Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Hint required - Find values of $a$ and $b$ such that the limit equals $\frac{5}{12}$. Find the values of $a$ and $b$ such that $$\lim_{x\to 0} \frac{\sqrt[3]{ax+b}-2}{x}= \frac{5}{12}.$$
Can someone please provide me with a hint on how to approach this question?
| $$\lim_{x\to 0} \frac{\sqrt[3]{ax+b}-2}{x}= \frac{5}{12}$$
An approach without using L'Hôpital's rule:
Since the denominator approaches zero as $x$ approaches zero, the numerator must also approach zero as $x$ approaches zero for the limit to be finite, i.e.,
$$\tag{1} \lim_{x\to0}\sqrt[3]{ax+b}-2=0
\implies b^\frac13 ... | {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate $\lim\limits_{n\rightarrow \infty }(n-\sum^n_{k=1}\cos\frac{2k}{n\sqrt{n}}) $? I tried this:
$\lim\limits_{n\rightarrow \infty }(n-\sum^n_{k=1}\cos\frac{2k}{n\sqrt{n}}) $=
= $ \lim\limits_{n\rightarrow \infty }(1-\cos\frac{2}{n\sqrt{n}}) $ + $ \lim\limits_{n\rightarrow \infty }(1-\cos\frac{4}{n\sqrt{n}... | You face the sum of cosines with angles in arithmetic progression.
So
$$\sum_{k=1}^n \cos \left(\frac{2 k}{n^{3/2}}\right)=\sin \left(\frac{1}{\sqrt{n}}\right) \cos \left(\frac{n+1}{n^{3/2}}\right) \csc
\left(\frac{1}{n^{3/2}}\right)$$
$$S_n=n-\sin \left(\frac{1}{\sqrt{n}}\right) \cos \left(\frac{n+1}{n^{3/2}}\right... | {
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"url": "https://math.stackexchange.com/questions/4440617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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find the maximum of $x+\frac{1}{2y}$ under the condition $(2xy-1)^2=(5y+2)(y-2)$ Assume $x,y \in \mathbb{R}^+$ and satisfy $$\left(2xy-1\right)^2 = (5y+2)(y-2)$$, find the maximum of the expression $x+\frac{1}{2y}$.
$\because (5y+2)(y-2) \geq 0$ , $\therefore y\geq 2$ or $y \leq -\frac{2}{5}$, and we have $$2xy = \sqrt... | Alternative solution:
Let $z = x + \frac{1}{2y}$.
We have $x = z - \frac{1}{2y}$ and
$2xy = 2yz - 1$.
Then, we have
$$(2yz - 2)^2 = (5y + 2)(y - 2)$$
or
$$(4z^2 - 5)y^2 + (-8z + 8)y + 8 = 0. \tag{1}$$
Since the quadratic equation in $y$ (1) has real roots, its discriminant is non-negative, i.e.
$$\Delta := -64z^2 - 128... | {
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Discriminant of depressed cubic If the cubic equation $x^3+px+q$ has roots $\alpha , \beta , \gamma $ then we know that $\alpha + \beta + \gamma =0 $, $\alpha \beta + \alpha \gamma + \beta \gamma =p $ and $\alpha \beta \gamma =-q $.
The discriminant is $\Delta = (\alpha - \beta )^2 (\alpha - \gamma )^2 (\beta - \gamma ... |
If the cubic equation $x^3+px+q$ has roots $\alpha , \beta , \gamma $ then we know that $\alpha + \beta + \gamma =0 $, $\alpha \beta + \alpha \gamma + \beta \gamma =p $ and $\alpha \beta \gamma =-q $.
The discriminant is $\Delta = (\alpha - \beta )^2 (\alpha - \gamma )^2 (\beta - \gamma ) ^2 $.
I know the answer s... | {
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Solving the system $a^3 + 15ab^2 = 9$, $\;\frac 35 a^2b + b^3 = \frac 45$ I have problem solving the following system of two cubic equations.
$$\begin{cases}
a^3 + 15ab^2 = 9 \\
\frac 35 a^2b + b^3 = \frac 45
\end{cases}
$$
I don't have any idea how to approach this kind of problem.
I'm looking for solutions included i... | Starting from
$$\begin{align}
a^3 + 15ab^2 = 9 \\
\frac 35 a^2b + b^3 = \frac 45
\end{align}$$
let $b=ax$ as obviously $b\neq0$:
$$\begin{align}
a^3 + 15x^2a^3 = 9 \\
3 xa^3 + 5x^3a^3 = 4
\end{align}$$
dividing both sides eliminates $a$:
$$
\frac{1+ 15x^2}{3 x + 5x^3} = \frac 94
$$
so you have a cubic in $x$, for which... | {
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Solve for x: $\lfloor x\rfloor \{\sqrt{x}\}=1$ where $\{x\}=x-\lfloor x\rfloor$ Solve for x:
$\lfloor x\rfloor \{\sqrt{x}\}=1$
where $\{x\}=x-\lfloor x\rfloor$
My Attempt:
I took intervals of $x$ as $x\in (2,3)$ so $\sqrt x\in (1,2)$. Due to which $\lfloor x\rfloor=2$ and $\{\sqrt x\}=\sqrt x-\lfloor \sqrt x\rfloor=\sq... | There is obviously no solution for x < 1. Let k, n be integers with k ≥ 1 and $k^2 ≤ n ≤ k^2 + 2k$, and look for solutions x in the interval [n, n+1); this covers all solutions.
In this interval, the equation is equivalent to $n \cdot (\sqrt x - k) = 1$ or $x = (k + 1 / n)^2$. Since $n ≥ k^2$, $x ≤ (k + 1 / k^2)^2 = k^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How to find $\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}$? By factorization:
$$\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}\tag{1}$$
$$=\lim_{x\to-\infty} \frac{x\sqrt{1+\frac{2}{x}}}{-x}$$
$$=\lim_{x\to-\infty}-\sqrt{1+\frac{2}{x}}$$
If I input $x=-\infty$, the limiting value seems to be $-1$. But according to desmos, ... | This answer is based on @KaviRamaMurthy's comments.
Alternative way to find the limit:
$$\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}$$
$$\text{[$\sqrt{x^2+2x}$ is positive. Moreover, $-x$ is also positive. So, $\frac{\sqrt{x^2+2x}}{-x}$ is positive.]}$$
$$=\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{|x|}$$
$$=\lim_{x\to-\i... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Why does the method of inclusion/exclusion give the wrong answer when finding the number of integers b/w 1 and 10 that are not divisible by 2,3 or 5? Let $S=\{1,2,\dots, 10\}$.
METHOD 1: I'm first counting the integers that are divisible by $2, 3$ or $5$ in $S$ and then subtracting from the total as follows:
Let $A, B... | In the second method you want to get $|A\cap B \cap C|$ not $|A \cup B \cup C|$.
| {
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"timestamp": "2023-03-29T00:00:00",
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The Maclaurin series of $1-(1-\frac{x^2}{2} + \frac{x^4}{24})^{2/3}$ has all coefficients positive It was shown in a previous post that the Maclaurin series of $1 - \cos^{2/3} x$ has positive coefficients. There @Dr. Wolfgang Hintze: has noticed that the truncation $1- \frac{x^2}{2} + \frac{x^4}{24}$ can be substituted... | Hint
Consider
$$f(a)=1-\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)^a$$ and use the binomial expansion.
It should write
$$f(a)=\frac a 2 \sum_{n=1}^\infty\frac {P_n(a)}{b_n} x^{2n}$$ You should see it very quickly.
Edit
After your edit and remarks, considering
$$f(k)=1-\big[ (1-ax)(1-b x)\big]^k$$ let $b=a c$ to obtain... | {
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fourth-order finite difference for $(a(x)u'(x))'$ Previously I asked here about constructing a symmetric matrix for doing finite difference for $(a(x)u'(x))'$ where the (diffusion) coefficient $a(x)$ is spatially varying. The answer provided there works for getting a second order accurate method. What about getting a f... | A bit of work with five-point Lagrange interpolants based on a uniform grid for $a$ and $u$ shows, with
\begin{align}
w_{-2} &:= \tfrac{1}{144}a_{-2} - \tfrac{1}{18} a_{-1} - \tfrac{1}{12}a_0 + \tfrac{1}{18} a_1 - \tfrac{1}{144} a_2, \\
w_{-1} &:= -\tfrac{1}{18}a_{-2} + \tfrac{4}{9} a_{-1} + \tfrac{4}{3} a_0 - \tfrac{4... | {
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Showing $\int_{0}^{2022}x^{2}-\lfloor{x}\rfloor\lceil{x}\rceil dx = 674$ It is from the 2022 MIT Integration Bee Question 3 states as follows:
$$\int_{0}^{2022}x^{2}-\lfloor{x}\rfloor\lceil{x}\rceil dx$$
I know that the answer is $674$, but I do not know the process and the steps to derive this solution. Can someone pl... | To keep things simple, let's limit the integration to the interval $(n, n+1)$. where $n \in \mathbb{Z}$.
$$\int_n^{n+1} x^2 - \lfloor x \rfloor \lceil x \rceil dx $$
$$= \int_n^{n+1} x^2 - n (n + 1) dx $$
$$= \left(\frac{1}{3}x^3 - n (n + 1)x\right)\Bigg|_{n}^{n+1} $$
$$= \left(\frac{1}{3}(n+1)^3 - n (n + 1)^2\right) -... | {
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"timestamp": "2023-03-29T00:00:00",
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What is the minimum value of the function $f(x)= \frac{x^2+3x-6}{x^2+3x+6}$? I was trying to use the differentiation method to find the minimum value of the person but it did not give any result, I mean when I differentiated this function and equated to zero for finding the value of $x$ when the function value would be... | You do not need calculus/derivatives for this. Let $u=x^2+3x+6=\left(x+\frac32\right)^2+\frac{15}{4}$, and note that you would like to minimize $\frac{u-12}{u}=1-\frac{12}{u}$.
$u$ can take values in $[15/4,\infty)$. And $1-\frac{12}{u}$ is an increasing function for positive $u$.
It follows that the minimum occurs whe... | {
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Single Solution for this Recurrence: $a(n)=3^n-a(n-1)+1$ I've solved this recurrence using the iteration method for even and odd values of $n$, but I cannot seem to find a singular explicit function that solves this recurrence for all values of $n$.
