Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Claculate limit $\lim_{x\to 0}\frac{1-(\cos(1-\sqrt{\frac{\sin(x)}{x}}))}{x^4}$ I have a problem to calculte this limit:
$$\lim_{x\to 0}\frac{1-(\cos(1-\sqrt{\frac{\sin(x)}{x}}))}{x^4}$$
I used Taylor expansion for $\sin(x), \cos(x)$ and considered also $1-\cos(\alpha)=2\sin^2(\frac{\alpha}{2})$ and $\alpha=2-2\sqrt{\f... | Hint
Compose Taylor series one piece at the time
$$\frac{\sin (x)}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right)$$
$$\sqrt{\frac{\sin (x)}{x}}=1-\frac{x^2}{12}+\frac{x^4}{1440}+O\left(x^6\right)$$
$$\cos \left(1-\sqrt{\frac{\sin (x)}{x}}\right)=1-\frac{x^4}{288}+\frac{x^6}{17280}+O\left(x^8\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3870497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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HINT: $a^2+b^2=c^2$ cannot hold for $a,b$ odd and $c$ even. (using congruences) Here is what I have: Suppose the contrary. Thus,
(2k+1)(2k+1) + (2j+1)(2j+1) = (2p)(2p)
Take mod 2 of both sides
[1][1]+[1][1]=[0]
[2]=[0]
[0]=[0]
No contradiction ... am I approaching this correctly? I want to figure this out myself but wo... | With $a$, $b$ odd we may write
$a = 2m + 1, \tag 1$
$b = 2n + 1; \tag 2$
then
$a^2 = 4m^2 + 4m + 1, \tag 3$
$b^2 = 4n^2 + 4n + 1, \tag 4$
$a^2 + b^2 = 4(m^2 + n^2 + m + n) + 2; \tag 5$
also,
$c = 2p, \tag 6$
whence
$c^2 = 4p^2; \tag 7$
if
$a^2 + b^2 = c^2, \tag 8$
then substituting in (3), (4) and (7) we find
$4(m^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3872457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Summation of $n$th partial products of the square of even numbers diverges, but for odd numbers they converge in this series I'm looking at. Why? So I have the two following series:
$$\sum_{n=1}^\infty \frac{\prod_{k=1}^n(2k)^2}{(2n+2)!}$$
$$\sum_{n=0}^\infty \frac{\prod_{k=0}^n(2k+1)^2}{(2n+3)!}$$
I figured out the $n... | Convergence
Using the asymptotic approximation given in inequality $(9)$ of this answer, we get
$$
\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}\tag1
$$
Therefore,
$$
\begin{align}
\frac{\prod\limits_{k=1}^n(2k)^2}{(2n+2)!}
&=\frac{4^nn!^2}{(2n)!(2n+1)(2n+2)}\\
&=\frac{\color{#090}{4^n}}{\color{#090}{\binom{2n}{n}}\color{#... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3875640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find the roots of the polynomial $x^3-2$. Find the roots of the polynomial $x^3-2$.
If $\alpha$ be the root of this polynomial i.e., $\alpha^3=2$,
then $(\zeta \alpha)^3=2$, where $\zeta$ is $3^{rd}$ root of unity.
Hence the solutions of this equations are
$$\zeta{2}^\frac{1}{3}$$, $\zeta$ a $p^{th}$ root of unity
And... | The best way to do it is to use the methods that other responses have detailed. I am going to show you an inferior method only because understanding the relationship between this inferior method and the better method (already described by others) will stretch your intuition.
Let $a$ denote $2^{(1/3)}.$
Given $0 = (x^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3876285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Find the image of the line $\frac{x}{3}=\frac{y}{2}=\frac{z-1}{1}$ in the plane $x+y+z=4$.
Find the image of the line $\frac{x}{3}=\frac{y}{2}=\frac{z-1}{1}$ with respect to plane $x+y+z=4$.
my method:
Any point in the given line will be $(3t,2t,t+1)$.
Thus the line meets the plane in $(3/2,1,3/2)$.
We now take a ran... | The line $r:\frac{x}{3}=\frac{y}{2}=\frac{z-1}{1}$ is the intersection of the planes $\alpha:\frac{x}{3}=\frac{y}{2};\;\beta:\frac{x}{3}=z-1$.
Any linear combination $\mathcal{P}$ of the two planes $\alpha:2x-3y=0$ and $\beta:x-3z+3=0$ passes through the line $r$
$$\mathcal{P}:h(2x-3y)+k(x-3z+3)=0;\;\forall h,k\in\math... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Partial fraction decomposition $\frac{1}{(x-y)^2}\frac{1}{x^2}$ I want to integrate $$\frac{1}{(x-y)^2}\frac{1}{x^2}$$ with respect to $x$.
I know that I have to apply partial fraction decomposition but which ansatz do I have to make to arrive at
$$\frac{1}{(x-y)^2}\frac{1}{x^2}=-\frac{2}{y^3(x-y)}+\frac{1}{y^2(x-y)^2}... | Here is the (comparatively) fast way to determine the partial fraction decomposition, using the poles of the fraction. To make the computation clearer, I'll denote temporarily $y$ as the constant $a$.
The decomposition has the form
$$\frac 1{x^2(x-a)^2}=\frac Ax+\frac B{x^2}+\frac C{x-a}+\frac D{(x-a)^2}.\tag 1$$
Mul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Integral $\int_0^\infty \frac{1-x(2-\sqrt x)}{1-x^3}dx$ vanishes Come across
$$I=\int_0^\infty \frac{1-x(2-\sqrt x)}{1-x^3}dx$$
and break the integrand as
$$\frac{1-x(2-\sqrt x)}{1-x^3}=\frac{1-x}{1-x^3}- \frac{x(1-\sqrt x)}{1-x^3}
$$
The first term simplifies and the second term transforms with $\sqrt x\to x$. Then, t... | I'm not sure if this is something that is of value to you since it uses your approach, but once you transform $\sqrt{x} \to x$ for the second integral you can use PFD to cancel out the first integral and be left with two simple integrals, instead of expressing it as you did:
$$-\int_0^{\infty} \frac{2x^3(1-x)}{1-x^6} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3879991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Trigonometric equation: $3\sin x = -\cot x \cdot \cot 2x \cdot (\tan^2 x + \tan 2x)$
Solve the following equation:
$$3\sin x = -\cot x \cdot \cot 2x \cdot (\tan^2 x + \tan 2x)$$
My attempt:
$$3\sin x = -\cot x \cdot \cot 2x \cdot (\tan^2 x + \tan 2x)$$
$$\implies 3\sin x = -\tan x \cdot \cot 2x - \cot x$$
$$\implies ... | I think that after writing that tou want to find the zeros of
$$f(x)=3\sin (x) + \dfrac{1 - \tan^2 (x)}{2} + \cot (x) $$ the problem becomes purely numerical.
If you plot the function, it is quite awful because of the discontinuities due to the tangent and cotangent. It would be better to remove them multipluing everyt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3880162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How to find a useful variable change for this integral I would like to find the area of the following region
$$
D=\left \{(x,y): -\sqrt{1+y^2}\leq x\leq \sqrt{1+y^2}; -1\leq y\leq (x+1)/2\right \}.
$$
I try to calculate the double integral brute force, but following this path
I came across some very unpleasant integral... |
See the diagram. You need to find the area of the region ABCO. If you integrate along $y$ axis taking strips of thickness $dy$ parallel to $x$ axis, you can see that from $y = -1$ to $y = 0$, both the left and right ends are bound by the hyperbola but for strips at $0 \leq y \leq \frac{4}{3}$, the left is bound by t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3882212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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(AIME 1994) $ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $
$($AIME $1994)$ Find the positive integer $n$ for which
$$ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \lfloor \log_2 3 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $$ where $\lfloor x \rflo... | Denote the sum by $S_{n}$. Note that for any $k\in\mathbb{N}$ there are $2^k$ positive integers $x$ for which $[\log_{2}(x)]=k$, and those are $x=2^{k},2^{k}+1,\ldots,2^{k+1}-1$. Thus we have $$S_{2^{k}-1}=0 + (1+1) + (2+2+2+2+) + \cdots + \bigl((k-1)+(k-1)+\cdots + (k-1)\bigr)$$ where there number of $(k-1)$ terms is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3884500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Limit multivarible How do I solve this limit?
$$\lim_{x,y,z) \to (0,0,0)} \frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2+xyz} $$
This is equal to $$\frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2}\times\frac1{1+\frac{xyz}{x^2+y^2+z^2}}$$
The first one is a standard limit with value one, but I'n not sure about the other term.
| Yes your idea is very good to put away the $\sin$ term, now we have that, for example by spherical coordinates
$$\frac{xyz}{x^2+y^2+z^2} =r \cos \theta\sin \theta\cos^2 \phi\sin \phi $$
or as an alternative by AM-GM we can use that
$$x^2+y^2+z^2 \ge 3 \sqrt[3]{x^2y^2z^2} \implies \left|\frac{xyz}{x^2+y^2+z^2}\right| \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3887468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find the product of all values of $(1+i\sqrt 3)^{\frac{3}{4}}$. Find the product of all values of $(1+i\sqrt 3)^{\frac{3}{4}}$.
My try:
$(1+i\sqrt 3)^{\frac{3}{4}}=\exp (\frac{3}{4}(Log(1+i\sqrt 3)))=\exp(\frac{3}{4}(\log2+i\frac{\pi}{3}+2n\pi i))$.
