Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Factorize $x^4 + y^4 -x^2 y^2$ over $\mathbb C$ As in the title, I'd like to factorize $x^4 + y^4 - x^2y^2$ into irreducible factors over $\mathbb C$ (i.e. linear factors).
Attempts:
First I tried doing
$$x^4 + y^4 - x^2y^2 = (x^2 - y^2)^2 + x^2y^2 \\ = (x^2 - y^2) - (ixy)^2 \\ = (x^2 - y^2 - ixy)(x^2 - y^2 + ixy) $$
A... | $$x^4-x^2y^2+y^4=(x^2+y^2)^2-3x^2y^2=(x^2-\sqrt3xy+y^2)(x^2+\sqrt3xy+y^2)=$$
$$=\left(\left(x-\frac{\sqrt3}{2}y\right)^2+\frac{y^2}{4}\right)\left(\left(x+\frac{\sqrt3}{2}y\right)^2+\frac{y^2}{4}\right)=$$
$$=\left(x-\frac{\sqrt3}{2}y+\frac{1}{2}yi\right)\left(x-\frac{\sqrt3}{2}y-\frac{1}{2}yi\right)\left(x+\frac{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Inverse of $\frac{x}{1-x^2}$ How did Munkres (page 104 – new international edition) invert the function
$$
F(x) = \frac{x}{1-x^2} \,, \quad x \in (-1,1)
$$
for
$$
G(y) = \frac{2y}{1 + \sqrt{1+4y^2}} \,, \quad y \in \mathbb{R} \,?
$$
Looks like some trig-substitution but cannot see it.
| The usual method to obtain the inverse is to let $F(x)=y$, interchange $x$ and $y$, and solve for $y$. Thus we solve
$$x=\frac{y}{1-y^2}$$
for $y$:
\begin{align*}
x&=\frac{y}{1-y^2}\\
\implies x-xy^2&=y\\
\implies xy^2+y-x&=0\\
\implies y^2+\tfrac1xy-1&=0\\
\implies(y+\tfrac1{2x})^2&=\frac{1+4x^2}{4x^2}\qquad\text{(co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
value of $2\tan^{-1}(\csc \alpha)+\tan^{-1}(2\sin \alpha\sec^2\alpha)$
If $ x^3+bx^2+cx+1=0$ has only real root $\alpha $.
Where $(b<c)$. Then $\displaystyle 2\tan^{-1}(\csc \alpha)+\tan^{-1}(2\sin \alpha\sec^2\alpha)$ is
Plan
$$\tan^{-1}\bigg(\frac{2\csc \alpha}{1-\csc^2\alpha}\bigg)+\tan^{-1}\bigg(2\sin \alpha\sec^... | Let $f(x)=x^3+bx^2+cx+1$
$f(0)=1>0$
$f(-1)=(b-c)<0$
So, $\alpha$ lies between $f(0)$ and $f(-1)$ which is $-1,0$
$2\tan^{-1}(\csc\alpha)+\tan^{-1}(2\sin\alpha\sec^2\alpha)$
$2\tan^{-1}\alpha\left(\dfrac{1}{\sin\alpha}\right)+\tan^{-1}\left(\dfrac{2\sin\alpha}{\cos^2\alpha}\right)$
$2\tan^{-1}\alpha\left(\dfrac{1}{\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3259906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $a,b,c>0$ and $ab+bc+ca=3$, prove that $\sum_{cyc} \frac{a}{\sqrt{a^3+5}} \leq \sqrt{6}/2$
If $a,b,c>0$ and $ab+bc+ca=3$, prove that $\displaystyle \sum_{cyc} \frac{a}{\sqrt{a^3+5}} \leq \sqrt{6}/2$.
My attempt was to use firstly AM-GM in the denominator, like $a^3+5 \geq 3a+3$ and the set $a=x^2-1$ but a lot of t... | Another AM-GM helps:
$$\sum_{cyc}\frac{a}{\sqrt{a^3+5}}\leq\sum_{cyc}\frac{a}{\sqrt{\frac{3a^2-1}{2}+5}}=\sqrt{\frac{2}{3}}\sum_{cyc}\frac{a}{\sqrt{a^2+3}}=$$
$$=\sqrt{\frac{2}{3}}\sum_{cyc}\frac{a}{\sqrt{(a+b)(a+c)}}\leq\frac{1}{2}\sqrt{\frac{2}{3}}\sum_{cyc}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{\sqrt6}{2}.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3260825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Formation of commissions with generating functions Representatives of three research institutes should form a commission of 9 researchers. How many ways can this committee be formed such that no institute should have an absolute majority in the group?
My partial solution:
I will use the exponential generating function ... | You should not use an exponential generating function here. You would use this if your were choosing an ordered lists of nine researchers. Here, we are just choosing how many representative appear from each institution, so ordinary generating functions are applicable.
The number of representatives can be anywhere betw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3261743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrating $\int^2_0 xe^{x^2}dx$ Well what I was thinking was to integrate the indefinite integral first.
$u=x^2$, $x=\sqrt u$
$du=2xdx = 2\sqrt {u} dx$
$dx= \frac{1}{2\sqrt{u}}du$
$\int xe^{x^2} dx = \int \sqrt{u}\frac{1}{2\sqrt{u}} du =\frac{1}{2}\int e^u du = \frac{1}{2}e^u =\frac{1}{2}e^{x^2} +C$
Now I can evalu... | You mean $\frac12 e^4-\frac12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
quadratic equation solving mistake I'm a student who started self learning quadratic equations for a youth university program. I'm busy at trying to solve such equation:
$$
(1 - 4x)^2 + 9x + 7 = 2(x+3)(1-x) + (x+4)^2
$$
this is my current progress:
\begin{align}
(1 - 4x)^2 + 9x + 7 &= 2(x+3)(1-x)+ (x+4)^2\\
(1 - 4x)(1 ... | Your working is all fine. Only the last step was remaining to be written down after factoring.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3269580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Circle and Locus _ ONLY PEN AND PAPER ALLOWED. Q) Let T be the line passing through the points P(–2, 7) and Q(2, –5). Let $F_{1}$ be the set of all pairs of circles $(S_{1}$, $S_{2}$) such that T is tangent to $S_{1}$ at P and tangent to $S_{2}$ at Q, and also such that $S_{1}$ and $S_{2}$ touch each other at a unique ... | As the diameter of the first is $PQ$ $$E_1=\{(x,y)| (x+2)(x-2)+(y-7)(y+5)=0\}-\{P,Q\}$$
and the diameter of the second is $DR$, with $D$ the midpoint of $PQ$
$$E_2=\{(x,y)| x(x-1)+(y-1)^2=0\}-\{(4/5,7/5),(36/37,43/37)\}$$
where the disallowed points stem from the intersections with $l_{PR}$ and $l_{QR}$.
Now
$$\#((V((x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $\lim\limits_{n\to \infty}\frac{1}{\sqrt n}\left|\sum\limits_{k=1}^n (-1)^k\sqrt k\right|= \frac{1}{2}$ I'm trying to show that $$\lim_{n\to \infty} x_n=\lim_{n\to \infty}\frac{1}{\sqrt n}\left|\sum_{k=1}^n (-1)^k\sqrt k\right|= \frac{1}{2}.$$
Assuming $\lim\limits_{n\to\infty} x_n=x$ exists, we have
$$x_{2n}=\fr... | $\displaystyle \lim_{n\to\infty}\frac{1}{\sqrt{n}}\left|\sum\limits_{k=1}^n (-1)^k\sqrt{k}\right|
= \lim_{n\to\infty}\frac{1}{\sqrt{2n}}\sum\limits_{k=1}^n \frac{1}{\sqrt{2k-1}+\sqrt{2k}} $
$\displaystyle \frac{1}{2\sqrt{2n}}\sum\limits_{k=1}^n\frac{1}{ \sqrt{2k} } < \frac{1}{\sqrt{2n}}\sum\limits_{k=1}^n\frac{1}{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3272758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
Stuck on this - In ABC right triangle AC= $2+\sqrt{3}$ and BC = $3+2\sqrt{3}$. circle touches point C and D, Find the Area of $AMD$ In ABC right triangle $AC= 2+\sqrt{3}$ and $BC = 3+2\sqrt{3}$. circle touches point C and D, Find the Area of $AMD$
Here's my strategy of solving this, I'm not sure if it's correct, if y... | Clearly, $\angle CAB = 60$ because $BC = 3+2\sqrt{3} = \sqrt{3} \cdot (2 + \sqrt{3}) = \sqrt{3} \cdot AC$. Now, we need to find the lengths $AM$ and $AD$. Let $r$ be the radius of the circle. It is clear that $OC = r$, and $AO = \sqrt{3}r$ because of the 30-60-90 triangle $AOD$, as $OD$ is a tangent to $AB$. Therefore,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3272856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
Find the limit of $\lim \limits_{x \to 1} \frac{\sqrt[\leftroot{0}\uproot{2}3]{5x-4} - 1}{\sqrt[\leftroot{0}\uproot{2}4]{5x-4} - 1}$ I need to find the limit of:$$\lim \limits_{x \to 1} \frac{\sqrt[\leftroot{0}\uproot{2}3]{5x-4} - 1}{\sqrt[\leftroot{0}\uproot{2}4]{5x-4} - 1}$$ without L'Hospital.
