Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find maximum of function $A=\sum _{cyc}\frac{1}{a^2+2}$ Let $a,b,c\in R^+$ such that $ab+bc+ca=1$. Find the maximum value of $$A=\frac{1}{a^2+2}+\frac{1}{b^2+2}+\frac{1}{c^2+2}$$
I will prove $A\le \dfrac{9}{7}$ and the equality occurs when $a=b=c=\dfrac{1}{\sqrt3 }$
$$\frac{\sum _{cyc}\left(b^2+2\right)\left(c^2+2\ri... | Your solution is right, but if you want to get all points for this problem, you need to explain, why it's enough to prove the starting inequality for equality case of two variables. It's not enough to say that it's true by uvw.
After words: $f$ decreases you can write the following.
Thus, it's enough to prove our ineq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3215787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to factorize the polynomial in the ring $\mathbb{Z}_5[x]$ I need to factor the polynomial $x^5+3x^4+x^3+x^2+3$ into the product of the irreducible ones in the ring $\mathbb{Z}_5[x]$.
The problem is I don't see any whole roots (I tried every possible divisor of 3) in the given polynomial. Does it mean that it is alr... | You need to make it more aggressively.
Check that $2$, $3$ and $4$ are roots, which gives the factorization:
$$x^5+3x^4+x^3+x^2+3\equiv x^5-2x^4+x^3+x^2-12=$$
$$=x^5-2x^4+x^3-2x^2+3x^2-6x+6x-12=(x-2)(x^4+x^2+3x+6)\equiv$$
$$\equiv(x-2)(x^4-9x^2-2x+6)=(x-2)(x^2(x-3)(x+3)-2(x-3))=$$
$$=(x-2)(x-3)(x^3+3x^2-2)\equiv(x-2)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Different methods of solving a linear system of first order DE?
$$ x'(t)=\begin{pmatrix} 1&\frac{-2}3\\3&4\end{pmatrix}x(t).$$
This is the method described in my book:
*
*finding the eigenvalues of matrix $A = \begin{pmatrix} 1&\frac{-2}3\\3&4\end{pmatrix}$:
$$ \begin{vmatrix} 1-\lambda&\frac{-2}3 \\3&4-\lambda\e... | The advantage of the exponential matrix method is that you do not have to solve for $c_1$ and $c_2$ to satisfy the initial conditions.
You just multiply your exponential matrix by the vector of initial condition and you get your solution.
$$ x(t)=e^{tA}x_0 = \dots = x_1\begin{pmatrix}2e^{2t}+\frac13e^{3t} \\-3e^{2t}-e^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve this quadratic congruent equation by inspection I found a systematic way (c.f. How to solve this quadratic congruence equation) to solve all congruent equations of the form of $ax^2+bx+c=0\pmod{p}$, or to determine that they have no solution.
But I wonder if there is some easy way to find solutions of simp... | The good old quadratic equation works just fine if $p\neq2$. If $a\not\equiv 0\pmod{p}$ and $b^2-4ac$ is a quadratic residue mod $p$, then the solutions to the quadratic congruence
$$ax^2+bx+c\equiv0\pmod{p},$$
are precisely
$$-\frac{b\pm\sqrt{b^2-4ac}}{2a}.$$
In this particular case we have $p=7$ and $a\equiv b\equiv1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3221002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$1/(51 +u^2) + 1/(51 +v^2) + 1/(51 +w^2)$ inequality I need to prove that $$1/(51 +u^2) + 1/(51 +v^2) + 1/(51 +w^2) \leq 1/20$$ given $u + v + w = 9$ and $u,v,w$ positive reals.
Using AM-HM inequality with $(51 +u^2, 51 +v^2, 51 +w^2)$ I arrive at
$(51 +u^2 + 51 +v^2 + 51 +w^2)/3 \geq 3/((1/(51 +u^2) +1/(51 +v^2) 1/(5... | You wrote the following:
"Using AM-HM inequality with $(51 +u^2, 51 +v^2, 51 +w^2)$ I arrive at
$(51 +u^2 + 51 +v^2 + 51 +w^2)/3 \geq 3/((1/(51 +u^2) +1/(51 +v^2) 1/(51 +w^2))$, $$or,$$
$1/(51 +u^2) +1/(51 +v^2)+1/(51 +w^2) \leq 9/(51 +u^2 + 51 +v^2 + 51 +w^2)$"
See please better, the last line should be:
$1/(51 +u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3222160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Different methods give different answers. Let A,B,C be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Let $A,B,C$ be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Find all possible values of $p$ such that $A,B$ and $C$ are angles of a triangle.
case 1- discriminant
We can re... | For $A,B,C$ to be the angles of a triangle, not only it shall be $A+B+C = \pi$,
but also $0 \le A,B,C$, or strictly greater than $0$ if you exclude the degenerate case.
So, the correct formulation of the problem is
$$
\left\{ \matrix{
A = \pi /4 \hfill \cr
\tan B\tan C = p \hfill \cr
A + B + C = \pi \hfill \cr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 3
} |
How to solve an equation with 6 degree polynomial? Does anyone have an idea to solve the following equation if it is possible? It's the best to get analytic solution, but if you can help me to show when equation has all real root with certain conditions of parameters, it is also alright.
Variable is $\Omega$, and para... | The following is a summary and expansion of some of my comments above, and by no means a complete answer to the question.
If $r<1$ then by Descartes' rule of signs $f$ has either $2$ or $0$ positive real roots, and either $4$, $2$ or $0$ negative real roots, and clearly $\Omega=0$ is not a root. It is not true in gen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $(\sum_{k=1}^{7} \tan^2(\frac{k\pi}{16})) - \tan^2(\frac{4\pi}{16})$ The question is:
Evaluate
$$\displaystyle \left(\sum_{k=1}^{7} \tan^2\left(\frac{k\pi}{16}\right)\right) - \left(\tan^2\frac{4\pi}{16}\right)$$
The given answer:$34$
What I've tried:
$$\displaystyle \left(\sum_{k=1}^{7} \tan^2\left(\f... | Using Expansion of $\tan(nx)$ in powers of $\tan(x)$,
$\tan\dfrac{k\pi}{2n},0\le k<2n,k\ne n$ are the roots of
$$\binom{2n}1t-\binom{2n}3t^3+\cdots+(-1)^{n-1}\binom{2n}{2n-1}t^{2n-1}=0$$
$$\implies \binom{2n}{2n-1}t^{2n-2}-\binom{2n}{2n-3}t^{2n-3}+\cdots+(-1)^{n-1}\binom{2n}1=0$$ will have the roots $\tan\dfrac{k\pi}n,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3226619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\frac{1}{AM \cdot BN} + \frac{1}{BN \cdot CP} + \frac{1}{CP \cdot AM} \le \frac{4}{3(R - OI)^2}$.
$(O, R)$ is the circumscribed circle of $\triangle ABC$. $I \in \triangle ABC$. $AI$, $BI$ and $CI$ intersects $AB$, $BC$ and $CA$ respectively at $M$, $N$ and $P$. Prove that $$\large \frac{1}{AM \cdot BN} +... | Firstly,
$$\frac{IM}{AM} + \frac{IN}{BN} + \frac{IP}{CP} = \frac{A_{CIB}}{A_{CAB}} + \frac{A_{AIC}}{A_{ABC}} + \frac{A_{BIA}}{A_{BCA}} = 1$$
where $A_m$ denotes the area of shape $m$.
$$\frac{AI}{AM} + \frac{BI}{BN} + \frac{CI}{CP} = 3 - \left(\frac{IM}{AM} + \frac{IN}{BN} + \frac{IP}{CP}\right) = 2$$
We have that $$\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3227215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Extracting integral solutions from a quartic equation The equation
\begin{equation*}
y^{4} + 4y^{3} + 10y^{2} + 12y - 27 = 0
\end{equation*}
has two integral roots. Without resorting to the quartic formula, how would one extract the roots from it?
| Here's one way we might proceed:
There is an old and simple result:
Let $R$ be a commutative unital ring and let
$p(x) = \displaystyle \sum_0^n p_i x^i \in R[x]; \tag 1$
then $1$ is a root of $p(x)$ if and only if the sum of the coefficients of $p(x)$ is $0$; that is,
$p(1) = 0 \Longleftrightarrow \displaystyle \sum_0^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
If $\space$ $\forall$ $x \in \Bbb R$, $\space$ $f(f(x))=x^2-x+1$. Find the value of $f(0)$. If $\space$ $\forall$ $x \in \Bbb R$, $\space$ $f(f(x))=x^2-x+1$. Find the value of $f(0)$.
I thought that making $f(x)=0$ implies that $f(0)= 0^2 - 0 + 1 = 1$, but i think that this isn't correct, because the $x$ in $f(f(x))$ i... | We first consider $f(1)$. Note that $f(f(1)) = 1$, so by substituting $x=f(1)$ we obtain $f(1) = f(f(f(1))) = f(1)^2 - f(1) + 1$. We therefore obtain the quadratic equation $f(1)^2 - 2f(1) + 1 = 0$ which implies $f(1) = 1$.
Also note that $f(f(0)) = 1$, so we have $f(1) = f(f(f(0))) = f(0)^2 - f(0) + 1$. So we obtain ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Fourier Series Representation for piecewise function I've been posed the following question:
$$
f(x)=
\begin{cases}
1-x^2, & 0 \leqslant|x|<1,\\
0, & 1\leqslant|x|<2\\
\end{cases}
$$
I'm trying to find the Fourier series representation but I am having trouble with $ a_n$ and $b_n$ coefficients.
For reference, I obta... | All the odd terms disappear.
Note that $f$ doesn't take value of $1-x^2$ from $1$ to $2$.
\begin{align}
\frac12 \int_{-2}^2 f(x) \cos \left( \frac{m\pi x}{2}\right) \, dx &= \int_0^1 (1-x^2) \cos \left( \frac{m\pi x}{2}\right) \, dx\\
&= \frac{2}{m \pi}\sin \left( \frac{m \pi x}{2} \right)(1-x^2)\big|_0^1 +\frac{4}{m ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$x_1,x_2,x_3$ the roots of $x^3+mx+n=0$ Find determinant of $A^{2}$ Let $m,n\in\mathbb{R}; x_1,x_2,x_3 $ the roots of $x^3+mx+n=0$ and the matrix $A=\begin{pmatrix}
1 & 1 &1 \\
x_1 &x_2 &x_3 \\
x^{2}_1 & x^{2}_2 & x^{2}_3
\end{pmatrix}$
I need to find determinant of $A^2$ which is $det(A)\cdot det(A)$
I got $det(A)... | An alternative method!