I am kinda stuck on how to find the product of all values of the abov... | $$1+i\sqrt3=2e^{i\pi/3}$$
The fourth roots of this number have common magnitude $2^{1/4}$ and arguments $\pi/12,7\pi/12,-5\pi/12,-11\pi/12$. The $\frac34$-powers thus have common magnitude $2^{3/4}$ and arguments $\pi/4,7\pi/4\equiv-\pi/4,-5\pi/4\equiv3\pi/4,-11\pi/4\equiv-3\pi/4$. The producf of all the four possible ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3888572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find the coefficient of $x^n$ in the generating functions: $g(x) = \frac{x^3}{(1+x)^5 (1−x)^6}$ One of the problems in my Discrete Math course states that we need to find the coefficient of $x^n$ in generating function $g(x) = \frac{x^3} {(1+x)^5 (1−x)^6}$
I separated $g(x) = \frac{x^3} {(1+x)^5 (1−x)^6}$ into $x^3$ * ... | $$
\begin{align}
\frac{x^3}{(1+x)^5 (1−x)^6}
&=\frac{x^3+x^4}{\left(1-x^2\right)^6}\tag1\\
&=\left(x^3+x^4\right)\sum_{k=0}^\infty(-1)^k\binom{-6}{k}x^{2k}\tag2\\
&=\left(x^3+x^4\right)\sum_{k=0}^\infty\binom{k+5}{5}x^{2k}\tag3\\
&=\sum_{k=0}^\infty\binom{k+5}{5}\left(x^{2k+3}+x^{2k+4}\right)\tag4\\
&=\sum_{k=2}^\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finite sum which sums to $x e^x$ In my analysis class, we have to prove this and another two related sums that I think I could prove if I knew this. However, I don't know how to begin solving this; any hints would be appreciated.
$$\sum_{n=0}^{\infty} \left( e^x-1-\frac{x}{1!}-\frac{x^2}{2!}-\frac{x^3}{3!}-\cdots-\frac... | \begin{align}
& \sum_{n=0}^\infty \left( e^x-1-\frac x {1!}-\frac{x^2}{2!}-\frac{x^3}{3!}-\cdots-\frac{x^n}{n!} \right) = \sum_{n=0}^\infty \sum_{k=n+1}^\infty \frac{x^k}{k!} \\[8pt]
& \begin{array}{cccccccccc}
= & x & + & \dfrac{x^2} 2 & + & \dfrac{x^3} 6 & + & \dfrac{x^4}{24} & + & \dfrac{x^5}{120} & + & \cdots \\[8p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3890345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Analysis - Find the Limit Let $P(x) = ax^3 + bx^2 + cx + d$, $a, b, c, d \in \mathbb{R}$, determine whether the sequence
$$
\left(
\frac{P(n)}
{P(n + 1)}
\right)
$$
it is convergent or not. If so, calculate the limit.
I managed to show that the sequence
$$\frac{an^3 + bn^2 + cn + d}{a(n+1)^3 + b(n+1)^2 + c(n+1) + d}$$
... | We can use that, for $a\neq 0$
$$\frac{an^3 + bn^2 + cn + d}{a(n+1)^3 + b(n+1)^2 + c(n+1) + d}=\frac{a + b\frac1{n} + c\frac1{n^2} + d\frac1{n^3}}{a\left(1+\frac1n\right)^3 + \frac b n\left(1+\frac1n\right)^2 + \frac c {n^2}\left(1+\frac1n\right) + \frac d{n^3}} \to 1$$
For $a=0$ we can divide by $n^2$ and so on.
The a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3894042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the solutions to an ODE are non-periodic oscillating. I have the ODE $$\ddot{x}+ \dot{x}+k^2x=0$$
where the dot means derivative with respect to $t$ ($x$ is a function of $t$) and $k$ is a constant such that $k>\frac{1}{2}$
prove that the solutions to the ODE are non-periodic and oscillating.
What I tried is... | Given the equation
$\ddot x + \dot x + k^2x = 0, \tag 1$
the characteristic equation is indeed
$\lambda^2 + \lambda + k^2 = 0, \tag 2$
but the "quadratic formula" used to find $\lambda_{1,2}$ is misstated; it should read
$\lambda_{1,2} = \dfrac{-1 \pm \sqrt{1 - 4k^2}}{2}; \tag 3$
from this we see that when $k > 1/2$ th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3895689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to simplify $\frac {\sin 3A - \cos 3A}{\sin A + \cos A} + 1$? So I started by using $\sin 3A$ and $\cos 3A$ identities and then I added the lone $1$ to the trigonometric term. (Done in the picture below)
But after this I don't have any clue on how to proceed.
$$=\frac{3 \sin \theta-4 \sin ^{3} \theta-\left(4 \cos ^... | $$\frac{\sin3x-\cos3x}{\sin x+\cos x}+1=\frac{\sin3x+\sin x+\cos x-\cos3x}{\sin x+\cos x}=\frac{\sin(2x+x)+\sin (2x-x)+\cos (2x-x)-\cos(2x+x)}{\sin x+\cos x}=\frac{\sin2x\cos x+\sin x\cos2x+\sin2x\cos x-\sin x\cos 2x+\cos 2x\cos x+ \sin 2x\sin x-(\cos 2x\cos x- \sin 2x\sin x)}{\sin x+\cos x}=\frac{2 \sin2x\cos x +2 \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3897329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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solve for the Laurent series of the function $f(z) = \frac{z-12}{z^2+z-6}$ that is valid for $1<|z-1|<4$. I am about to solve for the Laurent series of the function $f(z) = \frac{z-12}{z^2+z-6}$ that is valid for $1<|z-1|<4$.
What I did is I use the partial fraction decomposition to rewrite the $f(z)$ into $f(z) = \fra... | Your answer cannot possibly be an answer to the given question, since the answer should be a series of the form $\sum_{n=-\infty}^\infty a_n(z-1)^n$.
Note that\begin{align}\frac{z-12}{z^2+z-6}&=\frac3{z+3}-\frac2{z-2}\\&=\frac3{z-1+4}-\frac2{z-1-1}\\&=\frac34\frac1{1+\frac{z-1}4}+2\frac1{1-(z-1)}.\end{align}Now, since ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3901339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Probability of first to draw red ball from urn without replacement There are R red and G green balls in a box. Alice and Bob pick one ball from the box in turns without replacement. First one to draw a red ball wins. What is the probability that Alice wins if she draws first?
Here is my attempt - Could someone please a... | Let random variable $R_1$ be the number of draws to get the first red ball. For $k\in\{1,\dots,G+1\}$, we have
\begin{align}
P(R_1=k)
&= \left(\prod_{d=0}^{k-2} \frac{G-d}{R+G-d}\right)\frac{R}{R+G-(k-1)} \\
&= \frac{\binom{G}{k-1}}{\binom{R+G}{k-1}}\frac{R}{R+G-(k-1)}\\
&= \frac{\binom{R+G-(k-1)}{G-(k-1)}}{\binom{R+G... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3901828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Proving that if $(x_n) \to x$ then $p(x_n) \to p(x)$ in Real Analysis I am self-learning Real Analysis from Stephen Abott's Understanding Analysis. Exercise 2.3.8 asks to prove that when a polynomial function $p$ is applied to values of a convergent sequence $(x_n)$, then if the input sequence converges to $x$, the res... | If you are using the algebra of limits in the end of your proof, you might as well use it in the beginning. Considering $p(x)=a_0 + a_1 x + \cdots + a_k x^k$, you have that
$$
\lim p(x_n) = \lim(a_0 + a_1 x_n + \cdots + a_k x_n^k)= a_0 + a_1 \lim x_n + \cdots + a_k (\lim x_n)^k = p(x).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3902947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ without induction? I've been trying to solve the following problem:
Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1... | Let $S=1^2+3^2+5^2+\cdots+(2n+1)^2$$.
Next, consider the sums
$$S_1=2^2+4^2+6^2+\cdots+(2n)^2$$
$$S_2=1^2+2^2+3^2+\cdots+(2n+1)^2$$
Thus, we can calculate $S$ as $S=S_2-S_1$.
Let's evaluate $S_1$ and $S_2$ using the identity we are given.
Clearly,
$$S_2=\frac{1}{6}(2n+1)(2n+2)(4n+3)$$
and
$$S_1=2^2(1^2+2^2+\cdots+n^2)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3909607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Simultaneous algebra question Let $x$, $y$ be real numbers such that
$$x+3y=4$$$$x^2+3y^2=8$$
Find $$x^3+3y^3$$
Also, let $$ax+by=v_1$$$$ax^2+by^2=v_2$$
Find the value of $ax^3+by^3$ in terms of $v_1$, $v_2$
I've tried $(x+3y)(x^2+9y^2−3xy)−6y^3$ and was left with $48+20y^2−6y^3$
| $$x+3y=4 \implies x=4-3y \implies (4-3y)^2+3y^2=8 \implies 3y^2-6y+2=0 \implies y=1\pm \frac{1}{\sqrt{3}}$$
We get $x=1-\sqrt{3}$ for $y=1+\frac{1}{\sqrt{3}}$, then $F=x^3+3y^3=(1-\sqrt{3})^3+3(1+\frac{1}{\sqrt{3}})^3=16-\frac{8}{\sqrt{3}}.$
or $x=1+\sqrt{3}$ for $y=1-\frac{1}{\sqrt{3}}$
Finally, $F=x^3+3y^3=(1+\sqrt{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3910717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Systems of Congruences \begin{cases} \overline{xyz138} \equiv 0 \mod7 \\ \overline{x1y3z8} \equiv 5 \mod11 \\ \overline{138xyz} \equiv 6 \mod13 \end{cases}
I worked my way up to this:
\begin{cases} 2000x+200y+20z \equiv 2 \mod7 \\ 100000x+1000y+10z \equiv 4 \mod11 \\ 2000x+200y+20z \equiv 7 \mod 13 \end{cases}
I tried ... | There are some basic division tricks. $7*13*11=1001$ so to have $xyz138\equiv 0 \pmod 7$ implies that $xyz138 \equiv xyz138 - 1001*xyz = 138-xyz \equiv 0$ so $xyz \equiv 138\pmod 5\pmod 7$.
Likewise $138xyz \equiv 138xyz - 1001*138\equiv xyz -138\equiv 6 \pmod {13}$ so $xyz \equiv 144\equiv 1 \pmod {13}$.
Both togethe... | {
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How do you prove that $\int \frac{1}{\sqrt{x^2+r^2}} \,dx = \ln{\left\lvert x + \sqrt{x^2+r^2} \right\rvert + C}$? I would like to prove that
$$\int \frac{1}{\sqrt{x^2+r^2}} \,dx = \ln{\left\lvert x + \sqrt{x^2+r^2} \right\rvert} + C$$
I tried with applying derivative with respect to $x$ to $\ln{\left\lvert x + \sqrt{x... | Your calculation of the derivative is incorrect.