The recurrence is
$$a(n) =
\begin{cases}
1, & \text{if $n=0$} \\
3^n-... | Let $A(x)=\sum_{n \ge 0} a_n x^n$ be the ordinary generating function. The recurrence and initial condition imply that
\begin{align}
A(x)
&= a_0 x^0 + \sum_{n \ge 1} a_n x^n \\
&= 1 + \sum_{n \ge 1} (3^n - a_{n-1} + 1) x^n \\
&= 1 + \sum_{n \ge 1} (3x)^n - x \sum_{n \ge 1} a_{n-1} x^{n-1} + \sum_{n \ge 1} x^n \\
&= 1 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the area of a triangle on a unit circle Assume that $0< \theta < \pi$. For three points $A(1,0)$, $B(\cos(\theta),\sin(\theta))$ and $C(\cos(2\theta),\sin(2\theta))$ on a unit circle, the area of triangle $ABC$ is: ???
I drew out the unit circle and tried to get the dimensions. I ended up drawing a triangle und... | The three side lengths of $\triangle ABC$ are:
$$BC = AB = \sqrt{(\cos \theta - 1)^2 + (\sin \theta - 0)^2}$$
$$= \sqrt{\cos^2 \theta - 2 \cos \theta + 1 + \sin^2 \theta}$$
$$= \sqrt{2 - 2 \cos \theta}$$
$$AC = \sqrt{(\cos(2\theta) - 1)^2 + (\sin(2\theta)-0)^2}$$
$$= \sqrt{\cos^2(2\theta) - 2\cos(2\theta) + 1 + \sin^2(... | {
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The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$. Find the other two roots. The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$.
Find the other two roots.
$(1+i)^3 -(1+i)^2+2= (1-i-3+3i)-(1-1+2i) +2= (-2+2i)-(2i) +2= 0$.
The other two roots are found by division.
$$
\require{enclose}
\begin{array}{... |
$x^3-x^2+2= (x-1-i)\,x^2 + 2 + i\,x^2\;$ How to pursue by this or some other approach?
The posted answers cover the other approaches.
OP's approach is also salvageable, but first the polynomial division must be taken to completion:
$$
x^3-x^2+2= \left(x-1-i\right)\left(x^2 + ix - 1 + i\right)
$$
The quotient can be f... | {
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Determine the image of the unit circle $S^1$ by the action of the matrix $e^A$. We have:
$$e^{ \begin{pmatrix}
-5 & 9\\
-4 & 7
\end{pmatrix} }$$
I need to determine the image of the unit circle $S^1$ by the action of the matrix $e^A$.
I think that I know how to calculate $e^A$:
I get the Jordan decomposition:
$$A = \be... | The calculation won't be easy, but standard.
To understand the image of any invertible matrix $T$ acting on the unit circle, first do the SVD decomposition $$T=U\Sigma V = U\begin{pmatrix} \sigma_1 & \\ & \sigma_2\end{pmatrix}V$$ Note that $V$ send the circle back to itself, as it keeps norms. Now it should be clear th... | {
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If $x$ and $y$ are positive integers such that $5x+3y=100$, what is the greatest possible value of $xy$? I first wrote $y$ in terms of $x$. in $5x + 3y = 100$, I subtracted $5x$ from both sides to get $3y = 100 - 5x$. Therefore, $y = \frac{100 - 5x}{3}$. I substituted this into $xy$ to get $x(\frac{100 - 5x}{3}$) – You... | Since $$3y= 100-5x\implies 3\mid 20-x$$ we have $20-x =3t$ for some integer $t$. Then $x= 20-3t$ and $y= 5t$. So we have $$xy = 5t(20-3t) \leq \max \{20\cdot 8, 15\cdot 11\}$$
(Maximum value of quadratic function is at $10\over 3$ so in the set of integers it is at $3$ or at $4$.)
| {
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Possible growth rates of a matrix entry with respect to exponentiation Let $A = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$, so $A^n = \begin{pmatrix}1 & n \\ 0 & 1\end{pmatrix}$. Thus, $(A^n)_{1,1} = 1 = \Theta(1)$, and $(A^n)_{1,2} = n = \Theta(n)$.
Given a constant $c$, if $B = \begin{pmatrix} c/2 & c/2 \\ c/2 & c/... | Consider $A=I+S,$ where $S$ a matrix with entries $s_{i,i+1}=1$ and $0$ otherwise. If the dimension is equal $m,$ then $$A^n=I+ \sum_{k=1}^{m-1} {n\choose k}S^k$$ Thus $$(A^n)_{1,1+j}={n\choose j}\approx {n^j\over j!}\quad \qquad 1\le j\le m-1$$
| {
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Inequality about $x^p$ for $p\geq 2$ a real number I am stuck with the following inequality
$$g^2(n+2)+2g(n+2)g(n)-g^2(n+1)-2g(n+1)g(n+2)\geq 0,$$
for all integer $n\geq 1$. Here, $g(x)=x^p$ where $p\geq 2$ is a real number. I need help.
| It suffices to prove that
$$(n + 2)^{2p} + 2(n + 2)^p n^p - (n + 1)^{2p} - 2(n + 1)^p(n + 2)^p \ge 0$$
or
$$[(n + 2)^p - (n + 1)^p]^2
- 2(n + 1)^{2p} + 2(n + 2)^p n^p \ge 0$$
or (multiplying both sides by $(n + 1)^{-2p}$)
$$\left(\left(1 + \frac{1}{n + 1}\right)^p - 1\right)^2
- 2 + 2\left(1 - \frac{1}{(n + 1)^2}\righ... | {
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Find the limit $\lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4(x+1)}$ I need to find $\displaystyle \lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4\left(x+1\right)}$.
I tried using the following:
\begin{align*}
\ln(1+x)&\approx x,\\
\sin(x)&\approx x-\frac{x^3}{2},\\
\cos(x)&\approx 1-\frac{x... | You only need to add to your equalities the $\log$
$$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \textrm{h.o.t}\\
x \sin x = x^2 - \frac{x^4}{6} + \textrm{h.o.t}\\
\log(1+x) = x + \textrm{h.o.t, so} \\
\log^4(1+x) = x^4 + \textrm{h.o.t}$$
Now calculate carefully and see that
$$\frac{\cos x-1 - \frac{x}{2} \sin x}{\lo... | {
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"timestamp": "2023-03-29T00:00:00",
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Solving the integral $\int_{0}^{1} \frac{s \ln{s} - t \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}\,\mathrm{d}s$ For $t \in [0,1]$, let
$$f(t) = \int_{0}^{1} \frac{s \ln{s} - t \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}\,\mathrm{d}s.$$
Is it possible to evaluate this integral to obtain an explicit expression for $f(t)$? Mathemat... | Split the integral into two
\begin{align}
I=& \int_{0}^{1} \frac{s \ln{s} - t \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}ds=I_1+I_2
\end{align}
where, with $b = \sqrt{1-t^2}$
\begin{align}
I_1= & \int_{0}^{1} \frac{s \ln{s} - s \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}ds
\overset{x=\sqrt{1-s^2}}=\frac12 \int_{0}^{1}
\frac{\ln... | {
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"timestamp": "2023-03-29T00:00:00",
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Simplest proof of $|I_n+J_n|=n+1$ Answering another question, I realised that I "know" that the determinant of the sum of an identity matrix $I_n$ and an all-ones square matrix $J_n$ is $n+1$. I.e.
$$|I_n+J_n|=\left|\begin{pmatrix}
1 & 0 & 0 &\cdots & 0 \\
0 & 1 & 0 &\cdots & 0 \\
0 & 0 & 1 &\cdots & 0 \\
\vdot... | The eigenvalues of $J_n$ are $0$ with geometric multiplicity $n-1$ (because the rank of $J_n$ is one) and $n$ with algebraic multiplicity at least $1$ (an eigenvector is the “all $1$” column vector).
Thus the algebraic multiplicity of $0$ is $n-1$ and of $n$ is $1$. Hence the characteristic polynomial of $J_n$ is
$$
p(... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Real part of $ \quad 1- \frac{2}{\pi} \arctan(r^{\rho}e^{i\rho\theta}).$ To solve the Dirichlet problem using mellin transform, i needed to find the real part of $ \quad 1- \displaystyle\frac{2}{\pi} \arctan(r^{\rho}e^{i\rho\theta}).$
I already know the result will be
\begin{cases}
\quad 1- \displaystyle \frac{1}{\pi}... | For a complex number $z \ne \pm \mathrm{i}$, we have
$$\arctan z = \frac{1}{2\mathrm{i}}\ln \frac{\mathrm{i} - z}{\mathrm{i} + z}$$
where $\ln u$ is the principal branch of complex logarithm ($u\ne 0$)
$$\ln u = \ln |u| + \mathrm{i}\arg u, \quad -\pi < \arg u \le \pi.$$
(See, e.g. http://scipp.ucsc.edu/~haber/archives/... | {
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Evaluating $\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$ for $n \in \mathbb{N}$ For $n \in \mathbb{N}$, evaluate
$$\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$$
I could not use wolframalpha, I do not know the reason.
For $... | Hint: Notice that the sum is of consecutive odd powers of $x$ changing sign every two terms. What if you break it up into two sums, one of which is of the powers $4k+1$ and one of which is the powers of $4k+3$, both of alternating sign, and compare them to the Taylor series for $\frac{1}{1+y}$ for some suitable $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4481299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Assistance with solving the integral Can you give me an idea how to handle this integral?
$\int_{0}^{2\pi} (1+\cos{x})\sqrt{3+\cos{x}}\,dx$
| Since $1 - \cos(x) = 2 \sin^2\left(\frac{x}{2} \right) $ then
\begin{align}
\int_{0}^{2\pi} (1+\cos{x})\sqrt{3+\cos{x}}\, \mathrm{d}x & \overset{u= x -\pi}{=} 2\int_{0}^{\pi}2 \sin^2\left(\frac{u}{2} \right) \sqrt{2+2\sin^2\left(\frac{u}{2} \right)}\, \mathrm{d}u \\
& \overset{t = \frac{u}{2}}{=} 8 \sqrt{2} \int_0^{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the extrema of $f(x,y)=max(x,y)$ constrained to $\mathscr A=\{(x,y) ∈ \mathbb R^2 ∣ x^2+y^2=1\}$ I know the maxima occurs at $f(x,y) = 1$ at the points $(1,0)$ and $(0,1)$... but I can't seem to find all the minima.
