Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Linear Transformation in $\mathbb R_2[X]$ I'm learning linear algebra, specifically linear transformations, and need help with the following exercise:
Consider the linear transformation $f:\mathbb R_2[X] \rightarrow \mathbb R_2[X]$ satisfying the relations
\begin{align}
f(1+X+X^2)&=1+X+2X^2\\
f(X+X^2)&=1+3X+6X^2\\
f(... | I would say your solution for $\alpha$ is one of the equivalent ways of expressing $f$. Another would be to say that $$f(ax^2+bx+c) = a(4x^2+x) + b(2x^2+2x+1) + c(-4x^2-2x)$$
and then simplify that expression.
For $\beta$, you already have the matrix set up, so the question is for which value of $\mu$ does the equatio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all complex numbers satisfying the equation $\bar{z}+1=iz^2+|z|^2$ Find all complex numbers z satisfying $$\bar{z}+1=iz^2+|z|^2$$
where $i=\sqrt{-1}$
I only know one way i.e. assuming $z=x+iy$ but that process is very cumbersome. I don't know how to proceed otherwise with a shorter approach.
| The answer is$$z=i,i\omega,i\omega^2$$where $\omega,\omega^2$ are the two complex cube roots of unity.
Putting $z=x+iy$,we get
$$(x-y)^2=1+x$$and $$x^2-y^2=-y$$
Let $A=x+y$ and also let $B=x-y$
Substituting in $$(x-y)^2=1+x$$ we get $$B^2=\frac{A+B}{2}$$i.e.$$A=2B^2-B-2$$
Substituting in $$x^2-y^2=-y$$ we get $$AB=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Simplify $\frac {2^{n(1-n)}\cdot 2^{n-1}\cdot 4^n}{2\cdot 2^n\cdot 2^{(n-1)}}$ Simplify::
$$\frac {2^{n(1-n)}\cdot 2^{n-1}\cdot 4^n}{2\cdot 2^n\cdot 2^{(n-1)}}$$
My Attempt:
\begin{align}
&\frac {2^{n-n^2}\cdot 2^{n-1}\cdot 2^{2n}}{2\cdot 2^n\cdot 2^{n-1}}\\
&=\frac {2^{n-n^2+n-1+2n}}{2^{1+n+n-1}} \\
&=\frac {2^{4n-n^2... | $$\frac {2^{4n-n^2-1}}{2^{2n}}=2^{4n-n^2-1-2n}=2^{2n-n^2-1}=2^{-(n-1)^2}$$
where I have used $\frac{a^b}{a^c}=a^{b-c}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
$\binom{28}{2}+\binom{28}{5}+\binom{28}{8}+\cdots+\binom{28}{26}=\frac{2}{3}(2^{27}+1)$ I need to prove the following identity
$$
\binom{28}{2}+\binom{28}{5}+\binom{28}{8}+\cdots+\binom{28}{26} = \frac{2}{3}(2^{27}+1).
$$
I have tried to use the fact that $\binom{m}{n}=\binom{m}{m-n}$ but it doesn't help.
| Another solution(just for fun)
Let $F_j(x)= \sum_{x \in 0..n}\binom{n}{x} | x \mod 3 = j$
$F_0(x) + F_1(x) +F_2(x) =2^x$
$F_j(x)=F_j(x-1) + F_{(j-1) \mod 3}(x-1)$
Self-cast previous equation
$F_j(x) = F_j(x-2) + 2*F_{(j-1) \mod 3}(x-2) + F_{(j-2) \mod 3}(x-2)$ =
$2^{x-2} + F_{(j-1) \mod 3}(x-2)$
Another self-cast
$F_j... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
find range of $a^2+b^2$ without trigonometric substution given $7a^2-9ab+7b^2=9$ and $a,b$ are real no. then find range of $a^2+b^2$ without trigonometric substution
from $7a^2-9ab+7b^2=9$
$\displaystyle ab = \frac{7(a^2+b^2)-9}{9}$
put into inequality $\displaystyle a^2+b^2 \geq 2ab$
$\displaystyle a^2+b^2 \geq \frac{... | It seems that you've got $\frac{18}{23}$ as the minimum value incorrectly though the value itself is correct. (you have already corrected the error.)
Let $k=a^2+b^2$.
Then, we get
$$ab=\frac{7k-9}{9}\tag1$$
Also, from
$$k=(a+b)^2-2ab=(a+b)^2-2\times\frac{7k-9}{9}$$
we get
$$(a+b)^2=\frac{23k-18}{9}\tag2$$
Therefore, we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Probability that my roll on a die will be higher than yours: Why divide by 6? I have to work out a question where on two fair, $6$ sided dice, what is the probability that the second die gives me a higher number than the first die.
So I broke the question down the long way and said "If you roll a $1$, I have a $\frac{5... | Okay:
If the first die is a $1$ the probability of the second die being higher is $5/6$. But that is the probability only if the first die is $1$.
If the second die is $2$ the probability of the second die being higher is $2/3$. But again that is only the probability if the first die is $2$.
And so one. You have six... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2073994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 3
} |
Confusion in fraction notation $$a_n = n\dfrac{n^2 + 5}{4}$$
In the above fraction series, for $n=3$ I think the answer should be $26/4$, while the answer in the answer book is $21/2$ (or $42/4$). I think the difference stems from how we treat the first $n$. In my understanding, the first number is a complete part and ... | in elementary school math the fraction $x\frac{y}{z}$ usually means $x+\frac{y}{z}$ and is called a mixed fraction.
However these are almost never used after junior high.
Most of the time when you see $x\frac{y}{z}$ the two terms should be multiplied, so it is equal to $\frac{xy}{z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 7,
"answer_id": 2
} |
Generating functions on recursive We have a recursive sequence, defined by $r_0 = r_1 = 0$, and
$$r_n = 7r_{n-1} + 4r_{n-2} + n + 1,\quad \text{for } n\geq 2.$$
Express the generating function of this sequence as a quotient of polynomials or products of polynomials.
I have written the sequence
$$R(x) = r_0(0)+r_1(x... | Indeed, begin by letting
$$R(x) = r_0 + r_1 x + r_2 x^2 + \dots = \sum _{n \ge 0} r_n x^n$$
be a formal series.
Notice that
$$R(x) = \underbrace{r_0} _{=0} + \underbrace{r_1} _{=0} x + \sum _{n \ge 2} \big( 7 r_{n-1} + 4 r_{n-2} + (n+1) \big) x^n = \\
\sum _{m \ge 1} 7 r_m x^{m+1} + \sum _{p \ge 0} 4 r_p x^{p+2} + \sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A simpler proof of $(1-x)^n<\frac{1}{1+nx}$ I proved the following inequality:
Let $x\in\mathbb{R}, 0<x<1, n\in\mathbb{N}$, then $(1-x)^n<\frac{1}{1+nx}$
however, judging from the context in the exercise book, I feel like there is a much simpler way to prove it, but I can't see it. So I'm asking for that simpler alte... | $(1-x)^n = 1/(1-x)^{-n}=1/(1+nx+n(n-1)x^2/2 +\cdots) < 1/(1+nx)$, assuming $ 0 < x < 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
A system of Diophantine equations Is there any positive solution for the Diophantine system
$$\left\{\begin{array}{l}
x^2-4y=z^2 \\
y^2-4x=w^2
\end{array} \right. ,$$
except that $(x,y)=(4,4), (6,5)$ (up to symmetry)?
| $x^2-4y=z^2$
$y^2-4x=w^2$
$x^2-y^2-4(y-x)=z^2-w^2$
$(x-y)(x+y+4)=(z-w)(z+w)$
Since factors in LHS and also in RHS of this relation are co-prime then we can write:
$x-y=z-w$
$x+y+4=z+w$
$⇒ 2x+4=2z$ $⇒z=x+2$
$2w=2y+4$ $⇒w=y+2$
Plugging z or w in 1st or 2nd equation we get:
$x+y=-1$
Therefore:
$z+w=x+y+4=-1+4=3$
Now w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that that $a>\frac{3}{4}$ Let $x^4+2ax^3+x^2+2ax+1=0$ .
If this equation has at least two real negative roots then show that that $a>\frac{3}{4}$.
I tried to solve the equation first by using the sub $t=x+ \frac{1}{x}$.
Then I got $t^2+2at-1=0$.
Thereafter how can I proceed to get the required inequality ?
| If we solve $\quad x^4+2ax^3+x^2+2ax+1=0\quad$ for $a$ we get
\begin{equation}
a = -\frac{x^4 + x^2 + 1}{2 x^3 + x}\quad \text{for}\quad x^3 + x\ne0
\end{equation}
which suggests that $\quad a<0\quad $ but
$\quad x=1\implies a=-\frac{3}{4}\quad $ and $\quad x=-1\implies a=\frac{3}{4}$
If we simply try a range of valu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
(floor function) sum of x: $\left\lfloor{\frac{x}{5}}\right\rfloor - \left\lfloor{\frac{x}{9}}\right\rfloor = \frac{x}{15}$ Find the sum of integers $x$ such that
$$\left\lfloor{\frac{x}{5}}\right\rfloor - \left\lfloor{\frac{x}{9}}\right\rfloor = \frac{x}{15}$$
I don't have much experience in solving problems with th... | First, since the LHS is an integer, we know $x$ is a multiple of $15$. Thus, let $x = 15n$. Then our equation reduces to
$$n = 3n - \left\lfloor \frac{15n}{9}\right\rfloor \implies$$
$$2n = \left\lfloor \frac{15n}{9}\right\rfloor \leq \frac{15n}{9}$$
thus ruling out any positive solutions. Now note that
$$2n = \left\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Trignometric integration We have to solve the following integration.
$$\int\frac{\tan 2\theta}{\sqrt{\cos^6 \theta+\sin^6\theta}}\ d\theta$$
In this, I write the denominator as $\sqrt 1-3\sin^2\theta \cos^2\theta$
But now how to proceed?
| hint: Put $t = \cos (2\theta)\implies 1 - 3\sin^2\theta\cos^2\theta = 1 - \dfrac{3}{4}\sin^2(2\theta)= \dfrac{1}{4} + \dfrac{3}{4}\cos^2(2\theta)\implies I = -\displaystyle \int\dfrac{dt}{t\sqrt{1+3t^2}}$. From this put again $\sqrt{3}t= \tan(\phi)$. Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to find two variables $a,b \in {\bf Z}$ that the matrix $A$ is orthogonal I have to find two variables $a,b \in {\bf Z}$ that the given $n \times n$ matrix A becomes orthogonal.
\begin{equation*}
A =
\begin{pmatrix}
1&2 \\
a&b
\end{pmatrix}
\end{equation*}
I know that a $n \times n$ matrix is called orthogonal ... | There is no solution since the norm of the column vector of an orthogonal matrix must be $1$. (Note that the second column is $(2,b)^T$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Factorization of quartic polynomial. I want to know other ways of factorization to get quadratic factors in this polynomial:
$$x^4+2x^3+3x^2+2x-3$$
Thanks in advance for your suggestions.