I tried to rationalize... | Let $5x-4=t^{12}$, then we have
:$$\lim \limits_{x \to 1} \frac{\sqrt[\leftroot{0}\uproot{2}3]{5x-4} - 1}{\sqrt[\leftroot{0}\uproot{2}4]{5x-4} - 1}=\lim \limits_{t \to 1}\frac{t^4-1}{t^3-1}$$
and $$\lim \limits_{t \to 1}\frac{t^4-1}{t^3-1}=\lim \limits_{t \to 1}\frac{(t-1)(t+1)(t^2+1)}{(t-1)(t^2+t+1)}=\frac{4}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
} |
Evaluate $\int\limits_0^{1}\frac{\sqrt{1+x^2}}{1+x}dx$ Evaluate:
$I=\int\limits_0^1 \frac{\sqrt{1+x^2}}{1+x}dx$
My try:
Let $x=\tan y$ then $dx=(1+\tan^{2} y)dy$
As for the integration limits: if $x=0$ then $y=0$ and if $x=1$ then $y=\frac{π}{4}$
So:
$I=\int\limits_0^{\frac{π}{4}}\frac{1+\tan^{2} y}{(1+\tan y)\cos... | Substituting $x=\sinh t$, we get:
$$I=\int_0^{\sinh^{-1}(1)} \frac{\cosh^2 t}{1+\sinh t}dt$$
This integral can be dealt with using the substitution $t=\log u$:
$$I=\frac{1}{2}\int_1^{1+\sqrt{2}} \frac{(u+1/u)^2 du}{u(2+u-1/u)}=\frac{1}{2}\int_1^{1+\sqrt{2}} \frac{(u^2+1)^2 du}{u^2(2u+u^2-1)}= \\ = \frac{1}{2}\int_1^{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to integrate $\int_{0}^{2\pi} \frac{1}{\sin^4x + \cos^4 x} \,dx$ So I followed the explanations made in this post and I got that:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \frac{\sqrt{2}}{2}\arctan\left(\frac{\sqrt{2}}{2}\tan\left(2x\right)\right) + C$$
But when I try to use the Leibniz-Newton formula and evaluate... | The given antiderivative, $$\frac{1}{\sqrt 2} \arctan \left( \frac{1}{\sqrt 2} \tan 2 x \right)$$ is not defined everywhere on the interval $[0, 2 \pi]$ of integration---it is undefined at $\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$, so the usual hypotheses of the Fundamental Theorem of Calculus ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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Find $\,n\bmod 42\,$ given $\,7\mid 3^n+4n+1$ I have been working on a problem relating to remainders: "The number $3^n+4n+1$ is divisible by $7$. Find the remainder of n when it is divided by $42$." I have been unable to find a strategic way to solve this problem and so far, I have only found the remainders, $4$, $8$,... | Let's say that $n=42a+b$, with $a \in Z$ and $b \in \{0,1,2,3,...,41\}$, since $n\in Z$.
Substituing we get $ 3^n+4n+1=0 (mod 7) \Rightarrow 3^{42a+b}+4(42a+b)+1=0 (mod 7)$
$ \Rightarrow 3^b.(3^{42})^a+7.(24a)+4b+1=0 (mod 7) \Rightarrow 3^b.(3^{42})^a+4b+1=0 (mod 7)$
Note that $3^{42}=({3^2})^{21}=9^{21}=(7+2)^{21}=2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve the following equation: $\sqrt {\sin x - \sqrt {\cos x + \sin x} } = \cos x$ Solve the following equation:
\begin{array}{l}{\sqrt{\sin x-\sqrt{\cos x+\sin x}}=\cos x} \\ \text{my try as follows:}\\{\sin x-\sqrt{\cos x+\sin x}=\cos ^{2} x} \\ {\sin x-\cos ^{2} x=\sqrt{\cos x+\sin x}} \\ {\sin ^{2} x+\cos ^{4} x-2... | We see that $\sin{x}\geq0$ and $\cos{x}\geq0.$
Let $f(x)=\sqrt{\sin{x}-\sqrt{\sin{x}+\cos{x}}},$ where $x\in\left[0,\frac{\pi}{2}\right].$
Thus, $$f'(x)=\frac{\cos{x}-\frac{\cos{x}-\sin{x}}{2\sqrt{\sin{x}+\cos{x}}}}{2\sqrt{\sin{x}-\sqrt{\sin{x}+\cos{x}}}}>0$$ for $\cos{x}-\sin{x}<0.$
But for $\cos{x}-\sin{x}\geq0$ we o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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$\int_0^{\infty} \frac{\ln x}{x^6 + 1}dx$ using contour integration I am looking for the solution to this improper integral. This was my approach:
Step I. (We use the 'pie slice' $\frac{\pi}{3}$). We define our log with $\frac{-\pi}{2} < \arg z < \frac{3\pi}{2}$.
\begin{equation}
\int_C \frac{\log(z)}{z^6 + 1}dz = \int... | The formula you've obtained is correct. The formula contains imaginary units, but it's not yet the final answer. Except for assmuing that the formula is incorrect, you did nothing wrong.
You have (after multiplying by $e^{-\frac{i \pi}{6}}$):
$$\frac{\pi^2}{18} = -i \int_{0}^\infty \frac{\ln x}{x^6+1}dx + \frac{\pi(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Consider the polynomial expression $P(x) = (x+1)^5 + x$ Consider the polynomial expression $P(x) = (x+1)^5 + x$
*
*Find the quotient of $P(x)$ divided by $x^2 + x + 1$
*Prove that for each positive integer $n$, the integer number $P(n)$ is divisibile by at least two different prime numbers.
I've done the first tas... | $B$ and $C$ aren't necessarily prime, but they have distinct factors.
To show this, continue as follows:
$(x^3+4x^2+5x+1)=(x^2+x+1)(x+3)+x-2.$
$x^2+x+1=(x-2)(x+3)+7.$
Therefore, $\gcd(x^3+4x^2+5x+1,x^2+x+1)|7,$ so there are two possibilities.
Case ($1$): $\gcd(x^3+4x^2+5x+1,x^2+x+1)=1.$
In this case, since, for posi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How to choose your $u$ value properly to solve $\int\frac{1}{5+3\cos (x)} \cdot dx$. We are given the following facts:
*
*$\cos(x) = \frac{1-u^2}{1+u^2}$,
*$dx = \frac{2du}{1+u^2}$
Using 1. and 2. along with u substitution solve the following integral.
$$\int\frac{1}{5+3\cos (x)} \cdot dx$$
I re-wrote the integral... | You need a substitution such that $$ 4+u^2 = 4(1+v^2)$$
that is $u=2v$. That gives you $$ \int \frac{du}{4+u^2} = \int\frac{2 dv}{4(1+v^2)} = \frac12 \arctan v + C = \frac12 \arctan\frac u 2 + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Smallest integer which leaves remainder 3, 2, 1 when divided by 17, 15, 13 Find the smallest positive integer so that the remainder when it is divided by $17,15,13$ is $3,2,1$ respectively.
This question can be solved using Chinese remainder theorem, but the theorem gives any integer, not the smallest.
For example, we... | If $x\equiv3\mod 17$ and $x\equiv2\mod 15,$ then $17a+3\equiv 15b+2\mod 17\times15,$
so $17a+3\equiv2\mod 15,$ so $17a\equiv -1 \mod 15,$ so $2a\equiv -1\mod 15,$ so $a\equiv7 \mod 15,$
so $x=17(15c+7)+3=255c+122$.
If also $x\equiv1\mod 13$ then $255c+122\equiv13e+1$ so $255c+122\equiv1 \mod 13,$
so $255c\equiv-121... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find the minimum $K$ such that $\sum_{cyc} \frac{a}{\sqrt{a+b}} \leq K\sqrt{a+b+c}$ The question is to find the minimum $K$ such that $$\sum_{cyc} \frac{a}{\sqrt{a+b}} \leq K\sqrt{a+b+c}$$ holds true for all non-negative $a,b$ and $c$.
My attempt:
I used Cauchy-Schwarz to get
$$\left(\sum_{cyc} \frac{a}{\sqrt{a+b}}\ri... | Some thoughts
Let $x=b/a, y=c/b, z=a/c$. This transforms the inequality to
$$2≥f(x,y,z)≥1,f(x,y,z)=\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}$$
Under the constraint $xyz=1$ (as you have shown).
Now
\begin{align}
f(x,y,z)&=\frac{(1+y)(1+z)+(1+z)(1+x)+(1+x)}{(1+y)(1+x)(1+y)(1+z)}\\
&=\frac{3+2(x+y+z)+(xy+yz+zx)}{1+(x+y+z)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximize $\int_{0}^{1} f(x)^5 dx$ over all $f\colon[0,1]\to[-1,1]$ with $\int_{0}^{1} f(x)^3 dx=\int_{0}^{1} f(x) dx= 0$.
Maximize $\int_{0}^{1} f(x)^5 dx$ over all $f\colon[0,1]\to[-1,1]$ with $\int_{0}^{1} f(x)^3 dx=\int_{0}^{1} f(x) dx= 0$.
I'm not even sure where to start with this problem. Any hints would be app... | The maximum value of $\int_0^1 f(x)^5 \mathrm{d}x$ is $\frac{1}{16}$.
We first prove that $\frac{1}{16}$ is an upper bound.
Note that if $t\le 1$ then $t^5\le \frac{5}{4} t^3 - \frac{5}{16} t + \frac{1}{16}$. Indeed,
$$t^5 - \left(\frac{5}{4} t^3 - \frac{5}{16} t + \frac{1}{16}\right) = (t-1)\left(t^2+\frac t2 - \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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Sum of all the numbers in the grid.
A square containing numbers
$$
\begin{array}{|c|c|c|}
\hline 1 & 2 & 3 \\ \hline
1 & 2 & 2 \\ \hline
1 & 1 & 1 \\\hline
\end{array}
\qquad \qquad\qquad
\begin{array}{|c|c|c|c|}
\hline 1 & 2 & 3 & 4 \\ \hline
1 & 2 & 3 & 3 \\ \hline
1 & 2 & 2 & 2 \\ \hline
1 & 1 & 1 & 1 \\\hline
\e... | To be honest I am not sure what you are doing. Let me add my solution. One way to write down a formula for the $n \times n$ version is $$\sum_{i = 1}^n i(2(n+1-i) - 1).$$
You will find $2n-1$ times the number $1$, $2(n-1)-1$ times the number $2$ etc. This means we want to add these up and then also add the values to th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How to obtain the function knowing its higher derivatives at $0$ does some one knows how to obtain $f(x)$ knowing that in x=0 they have the following value
$f^{n}(0)= \frac{1}{n-s}$ if $ n=1,3,5,\cdots$ and $f^{n}(0)=0$ otherwise
| Using the derivatives to fill in a Taylor series expansion...
$$
f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \\
\frac{x}{1! (1-s)} + \frac{x^3}{3! (3 - s)} + \frac{x^5}{5! (5 - s)} + \cdots \\
\sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!(2n-1-s)}
$$
This would be notoriously difficult to manually eval... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How to define function $f(n)$ for the number of integers in $\mathbb{O}_n$ that are evenly divisible by 11? Consider the set $\mathbb{O}_n$ which contains the odd integers less than or equal to $n$ that are not divisible by 3, 5 or 7 but includes 3, 5 and 7.