I've come to the conclusion that the minima occurs at a point $(a,a)$ where $a ∈ \mathbb R_<0$, suc... | The original curve can be expressed by the following equation in polar coordinates $(\rho,\theta)$:
$$
\rho=1.
$$
$(1)$ When $y\ge x$, there is $\theta\in[\frac{\pi}{4},\frac{3\pi}{4}]$, we have $f(x,y)=y=\text{sin}(\theta)$, so $\min f(x,y)=-\frac{\sqrt{2}}{2}$ when $\theta=\frac{3\pi}{4}$ or $x=y=-\frac{\sqrt{2}}{2}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Irrational equation $\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$ I saw the problem from one math competition:
$$\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$$.
I tried to solve it this way:
\begin{align*}
& \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0\\
\Leftrightarrow\,\, & \left... | HINT
Based on the suggestion of @JohnOmielan in the comments, assuming that $|x| > 1$, it yields that
\begin{align*}
(x^{3} - 3x + 1)(\sqrt{x^{2} - 1} + x) = -1 & \Longleftrightarrow x^{3} - 3x + 1 = \sqrt{x^{2} - 1} - x\\\\
& \Longleftrightarrow x^{3} - 2x + 1 = \sqrt{x^{2} - 1}\\\\
& \Longleftrightarrow (x^{3} - 2x +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
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Prove, using a combinatorial argument, that $\binom{m}{n}\cdot\binom{n}{r}= \binom{m}{r}\cdot \binom{m-r}{n-r}$
Prove, using a combinatorial argument, that
$$\binom{m}{n}\cdot\binom{n}{r}= \binom{m}{r}\cdot \binom{m-r}{n-r}$$
attempt: In $m$, suppose that $n$ went for a ride, and out of those $n$, $r$ were awarded. T... | Solution:
For the left part, we have
$$
\left( \begin{array}{c}
m\\
n\\
\end{array} \right) \cdot \left( \begin{array}{c}
n\\
r\\
\end{array} \right) =\frac{m!}{n!\left( m-n \right) !}\cdot \frac{n!}{r!\left( n-r \right) !}=\frac{m!}{\left( m-n \right) !\left( n-r \right) !r!},
$$
and for the right part, we have
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $A+B+C+D=\pi$ then find $\sum\cos A\cos B-\sum\sin A\sin B$ Question:
If $A+B+C+D=\pi$ then find $\sum\cos A\cos B-\sum\sin A\sin B$
My Attempt:
$$\cos((A+B)+(C+D))=-1\\\implies\cos(A+B)\cos(C+D)-\sin(A+B)\sin(C+D)=-1\\\implies(\cos A\cos B-\sin A\sin B)(\cos C\cos D-\sin C\sin D)-(\sin A\cos B+\cos A\sin B)(\sin C\... | Using $\cos(x+y)=\cos x\cos y-\sin x\sin y$, we have
$$\cos A\cos B-\sin A\sin B=\cos(A+B),$$
$$\cos B\cos C-\sin B\sin C=\cos(B+C),$$
$$\cos C\cos D-\sin C\sin D=\cos(C+D),$$
$$\cos D\cos A-\sin D\sin A=\cos(D+A),$$
hence
$$\sum\cos A\cos B-\sum\sin A\sin B=\cos(A+B)+\cos(C+D)+\cos(B+C)+\cos(D+A).$$
Now using $\cos(\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to solve this function $f(x)$ for degree $3$ or $4$
If $f(x)=0$ is a polynomial whose coefficients are either $1$ or $-1$ and whose roots are all real, then the degree of $f(x)$ can be equal to$:$
$A$. $1$
$B$. $2$
$C$. $3$
$D$. $4$
My work$:$
For linear only four polynomials are possible which are $x+1$ ... | The cubic case is handled by noting that
$(x-1)^2(x+1)=x^3-x^2-x+1=0$
has all of its roots real.
For the quartic case, we do not have such a simply soluble case. But we can use Descartes' Rule of Signs, if we are careful with it.
For instance, when we have all coefficients positive, as in $x^4+x^3+x^2+x+1=0$, four nega... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to find $\int_0^1 x^4(1-x)^5dx$ quickly? This question came in the Rajshahi University admission exam 2018-19
Q) $\int_0^1 x^4(1-x)^5dx$=?
(a) $\frac{1}{1260}$
(b) $\frac{1}{280}$
(c)$\frac{1}{315}$
(d) None
This is a big integral (click on show steps):
$$\left[-\dfrac{\left(x-1\right)^6\left(126x^4+56x^3+21x^2+6x+... | If you don't have the beta function formula memorized, you can use integration by parts repeatedly
\begin{align*}
\int_{0}^{1}x^4(1-x)^5\,dx &= \left[x^4 \cdot -\dfrac{1}{6}(1-x)^6\right]_{0}^{1} - \int_{0}^{1}4x^3 \cdot -\dfrac{1}{6}(1-x)^6\,dx
\\
&= \dfrac{4}{6}\int_{0}^{1}x^3(1-x)^6\,dx
\\
&= \dfrac{4}{6}\left[x^3 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
} |
A golden question $\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$ How would you solve this problem for real $x$?
$$\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$$
It can be easily shown that both equations
$$x=\sqrt{2+\sqrt{2-x}}\tag{1}$$
and $$x=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}\tag{2}$$
h... | With some difficulty.
You can turn the equation into a polynomial through lots of squaring and rearranging, starting with:
$\begin{eqnarray} \sqrt{2+\sqrt{2-x}} & = & \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} \\
& = & \sqrt{\frac{x - 1}{x}}\left(\sqrt{x + 1} + 1\right) \\
2 + \sqrt{2 - x} & = & \frac{x - 1}{x} \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4495058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Evaluating $\lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}$ $$
\begin{align*}
&\text { Let } \mathrm{x}=2 \mathrm{y} \quad \because x \rightarrow 0 \quad \therefore \mathrm{y} \rightarrow 0\\
&\therefore \lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}\\
&=\lim _{y \rightarrow 0} \frac{12-6(2 y)^{... | While I'm not entirely sure what's going on with the answer you have provided (exponents seem to appear and disappear, multiplication turns into addition etc.), I shall offer you two methods to evaluate your given limit.
Method 1 : As both the numerator and denominator both tend to $0$ as $x\to0 $, we can use L'Hopita... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
On Réalis’s solution of the “cubic Markov equation” I am interested in the Diophantine equation
$$X^3+Y^3+Z^3=3XYZ$$
and its solutions. In Nouv. Corr. Math. 1879 (page 7), Réalis claims that, other than the trivial solution $x=y=z$, the complete solution in integers is given by
$$
x=(a-b)^3+(a-c)^3, \quad y=(b-c)^3+(b-... | This is quite wrong. We have
$$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x + y \omega + z \omega^2)(x + y \omega^2 + z \omega)$$
where $\omega = e^{ \frac{2 \pi i}{3} }$ is a primitive third root of unity. This identity comes from recognizing the LHS as the determinant of the circulant matrix $\left[ \begin{array}{ccc} x & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4499443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$\int_0^1 \frac{\arcsin x\arccos x}{x}dx$ Someone on Youtube posted a video solving this integral.
I can't find on Math.stack.exchange this integral using search engine https://approach0.xyz
It is related to $\displaystyle \int_0^\infty \frac{\arctan x\ln x}{1+x^2}dx$
Following is a solution that is not requiring the u... | Letting $y=\arcsin x$ transforms our integral
$$
\begin{aligned}
I &=\int_{0}^{\frac{\pi}{2}} \frac{y\left(\frac{\pi}{2}-y\right) \cos y d y}{\sin y} \\
&=\frac{\pi}{2} \underbrace{\int_{0}^{\frac{\pi}{2}} \frac{y \cos y}{\sin y} d y}_{J}-\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{y^{2} \cos y}{\sin y} d y}_{K}
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
Use the definition of limit to prove that $\lim_{x \to 0} \frac{x}{x+1}=0$ Use the definition of limit to prove that $\lim_{x \to 0} \dfrac{x}{x+1}=0$
My attempt:
Let $\epsilon >0$ then I need to find $\delta>0$ such that if $0<|x|<\delta$ then $\left| \dfrac{x}{x+1} \right| < \epsilon$
What I was thinking is that, to ... | Suppose $\delta < \min\{\frac12, \frac{\epsilon}{2}\}$. Then $|x + 1| \ge \min\{\frac12, 1 - \frac{\epsilon}{2}\} \ge \frac12$ so that $\frac{|x|}{|x+1|} \le 2\delta < \epsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Show that $\frac{\sin^3 \beta}{\sin \alpha} + \frac{\cos^3 \beta}{\cos \alpha} = 1$ with certain given $\alpha, \beta$ Let $$\frac{\sin (\alpha)}{\sin (\beta)} + \frac{\cos (\alpha)}{\cos (\beta)} = -1 \tag{$1$}$$ where $\alpha, \beta$ are not multiples of $\pi / 2$. Show that
$$\frac{\sin^3 (\beta)}{\sin (\alpha)} + \... | Multiplying by a common denominator gives:
$$\sin \alpha \cos \beta + \cos \alpha \sin \beta = -\cos \beta \sin \beta$$
which is the same as
$$\sin(\alpha + \beta) = -\cos \beta \sin \beta. \tag{1}$$
Now:
$$(\sin \alpha \cos \beta + \cos \alpha \sin \beta)(\sin^2 \beta + \cos^2 \beta) = -\cos \beta \sin \beta$$
$$\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How many method are there to handle the integral $\int \frac{\sin x}{1-\sin x \cos x} d x$ $$
\begin{aligned}
\int \frac{\sin x}{1-\sin x \cos x} d x =& \frac{1}{2} \int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{1-\sin x \cos x} d x \\
=& \int \frac{d(\sin x-\cos x)}{2-2 \sin x \cos x}-\int \frac{d(\sin x+\cos x)}{2-2 \si... | Letting $y=\frac{\pi}{2}-x$ converts the integral into
$$
I=\int \frac{\cos y}{1-\sin y \cos y} d y
$$
Using the Weierstrass substitution, $t =\tan \frac{y}{2} $, we simplify and get
$$
I=-2 \int \frac{t^{2}-1}{t^{4}-2 t^{3}+2 t^{2}-2 t+1} d t
$$
Playing a little trick on the $I$ gives
$$
I =-2 \int \frac{1-\frac{1}{t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Determination of order of cosets in a factor group of a finite abelian group. Consider the group $G=\frac{\mathbb{Z}_{3^{10}}\times \mathbb{Z}_{3^7} }{\langle (3^2, 3^3)\rangle}$. Let $a=(1,0)+\langle (3^2, 3^3)\rangle$ and $b=(0,1)+\langle (3^2, 3^3)\rangle$. Since the order of the group $G$ is $3^9$, so order of $a$ ... | Let $G_1=\Bbb{Z}_{3^{10}} $,$G_2=\Bbb{Z}_{3^{7}}$ , $H=\langle (3^2, 3^3)\rangle$
Then $|3^2|_{G_{1}}=3^8$ , $|3^3|_{G_{2}}=3^4$
Let $|a|=n$ where $a=(1,0)+H$
Then $n$ is the least positive integer such that $na=n(1, 0) +H=H$
$(n, 0) +H=H$ implies $(n, 0) \in H$
Hence $n$ is the least positive integer such that $(n, 0)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Coefficients of the polynomial? I have the following function
$$ p_n(x,y) = \sum_{i=0}^n c_{n}(i)~x^{n-i}y^i . $$
I know that first couple of $n$'s have a form
$$
n=1, ~~~ p_1(x,y) = 3 x + y\, , \\
n=2, ~~~ p_2(x,y) = 15 x^2 + 10 x y + 3 y^2\, , \\
n=3, ~~~ p_3(x,y) = 105 x^3 + 105 x^2 y + 63 x y^2 + 15 y^3\, , \\
n=4,... | Thanks to the information provided in OPs comment we can show
\begin{align*}
\color{blue}{c_n(i)=\frac{2n+1}{2i+1}\,\frac{(2n)!}{2^nn!}\binom{n}{i}\qquad\qquad 0\leq i\leq n}\tag{1}
\end{align*}
We are looking according to OPs comment for the coefficients $c_i(n)$ of the bivariate polynomial
\begin{align*}
p_n(x,y)=\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the sum of radicals without squaring, Is that impossible?