Note $$\begin{align}
\frac{d}{dx}\left[ x + \sqrt{x^2 + r^2} \right]
&= 1 + \frac{1}{2}\left(x^2 + r^2 \right)^{-1/2} \cdot \frac{d}{dx}\left[x^2 + r^2\right] \\
&= 1 + \frac{1}{2 \sqrt{x^2 + r^2}} \cdot 2x \\
&= 1 + \frac{x}{\sqrt{x^2 + r^2}}. \end{align}$$ You are mi... | {
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Is $n=26$ the only integer satisfying $n-1$ being a square and $n+1$ being a cube? I guess it is possible to solve it by ring theory, but I would prefer an elementary method instead. I can only figure out that $n$ is an even integer, otherwise there will be contradiction w.r.t. modulo $4$. I have also tried to modulo s... | Lemma.
Let $a$ and $b$ be coprime integers, and let $m$ and $n$ be positive integers such that $a^2+2b^2=mn$. Then there are coprime integers $r$ and $s$ such that $m=r^2+2s^2$ divides $br-as$. Furthermore, for any such choice of $r$ and $s$, there are coprime integers $t$ and $u$ such that $a=rt-2su$, $b=ru+st$, and $... | {
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Let $x,y \in R$ such that $|x+y| + |x-y| = 2$
Let $x,y \in R$ such that $|x+y| + |x-y| = 2$, then find maximum values of $x^2 - 6x + y^2$ and $x^2 + y^2 + 10y$.
How do I go about solving this question? Is it possible to find all real values of $x$ and $y$ from the first equation? Please help.
| Hint:
If $f(x,y)=|x+y|+|x-y|-2,$
observe that $f(x,y)=f(-x,y)=f(x,-y)=f(-x,-y)$
WLOG $x+y, x-y\ge0;$
$$x=1, -x\le y\le x, y^2\le x^2=1$$
Now,
$$x^2+y^2+10y=1+(y+5)^2-25$$
$-1\le y\le1\iff5-1\le5+y\le5+1\implies4^2\le(y+5)^2\le6^2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Function satisfying the relation $f(x+y)=f(x)+f(y)-(e^{-x}-1)(e^{-y}-1)+1$ Let f be the differentiable function satisfying the relation $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$; $\forall x,y \in R$ and $\mathop {\lim }\limits... | Let $g(x)=f(x)+e^{-x}$. Then given equation reduces to Cauchy's equation $g(x+y)=g(x)+g(y)$. If $f$ is assumed to be continuous (or at least measurable) then $g(x)=cx$ for some constant $c$ and $f(x)=-e^{-x}+cx$.
The given limit exists if and only if $c=1$ and this gives $\int_0^{1} f(x)dx=\frac 1 e-\frac 1 2$.
| {
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Prove that $\left(1-\frac{1}{n+1}\right)\left(1-\frac{1}{n+2}\right)\cdot\ldots\cdot\left(1-\frac{1}{2n}\right)=\frac{1}{2}$ Prove the following equality:$$\left(1-\frac{1}{n+1}\right)\left(1-\frac{1}{n+2}\right)\cdot\ldots\cdot\left(1-\frac{1}{2n}\right)=\frac{1}{2}$$
$$\text{for all }\;n\in\mathbb{N}\;.$$
Could you ... | HINT
Notice that each denominator cancels the next numerator, whence we get
$\begin{align}
&\left(1 - \frac{1}{n+1}\right)\left(1 - \frac{1}{n+2}\right)\cdot\ldots\cdot\left(1 - \frac{1}{2n}\right) =\\
&=\left(\frac{n}{n+1}\right)\left(\frac{n+1}{n+2}\right)\cdot\ldots\cdot\left(\frac{2n-1}{2n}\right) = \frac{n}{2n}=\f... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Pythagoras Theorem Proof Is my logic correct below to prove the Pythagoras Theorem? Thanks.
Area Rectangle R
\begin{align*}
R &= WL\\
&=(2a+b)(2b+a)\\
&=4ab+2a^2+2b^2+ab\\
&=5ab+2a^2+2b^2
\end{align*}
Total Area Yellow Triangles T
\begin{align*}
T &= 10(\frac{ab}{2})\\
&=5ab
\end{align*}
Calculate Area $c^2$
\beg... | Yes. This is correct provided you have shown that all yellow triangles are congruent.
| {
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Matrix exponential of a skew-symmetric Toeplitz matrix From today's exam:
Given the following matrix,
$$B = \begin{pmatrix} 0 & -2/3 & 1/3\\ 2/3 & 0 & -2/3\\ -1/3 & 2/3 & 0\end{pmatrix}$$ Prove that $$\exp(aB) = I + \sin(a) B + (1 -\cos(a)) B^2$$
I have tried expanding the exponential
$$\exp(aB)=I+aB+\frac{a^2}{2}B^... | The characteristic polynomial of the given matrix is $x^3 +x$. Hence by Cayley-Hamilton theorem,
\begin{align*}
B^3+B&=0\\
B^3 &= -B\\
B^4 &= -B^2\\
B^5 &= B\\
\vdots\\
\exp(aB)&= I + aB+\frac{a^2B^2}{2!}+\frac{a^3B^3}{3!}+\cdots\\
\exp(aB) &= I + \left(a-\frac{a^3}{3!}+\cdots \right) B + \left( \frac{a^2}{2!}-\frac{a^... | {
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Showing that if $P_n(X)=\frac{(X+i)^{2n+1} - (X-i)^{2n+1}}{2i} = 0$, then x verify $(\frac{x+i}{x-i})^{2n+1} = 1$ I need to show that if $P_n(X)=\frac{(X+i)^{2n+1} - (X-i)^{2n+1}}{2i} = 0$, then x verify $(\frac{x+i}{x-i})^{2n+1} = 1$.
Here is what i have done:
$$P_n(X)=\frac{(X+i)^{2n+1} - (X-i)^{2n+1}}{2i} = 0 \\ (... | Please see the discussion here where it is mentioned that $z=x+i$, $z^\ast=x-i$ and $z^{2n+1}={z^{\ast}}^{2n+1}$. Another way to write this is $$\left(\frac{x+i}{x-i}\right)^{2n+1} = 1.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a^2+b^2\leq 1$ and $c^2+d^2<1$, then $a^2+b^2+c^2+d^2\leq 1+a^2c^2+b^2d^2+b^2c^2+a^2d^2$ If $a^2+b^2\leq 1$ and $c^2+d^2<1$, then $a^2+b^2+c^2+d^2\leq 1+a^2c^2+b^2d^2+b^2c^2+a^2d^2$
Is this true? This simple algebra should hold in order to finish my problem on Complex Analysis. Computing few numbers suggests that t... | Hint: If $x$, $y\le1$ then $$0\le(1-x)(1-y)=1-x-y+xy$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $a_0$ and $a_1$ when given a generating function. The recursive sequence $a_n$ has the following generating function
$$\begin{align}
&f(x)={\frac{x}{1-2x}+{\frac{4}{1+3x}}}
\end{align}$$
I am to find $a_0$ and $a_1$:
Rule: ${\frac{1}{1-a}}=1+a+a^2+a^3+...+a^n$
So we have
${\frac{1}{1-2x}}=1+2x+4x^2+8x^3+...+2^{n-1... | \begin{align}f(x) &= x+2x^2 + 4x^3+ \dots + 4 - 12 x + 36 x^2 - 108x^3 + \dots \\&= 4-11x+38x^2-104x^3+\dots\\&=a_0+a_1x+a_2x^2+a_3x^3+\dots\end{align}
where the last line is precisely the definition of a generating function. There is no need to substitute any values of $x$.
Alternatively we have $a_n = \dfrac {f^{(n)}... | {
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$a + b + c, \ ab + bc + ca, \ abc$ Prove a, b and c are real. I have found this problem in a book. If a, b and c are complex numbers, such that $|a| \neq |b| \neq |c| \neq |a|$ and $a + b + c, \ ab + bc + ca, \ abc$ are all real numbers, we must prove $a, \ b, \ c$ are also real numbers. We can get $(a + b + c) ^ 2 \in... | By Viète's relations, $a,b,c$ are the roots of
$$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$$
All the coefficients of this polynomial are real. If the polynomial has complex roots, there must be one such pair of them related by conjugation, but conjugate roots have the same magnitude, which is disallowed by the question. Thus the ... | {
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"source": "stackexchange",
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Powers of complex numbers : Which is the general formula? Let $z=\frac{1}{2}\left (\sqrt{3}+i\right )\in \mathbb{C}$.
I want to calcualte the powers $z^n$ for $n\in \mathbb{Z}$.
I have done the following: \begin{align*}&z^1=\frac{1}{2}\left (\sqrt{3}+i\right ) \\ &z^2=\left (\frac{1}{2}\left (\sqrt{3}+i\right )\right )... | It is easiest to express the general formula using the polar representation of $z = r e^{i\theta}$
$$
z^n = r^n e^{in\theta} = r^n\left(\cos(n\theta) + i \sin(n\theta)\right).
$$
See also Euler formula and De Moivre's formula.
| {
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"source": "stackexchange",
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A question of Number Theory+Binomial Theorem Let the sum of all divisors of the form $2^{p} \cdot 3^{q}$ (p,q are positive integers) of the number $19^{88} - 1$ be $\lambda$ . Find unit digit of $\lambda$
I noticed one fact that $2^{p} \cdot 3^{q}$ means any number which is not of form $6k\pm 1$.
Tried to expand it $(1... | We just need to find the largest factor of $19^{88}-1$ of the form $2^p3^q$. The rest shall follow from considering its factors.
