3 + .. + 3}^{7 \space \text{terms}} + \overbrace{4 + .. + 4}^{9 \space \text{terms}} + .. + n
$$
That reduces to:
$$
3 \times 1 + 5 \times 2 + 7 \times 4 + .. + (2n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Spivak: How do we divide $1$ by $1+t^2$, to obtain $\frac{1}{1+t^2}=1-t^2+t^4-t^6+...+(-1)^nt^{2n}+\frac{(-1)^{n+1}t^{2n+2}}{1+t^2}$? In Chapter 20, "Approximation by Polynomial Functions" in Spivak's Calculus, there is the following snippet on page $420$
The equation
$$\arctan{x}=\int_0^x \frac{1}{1+t^2}dt$$
suggests... | Use geometric sum:
$$1-t^2+t^4-t^6+...+(-1)^nt^{2n}=\frac{1-(-t^2)^{n+1}}{1-(-t^2)}=\frac{1}{1+t^2}+\frac{(-1)^nt^{2n+2}}{1+t^2}$$
Move terms and we get:
$$\begin{align}\frac{1}{1+t^2}&=1-t^2+t^4-t^6+...+(-1)^nt^{2n}-\frac{(-1)^{n}t^{2n+2}}{1+t^2}\\
\\\frac{1}{1+t^2}&=1-t^2+t^4-t^6+...+(-1)^nt^{2n}+\frac{(-1)^{n+1}t^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Given $p(x)=x^2+ax+b$, where $a,b\in \mathbb{Z}$. If $p(2020 + p(2020))=1$, find $a,b$
Given $p(x)=x^2+ax+b$, where $a,b\in \mathbb{Z}$. If $p(2020 + p(2020))=1$,
find $a,b.$
My attempt:
Given $p(x) = x^2+ax+b$, then $p(x+p(x))=p(x^2+(a+1)x+b)=(x^2+(a+1)x+b)^2+a(x^2+(a+1)x+b)+b=1$
After a few pages of algebra, I get.... | weird fact about parabolas:
$$p(x)=x^2+ax+b\iff \boxed{p(n)p(n+1)}=p\Big(p(n)+n\Big)$$
$$\begin{cases}2020a+b+2020^2&=\pm1\\2021a+b+2021^2&=\pm1\end{cases}$$
$0=2021^2-2020^2+a \iff a=-4041$
$b=\pm 1-2021(-4041)-2021^2=4082420 \pm 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is there a closed form for $\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{(\sin x+\cos x)^{2 n+1}} d x$? When I encountered the integral $$I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{(\sin x+\cos x)^{2 n+1}} d x,$$
where n is a non-negative integer, I, as usual, used the substitution $x\mapsto \frac{\pi}{2}-x$ to transforms
$$I=... | Something which could be interesting
$$I_n=\frac 1 {2^n} \sum_{k=0}^{n-1}\binom{n-1}{k}\frac{1}{2 k+1}=2^{-n} \, \,_2F_1\left(\frac{1}{2},1-n;\frac{3}{2};-1\right)$$ generates the unknown (?) sequence
$$\left\{\frac{1}{2},\frac{1}{3},\frac{7}{30},\frac{6}{35},\frac{83}{630},\frac{73}{6
93},\frac{523}{6006},\frac{476... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Limit of $u_{n}=\frac{1}{n !} \sum_{k=0}^{n} k !$ I encountered a sequence $(u_n)_{n \in \mathbb{N}} $ defined as
$$ u_{n}=\frac{1}{n !} \sum_{k=0}^{n} k ! $$
And I wonder what is the limit. It seems to be 1 but even Wolfram Alpha cannot figure it out.
My first idea was to write $$ \frac{k!}{n!} = \frac{1}{(k+1)...(n-1... | To go beyond the limit itself, using the same approach as @Yanko, let
$$a_p=\frac 1{\prod_{i=0}^p (n-i)}$$ and consider
$$1+\sum_{p=0}^\infty a_p=1+\sum_{k=0}^\infty \frac {B_k}{n^{k+1}}=1+\frac{1}{n}+\frac{1}{n^2}+\frac{2}{n^3}+\frac{5}{n^4}+O\left(\frac{1}{n^5}
\right)$$ where appear Bell numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4515299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Prove $\int_{- \infty}^{\infty} \frac{\mathrm{dx}}{({x}^{2} + {b}^{2}) {({x}^{2} + {a}^{2})}^{2}} = \frac{\pi (2 a + b)}{2 {a}^{3} b {(a + b)}^{2}}$ I am in chapter 6 of Whittaker and Watson, and I am currently confused attempting to find the mistake I made in computing the following integral. The problem asks me to pr... | Without residues but partial fractions, the integrand is
$$-\frac{i}{2 b \left(b^2-a^2\right)^2 (z-i b)}+\frac{i}{2 b \left(b^2-a^2\right)^2
(z+i b)}+$$ $$\frac{1}{4 a^2 \left(a^2-b^2\right) (z-i a)^2}+\frac{1}{4 a^2
\left(a^2-b^2\right) (z+i a)^2}+$$ $$\frac{i \left(3 a^2-b^2\right)}{4 a^3
\left(a^2-b^2\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4515492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Can we write square roots in a fraction separately? eg, $\sqrt{\frac{9-x^2}{x-2}}$ vs $\frac{\sqrt{9-x^2}}{\sqrt{x-2}}$ I was doing some questions related to functions when I came across this question where they gave us two functions as
$$\sqrt{\frac{9-x^2}{x-2}}\quad\text{and}\quad\frac{\sqrt{9-x^2}}{\sqrt{x-2}}$$
I c... | Hint: $\sqrt{\frac{-3}{-2}}$ is defined, while $\frac{\sqrt{-3}}{\sqrt{-2}}$ is not (in case we work only with real numbers).
In your case, for $\sqrt{\frac{9-x^2}{x-2}}$ to be defined, you need $\frac{9-x^2}{x-2} \geq 0 \Leftrightarrow x \in (-\infty, -3] \cup [2, 3]$ and for $\frac{\sqrt{9-x^2}}{\sqrt{x-2}}$ you need... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show that $a_{n+1} = a_{n} + a_{n}^{\frac{1}{3}}$ satisfies $\lim_{n\to\infty}\frac{a_n}{pn^q}=1$ for some real valued $p, q$. Show that $a_{n+1} = a_{n} + a_{n}^{\frac{1}{3}}$ satisfies $\lim_{n\to\infty}\frac{a_n}{pn^q}=1$ for some real valued $p, q$.
I had hoped that I could use a strategy similar to Aproximation of... | If $a_0=0$ then the result is not true.
I'm assuming that $a_0>0$.
It's easy to check that $\{a_n\}$ is non-decreasing. So it either diverges to $+\infty$ or converges to a finite number $l>0$.
If the latter, then $l$ must verify $l=l+l^{\frac 1 3}$, which is a contradiction. Thus the sequence diverges to $+\infty$.
Le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Two types of days, expected number of days of one type Say I have two types of days, A and B. We have that:
*
*A is followed by A with probability $p$.
*A is followed by B with probability $1 - p$.
*B is followed by B with probability $q$.
*B is followed by A with probability $1 - q$.
