Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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What is the equation of the circle having a radius of square root of 85, through (5,9) and (1,-7)? In solving this problem, I first solved the slope of the perpendicular bisector and I solved the midpoint of the two points. I don't know what to do next. I hope you can help me.
| Let center to be $(a,b)$. $$(a-5)^2+(b-9)^2=85, (a-1)^2+(b+7)^2=85$$
$$a^2-10a+b^2-18b=-21,a^2-2a+b^2+14b=35$$
Subtracting gives $$8a+32b=56\Rightarrow a+4b=7\Rightarrow a=7-4b$$
$$35=a^2-2a+b^2+14b=49-56b+16b^2-14+8b+b^2+14b=35+17b^2-34b\Rightarrow b^2-2b=0$$
$$b=0\Rightarrow a=7-4b=7; b=2 \Rightarrow a=7-4b=-1$$
$$(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4442791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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$\forall a\ge 2, \exists n\ge 1$ such that $2n+1\mid a+n!$ Let $a\ge 2$ be an integer. Prove that $\exists n\ge 1$ such that $$2n+1\mid a+n! \text{ or }a-n!$$
I think we'll construct a solution for $n$ as a function of $a$. But I have no idea what that function looks like. I just took care of some trivial cases.
First ... | Note that if $2n+1$ is prime, then by Wilson's theorem:
$$-1 \equiv (2n)!\equiv (-1)^{n}(n!)^2 \mod (2n+1)$$
Thus $(n!)^2 \equiv (-1)^{n+1} \mod (2n+1)$
Thus if $2n+1$ is a prime of the form $1\text{mod}$ $ 4$ then $(n)!^2 \equiv -1 \mod(2n+1)$ (also some thanks to John Omielan).
Claim: For all $a \geq 2$, $a^2+1$ cont... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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A beautiful problem on the Pigeonhole Principle I recently solved this beautiful mathematical problem on Pigeonhole Principle from Romania TST 2000.
Let $S$ be the set of interior points of a sphere in three-dimensional space and let $D$ be the set of interior points of a disc on a plane. For any two points $X,$ and $... | We claim that such a function does not exist. To prove this claim, we will assume that there exists such a function $f$, and arrive at a contradiction.
Suppose that there exists such a function $f: S\rightarrow D,$ then we will consider a cube of side length $a$ that is completely in the sphere.
Fix a random positive i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4450772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Eigenvalue and spectral condition
Let $A=\begin{pmatrix} 1& 1 \\ a^2 &1 \end{pmatrix} \text{ with } a\in (0,\frac{1}{2}]$. Show $$cond_2(A)=||A||_2 \cdot ||A^{-1}||_2\leq 4(1-a^2)^{-1}$$ by first showing $||A||^2_2\leq||A||_1||A||_{\infty}$.
$||A||^2_2$ is the maximal eigenvalue of $A^TA$ and $||A||_1||A||_{\infty}=2... | If $\lambda_1,..., \lambda_n$ are the eigenvalues of an invertible $n\times n$ matrix $M$, then $\frac{1}{\lambda_1},\dots,\frac{1}{\lambda_n}$ are the eigenvalues of $M^{-1}$. Moreover, the eigenvalues of $M^T$ are the same as the eigenvalues of $M$.
Putting this together, we have that the eigenvalues of $(A^{-1})^{T}... | {
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"timestamp": "2023-03-29T00:00:00",
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How can $x=1$ be a solution to this equation? This is the equation I am supposed to solve:
$$\log_{3}(2x^{2} - x-1)\ -\ \log_{3}(\ x-1)\ = 2$$
The textbook gives the solutions, $x=1, x=4$, with the working out as shown below:
While I understand what they are doing, I don't completely understand why they work it out as... | $x=1$ cannot be a solution because the equation is undefined for $x=1$. The equation in reals is only meaningful if $x-1 > 0$ and $2x^2-x-1>0$, which means $x>1$. In that case one may rewrite the equation as follows:
$$\begin{align}
\log_3(2x^2-x-1) - \log_3(x-1)
&= \log_3\big((2x+1)(x-1)\big) - \log_3(x-1) \\
&= ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is $\tan^{-1}(0)=\pi$ or is $\tan^{-1}(0)=0$? This question came in the Dhaka University admission exam 2006-7
Q) The value of $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3$ is -
(a) $0$
(b) $\frac{\pi}{2}$
(c) $\pi$
(d) $2\pi$
My attempt:
$$\tan^{-1}1+\tan^{-1}2+\tan^{-1}3$$
$$=\tan^{-1}\frac{1+2+3-1\cdot2\cdot3}{1-1\cdot2-2\cdot... | Let's find the value of $\arctan 2 + \arctan 3$ first...
Using the formula $$\arctan x + \arctan y = \arctan \left(\dfrac {x+y}{1-xy}\right)$$ we have $$\arctan 2 + \arctan 3 = \arctan \left(\dfrac {2+3}{1-2 \cdot 3}\right) \\ = \arctan \dfrac {5}{-5} \\ = \arctan (-1) \\ = \dfrac {3 \pi}{4}$$ but since $\arctan 1 = \d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4459550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do you finish the equations for $S^2 + \frac{7}{4}S= \frac{1}{2}$? Complete the Square:
$$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}$$
THIS is where I get stuck. I know that the Square Root of $\frac{81}{64}$ is $\frac{... | When you complete the square, assuming the coefficient on the quadratic term is $1$, you take the coefficient on the linear term and divide by two and then square it. That gives you two numbers,
$b/2$ and $b^2/4$. Don't throw away the $b/2$. See where it comes in:
$$x^2+bx = c$$
$$x^2+bx+\frac{b^2}{4} = c+\frac{b^2}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Range of $f(x) = x+ \frac {1}{x}$ for $x<0$ One method to do this is by re-writing the function:
$$ x + \frac{1}{x} = (\sqrt{x})^2 + \left(\frac{1}{\sqrt{x}}\right)^2 $$
Then adding and subtracting 2 from RHS:
$$ = (\sqrt{x})^2 + \left(\frac{1}{\sqrt{x}}\right)^2 - 2 + 2 $$
$$=(\sqrt{x})^2 + \left(\frac{1}{\sqrt{x}}\r... | Since for if $x \neq 0, |x| > 0 \implies f(|x|) \ge 2 \implies f(x) = -f(|x|) \le -2$. This means $(1) \implies (2)$. Alternatively, set $x = -x' \implies x' > 0\implies f(x) = x+\dfrac{1}{x} = -\left(x'+\dfrac{1}{x'}\right)= -\left(\sqrt{x'}-\dfrac{1}{\sqrt{x'}}\right)^2-2\le -2$.
| {
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"url": "https://math.stackexchange.com/questions/4462829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving the integral of $\cos^2x\sin^2 x$ Solving the integral of $\cos^2x\sin^2 x$:
My steps are: $(\cos x\sin x)^2=\left(\frac{\sin(2x)}{2}\right)^2$. Now we know that
$$\sin^2(\alpha)=\frac{1-\cos (2x)}{2}\iff\left(\frac{\sin(2x)}{2}\right)^2=\frac 14\sin^2(2x)=\frac 14\cdot \frac{1-\cos (4x)}{2}=\frac 18(1-\cos (4x... | You can excpress $\cos^2 x$ as $1 - \sin^2 x$ and hence your primitive becomes:
$$\int \sin^2 x d x - \int \sin^4 x dx$$
which comes to the same answer in the end.
Full working:
$$\int \sin^2 a x \cos^2 a x \, \mathrm d x$$
$$=\int \sin^2 a x (1 - \sin^2 a x) \, \mathrm d x$$
$$=\int \sin^2 a x \, \mathrm d x - \int \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4465528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\int_0^\infty \frac{\arctan (x^2) \ln(1+x^4)}{x(x^8+x^4+1)} \mathrm dx$ So, my friend has challenged me to solve the following integral
$\displaystyle \tag*{} I=\int_0^\infty \frac{\arctan (x^2) \ln(1+x^4)}{x(x^8+x^4+1)} \mathrm dx$
I started by doing $x^2=t$, so
$\displaystyle \tag{1} I=\frac 12 \int_0^\i... | Note that
$$\Im\left(-\frac{2}{\sqrt{3}\left(x^2+\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)} \right) = \frac{1}{x^4+x^2+1}$$
Hence
$$ I=\frac 12 \int_0^\infty \frac{\arctan (x) \ln(1+x^2)}{x(x^4+x^2+1)} \mathrm dx = \Im \left( -\frac{1}{\sqrt{3}} \int_0^\infty \frac{\arctan (x) \ln(1+x^2)}{x\left(x^2+\frac{1}{2}+\frac{i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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Determine area bounded by $(x+y)^4 = ax^2y$ Determine the area of region $U$ bounded by the graphic of the curve:
$$(x+y)^4 = ax^2y,\ a > 0 \quad \text{(loop in the first quadrant)}$$
I used polar coordinates and I arrived to:
$$\iint r\,dr\,d\theta = \frac{a^2}{2}\int_0^{\frac{\pi}{2}}
\frac{\cos^4 \theta \,\sin^2 \th... | \begin{align}
&\int_0^{\pi/2}
\frac{\cos^4 \theta \,\sin^2 \theta} {(\cos \theta + \sin \theta)^8}
\,d\theta\\
=& \int_0^{\pi/2}
\frac{\tan^2 \theta \,\sec^2 \theta} {(1+ \tan \theta)^8}
\,d\theta
\overset{x=\tan\theta}=\int_0^{\infty}
\frac{x^2}{(1+x)^8}dx\\
=& \int_0^{\infty}
\left( \frac{1}{(1+x)^6}-\frac{2}{(1+x)^7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove close form $\int_a^1 \frac{x}{\sqrt{x^2-a^2}}\ln\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}dx=\frac{\pi}{2}(1-a)$ I found this integral in a textbook (stated without proof),
$$J(a)=\int_a^1 \frac{x}{\sqrt{x^2-a^2}}\ln\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}dx=\frac{\pi}{2}(1-a)$$
with $0\le\,a\... | Integrate by parts with $d\left( \sqrt{x^2-a^2}\right)= \frac{x}{\sqrt{x^2-a^2}}dx$
\begin{align}
&\int_a^1 \frac{x}{\sqrt{x^2-a^2}}\ln\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\ dx\\
=& \int_a^1\frac1x \sqrt{\frac{x^2-a^2}{1-x^2}}dx \overset{y^2=\frac{x^2-a^2}{1-x^2}}=\int_0^\infty
\frac{(1-a^2) y^2}{(y^2+a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4471072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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What are the eigenvalues of $\begin{pmatrix} A & B \\ B &-A \end{pmatrix}$ in terms of $A$ and $B$? It is known that the set of eigenvalues of the following block matrix
$$ C = \begin{pmatrix} A & B \\ B & A \end{pmatrix} $$
is the union of the eigenvalues of the matrices $A + B$ and $A - B$. I am interested in the mat... | Partial answer when $A$ has inverse $A^{-1}$. Use the formula for the determinant of block-partitioned matrices in terms of Schur's complement. If $(A-\lambda I)$ has an inverse then $\det(A-\lambda I)\neq 0$
\begin{align}
\det \left(\begin{bmatrix} A&B \\ B& -A \end{bmatrix}-\lambda\begin{bmatrix} I&0 \\ 0& I
\end{bm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4471978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Let $x^2 + 3x +1 = 0$. Is $x^{2048} + \dfrac{1}{x^{2048}}$ divisible by 3? Let $x^2 + 3x +1 = 0$. Solve for $x^{2048} + \dfrac{1}{x^{2048}}$. Is it divisible by 3?