Using the factorization $19^{88}-1 = (19^{44}+1)(19^{22}+1)(19^{11}+1)(19^{11}-1)$, we first check the value of $p$. Taking mod $4$:
$$19^{88}+1 \equiv (-1)^{88}+1 \equiv 2 \equiv 19^{44}-1 \... | {
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prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$ number theory prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$
attempt:
$$2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$$
we can rewrite the equation
$$(2^4)^{2^{6k}}\equiv 2^4\pmod{19}$$
How I can continue from the... | By Fermat's little theorem, we have
$$2^{18} \equiv 1 \pmod{19} \implies 2^{36} = (2^{4})^{9} \equiv 1 \pmod{19} \tag{1}\label{eq1A}$$
Since $2^6 = 64 \equiv 1 \pmod{9}$, we thus get $2^{6k} = (2^6)^k \equiv 1 \pmod{9}$, so there's an integer $j$ where $2^{6k} = 9j + 1$. Thus,
$$(2^4)^{2^{6k}} = (2^4)^{9j + 1} = ((2^4)... | {
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Limit of a cube root Given $\lim_{x \to +\infty} (\sqrt[3]{x^3+x^2+x+1}-\alpha x-\beta)=0$ for some $\alpha, \beta \in \mathbb{R}$. I have tried equating it as $\lim_{x \to +\infty}\alpha x+\beta=\lim_{x \to +\infty} (\sqrt[3]{x^3+x^2+x+1})$. Cubing both sides would give$$(\alpha x)^3+\beta^3+3\alpha^2x^2\beta+3\beta^2... | For large $x$, it is guaranteed that $\left|\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}\right| < 1$, and so we may use a Binomial expansion :
$$ \left(1 + \left( \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}\right)\right)^\frac13 = 1 + \frac13 \left( \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}\right) + O\left( \frac{1}{x... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$ Limit i want to solve: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
This is how i started solving this limit:
*
*$\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
*$\left(\frac {3x+x-x+2+1-1}{4x+3}\right)^x$
*$\left(\frac {4x+3}{... | Note that $$\lim_{u\to \infty}\left( 1-\frac{1}{u}\right)^{u}=\frac{1}{e}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3952733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Solving a 2-D SDE Need to solve the system of SDE
\begin{align*}
dX^1&=-X^2dt+dW^1\\
dX^2&=X^1dt+dW^2
\end{align*}
I wrote the first eq. in integral form as $$X_t^1=X_0^1-\int_0^tX^2(s)ds+W_t^1$$
Then I plugged it into the second eq. and took the integral form of that which turned out to be
$$X_t^2=X_0^2+X_0^1t-\int_0^... | Good question!
It may first be helpful to rewrite as
\begin{align*}
\begin{pmatrix} dX_{t}^{(1)} \\ dX_{t}^{(2)} \end{pmatrix}
=
\begin{pmatrix} 0 & - 1 \\ 1 & 0 \end{pmatrix}
\begin{pmatrix} X_{t}^{(1)} \\ X_{t}^{(2)} \end{pmatrix}\, dt
+ \begin{pmatrix} dW_{t}^{(1)} \\ dW_{t}^{(2)} \end{pmatrix}.
\end{align*}
This ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to integrate $\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$ The question given is to calculate
$$\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$$
My attempt
I managed to figure out that the denominator is given out as a pe... | Let $t=x-\frac\pi6$ to get
\begin{align}
& I= \int \frac{\cos x+ \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{d}x
=\int \frac{\frac{\sqrt3}2(2+\cos t)-\frac12\sin t }{(1+ 2 \cos t)^2}{d}t
\end{align}
Note that $\left( \frac{\sin t}{1+2\cos t}\right)’ = \frac{2+\cos t}{(1+2\cos t)^2}$ and $\left( \fra... | {
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"source": "stackexchange",
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Find the minimum value of the ratio $\frac{S_1}{S_2}$ using given data
3 points $O(0, 0) , P(a, a2 ) , Q(b, b2 )$ are on the parabola $y=x^2$. Let S1 be the area bounded by the line PQ and the parabola and let S2 be the area of the triangle OPQ, then find min of $\frac{S1}{S2}$
$$S_1 = \frac 12 (a+b)(a^2+b^2) -\int_{... | In general, if you have a fraction involving $a,b$ where each term in the numerator and denominator has the same degree $d$, you can divide through by say $b^d$ to get an expression in $\frac{a}{b}$. Here, this would look like $$\frac{S_1}{S_2} =\frac{ \frac 16 (a^3+b^3) +\frac 12 (ab^2+a^2b)}{\frac 12 (a^2b+ab^2)} =\... | {
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Calculate $X_A(x) $ and $m_A(x) $ of a matrix $A\in \mathbb{C}^{n\times n}:a_{ij}=i\cdot j$
*
*Caclulate the characteristic & the minimal polynomial of the matrix:
$$A\in\mathbb C^{n\times n}:a_{ij}=i\cdot j ,\forall i,j=1,..,n$$
$$\text{i.e for $n=3$, } $$
$$A=\left[\begin{matrix}1 & 2 & 3\\2 & 4 & 6\\3 & 6 & 9\en... | Let $\mathbf{v}:=(1,2,\ldots,n)$ and $\mathbf{w}_i:=(i,0,\ldots,0,-1,0,\ldots)$ for $i=2,\ldots,n$. These vectors are non-zero orthogonal and hence form a basis of $\mathbb{R}^n$. They are also eigenvectors of $A$ since $$A\mathbf{v}=|\mathbf{v}|^2\mathbf{v},\qquad A\mathbf{w}_i=\mathbf{0}$$ Note that $|\mathbf{v}|^2=\... | {
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"source": "stackexchange",
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Difference of equal derivative functions value $f_1(x) = \text{tan}^{-1}(x)$
$f_2(x) = \text{tan}^{-1}\big(\frac{x-1}{x+1}\big)$
Where
$f_1^\prime(x) = f_2^\prime(x)=\frac{1}{x^2+1}$
Therefore
$f_1(x) - f_2(x) = \theta$
How to find values of $\theta$ mathematically. I have solved it graphically and the answers are $\fr... | Hint:
Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
$$\arctan x+\arctan(-1)=\begin{cases} \arctan\frac{x-1}{1-(-1)x} &\mbox{if } x\cdot(-1)<1\\ \pi+\arctan\frac{x-1}{1-(-1)x} & \mbox{if } x\cdot(-1)>1\\\text{sign}... | {
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Given triangle ABC and its inscribed circle O, AO = 3, BO = 4, CO = 5, find the perimeter of ABC As title,
I could only figure out that
$$a^2 + r^2 = 9$$
$$b^2 + r^2 = 16$$
$$c^2 + r^2 = 25$$
$$r^2 = \frac{abc}{a+b+c}$$
But couldn't get how to derive $2(a+b+c)$.
| In a triangle $ABC$ with sides $a,b,c$ opposite to $A,B,C$, respectively, and with incenter $I$, one has $$AI^2 = bc\cdot \frac{b+c-a}{a+b+c}, \quad BI^2 = ca\cdot \frac{c+a-b}{a+b+c}, \quad CI^2 = ab\cdot \frac{a+b-c}{a+b+c}.$$
The conditions $AI=3$, $BI=4$ and $CI=5$ can be rewritten as
$$\lambda := \frac{2abc}{a+b+c... | {
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"source": "stackexchange",
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Is $x=2,y=13$ the unique solution? Problem:
Find all positive integers $x$ and $y$ satisfying:
$$12x^4-6x^2+1=y^2.$$
If $x=1, 12x^4-6x^2+1=12-6+1=7,$ which is not a perfect square.
If $x=2, 12x^4-6x^2+1=192-24+1=169=13^2$, which is a perfect square. Thus, $x=2,y=13$ is a solution to the given Diophantine equation.
Howe... | We try to get a perfect square term on the LHS.
Note that multiplying both sides by 4 keeps the right side as a square:
$$48x^4 - 24x^2 + 4 = (2y)^2$$
Now, completing the square yields: $$3(4x^2 -1)^2 +1 = (2y)^2$$
Try to now find the solutions to $3(2x-1)^2(2x+1)^2 = (2y-1)(2y+1).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 1
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Equation using $f(x)=\frac{\sin\pi x}{x^2} $ If $f(x)= \frac{\sin \pi x}{x^2}$, $x>0$
Let $x_1<x_2<x_3<\cdots<x_n<\cdots$ be all the points of local maximum of $f$.
Let $y_1<y_2<y_3<\cdots<y_n<\cdots$ be all the points of local minimum of $f$.
Then which of the options is(are) correct.
(A) $x_1<y_1$
(B) $x_{n+1}-x_n>2$... | Only the option $A)$ is correct.
$$x_1=\frac 12\;\;,\; y_1=\frac 32$$
Let $$f(x)=\frac{\sin(\pi x)}{x^2}$$
it is clear that
$$(\forall x>0)\;\; -\frac{1}{x^2}\le f(x)\le \frac{1}{x^2}$$
The local maximums are such that
$$f(x_i)=\frac{1}{x_i^2}$$
or
$$\sin(\pi x_i)=1$$
thus
$$x_i=\frac 12+2n$$
with $n\ge 0$.
So, $$x_i\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3964918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluate and explain $\int_{0}^{\frac{\pi}{2}}\arccos\left(\frac{\cos x}{\left(1+2\cos x\right)}\right)dx$ $$\int_{0}^{\frac{\pi}{2}}\arccos\left(\frac{\cos x}{1+2\cos x}\right)dx$$
I found this integral on a math discord server. I was unable to apply any standard integration technique to solve it (I am just an advance... | As @K.defaoite notes, this is Coxeter's integral, equal to $5\pi^2/24$. The usual way (see Inside Interesting Integrals Secs. 6.2 and 6.3, where this takes Nahin just over 11 pages) to evaluate it (whose details I've half-provided as they're worth working through to appreciate how hard this is!) is to prove it's four t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3966889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$a,b,c > 0$ and $a+b+c=1$. Prove that $\frac{1}{a+b^2}+\frac{1}{b+c^2}+\frac{1}{c+a^2}\geqslant \frac{13}{2(1-abc)} $ $a,b,c > 0$ and $a+b+c=1$. Prove that
$$\frac{1}{a+b^2}+\frac{1}{b+c^2}+\frac{1}{c+a^2}\geqslant \frac{13}{2(1-abc)} $$
This inequality can be solved by computer ( essentially, remove the constrain thro... | I would not call this an answer, but perhaps could be helpful. Observe that:
\begin{align*}
(1-a)(1-b)(1-c)&=\color{red}{1-a-b-c}+ab+ac+bc-abc\\
&=ab+ac+bc-abc \\
&=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}-abc \\
&=\frac{1-(a^2+b^2+c^2)}{2}-abc \\
&=1-abc-\frac{1+a^2+b^2+c^2}{2} \\
&=1-abc-\frac{a+b+c+a^2+b^2+c^2}{2}.