The original polynomial is $$x^6-2x^3-4x^2+8x-3$$ where the two found factors are $(x+1)$ and $(x-1)$ by synthetic division.
| Splitting up the x^2 term works well for factorising quartic but one has to be smart in grouping terms
In this , write 3x^2 as x^2 + 2x^2 and group terms as
= x^2(x^2+2x+1) +2x^2+2x-3 =[x(x+1)]^2 +2x(x+1) -3
= [x(x+1) +3][x(x+1)-1]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer When checked in calculator it is 1. But how to prove it?
Also it is not a normal addition like $x+ \frac1x$ which needs direct rationalization. So I just need something to proceed.
| We have $x- \left(2+\sqrt 5 \right)^{\frac{1}{3}} - \left(2-\sqrt 5 \right)^{\frac{1}{3}} = 0$
So $x^3 -(2-\sqrt 5)-(2+\sqrt 5) = 3x \left(2+\sqrt 5 \right)^{\frac{1}{3}}\left(2-\sqrt 5 \right)^{\frac{1}{3}}= -3x$
(invoking that $a^3+b^3+c^3 = 3abc$ when $a+b+c = 0$)
from which we see that $x$ is a root of $x^3+3x-4 =0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 1
} |
Prove the Identity $\dfrac{\tan\theta}{\cos\theta-\sec\theta} = -\csc\theta$ Been trying this question for over an hour and would like to know how it's done. Thanks.
| \begin{align}
& \frac{\frac{\sin\theta}{\cos\theta}}{\cos\theta - \frac{1}{\cos\theta}} \\[10pt]
= {} & \frac{\frac{\sin\theta}{\cos\theta}}{\frac{\cos^2\theta - 1}{\cos\theta}} \\[10pt]
= {} & \frac{\frac{\sin\theta}{\cos\theta}}{\frac{-(1 - \cos^2\theta)}{\cos\theta}} \\[10pt]
= {} & \frac{\frac{\sin\theta}{\cos\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Which integration formula for $\frac{1}{a^2-x^2}$ is correct? In my book integration formula for $\frac{1}{x^2-a^2}$ is given as $\frac{1}{2a}\ln(\frac{x-a}{x+a})$.
From the above formula we can write the formula for integration of $\frac{1}{a^2-x^2}$ as $-\frac{1}{2a}\ln(\frac{x-a}{x+a}) = \frac{1}{2a}\ln(\frac{x+a}... | We do have that $$I =\int \frac {1}{a^2-x^2} dx =\frac {1}{2a}\ln \frac {x+a}{x-a} $$ But however we apply the absolute value function to arguments of logarithm functions in order to extend the antiderivative's domain giving us: $$I =\int \frac {1}{a^2-x^2} dx =\frac {1}{2a} \ln |\frac {x+a}{x-a}|$$Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find $x^2+y^2+z^2$ if $x+y+z=0$ and $\sin{x}+\sin{y}+\sin{z}=0$ and $\cos{x}+\cos{y}+\cos{z}=0$ for any $x,y,z \in [-\pi,\pi]$ Find $x^2+y^2+z^2$ if $x+y+z=0$ and $\sin{x}+\sin{y}+\sin{z}=0$ and $\cos{x}+\cos{y}+\cos{z}=0$ for any $x,y,z \in [-\pi,\pi]$.
My attempt:I found one answer $x=0,y=\frac{2\pi}{3},z=-\frac{2\pi... | HINT.- $$\begin{cases}\sin{x}+\sin{y}+\sin{z}=0\\\cos{x}+\cos{y}+\cos{z}=0\end{cases}\iff\begin{cases}2\sin\dfrac{x+y}{2}\cos\dfrac{x-y}{2}=-\sin z\\2\cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2}=-\cos z\end{cases}\Rightarrow \tan\frac{x+y}{2}=\tan z$$
Hence, because of $x,y,z \in [-\pi,\pi]$, $$x+y=2z\iff x+y+z=3z\Rightarrow \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find an analytic function $w=u+iv$ such that $u=\frac{x(1-x^2-y^2)}{4x^2y^2+(1+y^2-x^2)^2}, w(0)=0$
Find an analytic function $w=u+iv$ such that
$$
u=\frac{x(1-x^2-y^2)}{4x^2y^2+(1+y^2-x^2)^2}, w(0)=0
$$
This is the assignment under b.) and on a.) the answer is given to use the substitutions:
$$x=\frac{z+\overline... | Simplifying the denominator gives
$$
u = \frac{(z + \bar{z})(z\bar{z}-1)}{2(z^2-1)(\bar{z}^2-1)}
$$
You can then separate the denominator to get
$$
u = \frac{1}{2}\left(\frac{z}{z^2-1} + \frac{\bar{z}}{\bar{z}^2-1}\right)
$$
This is clearly the real part of the analytic function $z/(z^2-1)$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int\sin^2(x)\cos^4(x)dx$ As in the title: how does one integrate $$ \int\sin^2(x)\cos^4(x)dx? $$
Any hints? Integration by parts doesn't seem to be a reasonable technique here.
| Or you could separate it into two integrals right from the beginning:
$$
\int \sin^2 x \cos^4 x \ dx = \int \cos^4 x \ dx - \int \cos^6 x \ dx.
$$
Then, we have
$$
\cos^4 x = \frac{3 + \cos 4x + 4\cos 2x}{8}
$$
and
$$
\cos^6 x = \frac{10 + \cos 6x + 6 \cos 4x + 15 \cos 2x}{32}.
$$
On completing the integration, the an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Determine whether the following sequence is increasing or decreasing $\frac{n^2+2n+1}{3n^2+n}$ Determine whether the following sequence is increasing or decreasing:
$$\frac{n^2+2n+1}{3n^2+n}$$
I'm not sure whether my solution is correct:
$$\frac{n^2+2n+1}{3n^2+n}=\frac{n(n+2)+1}{n(3n+1)}=\frac{n+2}{3n+1}+\frac{1}{n(3n... | Let $$a_n=\frac{n^2+2n+1}{3n^2+n}$$ Then
$$a_{n+1}-a_n=\frac{-5n^2-11n-4}{(3n^2+7n+4)(3n^2+n)}<0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Finding $B,C$ such that $B\left[\begin{smallmatrix}1&2\\4&8\end{smallmatrix}\right]C=\left[\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right]$
State $B,C$, such that $B\begin{bmatrix} 1 & 2 \\ 4 & 8 \end{bmatrix}C=\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}$
I tried some things, but I ended up with non inver... | $B,C$ are not unique.
$\begin {bmatrix} 1&2\\4&8 \end{bmatrix} = B^{-1}
\begin {bmatrix} 1&0\\0&0 \end{bmatrix}C^{-1}$
Suppose
$B^{-1} = \begin {bmatrix} b_{11}&b_{12}\\b_{21}&b_{22} \end{bmatrix}$ and
$C^{-1} = \begin {bmatrix} c_{11}&c_{12}\\c_{21}&c_{22} \end{bmatrix}$
$\begin {bmatrix} 1&2\\4&8 \end{bmatrix} = $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Evaluate $f^{2016}(0)$ for the function $f(x)=x^2 \ln(x+1)$
Question: Evaluate $f^{2016}(0)$ for the function $f(x)=x^2 \ln(x+1)$
Background: I know we can use taylor/mclaurin/power series to solve this but these were not teached to me by my teacher and not part of the syllabus so we are required to find the $nth$... | It is okay to use partial fractions and derive the result, but you should still eventually prove that $n$-th derivative formula is correct. You claim it is for $n\geq 3$, so why not to use induction for this?
For $n=3$ you can simply verify this (and you did that).
For induction step, you assume
$$f^{(n)}(x)=(-1)^{n-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find all $n \in \mathbb{N}$ for which $(2^n + n) | (8^n + n)$.
Find all $n \in \mathbb{N}$ for which $(2^n + n) | (8^n + n)$.
$n = 1, 2, 6$ are some solutions. Also, if the above holds then
$$(2^n + n) | 2^n(2^n-1)(2^n+1)$$
and
$$(2^n + n)| n(2^n+1)(2^n-1)$$
I've tried using cases when $n$ is even or odd and have tr... | More generally,
if
$(a^n+n) |((a^3)^n+n)$,
then,
since
$(a^3)^n+n^3
=(a^n)^3+n^3
=(a^n+n)((a^2)^n-na^n+n^2)
$,
we have
$(a^3)^n+n
=(a^3)^n+n^3-n^3+n
=(a^n+n)((a^2)^n-na^n+n^2)-n^3+n
$
so
$(a^n+n)
| (n^3-n)
$.
Therefore
$a^n+n \le n^3-n$,
which bounds $n$
as a function of $a$.