$x^2 + 1 = -3x \Rightarrow x+ \dfrac{1}{x} = -3$
$x^2 + \dfrac{1}{x^2} = (x + \dfrac{1}{x})^2 - 2 = 9 -2 = 7$
$x^4 + \dfrac{1}{x^4} = (x^2 + \dfrac{1}{x^2}... | Let $x$ be a root of $x^2 + 3x + 1$. Then $y = 1/x$ is also a root, since $$y^2 + 3y + 1 = (1/x)^2 + 3/x + 1 = \frac{1 + 3x + x^2}{x^2}$$ and $x \ne 0$. Moreover, for each positive integer $n$,
$$x^{n+1} + y^{n+1} = (x+y)(x^n + y^n) - xy(x^{n-1} + y^{n-1}).$$
Since $x + y = -3$ and $xy = 1$, it follows that
$$x^{n+1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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... | $a+b+c=3d, a^2+b^2+c^2=4d^2-2d+1$
Just set $(a, b, c)=(k_1d+l_1, k_2d+l_2, k_3d+l_3)$, since if we have quadratic polynomials for $a, b, c$, the 4th degree polynomials have to be in $a^2+b^2+c^2$.
$4d^2-2d+1=(k_1^2+k_2^2+k_3^2)d^2+2(k_1l_1+k_2l_2+k_3l_3)d+(l_1^2+l_2^2+l_3^2).$
$k_1^2+k_2^2+k_3^2=4, k_1l_1+k_2l_2+k_3l_3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding $\lim _{n\to \infty}n\sin (2\pi \sqrt{1+n^2})$ Question:
Find $\lim \limits _{n\to \infty}n\sin \left (2\pi \sqrt{1+n^2}\right )$, where $n\in \mathbb{N}$.
Context:
This was a question given in my workbook, where it mentioned a hint that $n$ belonging to natural numbers is an important piece of information.
App... | Note that $\sqrt{x^2+1}$ is very close to $x$. And $\sin$ has period $2\pi$.
Computations. As integer $x \to \infty$,
\begin{align}
\sqrt{x^2+1} &= x\sqrt{1+\frac{1}{x^2}}
= x\left(1+\frac{1}{2 x^2} + o(x^{-2})\right)
= x + \frac{1}{2 x} + o(x^{-1})
\\
\sin\left(2\pi\sqrt{x^2+1}\right) &=
\sin\left(2\pi x + \frac{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478975",
"timestamp": "2023-03-29T00:00:00",
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Optimize $xyz$ where $x+y+z=1$ and $x^2+y^2+z^2=1$? Im trying to optimize
$f(x,y,z)=xyz$ restricted to $g(x,y,z)=x+y+z=1$ and $h(x,y,z)=x^2+y^2+z^2=1$.
$∇f=(yz,xz,xy)$, $∇g=(1,1,1)$ and $∇h=(2x,2y,2z)$.
I tried using the determinant $det(∇f,∇g,∇h)=yz(2z-2y)-xz(2z-2x)+xy(2y-2x)=0$ which I dont know what to do with and ... | $z = 1-x-y\implies f(x,y,z) = xyz = xy(1-x-y)=xy-xy(x+y)=g(x,y)$, subject to: $x^2+y^2+(1-x-y)^2 = 1$, or $2(x^2+y^2)-2x-2y+2xy=0$, or $x^2+y^2-x-y+xy=0$, or $(x+y)^2-xy-(x+y)=0\implies g(x,y) = xy(1-(x+y))=((x+y)^2-(x+y))(1-(x+y))=(x+y)^2-(x+y)^3-(x+y)+(x+y)^2 = -t^3+2t^2-t,t = x+y$. Observe that: $(x+y)^2 - (x+y) = x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate the definite integral $I = \int_0^\pi \frac{\sin^2(\theta) d\theta}{10-6\cos(\theta)}$ I'm studying for my upcoming complex analysis qualifying exam by working through problems in past exams. For this problem, I'd like to know (1) if my answer is correct and complete (i.e. whether I've made any errors/omission... | I think is perfect. Another way is to use the Weierstrass substitution $u=\tan(x/2)$. But is cumbersome and also your way is the apropiate way for a complex analysis course
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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... | This recurrence is linear so we can solve it as follows
$$
\cases{
f_n = f_n^h+f_n^p\\
f_n^h = \frac{n-4}{n}f_{n-1}^h\\
f_n^p = \frac{n-4}{n}f_{n-1}^p + 1\\
}
$$
The solution for the homogeneous part is easy to determine
$$
f_n^h = \frac{c_0}{n(n-1)(n-2)(n-3)}
$$
now assuming $f_n^p = \frac{c_0(n)}{n(n-1)(n-2)(n-3)}$ a... | {
"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": 3
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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... | Some of the last several steps I took in the OP are unnecessary. The last obstacle is the sum
$$\sum_{(a,b)\in\Bbb N^2} \frac1{ab(a+b)} = \sum_{a=1}^\infty \sum_{b=1}^\infty \frac1{ab(a+b)} = \Gamma(3)\zeta(3) = 2\zeta(3)$$
Let
$$f(x) = \sum_{a=1}^\infty \sum_{b=1}^\infty \frac{x^{a+b}}{ab(a+b)}$$
Then for $|x|<1$,
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
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Finding roots of $\tan z=2i$ I'm trying to find the roots of $\tan(z) = 2i.$
At the moment I have
$$
\tan(z)=\frac{\sin 2x +i \sinh 2y }{\cos 2x+\cosh2y}=2i
$$
At a loss as to where I can go from here. Any advice would be appreciated.
| Recall that $\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}$, $\cos(z) = \frac{e^{iz} + e^{-iz}}{2}$, and $\tan(z) = \frac{\sin(z)}{\cos(z)} = \frac{e^{iz} - e^{-iz}}{i(e^{iz} + e^{-iz})}$. So your equation becomes:
$$\frac{e^{iz} - e^{-iz}}{i(e^{iz} + e^{-iz})} = 2i$$
$$\frac{e^{iz} - e^{-iz}}{e^{iz} + e^{-iz}} = -2$$
Make th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Help learning and understanding polynomial factorizations A number theory book I'm reading used this factorization as a main step for a proof:
Given $m > n$, integers, $$\left( a^{2^{m}} - 1 \right) = \left( a^{2^{m-1}} + 1 \right)\left( a^{2^{m-2}} + 1 \right) \left( a^{2^{m-3}} + 1 \right) \cdots \left( a^{2^n} + 1 ... | *
*
$(x^2 -1) = (x+1)(x-1)$
We know that through experience. We can explain it through "you see that $+$ in one term and the $-$ in the other... they cause all the middle terms in an expansion to cancel out". That is $(x+1)(x-1) = x(x-1) + 1(x-1) = x^2 \underbrace{\color{red}{- x + x}}_{\text{cancel out}} -1=x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"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
$$\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx=8\... | To answer your question on evaluating $I_1$, partial fraction decomposition yields a rather tedious result to work with. Instead, there is a nice substitution one can make by taking advantage of the symmetry within the integral. Consider the general case
$$I\equiv\int\limits_0^{+\infty}\frac {\mathrm dt}{t^4+\alpha t^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Are there more examples of this method in calculus? One of my absolute favorite derivations in math is the calculation of the sum of the alternating harmonic series, $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$, which is done by considering the function $f(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}... | $(1+x^2)^{-1}=1-x^2+x^4-x^6+\dots$,
$\arctan x=\int(1+x^2)^{-1}\,dx=x-(x^3/3)+(x^5/5)-(x^7/7)+\cdots$
$\pi/4=\arctan1=1-(1/3)+(1/5)-(1/7)+\cdots$.