Over a year with 365 days, w... | Here is the brute force method:
Notation: Let us give the state $A$ the value $1$ and the state $B$ the value $0$. Let us denote the state in time $n$ as $X_n$. $X_0$ will now denote the starting state. Now when we are in state $A$ at time $n$, we say $X_n = 1$ and when we are in state $B$ at that time we say $X_n = 0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4519636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the derivative of $y=\ln^3(3-x+x^2)$ Find the derivative of $$y=\ln^3(3-x+x^2)$$ We have to use the chain rule, but I am not sure how to think about it when we have more complex functions as in this problem. We can rewrite the function as $$y=\left[\ln(3-x+x^2)\right]^3,$$ so the outermost function is $g(x)=x^3$, ... | Another possible approach:
\begin{align*}
y = \ln^{3}(3 - x + x^{2}) & \Longleftrightarrow y^{1/3} = \ln(3 - x + x^{2})\\\\
& \Longleftrightarrow \frac{y'}{3y^{2/3}} = \frac{2x - 1}{3 - x + x^{2}}\\\\
& \Longleftrightarrow y' = \frac{3(2x - 1)\ln^{2}(3 - x + x^{2})}{3 - x + x^{2}}
\end{align*}
Hopefully this helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4520014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Determinant of block matrices. $$
X = \begin{pmatrix}
1+b_1 & 1 & 0 & 0 & 0 & \frac{1}{a_{6}} \\
1+b_2 & 1 & 1 & 0 & 0 & -\frac{a_1}{a_6} \\
b_3 & 1 & 1 & 1 & 0 & -\frac{a_2}{a_6} \\
b_4 & 0 & 1 & 1 & 1 & -\frac{a_3}{a_6} \\
b_5 & 0 & 0 & 1 & 1 & 1-\frac{a_4}{a_6} \\
b_6 & 0 & 0 & 0 & 1 & 1-\frac{a_5}{a_6}
\end{pm... | I think you have a partition as follows
$$ X=\left(\begin{array}{c|cccc|c}
1+b_{1} & 1 & 0 & 0 & 0 & \frac{1}{a_{6}}\\
\hline 1+b_{2} & 1 & 1 & 0 & 0 & -\frac{a_{1}}{a_{6}}\\
b_{3} & 1 & 1 & 1 & 0 & -\frac{a_{2}}{a_{6}}\\
b_{4} & 0 & 1 & 1 & 1 & -\frac{a_{3}}{a_{6}}\\
b_{5} & 0 & 0 & 1 & 1 & 1-\frac{a_{4}}{a_{6}}\\
\hl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit of $(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$ Find $a,b$ of:
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$$
I can't use L'hopital, I tried multiplying by the conjugate, and solving it,
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-(ax+b))\cdot \frac{\sqrt{4x^2+2x+1}+(ax+b)}{\sqrt{4x^2+2x+1}+(ax+b)}$$
$$=\li... | $$ L=\lim_{x\to \infty}(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$$
Let $x=\frac{1}{t}$, then
$$L=\lim_{t\to 0}\left( \frac{2}{t} (1+\frac{t}{2}+{\frac{t^2}{4}})^{\frac{1}{2}}-\frac{a}{t}-b\right)$$
Use $(1+z)^k=1+kz+O(z^2)$. Then
$$L=\lim_{t\to 0}\frac{2(1+\frac{t}{4}+O(t^2))-a-bt}{t}=\lim_{t\to 0} \frac{(2-a)+(\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
The equation with integer solution: $x^4+x-y^4+y=2x^2y^2(x^4-y^4+1)^3$ I am trying to solve this equation:
Given $x,y\in \text {Z}$: Solve : $x^4+x-y^4+y=2x^2y^2(x^4-y^4+1)^3$
Obviously, when $x=0$, it is easy to find that $y=0,y=1$ and $y=0$ then $x=0, x=-1$.
My question is how to find another solutions.
My try is fac... | Write $a=x+y$, $b=x-y$, $c=x^2+y^2$ then the equation becomes
$$2a(bc+1)=(c-ab)(c+ab)(abc+1)^3.$$
Now if $|a|>1$ and $|bc|>1$ then by the fundamental theorem of arithmetic either $a$ or $bc+1$ must contain three copies of all the prime factors of $abc+1$. However $a$ is coprime with $abc+1$, so $bc+1$ must contain thr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Geometric proof for derivative of $f(x)=\sqrt{x}$ Using a square of area $x$,
we obtain
$$dx = (2\sqrt{x})d\sqrt{x}+(d\sqrt{x})^2$$
$$\frac{d\sqrt{x}}{dx} = \frac{1}{2 \sqrt{x}} - \frac{(d\sqrt{x})^2}{2\sqrt{x}dx}$$
How do I show that $\frac{(d\sqrt{x})^2}{2\sqrt{x}dx} \rightarrow 0$ as $dx\rightarrow0$?
| \begin{eqnarray}
\Delta x &=& (2\sqrt{x})\Delta\sqrt{x}+(\Delta\sqrt{x})^2\\
\Delta x&=&(2\sqrt{x}+\Delta\sqrt{x})\Delta\sqrt{x}\\
(2\sqrt{x}+\Delta\sqrt{x})\Delta\sqrt{x}&=&\Delta x\\
\frac{\Delta\sqrt{x}}{\Delta x}&=&\frac{1}{2\sqrt{x}+\Delta\sqrt{x}}
\end{eqnarray}
As $\Delta x\to0$, then $\Delta\sqrt{x}\to0$.
So, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4524342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is this limit at infinity negative or positive 1
$\lim_{x \to -∞} \frac{x}{\sqrt{x^2+4}}$
I solved it:
$\lim_{x \to -∞} \frac{x}{\sqrt{x^2+4}} = \frac{x}{\sqrt{x^2}} = \frac{x}{x} = 1 , -1$ ?
Since its $x/x$, a negative infinity number divided by another negative infinity number is a positive infinity. But I am told ... | Note that
$$
\frac{x}{\sqrt{x^2+4}} = \frac{x}{\sqrt{x^2(1+4x^{-2}})} = \frac{x}{|x|}\frac{1}{\sqrt{1+4x^{-2}}}
$$
so
$$
\lim_{x\to -\infty}\frac{x}{\sqrt{x^2+4}} = \lim_{x\to -\infty}\frac{x}{|x|} \times\lim_{x\to -\infty}\frac{1}{\sqrt{1+4x^{-2}}} = -1\times 1 = -1
$$
since
$$
\frac{x}{|x|} = -1
$$
for $x<0$ and $(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
prove $\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\ge \frac{2}{n}-\frac{1}{n^2}$
Let $x_1,x_2,\cdots,x_n$ be $n$ non-negative numbers such that
$x_1+x_2+\cdots+x_n=1$. Show $$\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i
x_j}\ge \frac{2}{n}-\frac{1}{n^2}.$$
We may consider using Cauchy. That is
$$\sum_{i=1}^n\sum_{j=1}^i x... | Base Case:- $n=1$
$$\sum_{i=1}^1 \frac{x_i^2}{\sum_{j=1}^i x_j}\ge \frac{2}{1}-\frac{1}{1^2}$$
$$1 \ge 1$$
Thus the base case holds.
Hypothesis:-
$$\sum_{i=1}^{n+1} \frac{x_i^2}{\sum_{j=1}^i x_j}\ge \frac{2}{n+1}-\frac{1}{{(n+1)}^2}$$
And now for the step:-
$$\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j} + \frac{x_{n+1}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4537840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find all real values of x and y such that $\left(x+yi\right)^3$ is real and greater than $8$, represent these values in the xy-plane My solution
$\left(x+yi\right)^3>8$
$\left(x^3-3xy^2-8\right)+\left(3x^2y-y^3\right)i>0$
and now I'm stuck with these equations
$\left(x^3-3xy^2-8\right)>0,\left(3x^2y-y^3\right)>0$
| I don't know if my answer is correct or not but this is my idea:
$(x+yi)^3= r^3(\cos\phi+i\sin\phi)^3,$
where $r=\sqrt{x^2+y^2}$ and $\phi \in [-\pi,\pi)$.