In particular,
$a^n < n^3$,
so
$a < n^{3/n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $x-\sqrt{x^2-x+1}<\frac{1}{2} $ for every real number $x$, without using differentiation Let $f$ be a function defined by :
$$f(x)=x-\sqrt{x^2-x+1}$$
Show that:
$$\forall x\in\mathbb{R},\quad f(x)<\dfrac{1}{2} $$
without use notion of différentiable
let $x\in\mathbb{R}$
\begin{aligned}
f(x)-\dfrac{1}{2}&=x-\... | For $x\le 1/2$ is trivial. For $x > 1/2$:
$$x - 1/2 < \sqrt{x^2-x+1}\iff (x-1/2)^2< x^2-x+1\iff 1/4 < 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
The sum of series $1\cdot 3\cdot 2^2+2\cdot 4\cdot 3^2+3\cdot 5\cdot 4^2+\cdots \cdots n$ terms The sum of series $1\cdot 3\cdot 2^2+2\cdot 4\cdot 3^2+3\cdot 5\cdot 4^2+\cdots \cdots n$ terms
i have calculate $a_{k} = k(k+2)(k+1)^2$
so $\displaystyle \sum^{n}_{k=1}a_{k} = \sum^{n}_{k=1}k(k+1)^2(k+2)$
i wan,t be able g... | Hint :
Since we have
$$(k+3)(k+4)-(k-2)(k-1)=10k+10$$
$$\implies \frac{(k+3)(k+4)-(k-2)(k-1)}{10}=k+1$$
we get
$$\begin{align}k(k+2)(k+1)^2&=k(k+1)(k+2)\color{red}{(k+1)}\\\\&=k(k+1)(k+2)\cdot\color{red}{\frac{1}{10}((k+3)(k+4)-(k-2)(k-1))}\\\\&=\frac{1}{10}(k(k+1)(k+2)(k+3)(k+4)-(k-2)(k-1)k(k+1)(k+2))\\\\&=\frac1{10}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Inverse Mellin transform of $f(s)= 2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)}$ Given the function
$$f(s)= 2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)},$$
can we find and inverse Mellin transform for $f(s)$? T... | Let me offer a partial solution, which I might finish later. Repeatedly using the Gauss multiplication formula yields the following identity:
$$2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)}=\frac1{\sqrt{\pi } \sqrt[6]{2}}\frac{\Gamma \left(\frac{s}{6}+\frac{11}{12}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Other way to derive/prove $a_n = \frac{r^{n+1} - 1}{r-1}$ for sum of geometric sequence? The way I know, for a sequence $a_1 = 1 + r + r^2 + ... + r^n$, is to create another sequence $a_2 = r \cdot a_1 = r + r^2 + ... + r^{n+1}$, then subtract $a_1$ from $a_2$, to end up with $$a_2 - a_1 = r^{n+1}-1 = ra_1-a_1 = a_1(r-... | Do you remember fat fit identity ?$$(x-y)=(x-y).1\\
(x^2-y^2)=(x-y)(x+y)\\
(x^3-y^3)=(x-y)(x^2+xy+y^2)\\
(x^4-y^4)=(x^3+x^2y+xy^2+y^3)\\...\\
(x^n-y^n)=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+...x^{2}y^{n-3}+xy^{n-2}+y^{n-1})$$
so now :take $x=r ,y=1$
you will have
$$(x^n-y^n)=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+...x^{2}y^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $n=2^{31}\cdot 3^{19}$ then how many divisors of $n^2$ are smaller than $n$ but they aren't divisor of $n$? If $n=2^{31}\cdot 3^{19}$ then how many divisors of $n^2$ are smaller than $n$ but they aren't divisor of $n$?
It is clear that the power of $2$ or $3$ should be bigger than how much there are in $n$.For examp... | Notice that if $d|n^2$ and $d<n$ then $n^2/d>n$.
We conclude the number of divisors of $n^2$ less than $n$ are $\frac{\tau(n^2)-1}{2}$
So the answer is $\frac{\tau(n^2)-1}{2}-\tau(n)+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$ The inequality:
$$\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$$
Conditions:... | Let $\{a,b,c,d\}=\{x,y,z,t\}$, where $x\geq y\geq z\geq t$, $x+y+z+t=4$, $x+t=2u$ and $y+z=2v$.
Hence, $u+v=2$ and by Rearrangement we obtain: $$\sum\limits_{cyc}\frac{a^2}{(a+b+c)^2}\geq\frac{x^2}{(x+y+z)^2}+\frac{y^2}{(x+y+t)^2}+\frac{z^2}{(x+z+t)^2}+\frac{t^2}{(y+z+t)^2}=$$
$$=\frac{x^2}{(4-t)^2}+\frac{t^2}{(4-x)^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
How can I prove this inequality in a triangle: $\frac{1}{(b+c-a)} +\frac{1}{(c+a-b)} +\frac{1}{(a+b-c)} \gt \frac{1}{a} +\frac{1}{b} +\frac{1}{c}$?
If $a,b,c$ are the sides of a triangle then show that-
$\frac{1}{(b+c-a)} +\frac{1}{(c+a-b)} +\frac{1}{(a+b-c)} \gt \frac{1}{a} +\frac{1}{b} +\frac{1}{c}$
I got this probl... | hint: use $\dfrac{1}{x}+\dfrac{1}{y} \ge \dfrac{4}{x+y}$ $ $ $3$ times. The first time for the first two terms with $x = a+b-c, y = b+c-a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to find the limit $z\rightarrow i$? How to calculate the limit of $\frac{z^{10}+1}{z^6+1}$ as $z\rightarrow i$?
I tried to take the limit but the denominator becomes zero. Does this mean that the limit is infinite?
| To avoid L'Hopital rememeber that
$$\begin{align}x^{10}+1=&(x^2+1)(x^8-x^6+x^4-x^2+1)\\
x^6+1=&(x^2+1)(x^4-x^2+1)\end{align}$$
So the fraction simplifies to
$${x^8-x^6+x^4-x^2+1\over x^4-x^2+1}\to {5\over 3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find all reals $x$,$y$ satisfying the following equation:
Find all positive reals $x,y \in \mathbb{R}^+$ satisfying:
$$\frac{x^9}{y} + \frac{y^9}{x} = 10-\frac{8}{xy}$$
Since this involves higher exponents I am unable to tackle this problem. Please help me.
| Hint: Write it as:
$x^{10}+y^{10}=10xy-8$
Then use AM-GM:
$10xy-5=x^{10}+y^{10}+1+1+1\ge5\sqrt[5]{x^{10}y^{10}}=5x^2y^2$
Which gives $5(xy-1)^2\le0$, so $xy=1$
Then initial equation can be rewritten as $x^{10}+\frac{1}{x^{10}}=2$
Which by AM-GM again, (or by writing it as a $(x^{10}-1)^2=0$) has the solution, $x^{10}=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Computing $ \lim \limits_{x \rightarrow \infty} \left(1+\frac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}$ For $a \geq 0 $, the following limit
$$ L = \lim_{x \rightarrow \infty} \left(1+\dfrac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}$$
can be computed by applying L'Hopital rule as follows
$$L = \exp \left(\lim_{x \rightarrow \infty... | I would start with
$$\left(1+\frac1x\right)^{\sqrt{ax^2+bx+c}} = \left(1+\frac1x\right)^{\sqrt a x \cdot\sqrt{1+\frac{b}{ax} + \frac{x}{ax^2}}} =\left(\left(1 + \frac1x\right)^x\right)^{\sqrt{a}\cdot \sqrt{1+\frac{b}{ax} + \frac{x}{ax^2}}}$$
Now, use the fact that, if $f$ and $g$ are continuous and all limits exist, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2101388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Show the recursive sequence is increasing
How do I show that the recursive sequence
$$a_n = a_{\lfloor n/2 \rfloor} + a_{\lceil n/2 \rceil} +3n+1, \quad
n\geq 2, \phantom{x} a_1 = 3$$
is an increasing sequence?
1. attempt:
If I can show that $a_{n+1}-a_n>0$, I would be able to show it is increasing.
\begin{align*}
a... | Show $a_{n+1}>a_n$ be induction. Note that $\lceil(n+1)/2\rceil $ is either $\lceil n/2\rceil$ or $\lceil n/2\rceil+1$. Hence by induction hypothesis (which is applicable if $\lceil n/2\rceil<n$, i.e., for $n\ge2$), we may use that $a_{\lceil(n+1)/2\rceil}$ is $>$ or $=$ to $a_{\lceil n/2\rceil}$. The same works for fl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Express the $n$th smallest number in this set in terms of $n$
Suppose $n$ is a perfect square. Consider the set of all numbers which are the product of two numbers, not necessarily distinct, both of which are at least $n$. Express the $n$th smallest number in this set in terms of $n$.
If, for example, $n = 4$, then t... | Clearly there $w=(n+\sqrt n -1)^2$ is at least the $n$'th number in the list.
We must only prove that $w<ab$ if $a+b>2n+2\sqrt{n}-2$, with both $a$ and $b\geq n$.
Clearly this product is minimized in the case $a=n$, Which leaves $b>n+2\sqrt{n}-2$, since $b$ is an integer we can take the minimal case $b=n+\sqrt{2n}-1$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the matrix associated with a linear map
Find the matrix associated with the linear map $f:R^2 \rightarrow R^2$ defined by $f(x,y)=(3x-y,y-x)$ with respect to the ordered basis ${(1,0),(1,1)}$
Let the matrix be $A$ and let $f(x)=AX$ where
$$X=\begin{bmatrix}
x \\
y
\end{bmatrix}$$
and
$$A=\begin{bmatrix}
... | You can set up the matrix with respect to the standard basis and then convert it to the matrix with respect to the given basis by multiplying with the appropriote change of bases matrices.
Altneratively, you can get the matrix of the linear map with respect to the given basis directly. In general: the $i$-th column of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
If $2^{2017} + 2^{2014} + 2^n$ is a perfect square, find $n$. If $2^{2017} + 2^{2014} + 2^n$ is a perfect square, find $n$.
My first shot would be to assume the perfect square is $2^{2018}$, but how would I prove that? Even if it is, what is $n$? All help is appreciated.
| If you want an answer then $(a+b)^2 = a^2 + 2ab + b^2$ should suggest an obvious answer by setting $a^2 = 2^{2014}; 2ab=2^{2017}; 2^n = c^2$.
i.e $(2^{1007} + 2^{\frac n2})^2 = 2^{2014} + 2*2^{1007}*2^{\frac n2} + 2^n= 2^{2014} + 2^{1 + 1007 + \frac n2} + 2^n$. We just have to solve for $1 + 1007 + \frac n2 = 2017$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Evaluating $2x^3+2x^2-7x+72$ where $x= \frac{3-5i}{2} $ So I have to find the value of :
$$2x^3+2x^2-7x+72$$ at $$x= \frac{3-5i}{2} $$
Where $i$ stands for $\sqrt{-1}$.
I know the obvious approach would be to substitute the value of $x$ in the equation, but the equation gets extremely messy in that process. Is there ... | You have $\frac{3-5i}{2}$ is the solution of $2x^2-6x+17 = 0$.