Another example (that has come up often on this site): show that
$$
\sum_1^{\infty}{n\over2^n}={1\over2}+{2\over4}+{3\over8}+{4\over16}+{5\over32}+\cdots=2.
$$
Solution: let... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How does $\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}$ become $\sqrt{2(2+\sqrt{2})}$? I'd like to know how can one simplify the following expression
$$\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}$$
into
$$\sqrt{2(2+\sqrt{2})}.$$
Wolfram alpha suggests it as an alternative form, and numerically it's easy to verify, but I can't find the ... | Alternatively
We want
$\sqrt{2+\sqrt 2} + \sqrt {2-\sqrt 2}=\sqrt{something}$ so
$\sqrt{2+\sqrt 2} + \sqrt {2-\sqrt 2}=$
$\sqrt{(\sqrt{2+\sqrt 2} + \sqrt {2-\sqrt 2})^2}=$
$\sqrt{(2+\sqrt 2) + (2-\sqrt 2) + 2\sqrt{2+\sqrt 2}\sqrt {2-\sqrt 2}}=$
$\sqrt{4 + 2\sqrt{(2+\sqrt 2)(2-\sqrt 2)}}=$
$\sqrt{4 + 2\sqrt{4 - 2}}=$
$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 1
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$\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 $$\lim\limits_{x\to 0} \frac{x^2\cos x - 6 \ln(1+x^2) +5x^2}{(e^{... | The numerator is $\sim x^2(\cos(x)-1)+3x^4\sim \frac{5}{2}x^4$
The denominator is $\sim (e^{\sqrt[4]{1+4x^3+8x^4}}-e)x$
So you need to evaluate the limit for
$$\frac{e^{\sqrt[4]{1+4x^3+8x^4}}-e}{x^3}$$
You did $x^2$ instead of $x^3$, so you got $0$, and you can show this term
$$e^{\sqrt[4]{1+4x^3+8x^4}}-e\sim e\cdot x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there any other method to show that $\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x =-\frac{\pi^{2}}{8} \ln 2+\frac{7}{16}\zeta(3)?$ Noting that the evaluation of the integral can be simplified by the Fourier series of $\ln(\sin x)$,
$$\ln (\sin x)+\ln 2=-\sum_{k=1}^{\infty} \frac{\cos (2 k x)}{k}$$
Multiplying the equ... | \begin{align}J&=\int_0^{\frac{\pi}{2}}x\ln(\sin x)dx\\
K&=\int_0^{\frac{\pi}{2}}x\ln(\cos x)dx\\
J+K&=\int_0^{\frac{\pi}{2}}x\ln(\sin x\cos x)dx\\
&\overset{y=\frac{\pi}{2}-x}=\int_0^{\frac{\pi}{2}}\left(\frac{\pi}{2}-y\right)\ln(\sin y\cos y)dy\\
&\frac{\pi}{4}\int_0^{\frac{\pi}{2}}\ln(\sin y\cos y)dy\\
&=\frac{\pi}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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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... | Okay.... I'd say the issue of "overpowering" is that it doesn't work when there isn't something else acting an the thing doing the overpowering to something finite, so that the thing being overpowered is forced to be insignificant. If the thing doing the overpowering goes to infinity and is only countered by somethin... | {
"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": 3
} |
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... | Here's a proof using the benefits of the Iverson bracket:
\begin{align}
\sum_{k=1}^{n^2}\lfloor\sqrt{k}\rfloor &= \sum_{k=1}^{n^2}\sum_{j=1}^n[\;j \le \sqrt{k}\;]\\
&= \sum_{j=1}^n \sum_{k=1}^{n^2}[\;j \le \sqrt{k}\;]\\
&= \sum_{j=1}^n \sum_{k=1}^{n^2}[\;j^2 \le k\;]\\
&= \sum_{j=1}^n \sum_{k=j^2}^{n^2}1\\
&= \sum_{j=1... | {
"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": 1
} |
Showing that $\mathbf{X}^{2} + \mathbf{X} = \mathbf{A}$ has a solution
Show that there exists some $\epsilon >0$ s.t. for all $\mathbf{A}\in \mathbb{R}^{2\times 2} $
with $|( \mathbf{A})_{i, j}| < \epsilon $ for all $i, j$ (let the space of all such matrices be $E$) the equation
$$
\begin{align*}
\mathbf{X}^{2} + \mat... | Lemma: There exists $\epsilon>0$ such that if $B$ is a $2\times 2$ matrix with $\|B-\frac{1}{4}I \|_\infty<\epsilon$, then the equation $Y^2=B$ has a solution $Y$.
Proof: Since the eigenvalues of $S$ are given by
$$\lambda_{1,2}(S)=\frac{1}{2} \left(S_{11}+S_{22}\pm \sqrt{(S_{11}+S_{22})^2-4(S_{11}S_{22}-S_{12}S_{21})}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
How to find all the solutions to $a+2b+3c+\cdots+9i=45$ I want to find all the solutions to the equation $a+2b+3c+4d+5e+6f+7g+8h+9i=45$ where all of the variables can only be of $3$ values: $0,1,2$
I know there are a few solutions for example, set all of them to be $1$, or $a=1, b=2, c=0, d=1, e=0, f=2, g=1, h=1, i=1$.... | As @robjohn mentioned , you can find the number of all solutions using generating functions. If the possible values for the variables are $0,1,2$ ,we can say that
*
*$a:0,1,2$ . Hence , its generating function is $$x^0+x^1+x^2 = \frac{1-x^3}{1-x}$$
*If "b" can take the values of $0,1,2$ , then $2b:0,1,4$ . Hence , ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^2-y^2}$ What is the value of
$$\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^2-y^2}$$
I have already tried the two paths result and I think now that the limit does exist and it is equals to 0. But i really cannot prove this.
|
$$L=\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^2-y^2}$$
$$L=\lim_{(x, y) \to (0, 0)} \frac{xy^2-x^3+x^3}{x^2-y^2}=\lim_{x \to0} (-x)+\lim_{(x, y) \to (0, 0)} \frac{x^3}{x^2-y^2}=\lim_{(x, y) \to (0, 0)} \frac{x^3}{x^2-y^2}$$
Let $y=x-ax^2$
$$L=\lim_{x \to 0} \frac{x^3}{2ax^3-a^2x^4}=\lim_{x \to 0} \frac{1}{2a-a^2x}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding the minimum value of the expression $xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2$ From R. D. Sharma's Objective Mathematics,
Given that $x + y + z = 1$, ($x,y,z$ are positive real numbers) find the minimum value of $$A = xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2$$
My attempt:
By A.M.-G.M. inequality,
$$\begin{align}
(x+y)^2... | An approach with symmetric functions:
$$A=xy(1-z)^2+xz(1-y)^2+yz(1-x)^2\\
\quad =xy+xz+yz-6xyz+xyz(x+y+z)\\
\quad =xy+xz+yz-5xyz$$
the symmetric $\quad xyz\quad$ in terms of $\quad x^3+y^3+z^3:$
$(x+y+z)^3=p_3+3(1-x)(1-y)(1-z)=\cdots \\
xyz=\frac 1 3(x^3+y^3+z^3)+(xy+xz+yz)-\frac 1 3$
note also $\quad xy+xz+yz=\frac 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Calculating eigenvalues of a matrix I have this quadratic matrix:
\begin{align*}
A = \begin{bmatrix}
0 & -1 & -2\\
-1 & 0 & -2\\
-2 & -2 & -3
\end{bmatrix} \implies A - \lambda I = \begin{bmatrix}
0 - \lambda & -1 & -2\\
-1 & 0 - \lambda & -2\\
-2 & -2 & -3 - \lambda
\end{bmatrix}
\end{align*}
After forming the charact... | For matrices of $\mathbb{R}^{3 \times 3}$ and above, the method is to try to apply row operations and factorize. Avoid computing the determinant directly, or expanding the terms first.
*
*Apply row operations to $ |A - \lambda I|$.
$$
\begin{matrix}
\\
-R_1+R_2 \rightarrow\\
\\
\end{matrix}
\begin{vmatrix}
-\lambda &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Apparent inconsistency in geometric product associativity With a little bit of work, I have proven to myself that the geometric product between three vectors is associative ($a$, $b$, and $c$ are 1-vectors):
$$\begin{aligned}(ab)c &= a(bc) \\ &= (b \cdot c) a - (a \cdot c) b + (a \cdot b) c + a \wedge b \wedge c.\end{... | It looks like you are missing a factor in your expansion of $ \left( {a \wedge b} \right) \left( {c \wedge d} \right) $. In general, a product of bivectors $ A B $, should have scalar, bivector, and quadvector components:
$$A B = A \cdot B + {\left\langle{{ A B }}\right\rangle}_{2} + A \wedge B,$$
so for $ A = a \wedg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Gelfands Trigonometry $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha$ Trying Prove the identity $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha$ using the figure provided in Gelfands trigonometry.
What I have so far
$\sin(\alpha - \beta) = \frac{CD}{AC} = \frac{PQ}{AC} ... | Try using the whole angle $\alpha$ as in the proof for the sin and cos addition formulas:
$$\sin \alpha = \frac{BQ}{AB} = \frac{BP}{AB} + \frac{CD}{AB}$$
$$=\frac{BP}{BC} \frac{BC}{AB} + \frac{CD}{AC} \frac{AC}{AB}$$
$$=\cos(\alpha - \beta) \sin \beta + \sin(\alpha - \beta) \cos \beta$$
and also:
$$\cos \alpha = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4517102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Show that $a_{n+1} = a_{n} + a_{n}^{\frac{1}{3}}$ satisfies $\lim_{n\to\infty}\frac{a_n}{pn^q}=1$ for some real valued $p, q$. Show that $a_{n+1} = a_{n} + a_{n}^{\frac{1}{3}}$ satisfies $\lim_{n\to\infty}\frac{a_n}{pn^q}=1$ for some real valued $p, q$.
I had hoped that I could use a strategy similar to Aproximation of... | Hint: It suffices to show that there exists a real number $q$ such that $\lim_{n\rightarrow \infty}{{a_n}\over{n^q}}$ converges. Once this is done, call the value $p$, and then divide both sides by $p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving definite integral $\int_{0}^{1} \tan^{-1}(1-1/x)dx$ $$\int_{0}^{1} \tan^{-1}\left(1-\frac1x\right)dx$$
Here's what I have done so far. (the answer is given as $-\pi/4$)
Let
$$
I = \int_{0}^{1}\tan^{-1}\left(1-\frac1x\right)dx = \int_{0}^{1}\tan^{-1}\left(\frac{x-1}x\right)dx.
$$
Since, $\int_{0}^{1} f(x)dx = \i... | By integration by parts, one has
\begin{eqnarray}
\int_0^1 \arctan\left(1 - \frac1x\right) \, dx &=& -\int_0^1 \frac{x}{1+(1-\frac1x)^2}\frac{dx}{x^2}\\
&=& -\int_0^1 \frac{x}{2x^2-2x+1}dx\\
&=& \boxed{-\frac\pi4}
\end{eqnarray}
Update:
\begin{eqnarray}
\int_0^1 \frac{x}{2x^2-2x+1}dx&=&\int_0^1 \frac{2x}{(2x-1)^2+1}dx\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4519205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
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Two different absorbing matrices seemingly giving the same answer. Can someone please explain to me how these two absorbing matrices,
$
M_1 = \left[
\begin{array}{ccccc}
\frac{2}{3} & \frac{1}{3} & 0 & 0 & 0 \\
\frac{2}{3} & 0 & \frac{1}{3} & 0 & 0 \\
\frac{2}{3} & 0 & 0 & \frac{1}{3} & 0 \\
\frac{1}{3} & 0 & 0 & 0... | Suppose the die faces are A, B, C.