You found:
$$\det(A)=(x_2-x_1)(x_3-x_1)(x_3-x_2), \\
x_1+x_2+x_3=0, x_1x_2+x_1x_3+x_2x_3=m, x_1x_2x_3=-n$$
Note:
$$x_1+x_2+x_3=0 \Rightarrow x_1+x_2=-x_3 \Rightarrow x_1^2+x_2^2=x_3^2-2x_1x_2 \ \ (1)\\
$$
Also note the famous formula:
$$x^3+y^3+z^3=3xyz+(x+y+z)[(x+y+z)^2-3(xy+yz+zx)] \ (2) \ \ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Inverse of an upper bidiagonal Toeplitz matrix I have a matrix with the following structure
$$\left[\begin{array}{cccccc|c}
-1 & 1-b & 0 & \dots & 0 & 0 & b \\
0 & -1 & 1-b & \dots & 0 & 0 & b \\
\cdots \\
0 & 0 & 0 & \dots &-1 & 1-b & b \\
0 & 0 & 0 & \dots & 0 & -1 & 1 \\ \hline
... | You can notice that $A=-I + (1-b)N$ where $N$ is the standard nilpotent matrix, so
$$A^{-1} = - \left( I + (1-b)N + (1-b)^2 N^2 + \cdots + (1-b)^{n-1} N^{n-1} \right)$$
and the powers of $N$ are easily computed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
prove $\ln(1+x^2)\arctan x=-2\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{2n+1}x^{2n+1}$ I was able to prove the above identity using 1) Cauchy Product of Power series and 2) integration but the point of posting it here is to use it as a reference in our solutions.
other approaches would be appreciated.
| I think this is a much simpler proof.
\begin{align}
\tanh^{-1}x\ln(1-x^2)&=\frac12\{\ln(1+x)-\ln(1-x)\}\{\ln(1+x)+\ln(1-x)\}\tag1\\
&=\frac12\ln^2(1+x)-\frac12\ln^2(1-x)\tag2\\
&=\sum_{n=1}^\infty(-1)^n\frac{H_{n-1}}{n}x^n-\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^n\tag3\\
&=-2\sum_{n=1}^\infty\frac{H_{2n-2}}{2n-1}x^{2n-1}\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
I need help with this rate of change problem The solid shown in the figure below consists of a cylinder of the radius (r) and height (h)
and a hemispherical void of the radius (r). The dimensions at a given instant (t) are:
$$h(t)=3t^2+2;r(t)=8-\frac{t^2}{4}$$
Find the rate of change of the volume (V) and surface area ... | Volume-->
$$V=\pi r^2h-\frac{2}{3}\pi r^3$$
$$V=\pi (8-\frac{t^2}{4})^2(3t^2+2)-\frac{2}{3}\pi (8-\frac{t^2}{4})^3$$
$$\frac{dV}{dt}=\pi[(3t^2+2)*2(8-\frac{t^2}{4})*(\frac{-t}{2})+(8-\frac{t^2}{4})^2*6t]-[\frac{2}{3}\pi *3(8-\frac{t^2}{4})^2*(\frac{-t}{2})]$$
$$\frac{dV}{dt}=\pi[\frac{19}{16}t^5-\frac{103}{2}t^3+432t]$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Non-linear system of equations with three unknowns I have the following non-linear system of equations
\begin{cases}
\tau + e^{(m+\frac{s}{2})} &= a\,\quad(1)\\
\tau^2 + 2\tau\,e^{(m+\frac{s}{2})}+e^{(2m+2s)} &= b.\quad(2)\\
\tau^3+3\tau^2\, e^{(m+\frac{s}{2})}+3\tau\,e^{(2m+2s)}+e^{(3m+\frac{9s}{2})}&= c\quad (3)
\end... | \begin{cases}
t + xy &= a\,\quad(3)\\
t^2 + 2t xy+ x^2y^4 &= b\quad(4)\\
t^3 + 3t^2xy+ 3t\,x^2y^4+x^3y^9 &= c \quad(5)
\end{cases}
$xy=a-t\quad$ that we put into $(4)$ and $(5)$.
$\begin{cases}
t^2 + 2t (a-t)+ (a-t)^2y^2 &= b\quad(6)\\
\color{red}{t^3} + 3t^2(a-t)+ 3t\,(a-t)^2y^2+(a-t)^3y^6 &= c \quad(7)
\end{cases}$
f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Difference in my and wolfram's integration. Calculate
$$\int \frac{\sin ^3(x)+1}{\cos (x)+1} \, dx$$
Let $$u = \tan(x/2)$$
$\int \frac{\sin ^3(x)+1}{\cos (x)+1} \, dx = \int \frac{2\left(\frac{8u^3}{(u^2+1)^3}+1 \right)}{(u^2+1)\left( \frac{1-u^2}{u^2+1} + 1 \right)} \, du = \int \frac{8u^3}{(u^2+1)^3 } + 1 \, du$
$$ ... | Use $$\frac{\sin^3x+1}{\cos{x}+1}=\frac{8\sin^3\frac{x}{2}\cos^3\frac{x}{2}+1}{2\cos^2\frac{x}{2}}=4\sin^3\frac{x}{2}\cos\frac{x}{2}+\frac{1}{2\cos^2\frac{x}{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta -\sin^6 \theta)$ How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta -\sin^6 \theta)$
| Write $\cos2\theta=x$. We have
$$x=2\cos^2\theta-1=1-2\sin^2\theta$$
$$\frac{1+x}2=\cos^2\theta$$
$$\frac{1-x}2=\sin^2\theta$$
Thus the RHS becomes
$$4\left(\frac{(1+x)^3}8-\frac{(1-x)^3}8\right)=\frac12(1+3x+3x^2+x^3-(1-3x+3x^2-x^3))=x^3+3x=LHS$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$. This question is taken from book: Exercises and Problems in Calculus, by
John M. Erdman, available online, from chapter 1.1, question $4$.
Request help, as not clear if my approach is correct.
(4) Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$.
Ha... | $\mathrm{dist}(x^2,-2)=|x^2+2|=|x^2-11|=\mathrm{dist}(x^2,11)$. Therefore you want $x^2$ to be equidistant from $-2$ and $11$, i.e. $x^2=11-\frac{\mathrm{dist}(-2,11)}{2}=-2+\frac{\mathrm{dist}(-2,11)}{2}=\frac{9}{2}$. It follows that $x=\pm \frac{3}{\sqrt{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Find $9$'th derivative of $\frac{x^3 e^{2x^2}}{(1-x^2)^2}$ How can I find $9$'th derivative at $0$ of $\displaystyle \frac{x^3 e^{2x^2}}{(1-x^2)^2}$. Is there any tricky way to do that?
This exercise comes from discrete mathematic's exam, so I think that tools like taylor can't be used there.
| We know that
$$\frac{x}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n}$$
$x\rightarrow x^2$
$$\frac{x^2}{(1-x^2)^2}=\sum_{n=1}^{\infty}nx^{2n}$$
multiply by $x$
$$\frac{x^3}{(1-x^2)^2}=\sum_{n=1}^{\infty}nx^{2n+1}\tag1$$
$$e^x=\sum_{m=0}^{\infty}\frac{x^m}{m!}$$
$x\rightarrow 2x^2$
$$e^{2x^2}=\sum_{m=0}^{\infty}\frac{2^mx^{2m}}{m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Trigonometric series Cosecant^4 How to prove the following identity
$$
\text{cosec}^4{\frac{π}{n}}+\text{cosec}^4{\frac{2π}{n}}+\text{cosec}^4{\frac{3π}{n}} + \ldots+ \text{cosec}^4{\frac{(n-1)π}{n}} = \frac{(n^2+11)(n^2-1)}{45}
$$
I tried to simplify the sum but I finally get stuck with the same question I don't kno... | The best way is probably appealing to Chebyshev polynomials and Vieta's formula.
Note that $x=\sin(k\pi/n)$, $k=0,\dots,2n-1$, are the roots to $\cos(n\arccos(1-2x^2))=1$, i.e., $T_n(1-2x^2)=1$, where $T_n$ is the Chebyshev polynomial of the first kind.
Now using
$$
T_n(y)=n\sum_{k=0}^n(-2)^k\frac{(n+k-1)!}{(n-k)!(2k)!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{x\to\infty}1+2x^2+2x\sqrt{1+x^2}$ Consider the function
$$f(x)=1+2x^2+2x\sqrt{1+x^2}$$
I want to find the limit $f(x\rightarrow-\infty)$
We can start by saying that $\sqrt{1+x^2}$ tends to $|x|$ when $x\rightarrow-\infty$, and so we have that
$$\lim_{x\rightarrow-\infty}{(1+2x^2+2x|x|)}=\lim_{x\rightarrow-\... |
We can start by saying that $\sqrt{1+x^2}$ tends to $|x|$ when $x\rightarrow-\infty$,
Not exactly. Both $\sqrt{1+x^2}$ and $|x|$ tend to $\infty$ as $x \to -\infty$. I think what you mean that each is asymptotic to the other, because their ratio tends to $1$ as $x\to-\infty$. But that doesn't mean you can replace o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
A hard integral, can't seem to prove it How to show that,
$$\int_{-\infty}^{+\infty}\frac{(x-1)^2}{(2x-1)^2+\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}\right)^2}\mathrm dx=\pi$$
Let assume $a\ge1$
I have tried substitution but it leads to a more complicated integral.
| Noting that
\begin{eqnarray}
&&(2x-1)^2+\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}\right)^2\\
&=&\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}+(2x-1)i\right)\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}-(2x-1)i\right) \\
&=&\left(ax^2+(2i-1)x-\frac{(a-1)^2}{2(2a-1)}-i\right)\left(ax^2-(2i+1)x-\frac{(a-1)^2}{2(2a-1)}+i\right)
\end{eqnarray}
s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to calculate limit as $n$ tends to infinity of $\frac{(n+1)^{n^2+n+1}}{n! (n+2)^{n^2+1}}$? This question stems from and old revision of this question, in which an upper bound for $n!$ was asked for.