Then, $$2x^3 + 2x^2-7x+72 = x(2x^2-6x+17) + 4(2x^2-6x+17) + 4 = 4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the determinant of a 5x5 matrix Find the determinant of the following matrix:
$$\begin{bmatrix}
1& 1& 1& 1& 1\\
3 & 3 &3 &3 &2\\
4& 4& 4& 3& 3\\
5& 5& 4& 4& 4\\
6& 5& 5& 5 &5\end{bmatrix}$$
Laplace doesn't seem like the best method here, can we somehow turn this into a triangular matrix so that the determinant ... | Now you can expand by the first collumn, and you get
$$\begin{vmatrix} 0 & 0 &0 &-1 \\ 0 & 0 &-1 &-1 \\ 0 &-1 &-1 &-1 \\ -1 &-1&-1&-1 \end{vmatrix}$$
which is triangular, so the determinant equals $(-1)^4=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How can I evaluate this series? I have to evaluate the following sum:
$$\sum_{n=0}^{\infty} \frac{n^{2}-5n+2}{n!}$$
Could you recommend some guide for solving this? I have not found any information that I could use for this type of exercise.
| Let us make the problem more general considering $$F(x)=\sum_{n=0}^{\infty} \frac{an^{2}+bn+c}{n!}x^n$$ First, write $$an^2+bn+c=a(n(n-1)+n)+bn+c=an(n-1)+(a+b)n+c$$ $$F(x)=a\sum_{n=0}^{\infty} \frac{n(n-1)}{n!}x^n+(a+b)\sum_{n=0}^{\infty} \frac{n}{n!}x^n+c\sum_{n=0}^{\infty} \frac{x^n}{n!}$$ $$F(x)=ax^2\sum_{n=0}^{\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Simplifying fraction with factorials: $\frac{(3(n+1))!}{(3n)!}$ I was trying to solve the limit:
$$\lim_{n \to \infty} \sqrt[n]{\frac{(3n)!}{(5n)^{3n}}}$$
By using the root's criterion for limits (which is valid in this case, since $b_n$ increases monotonically):
$$L= \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \... | By Stirling's inequality the answer is clearly $\left(\frac{3}{5e}\right)^3$. To prove it, you may notice that by setting
$$ a_n = \frac{(3n)!}{(5n)^{3n}} $$
you have:
$$ \frac{a_{n+1}}{a_n} = \frac{(3n+3)(3n+2)(3n+1)(5n)^{3n}}{(5n+5)^{3n+3}} = \frac{\frac{3n+3}{5n+5}\cdot\frac{3n+2}{5n+5}\cdot\frac{3n+1}{5n+5}}{\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2109192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Solve using generating functions the following recurrence $F_0 = F_1 = 0$
$F_2 = 1$
$F_n = F_{n-1} + F_{n-2} + F_{n-3} , n \ge 3 $
Hint: You will have to write the recurrence in relation to the Fibonacci sequence. I seem to be stuck in this method since I am at a point where $G(z) = \frac{z}{1-z-z^2-z^3}$ and I do n... | The sequence
\begin{align*}
\left(F_n\right)_{n\geq 0}=(0,0,1,1,2,4,7,13,24,44,81,\ldots)
\end{align*}
has the corresponding generating function
\begin{align*}
G(z)=z^2+z^3+2z^4+4z^5+7z^6+\cdots
\end{align*}
Since the series expansion of $G(z)$ at $z=0$ starts with $z^2$ and the geometric series $\frac{1}{1-y}=1+y+y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
find the maximum and minimum value of the function $x^3+y^3-3x-12y+10$ find the maximum and minimum value of the function $x^3+y^3-3x-12y+10$? here the question is of $2$ variables and i am not able to solve that ,i know how to solve maximum and minimum question of 1 variable,so please help me to solve this
| For positive variables by AM-GM we obtain:
$$x^3+y^3-3x-12y+10=x^3+1+1+y^3+8+8-3x-12y-8\geq$$
$$\geq3\sqrt[3]{x^3\cdot1\cdot1}+3\sqrt[3]{y^3\cdot8\cdot8}-3x-12y-6=-8.$$
The equality occurs for $x=1$ and $y=2$, which says that
$$\min_{x>0,y>0}(x^3+y^3-3x-12y+10)=-8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
if : $abc=8 $ then : $(a+1)(b+1)(c+1)≥27$ if : $$abc=8 :a,b,c\in \mathbb{R}_{> 0}$$
then :
$$(a+1)(b+1)(c+1)\ge 27.$$
My try :
$$(a+1)(b+1)(c+1)=1+(a+b+c)+(ab+ac+bc)+abc$$
$$(a+1)(b+1)(c+1)=1+(a+b+c)+(ab+ac+bc)+8, $$
then?
| Using $\bf{A.M\geq G.M}$
$$\frac{a}{2}+\frac{a}{2}+1\geq 3\left(\frac{a^2}{4}\right)^{\frac{1}{3}}$$
Similarly $$\frac{b}{2}+\frac{b}{2}+1\geq 3\left(\frac{b^2}{4}\right)^{\frac{1}{3}}$$
$$\frac{c}{2}+\frac{c}{2}+1\geq 3\left(\frac{c^2}{4}\right)^{\frac{1}{3}}$$
So $$(a+1)(b+1)(c+1)\geq 27 \left(\frac{(abc)^2}{64}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Converting a power series to a function
$$\sum (n+1)^2x^n = {1+x\over (1-x)^3}$$
I tried to find a similar power series and start from there. I see $ x^n $ and it reminds me of geometric series, but $(n+1)^2$ confuses me. How do I prove this?
| First use the fact that
$$\sum_{n=0}^\infty x^{n+1} = {x\over 1-x}$$
now take the derivative to get
$$\sum_{n=0}^\infty (n+1)x^n = {1\over (1-x)^2}$$
Now multiply by $x$ and take the derivative again:
$$\sum_{n=0}^\infty (n+1)^2x^n = {d\over dx}\left({x\over (1-x)^2}\right) = {(1-x)^2+2x(1-x)\over (1-x)^3}$$
$$={1-x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2115855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is it possible to evaluate $\lim_{x\rightarrow 0}(-1+\cos x)^{\tan x}$? Let's say we have the following limit:
$$\lim_{x\rightarrow 0}(-1+\cos x)^{\tan x}$$
Would the following solution be correct?
The solution is incorrect, please see the correction of @YvesDaoust
\begin{align}
\lim_{x \rightarrow 0}(-1+\cos x)^{\ta... | $$\cos x-1<0$$ in the neighborhood of $0$ so that the function cannot be evaluated (at best values closer and closer to $\pm1$ for rational exponents).
Hence the limit does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
System of two cubic equations: $x+y^2 = y^3$, $y+x^2=x^3$ I got stuck on this system of equations. Could you help and tell me how should I approach this problem?
\begin{align*}
x+y^2 &= y^3\\
y+x^2 &= x^3
\end{align*}
These are the solutions:
\begin{align*}
(0, 0); \\((1+\sqrt{5})/2, (1+\sqrt{5})/2); \\((1-\sqrt... | Here is a dynamical systems approach assuming that we only look for real solutions. Consider the map: $$f(x)=x^3-x^2=x^2(x-1)$$
Solving $f(x)=y$ and $f(y)=x$ is equivalent to solving $f(f(x))=x$. We want to show that a solution of the latter (a periodic orbit of period 2) is necessarily a fixed point, i.e. $f(x)=x$ whi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Given positive integers $ x , y$ and that $3x^2 + x = 4y^2 +y$ Prove that $x-y$ is a perfect square This is some old Olympiad problem (I vaguely remember its Iranian )
I factorised it as
$ (x-y)(3x + 3y +1) = y^2 $
Then I fruitlessly tried to prove
$\gcd(x-y, 3x + 3y +1 ) = \gcd (6x+1,6y+1) = 1$
for proving that $x-y$... | Here is the method to approach this sort of problem. Note that $x>y$. Let $\gcd(x,y)=d$. Now, first note that $$x-y=4y^2-3x^2 \equiv 0 \pmod{d^2} \tag{1}$$
Now, $$4y^2-3x^2=4(y-x)(y+x)+x^2 \equiv 0 \pmod {x-y} $$
And $$4y^2-3x^2=3(y-x)(y+x)+y^2 \equiv 0 \pmod {x-y} $$
This gives us that $x-y$ is a common factor of $x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the integral of $\int \frac{x^4}{x-1}$ I was not happy with the responses so I went ahead and solved the integral to get feedback on the answer:
Dividing as the book shows us:
we get $x^4$ divided by $x-1$ which gives us the quotient plus the remainder over the divisor which is: $x^3 + \frac{x^3}{x-1}$
$\int x^3+\... | Substitute $\text{u}=x-1$:
$$\int\frac{x^4}{x-1}\space\text{d}x=\int\frac{\left(1+\text{u}\right)^4}{\text{u}}\space\text{d}\text{u}=\int\left(\text{u}^3+4\text{u}^2+6\text{u}+\frac{1}{\text{u}}+4\right)\space\text{d}\text{u}=$$
$$\int\text{u}^3\space\text{d}\text{u}+4\int\text{u}^2\space\text{d}\text{u}+6\int\text{u}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$ \lim_{n \rightarrow \infty} n \left( \frac{1}{(n+1)^2} + \frac{1}{(n+2)^2} + \cdots + \frac{1}{(2n)^2} \right)$ as Riemann sum? I am trying to evaluate the limit $$
\lim_{n \rightarrow \infty} n \left( \frac{1}{(n+1)^2} + \frac{1}{(n+2)^2} + \cdots +
\frac{1}{(2n)^2} \right).$$
I have been trying to convert it to th... | The polygamma functions are great to solve this kind of problems.
$$\sum_{k=1}^N \frac{1}{(x+k)^2}=\psi^{(1)}(x+1)-\psi^{(1)}(N+x+1)$$
$\psi^{(1)}(z)$ is the trigamma function, i.e.: the polygamma[1,z] function.
With $z=n$ and $N=n$ :
$$n\sum_{k=1}^n \frac{1}{(n+k)^2}=n\left(\psi^{(1)}(n+1)-\psi^{(1)}(2n+1) \right)$$
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How to prove that a Möbius image is a circle? Let $\mu(z) = \frac{z-ia}{z-ib}$, $z=x+iy$, where $x,y, a, b$ are real, with $a<b$. Define a line $l = x+ia$. We can see that the image of this line under $\mu$ is not a line:
$$\mu(l)=\frac{x}{x+i(a-b)}.$$
We also know that Möbius transformations map lines to lines or circ... | The circle is formed by $\Re(\mu(x))$ and $\Im(\mu(x))$, with $x$ acting as a parameter. To avoid confusion I will call the parameter $t$, and use $c = a - b$.