(1,5) of the second matrix is as you said, the chance that within n steps you have seen four As in a row or four Bs in a row.
(1,5) of the first matrix is the chance that within n steps that you have seen three As in a row, then something not an A.
In the second matrix, you have a 2/3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4520787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to derive the polar equation of a circle.
I have learnt that the standard equation of a circle whose centre is at (a,b) is
$$(x-a)^2 + (y-b)^2 = c^2
$$
I am trying to derive the polar equation of this circle but I am unfortunately stuck.
$$\begin{align}(x-a)^2+(y-b)^2&=c^2\\
(r\cos\theta-a)^2+(r\sin\theta-b)^2&=c^... | By to reach a similar result I understood that you are interested in an unparametrized form.
We obtain polar equation by continuing the derivation to its logical conclusion.
$$ r(r-2a\cos\theta-2b\sin\theta)=c^2-a^2-b^2$$
Distance to center from origin $OC=\sqrt {a^2+b^2}$ and power of circle /squared tangent length... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
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,
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-(ax+b))\cdot \frac{\sqrt{4x^2+2x+1}+(ax+b)}{\sqrt{4x^2+2x+1}+(ax+b)}$$
$$=\li... | A variation. We obtain
\begin{align*}
\color{blue}{-\frac{1}{2}}&=\lim_{x\to\infty}\left(\sqrt{4x^2+2x+1}-ax-b\right)\\
&=\lim_{x\to\infty}\left(\sqrt{\left(2x+\frac{1}{2}\right)^2+\frac{3}{4}}-ax-b\right)\\
&=\lim_{y:=2x+\frac{1}{2}\to\infty}\left(\sqrt{y^2+\frac{3}{4}}-a\left(\frac{y}{2}-\frac{1}{4}\right)-b\right)\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Geometric proof for derivative of $f(x)=\sqrt{x}$ Using a square of area $x$,
we obtain
$$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}$$
How do I show that $\frac{(d\sqrt{x})^2}{2\sqrt{x}dx} \rightarrow 0$ as $dx\rightarrow0$?
| By your geometric construction, and also by your first line, when $dx \rightarrow 0$, $d\sqrt x \rightarrow 0$.
$\frac {dx} {d\sqrt x} = 2\sqrt x + d\sqrt x$
So when $dx \rightarrow 0, \frac {dx} {d\sqrt x} \rightarrow 2\sqrt x$
So when $x\ne 0, \frac {(d\sqrt x)^2} {2\sqrt x dx}$ is equivalent when $dx \rightarrow 0$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4524342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding the range of a Quadratic divided by a Quadratic I need to find the range of a function:
$$f(x) = \frac{e^{2x} - e^x + 1}{e^{2x} + 3e^x - 7} $$
Substituting $e^x = t $ and equating $f(x)$ to $y$, I get
$$(y-1)t^2 + (3y+1)t - (7y+1) = 0$$
Since $t$ must be real and positive, I get the range of y to be $(-\infty, ... | Substitute $f(x)$ with $y$: $y=\frac{e^{2x}-e^x+1}{e^{2x}+3e^x-7}$.
Rewrite in terms of $e^x$:
$$e^{2x}y+3e^xy-7y=e^{2x}-e^x+1$$
$$(y-1)(e^x)^2+(3y+1)e^x-(7y+1)=0$$
This equation has roots (with respect to $x$), when there are roots with respect to $e^x$ (discriminant is non-negative) and at least one root is positive.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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What is the biggest possible value of $\frac 1a+\frac 1b$ given that $2 <\frac ab < 7$ where $a$ and $b$ are distinct integers and $a + b = 10$. Let $a$ and $b$ be distinct integers that satisfy $2 < \frac ab < 7$. If $a + b = 10$, then what is the biggest possible value of $\frac 1a +\frac 1b$?
It's easy to show that ... | If you assume that $a$ and $b$ are positive, then there are only nine possible combinations with $a + b = 10$, and of these, only two also meet $2 < \frac{a}{b} < 7$:
*
*$a = 7, b = 3 \implies \frac{1}{a} + \frac{1}{b} = \frac{10}{21} \approx 0.47619$
*$a = 8, b = 2 \implies \frac{1}{a} + \frac{1}{b} = \frac{5}{8} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solve in exact form: $x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1=0$ ( WolframAlpha failed)
Solve the polynomial in closed form:
$$x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1=0$$
WolframAlpha obviously failed.
I tried several ways:
*
*I tried the Rational Root Thereom, but there is no rational root.
*I tried possible ... | The problem can be rewritten as $$x((x-1)x+1)(x^3+3x-1)+1=0$$ Now solving this we get $6$ roots:
$1.$ $-0.1399-1.7659i$
$2.$ $-0.1399+1.7659i$
$3.$ $-0.04270-0.56305i$
$4.$ $-0.04270+0.56305i$
$5.$ $0.67660-0.73635i$
$6.$ $0.67660+0.73635i$
These are all approximations I got from a software. Strict factorisation $($wri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4531994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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Connection between thinking of an ellipse as a squished circle and the formal definition If I think of an ellipse fundamentaly as a squished circle. For exemple, I have a initial circle $x^2+y^2=1$ and I morph into $x^2+ 2y^2=1$.
How can I see, intuitively, that for all points on the curve, the sum of the two distanc... | What do we have?
$$x^2+2y^2=1$$
$$\frac{x^2}{1^2}+\frac{y^2}{(1/\sqrt{2})^2}=1$$
$$a=1, b=\frac{1}{\sqrt{2}}$$
$$c=\sqrt{a^2-b^2}=\frac{1}{\sqrt{2}}$$
We need to show
$$A=\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}={\rm const}$$
$$x^2+2y^2=1\Rightarrow y^2=\frac{1-x^2}{2}$$
$$(x-c)^2+y^2=\frac{x^2-2x\sqrt{2}+2}{2}=\frac{(x-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4534562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Olympiad Math Problem About Simultaneous Equations Given the question:
*
*$x + y = 1$
*$x^2 + y^2 = 2$
*$x^5 + y^5 = ?$
After a bunch of manipulation of the above equations we find the solution to be 19/4. Yet could the above be solved by simultaneous equations?
From $1)$ we can conclude that $x = 1 - y$.
Substitut... | $$1=(x+y)^2= x^2+2xy+y^2 = 2+2xy$$$$ \implies xy = -1/2$$
$$4=(x^2+y^2)^2 =x^4+2x^2y^2 +y^4$$$$\implies x^4+y^4= 7/2$$
and $$2=(x+y)(x^2+y^2) = x^3+xy(x+y)+y^3$$ $$\implies x^3+y^3 = 5/2
$$
So
$$7/2=(x+y)(x^4+y^4) = x^5+y^5 + xy(x^3+y^3)$$ $$\implies x^5+y^5 =19/4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4537015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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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.
$$\boxed{I_n=∫_0^∞ \tanh(2x)\ln^n(\tanh x)dx =\frac{(-1)^n n !}{2^... | If we consider
$$I_n=\int \tanh(nx)\,\log(\tanh(x))\,dx$$ there is a (nasty) antiderivative involving logarithms of complex arguments and polylogarithms.
Doing the same as you did and using multiple angles formulae
$$I_1=\int_0^1 \frac y{1-y^2}\log(y)\,dy=-\frac{\pi ^2}{24}$$
$$I_2=\int_0^1 \frac {2y}{1-y^4}\log(y)\,... | {
"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": 2
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How do I find the points of tangency given a 2 variable function and a normal vector? I'm given the two variable function $f(x,y) = 4x^2+7y^2+5xy+14$ and asked to find the (2) points on the surface where the vector $6\hat i + 69\hat j + 3\hat k$ is normal to the tangent plane. So far I have written a generalized linea... | $z = f(x,y) = 4x^2 + 7y^2+5xy + 14$
The normal to the tangent plane is along the gradient of $f$
$\nabla f = \begin{bmatrix} 8 x + 5 y \\ 14 y + 5 x \\-1 \end{bmatrix} $
If $\nabla f$ is parallel to $ (6, 69, 3) $, then their cross product is zero.