$\mathbb{O}_n$ = { 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 4... | Using Inclusion/Exclusion, I believe this would be:
$$g(n) = \left\lfloor \dfrac{n}{11} \right\rfloor - \left( \left\lfloor \dfrac{n}{22} \right\rfloor + \left\lfloor \dfrac{n}{33} \right\rfloor + \left\lfloor \dfrac{n}{55} \right\rfloor + \left\lfloor \dfrac{n}{77} \right\rfloor\right) + \left(\left\lfloor \dfrac{n}{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3284413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Inverse Function Verification
Are functions $f(x)=\frac{7x+4}{x+6}$ and $g(x)=\frac{6x-4}{7-x}$ inverses of each other?
I'm experiencing a very strange issue with this problem. When I find the inverse of $g$ I get $f(x)$. However, when I do the same thing for $f$ I do not get $g(x)$.
\begin{align*}
f(x)&=\frac{7x+4}... | You got the correct result for $f^{-1}$. Just multiply top and bottom by $-1$:
$$f^{-1}(x)=\frac{-6x+4}{x-7}=\frac{6x-4}{7-x}=g(x)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $y=\frac{1}{3}[\sin x+[\sin x+[\sin x]]]$ & $[y+[y]]=2\cos x$ $[x]$ represents the greatest integer function
$y=\frac{1}{3}[\sin x+[\sin x+[\sin x]]]$
&
$[y+[y]]=2 \cos x$
Find the number of solution
My approach is as follows
$\sin x \in (\pi,2\pi)$
$y=\frac{1}{3}[\sin x+[\sin x+[\sin x]]]$
$y=-1$
$[-1+[-1]]=2 \c... | I believe that your answer is correct. $[x+k]=[x]+k$ if $k$ is an integer, so the given pair of equations is just equivalent to $y=[\sin x]$ and $[y]=\cos x$.
This gives $\cos x=[[\sin x]]=[\sin x]$ which has no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3286583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding $k$ such that $(x^2 + kx + 1)$ is a factor of $(x^4 - 12 x^2 + 8 x + 3)$
$(x^2 + kx + 1)$ is a factor of $(x^4 - 12 x^2 + 8 x + 3)$ . Find $k$
....couldnt figure out how to find $k$
I tried assuming $(x^2 + kx + 1)= (x - 1)^2 $ where $k = (-2) $ comsidering that $(x-1) $ is a factor of the above polynomial...... | Let $x$ be a root of the polynomial $x^2+kx+1$, thus $x^2=-(kx+1)$. By substitute it into ${x}^{4}-12\,{x}^{2}+8\,x+3$ we get ${x}^{4}-12\,{x}^{2}+8\,x+3=\left( -{k}^{3}+14\,k+8 \right) x-{k}^{2}+16 \equiv 0$. Then solve system
\begin{cases}
-{k}^{3}+14\,k+8=0,\\
-{k}^{2}+16=0,
\end{cases}
we get that $k=4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3288779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Volume generated by rotating around y-axis, curve $y=x^3$ and the lines $y=0$ and $x=2$
Find the volume of the solid generated by revolving about the y-axis the region bounded by the curve $y=x^3$ and the lines $y=0$ and $x=2$
I first found what $x=2$ would be in terms of y.
$$y= (2)^3
\\ = 8$$
And in terms of y, the... | Washer method is quite easy. Express the washer area including its hole and extrude it along $y$.
$$
V=\pi\int_{0}^{8}\left(2^2-x^2\right)\,dy=\pi\int_{0}^{2}\left(2^2-x^2\right)\,3x^2 dx= $$
$$
V=3\pi\int_{0}^{2}\left(4-x^2\right)\, \cdot x^2 \, dx = ... = \frac{64 \pi}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3289234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $4 = \frac{3}{a_1} = \frac{3}{a_2}+a_1= \frac{3}{a_3} + a_2 = ... = \frac{3}{a_{n+1}} + a_n$ Prove that $a_n= \frac{3^{n+1} - 3}{3^{n+1}-1}$ If $$4 = \frac{3}{a_1} = \frac{3}{a_2}+a_1= \frac{3}{a_3} + a_2 = \space ... \space = \frac{3}{a_{n+1}} + a_n$$
Prove that $$a_n= \frac{3^{n+1} - 3}{3^{n+1}-1}$$
I tried to pro... | Proof by induction
Base case
$a_1 = \frac 43$
Suppose our proposition is true.
$\frac{3}{a_{n+1}} + a_n = 4\\
a_{n+1} = \frac{3}{4-a_n}\\
a_{n+1} = \frac{3}{4-\frac{3^{n+1} - 3}{3^{n+1} - 1}}$
By the inductive hypothesis
$a_{n+1} = \frac{3^{n+2} - 3}{3^{n+2}-1}$
| {
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Is there a closed-form solution for $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(3n+m)}$? I am seeking a closed-form solution for this double sum:
\begin{eqnarray*}
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(\color{blue}{3}n+m)}= ?.
\end{eqnarray*}
I will turn it into $3$ tough integrals in a moment. ... | $$\boxed{I=\int_0^1 \frac{\ln(1-x) \ln(1-x^3)}{x}dx=\frac53\zeta(3) +\frac{2\pi^3}{27\sqrt 3} -\frac{\pi}{9\sqrt 3}\psi_1\left(\frac13\right)}$$
As mentioned in the question we have:
$$I=\int_0^1 \frac{\ln^2(1-x)}{x}dx+\int_0^1 \frac{\ln(1-x)\ln(1+x+x^2)}{x}dx=2\zeta(3)+J$$
We can make use of the following series:
$$ ... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "31",
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Prove that if $x > 3$ and $y < 2$, then $x^2 − 2y > 5$
Suppose $x > 3$ and $y < 2$. Then $x^2 − 2y > 5$.
My attempt:
*
*$x>3 \implies x^2 > 9$
*$ y < 2 \implies 5 + 2y < 9$
It follows that $5+2y < 9 < x^2$. From this we can see that $5+2y < x^2 \implies x^2 -2y > 5$
Therefore, if $x>3$ and $y<2$, then $x^2 − 2... | Proof is fine:
Option:
1)$x >3$, then
$x^2= xx >3x >9$;
2) $y <2$, then
$x^2 - 2y >9-2y >9-2 \cdot 2=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Comparing two close numbers How to compare these two numbers without using a calculator ?
$A=\left(\dfrac{11}{10}\right)^{\sqrt{5}}$ and $\;B=\left(\dfrac{12}{11}\right)^{\sqrt{6}}$.
Thanks for your help !
Here is what I tried for example : $$\left(\frac{A}{B}\right)^{\sqrt6-\sqrt5}=\frac{11}{10^{\sqrt{30}−5}12^{6−\sq... | Let $f(x)=\frac{\sqrt{x+1}\ln(1+x)}{x},$ where $x>0$.
Thus, $$f'(x)=\frac{\left(\frac{\ln(1+x)}{2\sqrt{1+x}}+\frac{1}{\sqrt{1+x}}\right)x-\sqrt{1+x}\ln(1+x)}{x^2}=\frac{2x-(x+2)\ln(1+x)}{2x^2\sqrt{1+x}}\leq0$$ because
$$\left(\ln(1+x)-\frac{2x}{x+2}\right)'=\frac{x^2}{(x+1)(x+2)^2}\geq0.$$
Id est, $f$ decreases and for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$?
How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$?
im trying to use AM-GM
$\sqrt[n]{ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}}
\le ... | It's wrong. Try $x_1\rightarrow0^-$
For positive variables it's true by C-S:
$$\sum_{k=1}^nx_k\sum_{k=1}\frac{1}{x_k}\geq\left(\sum_{k=1}^n\sqrt{x_k\cdot\frac{1}{x_k}}\right)^2=n^2.$$
Also, AM-GM works:
$$\sum_{k=1}^nx_k\sum_{k=1}\frac{1}{x_k}\geq n\sqrt[n]{\prod_{k=1}^nx_k}\cdot n\sqrt[n]{\prod_{k=1}^n\frac{1}{x_k}}=n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the coefficient of $x^3$ in expansion of $(x^2 - x + 2)^{10}$
What is the coefficient of $x^3$ in expansion of $(x^2 - x + 2)^{10}$.
I have tried:
$$\frac{10!}{(3!\times7!)} \times (-x + 2)^7 \times (x^2)^3 $$
But got an incorrect answer $-15360$.
| You can get $x^3$ only by multiplying $x^2$ with $-x$, or by multiplying three $(-x)$'s together where the rest of the terms you are multiplying with should be $2$'s.
We can multiply $x^2$ with $(-x)$ in $2\cdot \binom{10}{2}$ ways (We multiply by $2$ to take into account the order of $x^2$ and $(-x)$). And we can mult... | {
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"timestamp": "2023-03-29T00:00:00",
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On the Clausen triple $8\rm{Cl}_2\left(\frac{\pi}2\right)+3\rm{Cl}_2\left(\frac{\pi}3\right)=12\,\rm{Cl}_2\left(\frac{\pi}6\right)$ While doing research on the Clausen function, I came across this nice identity,
$$8\operatorname{Cl}_2\left(\frac{\pi}2\right)+3\operatorname{Cl}_2\left(\frac{\pi}3\right)=12\operatorname{... | Let $~~m,n\in\mathbb{N}~~$ and $~~0<x<1~$ .
$\displaystyle \text{E}_m(x):=\text{Li}_m(e^{i2\pi x})~ , ~~ \text{Cl}_m(2\pi x)=\Im \text{E}_m(x)~ , ~~ \overline{\text{E}_m(x)}~$ is conjugated complex to $~\text{E}_m(x)$
We have:
$$ \text{E}_m(1-x) = \overline{\text{E}_m(x)}~~ , ~~~~ \sum\limits_{k=0}^{n-1}\text{E}_m \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find $n$ if $\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$ Find $n$ if $$\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$$
In this video they show a shortcut and say $n=-1/2$ without any explanation.
Key observation here is that the geometric mean of $9$ and $4$ is $6$.