Find the summation:
$$\sqrt {3-\sqrt 5}+\sqrt {3+\sqrt 5}$$
My attempts:
\begin{align*}
&A = \sqrt{3-\sqrt{5}}+ \sqrt{3+\sqrt{5}}\\
\implies &A^2 = 3-\sqrt{5} + 3 + \sqrt{5} + 2\sqrt{9-5}\\
\implies &A^2 = 6+4 = 10\\
\implies &A = \sqrt{10}
\end{align*}
S... | I want to solve the problem with a geometric approach. Consider a right triangle $ABC$ as below:
If we suppose that $BH = \sqrt{3 - \sqrt{5}}$ and $HC = \sqrt{3 + \sqrt{5}}$, then we have $AH^2 = BH.HC = \sqrt{9 - 5} = 2$. So $AH = \sqrt{2}$. Now use Pythagorean theorem:
\begin{align*}
\overset{\triangle}{ABH}&: AB^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 5
} |
If a and b are the distinct roots of the equation $x^2+3^{1/4}x + 3^{1/2} =0$ then find the value of $a^{96}(a^{12}-1)+b^{96}(b^{12}-1)$.
If $a$ and $b$ are the distinct roots of the equation $x^2+3^{1/4}x +
3^{1/2} =0$ then find the value of $a^{96}(a^{12}-1)+b^{96}(b^{12}-1)$
My attempt:
LHS = $(a^{108}+b^{108})-(... | As suggested in Eric Snyder's comment, let $c = \sqrt[4]{3}$. Then rearrange your equation to get
$$x^2 = -cx - c^2 \tag{1}\label{eq1A}$$
Multiply both sides by $x$ and use \eqref{eq1A} to get
$$\begin{equation}\begin{aligned}
x^3 & = -cx^2 - c^{2}x \\
& = -c(-cx - c^2) - c^{2}x \\
& = c^{2}x + c^3 - c^{2}x \\
& = c^3
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving floor function system of equations given $x \lfloor y \rfloor + y \lfloor x \rfloor =66$ and $x \lfloor x \rfloor + y \lfloor y \rfloor=144$
$x$ and $y$ are real numbers satisfying $x \lfloor y \rfloor + y \lfloor x \rfloor =66$ and $x \lfloor x \rfloor + y \lfloor y \rfloor=144.$ Find $x$ and $y.$
I first a... | Here is an idea: you already showed that $(x+y)(\lfloor x \rfloor + \lfloor y \rfloor)=210$. If you denote $k=\lfloor x \rfloor + \lfloor y \rfloor$ then $k\leq x+y<k+2$. So we have $k^2\leq k(x+y)<k(k+2)$. Since $k(x+y)=210$, we now know that $k^2\leq 210<k(k+2)$. Solving this inequality for integer $k$ yields $k=14$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sum of squares of ratios of diagonal of a regular heptagon A problem from the 1998 Lower Michigan Math Competition...
A regular heptagon has diagonals of two different lengths. Let $a$ be the length of a side, $b$ the length of a shorter diagonal, and $c$ the length of a longer diagonal. Prove that
$$
\frac{a^2}{b^2}... | Adding details to David's answer---too much algebra to fit in a comment. ;-)
Some milestones along the way:
*
*Taking the modulus-squared on both sides of David's expression for $\frac{1}{1+\omega}$ yields
$$
\frac{1}{|1+\omega|^2} = |1+\omega^2 + \omega^4|^2
= (1+\omega^2+\omega^4) \overline{(1+\omega^2+\omega^4)}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that for any two positive real numbers $a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$ Question:
Show that for any two positive real numbers $a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$.
So for this question, I began by expanding all terms and moving them all to one side. However, I do not know how to... | first let's do some cleaning:
$$\frac{a}{a+2b}+\frac{b}{b+2a} = \frac{2a^2 + 2ab + 2b^2}{2a^2 + 4ab + 2a^2} = \frac{2(a+b)^2 - 2ab}{2(a+b)^2} = 1-\frac{ab}{(a+b)^2}$$
and we have :
$$A = \frac{a}{a+2b}+\frac{b}{b+2a} = \frac{1}{1+\frac{2b}{a}} + \frac{1}{1+\frac{2a}{b}}$$
let $x = \frac{a}{b} $ then we have :
$$f(x) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Is there any method to compute $\frac{d^{n}}{d x^{n}}\left(\cos ^{k} x\right)$ other than complex numbers? In this couple of days, I need to know the high derivatives of $cos^kx$ whose power $k$ make the differentiation much harder. Then I attempt to
use the identity $$
\cos x=\frac{1}{2}\left(e^{x i}+e^{-x i}\right),... | Although essentially equivalent, you can say
$$\cos^k(x)=\frac1{2^k}\sum_{j=0}^k{k\choose j}\cos(k-2j)x,$$
which can be proved by induction on $k$, together with the sum formula. Now, differentiating is fairly straightforward, based on casework on $n\pmod 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4519048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Combinatorial Identity involving Bernoulli numbers For all $n\geq 1$ and $m\geq0$, I'm trying to prove that
$\sum_{k=0}^n\sum_{l=0}^m\binom{n}{k}\binom{m}{l}\frac{(n-k)!(m-l)!}{(n+m-k-l+1)!}(-1)^l B_{k+l}=0$
where $B_n$ are the Bernoulli numbers with $B_{1}=-\frac{1}{2}$.
I made a couple of attempts using the recursive... | With the work by @RenéGy and @epi163sqrt we are trying to prove the
statement
$$(-1)^n \sum_{g=0}^m \frac{B_{n+g+1}}{n+g+1} {m\choose g}
+ (-1)^m \sum_{g=0}^n \frac{B_{m+g+1}}{m+g+1} {n\choose g}
= - \frac{1}{n+m+1} {n+m\choose m}^{-1}$$
We prove this for $n\ge m$, it then follows by symmetry for $m\ge n.$
Using
$$B_n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4520057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 2
} |
Proving $(b + c) \cos \frac{B + C}{2} = a \cos \frac{B - C}{2}$ for any $\triangle ABC$ I found this problem on Mathematics, Textbook for Class XI by NCERT, ed. January 2021 that uses the sine and cosine formulas along with standard trigonometric identities for proving an identity.
By triangle ABC, the question assumes... | $$\sin(B)+\sin(C)=2\sin\left(\frac{B+C}{2}\right)\cos\left(\frac{B-C}{2}\right)$$
Multiply $\cos\left(\frac{B+C}{2}\right)$ on both sides, and use double angle formula for sine
$$(\sin(B)+\sin(C))\cos\left(\frac{B+C}{2}\right)=\sin(B+C)\cos\left(\frac{B-C}{2}\right)$$
$\sin(B+C)=\sin(\pi-A)=\sin(A)$
$$(\sin(B)+\sin(C))... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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why exactly doesn't logarithmic differentiation work here? As part of a larger problem, I have to differentiate $y^x$
so if we let $y=y^x$
we get $\ln(y)=x\ln(y)$
which means the differential is zero
alternatively, let $y=e^{\ln(x^y)}$
so $y=e^{x\ln(y)}$
which can be differentiated as follows
$y'=x^y\left(\ln(y)+y'\ \f... | Differentiating the function $f(x) = y^x$ (for some $y$ which is a function of $x$) is not the same thing as using implicit differentiation on $y=y^x$.
For example, if $y=e^x$, then $f(x)=y^x$ is the function $f(x)=(e^x)^x = e^{x^2}$; but $y=y^x$ gives $e^x=e^{x^2}$, which gives $e^{x^2-x}=1$, so the only values of $x$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Power series expansion of $f(x)=\frac{x}{2x^2+1}$ I was requested to find a series representation of $$f(x)=\frac{x}{2x^2+1}$$
and its convergence radius. I attempted a solution but it is wrong according to online calculators, and I can't find my mistake.