By de Moivre's formula, we have
$(x+yi)^3= r^3(\cos\phi+i\sin\phi)^3 = r^3(\cos(3\phi)+i\sin(3\phi)).$
Since this is real, then $\sin(3\phi)=0 \Leftrightarrow \phi ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4538996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Computing the exact value of the integral $∫_0^∞ \tanh(2x)\ln(\tanh x)dx.$ LATEST EDIT
Thanks to @FDP’s alternative method and @Claude Leibovici’s generalisation on the integral. Meanwhile, I had found a formula for the integral with power n in general.
$$\boxed{I_n=∫_0^∞ \tanh(2x)\ln^n(\tanh x)dx =\frac{(-1)^n n !}{2^... | \begin{align}J&=\int_0^\infty \tanh(2x)\ln (\tanh x)dx\\
&=\int_0^\infty \frac{\text{e}^{2x}-\text{e}^{-2x}}{\text{e}^{2x}+\text{e}^{-2x}}\ln\left(\frac{\text{e}^{x}-\text{e}^{-x}}{\text{e}^{x}+\text{e}^{-x}}\right)dx\\
&=\int_0^\infty \frac{1-\text{e}^{-4x}}{1+\text{e}^{-4x}}\ln\left(\frac{1-\text{e}^{-2x}}{1+\text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4540019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
complex numbers $z^n = 1$ complex plane I would like to know if I answered the question below correctly or if I need to change something.
Question: Take $n ∈ \mathbb N$. Find all solutions for $z^n=1$. Where can we find the solutions in the complex plane? Draw them in the complex plane for some values of $n$.**
$z^n ... | An easier way to solve would be:
$$
\begin{align*}
z^{n} &= 1 \Rightarrow z^{n} = |z^{n}|\\
z &= |z| \cdot \mathrm{e}^{\arg(z) \cdot \mathrm{i}}
(|z| \cdot \mathrm{e}^{\arg(z) \cdot \mathrm{i}})^{n} &= 1\\
|z|^{n} \cdot (\mathrm{e}^{\arg(z) \cdot \mathrm{i}})^{n} &= 1\\
|z|^{n} \cdot \mathrm{e}^{n \cdot \arg(z) \cdot \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4543557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int \sqrt{x^2 + a^2}\mathrm{d}x$ I tried using integration by parts and could get to this point:
\begin{align}
\int \sqrt{x^2 + a^2}\mathrm{d}x &= x\sqrt{x^2+a^2} - \int x\frac{2x}{\sqrt{x^2 + a^2}}\mathrm{d}x
\\\\ &= x\sqrt{x^2+a^2} -2\int \sqrt{x^2 + a^2}\mathrm{d}x + \int \frac{2a^2}{\sqrt{x^2 + a^2}}\ma... | You have a little mistake in the integration by parts, $\displaystyle(\sqrt{x^2 + a^2})' = \frac{x}{x^2 + a^2}$ therefore
$$
\int \sqrt{x^2 + a^2}\,dx = x\sqrt{x^2 + a^2} - \int \sqrt{x^2 + a^2}\,dx + a^2\int\frac{1}{\sqrt{x^2 + a^2}}.
$$
The latter integral is simply a can be calculated using the hyperbolic trig subst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4545307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Calculate the limit $\lim_{n\to \infty }\int_{1}^{n}(n^2+x^3)^{-1/2} dx$ \begin{align}
\lim_{n\to \infty }\int_{1}^{n}\frac{dx}{\sqrt{n^2+x^3}}& \overset{x=nt}{=}\int_{1/n}^{1}\frac{dt}{\sqrt{1+nt^3}} \\
& \overset{t^3=p}{=}\int_{1/n^3}^{1}\frac{1/3p^{-2/3}}{\sqrt{1+np}}dp\\
& \overset{pn=q}{=}\frac{n^{-1/3}}{3}\int_{1... | Your mistake is saying
$$\frac{n^{-\frac{1}{3}}}{3}\int_{\frac{1}{n^{2}}}^{n}\frac{t^{-\frac{2}{3}}}{\sqrt{t+1}}dt = 2\int_{0}^{\frac{\pi}{2}}\sin\left(\theta\right)^{-\frac{1}{3}}\cos\left(\theta\right)^{-\frac{2}{3}}d\theta.$$
I'm not too sure how you tried using the substitution $t=tg^2\theta$ since that second inte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve initial value problem using Laplace Transform Use Laplace transform to solve the following initial–value problems.
I'm confused with $f(t)$ defined by cases and both $x,y$ being differentiated as shown below.
*
*$(\frac{d^2}{dt^2}-4\frac{d}{dt}+4)y=f(t)$, subject to $y(0)=-2,y'(0)=1$, with
$$f(t)=\begin{cases}
... | The Laplace transform of $f(t)$ is given by
\begin{align}
f(t) &\doteqdot \int_{0}^{\infty} e^{-s t} \, f(t) \, dt \\
&\doteqdot \int_{0}^{3} e^{- s t} \, t \, dt + \int_{3}^{\infty} e^{-s t} \, (t + 2) \, dt \\
&\doteqdot \frac{1}{s^2} + \frac{2}{s} \, e^{- 3 s}.
\end{align}
Now the differential equation, when transfo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$ by partial fractions
Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$
I know that we can use substitution to make the expression simplier before solving it, but I am trying to solve this by using partial fractions only.
$\frac{x^2}{(x-3)(x+2)^2} = \frac{A}{x-3} + \frac{B}{... | From
$$\frac{x^2}{(x-3)(x+2)^2} = \frac{A}{x-3} + \frac{B}{x+2} + \frac{C}{(x+2)^2}$$
we evaluate the coefficients directly as follows
\begin{align}
&A=\left[\frac{x^2}{(x+2)^2} - \frac{B(x-3)}{x+2} + \frac{C(x-3)}{(x+2)^2} \right]_{x=3}=\frac9{25}\\
&B=\left[\frac{x^2}{(x-3)(x+2)} - \frac{A(x+2)}{x-3} -{\frac C{x+2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How is ${n \choose n/2}$ approximated in this manner? In this question, the author asked how to get an asymptotic growth of ${n \choose n/2}$, to which the answer is as follows. First:
$$ {n \choose n/2} = \frac{n!}{(n/2)!(n- n/2)!} = \frac{n!}{(\tfrac{n}{2}!)^2} $$
Now, to find an upper bound on this, take the upper b... | Using,as you did,
$$\sqrt{2 \pi n}\ \left(\frac{n}{e}\right)^n e^{\frac{1}{12n + 1}} < n! < \sqrt{2 \pi n}\ \left(\frac{n}{e}\right)^n e^{\frac{1}{12n}}$$
$$\frac{2^{n+\frac{1}{2}} e^{\frac{1}{12 n+1}-\frac{1}{3 n}}}{\sqrt{\pi n}
}\lt \binom{n}{\frac{n}{2}} \lt \frac{2^{n+\frac{1}{2}} e^{\frac{1}{12 n}-\frac{2}{6 n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving $\frac1{2^1}+\frac2{2^2}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}$ by induction I need to prove that
$$\frac1{2^1}+\frac2{2^2}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}$$ by induction.
The base case was fine, and after that we have the induction hypothesis:
$$\frac1{2^1}+\frac2{2^2}+\cdots+\frac{k}{2^k}=2-\frac{k+2}... | You forgot to take into account the fact that the
$\dfrac{2(k + 2)}{2^{k + 1}}$
term had a negative coefficient.