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Indefinite integral $\displaystyle \int \frac 1 {p \sin ax + q \cos ax + \sqrt {p^2 + q^2} }\mathrm{d}x$ Working on $\displaystyle \int \frac 1 {p \sin ax + q \cos ax + \sqrt {p^2 + q^2} }\mathrm{d}x$.
According to Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968) item $14.422$ it should work out ... | Integrate as follows
\begin{align}
& \int \frac {dx} {p \sin ax + q \cos ax + \sqrt {p^2 + q^2} }\\
=&\frac1{\sqrt {p^2 + q^2} } \int \frac {dx} {\cos(\tan^{-1}\frac pq -ax)+1 }\\
=&\frac1{\sqrt {p^2 + q^2} } \int \frac {dx} {\cos(\frac\pi2-ax -\tan^{-1}\frac qp )+1 }\\
=&\frac1{2\sqrt {p^2 + q^2} } \int \sec^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3968147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to calculate the second derivative of this example? I am trying to calculate the second derivative of the following equation:
$$f(x)=\frac{6}{(x^2+3)}$$
$$f'(x)=\frac{-12x}{(x^2+3)^2}$$
The correct answer given was $$f''(x) = \frac{36(x+1)(x-1)}{(x^2+3)^3}$$ but the second derivative that I have gotten was differen... | Keep going...
$$48x^2(x^2+3)^{-3}-12(x^2+3)^{-2}$$
$$=\frac{48x^2}{(x^2+3)^3}-\frac{12}{(x^2+3)^2}$$
$$=\frac{48x^2-12(x^2+3)}{(x^2+3)^3}$$
$$=\frac{36x^2-36}{(x^2+3)^3}$$
Factorise the numerator to match the given answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3969015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}\geqslant 3$ Let $a,b,c>0$ and $a^2+b^2+c^2=3$, prove
$$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}\geqslant 3$$
This inequality looks simple but I do not know how to solve it. The straightforward method is to bash the inequality wi... | If you are unable to solve this equation by other means, this may help
WolframAlpha has a solution
$$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}\ge 3
\implies(a,b,c)\\= (-3,3,3), (0,1,1), (1,0,1),(1,1,0), (1,1,1), (3,-3,3), (3,3,-3) $$
WolframAlpha has a solution
$$\frac{a^2+3b^2}{a+3b}+\frac{b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3970317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
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$3^{1234}$ can be written as $abcdef...qr$. What is the value of $q+r$? It was possible to find $3^{15} ≡ 7\pmod{100}$. Knowing that $7^{4k}≡1\pmod{100}, (3^{15})^{80}≡3^{1200}≡(7)^{80}≡1\pmod{100}$.
$(3^{15})(3^{15})3^{1200}≡1\cdot 7\cdot 7\pmod{100}, 3^{1230}\cdot 3^{4}≡49\cdot 81\pmod{100}≡69\pmod{100}$ to get $6+9=... | The Carmichael function value for $100$, $\lambda(100)=\text{lcm}(\lambda(25),\lambda(4)) = \text{lcm}(20,2) = 20$ gives us $1234 \equiv 14 \bmod \lambda(100) \implies 3^{1234} \equiv 3^{14} \bmod 100$.
Probably the easiest way to calculate $3^{14} \bmod 100$ is as $81^3 \cdot 9$ discarding higher digits, so $81^2 \equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3971549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
difference of recursive equations Lets have two recursive equations:
\begin{align}
f(0) &= 2 \\
f(n+1) &= 3 \cdot f(n) + 8 \cdot n \\ \\
g(0) &= -2 \\
g(n+1) &= 3 \cdot g(n) + 12
\end{align}
We want a explicit equation for f(x) - g (x).
I firstly tried to do in manually for first $n$ numbers
\begin{array}{|c|c|c|c|}... | I think you can just use an induction proof, since you already have an intuition of the result, it is much easier to check :
$$\text{Let : } H_n :"w_n=f(n)-g(n)=4-4n" $$
First, for $n=0$ :
$$w_0=4=4-4\times 0 $$
Hence, $H_0$ is true.
Let $n\in\mathbb{N} $, such that $H_n$ is true, let us show $H_{n+1}$
$$w_{n+1}=3w_n+8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Finding the derivative of $ \cos(\arcsin x)$ I study maths as a hobby. I am trying to find the derivative of $ \cos(\arcsin x)$
This is how I have been proceeding:
Let u = $\arcsin x$
Then $\sin u = x$
Differentiating:
\begin{align}
\cos u \frac{du}{dx} &= 1 \implies \frac{du}{dx} = \frac{1}{\cos u} \\[4pt]
\cos^2 u + ... | With your notation, $x = \sin u$, we have $$\frac{d}{dx}\left[\cos (\sin^{-1} x)\right] = -\sin (\sin^{-1} x) \cdot \frac{d}{dx} \left[ \sin^{-1} x \right] = -x \frac{d}{dx} \left[ \sin^{-1} x \right].$$ Since $$\frac{dx}{du} = \cos u,$$ we have $$\frac{du}{dx} = \frac{1}{\cos u} = \frac{1}{\sqrt{1 - \sin^2 u}} = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3973505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the volume of the set in $\mathbb{R}^3$ defined by the equation $x^2 + y^2 \leq z \leq 1 + x + y$. Let $D=\left\{\left(x,y,z\right) \in \mathbb{R}^3: x^2 + y^2 \leq z \leq 1 + x + y \right\}$. I tried to integrate with cartesian coordinates, so that
$$\iiint_Ddxdydz = \int_{x=\frac{1-\sqrt{5}}{2}}^{{\frac{1+\sqrt{... | Define $$x=r\cos(\theta)+\frac{1}{2}$$ $$y=r\sin(\theta)+\frac{1}{2}$$ $$z=z$$ The volume of your solid is the triple integral $$\int_0^{2\pi} \int_0^{\sqrt{3/2}} \int_{r^2+r\big(\sin(\theta)+\cos(\theta)\big)+\frac{1}{2}}^{r\big(\sin(\theta)+\cos(\theta)\big)+2}rdzdrd\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Since dividing $x=x^6$ by $x$ gives $1=x^5$, how can I get to $x=0$ as a root? this might sound like a stupid question,
bear with me it probably is.
I know the solutions for $x=x^6$ are 1 and 0.
Now, since $1 \cdot x = 1 \cdot x^6 $ and it follows $ 1 \cdot x = 1 \cdot x \cdot x^5$ and I divide both sides by $ 1 \cdot... | Don't divide. Factorise:
$x=x^6$
$x-x^6=0$
$x(1-x^5)=0$
Either $x=0$ or $1-x^5=0 \Rightarrow x=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3976354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Want to verify my approach
If repetition of digit is not allowed, then how many five digit numbers which are divisible by $3$ can be formed
using the digits?
My approach is that i made 3 sets of digits according to the remainder they give on dividing by $3$ -
$ A:(0,3,6,9)$ , $B:(1,4,7)$ , $C:(2,5,8)$, Now the follow... | The count can be simplified using probability as under:
Without zeroes, there will be $\binom9 5 = 126$ digit strings, and the same number with zeroes, at$\,\binom9 4 = 126$
$\frac 1 3$ of each, i.e. $42$ will be divisible by $3$, but since zeroes can only be allowed at $4$ of the $5$ digits, the multiplication factor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3980211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Transformation to complex spherical basis. According to this Wikipedia page about the complex spherical basis, it is stated that the spherical basis vectors are written in terms of the Cartesian basis vectors the following way:
$$
\left(
\begin{array}{c}
e_{+} \\
e_{-} \\
e_{0}
\end{array}
\right)
=
\left(
\begin{array... | Let $B_c$ denote the Cartesian basis as a column vector. Let $B_s$ be the spherical basis. Then we have $B_s=U B_c.$ If $v$ is a vector, let $v_c$ and $v_s$ be the coordinate representation in each basis. Then
$$
{B_c}^T v_c={B_s}^T v_s=(UB_c)^T v_s={B_c}^T U^T v_s.
$$
But because $B_c$ is a basis this gives $v_c=U^T v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3980394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove the polynomial $x^4+4 x^3+4 x^2-4 x+3$ is positive Given the following polynomial
$$
x^4+4 x^3+4 x^2-4 x+3
$$
I know it is positive, because I looked at the graphics
and I found with the help of Mathematica that the following form
$$
(x + a)^2 (x + b)^2 + c^2(x + d)^2 + e^2
$$
can represent the polynomial with t... | Here's a slightly more motivated way to find a sum-of-squares representation.
We should want the first term to be the square of a quadratic with leading coefficient $1$, and to cancel out the $x^3$ term we need it to be $(x^2+2x+a)^2$ for some real $a$. This will leave us to show that
$$-2ax^2-(4a+4)x+(3-a^2)\geq 0;$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3980896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Let $a$ and $n$ be integers with $n>0$.
Show that the additive order of $a$ modulo $n$ is $\frac{n}{\gcd(a,n)}$. Let $a$ and $n$ be integers with $n>0$.
Show that the additive order of $a$ modulo $n$ is $\frac{n}{\gcd(a,n)}$.
attempt:
Let $a$ and $n$ be integers with $n>0$.
Let $m=\frac{n}{\gcd(a,n)}$.
Then,
\begin{equ... | The additive order of $a$ modulo $n$ is the smallest positive integer $m$ such that $ma \equiv 0 \pmod{n}$.
Now, $ma = kn$ for some integer $k$.
Then, $m = \frac{kn}{a}$.
Let $\gcd(a,n) = d$. Then,
$m = k\frac{\frac{n}{d}}{\frac{a}{d}}$.
Since $\frac{a}{d}$ and $\frac{n}{d}$ are relative prime, then $\frac{a}{d} \mid k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3982080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$. If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$.
I found this in a Facebook group.
I start by doing the math in the LHS:
$8(1-x-y+xy-z+xz+yz-xyz) = 8(1-2+xy+xz+yz-xyz) = 8(xy+xz+yz-1-xyz)$.
Then we set $z=2-... | We let $a=1-x,b=1-y,c=1-z$ so that we have $a+b+c=1$ and $0 < a,b,c < 1$.