It seems numerator and denominator are partial sums of geometric ser... | Start by dividing top and bottom by $4^n$:
$$\frac{\frac{9^{n+1}}{4^n} + \frac{4^{n+1}}{4^n}}{\frac{9^n}{4^n} + 1} = 6.$$
Substitute $u = \frac{9^n}{4^n}$ to get
$$\frac{9u + 4}{u + 1} = 6 \iff 9u + 4 = 6(u + 1) \iff u = \frac{2}{3}.$$
Therefore,
$$\left(\frac{9}{4}\right)^n = \frac{2}{3} \implies n = -\frac{1}{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find $f(x) = x + \frac{2}{3}x^3 + \frac{2\cdot4}{3\cdot5}x^5 + \frac{2\cdot4\cdot6}{3\cdot5\cdot7}x^7+\cdots$ where $|x|<1$
Find $$f(x) = x + \frac{2}{3}x^3 + \frac{2\cdot4}{3\cdot5}x^5 + \frac{2\cdot4\cdot6}{3\cdot5\cdot7}x^7+\cdots +\infty\,,\quad|x|<1$$
My solution:
$$f'(x) = 1 + 2x^2 + \frac{2}{3}\cdot4x^4 + \fra... | The question asks for other ways to identity the function whose
power series expansion is
$$ f(x) := x + \frac{2}{3}x^3 + \frac{2\cdot4}{3\cdot5}x^5 +
\frac{2\cdot4\cdot6}{3\cdot5\cdot7}x^7+
\ldots .$$
There are many potential methods but maybe finding numerical values of
the function will help to identify it. For ... | {
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"timestamp": "2023-03-29T00:00:00",
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What $x$ solves $x^2 = R^2 - x^2 +x(R^2 - x^2)^\frac{1}{2}$? This is from "Calculus Made Easy", Exercise 9 problem 9c. The formula in the title is what I've gotten and I know from the answer in the back of the book that it reduces to $x = 0.8507R$, but I can't see how to reduce it further (other than grouping the two ... | Rearrange to get $$\frac{2x^2-R^2}{x} = (R^2-x^2)^{\frac{1}{2}}$$
Square both sides: $$\frac{4x^4-4x^2R^2+R^4}{x^2} = R^2-x^2$$
Multiply by $x^2$: $$4x^4-4x^2R^2+R^4 = x^2R^2-x^4$$
Rearrange some more: $$5x^4-5x^2R^2+R^4 = 0$$
This is a quadratic in $x^2$: $$x^2 = \frac{5R^2\pm \sqrt{25R^4-20R^4}}{10} = \frac{5R^2\pm R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306999",
"timestamp": "2023-03-29T00:00:00",
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Does the equation $a^n+b^n=c^n$ have positive integer solutions? We know that the equation $a^n+b^n=c^n$ has infinitely many positive integer solutions when $n=1,n=2$. Also, by Fermat's last theorem, we know that this equation has no positive integer solution for any integer value of $n$ greater than $2$.
My question i... | Well, [1] obviously not for every non-integer $n$ will have integer solutions, $a,b,c$.
But obviously [2] some (many) integer $a,b,c$ will have non-integer $n$ solutions.
[1]. It'd be rather bizarre and strange if none of the $n\in \mathbb Z; n> 2$ will have integer solutions to $a^n + b^n = c^n$ but that every $\x \n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3308524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Laurent Expansion Coefficients of $\exp\left(\frac{z}{1-z}\right)$ Given the function $f(z)=\exp\left(\frac{z}{1-z}\right)$, I want to find the coefficients $a_0$, $a_{-1}$, and $a_{-2}$ of the Laurent expansion $f(z)=\sum_{n=-\infty}^{\infty}a_n(z+1)^n$ about $z=-1$, on the annulus $\{z\in\mathbb{C}:|z+1|>2\}$.
We kn... | Similar to José Carlos Santos'answer
Let $$z=x-1\implies \frac z {1-z}=\frac{1-x}{x-2}=-\frac 12 + \sum_{n=1}^\infty 2^{-(n+1)} x^n$$ which make
$$e^{\frac{1-x}{x-2}}=\frac{1}{\sqrt{e}}\left(1+\frac{x}{4 }+\frac{5 x^2}{32 }+\frac{37 x^3}{384
}+\frac{361 x^4}{6144 }\right)+O\left(x^5\right)$$
The numerator of the co... | {
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"url": "https://math.stackexchange.com/questions/3310902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Interesting four-sum inequality $n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right) \ge...$
Prove that for all $n \in \mathbb{N}$ the inequality $$ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le n^2 \lef... | Just a partial approach.
For the different sums, we have
$$\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}=n^2+2 n-H_n$$
$$\sum \limits_{k=1}^n (2k-1)\frac k{k+1}=3 H_{n+1}+(n-3) (n+1)$$
$$\sum \limits_{k=1}^n \frac{k+1}{k}=H_n+n$$
$$\sum \limits_{k=1}^n \frac k{k+1}=n+1-H_{n+1}$$
So, we need to prove that
$$f(n)=n^2 \left(H... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Prove the following inequality for any positive real number $x,y$ Prove the following inequality for any positive real number $x,y$
$xy^3 \leq \frac14x^4 + \frac34y^4$
I have tried subtracting $y^4$ from both sides to get:
$xy^3 - y^4 \leq \frac14x^4 - \frac14y^4$
$y^3(x - y) \leq \frac14(x - y)(x + y)(x^2 + y^2)$
$y^3... | We need to prove that
$$x^4-4xy^3+3y^4\geq0$$ or
$$x^4-2x^3y+x^2y^2+2x^3y-4x^2y^2+2xy^3+3x^2y^2-6xy^3+3y^4\geq0$$ or
$$(x-y)^2(x^2+2xy+3y^2)\geq0,$$ which is true even for all reals $x$ and $y$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that for $\forall m \ge 2$, $\dfrac{1}{2}(a^4 + b^4 + c^4) \mid [(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m}]$.
Let $(a, b, c)$ be a Pythagorean triple. Prove that for $\forall m \ge 2$, $$\large \dfrac{1}{2}(a^4 + b^4 + c^4) \mid \left[(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m}\right]$$
Let $P_{n} = a^{2^n} + b^{2^n} + c^{... | As you did, this answer uses $P_m$ where
$$P_m=a^{2^m}+b^{2^m}+c^{2^m}$$
We may suppose that $c^2=a^2+b^2$.
Let $$Q_m=(ab)^{2^m}+(bc)^{2^m}+(ca)^{2^m},\qquad R=\frac{a^4+b^4+c^4}{2}$$
Now, let us prove by induction on $m$ that $R\mid P_m$ and $R\mid Q_m$.
We see that $R\mid P_2$ and $R\mid Q_2$ since
$$Q_2=(a^4+a^2b^2+... | {
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Indices - factorising Simplify: $$\frac {x^5y^2x^3 + x^4y^5 - y^5x^7y^4}{x^4y^3}$$
I know this is probably low level stuff but I need to be able to do this specific type of question and I have no way of checking my work. If anyone could offer a step by step working I'd be appreciative
| You have
\begin{align*}
\frac {x^5y^2x^3 + x^4y^5 - y^5x^7y^4}{x^4y^3}
&=\frac {x^8y^2 + x^4y^5 - x^7y^9}{x^4y^3} \\
&=\frac {x^4y^2\big(x^4 + y^3 - x^3y^7\big)}{x^4y^3} \\
&=\frac {x^4 + y^3 - x^3y^7}{y},\quad x\not=0.
\end{align*}
That's about as far as you can go.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the surface area of a region which is generated by revolving a curve around a line The following problem is from the book, Calculus and Analytical Geometer by Thomas and Finney. It is early on in the book
so I would expect / hope any integral would be easy to solve.
Problem:
Find the area of the surface gener... | Consider a scalar function $f(x)$: $R$ $\rightarrow$ $R$. Then the surface area of the solid formed by revoling $f(x)$ about $y=0$ is S=$2\pi$$\int_a^b$$f(x)ds$, where $ds$ represents an infintesimal arclength element of the curve $f(x)$.
To calculate the surface area of the solid formed by revolving $f(x)$ about y=$... | {
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"url": "https://math.stackexchange.com/questions/3316716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many real roots does $x^4 - 4x^3 + 4x^2 - 10$ have?
How many real roots does this polynomial have?
$$x^4 - 4x^3 + 4x^2 - 10$$
Because non-real roots come in pairs, it must have 4, 2 or 0 real roots.
Following Descartes' rules of signs, it either has one negative (real) number and one or three positive numbers.
... | As $x\to -\infty$ $f(x)\to +\infty>0$. And $f(-1)= -1$ so $f(x)$ "crosses" the $x$-axis at some $x < -1$. That's our one and only negative root.
Now $f(x) = x^2(x-2)^2 - 10$. If $x > 0$ then $x^2$ is strictly increasing and positive and $(x-2)^2$ is strictly increasing and non-negative so $f(x)$ is strictly increasi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3318485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
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Integrate ${\int\sqrt{1 + \sin\frac{x}2}\,\mathrm{d}x}$ So I was doing a integral question and I stumbled upon this question.
$\displaystyle{\int\sqrt{1 + \sin\left(\frac x2\right)}\,dx}$
In order to solve it I did the following:
I took $u = \frac12x$
Then $\frac {du}{dx}$
Which gave me $2 du = dx$.
After that I subst... | Obviously, knowing the trigonometric identities as already covered is the far superior and simpler method. Here is a more 'generalised' approach. I hope it serves some value.
Here I will address the integral:
\begin{equation}
I = \int \sqrt{1 + \sin\left(\frac{x}{2}\right)}\:dx
\end{equation}
First let $u = \frac{x}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Prove $(x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{2} \le (x^{6} + y^{6} + z^{6})[ (1+x)^{2} + (1+y)^{2} + (1+z)^{2} ] $ Prove
$$ (x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{2} \le (x^{6} + y^{6} + z^{6})[ (1+x)^{2} + (1+y)^{2} + (1+z)^{2} ] $$
if $x,y,z \ge 0$
I made this problem using Holders inequality, no... | This is just Cauchy inequality $$(a^2+b^2+c^2)(a'^2+b'^2+c'^2)\geq (aa'+bb'+cc')^2$$
where $a= 1+x$ and $a'=x^3$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proving $(1-\frac{1}{x-c})^{\epsilon x-c}\geq(1-\frac{1}{x})^{\epsilon x}$ How do I formally prove that $(1-\frac{1}{x-c})^{\epsilon x-c}\geq(1-\frac{1}{x})^{\epsilon x}$ which can be seen graphically. Here $\epsilon$ is a small positive constant $\ll1$, $x\gg0$ and $c>0$.
| Let us prove the inequality in a more precise manner.