Observe that
\begin{align*}
2 - \frac{2(k + 2)}{2^{k + 1}} + \frac{k + 1}{2^{k + 1}} & = 2 - \left[\frac{2(k + 2)}{2^{k + 1}} - \frac{k + 1}{2^{k + 1}}\right]\\
& = 2 - \left[\frac{2k + 4}{2^{k + 1}} + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4559625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Simplifying $\frac{1-4\cos80^\circ}{\tan20^\circ}$ I am working on simplifying the value
$$\frac{1-4\cos80^\circ}{\tan20^\circ}$$
I am looking for it to be expressed in terms of tangent, because the value seems to be $\tan40^\circ$. However, I can't find any obvious way to do so. How would simplify it to express like ... | Let $\theta = \pi/9 = 20^\circ$. We wish to show that $$1 - 4 \cos 4\theta = \tan \theta \tan 2\theta. \tag{1}$$ This suggests the tangent angle addition identity $$\tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \tag{2}$$ for the choice $$\alpha = \theta, \beta = 2\theta$$ which y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Prove that $(\sin x + \cos x)(6 - \sin x)<9$ Is there any elementary way to prove that $(\sin x + \cos x)(6 - \sin x)<9$?
I've noticed that $(\sin x + \cos x)$ has to be positive so $x \in\left(-\dfrac{\pi}{4}, \dfrac{3\pi}{4}\right)$ and then $(6 - \sin x)\in\left(6-\dfrac{\sqrt2}{2},6+ \dfrac{\sqrt2}{2}\right)$ but s... | Using the product-to-sum identity $\sin\theta\sin\phi = \frac{1}{2}(\cos(\theta - \phi) - \cos(\theta + \phi))$ to simplify:
\begin{align*}
& (\sin x + \cos x)(6 - \sin x) \\
= & \sqrt{2}\sin(x + \pi/4)(6 - \sin x) \\
= & 6\sqrt{2}\sin(x + \pi/4) - \sqrt{2}\sin x\sin(x + \pi/4) \\
= & 6\sqrt{2}\sin(x + \pi/4) - \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Using complex analysis, show that $\int_0^\infty\frac{\arctan(x)}{1+x^2}\,dx=\frac{\pi^2}8$ The result is of course confirmed by substituting $x\mapsto\tan(x)$:
$$I = \int_0^\infty \frac{\arctan(x)}{1+x^2} \, dx = \int_0^{\frac\pi2} x \, dx = \frac{\pi^2}8$$
Or integrating by parts:
$$I = \lim_{x\to\infty} \arctan^2(x)... | Complex integration can also be used to evaluate the integral; often, this is a very power tool for finding general solutions - if the system has an appropriate symmetry.
We will consider a general case:
$$I(a,b)=\int_0^\infty\frac{x^b\arctan x}{a^2+x^2}dx;\,b\in[0;1);\,a>0\tag{0}$$
We will also choose $a<1$ - it allow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4561442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the limit of a sequence by Riemann sums and mean value theorem Calculate:
$\displaystyle \lim_{n \rightarrow \infty} \left( \frac{n \pi}{4} - \left( \frac{n^2}{n^2+1^2} + \frac{n^2}{n^2+2^2} + \cdots \frac{n^2}{n^2+n^2} \right) \right)$.
I solved it by taking into account that $\displaystyle \int_0^{1} \frac{1... | Using
$$ \sum_{k=1}^{n} \frac{1}{k^2 +n^2} = \frac{i}{2 n} \, \left( \psi(1 - i n) - \psi(1 + i n) + \psi(1 + (1+i) \, n) - \psi(1 + (1-i) \, n) \right), $$
where $\psi(x)$ is the digamma function, and
$$ \psi(x) \approx \ln(x) + \frac{1}{2 x} - \sum_{j=1}^{\infty} \frac{B_{2j}}{2 \, j \, x^{2 j}}, $$
where $B_{n}$ are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4564940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Telescoping a finite sum Working through an upper-school pre-calculus book which starts with a bit of revision. Problem is that one of the questions in the development section of exercises is solvable using a technique they’ve not yet taught called ‘telescoping’.
If I process the equation by hand, I see the telescopin... | $$\frac{\sqrt{1}-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+...+\frac{\sqrt{15}-\sqrt{16}}{15-16}$$
Your work till here is correct. Let's evaluate each of this in turn. We notice that the deominantor of all the fractions is $-1$, so they evaluate to the following:
$$-(\sqrt{1}-\sqrt{2})-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4565446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating double integral on different domains $D$ $$\iint\limits_{D} \left(x^2+y^2\right)\mathrm{d}x \mathrm{d}y$$
where $D$ is given each time by
$D=x^2-y^2=1,\hspace{0.5cm} x^2-y^2=9,\hspace{0.5cm} xy=2,\hspace{0.5cm} xy=4$
I try to use Polar coordinate transformation
\begin{cases}
x &= \rho \cos \theta \\
y... | $D$ is made of two disconnected but symmetric regions. $xy$ is positive, so either both $x,y$ are positive or they are both negative. Let $D_+$ be the part of $D$ in the first quadrant. The other part is mirrored in the fourth quadrant. The integrand $x^2+y^2$ is symmetric about the origin, so its integral over $D$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4568221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If in a $\Delta ABC$, $\sqrt 3 \sin C = 2\sec A - \tan A$ and $\angle C = {\lambda ^0}$, find the value of $\lambda$ If in a $\Delta ABC$, $\sqrt 3 \sin C = 2\sec A - \tan A$ and $\angle C = {\lambda ^0}$, find the value of $\lambda$
My approach is as follow
$\sqrt 3 \sin C = 2\sec A - \tan A \Rightarrow \sqrt 3 \sin C... | Hint:
We have $$\sqrt3\sin C\cos A+\sin A=2$$
But $$\sqrt3\sin C\cos A+\sin A\le\sqrt{(\sqrt3\sin C)^2+1^2}=\sqrt{4-3\cos^2C}\le2$$
The equality occurs if $\cos C=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4568759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Determine the value of $\theta$. Given two random variables $X_1$ and $X_2$ with pdf
$$
f\left(x_i\right)= \begin{cases}\frac{1}{2 \theta}, & -\theta<x_i<\theta \\ 0, & x_i \text { otherwise }\end{cases}
$$
If it is known that $X_1$ and $X_2$ are independent and $\operatorname{Var}\left(X_1 X_2\right)=\frac{64}{9}$, de... | Note that $EX_1 = EX_2 = 0$, hence
$$
\begin{align}
Var (X_1 X_2) &= E(X_1X_2)^2 - \left(E\left(X_1X_2\right)\right)^2 \\
&= EX_1^2 EX_2^2 - (EX_1 EX_2)^2 \\
&=(EX_1^2)^2 \\
&= \left(\int_{-\theta}^\theta\frac{x^2}{2\theta} dx\right)^2 \\
&= \left(\frac{\theta^2}{3}\right)^2 \\
\end{align}
$$
Solve equatioin $\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4569593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Struggles solving : $\frac23xyy'=\sqrt{x^6-y^4}+y^2.$
Solve $\frac23xyy'=\sqrt{x^6-y^4}+y^2.$
(Hint: the equation can be boiled down to a homogeneous one.)
My attempt:
I didn't know how to obtain a homogeneous equation, so I'm going to show what I did instead.