Our new inequality now becomes $$8abc \leq (1-a)(1-b)(1-c)\\ \iff 8 \leq \left(\dfrac{1}{a}-1\right)\left(\dfrac{1}{b}-1\right)\left(\dfrac{1}{c}-1\right) \\ \iff 8 \leq \left(\dfrac{b+c}{a}\right)\left(\dfrac{c+a}{b}\right)\left(\dfrac{a+b}{c}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3982410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Analytic Proof of Wilson's Theorem I'm trying to find a reference for the following (analytic) proof of Wilson's Theorem. I would appreciate it immensely if some of you knew and told me the person who first came up with it. Here is the proof:
We start by considering the Maclaurin series of the function $f(x)=\ln \left(... | An alternative viewpoint.
*
*Analytic part
In $\Bbb{Q}[[x]]$ we have
$$\frac1{1-x}=\prod_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{x^{nk}}{n^k k!}$$
*
*Algebraic part
Let $R=\Bbb{Z}_{(p)}[x,x^p/p]+x^{p+1}\Bbb{Q}[[x]]$ the subring of formal series such that $p$ doesn't divide the denominator of the coefficients of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Finding integers $s$ and $t$ such that $s+t=\frac{\alpha}{b}+b$ and $st=\frac{\beta}{b}+b$ for integers $\alpha$, $\beta$, $b$ In my research I need to find integers $s$ and $t$ with the following properties (respect to $\alpha, \beta, b$) :
\begin{align}
s+t=\frac{\alpha}{b}+b\\
st=\frac{\beta}{b}+b,
\end{align}
where... | $$s+t=\frac{\alpha}{b}+b \qquad st=\frac{\beta}{b}+b\\
\implies {α + b^2 = b (s + t) \qquad b^2 + β = b s t}\\
\implies α = -b^2 + b s + b t \qquad β = b s t - b^2\qquad b\ne0
$$
We can see infinite solutions in terms of $\quad b, s, t\quad$ but it appears $\quad \alpha,\beta\subset\mathbb{Z}.$
Example of $(b,s,t,\al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3987883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the Maximum Trigonometric polynomial coefficient $A_{k}$
Let $n,k$ be given positive integers and $n\ge k$. Let $A_i, i=1, 2, \cdots, n$ be given real numbers. If for all real numbers $x$ we have $$A_{1}\cos{x}+A_{2}\cos{(2x)}+\cdots+A_{n}\cos{(nx)}\le 1$$
Find the maximum value of $A_{k}$.
I don't know if this ... | I'll just do a subcase of the $n=3$ case.
$A_{k}=1$ is the maximum for k=2, 3.
First notice that letting $A_{k}=1$ and taking the other values of $A_{j}=0$ we have
$$cos(kx)\leq 1$$
which holds for all x. So each $A_{k}$ is at least 1. Now we will show that each $A_{k}$ is at most 1 for $k=2,3$.
Plugging in $x=0$ gives... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3988744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 3
} |
Prove that $x/(1 + \frac{2}{\pi} \cdot x) < \arctan(x)$ $\forall x > 0$ The first thing that I tried to do is to differentiate both functions and try to see if there establishes inequality that we want (considering that they are equal when $x = 0$). This attempt failed because firstly it really is true but after that w... | With the idea of @trancelocation: the map $x\mapsto x/(1 + a x)= \frac{1}{a} ( 1 - \frac{1}{1+a x})$ is increasing on $(0, \infty)$ ( for $a>0$), so the inequality is equivalent to
$$x < \frac{ \arctan x}{ 1 - \frac{\arctan x}{\pi/2}} \textrm{ or } \tan t < \frac{t}{1- \frac{t}{\pi/2}}$$
We have the product expansi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$ subject to $x+y=23$
Solve the system of equations:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$, $x+y=23$
The obvious way to solve it is by substituting for one variable. However I was looking for a more clever solution and went ahead and plotted two graphs.
The first graph looks pretty wei... | Don't rush to square equations to get rid of square roots. It's almost always more work and you end up missing opportunities for simplification. Instead eliminate square roots by substitution. Here, let $a\ge0$ and $b\ge0$ be the values of the two square roots:
$$
\begin{cases}
x+y=23\\
a+b=33\\
a^2 = x^2 + 12y\\
b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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$\lim\limits_{n\to\infty}n\big(\sum_{k=1}^n\frac{k^2}{n^3+kn}-\frac{1}{3}\big)$? calculate $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right).$$
I got it $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\lim\limits_{n\to\infty}\dfrac{1}{n}\sum\limits_{k=1}^n\dfra... | Although $\sum\frac{k^2}{n^3+kn}$ and $\sum\frac{k^2}{n^3}$ have the same limit, they differ to $O(1/n)$. So when they are multiplied by $n$, they give different limits
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3996425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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How do we integrate the function $\frac{b\sqrt{[1+(\theta-1)2b]^{2} - 4\theta(\theta-1)b^{2}}}{4(\theta-1)}$? I came across the problem of determining the following integral:
\begin{align*}
I(\theta) = \int_{0}^{1}\dfrac{b\sqrt{[1+(\theta-1)2b]^{2} - 4\theta(\theta-1)b^{2}}}{4(\theta-1)}\mathrm{d}b
\end{align*}
where $... | Rearranging
\begin{equation}
[1+(\theta-1)2b]^{2} - 4\theta(\theta-1)b^2=(1-\theta)(2b-1)^2+\theta
\end{equation}
the integral can be written as
\begin{align}
I(\theta) &= \frac{1}{4(\theta-1)}\int_{0}^{1}b\sqrt{[1+(\theta-1)2b]^{2} - 4\theta(\theta-1)b^{2}}\,db\\
&=\frac{1}{4(\theta-1)}\int_{0}^{1}b\sqrt{(1-\theta)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3998258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Does the following lower bound improve on $I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q-1)}$, where $q^k n^2$ is an odd perfect number? Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Define the abundancy index
$$I(x)=\frac{\sigma(x)}{x}$$
where $\si... | Suppose to the contrary that there exists an integer $k \geq 1$ and a (special) prime $q \geq 5$ such that
$$\frac{q}{(q-1)(q^{k+1}+1)} \geq \frac{2(q-1)}{q^{k+2}}.$$
This inequality is equivalent to
$$q^{k+3} \geq 2(q-1)^2 (q^{k+1}+1) = 2q^{k+1} - 4q^{k+2} + 2q^{k+3} + 2q^2 - 4q + 2,$$
which in turn is equivalent to
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3998982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that $0<\sum_{i=2}^n\ (-1)^n \frac{x^{2n}}{2n!}$ for $x>0$ So far, I've tried using induction:
Let P(n): $0<\sum_{i=2}^n\ (-1)^i \frac{x^{2i}}{2i!}$
Let $n=2$. Thus, P(2): $\frac{x^4}{4!}>0$ for $x>0$. Then, let P(k) be true (Induction Hypothesis). Now, prove the validity of P for n=k+1:
$P(k+1) : \sum_{i=2}^{k+1... | $$S_n=\sum_{i=2}^n\ (-1)^n \frac{x^{2n}}{2n!}=\sum_{i=0}^n\ (-1)^n \frac{x^{2n}}{2n!}-\sum_{i=0}^2\ (-1)^n \frac{x^{2n}}{2n!}$$
$$S_n=\frac{x^2-1}2+\sum_{i=0}^n\ (-1)^n \frac{x^{2n}}{2n!}$$ If you are aware of the complete and incomplete gamma function
$$\sum_{i=0}^n\ (-1)^n \frac{x^{2n}}{2n!}=\frac{e^{-x^2}\,\, \Gamma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3999923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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(Proof) Equality of the distances of any point $P(x, y)$ on the isosceles hyperbola to the foci and center of the hyperbola I searched but couldn't find the proof.
Isosceles hyperbola equation:
$${H:x^{2}-y^{2} = a^{2}}$$
And let's take any point $P(x, y)$ on this hyperbola. Now, the product of the distances of this po... | $$PS_1.PS_2=\sqrt{[(x-ae)^2+y^2][(x+ae)^2+y^2]}$$
$$=\sqrt{(x^2-a^2e^2)^2+y^2(x^2+a^2e^2+2aex)+y^2(x^2+a^2e^2-2aex)+y^4} $$
$$=\sqrt{x^4+a^4e^4-2a^2e^2x^2+y^2x^2+y^2a^2e^2+2aexy^2+y^2x^2+y^2a^2e^2-2aexy^2}$$
$$=\sqrt{x^2+2x^2y^2+y^4+a^4e^4-2a^2e^2(x^2-y^2)}$$
$$=\sqrt{(x^2+y^2)^2+a^4e^4-2a^4e^2}$$
Since $e=\sqrt{2}$ fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Proving $\sum_{cyc}\frac{\sqrt{s(s-a)}}{a}\geq\frac{3\sqrt{3}}2$ for $a$, $b$, $c$, $s$ the sides and semi-perimeter of a triangle
If $a, b, c$ are the lengths of the sides of a triangle and s is the semiperimeter, prove that:
$$\sum_{cyc}\frac{\sqrt{s(s-a)}}{a}\geq \frac{3\sqrt{3}}{2}$$
My attempt: $$\sum_{cyc} cos ... | After using Ravi's substituition $a=x+y,b=y+z,c=x+z$ as nguyenhuyen_ag did we have to prove $$\sum \frac{\sqrt{z(x+y+z)}}{y+x} \geq \frac{3\sqrt 3}{2}.$$ Now we will use the method of Isolated fudging. We guess that we might be able to prove the follwing inequality $$\frac{\sqrt{z(x+y+z)}}{y+x}\ge \frac{3\sqrt{3}}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4005064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Three unit vectors whose sum is zero Let $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ be unit vector such that $\overrightarrow a + \overrightarrow b + \overrightarrow c = 0$. Which of the following is correct ?