Obtain real and imaginary parts as $$\frac{t}{t - ic} = \frac{t^2 + ict}{t^2 + c^2} = \frac{t^2}{t^2+c^2} + i\frac{ct}{t^2+c^2}$$ Now let $x = \frac{t^2}{t^2+c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2123158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $\text {limsup}_{x\to \infty}x\int_{x}^{x+1}\sin (t^2) \mathrm{d}t=1$ It is not difficult through the substitution $u=t^2$ to show that $$x\int_{x}^{x+1}\sin (t^2) \mathrm{d}t=\frac 12 \left(\cos(x^2)-\cos[(x+1)^2]\right)+r(x)$$
where $\lim_{x \to \infty} r(x) =0$, it seems intuitive that $\frac 12 \left(\cos(x... | Your initial idea of enforcing the substitution $t\to \sqrt{t}$ is a good one. Indeed we have
$$\int_x^{x+1} \sin(t^2)\,dt=\int_{x^2}^{(x+1)^2}\frac{\sin(t)}{2\sqrt{t}}\,dt$$
Now, let's integrate by parts with $u=\frac{1}{2\sqrt{t}}$ and $v=-\cos(t)$. Then, we have
$$\begin{align}
\int_x^{x+1} \sin(t^2)\,dt&=\frac{\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2123493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $(x-a)^2+y^2=r^2$, prove that $r=\sqrt{(x_1-a)^2+y_1^2}=\sqrt{(x_2-a)^2+y_2^2}$ and $a=\frac{x_2^2+y_2^2-x_1^2-y_1^2}{2(x_2-x_1)}$. Consider this is taking place in the hyperbolic half plane model
If $x_1 \not= x_2$ then the line $\overleftrightarrow{AB}$ has the following equation $(x-a)^2+y^2=r^2$. Prove that $r... | You have:
$$\sqrt{(x_1-a)^2+y_1^2}=\sqrt{(x_2-a)^2+y_2^2}$$
Squaring both the sides we get:
$$(x_1-a)^2+y_1^2=(x_2-a)^2+y_2^2$$
$$(x_1-a)^2-(x_2-a)^2=y_2^2-y_1^2$$
$$(x_1-a-x_2+a)(x_1-a+x_2-a)=y_2^2-y_1^2$$
$$(x_1-x_2)(x_1+x_2-2a)=y_2^2-y_1^2$$
$$x_1^2-x_2^2-2a(x_1-x_2)=y_2^2-y_1^2$$
$$2a(x_2-x_1)=x_2^2+y_2^2-x_1^2-y_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2123776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving trigonometrical identity from given identity.
Given $$\dfrac{\sin^4 A}{a}+\dfrac{\cos^4 A}{b}=\dfrac{1}{a+b}$$ prove that:$$\dfrac{\sin^8 A}{a^3}+\dfrac{\cos^8 A}{b^3}=\dfrac{1}{(a+b)^3}$$
Using given, I proved: $b\sin^2 A=a\cos^2 A$ Help me proceeding after this.
| $\dfrac{\sin^4A}{a}+\dfrac{\left(1-sin^2A\right)^2}{b}-\dfrac{1}{a+b}=0$
$\sin^4A\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{2\sin^2A}{b}+\dfrac{1}{b}-\dfrac{1}{a+b}=0$
$\left(\sin^2A-\dfrac{a}{a+b}\right)^2=0$
$\sin^2A=\dfrac{a}{a+b}$
$\cos^2A=\dfrac{b}{a+b}$
You can take it up from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Fourier Coefficients for $\frac{3}{5-4\cos{2\theta}}$ I want to compute the (even) fourier coefficients for
$\dfrac 3 {5-4\cos 2\theta}$ on the interval $[0, 2\pi]$.
Namely, I want to compute the integral:
$$b_n = \int_0^{2\pi} \cos(n\theta) \frac 3 {5-4\cos 2\theta } \frac{d\theta}{2\pi}$$
Integrating this in mathem... | Complex analysis approach:
Let $$f(z)=\frac{3}{5-2(z^2+z^{-2})}=\frac{-3z^2}{2z^4-5z^2+2}=\frac{1}{2z^2-1}-\frac{2}{z^2-2}$$ Then $f(e^{i\theta})=\frac{3}{5-4\cos 2\theta}$.
Now, for $|z|<\sqrt{2}$, we have:
$$\frac{-2}{z^2-2}=\frac{1}{1-z^2/2}=\sum_{k=0}^{\infty} \frac{1}{2^k}z^{2k}$$
and for $|z|>\frac{1}{\sqrt{2}}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $A^n$ for all $n \in \mathbb{N}$. I'm learning linear algebra and need help with the following problem:
Let $A = \begin{pmatrix}-2 & -3 & -3\\-1 & 0 & -1\\5 & 5 & 6\end{pmatrix} \in M_{3x3}(\mathbb{R}).$ Find $A^n$ for all $n \in \mathbb{N}$.
My first thought was to compute $A^2$, $A^3$, $A^4$ and see if a patte... | The matrix is similar to the diagonal matrix $D=diag(1,1,2)$. Hence we have
$$
A^n=(SDS^{-1})^n=SD^nS^{-1}
$$
with
$$
S=\begin{pmatrix} -5 & -4 & 3\cr -1 & -1 & 1\cr 6 & 5 & -5\end{pmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
if $z=2+4i$, find $\sqrt{z}$ when $\sqrt{z}=a+bi$ Question
if $z=2+4i$, find $\sqrt{z}$ when $\sqrt{z}=a+bi$
what I have so far
$\sqrt{z}=a+bi$, so I square both sides ->
$z=a^2+2abi-b^2$
we can substitute for z
$z=2+4i=a^2+2abi-b^2$
that means $a^2-b^2=2$ and $2abi=4i$
and ab=2 so then $b=\frac{a}{2}$
from then i dont... | You have
$$a^2-b^2=2\quad (1)\\
ab=2\to b=\frac{2}{a}\quad (2)$$
Plug $(2)$ in $(1)$
$$a^2-\frac{4}{a^2}=2\to a^4-2a^2-4=0$$
so
$$a^2=\frac{2\pm\sqrt{20}}{2}=1\pm\sqrt{5}$$
so,
$$a=\pm\sqrt{1+\sqrt{5}}\to b=\pm\frac{2}{\sqrt{1+\sqrt{5}}}=\pm \frac{\sqrt{\sqrt{5}-1}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding basis of subspace from homogenous equations (checking) The question;
$U = \{x |Ax = 0\}$ If $ A = \begin{bmatrix}1 & 2 & 1 & 0 & -2\\ 2 & 1 & 2 & 1 & 2\\1 & 1 & 0 & -1 & -2\\ 0 & 0 & 2 & 0 & 4\end{bmatrix}$
Find a basis for $U$.
To make it linearly independent, I reduce the rows of $A$;
$\begin{bmatrix}1 & 0 ... | $U$ is the set of vectors $\vec x$, whose image under $f_A$ is $\vec 0$.
Since your matrix $A$ is rank 4 and $f_A:K^5 \rightarrow K^4$, it must have a $1$ dimensional Kernel.
So your answer should be subspace of $K^5$ spanned by 1 vector. What vector could that be from your reduced matrix?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many numbers smaller than $10^6$ contain exactly three '$9$'s and have an odd sum of digits? i came up with an idea that i choose
Even Even Even 9 9 9 - and i sort it in ways such that
all even are the same - $\frac{6!}{3!*3!}$
two are the same - $\frac{6!}{3!*2!}$
all are different - $\frac{6!}{3!}$
other one is... | Assuming leading zeros allow us to assume all numbers have six digits you need.
A) 1 non-nine odd, two different evens, 3 nines
plus
B) 1 (non-nine odd), two same evens, 3 nines
plus
C) 3 non-nine odds, all different, 3 nines
plus
D) 3 non-nine odds, two the same, 3 nines
E) 3 non-nine odds, all the same, 3 nines.
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2129311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
Number of ways to get $xy^2$ from $(x+y+5)^5$ I am trying to understand a proof of counting number of trees using the conclusion of Cayley's formula.
At the end they got a $(x+y+5)^5$ and they say that the number of ways to get $xy^2$ from this is $5 \cdot \binom{4}{2} \cdot 5^2 = 750 $ and I can't figure how. Can some... | $$(x+y+z)^n \to \dfrac{n!}{k_1!k_2!k_3!}x^{k_1}y^{k_2}z^{k_3} ,k_1+k_2+k_3=n\\$$
$$(x+y+5)^5 \to \dfrac{5!}{k_1!k_2!k_3!}x^{k_1}y^{k_2}5^{k_3} ,k_1+k_2+k_3=5\\ \to xy^2 \to ,1+2+k_3=5 \to k_3=2\\ \dfrac{5!}{1!2!2!}x^{1}y^{2}5^{2} \\25\dfrac{5!}{1!2!2!}x^{1}y^{2}=25\times 30 xy^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If $\tan^3A+\tan^3B+\tan^3C=3\tan(A)\tan(B)\tan(C)$, prove triangle ABC is equilateral triangle If $\tan^3A+\tan^3B+\tan^3C=3\tan(A)\tan(B)\tan(C)$, prove triangle ABC is equilateral triangle
Now i remember a identity which was like if $a+b+c=0$,then $a^3+b^3+c^3=3abc$. So i have $\sum_{}^{} \tan(A)=0$. How do i procee... | $$a^3+b^3+c^3-3abc=(a+b)^3-3ab(a+b)+c^3-3abc=\{(a+b)+c\}\{(a+b)^2-(a+b)c+c^2\}-3ab\{(a+b)+c\}$$
$$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
But $2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2$
Now if $a^3+b^3+c^3-3abc=0$
either $a+b+c=0$
But with $A+B+C=m\pi,\tan A+\tan B+\tan C=\tan A\tan B\tan C$
As $0<A,B,C<\pi; \tan A\tan ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve the differential equation $2r(1+s^2)dr + (1+r^4)ds = 0$ $2r(1+s^2)dr + (1+r^4)ds = 0$
$\Rightarrow \frac {2r}{1+r^4}dr + \frac {ds}{1+s^2} = 0$ $(1)$
Integrate both sides:
For integration of $\frac {2r}{1+r^4}dr$ $(2)$
I put $1+r^4=t\Rightarrow 4r^3dr=dt\Rightarrow 2rdr = \frac {dt}{2r^2} = \frac {dt}{2\sqrt{t-1... | In addition to @Did's hint, there is a relatively easy way to handle that integral.