We can write $\frac23xyy'=|x|^3\sqrt{1-\frac{y^4}{x^6}}+y... | To make it homogeneous, we use the following variable change
$y=t^{a}$ so $~dy=at^{a-1}~dt$
$\frac{2}{3}ax t^{2a-1}~dt=(\sqrt{x^{6}-t^{4a}}+t^{2a})~dx$
To be homogeneous : $2a=3$ ,so $a=\frac{3}{2}$
$xt^{2}~dt=(\sqrt{x^{6}-t^{6}}+t^{3})~dx$ , $t=zx$
$z'x+z=\frac{\sqrt{1-z^{6}}+z^{3}}{z^{2}}$
$\int\frac{z^2}{\sqrt{1-z^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4577757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solve the equation $\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=|5-x|$ Solve the equation $$\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=|5-x|$$
$x=5$ is the only real solution
We can note that $x^2-9x+24>0$ and $6x^2-59x+149>0$ for all $x$. The first thing I decided to try: as $$|5-x|=\begin{cases}5-x,x\le5\\x-5,x>5\end{cases},$$ ... | Well we must have the following conditions.
$x^2 -9x +24 \ge 0$ (other wise the square root doesn't exist)
$6x^2 - 59x + 149 \ge 0$. (ditto)
$\sqrt{x^2 - 9x + 24} \ge \sqrt{6x^2 - 59x + 149}$ (because their difference is $|5-x|$ which is $\ge 0$)
$x^2 - 9x + 24 \ge 6x^2 - 59x + 149$ (because $0 \le a \le b \implies a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Why $\lim\limits_{n \to \infty} \frac{\frac1{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim\limits_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}$? I just need clarification about answer to this question:
$$\lim_{n \to \infty} \frac{\sum\limits_{i = 1}^{n} \frac{1}{\sqrt{i}}}{\sqrt{n}} = \lim_{n \to \infty} \frac{\s... | This equality is purely algebraic. Forget about the limits.
$$\frac{\frac1{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}}=\frac{\frac1{\sqrt{n+1}}\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}.$$The numerator equals $\frac{\sqrt{n+1}+\sqrt n}{\sqrt{n+1}}$ and the denominator equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove $\frac{1}{(y+z) x^4} + \frac{1}{(x+z) y^4} + \frac{1}{(y+x) z^4}\geq\frac{3}{2}$ for $x, y, z>0$, with $xyz=1$ How to prove that $\frac{1}{(y+z) x^4} + \cfrac{1}{(x+z) y^4} + \cfrac{1}{(y+x) z^4}\geq\frac{3}{2}$ for $x, y, z>0$...
I've tried substituting $a=1/x$, $b=1/y,$ $c=1/z$, and also writing the 1 that is b... | By arithmetic mean-geometric mean inequality: $y^2z^2 + z^2x^2 \geq 2z^2xy = 2z$. Also, we can use Sedrakyan's Form of Cauchy-Schwarz inequality and arithmetic mean-geometric mean inequality we get
$S= \cfrac{1}{(y+z) x^4} + \cfrac{1}{(x+z) y^4} + \cfrac{1}{(y+x) z^4} = \cfrac{(xyz)^4}{(y+z) x^4} + \cfrac{(xyz)^4}{(x+z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4581269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Verify that $abc^2x^2+c(3a^2+b^2)x+3a^2-ab+b^2=0$ has real roots As the title states, the goal is to verify that the quadratic equation: $abc^2x^2+c(3a^2+b^2)x+3a^2-ab+b^2=0$ has real roots. This problem comes from an interschool mathematics contest for High-schoolers, here's my (brute force) attempt:
Since we need to ... | It seems your approach is the most straightforward, and you are almost done. Observe that WLOG we may assume $a, b \ge0$, and we, by AM-GM, have:
$$9a^4+10a^2b^2+b^4 \ge (6a^4+6a^2b^2)+(4a^2b^2+b^4) \ge 12a^3b +4ab^3.$$
EDIT: Another easy way is: $D=c^2(3a^4+6(a^2-ab)^2+(b^2-2ab)^2),$ not using the AM-GM inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4582566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Trig and de Moivre's theorem
A) Use de Moivre's theorem to prove that $\cos^4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$
B) Therefore deduce that $\cos(\pi/8) = \left(\frac{2 + \sqrt{2}}{4}\right)^{1/2}$
C) and write down an expression for $\cos(3\pi/8)$.
I have proved the first part of the question but I am not sure... | As noted, the firt proposition should be
$\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$
What you show is fine.
What is $\cos \frac {\pi}{8}$?
$\cos 4\frac {\pi}{8} = 8\cos^4\frac {\pi}{8} - 8\cos^2\frac {\pi}{8} +1\\
\cos \frac {\pi}{2} = 0 =8\cos^4\frac {\pi}{8} - 8\cos^2\frac {\pi}{8} +1$
Let $u = \cos \frac{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4583737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
compute $\int_0^1 f(x)dx$
A continuous function $f:[0,1)\to [0,\infty)$ satisfies $f(x/2+1/2) = f(x)+1$ for $x\in[0,1)$ and $f(1-x) = 1/f(x)$ for all $x\in (0,1)$. Compute $\int_0^1 f(x)dx$.
This is question 11 from this problem set.
$f(x)$ cannot be zero for any $x>0$ since otherwise $f(1-x) = 1/f(x)$ would not ho... | The first thing to notice is that $f(x)+f(1/2-x)=1$ whenever $0<x<\frac{1}{2}$:
\begin{align*}
f(x)+f\left(\frac{1}{2}-x\right)&=\frac{1}{f(1-x)}+\frac{1}{f\left(\frac{1}{2}+x\right)}\\
&=\frac{1}{f(1-2x)+1}+\frac{1}{f(2x)+1}\\
&=\frac{1}{\frac{1}{f(2x)}+1}+\frac{1}{f(2x)+1}\\
&=\frac{f(2x)}{1+f(2x)}+\frac{1}{f(2x)+1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Diophantine equation ${x ^ 2}+{y ^ 2}+{z ^ 2}={m ^ 2}+{n ^ 2} $ has infinite solutions. What is its general solution formula? Diophantine equation ${x ^ 2}+{y ^ 2}+{z ^ 2}={m ^ 2}+{n ^ 2} $ has infinite solutions. For examples,
$\begin{array}{l}
{1^2}+ {3^2} + {4^2} = {1^2} + {5^2}\\
{1^2} + {2^2} + {6^2} = {4^2} + {5... | A very simple equation. You can write various combinations.
https://artofproblemsolving.com/community/c3046h1055645_quadratic_diophantine_equation
For the equation:
$$X_1^2+X_2^2=Y_1^2+Y_2^2+Y_3^2$$
Solutions have the form:
$$X_1=t^2+2(p+s-k)t+2k^2+2p^2+4ps-4pk-2sk$$
$$X_2=t^2+2(p+s-k)t+2k^2+2s^2+4ps-2pk-4sk$$
$$Y_1=t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to find the shape of $x^3 + y^3 - 3xy = 0$ using polar coordinates? There is a curve known as the Folium of Descartes, and it has the equation $x^3 + y^3 - 3axy = 0$. I wanted to know whether it was possible to determine the shape of the case a = 1, using polar coordinates. I have only determined its polar coordina... | You can write $r^2$ in terms of $\sin 2\theta$, which shows some of the symmetries of the curve: $$(s^3+c^3)^2 = (s^2+c^2)^3 +2s^3c^3 - 3s^2c^2(s^2+c^2)$$so $$r^2=\dfrac{\tfrac{9}{4}\sin^2 2\theta}{1+\tfrac{1}{4}\sin^3 2\theta-\tfrac{3}{4}\sin^2 2\theta}=\dfrac{9\sin^2 2\theta}{4+\sin^3 2\theta-3\sin^2 2\theta}=\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Right triangle with its perimeter and median and altitude A right triangle $ABC$ is given with $\measuredangle ACB=90^\circ$. If the perimeter of the triangle is $72$ and the difference between the lengths of the median and the altitude to the hypotenuse is $7$, find the area.