The original bound was incorrect. In fact, I want to show that the given expression divided by $n!$ goes to $0$ as $... | The easy part first: We have by basic analysis (where $x\in\Bbb R$)
\begin{equation}\tag 1\label 1\lim_{x\to\infty} \left(\frac{x+1}x\right)^x=\lim_{x\to\infty}(1+1/x)^x=e.\end{equation}
Now comes the harder part:
Note that
\begin{equation}\label 2\tag 2
\lim_{x\to\infty} e^x \left(\frac{x+1}{x+2}\right)^{(x^2)}
= \lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Asymptotic behavior of $a_{n+1}=\frac{a_n^2+1}{2}$ Define a sequence as follows:
$$a_0=0$$
$$a_{n+1}=\frac{a_n^2+1}{2}$$
I would like to know the asymptotic behavior of $a_n$. I already know (by roughly approximating $a_n$ with a differential equation) that
$$a_n\sim 1-\frac{2}{n}$$
as $n\to\infty$. However, my approxi... | Write $a_n = 1 - \epsilon_n$ and notice that $(\epsilon_n)$ solves
$$ \epsilon_{n+1} = \epsilon_n - \frac{\epsilon_n^2}{2}.$$
For the purpose of future use, we allow $\epsilon_0$ to take any value in $(0, 1]$. This type of sequence is well-studied, and here is a method of extracting asymptotic forms up to certain order... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Given $f(x)=ax^3-ax^2+bx+4$ Find the Value of $a+b$ Let $f(x)=ax^3-ax^2+bx+4$. If $f(x)$ divided by $x^2+1$ then the remainder is $0$. If $f(x)$ divided by $x-4$ then the remainder is $51$. What is the value of $a+b$?
From the problem I know that $f(4)=51$.
Using long division, I found that remainder of $\frac{ax^3-ax^... | One more way to see the conditions imposed on $f(x)$ are inconsistent/impossible:
Since $x^2 + 1$ divides $f(x)$ with remainder $0$, $f(x)$ factors as
$ax^3 - ax^2 + bx + 4 = (x^2 + 1)(cx + d) = cx^3 + dx^2 + cx + d; \tag 1$
comparing coefficients:
$a = c = -d, \tag 2$
$b = c, \tag 3$
$ d = 4; \tag 4$
thus,
$a = b = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Evaluating $\sum_{k=1}^{n} \frac{1}{k(k+1)}$ I have just started learning sums. I need to evaluate the following sum:
$$S_{n} = \sum_{k=1}^{n} \frac{1}{k(k+1)}$$
$$a_{1} = \frac{1}{2}, a_{2} = \frac{1}{6}, a_{3} = \frac{1}{12}, a_{4} = \frac{1}{20}, a_{5} = \frac{1}{30}, a_{6} = \frac{1}{42}, ...$$
I tried splitting it... | $$S_{n} = \sum_{k=1}^{n} \frac{1}{k}-\frac{1}{k+1}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1}$$
Check this: https://en.wikipedia.org/wiki/Telescoping_series
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Five different positive non-integer rational numbers such that increased by 1 the product of any two numbers is a square of some rational number Find at least one example of five rational numbers $x_1, \; x_2, ..., \; x_5$ such that
i) $x_k > 0$ for all $k=1,2,...,5$;
ii) $x_k$ is not an integer for all $k=1,2,...,5$;
... | See OEIS A192629 for
$$1,3,8,120,\frac {777480}{8288641}$$
and OEIS A030063 (which adds $0$ and deletes the fraction) for more information.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $\int_0^1 \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx$ I came across this integral while i was working on a tough series.
a friend was able to evaluate it giving:
$$\int_0^1 \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx=\frac{\pi^3}{16}\ln2-\frac{7\pi}{64}\zeta(3)-\frac{\pi^4}{96}+\frac1{768}\psi^{(3)}\left(\frac14\right)$$ usi... | solution by Kartick Betal.
\begin{align}
I&=\int_0^1 \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\underbrace{\int_1^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx}_{\displaystyle x\mapsto 1/x}\\
&=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\int_0^1 \frac{x\ln^2x\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3260176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
maximum and minimum value of $\frac{x^2+y^2}{x^2+xy+4y^2}$
If $x,y\in\mathbb{R}$ and $x^2+y^2>0.$ Then maximum and minimum value of $\displaystyle \frac{x^2+y^2}{x^2+xy+4y^2}$
Plan
Let $$K=\frac{x^2+y^2}{x^2+xy+4y^2}$$
$$Kx^2+Kxy+4Ky^2=x^2+y^2\Rightarrow (4K-1)y^2+Kxy+(K-1)x^2=0$$
put $y/x=t$ and equation is $(4K-1)t... | Let $v = (x,y)$,
$$
A = \begin{bmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 4 \end{bmatrix}
$$
and define the Rayleigh cuocient.
$$
R_A(v) = \frac{v^T A v}{v^Tv}.
$$
Then, it is clear that
$$
\frac{x^2+y^2}{x^2+xy+4y^2} = \frac{1}{R_A(v)}.
$$
It is well-known (Linear Algebra result) that $\operatorname{min}_{v \neq 0} R_A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3260474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Calculate local maximum and minimum of $f(x)=(2+\sin x)(5-\sin x)$ I am working on my scholarship exam practice which assumes high school or pre-university math knowledge. Could you please have a look on my approach?
The minimum of the function $f(x)=(2+\sin x)(5-\sin x)$ is ......
First, I began with some basic appr... | You cannot divide by $\cos x$ since it may be equal to zero.
It is true that
$$f(x) = (2 + \sin x)(5 - \sin x) = 10 + 3\sin x - \sin^2x$$
Therefore,
$$f'(x) = 3\cos x - 2\sin x\cos x$$
Setting the derivative equal to zero yields
\begin{align*}
f'(x) & = 0\\
3\cos x - 2\sin x\cos x & = 0\\
\cos x(3 - 2\sin x) & = 0
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3265097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If $\tan x=3$, then what is the value of ${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}$?
If $\tan x=3$, then what is the value of
$${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}$$
So what I did is change all the $\sin{2x}$ and $cos{2x}$ with double angle formulas, getting
$${3\cos^2{x}-3\sin^2{x}-4\sin{x}\cos{x}\o... | $$\tan2x=\dfrac{2\tan x}{1-\tan^2x}=-\dfrac34$$ then
$${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}=\dfrac{3-2\tan 2x}{4\tan2x+5}=\dfrac{9}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$ I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.
Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:
$\sqrt{3x+7}-\sqrt{x+2}=1$
$(3x+... | There are actually two solutions: $x = -1; x = -2$ when you continue with the ``Can you finish?'' step of Dr. S.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 5
} |
Simplifying $\sqrt\frac{\left(a^2\cos^2t+b^2\sin^2t\right)^3}{\left(b^2\cos^2t+a^2\sin^2t\right)^3}$ I am looking to simplify these term [ I forgot the 3 :( ]
$$\sqrt\frac{\left(a^2\cos^2t+b^2\sin^2t\right)^3}{\left(b^2\cos^2t+a^2\sin^2t\right)^3}$$
where $a$ and $b$ are two non-negative reals.
(This is not homewor... | Divide both the numerator and denominator by $a^2+b^2$, then use $\frac{a^2}{a^2+b^2}=\sin^2 u$. Your expression will become $$\sqrt{\frac{\sin^2 u\cos^2 t+\cos^2u\sin^2t}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}$$
Now use $\sin^2u+\cos^2 u=1$
$$\sqrt{\frac{\sin^2 u\cos^2 t+\cos^2u\sin^2t}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}=\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3277124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Problems with functions. I have been working on the following problem relating to functions:
"Consider a line $y=kx+b$ ($k<0$, $b>0$) which is tangent to the parabola $y=x^2−4x+4$.
Find the maximal value of the region bounded by the lines $y=kx+b$, $y=0$ and $x=0$".
The greatest region I found that followed such rest... | Let the parabola be $f(x) = x^2 -4x + 4$.
Let's let the point of the tangent point of the parabola be $(x, y)$
The slope of the tangent line is $k = f'(x) = 2x - 4$.
Given a right triangle made be the lines $y=kx + b$ and $y = 0$ and $x=0$ the base of the triangle will be from $(0,0)$ to the $x$-intercept of the line $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}dx$
How to prove $$\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}\ dx=\text{Im}\left(\operatorname{Li}_3(1+i)\right)-\frac{\pi^3}{32}-G\ln2 \ ?$$
where $\operatorname{Li}_3(x)=\sum\limits_{n=1}^\infty\frac{x^n}{n^3}$ is the trilogarithm and $G=\sum_{n=0}^\infty\frac{(-1)^n}{(2... | lets start with $\displaystyle\int_0^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx=2\Im\operatorname{Li}_3(1+i)\quad$ (proved here)
\begin{align}
\int_0^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx&=\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ d+\underbrace{\int_1^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx}_{\small\displaystyle x\mapsto1/x}\\
2\Im\operatorna... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}.$
Prove that $$S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}=\frac{e}{2}-1.$$
$$
S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}=\sum_{n=2}^\infty\frac{n!}{2(n-2)!(n+1)!}=\sum_{n=2}^\infty\frac{1}{2(n+1)(n-2)!}\\
=\frac{1}{2}\bigg[\frac{1}{3.0!}+\frac{1}{4.1!}+\frac{1}{5.2!}+\frac{... | $$S=\sum_{n=2}^\infty\frac{1}{2(n+1)(n-2)!}=\sum_{n=0}^\infty\frac{1}{2(n+3)(n)!}$$
$$=\frac{1}{2}\sum_{n=0}^\infty\int_{0}^{1}\frac{x^{n+2}}{n!}dx=\frac{1}{2}\int_{0}^{1}{x^2e^{x}}dx=\frac{e}{2}-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Calculate $\int_0^1\arctan x\left(\frac{3\ln(1+x^2)}{1+x}-\frac{2\ln(1+x)}{x}\right)\ dx$ Prove that
$$\int_0^1\arctan x\left(\frac{3\ln(1+x^2)}{1+x}-\frac{2\ln(1+x)}{x}\right)\ dx=\frac{3\pi}{8}\ln^22-\frac{\pi^3}{32}$$
I managed to prove the above equality using integral manipulation (solution be posted soon) , bu... | @Kemono Chen elegantly proved here
$$\int_0^y\frac{\ln(1+yx)}{1+x^2}dx=\frac12 \tan^{-1}(y)\ln(1+y^2)$$
By integration by parts we have
$$\int_0^y\frac{y\tan^{-1}(x)}{1+yx}dx=\frac12 \tan^{-1}(y)\ln(1+y^2)$$
divide both sides by $1+y$ and integrate from $y=0$ to $y=1$, we get,
\begin{align}
\frac12I&=\frac12\int_0^1\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding zero divisors in a polynomial quotient ring
Is $x^2+x+I$ a zero divisor in $\mathbb{Z}_7[x]/I$, where $I=(x^3+5x^2+2x+5)$?