We have $$ \frac {2r}{1+ r^4} \mathrm {d}r = - \frac {1}{1+ s^2 } \mathrm {d}s $$ $$\Rightarrow \int \frac {\mathrm {d}(r^2)}{1+(r^2)^2} = \int - \frac {1}{1+ s^2 } \mathrm {d}s$$ $$\Rightarrow \arctan r^2 = - \arctan s +c $$
Hope you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving $\sin 2x + \cos 2x =-1$
$$\sin 2x + \cos 2x =-1$$
How would I go about solving this equation? Some hint on how to start, so I can try to figure it out on my own.
Thank you
| Use sum to product for $\sin x + \sin y$ (it works also for $\cos x + \cos y$):
$$\begin{align}
\sin 2x + \cos 2x&=\sin 2x + \sin\left(\frac{\pi}{2}-2x\right)\\
&=2\sin\frac{2x+\pi/2-2x}{2}\cos\frac{2x-\pi/2+2x}{2}\\
&=2\sin\frac{\pi}{4}\cos\left(2x-\frac{\pi}{4}\right)\\
&=\sqrt 2\cos\left(2x-\frac{\pi}{4}\right)
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 3
} |
Solve the inequality $\sin(x)\sin(3x) > \frac{1}{4}$
Find the range of possible values of $x$ which satisfy the inequation $$\sin(x)\sin(3x) > \frac{1}{4}$$
SOURCE : Inequalities (PDF)( Page Number 6; Question Number 306)
One simple observation is that both $x$ and $3x$ have to positive or negative simultaneously. I ... | You are on the correct path.
Now you need to solve the following equation:
$$\sin^2(x)\cdot \big(3-4\sin^2(x)\big) =\frac{1}{4}$$
$$4\sin^2(x)\cdot\big(3-4\sin^2(x)\big)-1=0$$
$$12\sin^2x-16\sin^4x-1=0$$
$$16\sin^4x-12\sin^2x+1=0$$
So we get by solving, $$\sin^2x=\frac{12\pm \sqrt{144-4\cdot 16}}{2\cdot 16}=\frac{12\pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
What is the z-transform of $\frac{1}{n}$ given that $n \geq 1$ and $x(n) = \frac{1}{n}$, how can I get the z-transform of $x(n)$?
Thanks
| Good Question
$$\frac{u(n-1)}{n} \rightleftharpoons ??$$
We know that Z transform is defined as :
$$X(z) = \sum_{n=-\infty}^{\infty}x(n)z^{-n}$$
$$u(n) \rightleftharpoons \frac{z}{z-1}$$
By the use of time shifting property
$$x(n-1) \rightleftharpoons z^{-1} X(z)$$
$$u(n-1) \rightleftharpoons \frac{1}{z-1}$$
Differe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2137107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Need help understanding certain steps of the solution to a limit. I had trouble evaluating this limit and found this solution from another post of Stack Exchange. The point of this solution was trying to find the limit without using Taylor's series or L'Hospital's rule.
I just need help understanding certain steps:
... | If we have:
$$4L=\lim_{x\to 0} \frac{\frac12\tan 2x-x}{x^3} \qquad L=\lim_{x\to 0} \frac{\tan x-x}{x^3}$$
substracting we would get
$$4L-L=3L=\lim_{x\to 0} \frac{\frac12\tan 2x-\tan x}{x^3}$$
Now, use that $\tan(2x)=\frac{2\tan x}{1-\tan^2 x}$ to go on:
\begin{equation*} \begin{split} 3L&=\lim_{x\to 0} \frac{\frac{\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Coefficients of products of binomials What is the coefficient of the term $\prod_{k=1}^N{z_k^{N-1}}$ in the expansion of $\prod_{1\leq i<j\leq N}{(z_i+z_j)^2}$?
From the symmetry of the problem all that I can argue is that this coefficient shall be the largest of all the other coefficients.
| This is more of a comment but slightly too large to post as such. If you are merely interested in how your coefficient grows with $N$, here are two crude bounds:
Let $C_N^*$ denote the coefficient of $\prod_{k=1}^N{z_k^{N-1}}$ in the expansion of $\prod_{1\leq i<j\leq N}{(z_i+z_j)^2}$. First note that
\begin{equation}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is the product of two consecutive integers $1$ $\pmod n$? Is there a simple test for $n$ to determine if there exists an integer $x$ such that $x$($x+1$) $=$ $1$ $\pmod n$. For example, $n$ $=$ $3$ and $n$ $=$ $7$, there are no integers $x$ such that $x$($x+1$) $=$ $1$ $\pmod n$. But for $n$ $=$ $5$ and $n$ $=$ $11$, t... | First we rewrite the congruence:
$$4x(x+1)=4\pmod{4n}\quad\Leftrightarrow\quad (2x+1)^2=5\pmod{4n}\ .$$
Now write $y=2x+1$ and consider various cases.
*
*If $n$ is even then the above implies $y^2=5\pmod8$, which has no solution.
*If $5^2\mid n$ then we get
$$\eqalign{y^2=5\pmod{25}\quad
&\Rightarrow\quad 5\mid y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
solving a Diophantine system of two equations Find all triples $(a,b,c) \in \mathbb{N}$ satisfing the following equations:-
\begin{align}
a^2 + b^2 & = c^3 \\
(a + b)^2 & = c^4
\end{align}
Thank you for your help.
| First observe that $(a+b)^2=c^4 \implies a+b = c^2 , c>0$ Then we can combine that with $a^2+b^2=c^3$ to get $(a-b)^2=2c^3-c^4$. As $a-b \in N, (2c^3-c^4)^{1/2}= c(2c-c^2)^{1/2} \in N$ Which implies $c = 1,2$ . Checking the two values, $c=0$ has no natural solutions and $c=1$ has one natural solution of $a=2,b=2,c=2$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
finding ratio of two definite integration If $\displaystyle A = \int^{1}_{0}x^{\frac{7}{2}}(1-x)^{\frac{5}{2}}dx$ and $\displaystyle B = \int^{1}_{0}\frac{x^{\frac{3}{2}}(1-x)^{\frac{7}{2}}}{(x+3)^8}dx\;,$ then value of $AB^{-1} = $
Attempt: i have tried using gamma function $\displaystyle \int^{1}_{0}x^m(1-x)^ndx = \f... | We first prove a result : $F(a,b,p)=\displaystyle\int_0^1 \frac{x^{a-1}(1-x)^{b -1}}{(x+p)^{a+b}}\,dx = \left(\frac{\beta(a,b)}{p^b(1+p)^a}\right) \\$
Proof :
We start with the integral $\displaystyle \beta(a,b) = \int\limits_0^1 x^{a-1}(1-x)^{b-1}\; dx \\$
The substitution $y = \dfrac{(p+1)x}{p+x}$ is pretty strai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to deduce $\sin A+\sin B+\sin C=4\cos\frac A2\cos\frac B2\cos\frac C2$ from $A+B+C=\pi$?
If $A+B+C=\pi$,$$\sin A+\sin B+\sin C=4\cos\dfrac A2\cos\dfrac B2\cos\dfrac C2\tag1$$$$\sin A+\sin B-\sin C=4\sin\dfrac A2\sin\dfrac B2\cos \dfrac C2\tag2$$$$\cos A+\cos B+\cos C=4\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2+1\tag3... | $$4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}=2\cos\frac{A}{2}\left(\cos\frac{B+C}{2}+\cos\frac{B-C}{2}\right)=$$
$$=\cos\frac{A+B+C}{2}+\cos\frac{-A+B+C}{2}+\cos\frac{A+B-C}{2}+\cos\frac{A-B+C}{2}=$$
$$=\sin{A}+\sin{B}+\sin{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is this limit $\lim\limits_{x \to\, -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} = 0$? Is this limit $\lim\limits_{x \to\, -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} = 0$?
It's in indeterminant form, 0/0 when $x$ approaches $-8$. So I used LHopital's rule and got $$-\frac{3x^{\frac{2}{3}}}{2\sqrt{1-x}}$$ plug in $-8$ it is $-2(-1)... |
Wolfram Alpha is not correct.
You're almost correct, except that we have
$$\left(-8\right)^{2/3}=(64)^{1/3}=4$$ or alternatively $$\left(-8\right)^{2/3}=(-2)^{2}=4$$
Hence, applying L'Hospital's Rule
$$\begin{align}
\lim_{x\to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}&=\lim_{x\to -8}\left(-\frac{3x^{2/3}}{2\sqrt{1-x}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$S=1+2i+3i^2+4i^3+\dots+(n+1)i^n$; $4\mid n$; closed form for $S$ $$S=1+2i+3i^2+4i^3+\dots+(n+1)i^n$$
where $4\mid n$. How can I simplify this exprerssion so as to obtain a general expression?
| Consider functions $f(x) = 1 + 2x + 3x^2 +\ldots +(n+1)x^n$ and $F(x) = 1+ x + x^2 +\ldots + x^{n+1}$. We have $F'(x) = f(x)$. On the other hand, $F(x)(x-1) = x^{n+2} -1$ which implies $F(x) = \frac{x^{n+2}-1}{x-1}$ for $x\neq 1$. From this we have $$f(x) = F'(x) = \frac{(n+2)x^{n+1}(x-1)-(x^{n+2}-1)}{(x-1)^2}$$
All yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
On the limit of $\sin^2 (\pi\sqrt{n^2+n})$ What is the limit of the sequence $\sin^2 (\pi\sqrt{n^2+n})$ as $n$ tends to infinity?
My Attemp: I replace the square root with $n+\frac 12$ (its equivalent) and the rest is routine:
$\lim \sin^2 (\pi\sqrt{n^2+n}) = \lim \sin^2 (\pi n + \frac{\pi}2) = 1$
Is this correct?
| Hint. The sequence is convergent.