$ \bigg(\nabla f\bigg) \times (6, 69, 3) = \bigg( (14 y + 5x)(3) + 69 ,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4541978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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compute $\sum_{k=0}^{(p-1)/2} \dfrac{(p-1)!}{(k!)^2(p-1-2k)!}\mod p$
Let $p$ be an odd prime. Compute $\sum_{k=0}^{(p-1)/2} \dfrac{(p-1)!}{(k!)^2(p-1-2k)!}\mod p.$
For an odd prime p, let $S(p) = \sum_{k=0}^{(p-1)/2} \dfrac{(p-1)!}{(k!)^2(p-1-2k)!}$. Note that we may assume $p\equiv 1, 5\mod 6,$ since for $p=3$, we ... | $$\begin{aligned}
(2k)!&=(2k)(2k-2)\cdots2\cdot(2k-1)(2k-3)\cdots1\\
&=k!2^k(2k-1)(2k-3)\cdots1\\
&=k!(4k-2)(4k-6)\cdots2\\
&\equiv_pk!(4k-2-2p)(4k-6-2p)\cdots(2-2p)\\
&=k!(-4)^k(\frac{p-1}2-k+1)(\frac{p-1}2-k+2)\cdots \frac{p-1}2
\end{aligned}$$
The idea of the transformation above comes from this comment.
$$\begin{al... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Evaluate $\int \sqrt{x^2 + a^2}\mathrm{d}x$ I tried using integration by parts and could get to this point:
\begin{align}
\int \sqrt{x^2 + a^2}\mathrm{d}x &= x\sqrt{x^2+a^2} - \int x\frac{2x}{\sqrt{x^2 + a^2}}\mathrm{d}x
\\\\ &= x\sqrt{x^2+a^2} -2\int \sqrt{x^2 + a^2}\mathrm{d}x + \int \frac{2a^2}{\sqrt{x^2 + a^2}}\ma... | Those types of integrals usually take advantage of some properties of the hyperbolic sine and cosine functions - $$\sinh t=\frac{e^t-e^{-t}}{2}\\\cosh t=\frac{e^t+e^{-t}}{2}\\(\sinh t)'=\cosh t\\(\cosh t)'=-\sinh t$$ We solve by plugging in $x=a\sinh t$: $$\int\sqrt{x^2+a^2}dx=\int a\sqrt{a^2\sinh^2 t+a^2}\cosh tdt=\in... | {
"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": 1
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Finding the sum of Arithmetico-geometric series It was asked to find the sum of the AGP series $\left\{\dfrac{e^{\frac{1}{k}}+2 e^{\frac{2}{k}}+3 e^{\frac{3}{k}}+\cdots+k e^{\frac{k}{k}}}{k^2}\right\}$
I tried solving it with the formula for sum of $n^{th}$ terms of AGP series $\frac{a-[a+(n-1) d] r^n}{1-r}+\frac{d r\l... | $S=e^{\frac{1}{k}}+2 e^{\frac{2}{k}}+3 e^{\frac{3}{k}}+\cdots+k e^{\frac{k}{k}}=e^{\frac{1}{k}}\left(1+2 e^{\frac{1}{k}}+3 e^{\frac{2}{k}}+\cdots+k e^{\frac{k-1}{k}}\right)$
Use your formula, set $a=1, d=1, r=e^{\frac{1}{k}}, n=k$
$S=e^{\frac{1}{k}}\left(\frac{a-[a+(n-1) d] r^n}{1-r}+\frac{d r\left(1-r^{n-1}\right)}{(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4546332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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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... | I did direct substitution $x=n^{\frac{2}{3}}\tan^{\frac{2}{3}}\theta$ and $dx=\frac{2}{3}n^{\frac{2}{3}}\tan^{-\frac{1}{3}}\theta\sec^2\theta d\theta$ and got
$$\lim_{n\rightarrow\infty}\int_{\arctan(n^{-2/3})}^{\arctan(n^{1/3})} \frac{\frac{2}{3}n^{\frac{2}{3}}\tan^{-\frac{1}{3}}\sec^2\theta}{n\sec\theta}d\theta=(\lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Complex contour integral using Cauchy integral formula I am trying to get integral in contour $C=\left \{ 3\cos(t)+2i\sin(t) : 0\leq t\leq 2\pi \right \}$
$$\oint_{C} \frac{z}{(z+1)(z-1)^2}dz$$
What I tried was using partial fractions and integrate them separately
$$\oint_{C} \frac{z}{(z+1)(z-1)^2}dz =\oint_{C}... | The contour is simply an ellipse $\frac{x^2}{3^2}+\frac{y^2}{2^2}=1$ that encloses both the singularities viz. $z=-1$ and $z=1$ of $f(z)$.
Use deformation principle to see
$\oint_C f(z)dz=\oint_{C_1}\frac{z/(z+1)}{(z-1)^2}dz+\oint_{C_2}\frac{z/(z-1)^2}{(z+1)}dz$
where $C_1$ and $C_2$ are circles enclosing the singulari... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4551261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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How do I prove that the distance between two points is longer than one point in the normal and the other point in a plane? So, imagine $P$ is a plane with the normal $n$, and $A = (a1, a2, a3)$ is a point above the plane $P$ that does not lie on the plane. $B$ is a point collinear on the line $L$: $r = (a_1, a_2, a_3) ... | You have $\mathbf{B} = \mathbf{A} + \lambda \mathbf{n} $ for some $\lambda$. Point $\mathbf{C}$ is in the plane and $\mathbf{C} \ne \mathbf{B}$, therefore,
$ \mathbf{n} \cdot (\mathbf{C} - \mathbf{B}) = 0 $
Now
$ \mathbf{A} - \mathbf{C} = (\mathbf{A} - \mathbf{B}) + (\mathbf{B} - \mathbf{C}) = -\lambda \mathbf{n} + \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Value of a cyclic sum Let $a,b,c$ satisfy the equation $x^3+px^2+qx+r=0$. Is it possible to determine $\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{a+c}$ in terms of $p,q,r$? I stumbled upon this while thinking about an inequality problem. What i could do so far is this:
We have the relations $a+b+c=-p,ab+bc+ca=q,abc=-r$ fr... | Note that $\frac{ab}{a+b} = \frac{1}{\frac{1}{a}+\frac{1}{b}}$.
Suppose we have the monic cubic $f(x) = x^3+a_2x^2+a_1x+a_0$ with roots $\alpha, \beta, \gamma$. We can make use of the following two operations:
*
*$\frac{1}{a_0}x^3f(\frac{1}{x})$ is the unique monic cubic with roots $\frac{1}{\alpha}, \frac{1}{\beta},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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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}{... | $$B=1-A$$
Putting this in 2nd equation.
$$4A-1+A+C=0\\\implies C=1-5A$$
Now, we have both $B$ and $C$ in terms of $A$. Putting that in 3rd equation.
$$4A-6+6A-3+15A=0\\\implies 25A=9\\\implies A=\frac9{25}$$
| {
"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": 4
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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... | Hint: It seems the answerer didn't use the inequality chain stated in the Wiki-page
as starting point, but started instead with the asymptotic formula
\begin{align*}
\color{blue}{n!\sim\left(\frac{n}{e}\right)^n\sqrt{2\pi n}}\tag{1}
\end{align*}
We obtain from (1) assuming even $n$:
\begin{align*}
\color{blue}{\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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finding a coefficient to obtain a geometric sequence I have the following:
$$u_0=2 \\
u_{n+1} = \frac{2}{3}(u_n)-n-\frac{8}{3}\\
v_n=u_n + x\cdot n -1$$
I need to find the value of $x$ in order for $v_n$ to be a geometric sequence, which mean that I need to compute the following :
$$\frac{v_{n+1}}{v_n}$$
edit:
thanks f... | The hard part is to show that $$ u_{n} = \left(\frac{2}{3}\right)^n - 3 \, n + 1.$$
The remaining part is
\begin{align}
v_{n} &= u_{n} + x \, n - 1 \\
&= \left(\frac{2}{3}\right)^n + (x - 3) \, n.
\end{align}
To show the form of $u_{n}$ compute the first few terms to develop a pattern and then check:
\begin{align}
f_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4557707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Approximations of $\sum_{i=0}^k\binom{2k+1}{i}p^i(1-p)^{k-i}$ Given a biased coin with a probability $p$ of coming up heads, what is the probability that strictly more than half of $2k+1$ flips turn up heads?
This probability can be modelled as
$$\sum_{i=k+1}^{2k+1}\binom{2k+1}{i}p^{i}(1-p)^{2k+1-i} = p^k\sum_{i=0}^{k}... | It looks similar but it really isn't. The binomial identity sum is, of course, always $1$, whereas the probability you are interested in has behavior that differs dramatically depending on whether $p > \frac{1}{2}, p = \frac{1}{2}$, or $p < \frac{1}{2}$.
Instead we can use the Chernoff bound. Let $X \sim \text{Bin}(2k+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $(\sin x + \cos x)(6 - \sin x)<9$ Is there any elementary way to prove that $(\sin x + \cos x)(6 - \sin x)<9$?
I've noticed that $(\sin x + \cos x)$ has to be positive so $x \in\left(-\dfrac{\pi}{4}, \dfrac{3\pi}{4}\right)$ and then $(6 - \sin x)\in\left(6-\dfrac{\sqrt2}{2},6+ \dfrac{\sqrt2}{2}\right)$ but s... | Proof 1: Note that $\sin x + \cos x = \sqrt 2 \, \sin (x + \pi/4) \le \sqrt 2$
and $6 - \sin x > 0$.
If $\sin x \ge 0$, we have
$(\sin x + \cos x)(6 - \sin x) \le \sqrt 2 \cdot (6 - \sin x)
\le \sqrt 2 \cdot 6 < 9$.
If $\sin x < 0$, we have
$\sin x + \cos x \le \cos x \le 1$ and thus $(\sin x + \cos x)(6 - \sin x)
\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Minimal polynomial of $\cos\left(\frac{\pi}{13}\right)$ Find the Minimal polynomial of $\cos\left(\frac{\pi}{13}\right)$.
My try:
Let $$13 \theta=\pi \implies 9\theta=\pi-4\theta$$
$$\implies \cos(9\theta)=-\cos(4\theta)$$
Now Let $x=\cos(\theta)$
$$\implies 4(4x^3-3x)^3-3(4x^3-3x)=-(2(2x^2-1)^2-1) $$
$$\implies 256 x^... | Chebyshev polynomial of the first kind $T_p(x)$ satisfies $T_p(\frac{\zeta^{i}+\zeta^{-i}}{2})=1$ for all $0\leq i\leq p-1$ where $\zeta=e^{\frac{2\pi i}{p}}$.
So, the minimal polynomial of $\cos(\frac{2\pi i}{p})$ over integers is
$$m_p(x)=\sqrt{\frac{T_p(x)-1}{x-1}}.$$
But, I don't know its series expansion.