I know that $\gcd(x^3+5x^2+2x+5,x^2+x)=x+1$, and that it means that $x^2+x$ is indeed a zero divisor.
What I struggle with is finding $g(x)$ such that $$(f(x)+\langle x^3+5x^2+2x+5\rangl... | Hint:
Note that $6$ is a root of given polynomial. That is, in $\Bbb Z_7$, $$x^3+5x^2+2x+5=(x-6)(x^2+4x+5)=(x+1)(x^2+4x+5)$$ So take $f(x)=x+1$ and $g(x)=x^2+4x+5$ and let $I=\langle x^3+5x^2+2x+5 \rangle$. Then $f+I \neq I$ and $g+I \neq I$ but $$(f+I)(g+I)=fg+I=I$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $(abc+xyz) \left(\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}\right)\geq3$ Let $a,b,c,x,y,z\in\mathbb{R}_+$ such that $a+x=b+y=c+z=1.$ Prove the inequality
$$(abc+xyz)\left(\frac1{ay}+\frac1{bz}+\frac1{cx}\right)\geq3$$
I tried using AM-HM to get
$$(abc+xyz)\left(\frac1{ay}+\frac1{bz}+\frac1{cx}\right)\geq9\frac{... | By AM-GM
$$(abc+xyz)\sum_{cyc}\frac{1}{ay}=\sum_{cyc}\left(\frac{zx}{a}+\frac{bc}{y}\right)=\sum_{cyc}\left(\frac{xy}{b}+\frac{ab}{x}\right)=$$
$$=\sum_{cyc}\left(\frac{x(1-b)}{b}+\frac{(1-x)b}{x}\right)=\sum_{cyc}\left(\frac{x}{b}+\frac{b}{x}-x-b\right)=$$
$$=\sum_{cyc}\left(\frac{x}{b}+\frac{b}{x}\right)-3\geq\sum_{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $N \in \mathbb{N}$ so for $n \geq N$: $ \frac{n}{n-1} \frac{\log(n)}{\sqrt{2^n}} \leq \frac{4}{25}\frac{\log(n)^2}{n^{1/5}}. $ Show that there exists some $N \in \mathbb{N}$ so that for all $n \geq N$:
$$
\frac{n}{n-1} \frac{\log(n)}{\sqrt{2^n}} \leq \frac{4}{25}\frac{\log(n)^2}{n^{1/5}}.
$$
Numerically I know tha... | We have
$$\frac{n}{n-1} \frac{\log(n)}{\sqrt{2^n}} \frac{25}4\frac{n^{1/5}}{\log(n)^2}
\sim\frac{25}4\frac{n^{1/5}}{\log(n)}2^{-n/2}\xrightarrow{n\to\infty}0$$
hence, by definition of limit, there exists $N\in\Bbb N$ such that
$$\frac{n}{n-1} \frac{\log(n)}{\sqrt{2^n}} \frac{25}4\frac{n^{1/5}}{\log(n)^2}\leq 1$$
for ev... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
On the determinant of a Toeplitz-Hessenberg matrix I am having trouble proving that
$$\det
\begin{pmatrix}
\dfrac{1}{1!} & 1 & 0 & 0 & \cdots & 0 \\
\dfrac{1}{2!} & \dfrac{1}{1!} & 1 & 0 & \cdots & 0 \\
\dfrac{1}{3!} & \dfrac{1}{2!} & \dfrac{1}{1!} & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \ddots & \v... | Hint. In general, let $d_0=d_1=1$ and let $(a_k)_{k=1,2,\ldots}$ be any sequence of numbers. For every $n\ge2$, denote by $d_n$ the determinant of the $n\times n$ Toeplitz-Hessenberg matrix
$$
\begin{pmatrix}
a_1 &1 &0 &0 &\cdots &0\\
a_2 &a_1 &1 &0 &\cdots &0\\
a_3 &a_2 &a_1 &1 &\cdots &0\\
\vdots &\vdots &\vdots &\dd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3288246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Depression of a car
A person on the summit of mountain observes that the angles of depression of a car moving on a straight road at three consecutive mile stones are x, y and z respectively. Prove that the height of the mountain is $\sqrt{\frac{2}{\cot^2x-2\cot^2y+\cot^2z}}$
I am getting the height as $\frac{1}{\cot ... |
Let the distance between the foot of the mountain and the first mile stone be $a$ . Then the distances between the foot of the mountain and the consecutive milestones are $a+1$ and $a+2$.
Now $\cot x = \frac{a}{h} \ , \cot y = \frac{a+1}{h} \ , \cot z = \frac{a+2}{h}$.
$$\frac{2}{\cot^2x - 2\cot^2y + \cot^2z} = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3289036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The limit of $(1+x^2+y^2)^{1\over x^2 + y^2 +xy^2}$ as $(x,y)$ approaches $(0.0)$ Im having trouble solving the following limit problem:
Whats the limit of $(1+x^2+y^2)^{1\over x^2 + y^2 +xy^2}$ as $(x,y)$ approaches $(0.0)$.
I know that the first step is to change $(x,y)$ to polar coordinates. However after I've done... | Let $\epsilon>0$. Then for $|x|<\epsilon$:
$$
(1+x^2+y^2)^{\frac{1}{x^2+y^2+\epsilon y^2+\epsilon x^2}} \leq
(1+x^2+y^2)^{\frac{1}{x^2+y^2+x y^2}} \leq
(1+x^2+y^2)^{\frac{1}{x^2+y^2-\epsilon y^2-\epsilon y^x}}.
$$
Since
$$
(1+x^2+y^2)^{\frac{1}{x^2+y^2-\epsilon y^2-\epsilon y^x}}=
\Big((1+x^2+y^2)^{\frac{1}{x^2+y^2}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
minimum value of of $(5+x)(5+y)$
If $x^2+xy+y^2=3$ and $x,y\in \mathbb{R}.$
Then find the minimum value of $(5+x)(5+y)$
What I try
$$(5+x)(5+y)=25+5(x+y)+xy$$
$x^2+xy+y^2=3\Rightarrow (x+y)^2-3=xy$
I am finding $f(x,y)=22+5(x+y)+(x+y)^2$
How do I solve it? Help me please
| Using Rotation of axes
if $t$ is the rotation of axes, $\cot2t=0\iff\cos2t=0$
Set $2t=\dfrac\pi2$
$\sqrt2u=x+y,\sqrt2v=y-x$
$$3=\dfrac{(u-v)^2+(u+v)^2}2=u^2+v^2$$
WLOG $u=\sqrt3\cos t, v=\sqrt3\sin t$
$$(5+x)(5+y)=25+\sqrt2u+\dfrac{2(u^2-v^2)}4=\dfrac{50+2\sqrt2u+u^2-v^2}2$$
$2\sqrt2u+u^2-v^2=2\sqrt2(\sqrt3\cos t)+3(\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
How to isolate y in a potential function The following is a potential function:
$Φ(x,y)=x^4y^4+x^2e^xy^2-3lnx-\frac{3}{4}e^2=0$
I would like to isolate $y$ to give $y(x)=…$.
An online calculator gave $y=\pm \frac{\sqrt{-\sqrt{x^4(e^2x+12ln(x)+3e^2}+e^x}x^2}{\sqrt{2}}$, yet I would like to learn how to go about this man... | As suggested in the comments, the function $\Phi(x,y)$ is quadratic in $y^2$ as
$$\Phi(x,y)=x^4(y^2)^2+x^2e^xy^2-3\ln x-\tfrac{3}{4}e^2=0,$$
and so if $x\neq0$ the quadratic formula gives
\begin{eqnarray*}
y^2
&=&\frac{-x^2e^x\pm\sqrt{(x^2e^x)^2-4x^4(-3\ln x-\tfrac34e^2)}}{2x^4}\\
&=&\frac{-x^2e^x\pm\sqrt{x^4(e^{2x}+12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
2019 AIME I Problems/Problem 5 Regarding the AIME Problem:
Problem Link
Tried to solve backward - going from 4,4 and moving towards the origin and recording the number of ways to reach different points- then find the number of ways to each point of the coordinate axis and calculate the probability of hitting the origin... | If you had used probabilities rather than counts, it would have worked.
For example, the probability of reaching position $(3,4)$ is ${\large{\frac{1}{3}}}$, and the same for position $(4,3)$.
But the probability of reaching position $(3,3)$ is
$${\small{\frac{1}{3}}}+{\small{\frac{2}{9}}}={\small{\frac{5}{9}}}$$
wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How do you go from $\sum_{n \geq 3} \frac{1}{3} \big(\frac{1}{2}\big)^{n-2}s^n$ to $\frac{1}{6} s^3 \sum_{n \geq 0}\big(\frac{s}{2}\big)^{n}$? I'm summing:
$$\sum_{n \geq 3} \frac{1}{3} \bigg(\frac{1}{2}\bigg)^{n-2}s^n$$
Which is just an infinite series, so if I can get it in the form of $\sum_{n\geq 0}$, I can use $s_... | Hint : $$ \bigg(\frac{1}{2}\bigg)^{n-2}s^n = \bigg(\frac{1}{2}\bigg)^{1}\bigg(\frac{1}{2}\bigg)^{n-3} s^3 s^{n-3}$$ then apply the change of variables in the summand.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_{0}^{\infty} \frac{\sin x-x\cos x}{x^2+\sin^2x } dx$ The integral $$\int_{0}^{\infty} \frac{\sin x-x\cos x}{x^2+\sin^2x } dx$$ admits
a nice closed form. The question is: How to evaluate it by hand.
| Note that
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\log(x-i\sin(x))
&=\frac{1-i\cos(x)}{x-i\sin(x)}\\
&=\frac{x+\sin(x)\cos(x)}{x^2+\sin^2(x)}+i\frac{\sin(x)-x\cos(x)}{x^2+\sin^2(x)}
\end{align}
$$
Taking the imaginary part of both sides
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\frac1{2i}\log\left(\frac{x-i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Proving $\frac{\sqrt{k(k+1)} }{k-1} \leq1 + \frac{2}{k} + \frac{3}{4k^2}$ for $k \geq 3$. Could you please give me a hint on how to prove the inequality below for $k \geq 3$?
$$\frac{\sqrt{k(k+1)} }{k-1} \leq1 + \frac{2}{k} + \frac{3}{4k^2} $$
Thank you in advance.