Let $CD=x$ $(x>0)$, then $CM=x+7$. By the... | For the last three equations, there is a way to solve them.
$$a+b=72-c$$
$$a^2+2ab+b^2=5184-144c+c^2$$
$$2ab=5184-144c$$
Now, we know that:
$$\frac{c}{2}-\frac{ab}{c}=7$$
$$2ab=c^2-14c$$
$$c^2-14c=5184-144c$$
And that yields $c=32$ and $c=-162$, and obviously we'll use the positive value. Since we now have $c$, its ver... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4586439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Is it possible to simplify $n \equiv - \frac d 4\bigg(\frac 3 2\bigg)^{m-2} - 2 \pmod {3^m}$ further? Is it possible to simplify this further:
$$
n \equiv - \frac d 4\bigg(\frac 3 2\bigg)^{m-2} - 2 \pmod {3^m} \qquad \text{where } n, d \geq 0 \text{ and } m \geq 2 \text{ are integers}
$$
or to glean some info on what $... | Simplfy $n \equiv - \dfrac d 4\bigg(\dfrac 3 2\bigg)^{m-2} - 2 \pmod {3^m} \qquad \text{where } n, d \geq 0 \text{ and } m \geq 2 \text{ are integers}$.
Multiplying by $\dfrac94$ we get
$$\dfrac94n\equiv -\dfrac d4 \left(\dfrac 32\right)^m-\dfrac 92\pmod{3^m}\Rightarrow 9n\equiv \left(\dfrac{-d}{2^m}\right)3^m-18\pmod{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4588115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that the sums are equal to: a) If $\sum_{k=1}^∞ \frac{1}{k^2} = \frac{π^2}{6}$ show that $\sum_{k=1}^∞ \frac{1}{(2k+1)^2} = \frac{π^2}{8}$
b) Show that $\sum_{k=1}^∞ \frac{1}{k(k+1)(k+2)} = \frac{1}{4}$
Hint for b): Find A, B, C so that for ∀k:
$\frac{1}{k(k+1)(k+2)}$= $\frac{A}{k}$ +$\frac{B}{k+1}$+ $\frac{C}{k+... | If you find the constants $A,B,C$, they are $\frac{1}{2},-1,\frac{1}{2}$ respectively, then you term
$\frac{1}{k(k+1)(k+2)}=\frac{1}{2k}-\frac{1}{k+1}+\frac{1}{2(k+2)}=\frac{1}{2}(\frac{1}{k}-\frac{1}{k+1})+\frac{1}{2}(\frac{1}{k+2}-\frac{1}{k+1})$ if you take the series (Harmonic series) you limit is $\frac{1}{2}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4588721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Describe the set of real numbers $x$ which satisfy $2\log_{2x+3}x<1$. Describe the set of real numbers $x$ which satisfy $2\log_{2x+3}x<1$.
Here, I am really stuck . In this problem, I tried to compute by writing this expression as $(2x+3)^a=x^2$, where $a<1$. But, I don't have a clue what to do ? I am not quite getti... | General approach.
Notice that, the logarithm $\log_{2x+3}x$ is defined only by the following condition:
$$x>0\wedge 2x+3>0\wedge 2x+3≠1$$
This is equivalent to the condition $x>0$.
Remember that,
If $a>1$ and $b,c>0$ then the inequality $\log_a b<\log_a c$ is equivalent to
$b<c$ and if $0<a<1$ and $b,c>0$ then the ine... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4589033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Help w/$(\frac{a}{b})^4+(\frac{b}{c})^4+(\frac{c}{d})^4+(\frac{d}{e})^4+(\frac{e}{a})^4\ge\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{e}{d}+\frac{a}{e}$ How exactly do I solve this problem? (Source: 1984 British Math Olympiad #3 part II)
\begin{equation*}
\bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\fra... | Applying the AM-GM
$$LHS - \bigl(\frac{e}{a}\bigr)^4 = \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 \ge4 \cdot \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{d} \cdot\frac{d}{e} = 4\cdot\frac{a}{e} $$
Do the same thing for these 4 others terms, and mak... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4591726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The sum $\frac{1}{2\sqrt1+1\sqrt 2}+\frac{1}{3\sqrt2+2\sqrt 3}+\cdots +\frac{1}{100\sqrt{99}+99\sqrt{100}}$ is equal to: From an exercise of a textbook of an high school of 15 years old, I have this partial sum
$$S_n=\frac{1}{2\sqrt1+1\sqrt 2}+\frac{1}{3\sqrt2+2\sqrt 3}+\cdots +\frac{1}{100\sqrt{99}+99\sqrt{100}}$$
Con... | We can split the terms into two that can be eliminated between adjacent terms.
\begin{align}
\sum_{i=1}^{99}{\frac{1}{\left( i+1 \right) \sqrt{i}+i\sqrt{i+1}}}&=\sum_{i=1}^{99}{\frac{\left( i+1 \right) \sqrt{i}-i\sqrt{i+1}}{\left( i+1 \right) ^2i-i^2\left( i+1 \right)}}=\sum_{i=1}^{99}{\frac{\left( i+1 \right) \sqrt{i}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4591969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Need advice: what should be my next step for solving the derivative of $f(z)$ using the definition? What should be my next step for solving the derivative of $f(z)$ using the definition?
$$f(z) = (2{z^2} + 1) \cdot ({z^3} - \sqrt {z}) $$
$$f'(z) = \mathop {\lim }\limits_{\vartriangle z \to 0} \frac{{f(z + \vartriangle ... | After distributing the terms we have...
$$ f(z) = (2{z^2} + 1) \cdot ({z^3} - \sqrt {z}) = 2z^5 - 2z^{
\frac{5}{2}} + z^3 - z^{\frac{1}{2}} $$
by definition of derivative...
$$ f'(z) = \lim _{\Delta z \to 0} \frac{f(z+\Delta z) - f(z)}{\Delta z} $$
$$ = \lim _{\Delta z \to 0} \frac{2{(z+\Delta z)}^5 - 2(z+\Delta z)^{\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4592177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why does cancelling by $\sin x$ when solving $4\tan x = 5\sin x$ for $0\leq x < 2\pi $ miss solutions? So the solution to the problem is:
$$4 \tan x = 5 \sin x$$
$$4 \frac{\sin x}{\cos x} = 5 \sin x$$
$$4 \sin x = 5 \sin x \cos x$$
So, either $\sin x = 0$ or $\cos x = 4/5$. I'm fine with the logic of this. But is there... | We can’t cancel terms on both sides without knowing whether it is zero.
Just like solving
$x^2=x \Leftrightarrow x^2-x=0 \Leftrightarrow x(x-1)=0 \Leftrightarrow x=0 \textrm{ or }x=1, \tag*{} $
we shall miss the solution $x=0.$
Back to our equation, we have
$$
\begin{aligned}
& 4 \tan x=5 \sin x \\
\Leftrightarrow \qu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4592572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ I would like to solve the equation $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ analytically, without using Wolfram Alpha. I have tried several substitutions, including $x=\cos(t)$, $\sqrt{1-x^2}=t$, but they all fail and eventually result in a quartic polynomial which i... | Another variation.
We observe that, $\sqrt {1-x^2}≥0$ and $1+\sqrt {1-x^2}>0$ imply $x≥0$.