From WA:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Do we have a closed form for $\int_0^{\infty} \frac{\ln t}{\left(1+t^2\right)^n} d t $? Latest Edit
We are glad to see there are 4 alternative solutions which give the same closed form to the integral:
$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(a^2+t^2\right)^n} d t = \frac{a^{1-2n}\sqrt{\pi}\Gamma\left(n-\frac{1}{2}\... | The generalized Leibniz rule gives
$$\newcommand{\d}{\mathrm{d}}
\frac{\d^n}{\d x^n} \ln(x) x^{-1/2} \bigg|_{x=1}
= \sum_{k=0}^n \binom n k \left( \frac{\d^{n-k}}{\d x^{n-k}} \ln(x) \bigg|_{x=1} \right)\left( \frac{\d ^{k}}{\d x^{k}} x^{-1/2} \bigg|_{x=1} \right)$$
For the $(n-k)$th derivative of $\ln(x)$, it is easy t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4562577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Evaluate the limit of a sequence by Riemann sums and mean value theorem Calculate:
$\displaystyle \lim_{n \rightarrow \infty} \left( \frac{n \pi}{4} - \left( \frac{n^2}{n^2+1^2} + \frac{n^2}{n^2+2^2} + \cdots \frac{n^2}{n^2+n^2} \right) \right)$.
I solved it by taking into account that $\displaystyle \int_0^{1} \frac{1... | Let us use the following
Lemma: If $f\in C^1[0,1]$ then we have $$\lim_{n\to\infty} \sum_{k=1}^nf(k/n)-n\int_0^1 f(x) \, dx=\frac{f(1)-f(0)}{2}$$ Using $f(x) =1/(1+x^2)$ we get your desired limit as $1/4$. The lemma is proved using mean value theorem and matches your approach. An extension of the lemma is available her... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4564940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Let $n$ be a fixed positive integer such that $\sin(\frac\pi{2n})+\cos(\frac\pi{2n})=\frac{\sqrt n}2$ then find $n$.
Let $n$ be a fixed positive integer such that $\sin(\frac\pi{2n})+\cos(\frac\pi{2n})=\frac{\sqrt n}2$ then find $n$.
This question already exists here, but the answer mentioned there is incorrect. That... | I try to solve a more general equation $$
\sin \left(\frac{\pi}{2 x}\right)+\cos \left(\frac{\pi}{2 x}\right)=\frac{\sqrt{x}}{2} \textrm{ for }x\geq 1\tag{*} $$
Squaring both sides yields $$
\begin{gathered}
\sin ^2\left(\frac{\pi}{2 x}\right)+\cos^2\left(\frac{\pi}{2 x}\right)+2 \sin \left(\frac{\pi}{2 x}\right) \cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4567239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find all triples $(x,y,z)$ of positive integers such that $5(x²+2y²+z²)=2(5xy-yz+4xz)$... Find all triples of $(x,y,z)$ of positive integers such that $5(x^2+2y^2+z^2)=2(5xy-yz+4xz)$, and at least one of $x,y,z$ is prime
My approach: Put $x,y,z=2$ one by one then we get value as $(x,y,z)=(14/3,2,10/3)$ and $(14/5,6/5,2... | Solving as a quadratic in $x$, we have
$$
x = \pm \frac{1}{5}\sqrt{-(5y-3z)^2} + y + \frac{4z}{5}
$$
Since $(5y-3z)^2$ is non-negative, the only possibility is $5y-3z = 0$ or
$$y = \frac{3z}{5}$$
This gives
$$
x = 0 + \frac{3z}{5} + \frac{4z}{5} = \frac{7z}{5}
$$
Now substitute $x$ and $y$ in terms of $z$ in the origin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4568399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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... | Use Weierstrass substitution
$$\sqrt3\sin C=\dfrac{2(1+t^2)-2t}{1-t^2}$$
$$\iff t^2(2+\sqrt3\sin C)-2t+2-\sqrt3\sin C=0$$
The discriminant $=2^2-4(2^2-3\sin^2C)=-12\cos^2C$ which is $\le0$
But we need $-12\cos^2C\ge0\implies12\cos^2C=0$
| {
"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": 1
} |
On solutions to special eighth/fourth powers $a^8+b^4 = c^8+d^4$? By the Lander-Parkin conjecture, it is assumed there are no non-trivial integer solutions to either,
$$a^8+b^8+c^8 = d^8+e^8+f^8\tag1$$
$$a^8+b^8 = c^8+d^8\tag2$$
However, if we relaxed $(1)$ a bit, then,
$$1118^8 + 1937^8 + 2502045^4 = 455^8 + 1846^8 + ... | Solutions of (5) can easily be obtained from solutions of the 4-2-2 equation. Given any integers $A,B,C,D$ such that:
$$A^4+B^4=C^4+D^4$$
we have:
$$A^8+(AB)^4=(AC)^4+(AD)^4$$
Taking for example the smallest 4-2-2 solution:
$$59^4+158^4=133^4+134^4$$
we have:
$$59^8+9322^4=7847^4+7906^4$$
Unfortunately there doesn't a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4578655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove $\frac{1}{(y+z) x^4} + \frac{1}{(x+z) y^4} + \frac{1}{(y+x) z^4}\geq\frac{3}{2}$ for $x, y, z>0$, with $xyz=1$ How to prove that $\frac{1}{(y+z) x^4} + \cfrac{1}{(x+z) y^4} + \cfrac{1}{(y+x) z^4}\geq\frac{3}{2}$ for $x, y, z>0$...
I've tried substituting $a=1/x$, $b=1/y,$ $c=1/z$, and also writing the 1 that is b... | By Holder and AM-GM we obtain:
$$\sum_{cyc}\frac{1}{(y+z)x^4}=\frac{\sum\limits_{cyc}\frac{1}{(y+z)x^4}\sum\limits_{cyc}(y+z)x\sum\limits_{cyc}1}{6(xy+xz+yz)}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}\frac{1}{x}\right)^3}{6(xy+xz+yz)}=\frac{(xy+xz+yz)^2}{6}\geq\frac{9}{6}=\frac{3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4581269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Verify that $abc^2x^2+c(3a^2+b^2)x+3a^2-ab+b^2=0$ has real roots As the title states, the goal is to verify that the quadratic equation: $abc^2x^2+c(3a^2+b^2)x+3a^2-ab+b^2=0$ has real roots. This problem comes from an interschool mathematics contest for High-schoolers, here's my (brute force) attempt:
Since we need to ... | Since this problem was written for high-school contestants, I also have my doubts as to how many of them would spot the factorability of the discriminant. I decided to take the approach of writing the quadratic polynomial in "vertex form", thus
$$ abc^2·x^2 \ + \ c·( \ 3a^2 + b^2 \ )·x \ + \ ( \ 3a^2 - ab + b^2 \ ) $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4582566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Show that $2 \operatorname{Re}(z-1/z) > 1/|z|^2 - 1.$
Let $n$ be a positive integer and let $z$ be a root of $x^n-x-1.$ Show that $2 \operatorname{Re}(z-1/z) > 1/|z|^2 - 1$.
Solution. Write $z = re^{i\theta}, r > 0$ as $0$ is obviously not a root. Then $$1+2r\cos \theta = |z+1|^2-r^2 = r^{2n} - r^2.$$ We need to show... | $n>1$ and your last parenthesis is $(r^2-1)(r^{2n-2}-1),$ so you just have to prove that $r\ne1.$
Let us prove it by contradiction. If $r=1$ then (by what you already proved) $\cos\theta=-\frac12$ hence $z=e^{\pm i2\pi/3},$ but such a $z$ satisfies $z+1=-z^2\ne z^n.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4583050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ I would like to solve the equation $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ analytically, without using Wolfram Alpha. I have tried several substitutions, including $x=\cos(t)$, $\sqrt{1-x^2}=t$, but they all fail and eventually result in a quartic polynomial which i... | $$x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$$
$$x^2\left(1+2\sqrt {1-x^2}+1-x^2\right)=1-x^2$$
$$x^2\left(2+2\sqrt {1-x^2}-x^2\right)=1-x^2$$
Substitute $x^2=y$,
$$y\left(2+2\sqrt {1-y}-y\right)=1-y$$
$$2y+2y\sqrt {1-y}-y^2=1-y$$
$$2y\sqrt {1-y}=1-y+y^2-2y$$
$$4y^2(1-y)=(y^2-3y+1)^2$$
$$y^4-2y^3+7y^2-6y+1=0$$
$$(y^2-y+... | {
"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": 4
} |
Encountered $\int_0^{\infty} \frac{x^3}{e^{bx}+1}dx $ in statistical mechanics, curious how to solve it According to Wolfram Alpha, this integral = $\frac{7\pi^4}{120 b^4}$. I tried solving it using the residue theorem, but there are infinite residues which I believe means you can't apply the theorem.
Edit: I forgot to... | Using the power series
$$
\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^n x^n \text { for }| x |<1,
$$
we can transform the integrand into a power series too.
For $b>0$, we have
$$
\begin{aligned}
I & =\int_0^{\infty} \frac{x^3}{e^{b x}+1} d x \\
& =\int_0^{\infty} \frac{e^{-b x} x^3}{1+e^{-b x}} d x \\
& =\sum_{n=0}^{\infty}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the real roots of the equation: $\sqrt [4]{x}-\sqrt [4]{2-x}=1$
Find the real roots of the following equation: $$\sqrt [4]{x}-\sqrt [4]{2-x}=1$$
This is my textbook contest exercise.
$$\sqrt [4]{x}=\sqrt [4]{2-x}+1$$
$$x=\big(\sqrt [4]{2-x}+1\big)^4$$
I know that
$$(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$$
So, we wi... | In this answer, I wanted to construct the trigonometric method that requires significantly less computation.
Since $0<x<2$, making the substitution $x=2\sin^2\theta$, where $0<\theta<\frac {\pi}{2}$, we have:
$$\begin{align}&\sqrt {\sin\theta}-\sqrt {\cos\theta}=\frac {1}{\sqrt [4]{2}}\\
\implies&\sin\theta+\cos\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"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 $$\cos x \sin 2 n x=\frac{1}{2}[\sin (2 n+1) x+\sin (2 n-1) x],$$
we have
$$
\begin{aligned}
&\quad \int_0^{\frac{\pi}{2}} x \cos x \sin 2 n x d x\\& =\frac{1}{2}\left[\int_0^{\frac{\pi}{2}} x[\sin (2 n+1)+\sin (2 n-1) x] d x\right] \\
& =-\frac{1}{2} \int_0^{\frac{\pi}{2}} x d\left(\frac{\cos (2 n+1) x}{2 n+1}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding remainder of dividing $3^{2^n}$ by $2^{n+3}$ Find the remainder of dividing $3^{2^n}$ by $2^{n+3}$
I was trying to use $ord_{2^{n+3}}(3)$, but i don't see a future in looking for a constant when I change the value of n. However, I was looking for a less advanced solution, is it possible?
| You could show by induction that the remainder is $2^{n+2}+1$.