The inequality is written as
$$(\epsilon x - c)\ln \big(1 - \frac{1}{x-c}\big) \ge \epsilon x \ln \big(1 - \frac{1}{x}\big)$$
or
$$\epsilon \le \frac{-c \ln \big(1 - \frac{1}{x-c}\big)}{x \ln \big(1 - \frac{1}{x}\big) - x \ln \big(1 - \frac{1}{x-c}\big) }.\tag{1}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the integral $\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}$ One may take $x= \cos t$ and get
$$I=\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}= -\frac{1}{4}\int \csc^4(t/2)~ dt=-\frac{1}{4} \int [\csc^2(t/2) +\csc^2(t/2) \cot^2(t/2)]~ dt.$$
$$\Rightarrow I=\frac{1}{2} \left [\cot (t/2)] +\frac{1}{3}\cot^3(t/2)\right]=\frac{(2-x)}... | Let us use the Euler transformation $$z=\frac{1+x}{1-x}\Rightarrow x =\frac{z-1}{z+1} \Rightarrow dx=\frac{2 dz}{(1+z)^2}$$ then
$$I=\int \frac{dx}{(1-x)^2\sqrt{1-x^2}} dx = \frac{1}{4}\int (z^{-1/2}+z^{1/2}) dz=\frac{1}{2}z^{1/2}[1+\frac{z}{3}]=\frac{(2-x)}{3(1-x)} \sqrt{\frac{1+x}{1-x}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Laurent series of $~\frac{1}{z+2}+\frac{1}{z^2}, ~~~~~~~0<|z+2|<2~$ Laurent series of $$~\frac{1}{z+2}+\frac{1}{z^2}, ~~~~~~~0<|z+2|<2~$$
I can't find a way to represent $z^2$ in terms of $z+2$. I've tried to do $(z+2)(z-2)+4$, but I'm stuck with the $+4$.
| Thanks for the answer !
I've tried the Taylor series, but because I was always simplifying the terms $\frac{2}{8}= \frac{1}{4}, \frac{4}{32}= \frac{1}{8}$ etc. to get it to look like the one in wolframalpha. I couldn't see the pattern to make the series looks like : $$\sum_{n=0}^\infty \frac{(n+1)}{2^{n+2}}(z+2)^n$$
A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maclaurin expansion of $\arccos(1-2x^2)$ Maclaurin expansion of $\arccos(1-2x^2)$
This is what I tried.
$f'(x)=2(1-x^2)^{-1/2} \\
f''(x)=2(1-x^2)^{-3/2}+3 \cdot 2 x^2(1-x^2)^{-5/2} \\
f^{(3)}(x)=18x(1-x^2)^{-5/2}+2\cdot 3\cdot 5x^3(1-x^2)^{-7/2} \\
f^{(4)}(x)=18(1-x^2)^{-5/2}+180x^2(1-x^2)^{-7/2}+2\cdot 3\cdot 5\cdot 7... | If you check your differentiation very carefully, you will find that $f(x)=\arccos(1-2x^2)$ is not continuously differentiable at $x=0$, so the MacLaurin expansion does not exist.
Details: if $x\ne0$ we have
$$\frac{d}{dx}\arccos(1-2x^2)=\frac{4x}{\sqrt{1-(1-2x^2)^2}}
=\frac{2x}{|x|\sqrt{1-x^2}}\ .$$
So
$$\lim_{x\to0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3329518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can $S_n=n$ for the series $S_n=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\cdots$ How can $S_n=n$ for the series $$S_n=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\cdots$$
My try:
We can Re-write $S_n$ as
$$S_n=\frac{0+1}{2}+\frac{1+2}{2^2}+\frac{3+2^2}{2^3}+\frac{7+2^3}{2^4}+\cdots \frac{2^{n-1}-1+2^... | Your solution is not quite correct because $S_n$ has terms from $\frac{1}{2}$ to $\frac{2^n - 1}{2^n}$, but with your adjusted value, the terms on the right are missing the last value of $S_n$. Thus, the appropriate expression is$$\begin{equation}\begin{aligned}
S_n & = \frac{n}{2} + \frac{S_n - \frac{2^{n} - 1}{2^{n}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3331620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a,b \in \mathbb{R}$ and $a+b\ge 0$, prove that $(a^2+b^2)^3\ge 32(a^3+b^3)(ab-a-b)$
If $a,b \in \mathbb{R}$ and $a+b\ge 0$, prove that $(a^2+b^2)^3\ge 32(a^3+b^3)(ab-a-b)$
This question in my opinion is difficult and have tried many things. I tried to expand both sides but that will not help since I will not be a... | We need to prove that
$$(a^2+b^2)^3+32(a^3+b^3)(a+b)\geq32(a^3+b^3)ab,$$ which is obviously true for $ab<0$.
Let $ab\geq0.$
Thus, by AM-GM three times we obtain:
$$(a^2+b^2)^3+32(a^3+b^3)(a+b)\geq2\sqrt{(a^2+b^2)^3\cdot32(a^3+b^3)(a+b)}=$$
$$=8\sqrt{2(a^2-ab+b^2+ab)^3(a^3+b^3)(a+b)}\geq8\sqrt{2\left(2\sqrt{(a^2-ab+b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to factorize $\frac{4x^3+4x^2-7x+2}{4x^4-17x^2+4}$? How to factorize and simplify the following?
$$\frac{4x^3+4x^2-7x+2}{4x^4-17x^2+4}$$
I've tried everything I know. Trying to factorize the numerator first then denominator, but I get no where. Usual identities like $(x+y)^2=x^2+2xy+y^2$ don't work either, and ne... | By the rational root test, $4x^3+4x^2-7x+2=0$ has roots $\frac{1}{2}$ and $-2$, so that
$$
4x^3+4x^2-7x+2=(2x - 1)^2(x + 2).
$$
In the same way we see that
$$
4x^4-17x^2+4=(2x + 1)(2x - 1)(x + 2)(x - 2).
$$
Now we can form the quotient and see the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there a simple pattern to memorize the sine of $0^\circ$, $15^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, $75^\circ$, $90^\circ$? We know there is a nice pattern to memorize the sine of $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, $90^\circ$ as follows.
\begin{align}
\sin 0^\circ &= \tfrac12\sqrt0\\
\sin 30^\circ... | This answer is inspired from Oleg567's answer. (I should say, I will explain why does this beautiful sequence appears.)
Firstly we know that: (It should be the really amazing and beautiful sequence)
$$\cos 0^\circ=\dfrac{\sqrt{4}}{2}\quad\cos 30^\circ=\dfrac{\sqrt{3}}{2}\quad\cos 45^\circ=\dfrac{\sqrt{2}}{2}\\\cos 60^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3335923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Remainders of powers of 10 when divided by prime While doing a number theory question, I observed that the repeating digits of decimal expansions of $\frac{1}{7},\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7}$ form a pattern. They are $142857,285714,428571,571428,714285,857142$. I wondered if there is a li... | This works because $7$ is a factor of $10^{(7-1)}-1=10^6-1=999999$. We have that the period of the decimal expansion of $7$ will be a factor of $6$.
If we look at $999999=3^3\cdot 7\cdot 11 \cdot 13\cdot 37$ we find that
$3$ and $9$ are factors of $10^1-1=9$ and have period $1$
$11$ is a factor of $10^2-1=99$ and has p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the sum $ 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + ... + n \cdot n! $ Find the sum
$$ 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + ... + n \cdot n! $$
Attempt
$$ 2 \cdot 2! + 3 \cdot 3! = 2 \cdot 2! + 3^{2} \cdot 2! = 2! (2+3^{2})$$
$$ 4 \cdot 4! + 5 \cdot 5! = 4 \cdot 4! + 5^{2} \cdot 4! = 4! (4+5^{2})$$
$$ 6 \cdot 6!... | HINT:
$n\cdot n!=(n+1-1)n!=(n+1)!-n!$
Do you know about Telescoping Series?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $ If $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $
Attempt:
Notice that $$ \tan(a+b+c) = \frac{\tan(a+b) + \tan(c)}{1 - \tan(a+b)\tan(c)} $$... | Another concise option: since $\tan c=\frac{1}{\tan(a+b)}=\frac{1-\tan a\tan b}{\tan a+\tan b}$, $\tan b\tan c+\tan c\tan a=1-\tan a\tan b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Geometry - Rectangle ABCD with inside point E. Find the least possible value for sum of interger distances from E to 4 vertices.
The point E lies within the rectangle ABCD.
If the distances from the vertices to E are all distinct integers, what is the least possible value of AE + BE + CE + DE?
| Suppose we have such rectangle. Place it on a coordinate system so that the $E$ is on $(0,0)$ and $A = (-a, b); B= (-a,-c), C=(d,b); D=(d,-c)$ and we need for $\sqrt{a^2 + b^2}=m, \sqrt{a^2+c^2}=n, \sqrt{d^2 + b^2}=r, \sqrt{d^2+c^2}=s$ are all integers.