Letting $\sqrt {1-x^2}=u,~u≥0$ we have:
$$\begin{cases}
ux=u-x\\u^2+x^2=1\end{cases}$$
Then setting $u=a+b,~ x=a-b$ under the restriction $a≥b≥0$, we have:
$$\begin{align}&\begin{cases}a^2-b^2=2b\\a^2+b^2=\frac 12\end{cases}\\
\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4593991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
deriving pdf with two methods, with two different results I have this problem to determine a density function $f_{Y}(y)$ of $Y$ given $X$ as a random variable with uniform distribution on $[−1, 1]$ and $Y = X^2$. So: $f_X(x) = \frac{1}{2}$ for $x \in [-1, 1]$. I approached this problem with two different ways, 1) throu... | This idea that you proposed only works if the transform $g(\cdot)$ is one to one.
https://www.math.arizona.edu/~jwatkins/f-transform.pdf
This is why the following reasoning is incorrect, for instance, if $g(x)=x^2$
\begin{align}
f_Y(y)&=f_X(g^{-1}(y))\Big| \frac{d}{dy}g^{-1}(y) \Big| \\
&= \frac{1}{2} \Big| \frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate expression with 3 multiplications $\int_0^{\frac\pi4}{x\cos(x)\sin(2nx)}dx$ Specifically I want to integrate the following:
$$\int_0^{\frac\pi4}{x\cos(x)\sin(2nx)}$$
I do know that
$$\frac4{\pi}\int_0^{\frac\pi4}{x\cos(x)\sin(2nx)} = \dfrac{-16\cos(n\pi)}{(4n^2-1)^2{\pi}}$$
I just have no idea how to integrat... | Using prostapheresis formulas:
$$\sin(p)+\sin(q)=2\cdot\sin\left(\frac{p+q}{2}\right)\cdot\cos\left(\frac{p-q}{2}\right)$$
One can write:
$$\left\{\begin{matrix}
\frac{p+q}{2}=2nx
\\\frac{p-q}{2}=x
\end{matrix}\right.$$
Which leads to:
$$\left\{\begin{matrix}
q=(2n-1)x
\\p=(2n+1)x
\end{matrix}\right.$$
So:
$$\int_0^{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proof by induction that $17n^3+103n$ is a multiple of 6 Problem use induction to prove that $17n^3+103n$ is a multiple of 6.
Im new to learning proof by induction and was wondering if my proof would be acceptable for my maths test coming up.
Proof. base case $n=1 = 120$ $17k^3+103k=6a$ for some $a$. then we prove $k+1$... | Your base case is correct: $17(1)^3+103(1)=120$ which is divisible by $6$.
For the inductive step we assume $17k^3+103k$ is divisible by $6$ and consider $17(k+1)^3+103(k+1)$.
Expanding this we get $$17(k+1)^3+103(k+1)=17(k^3+3k^2+3k+1)+103k +103=(17k^3+103k)+51k^2+51k+120$$
We know that $17k^3+103k$ is divisible by $6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $\iint_{x^2-xy+y^2\leq1}(x-y)^2\mathrm{d}x\mathrm{d}y$ $$
I:=\iint_{x^2-xy+y^2\leq1}(x-y)^2\mathrm{d}x\mathrm{d}y
$$
I've got the following 2 colored equations but neither seems useful for evaluating this double integral.
$$\begin{align}
x^2-xy+y^2&= \color{blue}{\left(x- {1 \over 2 }y \right)^2+ {3 \ove... | Observe that $a(x+y)^2+b(x-y)^2=x^2+y^2-xy$ if and only if $a=\frac1{4}$ and $b=\frac{3}{4}$. Now let the change of variable $T(x,y):=\frac1{2}(x+y,\sqrt{3}(x-y))=(z,t)$ and notice that $T=\left[\begin{smallmatrix}1/2&1/2\\\sqrt{3}/2&-\sqrt{3}/2\end{smallmatrix}\right]$ is an invertible linear map, therefore
$$
\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4596255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Determinant of $A_{2n}$ Determine the determinant of $$A_{2n}:=\begin{pmatrix}
a &0 &\cdots & \cdots &0 &b \\
0& a & & & b&0 \\
\vdots &\vdots &\ddots &\ddots &\vdots &\vdots \\
\vdots&\vdots & \ddots & \ddots &\vdots &\vdots \\
0& b & & & a&0 \\
b&0 &\cdots &\cdots &0 &a
\end{pmatr... | We expand the determinant along the first row:
$$
\det A_{2n}=aM_{11}+(-1)^{2n+1}bM_{1,2n}
$$
Now we expand $M_{11}$ and $M_{1,2n}$ along the last row:
$$
\det A_{2n}=a^2\det A_{2n-2}-b^2\det A_{2n-2}=(a^2-b^2)\det A_{2n-2}.
$$
Now induction works:
$$
\det A_{2n}=(a^2-b^2)(a^2-b^2)^{n-1}=(a^2-b^2)^n.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4597826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find a permutation $X \in S_6$ which satisfies the equation $\begin{pmatrix}1&2&3&4&5&6\\ 4&3&1&5&2&6\end{pmatrix} \circ X = \begin{pmatrix}1&2&3&4&5&6\\ 3&2&4&5&1&6\end{pmatrix}$
I've tried this solution and checked this website, but none of them were helpful or my problem is little bit different and may require more ... | Hint:
$$X=\begin{pmatrix}1&2&3&4&5&6\\ 4&3&1&5&2&6\end{pmatrix}^{-1}\circ \begin{pmatrix}1&2&3&4&5&6\\ 3&2&4&5&1&6\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4598172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrals of the form $\int_0^1\left(\frac{1}{x+1}-\ln\left|\frac{x+1}{2}\right|-\ln\ln\left|\frac{x+1}{x-1}\right|\right)dx$
How could we evaluate the following integrals? (Integrands pictures above)
Blue:
\begin{align}
\int_0^1 \left(\frac{1}{x+1}-\ln\left|\frac{x+1}{2}\right|-\ln\ln\left|\frac{x+1}{x-1}\right|\righ... | A solution using Riemann $\zeta$, with $\zeta(0)=-1/2$ and $\zeta'(0)/\zeta(0)=\ln2\pi$. Let \begin{align}
F(a,b)&=\int_a^b\left(\frac1{x+1}+\ln\left|\frac{x-1}{2}\right|+\ln\ln\left|\frac{x+1}{x-1}\right|\right)dx,\\
G(a,b)&=\int_a^b\left(\frac1{x+1}-\ln\left|\frac{x+1}{2}\right|-\ln\ln\left|\frac{x+1}{x-1}\right|\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_0^\infty\frac{\sqrt[5]{x}}{(x^2+1)(x+1)^2}dx$ using Gamma and Beta functions I need to evaluate the following integral, using Gamma and Beta functions:
$$\int_0^\infty\frac{\sqrt[5]{x}}{(x^2+1)(x+1)^2}dx$$
I tried to use partial fractions, but I am not sure if that is correct, since doing that I get that... | We can make your approach work by modifying it slightly. Consider the integral
$$I(s)=\int_0^\infty\frac{x^{s-1}}{(1+x^2)(1+x)^2}\,\mathrm dx$$
instead. Applying your partial fraction expansion, an integral representation of the beta function $\text{B}(x,\,y)=\int_0^\infty\frac{t^{x-1}}{(1+t)^{x+y}}\,\mathrm dt$, and t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Two linear equations with three unknowns and a magnitude constraint Given $u,v,w,x,y,z$ is there a way to solve the following for $a,b,c$, assuming that there is a solution (e.g. $(u,v,w)$ and $(x,y,z)$ are not proportional):
$$\begin{aligned}
au+bv &= cw\\
ax+by &= cz\\
a^2 + b^2 &= 1
\end{aligned}$$
An equivalent pro... | The system should be digestible by a Grobner basis computation. I ran it in Mathematica V13 with respect to the monomial order $\{a, b, c\}$, with the variables $b$ and $c$ eliminated
GroebnerBasis[{a u + b v - c w, a x + b y - c z, a^2 + b^2 - 1}, {a, b, c}, {b, c}]
and obtained the following polynomial for $a$
$$a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4604461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Frullani like Trig integral $$\int_{0}^{\infty}\frac{\left|\cos\left ( x-\frac{\pi}{4} \right ) \right|- \left|\cos\left ( x+\frac{\pi}{4} \right ) \right| }{x}dx=\sqrt{2}\ln(1+\sqrt{2})$$ The above integral seems to look like a frullani type integral and has a closed form in terms of natural log. I tried to indefinite... | Let $$I = \int_{0}^{\infty} \frac{\left|\cos\left(x-\frac{\pi}{4}\right) \right|-\left|\cos\left(x+\frac{\pi}{4}\right) \right|}{x} \, \mathrm dx .$$
Then
$ \require{cancel} \begin{align}I &= \int_{0}^{\infty}\frac{\left|\cos\left(x-\frac{\pi}{4}\right) \right|-\left|\sin\left(x-\frac{\pi}{4} \right) \right|}{x} \, \ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
In a triangle $\triangle ABC$, the sum of two sides is $x$, and their product is $y$. If $x^2-c^2=y$, find $\frac{r}{R}$ in terms of $x, c, y$ This problem comes from a previous JEE Advanced examination, the problem is as follows:
In a triangle $\triangle ABC$, the sum of two sides is $x$, and their product is $y$. If... | Let’s assume $S$ is the Area of the triangle and $P$ is half of the perimeter.