Base case ($n=1$): $\;3^{2^n}=9=2^{n+2}+1\pmod {2^4=16}$.
Now assume $3^{2^n}\equiv 2^{n+2}+1\bmod 2^{n+3}$; i.e., $3^{2^n}=2^{n+2}+1+k\cdot2^{n+3}$ for some integer $k$.
Then $3^{2^{n+1}}=\left(3^{2^n}\right)^2=2^{2n+4}+1+k^2\cdot2^{2n+6}+2^{n+3}+k\cdot2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4597434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Determinant of $A_{2n}$ Determine the determinant of $$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... | $\vert A_{2n}\vert=\begin{vmatrix} B_{n} & C_{n} \\ C_{n} & B_{n} \end{vmatrix}$, Where $B_{n}=aI_{n}$ and $C_{n}=bI_{n}$. Because $B_{n}$ and $C_{n}$ non singular, then
$\vert A_{2n}\vert=\vert B_{n}^2-C_{n}^2\vert=\vert a^2I_{n}-b^2I_{n}\vert=\vert (a^2-b^2)I_{n}\vert=(a^2-b^2)^n.1=(a^2-b^2)^n.$
$\blacksquare$
| {
"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": 2
} |
Evaluating the limit $\lim_{x\to\infty}\left(\frac {x^5+\pi x^4+e}{x^5+ex^4+\pi}\right)^x$ How can I evaluate the following limit?
$$\lim_{x\to\infty}\left(\frac {x^5+\pi x^4+e}{x^5+ex^4+\pi}\right)^x$$
To solve this limit, I divided the numerator and denominator by $x^5$ and I got
$$\lim_{x\to\infty}\left(\frac {1+\fr... | So this is the typical $1^\infty$ type-limit. To find the limit I always start by adding and subtracting $1$ inside the parenthesis (you are "apparentlly" doing nothing), like so $$\lim_{x\rightarrow \infty}\Bigg(1-1+\frac{x^5+\pi x^4+e}{x^5+e x^4+\pi}\Bigg)^x=\lim_{x\rightarrow \infty}\Bigg(1+\frac{(\pi-e)(x^4-1)}{x^5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4599278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluating a simple looking integral While solving an problem of Reator design i have encountered a integral,
$$ I =\int \left(\frac{3x+1}{1-x}\right)^{1/2} dx$$
$0<x<1$
My Approach: Rationalization didn't helped much, so i have substituted $$\frac{3x+1}{1-x}=t $$
and $$dt =\frac{3(1-x)+(3x+1)}{(1-x)^2}dx= \frac{4}{(1-... | We may deal with the integral by a single substitution
$$
\sqrt{3} \tan \theta=\left(\frac{3 x+1}{1-x}\right)^{\frac{1}{2}},
$$
then
$$
\begin{aligned}
6 \tan \theta \sec ^2 \theta d \theta & =\frac{4}{(1-x)^2} d x \\
& =4\left(\frac{3+3 \tan ^2 \theta}{4}\right)^2 d x \\
& =\frac{9}{4} \sec ^4 \theta d x
\end{aligned}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4599859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Showing $\sum_{cyc}\frac{a^2+bc}{b+c}\geq a+b+c$ for positive $a$, $b$, $c$ The following is an inequality which I have trouble solving:
$$\frac{a^2+bc}{b+c}+\frac{b^2+ac}{c+a}+\frac{c^2+ab}{a+b}\geq a+b+c$$
($a, b, c>0$)
I tried multiplying LHS and RHS by 2 and then use
$${2(a+b+c)}={(a+b)+(b+c)+(c+a)}$$
and move to L... | Another way.
Since $(x+y+z)^2\geq3(xy+xz+yz),$ by C-S we obtain:
$$\sum_{cyc}\frac{a^2+bc}{b+c}=\sqrt{\left(\sum_{cyc}\frac{a^2+bc}{b+c}\right)^2}\geq\sqrt{3\sum_{cyc}\frac{(a^2+bc)(b^2+ac)}{(b+c)(a+c)}}=$$
$$=\sqrt{3(a^2+b^2+c^2)}\geq a+b+c.$$
The following inequality is also true.
Let $a$, $b$ and $c$ be positive nu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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For $a+b+c+d=16$, are there always four integers $x, y, z, w$ that satisfy some conditions including $10\mid x+2y+3z+4w$?
For $a, b, c, d \in \Bbb{N}, a+b+c+d=16$, do four integers $x, y, z, w$ always exist that satisfy the following three conditions?
*
*$\ 0\leq x\leq a, \ 0\leq y\leq b, \ 0\leq z\leq c, \ 0\leq w\... | zwim found by brute force that
no such $x,y,z,w$ exist for $(a,b,c,d)=(1,7,7,1)$.
Let me prove that fact by hand. Suppose integer $x,y,z,w$ satisfy condition 1 and 2 for $(a,b,c,d)=(1,7,7,1)$.
$x+2y+3z+4w=3(x+y+z+w)-2x-y+w\ge30-2\times1-7+0=21.$
$x+2y+3z+4w=2(x+y+z+w)-x+z+2w\le20-0+7+2\times1=29.$
The two inequalites a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"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... | Substitute $x=\tan u$
$$\begin{align}
& =\int_{0}^{\pi /2}{\frac{\sqrt[5]{\tan u}}{{{(\tan u+1)}^{2}}}du} \\
& =\int_{0}^{\pi /2}{\frac{\sqrt[5]{\frac{\sin u}{\cos u}}}{{{(\frac{\sin u}{\cos u}+1)}^{2}}}du} \\
& =\int_{0}^{\pi /2}{\frac{\sqrt[5]{\sin u}}{\sqrt[5]{\cos u}}\frac{{{\cos }^{2}}u}{{{(\sin u+\cos u)}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Asymptotic expansion of the solution of $\tan x=x$. I am trying to find an asymptotic expansion of the soltion $x_n$ of the equation $\tan x=x$ in the intervall $I_n= \left]-\frac{\pi}{2}+n\pi , \frac{\pi}{2} + n\pi\right[$. I have showed that
$$x_n = n\pi + \frac{\pi}{2}-\frac{1}{n\pi}+o\left(\frac{1}{n}\right)$$ But ... | Too long for a comment
Let's look for the solution in the form $x=\pi n+\frac{\pi}{2}-x_1+x_2+...$, where $x_2\ll x_1\ll1$.
We also suppose $\frac{x_2}{x_1}\sim x_1$
$$\tan x=x\,\Rightarrow\,\tan(\pi n+\frac{\pi}{2}-x_1+x_2+..)=\cot(x_1-x_2-..)=x=\pi n+\frac{\pi}{2}-x_1+x_2+..$$
$$\cot(x_1-x_2-..)=\frac{\cos()}{\sin()}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Let $f(x)=1+2\cos x +3 \sin x$. If real number $a, b, c$ are such that $af(x)+bf(x+c)=1$ holds for any $x \in \mathbb R$, then find $a, b, c$
Let $f(x)=1+2\cos x +3\sin x$. If real number $a, b, c$ are such that $af(x)+bf(x+c)=1$ holds for any $x \in \mathbb R$, then find $a, b, c$
Solution given as in book:
since ... | Using $\cos(\theta)=\sin\left(\theta+\frac{\pi}{2}\right)$, as well as several formulas for linear combinations of $\sin$ and $\cos$ (as suggested in Hypernova's comment), gives
$$\begin{equation}\begin{aligned}
1 &= af(x)+bf(x+c) \\
& = a(1+3\sin(x)+2\cos(x)) + b(1+3\sin(x+c)+2\cos(x+c)) \\
& = a + b + a\sqrt{13}\cos(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Computing $\lim\limits_{x \to 0} {\frac{\sin x-x}{x^3}}$ So I was solving the Cengage Mathematics Calculus book for JEE Adv. and I came across this question in the examples $$\lim\limits_{x \to 0} {\frac{\sin x-x}{x^3}}$$
The solution given in the book uses the expansion of $\sin x$ in solving the question, whereas Wol... | The limit $$\lim_{x\to 0}\left(\frac {\sin x}x \cdot \frac 1{x^2} - \frac 1{x^2}\right)$$ is of the indeterminate form $\infty - \infty$ so
$$\lim_{x\to 0}\left(\frac {\sin x}x \cdot \frac 1{x^2} - \frac 1{x^2}\right) \neq \lim_{x\to 0}\left(\frac {\sin x}x \cdot \frac 1{x^2}\right) - \lim_{x\to 0}\left(\frac 1{x^2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving the functional equations $f(g(x)) = \sin x$ and $g(f(x)) = \cos x$
Solving the functional equations $f(g(x)) = \sin x$ and $g(f(x)) = \cos x$ .
I've managed to reduce an expression for $f(x)$ as:
$f(\cos x) = \sin(f(x))$
But I have no idea how to proceed. I would appreciate some help !
| (Too long for a comment.)
If we assume that $g(x)$ is nice enough to have a Taylor series, then there is no solution.
Going in the other direction (i.e. replacing $x$ with $g(x)$ in the cosine equation and reduce using the sine equation), we get
$$ g(\sin x) = \cos g(x). $$
Putting $g(x) = a_0 + a_1 x + a_2 x^2 + \text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
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... | Goku's solution with some geometric point of view:
*
*Dropping a perpendicular from $O$, the circumcenter, to side $c$, we have $\frac{c}{2}=R\sin C=R\sin120^{\circ}$ and hence $R=\frac{c}{\sqrt 3}.$
*Droping perpendiculars from the incenter $I$, to the sides $a,b,c$ we find:
$a=r\cot\frac B2+r\cot \frac C2=r\cot\fr... | {
"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": 1
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Remainder of a function in two variables Supposed a function in two variables is defined as $f(a,b) = \frac{a(a-1)b^{2}}{2}$, and $a,b \in N$ where a is fixed on a certain value while b runs from 1 to n.