We must have $m^2 +s^2 = n^2 + r^2$ and that simply requires find... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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An application of the Banach Fixed Point Theorem to a system of equations I am solving a question whose first item is to demonstrate the Banach Fixed Point Theorem, and the second item is as follows:
Show that for any parameter $t \in \mathbb{R}$ the system
$$
\begin{cases}
x = \frac{1}{2}\sin(x+y) + t - 1 \\
y = \frac... | By the mean value theorem, for any $a,b \in \mathbb{R}$, $|\sin(a)-\sin(b)| \le |a-b|$. Therefore, $$|(\frac{1}{2}\sin(x+y)+t-1)-(\frac{1}{2}\sin(x'+y')+t-1)| = \frac{1}{2}|\sin(x+y)-\sin(x'+y')|$$ $$\le \frac{1}{2}|(x+y)-(x'+y')| = \frac{1}{2}|(x-x')+(y-y')|.$$ Similarly, $$|(\frac{1}{2}\cos(x-y)-t+\frac{1}{2})-(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3344182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to solve given differential equation $(x^3y^3+x^2y^2+xy+1)ydx+(x^3y^3-x^2y^2-xy+1)xdy=0$? The equation to solve is:
$$(x^3y^3+x^2y^2+xy+1)ydx+(x^3y^3-x^2y^2-xy+1)xdy=0$$
I tried putting $xy=t$ but that just gave me this:
$$\frac{t^3-t^2-t+1}{t^3+t^2+t+1}dt=\frac{dx}{x}$$
I suppose there must be some clever factorin... | The given equation is of the form $f_1(xy)ydx+f_2(xy)xdy=0.$
Here, $M=(x^3y^3+x^2y^2+xy+1)y$ & $N=(x^3y^3-x^2y^2-xy+1)x$
$$\therefore I.F.=\frac1{Mx-Ny}\\
=\frac1{2x^2y^2(xy+1)}$$
$\text{Multiplying the I.F. with the equation, we get}$
\begin{align}\\
&{\begin{aligned}\\
\frac{(x^3y^3+x^2y^2+xy+1)y}{2x^2y^2(xy+1)}dx &+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Calculate the integral value using residues Hello I'm trying to solve this integral : $\\$
$$\int_{0}^{2\pi} \frac {\cos^2(x)}{4+3\cos(x)} dx.$$
I want to solve this integral using the theorem:
$$\int_{0}^{2\pi} R\bigl(\cos(\alpha),\sin(\alpha)\bigr) d\alpha =2\pi i \sum_{|z_{k}|<1} \operatorname*{Res}_{z=z_k}f(z)$$
w... | $$I=\int_{0}^{2\pi}\frac{\cos^2(x)}{4+3\cos x}\,dx = \int_{0}^{2\pi}\frac{\left(\frac{e^{ix}+e^{-ix}}{2}\right)^2}{4+3\left(\frac{e^{ix}+e^{-ix}}{2}\right)}\,dx=\int_{0}^{2\pi}\frac{e^{2ix}+e^{-2ix}+2}{16+6e^{ix}+6e^{-ix}}\,dx $$
equals, via $e^{ix}\to z$,
$$ -i\oint_{|z|=1}\frac{(z^2+1)^2}{z^2(6z^2+16z+6)}\,dz. $$
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3347485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Determining $\lim_{h \to 0} \frac{f(x +h) - f(x)}{h}$ for the function $f(x) = 3x^2 + 2x$ Determine $\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}$ for the function $f(x)=3x^2+2x$.
Can anyone please confirm whether this is correct:
\begin{align*}
& = \frac{3x^2+6xh+3h^2+2x+2h-3x^2-2x}{h}\\
& = \frac{6xh+3h^2+2h}{h}\\
& = \... | Given $f(x)=3x^2+2x$:
\begin{align}
\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}&=\lim\limits_{h\to0}\frac{3(x+h)^2+2(x+h)-2x^2-2x}{h}\\
&=\lim\limits_{h\to0}\frac{6xh+3h^2+2h}{h}\\
&=\lim\limits_{h\to0}\frac{h(6x+3h+2)}{h}\\
&=\lim\limits_{h\to0}\ (6x+3h+2)\\
&=6x+2
\end{align}
You have approached the problem correctly, b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3348040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probabilities of selecting exactly $n$ white and $m$ black balls I stumbled upon a tricky problem that I don't know how to solve.
There $a$ white and $b$ black balls in an urn. We select one first ball at random (let's call its color $F$) and then return this ball back to the urn. We then add $c$ balls of the same colo... | Part 1)
Let's say we have urn with $a$ white balls, $b$ black balls. After first draw, we return ball back to the urn and add $c$ balls of the same colour. We're asked about probability of event $E - $exactly $n$ white, $m$ black balls (We draw exactly $n+m$ balls ).
Let $X$ - be first draw, that is $X =$ white with pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the derivative of a function of the form $u(x)^{v(x)}$? So I have a given lets say $(x+1)^{2x}$ in addition to $\frac{\mathrm dy}{\mathrm dx}a^u=a^u\log(a)u'$. I still have to multiply this by the derivative of the inside function $x+1$ correct?
| This is what logarithmic differentiation is for. You start with writing the function as an equation
$$y = (x + 1)^{2x},$$
then take the natural log of both sides:
$$\ln y = \ln\left[(x + 1)^{2x}\right] = 2x \ln(x+1).$$
We then implicitly differentiate both sides with respect to $x$. By chain rule (remember, $y$ is a fu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3353508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
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Expression of $x^n+\frac1{x^n}$ by $x+\frac1{x}$ where $n$ is a positive odd number. There was a problem in a book:
Denote that $y=x+\dfrac{1}{x}$, express $x^7+\dfrac{1}{x^7}$ using $y$.
It's not a hard question, but I find a special sequence:
$x+\dfrac{1}{x}=y\\x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\lef... | I don't see a clear pattern in the coefficients but there is a recurrence:
$$
y_{n+2} = y_2 y_n -y_{n-2} = (y^2-2)y_n -y_{n-2}
$$
where
$$
y_n = x^n+\dfrac{1}{x^n}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3353877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Calculate the limit without l'Hopital rule I have the following limit:
$$
\lim_{x\to 7}\dfrac{x^2-4x-21}{x-4-\sqrt{x+2}}
$$
I could easily calculate the limit = 12 using the l'Hopital rule.
Could you please suggest any other ways to solve this limit without using the l'Hopital rule?
Thank you
| The numerator can be factored as $(x-7)(x+3)$. Now consider
$$
\lim_{x\to7}\frac{x-4-\sqrt{x+2}}{x-7}=\lim_{x\to7}\frac{x-7-(\sqrt{x+2}-3)}{x-7}=
1-\lim_{x\to7}\frac{x+2-9}{(x-7)(\sqrt{x+2}+3)}=1-\frac{1}{6}=\frac{5}{6}
$$
So your limit is
$$
\lim_{x\to7}\frac{x-7}{x-4-\sqrt{x+2}}(x+3)=\frac{6}{5}\cdot10=12
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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If $a$, $b$, $c$ are three positive integers such that $a^3+b^3=c^3$ then one of the integer is divisible by $7$ Let on contrary that none of the $a$, $b$, $c$ is divisible by $7$. Then either $a^3\equiv b^3\pmod{7}$ or $b^3\equiv c^3\pmod{7}$ or $c^3\equiv a^3\pmod{7}$.
Now how to go further?
| If none of a,b,c are divisible by 7,
then each of a$^3$, b$^3$ and c$^3$ = either 1 or -1 (mod 7).
A contradiction ensues.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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About $e^{i z} = \cos z + i \sin z$ in Michael Spivak "Calculus 3rd Edition". I am reading "Calculus 3rd Edition" by Michael Spivak.
The author wrote as follows (p. 555):
Moreover, if we replace $z$ by $i z$ in the series for $e^z$, and make a rearrangement of the terms (justified by absolute convergence), something p... | You are right that absolute convergence, while present as "security blanket", is not really used. The only re-arrangement that happens is that the exponential series is organized in groups of two subsequent terms, so that then the addition rule for convergent series can be applied:
$\sum_{n=0}^\infty a_n+\sum_{n=0}^\... | {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Derivation of 2$\pi$ in the period of a simple pendulum Most of us know that the period of a simple pendulum is given by
$$2 \pi \sqrt{\dfrac{\ell}{g}}.$$
But how did the $2\pi$ term get into that argument. From dimensional analysis, we can find the period but not the constant.
Is there a calculus argument as some limi... | The pendulum movement equation is given by
\begin{align*}
\frac{\mathrm{d}^{2}\theta}{\mathrm{d}t^{2}} + \frac{g}{\ell}\sin(\theta) = 0
\end{align*}
For small values of $\theta$, we can make the approximation $\sin(\theta) \approx \theta$, from whence we obtain the equation
\begin{align*}
\ddot{\theta} + \frac{g}{\ell... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3358167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a+b+c=2$ where $0 \leq {a,b,c} \leq 1$ find the range of $\frac{a}{1-a}\cdot\frac{b}{1-b}\cdot\frac{c}{1-c}$
If $a+b+c=2$ where $0 < {a,b,c} <1$ find the range of $$\frac{a}{1-a}\cdot\frac{b}{1-b}\cdot\frac{c}{1-c}.$$
I have tried using AM-GM but I was not able to solve it . I also assumed the expression equal to... | Let $\frac{a}{1-a}=x$, $\frac{b}{1-b}=y$ and $\frac{c}{1-c}=z.$
Thus, $x$, $y$ and $z$ are positives, $a=\frac{x}{1+x},$ $b=\frac{y}{1+y},$ $c=\frac{z}{1+z}$ and the condition gives:
$$\sum_{cyc}\frac{x}{1+x}=2.$$
Id est, by AM-GM we obtain:
$$\prod_{cyc}\frac{x}{1+x}=\prod_{cyc}\left(2-\frac{y}{1+y}-\frac{z}{1+z}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3359503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is $\alpha^{4} + \beta^{4} + \gamma^{4}$? If
$$ \alpha + \beta + \gamma = 14 $$
$$ \alpha^{2} + \beta^{2} + \gamma^{2} = 84 $$
$$ \alpha^{3} + \beta^{3} + \gamma^{3} = 584 $$
What is $\alpha^{4} + \beta^{4} + \gamma^{4}$?
Attempt:
Notice that
$$ \alpha + \beta + \gamma = 14 \implies \alpha^{2} + \beta^{2} + \ga... | Let $x=\alpha+\beta+\gamma$, $y=\alpha^2+\beta^2+\gamma^2$ and $z=\alpha^3+\beta^3+\gamma^3$ for any real numbers $\alpha,\beta,\gamma$. Then we have the identity:
$$\alpha^4+\beta^4+\gamma^4=\frac16\left(x^4-6x^2y+3y^2+8xz\right).\tag{1}$$
You can verify this by expanding everything. In our case, $x=14,y=84$ and $z=58... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Contradiction when solving a linear system with Gauss-Jordan elimination Consider a linear system with an unknown constant $a$:
$$
\left\{
\begin{aligned}
x+(a-1)y+az&=1 \\
ax+ay+az&=1 \\
a^2x+y+z&=a
\end{aligned}
\right.
$$
This gives us an augmented matrix:
$$A= \left[
\begin{array}{ccc|c}
1&a-1&a&1\\
a&a&a&1\\... | In the case $a=1$, the rows 2 and 3 of the $3\times 3$ matrix are the same and so the matrix has rank 2 and there are infinitely many solutions (the solution set is a line).