| $$\dfrac{\sqrt{k(k+1)} }{k-1} \leq1 + \dfrac{2}{k} + \dfrac{3}{4k^2}=\dfrac{(2k+1)(2k+3)}{4k^2}$$
$$\sqrt{k(k+1)}\le\dfrac{(2k+1)(2k+3)(k-1)}{4k^2}=\frac{2k+1}{2}\times\dfrac{(2k+3)(k-1)}{2k^2}$$ We know that $\sqrt{k(k+1)}\le\dfrac{k+(k+1)}{2}$ (AM–GM inequality) and $\dfrac{(2k+3)(k-1)}{2k^2}=\dfrac{2k^2+k-3}{2k^2}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
1025th term of the sequence $ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $ Consider the following sequence - $$ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $$
In this sequence, what will be the $ 1025^{th}\, term $
So, when we write down the sequence and then write the value of $ n $ (Here, $n$ stands for the number of the below term) ... | $$ 2^0 + 2^1 + 2^2 + 2^3 + ... + 2^n < 1025 = 2^{10}+ 2^0 $$
$$ 2^1 + 2^2 + 2^3 + ... + 2^n < 2^{10} $$
$$ \frac{1}{2^9} + \frac{1}{2^8} + ... + \frac{1}{2^{10-n}} < 1 $$
In the left-hand side of the above inequality, we've acquired a geometrical sequence. The summation formula for geometrical series is;
$$ S = \frac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 5
} |
Finding the number of solutions to $\cos^4(2x)+2\sin^2(2x)=17(1+\sin 2x)^4$ for $x\in(0,2\pi)$
Number of solution of the equation
$\cos^4(2x)+2\sin^2(2x)=17(1+\sin 2x)^4\; \forall $ $x\in(0,2\pi)$
what i try
$\cos^4(2x)+2\sin^2 2x=17(1+\sin^2(2x)+2\sin 2x)^2$
$1+\sin^4 (2x)=17(1+\sin^4 2x+2\sin^2 2x+4\sin^24x+4\sin 2... | Hint: Your equation is equivalent to
$$2 (\sin (2 x)+2) (2 \sin (2 x)+1) (-7 \sin (2 x)+2 \cos
(4 x)-6)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3300465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Distance between two circles in a sphere I have a sphere with radius $R$ and $O$ is the origin.
Inside sphere there are 3 circles. The small circle in black colour is fixed with it's position defined by and $\alpha$ angle and among other two circles one is great circle and another small circle can rotate by maintaining... | The black circle is the intersection of a plane $P\cdot (0,0,1)=a$ (I don't think you've specified $a$) and the sphere $P\cdot P =1$.
The blue great circle is the intersection of $P\cdot (\cos \beta, 0, \sin \beta) = 0$ with the sphere, and the blue small circle is the intersection of $P\cdot (\cos \beta, 0, \sin \beta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to show that $\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
$\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
How to change $\sqrt{2+\sqrt{3}}$ into $\dfrac{\sqrt{6}+\sqrt{2}}{2}$
| You can use the identity
\begin{equation}
\sqrt{a+\sqrt{b}} = \sqrt{\frac{a+\sqrt{a^2-b}}{2}} + \sqrt{\frac{a-\sqrt{a^2-b}}{2}}.
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
$a, b$ are the integer part and decimal fraction of $\sqrt7$ find integer part of $\frac{a}{b}$
$a, b$ are the integer part and decimal fraction of $\sqrt7$
find integer part of $\frac{a}{b}$
using calculator :
$\sqrt 7$ = 2.645
$\frac{a}{b} = \frac{2}{0.6} = \frac{20}{6} = 3.333$
integer part of $\frac{a}{b} = 3$
ho... | $2.6^2=6.76<7<7.29=2.7^2.$
So $a=2$ and $0.6<b<0.7.$
So $4>10/3=2/0.6>a/b>2/0.7=20/7>2.$
So the integer part of $a/b$ is $2$ or $3.$
Now $$3\le a/b\iff 3b\le a\iff 3b\le 2 \iff b\le 2/3 \iff$$ $$\iff \sqrt 7=a+b=2+b\le 2+2/3=8/3\iff$$ $$\iff 7\le (8/3)^2=64/9 =7+1/9.$$ Since we used "$\iff$" throughout, and since i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Infinite Series $\sum_{n=1}^{\infty}\frac{4^nH_n}{n^2{2n\choose n}}$ I am trying to find a closed form for this infinite series:
$$ S=\sum_{n=1}^{\infty}\frac{4^nH_n}{n^2{2n\choose n}}$$
Whith $H_n=\sum\limits_{k=1}^{n}\frac{1}{k}$ the harmonic numbers.
I found this integral representation of S:
$$S=2\int_{0}^{1}\frac{... | $$S=2\int_{0}^{1}\frac{x}{1-x^2}\left(\frac{\pi^2}{2}-2\arcsin^2(x)\right)dx\overset{IBP}=-4\int_0^1 \frac{\arcsin x\ln(1-x^2)}{\sqrt{1-x^2}}dx$$
$$\overset{x=\sin t}=-8\int_0^\frac{\pi}{2} t \ln(\cos t)dt=8 \ln 2 \int_0^\frac{\pi}{2}t dt+8\sum_{n=1}^\infty \frac{(-1)^n}{n}\int_0^\frac{\pi}{2} t\cos(2n t)dt$$
$$={\pi^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Proof verification induction on sequence I am doing problems that involve inequalities. My understanding is that you through a string of inequalities show that one is less than the other. Kind of like the transitive property. For example:
$2n+1 \lt 2^n$ for $n=3,4,...$
Assume this is true for P(k):
Base case $k = 3$
$L... | You can not use induction on $\mathbb R$ as the induction step will prove if it is true for $x$ then it is true for $x+1$ but there is nothing that will assure that it is true for any $k; x < k < x+1$.
Are you trying to prove $2x + 1 < 2^x$ for all real $x\ge 1$? If so do the following.
Use induction to prove that for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3307704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to get the value of $A + B ?$ I have this statement:
If $\frac{x+6}{x^2-x-6} = \frac{A}{x-3} + \frac{B}{x+2}$, what is the value of $A+B$ ?
My attempt was:
$\frac{x+6}{(x-3)(x+2)} = \frac{A(x+2) + B(x-3)}{(x-3)(x+2)}$
$x+6=(x+2)A + B(x-3)$: But from here, I don't know how to get $A + B$, any hint is appreciated.
| If $x+6=(x+2)A+B(x-3)$ then $$x+6=x(A+B)+(2A-3B)\implies$$ $$A+B=1\text{ and } 2A-3B=6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find the point located on an offset ellipse I am attempting to find either the x & y coordinates of a point located on an offset ellipse. I have the following information to go on:
*
*The angle of a line which intersects the point from the origin
*The radii of the ellipse
*The offset of the ellipse
This image vi... | The intersections of
$$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$$ and $$y=mx$$
are given by the solutions of
$$b^2(x-h)^2+a^2(mx-k)^2-a^2b^2=0.$$
The coefficients are
$$\begin{cases}A=b^2+a^2m^2,\\B=-2b^2h-2a^2mk,\\C=b^2h^2+a^2k^2-a^2b^2.\end{cases}$$
Then the discriminant
$$\Delta'=a^2b^2(a^2m^2+b^2-(mh-k)^2)$$
and t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3314125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
In the exapnsion of $(1+x+x^3+x^4)^{10}$, find the coefficient of $x^4$ In the exapnsion of $(1+x+x^3+x^4)^{10}$, find the coefficient of $x^4$.
What's the strategy to approach such problems. Writing expansion seems tedious here.
| You can use the multinomial expansion:
$$
(a+b+c+d)^{10}=\sum_{p+q+r+s=10}\binom{10}{p\ q\ r\ s}a^pb^qc^rd^s
$$
where
$$
\binom{10}{p\ q\ r\ s}=\frac{10!}{p!q!r!s!}
$$
With $a=1$, $b=x$, $c=x^3$, $d=x^4$, we have
$$
a^pb^qc^rd^s=x^{q+3r+4s}
$$
We get $q+3r+4s=4$ if and only if $q=4$, $r=0$, $s=0$ or $q=1$, $r=1$, $s=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3314754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Modular arithmetic $(2n+1)x \equiv -7 \pmod 9$ Find a solution $(2n+1)x \equiv -7 \pmod 9$
I’m sure this is trivial but I still have doubts about it.
I know the equation has solution for certain $n \in \mathbb {Z}$. Actually I have tried a few and got a similar results (with Diophantine equations ). I wonder if there’s... | \begin{align}
(2n+1)x &\equiv 2 \pmod 9 \\
5(2n+1)x &\equiv 1 \pmod 9 \\
(10n + 5)x &\equiv 1 \pmod 9 \\
(n-4)x &\equiv 1 \pmod 9 \\
n-4 &\equiv x^{-1} \pmod 9 \\
n &\equiv 4 + x^{-1}
\end{align}
\begin{array}{c}
x & n \equiv x^{-1} + 4 \\
\hline
1 & 5 \\
2 & 0 \\
3 & \text{No solution.} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 6
} |
Computation Of Integrals Computer the Integral: $$\int\frac{2x+1}{(x-1)(x-2)}dx$$
Now using partial fraction we can write $$\frac{2x+1}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$$, So we get $$\frac{2x+1}{(x-1)(x-2)}=\frac{A(x-2)+B(x-1)}{(x-1)(x-2)}$$ Now for all $x$ not equal to $1, 2$ we can cancel out the denominator ... | While $\frac{2x+1}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$ is only valid for $x\not\in\{1,\,2\}$ for the values of $A,\,B$ we seek, $2x+1=A(x-2)+B(x-1)$ will be valid for all $x\in\Bbb R$ because its validity at other values implies validity on $x\in\{1,\,2\}$ by continuity. You could also find $A,\,B$ from simultaneo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Inequality $\frac{\ln(7a+b)}{7a+b}+\frac{\ln(7b+c)}{7b+c}+\frac{\ln(7c+a)}{7c+a}\leq \frac{3\ln(8\sqrt{3})}{8\sqrt{3}}$ I'm interested by the following problem :
Let $a,b,c>0$ such that $abc=a+b+c$ then we have :
$$\frac{\ln(7a+b)}{7a+b}+\frac{\ln(7b+c)}{7b+c}+\frac{\ln(7c+a)}{7c+a}\leq \frac{3\ln(8\sqrt{3})}{8\sqrt... | As pointed out by River Li, first show that for every $x>0$ holds the inequality
$$
\log x\leq\frac{x}{8\sqrt{3}}+\log\left(\frac{8\sqrt{3}}{e}\right).\tag 1
$$
The equality in $(1)$ for holds only for $x=8\sqrt{3}$.