For instance,
$$f(4,1) = \frac{4(4-1)1^{2}}{2} = 6$$
$$f(4,2) = \frac{4(4-1)2^{2}}{2} = 24$$
$$f(4,3) = \frac{4(4-1)... | When $a$ is odd, $\frac{a-1}2$ is an integer, so your fraction is $a\times\frac{a-1}2\times b^2$, which is an integer multiple of $a$, so it's equal to $0$ mod $a$.
When $a$ is even, $\frac a2$ is an integer, so your fraction is $\frac a2\times(a-1)b^2$. This is an integer multiple of $\frac a2$, so mod $a$ it can only... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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What is the concrete mathematical argument why $a^3+a=b^3+b\Rightarrow a=b$ is true in $\mathbb{R}$? The question arose from the following proof:
$f(x)=x^3+x+1$ is injective:
Let $a,b\in \mathbb{R}$ and $f(a)=f(b)$:
$f(a)=f(b)\Leftrightarrow a^3+a+1=b^3+b+1\Leftrightarrow a^3+a=b^3+b \Rightarrow a=b$
Is $a^3+a=b^3+b\Ri... | What do you mean by "rigorous enough"? No argument is in your step $a^3 + a = b^3 + b \Longrightarrow a = b$. Some reasoning should be included there.
Method 1. Use calculus to show $x^3 + x$ is an increasing function. That it should be an increasing function is obvious by staring at a graph of it, and calculus allow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Proving that $\lim_\limits{x \to \infty} x\cdot \ln\left(1 + \frac{1}{x}\right) = 1$ using Taylor series We need to prove that $\lim_\limits{x \to \infty} \ln((1 + \frac{1}{x})^x) = \lim_\limits{x \to \infty} x \cdot \ln(1 + \frac{1}{x}) = 1$ and would like to use Taylor series.
To do this, when we expand $\ln(1 + \fra... | It seems that you're making this a bit more complicated than is necessary.
For $\lvert u \rvert < 1$,
$$
\ln(1 + u) = u - \frac12 u^2 + \frac13 u^3 - \frac14 u^4
+ \cdots
$$
You can verify this Maclaurin series (Taylor series centered at $0$) by differentiating both sides to obtain the well-known geometric series
$$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all positive integers x, y Find all positive integers x, y such that $(x^2 + y)(x + y^2) = (xy)^3$
I get the problem from a practice problem set of a math camp which was organised for our national math olympiad.
I think that I can solve it if I multiply the left but after multipling I found no other way other th... | If $x,y$ are both odd, then $(x^2+y)(x+y^2)$ is even and $(xy)^3$ is odd.
If one of $x,y$ is even and the other is odd, then $(x^2+y)(x+y^2)$ is odd and $(xy)^3$ is even.
So $x,y$ must both be even and therefore $\geq 2$. Hence $\dfrac{x^2}{y} \leq \dfrac{x^2}{2}$ and $1 < \dfrac{x^2}{2}$ so that:
$$\dfrac{x^2}{y} + 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove a binomial identity: $\sum_{i=1}^n i \binom{2n}{n-i}=\frac12(n+1) \binom{2n}{n-1}$ I want to prove the product of even/odd power with a combinatorial number:
\begin{aligned}
\sum_{i=1}^n i \binom{2n}{n-i}&=\frac12(n+1) \binom{2n}{n-1}, \\
\sum_{i=1}^n i^2 \binom{2n}{n-i}&=2^{2n-2} n.
\end{aligned}
I am sure ... | I don't konw how to prove it, but for general case I found something:
Integral Representation and the Computation of Combinatorial Sums by G.P.EGORYCHEV Page77
says that:
verify the formula$\sum\limits_{i=1}^n i\left(\begin{array}{c}
2 n \\
n-i
\end{array}\right)=\frac{1}{2}(n+1)\left(\begin{array}{c}
2 n \\
n-1
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Limits problem (unable to solve further) Struggling to solve this problem,
$\displaystyle \lim\limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} +\dots+ \frac{1}{6n}\right)$
My approach:
$\displaystyle \lim\limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} +...+ \frac{1}{6n}\... | Let define $\displaystyle H_n=1+\frac {1}{2}+ \frac {1}{3}...+\frac {1}{n}
$ more precisely is the sum up to n $H_n=\sum_{k=0}^n=\frac {1}{k}$.
Using Euler-Maclaurin formula $\displaystyle H_n=ln(n)+\gamma+\frac{1}{2n}-\epsilon_n$ where $\gamma\approx0.5772...$and $0\le\epsilon_n\le\frac{1}{8n^2}$
$\displaystyle H_{6n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4612750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"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$... | By AM-GM, $\frac{a^4}{1+a^4}=\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}\ge \frac{3}{\sqrt[3]{(1+b^4)(1+c^4)(1+d^4)}}$, etc. Multiplying all these together we have $\frac{(abcd)^4}{\prod (1+a^4)}\ge \frac{3^4}{\prod (1+a^4)}$, i.e. $abcd\ge 3$.
| {
"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": 2
} |
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}}$$
A.M.-G.M. inequality makes this problem complex to solve.
It was true for $n=1,2,3,4$ then solving this became difficult. How can I solve this?
| I think it is a good idea to prove the given inequality with First Principle of Mathematical Induction.
1) Step-1:
To see if this inequality is true for $n=1$,
$$ \frac{1}{2^ \frac{3}{4}}>2- \sqrt{2}$$
Which is true (substitute $2^{ \frac{1}{4}}=k$ where and $k>1$ you will get result). Hence $P(1)$ is true.
2) Step-2
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
The real and imaginary parts of $\frac{z}{(1-z)^2}$ I have been trying to evaluate the real and imaginary parts of $\frac{z}{(1-z)^2}$ and I was wondering if my solution was correct, as it's quite not nice.
I first wrote out the z's as (x+iy)'s, turning the denominator into $1+x^2 -y^2 -2x +i(2y(1+x))$. Then I multipli... | By $w=\frac{z}{(1-z)^2}$, for the real part we have
$$2\Re(w)=w+\bar w=\frac{z}{(1-z)^2}+\frac{\bar z}{(1-\bar z)^2}=\frac{z(1-\bar z)^2+\bar z(1-z)^2}{(1-2\Re(z)+|z|^2)^2}=$$
$$=\frac{2\Re (z)-4|z|^2+2\Re(z)|z|^2}{\left(1-2\Re(z)+|z|^2\right)^2} \implies \Re \left(\frac{z}{(1-z)^2}\right) =\frac{x-2(x^2+y^2)+x(x^2+y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4615900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
how to solve $|z|^2 + z|z| + 2 = 2i Im(z)$? this is the process I follow:
*
*$z|z| = 2iImz -|z|^2- 2 $
*$z \sqrt{Re(z)^2 + Im(z)^2} = 2iy - (Re(z)^2 + Im(z)^2) - 2$
*$z\sqrt{x^2 + y^2} =2iy -(x^2+y^2)-2$
*$z=(2iy -(x^2+y^2)-2)/\sqrt{x^2 + y^2}$
then i get: $Im(z)=2y/\sqrt{x^2 + y^2}$ and $Re(z)=(-(x^2+y^2)-2)/\sq... | Write $z=re^{i\theta}=r(\cos\theta + i \sin\theta)$ with $r,\theta\in\Bbb{R}$; your equation becomes $r^2+r\times r(\cos\theta+i\sin\theta)+2=2ir\sin\theta$.
Identifying real and imaginary parts of the left & right members, you get:
$\begin{cases}
r^2(1+\cos\theta)+2=0&(1)\\[5pt]
r^2\sin\theta=2r\sin\theta&(2)
\end{cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4617200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Splitting six friends into two pairs and two singles Six friends agree to meet at the hotel Acropolis in Athens. It happens that there are four hotels with the same name. Each of the six friends picks one hotel at random and goes there.
What is the probability that two friends end up alone and the rest four in pairs?
M... | For another approach :
We know that $2$ hotels will take pairs and the rest $2$ will take only an individual. So , select which hotels will take pairs by $C(4,2)$.The rest automatically take one person.
Now , determine who will be alone and who will be pairs by $$\binom{6}{1}\times\binom{5}{1}\times\binom{4}{2}\times\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation $2^{2x}+2^{2x-1}=3^{x+0.5}+3^{x-0.5}$ Solve the equation $$2^{2x}+2^{2x-1}=3^{x+0.5}+3^{x-0.5}$$
The given equation is equivalent to $$2^{2x}+\dfrac12\cdot2^{2x}=3^x\sqrt3+\dfrac{3^x}{\sqrt3}$$ which is $$\dfrac32\cdot2^{2x}=3^x\left(\sqrt3+\dfrac{1}{\sqrt3}\right)$$ The last equation can be written ... | Since your last step then $(4/3)^{x}=(4/3)^{3/2}$ then $x=3/2$ as Sil noticed.
Alternatively, let $x$ be a real number, then
\begin{align*}
2^{2x}+2^{2x-1}&=3^{x+1/2}+3^{x-1/2}\\
\frac{1}{3^{x+1/2}}\left(2^{2x}+2^{2x-1} \right)&=\frac{1}{3^{x+1/2}}\left(3^{x+1/2}+3^{x-1/2} \right)\\
2^{2x}\cdot3^{-(x+1/2)}+2^{2x-1}\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Activation function defined as a convolution (proof verification) I would like to request a proof verification.
I am currently reading this paper, and I am trying to prove Eq. (3) from the paper.
The operator $\ast$ is the convolution operation and $\phi_n(x) = n\phi(nx)$ with $\phi(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac... | You’re right; the paper is wrong. This is easiest to see if you choose $\alpha=1$. Then the convolution of $x$ with a Gaussian should indeed yield $x$, not an additional Gaussian term. You don’t need to evaluate any integrals for that; it follows by symmetry, since a convolution of a symmetric and an antisymmetric func... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4619954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.