My approach was the standard procedure of rewriting $f(x)$ as $\... | Another way to do it
$$f(x)=\frac{x}{2x^2+1}\quad \implies \int f(x)\,dx=\frac{1}{4} \log \left(1+2 x^2\right)$$ Let $t=2x^2$
$$\log(1+t)=\sum_{n=1}^\infty (-1)^{n+1} \frac {t^n}n$$
$$\frac{1}{4} \log \left(1+2 x^2\right)=\sum_{n=1}^\infty (-1)^{n+1} \frac {2^{n-2}}n x^{2n}$$
Now, differentiate
$$f(x)=\frac{x}{2x^2+1}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation $x^6-2x^5+3x^4-3x^2+2x-1=0$ Solve the equation $$x^6-2x^5+3x^4-3x^2+2x-1=0$$ Let's divide both sides of the equation by $x^3\ne0$ (as $x=0$ is obviously not a solution, we can consider $x\ne0$). Then we have $$x^3-2x^2+3x-3\cdot\dfrac{1}{x}+2\cdot\dfrac{1}{x^2}-\dfrac{1}{x^3}=0\\\left(x^3-\dfrac{1}{x... |
$$\left(x^3-\dfrac{1}{x^3}\right)-2\left(x^2-\dfrac{1}{x^2}\right)+3\left(x-\dfrac{1}{x}\right)=0$$
There are two roots $\pm1$ associated with the factor of $x-1/x$. To find the other roots, divide by $x-1/x$ to get
$$
x^2+1+{1\over x^2}-2\left(x+{1\over x}\right)+3=0
$$
Noticing that
$(x+1/x)^2=x^2+2+1/x^2$ we have
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\ {g'(x)} & {x< k} \end{cases} $ , $\max (k)=?$ if $f$ is a differentiable function
If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\
{g'(x)} & {x< k} \end{cases} $. If $f(x)$ is a differentiable
function, what is the maximum value of $k... | Assume $a\neq 0$:
Statement 1:
$$ak^2+bk+c=2ak+b$$
$$ak^2+(b-2a)k+(c-b)=0$$
$$k=\frac{2a-b\pm\sqrt{b^2-4ab+4a^2+4ab-4ac}}{2a}$$
$$k=\frac{2a-b\pm\sqrt{b^2+4a^2-4ac}}{2a}$$
Statement 2:
$$2ak+b=2a$$
$$k=\frac{2a-b}{2a}$$
Statement 3, combining 1 and 2:
$$\frac{2a-b\pm\sqrt{b^2+4a^2-4ac}}{2a}=\frac{2a-b}{2a}$$
$$2a-b\pm\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Finding the value of $\mathop {\lim }\limits_{n \to \infty } f\left( n \right)$ of a series having the value "n" in both numerator and denominator If $f\left( n \right) = n + \frac{{16 + 5n - 3{n^2}}}{{4n + 3{n^2}}} + \frac{{32 + n - 3{n^2}}}{{8n + 3{n^2}}} + \frac{{48 - 3n - 3{n^2}}}{{12n + 3{n^2}}} + .. + \frac{{25 -... | $f(n) = \sum_{r =1}^n (1 + \frac{16r + (9-4r)n -3n^2}{4rn + 3n^2}) $
$ = \sum_{r =1}^n ( \frac{16r +9n}{4rn +3n^2} ) \\ $
$ \lim_{n\to\infty} f(n) = \lim_{n\to\infty} \frac{1}{n} ( \sum_{r =1}^n ( \frac{ 16(\frac{r}{n}) +9}{4(\frac{r}{n}) +3}) )$
$ = \int_{0}^1 \frac{16x +9}{4x+3} dx$
$ = 4 + \frac{3}{4} ln \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4524114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solve inequality $\sinh (2x) - 3 \sinh x \ > 0$
Solve inequality $\sinh (2x) - 3 \sinh x \ > 0$
Using hyperbolic trigo indenties:
$ 2 \sinh x \cosh x - 3 \sinh x \ > 0$
$\sinh x (2 \cosh x -3) \ > 0$
To solve this, we need to take
*
*$\sinh x (2 \cosh x -3) \ > 0$
*$\sinh x (2 \cosh x -3) \ < 0$
why is this the c... | $\sinh x = \frac {e^x - e^{-x}}{2}$
$\sinh 2x - 3\sinh x = \frac {e^{2x} - 3e^{x} + 3e^{-x} - e^{-2x}}{2}$
Let $u = e^x$
$\frac {u^2 - 3u + u^{-1} + u{-2}}{2}$
Multiply numerator and denominator by $u^2$
$\frac {u^4 - 3u^3 + 3u - 1}{2u^2}$
Factor the numerator
$\frac {(u-1)(u+1)(u - \frac {3 - \sqrt {5}}{2})(u - \frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4524597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How did people come up with the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$? Every resource that I've read proves the formula
$$ a^3 + b^3 = (a+b)(a^2-ab+b^2) \tag1$$
by just multiplying $(a+b)$ and $(a^2 - ab + b^2)$.
But how did people come up with that formula? Did they think like, "Oh, let's just multiply these polynomials... | In this answer I wanted to suggest a method where I get results by applying substitutions (based on the idea that it is the most basic algebraic technique).
Let, $a=m+n$ and $b=m-n$ then:
$$
\begin{aligned}a^3+b^3=m^3+3m^2n+3mn^2+n^3+m^3-3m^2n+3mn^2-n^3=2m^3+6mn^2=2m(m^2+3n^2)\end{aligned}
$$
and note that,
$$
\begin{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Can we show algebraically that this is true? This problem is part of a larger proof I've been working on. Long story short, I'm trying to prove a particular explicit formula for the recursive function
$$
t(p, 0) = 0 \\
t(p, k+1) = \left\lfloor \dfrac{t(p, k)}{3} \right\rfloor + p
$$
where $p=2n+1, n \in \mathbb{N}$. Th... |
Claim. Let $m,n$ be integers with $n>0$ and $x$ a real number. Then
$$\Bigl\lfloor \frac{-\lfloor x \rfloor+m}{n} \Bigr\rfloor = -\Bigl\lfloor \frac{x-m+n-1}{n} \Bigr\rfloor.$$
Proof.
We have
\begin{align}
\Bigl\lfloor \frac{-\lfloor x \rfloor+m}{n} \Bigr\rfloor
&=\Bigl\lceil \frac{-\lfloor x \rfloor+m-n+1}{n} \Bigr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Inequality $xy+yz+zx-xyz \leq \frac{9}{4}.$ Currently I try to tackle some olympiad questions:
Let $x, y, z \geq 0$ with $x+y+z=3$. Show that
$$
x y+y z+z x-x y z \leq \frac{9}{4}.
$$
and also find out when the equality holds.
I started by plugging in $z=3-x-y$ on the LHS and got
$$
3y-y^2+3x-x^2-4xy+x^2y+xy^2 = 3y-... | This is equivalent to $$(1-x)(1-y)(1-z)\leq \frac14.\tag 1$$
This is true if any $x,y,z$ is exactly one, since the the left side of $(1)$ is zero.
It is not possible for all $x,y,z$ to be greater than $1,$ nor for all $x,y,z$ to be less than $1,$ since $x+y+z=3.$
If only one of the values is $>1,$ then the left side of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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Integral of infinite G.P. : $ \int 1+ 2x+ 3x^2 + 4x^3 ...... \, dx $ In this $$\int 1+ 2x+ 3x^2 + 4x^3 ...... \, dx $$
textbook solution goes like let $$ S = 1+ 2x+ 3x^2 + 4x^3 ...... $$ (call this equation 1, then multiple both sides by x , we get $$ Sx= x + 2x^2 + 3x^3 + 4x^4 ...... $$ (call this equation 2), now su... | They are actually the same, up to the famous constant:
$$\frac{x}{1-x}=\frac{1}{1-x}-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\fr... | I found the solution sets in your cases:
Case-1) $((-\infty,4)\cup(5,\infty))\cap[8,\infty)=[8,\infty)$
Case-2) $((-\infty,1)\cup(4,\infty))\cap[-\infty,0)=(-\infty,0]$
Case-3) $((-\infty,\frac{10-2\sqrt{13}}{3})\cup(\frac{10+2\sqrt{13}}{3},\infty))\cap[0,8]=[0,\frac{10-2\sqrt{13}}{3})\cup(\frac{10+2\sqrt{13}}{3},8]$
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 11,
"answer_id": 2
} |
Is this proof of $\left\lfloor \frac{n}{m} \right\rfloor = \left\lceil \frac{n-m+1}{m} \right\rceil$ correct? I've been practicing proving things about floor and ceiling functions, so I thought I'd try to prove this well-known identity:
$$\left\lfloor \frac{n}{m} \right\rfloor = \left\lceil \frac{n-m+1}{m} \right\rceil... | Another succinct proof is
$$\left\lfloor\frac nm\right\rfloor = \left\lfloor\frac{n + \frac12}m\right\rfloor = \left\lceil\frac{n - m + \frac12}m\right\rceil = \left\lceil\frac{n - m + 1}m\right\rceil,$$
since $\frac{n + \frac12}m$ and $\frac{n - m + \frac12}m$ are always non-integers with a difference of $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4530070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
The locus of the intersection point of two perpendicular tangents to the curve $xy^2=1$. Find the locus of the intersection point of two perpendicular tangents to the curve $xy^2=1$.
I find out that the tangents to this curve with slope $m$ has this general form:
$y = mx+(-2m)^{\frac{1}{3}}-m \left( -\frac{1}{2m} \righ... | The curve is $x=1/t^2,y=t$ with dual curve (The dual curve lives in the dual plane of lines in the plane and is the set of tangents to the original curve. I use the parametric form $X=\frac{-y'}{xy'-yx'},Y=\frac{x'}{xy'-yx'}$) $X=-t^2/3,Y=-2/(3t)$ so two independent tangents are
$(-t^2/3)x -2y/(3t)+1=0 \\
(-s^2/3)x -2y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4532890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Representing $\frac{1}{x^2}$ in powers of $(x+2)$ I was asked to represent $f(x)=\frac{1}{x^2}$ in powers of $(x+2)$ using the fact that $\frac{1}{1 − x} = 1 + x + x^2 + x^3 + ...$. I am able to represent $\frac{1}{x^2}$ as a power series, but I am struggling withdoing it in powers of $(x+2)$. This is what I attempted.... | There is a better way.
$\frac 1 {x^{2}}=\frac 1 {((x+2)-2)^{2}}=\frac 1 4 \frac 1 {(1-\frac {x+2} 2)^{2} } =\frac 1 2\frac d {dx} \frac 1{1-\frac {x+2} 2}=\frac 1 2\frac d {dx} [\sum\limits_{n=0}^{\infty} (\frac {x+2} 2 )^{n}]$. Can you continue?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4534269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Find $f(7)$ given $f(x)f(y)=f(x+y)+f(x-y)$ and $f(1)=3$
Let $f:\Bbb R\to\Bbb R$ such that $f(1)=3$ and $f$ satisfies the functional equation
$$f(x)f(y) = f(x+y) + f(x-y)$$
Find the value of $f(7)$.