As $n$ tends to $+\infty$, we may write
$$
\begin{align}
u_n &:=\sin^2 \left( \pi \sqrt{n^2+n }\right)\\
&=\sin^2 \left( \pi n \:\sqrt{1+\frac{1}{n}}\right)\\
&=\sin^2 \left( \pi n \:\left(1+\frac{1}{2n}+\mathcal{O}\left(\frac{1}{n^2}\right)\right)\right)\\
&=\sin^2 \left( \pi n +\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2152726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to solve $x - 3\sqrt{\frac{5}{x}} = 8$ for $x - \sqrt{5x}$? I have a problem from my textbook. By using $x - 3\sqrt{\frac{5}{x}} = 8$ how can we find the value of $x - \sqrt{5x}$. I have derived the equation that's given so much, but i couldn't find the answer. Solvings or hints are appreciated.
| The function $f(x)=x-3\sqrt{5\over x}$ increases from $-\infty$ at $x=0$ to $+\infty$ as $x\to\infty$, so the equation $x-3\sqrt{5\over x}=8$ has exactly one solution for $x\gt0$. Multiplying through by $x$ and moving everything to the left hand side, we have
$$x^2-8x-3\sqrt{5x}=0$$
Now let $u=x-\sqrt{5x}$. We can re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2153038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$ Given that $a, b$ and $c$ are the sides of a triangle.
How to prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$?
Maybe any hint? Am I going to wrong direction?
$$2(ab + bc + ca)-a^2 + b^2 + c^2>0$$
$$2ab + 2bc + 2ca-a^2 + b^2 + c^2>0$$
$$2b(a+c) + 2ca-a^2 + b^2 + c^2>0$$
... | Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, $x$, $y$ and $z$ are positives and
$$\sum_{cyc}(2ab-a^2=\sum_{cyc}(2(x+z)(y+z)-(y+z)^2)=$$
$$=\sum_{cyc}(2x^2+6xy-2x^2-2xy)=4(xy+xz+yz)>0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Solve $\lim_{x \rightarrow 0} \frac{e^x+e^{-x}-2}{x^2+2x}$ without using L'Hopital's rule I tried:
$$\lim_{x \rightarrow 0} \frac{e^x+e^{-x}-2}{x^2+2x} = \\
\frac{e^{x}(1+e^{-2x}-\frac{2}{e^x})}{x(x-2)} = \frac{e^x(1+e^{-2x})-2}{x(x-2)} = \frac{e^x(1+e^{-2x})}{x(x-2)} - \frac{2}{x(x-2)} = \\
\frac{1+e^{-2x}}{x} \cdot ... | You can use hyperbolics: $2\cosh x=e^x+e^{-x}$ and so the limit expression can be written as $$\frac{2}{x+2}\cdot\frac{\cosh x-1}{x-0}$$ and use the limit definition of the derivative on $y=\cosh x$ taken at $x=0$. Now give it a try from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2155353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Definite integration with variable substitution gives wrong result I have the following definite integral:
$$\int^{5}_{-5}{\sqrt{25-x^2} dx}$$
I do a variable substitution:
$$y = 25 - x^2$$
$$x = \sqrt{25 - y}$$
$$dy = -2x~dx$$
$$dx = \frac{dy}{-2x} = -\frac{1}{2\sqrt{25-y}}$$
I get the new integral:
$$-\frac{1}{2}\int... | The "mistake" is $x=\sqrt{25 - y}$. What happens if $x < 0$? One needs to be careful when making non-injective substitutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$ How to prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$? Factoring $504$ yields $2^3 \cdot 3^2 \cdot 7$. Divisibility by $8$ can be easily seen, because if $n$ is even then $8 | n^3$, else $(n^3 - 1)$ and $(n^3 + 1)$ are both even and one of these is divi... | $(n^3-1)(n^3)(n^3+1) = n^9-n^3$
$n^6 \equiv 1 \pmod 7$ This is Fermat's little theorem. If $p$ prime $n^{p-1}\equiv 1\pmod p$
$n^9-n^3 \equiv n^3-n^3\equiv 0 \pmod 7$
$n^6 \equiv 1 \pmod 9$
If this is not obvious:
$n^2 \equiv 1 \pmod 3\\
n^2 = (3k+1)\\
n^6 = (3k+1)^3 = (27k^3 + 27k^2 + 9k + 1)$
$n^9-n^3 \equiv 0 \pmod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Finding $ E ( X Y \mid X + Y = w ) $ when $ X $ and $ Y $ are i.i.d $\mathcal N ( 0 , 1 ) $ variables $ X $ and $ Y $ are both independent and identically distributed random variables with normal distributions $ \mathcal N ( 0 , 1 ) $. What is $ E ( X Y \mid X + Y = w ) $?
I know this means that $ W=X+Y $ must be norma... | Letting $W = X + Y$ then you can write out the joint Gaussian as
\begin{align*}
\begin{bmatrix}
X \\ Y \\ W
\end{bmatrix} \sim \mathcal{N} \left(
\begin{bmatrix}
0 \\ 0 \\0
\end{bmatrix},
\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 2
\end{bmatrix}
\right)
\end{align*}
without worrying too much that this covari... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $I_{n}=\int\frac{\cos nx}{5-4\cos x}=$ I stumbled across the following integral:-
$$I_{n}=\int\frac{\cos nx}{5-4\cos x}dx$$
where $n$ is a positive integer.
I have no idea how to proceed....I tried integration by parts and even writing
$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$
I couldn't make much headway.....
Any i... | Hint:
It is just a piece of cake of creating the reduction formula:
$\int\dfrac{\cos nx}{5-4\cos x}~dx$
$=\int\dfrac{2\cos x\cos((n-1)x)}{5-4\cos x}~dx-\int\dfrac{\cos((n-2)x)}{5-4\cos x}~dx$ (according to http://mathworld.wolfram.com/Multiple-AngleFormulas.html)
$=\dfrac{1}{2}\int\dfrac{4\cos x\cos((n-1)x)}{5-4\cos x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing $3^n=\sum_{k=0}^n{(-1)^k\binom{n}{k}4^{n-k}}$. This is Exercise 1.9.3 of F. M. Goodman's "Algebra: Abstract and Concrete".
Show $$3^n=\sum_{k=0}^n{(-1)^k\binom{n}{k}4^{n-k}}.$$
My Attempt:
I tried induction on $n$ as follows.
Base: Try when $n=1$. Then $LHS=3^1=3$ and
$$\begin{align}
RHS&=\sum_{k=0}^1{(-1)^k\... | Start with
$$ \sum_{k=0}^{r+1} (-1)^k\binom{r+1}{k}4^{r+1-k} = 4^{r+1} + \sum_{k=1}^{r+1} (-1)^k\Bigg( \binom{r}{k} + \binom{r}{k-1}\Bigg)4^{r+1-k}$$
$$ = 4^{r+1} + \sum_{k=1}^{r+1} (-1)^k\binom{r}{k}4^{r+1-k} + \sum_{k=1}^{r+1} (-1)^k\binom{r}{k-1}4^{r+1-k}.$$
In the first sum the term $k=r+1$ is zero and you can a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove $\frac12+\frac16+...+\frac{1}{n(n+1)}=\frac{n}{n+1}$ for $n \in \mathbb{N}$ I am using Induction:
Base Case $n=1$ holds ; $\frac12$= $\frac{1}{(1)+1}$
Assume $\frac{n}{n+1}$ is true from some $n \in \mathbb{N}$.
Then $\frac12+\frac16+...+\frac{1}{n(n+1)}+ \frac{1}{(n+1)((n+1)+1)}=\frac{n}{n+1}+ \frac{1}{(n+1)((n+... | Well, I will tell you a method without having to use induction. We have, $$\frac {1}{2} = \frac {2-1}{2\times 1} = \frac {1}{1}-\frac {1}{2} $$ $$\frac {1}{6} = \frac {3-2}{3\times 2} = \frac {1}{2}-\frac {1}{3} $$ $$\vdots $$ $$\frac {1}{n (n+1)} = \frac {n+1-n}{n (n+1)} = \frac {1}{n}-\frac {1}{n+1} $$
Adding, we get... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Question from applications of derivatives. Prove that the least perimeter of an isoceles triangle in which a circle of radius $r$ can be inscribed is $6r\sqrt3$.
I have seen answer online on two sites. One is on meritnation but the problem is that answer is difficult and bad formatting. Other answer on topperlearning b... | Let $\Delta ABC$ be our triangle, where $AB=BC$ and $\measuredangle ABC=2x$.
Hence, $P_{\Delta ABC}=2r\left(\cot{x}+2\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right)$ and we need to prove that $\min\limits_{\left(0,\frac{\pi}{2}\right)}f=3\sqrt3$, where
$$f(x)=\cot{x}+2\tan\left(\frac{\pi}{4}+\frac{x}{2}\right).$$
Ind... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Finding $\lim_{n\to\infty}(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}})$ I'm trying to find $\lim_{n\to\infty}(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}})$.
*
*I tried to use the squeeze theorem, failed.
*I tried to use a sequence defined recurs... | It looks squeezable.
\begin{align}
\frac{n}{\sqrt{n^2+n}} \le \sum_{k=1}^n\frac{1}{\sqrt{n^2+k}} \le \frac{n}{\sqrt{n^2+1}}
\\
\\
\frac{1}{\sqrt{1+\frac{1}{n}}} \le \sum_{k=1}^n\frac{1}{\sqrt{n^2+k}} \le \frac{1}{\sqrt{1+\frac{1}{n^2}}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that the sum of pythagorean triples is always even Problem: Given $a^2 + b^2 = c^2$ show $a + b + c$ is always even
My Attempt, Case by case analysis:
Case 1: a is odd, b is odd. From the first equation,
$odd^2 + odd^2 = c^2$
$odd + odd = c^2 \implies c^2 = even$
Squaring a number does not change its congruence m... | Notice that $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=2c^2+2(ab+bc+ac)$, so the square of $a+b+c$ is even and thus $a+b+c$ is also even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 6,
"answer_id": 4
} |
$5x^2−10x+7$ in completed square form? I was studying quadratic equations and practicing to solve them using the technique of completing the squares.