From $(1)$ we have
$$
\sum_{cyc}\frac{\log(7a+b)}{7a+b}\leq \frac{\sqrt{3}}{8}+\sum_{cyc}\frac{1}{7a+b}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $\sec A-\cos A=1$, then determine the value of $\tan^2\frac A2$ This is what I tried
$\sec A=\frac{1}{\cos A}$, so the equation becomes
$1-\cos^2A=\cos A$
If we solve the above quadratic equation, we the values of $\cos A$ as $\frac{-1\pm \sqrt5}{2}$
Therefore, $\tan\frac A2$ becomes
$$\sqrt \frac{3-\sqrt 5}{1+\sq... | Just notice that
$$
\frac{3-\sqrt 5}{1+\sqrt 5}=\frac{(3-\sqrt 5)(1-\sqrt{5})}{(1+\sqrt 5)(1-\sqrt{5})}=\frac{-4\sqrt{5}+8}{-4}=\sqrt{5}-2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that number of partitions $n^2$ on $n$ distinct parts $=$ partitions of $\binom{n}{2}$ on at most $n$ parts
Prove that number of partitions $n^2$ on $n$ distinct parts $=$ partitions of $\binom{n}{2}$ on at most $n$ parts
I tried to find bijection.
We can noticed that
$$ \binom{n}{2} = \frac{n^2-n}{2} $$
which ... | Example for $n=3$ (rather easily generalized):
Take $9$ dots and arrange in three rows so that each row has more dots than the one below it. For instance,
$$
\begin{array}{ccccc}
\bullet&\bullet&\bullet&\bullet&\bullet\\
\bullet&\bullet&\bullet\\
\bullet
\end{array}
$$
Each partition of $9$ into $3$ distinct parts corr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find "A" in this equation. $$ \sqrt[3] {A-15√3} + \sqrt[3] {A+15√3} = 4 $$
Find "A" ?
The way of exponentiation took too much time, is there any easier method?
| Let $x=A-15\sqrt 3$ and $y=A+15\sqrt 3,$ then the equation becomes $$x^{1/3}+y^{1/3}=4.$$ Taking cubes of both sides gives $$x+y+3(xy)^{1/3}(x^{1/3}+y^{1/3})=4^3.$$ Now since $x+y=2A,$ and $x^{1/3}+y^{1/3}=4,$ the equation becomes $$2A+3(xy)^{1/3}(4)=4^3,$$ which gives $$6(xy)^{1/3}=32-A.$$ Now, cubing and substituting... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3320229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Determine points on line with specific distance from plane There is a line $$p: \dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z+1}{-1}$$
and plane $$\pi : x+y+2z-3=0.$$
I need to find points $T_1, T_2 \in p$. Requirement when finding those points are that they have a set distance from plane $\pi$, with distance being $\sqrt... | If you are going to work with parallel planes, you can use the fact that they have the same normal vector. This means you only need to change the $d$ in the plane equation. So you want to find a plane $x+y+2z+d=0$ such that its distance to $\pi$ is $\sqrt{6}$.
The distance between two planes with the same normal vector... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3320539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
if $2^n+1=xy$, show $2^a|(x-1) \iff 2^a|(y-1)$
Suppose that $2^n+1=xy$, where $x$ and $y$ are integers greater than $1$ and $n>0$. Show that $2^a|(x-1)$ if and only if $2^a|(y-1)$.
What I noticed is that $2^n+1$ is an odd number for any $n$, so $x$ and $y$ are both odd numbers, and hence $x-1$ and $y-1$ will always ... | You can rewrite $$2^n+1=xy$$ as $$2^n=(x-1)(y-1)+(x-1)+(y-1)$$
With this equation, we can easily see that for every integer $a$ with $\ 1\le a\le n\ $ , $\ x-1\ $ is divisible by $\ 2^a\ $ if and only if $\ y-1\ $ is divisble by $\ 2^a\ $ , which is what has to be proven.
With some additional considerations, we can al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Computing $\int\frac {du}{\sqrt{u^2 + s^2}} =\log \lvert (u + \sqrt{u^2 + s^2}) \rvert$ with a substitution Can someone please show me where I am going wrong? It seems there is a contradiction in the formula for the the following integral.
$$\int\frac {du}{\sqrt{u^2 + s^2}} = \log \bigl\lvert u + \sqrt{u^2 + s^2}\bigr... | Both are not equal . But you'll get a term which will get added to the arbitrary constant and get cancelled in definite integral.
$$I=-\log|-u+\sqrt{u^2+s^2}|+c = \log\bigg\vert\frac{1}{-u+\sqrt{u^2+s^2}}\bigg\vert+c$$
$$I = \log\bigg\vert\frac{u+\sqrt{u^2+s^2}}{-u^2+u^2+s^2}\bigg\vert+c=\log\vert u+\sqrt{u^2+s^2}\ver... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3322560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that for $n\in\mathbb{N}$, $\sum_{k=1}^{\infty}\left[\frac{n}{5^{k}}\right]=15\iff\left[\frac{n}{5}\right]=13.$ How to show that the following relation? : for $n\in\mathbb{N}$, $$\sum_{k=1}^{\infty}\left[\frac{n}{5^{k}}\right]=15\iff\left[\frac{n}{5}\right]=13.$$ It's not obvious to me. Can anyone help me? Thank ... | Observe that $\sum_{k=1}^{\infty}\left\lfloor\frac{n}{5^k}\right\rfloor=\nu_5(n!)$, where $\nu_5(n!)$ is the exponent of the largest power of $5$ that divides $n!$ (if you are not familiar with this, then see Legendre's formula).
Since this sum is $15$, thus the highest power of $5$ that divides $n!$ is $15$.
Conside... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3322763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
equation simplification. $(5y-1)/3 + 4 =(-8y+4)/6$ Simplification of this equation gives two answers when approched by two different methods.
Method 1 Using L.C.M( least common multiple)
$(5y-1)/3 + 4 =(-8y+4)/6$
$(5y-1+12)/3 = (-8y+4)/6$
$5y-11 = (-8y+4)/2$
$(5y-11)2= (-8y+4)$
$10y-22 = -8y+4$
$18y=26$
$y = 26/18=13/... | When you do two algebraic manipulations simultaneously, like here:
$$(5y-1+12)/3 = (-8y+4)/6$$
$$5y-11 = (-8y+4)/2$$
i.e. multiply through by $3$ AND simplify the left bracket, you have confusion. It should be $5y+11$. And you could always multiply through by $6$ instead of $3$ ($6=\operatorname{lcm}(3,6)$).
I would:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Uniform convergence of $f_n(x) = \cos^n(x/ \sqrt{n})$ . I'm studying the uniform convergence of the following sequence :
\begin{equation*}
f_n(x) =
\left\{
\begin{split}
& \ \cos^n\left(\frac{x}{\sqrt{n}}\right) \ \ \textrm{if} \ x \in \left[0, \frac{\sqrt{n} \pi}{2} \right] \\
& 0 \ \ \tex... | Too long for comments.
Since composition of Taylor series is one of my hobbies, continuing your work, we have
$$\cos^n\left(\frac{x}{\sqrt{n}}\right)=e^{-\frac{x^2}{2}}\left(1-\frac{x^4}{12 n}+\frac{x^6 \left(5 x^2-32\right)}{1440 n^2}+O\left(\frac{1}{n^3}\right) \right)$$ We have also the "nice"
$$\text{sinc}^n\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is a (short and) beautiful proof for symmetric inequality must exist always? There are several and almost similar inequalities in MSE that some of them can be proved in long page. some of these questions listed below:
*
*For $abc=1$ prove that $\sum\limits_{cyc}\frac{a}{a^{11}+1}\leq\frac{3}{2}.$
*For positive $a$,... | It not always exists, but very very wanted that we can find a nice solution.
I'll give one example.
In 1988 Walther Janous proposed the following problem (Crux 1366).
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq\frac{\sqrt{a}+\sqrt{b}+\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inductively simplify specific Vandermonde determinant From Serge Lang's Linear Algebra:
Let $x_1$, $x_2$, $x_3$ be numbers. Show that:
$$\begin{vmatrix} 1 & x_1 & x_1^2\\ 1 &x_2 & x_2^2\\ 1 & x_3 &
x_3^2 \end{vmatrix}=(x_2-x_1)(x_3-x_1)(x_3-x_2)$$
The matrix presented above seems to be the specific case of Vander... | "Since determinant is a multilinear alternating function, it can be seen that adding a scalar multiple of one column (resp. row) to other column (resp. row) does not change the value (I omitted the proof to avoid too much text)
" is right. But
$$
\begin{vmatrix} 1 & x_1 & x_1^2\\ 1 &x_2 & x_2^2\\ 1 & x_3 &
x_3^2 \e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
find min of $a+b$ given sum of $a\geq b\geq c\geq d$ is 9 and square sum is 21 Suppose $a\geq b\geq c\geq d>0$ and all are real numbers, and $a+b+c+d=9,a^2+b^2+c^2+d^2=21$, how to find the minmum of $a+b$?