Attempt:
If $x=1$ and $y=0$, we find
$$f(1) f(0) = 2f(1) \implies f(0) = 2$$
If we fix $y=1$, we get the recurrence rela... | Well, is not too fancy, but you could do
$f(7)=f(4+3)=f(4)f(3)-f(1),$
and using the fact that
$f(4)=f(3)f(1)-f(2),$
you could compute $f(7)$ by knowing only $f(1),\;f(2),$ and $f(3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4536280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Roll two balanced dice until the sum of the faces equals 7 appears for the first time. Determine the expected value of tosses in this experiment. Roll two balanced dice until the sum of the faces equals 7 appears for the first time. After that, roll the same two dice until some face 3 appears for the first time. Determ... | Let us represent
*
*$X$: number of tosses until dice sum is 7, $X\sim$ Geometric$(p=1/6)$, $E(X)=1/p=6$.
*$Y$: number of tosses until some face is 3, $Y\sim$ Geometric$(q=11/36)$, $E(Y)=1/q=36/11$.
*$X+Y$: total number of tosses
It is true that $E(X+Y)=E(X)+E(Y)=6+36/11=102/11.$
Therefore the expected number of to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4536439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Why am I getting $\beta=90^{\circ}$ Consider the geometry below, where the small circle is touching both semi circles of radius $5$ and the side of the square. Find the radius of the small circle.
My try: $M$ and $N$ are centers of semicircles and $G$ is center of small circle indicated below.Let $r$ be the radius of ... | The strategy is fine, but you have made an algebraic mistake: $\frac{(5-2 r)+\sqrt{-2 r^2+25}}{\sqrt{2}(5-r)}$ is only $\cos \alpha + \sin \alpha$, not $\frac{\cos \alpha + \sin \alpha}{\sqrt 2}$. Solving $$\frac{(5-2 r)+\sqrt{-2 r^2+25}}{2(5-r)} = \frac{r}{5-r}$$ instead gives $r = \frac{20}{9}$.
A simpler approach is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4545101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Deriving exact value of $\sin \pi/12$ using double angle identity
Deriving the exact value of $\sin \pi/12$ using double angle identity
Double angle identity - $\sin 2A = 2\sin A \cos A$
So, $\sin (2 \frac{\pi}{12}) = 2 \sin \frac{\pi}{12} \cos \frac{\pi}{12} $
$\sin^2 A + \cos^2 A = 1 $ so, $\cos \frac{\pi}{12} = \... | Let $x := \sin^2\frac{\pi}{12}$. Then your last equality is
$$\frac{1}{4} = 4x(1-x) = 4x - 4x^2.$$
The solutions of this quadratic equation are $x_{1,2} = \frac{1}{2} \pm \frac{\sqrt{3}}{4}$. Clearly, $x$ is smaller than $\frac{1}{2}$, so we must have $x = \frac{1}{2} - \frac{\sqrt{3}}{4}$. Taking the square root gives... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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show that $f(x)=x$ where $\,\forall x,y\in\mathbb{R}^+, f(x+f(y+xy)) = (y+1)f(x+1)-1$
Let $f:\mathbb{R}^+\to\mathbb{R}^+$ be a function such that $\,\forall x,y\in\mathbb{R}^+, f(x+f(y+xy)) = (y+1)f(x+1)-1$. Prove that $f(x)=x$ for all real numbers $x>0$.
I think it might be useful to prove that f is injective and t... | Let $P(x,y)$ be the assertion, as usual.
$P(1,x/2)\implies f(1+f(x))=(1+x/2)f(2)-1$. Thus if $f(a)=f(b)$, $a=b$ i.e. $f$ is injective.
Now we put the fairly obvious $P(x,\frac{1}{f(x+1)})$ to get $f(x+f\left(\frac{x+1}{f(x+1)}\right))=f(x+1)$ which, by injectivity, implies $f\left(\frac{x+1}{f(x+1)}\right)=1$. Again by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4549625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Being given a cyclic summation on 3 letters equal to $1$, deduce the value of another cyclic summation If
$$
\dfrac{a}{b+c} + \dfrac{b}{a+c} + \dfrac{c}{a+b} = 1
$$
then
$$
\dfrac{a^2}{b+c} + \dfrac{b^2}{a+c} + \dfrac{c^2}{a+b} = \;?
$$
I tried to manipulated the equation above using some properties as
$(a+b+c)^2 = a^2... | By assumption we have
$$
\frac{a^3 + abc + b^3 + c^3}{(a + b)(a + c)(b + c)}=0.
$$
The second term equals
$$
\dfrac{a^2}{b+c} + \dfrac{b^2}{a+c} + \dfrac{c^2}{a+b} =\frac{(a^3 + abc + b^3 + c^3)(a+b+c)}{(a + b)(a + c)(b + c)}
$$
So it is equal to zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation $(x-1)^5+(x+3)^5=242(x+1)$ Solve the equation $$(x-1)^5+(x+3)^5=242(x+1)$$ My idea was to let $x+1=t$ and use the formula $$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4),$$ but I have troubles to implement it. The equation becomes $$(t-2)^5+(t+2)^5=242t\\(t-2+t+2)\left[(t-2)^4-(t-2)^3(t+2)+(t-2)^2(t+2)^2-\... | You could also solve it in a much simpler way, like this:
$$
\begin{align*}
(x - 1)^{5} + (x + 3)^{5} &= 242 \cdot (x + 1) \quad\mid\quad -(242 \cdot (x + 1))\\
(x - 1)^{5} + (x + 3)^{5} - 242 \cdot (x + 1) &= 0\\
2 \cdot x^{5} + 10 \cdot x^{4} + 100 \cdot x^{3} + 260 \cdot x^{2} + 168 \cdot x &= 0 \quad\mid\quad \div ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability that 2 out of 3 balls will be of the same colour if the first one drawn is red Three balls are drawn from a box containing 3 red, 4 black and 5 white balls. Find the probability that 2 out of 3 balls will be of the same colour (event A) if the first one drawn is red (event B).
I tried to solve this question... |
First, I calculated $P(A \cap B) = \frac{{3\choose1}{4\choose2}+{3\choose1}{5\choose2}+{3\choose2}{4\choose1}+{3\choose2}{5\choose1}} {12\choose3} = 15/44 $
This is the probability that exactly two among the three are the same colour and at least one among the three is red. That is not the required event.
Event $B$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is it possible for two non real complex numbers a and b that are squares of each other? ($a^2=b$ and $b^2=a$)? Is it possible for two non real complex numbers a and b that are squares of each other? ($a^2=b$ and $b^2=a$)?
My answer is not possible because for $a^2$ to be equal to $b$ means that the argument of $b$ is t... | It is entirely possible. Consider the complex conjugate cube roots of one. $\omega = -\frac 12 + \frac{\sqrt 3}{2}i$ and $\overline \omega = -\frac 12 - \frac{\sqrt 3}{2}i$
Note that $\omega = (\overline \omega)^2$, but also $\overline \omega = \omega^2$.
There is no contradiction here, because from the two equations, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4563725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Is this proof (scratch work) of $\lim_{x\rightarrow c}x^3=c^3$ correct? Is this scratch work for the proof of $\lim_{x\rightarrow c}x^3=c^3$ correct?
We have that $|x^3-c^3|=|x-c||x^2+xc+c^2|$.
If $|x-c|<|c|$ then $|x|=|x-c+c|\leq |x-c|+|c|<2|c|$, it follows that $|x^2+xc+c^2|\leq |x|^2+|x||c|+|c|^2< 4|c|^2+2|c|^2+|c|... | (After reading the comments.) With a little more care, you can include the case $c=0$:
If $|x-c|\le\delta\le1$ then $|x|=|x-c+c|\leq |x-c|+|c|\le1+|c|,$ so
$|x^3-c^3|\le\delta\left(3(1+|c|)^2\right),$
which is $\le\epsilon$ if $\delta$ is not only $\le1$ but also $\le\frac\epsilon{3(1+|c|)^2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4566047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why is $ab+\frac{1}{2} - \frac{a+b}{2} >0$ for $0I am trying to show that the function $$ab+\frac{1}{2}-\frac{a+b}{2}$$ is positive when $0<a,b<1.$
Here is what I have done.
*
*The arithmetic-geometric mean inequality doesn't seem like the way to go because it implies $$ab < \sqrt{ab} \le \frac{a+b}{2}.$$
This implie... | As a rule, when you are dealing with numbers in $[0,1]$, you can write them as $a = \frac{u}{u+v}$ and $b= \frac{p}{p+q}$, where $u$, $v$, $p$, $q$ are positive. Substituting we get
$$a b + \frac{1}{2} - \frac{a+b}{2} = \frac{p u + q v}{2 (p + q) (u + v)}$$
clearly positive. We can substitute back if we wish, $u+v= p+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\lim_{n \to \infty} \prod_{r=1}^n (n^2+r^2)^{\frac{1}{n}}/n^2$
Find $\lim_{n \to \infty} \prod_{r=1}^n \frac{(n^2+r^2)^{\frac{1}{n}}}{n^2}$
Let $L = \lim_{n \to \infty} \prod_{r=1}^n \frac{(n^2+r^2)^{\frac{1}{n}}}{n^2}$
Then $\log(L) = \lim_{n \to \infty} \log(\prod_{r=1}^n \frac{(n^2+r^2)^{\frac{1}{n}}}{n^2})$... | Notice that:
$$\prod\nolimits_{r = 1}^{n} \frac {(n^2+r^2)^{\frac{1}{n}}}{n^2}\le \prod\nolimits_{r = 1}^{n}\frac {(n^2+n^2)^{\frac{1}{n}}}{n^2}\le \frac {2n^2}{n^{2n}};$$
hence $L=0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The least value of $f(x)$=$\sqrt{x^2-2x+2}+\sqrt{x^2-4x+29}$ If $m$ is the least value of $f(x)$=$\sqrt{x^2-2x+2}+\sqrt{x^2-4x+29}$, occur at $x = α$ , then $[m]+[α]$ is equal to (where [.] denotes greatest integer function)
(A)6
(B)7
(C)5
(D)4
My approach is as follow
$f(x)=\sqrt{(x-1)^2+1}+\sqrt{(x-2)^2+5^2}$
$\alpha... |
In the figure, let $X(x,0)$ be any point on the $x$-axis.
Note that $AX=\sqrt{(x-1)^2+1}$ and $BX=\sqrt{(x-2)^2+5^2}$.