(A) $\overrightarrow a \times \overrightarrow b = \overrightarrow b \times \overrightarr... | (A) would imply the vectors are pairwise parallel or antiparallel, so their sum is of length $1$ or $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Calculating the improper integral $\int _{1}^{\infty}\frac{e^{\sin x}\cos x}{x}\,dx$ I want to calculate the value of $$\int_{1}^{\infty}\frac{e^{\sin x}\cos x}{x}\,dx$$
I was able to prove using Dirichlet's test that it does converge, but how can I calculate its value?
| We have that, integrating by parts
$$
\eqalign{
& I = \int_1^\infty {e^{\,\sin x} {{\cos x} \over x}dx}
= \int_1^\infty {{1 \over x}d\left( {e^{\,\sin x} } \right)} = \cr
& = \left. {{{e^{\,\sin x} } \over x}} \right|_1^\infty
- \int_1^\infty {e^{\,\sin x} d\left( {{1 \over x}} \right)}
= - e^{\,\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4007090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
doubt about the calculation of the rest for a taylor's series I'm trying to calculate some numbers with the help of series.
My problem is the calculation of the rest of the series.
For example I have to calculate $log 2$.
I can consider the function $$log (\frac{1+x}{1-x})$$ in which I can put $x= \frac{1}{3}$.
The ser... | For $k \ge n+1$, $\vert x \vert^{2k+1} = \vert x \vert^{2n+3} x^{2(k-n-1)}$ and $\frac{1}{2k+1} \le \frac{1}{2n+3}$. Hence
$$2\sum_{k=n+1}^ \infty \frac{x^{2k+1}}{2k+1} \le \frac{2 |x|^{2n+3}}{2n+3}\sum_{h=0}^ \infty x^{2h}.$$
Finally $\sum_{h=0}^ \infty x^{2h}$ is the sum of the terms of a geometric sequence and there... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4007271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions over $\mathbb{R}$ The problem goes as follows:
Using elementary methods prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions for $x \in \mathbb{R}$.
I first came across the problem in a school book targeted towards students just expo... | Define the real polynomial
$$ p(x)\!:=\!x^8\! + \!x^7\! + \!x^6\! - \!5x^5\!-
\!5x^4\!- \!5x^3\! + \!7x^2\! + \!7x\!+\!7. \tag{1} $$
By observation, the polynomial easily factors as the product
$$ p(x) = (x^2+x+1)(x^6-5x^3+7) = p_1(x)p_2(x^3). \tag{2} $$
If we can prove for all real $\,x\,$ that $\,p(x)>1\,$ then
also... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4007456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Inequality $a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0$ Let $a,b,c$ be the lengths of the sides of a triangle. Prove that:
$$a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0.$$
Now, I am supposed to solve this inequality by applying only the Rearrangement Inequality.
My Attempt:
W.L.O.G. let $a \geq b \geq c.$
Then, L.H.S. = $a ... | For $a,b,c$ sides of a triangle, let $$\begin{matrix}a=y+z\\b=z+x\\c=x+y\end{matrix}$$
Then the inequality becomes $$\frac{y^2}{z}+\frac{x^2}{y}+\frac{z^2}{x}\ge x+y+z$$ This is true by Cauchy's inequality: $$x+y+z\le\sqrt{\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}}\sqrt{y+z+x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Prove by induction that $(n+1)(n+2)(n+3)(n+4)(n+5)$ is divisible by $120$.
Prove by induction that $$(n+1)(n+2)(n+3)(n+4)(n+5)$$ is divisible by $120$.
I tried to solve it, but could do so only till the inductive step.
I assumed: $$p(k)=(k+1)(k+2)(k+3)(k+4)(k+5), and $$ $$p(k+1)=(k+2)(k+3)...(k+6)$$
Then, I distribut... | $120=2\cdot 3\cdot 4\cdot 5$. $\Rightarrow 120|1\cdot 2\cdot 3\cdot 4\cdot 5$ and $120|2\cdot 3\cdot 4\cdot 5\cdot 6$.
Now, if $120|k(k+1)(k+2)(k+3)(k+4)$ and $120|(k+1)(k+2)(k+3)(k+4)(k+5)$, then,
$120|(k+1)(k+2)(k+3)(k+4)\{(k+5)-k\}$
$\Rightarrow 24|(k+1)(k+2)(k+3)(k+4)$
Now, use normal induction and proceed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4009762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to $2^n/n! \leq 4/n$ by induction I'm trying to prove that $\frac{2^n}{n!} \leq \frac{4}{n}$ for all positive integers $n$.
It's obvious this holds for $n=1$.
My inductive hypothesis is that $\frac{2^n}{n!} \leq \frac{4}{n}$.
I want to show that $\frac{2^{n+1}}{(n+1)!} \leq \frac{4}{(n+1)}$.
I can see that $\frac{2... | $\frac{4\cdot 2}{n(n+1)} = \frac{2}{n}\frac{4}{n+1}\leq \frac{4}{n+1}$, since $\frac{2}{n}\leq 1$ for $n\geq 2$.
In the case $n=1$, $2/1 \leq 4/1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving a positive integer Suppose that a positive integer $n$ ends with $k$ $9\text{'s}$. How can we prove that the integer $3n^4+4n^3$ ends with at least $2k$ $9\text{'s}$?
| Firstly note that saying a number ends with $k$ $9$s is equivalent to saying that it is congruent $-1\ (\text{mod }10^k)$, as for example $99$ ends with two $9$s and $99\equiv -1\ (\text{mod }100)$.
So write $n=A\cdot 10^k -1$, for an integer $A$. Then note:
\begin{align}
n^3&\equiv 3A\cdot10^k-1\ \quad&(\text{mod }10^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Positive integer solutions of equations of the form $ x^4+bx^2y^2+dy^4=z^2$ A method used to answer the post Does the equation $y^2=3x^4-3x^2+1$ have an elementary solution? showed that the only positive integer solutions of the equations $$ x^4-3x^2y^2+3y^4=z^2$$ $$X^4+6X^2Y^2-3Y^4=Z^2,$$ are $(1,1,1)$ and $(1,1,2)$,... | SOME PROGRESS
Let the loop of solutions have period $2n$. We can suppose that successive values taken by $x$ and $X$ are $a_1,a_2,...a_n$ and $A_1,A_2,...A_n$, respectively, where for each $i$ there are integers $c_i$ and $C_i$ such that $d=a_i^2c_i$ and $D=A_i^2C_i$.
Then the conditions given in the post simplify to:-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove that $\angle AEF =90^\circ$ given a square $ABCD$ Let $ABCD$ be a square and $E\in CD$ such that $DE=EC$. Let $F\in BC$ such that $BC=4FC$. Prove that $\angle AEF =90^\circ$.
My attempt:
Proving that $\angle AEF =90^\circ$ is the same as proving that $\triangle AEF$ is a right triangle. In other words, we wish t... | Let $AB = BC = CD = DA = 4x$.
Then: $DE = EC = 2x$, $CF = x$, and $FB = 3x$.
Now,
*
*In triangle $ABF$, we have $AF^2 = AB^2 + FB^2 = (4x)^2 + (3x)^2 \Rightarrow AF = 5x$.
*In triangle $ADE$, we have $AE^2 = DA^2 + ED^2 = (4x)^2 + (2x)^2 \Rightarrow AE = 2\sqrt{5}x$.
*In triangle $ECF$, we have $EF^2 = EC^2 + CF^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
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These functions are very similar. Is there an explanation? Blue: $x \ln x-x$
Brown: $\ln \frac{\Gamma(x+1/2)}{\sqrt{\pi}}$
| Using Sterling’s approximation for the Gamma function, $$\Gamma(z)\sim \sqrt{\frac{2\pi}{z}}\left(\frac{z}{e}\right)^z$$
In this case \begin{align*}
\Gamma\left(x+\frac{1}{2}\right)&\sim \sqrt{\frac{2\pi}{x+\frac{1}{2}}}\left(\frac{x+\frac{1}{2}}{e}\right)^{x+\frac{1}{2}}\\
\ln\Gamma\left(x+\frac{1}{2}\right)&\sim \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solving recurrence with substitution I am trying to solve this recurrence $T(n) = 4T(n − 2) + 2^{2n}$ by using substitution and knowing $T(1)=1,T(2)=2$ and here is my attempt:Expanding $T(n)=4T(n-2)+2^{2n}$ using $T(n-2)=4T(n-4)+2^{2(n-2)}$, we get $T(n)=4(4T(n-4)+2^{2(n-2)})+2^{2n}=4^2T(n-4)+2^{2n-2}+2^{2n}=4^2(4T(n-6... | This is a linear difference equation. You can get the solution as follows:
*
*Compute the general solution of the homogenous equation $T_n = 4 T_{n-2}$, which is $H_n = c_1 \cdot 2^n +c_2 \cdot (-2)^n$.
*Obtain a particular solution of the complete equation by trying something similar to the RHS, i.e. $P_n = k \cdo... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the volume of a solid through 2$\pi$ radians With reference to this question
Curve $C$ with the equation:
$$y = (2x-1)^{\frac{3}{4}}, \quad x \ge\frac{1}{2}$$
The finite region $S$, is bounded by the curve $C$, the $x$-axis. The $y$-axis and the line $y = 8$. This region is rotated through $2\pi$ radians about ... | Curve $C$ is given by $y = (2x-1)^{\frac{3}{4}}$ for $x \ge\frac{1}{2}$.
Note that the region S, is bounded by the curve C, the x-axis, the y-axis and the line $y = 8$. This region is rotated through $2\pi$ about the x-axis.
a) If we find the volume using shell method, it is one integral.
$\displaystyle y = (2x-1)^{\fr... | {
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"url": "https://math.stackexchange.com/questions/4031794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the zeros of the polynomial $P(z)=z^7+7z^4+4z+1$ lie inside the disk of radius $2$ centered at the origin.
Prove that the zeros of the polynomial $P(z)=z^7+7z^4+4z+1$ lie inside the disk of radius $2$ centered at the origin.