What I attempted:
I can show $b\geq 1.5$ and $a\leq 3$.for $r\geq 0$, I consider $\sum(a-r)^2=\sum a^2-2r\sum a+4r... | Set $c+d=x,c^2+d^2=y$ then we have
$$\begin{cases}c=\frac{x+\sqrt{2y-x^2}}{2}\\ d=\frac{x-\sqrt{2y-x^2}}{2}\end{cases}$$
Also it is easy to find out that
$$b=\frac{9-x-\sqrt{42-2y-81+18x-x^2}}{2},$$
Note that $b\geq c$ we have
$$(9-2x)^2\geq 18x-2x^2-39+2\sqrt{(18x-x^2-2y-39)(2y-x^2)}$$
which also implie that
$$6x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Range of $f(x,y)=\frac{4x^2+(y+2)^2}{x^2+y^2+1}$ I am trying to find the range of this function:
$$f(x,y)=\frac{4x^2+(y+2)^2}{x^2+y^2+1}$$
So I think that means I have to find minima and maxima. Using partial derivatives gets messy, so I was wondering if I could do some change of variables to make it easier computatio... | For
$$f(x,y)=\frac{4x^2+(y+2)^2}{x^2+y^2+1}$$
partial derivatives are not so terrifying:
\begin{align}
D_1f(x,y)&=\frac{8x(x^2+y^2+1)-2x(4x^2+(y+2)^2)}{(x^2+y^2+1)^2}\\[4px]
&=\frac{2xy(3y-4)}{(x^2+y^2+1)^2}
\\[8px]
D_2f(x,y)&=\frac{2(y+2)(x^2+y^2+1)-2y(4x^2+(y+2)^2)}{(x^2+y^2+1)^2}\\[4px]
&=\frac{2(-3x^2y+2x^2-2y^2-3y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3329898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\int _0^1\int _0^1\int _0^1\int _0^1\sqrt{(z-w)^2+(x-y)^2} \, dw \, dz \, dy \, dx$ This page contains an interesting identity$$\int _0^1\int _0^1\int _0^1\int _0^1\sqrt{(z-w)^2+(x-y)^2} \, dw \, dz \, dy \, dx=\frac{1}{15} \left(\sqrt{2}+2+5 \log \left(\sqrt{2}+1\right)\right)$$
Which calculates the average ... | By symmetry:
$$I=4\int _0^1\int _0^1\int _0^x\int _0^z\sqrt{(z-w)^2+(x-y)^2} \, dw \, dy \, dx dz$$
Substituting $w \to z w, \quad y \to x y$:
$$I=4\int _0^1\int _0^1\int _0^1\int _0^1\sqrt{z^2(1-w)^2+x^2(1-y)^2}~ x z \, dw \, dy \, dx dz$$
Substituting $w \to 1-w, \quad y \to 1-y$:
$$I=4\int _0^1\int _0^1\int _0^1\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Non-graphical solution to $5\log_{4}a\ + 48\log_{a}4 = \frac{a}{8}$ It is required to solve
$$5 \log_{4}a\ + 48\log_{a}4 = \frac{a}{8}$$
Here is my attempt,
Let $$ x = \log_{4}a$$, then $$a = 2^{2x}$$
And our equation becomes
$$ 5x^{2} - x\cdot 2^{2x-3} + 48 = 0$$
But this is as far as I can go. I've tried several s... | To localize the solution x=4 you can write
$$
5x^2 - x \cdot 2^{2x - 3} + 48 = 0
$$
as
$$
5x + \frac{{48}}
{x} = 2^{2x - 3}
$$
This tell you that x must be positive. Now, first search for integer solutions. It is obvious that $$x=1$$ is not a solution and therefore you can imagine that $$x \geq 2$$. In this case the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Regarding perfect squares Is there any positive integer $n > 2$ such that $(n - 1)(5n - 1)$ is a perfect square? It is observed that $(n - 1)(5n - 1)$ is of the form $4k$ or $4k+ 1$.
Affirmative answers were given by Pspl and Mindlack (by providing some examples).
Now my question is the following:
Is there any charact... | As $(n-1)(5n-1)=m^2$ for some $m\in\mathbb{N}$, then
$$\begin{split}(3n-1)^2-(2n)^2&=m^2\\m^2+(2n)^2&=(3n-1)^2\end{split}$$
By the formula for Pythagorean Triple, we can use it.
Let $m=x^2-y^2, 2n=2xy, 3n-1=x^2+y^2$ and assume $x>y>0$. Then
$$\begin{cases}
(x + y)^2 =5n-1 \\
(x-y)^2 = n-1
\end{cases}$$
By the assumptio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3337907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Simple sum with $i = 2$ step I need to calculate the following sum:
$$S_{n} = \sum_{2 \leq i \leq n } (3i - 2)$$
(two more conditions for the above sum: $n$ is even and $i$ with step $2$ (not sure how to do multi-line))
Adding image of the task with multi-line visible:
I wrote down the first few terms and it looks li... | Make the substitution $j=i-2$: $$S_n=\sum_{i=2}^{n}(3i-2) = \sum_{j=0}^{n-2}[3(j+2)-2] = \sum_{j=0}^{n-2}(3j+4) = 3 \cdot \sum_{j=0}^{n-2} j + 4 \cdot \sum_{j=0}^{n-2} 1 .$$ Now use the well known formula $\sum_{j=0}^{m} j = \frac{m(m+1)}{2}$ and get
$$ S_n = 3 \cdot \frac{(n-2)(n-1)}{2} + 4 \cdot (n-1) = \frac{3n^2-9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Inverse of a skew-symmetric matrix For $a, x, y, z \in \mathbb R$, let $$M= \left(
\begin{array}{cccc}
\cos(a) & \sin(a) \, x & \sin(a)\, y & \sin(a) \, z \\
-\sin(a) \, x & \cos(a) & \sin(a) \,z & -\sin(a)\, y \\
-\sin(a) \, y & -\sin(a) \, z & \cos(a) & \sin(a) \, x \\
-\sin(a) \, z & \sin(a) \, y & ... | Here's an approach which is relatively light on computation.
As in the comments, we observe that $A^2 = -(x^2 + y^2 + z^2)I$. It follows that any rational function of $A$ can be written in the form $p I + q A$ for some $p,q$. In other words, there necessarily exist numbers $p,q$ such that
$$
(M-I)^{-1} = ([\cos(a) - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Quartic polynomial roots The equation
$x^4 - x^3-1=0$
has roots α, β, γ, δ. By using the substitution $y=x^3$ find the exact value of $α^6+β^6+γ^6+δ^6$ .
The solution is
$x=y$ (1/3)
$y^4=(1+y)^3$
$y^4 -y^3 -3y^2-3y^2-1=0$
$S$N+4 $=$ $S$N + $S$N+3
$S$-1 = $\frac {0}{1} =0$
$S$2 = $ 1^2 -2*0 =1$
$S$3 = $0+1 =1$
$S$4 = $... | Multiply each term by $x^n$
$\pmb S_n =x^n$
Then it becomes
$S$N+4 $=$ $S$N + $S$N+3
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3341019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Big Factorials And Rules With Their Digits This is the issue I have.
Given that $34! = 295,232,799,**cd**9,604,140,847,618,609,643,5**ab**,000,000$ determine the digits $a$, $b$, $c$, $d$.
As far as I've gotten is $34!$ contains these multiples of $5$. $5$, $10$, $15$, $20$, $25$, $30$. So $34!$ is divisible by $5^7$... | Yes, $b=0$ for the reason you stated ($34!$ is divisible by $10^7$ but not $10^8$ so $b=0$ but $a\neq0$). That's step $1$.
Step $2$. Now say you divided $34!$ by the $10^7$ so you take away seven products of $2$ and seven product of $5$ (speaking in terms of prime factors). What you're left with has at least three prod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A matrix of order 8 over $\mathbb{F}_3$ What is an example of an invertible matrix of size 2x2 with coefficients in $\mathbb{F}_3$ that has exact order 8?
I have found by computation that the condition that the 8th power of a matrix
$\begin{bmatrix}a & b\\c & d\end{bmatrix}$ is the identity is
$$
b c (a + d)^2 (a^2 +... | Using a similar strategy to Travis, we can actually answer this question quite nicely:
$$A \text{ has order 8 if and only if }\det(A) = 2 \text{ and } \text{tr}(A) \neq 0.$$ We can do this by looking at the characteristic polynomial $p(x)$ of $A$, with which we can utilize the fact that $p(A) = 0$ as shown below.
Sinc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
For $z \in \mathbb{C}$, if $(z^a,z^c,z^b \bar{z}^d,z^{b+d})=(1,1,1,1)$, then $z=1$ under certain gcd conditions for $a,b,c,d$.
Let $z \in \mathbb{C}$. I would like to show that if $(z^a,z^c,z^b \bar{z}^d,z^{b+d})=(1,1,1,1)$ and gcd$(a,b,c,d)=1$ and gcd$(a^2-c^2,b^2-d^2)=4$, then $z=1$.
Note that I am not sure whethe... | The given condition is false. Take $z = -1$, and $a=4,c=2$, with $b=3,d=1$. Then, of course $\mbox{gcd}(a,b,c,d) = 1$, and $a^2-c^2 = 12$ with $b^2-d^2 = 8$ whose $\gcd$ is $4$. Furthermore, $\bar z = z$, so $z^a = z^c= 1$ and $z^b\bar z^d = z^bz^d=z^{b+d} = z^4=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $a$, $b$, $c$ are sides of a triangle, prove $2(a+b+c)(a^2+b^2+c^2)\geq3(a^3+b^3+c^3+3abc)$
$a$, $b$, $c$ are sides of a triangle, prove:
$$2(a+b+c)(a^2+b^2+c^2)\geq3(a^3+b^3+c^3+3abc)$$
What I have tried:
$$
⇔2\sum (a+b)ab\geq \sum a^3+9abc
$$
so I can't use
$$\sum a^3+3abc\geq \sum ab(a+b)$$
| Also, we can use SOS here.
Indeed, by your work we need to prove that
$$\sum_{cyc}(2a^2b+2a^2c-a^3-3abc)\geq0.$$
Now, let $a\geq b\geq c$.
Thus, since $$3b-a-c=b+c-a+2b-2c>0,$$ we obtain:
$$\sum_{cyc}(2a^2b+2a^2c-a^3-3abc)=\sum_{cyc}(2a^2b+2a^2c-4abc-a^3+abc)=$$
$$=\sum_{cyc}\left(2c(a-b)^2-\frac{1}{2}(a+b+c)(a-b)^2\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integral solutions of $a^{2} + a = b^{3} + b$ Find all pairs of coprime positive integers $(a,b)$ such that
$b<a$ and $a^2+a=b^3+b$
My approach:
$a(a+1)=b(b^2+1)$ so $a|(b^2+1)$ and $b|(a+1)$
Now after this I am not able to do anything.pls help.
| As you already note, if $a$ and $b$ are coprime and
$$a(a+1)=a^2+a=b^3+b=b(b^2+1),$$
then it follows that $b$ divides $a+1$. Then $a=bc-1$ for some integer $c$, where $c>1$ because $a>b$. Then
$$b^3+b=a^2+a=(bc-1)^2+(bc-1)=c^2b^2-cb,$$
and since $b$ is positive we can divide both sides by $b$ and rearrange to get the q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3346708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Prove that $0<\sum_{k=1}^n \frac{g(k)}{k} - \frac{2n}{3} < \frac{2}{3}$ Prove
$$0<\sum_{k=1}^n \frac{g(k)}{k} - \frac{2n}{3} < \frac{2}{3}$$
where $g(k)$ is the greatest odd divisor of k
Please Find Holes in my Proof.