Therefore our job is to minimize $AX+BX$ where $X$ is an arbitrary point on the $x$-axis
Let $A'(1,-1)$ be the mirror image of $A(1,1)$ with respect to the $x$-axis.
From perpendicular bisector theore... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4570233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For $5b^2+4=n^2$, prove that $b=\frac{1}{5}(\lambda_2-\lambda_1)(\lambda_2^k-\lambda_1^k)$ where $\lambda_1, \lambda_2$ are roots of $x^2-3x+1=0$. If $b\in \mathbb{Z}_{>0}$ and $5b^2+4$ is a perfect square, then $b$ can be written in the form $b=\frac{1}{5}(\lambda_2-\lambda_1)(\lambda_2^k-\lambda_1^k)$, where $\lambda... | COMMENT.-Your equation $5b^2+4=k^2$ is equivalent to $$\left(\dfrac k2\right)^2-5\left(\dfrac b2\right)^2=1$$ and the Pell-Fermat equation $$x^2-5y^2=1$$ has the infinite set of integer solutions where $n$ is natural integer.
$$x=\frac{(9+4\sqrt5)^n+(9-4\sqrt5)^n}{2}\\y=\frac{(9+4\sqrt5)^n-(9-4\sqrt5)^n}{2\sqrt5}$$ so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4574787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does $\sum_{j=0}^n\sum_{k=0}^n\binom{j}{a}\binom{k}{b}\binom{n-j-k}{c}=\binom{n+1}{a+b+c+2}?$ I'm working on a problem,
(a) Let $a,b,n\geq1$ with $a+b\leq n$. By considering choosing $a+b+1$ numbers from the set $\{0,1,...,n\}$, and the possibilities for the number in position $a+1$ when the chosen numbers are listed ... | We can also derive a closed formula of the double sum by applying the identity
\begin{align*}
\sum_{k=0}^n\binom{k}{a}\binom{n-k}{b}=\binom{n+1}{a+b+1}\tag{1}
\end{align*}
twice.
We obtain
\begin{align*}
\color{blue}{\sum_{j=0}^n}&\color{blue}{\sum_{k=0}^{n-j}\binom{j}{a}\binom{k}{b}\binom{n-j-k}{c}}\tag{2}\\
&=\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4576083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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How can we guess one solution of the equation $y''+\frac{2}{x}y'+y=0$? I want to solve the equation $$\frac{1}{x^2} \, \frac{d}{dx}\left(x^2\frac{dy}{dx}\right)=-y.$$ I converted it to this equation $y''+\frac{2}{x}y'+y=0$. How can we guess one solution of the equation $y''+\frac{2}{x}y'+y=0$? Is there any way to solve... | One thing to try for non-constant coefficients second order linear ode's is the Liouville transformation. This involves no guessing and is one of the methods to try for such ode's if other methods fail.
Solve
\begin{align*}
y''+\frac{2 y'}{x}+y&=0\\ \tag{1}
A y''+ B y' + C y&=0
\end{align*}
Hence $A=1,B=\frac{1}{2},C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4576195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Show that $\ln(1/1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$ for $-1
Show that $\ln(1/1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$ for $-1<x<1$ using the power series $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$
I did this using term by term integration
$\int \frac{1}{1-x} dx = (-1) \ln (1-x) + C = \sum_{n=0}^{\infty} \frac{x^... | You are on the right track. For $ |x|<1$ let
$$f(x)= \sum_{n=1}^{\infty}\frac{x^n}{n}.$$
Then
$$f'(x)= \sum_{n=1}^{\infty}x^{n-1}= \sum_{n=0}^{\infty}x^{n}= \frac{1}{1-x}.$$
Hence
$$f(x)= - \ln (1-x)+c.$$
With $x=0$ we see that $c=0.$
Therefore
$$f(x)= - \ln (1-x)= \ln (1)-\ln (1-x)= \ln ( \frac{1}{1-x}).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Laurent Series of f On Given Annulus I am attempting to solve the following question and I am encountering some difficulties:
Expand the function:
$$
f(z) = \frac{z}{z^2 + 2z -3}
$$
in powers of z to find a series that is valid for an annulus containing z=2. For what values of z does this series converge?
My attempt at... | We want to expand a series valid for an annulus containing $z=2$. When looking at
\begin{align*}
\color{blue}{f(z)=\frac{3}{4}\left(\frac{1}{z+3}\right)-\frac{1}{4}\left(\frac{1}{1-z}\right)}\tag{1}
\end{align*}
*
*the left summand of (1) can be expanded as usual. We derive using the geometrics series expansion
\begi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\lim_{n \to \infty} \frac{1}{\sqrt{n^k}} (1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}})^k$ by Stolz–Cesàro $$\lim_{n \to \infty} \frac{1}{\sqrt{n^k}} (1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}})^k, k \in \mathbb{N}$$
Putting $\sqrt{n^k}$ in denominator, we ge... | There is a nice proof already provided by iskander using Stolz–Cesàro.
Also, using definite integrals it can be shown the limit is $2^k$
$\left( \lim_{n \to \infty} \frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\sqrt{i}} \right)^k=\left( \lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^{n}\sqrt{\frac{n}{i}} \right)^k=\left( \int_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof verification: $\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$ The question states:
Let $a, b, c$ be real numbers such that $$\frac{1}{(bc-a^2)}+\frac{1}{(ca-b^2)}+\frac{1}{(ab-c^2)}=0$$ Prove that $$\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$$
Now this can be solved with... | Your approach is not valid, and by coincidence, you don't arrive at an obvious contradiction because the final answer is zero too.
Let's examine a numerical example.
For $a=1, b=1, c=\frac{-1}{2}$, we get $\frac {1}{bc-a^2}+\frac {1}{ca-b^2}+\frac {1}{ab-c^2}=0$. If your approach was supposed to be correct, in a simila... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integration of Root What will be the integral
$$\int_0^1\sqrt{1-y^2}dy$$
I initially substituted $y =\sin t$ to solve (but couldn't since I'm bad at trigonometry).
| If you are not able to work around trig substitutions, you can use other elementary functions to substitute as well.
$$\sqrt{1 - y^2} \to y\sqrt{\frac{1}{y^2} - 1}$$
Let $u = y^2, du = 2y\cdot dy$
$u(0) = 0^2 = 0$
$u(1) = 1^2 = 1$
Hence, we have
$$y\sqrt{\frac{1}{y^2} - 1} \cdot dy \overset{u = y^2}{\to} \frac{1}{2}\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4582636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of: $\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{a+c}$
Let $a,b,c$ be roots of the cubic
$$x^3-x^2-2x+1=0$$
Then, find the value of:
$$\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{a+c}$$
My attempt.
I used the substitutions $$a+b=x^3, b+c=y^3, a+c=z^3$$
$$x^3+y^3+z^3=2(a+b+c)=2$$
Then I used the identity $$x^3+y^3+z^... | From the given cubic, we know that $a+b+c=1$, therefore
$$\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{a+c}=\sqrt[3]{1-c}+\sqrt[3]{1-a}+\sqrt[3]{1-b}=\sqrt[3]{A}+\sqrt[3]{B}+\sqrt[3]{C}$$
where $A,B,C$ are the roots of
$(1-x)^3-(1-x)^2-2(1-x)+1=0$, that is $x^3-2 x^2-x+1=0$.
It can be shown (see for example this link) that
$$x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How should I go about proving this identity? Found this on a math worksheet but wasn't able to prove it.
$$4(\cos^6x+\sin^6x)=1+3\cos^2(2x)$$
| Looking at the LHS of this equation, we can lower the exponent as follow:
$4(cos^6( x) +sin^6( x ) )= 4(cos^{2\cdot 3}(x) + sin^{2\cdot 3}(x) )= 4[(\dfrac{1+cos(2x)}{2})^3+(\dfrac{1-cos(2x)}{2})^3] = 4 \cdot \dfrac{2 + 2 cos^2(2x) + 4 cos^2(2x)}{2^3}= 4 \cdot \dfrac{2 + 6cos^2(2x)}{8} = \dfrac{2 + 6 cos^2(2x)}{2} = 1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Solve the radical equation for all reals: $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$ Question:
Solve the radical equation for all reals: $$x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$$
My approach:
$$1+\sqrt {1-x^2}=\frac {\sqrt {1+x^2}}{x}\\1+2\sqrt {1-x^2}+1-x^2=\frac{1+x^2}{x^2}\\4(1-x^2)=\left(\frac{1+x^2}{x^2}+x^... | Starting from: $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$ then
$$ x^2 \, (2 - x^2 + 2 \, \sqrt{1-x^2}) = 1 + x^2 \\
2 \, x^2 \, \sqrt{1-x^2} = 1 - x^2 + x^4 \\
4 \, x^4 \, (1-x^2) = 1 - 2 x^2 + 3 x^4 - 2 x^6 + x^8 \\
x^8 + 2 x^6 - x^4 - 2 x^2 + 1 = 0 \\
(x^4 + x^2 - 1)^2 = 0 \\
x^4 + x^2 - 1 = 0 \\
(x^2 + \alpha)(x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Basic question about floor function and limit ( $\lim \limits_{x \to 0} \lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor$)
$\lim \limits_{x \to 0} \lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor$
calculate the limit if it exists if not then prove it does not exist
I tried approaching by squeeze theorem and floor function... | As $x$ approaches $0$ from above (i.e. the right side limit), you have that
$\lfloor x-2\rfloor~$ stays at $~-2~$ and
$\lfloor x+3\rfloor~$ stays at $~3.~$
Therefore, the product stays at $~-6.~$
As $x$ approaches $0$ from below (i.e. the left side limit), you have that
$\lfloor x-2\rfloor~$ stays at $~-3~$ and
$\lf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I find the equation of an ellipse formed by a projection of a tilted rotating circle? Suppose I have a circle with radius $R$ on $xy$ plane with it's center at $(0,0,0)$. Then I rotate this circle about the y-axis by $0 \le \varphi \le {\pi \over 2}$, and then start rotating it about z-axis by $0 \le \alpha \le ... | The circle is initially (before any rotation) spanned by two vectors
$ u_1 = (R, 0, 0) $ and $u_2 = (0, R, 0) $
So that points on the circle are parametrically given by
$ P(t) = \cos t u_1 + \sin t u_2 $
Apply the rotation about the $y$ axis gives
$ Q(t) = \cos t R_y(\varphi) u_1 + \sin t R_y(\varphi) u_2 $
where
$ R_y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4597523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Representation of number as a sums and differences of natural numbers Lets consider all the combinations of:
$$1+2+3+4=10,\ \ 1+2+3-4=2,\ \ 1+2-3+4=4,\ \ 1+2-3-4=-4, $$
$$1-2+3+4=6,\ \ 1-2+3-4=-2,\ \ 1-2-3+4=0,\ \ 1-2-3-4=-8,$$
$$-1+2+3+4=8,\ \ -1+2+3-4=0,\ \ -1+2-3+4=2,\ \ -1+2-3-4=-6,$$
$$-1-2+3+4=4,\ \ -1-2+3-4=-4,\... | I believe you just need
$$ \prod_{k=1}^n (q^k - q^{-k}).$$
For example, $$(q-q^{-1})(q^2-q^{-2})(q^3-q^{-3})(q^4-q^{-4}) = q^{10} - q^8 - q^6 + 2 - q^{-6} - q^{-8} + q^{-10}$$
which corresponds to your final list of ordered pairs.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Algebraic calculation with polynomial and complex root.