Assuming the contrary that there exists an $|z|\geqslant2$ such that $P(z)=0$ I ha... | If there is a solution, $|z^7|=|7z^4+4Z+1|$ Suppose that $|z|\geq 2$,
$2^7=128>7 \times 2^4 +4 \times 2+1=127$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4032196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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calculate the limit $\lim\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$ I'm working on finding the limit for this equation, and would kindly welcome your support to my solution:
$$\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$$
These are my steps in hopefully deriving the correct re... | You are almost done!
$$\frac{1-\cos \theta}{\theta \sin \theta} = \frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin\theta}=\frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin \theta}=$$
$$=\frac{\sin\theta}{\theta}\cdot\frac{1}{1+\cos\theta} \to 1\cdot \frac{1}{1+1} = \frac{1}{2}$$
Note (and prove) that indeed
$$\lim_{\theta \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4034360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A travelled inequality found by discriminant
Given three real numbers $x, y, z$ so that $1\leq x, y, z\leq 8.$ Prove that
$$\sum\limits_{cyc}\frac{x}{y}\geq\sum\limits_{cyc}\frac{2x}{y+ z}$$
I found this inequality by discriminant, we realised the homogenous, the generality of this problem. We assume $y= 1$ that lead... | Eliminating denominators, we arrive to the inequality $\;f(x,y,z)\ge 0,\;$ where
$$\begin{align}
&f(x,y,z) = x^3 y^3 + y^3 z^3 + x^3 z^3 - 3 x^2 y^2 z^2
\\[4pt]&
+ x^2 y^4 + y^2 z^4 + z^2 x^4 - x^4 y z - x y^4 z - x y z^4.
\end{align}\tag1$$
Since $(1)$ is homogenius and has the rotational symmetry, then WLOG there ar... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Faster way to calculate $\frac{\sqrt8+\sqrt{27}}{5-\sqrt6}-2(\sqrt[4]9-1)^{-1}$ What is the value of $\cfrac{\sqrt8+\sqrt{27}}{5-\sqrt6}-2(\sqrt[4]9-1)^{-1}$ ?
$$1)1+\sqrt3\quad\quad\quad\quad\quad\quad2)-1+\sqrt2\quad\quad\quad\quad\quad\quad3)1-\sqrt2\quad\quad\quad\quad\quad\quad4)\sqrt2-2\sqrt3$$
It was one of the... | $$\begin{align}
\frac{\sqrt{8}+\sqrt{27}}{5-\sqrt{6}}
&= \frac{(\sqrt{8} + \sqrt{27})(5 + \sqrt{6})}{25-6} \\
&= \frac{5 \sqrt{8} + \sqrt{48} + 5\sqrt{27} + \sqrt{81(2)}}{19} \\
&= \frac{10\sqrt{2} + 4\sqrt{3} + 15\sqrt{3} + 9\sqrt{2}}{19} \\
&= \sqrt{2} + \sqrt{3}.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4037499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Comparison test proof of convergence I want to prove that $\sum_{n=1}^\infty\frac{3^n+4^n}{3^n+5^n}$ converges. To do this, I said that $\sum_{n=1}^\infty\frac{3^n+4^n}{3^n+5^n}<3\sum_{n=1}^\infty (\frac{4}{5})^n$, which indeed converges. Hence by the comparison test, the series we are given converges. Is this valid?
| Hint:
$$\frac{3^n+4^n}{3^n+5^n} \le \frac{3^n+4^n}{5^n} = \left(\frac 35 \right)^n + \left(\frac 45 \right)^n \le 2\left(\frac 45 \right)^n $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4038446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the probability of placing $5$ dots on an $8 \times 8$ grid. $5$ dots are placed at random on an $8 \times 8$ grid such that no cell has more than $1$ dot.
What is the probability that no row or column has more than $1$ dot?
I thought about this in the following way, the number of ways to place $5$ dots on $64$ ce... | ${64 \choose 5}$ looks OK for the denominator
For distinct rows, there are ${8 \choose 5}$ ways of choosing the five rows.
*
*The dot in the first occupied row can be in any of the $8$ columns
*The dot in the second occupied row can be in any of the $7$ remaining columns
*The dot in the third occupied row can be in... | {
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Find the minimum of $\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$ Find the minimum of $$f(x,y)=\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$$ It seems that $f_x=f_y=0$ is very hard to compute. Is there any easier idea?
| We have
\begin{align}
f(x, y) &= \sqrt{4y^{2}- 12y+ 10}+ \sqrt{18x^{2}- 18x+ 5} \\
&\qquad + \sqrt{18x^{2}+ 4y^{2}- 12xy+ 6x- 4y+ 1} \\[6pt]
&= \sqrt{\frac{ ( 2y- 6 )^{2} + ( 6y- 8 )^{2}}{10}} + \sqrt{\frac{( 6x- 5 )^{2}+ ( 12x- 5 )^{2}}{10}} \\[6pt]
&\qquad + \sqrt{\frac{ ( 6x+ 2y- 1 )^{2}+ ( 12x- 6y+ 3 )^{2}}{10}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 1
} |
Find the volume of the solid in the first octant bounded by the three surfaces $z = 1-y^2$, $y=2x$, and $x=3$ I want to find the volume of the solid in the first octant bounded by the three surfaces $z = 1-y^2$, $y=2x$, and $x=3$. It seems that would simply be to calculate the following triple integral:
$\int_0^3 \int_... | I ended up doing the following (omitting some algebra steps):
$1 - y^2 \geq 0$ when $y \leq 1$ which gives an integration interval $0 \leq y \leq 1$. By dividing the integral into two parts with the intervals $0 \leq x \leq \frac{1}{2}$ and $\frac{1}{2} \leq x \leq 3$ we get the two simpler integrals:
$V =
V_1 + V_2 =
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4044263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 12$ if $ab+bc+ca=3$ Let $a,b,c$ be real positive number, $ab+bc+ca=3$.
Prove that $$P=(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 12$$
My attempt: $$P=(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 3\sqrt[3]{(a+b)(a+bc)(b+c)(b+ca)(c+a)(c+ab)}$$
$$P \ge 3\sqrt[3]{\frac{8}{9}\cdo... | Your last inequlaity is equivalent to
$$
a^2b^2c^2+abc(a^2+b^2+c^2)+a^2b^2+b^2c^2+c^2a^2+abc\ge 8.
$$
Since
$$
a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=(a+b+c)^2-6~\text{and}~
\\
a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)=9-2abc(a+b+c),
$$
the previous inequlaity is equivalent to
$$
a^2b^2c^2+abc(a+b+c)^2-6abc+9-2abc(a+b+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4052142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculating the following integral $\int^{\infty}_{2}\frac{x-1}{x^3+4x+7}$ I'm trying to find the solution of the following integral, to check for convergence or divergence. Though, I'm finding the integral itself difficult to resolve. I welcome any feedback towards my approach.
$$\int^{\infty}_{2}\frac{x-1}{x^3+4x+7}$... | Write $x^3+4x+7=(x-a)((x-b)^2+c^2)$ with $a\approx-1.25538,\,b\approx0.627692,\,c\approx2.2764$. Then$$\begin{align}\frac{x-1}{x^3+4x+7}&\equiv\frac{A}{x-a}+\frac{B(x-b)+C}{(x-b)^2+c^2}\\\iff x-b+b-1&\equiv A((x-b)^2+c^2)+(x-b+b-a)(B(x-b)+C)\\&=(A+B)(x-b)^2+((b-a)B+C)(x-b)+Ac^2+(b-a)C\\\iff A=-B&=\frac{a-1}{(b-a)^2},\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4053864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$ Prove that $\forall x,y\in \mathbb{R}$ the inequality $x^2+y^2+1 \geq xy+x+y$ holds.
Attempt
First attempt: I was trying see the geometric meaning, but I´m fall.
Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$
and then compare ... | We can rewrite $x^2+y^2+1 \geq xy+x+y \ $ as
$2x^2+2y^2+2 - 2xy - 2x - 2y \geq 0$
or as $(x-y)^2 + (x-1)^2 + (y-1)^2 \geq 0$
which holds for all $x, y \in \mathbb{R}$
Or start from $(x-y)^2 + (x-1)^2 + (y-1)^2 \geq 0$ and expand to show that $x^2+y^2+1 \geq xy+x+y \ $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 0
} |
If $x\geq 0,$ what is the smallest value of the function $f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$ If $x\geq 0,$ what is the smallest value of the function
$$f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$$
I tried doing it by completing the square in numerator and making it of the form
$$\frac{4(x+ 1)^2+ 9}{6(1+ x)}$$
and then, I pu... | Let $x+1=y^2$ as $x\ge0, y^2\ge1$
$$\dfrac{4x^2+8x+13}{1+x}=\dfrac{4(y^2-1)^2+8(y^2-1)+13}{y^2}=4y^2+\dfrac9{y^2}=\left(2y-\dfrac3y\right)^2+2\cdot2y\cdot\dfrac3y\ge12$$ the equality occurs if $2y-\dfrac3y=0\iff y^2=\dfrac32\iff x=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $2^n > n^2 + n + 1$ I did this but I'm pretty sure it's wrong
$P(5) $ is true
Assume $P(k)$ is true
for some $k \ge 5$, ie. $2^k > k^2 + k + 1$
$$2^{k+1} = (k + 1)^2 + (k + 1) + 1$$
$$2^{k+1} = k^2 + 3k + 3$$
Then:
$$2^{k+1} > (k^2 + k + 1) \cdot 2 > k^2 + 3k + 3$$
$$2^{k+1} > k^2 > k + 1$$
From the onset we know... | You started well but the end of your proof is not right and it is unclear when you write 2 inequalities.
We want to prove $2^{k+1}>(k+1)^2+(k+1)+1$ knowing already that $2^k>k^2+k+1$.
$2^{k+1}=2\times2^k>2\times(k^2+k+1)=2k^2+2k+2$ but we wanted to prove $2^{k+1}>(k+1)^2+(k+1)+1=k^2+3k+3$ so it is enough to prove that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4060214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Long Division of polynomials How can we prove that $x^n$ isn't divisible by $g(x)=x^4+x+1$ without remainders?
I understand why for all $n<4$ it's working, but how with $n \ge 4$? How we can prove it formally?
Note: $x^n,\:g\left(x\right)\in \mathbb{Z}_2\left[x\right]$
| Write $x^n = q_n(x) (x^4+x+1) + r_n(x)$.
Then $$x^{n+4} = x^4 q_n(x) (x^4+x+1) + x ^4 r_n(x) = x^4 q_n(x) (x^4+x+1) + (x+1) r_n(x)$$ and so $r_{n+4} \equiv (x+1) r_n \bmod (x^4+x+1)$.
The result follows by induction because $r_n = x^n \ne 0 $ for $n=0,1,2,3$ and $x+1$ is a unit mod $(x^4+x+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4065341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
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