Let $k=2m+1$ if we show that the right hand side of the equation is true for odd numbers, then it is t... | Let me prove a slightly sharper inequality. While it looks slightly more complicated it gives a better insight:
$$ \frac 1{3\cdot 2^{\lfloor \log_2(n)\rfloor}}\,\ \le
\,\ \sum_{k=1}^n\frac{g(k)}k\ -\ \frac 23\cdot n\,\ \le
\,\ \frac 23 - \frac 1{3\cdot 2^{\lfloor \log_2(n)\rfloor}}
$$
Even better, the above inequ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
proof based on cauchy schwartz inequality If a,b,c are positive real numbers,prove that
$$ \frac{a}{b+2c} + \frac{b}{c+2a} + \frac{c}{a+2b} \ge 1 $$
I tried solving and i have no idea how to proceed I mechanically simplified it it looks promising but im still stuck. This is from the excersice on Cauchy Schwartz Inequal... | As we have, $2+a^3=a^3+1+1\geqslant 3a$, $b^2+1\geqslant 2b$, thus$$\dfrac{a}{a^3+b^2+c}=\frac{a}{3+a^3+b^2-a-b}\leqslant\frac{a}{3a+2b-a-b}=\frac{a}{2a+b}.$$Similarly, we can get $$\dfrac{b}{b^3+c^2+a}\leqslant\frac{b}{2b+c},\,\,\,\,\,\dfrac{c}{c^3+a^2+b}\leqslant\frac{c}{2c+a}.$$It suffices to show$$\frac{a}{2a+b}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The sum of infinite series $(1/2)(1/5)^2 + (2/3)(1/5)^3 + (3/4)(1/5)^4..................$ Initially, I broke this series as $1/2 = 1-1/2, 2/3 = 1-1/3, 3/4 = 1-1/4$, then I got two series as it is
$$= (1/5)^2 + (1/5)^3 + (1/5)^4 +.......-[1/2(1/5)^2 + 1/3(1/5)^3 + 1/4(1/5)^4 +.....]$$
After solving the first part, we re... | Thanks all who have given the answers. Some people use "Lerch Transenden" methods to solve some uses "integration", "convergence" etc. But I was looking to solve it as using 10+2 level math techniques.
Finally I got success in solving it. See my solution below -
= (1/2)(1/5)^2 + (2/3)(1/5)^3 + (3/4)(1/5)^4 +............ | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How many $14$-digit even numbers can be formed using $0,1,1,2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5$? What is a good way to approach this question?
How many $14$-digit even numbers can be formed using $0,1,1,2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5$?
I know that I can list the possible outcomes with different ending numbers and then ... | Case 1: The number ends in $0$. Considering all possible orderings of the other $13$ numbers we get $$\frac{13!}{2!2!3!3!3!}$$ possible outcomes (since we have two of $1$ and $2$ each and three of $3,4,5$ each, we divide by $2!2!3!3!3!$) to remove duplicates.
Case 2: The number does not end in $0$. We have $5$ possible... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3351314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Relate to Dirichlet's theorem Diophantine approximation I have a problem:
Let $a \in \mathbb{Z}$, $a \geq 3$, and set $\xi= \sum_{n=0}^\infty
10^{-a^{2n}}>0$. Then the inequality
$$\Big|\,\xi - \dfrac{x}{y}\,\Big| \leq \dfrac{1}{y^a}$$ has infinitely many solutions with $x,y \in \mathbb{Z}$, $y>0$ and $\gcd(x,y)=1$.
... | With regards to other ideas, just straight attack. Noting $y=10^{a^{2n}}$ then
$$\sum\limits_{k=0}^{n}\frac{1}{10^{a^{2k}}}=\frac{\sum\limits_{k=0}^{n}10^{a^{2n}-a^{2k}}}{y}$$
where $x=\sum\limits_{k=0}^{n}10^{a^{2n}-a^{2k}}=\sum\limits_{k=0}^{\color{red}{n-1}}10^{a^{2n}-a^{2k}}\color{red}{+1}=10\cdot Q\color{red}{+1}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3352604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is my summation result correct? $\sum_{i=0}^{n-1} \sum_{j=0}^{i-1} (i+j) =\cdots= \frac{n(n-1)^2}{2}$ Is this result correct?
$$\begin{align}
\sum_{i=0}^{n-1} \sum_{j=0}^{i-1} (i+j)
&= \sum_{i=0}^{n-1}\left( i^2 + \frac{i(i-1)}{2} \right)\\
&= \frac{3}{2}\sum_{i=0}^{n-1}i^2 - \frac{1}{2}\sum_{i=0}^{n-1}i \\
&= \frac{3... | You are almost correct. It should be
\begin{align}
\sum_{i=0}^{n-1} \sum_{j=0}^{i-1} (i+j)
&= \frac{3}{2}\sum_{i=0}^{n-1}i^2 - \frac{1}{2}\sum_{i=0}^{n-1}i \\
&= \frac{3}{2}\cdot \frac{n(n-1)(2n-1)}{\color{red}{6}} -\frac{1}{2}\cdot \frac{n(n-1)}{2}\\
&=\frac{n(n-1)((2n-1)-1)}{4}=\frac{n(n-1)^2}{2}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3352845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Olympiad Inequality $\frac{ab+bc+ca}{a+b+c}(\sum_{cyc}\frac{1}{2a+b})\leq 1$ with proof (long post) It takes me a lot a time to solve this :
Let $a,b,c>0$ with $a\geq b\geq c$ then we have :
$$\frac{ab+bc+ca}{a+b+c}(\sum_{cyc}\frac{1}{2a+b})\leq 1$$
My proof :
We need some lemmas to begin :
First lemma :
Let $x,... | After full expanding we need to prove that
$$\sum_{cyc}(2a^3b-a^2b^2-a^2bc)\geq0$$ or
$$\sum_{cyc}(a^3b+a^3c-a^2b^2-a^2bc)\geq\sum_{cyc}(a^3c-a^3b)$$ or
$$\sum_{cyc}(a^3b+a^3c-a^2b^2-a^2bc)\geq(a-b)(b-c)(c-a)(a+b+c),$$ which is true because by Muirhead
$$\sum_{cyc}(a^3b+a^3c-a^2b^2-a^2bc)\geq0$$ and by the given $$(a-b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3354020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Determining The $I^{th}$ Number In The Sequence I've recently come across a sequence of numbers but I can't determine any sensible way to say what the $I^{th}$ number in the pattern will be. The pattern is as follows:
$\frac{1}{2}, \\\frac{1}{4}, \frac{3}{4}, \\\frac{1}{8}, \frac{5}{8}, \frac{3}{8}, \frac{7}{8}, \\\fr... | Ignore the denominator. For the $i^\text{th}$ number, convert that number to binary, reverse the order of the digits, and then convert that back to a base 10 number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3354574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Hard inequality :$\Big(\frac{1}{a^2+b^2}\Big)^2+\Big(\frac{1}{b^2+c^2}\Big)^2+\Big(\frac{1}{c^2+a^2}\Big)^2\geq \frac{3}{4}$ I have a hard problem this is it :
Let $a,b,c>0$ such that $a^ab^bc^c=1$ then we have :
$$\Big(\frac{1}{a^2+b^2}\Big)^2+\Big(\frac{1}{b^2+c^2}\Big)^2+\Big(\frac{1}{c^2+a^2}\Big)^2\geq \frac{3}... | Define: $$F(a,b,c)=\left(\frac{1}{a^2+b^2}\right)^2+\left(\frac{1}{b^2+c^2}\right)^2+\left(\frac{1}{c^2+a^2}\right)^2$$
By the method of Lagrange multipliers, the minimizer of $F$ under the constraint $a^ab^bc^c=1$ must be a critical point of the Lagrangian
$$L(a,b,c,\lambda)=\left(\frac{1}{a^2+b^2}\right)^2+\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3355161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
} |
partial fraction decomposition in integral I have a question that asks to calculate the following integral:
$$ \int_0^\infty {\frac{w\cdot \sin w}{4a^2+w^2}dw} $$
In the official solution they used partial fraction decomposition in order to later use Plancherel's identity:
$$ \frac{w\cdot \sin w}{4a^2+w^2} = $$
$$ \fr... | Consider the function:
$$I(a)=\int_0^\infty {\frac{w\cdot \sin (w)}{w^2+4\cdot a^2}\text{d}w},\space a>0$$
Rewrite the integrand as follows:
$$\frac{w\cdot \sin (w)}{w^2+4\cdot a^2}=\frac{w}{w}\cdot\frac{w\cdot \sin (w)}{w^2+4\cdot a^2}=\frac{\sin(w)}{w}\cdot\frac{w^2}{w^2+4\cdot a^2}=\frac{\sin(w)}{w}\cdot\frac{w^2+4\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How many triplets $(p,q,r)$ of integers between $1$ and $1000$ satisfy $p^2\sin^2x+q\sin2x+r^2\cos^2x>1$, if $\sin x$ and $\cos x$ are non-zero?
We have an inequality of the form
$$p^2\sin^2 x + q\sin2x + r^2\cos^2x > 1$$
for integers $p$, $q$, $r$ with $1\leq p,q,r \leq 1000$, and such that $x$ cannot be any val... | Divide both sides by $\cos^2x$ and writing $\tan x=t$
$$p^2t^2+2qt+r^2>1+t^2$$
$$t^2(p^2-1)+2qt+r^2-1>0$$
Now the roots of $$t^2(p^2-1)+2qt+r^2-1=0$$ are $t=\dfrac{-q\pm\sqrt{q^2-(p^2-1)(r^2-1)}}{p^2-1}$
So, either $t>$max $\left(\dfrac{-q+\sqrt{q^2-(p^2-1)(r^2-1)}}{p^2-1},\dfrac{-q-\sqrt{q^2-(p^2-1)(r^2-1)}}{p^2-1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3357097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $\frac{a}{b^{2}+1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2} + 1} \ge \frac{3}{2}$ $a,b,c > 0$ and $a+b+c=3$, prove
$$ \frac{a}{b^{2} + 1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2}+1} \ge 3/2 $$
Attempt:
Notice that by AM-Gm
$$\frac{a}{b^{2} + 1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2}+1} \ge 3\frac{\sqrt[3]{abc}}{\sqrt[3]{... | Since for $x>0$ we have (just draw a graph for ${1\over 1+x^2}$ and a tangent at $x=1$) $${1\over 1+x^2}\geq -{1\over 2}x+1$$ it is enough to check if $$-{1\over 2}(ab+bc+ca)+3\geq {3\over 2}$$ i.e. $$3\geq ab+bc+ca$$ is true?
Since $$a^2+b^2+c^2\geq ab+bc+ca$$ that is easy to verify. :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
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.