Let $f=X^{3}-7 X+7$ be in $\mathbb{Q}[X]$.
Let $\alpha \in \mathbb{C}$ be a root of $f$ and hence $1, \alpha, \alpha^{2}$ be a basis of the $\mathbb{Q}$ vector space $\mathbb{Q}(\alpha)$. Let $\beta=3 \alpha^{2}+4 \alpha-14$. Write $\beta^{2}$ and $\beta^{3}$ as ... | $\beta$ cannot equal $\alpha$. If so, then $\alpha$ would be a root of the polynomial $3x^2+3x-14$. But your starting polynomial $x^3-7x+7$ is irreducible over $\mathbb{Q}[x]$ (use Eisenstein criterion for instance), so $x^3-7x+7$ would divide $3x^2+3x-14$, which is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Is this quadratic polynomial monotone at solutions of this cubic? Let $0<s < \frac{4}{27}$. The equation $x(1-x)^{2}=s$ admits exactly two solutions in $(0,1)$: Denote by $a,b$ be these solutions, and suppose that $a<b$.
Does
$$
(1-a)^2+2a^2<(1-b)^2+2b^2
$$
hold?
The limitation on the range of $s$ is due to $$\max_{a \... | The given inequality
$$(1-a)^2 + 2a^2 < (1-b)^2 + 2b^2$$
is equivalent to
$$0 < (b-a)\cdot(3(b+a) - 2)$$
i.e.
$$ b+a > \frac{2}{3}.$$
If $c$ is the third root of the equation $x(1-x)^2 = s$, then Vieta's formula says $a+b+c = 2$, so the inequality is equivalent to
$$ c < \frac{4}{3}.$$
But if $c \ge \frac{4}{3}$, then
... | {
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"url": "https://math.stackexchange.com/questions/4600892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $(x+1)^{p}x^{1-p}-(x+1)^{1-p}x^{p}$ is strictly increasing Let $1/2<p< 1$. I am asked to show that
$$f(x)=(x+1)^{p}x^{1-p}-(x+1)^{1-p}x^{p}$$
is strictly increasing for $x\geq 0$ and to compute $\lim_{x\to\infty} f(x)$.
I first computed the derivative, but I don't see why it must be positive:
$$\frac{d f(x)}{... | Rather a long comment than a solution:
My idea was to set $t=p-\frac 12\in(0,\frac 12)$ to introduce symmetry in the formulas, and then rewrite the expression as hyperbolic trig.
$\begin{align}f(x)
&=(x+1)^{\frac 12+t}x^{\frac 12-t}-x^{\frac 12+t}(x+1)^{\frac 12-t}\\\\
&=\sqrt{x}\sqrt{x+1}\,\Big(\tfrac{(x+1)^t}{x^t}-\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4604904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Is there another simpler method to evaluate the integral $\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x , \textrm{ where } \theta \in (0, \pi)?$ Using ‘rationalization’, we can split the integral into two manageable integrals as:
$\displaystyle \begin{aligned}\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x = & \i... | Utilize the Fourier series below\begin{equation}
\frac{\sin\theta}{1+\cos\theta\cos x}=1+2\sum_{k=1}^\infty \left(\frac{\sin\theta-1}{\cos\theta}\right)^k\cos kx
\end{equation}
and recognize that only the constant term survives upon integration, i.e.
$$\int_0^{2 \pi} \frac{\sin\theta}{1+\cos \theta \cos x} d x=2\pi
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Find the greatest integer in the expression below as a function of the given conditions Find the greatest integer less than $\sqrt{2^{100}+10^{10}}.$
Answer: $2^{50}$
I tried, but I can't finish:
$\sqrt{2^{100}+(2\cdot5)^{10}}=\sqrt{2^{10}(2^{90}+5^{10})}=2^5\sqrt{2^{90}+5^{10}}$
| A little GM-AM can also be helpful:
\begin{eqnarray*} \sqrt{2^{100}+10^{10}}
& = & 2^{50} \sqrt{ 1+\frac{10^{10}}{2^{100}}} \\
& \stackrel{GM-AM}{<} & 2^{50}\frac{1+1+\frac{10^{10}}{2^{100}}}2 \\
& = & 2^{50} + \frac{10^{10}}{2^{51}} \\
& < & 2^{50} + \frac{10^{10}}{2^{50}} \\
& = & 2^{50} + \left( \frac{10}{2^5}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How To Prove $a^3b＋b^3c＋c^3a>a^2b^2＋b^2c^2＋a^2c^2 $ if $ a > b > c > 0\,$? How to prove this :
$$a^3b＋b^3c＋c^3a>a^2b^2＋b^2c^2＋a^2c^2 $$
if we know: $$ a > b > c > 0 $$
My attempt:
$$\frac {a^3b＋b^3a}{2}>a^2b^2
...(1)$$
$$\frac {b^3c＋c^3b}{2}>c^2b^2...(2)$$
$$\frac {a^3c＋c^3a}{2}>a^2c^2...(3)$$
(1) + (2)+ (3) :
$$\... | Here is a no-brainer approach suggested by Calvin Lin, compared to Michaels Rozenberg's magical answer.
We will use a positive variable to stand for the difference between two given variables and then eliminate the larger one of the two given variables.
For example since $b>c$, we let $b=c+x$ where $x>0$. We will su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Indefinite integral $\int \frac{x^2}{\sqrt{x^2 + x^4}} \, dx $ involving $\mathrm{sgn}()$ function How to solve integrals like
$$\int \frac{x^2}{\sqrt{x^2 + x^4}} \, dx $$
If we factor out $x^2$ from denominator it become
$=\int \frac{x}{\left| x \right|} \frac{x}{\sqrt{1 + x^2}} \, dx $
$=\int \mathrm{sgn}(x) \frac... | Too long for a comment.
The integrand is well defined on $\mathbb{R}\backslash\{0\}$. This is the crucial point then. So how do you work with such cases?
You observe that indeed
$$\int \dfrac{x^2}{\sqrt{x^2+x^4}}\ \text{d}x = \int \dfrac{\vert x\vert }{x} \dfrac{x}{\sqrt{x^2+1}}\ \text{d}x = \int \dfrac{\vert x\vert }{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculating double integral using variables substitution $\displaystyle
D = \left\lbrace \left. \rule{0pt}{12pt} (x,y) \; \right| \; 3 x^2 + 6 y^2 \leq 1 \right\rbrace$
Calculate $\displaystyle
\iint_D \frac{ x^2 }{ ( 3 x^2 + 6 y^2 )^{ 3/2 } } \; dx dy{}$.
Attempt:
$x=\frac{r}{\sqrt3}cost,y=\frac{r}{\sqrt6}sin... | $$J = \begin{bmatrix} x_r & x_t \\ y_r & y_t \end{bmatrix} = \begin{bmatrix} \frac1{\sqrt3}\cos(t) & -\frac r{\sqrt3} \sin(t) \\ \frac1{\sqrt6}\sin(t) & \frac r{\sqrt 6}\cos(t) \end{bmatrix}$$
$$\implies \det(J) = x_r y_t - x_t y_r = \frac r{\sqrt{18}} \cos^2(t) + \frac r{\sqrt{18}} \sin^2(t) = \frac r{\sqrt{18}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $A$ is any matrix and $n$ is an integer, what is $A^n$? If $A$ is any matrix and $n$ is an integer, what is $A^n$? What I know is $A_n$ is a matrix of order $n$. It seems that the superscript symbol (integer) is not common. What does it imply?
| The superscript symbol typically means the product of $A$ with itself $n$-many times. So $$A^3 = A AA$$ for instance, in the same way that $$5^3 = 5 \cdot 5 \cdot 5$$ If negative and $A$ is invertible, then it is the inverse being raised to $|n|$; for instance, if $A$ is invertible, then $A^{-4}$ means $$A^{-4} = \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$ Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$.
My effort: I tried using the fact $9^{\sqrt{2}}<9^{1.5}=27.$
Also We have $512 <729 \Rightarrow 2^9<27^2 \Rightarrow 2^{\frac{9}{2}}<27 \Rightarrow \sqrt{2}^9=2^{4.5}<27$. But both are below $27$.
| What you want is a suitable rational upper bound to $\sqrt{2} \approx 1.4142$ such as $\frac{17}{12} \approx 1.4167$ which you can find for example with continued fractions. As a check, $\left(\frac{17}{12}\right)^2=\frac{289}{144}>2$.
Then you can say $$9^{\sqrt{2}} \lt 9^{17/12}=3^{17/6} = 129140163^{1/6} < 13421772... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 4
} |
Find $a>0$ if the floor of $(n^{2}-n)(\sqrt[n]{a}-1)$ is equal to $n-1$. Find $a>0$ knowing that for every n non-zero natural number, the floor of $(n^{2}-n)(\sqrt[n]{a}-1)$ is equal to $n-1$.
I know a is e, because taking the limit of the expresion we find that a is e. But i dont know how to prove that this works for ... | We have that $$ \lfloor (n^2 - n)(a^{1/n} - 1)\rfloor = n-1$$
Since $x-1< \lfloor x\rfloor \le x,$ we can conclude that for every $n$, $$ n -1 \le (n^2-n)(a^{1/n} - 1) \le n \\ \iff \left(1 + \frac{1}{n}\right)^n \le a \le \left(1 + \frac{1}{n-1}\right)^n.$$ Now taking the limit shows that if $a$ exists then $a = e$.
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.