If $a\ne 1$, the $3\times 3$ matrix has rank 3 and there is unique solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove the sequence of $p_n(x)$ are all integer polynomials on x? for $x \in \mathbb R ,|x|< \frac12$
define:
$A_n(x)$
solution of the relation:
$A_{n+1}(x) = \frac{x^2}{1-A_n(x)}$
$A_0(x) = 0$
(I found that $A_n(x) = 2x^2 \cdot \frac{(1+\sqrt{1-4x^2})^n - (1 - \sqrt{1-4x^2})^n}{(1+\sqrt{1-4x^2})^{n+1} - (1 - \sq... | I will sketch a solution assuming the relation given by the OP for $A_n$ which looks correct. First some notations to simplify things: let $c=\sqrt{1-4x^2}$ and $Y_{1,2}=\frac{1 \pm c}{2}$. We will first prove that:
$Y_1^n-Y_2^n=cQ_n(x)$ where $Q_n(x), n \ge 0$ is an integral polynomial in $x$
Since $Y_{1,2}$ are the r... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A pattern on finding squares for which their sum is $\Big(\sum\limits_{i=0}^j x^i\Big)^2$ Apologies for the inactivity. I haven't been doing so well in life, lately, but I'm glad to be back at some maths! Here's a pattern I discovered. I'm not good at explaining with words, so hope you get it. It shouldn't be too diffi... | Induction.
We want to show that $$\left(\sum\limits_{i=0}^{2k+1}x^i\right)^2=(k+1)(x^{k+1}+x^k)^2+\sum_{j=1}^kj[(x^j+x^{j-1})^2+(x^{2k+2-j}+x^{2k+1-j})^2]$$ for all $k\ge1$. As you have shown the base case, suppose this is true for some $k\in\Bbb N$. Then consider $k+1$. \begin{align}\left(\sum\limits_{i=0}^{2k+3}x^i\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all natural numbers $n$ such that $2n+1$ divides $n^{4}+n^{2}$.
Find all natural numbers $n$ such that $2n+1$ divides $n^{4}+n^{2}$.
My attempt:-
We need to find natural numbers, $n$, for which $2n+1\mid n^{4}+n^{2}$. We write, $$n^{4}+n^{2}=n^{2}(n^{2}+1)$$. It can be easily proved that $(n^{2},2n+1)=1$(so I ... | Divide $$ \frac{n^4 + n^2}{2n+1} = \frac{n^3}{2} - \frac{n^2}{4} + \frac{5n}{8} - \frac{5}{16} + \frac{5}{8 (2n+1)}$$ and note that the $\frac{5}{8 (2n+1)}$ term, if rewritten in lowest terms, will have an odd factor in the denominator unless $n = 2$, so the whole expression cannot be an integer. Then all that remains ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3369622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the sum to n terms $$S=1^2+3^2+6^2+10^2+15^2+.......$$
My attempt is as follows:
$$T_n=\left(\frac{n\cdot\left(n+1\right))}{2}\right)^2$$
$$T_n=\frac{n^4+n^2+2\cdot n^3}{4}$$
$$S=\frac{1}{4}\cdot\sum_{n=1}^{n}\left(n^4+n^2+2\cdot n^3\right)$$
Now to solve this one has to calculate $\sum_{n=1}^{n}n^4$ which will b... | I change a little the notations from your original wording.
$T(n)=\dfrac{n(n+1)}{2}$
$S(n)=\sum\limits_{k=1}^nT(k)^2$
Cheating on the resulting formula for $f(n)=\dfrac{n(n+1)(n+2)(3n^2+6n+1)}{60}$
we notice that $f(0)=f(-1)=f(-2)=0$.
Is there a way to exploit these negative indices in order to find the $(3n^2+6n+1)$... | {
"language": "en",
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"source": "stackexchange",
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For what values of $a$ does $b\in\mathbb{N}$? I've the following function:
$$b=\frac{1}{2}-\frac{1}{r-2}+\sqrt{\frac{1}{4}-\frac{r-3}{(r-2)^2}+\frac{a(1+a)}{r-2}+\frac{a(a^2-1)}{3}}$$
I know that:
*
*$\text{a}\ge1$ and $a\in\mathbb{N}$;
*$\text{r}\ge3$ and $r\in\mathbb{N}$.
If I set the value of $r$, what does $... | Let $s = r - 2$. Then $$b=\frac{1}{2}-\frac{1}{s}+\sqrt{\frac{1}{4}-\frac{s - 1}{s^2}+\frac{a(1+a)}{r-2}+\frac{a(a^2-1)}{3}}$$ $$b = \frac{1}{2}-\frac{1}{s}+\sqrt{\frac{1}{4}-\frac{1}{s}+\frac{1}{s^2}+\frac{a(1+a)}{s}+\frac{a(a^2-1)}{3}}$$ $$b = \frac{1}{2}-\frac{1}{s}+\sqrt{\left(\frac{1}{2}-\frac{1}{s}\right)^2+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3375647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving equality of $3\times3$ determinant In Basic Mathematics, Lang, exercise of section 17.5...
Gives this answer...
My own calculations gave...
$(x_2-x_1)(x_3^2-x_1^2)-(x_3-x_1)(x_2^2-x_1^2)$
Is it correct?
| Personally, I would calculate in a slightly different way, in order to obtain directly a formula easier to memorise:
\begin{align}
\begin{vmatrix}
1&x_1&x_1^2 \\
1&x_2&x_2^2 \\
1&x_3&x_3^2 \\
\end{vmatrix}
&=
\begin{vmatrix}
0&x_1-x_2&x_1^2-x_2^2 \\
0&x_2-x_3&x_2^2-x_3^2 \\
1&x_3&x_3^2 \\
\end{vmatrix}
=(x_1-x_2)(x_2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Area of a triangle ABC Triangle's $ABC$ angle $C=60°$, $AB = AC+2 = BC-1$.
Find the area of this triangle.
I've tried writing $AB$ as $x$ so $AC= x-2$ and $BC = x+1$.
Then i calculated the area with
$$\frac12\sin(60°)\cdot(x-2)\cdot(x+1)=\frac12\sin(60°)\cdot(x^2-x-2)$$
Now i have no idea what to do next.
| With the theorem of cosines we get
$$c^2=a^2+b^2-2ab\cos(\frac{\pi}{3})$$
substituting $$c=b+2,a=b+3$$ we get an equation for $b$:
$$(b+2)^2=(b+3)^2+b^2-2(b+3)b\cos(\frac{\pi}{3})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3378392",
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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A basket has $10$ green, $10$ blue, and $10$ red balls. In how many ways can $10$ balls be selected from the basket if the number of red balls is odd? I tried to learn generating functions, but I am a way out of it.
And I know a way to solve problems like that by $S_0 - S_1 + S_2 +\cdots$
Please show me in this way bec... | Let $b$, $g$, and $r$ represent, respectively, the number of blue, green, and red balls that are selected from the basket. We know that
$$b + g + r = 10$$
However, we require that $r$ is odd, so there are five cases.
$r = 1$: Then
\begin{align*}
b + g + 1 & = 10\\
b + g & = 9
\end{align*}
which is an equation in th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all integers such that $2n-1 | n^3 +1$ I try
$$2n-1| n^3 +1$$
$$\therefore 2n11 | 2n^3 + 2 -n^2(2n-1)$$
$$\therefore 2n-1| 4n-2$$
But but this is valid for all $n$. How to proceed? Thanks in advance
and I'm sorry if this is a duplicate, I don't see any similar questions
| Problem. Find all integers $n$ such that $2n - 1\, |\, n^3 + 1$.
Sol. Let $(a, b)$ denote the greatest common divisor (gcd) of $a$ and $b$. Then
$$\begin{array}{lll}
(2n - 1, n^3 + 1) &= (2n - 1, n^3 + 2n) \\
&= (2n - 1, n^2 + 2) \mbox{ since } (2n - 1, n) = 1 \\
&= (2n - 1, n^2 + 2 + 4n - 2) = (2n - 1, n(n + 4)) \\
... | {
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"url": "https://math.stackexchange.com/questions/3382247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Which is bigger, $3$ or $\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$? Which is bigger, $3$ or $\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$
I am not sure how to do it. I thought of squaring both sides and moving the $2$ to the other side, then squaring again etc. But ... | Actually, there is a rather simple method to do this:
We can rewrite
$$a=\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$$
into
$$b=\sqrt{2 + \sqrt{3 + \sqrt{100 + \sqrt{100 + \sqrt{100+\cdots}}}}}$$
And we can clearly see that
$$b>a$$
There is a classic method to evaluate continued root. Let
$$x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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Solve in $\mathbb N^{2}$ the following equation : $5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$ Question :
Solve for natural number the equation :
$5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$
My try :
Let : $X=5^{x}$ and $Y=2^{y}$ so above equation
equivalent :
$2X^{2}+(Y-4)X-6Y^{2}-Y+2... | Different substitutions, let
$$ u = 5^x \; , \; \; v = 2^{y-1} $$
Then
$$ \frac{1}{4} \left( (2u+v-2)^2 - 49 v^2 \right) = 0 \; , \; \; $$
$$ \left( 2u+8v-2 \right) \left( 2u-6v-2 \right) = 0 \; , \; \; $$
$$ \left( u+4v-1 \right) \left( u-3v-1 \right) = 0 \; , \; \; $$
Both $u,v > 0$ so we are left with
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3383436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all $n$ natural numbers such that $10\mid n^{10}+1$ Determine all natural numbers $n$ such that :
$10$ divisor of $n^{10}+1$
My attempt :
Let $n=r(\mod{10})$ so $n^{10}+1=(r^{10}+1)(\mod{10})$
This mean : $r^{10}+1=0(\mod{10})$
Now $r\in {0,1,2,3,4,5,6,7,8,9}$ after try I get $r=3,7$
So : $n=10k+3,10k+7$
Is... | You are correct but a few things to make you calculations much fewer.
1) if $n$ and $10$ are not relatively prime then any common divisor $a;a\ne 1$ will not divide $n^{10}+1$ so $10\not \mid n^{10}+1$.
So we need only test $1,3,7,9$
2) $(10-i)^{10}\equiv i^{10}\pmod{10}$ so we only check $1,3$.
3) Eulers Thereom says... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3384211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Cauchy equation $ f ( x + y ) = f ( x ) + f ( y ) $ with an additional condition $ f \left( \frac 1 x \right) = \frac 1 { x ^ 2 } f ( x ) $
Let $ f : \mathbb R \to \mathbb R $ satisfy the following:
*
*$ f ( 1 ) = 1 $;
*$ f ( x + y ) = f ( x ) + f ( y ) $, $ \forall x , y \in \mathbb R $;
*$ f \left( \frac 1 x \ri... | Let $(1)$ denote the additive condition, and $(2)$ denote the condition that $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^2} $. Now note that for all $x\notin \{-1,0\}$, $$\begin{align*}\frac{f\left(x^2+x\right)}{(x^2+x)^2}\stackrel{(2)}=f\left(\frac{1}{x^2+x}\right)&\stackrel{(1)}=f\left(\frac{1}{x}\right)-f\left(\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3384486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the explicit form of $ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1} $. Find the explicit form of
$$
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1}.
$$
Let $S(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1}$. It has radius of convergence $1$.