My answer was as follows:
$$5x^2−10x=−7$$
$$5x^2−10x+25=18$$
$$x(x-5)^2−5(x-5)=18$$
$$\boldsymbol{(x-5)^2-18=0}$$
The answer in the book is:
$$\boldsymbol{5(x-1)^2+2=... | We know that $ax^2+bx+c=0$ and we want to get this form $a(x+d)^2+e=0$.
This is a shortcut: $d=\frac{b}{2a}$ and $e=c-\frac{b^2}{4a}$.
Then we know that $5x^2-10x+7=0$, here we can see that $a=5,b=-10,$ and $c=7$.
So we simply substitute and we get $d=\frac{-10}{2(5)}$ and $e= 7-\frac{(-10)^2}{4(5)}$.
Now we simplify a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2164443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
How can I calculate the limit without using the L'Hopital's rule $$ \lim_{x \to 2} \frac{1-\sqrt{1-\sqrt{4x-8}}}{1-\sqrt{1-\sqrt{\frac{x-2}{x+2}}}} $$
I have tried to multiply with conjugates or use auxiliary variables but it did not arrive at all simple
| $$\lim_{x \to 2} \frac{1-\sqrt{1-\sqrt{4x-8}}}{1-\sqrt{1-\sqrt{\frac{x-2}{x+2}}}}$$
Let's let $z = x-2$ (or $x = z + 2$). Then we get
$$\lim_{z \to 0} \frac{1-\sqrt{1-2\sqrt z}}{1-\sqrt{1-\sqrt{\frac{z}{z+4}}}}$$
\begin{align}
1-\sqrt{1-2\sqrt z}
&= \dfrac{1-(1-2\sqrt z)}{1+\sqrt{1-2\sqrt z}} \\
&= \dfrac{2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2170336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
Find a positive integer $a$ such that $4 \mid a$ and $9 \mid a+1$ and $25 \mid a+2$ I need to find a positive integer $a$ such that $4 \mid a$ and $9 \mid a+1$ and $25 \mid a+2$.
I tried converting this to a system of congruences, and I got the following.
$a \equiv 0$ mod $4$
$(a+1) \equiv 0$ mod $9$
$(a+2) \equiv 0$ m... | $a \equiv 0 \pmod{4}$, so $a = 4k$, and $a+1 \equiv 0 \pmod{9}$, it means $4k+1 \equiv 0 \pmod{9}$, the minimum $k$(here $k \ge 0$) is $k = 2$. so, when $a \equiv 0 \pmod{4}$ and $a+1 \equiv 0 \pmod{9}$, the minimum $a$ is $a = 4k = 8$, all the $a$ is $8+(4 \cdot 9)m = 8+36m$.
then, we have $a+2 \equiv 0 \pmod{25}$, it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $a^2+b^2 = c^2 + \sqrt{3}$, find value of $\angle C$.
In a triangle with sides $a$, $b$, $c$, consider this relationship:
$$a^2+b^2 = c^2 +\sqrt 3$$
Now find value of $\measuredangle C$ .
My try: I tried to use Laws of Cosines and Law of Sinee, but I didn't get any result.
| Note: In what follows I am interpreting $a^2+b^2 = c^2 +\sqrt 3$ as meaning that $c^2$ is larger than what would be obtained using the Pythagorean theorem by $\sqrt 3$.
Since you must have $$a^2+b^2 = c^2 +\sqrt 3$$ the law of cosines states that in this case $$- 2\ a\ b\ cos (\measuredangle C) = \sqrt 3$$ Therefore t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Proving: $\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2\ge\frac{25}2$
For $a>0,b>0$, $a+b=1$ prove:$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge\dfrac{25}{2}$$
My try don't do much, tough
$a+b=1\implies\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{ab}$
Expanding: $\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfra... | Since $f(x)=\left(x+\frac{1}{x}\right)^2$ is a convex function, your inequality follows from Jensen:
$$LS\geq2\left(\frac{a+b}{2}+\frac{1}{\frac{a+b}{2}}\right)^2=\frac{25}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Is this elementary proof of FLT correct? Consider the classic FLT for $n=3$, $x^3+y^3=z^3 $
Without loss of generality,we can rewrite as $a^3+(b-a)^3=c^3…Equation (1)$
We can also assume there is no common factor between $a,b-a$ and $c …Assumption (1)$
Since $m+n$ is a factor of $m^3+n^3$, $b$ divides $c^3$.
Let $p$ b... | The proof is incorrect. The error is in the statement
$$``\text{$p^2$ divides the first term $3b^{2}a$ but not $3a^{2}b$}."$$
In case 1, when $p$ is different than $3$, you know that $p \nmid a$, $p \nmid 3$, and $p\mid b$. But, you do not know that $p^{2} \nmid b$. This is because you chose $p$ to be any prime factor ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Does there exist $a \in \mathbb{Q}$ such that $a^2 - a + 1$ is a square? Does there exist $a \in \mathbb{Q}$ such that $a^2 - a + 1$ is a square?
Context for this question: For pedagogical purposes I was trying to create an example of a cubic polynomial with both its critical values and its zeros all at integers. Aft... | $$a^2-a+1=(a-\frac{1}{2})^2+\frac{3}{4}$$
So what you ask are there any rational numbers $u=a-\frac{1}{2}$ and $v$ such that,
$$u^2+\frac{3}{4}=v^2$$
$$v^2-u^2=\frac{3}{4}$$
$$(v-u)(v+u)=\frac{3}{4}$$
Let $x$ and $y$ be two real numbers that multiply to $\frac{3}{4}$, with the condition that $\frac{x+y}{2}$ and $\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Abstract Algebra (Multiplicative Inverse using the Euclidean Algorithm) The question I am trying to solve is to find the multiplicative inverse of
$2-\sqrt[3]{2}+\sqrt[3]{4}$.
So far I am come up with the following.
Let $\alpha = 2-\sqrt[3]{2}+\sqrt[3]{4}$.
Then $s(x)=2-x+x^2$ and $x=\sqrt[3]{2}.$
Then $t(x)=x^3-2$.
T... | $\mathbb Q(\sqrt[3]{2} )$ is a field. Since $1, \sqrt[3]{2} , \sqrt[3]{4} $ form a basis for $\mathbb Q(\sqrt[3]{2} )$ over $\mathbb Q$, you know that
$$ (2-\sqrt[3]{2}+\sqrt[3]{4} )^{-1} = a + b\sqrt[3]{2}+ c\sqrt[3]{4} $$
for some $a, b, c \in \mathbb Q$.
One natural approach would be write the equation
$$ (2-\sqrt[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Infinite series of formula How do I compute this infinite series
$$ \sum_{k=0}^{\infty}{\frac{(7n+32)(3^n)}{n(n+2)(4^n)}}$$
I believe partial fraction decomposition is part of the solving method but am a little stuck because the $3^n$ and the $4^n$ make the decomposition method a bit strange:
${\frac{(7n+32)(3^n)}{n(n+... | HINT:
$$\frac{(7n+32)(3^n)}{n(n+2)(4^n)}=\frac{(4^2(n+2)-3^2)(3^n)}{n(n+2)(4^n)}$$
$$=\dfrac{3^n}{n4^{n-2}}-\dfrac{3^{n+2}}{(n+2)4^n}$$
$$=16\left(\dfrac{(3/4)^n}n-\dfrac{(3/4)^{n+2}}{n+2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2175558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Gain matrices of numerical schemes Introduction
Say I have a linear multiple degree of freedom system, as below:
\begin{equation*}
\frac{dx}{dt}=ax+by
\end{equation*}
\begin{equation*}
\frac{dy}{dt}=dx+ey
\end{equation*}
This can also be written as $\frac{dr}{dt}=Ar$ with $A$ given by
\begin{equation*}
A=
... | You just insert the $f$ of the homogeneous linear equation,
\begin{alignat}{3}
k_1&=f(y)&&&&=Ay
\\
k_2&=f(y+0.5Δt·k_1)&&=A(y+0.5Δt·k_1))&&=Ay+0.5Δt·A^2y
\\
k_3&=f(y+0.5Δt·k_3)&&=A(y+0.5Δt·k_3))&&=Ay+0.5Δt·A^2y+0.25Δt^2·A^3y
\\
k_4&=f(y+Δt·k_4)&&=A(y+Δt·k_4)&&=Ay+Δt·A^2y+0.5Δt^2·A^3y+0.25Δt^3·A^4y
\\\hline
&k_1+2k_2+2k_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2177590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Test convergence of the series and find its sum: $\sum_{n=0}^{\infty}\left(\frac{-1-2i}{2+3i}\right)^n$ Test convergence of the series and find its sum: $$\sum_{n=0}^{\infty}\left(\frac{-1-2i}{2+3i}\right)^n$$
My try:
This is simplified to $$\sum_{n=0}^{\infty}\left(\frac{-i-8}{13}\right)^n$$
Then how to proceed?
| $\sum_{n=0}^{\infty}\left(\frac{-1-2i}{2+3i}\right)^n
$
Since
$\frac{-1-2i}{2+3i}
=\frac{-1-2i}{2+3i}\frac{2-3i}{2-3i}
=\frac{-8-i}{13}
$
and
$|\frac{-8-i}{13}|^2
=\frac{65}{169}
\lt 1
$,
the sum converges.
Applying the standard formula,
the sum is
$\begin{array}\\
\dfrac{1}{1-\frac{-8-i}{13}}
&=\dfrac{13}{13+(8+i)}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2179818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove the inequality, power. $\{ x,y \in\Bbb R\ \}$ If $x+y = 2$ then prove the inequality:
$x^4 + y^4 \ge 2$
How I started
*
*$(x+y)^2 = 4$
*$x^2 + y^2 = 4 - 2xy$
*$(x^2+y^2)^2 - 2(xy)^2 \ge 2$
*$(4-2xy)^2 - 2(xy)^2 \ge 2$
*$16-16xy + 4(xy)^2 -2(xy)^2 - 2 \ge 0$
*$2(xy)^2 - 16xy + 14 \ge 0$
*for $t=xy$
*$2t^... | Using Holder's inequality we get that
$$(1+1+1+1)^{3/4}(x^4+y^4+1+1)^{1/4} \geq |x|+|y| +1+1 \geq4 $$
Therefore,
$$(x^4+y^4+1+1)^{1/4}\geq 4^{1/4}$$
We conclude,
$$x^4+y^4\geq 4-2=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.