If $x=a+b$ and $y=ab$ $\\$ then $(a-b)^2=(a+b)^2-4ab=x^2-4y$.
We know that $S=\frac{abc}{4S}$ and $r=\frac{S}{P}$ then using the Heron’s formula we have:$$\frac{r}{R}=\frac{\frac{S}{P}}{\frac{abc}{4S}}=\frac{4S^2}{Pabc}=\frac{4(P-a)(P-b)(P-c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Help Solve $\int \sqrt{x^{2}+x-2}\,dx$ I have this integral $\int \sqrt{x^{2}+x-2}\,dx=\int |x-1|\sqrt{\frac{x+2}{x-1}}\,dx$.
My textbook advices to use the substitution $t^{2}=\frac{x+2}{x-1}$.
Observation: This integral should be solvable without using integral of modules as it is placed before moduled function integ... | $$
\begin{aligned}
I &= \int{\sqrt{x^2 + x - 2}dx} = \int{\underbrace{\sqrt{x^2 + x - 2}}_{u}\underbrace{dx}_{dv}} = |\text{ integrating by parts }| = \\
&= x\sqrt{x^2+x-2}-\int{xd\sqrt{x^2+ x-2}} = \\
&= x\sqrt{x^2+x-2}-\int{x\frac{2x+1}{2\sqrt{x^2+x-2}}dx} = \\
&= x\sqrt{x^2+x-2}-\int{\frac{2\left(x^2+x-2\right)-\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Any better way to find $\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)$
Find the value of $\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)$
My Method:
I used the following Identities:
\begin{aligned}
& \sin A \cos B-\cos A \sin B=\sin (... | Yet an other way to proceed, which arguably explains why we get a "nice value" for the sum. Let $c_1,c_2,c_3$ be the three different values of the cotangent function computed in $x_1,x_2,x_3$, which are respectively
$10^\circ$, $70^\circ$, $130^\circ$. These angles are of the shape $10^\circ+k\cdot 60^\circ$, so taking... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What is the concrete mathematical argument why $a^3+a=b^3+b\Rightarrow a=b$ is true in $\mathbb{R}$? The question arose from the following proof:
$f(x)=x^3+x+1$ is injective:
Let $a,b\in \mathbb{R}$ and $f(a)=f(b)$:
$f(a)=f(b)\Leftrightarrow a^3+a+1=b^3+b+1\Leftrightarrow a^3+a=b^3+b \Rightarrow a=b$
Is $a^3+a=b^3+b\Ri... | If $a^3+a=b^3+b$, then $$(a^3-b^3)+(a-b)=0 \quad \Leftrightarrow $$
$$(a-b)(a^2+ab+b^2+1)=0.$$
However, $a^2+ab+b^2+1=(a+\frac{b}{2})^2+\frac{3b^2}{4}+1>0$. This immediately implies that $a-b=0$, i.e. $a=b$. Please let me know does it make sense.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the symmetric matrix given its eigenvalues and eigenvector.
$A$ is a $3 \times 3$ symmetric matrix. It has the eigenvalue $\lambda_1 = 3$ with the eigenvector
$$
\begin{bmatrix}
1 \\ 0 \\ -1
\end{bmatrix}
$$
and a double eigenvalue $\lambda_2 = -1$.
Find the matrix $A$.
Since $A$ is symmetric, it can be diagon... | Since $3, -1, -1$ are all eigenvalues of $A$, $4, 0, 0$ are all eigenvalues of the matrix $A + I$, which means the rank of the symmetric matrix $A + I$ is $1$, hence $A + I = vv^T$ for some length-$3$ vector $v$.
Denote $\begin{bmatrix} 1 & 0 & -1 \end{bmatrix}^T$ by $w$, $Aw = 3w$ and $A + I = vv^T$ implies that
\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1$, prove $abcd \geq 3$. Given four positive real numbers $a,b,c,d$. It is given that $\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1$. Prove $abcd \geq 3$.
I've applied AM-HM inequality and got the result, $a^4+b^4+c^4+d^4 \geq 12$... | Let
$$w=\frac{1}{1+a^4}, x=\frac{1}{1+b^4}$$$$ y=\frac{1}{1+c^4}, z=\frac{1}{1+b^4}$$
So, $$a^4=\frac{1-w}{w}=\frac{x+y+z}{w}$$$$ b^4=\frac{1-x}{x}=\frac{w+y+z}{x}$$$$ c^4=\frac{1-y}{y}=\frac{w+x+z}{y}$$$$ d^4=\frac{1-z}{z}=\frac{w+x+y}{z}$$
$$a^4b^4c^4d^4=(\frac{x+y+z}{w}) (\frac{w+y+z}{x})( \frac{w+x+z}{y} )(\frac{w+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4612949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Prove $\sum_{i=1}^{n} \frac{1}{(i²+i)^\frac{3}{4}}>2- \frac{2}{\sqrt{n+1}}$
Let $i\in\mathbb{N}$ $$\sum_{i=1}^{n} \frac{1}{(i²+i)^\frac{3}{4}}>2-\frac{2}{\sqrt{n+1}}$$
A.M.-G.M. inequality makes this problem complex to solve.
It was true for $n=1,2,3,4$ then solving this became difficult. How can I solve this?
| Note that since
$$
(i^2 + i)^{\frac34} = (i^2 + i)^{\frac12} \cdot i^\frac14 \cdot (i+1)^\frac14 < \sqrt{i} \cdot \sqrt{i+1} \cdot \frac{\sqrt{i} + \sqrt{i+1} }{2} = \frac12 \cdot \frac{\sqrt{i} \cdot \sqrt{i+1}}{\sqrt{i+1} - \sqrt{i}},
$$
we have
$$
\frac{1}{(i^2 + i)^{\frac34}} > \frac{2(\sqrt{i+1} - \sqrt{i})}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"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.