Let $S_1(x)=xS(x)$. Then $S_1'(x)=\sum_{n=1}^{\infty}\... | $$\log(1+x)=\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n}~~~(1)$$
Multipluy by $x$ on both sides and integrate w.r.t. $x$
$$\int x \log(1+x) dx= \sum_{n=1}^{\infty} (-1)^{n-1}\int \frac{x^{n+1}}{n} dx= \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{n+2}}{n(n+2)} .$$
$$\implies S(x)=\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{n-1}}... | {
"language": "en",
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Express the other roots with the powers of $\alpha$. Let $f(x)=x^4+x+1 \in Z_2[x]$ and $\alpha$ one of its roots'.
Express the other roots using the powers of $\alpha$.
I believe I have to find the exact elements of its Galois group, and use them to express the other roots. The problem is, I could only find the identit... | Hint: Let $\alpha$ be a root of $x^4+x+1$ in $GF(16)$, the splitting field of the polynomial over $GF(2)$. Then $\alpha^4=\alpha+1$ in $GF(16)$ and so each nonzero element of $GF(16)$ can be expressed as a power of $\alpha$:
$$1, \alpha, \alpha^2, \alpha^3,\alpha^4=\alpha+1, \alpha^5=\alpha^2+\alpha, \alpha^6=\alpha^3+... | {
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Find the smallest number for which the number divides completely into 101
Find the smallest number $n,(n>4)$,
$A=\binom{3n-1}{11}+\binom{3n-1}{12}+\binom{3n}{13}+\binom{3n+1}{14}$
for which the number divides completely into $101$.
My solution:
$\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}\\\binom{3n-1}{11}+\bino... | Hint: You're almost there. Since $101$ is a prime number, it will divide $\ \frac{(3n+2)!}{14!(3n-12)!}\ $ evenly if and only if it divides one of the numbers $\ 3n+2,3n+1, 3n, \dots,3n-11\ $.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute the value of the $30$th derivative of the function $g(x) = \sinh{\left(x^4\right)}$ at the origin, i.e. $g^{(30)}(0)$. Compute the value of the $30$th derivative of the function $g(x) = \sinh{\left(x^4\right)}$ at the origin, i.e. $g^{(30)}(0)$.
So we have the (Macluarin) series
\begin{equation*}
\sinh{(x)} = \... | Definitely on the right track. Last thing you need to know is that the coefficient of $x^n$ in the Maclaurin series is given by $a_n=\frac{f^{n}(0)}{n!}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If each pair of equations $x^2=b_1x+c_1=0,x^2=b_2x+c_2 \text{ and } x^2+b_3x=c_3$ have a common root, prove following If each pair of equations $x^2=b_1x+c_1=0,x^2=b_2x+c_2 \text{ and } x^2+b_3x=c_3$ have a common root, prove that
$(b_1+b_2+b_3)^2=4(c_1+c_2+c_3+b_1b_2)$
My attempt is as follows:
For equations $x^2=b_1x... | I finally got this, thanks to @lab bhattacharjee
There are three quadratic equations
\begin{equation}
x^2-b_1x-c_1=0\tag{1}
\end{equation}
\begin{equation}
x^2-b_2x-c_2=0\tag{2}
\end{equation}
\begin{equation}
x^2+b_3x-c_3=0\tag{3}
\end{equation}
Suppose $(1)$ and $(2)$ have a common root as $p$, $(2)$ and $(3)$ have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3397072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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What is $\frac{d^2y}{dx^2}$ of $(y^5) + (x^5) = 8$ when you solve using implicit differentiation?
Find the Second derivative with respect to $x$ of :
$$ y^5 + x^5 = 8$$
When I solved it I got $\frac yx$ as the second derivative but I don’t think that I am right. Could someone explain it to me.
| Starting with
$y^5 + x^5 = 8, \tag 1$
we differentiate once to obtain
$5y^4 y' + 5x^4 = 0, \tag 2$
from which we may isolate $y'$:
$y' = -\dfrac{x^4}{y^4}; \tag 3$
differentiating this expression using the quotient rule yields
$y'' = -\dfrac{4x^3y^4 - 4x^4y^3y'}{y^8} = -\dfrac{4x^3y - 4x^4y'}{y^5}; \tag 4$
substitute (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3397803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Showing continuity of $g(x, y)$ Let $g : \mathbb{R}^2 \to \mathbb{R}$. Let the domain be $y^2 \leq \frac{x}{3}$.
$g(x, y) = \begin{cases}
\frac{x^3}{y^2} \left (\sqrt{x} - \sqrt{y^2 + x} \right ) & y \neq 0\\
0 & y = 0
\end{cases} $
I want to show $g$ is continuous at $(0, 0)$.
We have $|g(x, y) - g(0,0)| = \frac{x^3}... | $|\frac{x^3}{y^2} \left ( \sqrt{y^2 + x} - \sqrt{x} \right )|=|\frac{x^3}{\sqrt{x}+\sqrt{y^2+x}}| \leq |x|^{\frac{5}{2}}$
So the limit of the function at zero is zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3398003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Introduction of a phase term in the solution of the wave equation I would like to understand this step in the solution of the wave equation as presented here. The wave equation is formulated as
$$u_{tt}=c^2\Delta u$$
where $u$ is a function of time $t$ and space coordinates; $c^2$ is a positive constant, and $\Delta u$... | Rewrite that expression in the following way:
$$A\sin(\omega t) + B\cos(\omega t) = \sqrt{A^2+B^2}\left(\frac{A}{\sqrt{A^2+B^2}}\sin(\omega t) + \frac{B}{\sqrt{A^2+B^2}}\cos(\omega t) \right)$$
Notice that the coefficients are not only numbers that have an absolute value less than $1$, they are also conveniently alread... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3398415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove inequalities $\frac {36}{25} \le A(a) < 2$ Given the expression,
$$ A(a) = \frac{\left( 1 +a +\frac 1a\right)^2 }{\left(\frac12+a+a^2\right)\left(\frac12+\frac 1a + \frac{1}{a^2}\right)} $$
with $a > 0$.
Prove the following inequalities:
$$\frac {36}{25} \le A(a) < 2$$
Note that the bounds are rather tight. I en... | In the form given, $A(a)$ is pretty hard to reason around. As a first step let's factor out a factor of $\frac{1}{a^2}$ and see if we can coerce the denominator to look more friendly.
$$
A(a) = \frac{\frac{1}{a^2} {(a^2 + a + 1)}^2}{(a^2 + a + 1 - \frac{1}{2})\frac{1}{a^2}(\frac{1}{2}a^2 + a + 1)} = \frac{{(a^2 + a + 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3399839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Can anyone prove this function is concave? Can anyone help me prove the following function $$f(x) = \frac{ax}{1-x^a}-\frac{x}{1-x}$$ is concave for any parameter $a \geq 1$ when $0<x<1$ ?
I have tried to prove its second derivative is negative, but it becomes another difficult problem.
| $f(x) = \frac{ax}{1-x^a} - \frac{x}{1-x} $ , $a\geq 1 , 0<x<1$
$$f'(x) =\frac{a}{1-x^a} + \frac{ax(ax^{a-1})}{(1-x^a)^2} - \frac{1}{(1-x)^2} $$
$$f'(x) = \frac{a}{1-x^a} + \frac{a^2 x^a}{(1-x^a)^2} - \frac{1}{(1-x)^2}$$
$$f''(x) = \frac{a^2x^{a-1}}{(1-x^a)^2} + \frac{a^3x^{a-1}}{(1-x^a)^2} + \frac{2a^3x^{2a-1}}{(1-x^a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Maximum value of $a+b+c$ in an inequality Given that $a$, $b$ and $c$ are real positive numbers, find the maximum possible value of $a+b+c$, if
$$a^2+b^2+c^2+ab+ac+bc\le1.$$
From the AM-GM theorem, I have
$$a^2+b^2+c^2+ab+ac+bc\geq 6\sqrt[6]{a^4b^4c^4} = 6\sqrt[3]{a^2b^2c^2} \\
6\sqrt[3]{a^2b^2c^2} \le1 \\
a^2b^2c^2 \l... | Hint: the inequality is equivalent to
$$
(a+b)^{2} + (b+c)^{2} + (c+a)^{2} \leq 2.
$$
Now put $x = b+c, y = c+a, z = a +b$ and then we need to find maximum of $a + b + c = \frac{1}{2}(x+y+z)$ under the condition $x^{2} + y^{2} + z^{2} \leq 2$ and you may use Cauchy-Schwartz inequality. Note that you need to check $x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove that $\angle ABC <$ $ 84^\circ$. Angles A, B and C meet the circle at P, Q and R respectively.
I made the picture below, but what would the solution be?
| $\begin{array}{} \text{Euler's relation} & IO=\sqrt{R(R-2r)} & IO=r=1 & R=1+\sqrt{2} \end{array}$
$\begin{array}{} sin(\frac{ π }{6})=\frac{1}{2} & AI=2 & AI^2=1+y_{A}^2 & A=(1,y_{A}=\sqrt{3}) \end{array}$
$\begin{array}{} {O}=λ ∩ ω & O=(h,k) \end{array}$
$\left\{ \begin{array}{} x^2+y^2=1 \\ (x-1)^2+(y-\sqrt{3})^2=(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3403579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Prove with the help of the $\varepsilon$-$n_0$ definition, that $\frac 1 4$ is the limit of the sequence $a_n=\frac{n^2-1}{4n^2+1}$
Prove with the help of the $\varepsilon$-$n_0$ definition, that $\frac 1 4$ is the limit of the number-sequence $a_n=\frac{n^2-1}{4n^2+1}$
Let $\varepsilon>0$. We need to show, that $\fr... | We have $$\left|\frac{n^2-1}{4n^2+1}-\frac{1}{4}\right|=\left|\frac{4n^2-4-4n^2-1}{4(4n^2+1)}\right|=\frac{5}{4(4n^2+1)}=\frac{5}{4(4n^2+1)}<\frac{5}{16n^2}$$
I hope you can finish now!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Fractions in Questions and Answers
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